Approximation by Complex Bernstein and Convolution Type Operators

APPROXIMATION BY COMPLEX BERNSTEIN AND CONVOLUTION TYPE OPERATORS SERIES ON CONCRETE AND APPLICABLE MATHEMATICS ISSN:...

Author: Sorin G. Gal

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APPROXIMATION BY COMPLEX BERNSTEIN AND CONVOLUTION TYPE OPERATORS

SERIES ON CONCRETE AND APPLICABLE MATHEMATICS ISSN: 1793-1142 Series Editor: Professor George A. Anastassiou Department of Mathematical Sciences The University of Memphis Memphis, TN 38152, USA

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Series on Concrete and Applicable Mathematics – Vol. 8

APPROXIMATION BY COMPLEX BERNSTEIN AND CONVOLUTION TYPE OPERATORS

Sorin G Gal University of Oradea, Romania

World Scientific NEW JERSEY

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Preface

The Bernstein polynomials attached to f : [0, 1] → R and given by n X k n k Bn (f )(x) = pn,k (x)f ( ), pn,k (x) = x (1 − x)n−k , x ∈ [0, 1], n k k=0

probably are the most famous algebraic polynomials in Approximation Theory and were introduced in 1912 by S.N. Bernstein [38] in order to give the first constructive (and simple) proof to the Weierstrass’ approximation theorem. Many books and papers were dedicated to their study, probably the most known being the book of Lorentz [125]. The importance of Bernstein polynomials also consists in the fact that their form has suggested and still continues to suggest to mathematicians the construction of a great variety of other approximation operators, like the Schurer polynomials, Kantorovich polynomials, Stancu polynomials, q-Bernstein polynomials, Durrmeyer polynomials, Favard-Sz´ asz-Mirakjan operators, Baskakov operators, and the list can continue with very many others. A natural question concerning the Bernstein polynomials of a real variable (and by analogy, concerning any Bernstein-type operator of a real variable x) is the following : if in the expression of Bn (f )(x) one replaces x ∈ [0, 1] by z in some regions in C (containing [0, 1]) where f is supposed to be analytic, (a process we call complexification), then what convergence properties have the complex Bernstein polynomials n X n k k z (1 − z)n−k , z ∈ C ? Bn (f )(z) = pn,k (z)f ( ), pn,k (z) = k n k=0

In other words, the problem is to study the overconvergence phenomenon (not in the sense known as Wash’s overconvergence in the interpolation of functions !) for the Bernstein polynomials, that is to extend their convergence properties and orders of these convergencies to larger sets in the complex plane than the real interval [0, 1]. The first goal of the present book is to give some answers in Chapter 1 to the above question of overconvergence, for several classes of Bernstein-type operators. In essence, it will be shown that for all the Bernstein-type operators, the orders of approximation from the real axis are preserved in complex domains too. vii

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Approximation by Complex Bernstein and Convolution Type Operators

We recall that concerning the complex Bernstein polynomials, Wright [199], Kantorovich [113], Bernstein [39; 40; 41], Lorentz [125] and Tonne [190] have given interesting answers to this question. It is worth noting that an entire Chapter 4 of 38 pages is dedicated to it in the book of Lorentz [125]. In that book interesting convergence properties of Bn (f )(z) and of its so-called degenerate form, in various domains in C, like compact disks, ellipses, loops, autonomous sets are presented. Notice that in the above mentioned papers no quantitative estimates of these convergence results were obtained. Also, convergence results without any quantitative estimate were obtained for the complex Favard-Sz´ asz-Mirakjan operators by Dressel-Gergen and Purcell [65] and for the complex Jakimovski-Leviatan operators by Wood [200]. The above qualitative results are theoretically based on the ”bridge” made by the classical result of Vitali (see Theorem 1.0.1), between the (well-established) approximation results for the Bernstein-type operators of real variable and those for the Bernstein-type operators of complex variable. It is worth noting that in the other books or long surveys dealing with complex approximation, like those of Sewell [162], Dzjadyk [69], Gaier [76], Suetin [186], Andrievskii-Belyi-Dzjadyk [26], [27], or the surveys of Dyn’kin [62] and Andrievskii [25], the topic of the present book is not considered. Also, in other books like Lorentz [125] (Chapter 4) and Gal [77] (Chapters 3 and 4), the topic of the present book is attended in a tangential (and somehow different) way only. In the Preface of their important book [60] in 1993 concerning the Approximation Theory of functions of real variable, DeVore and Lorentz note that, I cite ”the Approximation Theory of functions of complex variables would require new books”. The present book seeks to be one among the answers to this requirement and can briefly be described as follows. In Chapter 1 one deepens the study of the approximation properties for the complex Bernstein polynomials Bn (f )(z) in compact disks and in some special compact subsets of C. In addition, similar results for other Bernstein-type polynomials/operators including those mentioned above are presented. In detail, Chapter 1 can be described as follows : – Section 1.0 contains the main results and concepts in complex analysis required for the proofs of the results in this book. For example, we mention here the Vitali’s theorem, Cauchy’s formula, Bernstein’s inequality, Faber polynomials associated to a domain in C, Faber series, Faber coefficients, Faber mapping. – in Section 1.1 the exact orders in simultaneous approximation by Bn (f )(z) and its derivatives, Voronovskaja’s result with quantitative upper estimate and shape preserving properties of Bn (f )(z) are obtained ; Subsection 1.1.1 contains the results on compact disks centered at origin while Subsection 1.1.2 contains some approximation results on compact sets in C for the so-called Bernstein-Faber polynomials ; – Section 1.2 contains convergence results with quantitative estimates of the iterates of Bn (f )(z), connected with the theory of the semigroups of operators and

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Preface

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ix

the shape preserving properties of these iterates, in the sense that beginning with an index they preserve some properties of f in Geometric Function Theory, like the starlikeness, convexity and spirallikeness ; – in Section 1.3 the exact order in the generalized Voronovskaja’s theorem for Bn (f )(z) is obtained ; – Section 1.4 presents the exact orders of approximation by Butzer’s linear combinations of complex Bernstein polynomials and of Bernstein-Faber polynomials in compact disks and in compact Faber sets, respectively ; – in Sections 1.5, 1.6, 1.7, 1.8, 1.9 and 1.10 we prove some similar properties for the complex q-Bernstein polynomials, Bernstein-Stancu polynomials, BernsteinKantorovich polynomials, Favard-Sz´ asz-Mirakjan operators, Baskakov operators and Bal´ azs-Szabados operators, respectively. Besides the approximation results in compact disks for all these complex Bernstein-type operators, it is worth mentioning here the approximation results for the Bernstein-Stancu-Faber polynomials in compact sets in C, a weighted-kind approximation result for the Favard-Sz´ asz-Mirakjan operator in strips and the study of two kinds of complex Baskakov operators generated by the real one ; – Section 1.11 contains bibliographical notes and some open problems. The open problems mainly consist in proposals of similar studies for other types of complex Bernstein-type operators too. In Chapter 2 we extend some of the results in Chapter 1 to the case of several complex variables. In Section 2.1 the concepts and results in the complex analysis of functions of several complex variables that we need for this chapter are presented. Section 2.2 deals with the approximation by two kinds of bivariate complex Bernstein polynomials, while Sections 2.3 and 2.4 treat the bivariate case of complex Favard-Sz´ asz-Mirakjan and Baskakov operators, respectively. All the approximation results are obtained in compact polydisks. Section 2.5 contains some bibliographical notes and open problems. Chapter 3 deals with the approximation and geometric properties of several types of complex convolutions. Section 3.1 contains the approximation properties of some complex linear convolution polynomials : of de la Vall´ee Poussin, Fej´er, Riesz-Zygmund, Jackson and Rogosinski kinds. More exactly, for these complex linear convolutions Voronovskaja-type results and the exact orders of approximation in compact disks are proved. Section 3.2 studies several kinds of linear non-polynomial convolutions. Thus, in Subsection 3.2.1 one studies the approximation properties (including the exact orders of approximation) in compact disks and compact sets in C of the non-polynomial complex convolutions of Picard, Poisson-Cauchy, and Gauss-Weierstrass. Also, their geometric properties are studied and applications to PDE in complex setting (i.e to heat and Laplace equations of complex spatial variable) are presented. In the Subsections 3.2.2, 3.2.3, 3.2.4 and 3.2.5 the approximation and geometric properties in compact disks of the complex q-Picard and q-Gauss-Weierstrass convolutions, Post-Widder complex convolution, rotation-

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Approximation by Complex Bernstein and Convolution Type Operators

invariant complex convolution and Sikkema complex convolution, respectively are presented. Section 3.3 contains the approximation and geometric properties of a nonlinear-type complex convolution in compact disks. Finally, in Chapter 4 one presents several related topics : approximation by Bernstein polynomials of quaternion variable in Section 4.1, approximation of vector-valued functions of real and complex variables by operators of the type introduced in the previous chapters in Section 4.2 and strong approximation by Taylor series in the unit disk in Section 4.3. Let us mention that most of the results in this book have been obtained by the author of this monograph : in a series of papers, single or jointly written with other researchers (as can be seen in the bibliography) and as new results that appear for the first time here. It is important to note that the present book suggests for further research similar studies for other complex linear and nonlinear convolutions and for the complex forms of other Bernstein-type operators in approximation theory, like those of Durrmeyer-type, Meyer-K¨ onig-Zeller-type, Jakimovski-Leviatan-type, BleimannButzer-Hahn-type, Gamma-type, beta-type, to mention only a few. The book mainly is addressed to researchers in the fields of complex approximation of functions and its applications, mathematical analysis and numerical analysis. Also, since most of the proofs use elementary complex analysis, it is accessible to graduate students and suitable for graduate courses in the above domains. Sorin G. Gal Department of Mathematics and Computer Science University of Oradea Romania

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Contents

Preface

vii

1. Bernstein-Type Operators of One Complex Variable 1.0 1.1

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

1

Auxiliary Results in Complex Analysis . . . . . . . . . . . . . . Bernstein Polynomials . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Bernstein Polynomials on Compact Disks . . . . . . . . 1.1.2 Bernstein-Faber Polynomials on Compact Sets . . . . . Iterates of Bernstein Polynomials . . . . . . . . . . . . . . . . . Generalized Voronovskaja Theorems for Bernstein Polynomials Butzer’s Linear Combination of Bernstein Polynomials . . . . . q-Bernstein Polynomials . . . . . . . . . . . . . . . . . . . . . . Bernstein-Stancu Polynomials . . . . . . . . . . . . . . . . . . . Bernstein-Kantorovich Type Polynomials . . . . . . . . . . . . Favard-Sz´ asz-Mirakjan Operators . . . . . . . . . . . . . . . . . Baskakov Operators . . . . . . . . . . . . . . . . . . . . . . . . Bal´ azs-Szabados Operators . . . . . . . . . . . . . . . . . . . . Bibliographical Notes and Open Problems . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

2. Bernstein-Type Operators of Several Complex Variables 2.1 2.2 2.3 2.4 2.5

Introduction . . . . . . . . . . . . . . . . . Bernstein Polynomials . . . . . . . . . . . Favard-Sz´ asz-Mirakjan Operators . . . . . Baskakov Operators . . . . . . . . . . . . Bibliographical Notes and Open Problems

. . . . .

. . . . .

. . . . .

. . . . .

155 . . . . .

. . . . .

. . . . .

. . . . .

3. Complex Convolutions 3.1 3.2

1 6 6 19 26 35 42 50 67 96 103 124 139 149

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

155 156 166 172 179 181

Linear Polynomial Convolutions . . . Linear Non-Polynomial Convolutions 3.2.1 Picard, Poisson-Cauchy and Convolutions . . . . . . . . . xi

. . . . . . . . . . . . . . . . . . . . . . Gauss-Weierstrass . . . . . . . . . . .

. . . . . . 181 . . . . . . 204 Complex . . . . . . 205

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xii

3.2.2

3.3 3.4

Complex q-Picard and q-Gauss-Weierstrass Integrals . . . . . . . . . . . . . . . . . . . 3.2.3 Post-Widder Complex Convolution . . . . 3.2.4 Rotation-Invariant Complex Convolutions 3.2.5 Sikkema Complex Convolutions . . . . . . Nonlinear Complex Convolutions . . . . . . . . . . Bibliographical Notes and Open Problems . . . . .

Singular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

4. Appendix : Related Topics 4.1 4.2

4.3 4.4

Bernstein Polynomials of Quaternion Variable . . Approximation of Vector-Valued Functions . . . 4.2.1 Real Variable Case . . . . . . . . . . . . 4.2.2 Complex Variable Case . . . . . . . . . . Strong Approximation by Complex Taylor Series Bibliographical Notes and Open Problems . . . .

257 264 269 279 286 294 295

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

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. . . . . .

295 299 300 314 321 324

Bibliography

327

Index

337

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Chapter 1

Bernstein-Type Operators of One Complex Variable

In the Sections 1.1-1.4 first we obtain the exact degrees in approximation of analytic functions in compact disks by complex Bernstein polynomials and their Butzer’s linear combination and in generalized Voronovskaja’s results. These sections include approximation results on compact sets in C for the so-called Bernstein-Faber polynomials and their Butzer’s linear combination in compact Faber sets. Convergence results on compact disks for the iterates of Bn (f )(z) connected with the theory of the semigroups of operators and shape preserving properties of these iterates (in the sense that beginning with an index they preserve some properties of f in Geometric Function Theory, like the starlikeness, convexity and spirallikeness) also are proved. Then in the next Sections 1.5-1.10 some similar properties for the complex q-Bernstein polynomials, Bernstein-Stancu polynomials, Bernstein-Kantorovich polynomials, Favard-Sz´ asz-Mirakjan operators, Baskakov operators and Bal´ azsSzabados operators are obtained. For all kinds of Bernstein operators, the exact degrees of approximation mainly are obtained by three steps : 1) upper estimates ; 2) quantitative Voronovskaja-type formula ; 3) lower estimates by using step 2. 1.0

Auxiliary Results in Complex Analysis

In order to make the book more self-contained, in this section we briefly present the main known results and methods in Complex Analysis we use in our study. The first one is called Vitali’s theorem and can be stated as follows. Theorem 1.0.1. (Vitali, see e.g. Kohr-Mocanu [118], p. 112, Theorem 3.2.10) Let Ω be a domain in C and F ⊂ Ω a set having at least one accumulation point in Ω. If the sequence (fn )n∈N of analytic functions in Ω is bounded in each compact in Ω and (fn (z))n is convergent for any z ∈ F , then (fn )n∈N is uniformly convergent in any compact of Ω. In our applications, in general Ω = DR = {z ∈ C; |z| < R} with R > 1, F is a segment included in DR and the compact subsets considered will be the closed disks Dr = {z ∈ C; |z| ≤ r} with 1 ≤ r < R. 1

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The second important result in Complex Analysis we use is the Cauchy’s formula for disks. Theorem 1.0.2. (Cauchy, see e.g. Kohr-Mocanu [118], p. 28, Theorem 1.2.20) Let r > 0 and f : Dr → C be analytic in Dr and continuous in Dr . Then, for any p ∈ {0, 1, 2, ..., } and all |z| < r we have Z p! f (u) f (p) (z) = du, 2πi Γ (u − z)p+1 where Γ = {z ∈ C; |z| = r} and i2 = −1. An immediate consequence of the Cauchy’s formula is the so-called Weierstrass’s theorem used in the proofs of shape preserving properties. Theorem 1.0.3. (Weierstrass, see e.g. Kohr-Mocanu [118], p. 18, Theorem 1.1.6) Let G ⊂ C be an open set. If the sequence (fn )n∈N of analytic functions on G converges to the analytic function f , uniformly in each compact in G, then for any (p) p ∈ N, the sequence of pth derivatives (fn )n∈N converges to f (p) uniformly on compacts in G. Indeed, note that by the above Cauchy’s formula we can write Z fn (u) − f (u) p! du, fn(p) (z) − f (p) (z) = 2πi Γ (u − z)p+1 from which by passing to modulus the theorem easily follows. In our applications, G = DR with R > 1 and the compact subsets in G are Dr with 1 ≤ r < R. Another well-known result used in the proof of shape preserving properties is the following. Theorem 1.0.4. (see e.g. Graham-Kohr [105], Theorem 6.1.18) If fn , f : Ω → C, n ∈ N are analytic in the domain Ω, f is univalent in Ω and fn → f uniformly in the compact K ⊂ Ω, then there exists n0 (K) such that for all n ≥ n0 , fn is univalent in K. The classical so called Maximum Principle (or Maximum Modulus Theorem) will be frequently used in the proofs of error estimates. Theorem 1.0.5. (see e.g. Kohr-Mocanu [118], p. 2, Corollary 1.1.20) If Ω ⊂ C is a bounded domain and f : Ω → C is analytic in Ω and continuous in Ω, then denoting by Γ the boundary of Ω we have max{|f (z)|; z ∈ Ω} = max{|f (z)|; z ∈ Γ}. For our applications again Ω will be an open disk centered at origin. Useful in some of our proofs will be the well-known so called theorem on the zeroes of analytic functions, which in essence says that the zeroes of an analytic function (non-identical null) necessarily are isolated points. More exactly we can state the following.

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Bernstein-Type Operators of One Complex Variable

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Theorem 1.0.6. (see e.g. Kohr-Mocanu [118], p. 20, Theorem 1.1.12) Suppose that f is analytic in the domain Ω and that f is not identical null in Ω. If a is a zero for f then there exists r = r(a) > 0 such that D(a, r) = {z ∈ C; |z − a| < r} ⊂ Ω and f (z) 6= 0, for all z ∈ D(a, r) \ {a}. Also, we will use the classical so called theorem on the identity of analytic functions. Theorem 1.0.7. (see e.g. Kohr-Mocanu [118], p. 21, Theorem 1.1.14) Let Ω ⊂ C be a domain. If f, g : Ω → C are analytic in Ω then f ≡ g on Ω is equivalent with the fact that the set {z ∈ Ω; f (z) = g(z)} has at least one accumulation point in Ω. Finally, we state a basic result very useful in the proofs of the approximation results and called Bernstein’s inequality for complex polynomials in compact disks. Theorem 1.0.8. (Bernstein [43], p. 45, relation (80) for general r > 0, see also e.g. Pn Lorentz [126], p. 40, Theorem 4, for r = 1) Let P (z) = k=0 ak z k be with ak ∈ C, for all k ∈ {0, 1, 2, ..., } and for r > 0 denote kPn kr = max{|Pn (z)|; |z| ≤ r}. (i) For all |z| ≤ 1 we have |Pn0 (z)| ≤ nkPn k1 ; (ii) If r > 0 then for all |z| ≤ r we have |Pn0 (z)| ≤ nr kPn kr . One observes that (ii) immediately follows from (i). Indeed, denoting Qn (z) = Pn (rz), |z| ≤ 1, by (i) applied to Qn (z) it easily follows r|Pn0 (rz)| ≤ nkPn kr , for all |z| ≤ 1, which proves (ii). Concerning the approximation of analytic functions by sequences of complex polynomials, as it will be seen in the next sections of this chapter and in the next chapters, the main results one refer to approximation in compact disks centered at origin (in particular in the compact unit disk). The advantage consists in the fact that in these kinds of disks constructive methods can be indicated. But of course that it is very important to obtain approximation results in more general domains in the complex plane. In what follows we briefly present the standard method based on the so-called Faber polynomials introduced by Faber [70], which allows to extend all the constructive methods from the closed unit disk to more general domains. The method is less constructive because a generally unknown mapping function (generated from the Riemann’s mapping theorem) enters into considerations. For all the details below on this method see e.g. the book of Gaier [76], pp. 42-54. Also, for other important contributions to the topic of constructive complex approximation see the book of Dzjadyk [69]. Definition 1.0.9. (i) γ : [a, b] → C is called Jordan curve if it is closed (i.e. γ(a) = γ(b)) and simple (i.e. injective). The length of the curve γ is defined by L(γ) = sup{

n X i=1

|γ(ti ) − γ(ti−1 )|; n ∈ N, a = t0 < ... < tn = b}.

γ is called rectifiable if L < +∞. The interior of a Jordan curve is called Jordan domain and the curve is called boundary curve of that domain.

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Approximation by Complex Bernstein and Convolution Type Operators

(ii) (Radon [158]) Suppose that γ : [a, b] → C is a rectifiable Jordan curve. Because L < +∞, it is known that γ has a tangent γ 0 almost everywhere. Then γ is called of bounded rotation if γ 0 can be extended to a function of bounded variation on the whole curve. Remark. Simple examples of Jordan curve of bounded rotation can be made up of finitely many convex arcs (where corners are permitted). Now, if G is a Jordan domain, then (by the Riemann’s mapping theorem) let us denote by Ψ the conformal mapping of C \ D1 onto C \ G, normalized at ∞, that is 0 < limw→∞ Ψ(w) w < ∞. Also, denote by Φ the inverse function of Ψ. Obviously that Ψ and Φ depend on G, but for the simplicity of notation we will not write them as ΨG and ΦG , considering in our presentation that G is arbitrary but fixed. For a Jordan domain G, denote by A(G) the class of all functions continuous in G and analytic in G. In what follows we sketch a method by which any f ∈ A(G) can be approximated by polynomials. For our considerations, it is sufficient to suppose that the boundary curve of G is rectifiable and of bounded rotation. First, one considers the Laurent expansion of [Φ(z)]n , n ∈ N ∪ {0}, valid for large z (n)

[Φ(z)]n = a0

n + ... + a(n) n z +

∞ X

k=1

(n)

a−k /z k .

(n) (n) Definition 1.0.10. (Faber [70]) (i) The polynomial Fn (z) = a0 + ... + an z n , n ∈ N ∪ {0} is called the Faber polynomial of degree n attached to the domain G. (Note that for z ∈ DR , R > 1 we can write Z [Φ(u)]n 1 du.) Fn (z) = 2πi |u|=R u − z

(ii) If f ∈ A(G) then Z Z π 1 f [Ψ(u)] 1 an (f ) = f [Ψ(eit )]e−int dt, n ∈ N ∪ {0} du = 2πi |u|=1 un+1 2πi −π P∞ are called the Faber coefficients of f and n=0 an (f )Fn (z) is called the Faber series attached to f on G. (Here i2 = 1.) The Faber series represent a natural generalization of Taylor series when the unit disk is replaced by an arbitrary simply connected domain bounded by a ”nice” curve. Pn (iii) The mapping T defined by T [Pn ](z) = k=0 ck Fk (z), where Pn (w) = Pn k c w is called the Faber mapping. k k=0

Remark. By Definition 1.0.10, (iii), the Faber mapping T is linear and defined on the set of all polynomials P defined on D1 and with values in the set of polynomials P defined on G. In some cases it can be extended as a linear and bounded mapping between the Banach spaces A(D1 ) and A(G) (both endowed with the corresponding uniform norms). Below we briefly point out this extension (for full details see e.g.

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Bernstein-Type Operators of One Complex Variable

5

the book of Gaier [76], pp. 48-49) under the hypothesis that the boundary of G is a rectifiable Jordan curve of bounded rotation. In this case for G, first it follows that kT (P )k ≤ CkP k for each P ∈ P, where C > 0 depends only on G. Then T can be extended to the closure of P and since P = A(D1 ), T can be extended as a linear and bounded operator from A(D1 ) into A(G), with the property that kT (f )k ≤ Ckf k for each f ∈ A(D1 ). Now, since the Faber mapping has the integral representation Z Pn [Φ(u)] 1 du, T [Pn ](z) = 2πi C u − z valid for each polynomial Pn , by passing to limits we obtain the formula Z F [Φ(u)] 1 du, z ∈ G, F ∈ A(D 1 ). T [F ](z) = 2πi C u − z

Also, the converse formula

1 F (w) = 2πi

Z

|u|=1

T [F ](Ψ(u)) du, w ∈ D1 u−w

holds. The following two known results are of great importance for approximation. Theorem 1.0.11. (more precisely see e.g. Gaier [76], p. 50, Theorem 3) If P∞ F ∈ A(D1 ), F (w) = n=0 cn wn then the Faber coefficients of T [F ] are cn .

Theorem 1.0.12. (more precisely see e.g. Theorem 4 in Gaier [76], p. 51) Suppose that the boundary of G is a rectifiable Jordan curve of bounded rotation and let f ∈ A(G). There exists F ∈R A(D1 ) with f = T [F ] if and only if as function of w ∈ D1 , the Cauchy integral |u|=1 f [Ψ(u)] u−w du belongs to A(D1 ) and in this case we have Z 1 f [Ψ(u)] du, w ∈ D1 , F (w) = 2πi |u|=1 u − w (F is extended by continuity on ∂D1 ). Remark. Theorem 1.0.12 allows to reduce the approximation of f ∈ A(G) to the approximation of F ∈ A(D1 ). Indeed, let (Sn (F )(w))n∈N , Sn (F )(w) = P mn k Then T [Sn (F )](z) = k=0 ak (F )w , be an approximation sequence for F . Pm n a (F )F (z), n ∈ N will represent an approximation sequence for f in the k k=0 k set G (here Fk (z), k ∈ N denote the Faber polynomials attached to G). Indeed, denoting the uniform norms by k · k, this follows from the relation (6.19), p. 51 in Gaier [76], kf −

mn X k=0

ak (F )Fk kG = kT (F −

mn X

k=0

ak (F )ek )kG ≤ k|T k| · kF −

mn X

k=0

ak (F )ek kD1 ,

where ek (w) = wk and k|T k| ≤ M < ∞, because of the hypothesis on the boundary of G.

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Bernstein Polynomials

In this section, we find the exact orders in simultaneous uniform approximation of analytic functions by complex Bernstein polynomials in closed disks, an upper estimate in Voronovskaja’s result and we prove that the complex Bernstein polynomials attached to an analytic function, preserve the univalence, starlikeness, convexity and spirallikeness. Also, to Jordan domains Bernstein-type polynomials are attached and approximation results on connected compact sets with estimates are obtained. 1.1.1

Bernstein Polynomials on Compact Disks

Concerning the approximation properties (uniform convergence), the results in Wright [199], Kantorovich [113], Bernstein [39; 40; 41], Lorentz [125] and Tonne [190] are well-known. It is worth nothing that an entire Chapter 4 of 38 pages is dedicated to this topic in the book of Lorentz [125]. In that book interesting convergence properties of Bn (f )(z) and of its so-called degenerate form, in various domains in C, like compact disks, ellipses, loops, autonomous sets are presented. For example, the following three approximation results due to Bernstein, Tonne and Kantorovich concerning the uniform approximation of Bernstein polynomials in the unit disk and in an ellipse hold. Theorem 1.1.1. (i) (Bernstein, see e.g. Lorentz [125], p. 88) For the open G ⊂ C, such that D1 ⊂ G and f : G → C is analytic in G, the complex Bernstein Pn polynomials Bn (f )(z) = k=0 nk z k (1 − z)n−k f (k/n), uniformly converge to f in D1 . Here D1 denotes the open unit disk. P∞ (ii) (Tonne [190]) If f (z) = k=0 ck z k is analytic in the open unit disk D1 , f (1) is a complex number and there exist M > 0 and m ∈ N such that |ck | ≤ M (k + 1)m , for all k = 0, 1, 2, ...., then Bn (f )(z) converges uniformly (as n → ∞) to f on each closed subset of D1 . (iii) (Kantorovich, see e.g. Lorentz [125], p. 90) If f is analytic in the interior of an ellipse of foci 0 and 1, then Bn (f )(z) converges uniformly to f (z) in any closed set contained in the interior of ellipse. But in all the previous mentioned work no quantitative estimates of these convergence results were obtained. In what follows, first we obtain upper quantitative estimates on compact disks. For this purpose, denote DR = {z ∈ C; |z| < R}. Theorem 1.1.2. (Gal [77], p. 264, Theorem 3.4.1, (iii)-(v)) Suppose that R > 1 P∞ and f : DR → C is analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . (i) Let 1 ≤ r < R be arbitrary fixed. For all |z| ≤ r and n ∈ N, we have |Bn (f )(z) − f (z)| ≤ where 0 < Cr (f ) =

3r(1+r) 2

P∞

j=2

Cr (f ) , n

j(j − 1)|cj |rj−2 < ∞.

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(ii) For the simultaneous approximation by complex Bernstein polynomials, we have : if 1 ≤ r < r1 < R are arbitrary fixed, then for all |z| ≤ r and n, p ∈ N, |Bn(p) (f )(z) − f (p) (z)| ≤

Cr1 (f )p!r1 , n(r1 − r)p+1

where Cr1 (f ) is given as at the above point (i). Proof. (i) Denoting ek (z) = z k and πk,n (z) = Bn (ek )(z), we evidently have P∞ Bn (f )(z) = k=0 ck πk,n (z) and we get |Bn (f )(z) − f (z)| ≤

∞ X k=0

|ck | · |πk,n (z) − ek (z)|,

so that we need an estimate for |πk,n (z) − ek (z)|. For this purpose we use the recurrence proved for the real variable case in Andrica [24] πk+1,n (z) =

z(1 − z) 0 πk,n (z) + zπk,n (z), n

for all n ∈ N, z ∈ C and k = 0, 1, .... Since the relationship in Andrica [24] proved for the real case is a simple algebraic manipulation, it is valid for complex variable as well. Taking into account that the paper Andrica [24] is less accessible, let us reproduce here the idea of proof. It consists of the simple algebraic relationship 0 Sk,n (z) =

Sk+1,n (z) Sk,n (z) −n , z(1 − z) 1−z

which is divided by nk , where Sk,n (z) =

n X j=0

jk

n j z (1 − z)n−j . j

(Note that the cases z = 0 and z = 1 are trivial in the recurrence for πk,n (z).) From this recurrence, we easily obtain that degree (πk,n (z)) = min{n, k} ≤ k. Also, it easily implies the next recurrence πk,n (z) − z k =

(k − 1)z k−1 (1 − z) z(1 − z) [πk−1,n (z) − z k−1 ]0 + + z[πk−1,n (z) − z k−1 ]. n n

Denoting with k · kr the norm in C(Dr ), where Dr = {z ∈ C; |z| ≤ r}, one observes that by a linear transformation the Bernstein’s inequality in the closed unit disk becomes |Pk0 (z)| ≤ kr kPk kr , for all |z| ≤ r, r ≥ 1, where Pk represents an algebraic

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polynomial of degree ≤ k. Therefore, from the above recurrence we get

r(1 + r) kπk−1,n − ek−1 (z)kr rn rk−1 (1 + r)(k − 1) + + r|πk−1,n (z) − ek−1 (z)| n (1 + r) · [kπk−1,n kr + kek−1 kr ] ≤ (k − 1) n rk−1 (1 + r)(k − 1) + + r|πk−1,n (z) − ek−1 (z)| n ≤ r|πk−1,n (z) − ek−1 (z)| k−1 . +[2(1 + r)rk−1 + (1 + r)rk−1 ] n

|πk,n (z) − ek (z)| ≤ (k − 1)

Above we used that for all k, n ∈ N and |z| ≤ r, r ≥ 1, we have |πk,n (z)| ≤ rk (see relation (4) in the proof of Theorem 4.1.1 in Lorentz[125], p. 88) and |ek (z)| ≤ rk . Now, by taking k = 1, 2, ..., in the inequality |πk,n (z) − ek (z)| ≤ r|πk−1,n (z) − ek−1 (z)| + 3(1 + r)r k−1

k−1 , n

by recurrence we easily obtain the required inequality 3(1 + r) k−1 [r + 2rk−1 + ... + (k − 1)r k−1 ] n 3r(1 + r) 3(1 + r) k(k − 1) k−1 · r ≤ · k(k − 1)rk−2 . = n 2 2n This immediately implies the estimate in (i). P∞ k Note that since by hypothesis, f (z) = k ck z is absolutely and uniformly convergent in |z| ≤ r, for any 1 ≤ r < R, it follows that the power series obtained by P∞ differentiating twice, i.e. f 00 (z) = k=2 k(k−1)ck z k−2 , also is absolutely convergent P∞ for |z| ≤ r, which implies k=2 k(k − 1)|ck |rk−2 < +∞. (ii) Denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z p! Bn (f )(v) − f (v) (p) (p) dv |Bn (f )(z) − f (z)| = 2π γ (v − z)p+1 |πk,n (z) − ek (z)| ≤

≤

which proves the theorem.

Cr1 (f ) p! Cr (f ) 2πr1 p!r1 = 1 , n 2π (r1 − r)p+1 n (r1 − r)p+1

Remarks. 1) An analogue to Theorem 1.1.2, (i), case r = 1, has been obtained by a different method in Ostrovska [146]. P∞ 2) Let us give a proof of the relationship Bn (f )(z) = k=0 ck Bn (ek )(z) used Pm at the beginning of the proof of Theorem 1.1.2, (i). Denoting fm (z) = j=0 cj z j , |z| ≤ r, m ∈ N, since from the linearity of Bn we obviously have Bn (fm )(z) =

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Pm

k=0 ck Bn (ek )(z), it suffices to prove that for any fixed n ∈ N and |z| ≤ r with r ≥ 1, we have limm→∞ Bn (fm )(z) = Bn (f )(z). But this is immediate from limm→∞ kfm − f kr = 0 and from the inequality n X n |z k (1 − z)n−k | · kfm − f kr ≤ Mr,n kfm − f kr , |Bn (fm )(z) − Bn (f )(z)| ≤ k k=0

valid for all |z| ≤ r. In what follows we present the Voronovskaja-type formula with a quantitative upper estimate.

Theorem 1.1.3. (Gal [78]) Let R > 1 and suppose that f : DR → C is analytic in P∞ DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . (i) The following Voronovskaja-type result in the closed unit disk holds Bn (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ |z(1 − z)| · 10M (f ) , 2n 2n n P∞ for all n ∈ N, z ∈ D1 , where 0 < M (f ) = k=3 k(k − 1)(k − 2)2 |ck | < ∞. (ii) Let r ∈ [1, R). Then for all n ∈ N, |z| ≤ r, we have 2 Bn (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ 5(1 + r) · Mr (f ) , 2n 2n n P∞ 2 k−2 where Mr (f ) = k=3 |ck |k(k − 1)(k − 2) r < ∞.

Proof. (i) Denoting ek (z) = z k , k = 0, 1, ..., and πk,n (z) = Bn (ek )(z), we can P∞ write Bn (f )(z) = k=0 ck πk,n (z), which immediately implies Bn (f )(z) − f (z) − z(1 − z) f 00 (z) 2n ∞ X z k−1 (1 − z)k(k − 1) ≤ |ck | · πk,n (z) − ek (z) − , 2n k=3

for all z ∈ D1 , n ∈ N. In what follows, we will use the recurrence obtained in the proof of Theorem 1.1.2, (i) z(1 − z) 0 πk+1,n (z) = πk,n (z) + zπk,n (z), n for all n ∈ N, z ∈ C and k = 0, 1, .... If we denote z k−1 (1 − z)k(k − 1) Ek,n (z) = πk,n (z) − ek (z) − , 2n then it is clear that Ek,n (z) is a polynomial of degree ≤ k and by a simple calculation and the use of the above recurrence we obtain the following relationship z(1 − z) 0 Ek,n (z) = Ek−1,n (z) + zEk−1,n (z) n z k−2 (1 − z)(k − 1)(k − 2) + [(k − 2) − z(k − 1)], 2n2

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for all k ≥ 2, n ∈ N and z ∈ D1 . 0 According to the Bernstein’s inequality kEk−1,n k ≤ (k − 1)kEk−1,n k, the above relationship implies for all |z| ≤ 1, k ≥ 2, n ∈ N that |z| · |1 − z| 0 [2kEk−1,n k] 2n |z| · |1 − z| |z|k−3 (k − 1)(k − 2) +|Ek−1,n (z)| + · (2k − 3) 2n n 2k(k − 1)(k − 2) |z| · |1 − z| 0 2kEk−1,n k+ ≤ |Ek−1,n (z)| + 2n n |z| · |1 − z| 2k(k − 1)(k − 2) ≤ |Ek−1,n (z)| + 2(k − 1)kEk−1,n k + 2n n |z| · |1 − z| ≤ |Ek−1,n (z)| + [2(k − 1)kπk−1,n − ek−1 k 2n

(k − 1)(k − 2) [ek−2 − ek−1 ] 2k(k − 1)(k − 2)

+ , + 2(k − 1)

2n n |Ek,n (z)| ≤

where k · k denotes the uniform norm in C(D1 ). Also, taking r = 1 in the inequality obtained in the proof of Theorem 1.1.2, (i), it follows 3 kπk,n − ek k ≤ (k − 1)k. n As a consequence, we get |z| · |1 − z| 3(k − 1)(k − 2) |Ek,n (z)| ≤ |Ek−1,n (z)| + 2(k − 1) 2n n

(k − 1)(k − 2)[ek−2 − ek−1 ] 2k(k − 1)(k − 2)

+ + 2(k − 1) ,

2n n

which by simple calculation implies

|z| · |1 − z| 10 · · k(k − 1)(k − 2). 2n n Since E0,n (z) = E1,n (z) = E2,n (z) = 0, for any z ∈ C, it follows that the last inequality is trivial for k = 0, 1, 2. By writing the last inequality for k = 3, 4, ..., we easily obtain, step by step the following |Ek,n (z)| ≤ |Ek−1,n (z)| +

|Ek,n (z)| ≤

k |z| · |1 − z| 10 X |z| · |1 − z| 10 · · j(j − 1)(j − 2) ≤ · · k(k − 1)(k − 2)2 . 2n n j=3 2n n

In conclusion, Bn (f )(z) − f (z) − z(1 − z) f 00 (z) 2n ∞ ∞ X |z| · |1 − z| 10 X ≤ |ck | · |Ek,n (z)| ≤ · · |ck |k(k − 1)(k − 2)2 . 2n n k=3

k=3

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P Note that since f (4) (z) = ∞ 2)(k − 3)z k−4 , and the series is k=4 ck k(k − 1)(k −P 2 absolutely convergent in D1 , it easily follows that ∞ k=3 |ck |k(k − 1)(k − 2) < ∞. (ii) We will use the relationship obtained in the proof of Theorem 1.1.2, (i) |πk,n (z) − ek (z)| ≤

3r(1 + r) · k(k − 1)rk−2 , 2n

for all k, n ∈ N, |z| ≤ r, with 1 ≤ r. Let us consider the relationship proved at the above point (i) given by Ek,n (z) =

z(1 − z) 0 Ek−1,n (z) + zEk−1,n (z) n z k−2 (1 − z)(k − 1)(k − 2) + [(k − 2) − z(k − 1)], 2n2

for all k ≥ 2, n ∈ N and z ∈ C, and let us restrict it only for |z| ≤ r. For all k, n ∈ N, k ≥ 2 and |z| ≤ r, it implies |Ek,n (z)| ≤

r(1 + r) 0 |Ek−1,n (z)| + r|Ek−1,n (z)| n (1 + r)rk−2 (k − 1)(k − 2) [(k − 2) + r(k − 1)]. + 2n2

0 Now we will estimate |Ek−1,n (z)|, for k ≥ 3. Taking into account that Ek−1,n (z) is a polynomial of degree ≤ (k − 1), we obtain

k−1 0 |Ek−1,n (z)| ≤ kEk−1,n (z)kr r

(k − 1)(k − 2)(ek−1 − ek−2 ) k−1

≤ kπk−1,n − ek−1 kr +

r 2n r k − 1 3r(1 + r)(k − 1)(k − 2)r k−3 rk−2 (r + 1)(k − 1)(k − 2) ≤ + r 2n 2n k(k − 1)(k − 2) ≤ 3(1 + r)rk−3 + rk−3 (r + 1) 2n 2k(k − 1)(k − 2)(1 + r)r k−3 . ≤ n This implies r(1 + r) 0 2r(1 + r)2 k(k − 1)(k − 2)r k−3 |Ek−1,n (z)| ≤ , n n2 and 2r(1 + r)2 k(k − 1)(k − 2)r k−3 n2 k−2 (1 + r)(k − 1)(k − 2)r + [(k − 2) + r(k − 1)] = r|Ek−1,n (z)| 2n2 (1 + r)(k − 1)(k − 2)r k−2 + [4k(1 + r) + (k − 2) + r(k − 1)] 2n2

|Ek,n (z)| ≤ r|Ek−1,n (z)| +

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(1 + r)(k − 1)(k − 2)r k−2 [(5k − 2) + r(5k − 1)] 2n2 (1 + r)(k − 1)(k − 2)r k−2 5k(1 + r) ≤ r|Ek−1,n (z)| + 2n2 2 5(1 + r) k(k − 1)(k − 2)r k−2 = r|Ek−1,n (z)| + . 2n2 But E0,n (z) = E1,n (z) = E2,n (z) = 0, for any z ∈ C. By writing the last inequality for k = 3, 4, ..., we easily obtain, step by step the following   k 5(1 + r)2 rk−2 X j(j − 1)(j − 2) |Ek,n (z)| ≤ 2n2 j=3 = r|Ek−1,n (z)| +

≤

5(1 + r)2 k(k − 1)(k − 2)2 rk−2 . 2n2

As a conclusion, we obtain X ∞ Bn (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ |ck | · |Ek,n (z)| 2n k=3

∞ 5(1 + r)2 X ≤ |ck |k(k − 1)(k − 2)2 rk−2 . 2n2 k=3 P (4) Note that since f (z) = ∞ 3)z k−4 , and the series k=4 ck k(k − 1)(k − 2)(k − P ∞ is absolutely convergent in |z| ≤ r, it easily follows that k=3 |ck |k(k − 1)(k − 2 k−2 2) r < ∞. Therefore the theorem has been proved.

Remark. By Gonska-Pit¸ul-Ra¸sa [102], p. 68, Proposition 7.2, for the real Bernstein polynomials n X n k Bn (f )(x) = x (1 − x)n−k f (k/n), x ∈ [0, 1] k k=0

attached to a function f ∈ C 2 [0, 1], for all x ∈ [0, 1] and n ∈ N it holds 1 Bn (f )(x) − f (x) − x(1 − x) f 00 (x) ≤ x(1 − x) ω ˜ 1 (f 00 ; √ ), 2n 2n 3 n

where ω ˜ 1 denotes the least concave majorant of the modulus of continuity ω1 and C 2 [0, 1] = {f : [0, 1] → R; f is twice continuously differentiable on [0, 1]}. Now, if f ∈ C 3 [0, 1] then we immediately get that the best quantitative uniform estimate in the real Voronovskaja’s result is of order O(1/n3/2 ), which is essentially worst that the order O(1/n2 ) in Theorem 1.1.3. This suggest that in the real case, the order of approximation could be improved, for example that maybe ω ˜ 1 could be replaced by ω2 .

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In what follows we will prove that the orders of approximation in Theorem 1.1.2, (i) and (ii) are exactly n1 . We present Theorem 1.1.4. (Gal [79]) Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose P∞ that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . If f is not a polynomial of degree ≤ 1, then for any r ∈ [1, R) we have kBn (f ) − f kr ≥

Cr (f ) , n ∈ N, n

where kf kr = max{|f (z)|; |z| ≤ r} and the constant Cr (f ) depends only on f and r. Proof.

For all z ∈ DR and n ∈ N we have

Bn (f )(z) − f (z) 1 z(1 − z) 00 1 2 z(1 − z) 00 = n Bn (f )(z) − f (z) − f (z) + f (z) . n 2 n 2n Since by hypothesis f 00 (z) is not identical zero in DR , there exists 0 < r0 < 1 such that M0 = inf |z|=r0 |f 00 (z)| > 0. Indeed, let us suppose the contrary. Then, choosing a sequence 0 < rn < 1, n ∈ N such that rn & 0, the continuity of f 00 on the compact set {z ∈ C; |z| = rn }, implies that there exists zn with |zn | = rn and f 00 (zn ) = 0. It follows zn → 0 and by the continuity of f 00 in DR we get f 00 (0) = 0. The analyticity of f 00 implies that 0 is an isolated zero, therefore there exists r 0 > 0 such that f 00 (z) 6= 0 for all z ∈ Dr0 , z 6= 0. But this is a contradiction because for sufficiently large n we have zn ∈ Dr0 . Now let r ≥ 1 be arbitrary. We obviously have kBn (f ) − f kr ≥ kBn (f ) − f kr0 and by the Maximum Principle, there exists a point z0 (depending on n, f and r0 ) with |z0 | = r0 , such that kBn (f ) − f kr0 = |Bn (f )(z0 ) − f (z0 )|. We get 1 z0 (1 − z0 ) 00 f (z0 ) kBn (f ) − f kr ≥ |Bn (f )(z0 ) − f (z0 )| = n 2 z0 (1 − z0 ) 00 1 2 f (z0 ) n Bn (f )(z0 ) − f (z0 ) − + n 2n 1 2 1 z0 (1 − z0 ) 00 z0 (1 − z0 ) 00 ≥ n Bn (f )(z0 ) − f (z0 ) − f (z0 ) − f (z0 ) . n 2 n 2n 0 ) 00 0) But z0 (1−z f (z0 ) ≥ r0 (1−r M0 > 0 and by Theorem 1.1.3 we have 2 2 z0 (1 − z0 ) 00 2 f (z0 ) n Bn (f )(z0 ) − f (z0 ) − 2n e1 (1 − e1 ) 00 ≤ n2 kBn (f ) − f − f kr 2n 5Kr (f )(1 + r)2 5Kr (f )(1 + r)2 ≤ n2 = . 2n2 2

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Therefore, there exists an index n0 depending only on f and r, such that for all n ≥ n0 we have z0 (1 − z0 ) 00 1 2 z0 (1 − z0 ) 00 n f (z ) − f (z ) B (f )(z ) − f (z ) − 0 0 n 0 0 2 n 2n r0 (1 − r0 ) ≥ M0 > 0, 4 which immediately implies kBn (f ) − f kr ≥

1 r0 (1 − r0 ) · M0 , ∀n ≥ n0 . n 4 M

(f )

For n ∈ {1, 2, ..., n0 − 1} we obviously have kBn (f ) − f kr ≥ r,n with Mr,n (f ) = n Cr (f ) n · kBn (f ) − f kr > 0, which finally implies kBn (f ) − f kr ≥ n for all n ∈ N, 0) where Cr (f ) = min{Mr,1 , Mr,2 (f ), ..., Mr,n0 −1 (f ), r0 (1−r M0 }. 4 Combining now Theorem 1.1.4 with Theorem 1.1.2, (i), we immediately get the following. Corollary 1.1.5. (Gal [79]) Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If f is not a polynomial of degree ≤ 1, then for any r ∈ [1, R) we have kBn (f ) − f kr ∼

1 , n ∈ N, n

where the constants in the equivalence depend on f and r. In the case of simultaneous approximation we present the following. Theorem 1.1.6. (Gal [79]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR . Also, let 1 ≤ r < r1 < R and p ∈ N be fixed. If f is not a polynomial of degree ≤ max{1, p − 1}, then we have kBn(p) (f ) − f (p) kr ∼

1 , n

where the constants in the equivalence depend on f , r, r1 and p. Proof. Taking into account the upper estimate in Theorem 1.1.2, (ii), it remains (p) to prove the lower estimate for kBn (f )−f (p) kr . Firstly, denoting by Γ the circle of radius r1 > and center 0 (where r1 > r ≥ 1), we have the inequality |v − z| ≥ r1 − r valid for all |z| ≤ r and v ∈ Γ. As in the proof of Theorem 1.1.4, for all v ∈ Γ and n ∈ N we have Bn (f )(v) − f (v) v(1 − v) 00 1 v(1 − v) 00 1 2 n Bn (f )(v) − f (v) − = f (v) + f (v) , n 2 n 2n

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which replaced in the Cauchy’s formula for derivatives implies Z v(1 − v)f 00 (v) p! 1 (p) (p) dv Bn (f )(z) − f (z) = n 2πi Γ 2(v − z)p+1  Z n2 Bn (f )(v) − f (v) − v(1−v) f 00 (v)  2n 1 p! + · dv p+1  n 2πi Γ (v − z) ( (p) z(1 − z) 00 1 f (z) = n 2  Z n2 Bn (f )(v) − f (v) − v(1−v) f 00 (v)  2n 1 p! dv . + ·  n 2πi Γ (v − z)p+1 Passing now to absolute value, for all |z| ≤ r and n ∈ N it follows ( (p) z(1 − z) 1 |Bn(p) (f )(z) − f (p) (z)| ≥ f 00 (z) n 2   Z n2 Bn (f )(v) − f (v) − v(1−v) f 00 (v) 2n 1 p! dv , − p+1 n 2π Γ (v − z)  where by using Theorem 1.1.3, (ii), for all |z| ≤ r and n ∈ N we get Z n2 Bn (f )(v) − f (v) − v(1−v) f 00 (v) p! 2n dv 2π (v − z)p+1 Γ

2πr1 n2 e1 (1 − e1 ) 00 p! · kBn (f ) − f − f k r1 p+1 2π (r1 − r) 2n p!r1 5Kr1 (f )(1 + r1 )2 · . ≤ 2 (r1 − r)p+1 h i(p) 00 Denoting now Fp (z) = z(1−z) , by the hypothesis on f it follows that Fp f (z) 2 is analytic and is not identically zero in DR . Reasoning exactly as in the proof of Theorem 1.1.4, there exists 0 < r0 < 1 such that C0 = inf |z|=r0 |Fp (z)| > 0. Continuing exactly as in the proof of Theorem 1.1.4 (with kBn (f ) − f kr replaced (p) by kBn (f ) − f (p) kr ), finally there exists an index n0 ∈ N depending on f , r, r1 and p, such that for all n ≥ n0 we have 1 C0 kBn(p) (f ) − f (p) kr ≥ · . n 2 The cases when n ∈ {1, 2, ..., n0 − 1} are similar with those in the proof of Theorem 1.1.4. ≤

Remark. Let us suppose that f (p) ∈ C[0, 1], p ∈ N. By taking r = 1 and λ = 1 in Xie [203], Theorem 2, we immediately obtain the following upper estimate for the derivatives of the real Bernstein polynomials attached to f , valid for all n ≥ n p √ kBn(p) (f ) − f (p) k ≤ Ap [ω1 (f (p) ; 1/n) + ω2ϕ (f (p) ; 1/ n) + kf (p) k/n],

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where k · k denotes the uniform norm on C[0, 1], np p ∈ N depends only on p, ω1 denotes the uniform modulus of continuity, ϕ(x) = x(1 − x) and ω2ϕ denotes the Ditzian-Totik second order modulus of smoothness defined in Ditzian-Totik [64]. Then, the above Theorem 1.1.6 suggests the following open question : for any p ∈ N, there exist the positive constants Cp and np depending only on p, such that for all n ≥ np √ Cp [ω1 (f (p) ; 1/n) + ω2ϕ (f (p) ; 1/ n) + kf (p) k/n] ≤ kBn(p) (f ) − f (p) k. The geometric properties of Bernstein polynomials are consequences of Theorem 1.1.2 and can be expressed by the following.

Theorem 1.1.7. (Gal [77], pp. 268-269, Theorem 3.4.2) Let us suppose that G ⊂ C is open, such that D1 ⊂ G and f : G → C is analytic in G. (i) If f is univalent in D1 , then there exists an index n0 depending on f , such Pn that for all n ≥ n0 , the complex Bernstein polynomials Bn (f )(z) = k=0 nk z k (1 − z)n−k f (k/n) are univalent in D1 . (ii) If f (0) = f 0 (0) − 1 = 0 and f is starlike in D1 , that is 0 zf (z) Re > 0, for all z ∈ D1 , f (z) then there exists an index n0 depending on f , such that for all n ≥ n0 , the complex Bernstein polynomials are starlike in D1 . If f (0) = f 0 (0) − 1 = 0 and f is starlike only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, Dr ), such that for all n ≥ n0 , the complex Bernstein polynomials Bn (f )(z) are starlike in Dr , that is, 0 zBn (f )(z) Re > 0, for all z ∈ Dr . Bn (f )(z) (iii) If f (0) = f 0 (0) − 1 = 0 and f is convex in D1 , that is 00 zf (z) Re + 1 > 0, for all z ∈ D1 , f 0 (z) then there exists an index n0 depending on f , such that for all n ≥ n0 , the complex Bernstein polynomials are convex in D1 . If f (0) = f 0 (0) − 1 = 0 and f is convex only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, Dr ), such that for all n ≥ n0 , the complex Bernstein polynomials Bn (f )(z) are convex in Dr , that is, 00 zBn (f )(z) Re + 1 > 0, for all z ∈ Dr . Bn0 (f )(z) (iv) If f (0) = f 0 (0) − 1 = 0, f (z) 6= 0, for all z ∈ D1 \ {0} and f is spirallike of type γ ∈ (−π/2, π/2) in D1 , that is zf 0 (z) Re eiγ > 0, for all z ∈ D1 , f (z)

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then there exists an index n0 depending on f and γ, such that for all n ≥ n0 we have Bn (f )(z) 6= 0, for all z ∈ D1 \ {0}, and Bn (f )(z) are spirallike of type γ in D1 . If f (0) = f 0 (0) − 1 = 0, f (z) 6= 0 for all z ∈ D1 \ {0} and f is spirallike of type γ only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, Dr , γ), such that for all n ≥ n0 , the Bernstein polynomials Bn (f )(z) 6= 0 for all z ∈ Dr \ {0} and they are spirallike of type γ in Dr , that is, zB 0 (f )(z) Re eiγ n > 0, for all z ∈ Dr . Bn (f )(z) Proof. (i) It is immediate from the uniform convergence in Theorem 1.1.2 and a well-known result concerning sequences of analytic functions converging locally uniformly to an univalent function (see e.g. Kohr-Mocanu [118], p. 130, Theorem 4.1.17 or Graham-Kohr [105], Theorem 6.1.18). For the proof of next points (ii), (iii) and (iv), let us observe that by Theorem 1.1.2, (i) and (ii) we get that for n → ∞, we have Bn (f )(z) → f (z), Bn0 (f )(z) → f 0 (z) and Bn00 (f )(z) → f 00 (z), uniformly in D1 . In all what follows, )(z) denote Pn (f )(z) = Bnfn (f (1/n) . By f (0) = f 0 (0)−1 = 0 and the univalence of f , we get nf (1/n) 6= 0, Pn (f )(0) = 0 Bn (f )(0) f (0) f (1/n)−f (0) 0 converges to nf (1/n) = 0, P (f )(0) = nf (1/n) = 1, n ≥ 2, nf (1/n) = 1/n 0 f (0) = 1 as n → ∞, which means that for n → ∞, we have Pn (f )(z) → f (z), Pn0 (f )(z) → f 0 (z) and Pn00 (f )(z) → f 00 (z), uniformly in D1 . (ii) By hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Writing Pn (f )(z) in the form Pn (f )(z) = zQn (f )(z), obviously Qn (f )(z) is a polynomial of degree ≤ n − 1. Let |z| = 1. We have |f (z) − Pn (f )(z)| = |z| · |g(z) − Qn (f )(z)| = |g(z) − Qn (f )(z)|,

which by the uniform convergence in D1 of Pn (f ) to f and by the maximum modulus principle, implies the uniform convergence in D1 of Qn (f )(z) to g(z). Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index n1 ∈ N and a > 0 depending on g, such that |Qn (f )(z)| > a > 0, for all z ∈ D1 and all n ≥ n0 . Also, for all |z| = 1, we have |f 0 (z) − Pn0 (f )(z)| = |z[g 0 (z) − Q0n (f )(z)] + [g(z) − Qn (f )(z)]|

≥ | |z| · |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| | = | |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| |,

which from the maximum modulus principle, the uniform convergence of Pn0 (f ) to f 0 and of Qn (f ) to g, evidently implies the uniform convergence of Q0n (f ) to g 0 .

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Then, for |z| = 1, we get

z[zQ0n(f )(z) + Qn (f )(z)] zPn0 (f )(z) = Pn (f ) zQn (f )(z) 0 zQn (f )(z) + Qn (f )(z) zg 0 (z) + g(z) = → Qn (f )(z) g(z) 0 0 f (z) zf (z) = = , g(z) f (z)

which again from the maximum modulus principle, implies

Since Re

zf 0 (z) f (z)

zPn0 (f )(z) zf 0 (z) → , uniformly in D1 . Pn (f ) f (z) is continuous in D1 , there exists α ∈ (0, 1), such that 0 zf (z) Re ≥ α, for all z ∈ D1 . f (z)

Therefore Re

0 zPn0 (f )(z) zf (z) → Re ≥α>0 Pn (f )(z) f (z)

uniformly on D1 , i.e. for any 0 < β < α, there is n0 such that for all n ≥ n0 we have 0 zPn (f )(z) Re > β > 0, for all z ∈ D1 . Pn (f )(z) Since Pn (f )(z) differs from Bn (f )(z) only by a constant, this proves the first part in (ii). For the second part, the proof is identical with the first part, with the only difference that instead of D1 , we reason for Dr . (iv) Obviously we have zP 0 (f )(z) zf 0 (z) Re eiγ n → Re eiγ , Pn (f )(z) f (z) uniformly in D1 . We also note that since f is univalent in D1 , by the above point (i), there exists n1 such that Bn (f )(z) is univalent in D1 for all n ≥ n1 , which by Bn (f )(0) = 0 implies Bn (f )(z) 6= 0, for all z ∈ D1 \ {0}, n ≥ n1 . For the rest, the proof is identical with that from the above point (ii). (iii) For the first part, by hypothesis there is α ∈ (0, 1), such that 00 zf (z) Re + 1 ≥ α > 0, f 0 (z) uniformly in D1 . It is not difficult to show that this is equivalent with the fact that for any β ∈ (0, α), the function zf 0 (z) is starlike of order β in D1 (see e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138], p. 77), which implies f 0 (z) 6= 0, for all z ∈ D1 ,

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i.e. |f 0 (z)| > 0, for all z ∈ D1 . Also, by the same type of reasonings as those from the above point (ii), we get 00 00 zPn (f )(z) zf (z) Re + 1 → Re + 1 ≥ α > 0, Pn0 (f )(z) f 0 (z) uniformly in D1 . As a conclusion, for any 0 < β < α, there is n0 depending on f , such that for all n ≥ n0 we have 00 zPn (f )(z) Re + 1 > β > 0, for all z ∈ D1 . Pn0 (f )(z) The proof of second part in (iii) is similar, which proves the theorem. 1.1.2

Bernstein-Faber Polynomials on Compact Sets

˜ \ G is In this subsection G ⊂ C will be considered a compact set such that C connected. In this case, according to the Riemann Mapping Theorem, a unique 0 ˜ ˜ conformal mapping Ψ of C\D 1 onto C\G exists so that Ψ(∞) = ∞ and Ψ (∞) > 0. By using the Faber polynomials Fp (z) attached to G (see Definition 1.0.10), for f ∈ A(G) we can introduce the Bernstein-Faber polynomials given by the formula n X n Bn (f ; G)(z) = ∆p1/n F (0) · Fp (z), z ∈ G, n ∈ N, p p=0 where ∆ph F (0)

=

p X k=0

(−1)

p−k

Z f (Ψ(u)) p 1 du, w ∈ D1 . F (kh), F (w) = 2πi |u|=1 u − w k

Here, since F (1) is involved in ∆n1/n F (0) and therefore in the definition of Bn (f ; G)(z) too, in addition we will suppose that F can be extended by continuity on the boundary ∂D1 . Remarks. 1) For G = D1 it is easy to see that the above Bernstein-Faber polynomials one reduce to the classical complex Bernstein polynomials given by n n X X n n p Bn (f )(z) = ∆p1/n f (0)z p = z (1 − z)n−p f (p/n). p p p=0 p=0

R 1 ω (f ◦Ψ;u) 2) It is known that, for example, 0 p u ∂D1 du < ∞ is a sufficient condition for the continuity on ∂D1 of F in the above definition of the Bernstein-Faber polynomials (see e.g. Gaier [76], p. 52, Theorem 6). Here p ∈ N is arbitrary fixed. The first main result one refers to approximation on compact sets without any restriction on their boundaries and can be stated as follows. Theorem 1.1.8. Let G be a continuum (that is a connected compact subset of C) and suppose that f is analytic in G, that is there exists R > 1 such that f is analytic in GR . Here recall that GR denotes the interior of the closed level curve ΓR given

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by ΓR = {z; |Φ(z)| = R} = {Ψ(w); |w| = R} (and that G ⊂ Gr for all 1 < r < R). Also, we suppose that F given in the definition of Bernstein-Faber polynomials can be extended by continuity on ∂D1 . For any 1 < r < R the following estimate |Bn (f ; G)(z) − f (z)| ≤

C , z ∈ Gr , n ∈ N, n

holds, where C > 0 depends on f , r and Gr but it is independent of n. ˜ \ G is Proof. First we note that since G is a continuum then it follows that C simply connected. By the proof of Theorem 2, p. 52 in Suetin [186], for any fixed P∞ 1 < β < R we have f (z) = k=0 ak (f )Fk (z) uniformly in Gβ , where ak (f ) are R f [Ψ(u)] 1 the Faber coefficients and are given by ak (f ) = 2πi |u|=β uk+1 du. Note here that G ⊂ Gβ . First we will prove that

Bn (f ; G)(z) =

∞ X

ak (f )Bn (Fk ; G)(z),

k=0

for all z ∈ G. (Note here that by hypothesis we have G = G). For this purpose, Pm denote fm (z) = k=0 ak (f )Fk (z), m ∈ N. Since by the linearity of Bn we easily get Bn (fm ; G)(z) =

m X k=0

ak (f )Bn (Fk ; G)(z), for all z ∈ G,

it suffices to prove that limm→∞ Bn (fm ; G)(z) = Bn (f ; G)(z), for all z ∈ G and n ∈ N. First we have n X n Bn (fm ; G)(z) = ∆p1/n Gm (0)Fk (z), p p=0

R R fm (Ψ(u)) f (Ψ(u)) 1 1 where Gm (w) = 2πi u−w du and F (w) = 2πi |u|=1 u−w du. |u|=1 Note here that since by Gaier [76], p. 48, first relation before (6.17), we have Z 1 Fk (Ψ(u)) Fk (w) = du = wk , for all |w| < 1, 2πi |u|=1 u − w

evidently that Fk (w) can be by continuity on ∂D1 . This also immediately R extended fm (Ψ(u)) 1 implies that Gm (w) = 2πi u−w du can be extended by continuity on ∂D1 , |u|=1 which means that Bn (Fk ; G)(z) and Bn (fm ; G)(z) are well defined. Now, taking into account the Cauchy’s theorem we also can write Z Z 1 1 fm (Ψ(u)) f (Ψ(u)) du and F (w) = du. Gm (w) = 2πi |u|=β u − w 2πi |u|=β u − w

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For all n, m ∈ N and z ∈ G it follows |Bn (fm ; G)(z) − Bn (f ; G)(z)| n X n ≤ |∆p1/n (Gm − F )(0)| · |Fk (z)| p p=0 p n X X n p ≤ |(Gm − F )((p − j)/n)| · |Fk (z)| p j p=0 j=0 p n X X n p ≤ Cj,p,β kfm − f kGβ · |Fk (z)| p j p=0 j=0

≤ Mn,p,β,Gβ kfm − f kGβ ,

which by limm→∞ kfm − f kGβ = 0 (see e.g. the proof of Theorem 2, p. 52 in Suetin [186]) implies the desired conclusion. Here kfm − f kGβ denotes the uniform norm of fm − f on Gβ . Consequently we obtain ∞ X |Bn (f ; G)(z) − f (z)| ≤ |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| =

k=0 n X

k=0

+

|ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ∞ X

k=n+1

|ak (f )| · |Bn (Fk ; G)(z) − Fk (z)|.

Therefore it remains to estimate |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)|, firstly for all 0 ≤ k ≤ n and secondly for k ≥ n + 1, where n X n Bn (Fk ; G)(z) = [∆p1/n Fk (0)] · Fp (z). p p=0

First it is useful to observe that by Gaier [76], p. 48, combined with the Cauchy’s theorem, for any fixed 1 < β < R we have Z 1 Fk [Ψ(u)] Fk (w) := du = wk = ek (w), for all |w| < β. 2πi |u|=β u − w Denote

Dn,p,k It follows

n n p = ∆1/n ek (0) = [0, 1/n, ..., p/n; ek ] · (p!)/np . p p Bn (Fk ; G)(z) =

n X p=0

Dn,p,k · Fp (z).

Since for the classical complex Bernstein polynomials attached to a disk of center Pn in origin we can write Bn (ek )(z) = p=0 Dn,p,k z p , since each ek is convex of any

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order and Bn (ek )(1) = ek (1) = 1 for all k, it follows that all Dn,p,k ≥ 0 and Pn n(n−1)...(n−k+1) . p=0 Dn,p,k = 1, for all k and n. Also, note that Dn,k,k = nk In the estimation of |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| we distinguish two cases : 1) 0 ≤ k ≤ n ; 2) k > n. Case 1. We have |Bn (Fk ; G)(z) − Fk (z)| ≤ |Fk (z)| · |1 − Dn,k,k | +

k−1 X p=0

Dn,p,k · |Fp (z)|.

Fix now 1 < r < β. By the inequality (13), p. 44 in Suetin [186] we have |Fp (z)| ≤ C(r)rp , for all z ∈ Gr , p ≥ 0, which immediately implies |Bn (Fk ; G)(z) − Fk (z)| ≤ 2C(r)[1 − Dn,k,k ]rk ≤ C(r) for all z ∈ Gr . Here we used the inequality 1 − Πki=1 xi ≤ xi ∈ [0, 1]) which implies the inequality

Pk

k(k − 1) k r , n

i=1 (1

− xi ) (valid if all

n(n − 1)...(n − k + 1) k−1 n − i = 1 − Πi=1 k n n k−1 k−1 X X 1 k(k − 1) ≤ (1 − (n − i)/n) = i= . n 2n i=1 i=0

1 − Dn,k,k = 1 −

Also by the above formula for ak (f ) we easily obtain |ak (f )| ≤ Note that C(r), C(β, f ) > 0 are constants independent of k. For all z ∈ Gr and k = 0, 1, 2, ...n it follows |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ≤

C(β,f ) , βk

for all k ≥ 0.

k r C(r, β, f ) k(k − 1) , n β

that is n n X C(r, β, f ) X k(k − 1)dk , for all z ∈ Gr , |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ≤ n k=2 k=0 Pn P k where 0 < d = r/β < 1. Also, clearly we have k=2 k(k−1)dk ≤ ∞ k=2 k(k−1)d < ∞ which finally implies that n X

k=0

|ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ≤

C ∗ (r, β, f ) . n

Case 2. We have ∞ ∞ X X |ak (f )| · |Bn (Fk ; G)(z)| |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ≤ k=n+1

k=n+1 ∞ X

+

k=n+1

|ak (f )| · |Fk (z)|.

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By the estimates mentioned in the case 1), we immediately get ∞ X

k=n+1

|ak (f )| · |Fk (z)| ≤ C(r, β, f )

k=n+1

with d = r/β. Also, ∞ X

k=n+1

|ak (f )| · |Bn (Fk ; G)(z)| = ≤

∞ X

∞ X

k=n+1 ∞ X

k=n+1

dk , for all z ∈ Gr ,

n X |ak (f )| · Dn,p,k · Fp (z) p=0

|ak (f )| ·

n X p=0

Dn,p,k · |Fp (z)|.

But for p ≤ n < k and taking into account the estimates obtained in the Case 1) we get |ak (f )| · |Fp (z)| ≤ C(r, β, f )

rk rp ≤ C(r, β, f ) , for all z ∈ Gr , βk βk

which implies k r Dn,p,k |ak (f )| · |Bn (Fk ; G)(z) − Fk (z)| ≤ C(r, β, f ) β k=n+1 p=0 k=n+1 ∞ k X r = C(r, β, f ) β ∞ X n X

∞ X

k=n+1 n+1

= C(r, β, f )

d , 1−d

with d = r/β. In conclusion, collecting the estimates in the Cases 1) and 2) we obtain |Bn (f ; G)(z) − f (z)| ≤ This proves the theorem.

C C1 + C2 dn+1 ≤ , z ∈ Gr , n ∈ N. n n

Remarks. 1) Simultaneous upper and lower estimates in Theorem 1.1.8 (that is a similar result to Corollary 1.1.5) can easily be obtained under some restrictions on the boundaries. For that purpose firstly we recall some useful concepts and results. The first important concept is that of Faber set. Thus, suppose that G ⊂ C is compact. If the Faber mapping T (given by Definition 1.0.10, (ii) ) defined on the set of all polynomials P(D1 ) and with values in the set of polynomials P(G) is continuous, then G is called a Faber set. In this case, T admits a unique extension to a linear and bounded mapping between the Banach spaces A(D 1 ) and A(G) (see the Remark after Definition 1.0.10, or for full details see e.g. the book of Gaier [76], pp. 48-49). For example, if G is a compact set which is a Jordan domain whose boundary Γ is a rectifiable curve of bounded rotation, then G is a Faber set (see

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e.g. Gaier [75], p. 51, Theorem 2). Also, Theorem 1 in Frerick-M¨ uller [74] gives other sufficient conditions on the boundary of G which assures that the compact set G is a Faber set. As a consequence, a compact set G whose boundary consists of piecewise convex curves also is a Faber set (see Frerick-M¨ uller [74], p. 429). By Lemma 1 in Frerick-M¨ uller [74], if G is a compact Faber set then the Faber mapping T : A(D1 ) → A(G) is injective. 2) Another important concept is that of inverse Faber set. Thus, according to Anderson-Clunie [22], p. 546, a Faber set G is called inverse Faber set if the Faber operator T is bijective, which implies that T −1 : A(G) → A(D1 ) given by (see Theorem 1.0.12 or e.g. Anderson-Clunie [22], relation (1.2) ) T

−1

1 (f )(ξ) = 2πi

Z

|w|=1

f [Ψ(w)] dw, w−ξ

also is linear and bounded. An important result is Theorem 2 in Anderson-Clunie [22], p. 548, which says that if G is the closure of a Jordan domain whose boundary Γ is rectifiable and of boundary rotation, and in addition Γ is free of cups, then G is an inverse Faber set. Let us also recall that if the compact set G is a Faber set, then for any 1 < r, Gr is an inverse Faber set, where Gr denotes the closure of the Jordan domain bounded by the analytic simple curve Γr = {Ψ(w); |w| = r} (see the Remark on page 434 in Frerick-M¨ uller [74] or Anderson-Clunie [22]). Also, in this case by Theorem 3 in Frerick-M¨ uller [74], for f ∈ A(Gr ) we have T −1 (f ) ∈ A(Dr ). As a consequence of the considerations in the above two remarks, we can state the following result. ˜ \ G is simply connected. Theorem 1.1.9. Let G be a compact Faber set such that C If f is analytic on G, that is there exists R > 1 such that f is analytic in G R and if f is not a polynomial of degree ≤ 1, then for any 1 < r < R we have kBn (f ; G) − f kGr ∼

1 , n ∈ N, n

where the constants in the equivalence depend on f , r and Gr but are independent of n. Here kf kGr = supz∈Gr |f (z)|. Proof. According to the above considerations, there exists g analytic in D r such that f = T (g), that is g = T −1 (f ) (therefore F can be extended by continuity on ∂D1 . By hypothesis on f it follows that f cannot be of the form f (z) = c0 F0 (z) + c1 F1 (z) where F0 and F1 are the Faber polynomials of degree 0 and 1 respectively and c0 , c1 ∈ C. This immediately implies that g is not a polynomial of degree ≤ 1. First we have Bn (T −1 (f )) = T −1 [Bn (f ; G)]. Indeed, Bn (T −1 (f ))(z) =

n X n p=0

p

∆p1/n T −1 (f )(0)z p =

n X n p=0

p

∆p1/n F (0)z p ,

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since T −1 (f )(ξ) = T

−1

1 2πi

R

|w|=1

f [Ψ(w)] w−ξ dw

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25

= F (ξ), and

Z

Bn (f ; G)[Ψ(w)] dw w−z |w|=1 Z n X 1 n Fp [Ψ(w)] ∆p1/n F (0) = dw p 2πi w−z |w|=1 p=0 n X n = ∆p1/n F (0)z p , p p=0

1 [Bn (f ; G)](z) = 2πi

since according to Gaier [76], p. 48, first relation before (6.17), we have Z 1 Fp [Ψ(w)] dw = z p . 2πi |w|=1 w − z Then by Corollary 1.1.5 and by the linearity and continuity of T −1 we get C ≤ kBn (g) − gkr = kBn (g) − T −1 (f )kr = kT −1 [Bn (f ; G)] − T −1 (f )kr n ≤ k|T −1 k| · kBn (f ; G) − f kGr ≤ M kBn (f ; G) − f kGr , which proves the lower estimate. On the other hand we have T [Bn (g)] = Bn (T (g); G). Indeed, T [Bn (g)](z) =

n X

∆p1/n g(0)Fp (z),

p=0

and Bn (T (g); G)(z) =

n X n p=0

p

∆p1/n H(0)Fp (z),

where according to Gaier [76], p. 49. relation (6.17’) we have Z T (g)[Ψ(u)] 1 du = g(w). H(w) = 2πi |u|=1 u − w Therefore by the same Corollary 1.1.5 and by the linearity and continuity of T we obtain kBn (f ; G) − f kGr = kBn (T (g); G) − T (g)kGr = kT [Bn (g)] − T (g)kGr C ≤ k|T k| · kBn (g) − gkr ≤ , n which proves the upper estimate and the theorem.

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1.2

Iterates of Bernstein Polynomials

First we deal with the approximation properties of the iterates of complex Bernstein polynomials and their relationship with the theory of the semigroups of operators. For R > 1, let us define by AR the space of all functions defined and analytic in the open disk of center 0 and radius R denoted by DR . Denoting rj = R − R−1 j , j ∈ N and for f ∈ AR , kf kj = max{|f (z)|; |z| ≤ rj }, since r1 = 1 and rj % R, it is well-known that {k · kj , j ∈ N} is a countable family of increasing semi-norms on AR and that AR becomes a metrizable complete locally convex space (Fr´echet space), with respect to the metric d(f, g) =

∞ X kf − gkj 1 · , f, g ∈ AR . j 1 + kf − gk 2 j j=1

It is well-known that limn→∞ d(fn , f ) = 0 is equivalent to the fact that the sequence (fn )n∈N converges to f uniformly on compacts in DR . Details about the space AR and the metric d can be found in e.g. Kohr-Mocanu [118], pp. 104-107. P∞ Now, for f ∈ AR , that is of the form f (z) = k=0 ck z k , for all z ∈ DR , let (1) us define the iterates of complex Bernstein polynomial Bn (f )(z), by Bn (f )(z) = (m) (m−1) Bn (f )(z) and Bn (f )(z) = Bn [Bn (f )](z), for any m ∈ N, m ≥ 2. Since we have (see e.g. Lorentz [125], p. 88, proof of Theorem 4.1.1 ), Bn (f )(z) = P∞ (m) ck Bn (ek )(z), by recurrence for all m ≥ 1, we easily get that Bn (f )(z) = Pk=0 (m) ∞ k k=0 ck Bn (ek )(z), with ek (z) = z . The first main result of this section is the following. P∞ Theorem 1.2.1. (Gal [78]) Let f ∈ AR with R > 1, that is f (z) = k=0 ck z k , for all z ∈ DR . (i) For any n ∈ N, we have lim d[Bn(m) (f ), B1 (f )] = 0;

m→∞

(ii) If limn→∞

mn n

= 0, then lim d[Bn(mn ) (f ), f ] = 0.

n→∞

Moreover, for any fixed q ∈ N, the following estimates hold ∞ mX |ck |k(k − 1)rqk , kBn(m) (f ) − f kq ≤ n k=2

and

d[Bn(m) (f ), f ] ≤ P∞

1)rqk

where k=2 |ck |k(k − < ∞. (iii) If limn→∞ mnn = ∞, then

∞

mX 1 |ck |k(k − 1)rqk + q , n 2 k=2

lim d[Bn(mn ) (f ), B1 (f )] = 0;

n→∞

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(iv) If limn→∞

mn n

27

= t ∈ (0, ∞), then lim d[Bn(mn ) (f ), T (t)(f )] = 0,

n→∞

where L(f )(z) = (1 − z)f (0) + zf (1), z ∈ DR , Z 1 T (t)(f )(z) = L(f )(z) + z(1 − z) Gt (z, y)[f (y) − L(f )(y)]dy, z ∈ DR , 0

Gt (z, y) =

∞ X k(2k − 1) k=2

k−1

(1,1)

(1,1)

e−k(k−1)t/2 Pk−2 (2z − 1)Pk−2 (2y − 1),

(1,1)

z ∈ DR , y ∈ [0, 1], and Pk−2 (z), |z| < R, are the Jacobi polynomials normalized to be k − 1 at z = 1. Proof. (i) Since from Karlin-Ziegler [114] it is known that if m → ∞, then (m) Bn (f )(x) → B1 (f )(x), uniformly on the interval [0, 1], according to the classical Vitali’s result (see e.g. Kohr-Mocanu [118], p. 112, Theorem 3.2.10], it suffices (m) to show that for any fixed n ∈ N, the iterate sequence of polynomials Bn (f )(z), m = 1, 2, ..., is uniformly bounded with respect to m ∈ N in each Dr with 1 ≤ r < R. Let n ∈ N be fixed. According to He [107], p. 580, relationship (7), we can write Bn (ek )(z) =

k X

S(k, j)

j=1

n(n − 1)...[n − (j − 1)] j z , nk

where S(k, j) are the Stirling numbers of second kind. It is well-known that these numbers satisfy S(k, j) ≥ 0, for all j, k ∈ N and k X j=1

S(k, j)n(n − 1)...[n − (j − 1)] = nk , for k, n ∈ N.

Let |z| ≤ r with r ≥ 1. Since S(k, j)n(n − 1)...[n − (j − 1)] ≥ 0, for all k, n, j ∈ N with 1 ≤ j ≤ k, it follows |Bn (ek )(z)| ≤ ≤ ≤

k X

S(k, j)

n(n − 1)...[n − (j − 1)] |ej (z)| nk

S(k, j)

n(n − 1)...[n − (j − 1)] j r nk

S(k, j)

n(n − 1)...[n − (j − 1)] k r = rk . nk

j=1

k X j=1

k X j=1

Applying now Bn to the above equality, we obtain Bn(2) (ek )(z) =

k X j=1

S(k, j)

n(n − 1)...[n − (j − 1)] Bn (ej )(z), nk

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which from the last inequality implies |Bn(2) (ek )(z)| = ≤

k X

S(k, j)

n(n − 1)...[n − (j − 1)] |Bn (ej )(z)| nk

S(k, j)

n(n − 1)...[n − (j − 1)] j r ≤ rk . nk

j=1

k X j=1

Reasoning by recurrence, we easily get |Bn(m) (ek )(z)| ≤ rk , for all k, n, m ∈ N and z ∈ Dr . This implies |Bn(m) (f )(z)| ≤

∞ X

k=0

|ck |rk < +∞,

for all m, n ∈ N and z ∈ Dr , which proves (i). (ii) Since from the last inequality in (i), for each r ∈ [1, R) the sequence (m) Bn (f )(z), m, n = 1, 2, ..., is in fact uniformly bounded in Dr with respect to (m ) both m, n ∈ N, and since by Kelisky-Rivlin [115], we have Bn n (f )(x) → f (x) as n → ∞, uniformly for x ∈ [0, 1], it follows that the Vitali’s convergence theorem implies the first convergence in (ii). P∞ (m) (m) Since Bn (f )(z) = k=0 ck Bn (ek )(z), with ek (z) = z k , we get |Bn(m) (f )(z) − f (z)| ≤

∞ X

k=2

|ck | · |Bn(m) (ek )(z) − ek (z)|.

But according to He [107], we have Bn (ek )(z) − ek (z) =

k X

S(k, j)

j=1

n(n − 1)...[n − (j − 1)] ej (z) − ek (z). nk

Therefore, Bn (ek )(z) − ek (z) =

k−1 X

S(k, j)

j=1

n(n − 1)...[n − (j − 1)] ej (z) nk

+[(1 − 1/n)...(1 − (k − 1)/n) − 1]ek (z),

which immediately implies Bn(p) [Bn (ek )(z) − ek (z)] =

k−1 X j=1

S(k, j)

n(n − 1)...[n − (j − 1)] (p) Bn (ej )(z) nk

+[(1 − 1/n)...(1 − (k − 1)/n) − 1]Bn(p) (ek )(z).

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29

(p)

Taking into account that by the proof of the above point (i), we have |Bn (ej )(z)| ≤ rj , for all p, n, j ∈ N and |z| ≤ r, it follows |Bn(p) [Bn (ek ) − ek ](z)| ≤

k−1 X

S(k, j)

j=1

n(n − 1)...[n − (j − 1)] (p) |Bn (ej )(z)| nk

+[1 − (1 − 1/n)...(1 − (k − 1)/n)]|Bn(p) (ek )(z)|

≤

k−1 X

S(k, j)

j=1

n(n − 1)...[n − (j − 1)] j r nk

+[1 − (1 − 1/n)...(1 − (k − 1)/n)]r k

≤ 2[1 − (1 − 1/n)...(1 − (k − 1)/n)]r k . But Bn(m) (ek )(z)

− ek (z) =

m−1 X p=0

Bn(p) [Bn (ek )(z) − ek (z)],

which implies for all |z| ≤ r that |Bn(m) (ek ) − ek | ≤

m−1 X p=0

|Bn(p) [Bn (ek ) − ek ](z)|

≤ 2m[1 − (1 − 1/n)...(1 − (k − 1)/n)]r k , and finally |Bn(m) (f )(z) − f (z)| ≤ 2m

∞ X

k=0

|ck | · [1 − (1 − 1/n)...(1 − (k − 1)/n)]r k .

Since [1 − (1 − 1/n)...(1 − (k − 1)/n)] ≤ n1 [1 + ... + (k − 1)] = k(k−1) 2n , for all k ∈ N, k ≥ 2, by choosing r = rq , we get the first required inequality in the statement. P P k−2 k−2 Note that ∞ < ∞, since we have f 00 (z) = ∞ , k=2 |ck |k(k − 1)r k=2 ck k(k − 1)z for all |z| ≤ r. The second estimate in (ii) is a direct consequence of the inequality d(f, g) = ≤ ≤ (m )

q ∞ X X kf − gkj 1 kf − gkj 1 · + · j j 1 + kf − gk 2 1 + kf − gk 2 j j j=q+1 j=1 q ∞ X 1 kf − gkq X 1 + j 1 + kf − gkq j=1 2 2j j=q+1

kf − gkq 1 1 + q ≤ kf − gkq + q . 1 + kf − gkq 2 2

(iii) Since Bn n (f )(x) → B1 (f )(x), as n → ∞, uniformly for x ∈ [0, 1] (see Kelisky-Rivlin [115]) the proof is similar with that of (ii).

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(iv) First we prove that for any t > 0, the complex series Gt (z, y) is uniformly and absolutely convergent for |z|, |y| ≤ r, with r ≥ 1. Indeed, from the representation formula n 1 X n+1 n+1 (z + 1)k (z − 1)n−k , |z| ≤ r, Pn(1,1) (z) = n n−k k 2 k=0

we get

n X n

n n+1 2n + 2 r+1 ≤ = 2 n−k n k=0 n n n (2n + 2)e r+1 r+1 ≤ (4e)n ≤ [6(r + 1)]n . ≤ 2 n 2 m Pj n n+m We used above the Vandermond’s equality and the ink=0 k j−k = j n ne k equality k ≤ k . Now, since for |z|, |y| ≤ r we get |2z − 1|, |2y − 1| ≤ 2r + 1, denoting ρ = 2r + 1, it follows ∞ X k(2k − 1) −k(k−1)t/2 (1,1) (1,1) e |Pk−2 (2z − 1)| · |Pk−2 (2y − 1)| |Gt (z, y)| ≤ k−1 |Pn(1,1) (z)|

≤

r+1 2

k=2 ∞ X

k=2

n+1 k

k(2k − 1) −k(k−1)t/2 e [6ρ]2k−4 . k−1

By applying the ratio test, the last series (of positive numbers) is convergent. This shows that T (t)(f )(z) is well defined for any t > 0 and all |z| < R. In what follows, it suffices to prove that for any fixed r ∈ [1, R) we have lim kBn(mn ) (f ) − T (t)(f )kr = 0.

n→∞

Since from Karlin-Ziegler [114], for t > 0 we have Z 1 lim Bn(mn ) (f )(x) = L(f )(x) + x(1 − x) Gt (x, y)[f (y) − L(f )(y)]dy, n→∞

0

uniformly with respect to x ∈ [0, 1], according to the Vitali’s theorem, it suffices (m ) to prove that the sequence (Bn n (f )(z))n∈N is uniformly bounded in Dr . However this fact was proved by the last inequality at the above point (i) (see also the remark at the beginning of point (ii)). Therefore the theorem has been proved. Remarks. 1) The property (iv) in Theorem 1.2.1 suggests that for f ∈ AR , the (m ) limit of the iterates Bn n (f )(z) represents the semigroup of operators T (t)(f )(z) defined on the locally convex space (Fr´echet) AR . 2) The results in Theorem 1.2.1 extend some related results in the case of iterates of real Bernstein polynomials on [0, 1] (see Karlin-Ziegler [114], Kelisky-Rivlin [115]). In what follows we prove that the shape preserving properties for complex Bernstein polynomials in Theorem 1.1.7 hold for their iterates too.

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31

In the proofs of these properties we need the following two auxiliary lemmas. P∞ Lemma 1.2.2. (Gal [78]) Let f ∈ AR with R > 1, that is f (z) = k=0 ck z k , for all z ∈ DR . For any m ∈ N, we have lim d[Bn(m) (f ), f ] = 0,

n→∞ (m)

that is the sequence (Bn (f ))n∈N uniformly converges to f on compacts disks in DR . Proof. Note that Lemma 1.2.2 is a particular case of Theorem 1.2.1, (ii), for the constant sequence mn ≡ m. Lemma 1.2.3. (Gal [78]) Let f ∈ AR , R > 1, be satisfying f (0) = 0. For all m, n ∈ (m) (m−1) (j) N we have [Bn (f )]0 (0) = nBn (f )(1/n) and limn→∞ [n · Bn (f )(1/n)] = f 0 (0), (0) (m) for any fixed j, where by convention, Bn (f ) = f and [Bn (f )]0 (0) denotes the (m) first derivative of Bn (f )(z) at 0. Proof.

We have Bn(m) (f )(z) = Bn [Bn(m−1) (f )](z),

which by Bn0 (f )(0) = nf (1/n) implies [Bn(m) (f )]0 (0) = nBn(m−1) (f )(1/n). For the second part of lemma, let us observe that it suffices to prove it for real functions f (x), x ∈ [0, 1], in C 2 [0, 1]. Indeed, by f (z) = U (x, y)+iV (x, y), z = x+iy, where U and V have partial derivatives of any order, we get f (x) = U (x, 0) + iV (x, 0) := g(x) + ih(x), for all x ∈ [0, 1], where g, h are continuously differentiable (m) of any order. Also, we take into account that Bn (·) is a linear operator on C[0, 1], for any n, m ∈ N. (0) We obviously have nf (1/n) = f (1/n)−f → f 0 (0), as n → ∞. (1/n) In what follows, we will use the well-known pointwise estimate for Bernstein polynomials when f ∈ C 2 [0, 1], given by x(1 − x) , for all x ∈ [0, 1], n ∈ N, n where k · k denotes the uniform norm in C[0, 1]. We get |Bn (f )(x) − f (x)| ≤ Ckf 00 k

Bn (f )(1/n) − Bn (f )(0) (1/n) Bn (f )(1/n) − f (1/n) f (1/n) − f (0) = + → f 0 (0), (1/n) (1/n)

nBn (f )(1/n) =

as n → ∞, since by the above estimate we have Bn (f )(1/n) − f (1/n) Ckf 00 k ≤ → 0, for n → ∞. (1/n) n

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Then, we get nBn(2) (f )(1/n) Bn [Bn (f )](1/n) − Bn (f )(1/n) Bn (f )(1/n) − Bn (f )(0) = + → f 0 (0), (1/n) (1/n) as n → ∞. Indeed, by applying again the above pointwise estimate for f replaced by Bn (f ) and taking into account the inequality kBn00 (f )k ≤ kf 00 k (which easily follows from e.g. Lorentz [125], p. 12, relation (2)), we obtain Bn [Bn (f )](1/n) − Bn (f )(1/n) CkBn00 (f )k Ckf 00 k ≤ ≤ → 0, (1/n) n n

as n → ∞. (j) But limn→∞ [n · Bn (f )(1/n)] = f 0 (0), for any j, easily follows by mathematical induction, which proves the lemma. The main result is the following.

Theorem 1.2.4. (Gal [78]) Let us suppose that G ⊂ C is open, so that D1 ⊂ G and f : G → C be analytic in G. Also, let m ∈ N be fixed. (i) If f is univalent in D1 , then there exists an index n0 depending on f and m, (m) so that the mth iterates Bn (f )(z), be univalent in D1 , for all n ≥ n0 . (ii) If f (0) = f 0 (0) − 1 = 0 and f is starlike in D1 , that is 0 zf (z) > 0, for all z ∈ D1 , Re f (z)

then there exists an index n0 depending on f and m, so that the mth iterates (m) Bn (f )(z), be starlike in D1 , for all n ≥ n0 . If f (0) = f 0 (0) − 1 = 0 and f is starlike only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, m, Dr ), so (m) that the mth iterates Bn (f )(z), be starlike in Dr for all n ≥ n0 , that is, ! (m) z[Bn ]0 (f )(z) > 0, for all z ∈ Dr . Re (m) Bn (f )(z) (m)

(m)

(Here [Bn ]0 (f )(z) denotes the first derivative of Bn (f )(z).) (iii) If f (0) = f 0 (0) − 1 = 0 and f is convex in D1 , that is 00 zf (z) + 1 > 0, for all z ∈ D1 , Re f 0 (z)

then there exists an index n0 depending on f and m, so that the mth iterates (m) Bn (f )(z), be convex in D1 , for all n ≥ n0 . If f (0) = f 0 (0) − 1 = 0 and f is convex only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, m, Dr ), so (m) that for all n ≥ n0 , the mth iterates Bn (f )(z) be convex in Dr , that is, ! (m) z[Bn ]00 (f )(z) + 1 > 0, for all z ∈ Dr . Re (m) [Bn ]0 (f )(z)

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(iv) If f (0) = f 0 (0) − 1 = 0, f (z) 6= 0, for all z ∈ D1 \ {0} and f is spirallike of type γ ∈ (−π/2, π/2) in D1 , that is 0 iγ zf (z) > 0, for all z ∈ D1 , Re e f (z) then there exists an index n0 depending on f, m and γ, so that the mth iterates (m) (m) Bn (f )(z) 6= 0, for all z ∈ D1 \ {0}, and Bn (f )(z) be spirallike of type γ in D1 , for all n ≥ n0 . If f (0) = f 0 (0) − 1 = 0, f (z) 6= 0 for all z ∈ D1 \ {0} and f is spirallike of type γ only in D1 , then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, m, Dr , γ), so that for all n ≥ n0 , the mth iterates (m) Bn (f )(z) 6= 0 for all z ∈ Dr \ {0} be spirallike of type γ in Dr , that is, ! (m) 0 iγ z[Bn ] (f )(z) Re e > 0, for all z ∈ Dr . (m) Bn (f )(z) Proof. (i) It is immediate from the uniform convergence in Lemma 1.2.2 and a well-known result concerning sequences of analytic functions converging locally uniformly to an univalent function (see e.g. Kohr-Mocanu [118], p. 130, Theorem 4.1.17). For the proofs of the next points (ii), (iii) and (iv), let us make some general useful considerations. According to Lemma 1.2.2, combined with the Weierstrass’s well-known result, it follows that as n → ∞, uniformly in D1 we have Bn(m) (f )(z) → f (z), [Bn(m) ]0 (f )(z) → f 0 (z) and [Bn(m) ]00 (f )(z) → f 00 (z). (m)

(m)

B (m) (f )(z)

In what follows we denote Cn,m = [Bn (f )]0 (0) and Pn (f )(z) = nCn,m . Note here that f 0 (0) = 1 combined with Lemma 1.2.3, implies that for any m ∈ N, there exists n(m, f ) so that Cn,m > 0, for all n ≥ n(m, f ), (in fact we have limn→∞ Cn,m = 1). (m) (m) From f (0) = 0 we get Bn (f )(0) = 0 and Pn (f )(0) = 0, while from the (m) (m) 0 definition of Pn we obtain [Pn ] (f )(0) = 1. By combining all of these facts with Lemma 1.2.2, we obtain that for n → ∞, (m) (m) (m) we have Pn (f )(z) → f (z), [Pn ]0 (f )(z) → f 0 (z) and [Pn ]00 (f )(z) → f 00 (z), uniformly in D1 . (ii) By hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . (m) (m) By writing Pn (f )(z) in the form Pn (f )(z) = zQn,m (f )(z), it is obvious that Qn,m (f )(z) is a polynomial of degree ≤ n − 1. Let |z| = 1. We have |f (z) − Pn(m) (f )(z)| = |z| · |g(z) − Qn,m (f )(z)| = |g(z) − Qn,m (f )(z)|, (m)

which by the uniform convergence in D1 of Pn (f ) to f and by the maximum modulus principle implies the uniform convergence in D1 of Qn,m (f )(z) to g(z).

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Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index n1 ∈ N and a > 0 depending on g, so that |Qn,m (f )(z)| > a > 0, for all z ∈ D1 and all n ≥ n0 . Also, for all |z| = 1, we have |f 0 (z) − [Pn(m) ]0 (f )(z)| = |z[g 0 (z) − Q0n,m (f )(z)] + [g(z) − Qn,m (f )(z)]|

≥ | |z||g 0 (z) − Q0n,m (f )(z)| − |g(z) − Qn,m (f )(z)| |

= | |g 0 (z) − Q0n,m (f )(z)| − |g(z) − Qn,m (f )(z)| |

(m)

which from the maximum modulus principle, the uniform convergence of [Pn ]0 (f ) to f 0 and of Qn,m (f ) to g, evidently implies the uniform convergence of Q0n,m (f ) to g0 . Then, for |z| = 1, we get (m) 0

] (f )(z)

z[Pn

(m)

Pn

=

(f )

z[zQ0n,m(f )(z) + Qn,m (f )(z)] zQn,m(f )(z)

zQ0n,m(f )(z) + Qn,m (f )(z) zg 0 (z) + g(z) → Qn,m (f )(z) g(z) 0 0 f (z) zf (z) = = , g(z) f (z) which again from the maximum modulus principle, implies =

(m) 0

z[Pn

] (f )(z)

(m)

Since Re

0

zf (z) f (z)

Therefore, Re

Pn

(f )

→

zf 0 (z) , uniformly in D1 . f (z)

is continuous in D1 , there exists α ∈ (0, 1), so that 0 zf (z) ≥ α, for all z ∈ D1 . Re f (z) "

(m) 0

z[Pn

(m)

Pn

] (f )(z)

(f )(z)

#

zf 0 (z) → Re ≥α>0 f (z)

uniformly on D1 , i.e. for any 0 < β < α, there is n0 so that for all n ≥ n0 we have " # (m) z[Pn ]0 (f )(z) Re > β > 0, for all z ∈ D1 . (m) Pn (f )(z) (m)

(m)

Since Pn (f )(z) differs from Bn (f )(z) only by a constant, this proves the first part in (ii). For the second part, the proof is identical with the first part, with the only difference that instead of D1 , we reason for Dr . (iv) Obviously we have " # (m) 0 0 iγ zf (z) iγ z[Pn ] (f )(z) → Re e , Re e (m) f (z) Pn (f )(z)

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uniformly in D1 . We also note that since f is univalent in D1 , according to Lemma 1.2.2, there (m) exists n1 (m, f ) so that Bn (f )(z) be univalent in D1 for all n ≥ n1 (m, f ). There(m) (m) fore, Bn (f )(0) = 0 implies Bn (f )(z) 6= 0, for all z ∈ D1 \ {0}, n ≥ n1 (m, f ). For the rest, the proof is identical with that from the above point (ii). (iii) For the first part, by hypothesis there is α ∈ (0, 1), so that 00 zf (z) + 1 ≥ α > 0, Re f 0 (z)

uniformly in D1 . It is not difficult to show that this is equivalent with the fact that for any β ∈ (0, α), the function zf 0 (z) is starlike of order β in D1 (see e.g. MocanuBulboac˘ a-S˘ al˘ agean [138], p. 77), which implies that f 0 (z) 6= 0, for all z ∈ D1 , i.e. 0 |f (z)| > 0, for all z ∈ D1 . Also, by using the same type of reasonings as those mentioned in the above point (ii), we get # " 00 (m) z[Pn ]00 (f )(z) zf (z) + 1 → Re + 1 ≥ α > 0, Re (m) 0 f 0 (z) [Pn ] (f )(z) uniformly in D1 . As a conclusion, for any 0 < β < α, there is n0 depending on f , so that for all n ≥ n0 we have " # (m) z[Pn ]00 (f )(z) + 1 > β > 0, for all z ∈ D1 . Re (m) [Pn ]0 (f )(z) The proof of second part in (iii) is similar, which proves the theorem.

1.3

Generalized Voronovskaja Theorems for Bernstein Polynomials

It is well-known the fact that the classical Voronovskaja’s theorem for real variable was generalized by Bernstein [42] as follows. Theorem 1.3.1. (see e.g. Lorentz [125], p. 22-23) If f is defined and bounded on [0, 1] and the derivative f (2p) (x) exists at x, then we can write Bn (f )(x) = f (x) + Pn

where Bn (f )(x) = P Tn,j (x) = ni=0 (i −

2p X f (j) (x) j=1

j!

n−j Tn,j (x) +

εn , np

n k k f (k/n) denotes the Bernstein k=0 k x (1 − x) j n i n−i nx) i x (1 − x) and εn → 0 as n → ∞.

polynomials,

Remark. The classical Voronovskaja’s theorem is recaptured for k = 1. The goal of this section is to obtain a similar result to Theorem 1.3.1 for the complex Bernstein polynomials attached to analytic functions in compact disks with the centers in origin and radii ≥ 1. For the particular case p = 1 one recapture Theorem 1.1.3. Moreover, the analyticity of f will imply exact orders of approximation in the generalized Voronovskaja’s theorems.

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The first main result of this section is the following. Theorem 1.3.2. (Gal [89]) Let R > 1 and let f : DR → C be an analytic function, P k that is f (z) = ∞ k=0 ck z . Then for any 1 ≤ r < R and any natural number p there exists a constant Cp > 0 such that the following estimate 2p (j) X Cp,r (f ) f (z) −j Bn (f ) (z) − f (z) − ≤ n T (z) n,j j! np+1 j=1

holds for all z ∈ C with |z| ≤ r and for all n ∈ N, where Cp,r (f ) = Cp ·

∞ X

k=2p+1

|ck |

k! · (k − 2p) k r < ∞. (k − 2p − 1)!

Remark. Theorem 1.3.2 is the complex analogue of the Bernstein’s result, with the quantitative estimate εn ≤ 1/n. Proof of Theorem 1.3.2. Let ek be the function defined by ek (z) = z k . Let us put πk,n = Bn (ek ) . By Lemma 2.2 in Pop [155] one has πk,n (z) = where z k

(j)

k X (j) 1 T (z) · z k , j n,j j!n j=0

is the j-th derivative of the function ek (z) = z k . Note that (j) zk = k (k − 1) ... (k − j + 1) z k−j .

(1.1)

(1.2)

Let us introduce the polynomial Ek,n,p by defining Ek,n,p (z) = πk,n (z) − ek (z) −

2p X (j) 1 Tn,j (z) · z k . j j!n

(1.3)

j=1

Since Tn,0 (z) = 1 we see by (1.1) and (1.2) that for k ≥ 2p + 1 we have k k X X 1 1 k k−j k (j) Ek,n,p (z) = z Tn,j (z) . T (z) · z = n,j j!nj nj j j=2p+1 j=2p+1

(1.4)

First we need the following auxiliary result.

Lemma 1.3.3. (Gal [89]) For given p ∈ N and r ≥ 1 there exists a constant Cp > 0 such that for all k ≥ 2p + 1, |z| ≤ r and n ∈ N the following estimate |Ek,n,p (z)| ≤ Cp rk holds.

np+1

k! 2 (k − 2p) (k − 2p)!

Proof of Lemma 1.3.3. We shall use mathematical induction over p. For p = 1 this inequality has been proved in the proof of Theorem 1.1.3, (ii). Now suppose

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37

that the inequality is valid for p, and we want to prove it for p + 1. At first we observe from (1.4) that Ek,n,p+1 (z) is equal to k−(2p+1) k−(2p+2) z z k k Ek,n,p (z) T (z) Tn,2p+2 (z) . (1.5) n,2p+1 2p + 1 2p + 2 n2p+1 n2p+2 Let us define Rk,n,p (z) = Ek,n,p+1 (z) − zEk−1,n,p+1 (z) −

z (1 − z) 0 Ek−1,n,p+1 (z) . n

Using (1.4), a simple computation shows that Rk,n,p (z) is equal to k k−1 X X 1 k k−j 1 k − 1 k−1−j z Tn,j (z) − z z Tn,j (z) nj j nj j j=2p+3 j=2p+3 −

k−1 z (1 − z) X 1 k − 1 d k−1−j z Tn,j (z) . j n n dz j j=2p+3

Now let us rewrite (1.1) in the following trivial way 2p+2 k X X 1 k 1 k k−j z Tn,j (z) = πk,n (z) − z k−j Tn,j (z) j j j j n n j=2p+3 j=0

(1.6)

(1.7)

(1.8)

(1.9)

and replace the summands in (1.7) and (1.8) by the corresponding expressions induced by (1.9). Then 2p+2 X 1 k z k−j Tn,j (z) − zπk−1,n (z) Rk,n,p (z) = πk,n (z) − j j n j=0 +

2p+2 X j=0

+

1 k − 1 k−j z (1 − z) 0 πk−1,n (z) z Tn,j (z) − nj j n

2p+2 0 z (1 − z) X 1 k − 1 d k−1−j z Tn,j (z) j n n j dz j=0

:= S1 − zS2 −

z(1 − z) S3 . n

Note that D. Andrica [24] has proved (see also the proof of Theorem 1.1.2, (i)) that πk,n (z) = zπk−1,n (z) +

z (1 − z) 0 πk−1,n (z) , n

which simplifies the above expression for Rk,n,p (z) in an obvious way. We want to deduce from the above formula that 1 |Rk,n,p (z)| ≤ Cp∗ p+2 rk k (k − 1) ... (k − 2p − 2) , n for all |z| ≤ r, n ∈ N and k ≥ 2p + 3.

(1.10)

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For this purpose, we observe that in the above expression of Rk,n,p (z) (with respect to S1 , S2 , S3 ), the expression S1 is equal to the left-hand side in (1.9), S2 is equal to the left-hand side in (1.9) written for k − 1 and S3 is equal to the derivative with respect to z of the left-hand side in (1.9) written for k − 1. But since by Lorentz [125], p. 14, Theorem 1.5.1, Tn,j (z) is a polynomial of degree [j/2] with respect to n, it is clear from its form that the left-hand side in (1.9) contains only terms having at denominator nj−[j/2] , j ≥ 2p + 3, that is only terms having at the denominator nj with j ≥ p + 2. This immediately implies that Rk,n,p (z) contains only terms having at denominator nj with j ≥ p + 2. Now, since by the same Lorentz [125], p. 14, Theorem 1.5.1, Tn,j (z) is a polynomial of degree j with respect to z, (by using again (1.9)) this immediately implies that for all |z| ≤ r 1 . we have an estimate of the form |Rn,k,p (z)| ≤ Cp,k rk np+2 Therefore it remains to find out the form of the constant Cp,k . By the recurrence formula for Tn,j (z) in Lorentz [125], p. 14, relation (3), it follows that Tn,j (z) is a polynomial in n of degree ≤ [j/2] with at most [j/2] terms containing the powers of j [j/2] n (where j ≤ 2p + 2), satisfying the , for all |z| ≤ r. estimate |Tn,j (z)| ≤ r Aj n k Combining with the estimates j ≤ k(k − 1)...(k − (2p + 2) + 1) ≤ k(k − 1)...(k − 2p−2), k−1 ≤ (k −1)(k −2)...(k −(2p+2)) ≤ k(k −1)...(k −2p−2), j = 1, , , 2p+2, j it easily follows that the modulus of all the nominators of the terms having at the denominators nj with j ≥ p + 2, can be bounded by Cp rk k(k − 1)...(k − 2p − 2), with a suitable chosen constant Cp depending only on p. Now we shall estimate Ek,n,p+1 (z) by using (1.10). Indeed, by (1.6) we have z (1 − z) 0 Ek−1,n,p+1 (z) + Rk,n,p (z) . n Let us denote kf kr = sup|z|≤r |f (z)| and let us recall Bernstein’s inequality

0

Pj ≤ j kPj k , r r r valid for any polynomial Pj of degree ≤ j. Since |z (1 − z)| ≤ r (1 + r) ≤ 2r 2 for all |z| ≤ r (recall that 1 ≤ r) and Ek−1,n,p+1 (z) is a polynomial of degree ≤ k we conclude that for |z| ≤ r 2r |Ek,n,p+1 (z)| ≤ r |Ek−1,n,p+1 (z)| + k |Ek−1,n,p+1 (z)| + |Rk,n,p (z)| . (1.11) n By equation (1.5) (applied to k − 1 instead of k) one obtains k−1 1 k−1−(2p+1) T (z) |Ek−1,n,p+1 (z)| ≤ |Ek−1,n,p (z)| + z n,2p+1 2p + 1 n2p+1 k−1 1 k−1−(2p+2) + T (z) . z n,2p+2 2p + 2 n2p+2 We use the last inequality in order to estimate the middle term in (1.11) in the following way: 2rk (k − 1)...(k − 2p)(k − 2p − 1)2 2r Cp rk−1 k |Ek−1,n,p+1 (z)| ≤ n n np+1 Ek,n,p+1 (z) = zEk−1,n,p+1 (z) +

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+ A2p+1 r

k−1 (k

39

− 1)...(k − (2p + 1)) k−1 (k − 1)...(k − (2p + 2)) + A2p r . np+1 np+2

From this we conclude that

|Ek,n,p+1 (z)| ≤ r |Ek−1,n,p+1 (z)| + Cp rk

1 np+2

k (k − 1) ... (k − 2p − 2) .

Since by Lemma 2.2 in Pop [155] we have E2p+2,n,p+1 (z) = 0, from the above inequality by an inductive argument applied for k = 2p + 3, ... we finally obtain 1 2 |Ek,n,p+1 (z)| ≤ Cp rk p+2 k (k − 1) ... (k − 2p − 1) (k − 2p − 2) , n which proves the lemma. Now we are in position to prove Theorem 1.3.2. Indeed, taking into account that by the estimate in Lemma 1.3.3 we have E0,n,p (z) = E1,n,p (z) = ... = E2p,n,p (z) = 0, it follows 2p (j) X f (z) −j Bn (f )(z) − f (z) − n T (z) n,j j! j=1 ≤

≤

∞ X

k=2p+1

Cp ·

|ck | · |Ek,n,p (z)|

P∞

k=2p+1

|ck |k(k − 1)...(k − 2p + 1)(k − 2p)2 rk np+1

which proves the theorem.

,

Remark. Analysing the proof of Lemma 1.3.3 and the reasonings for the proof of Theorem 1.3.2, in a similar way we can prove that in the case when r = 1 the pointwise estimate 2p (j) X |z| · |1 − z|Cp,1 (f ) f (z) −j Bn (f )(z) − f (z) − ≤ n T (z) , ∀|z| ≤ 1, n,j j! np+1 j=1 P 2 holds, where Cp,1 (f ) = Cp · ∞ k=2p+1 |ck |k(k − 1)...(k − 2p + 1)(k − 2p) < ∞. Unlike the real case, for complex analytic functions the order of approximation 1 in Theorem 1.3.2 is exactly np+1 . More exactly, the second main result of this paper is the following. Corollary 1.3.4. (Gal [89]) Let R > 1 and let f : DR → C be an analytic P∞ k function, say f (z) = k=0 ck z . If f is not a polynomial of degree ≤ 2p then for any 1 ≤ r < R and any natural number p we have

2p (j) X

f 1 −j

Bn (f ) − f − n Tn,j ∼ p+1 , n ∈ N,

j! n

j=1 r

where the constants in the equivalence depend only on f , r and p and are independent of n. Here kf kr = sup|z|≤r |f (z)|.

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Proof. Taking into account Theorem 1.3.2, it remains to obtain the lower estimate for the quantity in the statement of Corollary 1.3.4. Thus, suppose that f is not a polynomial of degree ≤ 2p. Keeping the notations in the previous section, since P∞ f (s) (z) = k=s ck k(k − 1)...(k − s + 1)z k−s , by using (1.4) and simple calculations we easily obtain the identity Bn (f )(z) − f (z) − =

∞ X

2p X f (j) (z) j=1

j!

n−j Tn,j (z)

ck Ek,n,p (z)

k=2p+1

 ∞  X

 k  X k k−j p+1−j ck  z n Tn,j (z)  j j=2p+1 

1 np+1  k=2p+1 1 Tn,2p+1 (z) (2p+1) Tn,2p+2 (z) (2p+2) = p+1 f (z) + p+1 f (z) n np (2p + 1)! n (2p + 2)!   ∞  X 1 + np+2 ck Ek,n,p+1 (z) .  n =

k=2p+3

By the recurrence formula for Tn,j (z) in Lorentz [125], p. 14, relation (3), it follows that Tn,j (z) is a polynomial in n of degree ≤ [j/2] with at most [j/2] terms containing the powers of n (where j ≤ 2p + 2). Also, by Lorentz [125], p. 14, The(2p+2)! p+1 , while orem 1.5.1, the coefficient of np+1 in Tn,2p+2 (z) is 2p+1 (p+1)! · [x(1 − x)] from the recurrence formula in Lorentz [125], p. 14, relation (3), it easily follows that the coefficient of np in Tn,2p+1 (z) is of the form ap (1 − 2z)[z(1 − z)]p with the constant ap > 0 (ap depends only on p). Therefore, it is easy to see that the sum Tn,2p+1 (z) (2p+1) T (z) (2p+2) 1 (z) + np+1 · n,2p+2 (z) can be written in the form np (2p+1)! f (2p+2)! f 1 Tn,2p+2 (z) (2p+2) Tn,2p+1 (z) (2p+1) f (z) + p+1 · f (z) p n (2p + 1)! n (2p + 2)! ap (1 − 2z)[z(1 − z)]p f (2p+1) (z) = (2p + 1)! [z(1 − z)]p+1 (2p+2) + p+1 f (z) 2 (p + 1)! 1 1 + F (z)f (2p+1) (z) + G(z)f (2p+2) (z), n n 1 where the polynomials F (z) := P1 (z) + n1 P2 (z) + ... + np−1 Pp (z) and G(z) := 1 1 Q1 (z)+ n Q2 (z)+...+ np Qp+1 (z) are bounded in any closed disk |z| ≤ r by constants depending on r and p but independent of n. Replacing this form in the above identity and taking into account the inequalities

kh + gkr ≥ | khkr − kgkr | ≥ khkr − kgkr ,

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we obtain

≥

1 np+1

2p (j) X

f −j

Bn (f ) − f −

n T n,j

j!

j=1

41

r

ap [e1 (1 − e1 )]p+1 (2p+2) p (2p+1)

(1 − 2e )[e (1 − e )] f + f 1 1 1

(2p + 1)!

2p+1 (p + 1)! r

 

 ∞

1 p+2 X (2p+1) (2p+2)  −  n c E + F f + Gf k k,n,p+1

 n

k=2p+3

:=

1 np+1

kU kr −

r

1 [kV kr ] n

≥

1 1 · kU kr , np+1 2

for all n ≥ n0 (n0 depends on f , p and r), under the conditions that kU kr > 0 and if kV kr is bounded by a constant depending only on f , p and r. But this is exactly what happens, because from Theorem 1.3.2 (written for p + 1) and from the above considerations on F and G it is immediate that kV kr is bounded by a constant depending only on f , p and r while by the fact that f is not a polynomial of degree ≤ 2p it follows kU kr > 0. Indeed, for the last fact let us suppose the contrary. It follows that f must satisfy the differential equation (here recall that ap > 0) [z(1 − z)]p+1 (2p+2) ap (1 − 2z)[z(1 − z)]p f (2p+1) (z) + p+1 f (z) = 0, |z| ≤ r. (2p + 1)! 2 (p + 1)! Making the substitution f (2p+1) (z) := y(z) it follows that y(z) necessarily is analytic in DR (since f is supposed analytic there) and is solution of the first order differential equation ap [z(1 − z)]p+1 0 (1 − 2z)[z(1 − z)]p y(z) + p+1 y (z) = 0, |z| ≤ r. (2p + 1)! 2 (p + 1)! After simplification with [z(1 − z)]p , we get that y(z) is an analytic function in DR satisfying the differential equation ap z(1 − z) (1 − 2z)y(z) + p+1 y 0 (z) = 0, |z| ≤ r, z 6= 0, z 6= 1. (2p + 1)! 2 (p + 1)! P∞ Writing y(z) in the form y(z) = k=0 bk z k , by comparison of coefficients, we easily obtain that bk = 0, for all k = 0, 1, ..., which implies that y(z) is identical zero in Dr \ {0, 1}. Since y is analytic, it is continuous and therefore y(0) = y(1) = 0, which implies that y(z) = 0, for all |z| ≤ r. But from the identity’s theorem of analytic functions, it necessarily follows that y(z) = 0 for all |z| < R, obviously in contradiction with the hypothesis that f is not a polynomial of degree ≤ 2p in DR . For n ∈ {1, ..., n0 − 1} we obviously have

2p (j) X

Mr,n (f ) f −j

Bn (f ) − f − n Tn,j

≥ np+1 , j!

j=1 r

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P2p (j)

with Mr,n (f ) = np+1 · Bn (f ) − f − j=1 fj! n−j Tn,j > 0, which finally implies r

2p X f (j)

Cp,r (f )

Bn (f ) − f − n−j Tn,j ≥ , for all n ∈ N,

j! np+1

j=1 r

where Cp,r (f ) = min{Mr,1 (f ), ..., Mr,n0 −1 (f ), 12 kU kr }. This completes the proof. 1.4

Butzer’s Linear Combination of Bernstein Polynomials

[q] In the paper of Butzer [51], were inductively introduced the operators Ln (x) of [0] real variable x ∈ [0, 1] by setting Ln (f ) (x) := Bn (f )(x) and [2q−2]

q (2q − 1) L[2q] n (f ) (x) = 2 L2n

(f ) (x) − L[2q−2] (f ) (x), n

for q ∈ N. For example, for q = 1 one obtains [0]

[0] L[2] n (f ) (x) := 2L2n (f ) (x) − Ln (f ) (x) = 2B2n (f )(x) − Bn (f )(x).

In the same paper of Butzer [51], by using the generalized Voronovskaja’s theorem of Bernstein [42] (that is Theorem 1.3.1), he proved that |L[2q−2] (f )(x) − f (x)| = O(n−q ). n The first main result of this section is the extension of Butzer’s result to the case of complex variable and can be stated as follows. Theorem 1.4.1. (Gal [89]) For any analytic function f : DR → C with R > 1, for each 1 ≤ r < R and given natural number q there exists a constant dq,r (f ) > 0 such that the following estimate d (f ) [2q−2] q,r , (f ) (z) − f (z) ≤ Ln nq is valid for all |z| ≤ r and n ∈ N. Proof. The proof of Theorem 1.4.1 is simple. Indeed, let us consider Butzer’s linear combination of complex Bernstein polynomials defined by the recurrence [2q−2]

q [2q] q L[0] n (f )(z) = Bn (f )(z), (2 − 1)Ln (f )(z) = 2 L2n

(f )(z) − Ln[2q−2] (f )(z),

where z ∈ C, q = 1, 2, .... Analysing the proofs of Lemma 1, Lemma 2 and Theorem 1 in Butzer [51] and taking into account Theorem 1.3.2, it is easy to see that their reasonings can analogously be used for the above linear combinations of complex Bernstein polynomials, so that finally we get the statement of Theorem 1.4.1. Remarks. 1) By the Remark after the proof of Lemma 1.3.3, it easily follows that in a similar way we get the pointwise estimate |L[2q−2] (f )(z) − f (z)| ≤ dq,1 (f ) · n

|z| · |1 − z| , for all |z| ≤ 1. nq

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2) For q = 2, the estimate in Theorem 1.4.1 can easily be obtained with an explicit constant dq,r (f ), by using the following estimate in Theorem 1.1.3, (ii) : z(1 − z) 00 5Mr (f )(1 + r)2 |Bn (f )(z) − f (z) − f (z)| ≤ , r ≥ 1, |z| ≤ r, n ∈ N, 2n 2n2 P∞ where Mr (f ) = k=3 |ck |k(k − 1)(k − 2)r k−2 . [2] Indeed, taking into account that Ln (f )(z) = B2n (f )(z) − Bn (f )(z) and that we can write the identity z(1 − z) 00 f (z) L[2] (f )(z) − f (z) = 2 B (f )(z) − f (z) − 2n n 4n z(1 − z) 00 + f (z) − Bn (f )(z) + f (z) , 2n we immediately obtain |L[2] n (f )(z) − f (z)| z(1 − z) 00 z(1 − z) 00 f (z) + Bn (f )(z) − f (z) − f (z) ≤ 2 B2n (f )(z) − f (z) − 4n 2n 5Mr (f )(1 + r)2 15(1 + r)2 Mr (f ) 5Mr (f )(1 + r)2 + = . ≤2 8n2 2n2 4n2 3) For q = 3, in Theorem 1.4.1 similar reasonings can be applied. Indeed, first [4] it easily follows that Ln (f )(z) = 83 B4n (f )(z) − 2B2n (f )(z) + 13 Bn (f )(z) and that we can write the identity L[4] n (f )(z) − f (z)   4 (j) X f (z) 8 (4n)−j T4n,j (z) = B4n (f )(z) − f (z) − 3 j! j=1   4 (j) X f (z) −2 B2n (f )(z) − f (z) − (2n)−j T2n,j (z) j! j=1   4 X f (j) (z) −j 1 Bn (f )(z) − f (z) − (n) Tn,j (z) + 3 j! j=1

f (4) (z)[1 − 6z(1 − z)]z(1 − z) . 4! · 8 · n3 This identity follows from the identities Tn,1 (z) = 0 and f 00 (z) 8 T4n,2 (z) T2n,2 (z) 1 Tn,2 (z) − · = 0, + 2 · − · 2! 3 16n2 4n2 3 n2 8 T4n,3 (z) T2n,3 (z) 1 Tn,3 (z) f 000 (z) − · = 0, +2· − · 3! 3 64n3 8n3 3 n3 8 T4n,4 (z) T2n,4 (z) 1 Tn,4 (z) f (4) (z) − · + 2 · − · 4! 3 256n4 16n4 3 n4 +

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=−

f (4) (z)nz(1 − z)[1 − 6z(1 − z)] . 4! · 8 · n4

As a conclusion, an estimate with explicit constant in Theorem 1.3.2 for p = 2 (which can be obtained by following the reasonings in the proof of Lemma 1.3.3 for p = 2, but with explicit constants), will immediately give an explicit constant d for [4] the estimate |Ln (f )(z) − f (z)| ≤ d/n3 . 4) A nice consequence of Corollary 1.3.4 and of Theorem 1.4.1 is that if f is not a polynomial of degree ≤ q then the order of approximation in Theorem 1.4.1 is exactly n1q . For simplicity first we illustrate the particular cases q = 1, 2, 3. Indeed, the case q = 1 is contained in Corollary 1.1.5. In the q = 2 case, taking into account the identity from the above Remark 2 and applying the reasonings in the proof of [2] Corollary 1.3.4 for the case p = 1, it follows that Ln (f )(z) − f (z) can be written in the form L[2] n (f )(z) − f (z) 1 T2n,3 (z) (3) T2n,4 (z) (4) 1 A (f )(z) + f (z) + f (z) =2 1,n (2n)2 2n 3!(2n) 4!(2n)2 1 1 Tn,3 (z) (3) Tn,4 (z) (4) − 2 A2,n (f )(z) + f (z) + f (z) n n 3!n 4!n2 1 1 1 1 T2n,3 (z) (3) = 2 A1,n (f )(z) − A2,n (f )(z) + · f (z) n 4n n 2 3!(2n) Tn,3 (z) (3) Tn,4 (z) (4) 1 T2n,4 (z) (4) f (z)− f (z) − f (z) + · 2 4!(2n)2 3!n 4!n2 1 1 (1 − 2z)z(1 − z)f (3) (z) 1 3z 2 (1 − z)2 f (4) (z) = 2 , A3,n (f )(z) − + · n n 2 · 3! 2 2 · 4! where for all n ∈ N kA1,n (f )kr , kA2,n (f )kr , kA3,n (f )kr ≤ Cr (f ), (Cr (f ) is independent of n) and we used the formulas in Lorentz [125], p. 14, Tn,3 (z) = n(1 − 2z)z(1 − z), Tn,4 (z) = nz(1 − z)[3nz(1 − z) + (1 − 6z(1 − z))]. By using for −

(1 − 2z)z(1 − z)f (3) (z) z 2 (1 − z)2 f (4) (z) + 12 32

similar reasonings with those in the proof of Corollary 1.3.4, case p = 1, finally we easily obtain kL[2] n (f )(z) − f (z)kr ∼

1 , n ∈ N, n2

where the constants in the equivalence depend only on f and r. 5) For the q = 3 case we can apply similar reasonings. Indeed, applying to the [4] three expressions between the brackets in the formula for Ln (f )(z) − f (z) from the

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above Remark 3, the reasonings in the proof of Corollary 1.3.4, the case p = 2, we obtain L[4] n (f )(z) − f (z) 1 1 8 T4n,5 (z) (5) 8 T4n,6 (z) (6) = 3 A1,n (f )(z) + · f (z) + · f (z) n n 3 · 43 5! · (4n)2 3 · 43 6! · (4n)3 1 T2n,5 (z) (5) 1 T2n,6 (z) (6) −2· 3 · f (z) − 2 · 3 · f (z) 2 5! · (2n)2 2 6! · (2n)3 1 Tn,5 (z) (5) 1 Tn,6 (z) (6) f (4) (z)[1 − 6z(1 − z)]z(1 − z) + · f (z) + · f (z) + 3 5! · n2 3 6! · n3 4! · 8 · n3 2 2 (5) 3 10z (1 − z) (1 − 2z)f (z) 15z (1 − z)3 f (6) (z) 1 1 An (f )(z) + + = 3 n n 8 · 5! 8 · 6! z(1 − z)(1 − 6z(1 − z))f (4) (z) , + 8 · 4!

where kA1,n (f )kr , kAn (f )kr ≤ Cr (f ) for all n ∈ N (Cr (f ) is independent of n) and we used the formula in Lorentz [125], p. 14, Tn,5 (z) = (1 − 2z)[10n2 z 2 (1 − z)2 + n(z(1 − z) − 12z 2(1 − z)2 )] and the Theorem 1.5.1 in Lorentz [125], p. 14, for the formula of n3 in the polynomial Tn,6 (z). Now, by using for 10z 2(1 − z)2 (1 − 2z)f (5) (z) 15z 3(1 − z)3 f (6) (z) + 8 · 5! 8 · 6! z(1 − z)(1 − 6z(1 − z))f (4) (z) + 8 · 4! similar reasonings with those for U (z) in the proof of Corollary 1.3.4, finally we easily obtain 1 kL[4] n (f )(z) − f (z)kr ∼ 3 , n where the constants in the equivalence depend only on f and r. In what follows we present a proof of the equivalence result in approximation [2q−2] by Ln (f )(z) for general q ≥ 3. Taking into account Theorem 1.4.1, therefore it suffices to prove the following lower estimate for general q. G(z) =

Theorem 1.4.2. Let DR = {z ∈ C; |z| < R} be with R > 1, f : DR → C analytic P∞ in DR , i.e. f (z) = j=0 ck z k , z ∈ DR , 1 ≤ r < R and q ≥ 3. If f is not a polynomial of degree ≤ q then for all n ∈ N we have C kL[2q−2] (f ) − f kr ≥ q , n n where the constant C is independent of n and depends on f , r and q. The proof of Theorem 1.4.2 requires the following two auxiliary results. Lemma 1.4.3. Let DR = {z ∈ C; |z| < R} be with R > 1, f : DR → C analytic in [q] DR , and 1 ≤ r < R. Also, let Ln (f )(z) be the Butzer’s polynomials defined by the recurrence in the previous section, q ≥ 2.

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(i) We have the representation formula L[2q−2] (f )(z) = n

q−1 X

αi,q−1 B2i n (f )(z),

i=0

k where αi,q−1 = (−1)q−1−i Si,q−1 /Πq−1 k=1 (2 − 1), S0,q−1 = 1 by convention and X 2j1 2j2 ...2ji , for all i ∈ {1, ..., q − 1}. Si,q−1 = j1 ,j2 ,...,ji =1, j1 <...<ji

(ii) The sums Si,q satisfy the recurrence formula Si,q = Si,q−1 + 2q Si−1,q−1 , for all i = 1, ..., q − 1. Pq−1 (iii) The numbers αi,q−1 satisfy i=0 αi,q−1 = 1 and the recurrence αi,q =

where α0,q−1 =

2q 1 αi−1,q−1 − q αi,q−1 , for all i = 1, q − 1, −1 2 −1

2q

(−1)q−1 . k Πq−1 k=1 (2 −1)

(iv) For all i = 0, ..., q we have αi,q = (−1)

q−i 2

i(i+1)/2 q Y

j=1

q

[2( i ) − 1]

.

j

(2 − 1)

Proof. The proofs of (i) and (ii) can easily be done by mathematical induction after q. Also, (iii) is an immediate consequence of (ii) and (iv) is an immediate consequence of (iii). Lemma 1.4.4. Let q ≥ 3. (i) For all j = 1, ..., q, z ∈ C and n ∈ N we have q−1 X

αi,q−1 (2i )−j T2i n,j (z) = 0.

i=0

(ii) For all j = q + 1, ..., 2q − 2, z ∈ C and n ∈ N we have q−1 X

αi,q−1 (2i )−j T2i n,j (z) = nBq−1,j (z),

i=0

where Bq−1,j (z) is a polynomial of degree ≤ j in z. Proof. By using the recurrence formulas in Lemma 1.4.3, (iii), the proofs of (i) and (ii) are immediate by mathematical induction with respect to q. The degree of the polynomials Bq−1,j (z) with respect to z follows from Lorentz [125], p. 14, Theorem 1.5.1.

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Proof of Theorem 1.4.2. Taking into account the representation formula in Lemma 1.4.3 (i), we can write (f )(z) L[2q−2] n q−1 X i=0

where

− f (z) =



αi,q−1 B2i n (f )(z) − f (z) −

q−1 X i=0

2q−2 X j=1



αi,q−1 [B2i n (f )(z) − f (z)] =

f

(j)



(z) i −j (2 ) T2i n,j (z) + Rn,q−1 (f )(z), j!

 (j) f (z) αi,q−1  Rn,q−1 (f )(z) = (2i n)−j T2i n,j (z) j! j=1 i=0 "q−1 # 2q−2 (j) X X f (z) i −j −j αi,q−1 (2 ) T2i n,j (z) . n = j! i=0 j=1 q−1 X

2q−2 X

Taking into account Lemma 1.4.4, (i) we have # "q−1 2q−2 X f (j) (z) X i −j −j αi,q−1 (2 ) T2i n,j (z) , n Rn,q−1 (f )(z) = j! j=q+1 i=0

which combined with Lemma 1.4.4, (ii) immediately implies   2q−2 1  X f (j) (z) nBq−1,j (z)  Rn,q−1 (f )(z) = q · n j=q+1 j! nj−q 1 f (q+1) (z) 1 = q Bq−1,q+1 (z) + Mq (f )(z) , n (q + 1)! n

where kMq (f )kr ≤ C, with C independent of n. Also, since from Theorem 1.5.1, p. 14 in Lorentz [125] we have that each T2i n,j (z) is a polynomial in z(1−z), this implies that Bq−1,q+1 (z) is of the form Bq−1,q+1 (z) = z(1 − z)Pq (z), with degree (Pq (z)) ≤ q − 1 and Pq (0) 6= 0. On the other hand, reasoning exactly as in the proof of Corollary 1.3.4, case q = p − 1, we get 2(q−1)

B2i n (f )(z) − f (z) −

X f (j) (z) (2i n)−j T2i n,j (z) j! j=1

T2i n,2q−1 (z) T2i n,2q (z) (2q) 1 = i q f (2q−1) (z) + f (z) (2 n) (2i n)q−1 (2q − 1)! (2q)!   ∞  X 1 + (2i n)q+1 ck Ek,2i n,q (z) .  n k=2q+1 P ∞ Clearly hat the expressions (2i n)q+1 k=2q+1 ck Ek,2i n,q (z) are bounded in Dr with bounds independent of n (see Corollary 1.3.4).

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Also, since by Theorem 1.5.1, p. 14 in Lorentz [125] the coefficient of (2i n)q in q T2i n,2q (z) is 2(2q)! q ·q! [z(1 − z)] , while from the recurrence formula in Lorentz [125], p. 14, relation (3), it easily follows that the coefficient of (2i n)q−1 in T2i n,2q−1 (z) is of the form aq−1 (1 − 2z)[z(1 − z)]q−1 with aq−1 > 0. Therefore, reasoning exactly as in the proof of Corollary 1.3.4 we can write T2i n,2q−1 (z) T2i n,2q (z) (2q) f (2q−1) (z) + f (z) (2i n)q−1 (2q − 1)! (2q)! aq−1 [z(1 − z)]q (2q) = (1 − 2z)[z(1 − z)]q−1 f (2q−1) (z) + f (z) (2q − 1)! 2q (q)! 1 1 + Fi (z)f (2q−1) (z) + Gi (z)f (2q) (z), n n where Fi (z) and Gi (z) are polynomials bounded in Dr by constants independent of n. Collecting all the above considerations in conclusion we obtain L[2q−2] (f )(z) − f (z) n (q+1) (z) 2q − 1 1 f z(1 − z)Pq (z) + q−1 aq−1 (1 − 2z)[z(1 − z)]q−1 f (2q−1) (z) = q n (q + 1)! 2 1 2q − 1 [z(1 − z)]q (2q) f (z) + Kq (f )(z) , + q−1 · 2 2q q! n where kKq (f )kr ≤ C with C independent of n. Denoting Hq (f )(z) =

f (q+1) (z) z(1 − z)Pq (z) (q + 1)!

2q − 1 aq−1 (1 − 2z)[z(1 − z)]q−1 f (2q−1) (z) 2q−1 2q − 1 [z(1 − z)]q (2q) + q−1 · f (z), 2 2q q! +

if kHq (f )kr > 0 then the expected lower estimate follows from the inequalities kL[2q−2] (f ) − f kr n 1 1 ≥ q kHq (f )kr − kKq (f )kr n n 1 1 kHq (f )kr ≥ q kHq (f )kr − kKq (f )kr ≥ , n n 2nq for all n ≥ n0 . From the reasonings in the same Corollary 1.3.4 we get the lower estimate of the same order for all n ∈ N. Now, to finish the proof it will be enough to show that if f is not a polynomial of degree ≤ q then we have kHq (f )kr > 0. For this purpose, it will be enough to prove that the differential equation Hq (f )(z) = 0 for z ∈ Dr implies that f is

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a polynomial of degree ≤ q. Making the substitution f (q+1) (z) = y(z) it will be enough to prove that the differential equation in z ∈ Dr y(z)z(1 − z)Pq (z) + Aq (1 − 2z)[z(1 − z)]q−1 y (q−2) (z) + Bq [z(1 − z)]q y (q−1) (z) = 0,

has as analytic solution only y(z) = 0. Here Aq , Bq > 0 and Pq (0) 6= 0. Simplifying with z(1 − z) supposed to be 6= 0, it follows the differential equation in z ∈ Dr1 \ {0} y(z)Pq (z) + Aq (1 − 2z)[z(1 − z)]q−2 y (q−2) (z) + Bq [z(1 − z)]q−1 y (q−1) (z) = 0,

with r1 < 1. Passing now with z → 0 in this equation we immediately obtain y(0) = 0. This means that we can write y(z) = zh(z), with h analytic in Dr1 . Calculating y 0 (z) = h(z)+zh0(z) and y 00 (z) = 2h0 (z)+zh00(z), replacing in the above differential equation, simplifying with z 2 and then passing to limit in the simplified differential equation with z → 0, we immediately obtain h(0) = 0. Therefore we can write h(z) = zu(z) and y(z) = z 2 u(z), that is y 0 (0) = 0. Repeating the same reasonings for y(z) written in this form, we arrive at the form y(z) = z 3 v(z), that is y 00 (0) = 0. Step by step by this kind of reasoning we will obtain y (k) (0) = 0 for all k = 0, 1, 2, ...,. In conclusion we obtain y(z) = 0 for all z ∈ Dr1 , which proves the Theorem 1.4.2. Remarks. 1) Let us suppose that f ∈ C[0, 1]. By taking λ = 1 in Guo-Li-Liu [106], Theorem 2, we immediately obtain the following upper estimate valid for all n∈N √ kf k ϕ [2q−2] + ω2q (f ; 1/ n) , kLn (f ) − f k ≤ C nq where k · k denotes the uniform norm on C[0, 1], C > 0 is an absolute constant ϕ and ω2q (f ; 1/δ) denotes the Ditzian-Totik modulus of smoothness of order 2q with p respect to the weight ϕ(x) = x(1 − x). Then, the equivalence results in the above Remarks 4 and 5 and Theorem 1.4.1 and 1.4.2 suggest the following open question : there exists an absolute constant C 0 > 0 such that for all n ∈ N we have √ kf k ϕ + ω (f ; 1/ n) ≤ kL[2q−2] (f ) − f k. C0 n 2q nq 2) Analogously, Corollary 1.3.4 suggests equivalence result with respect to some suitable expressions involving Ditzian-Totik moduli of smoothness for the generalized Voronovskaja’s theorem in the case of functions of real variable. [q] For G ⊂ C a compact set and q ∈ N, define Ln (z), z ∈ G by setting [0] Ln f ; G (z) := Bn (f ; G)(z) and [2q−2] (2q − 1) L[2q] f ; G (z) − L[2q−2] f ; G (z) = 2q L2n f ; G (z). n n

Taking into account that these Butzer kind polynomials are linear combinations of Bernstein polynomials attached to G, by the above Theorems 1.4.1 and 1.4.2 and

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by similar reasonings with those in the proof of Theorem 1.1.9 (since T and T −1 are linear), we immediately obtain the following result. ˜ \ G is simply connected. Theorem 1.4.5. Let G be a compact Faber set such that C If f is analytic on G, that is there exists R > 1 such that f is analytic in G R and if is not a polynomial of degree ≤ q then for any 1 < r < R we have

C , for all n, q ∈ N, nq where the constants in the equivalence depend on f , q, r and Gr but are independent of n. Here kf kGr = supz∈Gr |f (z)|. kL[2q−2] (f ; G) − f kGr ∼ n

1.5

q-Bernstein Polynomials

In this section we present the approximation and shape preserving properties of the complex q-Bernstein polynomials. First we present upper estimates in approximation and we prove the Voronovskaja’s convergence theorem in compact disks in C, centered at origin, with quantitative estimate of this convergence. These results alow us to obtain the exact degrees in simultaneous approximation by complex q-Bernstein polynomials and their derivatives. Then we study the approximation properties of their iterates and finally we prove that the complex q-Bernstein polynomials preserve in the unit disk (beginning with an index) the starlikeness, convexity and spirallikeness. For q = 1, all these results become those proved for complex Bernstein polynomials in Sections 1.1 and 1.2. Let q > 0. For any n = 1, 2, ..., define the q-integer [n]q := 1 + q + ... + q n−1 , [0]q := 0 and the q-factorial [n]q ! := [1]q [2]q ...[n]q , [0]q ! := 1. For q = 1 we obviously get [n]q = n. For integers 0 ≤ k ≤ n, define [n]q ! n . := [k]q ![n − k]q ! k q Evidently, for q = 1 we get [n]1 = n, [n]1 ! = n! and nk 1 = nk . Now, for f : [0, 1] → C, the complex q-Bernstein polynomials are defined simply replacing z by x in the Phillips definition in [149], that is n−1−k n Y X n [k]q k f z (1 − q s z), n ∈ N, z ∈ C. Bn,q (f )(z) = k q [n]q s=0 k=0

Here conventionally, the empty product is equal to 1. Also, note that for q = 1 we recapture the classical complex Bernstein polynomials. First let us briefly expose the present situation of the main approximation results for the complex q-Bernstein polynomials. Thus, concerning the estimates in the convergence of complex q-Bernstein polynomials attached to analytic functions, by the next theorem and remarks we mention the following known results.

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Theorem 1.5.1. (Ostrovska [146], Gal [87]) Let q > 0, R > 1, DR = {z ∈ C; |z| ≤ R} and let us suppose that f : DR → C is analytic in DR , that is we can write P∞ f (z) = k=0 ck z k , for all z ∈ DR . Then for the complex q-Bernstein polynomials we have the estimate |Bn,q (f )(z) − f (z)| ≤

Mr,q (f ) , for all n ∈ N, [n]q

valid for all n ∈ N and |z| ≤ r, with 1 ≤ r < R, where 0 < Mr,q (f ) = 2 P∞

∞ X

k=2

(k − 1)[k − 1]q |ck |rk .

Moreover, Mr,q (f ) ≤ 2 k=2 (k − 1)k|ck |rk := Mr (f ) < ∞, for all r ∈ [1, R) and q ∈ (0, 1], while if q > 1, then Mr,q (f ) < ∞, for all q < R and r ∈ [1, Rq ). Proof. Since only the case q ≥ 1 (and |z| ≤ 1) was stated explicitly in Ostrovska [146], let us indicate below the proof in its full generality by using some results already proved in Ostrovska [146]. Indeed, first one easily observe that Lemma 3 in Ostrovska [146] is valid for all q > 0. Also, analysing the proof of Theorem 5 in Ostrovska [146] (which use Lemma 3), again it easily follows that its estimate is valid for any q > 0, and not only for q ≥ 1. That is, for any q > 0, denoting ek (z) = z k , k = 1, 2..., z ∈ C, for all k, n ∈ N and |z| ≤ r, we get the kind of estimate in Theorem 5 in Ostrovska [146], i.e. |Bn,q (ek )(z) − ek (z)| ≤ 2rk

(k − 1)[k − 1]q , [n]q

P∞ which by Bn,q (f )(z) − f (z) = k=0 ck (Bn,q (ek )(z) − ek (z)), immediately implies the estimate. P∞ Now, if 0 < q ≤ 1, since [k − 1]q ≤ k, it is immediate that Mr,q (f ) = 2 k=2 (k − P∞ 1)[k − 1]q |ck |rk ≤ 2 k=2 (k − 1)k|ck |rk < ∞. qk If q > 1, then by the estimates [k − 1]q ≤ [k]q ≤ q−1 and Mr,q (f ) = 2

∞ X

k=2

(k − 1)[k − 1]q |ck |rk ≤

∞

2 X (k − 1)|ck |rk q k , q−1 k=2

it follows that Mr,q (f ) < ∞ for rq < R, which proves the theorem.

Remarks. 1) Let 0 < q ≤ 1 be fixed. Since we have [n]1 q → 1 − q as n → ∞, by passing to limit with n → ∞ in the estimate in Theorem 1.5.1 we don’t obtain convergence of Bn (f )(z) to f (z). But this situation can be improved by choosing 0 < qn < 1 with qn % 1 as n → ∞. Indeed, since in this case [n]1q → 0 as n → ∞ n (see Videnskii [194], formula (2.7)), from Theorem 1.5.1 we get that Bn,qn (f )(z) → f (z), uniformly for |z| ≤ r, for any 1 ≤ r < R. 2) If q > 1, since [n]1 q ≤ n1 , then by Theorem 1.5.1 it follows that for any r ≥ 1 with rq < R, we have Bn,q (f )(z) → f (z) as n → ∞, uniformly for |z| ≤ r. In

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fact, in this case by Theorem 6 in Ostrovska [146] (for upper estimate) and by Corollary 1 in Wang-Wu [198] we know much more : if f is not a linear function then kBn,q (f ) − f kr ∼ q −n , for any 0 < r < R/q. Here kf kr = sup{|f (z)|; |z| ≤ r}. 3) It is worth mentioning other two interesting papers in the topic : approximation by complex q-Bernstein polynomials of the Cauchy kernel 1/(z − a) (see Ostrovska [147]) and of the logarithmic function (see Ostrovska [148]). Now, concerning Voronovskaja-type results and approximation by iterates we can mention the following known results. If q ≥ 1 then qualitative Voronovskaja-type and saturation-type results for complex q-Bernstein polynomials were obtained in Wang-Wu [198]. If 0 < q < 1 then for the real q-Bernstein polynomials, qualitative Voronovskajatype and saturation results (see Wang [197]) and quantitative Voronovskaja’s result (see Videnskii [195]) were recently obtained. In this section we fulfil this gap for the complex case and obtain a Voronovskajatype result with quantitative estimate for complex q-Bernstein polynomials with 0 < q < 1. Compared with the quantitative result proved for the real q-Bernstein polynomials in Videnskii [195], our result is essentially better. Also, as an application of our quantitative Voronovskaja’s result, the exact order in approximation by complex qn -Bernstein polynomials with 0 < qn ≤ 1 and limn→∞ qn = 1 is obtained. Taking into account that until present only iterates for real q-Bernstein polynomial were studied (see Ostrovska [146], Xiang-He-Yang [202]), also we fulfil this gap for the complex case, by obtaining approximation results for the iterates of complex q-Bernstein polynomials with q > 0. Also, suggested by the fact that for n sufficiently large the classical complex Bernstein polynomial Bn (f )(z) preserves in the unit disk the starlikeness, convexity and spirallikeness, we will extend these kind of results to complex qn -Bernstein polynomials, Bn,qn (f )(z), n ∈ N, with 0 < qn < 1 and qn → 1. First we present the Voronovskaja-type formula. Theorem 1.5.2. (Gal [87]) Let 0 < q < 1, R > 1, DR = {z ∈ C; |z| < R} and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . (i) The following estimate holds : Bn,q (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ |z(1 − z)| · 9M (f ) , 2[n]q 2[n]q [n]q P∞ for all n ∈ N, z ∈ D1 , where 0 < M (f ) = k=3 |ck |k(k − 1)(k − 2)2 < ∞. (ii) Let r ∈ [1, R). Then Bn,q (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ (1 + r) · 9Kr (f ) , 2[n]q 2[n]q [n]q

for all n ∈ N, |z| ≤ r, where Kr (f ) =

P∞

k=3

|ck |k(k − 1)(k − 2)2 rk < ∞.

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Proof. (i) Denoting ek (z) = z k , k = 0, 1, ..., and πk,n,q (z) = Bn,q (ek )(z), we can P∞ write Bn,q (f )(z) = k=0 ck πk,n,q (z), which immediately implies Bn,q (f )(z) − f (z) − z(1 − z) f 00 (z) 2[n]q ∞ X z k−1 (1 − z)k(k − 1) ≤ |ck | · πk,n,q (z) − ek (z) − , 2[n]q k=2

for all z ∈ D1 , n ∈ N. Denote Dq (f )(z) = formula

f (z)−f (qz) , z−qz

q 6= 1. In what follows, we prove the recurrence

z(1 − z) Dq [πk,n,q ](z) + zπk,n,q (z), [n]q for all n ∈ N, z ∈ C and k = 0, 1, 2, .... For z = 0 and z = 1, this recurrence is obviously satisfied. Therefore let us suppose z 6= 0 and z 6= 1. Denoting n X n [j]kq z j Πn−1−j (1 − q s z), Sk,n,q (z) = s=0 j q j=0 πk+1,n,q (z) =

n−j

and taking into account the formulas q j 1−q 1−q

= [n]q − [j]q and

Dq [f · g](z) = g(z)Dq (f )(z) + f (qz)Dq (g)(z), Dq (z j )(z) = [j]q z j−1 ,

we obtain

Dq [Sk,n,q ](z) n−1−j n X Y n = [j]kq {Dq (z j )(z) (1 − q s z) + q j z j Dq (Πn−1−j (1 − q s z))} s=0 j q s=0 j=0

n−1−j n Y qj 1 − q n−j Sk+1,n,q (z) X k n j − [j]q (1 − q s z) · z = z 1−z 1−q j q s=0 j=0 =

Sk+1,n,q (z) [n]q Sk+1,n,q (z) Sk+1,n,q (z) [n]q − Sk,n,q (z) + = − Sk,n,q (z). z 1−z 1−z z(1 − z) 1−z

Dividing now by [n]k+1 , the recurrence formula for πk,n,q (z) is immediate. Note q that from this recurrence we easily obtain that degree (πk,n,q (z)) = k. Now, let us denote Ek,n,q (z) = πk,n,q (z) − ek (z) −

z k−1 (1−z)k(k−1) . 2[n]q

For all k ≥ 2, n ∈ N and z ∈ D1 , the above recurrence leads us to z(1 − z) Ek,n,q (z) = Dq [Ek−1,n,q ](z) + zEk−1,n,q (z) + Gk,n,q (z), [n]q where z k−2 (1 − z) (k − 1)(k − 2)[k − 2]q Gk,n,q (z) = · 2[n]q [n]q k−2 z (1 − z) (k − 1)[k − 1]q (k − 2) − ·z + 2(k − 1) − 2[k − 1]q , 2[n]q [n]q

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Taking into account that by the mean value theorem in complex analysis we have |Dq (f )(z)| ≤ kf 0 k1 , where k · k1 denotes the uniform norm in C(D1 ), and by using the relationships 0 < 1 − q < [n]1 q and (k − 1) − [k − 1]q = (1 − q) + ... + (1 − q k−2 )

= (1 − q)[1 + (1 + q) + ... + (1 + q + ...q k−3 )] 1 (k − 2)(k − 1) ≤ (1 + 2 + ... + k − 2) = , [n]q 2[n]q

we obtain, for all |z| ≤ 1, k ≥ 2, n ∈ N, |z| · |1 − z| · |z|k−3 |z| · |1 − z| 0 [2kEk−1,n,q k1 ] + |Ek−1,n,q (z)| + |Ek,n,q (z)| ≤ 2[n]q 2[n]q (k − 1)(k − 2)[k − 2]q (k − 1)(k − 2)[k − 1]q · + + 2(k − 1) − 2[k − 1]q [n]q [n]q |z| · |1 − z| ≤ |Ek−1,n,q (z)| + 2[n]q (k − 2)(k − 1) 2(k − 1)(k − 2)[k − 1]q 0 + · 2kEk−1,n,q k1 + [n]q [n]q |z| · |1 − z| ≤ |Ek−1,n,q (z)| + 2[n]q (k − 2)(k − 1) 2(k − 1)(k − 2)[k − 1]q + · 2(k − 1)kEk−1,n,q k1 + [n]q [n]q |z| · |1 − z| ≤ |Ek−1,n,q (z)| + [2(k − 1)kπk−1,n,q − ek−1 k1 2[n]q 2(k − 1)(k − 2)[k − 1]q kek−2 (z) − ek−1 (z)k1 + + + 2(k − 1)2 (k − 2) 2[n]q [n]q (k − 2)(k − 1) + ≤ (by the proof of Theorem 1.5.1) [n]q |z| · |1 − z| 2(k − 2)[k − 2]q ≤ |Ek−1,n,q (z)| + 2(k − 1) 2[n]q [n]q 2 (k − 2)(k − 1) 2(k − 1) (k − 2) 2(k − 1)(k − 2)[k − 1]q + + + [n]q [n]q [n]q 9|z| · |1 − z|k(k − 1)(k − 2) ≤ |Ek−1,n,q (z)| + . 2[n]2q Therefore, we have obtained |Ek,n,q (z)| ≤ |Ek−1,n,q (z)| +

9|z| · |1 − z|k(k − 1)(k − 2) . 2[n]2q

The last inequality is trivial for k = 1, 2, since E1,n,q (z) = E2,n,q (z) = 0, for any z ∈ C.

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Writing now the last inequality for k = 3, 4, ..., step by step we easily obtain |Ek,n,q (z)| ≤ ≤

k |z| · |1 − z| 9 X · · j(j − 1)(j − 2) 2[n]q [n]q j=3

|z| · |1 − z| 9 · · k(k − 1)(k − 2)2 . 2[n]q [n]q

In conclusion, ∞ X Bn,q (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ |ck | · |Ek,n,q (z)| 2[n]q k=3

∞ X

9 |z| · |1 − z| |ck |k(k − 1)(k − 2)2 . · · 2[n]q [n]q k=3 P∞ (4) Note that since f (z) = k=4 ck k(k − 1)(k − 2)(k − 3)z k−4 and the series is P∞ absolutely convergent in D1 , it easily follows that k=3 |ck |k(k − 1)(k − 2)2 < ∞, which proves (i). (ii) First we use the relationship in the proof of Theorem 1.5.1 ≤

|πk,n,q (z) − ek (z)| ≤ 2rk

(k − 1)[k − 1]q , [n]q

for all k, n ∈ N, |z| ≤ r, with 1 ≤ r < R. Denoting with k · kr the norm in C(Dr ), where Dr = {z ∈ C; |z| ≤ r}, one observes that by a linear transformation, the Bernstein’s inequality in the closed unit disk becomes |Pk0 (z)| ≤ kr kPk kr , for all |z| ≤ r, r ≥ 1, which combined with the mean value theorem in complex analysis, implies |Dq (Pk )(z)| ≤ kPk0 kr ≤ kr kPk kr , for all |z| ≤ r, where Pk (z) is a complex polynomial of degree ≤ k. Now, taking into account the formula proved at the above point (i), given by Ek,n,q (z) =

z(1 − z) Dq [Ek−1,n,q ](z) + zEk−1,n,q (z) [n]q z k−2 (1 − z) (k − 1)(k − 2)[k − 2]q + 2[n]q [n]q (k − 1)[k − 1]q (k − 2) −z , + 2(k − 1) − [k − 1]q [n]q

it follows for all k, n ∈ N, k ≥ 2 and |z| ≤ r, |Ek,n,q (z)| ≤

r(1 + r) (1 + r)rk−2 |Dq [Ek−1,n,q ](z)| + r|Ek−1,n,q (z)| + · [n]q 2[n]q

(k − 1)(k − 2)[k − 2]q +r [n]q

≤ r|Ek−1,n,q (z)| +

(k − 1)(k − 2)[k − 1]q (k − 2)(k − 1) + [n]q [n]q

r(1 + r) 3(1 + r)rk−1 k(k − 1)(k − 2) |Dq [Ek−1,n,q ](z)| + . [n]q 2[n]2q

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Now, we will estimate |Dq [Ek−1,n,q ](z)|, for k ≥ 3. Taking into account that Ek−1,n,q (z) is a polynomial of degree ≤ (k − 1), we obtain k−1 kEk−1,n,q (z)kr |Dq [Ek−1,n,q ](z)| ≤ r k−1 (k − 1)(k − 2)(ek−1 − ek−2 ) kr ≤ kπk−1,n,q − ek−1 kr + k r 2[n]q k − 1 2(k − 2)[k − 2]q rk 2rk−1 (k − 1)(k − 2) ≤ + r [n]q 2[n]q ≤

3rk−1 k(k − 1)(k − 2) . [n]q

This implies 3(1 + r)k(k − 1)(k − 2)r k r(1 + r) |Dq [Ek−1,n,q ](z)| ≤ , [n]q [n]2q and |Ek,n,q (z)| ≤ r|Ek−1,n,q (z)| +

3(1 + r)k(k − 1)(k − 2)r k−1 3(1 + r)k(k − 1)(k − 2)r k + [n]2q 2[n]2q

≤ r|Ek−1,n,q (z)| +

9(1 + r)k(k − 1)(k − 2)r k . 2[n]2q

But E0,n (z) = E1,n (z) = E2,n (z) = 0, for any z ∈ C. Writing now the last inequality for k = 3, 4, ..., step by step we easily obtain   k 9(1 + r)rk X |Ek,n,q (z)| ≤ j(j − 1)(j − 2) 2[n]2q j=3 ≤

9(1 + r)k(k − 1)(k − 2)2 rk . 2[n]2q

As a conclusion, we obtain X ∞ Bn,q (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ |ck | · |Ek,n (z)| 2[n]q k=3

∞ 9(1 + r) X ≤ |ck |k(k − 1)(k − 2)2 rk . 2[n]2q k=3 P∞ (4) Note that since f (z) = k=4 ck k(k − 1)(k − 2)(k − 3)z k−4 and the series is P∞ absolutely convergent in |z| ≤ r, it easily follows that k=3 |ck |k(k − 1)(k − 2)2 rk < ∞, which proves (ii).

Remarks. 1) In the hypothesis on f in Theorem 1.5.2 and choosing 0 < qn < 1 with qn % 1 as n → ∞, it follows that z(1 − z)f 00 (z) , lim [n]qn [Bn,qn (f )(z) − f (z)] = n→∞ 2

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uniformly in any compact disk included in the open disk of center 0 and radius R. 2) In Videnskii [194], Theorem 5.1, estimate (5.7), for the real q-Bernstein polynomials it is proved that for f ∈ C 2 [0, 1], x ∈ [0, 1], and 0 < qn < 1 with limn→∞ qn = 1, we have 00 Bn,qn (f )(x) − f (x) − f (x) · x(1 − x) ≤ Kx(1 − x) ω1 (f 00 ; [n]q−1/2 ), n 2 [n]qn [n]qn

where ω1 denotes the modulus of continuity. Obviously that the best order of −3/2 approximation that can be obtained from this estimate is O(1/[n]qn ) (for f ∈ C m [0, 1] with m ≥ 3), while the order given by our Theorem 1.5.2 is O(1/[n]−2 qn ), which is essentially better taking into account that as n → ∞ we have [n]qn → ∞. In what follows we obtain the exact orders in approximation by complex qBernstein polynomials and their derivatives on compact disks. In this sense, the first result is the following. Theorem 1.5.3. (Gal [87]) Let 0 < qn ≤ 1 be with limn→∞ qn = 1, R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR , that P∞ is we can write f (z) = k=0 ck z k , for all z ∈ DR . If f is not a polynomial of degree ≤ 1, then for any r ∈ [1, R) we have kBn,qn (f ) − f kr ≥

Cr (f ) , n ∈ N, [n]qn

where kf kr = max{|f (z)|; |z| ≤ r} and the constant Cr (f ) > 0 depends on f , r and on the sequence (qn )n∈N but it is independent of n. For all z ∈ DR and n ∈ N we have z(1 − z) 00 1 f (z) Bn,qn (f )(z) − f (z) = [n]qn 2 1 z(1 − z) 00 2 + [n]qn Bn,qn (f )(z) − f (z) − f (z) . [n]qn 2[n]qn

Proof.

We will apply to this identity the following obvious property : kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . Denoting e1 (z) = z it follows kBn,qn (f ) − f kr ≥

1 [n]qn

e1 (1 − e1 ) 00

1 e1 (1 − e1 ) 00 2

f − f [n]qn Bn,qn (f ) − f − .

2 [n] 2[n] q q n n r r

Taking into DR , we get z(1−z) 00 f (z) 2

account that

by hypothesis f is not a polynomial of degree ≤ 1 in

e1 (1−e1 ) 00 f > 0. Indeed, supposing the contrary it follows that

2 r

= 0 for all z ∈ Dr , which implies f 00 (z) = 0 for all z ∈ Dr \ {0, 1}. Since f is supposed to be analytic, from the identity’s theorem of analytic (holomorphic)

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functions this necessarily implies that f 00 (z) = 0, for all z ∈ DR , i.e. that f is a polynomial of degree ≤ 1, which is a contradiction. But by Theorem 1.5.2 we have

e1 (1 − e1 ) 00

≤ ([n]qn )2 9Kr (f )(1 + r) ([n]qn )2 f (f ) − f − B

n,qn

2[n]qn 2([n]qn )2 r 9Kr (f )(1 + r) = . 2 Since by the Remark after the proof of Theorem 1.5.1 we have [n]1q → 0 as n → ∞, n it follows that there exists an index n0 depending only on f , r and on the sequence (qn )n , such that for all n ≥ n0 we have

e1 (1 − e1 ) 00 e1 (1 − e1 ) 00 1 2

([n]qn ) Bn,qn (f ) − f − f − f

2 [n]qn 2[n]qn r r

e1 (1 − e1 ) 00 f ≥

> 0,

2 r which immediately implies

1 kBn,qn (f ) − f kr ≥ [n]qn For 1 ≤ n ≤ n0 − 1 we obviously have

e1 (1 − e1 ) 00

· f

, ∀n ≥ n0 . 4 r

kBn,qn (f ) − f kr ≥

Mr,n (f ) , [n]qn

with Mr,n (f ) = [n]qn · kBn,qn (f ) − f kr > 0, which finally implies Cr (f ) , for all n ∈ N, [n]qn

o n

1 ) 00 where Cr (f ) = min Mr,1 , Mr,2 (f ), ..., Mr,n0 −1 (f ), e1 (1−e f

. 4 kBn,qn (f ) − f kr ≥

r

Combining Theorem 1.5.3 with Theorem 1.5.1 we get the following.

Corollary 1.5.4. (Gal [87]) Let Let 0 < qn ≤ 1 be with limn→∞ qn = 1, R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If f is not a polynomial of degree ≤ 1, then for any r ∈ [1, R) we have kBn,qn (f ) − f kr ∼

1 , n ∈ N, [n]qn

where the constants in the equivalence depend on f , r and on the sequence (q n )n but are independent of n. Proof. Since qn ≤ 1 for all n ∈ N, by Theorem 1.5.1 it follows the upper estimate with the constant depending only on f and r (independent of the sequence (qn )n ). Theorem 1.5.3 assures the lower estimate with the constant depending on f , r and on the sequence (qn )n , but independent of n.

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Remark. Theorem 1.5.3 and Corollary 1.5.4 in the case when qn = 1 for all n ∈ N were obtained by Theorem 1.1.4 and Corollary 1.1.5. In the case of approximation by the derivatives of complex q-Bernstein polynomials we can present the following new result which appears for the first time here. Theorem 1.5.5. Let 0 < qn ≤ 1 be with limn→∞ qn = 1, R > 1, DR = {z ∈ C; |z| < P∞ R} and let us suppose that f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR . Also, let 1 ≤ r < r1 < R and p ∈ N be fixed. If f is not a polynomial of degree ≤ max{1, p − 1}, then we have 1 (p) kBn,q (f ) − f (p) kr ∼ , n [n]qn where the constants in the equivalence depend on f , r, r1 , p and on the sequence (qn )n , but are independent of n. Proof. Denoting by Γ the circle of radius r1 > and center 0 (where r1 > r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N we have Z Bn,qn (f )(v) − f (v) p! (p) (p) dv, Bn,qn (f )(z) − f (z) = 2πi Γ (v − z)p+1 which by Theorem 1.5.1 and by the inequality |v − z| ≥ r1 − r valid for all |z| ≤ r and v ∈ Γ, immediately implies 2πr1 p! (p) · kBn,qn (f ) − f kr1 kBn,q (f ) − f (p) kr ≤ n 2π (r1 − r)p+1 p!r1 ≤ Mr1 (f ) . [n]qn (r1 − r)p+1 (p)

It remains to prove the lower estimate for kBn,qn (f ) − f (p) kr . For this purpose, as in the proof of Theorem 1.5.3, for all v ∈ Γ and n ∈ N we have 1 v(1 − v) 00 Bn,qn (f )(v) − f (v) = f (v) [n]qn 2 v(1 − v) 00 1 2 f (v) , ([n]qn ) Bn,qn (f )(v) − f (v) − + [n]qn 2[n]qn which replaced in the above Cauchy’s formula implies Z 1 p! v(1 − v)f 00 (v) (p) (p) (f )(z) − f (z) = Bn,q dv n [n]qn 2πi Γ 2(v − z)p+1  Z n2 Bn,q (f )(v) − f (v) − v(1−v) f 00 (v)  n 2[n]qn p! 1 · + dv  [n]qn 2πi Γ (v − z)p+1 ( (p) z(1 − z) 00 1 = f (z) [n]qn 2  Z ([n]q )2 Bn,q (f )(v) − f (v) − v(1−v) f 00 (v)  n n 2[n]qn 1 p! + · dv .  [n]qn 2πi Γ (v − z)p+1

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Passing now to k · kr it follows

(

e (1 − e ) (p)

1

1

(p)

1 (p) 00 f

Bn,qn (f ) − f ≥

[n]qn 2 r r 

 Z ([n]q )2 Bn,q (f )(v) − f (v) − v(1−v) f 00 (v)

n n 2[n]qn p! 1

dv −

, p+1 [n]qn 2π Γ (v − e1 )

r

where by using Theorem 1.5.2 and denoting e1 (z) = z we get

Z ([n]q )2 Bn,q (f )(v) − f (v) − v(1−v) f 00 (v)

p! n n 2[n]qn

dv

2π p+1 (v − e1 ) Γ

r

e1 (1 − e1 ) 00 p! 2πr1 ([n]qn )2 f k r1 · kBn,qn (f ) − f − ≤ p+1 2π (r1 − r) 2[n]qn 5Kr1 (f )(1 + r1 )2 p!r1 ≤ · . 2 (r1 − r)p+1

h

e1 (1−e1 ) 00 i(p)

> 0. Indeed, supposing the

f But by hypothesis on f we have 2

r

f 00 (z) is a polynomial of degree ≤ p − 1. Now, if contrary it follows that z(1−z) 2 p = 1 and p = 2 then the analyticity of f obviously implies that f necessarily is a polynomial of degree ≤ 1 = max{1, p − 1}, which contradicts the hypothesis. If p > 2 then the analyticity of f obviously implies that f necessarily is a polynomial of degree ≤ p − 1 = max{1, p − 1}, which again contradicts the hypothesis. In continuation reasoning exactly as in the proof of Theorem 1.5.3, we immediately get the desired conclusion.

Remark. Theorem 1.5.5 in the case when qn = 1 for all n ∈ N was obtained by Theorem 1.1.6. In what follows we consider the approximation properties for iterates. First we recall some considerations in Section 1.2. For R > 1 let us define by AR the space of all functions defined and analytic in the open disk of center 0 and radius R denoted by DR . Denoting rj = R− R−1 j , j ∈ N and for f ∈ AR , kf kj = max{|f (z)|; |z| ≤ rj }, since r1 = 1 and rj % R as j → ∞, it is well-known that {k · kj , j ∈ N} it is a countable family of increasing semi-norms on AR and that AR becomes a metrizable complete locally convex space (Fr´echet space), with respect to the metric ∞ X kf − gkj 1 · , f, g ∈ AR . d(f, g) = j 1 + kf − gk 2 j j=1

It is well-known that limn→∞ d(fn , f ) = 0 is equivalent to the fact that the sequence (fn )n∈N converges to f uniformly on compacts in DR . P∞ Now, for f ∈ AR , that is of the form f (z) = k=0 ck z k , for all z ∈ DR , let us (1) define the iterates of complex q-Bernstein polynomial Bn,q (f )(z), by Bn,q (f )(z) =

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(m−1)

Bn,q (f )(z) and Bn,q (f )(z) = Bn,q [Bn,q (f )](z), for any m ∈ N, m ≥ 2. Since we P have Bn,q (f )(z) = ∞ k=0 ck Bn,q (ek )(z), by recurrence for all m ≥ 1, it easily follows P (m) (m) k Bn,q (f )(z) = ∞ c k=0 k Bn,q (ek )(z), with ek (z) = z . The main result is the following. P∞ Theorem 1.5.6. (Gal [87]) Let f ∈ AR , that is f (z) = k=0 ck z k , for all z ∈ DR . (i) Let q ∈ (0, 1). If limn→∞ mn = +∞, then (mn ) (f ), L1 (f )] = 0. lim d[Bn,q

n→∞

(ii) If q ∈ (0, 1] then for any fixed s ∈ N, the following estimates hold (m) kBn,q (f ) − f ks ≤

∞ 2m X |ck |k(k − 1)rsk , [n]q k=2

and (m) d[Bn,q (f ), f ] ≤

P∞

1)rsk

where k=2 |ck |k(k − < ∞. (iii) Let q ∈ (1, ∞). If limn→∞

∞ 2m X 1 |ck |k(k − 1)rsk + s , [n]q 2 k=2

mn [n]q

= 0, then

(mn ) lim d[Bn,q (P ), P ] = 0,

n→∞

for any polynomial P . (iv) Let q ∈ (1, ∞). If 1 ≤ r < R, then the following estimate holds for all |z| ≤ r k ∞ q − 1 − k(q − 1) 2m X (m) |ck | + k(k − 1) rk . |Bn,q (f )(z) − f (z)| ≤ n (q − 1)2 k=2 i h k P∞ q −1−k(q−1) + k(k − 1) < ∞, we If, in addition, q < R, then since |c | 2 k k=2 (q−1) (m )

obtain Bn,qn (f )(z) → f (z), uniformly in D1 , for mn = ∞, then (v) Let q ∈ (1, ∞). If limn→∞ [n] q

mn n

→ 0 as n → ∞.

(mn ) lim d[Bn,q (P ), L1 (P )] = 0,

n→∞

for any polynomial P . Proof. (i) Let 0 < q < 1 and n ∈ N be fixed. Denoting s = min{n, k}, by Ps j Lemma 3 in Ostrovska [146] we can write Bn,q (ek )(z) = j=1 αj,k,n,q z , where Ps α( j, k, n, q) ≥ 0, for all j, k, n ∈ N and j=1 αj,k,n,q = 1, for k, n ∈ N. Let |z| ≤ r with r ≥ 1. It follows |Bn,q (ek )(z)| ≤

s X j=1

αj,k,n,q |ej (z)| ≤

s X j=1

αj,k,n,q rj ≤

s X j=1

αj,k,n,q rs = rs ≤ rk .

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Applying now above Bn,q , we obtain (2) Bn,q (ek )(z) =

s X

αj,k,n,q Bn,q (ej )(z),

j=1

which from the last inequality implies (2) |Bn,q (ek )(z)| ≤

≤

s X

j=1 s X j=1

αj,k,n,q |Bn,q (ej )(z)| αj,k,n,q rj ≤

Reasoning by recurrence, we easily get z ∈ Dr . This implies (m) |Bn,q (f )(z)| = |

∞ X k=0

s X j=1

αj,k,n,q rs = rs ≤ rk .

(m) |Bn,q (ek )(z)|

(m) ck Bn,q (ek )(z)| ≤

≤ rk for all k, n, m ∈ N and

∞ X k=0

|ck |rk < +∞, (m)

for all m, n ∈ N and z ∈ Dr , that is for each r ∈ [1, R), the sequence Bn,q (f )(z), m, n = 1, 2, ..., is uniformly bounded in Dr with respect to both m, n ∈ N. (m ) Therefore, the sequence Bn,qn (f )(z), n = 1, 2, ..., is uniformly bounded in Dr with respect to n ∈ N. Since by Theorem 8 in Ostrovska [146], for each q ∈ (0, 1) and for n → ∞ we (m ) have Bn,qn (f )(x) → L1 (f )(x), for x ∈ [0, 1], (even uniformly), the classical Vitali’s convergence theorem implies the uniform convergence (as n → ∞) on compacts in (m ) DR of the sequence Bn,qn (f )(z) to L1 (f )(z). Taking into account that the uniform convergence on compacts is equivalent to the convergence with respect to the metric d, (i) is proved. P∞ (m) (m) (ii) Since Bn,q (f )(z) = k=0 ck Bn,q (ek )(z), with ek (z) = z k , we get (m) |Bn,q (f )(z)

− f (z)| ≤

∞ X

k=2

(m) |ck | · |Bn,q (ek )(z) − ek (z)|.

(m) |Bn,q (ek )(z)

To estimate − ek (z)|, we have two possibilities : 1) 0 ≤ k ≤ n ; 2) k > n ≥ 1. Case 1). According to Lemma 3 in Ostrovska [146], we have Bn,q (ek )(z) − ek (z) =

k X j=1

αj,k,n,q ej (z) − ek (z).

Therefore, Bn,q (ek )(z) − ek (z) =

k−1 X j=1

αj,k,n,q ej (z) + [αk,k,n,q − 1]ek (z),

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which immediately implies (p) Bn,q [Bn,q (ek )(z)

− ek (z)] =

k−1 X j=1

(p) (p) αj,k,n,q Bn,q (ej )(z) + [αk,k,n,q − 1]Bn,q (ek )(z). (p)

Taking into account that by the proof of (i) we have |Bn,q (ej )(z)| ≤ rj , for all p, n, j ∈ N and |z| ≤ r with r ≥ 1, it follows (p) |Bn,q [Bn,q (ek ) − ek ](z)|

≤ ≤

k−1 X j=1

k−1 X j=1

(p) (p) αj,k,n,q |Bn,q (ej )(z)| + |1 − αk,k,n,q | · |Bn,q (ek )(z)|

αj,k,n,q rj + |1 − αk,k,n,q |rk ≤ 2|1 − αk,k,n,q |rk .

But (m) Bn,q (ek )(z)

− ek (z) =

m−1 X p=0

(p) Bn,q [Bn,q (ek )(z) − ek (z)],

which implies, for all |z| ≤ r (m) |Bn,q (ek )

− ek | ≤

m−1 X p=0

(p) |Bn,q [Bn,q (ek ) − ek ](z)| ≤ 2m|1 − αk,k,n,q |rk .

Since by the same Lemma 3 in Ostrovska [146], we have 1 [k − 1]q αk,k,n,q = 1 − ... 1 − , [n]q [n]q by using the inequality 1−

k Y

j=1

xj ≤

k X j=1

(1 − xj ), 0 ≤ xj ≤ 1, j = 1, ..., k,

it follows |1 − αk,k,n,q | ≤ = =

k−1 X j=1

k−1 [j]q 1 X = [j]q [n]q [n]q j=1

k−2 k−2 1 X j 1 X q [k − (j + 1)] ≤ [k − (j + 1)] [n]q j=0 [n]q j=0

1 k(k − 1) [k(k − 1) − k(k − 1)/2] = . [n]q 2[n]q

Therefore (m) |Bn,q (ek ) − ek | ≤

m k(k − 1)rk . [n]q

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Case 2). For 1 ≤ r < R, |z| ≤ r and k > n ≥ 1, we get (m) (m) |Bn,q (ek )(z) − ek (z)| ≤ |Bn,q (ek )(z)| + |ek (z)|

≤ 2rk ≤ 2

k(k − 1) k k(k − 1) k r ≤2 r , n [n]q

since for q ∈ (0, 1] we have [n]q ≤ n. As a conclusion, for both Cases 1) and 2), for |z| ≤ r we obtain (m) |Bn,q (f )(z) − f (z)| ≤

=

∞ X k=1

n X k=1

+ ≤

(m) |ck | · |Bn,q (ek )(z) − ek (z)| (m) |ck | · |Bn,q (ek )(z) − ek (z)| ∞ X

k=n+1 ∞ X

2m [n]q

k=2

(m) |ck | · |Bn,q (ek )(z) − ek (z)|

|ck |k(k − 1)rk .

Now, by choosing r = rs , we get the first required inequality in the statement. Note P∞ P∞ that k=2 |ck |k(k − 1)rk−2 < ∞, since we have f 00 (z) = k=2 ck k(k − 1)z k−2 , for all |z| ≤ r. The second estimate in (ii) is a direct consequence of the inequality d(f, g) =

∞ s X X kf − gkj 1 kf − gkj 1 · + · j j 2 1 + kf − gkj j=s+1 2 1 + kf − gkj j=1

s ∞ X kf − gks X 1 1 ≤ + 1 + kf − gks j=1 2j j=s+1 2j

≤

kf − gks 1 1 + s ≤ kf − gks + s . 1 + kf − gks 2 2

This proves (ii). (iii) The proof is similar with that of the point (i), by taking into account that from Theorem 10 in Ostrovska [146], for each q ∈ (1, ∞), any polynomial P and for (m ) mn = 0 we have limn→∞ Bn,qn (P )(x) = P (x), x ∈ [0, 1]. limn→∞ [n] q (iv) Let us suppose that q ∈ (1, ∞) and |z| ≤ r, with 1 ≤ r < R. We reason (m) exactly as in the proof of the point (ii). First, to estimate |Bn,q (ek )(z) − ek (z)|, again we have two possibilities : 1) 0 ≤ k ≤ n ; 2) k > n ≥ 1. Case 1). Reasoning exactly as at the proof of the above point (ii), Case 1), by simple calculation we obtain (m) (ek )(z) − ek (z)| ≤ |Bn,q

k−2 2m X j 2m q k − 1 − k(q − 1) k q [k − (j + 1)]rk = · r . [n]q j=0 [n]q (q − 1)2

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Since for q ∈ (1, ∞) we have n ≤ [n]q , it follows

2m q k − 1 − k(q − 1) k · r . n (q − 1)2 Case 2). Reasoning exactly as the proof of the above point (ii), Case 2, we get k(k − 1) k (m) |Bn,q (ek )(z) − ek (z)| ≤ 2 r . n As a conclusion, collecting the estimates in the Cases 1) and 2), we get ∞ X (m) (m) |Bn,q (f )(z) − f (z)| ≤ |ck | · |Bn,q (ek )(z) − ek (z)| (m) |Bn,q (ek )(z) − ek (z)| ≤

k=1

=

n X k=2

+ ≤

(m) |ck | · |Bn,q (ek )(z) − ek (z)| ∞ X

k=n+1 n X

2m n +

2 n

(m) (ek )(z) − ek (z)| |ck | · |Bn,q

|ck | ·

k=2 ∞ X

k=n+1 ∞ X

q k − 1 − k(q − 1) k r (q − 1)2

|ck | · k(k − 1)rk

q k − 1 − k(q − 1) + k(k − 1) rk . |ck | (q − 1)2 k=2 P∞ If, in addition we take r = 1 and q < R, then obviously k=2 |ck |q k < ∞ and P∞ mn k=2 |ck |k(k − 1) < ∞, which for n → 0 as n → ∞, by the above inequality (mn ) implies that Bn,q (f )(z) → f (z), uniformly in D1 . (v) Let q ∈ (1, ∞) and 1 ≤ r < R be arbitrary. Taking into account that for any mn = ∞ as n → ∞ polynomial P , by Theorem 10 in Ostrovska [146], for limn→∞ [n] q 2m ≤ n

(m )

we have Bn,qn (P )(x) → L1 (P )(x), uniformly for x ∈ [0, 1] and since by the above (m ) point (i), it is immediate that Bn,qn (P )(z), n = 1, 2, ..., is uniformly bounded in Dr with respect to n ∈ N, by Vitali’s theorem it follows the uniform convergence (m ) on Dr to L1 (P )(z) of Bn,qn (P )(z) (as n → ∞). Because r is arbitrary, it follows the convergence in d too (see the considerations at the beginning of this section). Note that for the Vitali’s convergence result, the uniform convergence on [0, 1] is not necessary, it suffices to have only pointwise convergence there. The theorem is proved. Finally we present the geometric properties for the complex q-Bernstein polynomials Bn,qn (f )(z), with 0 < qn < 1, and qn → 1 as n → ∞. Theorem 1.5.7. (Gal [87]) Let us suppose that G ⊂ C is open, such that D1 ⊂ G and f : G → C is analytic in G. Also, let us consider Bn,qn (f )(z), with 0 < qn < 1, and qn → 1 as n → ∞.

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If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) in D1 , that is for all z ∈ D1 (see e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138]) 0 00 0 zf (z) zf (z) iη zf (z) Re > 0 Re + 1 > 0, Re e > 0, resp. , f (z) f 0 (z) f (z) then there exists an index n0 depending on f (and on η for spirallikeness), such that for all n ≥ n0 , Bn,qn (f )(z), are starlike (convex, spirallike of type η, respectively) in D1 . If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) only in D1 (that is the corresponding inequalities hold only in D1 ), then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, Dr ) (n0 depends on η too in the case of spirallikeness), such that for all n ≥ n 0 , Bn,qn (f )(z), are starlike (convex, spirallike of type η, respectively) in D r (that is, the corresponding inequalities hold in Dr ). Proof. By Theorem 2 in Phillips [149] and by the classical Vitali’s theorem, it follows that we have Bn,qn (f )(z) → f (z), uniformly for |z| ≤ 1, which by the wellknown Weierstrass’s theorem implies [Bn,qn (f )]0 (z) → f 0 (z) and [Bn,qn (f )]00 (z) → f 00 (z), for n → ∞, uniformly in D1 . In all what follows, denote Pn (f )(z) = Bn,qn (f )(z) [Bn,qn (f )]0 (0) , well defined for sufficiently large n. We easily get Pn (f )(0) = 0, Pn0 (f )(0) = 1 for sufficiently large n, and Pn (f )(z) → f (z), Pn0 (f )(z) → f 0 (z) and Pn00 (f )(z) → f 00 (z), uniformly in D1 . Suppose first that f is starlike in D1 . Then, by hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Writing Pn (f )(z) in the form Pn (f )(z) = zQn (f )(z), obviously Qn (f )(z) is a polynomial of degree ≤ n − 1. Also, for |z| = 1 we have |f (z) − Pn (f )(z)| = |z| · |g(z) − Qn (f )(z)| = |g(z) − Qn (f )(z)|, which by the uniform convergence in D1 of Pn (f ) to f and by the maximum modulus principle, implies the uniform convergence in D1 of Qn (f )(z) to g(z). Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index n1 ∈ N and a > 0 depending on g, such that |Qn (f )(z)| > a > 0, for all z ∈ D1 and all n ≥ n0 . Also, for all |z| = 1, we have |f 0 (z) − Pn0 (f )(z)| = |z[g 0 (z) − Q0n (f )(z)] + [g(z) − Qn (f )(z)]|

≥ | |z| · |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| | = | |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| |,

which from the maximum modulus principle, the uniform convergence of Pn0 (f ) to f 0 and of Qn (f ) to g, evidently implies the uniform convergence of Q0n (f ) to g 0 .

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Then, for |z| = 1, we get

z[zQ0n(f )(z) + Qn (f )(z)] zPn0 (f )(z) = Pn (f ) zQn (f )(z) zg 0 (z) + g(z) zQ0n (f )(z) + Qn (f )(z) → = Qn (f )(z) g(z) zf 0 (z) f 0 (z) = , = g(z) f (z)

which again from the maximum modulus principle, implies zPn0 (f )(z) zf 0 (z) → , uniformly in D1 . Pn (f ) f (z) Since Re

Therefore

zf 0 (z) f (z)

is continuous in D1 , there exists ε ∈ (0, 1), such that 0 zf (z) Re ≥ ε, for all z ∈ D1 . f (z)

0 zPn0 (f )(z) zf (z) Re → Re ≥ε>0 Pn (f )(z) f (z)

uniformly on D1 , i.e. for any 0 < ρ < ε, there is n0 such that for all n ≥ n0 we have 0 zPn (f )(z) Re > ρ > 0, for all z ∈ D1 . Pn (f )(z) Since Pn (f )(z) differs from Bn,qn (f )(z) only by a constant, this proves the starlikeness of Bn,qn (f )(z), for sufficiently large n. If f is supposed to be starlike only in D1 , the proof is identical, with the only difference that instead of D1 , we reason for Dr . The proofs in the cases when f is convex or spirallike of order η are similar and follow from the following uniform convergency (on D1 or on Dr ) 00 00 zPn (f )(z) zf (z) Re + 1 → Re +1 0 Pn (f )(z) f 0 (z) and

0 0 iη zPn (f )(z) iη zf (z) Re e → Re e . Pn (f )(z) f (z)

The theorem is proved. 1.6

Bernstein-Stancu Polynomials

In this section for two kinds of complex Bernstein-Stancu polynomials we study similar properties with those for the classical complex Bernstein polynomials and

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q-Bernstein polynomials in Sections 1.1, 1.2 and 1.5. More exactly we consider the following two kinds of polynomials : n X n k (α,β) z (1 − z)n−k f [(k + α)/(n + β)], z ∈ C, Sn (f )(z) = k k=0

where 0 ≤ α ≤ β are independent of n, (introduced and studied for the case of real variable in Stancu [173]) and n X Sn<γ> (f )(z) = p<γ> n,k (z)f (k/n), z ∈ C, k=0

(introduced and studied for the case of real variable in Stancu [174]) where γ ≥ 0 may to depend on n and p<γ> n,k (z) n z(z + γ)...(z + (k − 1)γ)(1 − z)(1 − z + γ)...(1 − z + (n − k − 1)γ) = . (1 + γ)(1 + 2γ)...(1 + (n − 1)γ) k (0,0)

Note that Sn (f )(z) = Sn<0> (f )(z) = Bn (f )(z). (α,β) We begin our study with Sn (f )(z). First we present upper estimates in simultaneous approximation. Theorem 1.6.1. (Gal [80]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR . Suppose 0 ≤ α ≤ β and 1 ≤ r < R are arbitrary fixed. For all |z| ≤ r and n ∈ N, we have (β)

M2,r (f ) , n+β P∞ P∞ (β) where 0 < M2,r (f ) = 2r2 j=2 j(j − 1)|cj |rj−2 + 2βr j=1 j|cj |rj−1 < ∞. Also, if 1 ≤ r < r1 < R, then for all |z| ≤ r and n, p ∈ N, we have |Sn(α,β) (f )(z) − f (z)| ≤

(β)

|[Sn(α,β) (f )](p) (z) − f (p) (z)| ≤ Proof.

(α,β)

Denoting ek (z) = z k , we get Sn |Sn(α,β) (f )(z) − f (z)| ≤ (α,β) |Sn (ek )(z) − ek (z)|

∞ X

k=0

M2,r1 (f )p!r1 . (n + β)(r1 − r)p+1

(f )(z) =

P∞

(α,β) (ek )(z) k=0 ck Sn

and

|ck | · |Sn(α,β) (ek )(z) − ek (z)|.

To estimate for fixed n ∈ N, we consider two possible cases : 1) 0 ≤ k ≤ n ; 2) k > n. Denoting by ∆k the finite difference of order k, we will use the representation formula (see Stancu [173]) n X n Sn(α,β) (f )(z) = ∆p1/(n+β) f (α/(n + β))ep (z). p p=0

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(α,β)

Case 1). If k = 0, then obviously we have Sn (ek )(z) − ek (z) = 0. Therefore, let us suppose that 1 ≤ k ≤ n and denote n (α,β) Cn,p,k = ∆p1/(n+β) ek (α/(n + β)) p n = [α/(n + β), (α + 1)/(n + β), ..., (α + p)/(n + β); ek ](p!)/(n + β)p . p (α,β)

Since ek is convex of any order, it follows Cn,p,k ≥ 0 and since Sn k Pn (α,β) ≤ 1. f [(n + α)/(n + β)], we get p=0 Cn,p,k = (n+α) (n+β)k For any |z| ≤ r with 1 ≤ r < R, we can write |Sn(α,β) (ek )(z) − ek (z)| =|

k X p=0

(α,β)

(α,β)

Cn,p,k ep (z) − ek (z)| = |[Cn,k,k − 1]ek (z) +

k−1 X

(f )(1) =

(α,β)

Cn,p,k ep (z)|

p=0

n(n − 1)...(n − (k − 1)) k r ≤ 1− (n + β)k (n + α)k n(n − 1)...(n − (k − 1)) k + − r (n + β)k (n + β)k (n + α)k n(n − 1)...(n − (k − 1)) k r + − 1 rk =2 1− (n + β)k (n + β)k 1 n(n − 1)...(n − (k − 1)) k [2βk + k(k − 1)] r k . r ≤ ≤2 1− k (n + β) n+β

Here we have applied the formula (easily proved by mathematical induction) 1 − Πkj=1 xj ≤

k X j=1

(1 − xj ), 0 ≤ xj ≤ 1, j = 1, ..., k.

Case 2). For 1 ≤ r < R, |z| ≤ r and k > n ≥ 1, we get |Sn(α,β) (ek )(z) − ek (z)| ≤ |Sn(α,β) (ek )(z)| + rk n X (α,β) ≤ Cn,p,k rp + rk ≤ rn + rk ≤ 2rk p=0

k+β k r n+β 2k(k − 1) + 2β(k − 1) k 2k(k − 1) + 2βk k = r ≤ r . n+β n+β ≤ 2nrk ≤ 2(k − 1) ·

Combining it with the above Case 1, we get the desired inequality. For the simultaneous approximation, denoting by Γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ Γ, we have |v − z| ≥ r1 − r, by the

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70

Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z (α,β) p! Sn (f )(v) − f (v) (α,β) (p) (p) |[Sn (f )] (z) − f (z)| = dv 2π Γ (v − z)p+1 (β)

M2,r1 (f ) p! 2πr1 ≤ n + β 2π (r1 − r)p+1 (β)

M2,r1 (f ) p!r1 = . n + β (r1 − r)p+1 P k Finally, since by hypothesis, f (z) = ∞ k ck z is absolutely and uniformly con(β) vergent in |z| ≤ r, for any 1 ≤ r < R, it is clear that M2,r (f ) < ∞. Remark. For α = β = 0 in Theorem 1.6.1 the estimates in Theorem 1.1.2 for classical Bernstein polynomials are obtained. A quantitative Voronovskaja-type formula follows. Theorem 1.6.2. (Gal [80]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . Let 0 ≤ α ≤ β. For all |z| ≤ 1 and n ∈ N, we have (α,β) Sn (f )(z) − f (z) + βz − α f 0 (z) − nz(1 − z) f 00 (z) n+β 2(n + β)2 (α,β)

≤

(α,β)

where 0 < M1 (α,β)

M1

|z| · |1 − z| (α,β) M (f ) M1 (f ) + 2 , (n + β)2 (n + β)2

(f ),

(f ) =

∞ X k=2

+

|ck |

9(k − 1)3 (k − 2) (k − 1)2 (k − 2)2 + + 4β(k − 1)3 2 2

3β(k − 1)2 (k − 2) 3α(k − 1)2 (k − 2) + + βk(k − 1)2 (k − 2) < ∞, 2 2 (α,β)

0 < M2

(f ) = (α + β)2

∞ X

k=2

|ck |

k(k − 1) < ∞. 2

(α,β)

Proof. Denoting ek (z) = z k and πn,k (z) = Sn (ek )(z), we obtain (α,β) Sn (f )(z) − f (z) + βz − α f 0 (z) − nz(1 − z) f 00 (z) 2 n+β 2(n + β) ∞ X z k−1 (βz − α)k nz k−1 (1 − z)k(k − 1) ≤ |ck | πn,k (z) − ek (z) + − . n+β 2(n + β)2 k=1

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P Differentiating the sum sn,k (z) = nj=0 (j + α)k nj z j (1 − z)n−j and then dividing the formula by (n + β)k+1 , by simple calculation we get the recurrence formula πn,k+1 (z) =

z(1 − z) 0 α + nz π (z) + πn,k (z), z ∈ C. n + β n,k n+β

Denoting Gn,k (z) = πn,k (z) − ek (z) + recurrence for πn,k (z) implies Gn,k (z) =

z k−1 (βz−α)k n+β

−

nz k−1 (1−z)k(k−1) , 2(n+β)2

the above

z(1 − z) 0 α + nz π (z) + πn,k−1 (z) − ek (z) n + β n,k−1 n+β z k−1 (βz − α)k nz k−1 (1 − z)k(k − 1) + − , n+β 2(n + β)2

which by simple calculation implies the following recurrence formula for Gn,k (z) (valid for all k ≥ 2 since Gn,0 (z) = Gn,1 (z) = 0) Gn,k (z) =

α + nz z(1 − z) 0 G (z) + Gn,k−1 (z) + A, n + β n,k−1 n+β

where z k−1 (1 − z)(k − 1) z k−2 (1 − z)(k − 1)[(k − 2)(βz − α) + βz] − n+β (n + β)2 z k−2 (1 − z)(k − 1)(k − 2)[(k − 2) − (k − 1)z] n+β β + − 2(n + β) (n + β)2 (n + β)2 α + nz k−1 α + nz k−2 + z − z (βz − α)(k − 1) n+β (n + β)2 n+β β α + nz k−2 n+β k + z (1 − z)(k − 1)(k − 2) − z − 2 2 2(n + β) (n + β) (n + β) n+β A :=

+

z k−1 (βz − α)k z k−1 (1 − z)k(k − 1) βz k−1 (1 − z)k(k − 1) − + . n+β 2(n + β) 2(n + β)2

In what follows, we will write the expression A in the form A := T1 + T2 + T3 , where T1 is the sum of all the terms containing (n + β) at the denominator, T2 is the sum of all the terms containing (n + β)2 at the denominator and T3 is the sum of all the terms containing (n + β)3 at the denominator. Therefore, by writing (for T2 ) α + nz = (α − βz) + (n + β)z, we obtain T1 =

T2 =

z k−1 [(1 − z)(k − 1) + (α + nz) − z(n + β) + k(βz − α) n+β z k−1 (k − 1) [z(k/2 + β − 1) + 1 − α − k/2], −k(k − 1)(1 − z)/2] = n+β

z k−2 (k − 1) (1 − z)(k − 2)z(n + β) (1 − z)(k − 2)z(−2β − (k − 1)) + (n + β)2 2 2

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(1 − z)(k − 2)[3α + (k − 2)] + − (βz − α)(n + β)z + (βz − α)2 2 z k−2 (k − 1) [(1 − z)z(k − 2)[−2β − (k − 1)] 2(n + β)2 + (1 − z)(k − 2)[3α + (k − 2)] + 2(βz − α)2

= −T1 +

= −T1 + +

z k−1 (1 − z)(k − 1)(k − 2) [−2β − (k − 1)] 2(n + β)2

z k−2 (1 − z)(k − 1)(k − 2)[3α + (k − 2)] z k−2 (k − 1) + (βz − α)2 , 2(n + β)2 (n + β)2 βz k−2 (1 − z)(k − 1)(k − 2)[(k − 2) − (k − 1)z] 2(n + β)3 β(α + nz)z k−2 (1 − z)(k − 1)(k − 2) − 2(n + β)2

T3 = −

= −

βz k−2 (1 − z)(k − 1)(k − 2) [z(n − (k − 1)) + α + (k − 2)]. 2(n + β)3

First, we observe that in the sum T1 + T2 + T3 , the terms containing (n + β) at the denominator cancel. Now, we will estimate |A| = |T1 + T2 + T3 | for |z| ≤ 1. For all k ≥ 2 we obtain |z| · |1 − z|(k − 1)(k − 2)[2β + (k − 1)] 2(n + β)2 |z| · |1 − z|(k − 1)(k − 2)[3α + (k − 2)] (k − 1)(β + α)2 + + 2(n + β)2 (n + β)2 |z| · |1 − z|β(k − 1)(k − 2)[n + α + 2k − 3] + 2(n + β)3

|A| ≤

(k − 1)(k − 2)[2β + (k − 1)] (k − 1)(k − 2)[3α + k − 2] + 2(n + β)2 2(n + β)2 β(k − 1)(k − 2)[n + α + 2k − 3] (k − 1)(β + α)2 + + 3 2(n + β) (n + β)2

= |z| · |1 − z|

≤

|z| · |1 − z| (k − 1)2 (k − 2)/2 + 2β(k − 1)(k − 2)/2 (n + β)2 3α(k − 2) β(k − 1)(k − 2) + (k − 1) (k − 2)2 /2 + + 2 2 β(α − β)(k − 1)(k − 2) (k − 1)(β + α)2 2βk(k − 1)(k − 2) + + + 2(n + β) 2(n + β) (n + β)2

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≤

73

|z| · |1 − z| (k − 1)2 (k − 2)/2 + β(k − 1)(k − 2) + (k − 1)(k − 2)2 /2 (n + β)2 β(k − 1)(k − 2) + 3α(k − 1)(k − 2)/2 + + βk(k − 1)(k − 2) 2 2 |z| · |1 − z| (k − 1)(α + β) 3β(k − 1)(k − 2) + = (k − 1)2 (k − 2)/2 + (n + β)2 (n + β)2 2 2 + (k − 1)(k − 2) /2 + 3α(k − 1)(k − 2)/2 + βk(k − 1)(k − 2)

+

(k − 1)(α + β)2 . (n + β)2

Therefore, denoting by k · k the uniform norm in the closed unit disk and estimating for |z| ≤ 1 the absolute value of Gn,k (z) in the formula of recurrence, also by using the Bernstein’s inequality (since Gn,k (z) is a polynomial of degree k) and the estimate for kπn,k − ek k in the proof of Theorem 1.6.1, we obtain n+α |z| · |1 − z| 0 kGn,k−1 k + |Gn,k−1 (z)| + |A| n+β n+β |z| · |1 − z|(k − 1) ≤ |Gn,k−1 (z)| + kGn,k−1 k + |A| n+β |z| · |1 − z|(k − 1) ≤ |Gn,k−1 (z)| + [kπn,k−1 − ek−1 k n+β (k − 1)(α + β) n(k − 1)(k − 2) + + + |A| ≤ |Gn,k−1 (z)| n+β (n + β)2 |z| · |1 − z|(k − 1) 2(k − 1)(k − 2) + 2β(k − 1) + n+β n+β (k − 1)(α + β) (k − 1)(k − 2) + + |A| ≤ |Gn,k−1 (z)| + n+β n+β |z| · |1 − z|(k − 1)2 2(k − 2) + 2β + (α + β) + (k − 2) · + n+β n+β 4|z| · |1 − z|(k − 1)2 (k − 2 + β) + |A| +|A| ≤ |Gn,k−1 (z)| + (n + β)2 4|z| · |1 − z|(k − 1)2 ≤ |Gn,k−1 (z)| + (k − 2 + β) (n + β)2 3β(k − 1)(k − 2) |z| · |1 − z| (k − 1)2 (k − 2)/2 + + (n + β)2 2

|Gn,k (z)| ≤

+ (k − 1)(k − 2)2 /2 + 3α(k − 1)(k − 2)/2 + βk(k − 1)(k − 2)

+

(k − 1)(α + β)2 := |Gn,k−1 (z)| + A(n, k, α, β)(z). (n + β)2

That is, in conclusion for all |z| ≤ 1 we can write |Gn,k (z)| ≤ |Gn,k−1 (z)| + A(n, k, α, β)(z).

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74

Since Gn,1 (z) = 0, reasoning by recurrence for k = 2, 3, ..., we immediately obtain |Gn,k (z)|

|z| · |1 − z| Xh 9(j − 1)2 (j − 2)/2 + 4β(j − 1)2 + (j − 1)(j − 2)2 /2 (n + β)2 j=2 i 3β(j − 1)(j − 2) + 3α(j − 1)(j − 2)/2 + βj(j − 1)(j − 2) + 2 k 2 X (α + β) (j − 1) + (n + β)2 j=2 |z| · |1 − z| h ≤ 9(k − 1)3 (k − 2)/2 + (k − 1)2 (k − 2)2 /2 + 4β(k − 1)3 (n + β)2 i 3β(k − 1)2 (k − 2) + 3α(k − 1)2 (k − 2)/2 + βk(k − 1)2 (k − 2) + 2 k(k − 1)(α + β)2 + . 2(n + β)2 As a conclusion, the desired estimate is immediate from X ∞ (α,β) Sn (f )(z) − f (z) + βz − α f 0 (z) − nz(1 − z) f 00 (z) ≤ |ck | · |Gn,k (z)|. n+β 2(n + β)2 k=1 k

≤

Remarks. 1) Taking now α = β = 0 in Theorem 1.6.2 we get the Voronovskaja’s theorem with an upper estimate for the classical Bernstein polynomials in Theorem 1.1.3. 2) Following exactly the lines in the proof of Theorem 1.6.2 it is immediate that in fact for any 1 ≤ r < R we have an upper estimate of the form

(α,β)

(α,β)

(f )

Sn (f ) − f + βe1 − α f 0 − ne1 (1 − e1 ) f 00 ≤ Mr ,

n+β 2(n + β)2 (n + β)2 r

(α,β) Mr (f )

where the constant > 0 is independent of n and depends on f , r, α and β. In what follows we prove that the degrees of approximation in Theorem 1.6.1 in fact are exact. Since the particular case α = β = 0 (that is the case of classical Bernstein polynomials) was already considered by Theorem 1.1.4, Corollary 1.1.5 and Theorem 1.1.6, in the next Theorem 1.6.3, Corollary 1.6.4 and Theorem 1.6.5 it will be excluded. First we present :

Theorem 1.6.3. (Gal [81]) Let R > 1, 0 ≤ α ≤ β with α + β > 0, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR , that is we can P∞ write f (z) = k=0 ck z k , for all z ∈ DR . If f is not a polynomial of degree 0 and 1 ≤ r < R, then we have (α,β)

kSn(α,β) (f ) − f kr ≥

Cr (f ) , n ∈ N, n+β

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where the constant Cr

bernstein

75

(f ) depends only on f , r, α and β.

For all z ∈ DR and n ∈ N we have 1 z(1 − z) 00 Sn(α,β) (f )(z) − f (z) = −(βz − α)f 0 (z) + f (z) n+β 2

Proof.

nz(1 − z) 00 βz − α 0 1 2 (α,β) f (z) − f (z) (n + β) Sn (f )(z) − f (z) + + n+β n+β 2(n + β)2 −

βz(1 − z) 00 f (z) 2

.

Note that in the case α = β = 0 in Corollary 1.1.5, necessarily f was supposed to be not a polynomial of degree ≤ 1. In what follows we will apply to the above identity the following obvious property : kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . It follows

e1 (1 − e1 ) 00 1 0

f −(βe1 − α)f + − f kr ≥

n+β 2 r

1 βe − α ne (1 − e1 ) 00 1 1 (α,β) 0 2

− · (n + β) Sn (f ) − f + f − f n+β n+β 2(n + β)2

βe1 (1 − e1 ) 00 − f .

2 r kSn(α,β)(f )

Since by Remark 2 after the proof of Theorem 1.6.2 we have

(n + β)2 S (α,β) (f ) − f + βe1 − α f 0 − ne1 (1 − e1 ) f 00 − βe1 (1 − e1 ) f 00 n

n+β 2(n + β)2 2 r ≤ Mr(α,β) (f ) + βkf 00 kr ,

and denoting H(z) = −(βz − α)f 0 (z) + z(1−z) f 00 (z), if we prove that kHkr > 0, 2 then it is clear that there exists an index n0 depending only on f , α and β, such that kHkr 1 · , ∀n ≥ n0 . kSn(α,β)(f ) − f kr ≥ n+β 2 (α,β)

For n ∈ {1, 2, ..., n0 − 1} we have kSn (α,β)

(n + β) · kSn

A(α,β) (α,β) n,r (f ) with An,r (f ) = n+β (α,β) C (α,β) (f ) kSn (f ) − f kr ≥ rn+β for

(f ) − f kr ≥

(f ) − f kr > 0, which finally implies o n (α,β) (α,β) (α,β) (α,β) r . all n ∈ N, with Cr (f ) = min A1,r , A2,r (f ), ..., An0 −1,r (f ), kHk 2 Therefore it remains to show that kHkr > 0. Indeed, suppose that kHkr = 0. We have two possibilities : 1) 0 = α < β or 2) 0 < α ≤ β.

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f 00 (z) = 0, for all |z| ≤ r and deCase 1). We obtain H(z) = −βzf 0 (z) + z(1−z) 2 0 noting y(z) = f (z), it follows that y(z) is an analytic function in DR , solution of the y 0 (z) = 0, |z| ≤ r, which after simplification differential equation −βzy(z) + z(1−z) 2 (1−z) 0 with z 6= 0 becomes −βy(z) + 2 y (z) = 0, |z| ≤ r. Now, seeking y(z) in the form P∞ y(z) = k=0 bk z k and replacing it in the differential equation, by the identification of the coefficients we easily obtain bk = 0 for all k = 0, 1, ...,. Therefore y(z) = 0 for all |z| ≤ r, which by the identity’s theorem on analytic (holomorphic) functions implies y(z) = 0 for all z ∈ DR and the contradiction that f is a polynomial of degree ≤ 0. Case 2). Denoting y(z) = f 0 (z) by hypothesis it follows that y(z) is an analytic y 0 (z) = function in DR solution of the differential equation (−βz + α)y(z) + z(1−z) 2 0, |z| ≤ r. Taking z = 0 it follows αy(0) = 0, which means y(0) = 0. Seeking y(z) in P∞ k the form y(z) = k=1 bk z and replacing it in the differential equation, by the identification of the coefficients we easily obtain bk = 0 for all k = 1, 2, ...,, which finally leads to the contradiction that f is a constant. Combining now Theorem 1.6.3 with Theorem 1.6.1 we immediately get the following. Corollary 1.6.4. (Gal [81]) Let R > 1, 0 ≤ α ≤ β with α + β > 0, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If f is not a polynomial of degree 0 and 1 ≤ r < R, then we have 1 , n ∈ N, kSn(α,β)(f ) − f kr ∼ n+β where the constants in the equivalence depend on f , r, α and β. In the case of simultaneous approximation we present : Theorem 1.6.5. (Gal [81]) Let DR = {z ∈ C; |z| < R} be with R > 1, 0 ≤ α ≤ β with α + β > 0 and let us suppose that f : DR → C is analytic in DR , i.e. f (z) = P∞ k k=0 ck z , for all z ∈ DR . Also, let 1 ≤ r < r1 < R and p ∈ N be fixed. If f is not a polynomial of degree ≤ p − 1, then we have k[Sn(α,β) (f )](p) − f (p) kr ∼

1 , n+β

where the constants in the equivalence depend on f , α, β, r, r1 and p. Proof. Taking into account Theorem 1.6.1, it remains to prove the lower estimate (α,β) for k[Sn (f )](p) − f (p) kr only. Denoting by Γ the circle of radius r1 > r (with r ≥ 1) and center 0, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N we have Z (α,β) p! Sn (f )(v) − f (v) (α,β) (p) (p) [Sn (f )] (z) − f (z) = dv, 2πi Γ (v − z)p+1

where we have the inequality |v − z| ≥ r1 − r valid for all |z| ≤ r and v ∈ Γ.

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As in the proof of Theorem 1.6.3 (keeping the notation for H), for all v ∈ Γ and n ∈ N we have 1 Sn(α,β) (f )(v) − f (v) = {H(v) n+β 1 βv − α 0 nv(1 − v) 00 2 (α,β) (n + β) Sn (f )(v) − f (v) + + f (v) − f (v) n+β n+β 2(n + β)2 βv(1 − v) 00 f (v) , − 2 which replaced in the above Cauchy’s formula implies 1 1 [Sn(α,β) (f )](p) (z) − f (p) (z) = H (p) (z) + · n+β n+β  Z (n + β)2 Sn(α,β) (f )(v) − f (v) + βv−α f 0 (v) − nv(1−v)2 f 00 (v) n+β 2(n+β) p!  dv 2πi Γ (v − z)p+1 Z βv(1 − v) 00 p! f (v)dv . − 2πi Γ 2(v − z)p+1

Passing now to absolute value, for all |z| ≤ r and n ∈ N it follows 1 1 |[Sn(α,β) (f )](p) (z) − f (p) (z)| ≥ · |H (p) (z)| − n+β n+β  Z (n + β)2 Sn(α,β) (f )(v) − f (v) + βv−α f 0 (v) − nv(1−v)2 f 00 (v) p! n+β 2(n+β)  dv 2πi (v − z)p+1 Γ Z βv(1 − v) 00 p! f (v)dv , − 2πi Γ 2(v − z)p+1

where by using the Remark 2 after the proof of Theorem 1.6.2, for all |z| ≤ r and n ∈ N we get Z (n + β)2 Sn(α,β) (f )(v) − f (v) + βv−α f 0 (v) − nv(1−v)2 f 00 (v) p! n+β 2(n+β) dv− 2πi p+1 (v − z) Γ p! 2πi

Z

Γ

(α,β) βv(1 − v) 00 p! 2πr1 βr1 (1 + r1 )kf 00 kr1 ≤ p! · 2πr1 Mr1 f (v)dv + · . 2(v − z)p+1 2π (r1 − r)p+1 2π 2(r1 − r)p+1

Denoting now Fp (z) = H (p) (z), we prove that kFp kr > 0. Indeed, if we suppose that kFp kr = 0 then it follows that f satisfies the differential equation

z(1 − z) 00 f (z) = Qp−1 (z), ∀|z| ≤ r, 2 where Qp−1 (z) is a polynomial of degree ≤ p − 1. Simplifying with z, making the P∞ substitution y(z) = f 0 (z), searching y(z) in the form y(z) = k=0 bk z k and then replacing in the differential equation, by simple calculations we easily obtain that −βzf 0 (z) +

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bk = 0 for all k ≥ p − 1, that is y(z) is a polynomial of degree ≤ p − 2. This implies the contradiction that f is a polynomial of degree ≤ p − 1. (α,β) Continuing exactly as in the proof of Theorem 1.6.3 (with kSn (f ) − f kr (α,β) replaced by k[Sn (f )](p) − f (p) kr ), finally there exists an index n0 ∈ N depending on f , r, r1 and p, such that for all n ≥ n0 we have 1 C0 k[Sn(α,β)(f )](p) − f (p) kr ≥ · . n 2 Also, the cases when n ∈ {1, 2, ..., n0 − 1} are similar with those in the proof of Theorem 1.6.3. Now defining the m-th iterates by

m

(α,β)

Sn

(f )(z), first we prove the following

Theorem 1.6.6. (Gal [80]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . Let 0 ≤ α ≤ β and 1 ≤ r < R. Then, uniformly in |z| ≤ r, ∀n ∈ N, we have m

lim

m→∞

Sn(0,β) (f )(z) = f (0),

lim

m

m→∞

Sn(α,α) (f )(z) = f (1),

(α,β)

limm→∞ m Sn

(f )(z) = b0 , where b0 is of the form n n X X b0 = dj f ((j + α)/(n + β)), with dj ≥ 0, j = 0, ..., n, dj = 1, j=1

j=1

all the values dj , j = 0, ..., n being independent of f . Proof.

By Theorem 2 in Gonska-Pit¸ul-Ra¸sa [103], for any n ∈ N we have lim

m

m→∞

Sn(0,β) (f )(x) = f (0), lim

m→∞

lim

m→∞

m

Sn(α,β) (f )(x) =

n X

m

Sn(α,α) (f )(x) = f (1),

dj f [(j + α)/(n + β)],

j=1

uniformly with respect to x ∈ [0, 1]. From the classical Vitali’s result, it suffices (α,β) to show that for any fixed n ∈ N, the sequence (m Sn (f )(z))m∈N is uniformly bounded for |z| ≤ r. P∞ (α,β) (α,β) We obviously have m Sn (f )(z) = k=0 ck · m Sn (ek )(z). But from the proof of Theorem 1.6.1 (both Cases 1) and 2) ) it easily follows (α,β) that |Sn (ek )(z)| ≤ rk , for all k, n ∈ N, |z| ≤ r, which implies min{n,k} X (α,β) (α,β) 2 (α,β) Cn,p,k Sn (ep )(z) ≤ rk , | Sn (ek )(z)| = p=0 (α,β)

and by recurrence it easily follows |m Sn (ek )(z)| ≤ rk , for all m, k, n ∈ N. This implies that ∞ ∞ X X |m Sn(α,β) (f )(z)| ≤ |ck | · |m Sn(α,β) (ek )(z)| ≤ |ck |rk < ∞, k=0

for all m, n ∈ N, which proves the theorem.

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Also, the following quantitative result holds. Theorem 1.6.7. (Gal [80]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P k suppose that f : DR → C is analytic in DR , that is we can write f (z) = ∞ k=0 ck z , for all z ∈ DR . Let 0 ≤ α ≤ β and 1 ≤ r < R. Then, for all |z| ≤ r, we have |m Sn(α,β) (f )(z) − f (z)| ≤ Proof.

∞ 2m X |ck | · [βk + k(k − 1)]r k . n+β k=1

We easily see that |m Sn(α,β) (f )(z) − f (z)| ≤

∞ X

k=1

|ck | · |m Sn(α,β) (ek )(z) − ek (z)|.

We have two possibilities : 1) 0 ≤ k ≤ n ; 2) k > n. Case 1). We successively get Sn(α,β) (ek )(z) − ek (z) = = p

k X j=1

(α,β)

Cn,j,k ej (z) − ek (z)

(α,β) [Cn,k,k

− 1]ek (z) +

k−1 X

(α,β)

Cn,j,k ej (z),

j=1

(α,β)

Sn(α,β) [Sn(α,β) (ek )(z) − ek (z)] = [Cn,k,k − 1] ·p Sn(α,β) (ek )(z) +

k−1 X j=0

(α,β)

Cn,j,k ·p Sn(α,β) (ej )(z),

(α,β)

|p Sn(α,β) [Sn(α,β) (ek )(z) − ek (z)]| ≤ |1 − Cn,k,k | · |p Sn(α,β) (ek )(z)| (α,β)

+|1 − Cn,k,k | ·

max

j=0,...,k−1

{|p Sn(α,β) (ej )(z)|}.

But by the proofs of Theorems 1.6.1 and 1.6.6, for all p, n, k ∈ N we have k(k − 1) 1 (α,β) + βk , |p Sn(α,β) (ek )(z)| ≤ rk , |1 − Cn,k,k | ≤ n+β 2

which implies

(β)

|p Sn(α,β) [Sn(α,β) (ek )(z) − ek (z)]| ≤ 2|1 − Cn,k,k |rk 1 = [2βk + k(k − 1)]r k , n+β and |

m

Sn(α,β) (ek )(z)

m−1 X p (α,β) (α,β) − ek (z)| = · Sn [Sn (ek )(z) − ek (z)] p=0

≤

m [2βk + k(k − 1)]r k . n+β

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Case 2). As in the proof of Theorem 1.6.1, Case 2), for all k > n we get |m Sn(α,β) (ek )(z) − ek (z)| ≤ 2rk ≤

2k(k − 1) + 2βk k r . n+β

As a conclusion, from both Cases 1) and 2), we obtain |m Sn(α,β) (f )(z) − f (z)| ≤ =

∞ X k=1

n X k=1

+ ≤

|ck | · |m Sn(α,β) (ek )(z) − ek (z)| ∞ X

k=n+1 n X k=1

+ ≤

|ck | · |m Sn(α,β) (ek )(z) − ek (z)|

|ck |

∞ X

|ck | · |m Sn(α,β) (ek )(z) − ek (z)|

m [2βk + k(k − 1)]r k n+β |ck |

k=n+1 ∞ X

2m n+β

k=1

2βk + 2k(k − 1) k r n+β

|ck | · [βk + k(k − 1)]r k ,

which proves the theorem.

Corollary 1.6.8. (Gal [80]) Suppose mn

mn n

→ 0 when n → ∞. Then

Sn(α,β) (f )(z) → f (z),

uniformly with respect to |z| ≤ r, for any 1 ≤ r < R. Proof.

It is immediate by passing to limit with n → ∞ in Theorem 1.6.7.

Remark. Theorem 1.6.7 and Corollary 1.6.8 are new even for the case of real functions of one real variable, since they are not covered by Gonska-Kacs´ o-Pit¸ul [101], Gonska-Pit¸ul-Ra¸sa [103]. In what follows we present some similar properties for the complex BernsteinStancu polynomials Sn<γ> (f )(z). The first results concern the approximation properties. Theorem 1.6.9. (Gal [82]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . Let 0 ≤ γ which can be dependent on n and 1 ≤ r < R. Then, for all |z| ≤ r and n ∈ N, we have <γ> (f ), |Sn<γ> (f )(z) − f (z)| ≤ M2,r,n

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P∞ P∞ <γ> j where 0 < M2,r,n (f ) = n2 j=2 j(j − 1)|cj |rj + γ(r+1) j=2 j(j − 1)(2j − 1)|cj |r < 6r ∞. Also, if 1 ≤ r < r1 < R, then for all |z| ≤ r and n, p ∈ N, we have |[Sn<γ> (f )](p) (z) − f (p) (z)| ≤ Proof.

Since Sn<γ> (f )(z) = |Sn<γ> (f )(z) −

<γ> M2,r (f )p!r1 1 ,n . (r1 − r)p+1

P∞

<γ> (ek )(z), we get k=0 ck Sn ∞ X f (z)| ≤ |ck | · |Sn<γ> (ek )(z) k=0

− ek (z)|.

To estimate |Sn<γ> (ek )(z) − ek (z)| for any fixed n ∈ N, we will consider two possible cases : 1) 0 ≤ k ≤ n ; 2) k > n. We will use the well-known representation (see Stancu [174]) n X n z(z + γ)...(z + (p − 1)γ) p <γ> ∆ f (0). Sn (f )(z) = p (1 + γ)...(1 + (p − 1)γ) 1/n p=0

Denoting

n n [0, 1/n, ..., p/n; ek ](p!)/np , ∆p1/n ek (0) = p p since ek is convex of any order, it follows that all Dn,p,k ≥ 0 and Dn,p,k =

min{n,k}

X

z(z + γ)...(z + (p − 1)γ) . (1 + γ)...(1 + (p − 1)γ) p=0 P Pmin{n,k} Also, since Sn<γ> (f )(1) = f (1), we get np=0 Dn,p,k = p=0 Dn,p,k = 1. r+jγ Note that since for any j = 0, 1, ..., we have 1+jγ ≤ r, for all 0 ≤ p ≤ min{n, k} ≤ k and |z| ≤ r we obtain r+γ r + (p − 1)γ |z(z + γ)...(z + (p − 1)γ)| ≤r · ... ≤ rp ≤ rk , (1 + γ)...(1 + (p − 1)γ) 1+γ 1 + (p − 1)γ which for all |z| ≤ r and n, k ∈ N, immediately implies Sn<γ> (ek )(z)

=

Dn,p,k

min{n,k}

|Sn<γ> (ek )(z)| ≤ rk

X

Dn,p,k = rk .

p=0

Case 1). If k = 0, then obviously we have Sn<γ> (ek )(z) − ek (z) = 0. Therefore, let us suppose that 1 ≤ k ≤ n. By using the representation in Stancu [174], we obtain |Sn<γ> (ek )(z) − ek (z)| n(n − 1)...(n − (k − 1)) z(z + γ)...(z + (k − 1)γ) k ≤ · −z k n (1 + γ)...(1 + (k − 1)γ) k−1 X z(z + γ)...(z + (p − 1)γ) Dn,p,k + (1 + γ)...(1 + (p − 1)γ) p=0 <γ> <γ> := En,k (z) + Fn,k (z).

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For |z| ≤ r it follows <γ> Fn,k (z) ≤ rk

k−1 X p=0

Dn,p,k = rk [1 − Dn,k,k ]

k(k − 1) n(n − 1)...(n − (k − 1)) ] ≤ rk . nk P 2n Here we have applied the inequality 1 − Πxi ≤ (1 − xi ), with all 0 ≤ xi ≤ 1. Also, n(n − 1)...(n − (k − 1)) z(z + γ)...(z + (k − 1)γ) <γ> En,k (z) ≤ nk (1 + γ)...(1 + (k − 1)γ) z(z + γ)...(z + (k − 1)γ) z(z + γ)...(z + (k − 1)γ) k −z − + (1 + γ)...(1 + (k − 1)γ) (1 + γ)...(1 + (k − 1)γ) z(z + γ)...(z + (k − 1)γ) n(n − 1)...(n − (k − 1)) · 1− ≤ (1 + γ)...(1 + (k − 1)γ) nk z(z + γ)...(z + (k − 1)γ) + − z k (1 + γ)...(1 + (k − 1)γ) k(k − 1) z(z + γ)...(z + (k − 1)γ) k k ≤ r + − z . 2n (1 + γ)...(1 + (k − 1)γ) = rk [1 −

For any fixed |z| ≤ r, let us denote gk (α)(z) = z(z+α)...(z+(k−1)α) (1+α)...(1+(k−1)α) , where α ≥ 0. Then, by applying the mean value theorem, there is ξ ∈ [0, γ] such that dgk (ξ)(z) z(z + γ)...(z + (k − 1)γ) k . (1 + γ)...(1 + (k − 1)γ) − z = |gk (γ)(z) − gk (0)(z)| ≤ γ · max dα k−1 But denoting uj (α)(z) = z+jα 1+jα , we have gk (α)(z) = zΠj=1 uj (α)(z) and k−1 k−1 X z + jα 0 Y z + iα dgk (α)(z) =z · dα 1 + jα α 1 + iα j=1

=z

k−1 X j=1

Since

j (1+jξ)2

2

i=1,i6=j

j(1 − z) (1 + jα)2

k−1 Y

i=1,i6=j

z + iα . 1 + iα

≤ j , passing to modulus (for 0 ≤ ξ ≤ γ and |z| ≤ r), we obtain k−1 X dgk (ξ)(z) k(k − 1)(2k − 1) ≤ r(r + 1) . j 2 rk−2 = (r + 1)rk−1 dα 6 j=1

It follows

k(k − 1) k(k − 1)(2k − 1) + γ(r + 1)rk−1 . 2n 6 Collecting all the above estimates, we get for all |z| ≤ r k(k − 1) k(k − 1) |Sn<γ> (ek )(z) − ek (z)| ≤ rk + rk 2n 2n k(k − 1)(2k − 1) +γ(r + 1)rk−1 6 r + 1 k(k − 1)(2k − 1) k k(k − 1) . = r +γ· · n r 6 <γ> En,k (z) ≤ rk

bernstein

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83

Case 2). We have |Sn<γ> (ek )(z) − ek (z)| ≤ |Sn<γ> (ek )(z)| + |ek (z)| n X z(z + γ)...(z + (p − 1)γ) + |ek (z)|. ≤ Dn,p,k (1 + γ)...(1 + (p − 1)γ) p=0

Reasoning as in the above Case 1), we get

2(k − 1)k k r . n Collecting all the results in the Cases 1) and 2), we immediately obtain for all |z| < r and k = 0, 1, 2, ..., r + 1 k(k − 1)(2k − 1) 2k(k − 1) , +γ· · |Sn<γ> (ek )(z) − ek (z)| ≤ rk n r 6 |Sn<γ> (ek )(z) − ek (z)| ≤ rn + rk ≤ 2rk ≤

which implies the corresponding estimate in statement. For the simultaneous approximation, denoting by Γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ Γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z p! Sn<γ> (f )(v) − f (v) dv |[Sn<γ> (f )](p) (z) − f (p) (z)| = 2π Γ (v − z)p+1 p! 2πr1 <γ> ≤ M2,r (f ) 1 ,n 2π (r1 − r)p+1 p!r1 <γ> = M2,r (f ) . 1 ,n (r1 − r)p+1 Remark. For γ = 0 we get the results for the classical complex Bernstein polynomials in Theorem 1.1.2. Now, defining the m-th iterates by m Sn<γ> (f )(z), first we prove the following qualitative result. Theorem 1.6.10. (Gal [82]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . Let 0 ≤ γ. Uniformly in |z| ≤ r, where 1 ≤ r < R, we have lim

m→∞

m

Sn<γ> (f )(z) = (1 − z)f (0) + zf (1), ∀n ∈ N.

Proof. From Agratini-Rus [4], Remark 2 after Theorem 9, p. 165, for any n ∈ N, we have limm→∞ m Sn<γ> (f )(x) = (1 − x)f (0) + xf (1), uniformly with respect to x ∈ [0, 1]. From the classical Vitali’s result, it suffices to show that for any fixed n ∈ N, the sequence (m Sn<γ> (f )(z))m∈N is uniformly bounded for |z| ≤ r. P∞ We have m Sn<γ> (f )(z) = k=0 ck · m Sn<γ> (ek )(z). We will prove that for all n, m, k ∈ N and |z| ≤ r, we have |m Sn<γ> (ek )(z)| ≤ rk .

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Indeed, for m = 1 it easily follows (also see the proof of Theorem 1.6.9) from the representation formula n X z(z + γ)...(z + (j − 1)γ) Dn,j,k Sn<γ> (ek )(z) = (1 + γ)...(1 + (j − 1)γ) j=0 min{n,k}

=

X

Dn,j,k

j=0

with Dn,j,k ≥ 0 and

Pn

Pmin{n,k}

Dn,j,k = 1. Pj (j) (j) (j) Denote hj (z) = z(z +γ)...(z +(j −1)γ) = i=0 ci ei (z), where ci ≥ 0, cj = 1 Pj (j) and i=0 ci = hj (1) = (1 + γ)...(1 + (j − 1)γ). By the linearity of Sn<γ> , we get min{n,k} j X X 1 (j) Dn,j,k ci Sn<γ> (ei )(z) · |2 Sn<γ> (ek )| = (1 + γ)...(1 + (j − 1)γ) i=0 j=0 j=0

Dn,j,k =

z(z + γ)...(z + (j − 1)γ) , (1 + γ)...(1 + (j − 1)γ)

min{n,k}

≤

X

Dn,j,k

j=0

j=0

j X 1 (j) c rj ≤ rk , · (1 + γ)...(1 + (j − 1)γ) i=0 i

and by mathematical induction it easily follows that for all n, m, k ∈ N we have This implies that

|m Sn<γ> (ek )(z)| ≤ rk , for all |z| ≤ r.

|m Sn<γ> (f )(z)| ≤

∞ X k=0

|ck | · |m Sn<γ> (ek )(z)| ≤

for all m, n ∈ N, which proves the theorem.

∞ X

k=0

|ck |rk < ∞,

The following quantitative result is not correspondent to the above qualitative one. Theorem 1.6.11. (Gal [82]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . Let 0 ≤ γ, 1 ≤ r < R and Dn,k,k = n(n−1)...(n−(k−1)) . Then, for all |z| ≤ r we nk have |m Sn<γ> (f )(z) − f (z)| ∞ X Dn,k,k 2k(k − 1) + 1− + γ(k − 1)2 rk . ≤m |ck | n (1 + γ)...(1 + (k − 1)γ) k=2

Proof. From the proof of Theorem 1.6.10, it follows that for all n, m, k ∈ N and |z| ≤ r, we have |m Sn<γ> (ek )(z)| ≤ rk . Also ∞ X |m Sn<γ> (f )(z) − f (z)| ≤ |ck | · |m Sn<γ> (ek )(z) − ek (z)|. k=2

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We have two possibilities : 1) 2 ≤ k ≤ n ; 2) k > n. Case 1). With the notations for gj (α)(z) in the proof of Theorem 1.6.9 and for (j) hj (z), ci in the proof of Theorem 1.6.10, we can write |m S <γ> (ek )(z) − ek (z)| n m−1 X p <γ> <γ> = Sn [Sn (ek )(z) − ek (z)] p=0   m−1 k X p <γ> X   Sn Dn,j,k · gj (γ)(z) − ek (z) = p=0 j=1   m−1 k X X p <γ> p <γ>   = Dn,j,k · Sn (gj (γ))(z) − Sn (ek )(z) p=0 j=1 ≤

m−1 X k−1 X p=0 j=1

+

m−1 X p=0

=

|Dn,k,k ·p Sn<γ> (gk (γ))(z) −p Sn<γ> (ek )(z)|

m−1 X k−1 X p=0 j=1

+

k X Dn,k,k (k) p <γ> p <γ> · Sn [ ci ei (z)] − Sn (ek )(z) (1 + γ)...(1 + (k − 1)γ)

m−1 X k−1 X p=0 j=1

+

i=0

Dn,j,k |p Sn<γ> (gj (γ))(z)|

m−1 X p=0

+

Dn,j,k |p Sn<γ> (gj (γ))(z)|

m−1 X p=0

≤

Dn,j,k |p Sn<γ> (gj (γ))(z)|

m−1 X p=0

Dn,k,k p <γ> p <γ> · S (e )(z) − S (e )(z) k k n n (1 + γ)...(1 + (k − 1)γ) k−1 Dn,k,k X (k) p <γ> ci · Sn (ei )(z) (1 + γ)...(1 + (k − 1)γ) i=0

:= T1 + T2 + T3 .

Reasoning exactly as in the proof of Theorem 1.6.10, we easily get for all j, p and |z| ≤ r that |p Sn<γ> (gj (γ))(z)| ≤ rj .

Taking into account the formula for 1 − Dn,k,k in the proof of Theorem 1.6.9, we get T1 ≤

m−1 X k−1 X p=0 j=1

rk Dn,j,k = mrk [1 − Dn,k,k ] ≤ mrk

k(k − 1) . 2n

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Also, m−1 X

Dn,k,k T2 = | 1− (1 + γ)...(1 + (k − 1)γ) p=0 Dn,k,k ≤ mrk 1 − . (1 + γ)...(1 + (k − 1)γ)

Finally,

p

Sn<γ> (ek )(z)|

m−1 X

Dn,k,k [(1 + γ)...(1 + (k − 1)γ) − 1]r k (1 + γ)...(1 + (k − 1)γ) p=0 1 k . = mr Dn,k,k 1 − (1 + γ)...(1 + (k − 1)γ) But, taking into account the inequalities Dn,k,k ≤ 1 and k−1 X k−1 (1 − xj ), 0 ≤ xj ≤ 1, j = 1, ..., k − 1, 1 − Πj=1 xj ≤ T3 ≤

j=1

applied for xj = Dn,k,k 1 −

1 1+jγ ,

we obtain

X k−1 k−1 X jγ 1 [1 − 1/(1 + jγ)] = ≤ (1 + γ)...(1 + (k − 1)γ) 1 + jγ j=1 j=1 ≤ (k − 1) ·

γ(k − 1) ≤ γ(k − 1)2 . 1 + γ(k − 1)

Collecting all these inequalities, we obtain |m Sn<γ> (ek )(z) − ek (z)| Dn,k,k k(k − 1) + 1− + γ(k − 1)2 . ≤ mrk 2n (1 + γ)...(1 + (k − 1)γ) Case 2). We get |m Sn<γ> (ek )(z) − ek (z)| ≤ |m Sn<γ> (ek )(z)| + |ek (z)| 2k(k − 1) k ≤ 2rk ≤ r . n As a conclusion, from both Cases 1) and 2), we obtain |m Sn<γ> (f )(z) − f (z)| ∞ X ≤ |ck | · |m Sn<γ> (ek )(z) − ek (z)| k=2

=

n X k=2

≤

n X k=2

+ ≤m

|ck | · |m Sn<γ> (ek )(z) − ek (z)| + |ck |mrk ∞ X

k=n+1 ∞ X k=2

k=n+1

|ck | · |m Sn<γ> (ek )(z) − ek (z)|

Dn,k,k k(k − 1) + 1− 2n (1 + γ)...(1 + (k − 1)γ)

|ck |rk

|ck |rk

∞ X

+ γ(k − 1)2

2k(k − 1) n

2k(k − 1) Dn,k,k + γ(k − 1)2 rk , + 1− n (1 + γ)...(1 + (k − 1)γ)

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which proves the theorem.

87

Remark. For γ = 0 we get some results for classical complex Bernstein polynomials in Section 1.2. Corollary 1.6.12. (Gal [82]) (i) Let 1 ≤ r < R. For γ := γn = 1/n and |z| ≤ r we have the estimate ∞ mX |ck | 2k(k − 1) + 2(k − 1)3 + (k − 1)2 rk . |m Sn<γn > (f )(z) − f (z)| ≤ n k=2

(ii) If γ := γn = 1/n and uniformly with respect |z| ≤ r. Proof.

mn n

→ 0 as n → ∞, then

mn

Sn<γn > (f )(z) → f (z),

(i) Taking γ = 1/n we obtain for all k ≥ 2 1−

Dn,k,k k−1 n − j = 1 − Πj=1 (1 + γ)...(1 + (k − 1)γ) n+j k−1 k−1 X j X n−j 1− =2 ≤ n+j j +n j=1 j=1 ≤ 2(k − 1)

(k − 1)3 k−1 ≤2 , n + (k − 1) n

which replaced in Theorem 1.6.11 gives |m Sn<γn > (f )(z) − f (z)| ≤ (i).

∞

mX |ck | 2k(k − 1) + 2(k − 1)3 + (k − 1)2 rk . n k=2

(ii) It is immediate by passing to limit with n → ∞ in the estimate proved in

Remark. The results in Theorem 1.6.11 and Corollary 1.6.12 are new even for the case of real functions of one real variable, since they are not covered by those in Gonska-Kacs´ o-Pit¸ul [101] or Gonska-Pit¸ul-Ra¸sa [103], whose estimates one refer to the difference |m Ln (f )(x) − B1 (f )(x)|, with B1 (f )(x) = f (0) + [f (1) − f (0)]x and m Ln (f ) representing the mth iterate of the positive linear operator Ln (f ). Finally we present the geometric properties of Sn<γ> (f )(z). Theorem 1.6.13. (Gal [82]) Let us suppose that G ⊂ C is open, such that D1 ⊂ G <γ(n)> and f : G → C is analytic in G. Also, let us consider (Sn (f )(z))n∈N , where we suppose that limn→∞ γ(n) = 0. If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) in D1 , that is for all z ∈ D1 (see e.g. Mocanu, P. T., Bulboac˘ a, T. and S˘ al˘ agean [138]) 00 0 0 zf (z) zf (z) iη zf (z) > 0 Re + 1 > 0, Re e > 0, resp. , Re f (z) f 0 (z) f (z)

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then there exists an index n0 depending on f (and on η for spirallikeness), such that <γ(n)> for all n ≥ n0 , Sn (f )(z), are starlike (convex, spirallike of type η, respectively) in D1 . If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) only in D1 (that is the corresponding inequalities hold only in D1 ), then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists an index n0 = n0 (f, Dr ) (n0 depends on η too in the case of spirallikeness), such that for all n ≥ n 0 , <γ(n)> Sn (f )(z), are starlike (convex, spirallike of type η, respectively) in D r (that is, the corresponding inequalities hold in Dr ). <γ(n)>

(f )(z) → f (z), uniProof. By Theorem 1.6.9 it follows that we have Sn formly for |z| ≤ 1, which by the well-known Weierstrass’s theorem implies <γ(n)> <γ(n)> [Sn (f )]0 (z) → f 0 (z) and [Sn (f )]00 (z) → f 00 (z), for n → ∞, uniformly

in D1 . In all what follows, denote Pn (f )(z) =

<γ(n)> Sn (f )(z) <γ(n)>

, well defined for suf-

[Sn (f )]0 (0) We easily get Pn (f )(0) = 0, Pn0 (f )(0) = 1 for sufficiently f (z), Pn0 (f )(z) → f 0 (z) and Pn00 (f )(z) → f 00 (z), uniformly

ficiently large n. large n, and Pn (f )(z) → in D1 . Suppose first that f is starlike in D1 . Then, by hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Writing Pn (f )(z) in the form Pn (f )(z) = zQn (f )(z), obviously Qn (f )(z) is a polynomial of degree ≤ n − 1. Also, for |z| = 1 we have |f (z) − Pn (f )(z)| = |z| · |g(z) − Qn (f )(z)| = |g(z) − Qn (f )(z)|, which by the uniform convergence in D1 of Pn (f ) to f and by the maximum modulus principle, implies the uniform convergence in D1 of Qn (f )(z) to g(z). Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index n1 ∈ N and a > 0 depending on g, such that |Qn (f )(z)| > a > 0, for all z ∈ D1 and all n ≥ n0 . Also, for all |z| = 1, we have |f 0 (z) − Pn0 (f )(z)| = |z[g 0 (z) − Q0n (f )(z)] + [g(z) − Qn (f )(z)]|

≥ | |z| · |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| |

= | |g 0 (z) − Q0n (f )(z)| − |g(z) − Qn (f )(z)| |,

which from the maximum modulus principle, the uniform convergence of Pn0 (f ) to f 0 and of Qn (f ) to g, evidently implies the uniform convergence of Q0n (f ) to g 0 . Then, for |z| = 1, we get z[zQ0n(f )(z) + Qn (f )(z)] zPn0 (f )(z) = Pn (f ) zQn (f )(z) 0 zg 0 (z) + g(z) zQn (f )(z) + Qn (f )(z) → = Qn (f )(z) g(z) 0 0 zf (z) f (z) = , = g(z) f (z)

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which again from the maximum modulus principle, implies zf 0 (z) zPn0 (f )(z) → , uniformly in D1 . Pn (f ) f (z) 0 (z) is continuous in D1 , there exists ε ∈ (0, 1), such that Since Re zff (z) 0 zf (z) Re ≥ ε, for all z ∈ D1 . f (z) Therefore 0 0 zf (z) zPn (f )(z) → Re ≥ε>0 Re Pn (f )(z) f (z)

uniformly on D1 , i.e. for any 0 < ρ < ε, there is n0 such that for all n ≥ n0 we have 0 zPn (f )(z) > ρ > 0, for all z ∈ D1 . Re Pn (f )(z) <γ(n)>

Since Pn (f )(z) differs from Sn (f )(z) only by a constant, this proves the star<γ(n)> (f )(z), for sufficiently large n. likeness of Sn If f is supposed to be starlike only in D1 , the proof is identical, with the only difference that instead of D1 , we reason for Dr . The proofs in the cases when f is convex or spirallike of order η are similar and follow from the following uniform convergences (on D1 or on Dr ) 00 00 zf (z) zPn (f )(z) + 1 → Re +1 Re Pn0 (f )(z) f 0 (z) and zP 0 (f )(z) zf 0 (z) Re eiη n → Re eiη . Pn (f )(z) f (z) Remark. If f is univalent in D1 , then from the uniform convergence in Theorem 1.6.9 and a well-known result in complex analysis, concerning sequences of analytic functions converging locally uniformly to an univalent function, it is immediate that <γ(n)> for sufficiently large n, the complex polynomials Sn (f )(z) (where γ(n) → 0, for n → ∞), must be univalent in D1 . At the end of this section we will extend the Bernstein-Stancu polynomials (α,β) Sn (f )(z) and some of their approximation results to compact subsets G ⊂ C. For this purpose, in what follows G ⊂ C we will be considered a compact set such ˜ \ G is connected. In this case, according to the Riemann Mapping Theorem, that C ˜ \ D1 onto C ˜ \ G exists so that Ψ(∞) = ∞ and a unique conformal mapping Ψ of C 0 Ψ (∞) > 0. By using the Faber polynomials Fp (z) attached to G (see Definition 1.0.10), for f ∈ A(G) and 0 ≤ α ≤ β we can introduce the Bernstein-Stancu-Faber polynomials given by the formula n X n Sn(α,β) (f ; G)(z) = ∆p1/(n+β) F (α/(n + β)) · Fp (z), z ∈ G, n ∈ N, p p=0

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where 0 ≤ α ≤ β, F (w) =

1 2πi

R

|u|=1

∆ph F (α/(n + β)) =

f (Ψ(u)) u−w du,

p X

k=0

w ∈ D1 , and p (−1)p−k F (α/(n + β) + kh). k

Here, in the case when α = β since F (1) is involved in ∆n1/(n+β) F (α/(n + β)) and (α,β)

therefore in the definition of Sn (f ; G)(z) too, in addition we will suppose that F can be extended by continuity on the boundary ∂D1 . Remarks. 1) For G = D1 it is easy to see that the above Bernstein-Stancu-Faber polynomials one reduce to the classical complex Bernstein-Stancu polynomials given by n X n Sn(α,β) (f )(z) = ∆p1/(n+β) f (α/(n + β))z p p p=0 n X n p z (1 − z)n−p f [(p + α)/(n + β)]. = p p=0

R 1 ω (f ◦Ψ;u) 2) It is known that, for example, 0 p u ∂D1 du < ∞ is a sufficient condition for the continuity on ∂D1 of F in the above definition of the Bernstein-Stancu-Faber polynomials (see e.g. Gaier [76], p. 52, Theorem 6). Here p ∈ N is arbitrary fixed. (α,β) 3) In the case when α = β = 0, Sn (f ; G)(z) becomes Bn (f ; G)(z). The first main result one refers to approximation on compact sets without any restriction on their boundaries and can be stated as follows.

Theorem 1.6.14. Let G be a continuum (that is a connected compact subset of C) and suppose that f is analytic in G, that is there exists R > 1 such that f is analytic in GR . Here recall that GR denotes the interior of the closed level curve ΓR given by ΓR = {z; |Φ(z)| = R} = {Ψ(w); |w| = R} (and that G ⊂ G r for all 1 < r < R). Also, we suppose that F given in the definition of Bernstein-StancuFaber polynomials can be extended by continuity on ∂D1 . For any 1 < r < R the following estimate C |Sn(α,β) (f ; G)(z) − f (z)| ≤ , z ∈ Gr , n ∈ N, n holds, where C > 0 depends on f , α, β, r and Gr but it is independent of n. ˜ \ G is Proof. First we note that since G is a continuum then it follows that C [ ] simply connected. By the proof of Theorem 2, p. 52 in Suetin 186 , for any fixed P∞ 1 < η < R we have f (z) = in Gη , where ak (f ) are k=0 ak (f )Fk (z) uniformly R f [Ψ(u)] 1 the Faber coefficients and are given by ak (f ) = 2πi |u|=η uk+1 du. Note here that G ⊂ Gη . First we will prove that

Sn(α,β) (f ; G)(z) =

∞ X k=0

ak (f )Sn(α,β) (Fk ; G)(z),

bernstein

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for all z ∈ G. (Note here that by hypothesis we have G = G). For this purpose, Pm denote fm (z) = k=0 ak (f )Fk (z), m ∈ N. Since by the linearity of Sn (α, β) we easily get Sn(α,β) (fm ; G)(z) =

m X k=0

ak (f )Sn(α,β) (Fk ; G)(z), for all z ∈ G, (α,β)

(α,β)

(f ; G)(z), for all z ∈ G it suffices to prove that limm→∞ Sn (fm ; G)(z) = Sn and n ∈ N. First we have n X n (α,β) ∆p1/(n+β) Gm (α/(n + β))Fk (z), Sn (fm ; G)(z) = p p=0 R R fm (Ψ(u)) f (Ψ(u)) 1 1 where Gm (w) = 2πi u−w du and F (w) = 2πi |u|=1 u−w du. |u|=1 Note here that since by Gaier [76], p. 48, first relation before (6.17), we have Z 1 Fk (Ψ(u)) Fk (w) = du = wk , for all |w| < 1, 2πi |u|=1 u − w

evidently that Fk (w) can be by continuity on ∂D1 . This also immediately R extended fm (Ψ(u)) 1 implies that Gm (w) = 2πi u−w du can be extended by continuity on ∂D1 , |u|=1 (α,β)

(α,β)

which means that Sn (Fk ; G)(z) and Sn (fm ; G)(z) are well defined. Now, taking into account the Cauchy’s theorem we also can write Z Z 1 fm (Ψ(u)) f (Ψ(u)) 1 du and F (w) = du. Gm (w) = 2πi |u|=η u − w 2πi |u|=η u − w

For all n, m ∈ N and z ∈ G it follows |Sn(α,β) (fm ; G)(z) − Sn(α,β) (f ; G)(z)| n X n ≤ |∆p1/(n+β) (Gm − F )(α/(n + β))| · |Fk (z)| p p=0 p n X X p n |(Gm − F )(α/(n + β) + (p − j)/(n + β))| · |Fk (z)| ≤ j p p=0 j=0 p n X X n p ≤ Cj,p,η,α,β kfm − f kGη · |Fk (z)| p j p=0 j=0

≤ Mn,p,η,α,β,Gη kfm − f kGη ,

which by limm→∞ kfm − f kGη = 0 (see e.g. the proof of Theorem 2, p. 52 in Suetin [186]) implies the desired conclusion. Here kfm − f kGη denotes the uniform norm of fm − f on Gη .

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Consequently we obtain |Sn(α,β) (f ; G)(z) − f (z)| ≤ = +

∞ X

k=0 n X

|ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)| |ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)|

k=0 ∞ X

k=n+1

|ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)|. (α,β)

Therefore it remains to estimate |ak (f )| · |Sn (Fk ; G)(z) − Fk (z)|, firstly for all 0 ≤ k ≤ n and secondly for k ≥ n + 1, where n X n (α,β) Sn (Fk ; G)(z) = [∆p1/(n+β) Fk (α/(n + β))] · Fp (z). p p=0 First it is useful to observe that by Gaier [76], p. 48, combined with the Cauchy’s theorem, for any fixed 1 < η < R we have Z 1 Fk [Ψ(u)] du = wk = ek (w), for all |w| < η. Fk (w) := 2πi |u|=η u − w Denote (α,β)

Dn,p,k =

n ∆p1/(n+β) ek (α/(n + β)) p

n = [α/(n + β), (α + 1)/(n + β), ..., (α + p)/(n + β); ek ] · (p!)/(n + β)p . p It follows Sn(α,β) (Fk ; G)(z) =

n X p=0

(α,β)

Dn,p,k · Fp (z).

(α,β)

Since Sn (f )(1) = f [(n + α)/(n + β)] and since each ek is convex of any order, it k P (α,β) (α,β) follows Dn,p,k > 0 and, by taking f (z) = ek (z) we get np=0 Dn,p,k = (n+α) ≤ 1, (n+β)k for all k and n. (α,β) In the estimation of |ak (f )| · |Sn (Fk ; G)(z) − Fk (z)| we distinguish two cases : 1) 0 ≤ k ≤ n ; 2) k > n . Case 1. We have (α,β)

|Sn(α,β) (Fk ; G)(z) − Fk (z)| ≤ |Fk (z)| · |1 − Dn,k,k | +

k−1 X p=0

(α,β)

Dn,p,k · |Fp (z)|.

Fix now 1 < r < η. By the inequality (13), p. 44 in Suetin [186] we have |Fp (z)| ≤ C(r)rp , for all z ∈ Gr , p ≥ 0,

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which immediately implies |Sn(α,β) (Fk ; G)(z) − Fk (z)| n(n − 1)...(n − (k − 1)) ≤ 1− C(r)rk (n + β)k n(n − 1)...(n − (k − 1)) (n + α)k C(r)rk − + (n + β)k (n + β)k n(n − 1)...(n − (k − 1)) (n + α)k k = 2 1− C(r)r + − 1 C(r)rk (n + β)k (n + β)k 1 n(n − 1)...(n − (k − 1)) C(r)rk ≤ ≤ 2 1− [2βk + k(k − 1)] C(r)r k . (n + β)k n+β Pk Here we used the inequality 1 − Πki=1 xi ≤ i=1 (1 − xi ), valid if all xi ∈ [0, 1]. Also ) by the above formula for ak (f ) we easily obtain |ak (f )| ≤ C(η,f η k , for all k ≥ 0. Note that C(r), C(η, f ) > 0 are constants independent of k. For all z ∈ Gr and k = 0, 1, 2, ...n it follows k r C(r, η, f ) (α,β) [2βk + k(k − 1)] · , |ak (f )| · |Sn (Fk ; G)(z) − Fk (z)| ≤ n+β η that is n X k=0

|ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)| ≤

n C(r, η, f ) X [2βk + k(k − 1)]dk , n+β k=1

for all z ∈ Gr , where 0 < d = r/η < 1. P P k Also, clearly we have nk=1 [2βk + k(k − 1)]dk ≤ ∞ k=1 [2βk+]k(k − 1)d < ∞ which finally implies that n X k=0

|ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)| ≤

C ∗ (r, η, β, f ) . n

Case 2. We have ∞ X

k=n+1

|ak (f )| ·

|Sn(α,β) (Fk ; G)(z)

− Fk (z)| ≤ +

∞ X

k=n+1 ∞ X

k=n+1

|ak (f )| · |Sn(α,β) (Fk ; G)(z)| |ak (f )| · |Fk (z)|.

By the estimates mentioned in the case 1), we immediately get ∞ X

k=n+1

with d = r/β.

|ak (f )| · |Fk (z)| ≤ C(r, η, f )

∞ X

k=n+1

dk , for all z ∈ Gr ,

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Also, ∞ X

k=n+1

|ak (f )| ·

|Sn(α,β) (Fk ; G)(z)|

= ≤

∞ X

k=n+1 ∞ X

k=n+1

n X (α,β) Dn,p,k · Fp (z) |ak (f )| · |ak (f )| ·

p=0 n X p=0

(α,β)

Dn,p,k · |Fp (z)|.

But for p ≤ n < k and taking into account the estimates obtained in the Case 1) we get |ak (f )| · |Fp (z)| ≤ C(r, η, f )

rp rk ≤ C(r, β, f ) , for all z ∈ Gr , ηk ηk

which implies ∞ X

k=n+1

|ak (f )| · |Sn(α,β) (Fk ; G)(z) − Fk (z)| ≤ C(r, η, f ) = C(r, β, f )

∞ X n X

k=n+1 p=0 ∞ k X k=n+1 n+1

= C(r, β, f )

(α,β)

Dn,p,k

k r β

r β

d , 1−d

with d = r/β. In conclusion, collecting the estimates in the Cases 1) and 2) we obtain |Sn(α,β) (f ; G)(z) − f (z)| ≤

C1 C + C2 dn+1 ≤ , z ∈ Gr , n ∈ N. n+β n

This proves the theorem.

Remark. In the case when α = β = 0 we recapture Theorem 1.1.8. As a consequence of the Remarks 1 and 2 after the proof of Theorem 1.1.8, we can state the following result. ˜ Theorem 1.6.15. Let G be a compact Faber set such that C\G is simply connected. If f is analytic on G, that is there exists R > 1 such that f is analytic in G R and if f is not a polynomial of degree ≤ 1, then for any 1 < r < R we have

1 , n ∈ N, n where the constants in the equivalence depend on f , α, β, r and Gr but are independent of n. Here kf kGr = supz∈Gr |f (z)|. kSn(α,β) (f ; G) − f kGr ∼

Proof. According to Remark 2 after the proof of Theorem 1.1.8, there exists g analytic in Dr such that f = T (g), that is g = T −1 (f ) (therefore F can be extended by continuity on ∂D1 . By hypothesis on f it follows that f cannot be of the form f (z) = c0 F0 (z)+c1 F1 (z) where F0 and F1 are the Faber polynomials of degree 0 and

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1 respectively and c0 , c1 ∈ C. This immediately implies that g is not a polynomial of degree ≤ 1. (α,β) (α,β) First we have Sn (T −1 (f )) = T −1 [Sn (f ; G)]. Indeed, n X n Sn(α,β) (T −1 (f ))(z) = ∆p1/(n+β) T −1 (f )(α/(n + β))z p p p=0 n X n = ∆p1/(n+β) F (α/(n + β))z p , p p=0 R f [Ψ(w)] 1 −1 since T (f )(ξ) = 2πi |w|=1 w−ξ dw = F (ξ), and T −1 [Sn(α,β) (f ; G)](z) Z (α,β) 1 Sn (f ; G)[Ψ(w)] = dw 2πi |w|=1 w−z Z n X n 1 Fp [Ψ(w)] p dw = ∆1/(n+β) F (α/(n + β)) 2πi w−z p |w|=1 p=0 n X n = ∆p1/(n+β) F (α/(n + β))z p , p p=0

since according to Gaier [76], p. 48, first relation before (6.17), we have Z Fp [Ψ(w)] 1 dw = z p . 2πi |w|=1 w − z

Then by Theorem 1.6.3 (see also Corollary 1.6.4) and by the linearity and continuity of T −1 we get C ≤ kSn(α,β) (g) − gkr = kSn(α,β) (g) − T −1 (f )kr n = kT −1 [Sn(α,β) (f ; G)] − T −1 (f )kr ≤ k|T −1 k| · kSn(α,β) (f ; G) − f kGr ≤ M kSn(α,β) (f ; G) − f kGr ,

which proves the lower estimate. (α,β) (α,β) On the other hand we have T [Sn (g)] = Sn (T (g); G). Indeed, n X T [Sn(α,β)(g)](z) = ∆p1/(n+β) g(α/(n + β))Fp (z), p=0

and

Sn(α,β) (T (g); G)(z)

=

n X n p=0

p

∆p1/(n+β) H(α/(n + β))Fp (z),

where according to Gaier [76], p. 49. relation (6.17’) we have Z 1 T (g)[Ψ(u)] du = g(w). H(w) = 2πi |u|=1 u − w

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Therefore by the same Corollary 1.6.4 and by the linearity and continuity of T we obtain kSn(α,β) (f ; G) − f kGr = kSn(α,β) (T (g); G) − T (g)kGr = kT [Sn(α,β)(g)] − T (g)kGr

≤ k|T k| · kSn(α,β) (g) − gkr ≤

which proves the upper estimate and the theorem.

C , n

Remark. For α = β = 0, we recapture Theorem 1.1.9.

1.7

Bernstein-Kantorovich Type Polynomials

In this section we extend some results in the previous sections to the following complex Kantorovich variants of these polynomials defined (for the case of real variable) by Kantorovich [112] Z (k+1)/(n+1) n X Kn (f )(z) = (n + 1) pn,k (z) f (t)dt, k=0

k/(n+1)

and (for the case of real variable) B˘ arbosu [36] Z n X (α,β) Kn (f )(z) = (n + 1 + β) pn,k (z) k=0

(k+1+α)/(n+1+β)

f (t)dt. (k+α)/(n+1+β)

For our purpose will be useful the results expressed by the following. Rz Theorem 1.7.1. Let F (z) = 0 f (t)dt. (i) (see e.g. Lorentz [125], p. 30) Denoting by Bn (f )(z) the Bernstein polynomials, we have 0 Kn (f )(z) = Bn+1 (F )(z), z ∈ C. P (α,β) [ ] (ii) (Gal 91 ) Denoting Sn (f )(z) = nk=0 nk z k (1 − z)n−k f [(k + α)/(n + β)], z ∈ C, the Bernstein-Stancu polynomials studied in Section 1.6, where 0 ≤ α ≤ β are independent of n, we have i0 n + 1 + β h (α,β) Sn+1 (F ) (z), z ∈ C. Kn(α,β) (f )(z) = n+1

Proof.

(ii) It is immediate by the formula (α,β) [Sn+1 (F )]0 (z) n X k+α k+α+1 −F = (n + 1 + β) pn,k (z) F n+β+1 n+1+β k=0 n X k+α+1 k+α −β pn,k (z) F −F n+β+1 n+1+β k=0

=

Kn(α,β) (f )(z)

−

β K (α,β) (f )(z). n+1+β n

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Now, as a consequence of Theorem 1.7.1, (i) and Theorem 1.1.6, we immediately get the following. Corollary 1.7.2. (Gal [91]) Let f : DR → C be analytic in DR with R > 1 and 1 ≤ r < R. (i) If f is not a polynomial of degree ≤ 0 then for all n ∈ N we have 1 kKn (f ) − f kr ∼ , n where the constants in the equivalence depend only on f and r. (ii) If f is not a polynomial of degree ≤ max{1, p − 1} then for all p, n ∈ N we have 1 kKn(p) (f ) − f (p) kr ∼ , n with the constants in the equivalence depending only on f , r and p. Proof. We combine Theorem 1.7.1, (i) with Theorem 1.1.6. (i) We get 1 0 kKn (f ) − f kr = kBn+1 (F ) − F 0 kr ∼ , n+1 if F is not a polynomial of degree ≤ max{1, 1} = 1, which ends the proof. (ii) We obtain 1 (p+1) kKn(p) (f ) − f (p) kr = kBn+1 (F ) − F (p+1) kr ∼ , n+1 if F is not a polynomial of degree ≤ max{1, p} = p, which ends the proof.

As a consequence of Theorems 1.7.1, (ii) and Theorem 1.6.5, we also get the following. Corollary 1.7.3. (Gal [91]) Let f : DR → C be analytic in DR with R > 1, 1 ≤ r < R and 0 ≤ α ≤ β, α + β > 0. (i) If f is not identical 0, then for all n ∈ N we have 1 , kKn(α,β) (f ) − f kr ∼ n+β where the constants in the equivalence depend only on f , r, α and β. (ii) If f is not a polynomial of degree ≤ p − 1 then for all p, n ∈ N we have 1 k[Kn(α,β) (f )](p) − f (p) kr ∼ , n with the constants in the equivalence depending only on f , r, α, β and p. Proof. We combine Theorem 1.7.1, (ii) with Theorem 1.6.5. (i) We get 1 (α,β) kKn(α,β) (f ) − f kr = k[Sn+1 (F )]0 − F 0 kr ∼ , n+β if F is not a polynomial of degree ≤ 0, which ends the proof. (ii) We obtain (α,β)

k[Kn(α,β)(f )](p) − f (p) kr = k[Sn+1 (F )](p+1) − F (p+1) kr ∼ if F is not a polynomial of degree ≤ p, which ends the proof.

1 , n+β

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Upper estimates with explicit constants in approximation by these kind of polynomials and in Voronovskaja-type formula can be derived as follows. First we consider the case of Kn (f )(z) polynomials. Theorem 1.7.4. (Gal [91]) Let f : DR → C be analytic in DR = {z ∈ C; |z| < R} P k with R > 1, i.e. f (z) = ∞ k=0 ck z , for all z ∈ DR . Suppose 1 ≤ r < r1 < R. Then for all |z| ≤ r and n, p ∈ N, we have : (i) |Kn(p) (f )(z) − f (p) (z)| ≤ where 0 < C2,r1 (f ) = 2 (ii)

P∞

j=2 (j

C2,r1 (f )(p + 1)!r1 , (n + 1)(r1 − r)p+2

− 1)|cj−1 |r1j < ∞ ;

1 − 2z z(1 − z) 00 · f 0 (z) − · f (z)| 2(n + 1) 2(n + 1) r1 ≤ Cr1 ,n+1 (f ) · , (r1 − r)2

|Kn (f )(z) − f (z) −

where Cr1 ,n (f ) =

Proof.

5(1 + r1 )2 · 2n

P∞

k=3

|ck−1 |(k − 1)(k − 2)2 r1k−2 . n

(i) Combining Theorem 1.7.1, (i) with Theorem 1.1.2, (ii), we obtain

M2,r1 (F )(p + 1)!r1 , (n + 1)(r1 − r)p+2 P∞ P∞ where 0 < M2,r1 (F ) = 2 j=2 j(j − 1)|Cj |r1j < ∞ and F (z) = k=0 Ck z k , z ∈ DR . But we also get Z zX ∞ ∞ ∞ X ck k+1 X ck−1 k z = z , F (z) = [ ck tk ]dt = k+1 k 0 (p+1)

|Kn(p) (f )(z) − f (p) (z)| = |Bn+1 (F )(z) − F (p+1) (z)| ≤

k=0

k=1

k=0

P∞

ck−1 k

which implies Ck = and C2,r1 (f ) = 2 j=2 (j − 1)|cj−1 |r1j . (ii) Replacing in Theorem 1.1.3, (ii), n by n + 1, r by r1 and f by F , for all |z| ≤ r1 and n ∈ N, we obtain 2 Bn+1 (F )(z) − F (z) − z(1 − z) F 00 (z) ≤ 5(1 + r1 ) · Mr1 (F ) , 2(n + 1) 2(n + 1) n+1 where

Mr1 (F ) =

∞ X k=3

|Ck |k(k − 1)(k − 2)2 r1k−2 =

:= Ar1 (f ). Here again we wrote F (z) =

P∞

k=0

∞ X k=3

|ck−1 |(k − 1)(k − 2)2 r1k−2

Ck z k , for all z ∈ DR .

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99

(f )

1) · r1n , by Γ the circle of radius r1 > r and Now, denoting Cr1 ,n (f ) = 5(1+r 2n z(1−z) 00 center 0, and En (F )(z) = Bn+1 (F )(z) − F (z) − 2(n+1) F (z), since for any |z| ≤ r and v ∈ Γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formula it follows that for all |z| ≤ r and n ∈ N, we obtain Z 1 2πr1 1 En (f )(z) 0 dv ≤ Cr1 ,n+1 (f ) |En (F )(z)| = 2 2π Γ (v − z) 2π (r1 − 1)2 r1 = Cr1 ,n+1 (f ) · . (r1 − r)2

But by Theorem 1.7.1, (i) we obtain En0 (F )(z) = Kn (f )(z) − f (z) −

z(1 − z) 00 1 − 2z · f 0 (z) − · f (z), 2(n + 1) 2(n + 1)

which proves the theorem. (α,β)

In the case of Kn

(f )(z) polynomials we have the following.

Theorem 1.7.5. (Gal [91]) Let f : DR → C be analytic in DR = {z ∈ C; |z| < R} P∞ with R > 1, i.e. f (z) = k=0 ck z k , for all z ∈ DR . Suppose 1 ≤ r < r1 < R. Then for all |z| ≤ r and n, p ∈ N, we have : (i) (β)

C2,r1 (f )(p + 1)!r1 β + kf kr , − f (z)| ≤ (n + 1)(r1 − r)p+2 n+1 P∞ P∞ (β) where 0 < C2,r1 (f ) = 2 j=2 (j − 1)|cj−1 |r1j + 2β j=1 |cj−1 |r1j < ∞ ; (ii) (α,β) 1 − 2z Kn (f )(z) − f (z) + βz − α − f 0 (z) n+1 2(n + β + 1) |[Kn(α,β) (f )](p) (z)

−

(p)

C(f, r1 , α, β) r1 z(1 − z) f 00 (z) ≤ · , 2(n + β + 1) (n + 1)(n + β + 1) (r1 − r)2

where C(f, r1 , α, β) is a positive constant depending only on f , r1 , α and β. Proof. obtain

(i) Combining Theorem 1.7.1, (ii) with Theorem 1.6.1, for all |z| ≤ r we

n + 1 + β (α,β) |[Kn(α,β) (f )](p) (z) − f (p) (z)| = [Sn+1 (F )](p+1) (z) − F (p+1) (z) n+1 n + 1 + β (α,β) β (p+1) (p+1) (p+1) ≤ |[Sn+1 (F )] |F (z)| (z) − F (z)| + n+1 n+1 (β)

≤

M2,r1 (F )(p + 1)!r1 β n+1+β · + · |f (p) (z)| n+1 (n + β + 1)(r1 − r)p+2 n+1

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M2,r1 (F )(p + 1)!r1 β ≤ + · kf (p) kr , (n + 1)(r1 − r)p+2 n+1

where kf (p) kr = sup{|f (p) (z)|; |z| ≤ r} and reasoning exactly as in the proof of Theorem 1.7.4, (i), we get (β)

M2,r1 (F ) = 2

∞ X j=2

=2

∞ X j=2

j(j − 1)|Cj |r1j + 2β (j − 1)|cj−1 |r1j + 2β

∞ X

j=1 ∞ X j=1

j|Cj |r1j (β)

|cj−1 |r1j := C2,r1 (f ).

(ii) Replacing in Remark 2 after Theorem 1.6.2 n by n + 1, r by r1 and f by F , for all |z| ≤ r1 and n ∈ N, we obtain (α,β) (n + 1)z(1 − z) 00 βz − α 0 S n+1 (F )(z) − F (z) + n + β + 1 F (z) − 2(n + β + 1)2 F (z) ≤

C(f, r1 , α, β) , (n + β + 1)2

where the positive constant C(f, r1 , α, β) depends only on f, r, α and β. Let us denote (n + 1)z(1 − z) 00 βz − α 0 (α,β) F (z) − F (z). En (F )(z) = Sn+1 (F )(z) − F (z) + n+β+1 2(n + β + 1)2 If Γ is the circle of radius r1 > r and center 0, and since for any |z| ≤ r and v ∈ Γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formula it follows that for all |z| ≤ r and n ∈ N, we obtain as in the proof of Theorem 1.7.4, (ii) |En0 (F )(z)| ≤ C(f, r1 , α, β) · But by Theorem 1.7.1, (ii) we obtain

1 r1 · . 2 (r1 − r) (n + β + 1)2

n+1 1 Kn(α,β) (f )(z) − f (z) + [(βz − α)f (z)]0 n+1+β n+β+1 n+1 n+1 − [(z − z 2 )f 0 (z)]0 = 2(n + β + 1)2 n+β+1 1 − 2z βz − α (α,β) 0 − · Kn (f )(z) − f (z) + f (z) f 0 (z) n+1 2(n + β + 1) z(1 − z) 00 − f (z) , 2(n + β + 1)

En0 (F )(z) =

which immediately proves the theorem.

In what follows we will prove an equivalence result for approximation in Voronovskaja’s theorem in the case of complex Bernstein-Kantorovich polynomials Kn (f )(z), analogous to that for complex Bernstein polynomials contained by Corollary 1.3.4.

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Theorem 1.7.6. Let R > 1 and let f : DR → C be an analytic function, say P∞ f (z) = k=0 ck z k . If f is not a polynomial of degree ≤ 2p−1 then for any 1 ≤ r < R and any natural number p we have

2p −j X

1 (n + 1) (j) (j−1) 0

Kn (f ) − f − ∼ p+1 , n ∈ N, Tn+1,j ] [f Tn+1,j + f

j! n

j=1 r

where the constants in the equivalence depend only on f , r and p and are independent Pn+1 of n. Recall here that Tn+1,j (z) = i=0 (i − nz)j n+1 z i (1 − z)n+1−i . i Rz Proof. Denoting F (z) = 0 f (t)dt, by Theorem 1.3.2 we get

2p (j) X

Cp,r (f ) F −j

Bn+1 (F ) − F − ≤ (n + 1) Tn+1,j .

j! (n + 1)p+1

j=1 r

Let 1 ≤ r < r1 < R and denote by γ the circle of radius r1 > r and center 0 and H(z) = Bn+1 (F )(z) − F (z) −

2p X F (j) (z) j=1

j!

(n + 1)−j Tn+1,j (z).

Since for any |z| ≤ Rr and v ∈ γ we have |v − z| ≥ r1 − r, by the Cauchy’s formula H(v) 1 dv, which implies we get H 0 (z) = 2πi γ (v−z)2 kH 0 kr ≤

2πr1 Cp,r (F )r1 1 Cp,r,r1 (F ) kHkr1 · ≤ · ≤ , 2π (r1 − r)2 (r1 − r)2 (n + 1)p+1 np+1

which is exactly the upper estimate in the statement of Theorem 1.7.6. Note that P∞ if f (z) = k=0 ck z k then F (z) =

∞ ∞ ∞ X X ck k+1 X cj−1 j z = z := Cj∗ z j , k+1 j j=1 j=1 k=0

c

where Cj∗ = j−1 j , for all j = 1, 2, ...,. So it remains to prove the lower estimate. In the proof of Corollary 1.3.4, write the first identity for Bn+1 (F ) and F and then take the first derivative. It follows Kn (f )(z) − f (z) −

2p X (n + 1)−j j=1

j!

0 [f (j) (z)Tn+1,j + f j−1 (z)Tn+1,j (z)]

0 Tn+1,2p+1 (z) 1 Tn+1,2p+1 (z) = F (2p+1) (z) + F (2p+2) (z) p+1 p (n + 1) (n + 1) (2p + 1)! (n + 1)p (2p + 1)! 0 Tn+1,2p+2 (z) Tn+1,2p+2 (z) F (2p+2) (z) + F (2p+3) (z) + p+1 (n + 1) (2p + 2)! (n + 1)p+1 (2p + 2)!   ∞  X 1  0 + (n + 1)p+2 Ck∗ Ek,n+1,p+1 (z) .  n+1 k=2p+3

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Here Ek,n+1,p (z) does not depend on F and it is a polynomial of degree ≤ k in z. By the Bernstein’s inequality and by Lemma 1.3.3 it follows k (k + 1)!(k − 2(p + 1))2 0 kEk,n+1,p+1 kr ≤ kEk,n+1,p+1 kr ≤ Cp rk−1 . r (n + 1)p+2 (k − 2(p + 1))! P ∞ 0 By the same Lemma 1.3.3 we get that (n + 1)p+2 k=2p+3 Ck∗ Ek,n+1,p+1 (z) is bounded in Dr by a constant independent of n. It remains to deal with the expression 0 Tn+1,2p+1 (z) Tn+1,2p+1 (z) F (2p+1) (z) + F (2p+2) (z) A := p (n + 1) (2p + 1)! (n + 1)p (2p + 1)! 0 Tn+1,2p+2 (z) Tn+1,2p+2 (z) + F (2p+2) (z) + F (2p+3) (z). p+1 (n + 1) (2p + 2)! (n + 1)p+1 (2p + 2)! By using the recurrence formula in Lorentz [125], p. 14, relation (3) 0 Tn+1,j+1 (z) = z(1 − z)[Tn+1,j (z) + (n + 1)jTn+1,j−1 (z)], for j = 2p + 1 and j = 2p + 2, we obtain Tn+1,2p+2 (z) 0 − (n + 1)(2p + 1)Tn+1,2p (z), Tn+1,2p+1 (z) = z(1 − z) and Tn+1,2p+3 (z) 0 Tn+1,2p+2 (z) = − (n + 1)(2p + 2)Tn+1,2p+1 (z). z(1 − z) Replacing these in A, exactly as in the proof of Corollary 1.3.4 it follows that in the expression of A only the terms independent of n matter for the lower estimate. Simple calculation shows that these terms are given by ap [(1 − 2z)(z(1 − z))p ]0 f (2p) (z) G(f )(z) = (2p + 1)! ap+1 (1 − 2z)[z(1 − z)]p+1 (2p+1) + f (z) z(1 − z)(2p + 2)! [z(1 − z)]p+1 (2p+2) + p+1 f (z), 2 (p + 1)! 1 , reasoning where ap , ap+1 > 0. In order to obtain the lower estimate of order np+1 as in the proof of Corollary 1.3.4 it suffices to prove that if f is not a polynomial of degree ≤ 2p − 1 then kG(f )kr > 0. Making the substitution f (2p) (z) = y(z) it follows that it suffices to prove that if y(z) is not identical zero then kG(y)k r > 0. For this purpose we reason as in the proof of Corollary 1.3.4. Thus we can show that the only solution of the differential equation ap [(1 − 2z)(z(1 − z))p ]0 y(z) ap+1 (1 − 2z)[z(1 − z)]p+1 0 + y (z) (2p + 1)! z(1 − z)(2p + 2)! [z(1 − z)]p+1 00 + p+1 y (z) = 0 2 (p + 1)! P∞ is y(z) = 0 for all z ∈ Dr . Writing y(z) = k=0 bk z k and reasoning as in the proof of Corollary 1.3.4 , we easily obtain step by step that b0 = 0, b1 = 0, so on, bk = 0 for all k. We omit here the calculation details which are simple. The theorem is proved.

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At the end of this section, concerning the mth iterates prove the following result.

m

103

(α,β)

Kn

(f )(z), we can

Theorem 1.7.7. (Gal [91]) Let f : DR → C be analytic in DR = {z ∈ C; |z| < R} P∞ with R > 1, i.e. f (z) = k=0 ck z k , for all z ∈ DR . Suppose 1 ≤ r < r1 < R. Then for all |z| ≤ r and n, p ∈ N, we have |[m Kn(α,β) (f )](p) (z) − f (p) (z)| ∞ X (p + 1)!r1 2m |ck−1 | · |β + (k − 1)|rk · ≤ . n+1+β (r1 − r)p+1 k=1

Proof.

First we easily observe that

d m (α,β) [ Sn+1 (F )](z), Kn(α,β) (f )(z) = dz Rz P∞ where F (z) = 0 f (t)dt = k=0 Ck z k . Taking into account Theorem 1.6.7, the Cauchy’s formula and reasoning exactly as in the proofs of Theorem 1.7.4, (i) and 1.7.5, (i), it follows m

|[m Kn(α,β) (f )](p) (z) − f (p) (z)| (α,β)

= |[m Sn+1 (F )](p+1) (z) − F (p+1) (z)| ∞ X (p + 1)!r1 2m |Ck | · |βk + k(k − 1)|r k · ≤ n+1+β (r1 − r)p+1 =

2m n+1+β

which proves the theorem.

k=1 ∞ X k=1

|ck−1 | · |β + (k − 1)|rk ·

(p + 1)!r1 , (r1 − r)p+1

Remark. 1) For β = 0 in Theorem 1.7.7 we get corresponding results for the iterates of classical complex Kantorovich polynomials. Note that in the real case, some asymptotic results for iterates of Kantorovich polynomials were obtained in Nagel [143]. 2) If mnn → 0 when n → ∞, then by Theorem 1.7.7 it is immediate that [mn Kn(α,β) (f )](p) (z) → f (p) (z),

uniformly with respect to |z| ≤ 1, for any 1 ≤ r < R. 1.8

Favard-Sz´ asz-Mirakjan Operators

In this section we obtain quantitative estimates of the convergence and of the Voronovskaja’s theorem in compact disks, for complex Favard-Sz´ asz-Mirakjan operators attached to an analytic function in a disk of radius R > 1 and center 0. The section is divided in two parts. In the first part of it, the analytic function satisfies

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some suitable exponential-type growth condition, while in the second part it does not satisfy such of conditions. But in the second case, the price paid is that the uniform convergence and the estimates hold in closed disks of radii < R2 only. Also, we will prove that beginning with an index, these operators preserve the starlikeness, convexity and spirallikeness in the unit disk. If f : [0, ∞) → R then it is well-known that the Favard-Sz´ asz-Mirakjan operators are given by (see Favard [72], Sz´ asz [188], Mirakjan [137]) Sn (f )(x) = P (nx)j e−nx ∞ j=0 j! f (j/n), x ∈ [0, ∞), where for the convergence of Sn (f )(x) to f (x), usually f is supposed to be of exponential growth, that is |f (x)| ≤ CeBx , for all x ∈ [0, +∞), with C, B > 0 (see Favard [72]). Also, concerning quantitative estimates in approximation of f (x) by Sn (f )(x), in e.g. Totik [191], it is proved that under some additional assumptions on f , we actually have |Sn (f )(x) − f (x)| ≤ C n, for all x ∈ R+ , n ∈ N. The complex Favard-Sz´ asz-Mirakjan operator is obtained from the real version, simply replacing the real variable x by the complex one z, that is ∞ X (nz)j f (j/n). Sn (f )(z) = e−nz j! j=0 Let us note that in our results, the domain of definition of the approximated S function f : DR [R, ∞) → C seem to be rather strange. However, the analyticity P k of f on R on DR assures the representation f (z) = ∞ k=0 ck z , which is essential in the proof of quantitative estimates in any Dr with 1 ≤ r < R (while on [0, ∞) the well known estimates in the case of real variable can be used). Probably a more natural domain of definition for f would be a strip around the P∞ k OX-axis, but in this case the representation f (z) = k=0 ck z fails, fact which produces the failure of the methods of proofs in this case. In this first part, supposing that f : [0, +∞) → C of exponential growth, can be prolonged to an analytic function in an open disk (with center in origin) by keeping exponential growth, we obtain quantitative estimates in closed disks with center in origin, similar in form with that in the real case in Totik [191] mentioned above. Also, we recall that the first result concerning the convergence of complex Sn (f )(z) to f (z) belonging to a class of analytic functions satisfying a suitable exponential-type growth condition in a parabolic domain, was proved in DresselGergen-Purcell [65], but without any estimate of the approximation error. The first main result of this section can be summarized by the following.

Theorem 1.8.1. (Gal [83]) Let DR = {z ∈ C; |z| < R} be with 1 < R < +∞ and suppose that f : [R, +∞) ∪ DR → C is continuous in [R, +∞) ∪ DR , analytic P∞ in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR , and that there exist M, C, B > 0 k 1 and A ∈ ( R , 1), with the property |ck | ≤ M Ak! , for all k = 0, 1, ..., (which implies |f (z)| ≤ M eA|z| for all z ∈ DR ) and |f (x)| ≤ CeBx , for all x ∈ [R, +∞). 1 be arbitrary fixed. For all |z| ≤ r and n ∈ N, we have (i) Let 1 ≤ r < A Cr,A |Sn (f )(z) − f (z)| ≤ , n

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P∞ k where Cr,A = M k=2 (k + 1)(rA) < ∞. 2r 1 (ii) If 1 ≤ r < r1 < A are arbitrary fixed, then for all |z| ≤ r and n, p ∈ N, |Sn(p) (f )(z) − f (p) (z)| ≤

p!r1 Cr1 ,A , n(r1 − r)p+1

where Cr1 ,A is given as at the above point (i). Proof.

(i) According to Theorem 2 in Lupa¸s [127], we can write Sn (f )(z) =

∞ X

[0, 1/n, ..., j/n; f ]z j ,

j=0

where [0, 1/n, ..., j/n; f ] denotes the divided difference of f on the knots 0, 1/n, ..., j/n. Note that the above formula was proved in Lupa¸s [127] for functions of real variable, but the formula holds in complex setting too, since only algebraic calculations were used (see the proof of Theorem 2 in Lupa¸s [127]). Taking in this representation formula ek (z) = z k , we get that Tn,k (z) := Sn (ek )(z) is a polynomial of degree ≤ k, k = 0, 1, 2, ..., and Tn,0 (z) = 1, Tn,1 (z) = z, for all z ∈ C. Also, differentiating Tn,k (z) with respect to z 6= 0, we get ∞ j X (nz)j−1 jk −nz (nz) −nz 0 −ne +e jn Tn,k (z) = nk j! j! j=0 = −nTn,k (z) + = −nTn,k (z) +

∞ X j k+1 −nz n (nz)j e nk+1 z j! j=0

n Tn,k+1 (z), z

which implies Tn,k+1 (z) =

z 0 T (z) + zTn,k (z), n n,k

for all z ∈ C, k ∈ {0, 1, 2, ..., }, n ∈ N. From this it is immediate the recurrence formula Tn,k (z) − z k =

z k − 1 k−1 [Tn,k−1 (z) − z k−1 ]0 + z[Tn,k−1 (z) − z k−1 ] + z , n n

for all z ∈ C, k, n ∈ N. Now, let 1 ≤ r < R. Denoting with k · kr the norm in C(Dr ), where Dr = {z ∈ C; |z| ≤ r}, by a linear transformation, the Bernstein’s inequality in the closed unit disk becomes |Pk0 (z)| ≤ kr kPk kr , for all |z| ≤ r, where Pk (z) is a polynomial of degree ≤ k. Therefore, from the above recurrence formula, we get kTn,k − ek kr ≤

r k−1 · kTn,k−1 − ek−1 kr n r rk−1 (k − 1) +rkTn,k−1 − ek−1 kr + , n

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which implies the recurrence k−1 rk−1 (k − 1) . kTn,k − ek kr ≤ r + kTn,k−1 − ek−1 kr + n n In what follows we prove by mathematical induction with respect to k (with n ≥ 1 supposed to be fixed, arbitrary), that this recurrence implies (k + 1)! k−1 r , for all k ≥ 2, n ≥ 1. kTn,k − ek kr ≤ 2n Indeed, for k = 2 and n ∈ N, the left-hand side is nr while the right-hand side is 3r n . Supposing now that it is true for k, the above recurrence implies k (k + 1)! k−1 rk k kTn,k+1 − ek+1 kr ≤ r + r + . n 2n n It remains to prove that k (k + 1)! k−1 rk k (k + 2)! k r+ r + ≤ r , n 2n n 2n or, after simplifications, equivalently to k (k + 1)! + 2rk ≤ (k + 2)!r. r+ n It is easy to see that this last inequality holds true for all k ≥ 2 and n ∈ N. Now, from the hypothesis on f (that is |f (x)| ≤ max{M, C}emax{A,B}x, for all x ∈ R+ ), it follows that ( see e.g. Dressel-Gergen-Purcell [65], pp. 1171-1172 and p. 1178 ) Sn (f )(z) is analytic in DR . Therefore, it is easy to see that we can write ∞ ∞ X X Sn (f )(z) = ck Sn (ek )(z) = ck Tn,k (z), for all z ∈ DR , k=0

k=0

which from the hypothesis on ck , immediately implies for all |z| ≤ r ∞ ∞ X X Ak (k + 1)! k−1 |Sn (f )(z) − f (z)| ≤ |ck | · |Tk,n (z) − ek (z)| ≤ M r k! 2n k=2

P

k=2

∞ Cr,A M X (k + 1)(rA)k = , = 2nr n

∞ M k=2 (k 2r P ∞ k+1 k=2 u

k=2

1 , taking into account where Cr,A = + 1)(rA)k < ∞, for all 1 ≤ r < A P∞ that the series and therefore its derivative k=2 (k +1)uk , are uniformly and absolutely convergent in any compact disk included in the open unit disk. (ii) Denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z p! Sn (f )(v) − f (v) Cr1 ,A p! 2πr1 |Sn(p) (f )(z) − f (p) (z)| = dv ≤ n 2π (r1 − r)p+1 2π γ (v − z)p+1 Cr1 ,A p!r1 = , n (r1 − r)p+1 which proves (ii) and the theorem.

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Remark. Let us show in detail the relationship ∞ X Sn (f )(z) = ck Sn (ek )(z), k=0

used in the proof of Theorem 1.8.1, (i) as follows. For this purpose, for any m ∈ N let us define m X fm (z) = cj z j if |z| ≤ r and fm (x) = f (x) if x ∈ (r, +∞). j=0

From the hypothesis on f it is clear that for any m ∈ N it follows |fm (x)| ≤ Cm eBm x , for all x ∈ [0, +∞). This implies that for each fixed m, n ∈ N and z, |Sn (fm )(z)| ≤ |e−nz |

∞ X (n|z|)j j=0

≤ Cm |e−nz |

j!

|fm (j/n)|

∞ X (n|z|)j j=0

j!

eBm j/n < ∞,

since by the ratio criterium the last series is convergent. Therefore Sn (fm )(z) is well-defined. Denoting f (x) if x ∈ (r, ∞), m+1 it is clear that each fm,k is of exponential growth on [0, ∞) and that fm (z) = Pm k=0 fm,k (z). Since from the linearity of Sn we have fm,k (z) = ck ek (z) if |z| ≤ r and fm,k (x) =

Sn (fm )(z) =

m X k=0

ck Sn (ek )(z), for all |z| ≤ r,

it suffices to prove that limm→∞ Sn (fm )(z) = Sn (f )(z) for any fixed n ∈ N and |z| ≤ r. But this is immediate from limm→∞ kfm −f kr = 0, from kfm −f kB[0,+∞) ≤ kfm − f kr and from the inequality |Sn (fm )(z) − Sn (f )(z)| ≤ |e−nz | · en|z| · kfm − f kB[0,∞) ≤ Mr,n kfm − f kr ,

valid for all |z| ≤ r. Here k · kB[0,+∞) denotes the uniform norm on C[0, +∞)-the space of all real-valued bounded functions on [0, +∞). In what follows we obtain the Voronovskaja-type formula with a quantitative estimate for the complex Favard-Sz´ asz-Mirakjan operator. Theorem 1.8.2. (Gal [83]) Suppose that the hypothesis on the function f and the constants R, M, C, B, A in the statement of Theorem 1.8.1 hold and let 1 ≤ r < A1 be arbitrary fixed. (i) The following upper estimate in the Voronovskaja-type formula holds ∞ z 00 3M A|z| X (k + 1)(rA)k−1 , f (z) ≤ Sn (f )(z) − f (z) − 2n r 2 n2 k=2

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for all n ∈ N, |z| ≤ r. (ii) We have the following equivalence in the Voronovskaja’s formula

1 e1 00

f ∼ 2,

Sn (f ) − f − 2n n r

where the constants in the equivalence depend on f and r but are independent of n. Proof. (i) Denoting ek (z) = z k , k = 0, 1, ..., and Tn,k (z) = Sn (ek )(z), by the P∞ proof of Theorem 1.8.1, (i), we can write Sn (f )(z) = k=0 ck Tn,k (z), which immediately implies ∞ z 00 X z k−1 k(k − 1) f (z) ≤ |ck | · Tn,k (z) − ek (z) − Sn (f )(z) − f (z) − 2n 2n k=2

for all z ∈ DR , n ∈ N. By the recurrence relationship in the proof of Theorem 1.8.1, (i), satisfied by k−1 k(k−1) , we immediately get the Tn,k (z), denoting Ek,n (z) = Tn,k (z) − ek (z) − z 2n new recurrence z 0 z k−2 (k − 1)(k − 2)2 Ek−1,n (z) + zEk−1,n (z) + , n 2n2 for all k ≥ 2, n ∈ N and z ∈ DR . This implies, for all |z| ≤ r, k ≥ 2, n ∈ N, Ek,n (z) =

|Ek,n (z)|

|z| rk−3 (k − 1)(k − 2)2 |z| 0 [2kEk−1,n kr ] + |z| · |Ek−1,n (z)| + · 2n 2n n k−3 2 |z| (k − 1)(k − 2) r 0 ≤ r|Ek−1,n (z)| + [2kEk−1,n kr + ] 2n n |z| 2(k − 1) rk−3 (k − 1)(k − 2)2 ≤ r|Ek−1,n (z)| + kEk−1,n kr + 2n r n |z| 2(k − 1) ≤ r|Ek−1,n (z)| + kTn,k−1 − ek−1 kr 2n r 2(k − 1) rk−2 (k − 1)(k − 2) rk−3 (k − 1)(k − 2)2 + ≤ r|Ek−1,n (z)| · + r 2n n |z| 2(k − 1) rk−2 k! 2(k − 1) rk−2 (k − 1)(k − 2) + · + · 2n r 2n r 2n k−3 2 r (k − 1)(k − 2) 3|z|rk−3 ≤ r|Ek−1,n (z)| + + (k − 1)k! n 2n2 ≤

≤ r|Ek−1,n (z)| +

3|z|rk−3 (k + 1)!, 2n2

that is |Ek,n (z)| ≤ r|Ek−1,n (z)| +

3|z|rk−3 (k + 1)!, for all |z| ≤ r. 2n2

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Taking k = 2, 3, ..., in this last inequality, step by step we obtain k+1 3|z|rk−3 X 3|z|rk−3 (k + 1)! |Ek,n (z)| ≤ , j! ≤ 2n2 j=3 n2

which implies ∞ z 00 X f (z) ≤ |ck | · |Ek,n (z)|| Sn (f )(z) − f (z) − 2n k=2

≤

∞

3M |z| X Ak (k + 1)!rk−3 n2 k! k=2 ∞ X

3M A|z| (k + 1)(rA)k−1 , r 2 n2 k=2 P∞ for all |z| ≤ r, where for rA < 1 we obviously have k=2 (k + 1)(rA)k−1 < ∞. (ii) Taking into account the above point (i) it will be enough to prove the lower estimate. For this purpose we will use the ideas in the proof of Corollary 1.3.4. More exactly, let us consider the expression which appear in the generalized Voronovskaja’s formula for the complex Favard-Sz´ asz-Mirakjan operators, that is ≤

2p X 1 A (z)f (j) (z), Qn,p (f )(z) = Sn (f )(z) − f (z) − j n,j j!n j=1

where An,j = nj Sn [(· − z)j ](z) = e−nz

∞ X (nz)k k=0

k!

(k − nz)j .

The idea in the proof of Corollary 1.3.4 is that in order to get the lower estimate C C we need first to prove the upper estimate kQn,p+1 (f )kr ≤ np+2 . kQn,p (f )kr ≥ np+1 In what follows for simplicity we will consider above the case p = 1. Therefore, first we need an upper estimate for |Qn,2 (f )(z)|. According to Lemma 1.2 in Pop [156], An,j (z) is a polynomial of degree [j/2]. Also, from Lemmas 1.2 and 1.3 and Consequence 1.2 in Pop [156] we easily get An,0 (z) = 1, An,1 (z) = 0, An,2 (z) = nz, An,3 (z) = nz, An,4 (z) = 3n2 z 2 + nz, which replaced in the expression of Qn,2 (f )(z) will mean that we need to prove an upper estimate of the form (valid for all |z| ≤ r) 2 Sn (f )(z) − f (z) − z f 00 (z) − z f (3) (z) − 3nz + z f (4) (z) ≤ C . n3 2 3 2n 6n 24n P∞ Since Sn (f )(z) = k=0 ck Tn,k (z) and denoting k(k − 1)(k − 2)z k−2 k(k − 1)z k−1 − 2n 6n2 k(k − 1)(k − 2)(k − 2) − (3nz 2 + z)z k−4 , 24n3

Ek,n,2 (z) = Tn,k (z) − ek (z) −

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we can write 2 Sn (f )(z) − f (z) − z f 00 (z) − z f (3) (z) − 3nz + z f (4) (z) 2n 6n2 24n3 ≤

∞ X

k=5

|ck | · |Ek,n,2 (z)|,

by taking into account that by simple calculation we get Ek,n,2 (z) = 0 for allk = 0, 1, 2, 3, 4. Also, clearly Ek,n,2 (z) is a polynomial of degree ≤ k.

Now, if we denote Ek,n,1 (z) = Tn,k (z) − ek − k(k−1)z 2n above point (i) we can write the recurrence formula Ek,n,1 (z) =

k−1

, by the proof of the

z k−2 (k − 1)(k − 2)2 z 0 Ek−1,n,1 (z) + zEk−1,n,1 (z) + . n 2n2

On the other hand, simple calculation lead us to the formula Ek,n,e (z) = Ek,n,1 (z) − Pk,n (z), where Pk,n (z) =

k(k − 1)(k − 2)(3k − 5) k−2 k(k − 1)(k − 2)(k − 3) k−3 z + z . 24n2 24n3

The recurrence formula for Ek,n,1 (z) implies the following recurrence formula for Ek,n,2 (z) Ek,n,2 (z) =

z 0 E (z) + zEk−1,n,2 (z) + Rk,n (z), n k−1,n,2

where Rk,n (z) =

(k − 1)(k − 2)(k − 3)(k − 4)(3k − 5) k−3 z 24n3 (k − 1)(k − 2)(k − 3)(k − 4)2 k−4 z . + 24n4

From this recurrence, applying the Bernstein’s inequality and the estimate for k−3 kEk−1,n,1 kr ≤ 3r n2 k! from the above point (i), for all |z| ≤ r, n ∈ N and k ≥ 5 we obtain kEk,n,2 kr

r kE 0 kr + kRk,n kr ≤ rkEk−1,n,2 kr n k−1,n,2 r k−1 r k−3 + · kEk−1,n,1 kr + · kPk,n kr + kRk,n kr ≤ rkEk−1,n,2 kr n r n r k − 1 3rk−3 k! k − 3 rk−3 + ·· + · (k − 1)(k − 2)(k − 3)(3k − 8) n n2 n 24n2 k − 3 rk−4 + · (k − 1)(k − 2)(k − 3)(k − 4) n 24n3

≤ rkEk−1,n,2 (z)kr +

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rk−3 (k − 1)(k − 2)(k − 3)(k − 4)(3k − 5) 24n3 rk−4 3rk−3 (k + 1)! + (k − 1)(k − 2)(k − 3)(k − 4)2 ≤ rkEk−1,n,2 kr + 4 24n n3 k−3 k−4 k−3 k−4 3r (k + 1)! r k! 3r k! r k! + + + + 24n3 24n4 24n3 24n4 3rk−3 (k + 1)! 3rk−3 (k + 1)! + ≤ rkEk−1,n,2 kr + n3 2n3 k−3 k−3 k−3 r k! 3r k! r k! + + + 3 3 3 2n 2n 2n 7rk−3 (k + 1)! . = rkEk−1,n,2 kr + n3 +

Therefore we have obtained that for all n ∈ N and k = 5, 6, ... kEk,n,2 kr ≤ rkEk−1,n,2 kr +

7rk−3 (k + 1)! . n3

Taking here step by step k = 5, 6, ... and taking into account that Ek,n,2 (z) = 0 for k = 0, 1, 2, 3, 4, we easily obtain kEk,n,2 kr ≤

k 7rk−3 (k + 2)! 7rk−3 X (j + 1)! ≤ , k = 5, 6, ..., 3 n j=5 n3

which implies

∞ X

Sn (f ) − f − e1 f 00 − e1 f (3) − 3ne2 + e1 f (4) ≤ |ck | · kEk,n,2 kr

2n 6n2 24n3 r k=5

≤

Cr (f ) . n3

Now, by similar reasonings with those in the case of Bernstein polynomials in Lemma 2.2 in Pop [155], we easily obtain Sn (ek )(z) =

k X 1 A (z)(z k )(j) . j n,j j!n j=0

Denoting for arbitrary p ∈ N Ek,n,p (z) = Sn (ek )(z) − ek (z) −

2p X 1 A (z)(z k )(j) , j j! n,j n j=1

the above formula for Sn (ek )(z) implies that Ek,n,p (z) =

k X

j=2p+1

1 k k−j z An,j (z), nj j

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and by direct calculation we arrive at Sn (f )(z) − f (z) −

2p X f (j) (z) j=1

j!

 ∞  X

n−j An,j (z) =

∞ X

Ek,n,p (z)

k=2p+1

 k  X k k−j p+1−j ck  z n An,j (z)  j j=2p+1 

1 np+1  k=2p+1 An,2p+1 (z) (2p+1) An,2p+2 (z) (2p+2) 1 f f (z) + p+1 (z) = p+1 p n n (2p + 1)! n (2p + 2)!   ∞  X 1 ck Ek,n,p+1 (z) . + np+2  n =

k=2p+3

By Corollary 1.3 in Pop [156], An,j (z) is a polynomial of degree ≤ [j/2] and by e.g. Agratini [3], p. 237, Lemma 3.9.4, is a polynomial of degree ≤ j in z. Therefore we can write An,2p+1 (z) (2p+1) An,2p+2 (z) (2p+2) f (z) + p+1 f (z) p n (2p + 1)! n (2p + 2)! = P2p+1 (z)f (2p+1) (z) + P2p+2 (z)f (2p+2) (z) +

1 1 F (z)f (2p+1) (z) + G(z)f (2p+2) (z), n n

where F (z) and G(z) are bounded polynomials on Dr by constants depending on r and p nut independent of n. In what follows we will find the form of P2p+1 (z) and P2p+2 (z). First, by taking p(z) = z in e.g. Lemma 1.3 in Pop [156] we get the recurrence formula An,j+1 (z) = z[A0n,j (z) + njAn,j−1 (z)], which immediately implies An,0 (z) = 1, An,1 (z) = 1, An,2 (z) = nz, An,3 (z) = nz, An,4 (z) = 3n2 z 2 +nz, An,5 (z) = 10n2 z 2 +nz, An,6 (z) = 15n3 z 3 +25n2z 2 , An,7 (z) = 105n3 z 3 + 56n2 z 2 , and so on. By mathematical induction it easily follows that the coefficient of n[(2p+1)/2] = np in An,2p+1 (z) is of the form cp z p with cp > 0, while the coefficient of n[(2p+2)/2] = np+1 in An,2p+2 (z) is of the form dp z p+1 with dp > 0. Therefore we have P2p+1 (z) = cp z p and P2p+2 (z) = dp z p+1 . Denoting now U (f )(z) = P2p+1 (z)f (2p+1) (z) + P2p+2 (z)f (2p+2) (z), we will prove that if f is not a polynomial of degree ≤ 2p then kU (f )kr > 0. Indeed, suppose that for such an f we have kU (f )kr = 0, that is the following differential equation holds P2p+1 (z)f (2p+1) (z) + P2p+2 (z)f (2p+2) (z) = 0, z ∈ Dr . Making the substitution f (2p+1) (z) = y(z) and replacing the above found form for P2p+1 (z) and P2p+2 (z), we obtain cp z p y(z) + dp z p+1 y 0 (z) = 0, z ∈ Dr .

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Simplifying with z p 6= 0 we obtain

cp y(z) + dp zy 0 (z) = 0, z ∈ Dr \ {0}. P∞ Passing here with z → 0 we get y(0) = 0. Writing y(z) = k=1 ak z k and replacing in the above differential, by the identification of the coefficients we easily obtain that ak = 0 for all k ≥ 1, that is y(z) = 0, for all z ∈ Dr and therefore f necessarily is a polynomial of degree ≤ 2p, a contradiction. Therefore for f a polynomial of degree ≤ 2p, the supposition that kU (f )kr = 0 is false, that is in this case we have kU (f )kr > 0. From this point, reasoning exactly as in the proof of Corollary 1.3.4 for p = 1 we obtain 2 X Cr (f ) f (j) −j n An,j kr ≥ , kSn (f ) − f − j! n2 j=1 which proves the point (ii) and the theorem.

The next result shows that the order of approximation in Theorem 1.8.1 is exactly n1 . Corollary 1.8.3. (Gal [83]) In the hypothesis of Theorem 1.8.1, if f is not a polynomial of degree ≤ 1 in the case (i) and if f is not a polynomial of degree ≤ p, (p ≥ 1) in the case (ii), then n1 is in fact the exact order of approximation. Applying the norm k · kr to the identity z 00 i 1 h 2 1 z 00 n Sn (f )(z) − f (z) − f (z) + f (z) , Sn (f )(z) − f (z) = n 2 n 2n it follows

i 1 h 2 e1 00 1

e1 00

. f n Sn (f ) − f − kSn (f ) − f kr ≥

f − n 2 n 2n r r

e 00 If f is not a polynomial of degree ≤ 1 then evidently 21 f r > 0, which combined with the estimate in Theorem 1.8.2 immediately implies that kSn (f ) − f kr ≥ C n , for all n ≥ n0 , with C > 0 independent of n. Since for n = 1, 2, ..., n0 − 1 the inequality kSn (f ) − f kr ≥ Cn1 is trivial with a constant C1 > 0 and taking into account the upper estimate in Theorem 1.8.1, (i), we get the desired conclusion. Now, replacing Sn (f )(z) − f (z) in the above identity, to the Cauchy formula in the proof of Theorem 1.8.1, (ii) and then applying the norm k · kr to the integral identity, we get Proof.

kSn(p) (f ) − f (p) kr

( Z 2 he i(p)

n Sn (f )(v) − f (v) − 1 1

p! 1 00

− f ≥

n 2 n 2π Γ (v − e1 )p+1 r

v 00 2n f (v)

)

dv ,

r

which combined again with Theorem 1.8.2 and taking into account that

e1 00 (p)

2f

> 0 (since f is not a a polynomial of degree ≤ p), as above leads r us to the same conclusion.

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In the second part of this section the growth conditions of exponential-type on f will be omitted. The only condition imposed to f is to be bounded on [0, ∞), case when it is clear that the complex Favard-Sz´ asz-Mirakjan operators given by P∞ (nz)j −nz Sn (f )(z) = e j=0 j! f (j/n) are well defined for all z ∈ C. In this sense, the first result is expressed by the following. Theorem 1.8.4. (Gal [84]) For 2 < R < +∞ let f : [R, +∞) ∪ DR → C be bounded P∞ on [0, +∞) and analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . (i) If 1 ≤ r < R2 then for all |z| ≤ r and n ∈ N it follows |Sn (f )(z) − f (z)| ≤

Cr,f , n

P∞ with Cr,f = 6 k=2 |ck |(k − 1)(2r)k−1 < ∞. (ii) If 1 ≤ r < r1 < R2 then for all |z| ≤ r and n, p ∈ N it follows |Sn(p) (f )(z) − f (p) (z)| ≤

p!r1 Cr1 ,f , n(r1 − r)p+1

where Cr1 ,f is as above. Proof.

(i) From the proof of Theorem 1.8.1, (i) we can write ∞ X

Sn (f )(z) =

[0, 1/n, ..., j/n; f ]z j ,

j=0

where [0, 1/n, ..., j/n; f ] denotes the divided difference of f on the knots 0, 1/n, ..., j/n. For f (z) = ek (z) = z k and applying the mean value theorem for divided differences, for all |z| ≤ r and k, n ∈ N it follows |Sn (ek )(z)| ≤ =

k X j=0

|[0, 1/n, ..., j/n; ek ]|rj

k X k(k − 1)...(k − j + 1)

j!

j=1

≤ rk

k X k j=0

j

rk−j rj

= (2r)k .

Denoting Tn,k (z) := Sn (ek )(z) clearly that it is a polynomial of degree ≤ k, k = 0, 1, 2, ...,, Tn,0 (z) = 1, Tn,1 (z) = z, for all z ∈ C and by the proof of Theorem 1.8.1, (i) the recurrence formula Tn,k (z) − z k =

k − 1 k−1 z [Tn,k−1 (z) − z k−1 ]0 + z[Tn,k−1 (z) − z k−1 ] + z , n n

holds for all z ∈ C, k, n ∈ N.

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Applying as in the proof of Theorem 1.8.1, (i) the Bernstein’s inequality, from the above recurrence formula, for all |z| ≤ r we get r k−1 rk−1 (k − 1) · kTn,k−1 − ek−1 kr + r|Tn,k−1 − ek−1 | + n r n k−1 k−1 r (k − 1) ≤ [kTn,k−1 kr + rk−1 ] + r|Tn,k−1 − ek−1 | + n n k−1 r (k − 1) k−1 [(2r)k−1 + rk−1 ] + r|Tn,k−1 − ek−1 | + ≤ n n 3(k − 1)(2r)k−1 ≤ r|Tn,k−1 − ek−1 | + . n

|Tn,k − ek | ≤

1 Since Tn,1 (z) = e1 (z), for k = 2 the above inequality implies |Tn,2 (z) − z 2 | ≤ 3r n2 , 2 for k = 3 it implies |Tn,3 (z) − z 3 | ≤ 3rn (1 · 21 + 2 · 22 ), and step by step for all |z| ≤ r we finally obtain   k−1 k−1 X 3rk−1 3r  j2j  = (k − 2)2k + 2 |Tn,k (z) − z k | ≤ n n j=1

≤

6(k − 1) (2r)k−1 . n

Pk−1 j Here the formula j=1 j2 = (k − 2)2k + 2 can easily be proved by mathematical induction. P∞ Since Sn (f )(z) = k=0 ck Tn,k (z), we get |Sn (f )(z) − f (z)| ≤

∞ X k=2

|ck | · |Tn,k (z) − z k | ≤

Cr,f , n

which proves (i). (ii) Denote by γ the circle of radius r1 > r and center 0. For any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r and by the Cauchy’s formulas for all |z| ≤ r and n ∈ N it follows Z p! Sn (f )(v) − f (v) Cr1 ,f p! 2πr1 |Sn(p) (f )(z) − f (p) (z)| = dv ≤ n 2π (r1 − r)p+1 2π γ (v − z)p+1 p!r1 Cr ,f , = 1 n (r1 − r)p+1 which proves (ii) and the theorem.

P∞ Remark. A proof of the relationship Sn (f )(z) = k=0 ck Sn (ek )(z) used in the proof of Theorem 1.8.4, (i) is as follows. For any m ∈ N define fm (z) =

m X j=0

cj z j if |z| ≤ r and fm (x) = f (x) if x ∈ (r, +∞).

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From the hypothesis on f it is clear that each fm is bounded on [0, +∞), which implies that ∞ X (n|z|)j |Sn (fm )(z)| ≤ |e−nz | M (fm ) = |e−nz | · en|z| M (fm ) < ∞, j! j=0 that is all Sn (fm )(z) are well-defined. Denoting f (x) if x ∈ (r, ∞), m+1 Pm it is clear that each fm,k is bounded on [0, ∞) and that fm (z) = k=0 fm,k (z). Since from the linearity of Sn we have m X Sn (fm )(z) = ck Sn (ek )(z), for all |z| ≤ r, fm,k (z) = ck ek (z) if |z| ≤ r and fm,k (x) =

k=0

it suffices to prove that limm→∞ Sn (fm )(z) = Sn (f )(z) for any fixed n ∈ N and |z| ≤ r. But this is immediate from limm→∞ kfm −f kr = 0, from kfm −f kB[0,+∞) ≤ kfm − f kr and from the inequality |Sn (fm )(z) − Sn (f )(z)| ≤ |e−nz | · en|z| · kfm − f kB[0,∞) ≤ Mr,n kfm − f kr ,

valid for all |z| ≤ r. Here k · kB[0,+∞) denotes the uniform norm on C[0, +∞)-the space of all real-valued bounded functions on [0, +∞). Also, the following Voronovskaja-type formula holds. Theorem 1.8.5. (Gal [84]) For 2 < R < +∞ let f : [R, +∞) ∪ DR → C be bounded P∞ on [0, +∞) and analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . Also, let 1 ≤ r < R2 . (i) For all |z| ≤ r and n ∈ N it follows

|z| z 00 f (z)| ≤ Mr,f · 2 , 2n n P∞ 2 k−3 with Mr,f = 26 k=3 |ck |(k − 1) (k − 2)(2r) < ∞. (ii) For all n ∈ N we have e1 00 1 kSn (f ) − f − f kr ∼ 2 , 2n n where the constants in the equivalence depend on f and r but are independent of n. |Sn (f )(z) − f (z) −

Proof.

(i) Keeping the notations in the proof of Theorem 1.8.2 (i) we have ∞ z 00 X |ck | · |Ek,n (z)|, f (z) ≤ Sn (f )(z) − f (z) − 2n k=2

where it is necessary to obtain a suitable estimate for |Ek,n (z)|. For this purpose we use the recurrence in the proof of Theorem 1.8.2 (i) given by Ek,n (z) =

z 0 z k−2 (k − 1)(k − 2)2 Ek−1,n (z) + zEk−1,n (z) + , n 2n2

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for all k ≥ 2, n ∈ N and z ∈ DR . This implies, for all |z| ≤ r, k ≥ 2, n ∈ N, |Ek,n (z)|

|z| |z| rk−3 (k − 1)(k − 2)2 0 [2kEk−1,n kr ] + |z| · |Ek−1,n (z)| + · 2n 2n n k−3 2 |z| r (k − 1)(k − 2) 0 2kEk−1,n kr + ≤ r|Ek−1,n (z)| ≤ r|Ek−1,n (z)| + 2n n rk−3 (k − 1)(k − 2)2 |z| 2(k − 1) kEk−1,n kr + ≤ r|Ek−1,n (z)| + 2n r n 2(k − 1) rk−2 (k − 1)(k − 2) |z| 2(k − 1) kTn,k−1 − ek−1 kr + · + 2n r r 2n k−3 2 r (k − 1)(k − 2) + ≤ r|Ek−1,n (z)| n |z| 2(k − 1) 6(k − 2) 2(k − 1) rk−2 (k − 1)(k − 2) + · · (2r)k−2 + · 2n r n r 2n k−3 2 2 26|z|(k − 1) (k − 2)(2r)k−3 r (k − 1)(k − 2) , + ≤ r|Ek−1,n (z)| + n 2n2 ≤

that is |Ek,n (z)| ≤ r|Ek−1,n (z)| +

13|z|(k − 1)2 (k − 2)(2r)k−3 . n2

For k = 1, 2 we get Ek,n (z) = 0, for k = 3 in this last inequality we obtain |E3,n (z)| ≤

13|z|r 0 n2 (3

− 1)2 (3 − 2)20 and for k = 4 it follows

13|z|r1 [(3 − 1)2 (3 − 2)20 + (4 − 1)2 (4 − 2)21 ]. n2 Then step by step finally we arrive at   k k−3 X 13|z|r  (j − 1)2 (j − 2)2j−3  |Ek,n (z)| ≤ n2 j=3 |E4,n (z)| ≤

≤

k X 13|z|rk−3 2 2j−3 · (k − 1) (k − 2) n2 j=3

13|z|rk−3 · (k − 1)2 (k − 2)(2k−2 − 1) n2 26|z|(2r)k−3 · (k − 1)2 (k − 2). ≤ n2 =

In conclusion it follows ∞ z 00 X |ck | · |Ek,n (z)| f (z) ≤ Sn (f )(z) − f (z) − 2n k=2

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≤

∞

26|z| X |z| |ck |(k − 1)2 (k − 2)(2r)k−3 = Mr,f · 2 , 2 n n k=3

which proves the point (i). (ii) The proof is similar to the proof of Theorem 1.8.2 (ii), with the only difference that for |Ek,n,1 (z)| := |Ek,n (z)| we use the upper estimate in the above point (i) and for the other terms appearing to be estimated we use upper estimates in accordance with this one. Now we are in position to prove that the order of approximation in Theorem 1.8.4 is exactly n1 . Thus we have : Theorem 1.8.6. (Gal [84]) Let 2 < R < +∞, 1 ≤ r < R2 and f : [R, +∞)∪DR → C P∞ be bounded on [0, +∞) and analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . If f is not a polynomial of degree ≤ 1, then the estimate Cr (f ) , n ∈ N, n holds, where the constant Cr (f ) depends on f and r but is independent of n. kSn (f ) − f kr ≥

Proof.

For all z ∈ DR and n ∈ N we get 1 = n

Sn (f )(z) − f (z)

z 00 i 1 h 2 z 00 n Sn (f )(z) − f (z) − f (z) + f (z) . 2 n 2n

We apply to this identity the following property :

kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . We get kSn (f ) − f kr ≥

1 n

i 1 h 2 e1 00

e1 00

f n Sn (f ) − f − .

f − 2 n 2n r r

Because

e 00 by hypothesis f is not a polynomial of degree ≤ 1 z in00 DR , it follows

1 f > 0. Indeed, supposing the contrary it follows that f (z) = 0 for all 2 2 r z ∈ Dr , that is f 00 (z) = 0 for all z ∈ Dr \ {0}. Since f is analytic, from the identity theorem on analytic functions this implies that f 00 (z) = 0, for all z ∈ DR , that is f is a polynomial of degree ≤ 1, which is contradiction with the hypothesis. Now, by Theorem 1.8.5 we have ∞

X e1 00

f ≤ 26r |ck |(k − 1)2 (k − 2)(2r)k−3 < ∞. n2 Sn (f ) − f − 2n r k=3

Consequently, there exists n1 (depending only on f and r) such that for all n ≥ n1 we have

e

i 1 e 1 h 2 e1 00

1 00

1 f ≥ f 00 , n Sn (f ) − f −

f − 2 n 2n 2 2 r r r

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1 1

e1 · f 00 , ∀n ≥ n1 . n 2 2 r

with Mr,n (f ) =

which implies kSn (f ) − f kr ≥

For n ∈ {1, ..., n1 − 1} we clearly have kSn (f ) − f kr ≥ n · kSn (f ) − f kr > 0, which finally implies

Mr,n (f ) n

Cr (f ) , n

for all n, with Cr (f ) = min{Mr,1 (f ), ..., Mr,n1 −1 (f ), 21 e21 f 00 r }. kSn (f ) − f kr ≥

From Theorem 1.8.6 and Theorem 1.8.4, (i) we immediately obtain the following consequence. Corollary 1.8.7. (Gal [84]) Let 2 < R < +∞ and f : [R, +∞) ∪ DR → C be P∞ bounded on [0, +∞) and analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . If 1 ≤ r < R2 is arbitrary fixed and if f is not a polynomial of degree ≤ 1, then the estimate 1 kSn (f ) − f kr ∼ , n ∈ N, n holds, where the constants in the equivalence depend only on f and r. Regarding the simultaneous approximation of the function and its derivatives we can present : Theorem 1.8.8. (Gal [84]) Let 2 < R < +∞ and f : [R, +∞) ∪ DR → C be P∞ bounded on [0, +∞) and analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR . If 1 ≤ r < r1 < R2 , p ∈ N and if f is not a polynomial of degree ≤ p, then we have 1 kSn(p) (f ) − f (p) kr ∼ , n where the constants in the equivalence depend only on f , r, r1 and p. Proof. The upper estimate is exactly Theorem 1.8.4, (ii), therefore it remains to prove the lower estimate. Denote by Γ the circle of radius r1 and center 0 (where R 2 > r1 > r ≥ 1). By the Cauchy’s formulas for all |z| ≤ r and n ∈ N it follows Z Sn (f )(v) − f (v) p! (p) (p) Sn (f )(z) − f (z) = dv, 2πi Γ (v − z)p+1

where |v − z| ≥ r1 − r for all |z| ≤ r and v ∈ Γ. As in the proof of Theorem 1.8.6, for all v ∈ Γ and n ∈ N we get

Sn (f )(v) − f (v)

v 00 i 1 v 00 1 h 2 n Sn (f )(v) − f (v) − f (v) + f (v) , n 2 n 2n which replaced in the Cauchy’s formula implies Z 1 p! vf 00 (v) Sn(p) (f )(z) − f (p) (z) = dv n 2πi Γ 2(v − z)p+1 =

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1 p! + · n 2πi 1 = n

(

hz

2

00

f (z)

i(p)

Z

Γ

n2 Sn (f )(v) − f (v) − (v − z)p+1

1 p! + · n 2πi

Z

Γ

v 00 2n f (v)

n2 Sn (f )(v) − f (v) − (v − z)p+1

dv

)

v 00 2n f (v)

dv

)

.

Passing to the norm k · kr , for all n ∈ N we obtain kSn(p) (f ) − f (p) kr

1 ≥ n

(

p! Z n2 S (f )(v) − f (v) −

h e1 00 i(p) 1 n

−n

2f

2π (v − z)p+1 Γ r

v 00 2n f (v)

)

dv ,

r

where by Theorem 1.8.5, for all n ∈ N it follows

p! Z n2 S (f )(v) − f (v) − v f 00 (v) n

2n dv

2π Γ

(v − z)p+1 r p! e1 00 2πr1 n2

≤ · f

Sn (f ) − f − 2π (r1 − r)p+1 2n r1 ∞ X p!r1 . ≤ 26r1 |ck |(k − 1)2 (k − 2)(2r1 )k−3 · (r1 − r)p+1 k=3

(p)

Now, by hypothesis on f we have e21 f 00

> 0. Indeed, supposing the contrary r

it follows that z2 f 00 (z) is a polynomial of degree ≤ p − 1, which by the analyticity of f obviously implies that f is a polynomial of degree ≤ p, in contradiction with the hypothesis. For the rest of the proof, reasoning exactly as in the proof of Theorem 1.8.6, we immediately get the required conclusion.

Remarks. 1) Since the boundedness of f on [0, ∞) in the Theorems 1.8.4, 1.8.5, 1.8.6, Corollary 1.8.7 and Theorem 1.8.8 is used only for the existence of the complex Favard-Sz´ asz-Mirakjan operator, taking into account the Remark after the proof of Theorem 1.8.1 it follows that in the above mentioned results it can replaced by the condition of exponential growth |f (z)| ≤ M eBx , for all x ∈ [0, ∞). 2) The domain of approximation [R, +∞) ∪ DR in the previous results of this section, seem less usual. More natural could be, for example, a strip of the form TR = {z = x + iy ∈ C; x ∈ R and |y| ≤ R}.

In what follows we will obtain a weighted approximation result by the complex Favard-Sz´ asz-Mirakjan operator Sn (f )(z) in such a strip TR . First we need the following. Lemma 1.8.9. For fixed arbitrary z0 ∈ C, let us denote e1 (z) = z and k ∞ X (nz)j j Tn,k,z0 (z) := Sn ((e1 − z0 )k )(z) = e−nz − z0 . j! n j=0

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(i) For all n ∈ N, k ∈ N

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S

{0} and z ∈ C we have z 0 Tn,k+1,z0 (z) = Tn,k,z (z) + (z − z0 )Tn,k,z0 (z). 0 n (ii) For all n ∈ N, k ∈ N with k ≥ 2 and z ∈ C we have 0 z Tn,k−1,z0 (z) − (z − z0 )k−1 Tn,k,z0 (z) − (z − z0 )k = n +(z − z0 )[Tn,k−1,z0 (z) − (z − z0 )k−1 ] k−1 z(z − z0 )k−2 . + n (iii) For all n, k ∈ N and z, z0 ∈ C with |z − z0 | ≤ r we have |Tn,k,z0 (z)| ≤ rk (3 + 2|z0 |)k .

Proof. (i) By differentiating Tn,k,z0 we easily get the required recurrence formula. (ii) It is an immediate consequence of (i). (iii) By the proof of Theorem 1.8.1 (i) we have the representation in Lupa¸s [127] ∞ X Sn (f )(z) = [0, 1/n, ..., j/n; f ]z j , j=0

which immediately implies

Sn ((e1 − z0 )p )(z) = =

p X j=0

 

p X k=j

For |z − z0 | ≤ r we obtain

p X j=0

[0, 1/n, ..., j/n; (e1 − z0 )p ]z j

 k [0, 1/n, ..., k/n; (e1 − z0 )p ] z k−j  (z − z0 )j . j 0

  p p X X p(p − 1)...(p − k + 1) k k−j p−k   rj |Sn ((e1 − z0 ) )(z)| ≤ |z0 | r j k! j=0 k=j   p p X X k p  ≤ rp |z0 |k−j  j k j=0 k=j   p p X X k p  (1 + |z0 |)p−j  ≤ rp j k j=0 p

k=j

p p−j p =r (1 + |z0 |) 2p−j = rp (3 + 2|z0 |)p , j j=0 p X

where we used the formula (see e.g. Tomescu [189], p. 11, Exercise 1.5, 2) ) p X k p p p−j = 2 . j k j k=j

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Corollary 1.8.10. (i) For all n, k ∈ N, z0 ∈ C and r ≥ 1 we have the estimate kTn,k,z0 − (e1 − z0 )k kD(z0 ,r) ≤

5 (k − 1)k(2k − 1)[r(3 + 2|z0 |)]k . 6nr

(ii) Let r ≥ 1. Suppose that f : Tr → C is analytic in the strip Tr and that f satisfies the conditions |f k (x0 )| ≤

M , k3

S for all k ∈ N {0} and all x0 ∈ R, where M > 0 is independent of x0 and k. Denoting the weight wr (x) = e−r(3+2|x|) , x ∈ R and the weighted norm on Tr by kf kTr ,wr = sup wr (x)kf − Sn (f )kD(x,r) , x∈R

we have kf − Sn (f )kTr ,wr ≤

15M , 6nr

for all n ∈ N. Proof. (i) First we estimate |Tn,k,z0 (z) − (z − z0 )k |. By the recurrence in Lemma 1.8.9 (i) it easily follows that Tn,k,z0 (z) is a polynomial in z of degree ≤ k. We will use the following generalization of the Bernstein’s inequality due to Pommerenke [151] kPn0 kK ≤

2n2 en2 kPn kK ≤ kPn kK , 2cap(K) cap(K)

where Pn (z) is a polynomial of degree n, K is a continuum in C and cap(K) denotes the capacity of K. By Lemma 1.8.9 (ii) and (iii) combined with the above Bernstein-type inequality applied for K = D(z0 , r) = {z ∈ C; |z − z0 | ≤ r}, since cap(K) = r we obtain kTn,k,z0 − (e1 − z0 )k kD(z0 ,r)

(r + |z0 |) 2(k − 1)2 kTn,k−1,z0 − (e1 − z0 )k−1 kD(z0 ,R) · n r k − 1 +rkTn,k−1,z0 − (e1 − z0 )k−1 kD(z0 ,r) + (r + |z0 |)rk−2 n i 2(k − 1)2 r + |z0 | h kTn,k−1,z0 kD(z0 ,R) + rk−1 ≤ n r k−1 k−1 +rkTn,k−1,z0 − (e1 − z0 ) kD(z0 ,r) + (r + |z0 |)rk−2 n 2(k − 1)2 r + |z0 | k−1 r (3 + 2|z0 |)k−1 + rk−1 ≤ n r k − 1 +rkTn,k−1,z0 − (e1 − z0 )k−1 kD(z0 ,r) + (r + |z0 |)rk−2 , n ≤

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which finally leads to the inequality kTn,k,z0 − (e1 − z0 )k kD(z0 ,r) ≤ rkTn,k−1,z0 − (e1 − z0 )k−1 kD(z0 ,r)

r(3 + 2|z0 |)2(k − 1)2 k−2 [2r (3 + 2|z0 |)k−1 ] n k − 1 k−2 + r(3 + 2|z0 |) r n 5(k − 1)2 [r(3 + 2|z0 |)]k ≤ rkTn,k−1,z0 − (e1 − z0 )k−1 kD(z0 ,r) + . nr For k = 1 this inequality obviously one reduces to 0 ≤ 0. Therefore, let k ≥ 2. For k = 2 we easily obtain 5 2 kTn,2,z0 − (e1 − z0 )2 kD(z0 ,r) ≤ {1 [r(3 + 2|z0 |)]2 }. nr For k = 3 it follows 5 2 5 · 22 [r(3 + 2|z0 |)]3 {1 [r(3 + 2|z0 |)]2 } + kTn,3,z0 − (e1 − z0 )3 kD(z0 ,r) ≤ r · nr nr 5 3 2 2 [r(3 + 2|z0 |)] (1 + 2 ), ≤ nr and reasoning by recurrence finally we arrive at 5 kTn,k,z0 − (e1 − z0 )k kD(z0 ,r) ≤ [r(3 + 2|z0 |)]k (12 + 22 + ... + (k − 1)2 ) nr 5 = (k − 1)k(2k − 1)[r(3 + 2|z0 |)]k , 6nr which proves (i). (ii) For arbitrary fixed x ∈ R, since f is analytic in the strip Tr , we have the Taylor expansion ∞ X f (k) (x) f (z) = (z − x)k , k! +

k=0

valid for all z ∈ C with |z − x| ≤ r. First we prove that ∞ X f (k) (x) Sn (f )(z) = Sn ((e1 − x)k )(z), k! k=0

P f (k) (x) (z− for all z ∈ C with |z−x| ≤ r. For this purpose let us define fm (z) = m k=0 k! S x)k , if |z − x| ≤ r and fm (z) = f (z) if z ∈ (−∞, x − r] [x + r, +∞). Since by hypothesis f is bounded on R, reasoning exactly as in the Remark after the proof of Theorem 1.8.4 we easily get the desired property. Therefore, taking into account the above point (i), for all |z − x| ≤ r we can write ∞ 5 X |f (k) (x)| |f (z) − Sn (f )(z)| ≤ · (k − 1)k(2k − 1)[r(3 + 2|x|)]k 6nr k! k=2

∞ 5M X (k − 1)k(2k − 1) [r(3 + 2|x|)]k · · ≤ 6nr k3 k!

≤

k=2 ∞ X

15M · 6nr

k=2

[r(3 + 2|x|)]k , k!

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124

that is kf − Sn (f )kD(x,r) ≤

∞ 15M X [r(3 + 2|x|)]k · . 6nr k! k=2

P∞

Multiplying this inequality by wr (x) = 1/{ k=0 [r(3 + 2|x|)]k /k!} and passing to supremum with x ∈ R we easily obtain the weighted inequality in the statement. The corollary is proved. 1.9

Baskakov Operators

The aim of the present section is to extend some kinds of results in the previous sections to complex Baskakov operators. For x real and ≥ 0, the original formula of the Baskakov operator is given by (see Baskakov [35]) k ∞ X n+k−1 x f (k/n). Zn (f )(x) = (1 + x)−n 1+x k k=0

Denoting Wn (f )(x) =

∞ X n(n + 1)...(n + j − 1)

nj

j=0

[0, 1/n, ..., j/n; f ]xj , x ≥ 0,

(where for j = 0 we take n(n + 1)...(n + j − 1) = 1), according to Lupa¸s [128], Theorem 2, Zn (f )(x) = Wn (f )(x), for all x ≥ 0 (under the hypothesis on f that Zn (f )(x) is well defined). But if x is not positive then Wn (f )(x) and Zn (f )(x) do not necessarily coincide. For example, if x = −1/2 then we easily get that for all n ∈ N, Zn (f )(−1/2) represents the sum of a divergent series even for the simplest function f (x) = 1, for all x, while clearly Wn (f )(−1/2) = 1, for f (x) = 1 and all n ∈ N. Consequently, the complex versions of these two operators denoted by k ∞ X n+k−1 z f (k/n), Zn (f )(z) = (1 + z)−n 1+z k k=0

and Wn (f )(z) =

∞ X n(n + 1)...(n + j − 1) j=0

nj

[0, 1/n, ..., j/n; f ]z j ,

do not necessarily coincide for all z ∈ C. Because of this reason in this section they will be studied separately, under different hypothesis on f and z ∈ C. Remarks. 1) A sufficient condition for the existence of the operator Wn (f )(z), z ∈ C, can be expressed by the fact that f has all its derivatives bounded in [0, ∞)

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by the same constant M > 0. Indeed, in this case, for r > 0, z ∈ C and |z| ≤ r, we get ∞ X n(n + 1)...(n + k − 1)r k |Wn (f )(z)| ≤ M , for all |z| ≤ r. nk k! k=0

k a (n,r) r(n+k) = n(1+k) . Since as k → ∞ , we have ak+1 Denoting ak (n, r) = n(n+1)...(n+k−1)r nk k! k (n,r) ak+1 (n,r) n we have ak (n,r) & nr , then for a fixed n0 ∈ N and r < 20 , there exists k0 such a (n0 ,r) 2r that for all k > k0 we have k+1 ak (n0 ,r) < n0 . By the ratio test and by the inequality ak+1 (n0 ,r) ak+1 (n,r) ak (n,r) < ak (n0 ,r) for all n > n0 , we immediately get that Wn (f )(z) is welldefined and analytic for all n > n0 and |z| ≤ n20 .

2) A sufficient condition for the existence of the complex operator Zn (f )(z) can be stated as follows. Suppose for example that z ∈ C satisfies Rez ≥ 0, |z| ≤ r and that |f (x)| ≤ M for all x ∈ [0, ∞). Then 1 + z 6= 0 and for all n ∈ N it follows k ∞ X |z| n+k−1 −n |Zn (f )(z)| ≤ M |1 + z| |1 + z| k k=0 ∞ X n+k−1 k ρ . ≤ M |1 + z|−n k k=0 q p r2 where ρ = 1+r x2 + y 2 ≤ r first 2 < 1. Indeed, for z = x + iy with x ≥ 0 and we get (since h(t) = t/(1 + t) is increasing for t ≥ 0) 2 |z| x2 + y 2 x2 + y 2 r2 = ≤ ≤ . 2 2 2 2 |1 + z| 1 + 2x + (x + y ) 1 + (x + y ) 1 + r2 k P∞ P∞ ρ := k=0 ak , we get that for By the ratio test applied to the series k=0 n+k−1 k 0 any fixed n ∈ N there exists k0 such that aak+1 = ρ k+n k+1 < ρ < 1, for all k ≥ k0 . k This implies that |Zn (f )(z)| < ∞ and therefore Zn (f )(z) is analytic as function of z as above. Also, as we will see later, for z ∈ C as above the operator Z n (f )(z) exists under more general conditions on f . 3) As in the case of complex Favard-Sz´ asz-Mirakjan operators, for the complex Baskakov operators too we note that the domain of definition of the approximated S function f : DR [R, ∞) → C seems to be rather strange. But the analyticity of f P∞ on DR assures the representation f (z) = k=0 ck z k , which is essential in the proof of quantitative estimates in any Dr with 1 ≤ r < R. On the other hand, on [0, ∞) the well-known estimates in the case of real variable can be used. A more natural domain of definition for f would be a strip around the OX-axis. P k Unfortunately, in this case the representation f (z) = ∞ k=0 ck z fails and we cannot use the methods of proofs in this case. Concerning the complex operator Wn (f )(z), upper estimates in simultaneous approximation, Voronovskaja’s result with a quantitative estimate and exact estimates in simultaneous approximation for these operators are obtained. The hypothesis on f in these cases consist in exponential growth in a compact disk and in

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the boundedness (by the same constant) on [0, ∞) of the derivatives of all orders of f. At the end of this section we present similar results for the complex operator Zn (f )(z), under different hypothesis on f and z. More exactly, the disks can be replaced by semidisks and the boundedness of all derivatives of f on [0, ∞) can be replaced by the weaker condition that f is of exponential growth on [0, ∞). For future research would be of interest to find larger classes of functions for which similar results with those stated in the next Theorems 1.9.1-1.9.9 hold. The first main result of this section can be summarized by the following. Theorem 1.9.1. (Gal [90]) For n0 ∈ N and R > 0 with 3 ≤ n0 < 2R < +∞ let f : [R, +∞) ∪ DR → C be with all its derivatives bounded in [0, ∞) by the same P∞ positive constant, analytic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR , and k 1 suppose that there exist M > 0 and A ∈ ( R , 1), with the property |ck | ≤ M Ak! , for all k = 0, 1, ..., (this implies |f (z)| ≤ M eA|z| for all z ∈ DR ). 1 }. For all |z| ≤ r and n > n0 the estimate (i) Let 1 ≤ r < min{ n20 , A |Wn (f )(z) − f (z)| ≤

Cr,A,M , n

P k holds, with Cr,A,M = 6M ∞ k=2 (k + 1)(k − 1)(rA) < ∞. (ii) In the case of simultaneous approximation by complex Baskakov operators, 1 } are fixed, then for all |z| ≤ r, p ∈ N and we have : if 1 ≤ r < r1 < min{ n20 , A n > n0 the estimate |Wn(p) (f )(z) − f (p) (z)| ≤

p!r1 Cr1 ,A,M , n(r1 − r)p+1

holds, with Cr1 ,A,M is given as above. Proof. (i) Denoting ek (z) = z k , Tn,k (z) := Wn (ek )(z), clearly that Tn,k (z) is a polynomial of degree ≤ k, k = 0, 1, 2, ..., and Tn,0 (z) = 1, Tn,1 (z) = z, for all z ∈ C. Also, for all z ∈ C and n, p ∈ N the following recurrence holds : Tn,p+1 (z) =

z(1 + z) 0 Tn,p (z) + zTn,p (z). n

Indeed, simple calculation shows that this recurrence is equivalent to [0, 1/n, ..., j/n; ep+1] =

j [0, 1/n, ..., j/n; ep] + [0, 1/n, ..., (j − 1)/n; ep ], n

which is an immediate consequence of the well-known relation (see e.g. Stancu [172], p. 256, Exercise 4.9) [x0 , ..., xm ; f · g] =

m X i=0

[x0 , ..., xi ; f ] · [xi , ..., xm ; g]

applied here for m = j, f = ep , g = e1 and xi = i/n.

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From the above recurrence we obtain z(1 + z) Tn,p (z) − z p = [Tn,p−1 (z) − z p−1 ]0 n z p−1 (1 + z)(p − 1) , +z[Tn,p−1 (z) − z p−1 ] + n for all z ∈ C, p, n ∈ N. In what follows we will use the Bernstein’s inequality on Dr = {z ∈ C; |z| ≤ r}. Thus, passing to modulus for |z| ≤ r, r ≥ 1, from the above recurrence formula, we obtain (p − 1)(1 + r) kTn,p−1 − ep−1 kr + r|Tn,p−1 (z) − ep−1 (z)| |Tn,p (z) − ep (z)| ≤ n rp−1 (p − 1)(1 + r) + ≤ r|Tn,p−1 (z) − ep−1 (z)| n rp−1 (p − 1)(1 + r) (p − 1)(1 + r) [kTn,p−1 kr + rp−1 ] + + n n 2(p − 1)r [kTn,p−1 kr + rp−1 ] ≤ r|Tn,p−1 (z) − ep−1 (z)| + n 2rp (p − 1) + . n Since for any p ∈ N we have Tn,p (z) =

p X n(n + 1)...(n + k − 1) [0, 1/n, ..., k/n; ep]z k , nk k=0

by using the mean value theorem in complex analysis we get

p X n(n + 1)...(n + k − 1) p(p − 1)...(p − k + 1) p−k k kTn,p (z)kr ≤ · r r nk k! k=1 p p X X n+k−1 p p n+1 · ... · ≤ rp k! = rp k n n k

= rp

k=1 p X

k=1

k=1

p(p − 1)...(p − k + 1) ≤ r p p · p! ≤ rp (p + 1)!.

Replacing in the above inequality, for all |z| ≤ r it follows

2(p − 1)r · [rp−1 p! + rp−1 ] n 2rp (p − 1) 6(p + 1)!rp + ≤ r|Tn,p−1 (z) − ep−1 (z)| + . n n Starting from p = 2, 3, ..., and reasoning by mathematical induction with respect to p we get   p 6rp X 6rp (p + 1)!(p − 1) kTn,p − ep kr ≤ (p + 1)! ≤ . n j=2 n |Tn,p (z) − ep (z)| ≤ r|Tn,p−1 (z) − ep−1 (z)| +

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From the next Remark 1 after the present proof we have Wn (f )(z) =

∞ X

ck Wn (ek )(z) =

k=0

∞ X

ck Tn,k (z),

k=0

which from the hypothesis on ck , implies that for all |z| ≤ r and n > n0 , where 1 }, we have 1 ≤ r < min{ n20 , A |Wn (f )(z) − f (z)| ≤ ≤ = P∞

∞ X

k=2 ∞ X k=2

|ck | · |Tk,n (z) − ek (z)| M

Ak 6rk (k + 1)!(k − 1) · k! n

∞ 6M X Cr,A,M (k + 1)(k − 1)(rA)k = , n n k=2

1 where Cr,A,M = 6M k=2 (k + 1)(k − 1)(rA)k < ∞, for all 1 ≤ r < A , because the P∞ k+1 P∞ k−1 series k=2 u and therefore its derivative k=2 (k + 1)ku too, are uniformly and absolutely convergent in any compact disk included in the open unit disk. (ii) Denote by γ the circle of radius r1 > r and center 0. Since for |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas for all |z| ≤ r and n ∈ N, we obtain Z p! Wn (f )(v) − f (v) (p) (p) dv |Wn (f )(z) − f (z)| = 2π (v − z)p+1 γ

≤

Cr ,A,M 2πr1 p!r1 Cr1 ,A,M p! = 1 , p+1 n 2π (r1 − r) n (r1 − r)p+1

which proves the theorem.

P∞

Remarks. 1) The proof of the relation Wn (f )(z) = k=0 ck Wn (ek )(z) used in the proof of Theorem 1.9.1, (i) is indicated bellow. First we show that under the P k hypothesis in the statement of Theorem 1.9.1 we can write f (x) = ∞ k=0 ck x , for all x ∈ [0, ∞), where the series is uniformly convergent in any compact interval [0, b]. Indeed, by hypothesis f is infinitely differentiable on [0, ∞) with all the derivatives bounded by the same constant. Then, if we consider the Taylor series of f at 0 (here it is not important that f : [0, ∞) → C since we can write the decomposition f (x) = F (x) + iG(x) with F, G : [0, ∞) → R and instead of f we reason on F P∞ (k) and G ), that is k=0 f k!(0) xk , with x ≥ 0, then by the Lagrange form of the P∞ (k) remainder it follows that f (x) = k=0 f k!(0) xk for all x ≥ 0, where the Taylor series uniformly converge on each compact interval [0, b]. Since for z ∈ DR we have P∞ P∞ f (z) = k=0 ck z k , obviously we obtain f (x) = k=0 ck xk , for all x ≥ 0. For m ∈ N define m m X X fm (z) = cj z j if |z| ≤ r and fm (x) = cj xj if x ∈ (r, +∞). j=0

j=0

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Clearly that on [0, ∞), each fm is infinitely differentiable with all its derivatives bounded by the same constant. Also, by the linearity of Wn we get m X ck Wn (ek )(z), for all |z| ≤ r. Wn (fm )(z) = k=0

For n ∈ N, reasoning as in the proof of Theorem 1.9.1, (i) it follows that for all |z| ≤ r we have |Wn (ek )(z)| ≤ rk (k + 1)!, which implies m ∞ X X |Wn (fm )(z)| ≤ |ck | · |Wn (ek )(z)| ≤ |ck | · |Wn (ek )(z)| k=0

≤M

∞ X

k=0

(k + 1)(rA)k := Mr (f ),

k=0

for all m ∈ N, |z| ≤ r, n > n0 with 1 ≤ r < min{ n20 , A1 }. From Vitali’s result (Theorem 1.0.1) it suffices to show that for any n > n0 ≥ 3 there exists 0 < x0 < 1 ( depending on n ) such that lim Wn (fm )(x) = Wn (f )(x), for x ∈ [0, x0 ]. m→∞

x0 ∈ [0, 1), Indeed, for x0 ≥ 0 we have Wn (f )(x0 ) = Zn (f )(x0 ) and denoting ρ0 = 1+x 0 by the hypothesis on f we get |Wn (f )(x0 ) − Wn (fm )(x0 )|   j k ∞ ∞ X X n + k − 1 k x 0   = (1 + x0 )−n cj k 1 + x n 0 j=m+1 k=0 ∞ ∞ j X n+k−1 X k ≤ (1 + x0 )−n ρk0 |cj | j n k j=m+1 k=0 ∞ ∞ 1 X n+k−1 k X kj = (1 + x0 )−n m+1 ρ0 |cj | j−m−1 n n k j=m+1 k=0 ∞ ∞ 1 X n+k−1 k X ≤ (1 + x0 )−n m+1 ρ0 |cj |k j n k j=m+1 k=0 ∞ ∞ 1 X n + k − 1 k X (kA)j ≤ M (1 + x0 )−n m+1 ρ0 n j! k j=m+1 k=0   ∞ m M (1 + x0 )−n X n + k − 1 k  kA X (kA)j  = ρ0 e − nm+1 j! k j=0 k=0

M (1 + x0 ) ≤ nm+1

∞ −n X

≤

M (1 + x0 )−n nm+1

=

M (1 + x0 )−n nm+1

n + k − 1 k ekA (kA)m+1 ρ0 k (m + 1)! k=0 ∞ X n + k − 1 k 2kA ρ0 e k k=0 ∞ X n+k−1 (ρ0 e2A )k . k k=0

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Choose x0 > 0 with ρ0 e2A := ρ(x0 ) < n1 , where n > n0 ≥ 3. Because the function t h(t) = 1+t is continuous and h(0) = 0, this is always possible. We get ∞ M X n+k−1 ρ(x0 )k (1 + x0 )n |Wn (f )(x0 ) − Wn (fm )(x0 )| ≤ m+1 k n k=0 ∞ M X n+k−1 1 . ≤ m+1 k n nk k=0 1 P∞ For n > n0 ≥ 3, by the ratio test we easily get that the series k=0 n+k−1 is k nk convergent, which implies lim Wn (fm )(x0 ) = Wn (f )(x0 ).

m→∞

But the above defined function h(t) also is increasing on [0, ∞), therefore for any x0 x x ∈ [0, x0 ] we get 1+x ≤ 1+x and ρ(x) ≤ ρ(x0 ) < n1 . Applying the above estimates 0 for x instead of x0 we get lim Wn (fm )(x) = Wn (f )(x),

m→∞

which proves our assertion. 2) Simple examples of functions f satisfying the hypothesis in Theorem 1.9.1 are f (z) = e−az , or f (z) = sin(az) with 0 < a < 1. Now, let us recall that in the case of real Baskakov operators the following Voronovskaja-type formula is known. Theorem 1.9.2. (Sikkema [163]) If f : [0, +∞) → R is twice continuous differentiable on [0, +∞), then uniformly in any compact subinterval of [0, ∞) we have x(1 + x) 00 f (x). lim n[Wn (f )(x) − f (x)] = n→∞ 2 In what follows we extend this result to the complex Baskakov operators obtaining, in addition, a quantitative estimate too. We have : Theorem 1.9.3. (Gal [90]) Suppose that the hypothesis on the function f and on the constants n0 , R, M, A in the statement of Theorem 1.9.1 hold and let 1 ≤ r < min{ n20 , A1 } be fixed. For all n > n0 , |z| ≤ r the following Voronovskaja-type result ∞ X Wn (f )(z) − f (z) − z(1 + z) f 00 (z) ≤ 16M (rA)k (k − 1)(k − 2)2 , 2n n2 k=3 P k holds, where for rA < 1 we have ∞ (rA) (k − 1)(k − 2) < ∞. k=3

Proof. Denote ek (z) = z k , k = 0, 1, ..., and Tn,k (z) = Wn (ek )(z). By the proof of P∞ Theorem 1.9.1, (i), we can write Wn (f )(z) = k=0 ck Tn,k (z), which implies Wn (f )(z) − f (z) − z(1 + z) f 00 (z) 2n ∞ X z k−1 (1 + z)k(k − 1) ≤ |ck | · Tn,k (z) − ek (z) − . 2n k=2

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By the recurrence in the proof of Theorem 1.9.1, (i), satisfied by Tn,k (z), denoting Ek,n (z) = Tn,k (z) − ek (z) −

z k−1 (1+z)k(k−1) , 2n

we obtain the new recurrence

Ek,n (z) = z k−2 (1 + z)(k − 1)(k − 2) z(1 + z) 0 Ek−1,n (z) + zEk−1,n (z) + [(k − 2) + z(k − 1)], n 2n2 for all k ≥ 2, n ∈ N. Therefore, for all |z| ≤ r, k ≥ 2, n ∈ N we obtain

r(1 + r) 0 |Ek−1,n (z)| + r|Ek−1,n (z)| n (1 + r)rk−2 (k − 1)(k − 2) + [(k − 2) + r(k − 1)]. 2n2 Since Ek−1,n (z) is a polynomial of degree ≤ (k − 1), by applying the Bernstein’s inequality we get |Ek,n (z)| ≤

k−1 kEk−1,n (z)kr r rk−2 (r + 1)(k − 1)(k − 2) k−1 ≤ kTn,k−1 − ek−1 kr + r 2n k − 1 6rk−1 k!(k − 2) rk−2 (r + 1)(k − 1)(k − 2) + ≤ r n 2n

0 |Ek−1,n (z)| ≤

7rk−2 k!(k − 1)(k − 2) . n Replacing above this inequality, for all |z| ≤ r we obtain ≤

14rk k!(k − 1)(k − 2) n2 (1 + r)rk−2 (k − 1)(k − 2) [(k − 2) + r(k − 1)] + 2n2 16rk k!(k − 1)(k − 2) ≤ r|Ek−1,n (z)| + . n2 Since E0,n (z) = E1,n (z) = E2,n (z) = 0, taking k = 3, 4, ..., in the last inequality, step by step finally we arrive at |Ek,n (z)| ≤ r|Ek−1,n (z)| +

|Ek,n (z)| ≤

k 16rk k!(k − 1)(k − 2)2 16rk X j!(j − 1)(j − 2) ≤ , n2 j=3 n2

which implies ∞ X Wn (f )(z) − f (z) − z(1 + z) f 00 (z) ≤ |ck | · |Ek,n (z)| 2n k=3

≤

∞ 16M X (rA)k (k − 1)(k − 2)2 . n2 k=3

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132

Here for rA < 1 we obviously have the theorem.

P∞

k=3 (rA)

k

(k − 1)(k − 2) < ∞, which proves

Remark. Under the hypothesis in Theorem 1.9.3 we have the equivalence

Wn (f ) − f (z) − e1 (1 + e1 ) f 00 ∼ 1 .

2n n2 r

For the proof of this equivalence we follow the ideas in the proof of Theorem 1.8.2 (ii), with the only differences that we use the upper estimate for |Ek,n,1 (z)| := |Ek,n (z)| in Theorem 1.9.3 and the recurrence formula satisfied in this case by An,j (z) := Wn [(· − z)j ](z), that is An,j+1 (z) = z(1 + z)[A0n,j (z) + njAn,j−1 (z)],

(the proof of this recurrence formula uses the same Lemma 1.3 in Pop [156]). Finally, everything one reduces to prove that the differential equation P2p+1 (z)f (2p+1) (z) + P2p+2 (z)f (2p+2) (z) = 0, z ∈ Dr , has the only solutions for f a polynomial of degree ≤ 2p. Here by the above recurrence formula we get P2p+1 (z) = cp [z(1 + z)]p(1 + 2z) and P2p+2 (z) = dp [z(1 + z)]p+1 , with cp , dp > 0. The proof is easy by writing y(z) in the form y(z) = P∞ k k=0 ak z . In what follows we obtain the exact degree in approximation by Wn (f )(z). The first main result is the following. Theorem 1.9.4. (Gal [90]) Suppose that the hypothesis on the function f and on the constants n0 , R, M, A in the statement of Theorem 1.9.1 hold and let 1 ≤ r < min{ n20 , A1 }. If f is not a polynomial of degree ≤ 1, then for all n > n0 the lower estimate Cr (f ) kWn (f ) − f kr ≥ n holds, where the constant Cr (f ) depends only on f (that is on A, M ) and r. Proof. For all |z| ≤ r and n > n0 we get

=

1 n

Wn (f )(z) − f (z) 1 2 z(1 + z) 00 z(1 + z) 00 f (z) + f (z) . n Wn (f )(z) − f (z) − 2 n 2n

We apply to this identity the following simple property :

kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . We get kWn (f ) − f kr

1 1 2 e1 (1 + e1 ) 00 e1 (1 + e1 ) 00

. n Wn (f ) − f − ≥ f − f

n 2 n 2n r r

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1 ) 00 Since f is not a polynomial of degree ≤ 1 in DR , we get e1 (1+e f

> 0. Indeed, 2 r

supposing the contrary it follows that z(1+z) f 00 (z) = 0 for all z ∈ Dr , which clearly 2 implies f 00 (z) = 0 for all z ∈ Dr \ {0, −1}. Since f is analytic, from the identity theorem on analytic functions this implies that f 00 (z) = 0, for all z ∈ DR , that is f is a polynomial of degree ≤ 1, in contradiction with the hypothesis. By Theorem 1.9.3 we have

∞ X

e1 (1 + e1 ) 00

f ≤ 16M (rA)k (k − 1)(k − 2)2 . n2 W (f ) − f −

n 2n r

k=3

Therefore, there exists n1 > n0 depending only on f and r, such that for any n ≥ n1 we have

e1 (1 + e1 ) 00

− 1 n2 Wn (f ) − f − e1 (1 + e1 ) f 00 f

2 n 2n r r

e (1 + e ) 1 1 1 00 f ≥

, 2 2 r which implies

1 1 e1 (1 + e1 ) 00

kWn (f ) − f kr ≥ · f

, ∀n ≥ n1 . n 2 2 r M

(f )

with Mr,n (f ) = n · kWn (f ) − For n ∈ {n0 + 1, ..., n1 } we get kWn (f ) − f kr ≥ r,n n Cr (f ) f kr > 0, which finally implies kWn (f for all n > n0 , with Cr (f ) = n

) − f kr ≥ o 1 e1 (1+e1 ) 00 min {Mr,n0 +1 (f ), ..., Mr,n1 −1 (f ), 2 f . This proves the theorem. 2 r From Theorem 1.9.4 and Theorem 1.9.1, (i), we get the following consequence.

Corollary 1.9.5. (Gal [90]) Suppose that the hypothesis on the function f and on the constants n0 , R, M, A in the statement of Theorem 1.9.1 hold and let 1 ≤ r < min{ n20 , A1 } be arbitrary fixed. If f is not a polynomial of degree ≤ 1, then for all n > n0 the estimate 1 kWn (f ) − f kr ∼ , n holds, where the constants in the equivalence depend only on f (i.e. on A, M ) and r. Regarding the simultaneous approximation we have the following result.

Theorem 1.9.6. (Gal [90]) Suppose that the hypothesis on the function f and on the constants n0 , R, M, A in the statement of Theorem 1.9.1 hold and let 1 ≤ r < 1 r1 < min{ n20 , A }, p ∈ N. If f is not a polynomial of degree ≤ max{1, p − 1}, then for all n > n0 the estimate 1 kWn(p) (f ) − f (p) kr ∼ , n holds, where the constants in the equivalence depend only on f (that is on A, M ), r, r1 and p.

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Proof. By Theorem 1.9.1, (ii) we have the upper estimate for kWn (f ) − f (p) kr , (p) therefore it remains to prove the lower estimate for kWn (f ) − f (p) kr . Denoting by n0 1 Γ the circle of radius r1 and center 0 (where min{ 2 , A } > r1 > r ≥ 1), we have the inequality |v − z| ≥ r1 − r for all |z| ≤ r and v ∈ Γ. By the Cauchy’s formula we obtain Z p! Wn (f )(v) − f (v) (p) (p) dv. Wn (f )(z) − f (z) = 2πi Γ (v − z)p+1

As in the proof of Theorem 1.9.1, (ii), for all v ∈ Γ and n > n0 we get

Wn (f )(v) − f (v) v(1 + v) 00 1 2 1 v(1 + v) 00 n Wn (f )(v) − f (v) − = f (v) + f (v) . n 2 n 2n

Replaced in the above Cauchy’s formula implies Z 1 v(1 + v)f 00 (v) p! (p) (p) Wn (f )(z) − f (z) = dv n 2πi Γ 2(v − z)p+1  Z n2 Wn (f )(v) − f (v) − v(1+v) f 00 (v)  2n 1 p! dv + ·  n 2πi Γ (v − z)p+1 ( (p) z(1 + z) 00 1 = f (z) n 2  Z n2 Wn (f )(v) − f (v) − v(1+v) f 00 (v)  2n 1 p! dv + ·  n 2πi Γ (v − z)p+1

and passing to the norm k · kr , for all n > n0 we obtain ( (p)

1

e1 (1 + e1 ) 00

(p) (p) kWn (f ) − f kr ≥ f

n 2

r



 Z n2 Wn (f )(v) − f (v) − v(1+v) f 00 (v)

2n p! 1

− dv

, p+1 n 2π Γ (v − z)

r

where by Theorem 1.9.3, for all n > n0 we have

Z n2 Wn (f )(v) − f (v) − v(1+v) f 00 (v)

p! 2n

dv

2π p+1 (v − z) Γ

r

p! 2πr1 n2 e (1 + e ) 1 1 00

Wn (f ) − f − ≤ · f

2π (r1 − r)p+1 2n r1 ≤ 16M

∞ X k=3

(r1 A)k (k − 1)(k − 2)2 ·

p!r1 . (r1 − r)p+1

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h

e1 (1+e1 ) 00 i(p)

> 0. Indeed, supposing the But by hypothesis on f , we have f 2

r

contrary it follows that z(1+z) f 00 (z) is a polynomial of degree ≤ p − 1. Now, if 2 p = 1 and p = 2 then the analyticity of f implies that f is a polynomial of degree ≤ 1 = max{1, p − 1}, in contradiction with the hypothesis. If p > 2 then the analyticity of f implies that f is a polynomial of degree ≤ p − 1 = max{1, p − 1}, which is again in contradiction with the hypothesis. Finally, reasoning exactly as in the proof of Theorem 1.9.4 we immediately obtain the required conclusion. At the end of this section we deal with the approximation properties of Zn (f )(z) given by k ∞ X n+k−1 z −n f (k/n), Zn (f )(z) = (1 + z) 1+z k k=0

mentioned and discussed at the beginning of this section. First we present some sufficient conditions on f for the analyticity of Zn (f )(z).

Theorem 1.9.7. (Gal [90]) Suppose that f : [0, ∞) → C is of exponential growth on [0, ∞), that is there exists M > 0 and A ≥ 0 such that |f (x)| ≤ M eAx for all x ∈ [0, ∞). (i) For any n ∈ N there exists a 0 < ρ < 1 (depending on n) such that Z n (f )(z) is well defined and analytic in the compact diskh Dρ/2 . i (ii) Let r ≥ 1 be fixed and denote n0 =

Zn (f )(z) is analytic in the compact semi-disk

2A ln(1+1/r 2 ) + Dr = {z ∈

+ 2. For all n ≥ n0 ,

C; |z| ≤ r, Rez ≥ 0}.

Proof. (i) It is clear that if 0 < ρ < 1 then for z ∈ Dρ/2 it follows 1 + z 6= 0 and therefore (1 + z)−n is analytic. We get k ∞ X n+k−1 |z| −n eak/n . |Zn (f )(z)| ≤ M |1 + z| k |1 + z| k=0 p 2 2 For z = x + iy with |x| ≤ x + y ≤ ρ/2 we obtain (since h(t) = t/(1 − ρ + t) is increasing for t ≥ 0) 2 |z| x2 + y 2 x2 + y 2 = = |1 + z| 1 + 2x + (x2 + y 2 ) 1 − ρ + (ρ + 2x) + (x2 + y 2 ) 2 2 (ρ/2)2 x +y ≤ . ≤ 1 − ρ + (x2 + y 2 ) 1 − ρ + (ρ/2)2 q (ρ/2)2 Denoting η := 1−ρ+(ρ/2) 2 < 1 we obtain ∞ ∞ X X n + k − 1 k ak/n |Zn (f )(z)| ≤ M |1 + z|−n η e := M |1 + z|−n ak . k k=0

k=0

ak+1 ak

a/n

k+n k+1 . a/n

We apply the ratio test to the last series. We have = ηe · ρ → 0 we obtain η → 0, for fixed n choose ρ so small that β := ηe

Since for n < 1. By

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a

k+n k+1

≤ n for all k ≥ 0 obviously that ak+1 ≤ β < 1, for all k ≥ 0, which proves that k P∞ the series k=0 ak is convergent and therefore Zn (f )(z) is analytic in Dρ/2 for ρ sufficiently small (depending on n) chosen as above. + (ii) Clearly that for z ∈ Dr we get 1 + z 6= 0 and therefore (1 + z)−n is analytic. r < 1, for all Reasoning exactly as at the above point (i) and denoting η = √1+r 2 +

z ∈ Dr we get |Zn (f )(z)| ≤ M |1 + z|−n = M |1 + z| Denote C =

2A ln(1+1/r 2 ) ,

−n

∞ X n+k−1

k=0 ∞ X

k=0

k

η k eAk/n

n+k−1 [ηeA/n ]k . k

n1 = [C] + 1 ≥ C. We get that for n ≥ n0 > n1 we have

ηeA/n ≤ ηeA/n0 < ηeA/n1 ≤ ηeA/C = 1. Denoting γ = ηeA/n0 < 1, for all n ≥ n0 we obtain ∞ X n+k−1 [ηeA/n ]k |Zn (f )(z)| ≤ M |1 + z|−n k k=0 ∞ ∞ X X n+k−1 k ≤ M |1 + z|−n γ := M |1 + z|−n ak . k k=0

k=0

= γ · k+n Let n ≥ n0 . By the ratio test applied to the last series we have aak+1 k+1 . k ak+1 k+n Since there exists a k0 with k+1 < 2 − γ for all k ≥ k0 , it follows that ak = < γ(2 − γ) < 1, for all k ≥ k0 . This implies the convergence of the series γ · k+n P∞k+1 a and therefore we obtain the analyticity of Zn (f )(z). k k=0 In what follows we show that Zn (f )(z) can be written under the form of divided differences. Corollary 1.9.8. (Gal [90]) (i) Suppose that f satisfies the hypothesis in Theorem 1.9.7. Then for any n ∈ N there exists a sufficiently small 0 < ρ < 1 (depending on n) such that for all z ∈ Dρ/2 we have ∞ X n(n + 1)...(n + j − 1) Zn (f )(z) = [0, 1/n, ..., j/n; f ]z j . j n j=0

(ii) Let p ∈ N ∪ {0} be fixed. For any n ∈ N there exists a sufficiently small 0 < ρ < 1 such that for all z ∈ Dρ/2 we have p X n(n + 1)...(n + k − 1) Zn (ep )(z) = [0, 1/n, ..., k/n; ep]z k . nk k=0 i h 2 + 2. For (iii) Let p ∈ N ∪ {0} and r ≥ 1 be fixed and denote n0 = ln(1+1/r 2) +

all n ≥ n0 and z ∈ Dr we have p X n(n + 1)...(n + k − 1) Zn (ep )(z) = [0, 1/n, ..., k/n; ep]z k . nk k=0

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Proof. (i) For fixed n let us define gk,n (z) = (1 + z)−n−k , k = 0, 1, 2, ...,. Since gk,n is analytic in Dρ/2 , we get gk,n (z) =

(p) ∞ X gk,n (0)

p!

p=0

zp =

∞ X

(−1)p

p=0

(n + k − 1 + p)! p z , (n + k − 1)!p!

which replaced in the expression of Zn (f )(z) implies ∞ ∞ X X (n + k − 1 + p)! p n+k−1 k z f (k/n) (−1)p Zn (f )(z) = z k (n + k − 1)!p! p=0 k=0

=

∞ X ∞ X

z

k+p

k=0 p=0

:=

∞ X ∞ X

f (k/n)(−1)

z k+p Ak,p =

k=0 p=0

Bj =

Aν,j−ν =

ν=0

z j Bj .

j=0

This implies the formulas j X

∞ X

+ k − 1 + p)! (n − 1)!k!p!

p (n

j X

f (ν/n)(−1)j−ν

ν=0

(n + j − 1)! (n − 1)!ν!(j − ν)!

j (n + j − 1)! X nj j−ν = f (ν/n)(−1) (n − 1)!nj ν=0 ν!(j − ν)!

(n + j − 1)! [0, 1/n, ..., j/n; f ], (n − 1)!nj which proves the corollary. (ii) It is immediate from the above point (i) since each ep satisfies the hypothesis in Theorem 1.9.7 with A = 1. (iii) By Theorem 1.9.7 (ii) we get that for all n ≥ n0 , Zn (ep )(z) is analytic in + Dr . On the other hand, by the above point (ii) , Zn (ep )(z) is a polynomial of degree ≤ p in a small disk Dρ/2 , that is its derivative of order p + 1 is zero in Dρ/2 . By the identity theorem on analytic functions it is clear that the derivative of order + p + 1 of Zn (ep )(z) also is zero in Dr , that is Zn (ep )(z) is a polynomial of degree + + ≤ p in Dr . Since Zn (f )(z) analytically extends the values in Dρ/2 to Dr , clearly that Zn (ep )(z) must be of the same form =

Zn (ep )(z) =

p X n(n + 1)...(n + k − 1) k=0

+ Dr .

nk

[0, 1/n, ..., k/n; ep]z k ,

for all z ∈ We are in position to prove the following result.

Theorem 1.9.9. (Gal [90]) For R > 1 suppose that f : [R, +∞) ∪ DR → C is P∞ k analytic in DR , that is f (z) = k=0 ck z , for all z ∈ DR , and that there exist

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M > 0 and A ∈ ( R1 , 1), with the property |ck | ≤ M Ak! , for all k = 0, 1, ..., (this implies |f (z)| ≤ M eA|z| for all z ∈ D+ R ). In addition, let us suppose that f is of exponential growth on [0, ∞) (for simplicity suppose that the exponent in the exponential growth also is A). Then the upper estimates in Theorems 1.9.1 (i) and 1.9.3 and the exact estimate in Corollary 1.9.5 hold forh Zn (f )(z) iin the compact +

semi-disk Dr , for any 1 ≤ r < R and n ≥ n0 with n0 =

2A ln(1+1/r 2 )

+ 2.

Proof. By Corollary 1.9.8 (iii), the recurrence for Zn (ep )(z) is similar to that in the + proof of Theorem 1.9.1, taking place for z in the compact semi-disk Dr = {z; |z| ≤ r, Rez ≥ 0}. The reasonings in the proofs of the corresponding results mentioned in the statement are similar. What remains to prove is the equality Zn (f )(z) =

∞ X

ck Zn (ek )(z),

k=0 +

for all z ∈ Dr and n ≥ n0 . For this purpose, for any m ∈ N let us define fm (z) =

m X j=0

+

cj z j if z ∈ Dr and fm (x) = f (x) if x ∈ (r, +∞).

From the hypothesis on the coefficients of f it is clear that for any m ∈ N we have |fm (x)| ≤

m X j=0

j

|cj |x ≤ M

m X (Ax)j j=0

j!

≤ M eAx , for all x ∈ [0, r],

which from the hypothesis on f on [0, ∞), immediately implies that for any m ∈ N we have |fm (x)| ≤ M eAx , for all x ∈ [0, ∞). By Theorem 1.9.7 (ii) we get that Zn (fm )(z), are well-defined and analytic in the + compact semi-disk Dr , for all m ∈ N and n ≥ n0 . Denoting +

fm,k (z) = ck ek (z) if z ∈ Dr and fm,k (x) =

f (x) if x ∈ (r, ∞), m+1

each fm,k is of exponential growth on [0, ∞) (with the same exponent A) and Pm fm (z) = k=0 fm,k (z). Since from the linearity of Zn we have Zn (fm )(z) =

m X

k=0

+

ck Zn (ek )(z), for all z ∈ Dr ,

it suffices to prove that limm→∞ Zn (fm )(z) = Zn (f )(z) for any fixed n ≥ n0 and + z ∈ Dr . But this is immediate from limm→∞ kfm −f kr = 0, from kfm −f kB[0,+∞) ≤

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kfm − f kr and from the inequality |Zn (fm )(z) − Zn (f )(z)| j ∞ X n+j−1 |z| −n kfm − f kB[0,∞) ≤ |1 + z| |1 + z| j j=0 j ∞ X n+j−1 r √ ≤ |1 + z|−n kfm − f kB[0,∞) , j 1 + r2 j=0 +

valid for all z ∈ Dr . Here, by using the ratio test it follows that the series P∞ n+j−1 r j √ is convergent by using the ratio test. Also, k · kB[0,+∞) j=0 j 1+r 2 denotes the uniform norm on C[0, +∞)-which represents the space of all real-valued bounded functions on [0, +∞). 1.10

Bal´ azs-Szabados Operators

The goal of the present section is to obtain similar type of results for the rational complex Bal´ azs-Szabados operators given by n X 1 n Rn (f )(z) = f (j/b ) (an z)j , n n (1 + an z) j=0 j where an = nβ−1 , bn = nβ , 0 < β ≤ 2/3, n ∈ N and z ∈ C, z 6= − a1n . The above complex form is obtained simply replacing x by z in the real form of rational operators introduced and studied in Bal´ azs [33] and Bal´ azs-Szabados [34]. Further studies on these operators in the case of real variable can be found in e.g. the paper Abel-Della Vecchia [1]. Remarks. 1) The complex operators Rn (f )(z) are well-defined and analytic for all n ≥ n0 and |z| ≤ r < n01−β . Indeed, in this case we easily obtain that z 6= − a1n , for all |z| ≤ r < n01−β and n ≥ n0 , which implies that (1+a1n z)n is analytic. 2) There exists a close connection between Rn (f )(z) and the classical complex Pn n j j Bernstein polynomials given by Bn (f )(z) = j=0 f (j/n) j z (1 − z) . Indeed, denoting ek (z) = z k , we easily get an z k(1−β) Rn (ek )(z) = n Bn (ek ) , 1 + an z valid for all n ≥ n0 , k ∈ N and |z| ≤ r < n01−β . This connection will be essential in our reasonings. First we will find some classes of analytic functions for which the uniform convergence of Rn (f )(z) to f (z) holds in some compact disks. As in the case of complex Favard-Sz´ asz-Mirakjan and complex Baskakov operators, for the complex Bal´ azsSzabados operators too let us note that in our results, the domain of definition of

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S the approximated function f : DR [R, ∞) → C seem to be rather strange. HowP∞ ever, the analyticity of f on R on DR assures the representation f (z) = k=0 ck z k , which is essential in the proof of quantitative estimates in any Dr with 1 ≤ r < R (while on [0, ∞) the well known estimates in the case of real variable can be used). Probably a more natural domain of definition for f would be a strip around the P∞ k OX-axis, but in this case the representation f (z) = k=0 ck z fails, fact which produces the failure of the proofs in this case. Because of the Remark 2 from the beginning of this section first we need to deal with the estimate of |Bn (ek )(z)| for |z| ≤ r with 0 < r < 1. Note that in the case when r ≥ 1 the estimate of |Bn (ek )(z)| is completely different and it was found in the proof of Theorem 1.1.2 (i). Lemma 1.10.1. Denote πk,n (z) = Bn (ek )(z), kf kr = sup|z|≤r {|f (z)|} and consider 0 < r < 1. S (i) For all |z| ≤ r, k ∈ N {0} and n ≥ 1 + r1 we have kπk,n kr ≤ k!(1 + r)r k .

(ii) For all |z| ≤ r < 1, k = 0, 1, 2, ... and n ∈ N we have kπk,n kr ≤ rk +

(1+r)k(k−1) . 2n

Proof. (i) We consider the following recurrence formula for Bernstein polynomials (see the proof of Theorem 1.1.2 (i)) πk+1,n (z) =

z(1 − z) 0 πk,n (z) + zπk,n (z), n

z ∈ C, k = 0, 1, 2, ...,, n ∈ N. We will use the mathematical induction. For k = 0 the inequality in statement is obvious. Suppose that it is true for k. By the above recurrence, by the Bernstein’s ≤ kr we inequality (since πk,n (z) is a polynomial of degree ≤ k) and since k(1+r) n obtain r(1 + r) k · · kπk,n kr + rkπk,n kr n r 1+r k ≤ k!(1 + r)r k · +r n

kπk+1,n kr ≤

≤ k!(1 + r)r k [(k + 1)r] = (k + 1)!(1 + r)r k+1 . (ii) Applying the Bernstein’s inequality to the above recurrence formula , for all |z| ≤ r we get |πk+1,n (z)| ≤

r(1 + r) k · kπk,n kr + r|πk,n (z)| ≤ r|πk,n (z)| n r (1 + r) (1 + r)k + · kkπk,n kr ≤ r|πk,n (z)| + , n n

since by the proof of Theorem 1.1.2 (i) we have kπk,n k1 ≤ 1, for all k, n ∈ N.

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Now, taking k = 1, 2, 3, ... in the above inequality, by recurrence we easily obtain for all |z| ≤ r (1 + r)k(k − 1) 1+r [1 + 2 + ... + (k − 1)] = r k + , n 2n which proves (ii) and the lemma. |πk,n (z)| ≤ rk +

Also we need the following. n1−β

Lemma 1.10.2. Let n0 ≥ 2, 0 < β ≤ 2/3 and 21 < r < 02 . If we denote 1 }, |z| ≤ r and k = 0, 1, 2, ...., we rk,n (z) = Rn (ek )(z) then for all n ≥ max{n0 , r1/β have |rk,n (z)| ≤ (k!) · (2r)k . Proof.

By Remark 2 from the beginning of this section it follows an z Rn (ek )(z) = nk(1−β) Bn (ek ) . 1 + an z n1−β

But for all n ≥ n0 ≥ 2 and |z| ≤ r < 02 it is easy to see that an z an r 1 + an z ≤ 1 − an r < 1.

an r Therefore, applying Lemma 1.10.1 (i) with 1−a instead of r, for all |z| ≤ nr S 1−an r 1 k ∈ N {0} and n ≥ 1 + an r = an r we get an z k(1−β) |rk,n (z)| = n Bn (ek ) 1 + an z k an r an r ≤ nk(1−β) k! 1 + . 1 − an r 1 − an r 1 an r

an r 1−an r ,

1 . Also, the r 1/β 1−β 1−β n n an r 1 conditions n ≥ {n0 , r1/β }, |z| ≤ r < 02 where 21 < r implies that 02 ≤ 1−a . nr 1−β n0 1−β an r Indeed, simple calculation shows that 2 ≤ 1−an r is equivalent to n0 ≤ 1−β 1−β n0 1−β 1 1 β−1 < 2 ≤ r, which implies n ≤ 1 ≤ r(2 + n0 ), rn (2 + n0 ). But 2+n1−β 0 1−β 1−β that is exactly n0 ≤ rnβ−1 (2 + n0 ). n1−β 1 }, |z| ≤ r < 02 and k = 0, 1, 2, ...., we In conclusion, for all n ≥ max{n0 , r1/β

But it is easy to see that the condition n ≥

is equivalent with n ≥

obtain

|rk,n (z)| ≤ nk(1−β) k!

nk(β−1) rk ≤ 2(k!)(2r)k , (1 − an r)k+1

if we can prove that 1−a1 n r < 2 for n ≥ n0 ≥ 2 and r < implies r < 2a1n for all n ≥ n0 , which is equivalent to proved.

1−β n1−β n1−β 0 0 < n2 2 . But r < 2 1 1−an r < 2. The lemma is

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Now we are in position to prove the following convergence result. n1−β

Theorem 1.10.3. Let n0 ≥ 2, 0 < β ≤ 2/3 and 21 < r < R ≤ 02 . Suppose S that f : DR [R, +∞) → C is uniformly continuous and bounded on [0, +∞), is P k analytic in DR , that is f (z) = ∞ j=0 ck z for all z ∈ DR and there exist M > 0, k

1 with |ck | ≤ M Ak! for all k = 0, 1, 2, ...., (which implies |f (z)| ≤ M eA|z| , 0 < A < 2r for all z ∈ DR ). Then the sequence (Rn (f )(z))n≥n0 is uniformly convergent to f in Dr . P∞ Proof. First we prove that Rn (f )(z) = j=0 Rn (ej )(z) for all z ∈ Dr , where ej (z) = z j . In this sense, for any m ∈ N define

fm (z) =

m X j=0

cj z j if |z| ≤ r and fm (x) = f (x) if x ∈ (r, +∞).

From the hypothesis on f it is clear that each fm is bounded on [0, +∞), which implies that n X 1 j n (a |z|) M (fm ) < ∞, |Rn (fm )(z)| ≤ n |1 + an z|n j=0 j that is all Rn (fm )(z) with n ≥ n0 , r < Denoting

n1−β 0 2 ,

m ∈ N are well-defined for z ∈ Dr .

f (x) if x ∈ (r, ∞), m+1 Pm it is clear that each fm,k is bounded on [0, ∞) and that fm (z) = k=0 fm,k (z). Since from the linearity of Sn we have fm,k (z) = ck ek (z) if |z| ≤ r and fm,k (x) =

Rn (fm )(z) =

m X k=0

ck Rn (ek )(z), for all |z| ≤ r,

it suffices to prove that limm→∞ Rn (fm )(z) = Rn (f )(z) for any fixed n ∈ N, n ≥ n0 and |z| ≤ r. But this is immediate from limm→∞ kfm − f kr = 0, from kfm − f kB[0,+∞) ≤ kfm − f kr and from the inequality |Rn (fm )(z) − Rn (f )(z)| ≤ Mr,n kfm − f kB[0,∞) ≤ Mr,n kfm − f kr , valid for all |z| ≤ r. Here k · kB[0,+∞) denotes the uniform norm on C[0, +∞)-the space of all real-valued bounded functions on [0, +∞). Therefore for all |z| ≤ r, n ≥ n0 we obtain |Rn (f )(z)| ≤

∞ X k=0

|ck | · |Rn (ek )(z)| ≤ 2

∞ X k=0

|ck | · k!(2r)k ≤ 2M

∞ X

k=0

(2rA)k < ∞.

Now, since we have limn→∞ Rn (f )(x) = f (x) for all x ∈ [0, r) (by Theorem 1 in Bal´ azs-Szabados [34]), by the classical Vitali’s theorem it follows that Rn (f )(z) uniformly converges to f (z) in Dr .

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As in the case of complex Bernstein polynomials, for an upper estimate in approximation by Rn (f )(z) we would need a recurrence formula. For this purpose, formally differentiating Rn (f )(z) we obtain n X nan 1 n 0 f (j/bn ) jan (an z)j−1 − Rn (f )(z) = Rn (f )(z). j (1 + an z)k j=1 1 + an z

Taking here f (z) = ek (z) by simple calculation we obtain Rn0 (ek )(z) = bn nan z Rn (ek+1 )(z) − 1+an z Rn (ek )(z), or denoting rk,n (z) = Rn (ek )(z) finally we easily arrive to the recurrence formula z z 0 (z) + rk,n (z), rk+1,n (z) = rk,n bn 1 + an z n1−β

valid for all k = 0, 1, 2, ..., |z| ≤ r < 02 and n ≥ n0 . Note that from this recurrence formula we easily get that rk,n (z) is a rational Pk,n (z) function of the form rk,n (z) = (1+a k with Pk,n (z) a polynomial of degree ≤ k in n z) z. Also, simple calculation leads us to the recurrence z z [rk,n (z) − z k ]0 + [rk,n (z) − z k ] rk+1,n (z) − z k+1 = bn 1 + an z an z 2 k . − +z k bn 1 + an z In order to use the kinds of reasonings in the case of complex Bernstein polynomials in Section 1.1, we need a Bernstein type inequality for rational functions. The following direct consequence of the Bernstein’s inequality in closed unit disk for rational functions in Borwein-Erd´elyi [47], Corollary 6, will be useful. Corollary 1.10.4. Let f (z) =

pk (z) , Πk j=1 (z−aj )

where pk (z) is a polynomial of degree

≤ k and we suppose that |aj | ≥ R > 1, for all j = 1, ..., k. If 1 ≤ r < R then for all |z| ≤ r we have |f 0 (z)| ≤ Proof.

Denote g(u) = f (ru), |u| ≤ 1. We get g(u) =

Since obtain

|aj | r

R+r k · kf kr . R−r r

≥

R r

pk (ru) pk (ru)/rk = , |u| ≤ 1. Πkj=1 (ru − aj ) Πkj=1 (u − aj /r)

> 1, we can apply Corollary 6 in Borwein-Erd´elyi [47] so that we

|g 0 (u)| ≤

R/r + 1 R+r R+r · kkgk1 = · kkgk1 = · kkf kr . R/r − 1 R−r R−r

But g 0 (u) = rf 0 (ru), which proves the corollary.

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Now we are in position to prove the following upper estimate in approximation by Rn (f )(z). n1−β

Theorem 1.10.5. Let n0 ≥ 2, 0 < β ≤ 2/3 and 21 < r < R ≤ 02 . Suppose S that f : DR [R, +∞) → C is uniformly continuous and bounded on [0, +∞), is P∞ analytic in DR , that is f (z) = j=0 ck z k for all z ∈ DR and there exist M > 0, k

1 with |ck | ≤ M Ak! for all k = 0, 1, 2, ...., (which implies |f (z)| ≤ M eA|z| , 0 < A < 2r 1 for all z ∈ DR ). Then for all n ≥ max{n0 , r1/β } and |z| ≤ r we have the upper estimate ∞ 1 1 8M X k(2rA)k . |Rn (f )(z) − f (z)| ≤ 1−β · 2M re2rA + β · n n r k=0

S n1−β Proof. Let |z| ≤ r < R ≤ 02 , n ≥ n0 and k ∈ N {0}. From the last proved recurrence, Corollary 1.10.4 and Lemma 1.10.2 we obtain |rk+1,n (z) − z k+1 |

r k an r 2 r 0 k−1 k k [|r (z)| + kr ]+ |rk,n (z) − z | + r + ≤ bn k,n 1 − an r bn 1 − an r " # krk r k n01−β + r r krk ≤ · 1−β · krk,n kr + + |rk,n (z) − z k | + b n r n0 − r bn 1 − an r bn + ≤

an rk+2 1 − an r

r 2k(k!) n1−β + r 2krk an rk+2 + |rk,n (z) − z k | + · (2r)k · 01−β + . 1 − an r bn bn 1 − an r n0 − r

But from the end of the proof of Lemma 1.10.2 we have condition r <

n1−β 0 2

is equivalent to

n1−β +r 0 n1−β −r 0

1 1−an r

< 2 while the

< 3, which immediately implies

|rk+1,n (z) − z k+1 | ≤ 2r|rk,n (z) − z k | +

6k(k!) 2krk (2r)k + + 2rk+2 an , bn bn

that is |rk+1,n (z) − z k+1 | ≤ 2r|rk,n (z) − z k | +

8k(k!) (2r)k + 2rk+2 an , bn

for all k = 0, 1, 2, .... Taking step by step k = 0, 1, 2, ... we easily obtain |rk,n (z) − z k | ≤ rk+1 an

k X j=1

2j +

k−1 8(2r)k−1 X · j(j!) ≤ 2rk+1 an (2k−1 − 1) bn j=1

8(2r)k−1 8(2r)k−1 + k(k!) ≤ (2ran )(2r)k + k(k!), bn bn Pk−1 taking into account that j=1 j(j!) ≤ k(k!).

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Taking into account the hypothesis in the coefficients of f we therefore obtain X |ck | · |rk,n (z) − z k | |Rn (f )(z) − f (z)| ≤ k=0

≤ 2M ran

where

P∞

k=0

∞ X (2rA)k k=0

k!

+

∞ 1 8M X k(2rA)k · bn r k=0

∞ 2M r 1 8M X k(2rA)k , = 1−β · e2rA + · n bn r k=0

k

k(2rA) < ∞ for 2rA < 1. The theorem is proved.

Remark. The upper estimate in Theorem 1.10.5 can obviously be written in the form 1 |Rn (f )(z) − f (z)| = O an + . bn The Voronovskaja-type result for Rn (f )(x), x ∈ [0, ∞) was obtained by Bal´ azs [33]. In the case of complex variable, the Voronovskaja-type result for Rn (f )(z) can be stated as follows. n1−β

Theorem 1.10.6. Let n0 ≥ 2, 0 < β ≤ 2/3 and 21 < r < R ≤ 02 . Suppose S that f : DR [R, +∞) → C is uniformly continuous and bounded on [0, +∞), is P k analytic in DR , that is f (z) = ∞ j=0 ck z for all z ∈ DR and there exist M > 0, k

1 0 < A < 2r with |ck | ≤ M Ak! for all k = 0, 1, 2, ...., (which implies |f (z)| ≤ M eA|z| , 1 for all z ∈ DR ). Then for all n ≥ max{n0 , r1/β } and |z| ≤ r we have 2 4 2 Rn (f )(z) − f (z) + an z f 0 (z) − an bn z + z f 00 (z) 2 1 + an z 2bn (1 + an z)

1 ≤ Cr (f ) an + bn

2

.

Proof. We clearly have ∞ 2 2 4 X Rn (f )(z) − f (z) + an z f 0 (z) − an bn z + z f 00 (z) ≤ |ck | · |Ek,n (z)|, 1 + an z 2bn (1 + an z)2 k=0

where

Ek,n (z) = rk,n (z) − ek (z) +

a2n bn z 4 + z an kz k+1 − k(k − 1)z k−2 . 1 + an z 2bn (1 + an z)2

Here we easily obtain E0,n (z) = E1,n (z) = 0 for all z. Now, from the recurrence formula for rk,n (z) obtained after the proof of Theorem 1.10.3, by long but simple calculation we obtain the recurrence for Ek,n (z) given by z 0 z Ek+1,n (z) = Ek,n (z) + Ek,n (z) + Ak,n (z), bn 1 + an z

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where Ak,n (z) =

ka2n z k+3 an 6an z + 2a2n z 2 + 5 − k + kz k+1 · · . 2 (1 + an z) bn 2(1 + an z)3

Under the hypothesis in the statement of Theorem 1.10.6, by the proof of Lemma 1.10.2 we have 1 1 1 + an z ≤ 1 − an r < 2, which immediately implies

an |Ak,n (z)| ≤ Cr k(k + 1)(k + 2)(max{r k+1 , rk+3 }) a2n + bn 2 1 . ≤ Cr k(k + 1)(k + 2)(2r)k an + bn

Therefore, from the recurrence we obtain r 0 (z)| + 2r|Ek−1,n (z)| + |Ak,n (z)|. |Ek,n (z)| ≤ |Ek−1,n bn n1−β

0 First we will estimate the quantity |Ek−1,n (z)|. Let r < r1 < R < 02 be with 1 r1 < 2A and denote by Γ the circle of center 0 and radius r1 . By the Cauchy’s theorem and taking into account the estimate for krk−1,n − ek−1 kr1 in the proof of Theorem 1.10.5, for all |z| ≤ r we obtain Z r 1 Ek−1,n (u)du 0 |Ek−1,n (z)| ≤ ≤ r1 − r kEk−1,n kr1 2π Γ u−z r 1 krk−1,n − ek−1 kr1 + Cr1 k(k − 1)(2r1 )k an + ≤ r1 − r bn 1 . ≤ Cr,r1 k(k − 1)k!(2r1 )k an + bn 2 By the inequality b1n an + b1n ≤ an + b1n , this immediately implies r 1 k |Ek,n (z)| ≤ 2r|Ek−1,n (z)| + Cr,r1 k(k − 1)k!(2r1 ) an + + |Ak,n (z)| bn bn 2 1 . ≤ 2r|Ek−1,n (z)| + Cr,r1 (k − 1)k(k + 1)k!(2r1 )k an + bn

Now taking in this inequality k = 1, 2, ..., by mathematical induction we easily arrive at   2 X k 1  (j − 1)j(j + 1)j!(2r1 )j  |Ek,n (z)| ≤ Cr,r1 an + bn j=2 ≤ Cr,r1

an +

1 bn

2

(k − 1)k(k + 1)(k + 2)k!(2r1 )k .

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Here the constants Cr,r1 can be different at each occurrence. Finally, taking into account the hypothesis too we obtain 2 4 2 Rn (f )(z) − f (z) + an z f 0 (z) − an bn z + z f 00 (z) 2 1 + an z 2bn (1 + an z) ∞ X ≤ |ck | · |Ek,n (z)| k=0

≤ Cr,r1 ≤ Cr,r1

an + an +

1 bn 1 bn

2 X ∞ · |ck |k!(k − 1)k(k + 1)(k + 2)(2r1 )k k=2

2 X ∞ k=2

(k − 1)k(k + 1)(k + 2)(2r1 A)k ,

which proves the theorem since 2r1 A < 1.

Now we are in position to obtain the exact degree of approximation for R n (f )(z). Theorem 1.10.7. Let n0 ≥ 2, 0 < β ≤ 2/3, β 6= 1/2 and 12 < r < R ≤ S n01−β [R, +∞) → C is uniformly continuous and bounded 2 . Suppose that f : DR P∞ on [0, +∞), is analytic in DR , that is f (z) = j=0 ck z k for all z ∈ DR and there k

1 exist M > 0, 0 < A < 2r with |ck | ≤ M Ak! for all k = 0, 1, 2, ...., (which implies A|z| |f (z)| ≤ M e , for all z ∈ DR ). If f is not a polynomial of degree ≤ 1 then for all 1 } we have n ≥ max{n0 , r1/β 1 kRn (f ) − f kr ∼ an + , bn

where the constants in the equivalence are independent of n. Proof.

We can write

Rn (f )(z) − f (z) 1 1 −an z 2 f 0 (z) 1 (a2 bn z 4 + z)f 00 (z) = an + · + · n bn an + 1/bn 1 + an z an + 1/bn 2bn (1 + an z)2 1 + an + bn 1 an z 2 f 0 (z) (a2n bn z 4 + z)f 00 (z) · − Rn (f )(z) − f (z) + . (an + 1/bn )2 1 + an z 2bn (1 + an z)2 Since an + b1n → 0 as n → ∞, taking into account the estimate in Theorem 1.10.6 and the reasonings in the cases of the previous complex Bernstein-type operators, it remains to show that for sufficiently large n and for all |z| ≤ r we have |T (z)| > ρ > 0, where ρ is independent of n and T (z) :=

1 −an z 2 f 0 (z) 1 (a2 bn z 4 + z)f 00 (z) · + · n . an + 1/bn 1 + an z an + 1/bn 2bn (1 + an z)2

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But since an = nβ−1 → 0 as n → ∞ and bn = nβ , we obtain n2β−1 1 −an z 2 f 0 (z) 1 ≥ · · |z 2 f 0 (z)| an + 1/bn 1 + an z 1 + n2β−1 1 + a1 r =

2β−1

|z 2 f 0 (z)| n2β−1 · . 1 + n2β−1 1 + a1 r

2

0

1 n z f (z) n · −a1+a → 0 as If 2β − 1 < 0 then 1+n 2β−1 → 0 and therefore the term a +1/b n n nz n → ∞, uniformly with respect to |z| ≤ r. In this case the term does not count for our estimate. n2β−1 If 2β − 1 > 0 then 1+n 2β−1 → 1 and therefore for sufficiently large n we have 1 −an z 2 f 0 (z) 1 |z 2 f 0 (z)| an + 1/bn · 1 + an z ≥ 2 · 1 + a1 r For the other term it follows (a2 bn z 4 + z)f 00 (z) n3β−2 z 4 f 00 (z) 1 · n = · an + 1/bn 2bn (1 + an z)2 1 + n2β−1 2(1 + an z)2 zf 00 (z) . + 2(1 + n2β−1 )(1 + an z)2 Here it is clear that if 2β − 1 > 0 then this term converges to 0 (uniformly with respect to |z| ≤ r) and therefore does not count for our estimate, while if 2β −1 < 0, zf 00 (z) then n2β−1 → 0 (as n → ∞) and here only the term 2(1+n2β−1 )(1+an z)2 counts for the estimate. Therefore, concluding all the above reasonings, there exists an index n1 depending on β, such that if 2β − 1 > 0 then for all n ≥ n1 and all |z| ≤ r we have 2 0 f (z)| , while if 2β − 1 < 0 then for all n ≥ n1 and all |z| ≤ r it follows |T (z)| ≥ 21 · |z 1+r 1 zf 00 (z) |zf 00 (z)| |T (z)| ≥ · . ≥ 2 2(1 + an z)2 4(1 + r)2

Since f is not a polynomial of degree ≤ 1, it easily follows that ke2 f 0 kr > 0 and ke1 f 00 kr > 0, which implies that in both cases 2β − 1 > 0 and 2β − 1 < 0 we have kT kr > ρ > 0, with ρ independent of n. In the case of 2β − 1 = 0, that is β = 1/2, we obtain n3β−2 z 4 f 00 (z) n1/2 z 4 f 00 (z) · = → ∞, 1 + n2β−1 2(1 + an z)2 4(1 + an z)2 as n → ∞, so that the case β = 1/2 remains unsettled. In conclusion, 1 1 1 kRn (f ) − f kr ≥ an + [kT kr − kGn kr ] ≥ an + · kT kr , bn bn 2 for all n ≥ n1 . (Here (Gn (z))n∈N is a sequence of analytic functions uniformly convergent to zero with respect to |z| ≤ r). In the case when n = 1, 2, ..., n1 − 1 the lower estimate is trivial. Finally, taking into account the upper estimate in Theorem 1.10.5 (in fact see the Remark after the proof of Theorem 1.10.5), our theorem is proved.

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1.11

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Bibliographical Notes and Open Problems

Theorems 1.1.8, 1.1.9, 1.4.2, Lemmas 1.4.3, 1.4.4, Theorems 1.4.5, 1.5.5, 1.6.14, 1.6.15, 1.7.6, Lemma 1.8.9, Corollary 1.8.10, Lemma 1.10.1, 1.10.2, Theorem 1.10.3, Corollary 1.10.4, Theorems 1.10.5, 1.10.6, 1.10.7 are new and appear for the first time here. Note 1.11.1. From a very long list, some references concerning the Bernstein-type operators of one real variable, different from those considered in this Chapter 1, for which would be possible to develop similar results are, for example : Altomare [8], Altomare-Campiti [9], Altomare-Mangino [10], Altomare-Ra¸sa [11], BleimannButzer-Hahn [46], Cimoca-Lupa¸s [55], Derriennic [56; 57; 59], Durrmeyer [67], Lupa¸s [129; 130; 131], Lupa¸s-M¨ uller [132], Meyer-K¨ onig and Zeller [136], Moldovan G. [140], Ra¸sa [159], Soardi [168], Stancu [176]. Open Problem 1.11.2. A Voronovskaja-type formula with a quantitative estimate and the exact orders of approximation in Theorem 1.6.9 for the complex polynomials Sn<γ> (f )(z) remain as open questions. Open Problem 1.11.3. It is well-known that if f : [a, b] → R, where a, b ∈ R, a < b, the Bernstein polynomials attached to f are given by the formula n X n 1 (x − a)k (b − x)n−k , f (ak ) Bn (f ; [a, b])(x) = (b − a)n k k=0

k b−a n .

ak = a + Now, if we consider the complexified form Bn (f ; [a, b])(z) for z belonging to a disk containing the real interval [a, b] and we suppose that f is analytic in that disk, then similar results with those for Bn (f ; [0, 1])(z) in Sections 1.1, 1.2, 1.3 and 1.4 can be obtained. But more interesting seems to be the case when [a, b] is a ”complex” interval in C, that is a, b ∈ C and [a, b] = {z = λa+(1−λ)b; λ ∈ [0, 1]}. In this case, for the complexified form Bn (f ; [a, b])(z) as above, it would be of interest to study its approximation properties in a disk containing the ”complex” interval [a, b], when f is supposed to be analytic in that disk. Open Problem 1.11.4. Let Ln : C[a, b] → C[a, b], n ∈ N, a, b ∈ R, a < b, be a sequence of positive and linear operators (of one real variable) attached to f ∈ C[a, b], satisfying for example the conditions in the classical Korovkin theorem, that is limn→∞ Ln (ek )(x) = ek (x), uniformly for x ∈ [a, b], for k = 0, 1, 2 (here we recall that ek (x) = xk ). Taking into account the results in this chapter, it is natural to ask if the convergence properties of the sequence (Ln (f ))n∈N remain valid if we complexify Ln (f )(x) (that is if we replace x ∈ [a, b] by z ∈ DR ), supposing in addition that f can be extended to an analytic function in the disk DR containing the real segment [a, b]. In general this does not happen. For example, if one considers as Ln (f )(x) the Hermite-Fej´er interpolation polynomials on an infinite interpolatory matrix in [−1, 1] consisting of the roots of orthogonal polynomials, or more general on an

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arbitrary infinite triangular matrix in [−1, 1], then for f (z) = z the polynomials L2n−1 (f )(z) diverges for all z ∈ C \ [−1, 1], see Brutman-Gopengauz [49] and Brutman-Gopengauz-V´ertesi [50]. However, for other sequences of positive and linear operators (possibly of noninterpolatory kind) the question remains open. More exactly, it would be interesting to see if the Korovkin theory for the complexified sequence of a sequence of positive and linear operators (possibly under some additional hypothesis) still holds, that is if under some possible additional hypothesis on Ln (e0 ), Ln (e1 ) and Ln (e2 ), the conditions lim Ln (f )(z) = ei (z), i = 0, 1, 2, uniformly in Dr ,

n→∞

would imply Ln (f )(z) → f (z) (as n → ∞), uniformly in Dr , for any analytic function in DR and r < R. Remark. From Vitali’s theorem it is immediate that if (Ln )n∈N is a sequence of positive and linear operators satisfying the Korovkin theorem and if in addition the complexified sequence satisfies |Ln (ek )(z)| ≤ Mr rk , for all |z| ≤ r, k ∈ N ∪ {0}, n ∈ N, then for any analytic function f we have limn→∞ Ln (f )(z) = f (z), uniformly in Dr . The problem is how to reduce the additional conditions on the set {Ln (e0 ), Ln (e1 ), Ln (e2 )} only. In the case of Bernstein-type operators this seems to be possible because of a recurrence formula with respect to k satisfied by Ln (ek ). Open Problem 1.11.5. It is left to the reader to prove that the upper estimates in the Voronovskaja-type results for q-Bernstein polynomials in Theorem 1.5.2 (ii) in fact hold with equivalence of order [n]1 2 . q More general, let us consider the exponential-type operators introduced in Rb May [134] by the general formula Mn (f )(x) = a Wn (x, t)f (t)dt, where the kernel Wn : I(a, b) × I(a, b) → R has the R b following properties : 1) Wn (x, t) ≥ 0 for all (x, t) ∈ I(a, b) × I(a, b) ; 2) a Wn (x, t)dt = 1, for all x ∈ I(a, b) ; ∂ Wn (x, t) = n(t−x) 3) ∂x p(x) Wn (x, t), where p(x) is a strictly positive polynomial for x ∈ I(a, b). Here I(a, b) can be of the form [a, b], (−∞, b], [a, +∞) or (−∞, +∞). Also, denote Z b An,m (x) = nm Wn (x, t)(t − x)m dt. a

Note that for a = 0, b = 1 and p(x) = x(1−x) we obtain the Bernstein polynomials, for a = 0, b = ∞ and p(x) = x we obtain the Favard-Sz´ asz-Mirakjan operators, for a = 0, b = ∞ and p(x) = x(1 + x) we obtain the Baskakov operators, for a = 0, b = ∞ and p(x) = x2 we obtain the Post-Widder operators, for a = −∞, b = ∞ and p(x) = 1 we obtain the Gauss-Weierstrass operators.

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In Pop [156] the following generalized Voronovskaja’s theorem was proved : if f is 2p continuous differentiable in I(a, b) then for all x ∈ I(a, b) we have |Mn (f )(x) −

2p X 1 A (x)f (i) (x)| = o(1/np ). i i! n,i n i=0

Our conjecture is that for the complex operators Z b Mn (f )(z) = Wn (z, t)f (t)dt, a

where f is supposed to be analytic in a suitable disk Dr depending on the operator (see the theorems mentioned at the beginning of this Open Problem), we have the equivalence kMn (f ) −

2p X 1 1 A f (i) kr ∼ p+1 . i i! n,i n n i=0

For general p ∈ N and general operators Mn (f ), it could be useful the ideas in the proof for the complex Bernstein polynomials (that is the proof of Corollary 1.3.4), see also the proof of Theorem 1.8.2 (ii) for the case of complex Favard-Sz´ aszMirakjan operators. Open Problem 1.11.6. The complex Bernstein-Stancu polynomials depending on the parameter γ ≥ 0 defined for disks of center in origin are given by the formula n X n z(z + γ)...(z + (p − 1)γ) p <γ> ∆ f (0), |z| ≤ r. Sn (f )(z) = p (1 + γ)...(1 + (p − 1)γ) 1/n p=0 Writing now p

z(z + γ)...(z + (p − 1)γ) X Ap,j z j , = (1 + γ)...(1 + (p − 1)γ) j=0 where by identification of the coefficients we evidently can explicitly find each Ap,j ∈ R, it follows   p n X X n  Sn<γ> (f )(z) = Ap,j z j  ∆p1/n f (0), |z| ≤ r. p p=0 j=0

˜ \ G is connected and by using the Then, for G ⊂ C a compact set such that C Faber polynomials Fp (z) attached to G (see Definition 1.0.10), for f ∈ A(G) we can introduce the Bernstein-Stancu-Faber polynomials depending on the parameter γ ≥ 0, given by the formula   p n X X n  Ap,j Fj (z) ∆p1/n F (0), z ∈ G, n ∈ N, Sn<γ> (f ; G)(z) = p p=0 j=0

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˜ \ D1 onto C ˜ \ G satisfying Ψ(∞) = where Ψ is the unique conformal mapping C R Ψ fof(Ψ(u)) 1 0 ∞ and Ψ (∞) > 0 and F (w) = 2πi |u|=1 u−w du, w ∈ D1 . Here, since F (1) is involved in ∆n1/n F (0) and therefore in the definition of Sn<γ> (f ; G)(z) too, in addition we will suppose that F can be extended by continuity on the boundary ∂D1 . It is an open problem to extend the results in Section 1.6 (corresponding there to Sn<γ> (f )(z), |z| ≤ r) to the Bernstein-Stancu-Faber polynomials Sn<γ> (f ; G)(z), z ∈ G (exactly as we did for the complex Bernstein polynomials and BernsteinStancu polynomials depending on two parameters 0 ≤ α ≤ β). Open Problem 1.11.7. It is known that the Kantorovich variant of the complex Bernstein polynomial for compact disks is given by Z z 0 Kn (f )(z) = Bn+1 (g)(z), g(z) = f (u)du, |z| ≤ r, 0 Pn+1 n+1 k k denotes the complex classical where Bn+1 (g)(z) = k=0 k ∆1/(n+1) g(0)z Bernstein polynomial of degree n + 1. This immediately implies the representation n X n+1 j (j + 1)∆j+1 Kn (f )(z) = 1/(n+1) g(0)z , |z| ≤ r, j + 1 j=0 and suggests the following expression for the Kantorovich-Faber polynomials attached to a set G ⊂ C n X n+1 Kn (f ; G)(z) = (j + 1)∆j+1 1/(n+1) F (0)Fj (z), z ∈ G, n ∈ N, j + 1 j=0

with F defined as in the above Open Problem 1.11.6 (with g instead of f ). Following the same procedure and taking into account that the complex StancuKantorovich polynomials depending on two parameters 0 ≤ α ≤ β are given by n n+1+β X n+1 j Kn(α,β) (f )(z) = ∆k+1 1/(n+β) g[α/(n + β)]z , |z| ≤ r, n+1 k+1 k=0

the expression of Stancu-Kantorovich-Faber polynomials attached to a set G ⊂ C, can be given by n n+1+β X n+1 (j + 1)∆j+1 Kn(α,β) (f ; G)(z) = 1/(n+β) F [α/(n + β)]Fj (z), n + 1 j=0 j + 1

z ∈ G, where F is defined as above. Remain as open questions the approximation properties of the polynomials (α,β) Kn (f ; G)(z) and Kn (f ; G)(z). Open Problem 1.11.8. In a similar manner, taking into account that the complex Favard-Sz´ asz-Mirakjan operators for compact disks can be written in the form ∞ X Sn (f )(z) = [0, 1/n, ..., j/n; f ]z j , |z| ≤ r, j=0

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for a set G ⊂ C we can formally define the Favard-Sz´ asz-Mirakjan-Faber operators by Sn (f ; G)(z) =

∞ X j=0

[0, 1/n, ..., j/n; f ]Fj (z), z ∈ G,

R f (Ψ(u)) 1 where F (w) = 2πi |u|=1 u−w du, w ∈ D1 and we suppose that F can be extended by continuity on the boundary ∂D1 . Let us observe that for any a with |a| > 1, F (a) is well-defined. It is an open question to find the approximation properties of the Favard-Sz´ aszMirakjan-Faber operators Sn (f ; G)(z). Open Problem 1.11.9. Prove Theorem 1.10.7 for the case β = 1/2. Open Problem 1.11.10. As in the case of Favard-Sz´ asz-Mirakjan operators, it is clear that in the case of Baskakov operators too, the domain of definition S [R, +∞) DR is rather unusual. Taking into account the form of complex Baskakov operators, more natural seems to consider and approximate the analytic function f on an annulus Ar,∞ = {z ∈ C; r ≤ |z + 1| < ∞}, with r > 0, where f can be represented as a Laurent series. The study of this problem is left as an open question. Open Problem 1.11.11. It is well known that in the case of Bernstein-type operators of real variables, starting with the paper of Altomare [7] a theory of strongly continuous contraction semigroups on Banach spaces obtained as a limit of the iterations of these operators is much developed. Then it would be interesting to consider the Altomare’s idea in the case of Bernstein-type operators of complex variables. Note that taking into account, for example, the considerations from the beginning of Section 1.2, in the complex case would correspond a Trotter’s theorem and a theory of limit semigroups on a Fr´echet space (i.e. a metrizable complete locally convex space) with respect to the topology of uniform convergence on compacts subsets of the open disk DR .

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Chapter 2

Bernstein-Type Operators of Several Complex Variables

The results for the complex univariate operators in Chapter 1 can be extended to the case of several complex variables. We illustrate here the theory on three important operators : Bernstein, Favard-Sz´ asz-Mirakjan and Baskakov. For simplicity, the results are presented for bivariate case, but from the proofs it is easy to see that they remain valid for several complex variables. 2.1

Introduction

In this section we present some requisites in the theory of holomorphic (analytic) functions of several complex variables. Definition 2.1.1. (see e.g. Andreian Cazacu [23] or Kohr [117]) Let Cp denote the space of p-complex variables z = (z1 , ..., zp ), zj ∈ C, j = 1, ..., p. (i) The open polydisk of center a = (a1 , ..., ap ) ∈ Cp and radius R = (R1 , ..., Rp ) ∈ R+ p is defined by P (a; R) = {z = (z1 , ..., zn ) ∈ C; |zj − aj | < Rj , ∀j = 1, ..., p}. (ii) Let Ω be a domain in Cp and f : Ω → C. We say that f is holomorphic (or analytic) on Ω if f is continuous in Ω and holomorphic (analytic) in each variable separately (when the others are fixed). Equivalently, f is holomorphic (analytic) in Ω, if for each point a = (a1 , ..., ap ) ∈ Ω, there exists a neighborhood of a such that we have f (z) =

∞ X

j1 ,...,jp =0

cj1 ,...,jp (z − a1 )j1 ...(z − ap )jp , ∀z ∈ Ω,

where the series converges absolutely and uniform on each compact subset of Ω. j By using the multi-index notations |j| = j1 + j2 + ... + jp , z j = z1j1 z2j2 ...zpp , j! = j1 !j2 !...jp !, for any j = (j1 , ..., jp ), jk ∈ {0, 1, ..., }, it is well known that for e.g.

a = 0, we can write cj1 ,...,jp = P∞ Taylor form f (z) = j1 ,...,jp =0

∂ |j| f (0)

1 j j ∂z11 ...∂zpp j! D j f (0) j z . j!

155

:= Dj f (0)/j! and therefore we get the

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156

Also, the following result called the Cauchy’s formula on polydisk is classical. Theorem 2.1.2. (see e.g. Kohr [117]) Let i2 = −1, a = (a1 , ..., ap ) ∈ Cp , R = (R1 , ..., Rp ), Rj > 0, j = 1, ..., p and f : P (a; R) → C be a holomorphic function in P (a; R) and continuous in P (a; R). Then for all z = (z1 , ..., zp ) ∈ P (a; R) and all S kj ∈ N {0}, j = 1, ..., p, we have : (i) Z Z f (u1 , ..., up )du1 ...dup 1 · · · . f (z1 , ..., zp ) = (2πi)p |up −zp |=Rp |u1 −z1 |=R1 (u1 − z1 )...(up − zp ) (ii) ∂ k1 +...+kp f k

∂z1k1 ...∂zp p (k1 !)...(kp !) = (2πi)p

Z

|up −zp |=Rp

···

Z

(z1 , ..., zp )

|u1 −z1 |=R1

f (u1 , ..., up )du1 ...dup . (u1 − z1 )k1 +1 ...(up − zp )kp +1

Remarks. 1) The Vitali’s theorem in univariate case (Theorem 1.0.1) holds for several complex variables (see e.g. Kohr [117], p. 29, Theorem 1.3.7.). 2) The Weierstrass’s theorem in univariate case (Theorem 1.0.3) holds for several complex variables, by replacing the derivatives with partial derivatives of any order (see e.g. Kohr [117], p. 26, Theorem 1.3.2). 3) The Maximum Modulus Theorem in univariate case (Theorem 1.0.5) holds for several complex variables on polydisks (see e.g. Kohr [117], p. 23, Corollary 1.2.5). 4) The identity’s theorem in univariate case (Theorem 1.0.7) holds in several complex variables in the following slightly modified form : if Ω ⊂ Cn is a domain and the analytic functions f, g : Ω → C coincide on a nonempty open set G ⊂ Ω, then they coincide on Ω (see e.g. Kohr [117], p. 20, Theorem 1.2.1). This chapter can be described as follows : - in Section 2.2 we study the complex Bernstein polynomials, both tensor product and non-tensor product types ; - in Section 2.3 the complex Favard-Sz´ asz-Mirakjan operator of tensor product kind, without exponential growth conditions on f is studied ; - Section 2.3 deals with the complex Baskakov operator of tensor product kind. 2.2

Bernstein Polynomials

For f (z1 , z2 ) analytic in the polydisc P (0; R) = DR1 × DR2 where R = (R1 , R2 ) and |z1 | ≤ r1 , |z2 | ≤ r2 , 1 ≤ r1 < R1 with 1 ≤ r2 < R2 , we can define two kinds of bivariate complex Bernstein polynomials : 1) the tensor product kind defined by

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Bn,m (f )(z1 , z2 ) =

n X m X

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pn,k (z1 )pm,j (z2 )f (k/n, j/m),

k=0 j=0

where pn,k (z) = nk z k (1 − z)n−k , and 2) the non-tensor product (or simplex) kind given by Bn (f )(z1 , z2 ) =

n n−k X X

pn,k,j (z1 , z2 )f (k/n, j/n),

k=0 j=0

where

n n−k k j z1 z2 (1 − z1 − z2 )n−k−j k j n! z k z j (1 − z1 − z2 )n−k−j . = k!j!(n − k − j)! 1 2

pn,k,j (z1 , z2 ) =

Note that in the case of several real variables, the tensor product Bernstein polynomial was first introduced and studied in Hildebrandt-Schoenberg [108] and Butzer [52], while the simplex kind Bernstein polynomial of several real variables was first introduced and studied in Dinghas [61], Lorentz [125], p. 51 and Stancu [175]. In this section, approximation properties of the above bivariate complex polynomials will be proved. We begin with the properties of the tensor-product kind. In this sense first we present : Theorem 2.2.1. Suppose that f : P (0; R) → C is analytic in P (0; R) = DR1 ×DR2 , P∞ P∞ that is f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , for all (z1 , z2 ) ∈ P (0; R), R = (R1 , R2 ). (i) For all |z1 | ≤ r1 , |z2 | ≤ r2 with 1 ≤ r1 < R1 , 1 ≤ r2 < R2 and n, m ∈ N we have |Bn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ Cr1 ,r2 ,n,m (f ), where Cr1 ,r2 ,n,m (f ) =

∞ ∞ 3r2 (1 + r2 ) X X · |ck,j | · j(j − 1)r2j−2 r1k 2m j=0

+

k=0 ∞ X ∞ X

3r1 (1 + r1 ) · 2n

k=0 j=0

|ck,j | · k(k − 1)r1k−2 r2j .

(ii) Let k1 , k2 ∈ N be with k1 + k2 ≥ 1 and 1 ≤ r1 < r1∗ < R1 , 1 ≤ r2 < r2∗ < R2 . Then for all |z1 | ≤ r1 , |z2 | ≤ r2 and n, m ∈ N we have ∂ k1 +k2 B (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 k k k k 1 2 ∂z1 1 ∂z2 2 ∂z1 ∂z2 ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

(r1∗

(k2 )! (k1 )! · ∗ . k +1 1 − r1 ) (r2 − r2 )k2 +1

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(i) Denote ek,j (z1 , z2 ) = ek (z1 ) · ej (z2 ), where ek (u) = uk . Clearly we get

Proof.

|Bn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤

∞ X ∞ X

k=0 j=0

|ck,j | · |Bn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )|.

But taking into account the estimates in the proof of Theorem 1.1.2, (i), for all |z1 | ≤ r1 and |z2 | ≤ r2 we obtain |Bn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )| = |Bn (ek )(z1 ) · Bm (ej )(z2 ) − z1k · z2j |

≤ |Bn (ek )(z1 ) · Bm (ej )(z2 ) − Bn (ek )(z1 )z2j | + |Bn (ek )(z1 )z2j − z1k z2j |

≤ |Bn (ek )(z1 )| · |Bm (ej )(z2 ) − z2j | + |z2j | · |Bn (ek )(z1 ) − z1k | 3r1 (1 + r1 ) 3r2 (1 + r2 ) j(j − 1)r2j−2 + r2j k(k − 1)r1k−2 , ≤ r1k · 2m 2n

which immediately implies the estimate in (i). (ii) Let 1 ≤ r1 < r1∗ < R1 , 1 ≤ r2 < r2∗ < R2 . By the Cauchy’s formula in Theorem 2.1.2, (ii), we get ∂ k1 +k2 f ∂ k1 +k2 Bn,m (f ) (z , z ) − (z1 , z2 ) 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2 =

(k1 !)(k2 !) (2πi)2

Z

|u2 −z2 |=r2∗

Z

|u1 −z1 |=r1∗

[Bn,m (u1 , u2 ) − f (u1 , u2 )]du1 du2 . (u1 − z1 )k1 +1 (u2 − z2 )k2 +1

Passing to absolute value with |z1 | ≤ r1 , |z2 | ≤ r2 and taking into account that |u1 − z1 | ≥ r1∗ − r1 , |u2 − z2 | ≥ r2∗ − r2 , by applying the estimate in (i) we easily obtain ∂ k1 +k2 B (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2 ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

(k2 )! (k1 )! · , (r1∗ − r1 )k1 +1 (r2∗ − r2 )k2 +1

which proves the theorem.

In what follows a Voronovskaja’s result for Bn,m (f ) is presented. It will be the product of the parametric extensions generated by the Voronovskaja’s formula in univariate case in Theorem 1.1.3. Indeed, for f (z1 , z2 ) defining the parametric extensions of the Voronovskaja’s formula by z1 Ln (f )(z1 , z2 )

:= Bn (f (·, z2 ))(z1 ) − f (z1 , z2 ) −

z1 (1 − z1 ) ∂ 2 f · (z1 , z2 ), 2n ∂z12

z2 Lm (f )(z1 , z2 )

:= Bm (f (z1 , ·))(z2 ) − f (z1 , z2 ) −

z2 (1 − z2 ) ∂ 2 f · (z1 , z2 ), 2m ∂z22

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their product (composition) gives ◦ z1 Ln (f )(z1 , z2 ) z1 (1 − z1 ) ∂ 2 f = Bm Bn (f (·, ·))(z1 ) − f (z1 , ·) − · 2 (z1 , ·) (z2 ) 2n ∂z1 z1 (1 − z1 ) ∂ 2 f z2 (1 − z2 ) − Bn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − · 2 (z1 , z2 ) − 2n ∂z1 2m 2 2 2 2 ∂ f ∂ f z1 (1 − z1 ) ∂ ∂ f (z1 , z2 ) · 2 · Bn ( 2 (·, z2 ))(z1 ) − 2 (z1 , z2 ) − ∂z2 ∂z2 2n ∂z1 ∂z22 := E1 − E2 − E3 . z2 Lm (f )(z1 , z2 )

After simple calculation evidently that we can write z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 )

2 z1 (1 − z1 ) ∂ f = Bn,m (f )(z1 , z2 ) − Bm (f (z1 , ·))(z2 ) − · Bm (z1 , ·) (z2 ) 2n ∂z12 z1 (1 − z1 ) ∂ 2 f −Bn (f (·, z2 ))(z1 ) + f (z1 , z2 ) + (z1 , z2 ) 2n ∂z12 2 z2 (1 − z2 ) ∂ f z2 (1 − z2 ) ∂ 2 f − · Bn (·, z ) (z ) + · 2 (z1 , z2 ) 2 1 2m ∂z22 2m ∂z2 ∂4f z1 (1 − z1 ) z2 (1 − z2 ) · · 2 2 (z1 , z2 ), + 2n 2m ∂z1 ∂z2 from which immediately can be derived the commutativity property z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 ) = z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 ).

The Voronovskaja-type formula can be stated as follows. Theorem 2.2.2. Suppose that f : P (0; R) → C is analytic in P (0; R) = DR1 ×DR2 , P∞ P∞ that is f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , for all (z1 , z2 ) ∈ P (0; R), R = (R1 , R2 ). For all |z1 | ≤ r1 , |z2 | ≤ r2 with 1 ≤ r1 < R1 , 1 ≤ r2 < R2 and n, m ∈ N we have 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ Cr1 ,r2 (f ) 2 + 2 , n m where  # " ∞ X ∞ X 2 2 k−2 j 5(1 + r1 )k(k − 1)(k − 2) r1 |ck,j |r2 , Cr1 ,r2 (f ) = max  2 j=0 k=2

" # ∞ X ∞ 2 2 k−2  X 5(1 + r1 )k(k − 1)(k − 2) r1 r2 (1 + r2 ) . |ck,j |j(j − 1)r2j−2 ·  2 2 j=2 k=2

P∞ k Proof. First by hypothesis we can write f (z1 , z2 ) = k=0 fk (z2 )z1 , where 2 P∞ P ∞ j ∂ f k−2 It follows ∂z2 (z1 , z2 ) = fk (z2 ) = and j=0 ck,j z2 . k=2 fk (z2 )k(k − 1)z1 1

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implies

P∞

∂ 2 fk ∂ 2 fk k k=0 ∂z22 (z2 )z1 , where ∂z22 (z2 ) P k Bn (f (·, z2 ))(z1 ) = ∞ k=0 fk (z2 )Bn (e1 )(z1 )

∂2f (z1 , z2 ) ∂z22

=

Bn (f (·, z2 ))(z1 ) − f (z1 , z2 ) −

=

∞ X k=2

=

P∞

j=2 ck,j j(j

and

− 1)z2j−2 . This

z1 (1 − z1 ) ∂ 2 f · (z1 , z2 ) 2n ∂z12

# k−1 z (1 − z )k(k − 1) 1 . fk (z2 ) Bn (ek1 )(z1 ) − ek1 (z1 ) − 1 2n "

Applying now Bm to the last expression with respect to z2 , we obtain " # ∞ X z1k−1 (1 − z1 )k(k − 1) k k E1 = Bm (fk )(z2 ) Bn (e1 )(z1 ) − e1 (z1 ) − 2n k=2

=

∞ X k=2

 " # ∞ k−1 X z (1 − z )k(k − 1) 1 j  . ck,j Bm (e1 )(z2 ) Bn (ek1 )(z1 ) − ek1 (z1 ) − 1 2n j=0

Passing now to absolute value with |z1 | ≤ r1 and |z2 | ≤ r2 and taking into account the estimates in the proofs of Theorem 1.1.2, (i) and Theorem 1.1.3, (ii), it follows " # ∞ X ∞ 2 2 k−2 X j 5(1 + r1 )k(k − 1)(k − 2) r1 |E1 | ≤ |ck,j |r2 . 2n2 j=0 k=2

Similarly, |E2 | ≤

≤

∞ X

k=2

∞ X ∞ X k=2 j=0

"

5(1 + r12 )k(k − 1)(k − 2)2 r1k−2 |fk (z2 )| · 2n2 |ck,j |r2j

"

Then Bn

= and

=

∞ X ∂ 2 fk

# 5(1 + r12 )k(k − 1)(k − 2)2 r1k−2 . 2n2

(z2 )Bn (ek1 )(z1 ) ∂z22 k=0

Bn (

#

∂2f (·, z2 ) (z1 ) ∂z22 =

∞ X ∞ X

k=0 j=2

ck,j j(j − 1)z2j−2 Bn (ek1 )(z1 ),

∂ 2f z1 (1 − z1 ) ∂ 2 ∂ 2 f ∂2f (z , z ) (·, z ))(z ) − (z , z ) − · 1 2 2 1 1 2 ∂z22 ∂z22 2n ∂z12 ∂z22

∞ ∞ X X

k=2 j=2

ck,j j(j −

1)z2j−2

"

Bn (ek1 )(z1 )

−

ek1 (z1 )

# z1k−1 (1 − z1 )k(k − 1) − , 2n

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which again by Theorem 1.1.3, (ii) implies # " ∞ ∞ 5(1 + r12 )k(k − 1)(k − 2)2 r1k−2 r2 (1 + r2 ) X X j−2 . |ck,j |j(j − 1)r2 · |E3 | ≤ 2m 2n2 j=2 k=2

Note that if we estimate now |z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 )|, then by reasons of symmetry we get a similar order of approximation, simply interchanging above the places of n and m. In conclusion, 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ |E1 | + |E2 | + |E3 | ≤ Cr1 ,r2 (f ) 2 + 2 , n m with Cr1 ,r2 (f ) given by the statement. Theorem 2.2.2 will be used to find the exact order in approximation by Bn,n (f ). In this sense we have the following. Theorem 2.2.3. Suppose that f : P (0; R) → C is analytic in P (0; R) = DR1 ×DR2 , P P∞ k j that is f (z1 , z2 ) = ∞ k=0 j=0 ck,j z1 z2 , for all (z1 , z2 ) ∈ P (0; R), R = (R1 , R2 ). Denoting kf kr1 ,r2 = sup{|f (z1 , z2 )|; |z1 | ≤ r1 ; |z2 | ≤ r2 }, if f is not a solution of the complex partial differential equation ∂2f ∂2f (z1 , z2 ) = 0, (z1 , z2 ) ∈ P (0, R), z1 (1 − z1 ) · 2 (z1 , z2 ) + z2 (1 − z2 ) · ∂z1 ∂z22 then we have Kr1 ,r2 ,f kBn,n (f ) − f kr1 ,r2 ≥ , for all n ∈ N, n where Kr1 ,r2 ,f is independent on n. Proof.

where

We can write Bn,n (f )(z1 , z2 ) − f (z1 , z2 ) 2 z1 (1 − z1 ) ∂ 2 f z2 (1 − z2 ) ∂ 2 f = · · 2 (z1 , z2 ) (z , z ) + 1 2 n 4 ∂z12 4 ∂z2 2 n2 (z2 Ln (f ) ◦ z1 Ln (f )) (z1 , z2 ) + Rn (f )(z1 , z2 ) , + n 4 Rn (f )(z1 , z2 ) n z2 (1 − z2 ) ∂ 2 f = Bn (f (z1 , ·))(z2 ) − f (z1 , z2 ) − · (z1 , z2 ) 2 2n ∂z22 z1 (1 − z1 ) ∂ 2 f n · 2 (z1 , z2 ) Bn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − + 2 2n ∂z1 2 z1 (1 − z1 ) ∂2f ∂ f + Bn (z1 , ·) (z2 ) − 2 (z1 , z2 ) 4 ∂z12 ∂z1 2 z2 (1 − z2 ) ∂2f ∂ f Bn (·, z ) (z ) − (z , z ) + 2 1 1 2 4 ∂z22 ∂z22 z1 (1 − z1 )z2 (1 − z2 ) ∂4f − · 2 2 (z1 , z2 ). 8n ∂z1 ∂z2

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By using Theorem 1.1.3 and the reasonings in the above Theorem 2.2.2, it is immediate that kRn (f )kr1 ,r2 → 0 as n → ∞. Also, by Theorem 2.2.2 we obtain Cr ,r (f ) n2 k(z2 Lm (f ) ◦ z1 Ln (f ))kr1 ,r2 ≤ 1 2 , 4 2

which implies

2

2 n

n 4 (z2 Ln (f ) ◦ z1 Ln (f )) + Rn (f ) 1) Denoting H(z1 , z2 ) = z1 (1−z · 4 account the inequalities

∂2f (z1 , z2 ) ∂z12

+

→ 0, as n → ∞.

r1 ,r2

z2 (1−z2 ) 4

·

∂2f (z1 , z2 ) ∂z22

and taking into

kF + Gkr1 ,r2 ≥ | kF kr1 ,r2 − kGkr1 ,r2 | ≥ kF kr1 ,r2 − kGkr1 ,r2 , it follows

2 ≥ n

(

kBn,n (f ) − f kr1 ,r2

kHkr1 ,r2

)

2

2 n

− (z Ln (f ) ◦ z1 Ln (f )) + Rn (f ) n 4 2 r1 ,r2

2 1 1 · kHkr1 ,r2 = · kHkr1 ,r2 , n 2 n for all n ≥ n0 , with n0 depending only on f , r1 and r2 . We used here that by hypothesis we have kHkr1 ,r2 > 0. For n ∈ {1, 2, ..., n0−1} we reason exactly as for the complex univariate Bernstein polynomials in the proof of Theorem 1.1.4. ≥

Combining Theorem 2.2.2 with Theorem 2.2.3 we immediately obtain the following. Corollary 2.2.4. Suppose that f : P (0; R) → C is analytic in P (0; R) = DR1 ×DR2 , P P∞ k j that is f (z1 , z2 ) = ∞ k=0 j=0 ck,j z1 z2 , for all (z1 , z2 ) ∈ P (0; R), R = (R1 , R2 ). If the Taylor series of f contains at least one term of the form ck,0 z1k with ck,0 6= 0 and k ≥ 2 or of the form c0,j z2j with c0,j 6= 0 and j ≥ 2, then we have kBn,n (f ) − f kr1 ,r2 ∼

1 , for all n ∈ N. n

Proof. It suffices to prove that under the hypothesis on f , it cannot be a solution of the complex partial differential equation z1 (1 − z1 ) ·

∂2f ∂2f (z1 , z2 ) + z2 (1 − z2 ) · (z1 , z2 ) = 0, (z1 , z2 ) ∈ P (0, R). 2 ∂z1 ∂z22

Indeed, suppose the contrary. Since simple calculation give z1 (1 − z1 ) ·

∞

∞

∞

X XX ∂2f = 2c2,j z1 z2j + z1k z2j [ck+1,j k(k + 1) − ck,j k(k − 1)] 2 ∂z1 j=0 j=0 k=2

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and z2 (1 − z2 ) ·

∞

∞

k=0

k=0

∞

X XX ∂2f = 2ck,2 z1k z2 + z1k z2j [ck,j+1 j(j + 1) − ck,j j(j − 1)] 2 ∂z2 j=2

by summing and equaling with zero, we obtain 2c0,2 z2 + 2c1,2 z1 z2 +

∞ X

k=2

+ +

∞ X

j=2 ∞ X j=2

z2j [c0,j+1 j(j + 1) − c0,j j(j − 1)] z1 z2j [c1,j+1 j(j + 1) − c1,j j(j − 1) + 2c2,j ]

+ 2c2,0 z1 + 2c2,1 z1 z2 +

∞ X

k=2

+

z1k z2 [ck+1,j k(k + 1) − ck,1 k(k − 1) + 2ck,2 ]

∞ X ∞ X k=2 j=2

z1k [ck+1,0 k(k + 1) − ck,0 k(k − 1)]

z1k z2j [ck+1,j k(k + 1) − ck,j k(k − 1)

+ ck,j+1 j(j + 1) − ck,j j(j − 1)] = 0. By the identification of coefficients, among others we immediately get c0,2 = c2,0 = 0 and from the terms under the second and fourth sign Σ, it follows c0,j+1 j(j + 1) − c0,j j(j − 1) = 0, ck+1,0 k(k + 1) − ck,0 k(k − 1) = 0, for all j = 2, 3, ..., k = 2, 3, ...,. From here by recurrence it is easy to deduce that c0,k = cj,0 = 0 for all j = 2, 3, ..., k = 2, 3, ..., which contradicts the hypothesis on f . Therefore the hypothesis and the lower estimate in Theorem 2.2.3 holds, which ends the proof. In what follows we study the approximation properties of the non-tensor product kind bivariate complex Bernstein polynomials Bn (f )(z1 , z2 ). Note that obviously we have Bn (f )(z1 , z2 ) 6= Bn,n (f )(z1 , z2 ). P∞ P∞ For f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , as in the case of tensor product Bernstein polynomials, denoting ek,j (z1 , z2 ) = ek (z1 ) · ej (z2 ) = z1k · z2j we can write Bn (f )(z1 , z2 ) − f (z1 , z2 ) =

∞ X ∞ X

k=0 j=0

ck,j [Bn (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )].

Therefore, in order to obtain approximation results a representation formula for Bn (ek,j )(z1 , z2 ) will be important. In this sense we present Theorem 2.2.5. (i) Bn (ep,q )(z1 , z2 ) = ep,q (z1 , z2 ), for all p, q with 0 ≤ p + q ≤ 1 and n ∈ N. Also, Bn (ep,0 )(z1 , z2 ) = Bn (ep )(z1 ) and Bn (e0,q )(z1 , z2 ) = Bn (eq )(z2 ),

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where Bn in the above right-hand sides denotes the Bernstein polynomial of one variable. (ii) Denoting m = min{n, p} and s = min{n, q}, for all n, p, q ∈ N we can write m X s X Bn (ep,q )(z1 , z2 ) = Cn,i,j,p,q z1i z2j , i=1 j=1

where

n n n[i+j] Cn,i,j,p,q = ∆i1/n ep (0) · ∆j1/n eq (0) · [i] [j] ≥ 0, i j n n [i] and n = n(n − 1)...(n − i + 1) ; (iii) Denoting m = min{n, p} and s = min{n, q}, for all n, p, q ∈ N we have p m X s n X X n n−k k Cn,i,j,p,q = ∆1/n eq (0) ≤ 1. k n i=1 j=1 k=0

Proof. (i) We have the possibilities p = q = 0, or p = 0, q = 1 or p = 1, q = 0, for which by simple calculation we obtain Bn (ep,q )(z1 , z2 ) = ep,q (z1 , z2 ). Also, the next two equalities follow by simple calculation. (ii) Denoting by S(p, i) and S(q, j) the Stirling numbers of second kind and taking into account that S(p, i) = S(q, j) = 0 for all i > p and j > q, by using Proposition 1 in Farca¸s [71], we have m s 1 X X [i+j] n S(p, i)S(q, j)z1i z2j Bn (ep,q )(z1 , z2 ) = p+q n i=1 j=1 = =

m X n[i] S(p, i) i=1 m X s X

np

z1i

s X n[j] S(p, j) j=1

nq

z2j

n[i+j] n[i] n[j]

Cn,i,j,p,q z1i z2j ,

i=1 j=1

where Cn,i,j,p,q is given by the formula in statement. Note that we used here the formula n n n[i] S(p, i) i = ∆ e (0) = [0, 1/n, ..., i/n; ep]i!/ni . 1/n p np i i Since ep is convex of any order, from the last formula it is clear that all Cn,i,j,p,q ≥ 0. (iii) From the definition formula for Bn (f )(z1 , z2 ), we obtain Bn (ep,q )(1, 1) n p n n−k p X X X nn − k jq n k n−k n−k−j k · = ∆ eq (0) = (−1) p nq k np 1/n n k j k=0 k=0 j=0 p X n n X n n−k n = ∆k1/n eq (0) ≤ ∆k1/n eq (0) = 1. k n k k=0

k=0

On the other hand, replacing x = y = 1 in the formula from the above point (ii), it follows exactly the required inequality.

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The approximation results are expressed by the following. Theorem 2.2.6. Suppose that f : P (0; R) → C is analytic in P (0; R) = DR1 ×DR2 , P∞ P∞ that is f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , for all (z1 , z2 ) ∈ P (0; R), R = (R1 , R2 ). (i) For all |z1 | ≤ r1 , |z2 | ≤ r2 with 1 ≤ r1 < R1 , 1 ≤ r2 < R2 and n ∈ N we have |Bn (f )(z1 , z2 ) − f (z1 , z2 )| ≤

Cr1 ,r2 (f ) , n

where Cr1 ,r2 (f ) = 2

∞ X ∞ X p=0 q=0

|cp,q |(p + q)(p + q − 1)r1p r2q .

(ii) Let k1 , k2 ∈ N be with k1 + k2 ≥ 1 and 1 ≤ r1 < r1∗ < R1 , 1 ≤ r2 < r2∗ < R2 . Then for all |z1 | ≤ r1 , |z2 | ≤ r2 and n ∈ N we have ∂ k1 +k2 B (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 k1 k2 ∂z1k1 ∂z2k2 ∂z1 ∂z2 ≤

Proof.

Cr1∗ ,r2∗ (f ) (k1 )! (k2 )! · ∗ · . n (r1 − r1 )k1 +1 (r2∗ − r2 )k2 +1

(i) It is immediate the inequality

|Bn (f )(z1 , z2 ) − f (z1 , z2 )| ≤

∞ X ∞ X p=0 q=0

|cp,q | · |Bn (ep,q )(z1 , z2 ) − ep,q (z1 , z2 )|.

To estimate |Bn (ep,q )(z1 , z2 ) − ep,q (z1 , z2 )| we have four possibilities : 1) 0 ≤ p ≤ n and 0 ≤ q ≤ n ; 2) 0 ≤ p ≤ n and q > n ; 3)p > n and 0 ≤ q ≤ n ; 4) p > n and q > n. Case 1). It follows m = p, s = q. By Theorem 2.2.5, (i) we may consider p ≥ 1 and q ≥ 1. Also, by Theorem 2.2.5, (ii), (iii), for all 1 ≤ |z1 | ≤ r1 and 1 ≤ |z2 | ≤ r2 we obtain |Bn (ep,q )(z1 , z2 ) − ep,q (z1 , z2 )| = |[Cn,p,q,p,q − 1]ep,q (z1 , z2 )] +

q p−1 XX

i j

Cn,i,j,p,q x y +

i=1 j=1

i=1 j=1

+[Bn (ep,q )(1, 1) −

q−1 p X X

q X j=1

Cn,p,j,p,q ]r1p r2q + [Bn (ep,q )(1, 1) −

≤ [1 − Cn,p,q,p,q ]r1p r2q + [1 − ≤ 3[1 −

Cn,p,q,p,q ]r1p r2q .

Cn,i,j,p,q xi y j | ≤ [1 − Cn,p,q,p,q ]r1p r2q

q X j=1

Cn,p,j,p,q ]r1p r2q + [1 −

p X i=1

p X

Cn,i,q,p,q ]r1p r2q

i=1

Cn,i,q,p,q ]r1p r2q

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But Cn,p,q,p,q which implies

n n p! q! n[p+q] n[p+q] n−j+1 = = · · = Πp+q , j=1 p q p+q [p] [q] q n n n n p n n 1 − Cn,p,q,p,q ≤

(p + q)(p + q − 1) . 2n

We used here the inequality 1− applied for yj =

p+q Y j=1

yj ≤

n−j+1 . n

p+q X j=1

(1 − yj ), 0 ≤ yj ≤ 1, j = 1, 2, ..., p + q,

This implies

3(p + q)(p + q − 1) p q r1 r2 . 2n Case 2). It follows m = p and s = n and again by Theorem 2.2.5, (i) we may consider p ≥ 1. By Theorem 2.2.5, (ii), (iii), for all 1 ≤ |z1 | ≤ r1 and 1 ≤ |z2 | ≤ r2 we obtain |Bn (ep,q )(z1 , z2 ) − ep,q (z1 , z2 )| ≤

|Bn (ep,q )(z1 , z2 ) − ep,q (z1 , z2 )| ≤ |Bn (ep,q )(z1 , z2 )| + |ep,q (z1 , z2 )| p X n X Cn,i,j,p,n r1p r2n + r1p r2q ≤ 2r1p r2q ≤ 2nr1p r2q ≤ 2(q − 1)r1p r2q ≤ i=1 j=1

(q − 1)q p q (p + q)(p + q − 1) p q r1 r2 ≤ 2 r1 r2 . n n Case 3). By reason of symmetry, the Case 3) is identical with the Case 2. Case 4). Identical with the Cases 2 and 3. (ii) By using the Cauchy’s formula in Theorem 2.1.2, (ii), we reason exactly as in the proof of Theorem 2.2.1, (ii). ≤2

2.3

Favard-Sz´ asz-Mirakjan Operators

In this section we consider the bivariate complex Favard-Sz´ asz-Mirakjan operators of tensor product kind given by ∞ X ∞ X Sn,m (f )(z1 , z2 ) = sn,k (z1 )sm,j (z2 )f (k/n, j/m), k=0 j=0

k

where sn,k (z) = e−nz (nz) k! , f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C is supposed to be bounded on [0, +∞) × [0, ∞) and analytic in DR1 × DR2 . Note that in the case of two real variables, these operators were first considered by Leonte-Virtopeanu [121]. We will use similar methods with those for complex bivariate Bernstein polynomials of tensor product kind in Section 2.2. The results in univariate case obtained for complex Favard-Sz´ asz-Mirakjan in Chapter 1 will be useful.

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First we present the following result. Theorem 2.3.1. Suppose 2 < R1 < ∞, 2 < R2 < ∞ and that f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C is bounded on [R1 , ∞) × [R2 , ∞) and analytic in P∞ P∞ DR1 × DR2 , i.e f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , for all |z1 | < R1 , |z2 | < R2 . (i) Let 1 ≤ r1 < R21 and 1 ≤ r2 < R22 . For all |z1 | ≤ r1 , |z2 | ≤ r2 and n, m ∈ N we have |Sn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ Cr1 ,r2 ,n,m (f ),

where

Cr1 ,r2 ,n,m (f ) =

∞ ∞ 6 XX |ck,j | · (j − 1)(2r1 )k (2r2j−1 ) · m j=0

+

k=0 ∞ X ∞ X

6 · n

k=0 j=0

|ck,j | · (k − 1)(2r1 )k−1 (2r2 )j .

(ii) Let k1 , k2 ∈ N be with k1 + k2 ≥ 1 and 1 ≤ r1 < r1∗ < R21 , 1 ≤ r2 < r2∗ < Then for all |z1 | ≤ r1 , |z2 | ≤ r2 and n, m ∈ N we have ∂ k1 +k2 S (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2 ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

R2 2 .

(k2 )! (k1 )! · , (r1∗ − r1 )k1 +1 (r2∗ − r2 )k2 +1

where Cr1∗ ,r2∗ ,n,m (f ) is given at the above point (i).

(i) Denote ek,j (z1 , z2 ) = ek (z1 ) · ej (z2 ), where ek (u) = uk . Clearly we get ∞ X ∞ X |Sn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ |ck,j | · |Sn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )|.

Proof.

k=0 j=0

But taking into account the estimates in the proof of Theorem 1.8.4, (i), for all |z1 | ≤ r1 and |z2 | ≤ r2 we obtain |Sn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )| = |Sn (ek )(z1 ) · Sm (ej )(z2 ) − z1k · z2j |

≤ |Sn (ek )(z1 ) · Sm (ej )(z2 ) − Sn (ek )(z1 )z2j | + |Sn (ek )(z1 )z2j − z1k z2j |

≤ |Sn (ek )(z1 )| · |Sm (ej )(z2 ) − z2j | + |z2j | · |Sn (ek )(z1 ) − z1k | 6(j − 1) 6(k − 1) ≤ (2r1 )k · (2r2 )j−1 + r2j · (2r1 )k−1 , m n which immediately implies the estimate in (i). (ii) Let 1 ≤ r1 < r1∗ < R21 , 1 ≤ r2 < r2∗ < R22 . By the Cauchy’s formula in Theorem 2.1.2, (ii), we obtain ∂ k1 +k2 Sn,m (f )

(z1 , z2 ) −

∂ k1 +k2 f

(z1 , z2 ) ∂z1k1 ∂z2k2 Z Z [Sn,m (u1 , u2 ) − f (u1 , u2 )]du1 du2 (k1 !)(k2 !) = . ∗ ∗ (2πi)2 (u1 − z1 )k1 +1 (u2 − z2 )k2 +1 |u2 −z2 |=r2 |u1 −z1 |=r1 ∂z1k1 ∂z2k2

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Passing to absolute value with |z1 | ≤ r1 , |z2 | ≤ r2 and taking into account that |u1 − z1 | ≥ r1∗ − r1 , |u2 − z2 | ≥ r2∗ − r2 , by applying the estimate in (i) we easily obtain ∂ k1 +k2 S (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2 ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

which proves the theorem.

(r1∗

(k2 )! (k1 )! · ∗ , k +1 1 − r1 ) (r2 − r2 )k2 +1

In what follows a Voronovskaja-type formula for Sn,m (f ) is presented. It will be the product of the parametric extensions generated by the Voronovskaja’s formula in univariate case in Theorem 1.8.5. Indeed, for f (z1 , z2 ) defining the parametric extensions of the Voronovskaja’s formula by z1 Ln (f )(z1 , z2 )

:= Sn (f (·, z2 ))(z1 ) − f (z1 , z2 ) −

z1 ∂ 2 f · (z1 , z2 ), 2n ∂z12

z2 Lm (f )(z1 , z2 )

:= Sm (f (z1 , ·))(z2 ) − f (z1 , z2 ) −

z2 ∂ 2 f · (z1 , z2 ), 2m ∂z22

their product (composition) give ◦ z1 Ln (f )(z1 , z2 ) z1 ∂ 2 f = Sm Sn (f (·, ·))(z1 ) − f (z1 , ·) − · 2 (z1 , ·) (z2 ) 2n ∂z1 z1 ∂ 2 f z2 − Sn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − · 2 (z1 , z2 ) − 2n ∂z1 2m ∂2f ∂2f z1 ∂ 2 ∂ 2 f · Sn ( 2 (·, z2 ))(z1 ) − 2 (z1 , z2 ) − · 2 (z , z ) 1 2 ∂z2 ∂z2 2n ∂z1 ∂z22 := E1 − E2 − E3 . z2 Lm (f )(z1 , z2 )

After simple calculation evidently that we can write z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 )

2 ∂ f z1 · Sm (z , ·) (z2 ) 1 2n ∂z12 z1 ∂ 2 f (z1 , z2 ) −Sn (f (·, z2 ))(z1 ) + f (z1 , z2 ) + 2n ∂z12 2 z2 ∂ f z2 ∂ 2 f − · Sn · (·, z ) (z ) + (z1 , z2 ) 2 1 2m ∂z22 2m ∂z22 ∂4f z1 z2 · · 2 2 (z1 , z2 ), + 2n 2m ∂z1 ∂z2

= Sn,m (f )(z1 , z2 ) − Sm (f (z1 , ·))(z2 ) −

from which immediately can be derived the commutativity property z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 ) = z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 ).

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The Voronovskaja’s result can be stated as follows. Theorem 2.3.2. Suppose 2 < R1 < ∞, 2 < R2 < ∞ and that f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C is bounded on [R1 , ∞) × [R2 , ∞) and analytic in P∞ P k j DR1 × DR2 , i.e f (z1 , z2 ) = ∞ j=0 ck,j z1 z2 , for all |z1 | < R1 , |z2 | < R2 . k=0 For all |z1 | ≤ r1 , |z2 | ≤ r2 with 1 ≤ r1 < R21 , 1 ≤ r2 < R22 and n, m ∈ N we have 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ Cr1 ,r2 (f ) 2 + 2 , n m where Cr1 ,r2 (f ) = max

∞ X ∞ X

k=2 j=2

 ∞ X ∞ X 

k=2 j=0

|ck,j |(2r2 )j 26r1 (2r1 )k−3 (k − 1)2 (k − 2) ,

|ck,j |j(j − 1)r2j−1 · 13r1 (2r1 )k−3 (k − 1)2 (k − 2)

  

.

P∞ k Proof. First by hypothesis we can write f (z1 , z2 ) = k=0 fk (z2 )z1 , where 2 P∞ P ∞ j k−2 It follows ∂∂zf2 (z1 , z2 ) = fk (z2 ) = and j=0 ck,j z2 . k=2 fk (z2 )k(k − 1)z1 1 P P 2 2 2 ∞ ∂ fk ∞ j−2 ∂ f ∂ fk k (z1 , z2 ) = . This k=0 ∂z22 (z2 )z1 , where ∂z22 (z2 ) = j=2 ck,j j(j − 1)z2 ∂z22 P∞ k implies Sn (f (·, z2 ))(z1 ) = k=0 fk (z2 )Sn (e1 )(z1 ) and z1 ∂ 2 f · (z1 , z2 ) Sn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − 2n ∂z12 # " ∞ k−1 X z k(k − 1) . = fk (z2 ) Sn (ek1 )(z1 ) − ek1 (z1 ) − 1 2n k=2

Applying now Sm to the last expression with respect to z2 , we obtain # " ∞ X z1k−1 k(k − 1) k k E1 = Sm (fk )(z2 ) Sn (e1 )(z1 ) − e1 (z1 ) − 2n k=2  " # ∞ ∞ k−1 X X z k(k − 1) j k k 1  = ck,j Sm (e1 )(z2 ) Sn (e1 )(z1 ) − e1 (z1 ) − . 2n j=0 k=2

Passing now to absolute value with |z1 | ≤ r1 and |z2 | ≤ r2 and taking into account the estimates in the proofs of Theorem 1.8.4, (i) and Theorem 1.8.5, it follows ∞ X ∞ X 26r1 (2r1 )k−3 (k − 1)2 (k − 2) |E1 | ≤ |ck,j |(2r2 )j . n2 j=0 k=2

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Similarly,

26r1 (2r1 )k−3 (k − 1)2 (k − 2) |fk (z2 )| · |E2 | ≤ n2 k=2 ∞ X ∞ X 26r1 (2r1 )k−3 (k − 1)2 (k − 2) ≤ |ck,j |r2j . n2 j=0 ∞ X

k=2

Then

Sn =

∞ X ∂ 2 fk k=0

and

∂2f (·, z2 ) (z1 ) ∂z22

∂z22

(z2 )Sn (ek1 )(z1 ) =

∞ X ∞ X k=0 j=2

ck,j j(j − 1)z2j−2 Sn (ek1 )(z1 ),

∂2f ∂ 2f z1 ∂ 2 ∂ 2 f Sn ( 2 (·, z2 ))(z1 ) − 2 (z1 , z2 ) − (z1 , z2 ) · ∂z2 ∂z2 2n ∂z12 ∂z22 # " ∞ X ∞ k−1 X z k(k − 1) j−2 , = ck,j j(j − 1)z2 Sn (ek1 )(z1 ) − ek1 (z1 ) − 1 2n j=2 k=2

which again by Theorem 1.8.5 implies ∞ ∞ 26r1 (2r1 )k−3 (k − 1)2 (k − 2) r2 X X |E3 | ≤ |ck,j |j(j − 1)r2j−2 · . 2m n2 j=2 k=2

Note that if we estimate now |z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 )|, then by reasons of symmetry we get a similar order of approximation, simply interchanging above the places of n and m. In conclusion, 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ |E1 | + |E2 | + |E3 | ≤ Cr1 ,r2 (f ) 2 + 2 , n m with Cr1 ,r2 (f ) given by the statement.

Theorem 2.3.2 will be used to find the exact order in approximation by Bn,n (f ). In this sense we have the following. Theorem 2.3.3. Suppose 2 < R1 < ∞, 2 < R2 < ∞ and that f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C is bounded on [R1 , ∞) × [R2 , ∞) and analytic in P∞ P∞ k j DR1 × DR2 , i.e f (z1 , z2 ) = j=0 ck,j z1 z2 , for all |z1 | < R1 , |z2 | < R2 . k=0 Denoting kf kr1 ,r2 = sup{|f (z1 , z2 )|; |z1 | ≤ r1 ; |z2 | ≤ r2 }, if f is not a solution of the complex partial differential equation z1 ·

∂2f ∂2f (z1 , z2 ) + z2 · 2 (z1 , z2 ) = 0, |z1 | < R1 , |z2 | < R2 , 2 ∂z1 ∂z2

then we have kSn,n (f ) − f kr1 ,r2 ≥

Kr1 ,r2 ,f , for all n ∈ N, n

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where Kr1 ,r2 ,f is independent on n. Proof.

We can write Sn,n (f )(z1 , z2 ) − f (z1 , z2 ) 2 z1 ∂ 2 f z2 ∂ 2 f = · (z1 , z2 ) + · (z1 , z2 ) 2 n 4 ∂z1 4 ∂z22 2 n2 (z Ln (f ) ◦ z1 Ln (f )) (z1 , z2 ) + Rn (f )(z1 , z2 ) , + n 4 2

where

Rn (f )(z1 , z2 ) z2 ∂ 2 f n Sn (f (z1 , ·))(z2 ) − f (z1 , z2 ) − · (z1 , z2 ) = 2 2n ∂z22 z1 ∂ 2 f n Sn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − · (z1 , z2 ) + 2 2n ∂z12 2 ∂ f ∂2f z1 Sn (z , ·) (z ) − (z1 , z2 ) + 1 2 4 ∂z 2 ∂z12 21 ∂2f ∂ f z2 (·, z2 ) (z1 ) − 2 (z1 , z2 ) Sn + 4 ∂z22 ∂z2 z1 z2 ∂ 4f − · (z1 , z2 ). 8n ∂z12 ∂z22 By using Theorem 1.8.5 and the reasonings in the above Theorem 2.3.2, it is immediate that kRn (f )kr1 ,r2 → 0 as n → ∞. Also, by Theorem 2.3.2 we obtain Cr ,r (f ) n2 k(z2 Lm (f ) ◦ z1 Ln (f ))kr1 ,r2 ≤ 1 2 , 4 2 which implies

2

2 n

→ 0, as n → ∞.

n 4 (z2 Ln (f ) ◦ z1 Ln (f )) + Rn (f ) r1 ,r2

Denoting H(z1 , z2 ) = inequalities it follows

z1 4

·

∂2f (z1 , z2 ) ∂z12

+

z2 4

·

∂2f (z1 , z2 ) ∂z22

and taking into account the

kF + Gkr1 ,r2 ≥ | kF kr1 ,r2 − kGkr1 ,r2 | ≥ kF kr1 ,r2 − kGkr1 ,r2 , kSn,n (f ) − f kr1 ,r2 ) (

2

2 n 2

≥ (z Ln (f ) ◦ z1 Ln (f )) + Rn (f ) kHkr1 ,r2 − n n 4 2 r1 ,r2

2 1 1 · kHkr1 ,r2 = · kHkr1 ,r2 , n 2 n for all n ≥ n0 , with n0 depending only on f , r1 and r2 . We used here that by hypothesis we have kHkr1 ,r2 > 0. For n ∈ {1, 2, ..., n0−1} we reason exactly as for the complex univariate Bernstein polynomials in the proof of Theorem 1.1.4. ≥

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Combining now Theorem 2.3.2 with Theorem 2.3.3 we immediately obtain the following. Corollary 2.3.4. Suppose 2 < R1 < ∞, 2 < R2 < ∞ and that f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C is bounded on [R1 , ∞) × [R2 , ∞) and analytic in P∞ P∞ DR1 × DR2 , i.e f (z1 , z2 ) = k=0 j=0 ck,j z1k z2j , for all |z1 | < R1 , |z2 | < R2 . If the Taylor series of f contains at least one term of the form ck,0 z1k with ck,0 6= 0 and k ≥ 2 or of the form c0,j z2j with c0,j 6= 0 and j ≥ 2, then we have 1 kSn,n (f ) − f kr1 ,r2 ∼ , for all n ∈ N. n Proof. It suffices to prove that under the hypothesis on f , it cannot be a solution of the complex partial differential equation ∂2f ∂2f z1 · 2 (z1 , z2 ) + z2 · 2 (z1 , z2 ) = 0, |z1 | < R1 , |z2 | < R2 . ∂z1 ∂z2 Indeed, suppose the contrary. Since simple calculation give ∞ X ∞ ∞ X ∞ X X ∂2f k−1 j z1 · = ck,j k(k − 1)z1 z2 = ck+1,j k(k + 1)z1k z2j ∂z12 j=0 j=0 =

k=0 ∞ X

k=1

ck+1,0 k(k + 1)z1k +

z2 ·

ck+1,j k(k + 1)z1k z2j ,

k=1 j=1

k=1

and

∞ X ∞ X

∞

∞

∞

XX X ∂2f j ck+1,j j(j + 1)z1k z2j , c j(j + 1)z + = 0,j+1 2 ∂z22 j=1 j=1 k=1

by summing and equaling with zero, by the identification of coefficients, we immediately obtain that c0,k = cj,0 = 0 for all j = 2, 3, ..., k = 2, 3, ..., which contradicts the hypothesis on f . Therefore the hypothesis and the lower estimate in Theorem 2.3.3 holds, which ends the proof. 2.4

Baskakov Operators

In this section we consider the bivariate complex Baskakov operators of tensor product kind given by ∞ X ∞ X n(n + 1)...(n + k − 1) m(m + 1)...(m + j − 1) Vn,m (f )(z1 , z2 ) = · nk mj j=0 k=0

· [0, 1/n, ..., k/n; [0, 1/n, ..., j/m; f (·, ·)]z2 ]z1 z1k z2j ,

where f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C has all the partial derivatives bounded in [0, ∞) × [0, ∞), by the same constant, is supposed to be analytic in DR1 × DR2 and satisfies some exponential growth conditions.

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Note that in the case of two real variables, these operators were first considered by Stancu [184]. We will use similar methods with those for complex bivariate Bernstein polynomials of tensor product kind in Section 2.2. The results in univariate case obtained for complex Baskakov operators in Chapter 1 will be useful. Theorem 2.4.1. Let n0 , m0 ∈ N and 3 ≤ n0 < 2R1 < ∞, 3 ≤ m0 < 2R2 < ∞. Suppose that f : ([R1 , +∞) ∪ DR1 ) × ([R2 , +∞) ∪ DR2 ) → C has all the partial derivatives bounded in [0, ∞) × [0, ∞) by the same constant, f is supposed to be P∞ P∞ k j analytic in DR1 × DR2 , i.e. f (z1 , z2 ) = all |z1 | < R1 , 1 z2 , for k=0 j=0 ck,j z 1 |z2 | < R2 and there exist the constants M > 0, Ai ∈ Ri , 1 , i = 1, 2, with Ak Aj

1 2 , for all k, j = 0, 1, 2, ..., (which implies |f (z1 , z2 )| ≤ the property |ck,j | ≤ M k!j! M eA1 |z1 |+A2 |z2 | , for all |z1 | < R1 , |z2 | < R2 ). (i) Let 1 ≤ r1 < min{ n20 , A11 }, 1 ≤ r2 < min{ m20 , A12 } be arbitrary fixed. For all |z1 | ≤ r1 , |z2 | ≤ r2 , n > n0 and m > m0 we have |Vn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ Cr1 ,r2 ,n,m (f ), where ∞ ∞ (r2 A2 )j 6M X X (r1 A1 )k (k + 1)(k − 1) Cr1 ,r2 ,n,m (f ) = n j! j=0

+

k=0 ∞ X ∞ X

6M m

k=0 j=0

(r1 A1 )k (r2 A2 )j (j + 1)(j − 1).

(ii) Let k1 , k2 ∈ N be with k1 + k2 ≥ 1 and 1 ≤ r1 < r1∗ < min{ n20 , A11 }, 1 ≤ r2 < r2∗ < min{ m20 , A12 } be arbitrary fixed. Then for all |z1 | ≤ r1 , |z2 | ≤ r2 , n > n0 and m > m0 we have ∂ k1 +k2 V (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2

(k2 )! (k1 )! · , (r1∗ − r1 )k1 +1 (r2∗ − r2 )k2 +1 where Cr1∗ ,r2∗ ,n,m (f ) is given as at the above point (i). ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

(i) Denote ek,j (z1 , z2 ) = ek (z1 ) · ej (z2 ), where ek (u) = uk . Clearly we get ∞ X ∞ X |Vn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ |ck,j | · |Vn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )|.

Proof.

k=0 j=0

But taking into account the estimates in the proof of Theorem 1.9.1, (i), for all |z1 | ≤ r1 and |z2 | ≤ r2 we obtain |Vn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )| = |Vn (ek )(z1 ) · Vm (ej )(z2 ) − z1k · z2j | ≤ |Vn (ek )(z1 ) · Vm (ej )(z2 ) − Vn (ek )(z1 )z2j | + |Vn (ek )(z1 )z2j − z1k z2j | ≤ |Vn (ek )(z1 )| · |Vm (ej )(z2 ) − z2j | + |z2j | · |Vn (ek )(z1 ) − z1k | ≤ r1k (k + 1)!

6r2j (j + 1)!(j − 1) 6rk (k + 1)!(k − 1) + r2j 1 , m n

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which from the conditions on the coefficients ck,j implies ∞ X ∞ X |Vn,m (f )(z1 , z2 ) − f (z1 , z2 )| ≤ |ck,j | · |Vn,m (ek,j )(z1 , z2 ) − ek,j (z1 , z2 )| ≤

k=0 j=0 ∞ X ∞ X

6M n

+

(r1 A1 )k

k=0 j=0 ∞ X ∞ X

6M m

k=0 j=0

(r2 A2 )j (k + 1)(k − 1) j!

(r1 A1 )k (r2 A2 )j (j + 1)(j − 1),

which proves (i). (ii) Let 1 ≤ r1 < r1∗ < R1 , 1 ≤ r2 < r2∗ < R2 . By the Cauchy’s formula in Theorem 2.1.2, (ii), we get ∂ k1 +k2 f ∂ k1 +k2 Vn,m (f ) (z , z ) − (z1 , z2 ) 1 2 ∂z1k1 ∂z2k2 ∂z1k1 ∂z2k2 Z Z (k1 !)(k2 !) [Vn,m (u1 , u2 ) − f (u1 , u2 )]du1 du2 = . 2 (2πi) (u1 − z1 )k1 +1 (u2 − z2 )k2 +1 |u2 −z2 |=r2∗ |u1 −z1 |=r1∗ Passing to absolute value with |z1 | ≤ r1 , |z2 | ≤ r2 and taking into account that |u1 − z1 | ≥ r1∗ − r1 , |u2 − z2 | ≥ r2∗ − r2 , by applying the estimate in (i) we easily obtain ∂ k1 +k2 V (f ) ∂ k1 +k2 f n,m (z , z ) − (z , z ) 1 2 1 2 k1 k2 ∂z1k1 ∂z2k2 ∂z1 ∂z2 ≤ Cr1∗ ,r2∗ ,n,m (f ) ·

which proves the theorem.

(k1 )! (k2 )! · , (r1∗ − r1 )k1 +1 (r2∗ − r2 )k2 +1

In what follows a Voronovskaja-type formula for Vn,m (f ) is presented. It will be the product of the parametric extensions generated by the Voronovskaja’s formula in univariate case in Theorem 1.9.3. Indeed, for f (z1 , z2 ) defining the parametric extensions of the Voronovskaja’s formula by z1 (1 + z1 ) ∂ 2 f · 2 (z1 , z2 ), z1 Ln (f )(z1 , z2 ) := Vn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − 2n ∂z1 z2 (1 + z2 ) ∂ 2 f · 2 (z1 , z2 ), z2 Lm (f )(z1 , z2 ) := Vm (f (z1 , ·))(z2 ) − f (z1 , z2 ) − 2m ∂z2 their product (composition) give z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 ) z1 (1 + z1 ) ∂ 2 f = Vm Vn (f (·, ·))(z1 ) − f (z1 , ·) − · 2 (z1 , ·) (z2 ) 2n ∂z1 z1 (1 + z1 ) ∂ 2 f z2 (1 + z2 ) · − Vn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − (z , z ) − 1 2 2n ∂z12 2m 2 2 2 2 ∂ f ∂ f z1 (1 + z1 ) ∂ ∂ f · Vn ( 2 (·, z2 ))(z1 ) − 2 (z1 , z2 ) − · 2 (z , z ) 1 2 ∂z2 ∂z2 2n ∂z1 ∂z22 := E1 − E2 − E3 .

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After simple calculation evidently that we can write z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 )

2 z1 (1 + z1 ) ∂ f (z , ·) · Vm (z2 ) 1 2n ∂z12 z1 (1 + z1 ) ∂ 2 f −Vn (f (·, z2 ))(z1 ) + f (z1 , z2 ) + (z1 , z2 ) 2n ∂z12 2 ∂ f z2 (1 + z2 ) ∂ 2 f z2 (1 + z2 ) · Vn (·, z ) (z1 ) + · (z1 , z2 ) − 2 2 2m ∂z2 2m ∂z22 z1 (1 + z1 ) z2 (1 + z2 ) ∂4f + · · 2 2 (z1 , z2 ), 2n 2m ∂z1 ∂z2 from which immediately can be derived the commutativity property = Vn,m (f )(z1 , z2 ) − Vm (f (z1 , ·))(z2 ) −

z2 Lm (f )(z1 , z2 )

◦ z1 Ln (f )(z1 , z2 ) = z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 ).

The Voronovskaja’s result can be stated as follows. Theorem 2.4.2. Suppose that the hypothesis on the function f and the constants n0 , m0 , R1 , R2 , M , A1 , A2 in the statement of Theorem 2.4.1 hold. Let 1 ≤ r1 < min{ n20 , A11 }, 1 ≤ r2 < min{ m20 , A12 } be arbitrary fixed. Then for all |z1 | ≤ r1 , |z2 | ≤ r2 , n > n0 and m > m0 we have 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ Cr1 ,r2 (f ) 2 + 2 , n m where ∞ X ∞ X (r1 A1 )k (r2 A2 )j (j + 1)(k − 1)(k − 2)2 . Cr1 ,r2 (f ) = 16M k=2 j=0

P∞ k Proof. First by hypothesis we can write f (z1 , z2 ) = k=0 fk (z2 )z1 , where P P∞ 2 ∞ j ∂ f k−2 and fk (z2 ) = It follows ∂z2 (z1 , z2 ) = k=2 fk (z2 )k(k − 1)z1 j=0 ck,j z2 . 1 P∞ ∂ 2 fk 2 P ∞ j−2 ∂2f ∂ f k k (z1 , z2 ) = . This k=0 ∂z22 (z2 )z1 , where ∂z22 (z2 ) = j=2 ck,j j(j − 1)z2 ∂z22 P∞ implies Vn (f (·, z2 ))(z1 ) = k=0 fk (z2 )Vn (ek1 )(z1 ) and z1 (1 + z1 ) ∂ 2 f · (z1 , z2 ) Vn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − 2n ∂z12 " # ∞ X z1k−1 (1 + z1 )k(k − 1) k k = fk (z2 ) Vn (e1 )(z1 ) − e1 (z1 ) − . 2n k=2

Applying now Vm to the last expression with respect to z2 , we obtain " # ∞ X z1k−1 (1 + z1 )k(k − 1) k k E1 = Vm (fk )(z2 ) Vn (e1 )(z1 ) − e1 (z1 ) − 2n k=2  " # ∞ ∞ k−1 X X z (1 + z )k(k − 1) 1 j  = ck,j Vm (e1 )(z2 ) Vn (ek1 )(z1 ) − ek1 (z1 ) − 1 . 2n j=0 k=2

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Passing now to absolute value with |z1 | ≤ r1 and |z2 | ≤ r2 and taking into account the estimates in the proofs of Theorem 1.9.1, (i) and Theorem 1.9.3, it follows ∞ X ∞ X 16r1k k!(k − 1)(k − 2)2 |E1 | ≤ |ck,j |r2j (j + 1)! n2 j=0 k=2

≤ Similarly,

∞ ∞ 16M X X [(r1 A1 )k (r2 A2 )j (j + 1)(k − 1)(k − 2)2 ]. n2 j=0 k=2

|E2 | ≤ ≤ ≤ Then Vn =

k=2

|fk (z2 )| ·

16r1k k!(k − 1)(k − 2)2 n2

∂z22

∞ ∞ 16 X X |ck,j |r2j r1k k!(k − 1)(k − 2)2 2 n j=0 k=2 ∞ X ∞ X

16M n2

(r1 A1 )k

k=2 j=0

∂2f (·, z ) (z1 ) 2 ∂z22

∞ X ∂ 2 fk

k=0

and

∞ X

(z2 )Vn (ek1 )(z1 ) =

(r2 A2 )j (k − 1)(k − 2)2 . j!

∞ X ∞ X k=0 j=2

ck,j j(j − 1)z2j−2 Vn (ek1 )(z1 ),

∂2f ∂2f z1 (1 + z1 ) ∂ 2 ∂ 2 f Vn ( 2 (·, z2 ))(z1 ) − 2 (z1 , z2 ) − (z , z ) · 2 1 2 ∂z2 ∂z2 2n ∂z1 ∂z22 # " ∞ X ∞ X z1k−1 (1 + z1 )k(k − 1) j−2 k k = ck,j j(j − 1)z2 , Vn (e1 )(z1 ) − e1 (z1 ) − 2n j=2 k=2

which again by Theorem 1.9.3, implies ∞ ∞ r2 (1 + r2 ) X X 16r1k k!(k − 1)(k − 2)2 j−2 |E3 | ≤ |ck,j |j(j − 1)r2 · 2m n2 j=2 ≤

k=2 ∞ X ∞ X

8(1 + r2 )M r 2 n2 m

(r1 A1 )k

k=2 j=2

(r2 A2 )j (k − 1)(k − 2)2 . (j − 2)!

Note that if we estimate now |z1 Ln (f )(z1 , z2 ) ◦ z2 Lm (f )(z1 , z2 )|, then by reasons of symmetry we get a similar order of approximation, simply interchanging above the places of n and m. In conclusion, 1 1 |z2 Lm (f )(z1 , z2 ) ◦ z1 Ln (f )(z1 , z2 )| ≤ |E1 | + |E2 | + |E3 | ≤ Cr1 ,r2 (f ) 2 + 2 , n m with Cr1 ,r2 (f ) given by the statement.

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Theorem 2.4.2 will be used to find the exact order in approximation by Vn,n (f ). In this sense we have the following. Theorem 2.4.3. Suppose that n0 = m0 and that the hypothesis on the function f and the constants n0 , m0 , R1 , R2 , M , A1 , A2 in the statement of Theorem 2.4.1 hold. Let 1 ≤ r1 < min{ n20 , A11 }, 1 ≤ r2 < min{ n20 , A12 } be arbitrary fixed. Denoting kf kr1 ,r2 = sup{|f (z1 , z2 )|; |z1 | ≤ r1 ; |z2 | ≤ r2 }, if f is not a solution of the complex partial differential equation z1 (1 + z1 ) ·

∂2f ∂2f (z1 , z2 ) + z2 (1 + z2 ) · 2 (z1 , z2 ) = 0, |z1 | < R1 , |z2 | < |R2 |, 2 ∂z1 ∂z2

then for all n > n0 we have kVn,n (f ) − f kr1 ,r2 ≥

Kr1 ,r2 ,f , n

where Kr1 ,r2 ,f is independent on n and m. Proof.

where

We can write Vn,n (f )(z1 , z2 ) − f (z1 , z2 ) z2 (1 + z2 ) ∂ 2 f 2 z1 (1 + z1 ) ∂ 2 f · (z , z ) + · 2 (z1 , z2 ) = 1 2 n 4 ∂z12 4 ∂z2 2 2 n + (z Ln (f ) ◦ z1 Ln (f )) (z1 , z2 ) + Rn (f )(z1 , z2 ) , n 4 2 Rn (f )(z1 , z2 ) z2 (1 + z2 ) ∂ 2 f n Vn (f (z1 , ·))(z2 ) − f (z1 , z2 ) − = · 2 (z1 , z2 ) 2 2n ∂z2 z1 (1 + z1 ) ∂ 2 f n Vn (f (·, z2 ))(z1 ) − f (z1 , z2 ) − · (z , z ) + 1 2 2 2n ∂z12 2 ∂ f ∂2f z1 (1 + z1 ) Vn (z1 , ·) (z2 ) − 2 (z1 , z2 ) + 2 4 ∂z ∂z1 21 z2 (1 + z2 ) ∂ f ∂2f Vn + (·, z ) (z ) − (z , z ) 2 1 1 2 4 ∂z22 ∂z22 ∂4f z1 (1 + z1 )z2 (1 + z2 ) · 2 2 (z1 , z2 ). − 8n ∂z1 ∂z2

By using Theorem 1.9.3 and the reasonings in the above Theorem 2.4.2, it is immediate that kRn (f )kr1 ,r2 → 0 as n → ∞. Also, by Theorem 2.4.2 we obtain n2 Cr ,r (f ) k(z2 Lm (f ) ◦ z1 Ln (f ))kr1 ,r2 ≤ 1 2 , 4 2

which implies

2

2 n

n 4 (z2 Ln (f ) ◦ z1 Ln (f )) + Rn (f )

r1 ,r2

→ 0, as n → ∞.

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178

1) · Denoting H(z1 , z2 ) = z1 (1+z 4 account the inequalities

∂2f (z1 , z2 ) ∂z12

+

z2 (1+z2 ) 4

·

∂2f (z1 , z2 ) ∂z22

and taking into

kF + Gkr1 ,r2 ≥ | kF kr1 ,r2 − kGkr1 ,r2 | ≥ kF kr1 ,r2 − kGkr1 ,r2 , it follows kVn,n (f ) − f kr1 ,r2 ( )

2

2 n

2

≥ kHkr1 ,r2 − (z Ln (f ) ◦ z1 Ln (f )) + Rn (f ) n n 4 2 r1 ,r2

1 2 1 · kHkr1 ,r2 = · kHkr1 ,r2 , n 2 n for all n ≥ n0 , with n0 depending only on f , r1 and r2 . We used here that by hypothesis we have kHkr1 ,r2 > 0. For n ∈ {1, 2, ..., n0−1} we reason exactly as for the complex univariate Bernstein polynomials in the proof of Theorem 1.1.4. ≥

Combining Theorem 2.4.2 with Theorem 2.4.3 we immediately obtain the following. Corollary 2.4.4. Suppose that the hypothesis in the statement of Theorem 2.4.3 hold. If the Taylor series of f contains at least one term of the form c k,0 z1k with ck,0 6= 0 and k ≥ 2 or of the form c0,j z2j with c0,j 6= 0 and j ≥ 2, then we have 1 kVn,n (f ) − f kr1 ,r2 ∼ , for all n > n0 . n Proof. It suffices to prove that under the hypothesis on f , it cannot be a solution of the complex partial differential equation ∂2f ∂2f (z1 , z2 ) + z2 (1 + z2 ) · 2 (z1 , z2 ) = 0, |z1 | < R1 , |z2 | < R2 . 2 ∂z1 ∂z2 Indeed, suppose the contrary. Since simple calculation give ∞ ∞ X ∞ X X ∂2f 2c2,j z1 z2j + z1k z2j [ck+1,j k(k + 1) + ck,j k(k − 1)] z1 (1 + z1 ) · 2 = ∂z1 j=0 j=0 z1 (1 + z1 ) ·

k=2

and

z2 (1 + z2 ) ·

∞

∞

k=0

k=0

∞

X XX ∂2f k = 2c z z + z1k z2j [ck,j+1 j(j + 1) + ck,j j(j − 1)] k,2 2 1 ∂z22 j=2

by summing and equaling with zero, we obtain ∞ X 2c0,2 z2 + 2c1,2 z1 z2 + z1k z2 [ck+1,j k(k + 1) + ck,1 k(k − 1) + 2ck,2 ] k=2

+

∞ X j=2

z2j [c0,j+1 j(j + 1) + c0,j j(j − 1)] +

∞ X j=2

z1 z2j [c1,j+1 j(j + 1) + c1,j j(j − 1) + 2c2,j ]

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+ 2c2,0z1 + 2c2,1 z1 z2 +

∞ X

k=2

+

∞ X ∞ X k=2 j=2

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z1k [ck+1,0 k(k + 1) + ck,0 k(k − 1)]

z1k z2j [ck+1,j k(k + 1) + ck,j k(k − 1) + ck,j+1 j(j + 1) + ck,j j(j − 1)] = 0.

By the identification of coefficients, among others we immediately get c0,2 = c2,0 = 0 and from the terms under the second and fourth sign Σ, it follows c0,j+1 j(j + 1) + c0,j j(j − 1) = 0, ck+1,0 k(k + 1) + ck,0 k(k − 1) = 0, for all j = 2, 3, ..., k = 2, 3, ...,. From here by recurrence it is easy to deduce that c0,k = cj,0 = 0 for all j = 2, 3, ..., k = 2, 3, ..., which contradicts the hypothesis on f . Therefore the hypothesis and the lower estimate in Theorem 2.4.3 holds, which ends the proof. 2.5

Bibliographical Notes and Open Problems

All the results in Sections 2.2, 2.3 and 2.4 are new and appear for the first time here. Note 2.5.1. From a long list, some references concerning Bernstein-type operators of two or several real variables, different from those considered in this Chapter 2, for which would be possible to develop similar results, are : Stancu [177; 178; 179; 180; 181; 182], Stancu-Vernescu [183], Vlaic [196]. Open Problem 2.5.2. Suppose that f : P (0; R) → C is analytic in P (0; R) = P∞ P∞ k j DR1 × DR2 , that is f (z1 , z2 ) = j=0 ck,j z1 z2 , for all (z1 , z2 ) ∈ P (0; R), k=0 R = (R1 , R2 ). Prove the following Voronovskaja-type formula for n n−k X X nn − k Bn (f )(z1 , z2 ) = z1k z2j (1 − z1 − z2 )n−k−j f (k/n, j/n) : k j j=0 k=0

for all |z1 | ≤ r1 , |z2 | ≤ r2 with 1 ≤ r1 < R1 , 1 ≤ r2 < R2 and n ∈ N we have 2 2 Bn (f )(z1 , z2 ) − f (z1 , z2 ) − z1 (1 − z1 ) · ∂ f (z1 , z2 ) + z1 z2 · ∂ f (z1 , z2 ) 2 2n ∂z1 n ∂z1 ∂z2 2 Kr1 ,r2 (f ) z2 (1 − z2 ) ∂ f · . (z1 , z2 ) ≤ − 2n ∂z22 n2

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Chapter 3

Complex Convolutions

This chapter deals with approximation of complex analytic functions by linear and non-linear complex convolutions, of polynomial or of non-polynomial kind.

3.1

Linear Polynomial Convolutions

It is known that upper estimates in terms of the moduli of smoothness of various orders in approximation in the unit disk by the complex convolution polynomials of de la Vall´ee Poussin, Fej´er, Riesz-Zygmund, Jackson and Beatson were obtained in Gaier [75] (see also Gaier [76], p. 53) and Gal [93]. Since most of these details can be found in the recent book Gal [77], we will not reproduce here the proofs of the above mentioned results and the geometric properties of these complex convolutions. Also, the saturation order of approximation for the complex Fej´er, RieszZygmund and Rogosinski convolutions even in more general Jordain domains in C were obtained in e.g. Bruj-Schmieder [48]. In this section first we present upper estimates (of the same order as those given by the moduli of smoothness) in approximation in compact disks with explicit constants depending on the coefficients of Taylor’s series of the approximated function f for the complex convolution polynomials of de la Vall´ee Poussin, Fej´er, Riesz-Zygmund, Jackson and Beatson. Then, in addition, as new results we prove Voronovskaja-type formulas with quantitative estimates which allow us to derive the exact orders of approximation not only for the convolutions but also for their derivatives of any order in compact disks. Therefore, as in the case of complex Bernstein-type operators in Chapter 1, the exact degrees of approximation for the above mentioned complex convolution polynomials will be obtained by three steps : 1) upper estimates ; 2) quantitative Voronovskaja’s theorems ; 3) lower estimates by using step 2. An exception is the case of complex convolutions of Beatson-type when the exact order of approximation is obtained by a different method. The classical de la Vall´ee Poussin convolution polynomials of real variable attached to a 2π-periodic function g are defined by (see e.g. Butzer-Nessel [53], 181

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pp. 299-300)

Z π 1 g(t)Kn (x − t)dt, x ∈ R, 2π −π where the kernel Kn (u) is given by n X (n!)2 u 2n (n!)2 eiju , i2 = −1, u ∈ R. Kn (u) = 2 cos = (2n)! 2 (n − j)!(n + j)! j=−n Pn∗ (g)(x) =

Their complex form attached to an analytic function f in a disk DR = {z ∈ C, |z| < R} and given by Z π n 1 X 1 2n f (zeiu )Kn (u)du = 2n cj Pn (f )(z) = zj 2π −π n + j n j=0 =

n X j=0

cj

(n!)2 zj , (n − j)!(n + j)!

have nice shape-preserving properties, that is preserve the starlikeness and convexity of f in the unit disk, for all n ∈ N (see P´ olya-Schoenberg [150]). For this reason the de la Vall´ee Poussin convolutions are considered as the ”trigonometric” analogues of the classical Bernstein polynomials. Also, in Gal [93] as an approximation order √ of f by Pn (f ) was found ω1 (f ; 1/ n)∂D1 . But probably it would be possible to √ obtain the approximation order in terms of ω2 (f ; 1/ n)∂D1 . Firstly, below upper estimates in approximation by Pn (f )(z) with explicit constants depending on the coefficients of Taylor’s series of the approximated function f (z) are presented. Theorem 3.1.1. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that P∞ k f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z , for all z ∈ DR . (i) For any r ∈ [1, R) we have Mr (f ) , n ∈ N, kPn (f ) − f kr ≤ n P∞ where kf kr = sup{|f (z)|; |z| ≤ r} and Mr (f ) = k=1 |ck |k 2 rk < ∞. (ii) If 1 ≤ r < r1 < R and p ∈ N then we have r1 p!Mr1 (f ) , n ∈ N. kPn(p) (f ) − f (p) kr ≤ (r1 − r)p+1 n Proof. (i) Denote ek (z) = z k . Since Pn (e0 )(z) = 1 and obviously we have P∞ P∞ Pn (f )(z) = k=0 ck Pn (ek )(z), it follows that kPn (f )(z) − f (z)kr ≤ k=1 |ck | · kPn (ek )(z) − ek (z)kr . But by the formula of Pn (f )(z) in Introduction, since we can write ek (z) = P∞ j j=0 cj z with ck = 1 and cj = 0 for all j 6= k, it is immediate that Pn (ek )(z) = 0 if k > n and k Y k (n!)2 ek (z) = ek (z) 1− , if k ≤ n. Pn (ek )(z) = (n − k)!(n + k)! n+j j=1

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Complex Convolutions

This immediately implies that kPn (ek ) − ek kr ≤ rk ≤ nk rk ≤ for k ≤ n we get k Y k k r kPn (ek )(z) − ek kr ≤ 1 − 1− n + j j=1 k X k rk ≤ 1− 1− n + j j=1

183

k2 k n r

if k > n, while

k X

k 2 rk k 2 rk 1 ≤ ≤ . n+j n+1 n j=1 Pk Here we have applied the inequality 1−Πki=1 xi ≤ i=1 (1−xi ), valid for 0 ≤ xi ≤ 1, i = 1, ..., k. In conclusion we have k2 k kPn (ek ) − ek kr ≤ r , for all k, n ∈ N, n which implies the estimate in (i). (ii) Denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z 2πr1 p! Pn (f )(v) − f (v) Mr1 (f ) p! (p) (p) dv ≤ , |Pn (f )(z) − f (z)| = 2π γ (v − z)p+1 n 2π (r1 − r)p+1 ≤ krk

which proves (ii) and the theorem.

The Voronovskaja-type formula for the real de la Vall´ee Poussin convolutions Pn∗ (g)(x) attached to an integrable 2π-periodic function g admitting a derivative of second order at x is due to Natanson (see e.g. Natanson [144], Chapter 10, Section 3, Satz 3) and is given by αn g 00 (x) = , Pn∗ (g)(x) − g(x) − n n where αn → 0 as n → ∞. In what follows we present its analogue for the complex convolutions Pn (f )(z), which will be used in the proof of the exactness degree of approximation. Theorem 3.1.2. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that P∞ k f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z , for all z ∈ DR . For any r ∈ [1, R) we have

00 0

Pn (f ) − f + e2 f + e1 f ≤ Ar (f ) , n ∈ N

n n r n2 P∞ where Ar (f ) = k=1 |ck |k 4 rk < ∞. Proof.

Denote ek (z) = z k and z k k(k − 1) z k k + . |Ek,n (z)| = Pn (ek )(z) − ek (z) + n n

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184

For all |z| ≤ r we get 00 0 Pn (f )(z) − f (z) + e2 (z)f (z)(z) + e1 f (z) n n ∞ n ∞ X X X ≤ |ck | · |Ek,n (z)| = |ck | · |Ek,n (z)| + |ck | · |Ek,n (z)| =

k=0 n X

k=1

|ck | · |Ek,n (z)| +

k=1 ∞ X

k=n+1

k=n+1

z k k(k − 1) z k k + . |ck | −z k + n n

But for |z| ≤ r we have ∞ ∞ X X k z k k(k − 1) z k k k(k − 1) k k + + ≤ = |ck |r −1 + |ck | −z + n n n n k=n+1

k=n+1 ∞ X

k=n+1

|ck |rk

∞ ∞ ∞ X k k2 1 X k2 1 X |ck | rk ≤ = 2 |ck |k 3 rk ≤ 2 |ck |k 4 rk . n n n n n k=n+1

k=n+1

k=n+1

Therefore, it remains to estimate |Ek,n (z)| for |z| ≤ r and 0 ≤ k ≤ n. Since it is immediate that E0,n (z) = 0, it suffices to consider 1 ≤ k ≤ n. We obtain z k k(k − 1) kz k (n!)2 zk − zk + + |Ek,n (z)| = (n − k)!(n + k)! n n (n!)2 k2 = |z|k − 1 + . (n − k)!(n + k)! n In what follows we prove by mathematical induction that for all k = 1, 2, ..., n and n ∈ N we have 0≤

(n!)2 k2 k4 −1+ ≤ 2. (n − k)!(n + k)! n n

(3.1)

Since (3.1) is immediate for k = n, we may suppose that 1 ≤ k ≤ n − 1. First we can write (k + 1)2 (n!)2 −1+ (n − k − 1)!(n + k + 1)! n k2 n−k (n!)2 n−k k2 −1+ + = · 1− (n − k)!(n + k)! n n+k+1 n+k+1 n k2 2k + 1 n−k k2 (n!)2 − 1− + · = −1+ n n (n − k)!(n + k)! n n+k+1 k2 (2k + 1) 2k + 1 + 1− − + n n+k+1 n n−k k2 (2k + 1)(k 2 + k + 1) (n!)2 = · −1+ + . (n − k)!(n + k)! n n+k+1 n(n + k + 1)

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Now suppose that both inequalities in (3.1) are valid for k. From the above relationship first it follows that the left-hand side in (3.1) is valid for k + 1. Also, passing above to the upper estimate, we obtain (k + 1)2 k4 n−k (2k + 1)(k 2 + k + 1) (n!)2 −1+ ≤ 2· + (n − k − 1)!(n + k + 1)! n n n+k+1 n(n + k + 1) ≤

k4 (2k + 1)(k 2 + k + 1) k 4 + 2k 3 + 3k 2 + 3k + 1 (k + 1)4 + = ≤ , n2 n2 n2 n2 4

which proves that (3.1) is valid for k + 1 too and that kEk,n kr ≤ rk nk 2 , for all 1 ≤ k ≤ n and n ∈ N. Replacing this in the first inequality of the proof we obtain the theorem. By using Theorems 3.1.1 and 3.1.2 now we are in position to obtain the exact degree of approximation by Pn (f )(z) and its derivatives. The first main result is a lower estimate in Theorem 3.1.1, (i). Theorem 3.1.3. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that P∞ k f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z , for all z ∈ DR . If f is not a polynomial of degree 0, then for any r ∈ [1, R) we have Cr (f ) , n ∈ N, n where the constant Cr (f ) depends only on f and r. kPn (f ) − f kr ≥

Proof.

For all z ∈ DR and n ∈ N we have

Pn (f )(z) − f (z) zf 0 (z) 1 2 1 z2 −[z 2 f 00 (z) + zf 0 (z)] + n Pn (f )(z) − f (z) + f 00 (z) + . = n n n n

In what follows we will apply to this identity the following obvious property : kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . It follows kPn (f ) − f kr

1 1 2 e2 f 00 e1 f 0 00 0

. ≥ ke2 f + e1 f kr − n Pn (f ) − f + + n n n n r

Taking into account that by hypothesis f is not a polynomial of degree 0 in DR , we get ke2 f 00 + e1 f 0 kr > 0. Indeed, supposing the contrary it follows that z 2 f 00 (z) + zf 0 (z) = 0 for all z ∈ Dr . But it is easy to see (by using the form of f as a power series and by identifying the coefficients) that the only analytic solution of this differential equation is f 0 (z) = 0, for all z ∈ Dr , which contradicts the hypothesis. But by Theorem 3.1.2 we have

e1 e2

n2 Pn (f ) − f + f 00 + f 0 ≤ Ar (f ). n n r

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Therefore, there exists an index n0 depending only on f and r, such that for all n ≥ n0 we have i e1 1 h 2 e2

ke2 f 00 + e1 f 0 kr − n Pn (f ) − f + f 00 + f 0 n n n r 1 ke2 f 00 + e1 f 0 kr , 2

≥

which immediately implies kPn (f ) − f kr ≥

1 1 · ke2 f 00 + e1 f 0 kr , ∀n ≥ n0 . n 2 M

(f )

For n ∈ {1, ..., n0 − 1} we obviously have kPn (f ) − f kr ≥ r,n with Mr,n (f ) = n n · kPn (f ) − f kr > 0, which finally implies kPn (f ) − f kr ≥ Crn(f ) for all n, where Cr (f ) = min{Mr,1 (f ), ..., Mr,n0 −1 (f ), 12 ke2 f 00 + e1 f 0 kr }. This completes the proof. Combining now Theorem 3.1.3 with Theorem 3.1.1, (i) we immediately get the following. Corollary 3.1.4. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If f is not a polynomial of degree 0, then for any r ∈ [1, R) we have 1 kPn (f ) − f kr ∼ , n ∈ N, n where the constants in the equivalence depend on f and r. In the case of approximation by the derivatives of Pn (f )(z) the following result holds. Theorem 3.1.5. Let DR = {z ∈ C; |z| < R} be with R > 1 and let us suppose that P∞ f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR . Also, let 1 ≤ r < r1 < R and p ∈ N be fixed. If f is not a polynomial of degree ≤ p − 1, then we have 1 kPn(p) (f ) − f (p) kr ∼ , n where the constants in the equivalence depend on f , r, r1 and p. Proof. Denoting by Γ the circle of radius r1 and center 0 (where r1 > r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N we have Z Pn (f )(v) − f (v) p! (p) (p) dv, Pn (f )(z) − f (z) = 2πi Γ (v − z)p+1

where the inequality |v − z| ≥ r1 − r is valid for all |z| ≤ r and v ∈ Γ. Taking into account Theorem 3.1.1, (ii), it remains to prove the lower estimate (p) for kPn (f ) − f (p) kr . For this purpose, as in the proof of Theorem 3.1.3, for all v ∈ Γ and n ∈ N we have Pn (f )(v) − f (v)

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1 = n

187

v 2 00 1 2 vf 0 (v) −[v f (v) + vf (v)] + n Pn (f )(v) − f (v) + f (v) + , n n n 2 00

0

which replaced in the above Cauchy’s formula implies Z p! 1 −[v 2 f 00 (v) + vf 0 (v)] (p) (p) Pn (f )(z) − f (z) = dv n 2πi Γ (v − z)p+1  Z n2 Pn (f )(v) − f (v) + v2 f 00 (v) + v f 0 (v)  n n 1 p! dv + · p+1  n 2πi Γ (v − z) (p) 1 n 2 00 −z f (z) − zf 0 (z) = n  Z n2 Pn (f )(v) − f (v) + v2 f 00 (v) + v f 0 (v)  n n 1 p! dv . + ·  n 2πi Γ (v − z)p+1 Passing now to k · kr it follows

1 n

(p) kPn(p) (f ) − f (p) kr ≥

[e2 f 00 + e1 f 0 ] n r 

 Z n2 Pn (f )(v) − f (v) + v2 f 00 (v) + v f 0

n n p! 1

− dv

, p+1 n 2π Γ (v − z)

r

where by using Theorem 3.1.2 we obtain

Z n2 Pn (f )(v) − f (v) + v2 f 00 (v) + v f 0

p!

n n

dv

2π

p+1 (v − z) Γ

p! 2πr1 n2 e2 00 e1 0

≤ · f + f (f ) − f +

P n 2π (r1 − r)p+1 n n r1 Ar1 (f )p!r1 . ≤ (r1 − r)p+1

(p) But by hypothesis on f we have [e2 f 00 + e1 f 0 ] > 0. r

r

Indeed, supposing the contrary it follows that z 2 f 00 (z) + zf 0 (z) = Qp−1 (z), for Pp−1 j all z ∈ Dr , where Qp−1 (z) = j=1 Aj z necessarily is a polynomial of degree ≤ p − 1, vanishing at z = 0. Denoting f 0 (z) = g(z) the differential equation becomes z 2 g 0 (z) + zg(z) = Qp−1 (z), for all z ∈ Dr . Seeking now the analytic P j solution in the form g(z) = ∞ j=0 αj z , replacing in the differential equation, by the identification of coefficients we easily obtain that g(z) necessarily is a polynomial of degree ≤ p − 2, which will imply that f (z) necessarily is a polynomial of degree ≤ p − 1, in contradiction with the hypothesis. Finally, reasoning exactly as in the proof of Theorem 3.1.3, we immediately get the desired conclusion.

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An important trigonometric mean of real variable in approximation theory is the Fej´er mean Fn∗ (g)(x), defined as the arithmetic mean of the sequence of partial sums of the Fourier series of g and given by Z π 1 ∗ Fn (g)(x) = g(t)Kn (x − t)dt, x ∈ R, 2π −π 2 . Concerning this mean where the kernel Kn (u) is given by Kn (u) = n1 sin(nu/2) sin(u/2)

Fej´er [73] proved that if f : R → C is a continuous function with period 2π, then the sequence (Fn (g)(x))n converges uniformly to g on [−π, π]. P∞ Its complex form attached to an analytic function f (z) = k=0 ck z k in a disk DR = {z ∈ C, |z| < R} and given by Z π 1 Fn (f )(z) = f (zeiu )Kn (u)du 2π −π 2 Z π n−1 X n−k 1 sin(nu/2) iu = f (ze ) zk, du = ck 2πn −π sin(u/2) n k=0

(for the last formula see e.g. Gal [86], Theorem 3.1) also has nice approximation properties, satisfying the estimate (see Gaier [75], Theorem 1) Cr (f ) M kf 0 kr 1 := , for all n ∈ N, kFn (f ) − f kr ≤ M ω1 f ; ≤ n Dr n n where M > 0, 0 < r < R, kf 0 kr = sup{|f (z)|; |z| ≤ r} and ω1 f ; n1 Dr is the classical modulus of continuity of f in Dr . In fact, from the saturation result in Bruj-Schmieder [48], pp. 161-162, it follows that if f is not a constant then the approximation order by Fn (f ) is exactly n1 . The Fej´er means, both trigonometric and complex cases were generalized by Riesz-Zygmund, their complex form being defined for any s ∈ N by s n−1 X k Rn,s (f )(z) = ck 1 − z k , n ∈ N. n k=0

For s = 1 one recapture the Fej´er means and for s = 2 one get the means introduced by Riesz. From the saturation result in Bruj-Schmieder [48], pp. 161-162, it follows that if f (z) is not a constant then the approximation order by Rn,s (f )(z) is exactly n1s . In what follows we complete this result by proving that the approximation order by the derivatives of Rn,s (f ) also is exactly n1s . Useful in the proof will be the Voronovskaja’s formula for Rn,s (f )(z) with a quantitative estimate. It is worth noting that our method is different from that in Bruj-Schmieder [1]. Also, in addition we obtain here a Voronovskaja result with a quantitative upper estimate and the exact orders in simultaneous approximation by derivatives. First we obtain an upper estimate in approximation by the derivatives of Rn,s (f )(z).

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Theorem 3.1.6. (Gal [85]) Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If 1 ≤ r < r1 < R and s, p ∈ N then we have (p) kRn,s (f ) − f (p) kr ≤

where Cr1 ,s (f ) = Proof.

P∞

k=0

r1 p!Cr1 ,s (f ) , n ∈ N, (r1 − r)p+1 ns

|ck |k s rk < ∞.

An estimate of the form kRn,s (f ) − f kr ≤

Cr,s (f ) , n ∈ N, ns

essentially follows from Bruj-Schmieder [48]. Below we reprove this estimate in a different and simple way with an explicit constant Cr,s (f ). Thus, denoting ek (z) = P z k and writing f (z) = ∞ k=0 ck ek (z), for all |z| ≤ r we obtain |Rn,s (f )(z) − f (z)| ≤

=

∞ X

k=0 n−1 X

|ck | · |Rn,s (ek )(z) − ek (z)|

|ck | · |Rn,s (ek )(z) − ek (z)|

k=0 ∞ X

+ ≤

|ck | · |Rn,s (ek )(z) − ek (z)|

k=n n−1 X

1 ns

P∞

k=0

|ck |k s rk +

∞ X

k=n

|ck |rk .

Here we used that ek (z) = j=0 cj z j , with ck = 1 and cj = 0 for all j 6= k. s Taking into account that for k ≥ n we have 1 ≤ nk s , from the previous inequality we get n−1 ∞ ∞ 1 X 1 X 1 X s k s k |Rn,s (f )(z) − f (z)| ≤ s |ck |k r + s |ck |k r = s |ck |k s rk , n n n k=0

k=n

P∞

k=0

s k

for all |z| ≤ r, therefore we can take Cr,s (f ) = k=0 |ck |k r < ∞. Now, denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z p! Rn,s (f )(v) − f (v) (p) (p) dv |Rn,s (f )(z) − f (z)| = 2π γ (v − z)p+1 ≤

which proves the theorem.

Cr1 ,s (f ) p! 2πr1 · · , s n 2π (r1 − r)p+1

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Useful in the proof of the exact degree of approximation will be the Voronovskaja’s formula for Rn,s (f )(z). For this purpose, we need the following simple lemma. P Lemma 3.1.7. (Gal [85]) Let k, s ∈ N. If we denote k s = sj=1 αj,s k(k − 1)...(k − (j − 1)), then the coefficients αj,s can be chosen independent of k and to satisfy α1,s = αs,s = 1 for all s ≥ 1 and αj,s+1 = αj−1,s + jαj,s , j = 2, ..., s, s ≥ 2. Proof.

We have

k s+1 s s+1 X X αj,s k(k − 1)...(k − (j − 1)) αj,s+1 k(k − 1)...(k − (j − 1)) = k = j=1

j=1

=

s X j=1

+ + +

s X

j=1 s X

j=1 s X j=2

αj,s (k − j + j)k(k − 1)...(k − (j − 1)) = jαj,s k(k − 1)...(k − (j − 1)) =

s+1 X j=2

s X j=1

αj,s k(k − 1)...(k − j)

αj−1,s k(k − 1)...(k − (j − 1))

jαj,s k(k − 1)...(k − (j − 1)) = αs,s +

s X j=2

αj−1,s k(k − 1)...(k − (j − 1))

jαj,s k(k − 1)...(k − (j − 1)) + α1,s ,

which implies s+1 X

+

j=1 s X j=2

αj,s+1 k(k − 1)...(k − (j − 1)) = α1,s+1 αj,s+1 k(k − 1)...(k − (j − 1))

+ αs+1,s+1 = αs,s +

s X j=2

+

s X j=2

αj−1,s k(k − 1)...(k − (j − 1))

jαj,s k(k − 1)...(k − (j − 1)) + α1,s ,

and proves the lemma.

Now, the Voronovskaja-type formula for Rn,s (f )(z) one states as follows. Theorem 3.1.8. (Gal [85]) Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose P∞ that f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z k , for all z ∈ DR . For any r ∈ [1, R) we have

s X

Ar,s (f ) (j)

Rn,s (f ) − f + 1 α e f j,s j

≤ ns+1 , n ∈ N s n

j=1 r

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where Ar,s (f ) = 3.1.7. Proof.

P∞

k=1

191

|ck |k s+1 rk < ∞, ej (z) = z j and αj,s are defined by Lemma

Denoting

Ek,n,s (z) = Rn,s (ek )(z) − ek (z) +

s 1 X αj,s k(k − 1)...(k − (j − 1))z k , ns j=1

we obtain X s ∞ X j j Rn,s (f )(z) − f (z) + 1 αj,s z f (z) ≤ |ck | · |Ek,n,s (z)| ns j=1 k=0 =

n−1 X k=1

|ck | · |Ek,n,s (z)| +

∞ X

k=n

|ck | · |Ek,n,s (z)|

s s X 1 k αj,s k(k − 1)...(k − (j − 1)) · |z|k −1+ s = |ck | · 1 − n n j=1 k=1 ∞ s X 1 X + |ck | · −1 + s αj,s k(k − 1)...(k − (j − 1)) · |z|k n j=1 k=n ∞ s X X 1 |ck | · −1 + s = 0+ αj,s k(k − 1)...(k − (j − 1)) · |z|k , n j=1 k=n n−1 X

where for the first sum we used Lemma 3.1.7. Ps Taking into account that by Lemma 3.1.7 we have j=1 αj,s n(n − 1)...(n − (j − 1)) = ns , for all |z| ≤ r it follows s X j j Rn,s (f )(z) − f (z) + 1 α z f (z) j,s ns j=1   ! ∞ s X X k(k − 1)...(k − (j − 1))  ≤ |ck |rk ·  αj,s s n j=1 k=n

∞ ∞ ∞ 1 X 1 X 1 X s k s+1 k = s |ck |k r ≤ s+1 |ck |k r ≤ s+1 |ck |k s+1 rk , n n n k=n

which proves the theorem.

k=n

k=1

By using Theorems 3.1.6. and 3.1.8 we are in position to obtain the exact degree of approximation by the derivatives of Rn,s (f )(z). Theorem 3.1.9. (Gal [85]) Let DR = {z ∈ C; |z| < R} be with R > 1 and let us P∞ suppose that f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR .

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Also, let 1 ≤ r < r1 < R and p, s ∈ N be fixed. If f is not a polynomial of degree ≤ p − 1 then we have

1 , ns where the constants in the equivalence depend on f , r, r1 , s and p. (p) (f ) − f (p) kr ∼ kRn,s

Proof. Denoting by Γ the circle of radius r1 > and center 0 (where r1 > r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N we have Z p! Rn,s (f )(v) − f (v) (p) dv, Rn,s (f )(z) − f (p) (z) = 2πi Γ (v − z)p+1 where the inequality |v − z| ≥ r1 − r is valid for all |z| ≤ r and v ∈ Γ. Taking into account Theorem 3.1.6, it remains to prove the lower estimate for (p) kRn,s (f ) − f (p) kr . For this purpose, for all v ∈ Γ and n ∈ N we have  s 1  X Rn,s (f )(v) − f (v) = s − αj,s v j f (j) (v) n  j=1     s  X 1 1 + ns+1 Rn,s (f )(v) − f (v) + s αj,s v j f (j) (v) ,  n n j=1

which replaced in the above Cauchy’s formula implies ( Ps Z − j=1 αj,s v j f (j) (v) 1 p! (p) (p) Rn,s (f )(z) − f (z) = s dv n 2πi Γ (v − z)p+1  Z ns+1 Rn,s (f )(v) − f (v) + 1s Ps αj,s v j f (j) (v)  j=1 n 1 p! dv + · p+1  n 2πi Γ (v − z)  (p)  s  X 1 = s − αj,s v j f (j) (v) n   j=1  Z ns+1 Rn,s (f )(v) − f (v) + 1s Ps αj,s v j f (j) (v)  j=1 n 1 p! dv . + ·  n 2πi Γ (v − z)p+1 Passing now to k · kr it follows

  (p)

 s



X 1 

(p) (p) (j)  kRn,s (f ) − f kr ≥ s αj,s ej f

n   j=1

r 

 Z ns+1 Rn,s (f )(v) − f (v) + 1s Ps αj,s v j f (j) (v)

j=1 n 1 p!

− dv

, n 2π Γ (v − z)p+1

r

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where by using Theorem 3.1.8 we obtain

Z ns+1 Rn,s (f )(v) − f (v) + 1s Ps αj,s v j f (j) (v)

p!

j=1 n

dv

2π

p+1 (v − z) Γ

r

s s+1 X

p! 2πr1 n Ar1 ,s (f )p!r1

Rn,s (f ) − f + 1 ≤ · . αj,s ej f (j) ≤

p+1 s 2π (r1 − r) n j=1 (r1 − r)p+1

r1

h i(p)

Ps

(j)

> 0. But by hypothesis on f we have j=1 αj,s ej f

Ps r Indeed, supposing the contrary it follows that j=1 αj,s z j f (j) (z) = Qp−1 (z), Pp−1 for all z ∈ Dr , where Qp−1 (z) = j=1 Aj z j necessarily is a polynomial of degree ≤ p − 1. P∞ Seeking now the analytic solution in the form f (z) = k=0 βk z k , replacing in the differential equation and taking into account again Lemma 3.1.7, by identification of the coefficients βk we easily obtain βk = 0, for all k ≥ p, that is f (z) necessarily is a polynomial of degree ≤ p − 1, in contradiction with the hypothesis. Therefore, there exists an index n0 depending only on f , s, p, r and r1 , such that for all n ≥ n0 we have

 (p)

s

X

 (j)  αj,s ej f

j=1 r

Z ns+1 Rn,s (f )(v) − f (v) + 1s Ps αj,s v j f (j) (v)

j=1 n 1 p! dv −

p+1 n 2π Γ (v − z)

r



 (p)

s

1

X

≥  αj,s ej f (j)  ,

2 j=1

r

which immediately implies

 (p)

s

1 1  X

(j)  (p) (p) kRn,s (f ) − f kr ≥ s · αj,s ej f

,

n 2 j=1

r

for all n ≥ n0 . (p) M (f ) For n ∈ {1, ..., n0 − 1} we obviously have kRn,s (f ) − f (p) kr ≥ r,s,p,n with n (p) (p) (p) (p) Mr,s,,p,n (f ) = n · kRn,s (f ) − f kr > 0, which finally implies kRn,s (f ) − f kr ≥ Cr,s,p (f ) for all n, where n  

 (p)



 s 



1 X

αj,s ej f (j)  . Cr,s,p (f ) = min Mr,s,p,1 (f ), ..., Mr,s,p,n0 −1 (f ), 

  2 j=1 



r

This completes the proof.

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In what follows we will make similar considerations on the complex convolutions of Jackson and of Beatson type given by Z 1 π Jn (f )(z) = f (zeit )Kn,2 (t)dt, π −π

and

Pn,2,p (f )(z) = respectively, where p ∈ N, 3 Kn,2 (t) = 2n(2n2 + 1)

Bn,2,p (u) =

1 π

Z

sin(nt/2) sin(t/2)

n 2π

Z

π

f (zeiu )Bn,2,p (u)du, −π

4

, Bn,2,1 (u) =

n 2π

Z

u+π/n

Kn,2 (t)dt, u−π/n

u+π/n

Bn,2,p−1 (t)dt, p = 2, 3, .... u−π/n

First we present without proofs the following results. Theorem 3.1.10. (Gal [93], Theorem 3.1.) We have : P (i) Kn,2 (t) = 21 + 2n−2 k=1 λk,n cos(kt), where λk,n =

4n3 − 6k 2 n + 3k 3 − 3k + 2n , if 1 ≤ k ≤ n, 2n(2n2 + 1)

(k − 2n) − (k − 2n)3 , if n ≤ k ≤ 2n − 2. 2n(2n2 + 1) p P2n−2 n (ii) Bn,2,p (t) = 21 + k=1 kπ sin(kπ/n) λk,n cos(kt), p = 1, 2, .... P∞ (iii) If f (z) = k=0 ck z k is analytic in |z| < R with R > 1 then λk,n =

Jn (f )(z) = c0 +

2n−2 X

ck λk,n z k ,

k=1

Pn,2,p (f )(z) = c0 +

2n−2 X

ck

k=1

ip h n sin(kπ/n) λk,n z k . kπ

(iv) (Gal [86], p. 423) If f is analytic in D1 and continuous in D1 then for all |z| ≤ 1 and n ∈ N we have |Jn (f )(z) − f (z)| ≤ Cω2 (f ; 1/n)∂D1 , where C > 0 is an absolute constant and ω2 (f ; δ)∂D1 = sup{|∆2u f (eit )|; |t| ≤ π, |u| ≤ δ},

∆2u g(t) = g(x) − 2g(x + u) + g(x + 2u). (v) (Gal [93], Corollary 2.4.) If f is analytic in D1 and continuous in D1 then for all |z| ≤ 1 and n ∈ N we have |Pn,2,p (f )(z) − f (z)| ≤ Cp ω2 (f ; 1/n)∂D1 ,

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where Cp > 0 is an absolute constant independent of n, f and z. Remarks. 1) The proofs of all the results in Theorem 3.1.10 can also be found in the recent book Gal [77]. 2) Analysing the proofs of Theorem 3.1.10, (iv), (v), it easily follows that for |z| ≤ r with 1 ≤ r < R, the approximation errors can be expressed in terms of ω2 (f ; 1/n)∂Dr = sup{|∆2u f (reit )|; |t| ≤ π, |u| ≤ 1/n}, with constants in front of ω2 depending on r. 3) We have λk,n ≥ 0 for all 1 ≤ k ≤ 2n − 2. Indeed, first let us suppose that 1 ≤ k < j ≤ n. We get λj,n − λk,n = = = ≤ =

−6j 2 n + 3j 3 − 3j + 6k 2 n − 3k 3 + 3k 4n3 + 2n 2 2 −6n(j − k ) + 3(j 3 − k 3 ) − 3(j − k) 4n3 + 2n (j − k) [−6n(j + k) + 3j 2 + 3jk + 3k 2 − 3] 4n3 + 2n (j − k) [−6j(j + k) + 3j 2 + 3jk + 3k 2 − 3] 4n3 + 2n (j − k) [−3j 2 − 3jk + 3k 2 − 3] ≤ 0, 4n3 + 2n

which means that for 1 ≤ k < j ≤ n, the sequence λk,n is decreasing, that is 0≤

n3 − n 4(n3 − n) = λ ≤ λ ≤ λ = < 1, for all 1 ≤ k ≤ n. n,n k,n 1,n 4n3 + 2n 4n3 + 2n

Now, let us suppose that 2 ≤ n ≤ k < j ≤ 2n − 2. We get

(j − 2n) − (j − 2n)3 − (k − 2n) + (k − 2n)3 4n3 + 2n (j − k) = [1 − (j − 2n)2 − (j − 2n)(k − 2n) − (k − 2n)2 ] ≤ 0, 4n3 + 2n

λj,n − λk,n =

which for all 2 ≤ n ≤ k ≤ 2n − 2 implies 0≤

4n3

n3 − n 6 = λ2n−2,n ≤ λk,n ≤ λn,n = 3 < 1. + 2n 4n + 2n

An immediate consequence of Theorem 3.1.10, (iv), (v), of the above Remark 2 and of the Cauchy’s formula is the following. P∞ Corollary 3.1.11. Let f (z) = k=0 ck z k be analytic in |z| < R with R > 1 and 1 ≤ r < R. (i) For all |z| ≤ r and n ∈ N we have |Jn (f )(z) − f (z)| ≤

∞ Cr X |ck |k(k − 1)rk−2 , n2 k=2

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|Pn,2,p (f )(z) − f (z)| ≤

∞ Cp,r X |ck |k(k − 1)rk−2 . n2 k=2

(ii) If 1 ≤ r < r1 < R then for all |z| ≤ r and n, p, q ∈ N we have |Jn(q) (f )(z) − f (q) (z)| ≤

r1 q!Mr1 (f ) , (r1 − r)q+1 n2

(q)

|Pn,2,p (f )(z) − f (q) (z)| ≤ where Mr1 (f ) = Cr1

∞ X k=2

r1 q!Mr1 ,p (f ) , (r1 − r)q+1 n2

|ck |k(k − 1)r1k−2 and Mr1 ,p (f ) = Cp,r1

∞ X

k=2

|ck |k(k − 1)r1k−2 .

Proof. (i) Since ∆2u g(t) = 2u2 [t, t + u, t + 2u; g], by the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get |∆2u f (reit )| ≤ u2 kf 00 kr , where kf 00 kr = sup{|f 00 (z)|; |z| ≤ r} ≤

∞ X

k=2

|ck |k(k − 1)rk−2 ,

which immediately proves (i). (ii) Denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N, we have Z q! Jn (f )(v) − f (v) Mr1 (f ) q! 2πr1 (q) (q) |Jn (f )(z) − f (z)| = dv ≤ . 2π (v − z)q+1 n2 2π (r1 − r)q+1 γ

The reasoning in the case of lary.

(q) Pn,2,p (f )(z)

is similar, which proves (ii) and the corol

For the complex Jackson polynomials Jn (f ), the following Voronovskaja’s result hold. P∞ Theorem 3.1.12. If f (z) = k=0 ck z k is analytic in |z| < R with R > 1 then for all |z| ≤ r, 1 ≤ r < R, we have 2 00 0 Jn (f )(z) − f (z) + 3z f (z) + 3zf (z) ≤ Ar (f ) , 2n2 2n2 n3 P ∞ 4 k where Ar (f ) = 15 k=0 |ck |k r . 2 P∞ Proof. Since we can write Jn (f )(z) = k=0 ck Jn (ek )(z), it follows that X ∞ 2 00 0 Jn (f )(z) − f (z) + 3z f (z) + 3zf (z) ≤ |ck | · |Ek,n (z)|, 2n2 2n2 k=0

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where Ek,n (z) = Jn (ek )(z) − z k + Since ∞ X k=0

|ck | · |Ek,n (z)| =

3k(k−1)z k 2n2

n X k=0

+

+

197

3kz k 2n2 .

|ck | · |Ek,n (z)| + ∞ X

k=2n−1

2n−2 X

k=n+1

|ck | · |Ek,n (z)|

|ck | · |Ek,n (z)| := A + B + C,

by Theorem 3.1.10, (i), (iii), we obtain A= 3 n X 4n − 6k 2 n + 3k 3 − 3k + 2n k 3k(k − 1)z k 3kz k k = |ck | · z − z + + 2n(2n2 + 1) 2n2 2n2 k=0 3 n 2 3 X 3k 2 k 4n − 6k n + 3k − 3k + 2n − 1 + 2 ≤ |ck |r · 2n(2n2 + 1) 2n k=0 n n n 3 2 X X 3n(k 3 − k) + 3k 2 X 2k 4 kk +k ≤ = |ck |rk · ≤ |ck |rk 3 , |c |r k 2 3 2n(2n + 1) n n k=0

k=0

k=0

(k − 2n) − (k − 2n)3 k 3k 2 z k k B= |ck | · z −z + 2n(2n2 + 1) 2n2 k=n+1 2n−2 3 X 3k 2 k (k − 2n) − (k − 2n) − 1 + 2 |ck |r ≤ 2n(2n2 + 1) 2n k=n+1 2n−2 3 2 2 3 2 X 4 k 4n − 12kn + n (12k − 4) + n(k − k ) + 3k = |ck |r 2n2 (2n2 + 1) 2n−2 X

k=n+1

≤ ≤

2n−2 X

k=n+1 2n−2 X

k=n+1

|ck |rk

4k 4 + 12k 4 + k 2 (12k 2 − 4) + k(k + k 3 ) + 3k 2 2n2 (2n2 + 1)

|ck |rk

2n−2 30k 4 15 X ≤ |ck |rk k 4 . 4n4 2n2 k=n+1

In the proof of inequality satisfied by B we applied the fact that n < k. Finally, ∞ X k 3z k k(k − 1) 3z k k + C= |ck | · −z + 2n2 2n2 k=2n−1 2 ∞ ∞ X X 3 k 3k ≤ |ck |r 2 − 1 ≤ |ck |rk k 2 2n 2n2 k=2n−1

3 ≤ 2n3

∞ X

k=2n−1

k=2n−1

3 |ck |rk k 3 ≤ 2n3

∞ X

k=2n−1

|ck |rk k 4 .

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In conclusion, ∞ X k=0

|ck | · |Ek,n (z)| = A + B + C ≤

which proves the theorem.

∞ 15 X |ck |rk k 4 , 2n3 k=0

By using Corollary 3.1.11 and Theorem 3.1.12 now we are in position to obtain the exact degree of approximation by Jn (f )(z) and its derivatives. The first main result is a lower estimate in Theorem 3.1.11, (i) for the complex Jackson polynomials. Theorem 3.1.13. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that P∞ k f : DR → C is analytic in DR , that is we can write f (z) = k=0 ck z , for all z ∈ DR . If f is not a polynomial of degree 0, then for any r ∈ [1, R) we have

Cr (f ) , n ∈ N, n2 where the constant Cr (f ) depends only on f and r. kJn (f ) − f kr ≥

Proof.

For all z ∈ DR and n ∈ N we have 2 00 1 3z f (z) 3zf 0 (z) Jn (f )(z) − f (z) = 2 − + n 2 2 3z 2 00 3zf 0 (z) 1 3 n Jn (f )(z) − f (z) + 2 f (z) + . + n 2n 2n2

In what follows we will apply to this identity the following obvious property : kF + Gkr ≥ | kF kr − kGkr | ≥ kF kr − kGkr . It follows kJn (f ) − f kr

0 00

3e1 f 0 1 3e2 f 00

− 1 n3 Jn (f ) − f + 3e2 f + 3e1 f + . ≥ 2

n 2 2 r n 2n2 2n2 r

Taking into account that by hypothesis polynomial of degree 0 in DR , by

f00is not a0

3e2 f 3e1 f the proof of Theorem 3.1.3, we get 2 + 2 > 0. r But by Theorem 3.1.12 we have

3e2 00 3e1 0

≤ Ar (f ). n3 J (f ) − f + f + f

n 2n2 2n2 r

Therefore, there exists an index n0 depending only on f and r, such that for all n ≥ n0 we have

3e2 f 00 3e1 f 0 1 3 3e2 00 3e1 0

2 + 2 − n n Jn (f ) − f + 2n2 f + 2n2 f r r

00 0 1 3e f 3e f 2 1 , ≥ + 2 2 2 r

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which immediately implies

3e1 f 0 3e2 f 00 1 1

, ∀n ≥ n0 . + kJn (f ) − f kr ≥ 2 · n 2 2 2 r M

(f )

For n ∈ {1, ..., n0 −1} we obviously have kJn (f )−f kr ≥ r,n with Mr,n (f ) = n2 · n2 Cr (f ) kJn (f )−f kr > 0, which finally implies kJn (f )−f

kr ≥ n2 for all n, where Cr (f ) = 3e1 f 0 1 3e2 f 00 min{Mr,1 (f ), ..., Mr,n0 −1 (f ), 2 2 + 2 }. This completes the proof. r

Combining now Theorem 3.1.13 with Corollary 3.1.11, (i) we immediately get the following.

Corollary 3.1.14. Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that f : DR → C is analytic in DR . If f is not a polynomial of degree 0, then for any r ∈ [1, R) we have kJn (f ) − f kr ∼

1 , n ∈ N, n2

where the constants in the equivalence depend on f and r. In the case of approximation by the derivatives of Jn (f )(z) the following result holds. Theorem 3.1.15. Let DR = {z ∈ C; |z| < R} be with R > 1 and let us suppose P k that f : DR → C is analytic in DR , i.e. f (z) = ∞ k=0 ck z , for all z ∈ DR . Also, let 1 ≤ r < r1 < R and q ∈ N be fixed. If f is not a polynomial of degree ≤ q − 1, then we have kJn(q) (f ) − f (q) kr ∼

1 , n2

where the constants in the equivalence depend on f , r, r1 and q. Proof. Denoting by Γ the circle of radius r1 and center 0 (where r1 > r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N we have Z Jn (f )(v) − f (v) q! (q) (q) Jn (f )(z) − f (z) = dv, 2πi Γ (v − z)q+1 where the inequality |v − z| ≥ r1 − r is valid for all |z| ≤ r and v ∈ Γ. Taking into account Corollary 3.1.11, (ii), it remains to prove the lower estimate (q) for kJn (f ) − f (q) kr . For this purpose, as in the proof of Theorem 3.1.13, for all v ∈ Γ and n ∈ N we have 2 00 3v f (v) 3vf 0 (v) 1 + Jn (f )(v) − f (v) = 2 − n 2 2 3v 2 00 1 3 3vf 0 (v) + n Jn (f )(v) − f (v) + 2 f (v) + , n 2n 2n2

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which replaced in the above Cauchy’s formula implies h 2 00 i  0  q! Z − 3v f2 (v) + 3vf2 (v) 1 Jn(q) (f )(z) − f (q) (z) = 2 dv n  2πi Γ (v − z)q+1  Z n3 Jn (f )(v) − f (v) + 3v22 f 00 (v) + 3v2 f 0 (v)  2n 2n 1 q! + · dv  n 2πi Γ (v − z)q+1 ( (q) 3z 2 f 00 (z) 3zf 0 (z) 1 − = 2 − n 2 2  Z n3 Jn (f )(v) − f (v) + 3v22 f 00 (v) + 3v2 f 0 (v)  2n 2n 1 q! dv . + ·  n 2πi Γ (v − z)q+1

Passing now to k · kr it follows kJn(q) (f )

−f

(q)

1 kr ≥ 2 n

( (q)

3e

3e1 0

2 00 f + f

2

2

Z n3 Jn (f )(v) − f (v) + 3v22 f 00 (v) +

2n 1 q! −

q+1 n 2π Γ (v − z)

r

3v 0 2n2 f





dv

,

r

where by using Theorem 3.1.12 we obtain

Z n3 Jn (f )(v) − f (v) + 3v22 f 00 (v) + 3v2 f 0

q!

2n 2n

dv

2π

(v − z)q+1 Γ

r

q! 2πr1 n3 3e 3e A 1 2 r1 (f )q!r1 00 0

Jn (f ) − f + ≤ · f + 2f . ≤ 2π (r1 − r)q+1 2n2 2n r1 (r1 − r)q+1

But by the hypothesis on f and by the proof of Theorem 3.1.5 we have

(q)

3e

3e1 0

2 00 f + f

> 0.

2

2 r

Finally, reasoning exactly as in the proof of Theorem 3.1.13, we immediately get the desired conclusion. In the case of Beatson-type convolutions Pn,2,p (f )(z), p ≥ 2, the following result concerning the exact degree of approximation holds. Theorem 3.1.16. Let DR = {z ∈ C; |z| < R} be with R > 1 and let us suppose P∞ that f : DR → C is analytic in DR , i.e. f (z) = k=0 ck z k , for all z ∈ DR . Also, S let 1 ≤ r < r1 < R, p ∈ N, p ≥ 2 even number and q ∈ N {0} be fixed. If f is not a constant for q = 0 and is not a polynomial of degree ≤ q − 1 for q ∈ N, then for all n ∈ N we have 1 (q) kPn,2,p (f ) − f (q) kr ∼ 2 , n

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where the constants in the equivalence depend on f , r, r1 , p and q. Proof. Denoting by Γ the circle of radius r1 and center 0 (where r1 > r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n, q ∈ N we have Z Pn,2,p (f )(v) − f (v) q! (q) (q) Pn,2,p (f )(z) − f (z) = dv, 2πi Γ (v − z)q+1 where the inequality |v − z| ≥ r1 − r is valid for all |z| ≤ r and v ∈ Γ. Taking into account Corollary 3.1.11, (ii), for q ∈ N and Corollary 3.1.11, (i) for (q) q = 0, in both cases it remains to prove the lower estimate for kPn,2,p (f ) − f (q) kr . We will use here a different method of proof, for which we don’t need a Voronovskaja-type formula. By Theorem 3.1.10, (iii), Pn,2,p (f )(z) is a polynomial of degree ≤ 2n − 2. Since P∞ f (q) (z) = k=q k(k − 1)...(k − q + 1)ck z k−q and by Theorem 3.1.10, (iii), we have (q)

Pn,2,p (f )(z) =

2n−2 X k=q

ck k(k − 1)...(k − q + 1)

ip hn sin(kπ/n) λk,n z k−q , kπ

with λk,n given by Theorem 3.1.10, (i), for z = reiϕ we can write (here i2 = −1) (q)

[Pn,2,p (f )(z) − f (q) (z)]e−isϕ

=

∞ X k=q

ck k(k − 1)...(k − q + 1)r k−q

=: En,p,q,s (z).

ip nh n o sin(kπ/n) λk,n − 1 eiϕ(k−q−s) kπ

Integrating from −π to π we immediately obtain Z π 1 En,p,q,s (z)dϕ 2π −π p n sin((q + s)π/n) λq+s,n − 1 . = cq+s (q + s)(q + s − 1)...(s + 1)r s (q + s)π Taking into account that by the Remark 3 after Theorem 3.1.10 we have 0 ≤ λ k,n ≤ 1 for all k, n, we immediately obtain (q)

kPn,2,p (f ) − f (q) kr p n (q + s)! s r 1− sin((q + s)π/n) λq+s,n ≥ |cq+s | s! (q + s)π (q + s)! s ≥ |cq+s | r [1 − λq+s,n ], s! h i p

n since 0 ≤ (q+s)π sin((q + s)π/n) ≤ 1 by the known inequality | sin(x)| ≤ x for all x ≥ 0. First consider q = 0 and denote Vn = inf 1≤s (1 − λs,n ). By Remark 3 after 3 1 4n3 −4n Theorem 3.1.10 we get Vn = 1 − 4n 3 +2n = 2n2 +1 ≥ n2 . By the above lower estimate for kPn,2,p (f ) − f kr , for all s ≥ 1 and n ∈ N it follows kPn,2,p (f ) − f kr kPn,2,p (f ) − f kr ≥ ≥ |cs |rs . n2 kPn,2,p (f ) − f kr ≥ Vn 1 − λs,n

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This implies that if there exists a subsequence (nk )k with limk→∞ nk = +∞ and such that limk→∞ n2k kPnk ,2,p (f ) − f kr = 0 then cs = 0 for all s ≥ 1, that is f is constant on Dr . Therefore, if f is not a constant then lim inf n2 kPn,2,p (f )−f kr > 0, which implies that there exists a constant Cr,p (f ) > 0 such that n2 kPn,2,p (f ) − f kr ≥ Cr,p (f ), for all n ∈ N, that is Cr,p (f ) , n ∈ N. n2 Now, consider q ≥ 1 and denote Vq,n = inf s≥0 (1 − λq+s,n ). Evidently that we have Vq,n ≥ inf s≥1 (1 − λs,n ) ≥ n12 . Reasoning as in the case of q = 0 we obtain kPn,2,p (f ) − f kr ≥

(q)

(q)

kPn,2,p (f ) − f (q) kr kPn,2,p (f ) − f (q) kr ≥ Vq,n 1 − λq+s,n (q + s)! s ≥ |cq+s | r , s! for all s ≥ 0 and n ∈ N. Now, if there exists a subsequence (nk )k with limk→∞ nk = (q) +∞ and such that limk→∞ n2k kPnk ,2,p (f ) − f (q) kr = 0 then cq+s = 0 for all s ≥ 0, that is f is a polynomial of degree ≤ q − 1 on Dr . (q) Therefore lim inf n2 kPn,2,p (f ) − f (q) kr > 0 when f is not a polynomial of degree ≤ q − 1, which implies that there exists a constant Cr,p,q (f ) > 0 such that (q) n2 kPn,2,p (f ) − f (q) kr ≥ Cr,p,q (f ), for all n ∈ N. That is (q)

n2 kPn,2,p (f ) − f (q) kr ≥

(q)

kPn,2,p (f ) − f (q) kr ≥

Cr,p,q (f ) , n ∈ N, n2

which proves the theorem.

Remarks. 1) The case when p is an odd number in Theorem 3.1.16 remains unsettled. 2) By using the considerations in Section 1.0, the above complex convolutions can be generalized to approximation in a Jordan domain G whose boundary is rectifiable and of bounded rotation. Indeed, keeping the notations there, let f ∈ A(G) and F ∈ A(D1 ) such that f = T [F ]. As in e.g. Gal [93], p. 318, Theorem 3.1, (i), P∞ attach to F (w) = k=0 ak wk , w ∈ D1 , the complex convolutions through a certain P mn trigonometric kernel O(t) = 12 + k=1 ρk,n cos(kt), Qn (F )(w) = a0 +

mn X

ak ρk,n wk .

k=1

Then, suggested by the Remark after Theorem 1.0.12, attach to f = T [F ] the complex polynomial Ln (f ) = T [Qn (F )] given by Ln (f )(z) =

mn X k=0

ak ρk,n Fk (z), z ∈ G,

where Fk (z) denotes the Faber polynomial of degree k attached to G.

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But according to Theorem 1.0.11, we have ak = ak (f )-the Faber coefficients of f on G and according to Definition 1.0.10, (ii), we can write Z Z π 1 1 f [Ψ(u)] f [Ψ(eit )]e−ikt dt, k ∈ N ∪ {0}. ak (f ) = du = 2πi |u|=1 uk+1 2πi −π Therefore, the complex convolutions attached to f in G can be written as mn X Ln (f )(z) = ak (f )ρk,n Fk (z), z ∈ G. k=0

The exact orders of approximation in the cases when the trigonometric kernels are those of Fej´er, Riesz-Zygmund and Rogosinski were obtained in Bruj-Schmieder [48], for details see Theorem 7 and its Remarks in Section 6 there. Below we present with proof a saturation result for the complex convolutions Ln (f )(z), n ∈ N, where for simplicity we take mn = n. Theorem 3.1.17. (Bruj-Schmieder [48], Theorem 4) Let G ⊂ C be a Jordan domain with rectifiable boundary and f ∈ A(G) a non-constant function. If ρ k,n 6= 1 for all n ∈ N, k = 1, 2, ..., n, then max |f (z) − Ln (f )(z)| 6= o( min |1 − ρk,n |). 1≤k≤n

z∈G

Recall here that by definition an = o(bn ) if limn→∞

an bn

= 0.

R f (Ψ(u)) 1 Proof. By Definition 1.0.10 (ii) we have am (f ) = 2πi |u|=1 um+1 du. Also, taking that by K¨ ovari-Pommerenke [120], p. 198, Lemma 2, the integral R intoFnaccount (Ψ(u)) 1 du is equal to zero for m 6= n and equal to 1 for m = n, we 2πi |u|=1 um+1 immediately obtain Z 1 f (Ψ(u)) − Ln (f )(Ψ(u)) du = (1 − ρm,n )am (f ), n ≥ m ≥ 0. 2πi |u|=1 um+1

This implies

|1 − ρm,n | · |am (f )| ≤ max |f (z) − Ln (f )(z)|. z∈G

Now, let us assume that the conclusion of Theorem 3.1.17 is false, that is max |f (z) − Ln (f )(z)| = o( min |1 − ρk,n |). 1≤k≤n

z∈G

It implies |1 − ρm,n | · |am (f )| = o( min |1 − ρk,n |) 1≤k≤n

and therefore lim

n→∞

|1 − ρm,n | · |am (f )| = 0, min1≤k≤n |1 − ρk,n |

which means am (f ) = 0, for all m ≥ 1. Then, by Theorem 1, p. 44 in Gaier [76] we get f (z) = a0 (f ) for all z ∈ G, a contradiction. In conclusion, the statement of Theorem 3.1.17 is valid.

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Remarks. 1) (Example A in Bruj-Schmieder [48], p. 161) If we take ρk,n := max{0, 1−k/(n+1)}, that is the Fej´er-type convolution, then Ln (f )(z) approximate f on G not better than the order min1≤m≤n |1 − ρm,n | = 1/(n + 1). 2) (Example B in Bruj-Schmieder [48], p. 161) If we take ρk,n := max{0, 1 − [k/(n + 1)]s }, where s ≥ 1 is fixed, that is the Riesz-Zygmund-type convolution, then Ln (f )(z) approximate f on G not better than the order min 1≤m≤n |1−ρm,n | = [1/(n + 1)]s . 3) For the so-called Rogosinki-type convolution, in Bruj-Schmieder [48], p. 162, similarly it is proved that approximate f on G not better than the order 1/n2 .

3.2

Linear Non-Polynomial Convolutions

In several recent papers, the geometric properties of complex linear integral transforms generated by the Hadamard product (convolution) of normalized analytic functions f in the unit disk with various kinds of kernels, when the kernels are special functions such as hypergeometric functions or Hurwitz kind functions were intensively studied, see e.g. Choi-Kim-Saigo [54], Dziok-Srivastava [68], Kanas-Srivastava [111], Murugusundaramoorthy-Magesh [142], Ponnusamy-Ronning [152], PrajapatRaina-Srivastava [157], R˘ aducanu-Srivastava [160], Srivastava-Attiya [169], Srivastava et al [170]. A common characteristic of these complex convolutions is that although they preserve the geometric properties of f like starlikeness or convexity, they do not have good approximation properties. In this section we study some complex linear integral convolutions of complex analytic f with various real non-trigonometric kernels, which besides the preservations of some geometric properties of f have also good approximation properties. More exactly, we will refer to the complex linear integral convolutions of Picard kind, Gauss-Weierstrass kind, Poisson-Cauchy kind, Post-Widder kind, rotationalinvariant kind, Sikkema kind and spline kind. Most of the results were obtained in Anastassiou-Gal [18], Anastassiou-Gal [16], Anastassiou-Gal [17], Anastassiou-Gal [19], Anastassiou-Gal [20], but also new results are presented. Applications to heat and Laplace equation with complex spatial variables are presented, see Gal-GalGoldstein [97]. The above mentioned results hold in compact disks of the complex plane, but extensions to Jordan domains whose boundaries are rectifiable and of bounded rotation also are presented. For D1 = {z ∈ C; |z| < 1} let us define A(D1 ) = {f : D1 → C; f is analytic on D1 , and continuous on D1 }, endowed with the uniform norm kf k1 = sup{|f (z)|; z ∈ D1 }. Is well-known that (A(D1 ), k · k1 ) is a Banach space. Therefore if f ∈ A(D1 ) then we have f (z) = P∞ k k=0 ak z , for all z ∈ D1 .

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Also, define the subclass of A(D1 ) of normalized functions by A∗ (D1 ) = {f ∈ A(D1 ); f (0) = 0, f 0 (0) = 1}.

For the geometric properties, the following classes of analytic functions will be of interest : 0 zf (z) S ∗ (D1 ) = {f ∈ A∗ (D1 ); Re > 0, for all z ∈ D1 }, f (z) zf 0 (z) Sγ∗ (D1 ) = {f ∈ A∗ (D1 ); Re eiγ > 0, for all z ∈ D1 }, |γ| < π/2, f (z) 00 zf (z) K(D1 ) = {f ∈ A∗ (D1 ); Re + 1 > 0, for all z ∈ D1 }, f 0 (z) ∞ ∞ X X ∗ k S1 = f ∈ A (D1 ); f (z) = z + ak z , k|ak | ≤ 1 , S2 =

k=2

f is analytic in D1 , f (z) =

k=2

∞ X

k=1 ∗

00

ak z k , z ∈ D 1 ,

S3 = {f ∈ A (D1 ); |f (z)| ≤ 1, ∀z ∈ D1 },

|a1 | ≥

∞ X

k=2

|ak | ,

P = {f : D1 → C; f is analytic on D1 , f (0) = 1, Re[f (z)] > 0, ∀z ∈ D1 },

R = {f ∈ A∗ (D1 ); Re[f 0 (z)] > 0, ∀z ∈ D1 },

SM = {f ∈ A∗ (D1 ); |f 0 (z)| < M, ∀z ∈ D1 },

M > 1.

According to e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138], p. 97, Exercise 4.9.1, if 00 (z) − 1 < 1, ∀z ∈ D1 and therefore f is starlike (and univalent) on f ∈ S1 then zff (z) D1 . According to Alexander [5], p. 22, if f ∈ S2 then f is starlike (and univalent) on D1 . By Obradovi´c [145], if f ∈ S3 then f is starlike (and univalent) on D1 . Also, it is well known that R is the class of functions with bounded turn (i.e. | arg f 0 (z)| < π2 , ∀z ∈ D1 ) and that f ∈ R implies the univalence of f on D1 . ] According to e.g. Mocanu-Bulboac˘ al˘ agean [138 a-S˘ , p. 111, Exercise 5.4.1, 1 f ∈ SM implies that f is univalent in z ∈ C; |z| < M . 3.2.1

Picard, Poisson-Cauchy and Gauss-Weierstrass Complex Convolutions

In this subsection for f ∈ A(D1 ), z ∈ D1 and ξ ∈ R, ξ > 0, we will study the approximation and geometric properties of the following complex linear convolutions Z +∞ 1 Pξ (f )(z) = f (zeiu )e−|u|/ξ du, 2ξ −∞ Z Z ξ π f (zeiu ) ξ +∞ f (ze−iu ) ∗ Qξ (f )(z) = du, Q (f )(z) = du, ξ π −π u2 + ξ 2 π −∞ u2 + ξ 2

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Z 2ξ 3 +∞ f (zeiu ) Rξ (f )(z) = du, π −∞ (u2 + ξ 2 )2 Z π 2 1 f (zeiu )e−u /ξ du, Wξ (f )(z) = √ πξ −π Z +∞ 2 1 Wξ∗ (f )(z) = √ f (ze−iu )e−u /ξ du, πξ −∞ and their Jackson-type generalizations # Z +∞ " n+1 X 1 iku −|u|/ξ k n+1 f (ze ) du, Pn,ξ (f )(z) = − e (−1) k 2ξ −∞ k=1

Qn,ξ (f )(z) = − 2 ξ

1 arctg

π · ξ

n+1 X

(−1)k

k=1

Z π n+1 f (zeiku ) du, 2 2 k −π u + ξ

Z π n+1 X 2 2 1 n+1 · (−1)k f (zeiku )e−u /ξ du, k 2C(ξ) −π k=1 Z n+1 +∞ X 2 2 n+1 1 ∗ (−1)k · Wn,ξ (f )(z) = − ∗ f (zeiku )e−u /ξ du, 2C (ξ) k −∞ Wn,ξ (f )(z) = −

k=1

Rπ

R∞ 2 2 where n ∈ N, C(ξ) = 0 e du and C ∗ (ξ) = 0 e−u /ξ du. Here Pξ (f ) is called of Picard–type, Qξ (f ), Q∗ξ (f ) and Rξ (f ) are called of Poisson–Cauchy–type, Wξ (f ), Wξ∗ (f ) are called of Gauss–Weierstrass–type, Pn,ξ (f )(z) is called of generalized Picard–type, Qn,ξ (f )(z) is called of generalized ∗ Poisson–Cauchy–type and Wn,ξ (f )(z), Wn,ξ (f )(z) are called of generalized Gauss– Weierstrass–type. The approximation properties of the Picard–type convolution Pξ (f ) are expressed by the following result. Note that while below Theorem 3.2.1, (i), (ii) and (iii) were obtained in Anastassiou–Gal [18], Theorem 3.2.1, (iv) is new. −u2 /ξ 2

Theorem 3.2.1. Let ξ ∈ R, ξ > 0. (i) If f ∈ A(D1 ) then Pξ (f ) ∈ A(D1 ), that is it is continuous on D1 and analytic P k on D1 . Moreover, if f (z) = ∞ k=0 ak z for all z ∈ DR , R ≥ 1, then Pξ (f )(z) =

∞ X k=0

ak zk, 1 + ξ 2 k2

for all z ∈ DR , if f (0) = 0 then Pξ (f )(0) = 0 and if f 0 (0) = 1 then Pξ0 (f )(0) = 1 1+ξ 2 6= 1, for all ξ > 0. (ii) If f is continuous on D1 then for all δ ≥ 0 and ξ > 0 ω1 (Pξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 , where ω1 (f ; δ)D1 = sup{|f (z1 ) − f (z2 )|; z1 , z2 ∈ D1 , |z1 − z2 | ≤ δ} ;

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(iii) If f ∈ A(D1 ) then for all z ∈ D1 and ξ > 0, |Pξ (f )(z) − f (z)| ≤ Cω2 (f ; ξ)∂D1 , where ω2 (f ; ξ)∂D1 = sup{|f (ei(x+u) ) − 2f (eiu ) + f (ei(x−u) )|; x ∈ R, |u| ≤ ξ}. P∞ (iv) Let us suppose that f (z) = k=0 ak z k for all z ∈ DR , R > 1. If f is not constant for q = 0 and not a polynomial of degree ≤ q − 1 for q ∈ N, then for all S 1 ≤ r < r1 < R, ξ ∈ (0, 1] and q ∈ N {0} we have (q)

kPξ (f ) − f (q) kr ∼ ξ 2 ,

where the constants in the equivalence depend only on f , q, r, r1 . Proof.

(i) Let z0 , zn ∈ D1 be with lim zn = z0 . We get n→∞

|Pξ (f )(zn ) − Pξ (f )(z0 )| ≤

1 2ξ

1 ≤ 2ξ =

1 2ξ

Z Z Z

+∞ −∞ +∞ −∞ +∞ −∞

|f (zn eiu ) − f (z0 eiu )|e−|u|/ξ du ω1 (f ; |zn eiu − z0 eiu |)D1 e−|u|/ξ du ω1 (f ; |zn − z0 |)D1 e−|u|/ξ du

= ω1 (f ; |zn − z0 |)D1 . Passing to limit with n → ∞, it follows that Pξ (f )(z) is continuous at z0 ∈ D1 , since f is continuous on D1 . It remains to prove that Pξ (f )(z) is analytic on D. P∞ For f ∈ A(D1 ), we can write f (z) = k=0 ak z k , z ∈ D1 . For fixed z ∈ D1 , we P∞ iku k get f (zeiu ) = z and since |ak eiku | = |ak |, for all u ∈ R and the k=0 ak e P∞ P∞ k series k=0 ak z is absolutely convergent, it follows that the series k=0 ak eiku z k is uniformly convergent with respect to u ∈ R. This immediately implies that the series can be integrated term by term, i.e. Z ∞ ∞ 1 X iku −|u|/ξ k e e du . ak z Pξ (f )(z) = 2ξ −∞ k=0

P∞

Now let f (z) = k=0 ak z k for all z ∈ DR , R ≥ 1. From the above reasonings we can write Z +∞ ∞ X 1 iku −|u|/ξ k e e du , Pξ (f )(z) = ak z 2ξ −∞ k=0

∀z ∈ D1 .

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But 1 2ξ

Z

+∞ −∞

eiku e−|u|/ξ du

Z 1 +∞ cos(ku) · e−|u|/ξ du = cos(ku)e−u/ξ du ξ 0 −∞ 1 −u/ξ − ξ cos(ku) + k sin(ku) ∞ 1 e 1 = = · . 1 2 ξ 1 + k2ξ2 0 ξ2 + k

=

1 2ξ

Z

+∞

1 Also, if a0 = 0 then Pξ (f )(0) = 0 and if a1 = 0 then Pξ (f )(0) = 1+ξ 2 , which proves (i). (ii) Let z1 , z2 ∈ D1 , |z1 − z2 | ≤ δ. We get Z +∞ 1 |Pξ (f )(z1 ) − Pξ (f )(z2 )| ≤ |f (z1 eiu ) − f (z2 eiu )|e−|u|/ξ du 2ξ −∞ ≤ ω1 (f ; |z1 − z2 |)D1 ≤ ω1 (f ; δ)D1 .

Passing to sup with |z1 − z2 | < δ, it follows the desired inequality. (iii) We have Z +∞ 1 Pξ (f )(z) − f (z) = [f (zeiu ) − f (z)]e−|u|/ξ du 2ξ −∞ Z ∞ 1 = [f (zeiu ) − 2f (z) + f (ze−iu )]e−u/ξ du, 2ξ 0 which implies Z ∞ 1 |f (zeiu ) − 2f (z) + f (ze−iu )|e−u/ξ du, |Pξ (f )(z) − f (z)| ≤ 2ξ 0

for all z ∈ D1 . By the maximum modulus principle we can take |z| = 1, case when |f (zeiu ) − 2f (z) + f (ze−iu )| ≤ ω2 (f ; u)∂D1 ,

which implies that for all z ∈ D1 we have Z +∞ 1 |Pξ (f )(z) − f (z)| ≤ ω2 (f ; u)∂D1 e−u/ξ du 2ξ 0 Z +∞ u 1 ω2 f ; · ξ = e−u/ξ du 2ξ 0 ξ ∂D1 ! 2 Z +∞ 1 u e−u/ξ du ω2 (f ; ξ)∂D1 ≤ 1+ 2ξ 0 ξ ≤ Cω2 (f ; ξ)∂D1 ,

(for the last inequalities see e.g. Gal [94], p. 252, proof of Theorem 2.1, (i)). (iv) Analysing the proofs of the above points (i)-(iii), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation error in (iii) can be expressed in terms of ω2 (f ; ξ)∂Dr = sup{|∆2u f (reit )|; |t| ≤ π, |u| ≤ ξ},

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with constants in front of ω2 depending on r ≥ 1. Since ∆2u g(t) = 2u2 [t, t + u, t + 2u; g], by the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get |∆2u f (reit )| ≤ u2 kf 00 kr , where 00

00

kf kr = sup{|f (z)|; |z| ≤ r} ≤ For all ξ ∈ (0, 1] from (iii) it follows

∞ X

k=2

|ak |k(k − 1)rk−2 .

kPξ (f ) − f kr ≤ Cr (f )ξ 2 .

Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and ξ ∈ (0, 1], we have Z q! Pξ (f )(v) − f (v) (q) dv |Pξ (f )(z) − f (q) (z)| = 2π γ (v − z)q+1 2πr1 q! ≤ Cr1 (f )ξ 2 2π (r1 − r)q+1 q!r1 = Cr1 (f )ξ 2 , (r1 − r)q+1 (q)

which proves the upper estimate in approximation by Pξ (f )(z). (q)

It remains to prove the lower estimate for kPξ (f ) − f (q) kr . For this purpose, S take z = reiϕ and p ∈ N {0}. We have 1 (q) (q) [f (z) − Pξ (f )(z)]e−ipϕ 2π ∞ 1 X 1 ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) 1 − = . 2π 1 + ξ 2 k2 k=q

Integrating from −π to π we obtain Z π 1 (q) [f (q) (z) − Pξ (f )(z)]e−ipϕ dϕ 2π −π = aq+p (q + p)(q + p − 1)...(p + 1)r p

ξ 2 (q + p)2 . 1 + ξ 2 (q + p)2

Passing now to absolute value we easily obtain |aq+p |(q + p)(q + p − 1)...(p + 1)r p

ξ 2 (q + p)2 (q) ≤ kf (q) − Pξ (f )kr . 1 + ξ 2 (q + p)2 2 2

ξ p First consider q = 0 and denote Vξ = inf 1≤p ( 1+ξ 2 p2 ). We get Vξ = 2 ξ ∈ (0, 1] it follows Vξ ≥ ξ /2.

ξ2 1+ξ 2

and for all

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By the above lower estimate for kPξ (f ) − f kr , for all p ≥ 1 and ξ ∈ (0, 1] it follows kPξ (f ) − f kr kPξ (f ) − f kr 2kPξ (f ) − f kr ≥ ≥ ≥ |ap |rp . 2 ξ 2 p2 ξ Vξ 2 2 1+ξ p

This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 kP (f )−f kr and such that limk→∞ ξk ξ2 = 0 then ap = 0 for all p ≥ 1, that is f is k

constant on Dr . Therefore, if f is not a constant then inf ξ∈(0,1]

there exists a constant Cr (f ) > 0 such that that is

kPξ (f )−f kr > 0, which implies that ξ2 kPξ (f )−f kr ≥ Cr (f ), for all ξ ∈ (0, 1], ξ2

kPξ (f ) − f kr ≥ Cr (f )ξ 2 , for all ξ ∈ (0, 1]. 2

2

ξ (q+p) Now, consider q ≥ 1 and denote Vq,ξ = inf p≥0 ( 1+ξ 2 (q+p)2 ). Evidently that we have 2 2

ξ p 2 Vq,ξ ≥ inf p≥1 ( 1+ξ 2 p2 ) ≥ ξ /2. Reasoning as in the case of q = 0 we obtain (q)

(q)

2kPξ (f ) − f (q) kr ξ2

≥

kPξ (f ) − f (q) kr Vq,ξ

≥ |aq+p |

(q)

≥

kPξ (f ) − f (q) kr ξ 2 (q+p)2 1+ξ 2 (q+p)2

(q + p)! p r , p!

for all p ≥ 0 and ξ ∈ (0, 1]. This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 (q)

and such that limk→∞

kPξ

k

(f )−f (q) kr ξk2

polynomial of degree ≤ q − 1 on Dr . kP

(q)

= 0 then aq+p = 0 for all p ≥ 0, that is f is a

(f )−f (q) kr

Therefore, inf ξ∈(0,1] ξ ξ2 > 0 when f is not a polynomial of degree ≤ q − 1, which implies that there exists a constant Cr,q (f ) > 0 such that (q)

kPξ

(f )−f (q) kr ξ2

≥ Cr,q (f ), for all ξ ∈ (0, 1], that is (q)

kPξ (f ) − f (q) kr ≥ Cr,q (f )ξ 2 , for all ξ ∈ (0, 1], which proves the theorem.

In what follows, we present some geometric properties of Pξ (f )(z). Theorem 3.2.2. (Anastassiou-Gal [18]) For all ξ > 0 we have Pξ (S2 ) ⊂ S2 Proof.

By Theorem 3.2.1, for f (z) =

and ∞ P

k=1

Pξ (P) ⊂ P. ak z k ∈ S2 , we get

X ∞ ∞ ∞ X ak |ak | |a1 | 1 + ξ2 1 X = |ak | ≤ · ≤ 1 + ξ 2 k2 1 + ξ 2 1 + ξ 2 k2 1 + ξ2 1 + ξ2

k=2

k=2

k=2

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P∞ ak k and since Pξ (f )(z) = k=0 1+ξ 2 k2 z , it follows Pξ (f ) ∈ S2 . P∞ Now, let f (z) = k=0 ak z k ∈ P, that is a0 = 1 and if f (z) = U (x, y) + iV (x, y), z = x + iy ∈ D1 , then U (x, y) > 0, ∀z = x + iy ∈ D1 . We get Pξ (f )(0) = a0 = 1 and Z +∞ 1 U (r cos(u + t), r sin(u + t))e−|u|/ξ du Pξ (f )(z) = 2ξ −∞ Z +∞ 1 + i· V (r cos(u + t), r sin(u + t))e−|u|/ξ du, 2ξ −∞ for all z = reit ∈ D1 , which immediately implies Z +∞ 1 Re[Pξ (f )(z)] = U (r cos(u + t), r sin(u + t))e−|u|/ξ du > 0, 2ξ −∞ that is Pξ (f ) ∈ P.

Theorem 3.2.3. (Anastassiou-Gal [18]) For all ξ > 0 we have (1+ξ 2 )Pξ (S1 ) ⊂ S1 , (1 + ξ 2 )Pξ (SM ) ⊂ SM (1+ξ2 ) and (1 + ξ 2 )Pξ (S3,ξ ) ⊂ S3 , where 1 00 , ∀z ∈ D ⊂ S3 . S3,ξ = f ∈ S3 ; |f (z)| ≤ 1 + ξ2 Proof.

Let f ∈ S1 . By Theorem 3.2.1, (i) we obtain (1 + ξ 2 )Pξ (f )(z) =

∞ X

ak

k=1

if f (z) =

∞ P

k=1

1 + ξ2 k z , 1 + ξ 2 k2

ak z k ∈ S1 . It follows (1 + ξ 2 )Pξ0 (f )(0) = a1 = 1, that is (1 + ξ 2 )Pξ (f )(z) = z +

∞ X

k=2

and ∞ X k=2 2

k|ak |

that is (1 + ξ )Pξ (f ) ∈ S1 . Let f ∈ SM . We get |(1 + ξ

2

)Pξ0 (f )(z)|

ak ·

1 + ξ2 k z 1 + ξ 2 k2

∞

X 1 + ξ2 ≤ k|ak | ≤ 1, 2 2 1+ξ k k=2

Z +∞ 1 0 iu iu −|u|/ξ f (ze )e e du = (1 + ξ ) · 2ξ −∞ Z +∞ 1 ≤ (1 + ξ 2 ) |f 0 (zeiu )|e−|u|/ξ du 2ξ −∞ 2

< M (1 + ξ 2 ),

z ∈ D1 .

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Also, Pξ (f )(0) = 0 and (1 + ξ 2 )Pξ0 (f )(0) = 1, which implies that (1 + ξ 2 )Pξ (f ) ∈ SM (1+ξ2 ) . Now, let f ∈ S3,ξ . We have Z +∞ 1 (1 + ξ 2 )Pξ00 (f )(z) = (1 + ξ 2 ) · f 00 (zeiu )e2iu e−|u|/ξ du, 2ξ −∞ which implies |(1 + ξ 2 )Pξ00 (f )(z)| ≤ (1 + ξ 2 ) 2

that is (1 + ξ )Pξ (f ) ∈ S3 .

1 · 2ξ

Z

+∞ −∞

|f 00 (zeiu )|e−|u|/ξ du ≤ 1,

Remarks. 1) Since the constant (1+ξ 2 ) does not influence the geometric properties of Pξ (f ), it follows that for all ξ > 0 we have: if f ∈ S1 then Pξ (f ) is starlike (and univalent) in D1 ; 1 if f ∈ SM then Pξ (f ) is univalent in z ∈ C; |z| < M (1+ξ 2) ; if f ∈ S3,ξ ⊂ S3 then Pξ (f ) is starlike and univalent in D1 . 2) Since Z +∞ 1 0 Pξ (f )(z) = f 0 (zeiu )eiu e−|u|/ξ du, 2ξ −∞ it is obvious that the condition Re[f 0 (z)] > 0, ∀z ∈ D1 , does not imply Re[Pξ0 (f )(z)] > 0 on D1 . In this case, we may follow the idea in e.g. Gal [93] Theorem 3.4, to construct anRz other singular integral, as follows: for f ∈ A∗ (D1 ), we define Sξ (f )(z) = 0 Qn (u) du with Z +∞ 1 f 0 (zeit )e−|t|/ξ dt. Qn (z) = 2ξ −∞ Then, it is an easy task to show that (1 + ξ 2 )Sξ (R) ⊂ R, for all ξ > 0 and the following estimate holds |Sξ (f )(z) − f (z)| ≤ Cω2 (f 0 ; ξ)∂D1 , ∀z ∈ D1 , ξ > 0. 1 Since inf 1+ξ = 12 , by Theorem 3.2.3 it is immediate the following 2 ; ξ ∈ [0, 1] result. Corollary 3.2.4. (Anastassiou-Gal [18]) Pξ S3, 12 ⊂ S3 and f ∈ SM implies that 1 Pξ (f ) is univalent in z ∈ C; |z| < 2M , for all ξ ∈ [0, 1]. 1 Remark. Of course that if we consider, for example, ξ ∈ 0, 21 , then inf 1+ξ 2; 1 4 4 x ∈ 0, 2 = 5 andby Theorem 3.2.3 get Pξ S13, 5 ⊂ S3 and f ∈ SM implies 4 , ∀ξ ∈ 0, 2 . Pξ (f ) is univalent in z ∈ C; |z| < 5M 1 4 Obviously S3, 21 ⊂ S3, 45 and z ∈ C; |z| < 2M ⊂ z ∈ C; |z| < 5M . In what follows the properties of Qξ (f ), Q∗ξ (f ) and Rξ (f ) will be studied.

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First we present their approximation properties. Note that Theorem 3.2.5, (i)(iii) were proved in Anastassiou-Gal [18], while Theorem 3.2.5, (iv) is new. P∞ Theorem 3.2.5. (i) If f (z) = k=0 ak z k is analytic in D1 , then for all ξ > 0, Qξ (f )(z), Q∗ξ (f )(z), Rξ (f )(z) are analytic in D1 and for all z ∈ D1 we have, Z ∞ X 2ξ π cos ku k ak bk (ξ)z , with bk (ξ) = Qξ (f )(z) = du, π 0 u2 + ξ 2 k=0

Q∗ξ (f )(z)

=

∞ X

ak b∗k (ξ)z k ,

with

k=0

Rξ (f )(z) =

∞ X

b∗k (ξ)

2ξ = π

ak ck (ξ)z k , with ck (ξ) =

k=0

Z

4ξ 3 π

+∞ 0

Z

∞ 0

cos ku du, u2 + ξ 2 cos ku du. (u2 + ξ 2 )2

Also, if f is continuous on D1 then Qξ (f ), Q∗ξ (f ) and Rξ (f ) are also continuous on D1 ; Here b1 (ξ) > 0, ∀ξ > 0, b∗k (ξ) = e−kξ , ck (ξ) = (1 + kξ)e−kξ , for all ξ > 0 and S k ∈ N {0}. (ii) If f is continuous on D1 then for all δ > 0 and ξ > 0 it follows ω1 (Q∗ξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 , ω1 (Qξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 , ω1 (Rξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 . (iii) Let f ∈ A(D1 ). For all z ∈ D1 and ξ > 0 it follows |Qξ (f )(z) − f (z)| ≤ C

ω2 (f ; ξ)∂D1 + Cξkf kD1 ξ

|Rξ (f )(z) − f (z)| ≤ Cω2 (f ; ξ)∂D1 .

If f ∈ Lipα (M ; D1 ) with 0 < α < 1, that is |f (u) − f (v)| ≤ M |u − v|α for all u, v ∈ D1 , then |Q∗ξ (f )(z) − f (z)| ≤ Cξ α , ∀|z| ≤ 1, ξ > 0,

where C > 0 is independent of ξ and z but depends on f . Moreover, if there exists an R > 1 such that f is analytic in DR then for any 1 ≤ r < R we have |Q∗ξ (f )(z) − f (z)| ≤ Cξ, ∀|z| ≤ r, ξ > 0,

where C > 0 is independent of ξ and z but depends on f and r. P∞ (iv) In addition, let us suppose that f (z) = k=0 ak z k for all z ∈ DR , R > 1. If f is not constant for q = 0 and not a polynomial of degree ≤ q − 1 for q ∈ N, S then for all 1 ≤ r < r1 < R, ξ ∈ (0, 1] and q ∈ N {0} we have k[Q∗ξ ](q) (f ) − f (q) kr ∼ ξ,

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(q)

kRξ (f ) − f (q) kr ∼ ξ 2 ,

where the constants in the equivalences depend only on f , q, r, r1 . P k Proof. (i) Let f (z) = ∞ k=0 ak z , z ∈ D1 . Reasoning as for the case of Picardtype integral in Theorem 3.2.1, (i), we obtain: Z π ∞ X ξ 1 eiku · 2 Qξ (f )(z) = ak z k du , π −π u + ξ2 k=0

where

Z Z ξ π cos ku ξ π sin ku 1 du = du + i du u2 + ξ 2 π −π u2 + ξ 2 π −π u2 + ξ 2 −π Z 2ξ π cos ku = du = bk (ξ). π 0 u2 + ξ 2 R∞ iu ) Since obviously we can write Q∗ξ (f )(z) = πξ −∞ fu(ze 2 +ξ 2 du, it follows Z +∞ ∞ X ξ 1 du , Q∗ξ (f )(z) = ak z k eiku · 2 π −∞ u + ξ2 ξ π

Z

π

eiku ·

k=0

where

ξ π

Z

Z 1 2ξ ∞ cos ku du = du = b∗k (ξ); 2 + ξ2 2 + ξ2 u π u −∞ 0 Z ∞ +∞ 3 X eiku 2ξ du , ak z k Rξ (f )(z) = π −∞ (u2 + ξ 2 )2 +∞

eiku ·

k=0

and where 2ξ 3 π

Z

+∞ −∞

eiku ·

1 4ξ 3 du = 2 2 2 (u + ξ ) π

Z

∞ 0

cos ku du. (u2 + ξ 2 )2

The continuity of f on D1 implies the continuity of Qξ (f ), Q∗ξ (f ) and Rξ (f ) as in the proof of Theorem 3.2.1, (i) for Pξ (f ). It remains to show that b1 (ξ) > 0, b∗k (ξ) = e−kξ and ck (ξ) = (1 + kξ)e−kξ , for all ξ > 0 and k ≥ 0. Indeed, firstly we have "Z # Z Z π π/2 2ξ π cos u 2ξ cos u cos u b1 (ξ) = du = du + du 2 2 π 0 u2 + ξ 2 π u2 + ξ 2 0 π/2 u + ξ "Z # Z π/2 π/2 cos u sin u 2ξ = du − du 2 π u2 + ξ 2 0 0 u + π2 + ξ 2 Z 2ξ π/2 cos u − sin u du > π 0 u2 + ξ 2 "Z # Z π/2 π/4 cos u − sin u 2ξ cos u − sin u = du + du π u2 + ξ 2 u2 + ξ 2 0 π/4 :=

2ξ [I1 + I2 ]. π

bernstein

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Z

215

Z π/4 cos u − sin u cos u − sin u du > du u2 + ξ 2 (π 2 /16) + ξ 2 0 0 √ π/4 16( 2 − 1) 16 sin u + cos u = = 2 . 0 π + 16ξ 2 π 2 + 16ξ 2

0 < I1 =

π/4

Also, I2 < 0 and Z π/2 Z π/2 sin u − cos u 1 |I2 | = −I2 = [sin u − cos u]du du ≤ 2 · u2 + ξ 2 (π /16) + ξ 2 π/4 π/4 √ π/2 16( 2 − 1) 16 = 2 − cos u − sin u = , π/4 π + 16ξ 2 π 2 + 16ξ 2

which implies I1 + I2 ≥ 0. Therefore, it follows b1 (ξ) > 2ξ π [I1 + I2 ] ≥ 0, for all ξ > 0. ∗ For bk (ξ) we obtain Z Z 2ξ +∞ dv 1 1 2ξ +∞ cos ku du = cos v b∗k (ξ) = π 0 u2 + ξ 2 π 0 v 2 /k 2 + ξ 2 k Z +∞ 1 2kξ cos v dv = e−kξ . = π 0 v2 + k2 ξ 2 iz

Applying now the classical residue’s theorem to f (z) = ze2 +1 , it is immediate R∞ π −ξ , which implies b∗1 (ξ) = π2 · π2 e−ξ = e−ξ , for all ξ > 0. For that 0 cos(uξ) u2 +1 du = 2 e Z Z ∞ cos ku 4ξ 3 4k 3 ξ 3 ∞ cos v ck (ξ) = · du = dv, 2 2 2 2 π (u + ξ ) π (v + k 2 ξ 2 )2 0 0 iz

applying the residue’s theorem to f (z) = (z2 e+η2 )2 , we immediately get Z ∞ π cos u du = 3 (1 + η)e−η , 2 + η 2 )2 (u 4η 0

that is replacing η = kξ it follows ck (ξ) = (1 + kξ)e−kξ , ∀ξ > 0. (ii) Let z1 , z2 ∈ D1 be with |z1 − z2 | ≤ δ. We get Z ξ +∞ |f (z1 e−iu ) − f (z2 e−iu )| |Q∗ξ (f )(z1 ) − Q∗ξ (f )(z2 )| ≤ du π −∞ u2 + ξ 2 Z ξ +∞ du ≤ ω1 (f ; |z1 − z2 |)D1 ≤ ω1 (f ; δ)D1 , π −∞ u2 + ξ 2

where from passing to supremum after z1 , z2 it follows ω1 (Q∗ξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 . Also Z ξ π |f (z1 eiu ) − f (z2 eiu )| |Qξ (f )(z1 ) − Qξ (f )(z2 )| ≤ du π −π u2 + ξ 2 Z π du ξ ≤ ω1 (f ; |z1 − z2 |)D1 · π −π u2 + ξ 2 Z ξ +∞ du = ω1 (f ; δ)D1 . ≤ ω1 (f ; δ)D1 · π −∞ u2 + ξ 2

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The reasonings for Rξ (f ) are similar, which proves (ii). (iii) We can write Z ξ π f (zeiu ) − 2f (z) + f (ze−iu ) du − f (z)E(ξ), Qξ (f )(z) − f (z) = π 0 u2 + ξ 2

where

|E(ξ)| = E(ξ) = 1 −

2ξ π

Z

π 0

2 π 2 du = 1 − arctg ≤ 2 ξ u2 + ξ 2 π ξ π

(for the last estimate |E(ξ)| ≤ π22 ξ see e.g. Gal [94], p. 257). Passing to modulus, it follows Z ξ π |f (zeiu ) − 2f (z) + f (ze−iu )| du + kf kD1 |E(ξ)| |Qξ (f )(z) − f (z)| ≤ π 0 u2 + ξ 2 Z ξ π ω2 (f ; u)∂D1 ≤ du + kf kD · |E(ξ)| π 0 u2 + ξ 2 2 Z π u 1 ξ 1+ du + Cξkf kD1 . ≤ C · ω2 (f ; ξ)∂D1 · 2 + ξ2 π ξ u 0

Reasoning as in the proof of Theorem 3.1, pp. 257–258 in Gal [94], we arrive at the desired estimate. For Q∗ξ (f )(z) we have Z ξ ∞ [f (ze−iu ) − f (z)] Q∗ξ (f )(z) − f (z) = du. π −∞ u2 + ξ 2 By f ∈ Lipα (M ; D1 ) and since by the maximum modulus theorem we can take |z| = 1, we obtain Z ξ ∞ |f (ze−iu ) − f (z)| ∗ |Qξ (f )(z) − f (z)| ≤ du π −∞ u2 + ξ 2 Z ∞ −iu Z |e − 1|α ξ ξ ∞ (2| sin(u/2)|)α ≤M du = M du π −∞ u2 + ξ 2 π −∞ u2 + ξ 2 Z ∞ uα ξ du ≤ 2M π 0 u2 + ξ 2 Z ∞ 2M α vα = ξ dv, π v2 + 1 0 R∞ α where it is easy to show that 0 v2v+1 dv < ∞. Now, if f is supposed to be analytic in DR with R > 1, we can write f (z) = P∞ k immediately implies that the series k=0 ak z , for all |z| ≤ r with 1 ≤ r < R, which P∞ P∞ is uniformly and absolutely convergent, that is k=0 |ak |·|z|k ≤ k=0 |ak |·rk < ∞. Taking into account the above point (i), since by the mean value theorem we have |e−kξ − 1| ≤ ξ for all k = 0, 1, ...,, for any |z| ≤ r we get ∞ ∞ X X ∗ −kξ k |Qξ (f )(z) − f (z)| = |ak |rk , ak [e − 1]z ≤ ξ k=0

k=0

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217

P∞ therefore by the above considerations we can take C = k=0 |ak |rk < ∞. For Rξ (f )(z) we obtain Z 2ξ 3 +∞ |f (zeiu ) − 2f (z) + f (ze−iu )| du |Rξ (f )(z) − f (z)| ≤ π 0 (u2 + ξ 2 )2 Z 2ξ 3 +∞ ω2 (f ; u)∂D1 ≤C du π 0 (u2 + ξ 2 )2 Z 2ξ 3 +∞ ω2 (f ; ξ(u/ξ))∂D1 du =C π 0 (u2 + ξ 2 )2 2 Z u 1 2ξ 3 ∞ 1+ · 2 ≤ Cω2 (f ; ξ)∂D1 · du π 0 ξ (u + ξ 2 )2 ≤ Cω2 (f ; ξ)∂D1 , since by easy calculation we get that 2 Z u 2ξ 3 ∞ 1 1+ du ≤ C, · 2 π 0 ξ (u + ξ 2 )2 where C > 0 is independent of ξ. (iv) Analysing the proofs of the above points (i)-(iii), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation errors in (iii) can be expressed in terms of ω2 (f ; ξ)∂Dr = sup{|∆2u f (reit )|; |t| ≤ π, |u| ≤ ξ}, with constants in front of ω2 depending on r ≥ 1. Since ∆2u g(t) = 2u2 [t, t + u, t + 2u; g], by the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get |∆2u f (reit )| ≤ u2 kf 00 kr , where ∞ X kf 00 kr = sup{|f 00 (z)|; |z| ≤ r} ≤ |ck |k(k − 1)rk−2 . k=2

For all ξ ∈ (0, 1] from (iii) it follows kQ∗ξ (f ) − f kr ≤ Cr (f )ξ. Also, from (iii) we immediately obtain kRξ (f ) − f kr ≤ Cr (f )ξ 2 , for all ξ ∈ (0, 1]. Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and ξ ∈ (0, 1], we have Z ∗ q! Qξ (f )(v) − f (v) dv |[Q∗ξ ](q) (f )(z) − f (q) (z)| = 2π γ (v − z)q+1 q! 2πr1 ≤ Cr1 (f )ξ 2π (r1 − r)q+1 q!r1 , = Cr1 (f )ξ (r1 − r)q+1

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which proves the upper estimate for k[Q∗ξ ](q) (f ) − f (q) kr . In a similar manner it follows an upper estimate of the form (q)

kRξ (f )(z) − f (q) (z)kr ≤ Cq,r,r1 (f )ξ 2 , for all ξ ∈ (0, 1]. (q)

It remains to prove the lower estimates for k[Q∗ξ ](q) (f )−f (q) kr and kRξ (f )−f (q) kr . First we deal with the Q∗ξ operator. For this purpose, take z = reiϕ and p ∈ S N {0}. We have 1 (q) [f (z) − (Q∗ξ )(q) (f )(z)]e−ipϕ 2π ∞ 1 X = ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − b∗k (ξ)] 2π =

1 2π

k=q ∞ X k=q

ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − e−kξ ].

Integrating from −π to π we obtain Z π 1 [f (q) (z) − (Q∗ξ )(q) (f )(z)]e−ipϕ dϕ 2π −π

= aq+p (q + p)(q + p − 1)...(p + 1)r p [1 − e−(q+p)ξ ].

Passing now to absolute value we easily obtain |aq+p |(q + p)(q + p − 1)...(p + 1)r p [1 − e−(q+p)ξ ] ≤ kf (q) − (Q∗ξ )(q) (f )kr . First consider q = 0 and denote Vξ = inf 1≤p (1 − e−pξ ). We get Vξ = 1 − e−ξ and by the mean value theorem applied to h(x) = e−x on [0, ξ] there exists η ∈ (0, ξ) such that for all ξ ∈ (0, 1] we have Vξ = h(0) − h(ξ) = (−ξ)h0 (η) = (ξ)e−η ≥ (ξ)e−ξ ≥ e−1 ξ ≥ ξ/3. By the above lower estimate for kQ∗ξ (f ) − f kr , for all p ≥ 1 and ξ ∈ (0, 1] it follows kQ∗ξ (f ) − f kr kQ∗ξ (f ) − f kr 3kQ∗ξ (f ) − f kr ≥ ≥ ≥ |ap |rp . ξ Vξ 1 − e−pξ This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 kQ∗ (f )−f kr

= 0 then ap = 0 for all p ≥ 1, that is f is and such that limk→∞ ξk ξk constant on Dr . kQ∗ (f )−f kr > 0, which implies that Therefore, if f is not a constant then inf ξ∈(0,1] ξ ξ there exists a constant Cr (f ) > 0 such that that is

kQ∗ ξ (f )−f kr ξ

≥ Cr (f ), for all ξ ∈ (0, 1],

kQ∗ξ (f ) − f kr ≥ Cr (f )ξ, for all ξ ∈ (0, 1]. Now, consider q ≥ 1 and denote Vq,ξ = inf p≥0 (1 − e−(p+q)ξ ). Evidently that we have Vq,ξ ≥ inf p≥1 (1 − e−pξ ) ≥ ξ/3.

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Reasoning as in the case of q = 0 we obtain 3k[Q∗ξ ](q) (f ) − f (q) kr k[Q∗ξ ](q) (f ) − f (q) kr k[Q∗ξ ](q) (f ) − f (q) kr ≥ ≥ ξ Vq,ξ 1 − e−(p+q)ξ (q + p)! p r , ≥ |aq+p | p! for all p ≥ 0 and ξ ∈ (0, 1]. This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 k[Q∗ ](q) (f )−f (q) kr

ξk and such that limk→∞ = 0 then aq+p = 0 for all p ≥ 0, that is f ξk is a polynomial of degree ≤ q − 1 on Dr .

k[Q∗ ](q) (f )−f (q) kr

ξ Therefore, inf ξ∈(0,1] > 0 when f is not a polynomial of deξ gree ≤ q − 1, which implies that there exists a constant Cr,q (f ) > 0 such that (q) k[Q∗ (f )−f (q) kr ξ] ξ

≥ Cr,q (f ), for all ξ ∈ (0, 1], that is

k[Q∗ξ ](q) (f ) − f (q) kr ≥ Cr,q (f )ξ, for all ξ ∈ (0, 1]. Now, let us consider the lower estimate in the case of Rξ operator. For this purpose, S take z = reiϕ and p ∈ N {0}. We have 1 (q) (q) [f (z) − Rξ (f )(z)]e−ipϕ 2π ∞ 1 X = ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − ck (ξ)] 2π =

1 2π

k=q ∞ X k=q

ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − (1 + kξ)e−kξ ].

Integrating from −π to π we obtain Z π 1 (q) [f (q) (z) − Rξ (f )(z)]e−ipϕ dϕ 2π −π

= aq+p (q + p)(q + p − 1)...(p + 1)r p [1 − (1 + (q + p)ξ)e−(q+p)ξ ].

Passing now to absolute value we easily obtain (q)

|aq+p |(q + p)(q + p − 1)...(p + 1)r p [1 − (1 + (q + p)ξ)e−(q+p)ξ ] ≤ kf (q) − Rξ (f )kr . First consider q = 0 and denote Vξ = inf 1≤p (1 − (1 + pξ)e−pξ ). We easily get Vξ = 1 − (1 + ξ)e−ξ and we have Vξ ≥

ξ2 for all ξ ∈ (0, 1]. 4 2

Indeed, denoting F (ξ) = 1 − (1 + ξ)e−ξ − ξ4 we get F 0 (ξ) = 1 − 2ξ + ξe−ξ > 0, for all ξ ∈ (0, 1]. That is F is increasing and since F (0) = 0 it follows the above inequality for Vξ .

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By the above lower estimate for kRξ (f ) − f kr , for all p ≥ 1 and ξ ∈ (0, 1] it follows kRξ (f ) − f kr kRξ (f ) − f kr 4kRξ (f ) − f kr ≥ ≥ ≥ |ap |rp . ξ2 Vξ 1 − (1 + pξ)e−pξ

This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 kR (f )−f kr = 0 then ap = 0 for all p ≥ 1, that is f is and such that limk→∞ ξk ξ2 k

constant on Dr . Therefore, if f is not a constant then inf ξ∈(0,1] that there exists a constant Cr,p (f ) > 0 such ξ ∈ (0, 1], that is

kRξ (f )−f kr > 0, which implies ξ2 kRξ (f )−f kr ≥ Cr,p (f ), for all that ξ2

kRξ (f ) − f kr ≥ Cr,p ξ 2 , for all ξ ∈ (0, 1].

Now, consider q ≥ 1 and denote Vq,ξ = inf p≥0 (1 − (1 + (q + p)ξ)e−(q+p)ξ ). Evidently that we have Vq,ξ ≥ inf p≥1 (1 − (1 + pξ)e−pξ ) ≥ ξ 2 /4. Reasoning as in the case of q = 0 we obtain (q)

(q)

4kRξ (f ) − f (q) kr

≥

ξ2

kRξ (f ) − f (q) kr

Vq,ξ (q + p)! p ≥ |aq+p | r , p!

(q)

≥

kRξ (f ) − f (q) kr

1 − (1 + (q + p)ξ)e−(q+p)ξ

for all p ≥ 0 and ξ ∈ (0, 1]. This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 (q)

and such that limk→∞

kRξ (f )−f (q) kr k

ξk2

polynomial of degree ≤ q − 1 on Dr . kR

(q)

= 0 then aq+p = 0 for all p ≥ 0, that is f is a

(f )−f (q) kr

Therefore, inf ξ∈(0,1] ξ ξ2 > 0 when f is not a polynomial of degree ≤ q − 1, which implies that there exists a constant Cr,q (f ) > 0 such that (q)

kRξ (f )−f (q) kr ξ2

≥ Cr,q (f ), for all ξ ∈ (0, 1], that is (q)

kRξ (f ) − f (q) kr ≥ Cr,q (f )ξ 2 , for all ξ ∈ (0, 1], which proves the theorem.

Remark. The estimates in approximation by Q∗ξ (f )(z) contained in Theorem 3.2.5 ω (f ;ξ)

(iii), correct the wrong estimate |Q∗ξ (f )(z) − f (z)| ≤ C 2 ξ ∂D1 obtained in the paper Anastassiou-Gal [18], Theorem 3.1, (ii) (and also stated without proof in the book Gal [77], p. 318, Theorem 5.4.2, 2) ). Also, note that Theorem 3.2 (ii) in Anastassiou-Gal [18] still remains valid since we can use there the second estimate for |Q∗ξ (f )(z) − f (z)| in Theorem 3.2.5 (iii), which implies the uniform convergence of Q∗ξ (f ) to f on D1 . In what follows we present some geometric properties of the complex Poisson– Cauchy convolutions.

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P∞ k Theorem 3.2.6. (Anastassiou-Gal [18]) (i) If f (z) = k=0 ak z , z ∈ D1 and P∞ k ∗ Tξ (f )(z) = k=0 Ak z , is any from Qξ (f ), Qξ (f ) and Rξ (f ), then P∞

|Ak | ≤ |ak |,

∀k = 0, 1, . . . , ;

(ii) If f (z) = k=1 ak z k , z ∈ D1 is univalent in D1 and f (D1 ) is convex, then for any ξ > 0, Qξ (f )(z) is close-to-convex on D1 ; (iii) For all ξ > 0 we have : Q∗ξ (P) ⊂ P, Rξ (P) ⊂ P; 1 1 · Qξ (S3,b1 (ξ) ) ⊂ S3 , ∗ · Q∗ξ (S3,b∗1 (ξ) ) ⊂ S3 , b1 (ξ) b1 (ξ) 1 1 · Rξ (S3,c1 (ξ) ) ⊂ S3 , Qξ (SM ) ⊂ SM/|b1 (ξ)| , c1 (ξ) b1 (ξ) 1 1 Q∗ (SM ) ⊂ SM/|b∗1 (ξ)| , Rξ (SM ) ⊂ SM/|c1 (ξ)| , b∗1 (ξ) ξ c1 (ξ) where S3,a = {f ∈ S3 ; |f 00 (z)| ≤ |a|} and SB = {f ∈ A∗ (D1 ); |f 0 (z)| < B, z ∈ D1 }. Proof. (i) With the notations in the statement of Theorem 3.2.5, (i), for all k = 0, 1, 2, . . ., we obtain Z Z 2ξ π | cos ku| 2ξ π du |bk (ξ)| ≤ du ≤ π 0 u2 + ξ 2 π 0 u2 + ξ 2 π 2ξ 1 2 u π = · arctg = arctg ≤ 1, π ξ ξ 0 π ξ ∞ 1 u 2ξ · arctg = 1, |b∗k (ξ)| ≤ π ξ ξ 0 3 Z ∞ 4ξ du |ck (ξ)| ≤ = 1, π 0 (u2 + ξ 2 )2 which immediately implies (i). (ii) First, it is immediate that we can write Z ξ π f (ze−iu ) du. Qξ (f )(z) = π −π u2 + ξ 2 1 Since h(u) = u2 +ξ 2 satisfies h(π) = h(−π), we may extend it by 2π-periodicity on the whole R, such that this extension is continuous on R. By h0 (u) = (u2−2u +ξ 2 )2 , it follows that h is non-decreasing on [−π, 0] and nonincreasing on [0, π]. Then by Suffridge [187], p. 799, Theorem 3, it follows that Qξ (f )(z) is close-to-convex on D1 . (iii) Let f ∈ P, f = U + iV , U > 0. Then by definitions, it easily follows that Qξ (f ), Q∗ξ (f ), Rξ (f ) ∈ P. We take here into account that by Theorem 3.2.5, (i), the condition a0 = f (0) = 1, implies Z ξ +∞ du Q∗ξ (f )(0) = a0 b∗0 (ξ) = b∗0 (ξ) = =1 π −∞ u2 + ξ 2

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and Rξ (f )(0) = a0 c0 (ξ) = Now, let f (z) =

∞ P

2ξ 3 π

Z

+∞ −∞

(u2

du = 1. + ξ 2 )2

ak z k , with a0 = 0, a1 = 1. First, by Theorem 3.2.5, (i), we get

k=0

1 1 1 Qξ (f )(0) = 0, Q0 (f )(0) = 1, ∗ Q∗ξ (f )(0) = 0, b1 (ξ) b1 (ξ) ξ b1 (ξ) 1 1 1 · [Q∗ξ (f )]0 (0) = 1, Rξ (f )(0) = 0, · Rξ0 (f )(0) = 1. ∗ b1 (ξ) c1 (ξ) c1 (ξ) Then Q00ξ (f )(z) =

ξ π

Z

π −π

f 00 (zeiu )e2iu ·

1 du, u2 + ξ 2

Z 1 ξ +∞ 00 −iu −2iu ∗ f (ze )e · 2 Qξ (f ) (z) = du, π −∞ u + ξ2 Z 00 1 2ξ 3 +∞ 00 iu 2iu f (ze )e · 2 du. Rξ (f ) (z) = π −∞ (u + ξ 2 )2

00

Let f ∈ S3,b1 (ξ) . We get Z 1 1 ξ π 00 iu 1 00 ≤ · Q (f )(z) · du |f (ze )| · 2 ξ |b1 (ξ)| π b1 (ξ) u + ξ2 −π Z π du 2 ξ π = arctg ≤ 1, ≤ π −π u2 + ξ 2 π ξ that is b11(ξ) · Qξ (f ) ∈ S3 . Let f ∈ S3,b∗1 (ξ) . We get Z 1 1 1 ξ +∞ 00 iu ∗ 00 |f (ze )| · 2 du b∗ (ξ) · [Q (f )] (z) ≤ |b∗ (ξ)| · π u + ξ2 −∞ 1 1 Z ξ +∞ du ≤ = 1, π −∞ u2 + ξ 2 that is b∗1(ξ) Q∗ξ (f ) ∈ S3 . The proof in the case of c11(ξ) · Rξ (f ) is similar. 1 Now, let f ∈ SM . It follows Z 1 1 1 ξ π 0 iu 0 |f (ze )| · 2 du b1 (ξ) Qξ (f )(z) ≤ |b1 (ξ)| π u + ξ2 −π M 2 π M < · arctg ≤ . |b1 (ξ)| π ξ |b1 (ξ)| The proofs in the case of the theorem.

1 b∗ 1 (ξ)

· Q∗ξ (f ) and

1 c1 (ξ)

· Rξ (f ) are similar, which proves

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Remarks. 1) Theorem 3.2.6 (iii) says that if f ∈ S3,b1 (ξ) then Qξ (f ) is starlike and univalent on D1 and if f ∈ SM/|b1 (ξ)| then Qξ (f ) is univalent in the disk |b1 (ξ)| 1 z ∈ C; |z| < ⊂ z ∈ C; |z| < . M M

Similar properties hold for Q∗ξ (f ), b∗1 (ξ), and Rξ (f ), c1 (ξ). 2) Let us denote B = inf{|b1 (ξ)|; ξ ∈ (0, 1]}. If B > 0, then by Theorem 3.2.6 (iii), the following properties hold: f ∈ S3,B implies Qξ (f ) ∈ S3 , for all ξ ∈ (0, 1], B , ∀ξ ∈ (0, 1]. Therefore it f ∈ SM , (M > 1) implies Qξ (f ) is univalent in |z| < M remains to calculate B, to check if B > 0, problems which are left to the reader as open questions. Now, since inf{|b∗1 (ξ)|; ξ ∈ (0, 1]} = inf{e−ξ ; ξ ∈ (0, 1]} = 1e and inf{|c1 (ξ)|; ξ ∈ (0, 1]} = inf{(1 + ξ)e−ξ ; ξ ∈ (0, 1]} = 2e (since h(ξ) = (1 + ξ)e−ξ is decreasing on [0, 1]), from Theorem 3.2.5 (i) and Theorem 3.2.6 (iii) we immediately get the following. Corollary 3.2.7. (Anastassiou-Gal [18]) (i) If f ∈ S3, 1e then Q∗ξ (f ) ∈ S3 , for all 1 ξ ∈ (0, 1] and if f ∈ SM , (M > 1), then Q∗ξ (f ) is univalent in z ∈ C; |z| < eM , for all ξ ∈ (0, 1]. (ii) If f ∈ S3, 2e then Rξ (f ) ∈ S3 for all ξ ∈ (0, 1] and if f ∈ SM then Rξ (f ) is 2 univalent in |z| < eM , for all ξ ∈ (0, 1]. The complex convolutions Wξ (f )(z) and Wξ∗ (f )(z) are studied in what follows. Concerning the Theorem 3.2.8 below, note here that (i), (ii) and the first estimate in (iii) were obtained in Anastassiou-Gal [18], while the second estimate (that for Wξ∗ (f )(z)) in (iii) and both estimates in (iv) are new. Theorem 3.2.8. (i) If f (z) =

∞ P

ak z k is analytic in D1 , then for all ξ > 0,

k=0

Wξ (f )(z) and Wξ∗ (f )(z) are analytic in D1 and for all z ∈ D1 we have Wξ (f )(z) =

∞ X

ak dk (ξ)z k ,

k=0

with 1 dk (ξ) = √ · πξ

Z

Wξ∗ (f )(z) =

π

e−u

2

/ξ

cos ku du

−π ∞ X

ak d∗k (ξ)z k ,

k=0

with d∗k (ξ)

1 =√ πξ

Z

+∞

e−u

2

/ξ

cos ku du.

−∞

Also, if f is continuous on D1 then Wξ (f ) and Wξ∗ (f ) are continuous on D1 . Here S 2 d1 (ξ) > 0 and d∗k (ξ) = e−k ξ/4 > 0 for all ξ > 0 and k ∈ N {0}.

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(ii) Let f be continuous on D1 . For all δ > 0 and ξ > 0 we have ω1 (Wξ∗ (f ); δ)D1 ≤ ω1 (f ; δ)D1 , ω1 (Wξ (f ); δ)D1 ≤ ω1 (f ; δ)D1 .

(iii) Let f ∈ A(D1 ). For all ξ ∈ (0, 1] and z ∈ D1 it follows p ω2 (f ; ξ)∂D1 |Wξ (f )(z) − f (z)| ≤ C + C ξkf kD1 , ξ p ∗ |Wξ (f )(z) − f (z)| ≤ Cω2 (f ; ξ)∂D1 . P∞ (iv) In addition, let us suppose that f (z) = k=0 ak z k for all z ∈ DR , R > 1. If f is not constant for q = 0 and not a polynomial of degree ≤ q − 1 for q ∈ N, S then for all 1 ≤ r < r1 < R, ξ ∈ (0, 1] and q ∈ N {0} we have k[Wξ∗ ](q) (f ) − f (q) kr ∼ ξ,

where the constants in the equivalences depend only on f , q, r, r1 . Proof.

(i) Reasoning as for the Pξ (f ) operator, we can write Z +∞ X ∞ 2 1 ak z k eiuk e−u /ξ du Wξ∗ (f )(z) = √ πξ −∞ k=0 Z +∞ ∞ X 2 1 [cos(ku) + i sin(ku)]e−u /ξ du = ak z k · √ πξ −∞ k=0 =

∞ X

ak d∗k (ξ)z k ,

k=0

where

d∗k (ξ)

1 =√ πξ

Z

+∞

cos(ku)e−u

2

/ξ

du.

−∞

The reasonings in the case of Wξ (f )(z) are similar. The proof of continuity on D1 of Wξ (f ) and Wξ∗ (f ) is similar to that for Pξ (f ) in the proof of Theorem 3.2.1, (i). 2

It remains to prove that d1 (ξ) > 0, ∀ξ > 0 and that d∗k (ξ) = e−k ξ/4 > 0, for all ξ > 0 and k ≥ 0. Indeed, firstly we have Z π Z π 2 2 1 2 d1 (ξ) = √ e−u /ξ cos u du = √ cos u e−(u/η) du, πη 0 πξ −π √ where η = ξ > 0. We obtain # "Z Z π π/2 2 −(u/η)2 −(u/η)2 cos u e du + cos u e du d1 (η) = √ · πη 0 π/2 "Z Z π/2 2 # π/2 2 − u+π/2 −(u/η)2 η du − sin u e = √ cos u e du πη 0 0 "Z # π/2 2 2 −(u/η)2 (cos u − sin u)e du := √ [I1 + I2 ], > √ πη 0 πη

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where I1 = and

Z

I2 = − It follows I1 > and

Z

π/4 0

π/4

2

(cos u − sin u)e−(u/η) du > 0

0

Z

225

π/2 π/4

2

(sin u − cos u)e−(u/η) du < 0.

√ 2 2 (cos u − sin u)e−(π/(4η)) du = ( 2 − 1)e−(π/(4η))

|I2 | = −I2 < e−(π/(4η))

2

Z

π/2 π/4

√ 2 (sin u − cos u) du = ( 2 − 1)e−(π/(4η)) .

Therefore,

√ √ 2 2 d1 (η) > I1 + I2 ≥ ( 2 − 1)e−(π/(4η)) − ( 2 − 1)e−(π/(4η)) = 0,

for any η > 0. Now, for d∗k (ξ) we have 1 d∗k (ξ) = √ πξ

Z

+∞

e−u

2

/(ξ)

cos ku du = e−k

2

ξ/4

,

−∞

for all ξ > 0 and k ≥ 0. (ii) For |z1 − z2 | < δ, we get

Z +∞ 2 1 |f (z1 e−iu ) − f (z2 e−iu )|e−u /ξ du |Wξ∗ (f )(z1 ) − Wξ∗ (f )(z2 )| ≤ √ · πξ −∞ ≤ ω1 (f ; |z1 − z2 |)D1 ≤ ω1 (f ; δ)D1 ,

and 1 |Wξ (f )(z1 ) − Wξ (f )(z2 )| ≤ √ πξ

Z

+π −π

|f (z1 eiu ) − f (z2 eiu )|e−u

2

/ξ

du

Z π 2 1 e−u /ξ du ≤ ω1 (f ; |z1 − z2 |)D1 · √ πξ −π Z +∞ 2 1 e−u /ξ du ≤ ω1 (f ; δ)D1 · √ πξ −∞ = ω1 (f ; δ)D1 .

(iii) We can write 1 Wξ (f )(z) − f (z) = √ πξ

Z

π

[f (zeiu ) − 2f (z) + f (ze−iu )]e−u 0 Z π 2 1 −f (z) 1 − √ e−u /ξ du . πξ −π

2

/ξ

du

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Here

bernstein

Z π Z π −u2 /ξ −u2 /ξ f (z) 1 − √1 · = f (z) 1 − √2 e du e du πξ −π πξ 0 Z ∞ Z π 2 2 2 2 = f (z) √ e−u /ξ du − √ e−u /ξ du πξ 0 πξ 0 Z ∞ 2 2 e−u /ξ du = |f (z)| · √ πξ π Z ∞ p 2 1 ξ du = 2kf kD1 ξ · √ . ≤ kf kD1 · √ 2 π π πξ π u

By the maximum modulus principle we can take |z| = 1, which implies Z π p 1 2 1 |Wξ (f )(z) − f (z)| ≤ √ ω2 (f ; u)∂D1 e−u /ξ du + 2kf kD1 ξ √ π π πξ 0 ≤ (reasoning as in Gal [94], p. 258) p 1 Cω2 (f ; ξ)∂D1 + 2kf kD1 ξ · √ . ≤ ξ π π R∞ R0 R∞ Also, writing −∞ = −∞ + 0 , we easily get Z ∞ 2 1 Wξ∗ (f )(z) − f (z) = √ [f (zeiu ) − 2f (z) + f (ze−iu )]e−u /ξ du, πξ 0 which for |z| = 1 implies |Wξ∗ (f )(z)

1 − f (z)| ≤ √ πξ

Z

+∞

ω2 (f ; u)∂D1 e−u

0

p

1 ξ)∂D1 √ πξ p ≤ Cω2 (f ; ξ)∂D1 , ≤ ω2 (f ;

since 1 √ πξ and 2 √ πξ

Z

∞ 0

Z

∞ 0

1 u2 −u2 /ξ e du = √ ξ π

2 u 2 p √ e−u /ξ du = √ ξ ξ πξ

Z

∞ 0

ve

Z

Z

∞ 0

∞ 0

−v 2

2

/ξ

du

u √ +1 ξ

2

e−u

2

/ξ

du

2

v 2 e−v dv < ∞

2 dv = √ π

Z

∞ 0

2

ve−v dv < ∞.

(iv) Analysing the proofs of the above points (i)-(iii), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation error in (iii) can be expressed in terms of p p ω2 (f ; ξ)∂Dr = sup{|∆2u f (reit )|; |t| ≤ π, |u| ≤ ξ},

with constants in front of ω2 depending on r ≥ 1.

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Since ∆2u g(t) = 2u2 [t, t + u, t + 2u; g], by the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get |∆2u f (reit )| ≤ u2 kf 00 kr ,

where 00

00

kf kr = sup{|f (z)|; |z| ≤ r} ≤ For all ξ ∈ (0, 1] from (iii) it follows

∞ X

k=2

|ak |k(k − 1)rk−2 .

kWξ∗ (f ) − f kr ≤ Cr (f )ξ.

Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and ξ ∈ (0, 1], we have Z q! Wξ∗ (f )(v) − f (v) ∗ (q) (q) |[Wξ ] (f )(z) − f (z)| = dv 2π γ (v − z)q+1 2πr1 q! ≤ Cr1 (f )ξ 2π (r1 − r)q+1 q!r1 = Cr1 (f )ξ , (r1 − r)q+1

which proves the upper estimate in approximation by [Wξ∗ ](q) (f )(z). It remains to prove the lower estimate for k[Wξ∗ ](q) (f )−f (q) kr . For this purpose, S take z = reiϕ and p ∈ N {0}. We have 1 (q) [f (z) − (Wξ∗ )(q) (f )(z)]e−ipϕ 2π ∞ 1 X ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − d∗k (ξ)] = 2π =

1 2π

k=q ∞ X k=q

ak k(k − 1)...(k − q + 1)r k−q eiϕ(k−q−p) [1 − e−k

Integrating from −π to π we obtain Z π 1 [f (q) (z) − (Wξ∗ )(q) (f )(z)]e−ipϕ dϕ 2π −π

= aq+p (q + p)(q + p − 1)...(p + 1)r p [1 − e−(q+p)

2

ξ/4

2

ξ/4

].

].

Passing now to absolute value we easily obtain

|aq+p |(q + p)(q + p − 1)...(p + 1)r p [1 − e−(q+p)

2

ξ/4

] ≤ kf (q) − (Wξ∗ )(q) (f )kr . 2

First consider q = 0 and denote Vξ = inf 1≤p (1 − e−p ξ/4 ). We get Vξ = 1 − e−ξ/4 and by the mean value theorem applied to h(x) = e−x/4 on [0, ξ] there exists η ∈ (0, ξ) such that for all ξ ∈ (0, 1] we have Vξ = h(0) − h(ξ) = (−ξ)h0 (η) = (ξ/4)e−η/4 ≥ (ξ/4)e−ξ/4 ≥

e−1/4 ξ ≥ ξ/8. 4

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By the above lower estimate for kWξ∗ (f ) − f kr , for all p ≥ 1 and ξ ∈ (0, 1] it follows 8kWξ∗ (f ) − f kr kWξ∗ (f ) − f kr kWξ∗ (f ) − f kr ≥ |ap |rp . ≥ ≥ ξ Vξ 1 − e−p2 ξ/4 This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 kW ∗ (f )−f kr

ξk and such that limk→∞ = 0 then ap = 0 for all p ≥ 1, that is f is ξk constant on Dr . kW ∗ (f )−f kr Therefore, if f is not a constant then inf ξ∈(0,1] ξ ξ > 0, which implies

that there exists a constant Cr (f ) > 0 such that ξ ∈ (0, 1], that is

kWξ∗ (f )−f kr ξ

≥ Cr (f ), for all

kWξ∗ (f ) − f kr ≥ Cr (f )ξ, for all ξ ∈ (0, 1]. Now, consider q ≥ 1 and denote Vq,ξ = inf p≥0 (1 − e−(p+q) 2 have Vq,ξ ≥ inf p≥1 (1 − e−p ξ/4 ) ≥ ξ/8. Reasoning as in the case of q = 0 we obtain

2

ξ/4

). Evidently that we

k[Wξ∗ ](q) (f ) − f (q) kr k[Wξ∗ ](q) (f ) − f (q) kr 8k[Wξ∗ ](q) (f ) − f (q) kr ≥ ≥ ξ Vq,ξ 1 − e−(p+q)2 ξ/4 (q + p)! p ≥ |aq+p | r , p! for all p ≥ 0 and ξ ∈ (0, 1]. This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 k[W ∗ ](q) (f )−f (q) kr

ξk = 0 then aq+p = 0 for all p ≥ 0, that is f and such that limk→∞ ξk is a polynomial of degree ≤ q − 1 on Dr .

k[W ∗ ](q) (f )−f (q) kr

ξ Therefore, inf ξ∈(0,1] > 0 when f is not a polynomial of deξ gree ≤ q − 1, which implies that there exists a constant Cr,q (f ) > 0 such that

k[Wξ∗ ](q) (f )−f (q) kr ξ

≥ Cr,q (f ), for all ξ ∈ (0, 1], that is

k[Wξ∗ ](q) (f ) − f (q) kr ≥ Cr,q (f )ξ, for all ξ ∈ (0, 1], which proves the theorem.

Concerning the geometric properties of the complex Gauss–Weierstrass convolutions, we present the following results. P∞ k Theorem 3.2.9. (Anastassiou-Gal [18]) (i) If f (z) = k=0 ak z , z ∈ D1 and P∞ Tξ (f )(z) = k=0 Ak z k is any from Wξ (f )(z) and Wξ∗ (f )(z) then P∞

|Ak | ≤ |ak |,

∀k = 0, 1, . . . ;

(ii) If f (z) = k=1 ak z k , z ∈ D1 is univalent in D1 and f (D1 ) is convex, then for any ξ > 0, Wξ (f )(z) is univalent in D1 and Wξ (f )(D1 ) is convex.

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Similarly, if f (z) is univalent in D1 and f (D1 ) is starlike with respect to the origin, then for any ξ > 0, Wξ (f )(z) is univalent in D1 and Wξ (f )(D1 ) is starlike with respect to the origin. (iii) For all ξ > 0 we have : 1 Wξ (S3,d1 (ξ) ) ⊂ S3 , d1 (ξ) 1 1 Wξ∗ (S3,d∗1 (ξ) ) ⊂ S3 , Wξ (SM ) ⊂ SM/|d1 (ξ)| , ∗ d1 (ξ) d1 (ξ) 1 W ∗ (SM ) ⊂ SM/|d∗1 (ξ)| . d∗1 (ξ) ξ

Wξ∗ (P) ⊂ P,

Proof.

(i) By Theorem 3.2.8 (i), we get

Z π 2 1 |ak dk (ξ)| ≤ |ak | · |dk (ξ)| ≤ |ak | · √ e−u /ξ | cos ku| du πξ −π Z +∞ 2 1 e−u /ξ du = |ak |, for all k = 0, 1, 2, . . . . ≤ |ak | · √ πξ −∞ Also, by the same theorem, we obtain |ak d∗k (ξ)| = |ak | · |d∗k (ξ)| ≤ |ak |,

for all k = 0, 1, 2, . . . .

Also, note that |d0 (ξ)| = d0 (ξ) ≤ 1 and |d∗0 (ξ)| = d0 (ξ) = 1. 2 (ii) Let g(u) = e−u /ξ , u ∈ [−π, π]. Since g(−π) = g(π), we can extend g(u) by 2π-periodicity on the whole R, such that the extension, denoted by h(u), is continuous on R. It is easy to check that log |h0 (u)| is concave in each interval [kπ, (k + 1)π], 0 h (u) = 0 if and only if u = 2kπ, k ∈ Z, and in uk = kπ, k ∈ Z, h takes its minimum and maximum values. Then, applying Ruscheweyh-Salinas [161], Theorem, p. 130, we get that h is PMP as in Ruscheweyh-Salinas [161], which implies that Wξ (f ) preserves the convexity of f . Also, by similar reasoning with those in P´ olya-Schoenberg [150], p. 321, Lemma 5 and Corollary 5, it follows that Wξ (f )(z) preserve the starlikeness of f (z) (with respect to origin) too. (iii) The proofs are similar to the proofs in Theorem 3.2.6 (iii), which proves Theorem 3.2.9 too. Remarks. 1) From the results presented above, it follows that Wξ (f )(z) has the best preservation property among the classes of complex singular integrals studied by the present book. 2) Let us denote D = inf{|d1 (ξ)|; ξ ∈ (0, 1]}. If D > 0 then by Theorem 3.2.9 (iii) we get: if f ∈ S3,D then Wξ (f ) ∈ S3 , for all ξ ∈ (0, 1],

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D , for all if f ∈ SM , (M > 1), then Wξ (f ) is univalent in z ∈ C; |z| < M ξ ∈ (0, 1]. Therefore it remains to calculate D, to check if D > 0, problems which are left to the reader as an open question. Since inf{|d∗1 (ξ)|; ξ ∈ (0, 1]} = 1/(e1/4 ), applying now Theorems 3.2.8 (i) and 3.2.9 (iii) to Wξ∗ (f )(z), we immediately get the following result. then Wξ∗ (f ) ∈ S3 , for all ξ ∈ (0, 1] and if f ∈ SM , (M > 1), then Wξ∗ (f ) is univalent in z ∈ C; |z| < M e11/4 , for all ξ ∈ (0, 1]. In what follows we study the Jackson-type generalizations of Picard-type Pn,ξ (f )(z), of Poisson–Cauchy–type Qn,ξ (f )(z), and of generalized Gauss– ∗ Weierstrass–type Wn,ξ (f )(z) and Wn,ξ (f )(z). The approximation results are expressed by the following theorem. Note that Theorem 3.2.11, (i), (ii) and (iii) were obtained in Anastassiou-Gal [16] while Theorem 3.2.11 (iv) is new. Corollary 3.2.10. (Anastassiou-Gal [18]) If f ∈ S3,

1 e1/4

Theorem 3.2.11. (i) Suppose that f is continuous on D1 . For all δ > 0, ξ > 0 and n ∈ N, we have ω1 Pn,ξ (f ); δ D ≤ (2n+1 − 1)ω1 (f ; δ)D1 , 1 ω1 Wn,ξ (f ); δ D1 ≤ (2n+1 − 1)ω1 (f ; δ)D1 , ∗ ω1 Wn,ξ (f ); δ D ≤ (2n+1 − 1)ω1 (f ; δ)D1 , 1 ω1 Qn,ξ (f ); δ D1 ≤ (2n+1 − 1)ω1 (f ; δ)D1 . (ii) Suppose that f ∈ A(D1 ). For all z ∈ D1 and ξ ∈ (0, 1] we have "n+1 # X n + 1 |Pn,ξ (f )(z) − f (z)| ≤ k! ωn+1 (f ; ξ)∂D1 , k k=0 R∞ 2 (1 + u)n+1 e−u du 0 Rπ , |Wn,ξ (f )(z) − f (z)| ≤ Cn ωn+1 (f ; ξ)∂D1 , Cn = e−u2 du 0 R∞ 2 (1 + u)n+1 e−u du ∗ ∗ ∗ 0 R∞ |Wn,ξ (f )(z) − f (z)| ≤ Cn ωn+1 (f ; ξ)∂D1 , Cn = , e−u2 du 0 R π/ξ (u+1)n+1 |Qn,ξ (f )(z) − f (z)| ≤ K(n, ξ)ωn+1 (f ; ξ)∂D1 , K(n, ξ) =

where

0

u2 +1 tan−1 πξ

du

,

ωn+1 (f ; ξ)∂D1 = sup |∆n+1 f (eix )|; |x| ≤ π, |u| ≤ ξ . u

(iii) If f (z) =

∞ P

ak z k is analytic in D1 and continuous in D1 , then Pn,ξ (f )(z),

k=0 ∗ Wn,ξ (f )(z), Wn,ξ (f )(z) and Qn,ξ (f )(z) are analytic in D1 and continuous in D1 , for all ξ > 0 and n ≥ 2.

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Also, we can write Pn,ξ (f )(z) =

∞ X

ap bp,n (ξ)z p ,

p=0

with bp,n (ξ) =

n+1 X

(−1)

k+1

k=1

Wn,ξ (f )(z) =

n+1 1 · 2 2 2 , ξ k p +1 k

∞ X

ap cp,n (ξ)z p ,

p=0

with Z π n+1 2 2 1 X k+1 n + 1 (−1) cos(kpu)e−u /ξ du, cp,n (ξ) = · C(ξ) k 0 k=1

∗ Wn,ξ (f )(z)

=

∞ X

ap c∗p,n (ξ)z p ,

p=0

with c∗p,n (ξ)

Z ∞ n+1 2 2 1 X k+1 n + 1 cos(kpu)e−u /ξ du, · (−1) = ∗ k C (ξ) 0 k=1

Qn,ξ (f )(z) =

∞ X

ap dp,n (ξ)z p ,

p=0

with dp,n (ξ) =

Z π n+1 X ξ cos(kpu) k+1 n + 1 (−1) du. arctg πξ u2 + ξ 2 k 0 k=1

Here for all ξ ∈ (0, ξn ] we have, n+1 X n+1 1 > 0, b1,n (ξ) = (−1)k+1 2 2 ξ k +1 k k=1 Z π n+1 X 2 2 1 k+1 n + 1 c1,n (ξ) = cos(ku)e−u /ξ du > 0, · (−1) C(ξ) k 0 k=1 Z n+1 ∞ 2 2 1 X n+1 (−1)k+1 c∗1,n (ξ) = ∗ cos(ku)e−u /ξ du C (ξ) k 0 k=1 n+1 X n + 1 −k2 ξ2 /4 (−1)k+1 = e > 0, k k=1

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232

and

Z π n+1 X ξ cos ku k+1 n + 1 (−1) d1,n (ξ) = du > 0, π · 2 2 k arctg ξ 0 u +ξ k=1

where 0 < ξn is independent of k and f (but may depend on n). P∞ k (iv) Let us suppose that f (z) = k z for all z ∈ DR , R > 1. For all k=0 aS 1 ≤ r < r1 < R, ξ ∈ (0, 1], n ∈ N and q ∈ N {0} we have (q)

kPn,ξ (f ) − f (q) kr ≤ C1 ξ n+1 , for all ξ ∈ (0, 1],

∗ (q) k[Wn,ξ ] (f ) − f (q) kr ≤ C2 ξ n+1 , for all ξ ∈ (0, 1],

where the constants C1 , C2 depend only on f , q, r, r1 and n. Proof.

(i) Let |z1 − z2 | ≤ δ, z1 , z2 ∈ D1 . We have

|Pn,ξ (f )(z1 ) − Pn,ξ (f )(z2 )| Z +∞ n+1 X n + 1 1 ≤ ω1 f ; |z1 − z2 | D1 · e−|u|/ξ du 2ξ −∞ k k=1 n+1 X n + 1 ≤ ω1 (f ; δ)D1 = 2n+1 − 1 ω1 (f ; δ)D1 . k k=1

As above, we obtain

Z π n+1 X n + 1 2 1 · e−u /ξ du · ω1 (f ; δ)D1 2C(ξ) −π k k=1 n+1 X n+1 ≤ ω1 (f ; δ)D1 = (2n+1 − 1)ω1 (f ; δ)D1 k

|Wn,ξ (f )(z1 ) − Wn,ξ (f )(z2 )| ≤

k=1

and analogously Finally,

∗ ∗ |Wn,ξ (f )(z1 ) − Wn,ξ (f )(z2 )| ≤ (2n+1 − 1)ω1 (f ; δ)D1 .

1 |Qn,ξ (f )(z1 ) − Qn,ξ (f )(z2 )| ≤ 2 π ξ arctg ξ = 2

n+1

Z

π −π

n+1 X n + 1 du · ω1 (f ; δ)D1 k u2 + ξ 2

− 1 ω1 (f ; δ)D1 .

k=1

Passing in all the above inequalities to sup with |z1 − z2 | ≤ δ, we obtain the required relations in (i). (ii) Let |z| = 1, ξ > 0 be fixed. Because of the maximum modulus principle, it suffices to estimate |Pn,ξ (f )(z) − f (z)|, for this |z| = 1, z = eix . We get Z +∞ 1 e−|u|/ξ du f (z) − Pn,ξ (f )(z) = f (z) · 2ξ −∞ # Z +∞ "n+1 X 1 k n+1 (−1) f (ei(x+ku) )e−|u|/ξ du + 2ξ −∞ k k=1 Z +∞ 1 = (−1)n+1 ∆n+1 f (eix )e−|u|/ξ du, u 2ξ −∞

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where from

Z +∞ 1 ωn+1 (f ; |u|)∂D1 e−|u|/ξ du 2ξ −∞ Z u 1 +∞ ωn+1 f ; · ξ = e−u/ξ du ξ 0 ξ ∂D1 n+1 Z u 1 +∞ e−u/ξ du 1+ ≤ ωn+1 (f ; ξ)∂D1 ξ 0 ξ = (reasoning exactly as in Gal [94], p. 254) n+1 X n + 1 = k!ωn+1 (f ; ξ)∂D1 . k

|f (z) − Pn,ξ (f )(z)| ≤

k=0

As above, we obtain

Z π 2 2 1 (−1)n+1 ∆n+1 f (eix )e−u /ξ du, f (z) − Wn,ξ (f )(z) = n 2C(ξ) −π which implies Z π 2 2 1 ωn+1 (f ; u)∂D1 e−u /ξ du |f (z) − Wn,ξ (f )(z)| ≤ C(ξ) 0 n+1 Z π 2 2 u 1 1+ ωn+1 (f ; ξ)∂D1 e−u /ξ du ≤ C(ξ) ξ 0 (reasoning exactly as in Gal [94], p. 260) R +∞ 2 [1 + u]n+1 e−u du 0 R ≤ · ωn+1 (f ; ξ)∂D1 . π −u2 e du 0 Similarly, Z +∞ 2 2 1 ∗ (−1)n+1 ∆n+1 f (eix )e−u /ξ du, f (z) − Wn,ξ (f )(z) = u ∗ 2C (ξ) −∞ which implies as above n+1 Z ∞ 2 2 u 1 ∗ 1+ e−u /ξ du |f (z) − Wn,ξ (f )(z)| ≤ ∗ ωn+1 (f ; ξ)∂D1 C (ξ) ξ 0 R +∞ n+1 −u2 [1 + u] e du R∞ ≤ 0 ωn+1 (f ; ξ)∂D1 . −u2 du 0 e Finally, by the relation Z π (−1)n+1 ∆n+1 f (eix ) 1 u du, f (z) − Qn,ξ (f )(z) = 2 π u2 + ξ 2 ξ arctg ξ −π

it follows (taking into account Anastassiou-Gal [18], p. 518 too) Z π ξ ωn+1 (f ; u)∂D1 |f (z) − Qn,ξ (f )(z)| ≤ du arctg πξ 0 u2 + ξ 2 n+1 Z π 1 u ξ · 2 +1 du ωn+1 (f ; ξ)∂D1 ≤ arctg πξ ξ u + ξ2 0 = K(n, ξ)ωn+1 (f ; ξ)∂D1

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which proves (ii). P∞ (iii) Let f (z) = p=0 ap z p , z ∈ D1 . For fixed z ∈ D1 , we can write f (zeiku ) = P∞ P∞ ikpu p z and since |ap eikpu | = |ap | for all u ∈ R and the series p=0 ap z p is p=0 ap e P∞ convergent, it follows that the series p=0 ap eikpu z p is uniformly convergent with respect to u ∈ R. Therefore the series can be integrated term by term (with respect to u), that is ∞ Z +∞ n+1 1 X n+1 X ap z p (−1)k eikpu e−|u|/ξ du. 2ξ k −∞ p=0

Pn,ξ (f )(z) = −

k=1

But −

1 2ξ

Z

+∞

eikpu e−|u|/ξ du Z ∞ 1 = − [cos(kpu) + i sin(kpu)]e−|u|/ξ du 2ξ −∞ Z 1 ∞ =− cos(kpu)e−u/ξ du ξ 0 1 −u/ξ − ξ cos(kpu) + k sin(kpu) +∞ 1 e = − · ξ k 2 p2 + ξ12 0 −∞

1

=

−ξ 1 1 · 2 2 2 · ξ2 = − 2 2 2 . ξ ξ k p +1 ξ k p +1

Therefore we can write Pn,ξ (f )(z) =

∞ X p=0

"

ap z · −

n+1 X

bp,n (ξ) =

n+1 X

p

k=1

(−1)

k

# ∞ X 1 n+1 = ap bp,n (ξ)z p , · 2 2 2 ξ k p +1 k p=0

with

k=1

(−1)k+1

n+1 1 · 2 2 2 , ξ k p +1 k

for all z ∈ D1 . For the continuity property, let z ∈ D1 and zm ∈ D1 , n ∈ N, with lim zm = z0 . m→∞ We have n+1 1 X n+1 |Pn,ξ (f )(zm ) − Pn,ξ (f )(z0 )| ≤ 2ξ k k=1 Z +∞ · |f (zm eiku ) − f (z0 eiku )|e−|u|/ξ du −∞

≤ (2n+1 − 1)ω1 (f ; |zm − z0 |)D1 .

Passing to limit with m → ∞, we get that Pn,ξ (f )(z) is continuous on D1 .

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∗ The proofs for the other operators Wn,ξ (f )(z), Wn,ξ (f )(z) and Qn,ξ (f )(z) are similar. The formulas for b1,n (ξ), c1,n (ξ) and d1,n (ξ) are immediate from above. √ R ∞ −u2 /ξ2 R ∞ −v2 ξ π ∗ Also, since C (ξ) = 0 e du = ξ 0 e dv = 2 and

Z

∞

cos(ku)e−u

2

/ξ 2

Z

du = ξ

0

∞

2

cos(kξv)e−v dv =

0

√ 2 2 ξ π · e−k ξ /4 , 2

we get c∗1,n (ξ)

√ n+1 X 2 ξ π −k2 ξ2 /4 k+1 n + 1 (−1) e = √ · · ξ π 2 k k=1 n+1 X 2 2 k+1 n + 1 e−k ξ /4 . = (−1) k k=1

Now, by n+1 X

0 = (−1 + 1)n+1 =

(−1)k

k=0

Pn+1 it follows k=1 (−1)k+1 Then, since

n+1 k

n+1 X n+1 n+1 , =1+ (−1)k k k k=1

= 1.

b1,n (ξ) =

n+1 X

(−1)k+1

k=1

n+1 1 ξ 2 k2 + 1 k

and c∗1,n (ξ)

=

n+1 X

(−1)

k+1

k=1

n + 1 −k2 ξ2 /4 e k

are obviously continuous functions of ξ ∈ R+ and b1,n (0) =

c∗1,n (0)

=

n+1 X

(−1)

k+1

k=1

n+1 k

= 1,

there exists ξn > 0 such that b1,n (ξ) > 0, c∗1,n (ξ) > 0, ∀ξ ∈ (0, ξn ]. Also, c1,n (ξ) and d1,n (ξ) are obviously continuous functions of ξ ∈ R+ \ {0}. Since Z π Z π/ξ 2 2 2 C(ξ) = e−u /ξ du = ξ e−v dv 0

0

and Z

π 0

cos(ku)e−u

2

/ξ 2

du = ξ

Z

π/ξ 0

2

cos(kξv)e−v dv,

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we get lim ξ↓0

Z

π

cos(ku)e 0

= lim ξ↓0

= lim ξ↓0

Z

Z

−u2 /ξ 2

du/C(ξ)

π/ξ

2

cos(kξv)e−v dv/ lim ξ↓0

0 π/ξ 0

2 = √ lim π ξ↓0

Z

2

cos(kξv)e−v dv/ π/ξ

Z

Z

∞

π/ξ

2

e−v dv 0 2

e−v dv

0

2

cos(kξv)e−v dv.

0

By the substitution ξv = u we get Z Z π/ξ 2 1 π −v 2 [1 − cos(ku)]e−(u/ξ) du, [1 − cos(kξv)]e dv = ξ 0 0 that is Z Z ∞ π/ξ −v 2 −v 2 cos(kξv)e dv − e dv 0 0 Z Z π/ξ π/ξ 2 2 e−v dv cos(kξv)e−v dv − ≤ 0 0 Z Z ∞ π/ξ 2 2 + e−v dv − e−v dv 0 0 Z Z π Z ∞ π/ξ 2 2 2 ≤ |1 − cos(ku)|ξ −1 e−(u/ξ) du + e−v dv − e−v dv . 0 0 0 Since

|1 − cos(ku)| = 2 sin2

2k 2 u2 k 2 u2 ku ≤ = , 2 4 2

we get 2

|1 − cos(ku)|ξ −1 e−(u/ξ) ≤

k 2 2 −1 −(u/ξ)2 u ξ e , 2

where 2 −1 −(u/ξ)2

lim u ξ ξ↓0

e

=

u2 ξ lim (u/ξ)2 ξ↓0 e 2

ξ→0

2

= lim

− uξ2

2 ξ↓0 − 2u3 eu2 /ξ 2 ξ

= lim

ξ

2 2 ξ↓0 eu /ξ

= 0,

that is |1 − cos(ku)|ξ −1 e−(u/ξ) −→ 0, uniformly with respect to u ∈ [0, π]. (We applied the l’Hospital’s rule). This immediately implies Z ∞ Z π/ξ 2 2 2 2 e−v dv = 1. lim cos(ku)e−u /ξ du/C(ξ) = √ ξ↓0 0 π 0

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Therefore, Rπ 2 2 cos(ku)e−u /ξ du n+1 0 lim c1,n (ξ) = (−1) lim ξ↓0 ξ↓0 C(ξ) k k=1 n+1 X n+1 = 1 > 0, = (−1)k+1 k n+1 X

k+1

k=1

which implies that there exists ξn > 0 such that c1,n (ξ) > 0, ∀ξ ∈ (0, ξn ]. In the case of d1,n (ξ), since 1 ξ = R π du = arctg πξ 0 u2 +u2

and Z

we get

π 0

1 cos ku du = u2 + ξ 2 ξ

ξ lim π · ξ↓0 arctg ξ

Z

π 0

Z

π/ξ 0

1 ξ

0

dv v 2 +1

cos(kξv) dv, v2 + 1 R π/ξ

cos(kξv) v 2 +1 dv R π/ξ dv v 2 +1 0 R π/ξ cos(kξv) limξ↓0 0 2 R ∞ dvv +1 0 v 2 +1 Z π/ξ

cos ku du = lim ξ↓0 u2 + ξ 2 =

1 R π/ξ

0

dv

2 cos(kξv) lim dv π ξ↓0 0 v2 + 1 Z ∞ 2 dv = · = 1. π 0 v2 + 1 =

Here, as in the above case, we write Z Z ∞ π/ξ cos(kξv) dv dv − 0 v2 + 1 v2 + 1 0 Z Z Z π/ξ Z ∞ π/ξ cos(kξv) dv π/ξ dv dv dv − − ≤ + 0 v2 + 1 v2 + 1 0 v2 + 1 v2 + 1 0 0 Z Z π/ξ Z ∞ π/ξ dv [1 − cos(kξv)] dv ≤ dv + − . 0 v2 + 1 v2 + 1 v2 + 1 0 0

But

Z

π/ξ

0

[1 − cos(kξv)] 1 dv = v2 + 1 ξ ≤

Z

k2 2

π

0

Z

(1 − cos ku) 1 du = u 2 ξ 1+ ξ

π 0

Z

π

0

u2 k2 1 du = · ξ 1+ u 2 2 ξ

2 sin2

1+ Z π 0

ξ

ku 2 u 2 ξ

du

u2 du. u2 + ξ 2

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2

Denote 0 ≤ gξ (u) = ξ u2u+ξ2 ≤ ξ. We obviously have lim gξ (u) = 0, uniformly with ξ↓0

respect to u ∈ [0, π], which implies Z Z π/ξ 2 2 ∞ dv cos(kξv) lim dv = = 1. π ξ↓0 0 v2 + 1 π 0 v2 + 1 Therefore, n+1 X n+1 = 1 > 0, (−1)k+1 lim d1,n (ξ) = ξ↓0 k k=1

which implies that there exists ξn > 0 such that d1,n (ξ) > 0, for all ξ ∈ (0, ξn ]. Obviously, we can choose the same ξn > 0 for all the four operators Pn,ξ (f ), ∗ Wn,ξ (f ), Wn,ξ (f ) and Qn,ξ (f ). (iv) Analysing the proofs of the above points (i)-(iii), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation error in (iii) can be expressed in terms of ωn+1 (f ; ξ)∂Dr = sup{|∆n+1 f (reit )|; |t| ≤ π, |u| ≤ ξ}, u

with constants in front of ω2 depending on r ≥ 1. By the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get where

|∆n+1 f (reit )| ≤ |u|n+1 kf (n+1) kr , u

kf (n+1) kr = sup{|f (n+1) (z)|; |z| ≤ r} ≤ Then by (ii) for all ξ ∈ (0, 1] it follows

∞ X

k=n+1

|ak |k(k − 1)...(k − n)r k−n−1 .

kPn,ξ (f ) − f kr ≤ Cr,n (f )ξ n+1 ,

∗ ∗ kWn,ξ (f ) − f kr ≤ Cr,n (f )ξ n+1 .

Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and ξ ∈ (0, 1], we have Z ∗ (f )(v) − f (v) q! Wn,ξ ∗ (q) (q) |[Wn,ξ ] (f )(z) − f (z)| = dv 2π (v − z)q+1 γ

2πr1 q! 2π (r1 − r)q+1 q!r1 , = Cr∗1 ,n (f )ξ n+1 (r1 − r)q+1

≤ Cr∗1 ,n (f )ξ n+1

(q)

which proves the upper estimate in approximation by Wn,ξ (f )(z). Similarly we get q!r1 (q) |Pn,ξ (f )(z) − f (q) (z)| ≤ Cr1 ,n (f )ξ n+1 , (r1 − r)q+1 which proves the theorem.

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The geometric properties of the Jackson-type generalized complex convolutions are contained in the following results. We present Theorem 3.2.12. (Anastassiou-Gal [16]) Let f ∈ SM . Then, for all ξ > 0, n ∈ N we have 1 Pn,ξ (SM ) ⊂ SM (2n+1 −1)/|b1,n (ξ)| , b1,n (ξ) 1 Wn,ξ (SM ) ⊂ SM (2n+1 −1)/|c1,n (ξ)| , c1,n (ξ) 1 W ∗ (SM ) ⊂ SM (2n+1 −1)/|c∗1,n (ξ)| , c∗1,n (ξ) n,ξ 1 Qn,ξ (SM ) ⊂ SM (2n+1 −1)/|d1,n (ξ)| . d1,n (ξ) Proof.

Let f (z) = z +

∞ X

k=2

ak z k ∈ A∗ (D1 ), |f 0 (z)| < M, ∀z ∈ D1 .

Since a0 = 0, by Theorem 3.2.11 (iii) we get ∗ Pn,ξ (f )(0) = Wn,ξ (f )(0) = Wn,ξ (f )(0) = Qn,ξ (f )(0) = 0.

Also, since a1 = 1, by Theorem 3.2.11 (iii) we get 1 1 0 · P 0 (f )(0) = · Wn,ξ (f )(0) b1,n (ξ) n,ξ c1,n (ξ) 1 1 ∗ · Q0n,ξ (f )(0) = 1, [Wn,ξ (f )]0 (0) = = ∗ c1,n (ξ) d1,n (ξ) which implies that 1 1 1 1 ∗ Pn,ξ (f ), Wn,ξ (f ), ∗ Qn,ξ (f ) ∈ A∗ (D1 ). · Wn,ξ (f ), b1,n (ξ) c1,n (ξ) c1,n (ξ) d1,n (ξ) Also, by 0 Pn,ξ (f )(z) = −

1 2ξ

Z

+∞ −∞

e−|u|/ξ

n+1 X k=1

(−1)k

n + 1 0 iku iku f (ze )e du, k

we obtain n+1 X 1 M n+1 M (2n+1 − 1) 0 k P (f )(z) < · |(−1) | = , n,ξ b1,n (ξ) |b1,n (ξ)| |b1,n (ξ)| k k=1

that is

1 · Pn,ξ (f )(z) ∈ SM (2n+1 −1)/|b1,n (ξ)| . b1,n (ξ)

The proofs for the other operators are similar, which proves the theorem.

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[138], p. 111, ExerRemarks. 1) Recall that by e.g. Mocanu–Bulboac˘ a–S˘ al˘ agean 1 ⊂ D1 . cise 5.4.1, f ∈ SM , M > 1, implies that f is univalent in z ∈ C; |z| < M Theorem 3.2.12 shows that f ∈ SM implies that Pn,ξ (f )(z) is univalent in |b1,n (ξ)| 1 z ∈ C; |z| < ⊂ z ∈ C; |z| < ⊂ D1 , M (2n+1 − 1) M since by Theorem 3.2.11, (iii), we have n+1 n+1 X n + 1 X n + 1 1 = 2n+1 − 1, ∀p = 0, 1, . . . . < · 2 2 2 |bp,n (ξ)| ≤ ξ k p +1 k k k=1

k=1

∗ For the operators Wn,ξ (f )(z), Wn,ξ (f )(z) and Qn,ξ (f )(z) similar conclusions hold by replacing above b1,n (ξ) by c1,n (ξ), c∗1,n (ξ) and d1,n (ξ), respectively. 2) For any fixed n ∈ N, let us denote

B1,n = inf{|b1,n (ξ)|; ξ ∈ (0, ξn ]}, C1,n = inf{|c1,n (ξ)|; ξ ∈ (0, ξn ]},

∗ C1,n = inf{|c∗1,n (ξ)|; ξ ∈ (0, ξn ]}, D1,n = inf{|d1,n (ξ)|; ξ ∈ (0, ξn ]}.

∗ If B1,n , C1,n , C1,n , D1,n > 0, then by Theorem 3.2.12 the following properties hold : B1,n f ∈ SM implies that Pn,ξ (f ) is univalent in z ∈ C; |z| < M (2n+1 −1) , for all x ∈ (0, ξn ], C1,n f ∈ SM implies that Wn,ξ (f ) is univalent in z ∈ C; |z| < M (2n+1 −1) , for all ξ ∈ (0, ξn ], ∗ C1,n ∗ f ∈ SM implies that Wn,ξ (f ) is univalent in z ∈ C; |z| < M (2n+1 −1) , for all ξ ∈ (0, ξn ], D1,n f ∈ SM implies that Qn,ξ (f ) is univalent in z ∈ C; |z| < M (2n+1 −1) , for all ξ ∈ (0, ξn ]. ∗ Therefore, it remains to calculate (for each fixed n ∈ N), B1,n , C1,n , C1,n , D1,n , ∗ to check if B1,n > 0, C1,n > 0, C1,n > 0, problems which are left to the reader as open questions. 3) It would be of interest to investigate for other geometric properties, for all the complex convolutions previously studied in Section 3.2. At the end of this subsection firstly we obtain some applications to PDE of complex variables and secondly we extend the above results in this subsection from compact disks with centers in origin to Jordan domains of rectifiable boundaries. Let f ∈ A(D1 ) and t ∈ (0, +∞). For reasons that come from physics, in the above expression of Wξ∗ (f )(z) let us replace ξ by 2t and for the simplicity of notation, ∗ (f )(z) denote instead of W2t Z +∞ 2 1 Wt∗ (f )(z) = √ f (ze−iu )e−u /(2t) du, z ∈ D1 . 2πt −∞

Indeed, the PDE in Theorem 3.2.13, (v) below represents the heat equation with complex spatial variable.

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Our first goal is to show that this complex convolution defines a contraction semigroup on the Banach space (A(D1 ), k·k1 ). Then applications to PDE equations with complex spatial variables and real time variable are considered. Theorem 3.2.13. (Gal-Gal-Goldstein [97]) Let f ∈ A(D1 ). (i) For all t > 0, Wt∗ (f ) ∈ A(D1 ) and Wt∗ (f )(z) =

∞ X

ak d∗k (t)z k ,

k=0

with d∗k (t) = √

1 2πt

Z

+∞

e−u

2

/(2t)

cos ku du = e−k

2

t/2

−∞

, k ≥ 1.

(ii) For all z ∈ D1 , t > 0, the following estimate holds : √ |Wt∗ (f )(z) − f (z)| ≤ Cω1 (f ; t)D1 . Here C > 0 is a constant independent of t and f . (iii) We have : √ √ |Wt∗ (f )(z) − Ws∗ (f )(z)| ≤ Cs | t − s|, for all z ∈ D1 , t ∈ Vs ⊂ (0, +∞), where Cs > 0 is a constant depending on f , independent of z and t and Vs is any neighborhood of s. (iv) The operator Wt∗ is contractive, that is kWt∗ (f )k1 ≤ kf k1 , for all t > 0, f ∈ A(D1 ). (v) (Wt∗ , t ≥ 0) is a (C0 )-contraction semigroup of linear operators on the Banach space (A(D1 ), k · k1 ) and the unique solution u(t, z) (that belongs to A(D 1 ), for each fixed t > 0) of the Cauchy problem 1 ∂2u ∂u (t, z) = (t, z), (t, z) ∈ (0, +∞) × D1 , z = reiϕ , z 6= 0, ∂t 2 ∂ϕ2 u(0, z) = f (z), z ∈ D1 , f ∈ A(D1 ), is given by u(t, z) = Wt∗ (f )(z) = √

1 2πt

Z

+∞

f (ze−iu )e−u

2

/(2t)

du.

−∞

Proof. (i) It is immediate by Theorem 3.2.8 (i). (ii) We obtain Z +∞ 2 1 |Wt∗ (f )(z) − f (z)| ≤ √ |f (ze−iu ) − f (z)|e−u /(2t) du 2πt −∞

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Z ∞ 2 1 ω1 (f ; |1 − e−iu |)D e−u /(2t) du 2πt −∞ Z +∞ 2 u 1 ω1 f ; 2 sin = √ e−u /(2t) du 2 D1 2πt −∞ Z +∞ 2 1 ω1 (f ; |u|)D1 e−u /(2t) du ≤ √ 2πt −∞ Z +∞ √ 2 1 |u| ω1 (f ; t)D1 √ + 1 e−u /(2t) du ≤ √ t 2πt −∞ √ Z √ ω1 (f ; t)D1 ∞ 2 = ω1 (f ; t)D1 + √ √ 2ue−u /(2t) du. t · 2πt 0 R∞ R∞ 2 Since 0 2ue−u /(2t) du = 2t 0 e−v dv = 2t, we infer i 2t h √ √ √ |Wt∗ (f )(z) − f (z)| ≤ ω1 (f ; t)D1 + ω1 (f ; t)D1 √ ≤ Cω1 (f ; t)D1 . t 2π (iii) We have Z 2 2 e−u /s kf k1 +∞ e−u /t ∗ ∗ |Wt (f )(z) − Ws (f )(z)| ≤ √ √ − √ du. s t 2π −∞ √ √ First, let us denote t = a, s = b. Applying now the mean value theorem, there exists a value c ∈ (a, b), such that 2 2 2 e−u2 /a2 e−u /b 1 −u2 /c2 2u − − 2 , = |a − b|e a b c4 c ≤ √

which together with the fact that Z +∞ Z 2 e−u /(2c) < ∞, −∞

+∞

−∞

u2 e−u

2

/(2c)

< ∞,

it immediately implies inequality for Wt∗ . R +∞the−udesired 2 1 /(2t) (iv) Since √2πt −∞ e du = 1, we deduce Z +∞ 2 1 |f (ze−iu )|e−u /(2t) du ≤ kf k, z ∈ D1 , |Wt∗ (f )(z)| ≤ √ 2πt −∞

which yields kWt∗ (f )k1 ≤ kf k1 . P∞ k (v) Let f ∈ A(D1 ), that is, f (z) = ∈ D1 . If z ∈ D1 , z = k=0 ak z , z P iϕ ∗ −k2 t/2 k kiϕ re , 0 < r < 1, then by (i), we can write Wt (f )(z) = ∞ r e . It k=0 ak e ∗ easily follows that Wt+s (f )(z) = Ws∗ [Wt∗ (f )](z), for all t, s > 0. If z is on the boundary of D1 , then we may take a sequence (zn )n∈N of points in D1 such that limn→∞ zn = z and we apply the above relationship and the continuity property from (i). Furthermore, denoting Wt∗ (f )(z) by T (t)(f ), it is easy to check that the property limt&0 T (t)(f ) = f , the continuity of T (·) and its contraction property follow from (ii), (iii) and (iv), respectively. Finally, all these facts together show that (Wt∗ , t ≥ 0) is a (C0 )-contraction semigroup of linear operators on A(D1 ).

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Consequently, since the above series representation for Wt∗ (f )(z) is uniformly convergent in any compact disk included in D1 , it can be differentiated term by term, ∂Wt∗ (f )(z) ∂ 2 Wt∗ (f )(z) with respect to t and ϕ. We then easily obtain that = (1/2) . ∂t ∂ϕ2 Finally, from the same series representation, it is easy to check that W0∗ (f )(z) = f (z), z ∈ D1 . We also note that in the differential equation we must take z 6= 0 simply because z = 0 has no polar representation, that is, z = 0 cannot be represented as function of ϕ. This completes the proof of the theorem. The next result shows that the solution of the above Cauchy problem in Theorem 3.2.13, (v), preserve some interesting geometric properties of the boundary function. Theorem 3.2.14. (Gal-Gal-Goldstein [97]) Let u(t, z) be the unique solution of the Cauchy problem in Theorem 3.2.13 (v). 1 then (i) As function of z, u(t, z) has the following properties : if f ∈ S3, 1/2 e u(t, z) ∈ S3 , and if f ∈ SM , (M > 1) then u(t, z) is univalent in z ∈ C; |z| < 1 , for all t ∈ (0, 1/2]. M e1/2 (ii) Assuming that f is analytic in an open set G including D1 , then for all t ∈ (0, tf ] with sufficiently small tf > 0 (depending on f ), the solution u(t, z) preserves as function of z, the starlikeness, convexity and spirallikeness of the boundary function f in D1 . Proof. (i) It is an immediate consequence of Corollary 3.2.10 for ξ = 2t. (ii) From Theorem 3.2.8 (i) and (iii), it is readily seen that Wt∗ (f )(z) is analytic in G and Wt∗ (f )(z) converges to f (z), (as t → 0) uniformly in D1 . This convergence together the well-known Weierstrass’s result yield that [Wt∗ (f )(z)]0 → f 0 (z) and [Wt∗ (f )(z)]00 → f 00 (z), uniformly in D 1 , as t → 0. In all what follows in the proof, Wt∗ (f )(z) from now on, Pt (f )(z) will denote b1 (t) , where b1 (t) is the coefficient of z, in the Taylor series representation of the analytic function Wt∗ (f )(z). We will also denote by Pt0 (f )(z) and Pt00 (f )(z), the corresponding first and second-order derivatives of Pt (f )(z) with respect to z. If f (0) = f 0 (0) − 1 = 0, it is not difficult to observe that Pt (f )(0) = bf1(0) (t) = 0, W 0 (f )(0)

Pt0 (f )(0) = tb1 (t) = 1 and b1 (t) converges to f 0 (0) = 1 as t → 0. This obviously implies that, for t → 0, we have Pt (f )(z) → f (z), Pt0 (f )(z) → f 0 (z) and Pt00 (f )(z) → f 00 (z), uniformly in D1 . First, suppose that f ∈ S ∗ (D1 ), that is f is starlike (and univalent) in D1 . The univalence implies that |f (z)| > 0 for all z ∈ D1 , z 6= 0 and that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 . Write Pt (f )(z) in the form Pt (f )(z) = zRt (f )(z). For |z| = 1, we have |f (z) − Pt (f )(z)| = |z| · |g(z) − Rt (f )(z)| = |g(z) − Rt (f )(z)|, which, by the uniform convergence in D1 , of Pt (f ) to f, and by the maximum modulus principle, yields the uniform convergence in D1 , of Rt (f )(z) to g(z), as

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t → 0. Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , then there exist an index t1 > 0 and a > 0 depending on g (that is, on f ), so that |Rt (f )(z)| > a > 0, for all z ∈ D1 and all t ∈ (0, t1 ). For all |z| = 1, we also have |f 0 (z) − Pt0 (f )(z)| = |z[g 0 (z) − Rt0 (f )(z)] + [g(z) − Rt (f )(z)]|

≥ ||z| · |g 0 (z) − Rt0 (f )(z)| − |g(z) − Rt (f )(z)|| = ||g 0 (z) − Rt0 (f )(z)| − |g(z) − Rt (f )(z)|| .

Clearly, the maximum modulus principle, the uniform convergence of Pt0 (f ) to f 0 and of Rt (f ) to g, imply the uniform convergence of Rt0 (f ) to g 0 , as t → 0. Then for |z| = 1, we obtain z[zRt0 (f )(z) + Rt (f )(z)] zPt0 (f )(z) = Pt (f ) zRt (f )(z) 0 zRt (f )(z) + Rt (f )(z) zg 0 (z) + g(z) = → Rt (f )(z) g(z) f 0 (z) zf 0 (z) = = . g(z) f (z) From this convergence and the maximum modulus principle, we infer zPt0 (f )(z) zf 0 (z) → , uniformly in D1 . Pt (f ) f (z) 0 (z) is continuous in D1 , there exists α ∈ (0, 1), so that Since Re zff (z) 0 zf (z) Re ≥ α, for all z ∈ D1 . f (z) Therefore 0 0 zf (z) zPt (f )(z) Re → Re ≥ α > 0, Pt (f )(z) f (z)

uniformly on D1 , that is, for any 0 < β < α there is tf > 0 so that for all t ∈ (0, tf ), we have 0 zPt (f )(z) Re > β > 0, for all z ∈ D1 . Pt (f )(z) Therefore, Pt (f ) ∈ S ∗ (D1 ), for all t ∈ (0, tf ) and since it only differs from Wt∗ (f ) by a constant, this proves the starlikeness in D1 of Wt∗ (f )(z), for all t ∈ (0, tf ). The proofs of the other cases, when f is convex or spirallike of order γ are similar and follow from the following uniform convergence on D1 : 00 00 zf (z) zPt (f )(z) + 1 → Re Re + 1, Pt0 (f )(z) f 0 (z) and

zP 0 (f )(z) zf 0 (z) Re eiγ nt → Re eiγ . Pt (f )(z) f (z)

The proof is complete.

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Replacing ξ by t (time) for reasons that come from physics, now we will consider the complex convolution operator of Poisson-Cauchy type Z t +∞ f (ze−iu ) Q∗t (f )(z) = du, z ∈ D1 . π −∞ u2 + t2 Theorem 3.2.15. (Gal-Gal-Goldstein [97]) Let f ∈ A(D1 ). (i) Then for all t > 0, Q∗t (f ) ∈ A(D1 ) and Q∗t (f )(z) =

∞ X

ak b∗k (t)z k ,

k=0

with b∗k (t) =

2t π

Z

+∞ 0

1 cos ku du = e−kt , k ≥ 1. u2 + t 2

(ii) For all z ∈ D1 , t ∈ Vs ⊂ (0, +∞), |Q∗t (f )(z) − Q∗s (f )(z)| ≤ Cs |t − s|, where Cs > 0 is a constant independent of z, t and f, and Vs is any neighborhood of (fixed) s > 0. (iii) The operator Q∗t is contractive, that is kQ∗t (f )k1 ≤ kf k1, for all t > 0, f ∈ A(D1 ). (iv) (Q∗t , t ≥ 0) is a (C0 )-contraction semigroup of linear operators on the Banach space (A(D1 ), k · k1 ) and the unique solution u(t, z) (that belongs to A(D 1 ), for each fixed t > 0) of the Cauchy problem ∂2u ∂2u (t, z) + (t, z) = 0, (t, z) ∈ D1 × (0, +∞) , z = reiϕ , z 6= 0, ∂t2 ∂ϕ2 u(0, z) = f (z), z ∈ D1 , f ∈ A(D1 ), is given by u(t, z) = Q∗t (f )(z) =

t π

Z

+∞ −∞

f (ze−iu )

du . t2 + u 2

Proof. (i) It is exactly Theorem 3.2.5 (i). (ii) We have Z kf k1 +∞ t s |Q∗t (f )(z) − Q∗s (f )(z)| ≤ − du . 2 2 2 2 π t +u s +u ∞ However, since

t2

s (t − s)(u2 − ts) t − 2 = 2 , 2 2 +u s +u (t + u2 )(s2 + u2 )

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integrating this relation with respect to u, for all t ∈ Vs , we obtain Z +∞ u2 − ts du 2 2 2 2 −∞ (t + u )(s + u ) Z =

Z +∞ 1 du 2 ≤ Cs < +∞, du − (t + ts) 2 2 2 2 2 2 −∞ s + u −∞ (u + t )(s + u ) R +∞ R +∞ 1 du since the integrals −∞ s2 +u 2 du and −∞ (u2 +t2 )(s2 +u2 ) are finite. This immediately implies (ii). R +∞ 1 (iii) Since πt −∞ t2 +u 2 du = 1, we deduce the following estimate : Z t +∞ 1 |Q∗t (f )(z)| ≤ |f (ze−iu )| 2 du ≤ kf k1 , z ∈ D1 . π −∞ t + u2 +∞

It is readily seen that kQ∗t (f )k1 ≤ kf k1 . (iv) The property Q∗t+s (f )(z) = Q∗t [Q∗s (f )] (z) , for all t, s > 0, f ∈ A(D1 ) and z ∈ D1 is immediate from the representation in (i). To prove that lim t&0 Q∗t (f ) = f , for any f ∈ A(D1 ), let f = U + iV , z = reix be fixed and denote F (v) = U [rcos(v), rsin(v)], G(v) = V [rcos(v), rsin(v)]. We can write Z Z t +∞ 1 1 t +∞ ∗ du + i du. F (x − u) 2 G(x − u) 2 Qt (f )(z) = π −∞ t + u2 π −∞ t + u2 From the maximum modulus principle, when estimating the quantity |Q∗t (f )(z) − f (z)|, we may take r = |z| = 1, that is, z = eix in the above formulas for F (v) and G(v). Therefore, passing to limit as t & 0 and taking into account the above property (see e.g. Goldstein [100], Exercise 2.18.8), we find Z +∞ t 1 ∗ lim |Qt (f )(z) − f (z)| ≤ lim F (x − u) 2 du − F (x) 2 t&0 t&0 π −∞ t +u Z +∞ t 1 = 0, G(x − u) 2 du − G(x) + lim 2 t&0 π −∞ t +u

which holds for z ∈ D1 . Combining this with the properties in (ii) and (iii), the first assertion of (iv) follows. Finally, from the series representation of Q∗t (f )(z) (cf. (i)) and reasoning exactly as in Theorem 3.2.13 (v), we can easily check the second assertion in (iv). This completes the proof of the theorem. Remark. The property limt&0 |Q∗t (f )(z) = f (z) for all z ∈ D1 proved in the above proof of Theorem 3.2.15 (iv), in fact holds uniformly with respect to z ∈ D1 . Indeed, since F and G in the proof of Theorem 3.2.15 (iv) are continuous and 2π-periodic on R, for all x ∈ R by Anghelutza [30] we have Z +∞ t 1 1 ≤ Cω1 (F ; ξ) log F (x − u) , du − F (x) π t2 + u 2 ω1 (F ; ξ) −∞

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and a similar estimate for G holds. Here log denotes the natural logarithm and C > 0 is independent of x and ξ. We can suppose that f is not a constant, that is that F and G are not constant (otherwise the trivial case Q∗t (f )(z) = f (z) for all z holds). Therefore, there exists a ξ0 > 0 such that ω1 (F ; ξ) > 0 for all 0 < ξ ≤ ξ0 . To simplify the notation denote H(ξ) = ω1 (F ; ξ), 0 ≤ ξ ≤ ξ0 . Clearly H(0) = 0, H(ξ) > 0 for ξ 6= 0 and H is continuous and increasing on [0, ξ0 ]. Therefore obviously that we can choose ξ0 > 0 such that in addition, H(ξ) < 1 for all ξ ∈ [0, ξ0 ]. We will show that limξ→0 H(ξ)[− log H(ξ)] = 0, which will imply the uniform convergence with respect to z ∈ D1 . For this purpose, let us suppose that would exists a sequence (ξn )n , ξn ∈ (0, ξ0 ], such that ξn → 0 as n → ∞ and limn→∞ H(ξn )[− log H(ξn )] = a > 0. This immediately implies that for any 0 < ε < a, there exists an n0 ∈ N such that |a − H(ξn )[− log H(ξn )]| < ε, for all n ≥ n0 , that is 0 < ρ = a − ε < H(ξn )[− log H(ξn )] for all n ≥ n0 . It follows ρ

1 1 ≤ log , for all n ≥ n0 , H(ξn ) H(ξn )

or denoting hn = H(ξn ) we get hn → ∞ and ρ≤

log hn , for all n ≥ n0 , hn

with ρ > 0 independent of n. But passing here to limit as n → ∞, since limn→∞ loghnhn = 0, we obtain the contradiction 0 < ρ = 0. In conclusion, for all the sequences (ξn )n , ξn ∈ (0, ξ0 ], such that ξn → 0 as n → ∞, we necessarily have limn→∞ H(ξn )[− log H(ξn )] = 0, which proves our assertion. The next result shows that the solution of the above Cauchy problem preserves some interesting geometric properties of the boundary function. Theorem 3.2.16. (Gal-Gal-Goldstein [97]) Let u(t, z) be the unique solution of the Cauchy problem in Theorem 3.2.15 (iv). (i) As function of z, u(t, z) has the following properties : if f ∈ S3, 1e , then u(t, z) ∈ S3 , and, if f ∈ SM , (M > 1) then u(t, z) is univalent in z ∈ C; |z| < 1 eM , for all t ∈ (0, 1] (ii) Assuming that f is analytic in an open set G including D1 , then, for all t ∈ (0, tf ] with a sufficiently small tf > 0 (depending on f ), the solution u(t, z) preserves as function of z, the starlikeness, convexity and spirallikeness of the boundary function f in D1 . Proof. (i) Both assertions are immediate consequences of Theorem 3.2.6 (iii). (ii) The proof is similar to that of Theorem 3.2.14 (ii), as a consequence of Theorem 3.2.5 (i) and (iii). We leave the rigorous details to the reader.

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In what follows we will extend some of the above results to more general domains in C. We will consider here the cases of the complex operators Pξ (f )(z) in Theorem 3.2.1, Q∗ξ (f )(z) in Theorem 3.2.5 and Wξ∗ (f )(z) in Theorem 3.2.8, which can be generalized for approximation of f ∈ A(G) in Jordan domains G ⊂ C, as follows. Keeping the notations in Section 1.0, let f ∈ A(G) and F ∈ A(D 1 ) be such that f = T [F ]. Here T denotes the Faber operator which is supposed to be continuous as mapping from A(D1 ) to A(G). S For ξ > 0, k ∈ N {0} and ρk,ξ ∈ (0, 1], attach to F ∈ A(D1 ), F (w) = P∞ k k=0 ak w , w ∈ D1 , the complex operators Oξ (F )(w) =

∞ X

ak ρk,ξ wk .

k=0

Then, suggested by the Remark after Theorem 1.0.12, attach to f = T [F ] the complex operator Lξ (f ) = T [Oξ (F )] given by Lξ (f )(z) =

∞ X k=0

ak ρk,ξ Fk (z), z ∈ G,

where Fk (z) denotes the Faber polynomial of degree k attached to G. But according to Theorem 1.0.11, we have ak = ak (f )-the Faber coefficients of f on G and according to Definition 1.0.10, (ii), we can write Z Z π 1 1 f [Ψ(u)] du = ak (f ) = f [Ψ(eit )]e−ikt dt, k ∈ N ∪ {0}. 2πi |u|=1 uk+1 2πi −π Therefore, the complex operators attached to an f ∈ A(G) can be formally written as ∞ X Lξ (f )(z) = ak (f )ρk,ξ Fk (z), z ∈ G. k=0

Remark. Of course that of interest are the cases when Lξ (f )(z) is well defined in G and belongs to A(G). For example, let us suppose that f ∈ A(G), where G is a bounded simply connected domain with the boundary Γ a regular analytic curve, that is Γ is an analytic mapping and its derivative never vanishes. In this case, by Theorem 1 and its proof, p. 51-52 in the book of Suetin [186] it follows that P∞ f (z) = k=0 ak (f )Fk (z), where the series is uniformly and absolutely convergent in any compact subset K of G. More exactly, applying the ideas in the proof of the general Theorem 3, p. 54 in Suetin [186], we easily obtain that there exists d ∈ (0, 1) such that |ak Fk (z)| ≤ Cdk , for all k = 0, 1, 2, ... and z ∈ K, which obviously implies P∞ that the series k=0 ak (f )Fk (z) is absolutely (and uniformly) convergent in K. As an immediate consequence, it follows that the series defining Lξ (f )(z) also is absolutely and uniformly convergent in K, that is Lξ (f )(z) is analytic in G. The two theorems below presents some approximation properties of Lξ (f ), ξ ∈ (0, 1] corresponding to the extensions of Pξ (f ), Q∗ξ (f ) and Wξ∗ (f ) when f ∈ A(G), in the light of the ideas mentioned in the above remark.

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Theorem 3.2.17. Let G ⊂ C be a bounded simply connected domain whose boundary Γ is a regular and analytic curve and for f ∈ A(G) define L ξ (f )(z) = P∞ k=0 ak (f )ρk,ξ Fk (z), where Fk (z) are the Faber polynomials attached to G, a k (f ) are the Faber coefficients of f and ρk,ξ ∈ (0, 1] for all ξ ∈ (0, 1], k ∈ N. Also, for any r > 0 define as Gr the interior of the (closed) level curve Γr given by Γr = {z; |Φ(z)| = r} = {Ψ(w); |w| = r}. (i) Then Lξ (f )(z) is analytic in G and if f is not a constant function on G and ρk,ξ 6= 1 for all ξ ∈ (0, 1] and k ∈ N, then there exists 0 < β0 < 1 such that for all β0 < β < 1 we have the saturation result max |f (z) − Lξ (f )(z)| 6= o(min |1 − ρk,ξ |), ξ ∈ (0, 1]. 1≤k

z∈Gβ

a

Recall here that by definition aξ = o(bξ ) if limξ→0 bξξ = 0. (ii) Take ρk,ξ = 1+ξ12 k2 , that is Lξ (f ) is the generalization of Pξ (f ). If f is not a constant function on G then there exists 0 < β0 < 1 such that for all β0 < β < 1 we have max |f (z) − Lξ (f )(z)| ∼ ξ 2 , ξ ∈ (0, 1],

z∈Gβ

where the constants in the equivalence depend only on f, β and Gβ (but are independent of ξ). (iii) Take ρk,ξ = e−kξ , that is Lξ (f ) is the generalization of Q∗ξ (f ). If f is not a constant function on G then there exists 0 < β0 < 1 such that for all β0 < β < 1 we have max |f (z) − Lξ (f )(z)| ∼ ξ, ξ ∈ (0, 1],

z∈Gβ

where the constants in the equivalence depend only on f, β and Gβ . 2 (iv) Take ρk,ξ = e−k ξ/4 , that is Lξ (f ) is the generalization of Wξ∗ (f ). If f is not a constant function on G then there exists 0 < β0 < 1 such that for all β0 < β < 1 we have max |f (z) − Lξ (f )(z)| ∼ ξ, ξ ∈ (0, 1],

z∈Gβ

where the constants in the equivalence depend only on f, β and Gβ . Proof. (i) First we use some ideas in the proof of Theorem 1, p. 51-52 in the book of Suetin [186]. Thus, since Γ is a regular analytic curve, then the mapping Φ(z) (and consequently its inverse Ψ(z)) can be analytically and univalently continued across the boundary Γ in the domain G. More exactly, there exists 0 < β0 < 1 such that Ψ(z) is analytic and univalent in Extβ0 (excepting the point ∞), where ˜ \ Gβ , with Gβ denoting the interior of the (closed) level curve Γβ Extβ0 = C 0 0 0 given by Γβ0 = {z; |Φ(z)| = β0 } = {Ψ(w); |w| = β0 }.

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Let β0 < r < β < 1. R f (Ψ(u)) 1 Now, by Definition 1.0.10 (ii) we have ak (f ) = 2πi |u|=1 uk+1 du. Also, by RK¨ ovari-Pommerenke [120], p. 198, Lemma 2, we have that the integral Fk (Ψ(u)) 1 2πi |u|=1 um+1 du is equal to zero for m 6= k and equal to 1 for m = k. But taking R f (Ψ(u)) 1 into account the Cauchy’s result, clearly we can write ak (f ) = 2πi |u|=β uk+1 du R Fk (Ψ(u)) 1 and that 2πi |u|=β um+1 du is equal to zero for m 6= k and equal to 1 for m = k. P∞ Let Lξ (f )(z) = k=0 ak (f )ρk,ξ Fk (z), z ∈ Gr , ξ ∈ (0, 1], with ρk,ξ ∈ (0, 1], for S all ξ ∈ (0, 1] and k ∈ N {0}. By the proof of Theorem 1, p. 52 in Suetin [186] we have |Fk (z)| ≤ C(r)rk , for all z ∈ Gr .

) . Note here Also, by the above formula for ak (f ) we easily obtain |ak (f )| ≤ C(β,f βk that C(r), C(β, f ) > 0 are independent of k. It follows k r |ak (f )ρk,ξ Fk (z)| ≤ C(r, β, f ) for all z ∈ Gr , k = 0, 1, 2, .... β

Since βr < 1, it follows that the series which defines Lξ (f )(z) is absolutely and uniformly convergent in the compact Gr since it is majorized by a geometric progression. Therefore Lξ (f ) is analytic in any Gr with β0 < r < 1, which immediately implies its analyticity in G. Note that by the above reasonings does not follow that Lξ (f )(z) is continuous on G. On the other hand we immediately obtain Z 1 f (Ψ(u)) − Lξ (f )(Ψ(u)) du = (1 − ρm,ξ )am (f ), m ≥ 0. 2πi |u|=β um+1 This implies |1 − ρm,ξ | · |am (f )| ≤ max |f (z) − Lξ (f )(z)|. z∈Gβ

Now, let us assume that the conclusion of Theorem 3.2.17 is false, that is there exists 0 < β < 1 such that max |f (z) − Lξ (f )(z)| = o(min |1 − ρk,ξ |). 1≤k

z∈Gβ

It implies |1 − ρm,ξ | · |am (f )| = o(min |1 − ρk,ξ |), for all m ≥ 1 1≤k

and therefore |1 − ρm,ξ | · |am (f )| = 0, ξ&0 min1≤k |1 − ρk,ξ | lim

which means am (f ) = 0, for all m ≥ 1. Then, by Theorem 1, p. 44 in Gaier [76] we get f (z) = a0 (f ) for all z ∈ Gβ . By the identity theorem on the analytic functions it follows that f is constant in G := G1 , a contradiction.

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(ii) By Theorem 3.2.1, (i) we get ρk,ξ = min(1 − ρk,ξ ) = 1≤k

1 1+ξ 2 k2

251

and

ξ2 ξ2 ≥ , for all ξ ∈ (0, 1]. 1 + ξ2 2

2 2

2

ξ p ξ Denote Vξ = inf 1≤p ( 1+ξ 2 p2 ). We get Vξ = 1+ξ 2 and for all ξ ∈ (0, 1] it follows Vξ ≥ ξ 2 /2. By the following lower estimate from the above point (i),

|1 − ρm,ξ | · |am (f )| ≤ max |f (z) − Lξ (f )(z)|, z∈Gβ

for all p ≥ 1 and ξ ∈ (0, 1] it follows 2 maxz∈Gβ |f (z) − Lξ (f )(z)| ξ2

≥ ≥

maxz∈Gβ |f (z) − Lξ (f )(z)|

Vξ maxz∈Gβ |f (z) − Lξ (f )(z)| ξ 2 p2 1+ξ 2 p2

≥ |ap (f )|.

This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 and such that limk→∞

maxz∈G |f (z)−Lξ (f )(z)| β

ξk2

f is constant on Gβ .

= 0 then ap = 0 for all p ≥ 1, that is maxz∈G |f (z)−Lξ (f )(z)|

β Therefore, if f is not a constant then inf ξ∈(0,1] implies that there exists a constant Cβ (f ) > 0 such that

maxz∈Gβ |f (z) − Lξ (f )(z)| ξ2

ξ2

> 0, which

≥ Cβ (f ) for all ξ ∈ (0, 1],

that is max |f (z) − Lξ (f )(z)| ≥ Cβ (f )ξ 2 , for all ξ ∈ (0, 1].

z∈Gβ

For the upper estimate, first by Theorem 1, p. 51 in Suetin [186], we have P∞ f (z) = k=0 ak (f )Fk (z), for all z ∈ G, where the series is absolutely and uniformly convergent inside G. Taking into account the estimate for |ak (f )Fk (z)|, z ∈ Gr from the above point (i) and denoting d = βr < 1, for all z ∈ Gr we obtain |f (z) − Lξ (f )(z)| ≤

∞ X

(1 − ρk,ξ )|ak (f )Fk (z)| =

k=0 ∞ X 2

≤ξ

k=1

k 2 |ak (f )Fk (z)| ≤ Cξ 2 P∞

∞ X

k=1 ∞ X

ξ 2 k2 |ak (f )Fk (z)| 1 + ξ 2 k2 k 2 dk .

k=0

1 , by differentiation with respect to But taking into account that k=0 d = 1−dP P∞ 1 2 k−1 k−2 d we easily obtain k=1 kd = (1−d)2 and ∞ = (1−d) 3 , which k=2 k(k − 1)d P∞ 2 k immediately implies k=0 k d < +∞ and therefore k

|f (z) − Lξ (f )(z)| ≤ M ξ 2 , for all z ∈ Gr , ξ ∈ (0, 1].

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Since r > β0 is arbitrary this proves (ii). (iii) By Theorem 3.2.5, (i) and (iv) we get ρk,ξ = e−kξ and min(1 − ρk,ξ ) = 1 − e−ξ ≥ ξ/3, for all ξ ∈ (0, 1]. 1≤k

Denote Vξ = inf 1≤p (1 − e−pξ ). We get Vξ = 1 − e−ξ and by the mean value theorem applied to h(x) = e−x on [0, ξ] there exists η ∈ (0, ξ) such that for all ξ ∈ (0, 1] we have Vξ = h(0) − h(ξ) = (−ξ)h0 (η) = (ξ)e−η ≥ (ξ)e−ξ ≥ e−1 ξ ≥ ξ/3. By the lower estimate |1 − ρm,ξ | · |am (f )| ≤ max |f (z) − Lξ (f )(z)|, z∈Gβ

for all p ≥ 1 and ξ ∈ (0, 1] it follows 3 maxz∈Gβ |f (z) − Lξ (f )(z)| ξ

≥ ≥

maxz∈Gβ |f (z) − Lξ (f )(z)| Vξ maxz∈Gβ |f (z) − Lξ (f )(z)| 1 − e−pξ

≥ |ap |.

Reasoning exactly as at the above point (ii), it follows that if f is not a constant then there exists a constant Cβ (f ) > 0 such that max |f (z) − Lξ (f )(z)| ≥ Cβ ξ, for all ξ ∈ (0, 1].

z∈Gβ

For the upper estimate we reason as at the above point (ii). By applying the mean value theorem to 1 − e−kξ = g(0) − g(k) with g(x) = e−ξx , there exists p ∈ (0, k) such that 1 − e−kξ = kξe−pξ ≤ kξ, which implies |f (z) − Lξ (f )(z)| ≤

∞ X

k=0 ∞ X 2

≤ξ P∞

(1 − e−kξ )|ak (f )Fk (z)| ≤

k=1

k|ak (f )Fk (z)| ≤ Cξ

∞ X

kξ|ak (f )Fk (z)|

k=1

∞ X

kdk ,

k=0

k

where k=0 kd < +∞. 2 (iv) By Theorem 3.2.8, (i) and (iv) we get ρk,ξ = e−k ξ/4 and min(1 − ρk,ξ ) = 1 − e−ξ/4 ≥ ξ/8, for all ξ ∈ (0, 1]. 1≤k

2

Denote Vξ = inf 1≤p (1 − e−p ξ/4 ). We get Vξ = 1 − e−ξ/4 and by the mean value theorem applied to h(x) = e−x/4 on [0, ξ] there exists η ∈ (0, ξ) such that for all ξ ∈ (0, 1] we have Vξ = h(0) − h(ξ) = (−ξ)h0 (η) = (ξ/4)e−η/4 ≥ (ξ/4)e−ξ/4 ≥

e−1/4 ξ ≥ ξ/8. 4

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As at the above points (ii) and (iii), for all p ≥ 1 and ξ ∈ (0, 1] it follows 8 maxz∈Gβ |f (z) − Lξ (f )(z)| ξ

≥ ≥

maxz∈Gβ |f (z) − Lξ (f )(z)| Vξ maxz∈Gβ |f (z) − Lξ (f )(z)| 1 − e−p2 ξ/4

≥ |ap |.

Again reasoning as the above points (ii) and (iii) it follows that if f is not a constant then there exists a constant Cβ (f ) > 0 such that max |f (z) − Lξ (f )(z)| ≥ Cβ ξ, for all ξ ∈ (0, 1].

z∈Gβ

For the upper estimate, we reason as at the above point (iii). By applying the 2 2 mean value theorem to 1 − e−k ξ/4 = g(0) g(x) = e−ξx /4 , there exists with − g(k) 2 2 2 e−p ξ/4 ≤ k2ξ , which implies p ∈ (0, k) such that 1 − e−k ξ/4 = (−k) − 2pξ 4 |f (z) − Lξ (f )(z)| ≤ ≤ where

P∞

k=0

2 k

∞ X

(1 − e−k

k=0 ∞ X

ξ 2

k=1

2

ξ/4

)|ak (f )Fk (z)| ≤

k 2 |ak (f )Fk (z)| ≤ Cξ

∞ X

∞ X k2 ξ

k=1

2

|ak (f )Fk (z)|

k 2 dk ,

k=0

k d < +∞. Theorem 3.2.17 is proved.

The results in Theorem 3.2.17 can be improved if we suppose that G is a continuum (that is a compact connected subset of C), without any requirement on its boundary. More exactly the following result holds. Theorem 3.2.18. Let G be a continuum and suppose that f is analytic in G, that is there exists R > 1 such that f is analytic in GR (which includes G). Here recall that GR denotes the interior of the closed level curve ΓR given by ΓR = {z; |Φ(z)| = P∞ R} = {Ψ(w); |w| = R}. Attach to f the operators Lξ (f )(z) = k=0 ak (f )ρk,ξ Fk (z), ξ ∈ (0, 1], where Fk (z) are the Faber polynomials attached to G, ak (f ) are the Faber coefficients of f and ρk,ξ ∈ (0, 1] for all ξ ∈ (0, 1], k ∈ N. (i) Then Lξ (f )(z) is analytic in G and if f is not a constant function on G and ρk,ξ 6= 1 for all ξ ∈ (0, 1] and k ∈ N, then for all 1 < β < R the saturation result max |f (z) − Lξ (f )(z)| 6= o(min |1 − ρk,ξ |), ξ ∈ (0, 1], 1≤k

z∈Gβ

holds. (ii) Take ρk,ξ = 1+ξ12 k2 , that is Lξ (f ) is the generalization of Pξ (f ). If f is not a constant function on G then for all 1 < β < R we have max |f (z) − Lξ (f )(z)| ∼ ξ 2 , ξ ∈ (0, 1],

z∈Gβ

where the constants in the equivalence depend only on f, β and Gβ (but are independent of ξ).

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(iii) Take ρk,ξ = e−kξ , that is Lξ (f ) is the generalization of Q∗ξ (f ). If f is not a constant function on G then for all 1 < β < R we have max |f (z) − Lξ (f )(z)| ∼ ξ, ξ ∈ (0, 1],

z∈Gβ

where the constants in the equivalence depend only on f, β and Gβ . 2 (iv) Take ρk,ξ = e−k ξ/4 , that is Lξ (f ) is the generalization of Wξ∗ (f ). If f is not a constant function on G then we have max |f (z) − Lξ (f )(z)| ∼ ξ, ξ ∈ (0, 1], z∈G

where the constants in the equivalence depend only on f, β and Gβ . Proof. (i) First we use some ideas in the proof of Theorem 3, p. 54 in the book of Suetin [186]. Since there exists R > 1 such that f is analytic GR and since G ⊂ GR , there exists r with 1 < r < R such that G ⊂ Gr . Let β satisfy the conditions 1 < r < β < R. R f (Ψ(u)) 1 du. Also, Now, by Definition 1.0.10 (ii) we have ak (f ) = 2πi |u|=1 uk+1 by RK¨ ovari-Pommerenke [120], p. 198, Lemma 2, we have that the integral Fk (Ψ(u)) 1 2πi |u|=1 um+1 du is equal to zero for m 6= k and equal to 1 for m = k. But taking R f (Ψ(u)) 1 into account the Cauchy’s result, clearly we can write ak (f ) = 2πi |u|=β uk+1 du R Fk (Ψ(u)) 1 du is equal to zero for m 6= k and equal to 1 for m = k. and that 2πi |u|=β um+1 P∞ Let Lξ (f )(z) = k=0 ak (f )ρk,ξ Fk (z), z ∈ Gr , ξ ∈ (0, 1], with ρk,ξ ∈ (0, 1], for S all ξ ∈ (0, 1] and k ∈ N {0}. By the inequality (13), p. 44 in Suetin [186] we have |Fk (z)| ≤ C(r)rk , for all z ∈ Gr .

) Also, by the above formula for ak (f ) we easily obtain |ak (f )| ≤ C(β,f β k . Note here that C(r), C(β, f ) > 0 are independent of k. It follows k r for all z ∈ Gr , k = 0, 1, 2, ..., . |ak (f )ρk,ξ Fk (z)| ≤ C(r, β, f ) β Since βr < 1, it follows that the series which defines Lξ (f )(z) is absolutely and uniformly convergent in the compact Gr since it is majorized by a geometric progression. Therefore Lξ (f ) is analytic in G. On the other hand we immediately obtain Z 1 f (Ψ(u)) − Lξ (f )(Ψ(u)) du = (1 − ρm,ξ )am (f ), m ≥ 0. 2πi |u|=β um+1

This implies

|1 − ρm,ξ | · |am (f )| ≤ max |f (z) − Lξ (f )(z)|. z∈Gβ

Now, let us assume that the conclusion of Theorem 3.2.17 is false, that is there exists 1 < β < R such that max |f (z) − Lξ (f )(z)| = o(min |1 − ρk,ξ |).

z∈Gβ

1≤k

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It implies |1 − ρm,ξ | · |am (f )| = o(min |1 − ρk,ξ |), for all m ≥ 1 1≤k

and therefore lim

ξ&0

|1 − ρm,ξ | · |am (f )| = 0, min1≤k |1 − ρk,ξ |

which means am (f ) = 0, for all m ≥ 1. Then, by Theorem 1, p. 44 in Gaier [76] we get f (z) = a0 (f ) for all z ∈ Gβ . Since G ⊂ Gβ it follows that f is constant in G, a contradiction. (ii) By Theorem 3.2.1, (i) we get ρk,ξ = 1+ξ12 k2 and min(1 − ρk,ξ ) = 1≤k

ξ2 ξ2 ≥ , for all ξ ∈ (0, 1]. 2 1+ξ 2

2 2

2

ξ p ξ Denote Vξ = inf 1≤p ( 1+ξ 2 p2 ). We get Vξ = 1+ξ 2 and for all ξ ∈ (0, 1] it follows Vξ ≥ ξ 2 /2. By the following lower estimate from the above point (i),

|1 − ρm,ξ | · |am (f )| ≤ max |f (z) − Lξ (f )(z)|, z∈Gβ

for all p ≥ 1 and ξ ∈ (0, 1] it follows 2 maxz∈Gβ |f (z) − Lξ (f )(z)| ξ2

≥ ≥

maxz∈Gβ |f (z) − Lξ (f )(z)|

Vξ maxz∈Gβ |f (z) − Lξ (f )(z)| ξ 2 p2 1+ξ 2 p2

≥ |ap (f )|.

This implies that if there exists a subsequence (ξk )k in (0, 1] with limk→∞ ξk = 0 and such that limk→∞

maxz∈G |f (z)−Lξ (f )(z)| β

ξk2

f is constant on Gβ .

= 0 then ap = 0 for all p ≥ 1, that is maxz∈G |f (z)−Lξ (f )(z)|

β Therefore, if f is not a constant then inf ξ∈(0,1] implies that there exists a constant Cβ (f ) > 0 such that

maxz∈Gβ |f (z) − Lξ (f )(z)| ξ2

ξ2

> 0, which

≥ Cβ (f ) for all ξ ∈ (0, 1],

that is max |f (z) − Lξ (f )(z)| ≥ Cβ (f )ξ 2 , for all ξ ∈ (0, 1].

z∈Gβ

For the upper estimate, first by Theorem 2, p. 52 in Suetin [186] we have P∞ f (z) = k=0 ak (f )Fk (z), for all z ∈ G, where the series is absolutely and uniformly

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convergent on the whole G. Taking into account the estimate for |ak (f )Fk (z)|, z ∈ Gr from the above point (i) and denoting d = βr < 1, for all z ∈ Gr we obtain |f (z) − Lξ (f )(z)| ≤

∞ X

(1 − ρk,ξ )|ak (f )Fk (z)| =

k=0 ∞ X 2

≤ξ

k=1

k 2 |ak (f )Fk (z)| ≤ Cξ 2 P∞

∞ X

k=1 ∞ X

ξ 2 k2 |ak (f )Fk (z)| 1 + ξ 2 k2 k 2 dk .

k=0

1 , by differentiation with respect to But taking into account that k=0 d = 1−dP P∞ 2 1 k−2 k−1 = (1−d) d we easily obtain k=1 kd = (1−d)2 and ∞ 3 , which k=2 k(k − 1)d P∞ 2 k immediately implies k=0 k d < +∞ and therefore |f (z) − Lξ (f )(z)| ≤ M ξ 2 , for all z ∈ Gr , ξ ∈ (0, 1]. k

Since r with 1 < r < R is arbitrary, this proves (ii). (iii) The proof follows word for word that for Theorem 3.2.17, (iii), with the unique difference that here β > 1. (iv) The proof follows word for word that for Theorem 3.2.17, (iv), with the unique difference that here β > 1. Theorem 3.2.18 is proved.

Remark. It is natural to ask for extensions of the Cauchy problems in Theorem 3.2.13 (v) and in Theorem 3.2.15 (iv) for domains in C. For example, let us suppose that f ∈ A(G), where G is a bounded simply connected domain with the boundary Γ a regular analytic curve, that is Γ is an analytic mapping and its derivative never vanishes. In this case, by Theorem 1 and its proof, p. 51-52 in the book of Suetin P∞ [186] it follows that f (z) = k=0 ak (f )Fk (z), where the series is uniformly and absolutely convergent in any compact subset K of G. Alternatively, we can suppose here that G is a continuum (that is a compact connected subset of C), without any requirement on its boundary. In this case too, from Theorem 2 in Suetin [186], p. P 52, it follows f (z) = ∞ k=0 ak (f )Fk (z), where the series is uniformly convergent in G. For this extension we need a suitable concept of derivative called Faber derivative and introduced in Bruj-Schmieder [48], Definition 3, p. 165. Taking into account the above hypothesis on G, it easily follows that its Faber derivative (of arbitrary order r) Rintroduced in Bruj-Schmieder [48], Definition 3 and denoted by F (r) (f )(z) π satisfies −π F (r) [ψ(eit )]dt = 0 and can be written as F (r) (f )(z) =

∞ X

ak (F (r) )Fk (z).

k=0

Therefore, taking into account Theorem 5, p. 166 in the same paper of BrujSchmieder [48], for the Faber derivative of order r of f formally it easily follows the relationship ∞ X F (r) (f )(z) = (ik)r ak (f )Fk (z), k=0

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for all z in any compact set included in G. Here i2 = −1. The following two corollaries are immediate. Corollary 3.2.19. Let us suppose that f ∈ A(G), where G ⊂ C is, for example a continuum (or a bounded simply connected domain with the boundary Γ a regular P∞ 2 analytic curve). Then Wt∗ (f )(z) = k=0 ak (f )e−k t/2 Fk (z) := u(f )(t, z), z ∈ G is a semigroup of linear operators on the Banach space (A(G), k·k C(G) ) and is solution (that belongs to A(G), for each fixed t > 0) of the ”Cauchy problem” ∂u 1 (t, z) = F 00 (u(f ))(t, z), (t, z) ∈ (0, +∞) × G, z = reiϕ , z 6= 0, ∂t 2 u(0, z) = f (z), z ∈ G, f ∈ A(G). Corollary 3.2.20. Let us suppose that f ∈ A(G), where G ⊂ C is, for example a continuum (or a bounded simply connected domain with the boundary Γ a regular P∞ analytic curve). Then Q∗t (f )(z) = k=0 ak (f )e−kt Fk (z) := u(f )(t, z), z ∈ G is a semigroup of linear operators on the Banach space (A(G), k · k C(G) ) and is solution (that belongs to A(G), for each fixed t > 0) of the ”Cauchy problem” ∂u (t, z) + F 00 (u(f ))(t, z) = 0, (t, z) ∈ (0, +∞) × G, z = reiϕ , z 6= 0, ∂t u(0, z) = f (z), z ∈ G, f ∈ A(G). Remark. For G = D1 , these equations become those in Theorem 3.2.13 (v) and Theorem 3.2.15 (iv).

3.2.2

Complex q-Picard and q-Gauss-Weierstrass Singular Integrals

In this subsection we extend the results in the case of classical complex Picard and Gauss-Weierstrass singular integrals proved in the previous subsection, to their q-analogues. P∞ S For f (z) = k=0 ak z k , z ∈ DR , λ ∈ R, λ > 0, 0 < q < 1, r ∈ N {0} and z ∈ DR , let us define the q-complex singular integrals Prλ (f ; q, z) ≡ Prλ (f ; z) Z ∞ f zei[k]q t r+1 (r−k+1)(r−k)/2 X (1 − q) q r+1 dt := − (−1)k −1 (r+1)r/2 (1−q)|t| k 2 [λ]q ln q q q −∞ Eq k=1 [λ] q

and

Wrλ (f ; q, z) ≡ Wrλ (f ; z) 1 := − q π [λ]q q 1/2 ; q 1/2 ·

r+1 X k=1

(−1)k

q

(r−k+1)(r−k)/2

q (r+1)r/2

r+1 k

Z q

∞ −∞

f zei[k]q t dt t2 Eq [λ] q

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called as the complex q- Jackson type generalization of the q-Picard and q-GaussWeierstrass singular integrals, respectively. For r = 0 we denote these singular integrals by Pλ (f ; q, z) ≡ Pλ (f ; z) and Wλ (f ; q, z) ≡ Wλ (f ; z), respectively. First we present the approximation properties. P k Theorem 3.2.21. (Aral-Gal [32]) Let f (z) = ∞ k=0 ak z , z ∈ DR be with a0 = 0, a1 = 1 and λ > 0, 0 < q < 1. We have : (i) Pλ (f ; q, z) := Pλ (f ; z) is continuous in DR , analytic in DR so that Pλ (f ; z) =

∞ X k=0

ck (λ, q) =

ak ck (λ, q)z k , z ∈ DR , Pλ (f ; 0) = 0 and

(1 − q) [λ]q ln q −1

Z

∞ 0

cos(ku) du, k = 0, 1, ...

Eq

(1−q)u [λ]q

Also, there exists qb ∈ (0, 1) such that for all q ∈ (b q , 1) we have c 1 (λ, q) > 0 and if we choose qλ such that 0 < qλ < 1 and qλ → 1 as λ → 0, then we have limλ→0 c1 (λ, qλ ) = 1 ; (ii) |Pλ (f ; z) − f (z)| ≤ (R + 1)(1 + q1 )ω1 (f ; [λ]q )DR , for all z ∈ DR , where ω1 (f ; δ)DR = sup{|f (z1 ) − f (z2 )|; z1 , z2 ∈ DR , |z1 − z2 | ≤ δ}.

Proof.

(i) Let z0 , zn ∈ DR be with lim zn = z0 . Since |eiu | = 1, we get n→∞

|Pλ (f ; zn ) − Pλ (f ; z0 )| (1 − q) ≤ 2 [λ]q ln q −1

Z

(1 − q) 2 [λ]q ln q −1

Z

≤

+∞ −∞

|f (zn eiu ) − f (z0 eiu )| ·

Eq

+∞ −∞

ω1 (f ; |zn − z0 |)DR ·

Eq

1 (1−q)|u| [λ]q

1 (1−q)|u| [λ]q

du

du

= ω1 (f ; |zn − z0 |)DR . Passing to limit with n → ∞, it follows that Pλ (f ; z) is continuous at z0 ∈ DR , since f is continuous on DR . It remains to prove that Pλ (f ; z) is analytic in P∞ k DR . We can write f (z) = k=0 ak z , z ∈ DR . For fixed z ∈ DR , we get P ∞ iku k iu f (ze ) = k=0 ak e z and since |ak eiku | = |ak |, for all u ∈ R and the series P∞ P∞ k iku k z is unik=0 ak z is absolutely convergent, it follows that the series k=0 ak e formly convergent with respect to u ∈ R. This immediately implies that the series

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can be integrated term by term, i.e.

  Z ∞ ∞ (1 − q) X 1 du Pλ (f ; z) = ak z k  eiku · (1−q)|u| 2 [λ]q ln q −1 −∞ E q k=0 [λ] q

=

∞ X

ak ck (λ, q)z k , where ck (λ, q)

k=0

(1 − q) = [λ]q ln q −1

Z

∞ 0

cos(ku) du. (1−q)u [λ]q

Eq

Since a0 = 0, we get Pλ (f ; 0) = 0. Then we have Z ∞ Z (1 − q) ∞ cos([λ]q u) (1 − q) cos(u) du = c1 (λ, q) = du. [λ]q ln q −1 0 E (1−q)u ln q −1 0 Eq ((1 − q) u) q

[λ]q

Now, if we choose qλ → 1 as λ → 0, then we get [λ]qλ → 0 (see Aral [31]). Since limq→1− Eq ((1 − q) t) = et (see Gasper-Rahman [99], p. 9, (1.3.16)) and limq→1− [λ]q = λ, by Lebesgue’s Dominated Convergence theorem, we obtain Z ∞ lim c1 (λ, qλ ) = e−t du = 1 and λ→0

lim c1 (λ, q) =

q→1−

Z

∞ 0

0

cos(λu) du > (by e.g. Anastassiou-Gal [18], p. 4) > 0. eu

Thus, there exists qb ∈ (0, 1) such that for all q ∈ (b q , 1) we have c1 (λ, q) > 0. (ii) By the Maximum Modulus Principle, it suffices to take |z| = R. Since |eiu − 1| ≤ 2|sin u2 | ≤ |u| for all u ∈ R, we easily get Z ∞ (1 − q) 1 du |Pλ (f ; z) − f (z)| ≤ ω1 (f ; |zeiu − z|)DR · (1−q)|u| 2 [λ]q ln q −1 −∞ Eq [λ]q Z ∞ (1 − q) 1 du ω1 (f ; R|u|)DR · ≤ (1−q)|u| 2 [λ]q ln q −1 −∞ Eq [λ]q Z ∞ 1 |u| (1 − q) du · ≤ ω1 (f ; [λ]q )DR (R + 1) 1 + −1 2 [λ]q ln q [λ]q −∞ Eq (1−q)|u| [λ]q 1 ≤ (by Aral [31]) ≤ (R + 1) 1 + ω1 (f ; [λ]q )DR . q P k Theorem 3.2.22. (Aral-Gal [32]) (i) If f (z) = ∞ k=0 ak z is analytic in DR , then for all λ > 0, 0 < q < 1, Wλ (f ; q, z) := Wλ (f ; z) is analytic in DR and we have in DR ∞ X Wλ (f ; z) = ak dk (λ, q)z k , k=0

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where dk (λ, q) =

π

q

2 [λ]q

q 1/2 ;

q

1/2

Z

∞ 0

cos(ku) du. u2 Eq [λ] q

Also, there exists qb ∈ (0, 1) such that for all q ∈ (b q , 1) we have d 1 (λ, q) > 0 and if we choose qλ such that 0 < qλ < 1 and qλ → 1 as λ → 0, then we have limλ→0 d1 (λ, qλ ) = 1. In addition, if f is continuous on DR then Wλ (f ; z) is continuous on DR . (ii) q q −1/2 1/2 (1 − q ) ω1 f ; [λ]q |Wλ (f ; z) − f (z)| ≤ (R + 1) 1 + q , DR

for all z ∈ DR .

(i) Reasoning as for the Pλ (f ) operator, we easily deduce Z +∞ X ∞ 1 1 du Wλ (f ; z) = q ak z k eiuk · u2 π [λ]q q 1/2 ; q 1/2 −∞ k=0 Eq [λ] q Z +∞ ∞ X 2 cos(ku) . = ak dk (λ, q)z k , where dk (λ, q) = q u2 π [λ]q q 1/2 ; q 1/2 0 Eq [λ] k=0

Proof.

q

Similar results with those for c1 (λ, q) (in Theorem 3.2.21), can be obtained for d1 (λ, q) too. Indeed, if we choose qλ such that 0 < qλ < 1 and qλ → 1 as λ → 0, then from Lebesgue’s Dominated Convergence theorem, Z ∞ we get 2 cos(u) du lim d1 (λ, qλ ) = lim q λ→0 λ→0 u2 1/2 0 π [λ]q q ; q 1/2 Eq [λ] q q Z ∞ cos( [λ] u) q 2 = lim du = 1, 2 1/2 λ→0 π q Eq (u ) ; q 1/2 0

(Here, for the last equality see e.g. Alvarez-Nodarse-Atakishiyeva-Atakishiyev [6], p. 132). Similarly we can see that limq→1− d1 (λ, q) > 0, which implies that there exists qb ∈ (0, 1) such that for all q ∈ (b q , 1) we have d1 (λ, q) > 0. The proof of continuity of Wλ (f ; z) is similar to that for Pλ (f ; z). (ii) Reasoning as in the case of Pλ (f ; z), we can write Z +∞ 1 1 |Wλ (f ; z) − f (z)| ≤ q |f (ze−iu ) − f (z)| 2 du u π [λ]q q 1/2 ; q 1/2 −∞ Eq [λ] q q 1 ≤ ω1 (f ; [λ]q )DR (R + 1) q π [λ]q q 1/2 ; q 1/2 ! Z +∞ |u| 1 du · 1+ p [λ]q Eq u2 −∞ [λ]q q q −1/2 1/2 ≤ (R + 1) 1 + q (1 − q ) ω1 f ; [λ]q . DR

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(For the last inequality see Aral [31].)

Theorem 3.2.23. (Aral-Gal [32]) Let R > 0, z ∈ DR , λ ∈ (0, 1], 0 < q < 1 and r ∈ N. For f analytic in DR and continuous on DR we have r+1 X r + 1 [k]q ! 1 ωr+1,q (f ; [λ]q )∂DR , |Prλ (f ; z) − f (z)| ≤ (r+1)r/2 k(k+1) k q q 2 k=0

|W(2r−1)λ (f ; z) − f (z)| ≤ 2 where

2r−1

q 2 − r2 1/2 q ; q ω2r,q f ; [λ]q 1+q r

∂DR

ωr,q (f ; δ)∂DR = sup |∆ru f (Reix )|; |x| ≤ π, |u| ≤ δ . Proof. Let z ∈ DR , |z| = R be fixed. Because of the Maximum Modulus Principle, it suffices to estimate |Prλ (f ; z) − f (z)|, for this |z| = R, z = Reix . Reasoning now exactly as in the proof of Theorem 3.2.21, we get r+1 Z ∞ ix ∆r+1 (1 − q) (−1) q,t f Re dt, f (z) − Prλ (f ; z) = 2 [λ]q ln q −1 q (r+1)r/2 −∞ Eq (1−q)|t| [λ] q

which implies

(1 − q) 1 |f (z) − Prλ (f ; z)| ≤ 2 [λ]q ln q −1 q (r+1)r/2 ≤ ωr+1,q f ; [λ]q

∂DR

Z

∞

ωr+1,q (f ; |t|)∂DR dt −∞ Eq (1−q)|t| [λ]q r+1 X 1 r + 1 [k]q ! . k(k+1) k q (r+1)r/2 q 2

The proof in the case of W(2r−1)λ (f ; z) is similar.

k=0

Remark. Since f is supposed to be analytic, by Theorems 3.2.21 and 3.2.22 it is immediate that p the orders of approximation by Pλ (f ; q, z) and Wλ (f ; q, z) are O([λ]q ) and O( [λ]q ), respectively. Also, from Theorem 3.2.23 it follows that for any r ∈ N, the orders of approximation by Prλ (f ; q; z) and W(2r−1)λ (f ; q; z) are O([λ]r+1 ) and O([λ]rq ), respectively. q The geometric properties are consequences of Theorems 3.2.21 and 3.2.22 and are expressed by the following. Theorem 3.2.24. (Aral-Gal [32]) Let us suppose that G ⊂ C is open, such that D1 ⊂ G and f : G → C is analytic in G. Denote by (Bλ (f )(z))λ>0 any from (Pλ (f ; q, z))λ>0 , (Wλ (f ; q, z))λ>0 , where we choose q := qλ such that 0 < qλ < 1 and qλ → 1 as λ → 0. (i) If f is univalent in D1 , then there exists λ0 > 0 sufficiently small (depending on f ), such that for all λ ∈ (0, λ0 ), Bλ (f )(z) are univalent in D1 .

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(ii) Let γ ∈ (−π/2, π/2). If f (0) = f 0 (0) − 1 = 0 (and f (z) 6= 0, for all z ∈ D1 \ {0} in the case of spirallikeness of order γ) and f is starlike (convex, spirallike of order γ, respectively) in D1 , that is for all z ∈ D1 0 00 0 zf (z) zf (z) iγ zf (z) Re > 0 Re + 1 > 0, Re e > 0, resp. , f (z) f 0 (z) f (z) then there exists λ0 > 0 sufficiently small (depending on f , and on f and γ in the case of spirallikeness), such that for all λ ∈ (0, λ0 ), Bλ (f )(z) are starlike (convex, spirallike of order γ, respectively) in D1 . If f (0) = f 0 (0) − 1 = 0 (and f (z) 6= 0, for all z ∈ D1 \ {0} in the case of spirallikeness of order γ) and f is starlike (convex, spirallike of order γ, respectively) only in D1 (that is the corresponding inequalities hold only in D1 ), then for any disk of radius 0 < ρ < 1 and center 0 denoted by Dρ , there exists λ0 > 0 sufficiently small (depending on f and Dρ , and in addition on γ for spirallikeness), such that for all λ ∈ (0, λ0 ), Bλ (f )(z) are starlike (convex, spirallike of order γ, respectively) in Dρ (that is, the corresponding inequalities hold in Dρ ). Proof. (i) Reasoning as in Aral [31], Theorem 2.3, we get uniform convergence (as λ → 0) in the Theorems 3.2.21 and 3.2.22, which together with a well-known results concerning sequences of analytic functions converging locally uniformly to an univalent function (see e.g. Kohr-Mocanu [118], p. 130, Theorem 4.1.17) implies the univalence of Bλ (f )(z) for sufficiently small λ. For the proof of the conclusions in (ii), let us make some general useful considerations. By Theorems 3.2.21 and 3.2.22 (reasoning again as in Aral [31], Theorem 2.3), it follows that for λ → 0, we have Bλ (f )(z) → f (z), uniformly in any compact disk included in G. By the well-known Weierstrass’ result (see e.g. Kohr-Mocanu [118] , p. 18, Theorem 1.1.6), this implies that Bλ0 (f )(z) → f 0 (z) and Bλ00 (f )(z) → f 00 (z), uniformly in any compact disk in G and therefore in D1 too, when λ → 0. In all (f )(z) , where b1 (λ, qλ ) > 0 (for λ sufficiently what follows, denote Pλ (f )(z) = Bb1λ(λ,q λ) small) is the coefficient of z in the Taylor series representing the analytic function Bλ (f )(z). (0) If f (0) = f 0 (0) − 1 = 0, then we get Pλ (f )(0) = b1f(λ,q = 0 and Pλ0 (f )(0) = λ) 0 Bλ (f )(0) b1 (λ,qλ )

= 1. Also, if f (0) = 0 and f 0 (0) = 1, then b1 (λ, qλ ) converges to f 0 (0) = 1 as λ → 0, which obviously implies that for λ → 0, we have Pλ (f )(z) → f (z), Pλ0 (f )(z) → f 0 (z) and Pλ00 (f )(z) → f 00 (z), uniformly in D1 . (ii) Suppose first that f is starlike in D1 . By hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Write Pλ (f )(z) in the form Pλ (f )(z) = zQλ (f )(z) . For |z| = 1 we have |f (z) − Pλ (f )(z)| = |z| · |g(z) − Qλ (f )(z)| = |g(z) − Qλ (f )(z)|,

which by the uniform convergence in D1 of Pλ (f ) to f and by the maximum modulus principle, implies the uniform convergence in D1 of Qλ (f )(z) to g(z), as λ → 0.

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Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index λ0 > 0 and a > 0 depending on g, such that |Qλ (f )(z)| > a > 0, for all z ∈ D1 and all λ ∈ (0, λ0 ). Also, for all |z| = 1, we have |f 0 (z) − Pλ0 (f )(z)| = |z[g 0 (z) − Q0λ (f )(z)] + [g(z) − Qλ (f )(z)]|

≥ | |z| · |g 0 (z) − Q0λ (f )(z)| − |g(z) − Qλ (f )(z)| |

= | |g 0 (z) − Q0λ (f )(z)| − |g(z) − Qλ (f )(z)| |,

which from the maximum modulus principle, the uniform convergence of Pλ0 (f ) to f 0 and of Qλ (f ) to g, evidently implies the uniform convergence of Q0λ (f ) to g 0 , as λ → 0. Then, for |z| = 1, we get zPλ0 (f )(z) z[zQ0λ(f )(z) + Qλ (f )(z)] = Pλ (f ) zQλ (f )(z) zQ0λ (f )(z) + Qλ (f )(z) zg 0 (z) + g(z) = → Qλ (f )(z) g(z) f 0 (z) zf 0 (z) = = , g(z) f (z) which again from the maximum modulus principle, implies zPλ0 (f )(z) zf 0 (z) → , uniformly in D1 . Pλ (f ) f (z) 0 (z) is continuous in D1 , there exists α ∈ (0, 1), such that Since Re zff (z) 0 zf (z) Re ≥ α, for all z ∈ D1 . f (z) Therefore 0 0 zPλ (f )(z) zf (z) Re → Re ≥ α > 0, Pλ (f )(z) f (z) uniformly on D1 , i.e. for any 0 < β < α, there is λ0 > 0 such that for all λ ∈ (0, λ0 ) we have 0 zPλ (f )(z) Re > β > 0, for all z ∈ D1 . Pλ (f )(z) Since Pλ (f )(z) differs from Bλ (f )(z) only by a constant, this proves the starlikeness in D1 . If f is supposed to be starlike only in D1 , the proof is identical, with the only difference that instead of D1 , we reason for Dρ . The proofs in the cases when f is convex or spirallike of order γ are similar and follows from the following uniform convergence (on D1 or on Dρ ) 00 00 zPλ (f )(z) zf (z) Re + 1 → Re + 1, Pλ0 (f )(z) f 0 (z) and 0 0 iγ zPnλ (f )(z) iγ zf (z) Re e → Re e . Pλ (f )(z) f (z) The proof is complete.

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Remark. By using Theorem 3.2.23 and reasoning as above, it is not difficult to prove that the geometric properties in Theorem 3.2.24 remain valid for Prλ (f ; z) and Wrλ (f ; z) too. 3.2.3

Post-Widder Complex Convolution

In the case of real functions, the Post-Widder operator is given by (see e.g. Ditzian [63]) Z 1 n n+1 ∞ −nu/x n Pn (f )(x) = e u f (u) du, x > 0, f ∈ C[0, +∞). n! x 0

Making the change of variable x1 = y, we obtain Z ∞ 1 1 = nn+1 y n+1 Pn (f ) e−nuy un f (u) du = (by uy = v) y n! 0 Z v 1 1 n+1 n+1 ∞ −nv v n y f dy e = n n n! y y y 0 Z nn+1 ∞ −nv n v dv. = e v f n! 0 y Denoting now

1 y

= w, we can write Z nn+1 ∞ −nv n Pn (f )(w) = e v f (vw) dv, n! 0

w > 0.

Now, let D1 = {z ∈ C; |z| < 1} and suppose

f ∈ A(D1 ) = {f ; analytic on D1 and continuous on D1 }. Suggested by the above form we propose the following complex Post-Widder operator by Z nn+1 ∞ −nv n Pn,α (f )(z) = e v f (zeiv/αn ) dv, z ∈ D1 , n! 0

where α = (αn )n , αn % +∞, is arbitrary, fixed. First a Jackson-type estimate in approximation of f ∈ A(D 1 ) by Pn,α (f )(z) and a global smoothness preservation property are proved. Secondly we prove some geometric properties of Pn,α (f )(z), in the sense that it preserves some sufficient conditions for starlikeness and univalence satisfied by f ∈ A∗ (D1 ). Concerning the approximation properties of Pn,α (f )(z) we present the following result. Note that Theorem 3.2.25 (i),(ii) and (iii) were proved in Anastassiou-Gal [17], while Theorem 3.2.25 (iv) is new. Theorem 3.2.25. Let f : D1 → C be continuous on D1 . For all δ ≥ 0, z ∈ D1 and n ∈ N we have: (i) ω1 (Pn,α (f ); δ)D1 ≤ ω1 (f ; δ)D1 .

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(ii) |Pn,α (f )(z) − f (z)| ≤ 3ω1 f ; α1n

D1

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.

P k (iii) If f (z) = ∞ k=0 ak z is analytic in D1 and continuous in D1 , then Pn,α (f ) is analytic in D1 and continuous in D1 for all n ∈ N and α = (αn )n . Also, we can write ∞ X Pn,α (f )(z) = Bk,α z k , ∀z ∈ D1 , k=0

where Bk,α = ak Ak,α with

Ak,α =

nn+1 n!

If αn = n, ∀n ∈ N, then

Z

∞

e−nv v n eikv/αn dv.

0

(n!)2 1 , |A1,α | := |A1,n | = √ · Q n √ 2 k4 + 1 k=1

that is A1,n 6= 0, for all n ∈ N.

P∞ (iv) In addition, let us suppose that f (z) = k=0 ak z k for all z ∈ DR , R > 1. For S all 1 ≤ r < r1 < R, n ∈ N and q ∈ N {0} we have (q) kPn,α (f ) − f (q) kr ≤

C , n ∈ N, αn

where the constant C depends only on f , q, r, r1 . (i) Let |z1 − z2 | ≤ δ, z1 , z2 ∈D1 . We have

Proof.

|z1 eiv/αn − z2 eiv/αn | = |z1 − z2 |, which implies |Pn,α (f )(z1 ) − Pn,α (f )(z2 )| Z nn+1 ∞ −nv n e v f (z1 eiv/αn ) − f (z2 eiv/αn ) dv ≤ n! 0 n+1 Z ∞ n −nv n e v dv ω1 (f ; δ)D1 . ≤ n! 0 Passing to supremum with |z1 − z2 | ≤ δ, it follows n

n+1

R∞

ω1 (Pn,α (f ); δ)D1 ≤ ω1 (f ; δ)D1 ,

∀δ ≥ 0.

e−nv v n dv = 1, we get Z nn+1 ∞ −nv n e v |f (zeiv/αn ) − f (z)| dv |Pn,α (f )(z) − f (z)| ≤ n! 0 Z nn+1 ∞ −nv n e v ω1 (f ; |z| · |eiv/αn − 1|)D1 dv. ≤ n! 0

(ii) Since

n!

0

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Combined with the inequality |z| · |e

iv/αn

it follows

v v − 1| ≤ 2 sin , ≤ 2αn αn

∀v > 0,

Z v nn+1 ∞ −nv n dv e v ω1 f ; n! 0 αn D 1 Z nn+1 ∞ −nv n 1 ≤ ω1 f ; · e v (v + 1) dv. αn D 1 n! 0

|Pn,α (f )(z) − f (z)| ≤

Since

Z

∞

e−nv v n+1 dv =

0

we arrive at |Pn,α (f )(z) − f (z)| ≤ ω1

1 f; αn

D1

(n + 1)! , nn+2

1 n+1 ≤ 3ω1 f ; 1+ . n αn D 1

(iii) Let z0 , zm ∈ D1 be with lim zm = z0 . We have m→∞

|Pn,α (f )(zm ) − Pn,α (f )(z0 )| nn+1 ≤ n!

Z

∞ 0

e−nv v n |f (zm eiv/αn ) − f (z0 eiv/αn )| dv

≤ ω1 (f ; |zm − z0 |)D1 ·

nn+1 n!

Z

∞ 0

e−nv v n dv = ω1 (f ; |zm − z0 |)D1 .

Passing to the limit with m → ∞, it follows the continuity of Pn,α (f ) at z0 ∈ D1 . P∞ Now, let f (z) = k=0 ak z k , z ∈ D1 , be analytic in D1 . For fixed z ∈ D1 , we get f (zeiv/αn ) =

∞ X

ak eikv/αn z k

k=0

P k | = |ak |, for all v ≥ 0 and the series ∞ and since |ak e k=0 ak z is convergent, P∞ it follows that the series k=0 ak eikv/αn z k is uniformly convergent with respect to v ≥ 0. This immediately implies that the series can be integrated term by term, that is Z ∞ ∞ X X nn+1 ∞ −nv n −ikv/αn e v e ak Ak,α z k , dv = Pn,α (f )(z) = ak z k n! 0 ikv/αn

k=0

k=0

where Ak,α

nn+1 = n!

Z

∞ 0

e−nv v n e−ikv/αn dv,

k = 0, 1, . . . .

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By using the substitution −nv = x and taking αn = n, ∀n ∈ N, we get Z Z 1 ∞ −x n −ix/n2 1 ∞ n −x(1+i/n2 ) e x e dx = x e dx. A1,α := A1,n = n! 0 n! 0

1 Denoting the last integral (without n! ) by In and integrating by parts, we immediately obtain 2 ∞ n n xn · e−x(1+i/n ) + · In−1 = In−1 , In = −(1 + i/n2 ) 0 (1 + i/n2 ) (1 + i/n2 )

for all n = 1, 2, . . . . Taking the modulus, it follows |In | = √ Here I0 =

Z

which implies |I0 | = √12 . Therefore |I1 | = √12 ·

∞

e

n3 · |In−1 |, n4 + 1

−x(1+i)

0

√1 2

∀n = 1, 2, . . . .

∞ e−x(1+i) 1 , dx = = −(1 + i) 0 1+i

= 12 . Writing |I1 | =

1 2

|I2 | = √

23 · |I1 | 24 + 1

.. . |In | = √

n3 · |In−1 |, n4 + 1

and taking the products of both members, we arrive at |In | =

√ (n!)3 1 · Q · 2. n √ 2 k4 + 1 k=1

It follows |A1,α | = |A1,n | =

1 1 (n!)2 · |In | = √ · Q 6= 0, for all n ∈ N, n √ n! 2 k4 + 1 k=1

that is A1,n 6= 0, ∀n ∈ N. (iv) Analysing the proofs of the above points (i)-(iii), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation error in (ii) can be expressed in terms of ω1 (f ; δ)Dr = sup{|f (u) − f (v)|; |u − v| ≤ δ, u, v ∈ Dr }, with constants in front of ω1 depending on r ≥ 1.

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By the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get ω1 (f ; δ)Dr ≤ kf 0 kr δ, where kf 0 kr = sup{|f 0 (z)|; |z| ≤ r} ≤ Then by (ii) for all n ∈ N it follows

∞ X k=1

|ak |krk−1 .

Cr (f ) . αn Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and ξ ∈ (0, 1], we have Z q! Pn,α (f )(v) − f (v) (q) (q) dv |Pn,α (f )(z) − f (z)| = 2π (v − z)q+1 kPn,α (f ) − f kr ≤

γ

Cr1 (f ) ≤ αn Cr1 (f ) = αn

2πr1 q! 2π (r1 − r)q+1 q!r1 · , (r1 − r)q+1 ·

(q)

which proves the upper estimate in approximation by Pn,α (f )(z).

Remark. If we take, for example, αn = np , then by Theorem 3.2.25 (ii), we get that the order of approximation by Pn,α (f )(z) will be ω1 f ; n1p D1 . Concerning the geometric properties of Pn,α (f )(z) the following result holds. ] Theorem 3.2.26. (Anastassiou-Gal [17 ) (i) |Bk,α | ≤ |ak |, 1 (ii) Pn,α (P) ⊂ P, A1,α Pn,α S3,A1,α ⊂ S3 and

∀k = 0, 1, . . . .

1 Pn,α (SM ) ⊂ SM/|A1,α | , A1,α

for α = (αn )n , αn = n, ∀n = 1, 2, . . ., where S3,a = f ∈ S3 ; |f 00 (z)| ≤ |a| ⊂ S3 , for |a| < 1 and

SB = f ∈ A(D1 ); |f 0 (z)| < B, ∀z ∈ D1 .

Proof. (i) We have |Bk,α | = |ak Ak,α | = |ak | · |Ak,α | ≤ |ak |, since |Ak,α | ≤ 1, ∀k = 0, 1, 2, . . . . (ii) By Theorem 3.2.25 (iii) we get Z nn+1 ∞ −nv n |Ak,α | ≤ e v dv = 1, ∀k = 0, 1, 2, . . . , n! 0

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and |A0,α | = |a0 |. Let f ∈ P, f = U + iV , U > 0 on D1 . Since a0 = f (0) = 1, we obtain Pn,α (f )(0) = a0 A0,α = 1. Also, by Z nn+1 ∞ −nv n Pn,α (f )(z) = e v U (r cos(x + v/αn ), r sin(x + v/αn )) n! 0 + iV (r cos(x + v/αn ), r sin(x + v/αn )) dv, ∀z = reix ∈ D1 , it follows

nn+1 Re[Pn,α (f )(z)] = n!

Z

∞

e−nv v n U (r cos(x + v/αn ), r sin(x + v/αn )) dv > 0,

0

that is Pn,α (f ) ∈ P. Let f ∈ A∗ (D1 ). It follows a0 = 0, a1 = 1, which by Theorem 3.2.25 (iii) implies 1 Pn,α (f )(0) = 0, A1,α

1 A1,α P 0 (f )(0) = = 1. A1,α n,α A1,α

If |f 00 (z)| < |A1,α | then Z 1 1 nn+1 ∞ −nv n 2iv/αn 00 e v |e | A1,α Pn,α (f )(z) ≤ |A1,α | · n! 0 · |f 00 (zeiv/αn )| dv ≤ 1,

1 which means that A1,α Pn,α (f ) ∈ S3 . Also, if |f 0 (z)| < M , ∀z ∈ D1 , then we obtain Z 1 1 nn+1 ∞ −nv n iv/αn 0 e v |e | · |f 0 (zeiv/αn )| dv A1,α · Pn,α (f )(z) ≤ |A1,α | · n! 0 Z M M nn+1 ∞ −nv n < e v dv = · , |A1,α | n! 0 |A1,α |

that is 3.2.4

1 A1,α Pn,α (f )

∈ SM/|A1,α | . Since |A1,α | ≤ 1, it follows that

M |A1,α |

≥ M.

Rotation-Invariant Complex Convolutions

For f ∈ A(D1 ), z ∈ D1 , let us consider the complex rotation-invariant integral operators given by Z +∞ Yk (f )(z) = 2k f (zeiv )ϕ(−2k v) dv, −∞

and the generalized complex rotation-invariant integral operators given by k Z 2k +∞ 2 Lk,j (f )(z) = `k (f )(2k zeiv )ϕ − v dv, k ∈ Z, j ∈ N. j −∞ j

Here i2 =R −1, ϕ is a real-valued function of compact support ⊆ [−a, a], a > 0, +∞ ϕ(x) ≥ 0, −∞ ϕ(x−u) du = 1, ∀x ∈ R, and {`k }k∈Z is a sequence of linear operators from A(D1 ) to A(D1 ) defined by recurrence as `k (f )(z) = `0 (fk )(z), z ∈ D, where z fk (z) = f 2k , z ∈ D and `0 : A(D1 ) → A(D1 ) is a linear operator.

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Also, let us consider the Jackson-type generalization of Lk,j (f )(z) given by q X j q (−1) Ik,q (f )(z) = − Lk,j (f )(z), ∀k ∈ Z, q ∈ N. j j=1 Note that the real variants (for real-valued functions of a real variable) of these operators were studied in Anastassiou-Gonska [21] and Anastassiou–Gal [12], [13]. In this subsection approximation and shape preserving properties (in geometric function theory) for the above complex rotation-invariant integral operators are presented. First the approximation properties are presented. Note that the results in Theorem 3.2.27 (i)-(v) were obtained in Anastassiou-Gal [19], while Theorem 3.2.27 (vi) is new. Theorem 3.2.27. Let f ∈ A(D1 ). (i) For all z ∈ D1 and k ∈ Z, we have

a |f (z) − Yk (f )(z)| ≤ 3ω1 f ; k ; 2 D1

(ii) For all z ∈ D1 , k ∈ Z, j ∈ N, we have mja + n , |f (z) − Lk,j (f )(z)| ≤ ω1 f ; 2k+r D1 where for fixed a > 0 it is assumed that sup z,y∈D1 |z−y|≤a

|`0 (f )(z) − f (y)| ≤ ω1

ma + n f; 2r

; D1

(iii) For the hypothesis in (ii) and q ∈ N, we have mqa + n q ; |f (z) − Ik,q (f )(z)| ≤ (2 − 1)ω1 f ; 2k+r D1 (iv) ω1 (Yk (f ); δ)D1 ≤ ω1 (f ; δ)D1 , δ > 0, k ∈ Z,

ω1 (Lk,j (f ); δ)D1 ≤ ω1 (f ; δ)D1 , δ > 0, k ∈ Z, j ∈ N,

ω1 (Ik,q (f ); δ)D1 ≤ (2q − 1)ω1 (f ; δ)D1 , δ > 0, k ∈ Z, q ∈ N,

in the hypothesis |`0 (f )(x − u + h) − `0 (f )(x − u)| ≤ ω1 (f ; h)D1 , ∀h > 0, ∀x, u ∈ D1 with x − u, x − u + h ∈ D1 . ∞ P (v) If f (z) = ap z p is analytic in D1 and continuous in D1 , then Yk (f )(z), p=0

Lk,j (f )(z) and Ik,q (f )(z) are analytic in D1 and continuous in D1 . The analyticity of Lk,j (f )(z) and Ik,q (f )(z) is proved here only for `0 (f ) ≡ f .

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Also we can write Yk (f )(z) =

∞ X

ap bp,k z p ,

p=0

where bp,k =

Z

+∞

cos −∞

If `0 (f ) ≡ f then Lk,j (f )(z) =

pu 2k

∞ X

ϕ(u) du,

ap bp,k,j z p ,

bp,k,j =

Z

∞

cos

−∞

p = 0, 1, . . . ,

k ∈ Z.

z ∈ D1 , k ∈ Z, j ∈ N

p=0

with

z ∈ D1 ,

pju 2k

ϕ(u) du = bpj,k ,

and Ik,q (f )(z) =

∞ X

ap cp,k,q z p ,

p=0

z ∈ D1 , k ∈ Z, q ∈ N

with cp,k,q =

q X j=1

(−1)

j+1

q bp,k,j . j

If ϕ(x) = 1 − x, x ∈ [0, 1], ϕ(x) = 1 + x, x ∈ [−1, 0], ϕ(x) = 0, x ∈ R \ (0, 1), then b1,k bj,k c1,k,q

1 1 − cos k > 0, ∀k ∈ Z, =2 2 2k+1 2 j = b1,k,j = 1 − cos k > 0, ∀k ∈ Z, j ∈ N, j2 2 2k+1 q X q 2 j = (−1)j+1 1 − cos , ∀k ∈ Z, q ∈ N, j2 2k j j=1 2k+1

and cp,k,q =

q X j=1

(−1)

j+1

2k+1 pj q 2 1 − cos k , 2 j p2 j 2

∀k ∈ Z, q ∈ N, p = 0, 1, 2, ....

(vi) Let us suppose that `0 (f ) ≡ f , the hypothesis in (ii) is satisfied for all z ∈ Dr P∞ and that f (z) = k=0 ak z k for all z ∈ DR , R > 1. For all 1 ≤ r < r1 < R, k ∈ Z, S j, q ∈ N and s ∈ N {0} we have (s)

kYk (f ) − f (s) kr ≤

C1 , 2k

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(s)

kLk,j (f ) − f (s) kr ≤ (s)

kIk,q (f ) − f (s) kr ≤

C2 , 2k+r C3 , 2k+r

where C1 depends on f , s, r, r1 and a, C2 depends on f , s, r, r1 , a, m and n, and C3 depends on f , s, r, r1 , a, m, q and n. Proof.

(i) Since 2k

Z

+∞

ϕ(−2k v) dv =

−∞

Z

+∞

ϕ(u) du = 1 −∞

we obtain Z +∞ k iv k |f (z) − Yk (f )(z)| = 2 [f (z) − f (ze )]ϕ(−2 v) dv −∞ Z +∞ |f (z) − f (zeiv )|ϕ(−2k v) dv ≤ 2k ≤ 2k ≤ 2k ≤ 2k

Z

−∞ +∞

−∞ Z +∞

Z

−∞ +∞ −∞

ω1 (f ; |z| · |1 − eiv |)D1 ϕ(−2k v) dv v ϕ(−2k v) dv ω1 f ; 2 sin 2 D1 ω1 (f ; |v|)D1 ϕ(−2k v) dv

Z +∞ k a 2 k |v| + 1 ϕ(−2k v) dv ≤ ω1 f ; k ·2 2 D1 a −∞ Z a 2k · 2k +∞ |v|ϕ(−2k v) dv . · 1+ = ω1 f ; k 2 D1 a −∞

But 2k · 2k a

Z

+∞

|v|ϕ(−2k v) dv = (by u = −2k v)

−∞ k k

Z +∞ Z |u| du 1 +∞ 2 ·2 · · ϕ(u) · = |u|ϕ(u) du k a 2k a −∞ −∞ 2 Z Z 1 a 1 a 2 = |u|ϕ(u) du ≤ |u| du = · a = 2, a −a a −a a =

which immediately proves (i). (ii) By k Z Z +∞ 2k +∞ 2 ϕ − v dv = ϕ(u) du = 1, j −∞ j −∞

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we get Z 2k +∞ 2k k iv Lk,j (f )(z) − f (z) = `k (f )(2 ze ) − f (z) ϕ − v dv j −∞ j k Z +∞ 2 2k = `0 (fk )(2k zeiv ) − fk (2k z) ϕ − v dv j −∞ j 2k by − v=u j Z +∞ j k i − 2k u `0 (fk ) 2 ze − fk (2k z) ϕ(u) du ≤ −∞ Z a j k i − 2k u − fk (2k z) ϕ(u) du. `0 (fk ) 2 ze = −a

But

k i 2 ze

−

j 2k

u

j ≤ 2k · j|u| = j|u| ≤ ja, u − 2k z ≤ 2k · 2 sin k 2·2 2k

for all |z| ≤ 1, k ∈ Z, j ∈ N, which implies (reasoning as in Anastassiou–Gal [12], p. 9) Z a j `0 (fk ) 2k zei − 2k u − fk (2k z) ϕ(u) du −a

Z

a

sup{|`0 (fk )(w) − fk (y)|; |w − y| ≤ ja}ϕ(u) du mja + n , ≤ ω1 f ; 2k+r D1

≤

−a

which proves (ii). (iii) By the relation −

q P

(−1)j

j=1

q j

= 1, we get

  X q X q q q j j   (−1) |Ik,q (f )(z) − f (z)| = − Lk,j (f )(z) − − (−1) f (z) j j j=1 j=1 X q j q [Lk,j (f )(z) − f (z)] = (−1) j j=1 ≤

q X q j=1 q X

j

· |Lk,j (f )(z) − f (z)|

mja + n q ω1 f ; ≤ 2k+r j D1 j=1 mqa + n ≤ (2q − 1)ω1 f ; , 2k+r D1

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which proves (iii) too. (iv) Let |z1 − z2 | ≤ δ, z1 , z2 ∈ D1 . We get Z +∞ |f (z1 eiv ) − f (z2 eiv )|ϕ(−2k v) dv |Yk (f )(z1 ) − Yk (f )(z2 )| ≤ 2k −∞

≤ ω1 (f ; |z1 − z2 |)D1 ≤ ω1 (f ; δ)D1 ,

where from passing to supremum with |z1 − z2 | ≤ δ, we obtain ω1 (Yk (f ) : δ)D1 ≤ ω(f ; δ)D1 ,

∀δ > 0, k ∈ Z.

Then, |Lk,j (f )(z1 ) − Lk,j (f )(z2 )| k Z 2 2k +∞ |`k (f )(2k z1 eiv ) − `k (f )(2k z2 eiv )|ϕ − v dv ≤ j −∞ j k 2 by − v = u j Z +∞ j j `0 (fk ) 2k z1 ei − 2k u − `0 (fk ) 2k z2 ei − 2k u ϕ(u) du ≤ −∞

≤ ω1 (f ; |z1 − z2 |)D1 ≤ ω1 (f ; δ)D1 ,

where from passing to supremum with |z1 − z2 | ≤ δ, we obtain ω1 (Lk,j (f ); δ)D1 ≤ ω1 (f ; δ)D1 . The inequality ω1 (Ik,q (f ); δ)D1 ≤ (2q − 1)ω1 (f ; δ)D1 follows immediately from the above inequality for Lk,j and from the relation q X q = 2q − 1, j j=1 which proves (iv). (v) Let z0 , zn ∈ D1 be with lim zn = z0 . We get (as in the proof of (iv)) n→∞

|Yk (f )(zn ) − Yk (f )(z0 )| ≤ ω1 (f ; |zn − z0 |)D1 ,

|Lk,j (f )(zn ) − Lk,j (f )(z0 )| ≤ ω1 (f ; |zn − z0 |)D1 , |Ik,q (f )(zn ) − Ik,q (f )(z0 )| ≤ ω1 (f ; |zn − z0 |)D1 ,

which proves the continuity of these operators in D1 . It remains to prove that Yk (f )(z), Lk,j (f )(z) and Ik,q (f )(z) are analytic in D1 . P p By hypothesis we have f (z) = ∞ p=0 ap z , z ∈ D1 . Let z ∈ D1 be fixed. We get f (zeiv ) =

∞ X p=0

ap eipv z p

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P∞ and since |ap eipv | = |ap | for all v ∈ R and the series p=0 ap z k is convergent, it P∞ follows that the series p=0 ap eipv z p is uniformly convergent with respect to v ∈ R. This immediately implies that the series can be integrated term by term, that is Z +∞ i − uk Yk (f )(z) = f ze 2 ϕ(u) du −∞

=

∞ X

ap

∞ X

ap

p=0

=

p=0

Z

Z

+∞

e

i − pu k 2

−∞ +∞ −∞

ϕ(u) du z p

∞ pu X ap bp,k z k , cos − k ϕ(u) du z p = 2 p=0

since cos is even function. If `0 (f ) ≡ f then `k (f )(2k zeiv ) = f (zeiv ) and we obtain k Z 2k +∞ 2 v Lk,j (f )(z) = dv f (zeiv )ϕ − j −∞ j and reasoning as for Yk (f )(z) we immediately obtain Lk,j (f )(z) =

∞ X

ap bp,k,j z p ,

p=0

with bk,p,j =

Z

+∞

cos −∞

pju 2k

z ∈ D1 ,

ϕ(u) du.

The development for Ik,q (f )(z) follows easily from above, which proves (v). For the particular choice of ϕ(x), we have : Z 1 Z 0 ju ju bj,k = b1,k,j = · (1 + u) du + cos · (1 − u) du cos k 2 2k −1 0 =2

Z

0

1

(1 − u) cos

k 1 Z 1 ju ju ju 2 − 2 u cos k du du = 2 sin · 2k 2k j 0 2 0

1 2k j ju 2k ju 2k+1 2 sin k − 2 cos k + u sin k = j 2 j2 2 j 2 0 j 22k+1 1 − cos > 0, ∀k ∈ Z, j ∈ N. = j2 2k For j = 1 we get b1,k,1 := b1,k = 22k+1 1 − cos 21k > 0. Therefore, q q X X q q 1 j c1,k,q = (−1)j+1 (−1)j+1 b1,k,j = 22k+1 1 − cos , j j j2 2k j=1 j=1

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and cp,k,q

q X q j+1 q (−1) (−1) = bp,k,j = bpj,k j j j=1 j=1 2k+1 q X pj q 2 1 − cos . (−1)j+1 = 2k j p2 j 2 j=1 q X

j+1

(vi) Analysing the proofs of the above points (i)-(v), it easily follows that they hold by replacing everywhere D1 with Dr , where 1 ≤ r < R. Therefore the approximation errors in (i)-(iii) can be expressed in terms of ω1 (f ; δ)Dr = sup{|f (u) − f (v)|; |u − v| ≤ δ}, with constants in front of ω1 depending on r ≥ 1. By the mean value theorem for divided differences in Complex Analysis (see e.g. Stancu [172], p. 258, Exercise 4.20) we get ω1 (f ; δ)Dr ≤ δkf 0 kr , where kf 0 kr = sup{|f 0 (z)|; |z| ≤ r} ≤ Then by (i), (ii) and (iii) it follows kYk (f ) − f kr ≤

∞ X

k=1

|ak |krk−n−1 .

C1 , 2k

kLk,j (f ) − f kr ≤

C2 , 2k+r

kIk,j (f ) − f kr ≤

C3 , 2k+r

respectively. Now denoting by γ the circle of radius r1 > 1 and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r we have Z s! Yk (f )(v) − f (v) (s) (q) |Yk (f )(z) − f (z)| = dv 2π (v − z)s+1 γ

C1 s! 2πr1 ≤ k · 2 2π (r1 − r)s+1 s!r1 C1 , = k · 2 (r1 − r)s+1 (s)

which proves the upper estimate in approximation by Yk (f )(z). (s) Similarly we get the upper estimates in approximation by Lk,j (f )(z) and (s)

Ik,q (f )(z), which proves the theorem.

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The geometric properties are expressed by the following results. Theorem 3.2.28. (Anastassiou-Gal [19]) It holds that Yk (P) ⊂ P, ∀k ∈ N, 1 1 Yk S3,b1,k ⊂ S3 , Yk (SM ) ⊂ SM/|b1,k | ∀k ∈ Z. b1,k b1,k If `0 (f ) ≡ f then 1 Lk,j (P) ⊂ P, Lk,j S3,b1,k,j ⊂ S3 , b1,k,j 1

Lk,j (SM ) ⊂ SM/|b1,k,j | , ∀k ∈ Z, j ∈ N. b1,k,j Here in all the cases we take ϕ(x) = 1 − x, x ∈ [0, 1], ϕ(x) = 1 + x, x ∈ [−1, 0], ϕ(x) = 0, x ∈ R \ (0, 1). Since from Theorem 3.2.27 (v) we have Z +∞ b0,k = b0,k,j = ϕ(u) du = 1, ∀k ∈ Z, j ∈ N,

Proof.

−∞

it follows Yk (f )(0) = Lk,j (f )(0) = a0 , that is if f ∈ P, f = U + iV then a0 = 1 and U > 0 on D1 , which implies Yk (f )(0) = Lk,j (f )(0) = 1, Z +∞ Re[Yk (f )(z)] = 2k U (r cos(x + v), r sin(x + v))ϕ(−2k v) dv > 0, ∀z = re

ix

−∞

∈ D1 , and for `0 (f ) ≡ f , k Z 2 v 2k +∞ dv > 0, U (r cos(x + v), r sin(x + v))ϕ − Re[Lk,j (f )(z)] = j −∞ j

for all z = reix ∈ D1 , that is Yk (P), Lk,j (P) ⊂ P. Let f (0) = f 0 (0) − 1 = 0. From Theorem 3.2.27, (v) we get 1 1 0 · Yk (f )(0) = B (f )(0) − 1 = 0 b1,k b1,k k and if `0 (f ) ≡ f then 1 1 Lk,j (f )(0) = · L0k,j (f )(0) − 1 = 0. b1,k,j b1,k,j Also, for f ∈ S3,b1 ,k we get Z 1 00 1 k +∞ 00 iv 2iv Y (f )(z) ≤ 2 |f (ze )e |ϕ(−2k v) dv |b1,k | b1,k k −∞ Z +∞ ≤ 2k ϕ(−2k v) dv = 1, −∞

1 Yk (f ) ∈ S3 , then for f ∈ SM it follows that is b1,k Z 1 0 1 k +∞ 0 iv iv M |f (ze )e |ϕ(−2k v) dv < , z ∈ D1 , b1,k Yk (f )(z) ≤ |b1,k | 2 |b 1,k | −∞ 1 that is b1,k Yk (f ) ∈ SM/|b1,k | . The proof in the case of Lk,j is similar, which proves the theorem.

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Remarks. 1) From the proof of Theorem 3.2.28 we obtain the following geometric properties: if f ∈ S3,b1,k then Yk (f ) is starlike (and univalent) on D1 , if f ∈ SM then Yk (f ) is univalent in |b1,k | 1 z ∈ C; |z| < ⊂ z ∈ C; |z| < , M M

and by Theorem 3.2.27 (v) Z +∞ Z +∞ u |b1,k | ≤ ϕ(u) du = 1; cos 2k ϕ(u) du ≤ −∞ −∞

if `0 (f ) ≡ f then f ∈ S3,b1,k,j implies that Lk,j (f ) is starlike (and univalent) on D1 and f ∈ SM implies that Lk,j (f ) is univalent in 1 |b1,k,j | ⊂ z ∈ C; |z| < , z ∈ C; |z| < M M

since by Theorem 3.2.27 (v) Z +∞ Z +∞ cos pju ϕ(u) du ≤ ϕ(u) du = 1. |b1,k,j | ≤ 2k −∞ −∞ 2) Let `0 (f ) ≡ f . If c1,k,q 6= 0 then similarly we get

1 Ik,q (SM ) ⊂ SM (2q −1)/|c1,k,q | , c1,k,q

that if f ∈ SM implies Ik,q (f ) is univalent in |c1,k,q | 1 z ∈ C; |z| < ⊂ z ∈ C; |z| < , M (2q − 1) M

since by Theorem 3.2.27 (v)

q X q j+1 |c1,k,q | = (−1) b1,k,j j j=1 q q X X q q ≤ |b1,k,j | ≤ = 2q − 1. j j j=1 j=1

3) For ϕ(x) = 1−x, ∀x ∈ [0, 1], ϕ(x) = 1+x, ∀x ∈ [−1, 0], ϕ(x) = 0, x ∈ R\(0, 1), let us consider 1 2k+1 1 − cos k ; k ∈ N , b1 = inf{|b1,k |; k ∈ N} = inf 2 2 b∗1 = inf{|b1,k,j |; k, j ∈ N, j ≤ 2k+1 } and c1,q = inf{|c1,k,q |; k ∈ N}.

We have: |b1,k | = 2 |b1,k,j | =

2k+1

22k+1 j2

2 1 1 1 2k+2 2 k+1 1 − cos k = 2 sin k+1 = 2 sin k+1 , 2 2 2 2 k+1 j 22k+2 j j 2 2 1 − cos k = , sin = · sin 2 j2 2k+1 j 2k+1

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which by the fact that f (t) = t sin 0 < b1 = Also, since 1 ≤ following.

2k+1 j ,

1 t

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279

is increasing for t ≥ 1, f (1) = sin 1, implies 2 1 1 4 sin = 16 sin2 . 4 4

j = 1, 2k+1 , we get b∗1 = sin2 1. Therefore, it is immediate the

Corollary 3.2.29. (Anastassiou-Gal [19]) (i) If f ∈ A∗ (D1 ) = {f ∈ A(D1 ); f (0) = f 0 (0) − 1 = 0},

|f 00 (z)| ≤ 16 sin2 14 , ∀z ∈ D1 then Yk (f ) ∈ S3 , for all k ∈ N and if f ∈ SM , M > 1, 16 sin2 1 then Yk (f ) is univalent in z ∈ C; |z| < M 4 , for all k ∈ N; (ii) If f ∈ A∗ (D1 ), |f 00 (z)| ≤ sin2 1, ∀z ∈ D1 , then Lk,j (f ) ∈ S3 and if f ∈ SM , 2 M > 1, then Lk,j (f ) is univalent in z ∈ C; |z| < sinM 1 , for all k, j ∈ N, j ≤ 2k+1 .

Remarks. 1) It would be of interest to find other geometric properties of the operators Yk , Lk,j and Ik,q . 2) Let f ∈ A∗ (D1 ) and define fα (z) := f (αz) for all α, z ∈ D1 . The operator Φ is called rotation invariant iff Φ(fα ) = (Φ(f ))α . We assume that `0 (f (2−k •))(az) = `0 (f (2−k α•))(z), k ∈ Z, a condition fulfilled trivially by Yk operators, case of `0 (f ) = f . Then easily one proves that `k (fα ) = (`k (f ))α and Yk (fα ) = (Yk f )α , (Lk,j (fα )) = (Lk,j (f ))α , Ik,q (fα ) = (Ik,q (f ))α . So all operators we are dealing with here are rotation invariant. 3.2.5

Sikkema Complex Convolutions

In a series of papers, Sikkema [164], [165], [166], [167] and Totik [193] studied the approximation properties of the convolution integral operators of real variable (for ρ → +∞), Z Z +∞ 1 +∞ ρ Uρ (f )(x) = f (x − t)β (t) dt, Iρ = β ρ (t) dt, Iρ −∞ −∞

where f, β : R → R satisfy some suitable properties. In this subsection we study approximation and geometric properties of the complexified version of the above operators, given by Z 1 +∞ f (ze−it )β ρ (t) dt, z ∈ Dr f ∈ A(Dr ), ρ ≥ 1, r ≥ 1, Lρ (f )(z) = Iρ −∞ where recall that

A(Dr ) = {f : Dr → C; f is analytic in Dr and continuous in Dr }. Let us suppose that β : R → R satisfies the following five properties:

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β(t) ≥ 0, ∀t ∈ R, β(0) = 1; ∀δ > 0, sup{β(t); |t| ≥ δ} < 1; β(t) is continuous at 0; t2 β(t) is Lebesgue integrable over R; β is even on R, i.e. β(−t) = β(t), ∀t ∈ R.

a) b) c) d) e)

First we present: P k Theorem 3.2.30. (Anastassiou-Gal [20]) If f (z) = ∞ k=0 ak z is analytic in D1 and continuous in D1 , then Lρ (f ) is analytic in D1 and continuous in D1 , for all ρ ≥ 1. Also, we can write Lρ (f )(z) =

∞ X

Ak (ρ)z k ,

k=0

where ak · Ak (ρ) = Iρ

Z

+∞

z ∈ D1 ,

cos(kt)β ρ (t) dt,

k = 0, 1, 2, . . . ,

−∞

and |Ak (ρ)| ≤ |ak |, Proof.

k = 0, 1, 2, . . . .

Let z0 , zn ∈ D1 be such that lim zn = z0 . We have n→∞

|Lρ (f )(zn ) − Lρ (f )(z0 )| ≤

1 Iρ

1 ≤ Iρ

Z Z

+∞ −∞ +∞

|f (zn e−it ) − f (z0 e−it )|β ρ (t) dt

ω1 f ; |e−it | · |zn − z0 | −∞ = ω1 f ; |zn − z0 | D ,

D1

β ρ (t) dt

1

which proves the continuity of Lρ (f ) in D1 . P k Now, let f (z) = ∞ k=0 ak z , z ∈ D1 , be analytic in D1 . For fixed z ∈ D1 , we get P ∞ −it −ikt k f (ze ) = k=0 ak e z , and since |ak e−ikt | = |ak |, for all t ∈ R and the series P∞ P∞ k −ikt k z is uniformly k=0 ak z is convergent, it follows that the series k=0 ak e convergent with respect to t ∈ R. This immediately implies that the series can be integrated term by term, that is Z ∞ X 1 +∞ [cos(kt) − i sin(kt)]β ρ (t) dt Lρ (f )(z) = ak z k · Iρ −∞ k=0 Z +∞ ∞ X 1 = ak cos(kt)β ρ (t) dt z k . Iρ −∞ k=0

Then, it is immediate that |Ak | ≤ |ak | ·

1 Iρ

Z

+∞ −∞

| cos(kt)|β ρ (t) dt ≤ |ak |,

k = 0, 1, . . .

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Remark. In the rest of the section those particular choices of β(t) for which R +∞ ρ (cos t)β (t) dt 6= 0 will be important. −∞ Concerning the approximation properties, we have

Theorem 3.2.31. (Anastassiou-Gal [20]) If f ∈ A(D1 ) then |Lρ (f )(z) − f (z)| ≤ 2 1 + Bρ (δ) ω1 (f ; δ)∂D1 , ∀z ∈ D1 , δ > 0, ρ ≥ 1, where

Bρ (δ) =

1 Iρ

Z

+∞ −∞

|t|δ −1 β ρ (t) dt.

Proof. By Theorem 3.2.30, Lρ (f ) is analytic in D1 and continuous on D1 , ∀ρ ≥ 1, f ∈ A(D1 ), so from the Maximum Modulus Principle, for the estimate |Lρ (f )(z) − f (z)| it suffices to take |z| = 1. For f ∈ A(D1 ) and |z| = 1, we can write f (z) = U (cos u, sin u) + iV (cos u, sin u),

∀z = eiu ∈ ∂D1 .

Denoting F (u) = U (cos u, sin u), G(u) = V (cos u, sin u), u ∈ R, by Sikkema [164], p. 356, Theorem 2, we get |Uρ (F )(u) − F (u)| ≤ 1 + Bρ (δ) ω1 (F ; δ)R , |Uρ (G)(u) − G(u)| ≤ 1 + Bρ (δ) ω1 (G; δ)R , ∀u ∈ R, δ > 0, ρ ≥ 1.

But, for |z| = 1, z = eiu , we have Lρ (f )(z) = Uρ (F )(u) + iUρ (G)(u) and for ω1 (f ; δ)∂D1 = sup{|f (eiu ) − f (eiv )|; u, v ∈ R, |u − v| ≤ δ},

it is easy to check the inequalities ω1 (F ; δ)R ≤ ω1 (f ; δ)∂D1 , ω1 (G; δ)R ≤ ω1 (f ; δ)∂D1 ,

δ > 0,

since |F (u) − F (v)| ≤ |f (eiu ) − f (eiv )|, |G(u) − G(v)| ≤ |f (eiu ) − f (eiv )|. In conclusion, for |z| = 1 we get

|Lρ (f )(z) − f (z)| ≤ 2 1 + Bρ (δ) ω1 (f ; δ)∂D1 ,

which proves the theorem.

∀δ > 0, ρ ≥ 1,

Remarks. 1) In the upper estimate in Theorem 3.2.31 it is important to make the best choice (depending of course on β) for δ := δ(ρ), in such a way that if ρ → ∞ then δ(ρ) → 0 in the fastest possible way and at the same time Bρ (δ) to remain bounded with respect to ρ > 0. Some well-known examples for β(t) 2 are as follows : β(t) = e−t (in this case Lρ (f )(z) is exactly Wξ∗ (f )(z) studied by Theorem 3.2.8), β(t) = e−|t| (in this case Lρ (f )(z) is exactly Q∗ξ (f )(z) studied by Theorem 3.2.1), β(t) = cos(πt/2) if |t| ≤ 1, β(t) = 0 if |t| > 1 (in this case Lρ (f )(z) is exactly Pn (f )(z) studied by Theorems 3.1.1–3.1.3 and by Corollary 3.1.4) and

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β(t) = 1 − |t|2p if |t| ≤ 1, β(t) = 0 if |t| > 1 (in this case we would obtain a complex version of the Landau operator). Also, for other choices for β(t) see e.g. Sikkema [164], pp. 358-359. 2) The lower estimate for approximation of f by Lρ (f )(z) can be obtained in a similar manner with the cases of the complex convolutions Wξ∗ (f )(z) (in the proof of Theorem 3.2.8) or Q∗ξ (f )(z) (in the proof of Theorem 3.2.1). Let us sketch here reasonings. Firstly let us observe that denoting dk (ρ) = R ∞the corresponding P∞ 1 ρ k k=0 ak z we get Iρ −∞ cos(kt) · β (t)dt and writing f (z) = Lρ (f )(z) =

Now take z = re

iϕ

and p ∈ N

S

∞ X

ak dk (ρ)z k .

k=0

{0}. We have

∞ 1 X 1 ak rk eiϕ(k−p) [1 − dk (ρ)]. [f (z) − Lρ (f )(z)]e−ipϕ = 2π 2π k=0

Integrating from −π to π we obtain Z π 1 [f (z) − Lρ (f )(z)]e−ipϕ dϕ = ap rp [1 − dp (ρ)]. 2π −π Passing now to absolute value we easily obtain |ap |rp |1 − dp (ρ)| ≤ kf − Lρ (f )kr .

Denote Vρ = inf 1≤p |1 − dp (ρ)|. By the above lower estimate for kLρ (f ) − f kr , for all p ≥ 1 and ρ > 0 it follows kLρ (f ) − f kr kLρ (f ) − f kr ≥ ≥ |ap |rp . Vρ |1 − dp (ρ)|

This implies that if there exists a subsequence (ρk )k with limk→∞ ρk = +∞ and kL (f )−f kr = 0 then ap = 0 for all p ≥ 1, that is f is constant such that limk→∞ ρk Vρ k

on Dr . Therefore, if f is not a constant then inf ρ>0 there exists a constant Cr (f ) > 0 such that is

kLρ (f )−f kr > 0, which implies Vρ kLρ (f )−f kr ≥ Cr (f ), for all ρ > 0, Vρ

that that

kLρ (f ) − f kr ≥ Cr (f )Vρ , for all ρ > 0. Now, concerning the geometric properties, first we present Theorem 3.2.32. (Anastassiou-Gal [20]) Let f ∈ A(D1 ) and Lρ (f )(z) = ∞ P Ak (ρ)z k . Suppose that β(t) is chosen such that A1 (ρ) 6= 0, for all ρ ≥ 1. k=0

Then, for all ρ ≥ 1 we have

1 1 Lρ S3,A1 (ρ) ⊂ S3 , Lρ (SM ) ⊂ SM/|A1 (ρ)| , A1 (ρ) A1 (ρ)

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where M > 1 and recall that S3 = f ∈ A(D1 ); |f 00 (z)| ≤ 1, ∀z ∈ D1 , S3,A1 (ρ) = f ∈ A(D1 ); |f 00 (z)| ≤ |A1 (ρ)|, ∀z ∈ D1 , SB = f ∈ A(D1 ); |f 0 (z)| < B, ∀z ∈ D1 . Proof.

Let f ∈ A(D1 ), f (z) =

∞ P

k=0

ak z k , z ∈ D1 . It follows a0 = 0, a1 = 1, which

by Theorem 3.2.30 immediately implies 1 · Lρ (f )(0) = 0, A1 (ρ)

1 A1 (ρ) L0ρ (f )(0) = = 1, A1 (ρ) A1 (ρ)

i.e. 1 Lρ (f ) ∈ A(D1 ). A1 (ρ) Then, by 1 1 L0ρ (f )(z) = A1 (ρ) Iρ

Z

1 1 L00ρ (f )(z) = A1 (ρ) Iρ

Z

and

+∞

e−it f 0 (ze−it )β ρ (t) dt,

−∞ +∞

−∞

e−2it f 00 (ze−it )β ρ (t) dt,

z ∈ D1 ,

z ∈ D1 ,

we get: f ∈ S3,A1 (ρ) implies |f 00 (z)| ≤ |A1 (ρ)|, ∀z ∈ D1 , i.e. Z 1 1 1 +∞ 00 −it −2it ρ 00 |f (ze )e |β (t) dt ≤ 1, ∀z ∈ D1 · |Lρ (f )(z)| ≤ |A1 (ρ)| |A1 (ρ)| Iρ −∞ and f ∈ SM implies |f 0 (z)| < M , ∀z ∈ D1 , i.e. Z M 1 1 1 +∞ 0 −it −it ρ |f (ze )e |β (t) dt < · |L0ρ (f )(z)| ≤ , |A1 (ρ)| |A1 (ρ)| Iρ −∞ |A1 (ρ)| which proves the theorem.

Remarks. 1) It is known (see e.g. Obradovici [145]) that f ∈ S3 implies that f is starlike (and univalent) in D1 and that f ∈ SM implies that f is univalent in 1 ⊂ D1 (see e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138], p. 111, Exercise z ∈ C; |z| < M 5.4.1). Since by Theorem 3.2.30 we have |A1 (ρ)| ≤ |a1 | = 1, it follows that S3,A1 (ρ) ⊂ S3 and |AM ≥ M > 1, i.e. if f ∈ S3,A1 (ρ) then Lρ (f ) remains starlike (and univalent) 1 (ρ)| in D1 and if f ∈ SM then Lρ (f ) is univalent in |A1 (ρ)| 1 z ∈ C; |z| < ⊂ z ∈ C; |z| < . M M

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2) Denote A = inf{|A1 (ρ)|; ρ ≥ 1}. If A > 0, then by Theorem 3.2.32 we get the following invariant geometric properties: if f ∈ S3,A then Lρ (f ) ∈ S3 for all ρ ≥ 1 and if f ∈ SM , M > 1, then Lρ (f ) is univalent in A z ∈ C; |z| < , ∀ρ ≥ 1. M Therefore, would remain to calculate A and to check that A > 0 for various choices of β(t), problems which are left to the reader as open questions. The second result concerning geometric properties of Lρ (f ) is the following. Theorem 3.2.33. (Anastassiou-Gal [20]) Let f ∈ A(Dr ), r > 1, and suppose that β(t) is such that for any bounded g : R → R, we have limρ→∞ Uρ (g) = g, uniformly in any compact interval of R. 0 (z) > 0, for all z ∈ D1 ), then there (i) If f is starlike in D1 (that is, Re zff (z) exists ρ0 > 0 (depending on f ), such that for all ρ ≥ ρ0 , Lρ (f )(z) are starlike in D1 . If f is starlike only in D1 , then for any disk of radius 0 < λ < 1 denoted by Dλ , there exists ρ0 (depending on f and Dλ ), such that for all ρ ≥ ρ0 , Lρ (f )(z) are zL0ρ (f )(z) Lρ (f )(z)

> 0, for all z ∈ Dλ ). 00 (z) + 1 > 0, for all z ∈ D1 ), then (ii) If f is convex in D1 (that is, Re zff 0 (z)

starlike in Dλ (that is, Re

there exists ρ0 (depending on f ), such that for all ρ ≥ ρ0 , Lρ (f )(z) are convex in D1 . If f is convex only in D1 , then for any disk of radius 0 < λ < 1 denoted by Dλ , there exists ρ0 (depending on f and Dλ ), such that for all ρ ≥ ρ0 , Lρ (f )(z) are

convex in Dλ (that is, Re

zL00 ρ (f )(z) L0ρ (f )(z)

+ 1 > 0, for all z ∈ Dλ ).

Proof. First let us make some general useful considerations. By hypothesis, it follows that for ρ → ∞, we have Lρ (f )(z) → f (z), uniformly in any compact disk included in Dr , that is in D1 too. Indeed, this is immediate from the relationship Lρ (f )(z) = Uρ (Fλ )(u) + iUρ (Gλ )(u), where Fλ (u) = U (λcos(u), λsin(u)), Gλ (u) = V (λcos(u), λsin(u)), f (z) = U (x, y) + iV (x, y), |z| = λ ∈ [0, r), z = λeiu = x + iy, u ∈ [0, 2π]. By the well-known Weierstrass’ result, this implies that L0ρ (f )(z) → f 0 (z) and 00 Lρ (f )(z) → f 00 (z), uniformly in any compact disk in Dr and therefore in D1 too, when ρ → ∞. Then, denoting by A1 (ρ) the coefficient of z in the Taylor series in Theorem 3.2.30 representing the analytic function Lρ (f )(z), since A1 (ρ) = L0ρ (0) and limρ→∞ L0ρ (0) = f 0 (0) = 1, it follows that limρ→∞ A1 (ρ) = 1 and for all ρ ≥ ρ0 we have A1 (ρ) > 0. Lρ (f )(z) Let us denote Pρ (f )(z) = A , for all ρ ≥ ρ0 . 1 (ρ) L0 (f )(0)

ρ 0 By f (0) = f 0 (0) − 1 = 0 we get Pρ (f )(0) = Af1(0) (ρ) = 0 and Pρ (f )(0) = A1 (ρ) = 0 0 00 1. Also, we obviously have Pρ (f )(z) → f (z), Pρ (f )(z) → f (z) and Pρ (f )(z) → f 00 (z), uniformly in D1 .

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(i) By hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Write Pρ (f )(z) in the form Pρ (f )(z) = zQρ (f )(z). Let |z| = 1. We have |f (z) − Pρ (f )(z)| = |z| · |g(z) − Qρ (f )(z)| = |g(z) − Qρ (f )(z)|, which by the uniform convergence in D1 of Pρ (f ) to f and by the Maximum Modulus Principle, implies the uniform convergence in D1 of Qρ (f )(z) to g(z), as ρ → ∞. Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist ρ0 and a > 0 depending on g, such that |Qρ (f )(z)| > a > 0, for all z ∈ D1 and all ρ ≥ ρ0 . Also, for all |z| = 1, we have |f 0 (z) − Pρ0 (f )(z)| = |z[g 0 (z) − Q0ρ (f )(z)] + [g(z) − Qρ (f )(z)]| ≥|

=|

|z| · |g 0 (z) − Q0ρ (f )(z)| − |g(z) − Qρ (f )(z)| | |g 0 (z) − Q0ρ (f )(z)| − |g(z) − Qρ (f )(z)| |,

which from the Maximum Modulus Principle, the uniform convergence of Pρ0 (f ) to f 0 and of Qρ (f ) to g, evidently implies the uniform convergence of Q0ρ (f ) to g 0 , as ρ → ∞. Then, for |z| = 1, we get z[zQ0ρ (f )(z) + Qρ (f )(z)] zPρ0 (f )(z) = Pρ (f ) zQρ (f )(z) zQ0ρ (f )(z) + Qρ (f )(z) zg 0 (z) + g(z) = → Qρ (f )(z) g(z) 0 0 f (z) zf (z) = = , g(z) f (z)

which again from the Maximum Modulus Principle, implies

Since Re

Therefore

zf 0 (z) f (z)

zPρ0 (f )(z) zf 0 (z) → , uniformly in D1 . Pρ (f ) f (z) is continuous in D1 , there exists α ∈ (0, 1), such that 0 zf (z) Re ≥ α, for all z ∈ D1 . f (z)

0 zPρ0 (f )(z) zf (z) Re → Re ≥ α > 0, Pρ (f )(z) f (z)

uniformly in D1 , i.e. for any 0 < β < α, there is ρ0 such that for all ρ ≥ ρ0 we have 0 zPρ (f )(z) Re > β > 0, for all z ∈ D1 . Pρ (f )(z)

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Since Pρ (f )(z) differs from Lρ (f )(z) only by a constant, this proves the first part in (i). For the second part, the proof is identical with the first part, with the only difference that instead of D1 , we reason for Dλ . (ii) For the first part, by hypothesis there is α ∈ (0, 1), such that 00 zf (z) Re + 1 ≥ α > 0, f 0 (z) uniformly in D1 . It is not difficult to show that this is equivalent with the fact that for any β ∈ (0, α), the function zf 0 (z) is starlike of order β in D1 (see e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138], p. 77), which implies f 0 (z) 6= 0, for all z ∈ D1 , i.e. |f 0 (z)| > 0, for all z ∈ D1 . Also, by the same type of reasonings as those from the above point (i), we get 00 00 zPρ (f )(z) zf (z) Re + 1 → Re + 1 ≥ α > 0, Pρ0 (f )(z) f 0 (z) uniformly in D1 . As a conclusion, for any 0 < β < α, there is ρ0 > 0 depending on f , such that for all ρ ≥ ρ0 we have 00 zPρ (f )(z) Re + 1 > β > 0, for all z ∈ D1 . Pρ0 (f )(z) The proof of second part in (ii) is similar, which proves the theorem.

3.3

Nonlinear Complex Convolutions

Geometric properties of nonlinear integral transforms generated by the (nonlinear) 2 Hadamard product (of fz(z) ) with special functions of hypergeometric or of Hurwitztype, were obtained in several papers, see e.g. Ponnusamy-Singh-Vasundhra [153], Ponnusamy-Vasundhra [154]. In this section we move in a different direction. Thus, we present convergence, shape preserving results and rate of approximation of analytic functions by some nonlinear complex integral convolution operators, related with the linear convolutions studied in the previous two sections. More exactly, we present approximation and shape preserving properties (that is preservation of univalence, starlikeness, convexity, etc) for the family of nonlinear complex convolution operators of the form Z π Tξ (f )(z) = Lξ (t)Hξ [f (ze−it )]dt, ξ > 0, z ∈ D1 , 2

−π

where i = −1, f is analytic in the open unit disk D1 and continuous in D1 , Lξ : R → R and Hξ : C → C. The results in this section extends those of the real case in Angeloni-Vinti [28], [29], with the addition that while the idea of shape preservation is absent in the real case, in the complex case it is present.

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First we recall some notations. For D1 = {∈ C; |z| < 1}, Dr = {z ∈ C; |z| < r}, Cr = {z ∈ C; |z| = r}, 0 < r < 1, let us consider A(D1 ) = {f : D1 → C; f is continuous in D1 and analytic in D1 }, and the total variation of f ∈ A(D1 ) on Cr , by Vr (f ) = V[−π,π] (Fr ), where Fr (u) = f (reiu ), u ∈ [−π, π] and V[−π,π] denotes the (usual) total variation of Fr on [−π, π]. Since for f ∈ A(D1 ) and r ∈ (0, 1), the modulus of derivative f 0 (z) is bounded on Dr , by the classical mean value theorem (inequality) in complex analysis it easily follows that Vr (f ) < ∞, for any r ∈ (0, 1). Denote by L12π , the class of all 2π-periodic functions f : R → R, which are Lebesgue integrable over [−π, π]. Everywhere in the section let us consider the family of kernel functions Kξ : R×C → C, ξ > 0, of the form Kξ (t, u) = Lξ (t)Hξ (u), with Hξ : C → C and Lξ : R → R are 2π-periodic functions, such that, in addition they satisfy the following properties : 1) LξR is Lebesgue measurable, Lξ ∈ L12π , kLξ kL1 ≤ A for all ξ > 0 and π limξ→∞ −π LξR(t)dt = 1 ; 2) limξ→∞ δ≤|t|≤π |Lξ (t)|dt = 0, for any fixed δ > 0 ; 3) Hξ (z), ξ > 0, are entire functions with Hξ (0) = 0, for all ξ > 0 and limξ→∞ Hξ (u) = u, for all u ∈ C ; 4) For every M > 0, there exists KM > 0 (independent of ξ), such that such that |Hξ (u) − Hξ (v)| ≤ KM |u − v|, for all u, v ∈ DM , ξ > 0, where DM = {z ∈ C; |z| ≤ M }. Remark. A simple example for Hξ (u) satisfying the above conditions is given by 1 Hξ (u) = u + ξ+1 g(u), where g : C → C is an entire function satisfying g(0) = 0. Indeed, since g is entire function, for any M > 0 it has bounded derivative on DM and let us denote that bound by BM . By the classical mean value inequality in complex analysis it follows |Hξ (u) − Hξ (v)| ≤ |u − v| +

1 |g(u) − g(v)| 1+ξ

≤ (1 + BM )|u − v|, for all u, v ∈ DM , ξ > 0, which means that condition 4) is satisfied. 1 Also, since Hξ (u) − u = ξ+1 g(u), this immediately implies 3) too. The first result shows the invariance property of the nonlinear convolutions Tξ , ξ > 0. Theorem 3.3.1. (Gal [88]) Suppose that the above conditions 1)-4) are fulfilled. If f ∈ A(D1 ) then Tξ (f ) ∈ A(D1 ), for all ξ > 0. Proof. Fix ξ > 0. First we show that Tξ (f ) is differentiable in any z0 ∈ D1 . Indeed, it is clear that there exists r < 1 such that z0 ∈ Dr . Let zn → z0 , as

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n → ∞, zn 6= z0 , for all n ∈ N. Obviously, without loss of generality, we can suppose that zn ∈ Dr , for all n ∈ N. We get Z π Tξ (f )(zn ) − Tξ (f )(z0 ) Hξ [f (zn e−it )] − Hξ [f (z0 e−it )] dt. = Lξ (t) zn − z 0 zn − z 0 −π

Denoting now

Fn (t) =

Hξ [f (zn e−it )] − Hξ [f (z0 e−it )] , zn − z 0

since H is entire function and f ∈ A(D1 ), it is immediate that for each t ∈ [−π, π], we have lim Fn (t) = Hξ0 [f (z0 e−it )] · f 0 (z0 e−it )[e−it ].

n→∞

This show that pointwise for t ∈ [−π, π], we have

lim L(t)ξ Fn (t) = Lξ (t)Hξ0 [f (z0 e−it )] · f 0 (z0 e−it )[e−it ].

n→∞

On the other hand, by the property 4) and by the classical mean value inequality in complex analysis, there is η ∈ Dr , such that Hξ [f (zn e−it )] − Hξ [f (z0 e−it )] ≤ |Lξ (t) · Fn (t)| ≤ |Lξ (t)| · zn − z 0 f (zn e−it ) − f (z0 e−it ) ≤ |Lξ (t)| · Mf · |f 0 (η)| ≤ |Lξ (t)| · Mf · kf 0 k . |Lξ (t)| · Mf · Dr zn − z 0

By the Lebesgue’s dominated convergence theorem, we can pass to limit under the T (f )(zn )−Tξ (f )(z0 ) exists, i.e. Tξ (f )(z) is integral sign, which shows that limn→∞ ξ zn −z0 analytic in D1 . For the continuity at z0 ∈ D1 , let zn ∈ D1 , with zn → z0 as n → ∞. We have Z π |Tξ (f )(zn ) − Tξ (f )(z0 )| = Lξ (t)[Hξ (f (zn e−it )) − Hξ (f (z0 e−it ))dt −π Z π ≤ |Lξ (t)| · |f (zn e−it ) − f (z0 e−it )|dt, −π

which by the continuity of f at z0 ∈ D1 , immediately implies the continuity of Tξ (f ) too at z0 . The next result represent an estimation in variation for the nonlinear convolutions Tξ , ξ > 0. Theorem 3.3.2. (Gal [88]) Suppose that the above conditions 1)-4) are fulfilled. Then, for any f ∈ A(D1 ), there exists a constant Cf > 0 (depending only on f ), such that Vr [Tξ (f )] ≤ Cf Vr [f ], for all 0 < r < 1, ξ > 0.

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Proof. For 0 < r < 1, f ∈ A(D1 ) and ξ > 0, denote Tr,ξ (u) = Tξ (f )(reiu ). Also, let s0 = −π < s1 < ... < sn = π be a partition of [−π, π] and denote zj = reisj , j = 0, ..., n. Denoting by Mf > 0, the smallest positive constant such that f [D1 ] ⊂ DMf and applying property 4), we get n X

≤

|Tr,ξ (sj ) j=1 n Z π X j=1

≤ Mf

−π

Z

− Tr,ξ (sj−1 )|

|Lξ (t)| · |Hξ [f (zj e−it )] − Hξ [f (zj−1 e−it )]|dt

π

−π

|Lξ (t)|

n X j=1

|f (zj e−it ) − f (zj−1 e−it )|dt ≤ A · Mf Vr (f ),

which proves the theorem.

Concerning the convergence, it holds Theorem 3.3.3. (Gal [88]) Suppose that the above conditions 1)-4) are fulfilled and f ∈ A(D1 ). Then limξ→∞ Tξ (f )(z) = f (z), uniformly in any compact disk included in D1 . Proof.

We can write Tξ (f )(z) − f (z) = =

Z

Z

π −π π −π

Z

Lξ (t)Hξ [f (ze−it )]dt − f (z) Lξ (t){Hξ [f (ze−it )] − Hξ [f (z)]}dt π

Lξ (t)[Hξ [f (z)] − f (z)]dt Z π Lξ (t)dt − 1 := E1 + E2 + E3 . +f (z)

+

−π

−π

We will estimate all Ek , k = 1, 2, 3. For this purpose, let z ∈ D1 and take ε > 0 arbitrary small, fixed. First, we note that there exists r < 1 such that z ∈ Dr . We have Z π |E3 | ≤ |f (z)| · Lξ (t)dt − 1 , −π

which by condition 1) implies that there exist ξ3 > 0, such that |E3 | ≤ ε · kf kD1 , for all ξ > ξ3 . Also, |E2 | ≤

Z

π −π

|Lξ (t)| · |Hξ [f (z)] − f (z)|dt,

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290

which by the conditions 3) and 1), implies that there exists ξ2 > 0, such that for all ξ > ξ2 we have Z π |E2 | ≤ ε · |Lξ (t)| ≤ εA. −π

Finally, by the conditions 4) and 2) we get Z π |E1 | ≤ |Lξ (t)| · |Hξ [f (ze−it )] − Hξ [f (z)]|dt −π Z π |Lξ (t)| · |f (zeit ) − f (z)|dt ≤ Mf −π Z π ≤ Mf |Lξ (t)|ω1 (f ; |z| · |eit − 1|)Dr dt −π Z π |Lξ (t)|ω1 (f ; 2|z| · |sin(t/2)|)Dr dt ≤ Mf −π Z π |Lξ (t)|ω1 (f ; |t|)Dr dt ≤ Mf −π Z Z ε ≤ Mf |Lξ (t)| · 2kf kD1 |Lξ (t)| · ω1 (f ; ε)Dr dt + Mf ε≤|t|≤π −ε Z ≤ Mf · A · ω1 (f ; ε)Dr + 2Mf · kf kD1 |Lξ (t)|dt ε≤|t|≤π

0

≤ Mf · A · εkf kDr + 2Mf kf kD1 ε, for all ξ > ξ1 .

Here ω1 (f ; ε)Dr := sup{|f (u) − f (v)|; u, v ∈ Dr , |u − v| ≤ ε}, and k · kDM denotes the uniform norm in DM . Collecting all the above estimates and denoting ξ0 = max{ξ1 , ξ2 , ξ3 }, for all ξ > ξ0 we get |Tξ (f )(z) − f (z)| ≤ |E1 | + |E2 | + |E3 |

≤ Mf · A · εkf 0 kDr + 2Mf kf kD1 ε + εA + ε · kf kD1 ,

which implies that for all z ∈ D1 , we have (pointwise) lim Tξ (f )(z) = f (z).

ξ→∞

On the other hand, we will show that (Tξ (f )(z))ξ>0 is uniformly bounded (that is independent of ξ) on each Dr . Indeed, taking into account that Hξ (0) = 0 and taking v = 0 in condition 4), for all ξ > 0 and |z| ≤ r we obtain Z

π −π

|Tξ (f )(z)| ≤ Z π |Lξ (t)| · |Hξ [f (ze−it )]|dt ≤ Kf |Lξ (t)| · |f (ze−it )|dt ≤ Kf · A · kf kDr . −π

Since by Theorem 3.3.1, it follows Tξ (f ) ∈ A(D1 ) for all ξ > 0, applying the classical Vitali’s result it follows the uniform convergence on each Dr , with 0 < r < 1.

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Corollary 3.3.4. (Gal [88]) Suppose that the conditions 1)-4) are fulfilled. If f ∈ A(DR ) with R > 1, then Tξ (f ) ∈ A(DR ), for all ξ > 0 and limξ→∞ Tξ (f )(z) = f (z), uniformly in any compact disk included in DR , that is in D1 too. Proof. The proof follows exactly the lines of the proofs for Theorem 3.3.1 and Theorem 3.3.3, but in this case for f ∈ A(DR ) with R > 1. The following result gives an estimate for the convergence result in Corollary 3.3.4 and it is a consequence of the estimates obtained in the proof of Theorem 3.3.3. For this purpose, we need some additional requirements, described below (see e.g. Angeloni-Vinti [28]). + Let h : R+ 0 → R0 with the properties : h is continuous atR0, h(0) = 0 and π h(u) > 0 for u > 0. One says that Lξ (t) is a h-singular kernel, if | −π Lξ (t)dt − 1| = O[h(1/ξ)] and for every δ > 0, Z |Lξ (t)|dt = O[h(1/ξ)], as ξ → ∞. δ≤|t|≤π

Corollary 3.3.5. (Gal [88]) Suppose that in addition to the above conditions 1)-4), the following conditions are fulfilled : a) Lξ (t) are h-singular kernels, for all ξ > 0 ; b) |Hξ (u) − u| = O[h(1/ξ)], as ξ → ∞, on each compact disk in C ; c) there exists δ > 0, such that Z |Lξ (t)| · |t|dt = O[h(1/ξ)], as ξ → ∞. 0≤|t|≤δ

Then, for any f ∈ A(DR ) with R > 1, and for sufficiently large ξ, we have kTξ (f ) − f kD1 = O[h(1/ξ)].

Proof.

From the proof of Theorem 3.3.3, for all |z| ≤ 1, it follows |Tξ (f )(z) − f (z)| ≤ |E1 | + |E2 | + |E3 |,

and (taking into account on the hypothesis too) Z π Z π |E3 | ≤ |f (z)| · Lξ (t)dt − 1 ≤ kf kD1 · Lξ (t)dt − 1 = O[h(1/ξ)], −π −π Z π |Lξ (t)| · |Hξ [f (z)] − f (z)|dt = O[h(1/ξ)], |E2 | ≤ Z

π

−π

|E1 | ≤ Mf |Lξ (t)| · ω1 (f ; |t|)D1 dt −π Z Z |Lξ (t)| · ω1 (f ; |t|)D1 dt + Mf = Mf |Lξ (t)|ω1 (f ; |t|)D1 dt δ≤|t|≤π 0≤|t|≤δ Z Z |Lξ (t)|dt + |Lξ (t)| · kf 0 kD1 · |t|dt = O[h(1/ξ)]. ≤ 2Mf kf kD1 δ≤|t|≤π

0≤|t|≤δ

All these imply the conclusion in Corollary 3.3.5.

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Remark. Concrete examples for various Lξ (t) satisfying the above results, can be found in e.g. Angeloni-Vinti [28], [29]. In what follows we present the geometric properties of the nonlinear complex operators Tξ (f )(z), ξ > 0. Theorem 3.3.6. (Gal [88]) Assume that the above conditions 1)-4) are fulfilled and let us suppose that f : DR → C is analytic in DR , where R > 1. If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) in D1 , that is for all z ∈ D1 (see e.g. Mocanu-Bulboac˘ a-S˘ al˘ agean [138]) 0 00 0 zf (z) zf (z) iη zf (z) Re > 0 Re + 1 > 0, Re e > 0, resp. , f (z) f 0 (z) f (z) then there exists ξ0 depending on f (and on η for spirallikeness), such that for all ξ ≥ ξ0 , Tξ (f )(z), are starlike (convex, spirallike of type η, respectively) in D 1 . If f (0) = f 0 (0)−1 = 0 and f is starlike (convex, spirallike of type η, respectively) only in D1 (that is the corresponding inequalities hold only in D1 ), then for any disk of radius 0 < r < 1 and center 0 denoted by Dr , there exists ξ0 = ξ0 (f, Dr ) (ξ0 depends on η too in the case of spirallikeness), such that for all ξ ≥ ξ 0 , Tξ (f )(z), are starlike (convex, spirallike of type η, respectively) in Dr (that is, the corresponding inequalities hold in Dr ). Proof. By Corollary 3.3.4, it follows that for ξ → ∞, we have Tξ (f )(z) → f (z), uniformly for |z| ≤ 1, which by the well-known Weierstrass’s theorem implies [Tξ (f )]0 (z) → f 0 (z) and [Tξ (f )]00 (z) → f 00 (z), for ξ → ∞, uniformly in D1 . In T (f )(z) all what follows, denote Pξ (f )(z) = [Tξξ(f )]0 (0) , well defined for sufficiently large ξ. 0 We easily get Pξ (f )(0) = 0, Pξ (f )(0) = 1, Pξ (f )(z) → f (z), Pξ0 (f )(z) → f 0 (z) and Pξ00 (f )(z) → f 00 (z), uniformly in D1 . Suppose first that f is starlike in D1 . Then, by hypothesis we get |f (z)| > 0 for all z ∈ D1 with z 6= 0, which from the univalence of f in D1 , implies that we can write f (z) = zg(z), with g(z) 6= 0, for all z ∈ D1 , where g is analytic in D1 and continuous in D1 . Write Pξ (f )(z) in the form Pξ (f )(z) = zQξ (f )(z). For |z| = 1 we have |f (z) − Pξ (f )(z)| = |z| · |g(z) − Qξ (f )(z)| = |g(z) − Qξ (f )(z)|, which by the uniform convergence in D1 of Pξ (f ) to f and by the maximum modulus principle, implies the uniform convergence in D1 of Qξ (f )(z) to g(z). Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist ξ1 > 0 and a > 0 depending on g, such that |Qξ (f )(z)| > a > 0, for all z ∈ D1 and all ξ ≥ ξ0 . Also, for all |z| = 1, we have |f 0 (z) − Pξ0 (f )(z)| = |z[g 0 (z) − Q0ξ (f )(z)] + [g(z) − Qξ (f )(z)]|

≥ | |z| · |g 0 (z) − Q0ξ (f )(z)| − |g(z) − Qξ (f )(z)| |

= | |g 0 (z) − Q0ξ (f )(z)| − |g(z) − Qξ (f )(z)| |,

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which from the maximum modulus principle, the uniform convergence of Pξ0 (f ) to f 0 and of Qξ (f ) to g, evidently implies the uniform convergence of Q0ξ (f ) to g 0 . Then, for |z| = 1, we get zPξ0 (f )(z) z[zQ0ξ (f )(z) + Qξ (f )(z)] = Pξ (f ) zQξ (f )(z) 0 zQξ (f )(z) + Qξ (f )(z) zg 0 (z) + g(z) = → Qξ (f )(z) g(z) zf 0 (z) f 0 (z) = , = g(z) f (z) which again from the maximum modulus principle, implies zPξ0 (f )(z) zf 0 (z) → , uniformly in D1 . Pξ (f ) f (z) Since Re

zf 0 (z) f (z)

is continuous in D1 , there exists ε ∈ (0, 1), such that Re

zf 0 (z) f (z)

≥ ε, for all z ∈ D1 .

Therefore Re

zPξ0 (f )(z) Pξ (f )(z)

→ Re

zf 0 (z) ≥ε>0 f (z)

uniformly on D1 , i.e. for any 0 < ρ < ε, there is ξ0 such that for all ξ ≥ ξ0 , we have 0 zPξ (f )(z) Re > ρ > 0, for all z ∈ D1 . Pξ (f )(z) Since Pξ (f )(z) differs from Tξ (f )(z) only by a constant, this proves the starlikeness of Tξ (f )(z), for sufficiently large ξ. If f is supposed to be starlike only in D1 , the proof is identical, with the only difference that instead of D1 , we reason for Dr with r < 1. The proofs in the cases when f is convex or spirallike of order η are similar and follow from the following uniform convergence (on D1 or on Dr ) as ξ → ∞ " # 00 zPξ00 (f )(z) zf (z) Re + 1 → Re +1 Pξ0 (f )(z) f 0 (z) and 0 zPξ0 (f )(z) iη zf (z) iη Re e → Re e . Pξ (f )(z) f (z)

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Bibliographical Notes and Open Problems

Theorems 3.1.1, 3.1.2, 3.1.3, Corollary 3.1.4, Theorem 3.1.5, Corollary 3.1.11, Theorems 3.1.12 and 3.1.13, Corollary 3.1.14, Theorems 3.1.15-3.1.16, 3.2.1 (iv), 3.2.5 (iv), 3.2.8 (iv), 3.2.11 (iv), 3.2.17, 3.2.18, Corollaries 3.2.19, 3.2.20, Theorems 3.2.25 (iv), 3.2.27 (iv) are new and appear for the first time here. Open Problem 3.4.1. Let f : DR → C be analytic on DR with R > 1 and suppose that 1 ≤ r < R and n ∈ N are fixed. Then for all ξ ∈ (0, 1] we have kQξ (f ) − f kr ≤ Cr (f )ξ, from Theorem 3.2.5, (iii) , kWξ (f ) − f kr ≤ Cr (f )

p ξ, from Theorem 3.2.8, (iii) ,

kWn,ξ (f ) − f kr ≤ Cn,r (f )ξ n+1 , from Theorem 3.2.11, (ii) . The exact order of approximations (with respect to ξ) remain as open questions. (q)

Open Problem 3.4.2. It remains to prove the lower estimates for kPn,ξ (f )−f (q) kr ∗ (q) and k[Wn,ξ ] (f ) − f (q) kr (for the upper estimates see Theorem 3.2.11, (iv)). (q)

Open Problem 3.4.3. It remains to prove the lower estimate for kPn,α (f )−f (q) kr (for the upper estimate see Theorem 3.2.25, (iv)). (s)

Open Problem 3.4.4. It remains to prove the lower estimates for kYk (f )−f (s) kr , (s) (s) kLk,j (f )−f (s) kr , kIk,q (f )−f (s) kr , (for the upper estimate see Theorem 3.2.27, (vi)). Open Problem 3.4.5. It would be of interest to study the complex Sikkema-type operators, for various kernels (different from those classical already mentioned in Subsection 3.2.4, Remark 1 after the proof of Theorem 3.2.31) introduced in the Sikkema’s papers [164], [165], [166], [167]. Open Problem 3.4.6. Taking into account the Remark after the proof of Theorem 3.2.23, it is an open question if for qn → 1, 0 < qn < 1, the approximation order by P1/n (f ; qn ; z) and W1/n (f ; qn ; z) are exactly [n]1q and √ 1 , respectively. n

[n]qn

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Chapter 4

Appendix : Related Topics

This chapter is a collection of several distinct directions of research generalizing and/or being related with those in the previous ones : approximation by quaternion Bernstein polynomials and approximation of vector-valued functions by Bernstein and convolution type operators. 4.1

Bernstein Polynomials of Quaternion Variable

As it is well-known, the field of complex numbers can be extended to more general algebraic structures (with several complex units) called hypercomplex numbers. These structures can be divided in two main classes : commutative hypercomplex structures which are rings with divisors of zeros and noncommutative hypercomplex structures which are fields (so without divisors of zero), the most known being the so-called quaternion numbers and Clifford algebras. Therefore, it is natural to see for extensions of the approximation results in the previous sections, to approximation by Bernstein-type operators of hypercomplex variables. In this section we limit our consideration to the case of Bernstein polynomials of quaternion variable. For this purpose first we make a short introduction. The quaternion field is defined by H = {q = x1 + x2 i + x3 j + x4 k; x1 , x2 , x3 , x4 ∈ R}, where the complex units i, j, k 6∈ R satisfy i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. It is a noncommutative field and since obviously C ⊂ H,pit extends the class of complex numbers. On H can be defined the norm kqk = x21 + x22 + x33 + x24 , for q = x1 + x2 i + x3 j + x4 k. If G ⊂ H then a function f : G → H can be written in the form f (q) = f1 (x1 , x2 , x3 , x4 ) + f2 (x1 , x2 , x3 , x4 )i + f3 (x1 , x2 , x3 , x4 )j + f4 (x1 , x2 , x3 , x4 )k, 295

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q = x1 + x2 i + x3 j + x4 k ∈ G, where fi are real valued functions, i = 1, 2, 3, 4. It is well-known the fact that a direct attempt to generalize the concept of differentiability for f as lim (q − q0 )−1 [f (q) − f (q0 )] ∈ H, q0 ∈ G,

q→q0

or as lim [f (q) − f (q0 )](q − q0 )−1 ∈ H, q0 ∈ G,

q→q0

necessarily implies that f is of the form f (q) = Aq + B (see Mejlihzon [135]). For this reason, the theory of holomorphic functions of quaternion variable can be constructed in several other ways, by producing different classes of holomorphic functions. We mention below only two ways. The first one is given by the following. Definition 4.1.1. (Moisil [139]) Let f = f1 + f2 i + f3 j + f4 k be such that each ∂ + fi has continuous partial derivatives of order one, i = 1, 2, 3, 4. Define F = ∂x 1 ∂ ∂ ∂ ∂x2 i + ∂x3 j + ∂x4 k. One says that f is left differentiable (monogenic) at q 0 if F f (q0 ) = 0. In this case, the derivative of f at q0 will be given f 0 (q0 ) = D(f )(q0 ), ∂ ∂ ∂ ∂ where F = ∂x − ∂x i − ∂x j − ∂x k. 1 2 3 4 If f is monogenic at each q, then it is called holomorphic. Remarks. 1) In the case of complex variable, the differential operator F one reduces to the areolar derivative of f and the operator F becomes the derivative of f. 2) f will be called right differentiable at q0 if f F (q0 ) = 0. The second kind of definition for holomorphy is suggested by the Weierstrass’s idea in the case of complex variable. Definition 4.1.2. Denoting DR = {q ∈ H; kqk < R}, one says that f : DR → H P∞ is left analytic in DR if f (q) = k=0 ck q k , for all q ∈ DR , where ck ∈ H for all P∞ k = 0, 1, 2, ...,. Also, f is called right analytic in DR if f (q) = k=0 q k ck , for all q ∈ DR . Remark. While in the case of complex variable the two concepts in Definitions 4.1.1. and 4.1.2 coincide, in the case of quaternion variable this does not happen. The most suitable concept for our purpose is that in Definition 4.1.2. Concerning the Bernstein polynomials, due to non-commutativity, for R > 1 to a function f : DR → H, three distinct Bernstein polynomials can be attached, as follows : n X n l l z (1 − z)n−l , z ∈ H, Bn (f )(z) = f n l l=0

Bn∗ (f )(z) =

n X n l=0

l

z l (1 − z)n−l f

l , z ∈ H, n

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Appendix : Related Topics

Bn∗∗ (f )(z) =

n X n l=0

l

z lf

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297

l (1 − z)n−l , z ∈ H. n

We may call them as the left Bernstein polynomials, right Bernstein polynomials and middle Bernstein polynomials, respectively. It is easy to show by a simple example that these kinds of Bernstein polynomials do not converge for any continuous function f . Indeed, if we take f (z) = izi, then we easily get |Bn (f )(z) − izi| = |Bn∗ (f )(z) − izi| = |Bn∗∗ (f )(z) − izi| = | − z − izi| = | − iz + zi| > 0, for all z 6= i.

However, for each kind of Bernstein polynomial there exists a suitable class of functions for which the convergence holds. To prove that we need some auxiliary results. Theorem 4.1.3. Suppose that f : DR → H has the property that f (z) ∈ R for all z ∈ [0, 1]. Then we have the representation formula n X n m Bn (f )(z) = ∆m 1/n f (0)z , for all z ∈ H. m m=0

Proof. Because of the hypothesis on f , the values f nl commutes with the other terms in the expression of Bn (f )(z), so that taking into account that z n+m = z n z m = z m z n (from associativity), z l (1 − z)n−l = (1 − z)n−l z l , αz = zα, for all α ∈ R, z ∈ H and that n−l X n−l s n−l (1 − z) = (−1) zs, s s=0 reasoning exactly as in the case of Bernstein polynomials of real variable (see e.g. Lorentz [125], p. 13) we obtain X n−l n n X X n l n l s n−l s m z (−1) z = ∆m Bn (f )(z) = f 1/n f (0)z , n l s m s=0 m=0 l=0

which proves the theorem.

Remark. It is clear that Theorem 4.1.3 holds for the right and middle Bernstein polynomials, Bn∗ (f )(z) and Bn∗∗ (f )(z) too. Now we are in position to state the following approximation result. Theorem 4.1.4. Suppose that f : DR → H is left analytic in DR , i.e. f (z) = P∞ p p=0 cp z , for all z ∈ DR , where cp ∈ H for all p = 0, 1, 2, ...,. Then for all 1 ≤ r < R, kzk ≤ r and n ∈ N we have ∞ 2X kcp kp(p − 1)r p . kBn (f )(z) − f (z)k ≤ n p=2

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Proof. First it is easy to see that Bn (f )(0) = f (0) and Bn (f )(1) = f (1). Then, for any f (z) = ep (z) = z p , taking into account Theorem 4.1.3 too we get n X n Bn (ep )(z) = ∆l1/n ep (0)z l , l l=0 P n n l and Bn (ep )(1) = 1 = l=0 l ∆1/n ep (0), where since ep is convex of any order it follows that nl ∆l1/n ep (0) ≥ 0 for all 0 ≤ l ≤ n. Moreover, taking into account the formula between the finite differences and divided differences, we can write n X n 1 Bn (ep )(z) = l! l [0, 1/n, ..., l/n; ep]z l n l l=0

=

n X n(n − 1)...(n − l + 1)

nl

l=0

[0, 1/n, ..., l/n; ep]z l .

On the other hand, Bn (f ) has the properties Bn (f + g) = Bn (f ) + Bn (g), Bn (αf ) = αBn (f ), α ∈ H, Bn (f α) 6= αBn (f ), α ∈ H \ R. P∞ Now let us prove the relationship Bn (f )(z) = k=0 ck Bn (ek )(z). Denoting fm (z) = Pm j ≤ r, m ∈ N, since from the above linearity of Bn we obviously have j=0 cj z , kzk Pm Bn (fm )(z) = k=0 ck Bn (ek )(z), it suffices to prove that for any fixed n ∈ N and kzk ≤ r with r ≥ 1, we have limm→∞ Bn (fm )(z) = Bn (f )(z). But this is immediate from limm→∞ kfm − f kr = 0 (where kf kr := sup{kf (z)k; kzk ≤ r} and from the inequality n X n kBn (fm )(z) − Bn (f )(z)k ≤ kz k (1 − z)n−k k · kfm − f kr k k=0

≤ Mr,n kfm − f kr ,

valid for all kzk ≤ r. Therefore we immediately get that kBn (f )(z) − f (z)k ≤

∞ X p=0

kcp k · kBn (ep )(z) − ep (z)k.

To estimate kBn (ep ) − ep k two possibilities exist : 1) 0 ≤ p ≤ n ; 2) p > n . Pp [0, 1/n, ..., l/n; ep]z l and deCase 1). We get Bn (ep )(z) = l=0 n(n−1)...(n−l+1) nl noting Cn,l = n(n−1)...(n−l+1) [0, 1/n, ..., l/n; ep] it follows nl Bn (ep )(z) − ep (z) =

1−

n(n − 1)...(n − (p − 1)) np

ep (z) +

p−1 X l=0

Cn,l z l .

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Passing to the norm k · k with kzk ≤ r and taking into account an inequality in the proof of Theorem 1.2.1, (ii), we obtain p−1 n(n − 1)...(n − (p − 1)) p X Cn,l rp kBn (ep )(z) − ep (z)k ≤ 1 − r + np l=0 p n(n − 1)...(n − (p − 1)) r = 2 1 − np p(p − 1) p r . ≤ n Case 2). For all kzk ≤ r, r ≥ 1, z ∈ H, p > n ≥ 1 we get kBn (ep )(z)k ≤ and therefore

p X n(n − 1)...(n − l + 1)

nl

l=0

[0, 1/n, ..., l/n; ep] · kz l k ≤ rn ,

kBn (ep )(z) − ep (z)k ≤ rn + rp ≤ 2rp ≤ 2rp n ≤ 2rp

p(p − 1) . n

In conclusion, combining both Cases 1 and 2 we obtain kBn (f )(z) − f (z)k ≤

∞ X p=0

kcp k · kBn (ep )(z) − ep (z)k ≤

∞

2X kcp kp(p − 1), n p=0

which proves the theorem.

In a similar manner we obtain the following. Corollary 4.1.5. Suppose that f : DR → H is right analytic in DR , i.e. f (z) = P∞ p p=0 z cp , for all z ∈ DR , where cp ∈ H for all p = 0, 1, 2, ...,. Then for all 1 ≤ r < R, kzk ≤ r and n ∈ N we have kBn∗ (f )(z)

∞

2X kcp kp(p − 1)r p . − f (z)k ≤ n p=2

Remark. It is not difficult to see that in the case of Bernstein-type polynomials Bn∗∗ (f )(z), an estimate of the form in Theorem 4.1.4 cannot be obtained, because in P∞ general we cannot write a formula of the type Bn∗∗ (f )(z) = p=0 cp Bn∗∗ (ep )(z) for P ∞ f left analytic or of the type Bn∗∗ (f )(z) = p=0 Bn∗∗ (ep )(z)cp for f right analytic. 4.2

Approximation of Vector-Valued Functions

By using a nice and powerful method based on a classical result in Functional Analysis, in this section we study the approximation of vector-valued functions of real variable and of complex variable, by the corresponding Bernstein-type or convolution-type operators.

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Approximation by Complex Bernstein and Convolution Type Operators

Real Variable Case

In this subsection we present some results concerning the approximation of functions f : I → X, where I is a subinterval of the real numbers R and X is a normed space over R.. The case of functions of one real variable is of intrinsic value and gives the main ideas of the method. Because of these reasons it is the first considered below in full details. Also, because we can take as X the space of all complex numbers C, it follows that this case one frames into the title of the present book. In essence, in this subsection we prove basic results in the approximation of vector-valued functions of real variable by polynomials with coefficients in normed spaces, called here generalized polynomials. Thus we obtain estimates in terms of Ditzian-Totik Lp -moduli of smoothness and inverse theorems for approximation by Bernstein, Bernstein-Kantorovich, Sz´ asz-Mirakjan, Baskakov generalized operators and their Kantorovich analogues, Post-Widder, Jackson-type generalized operators, etc. Some applications to approximation of random functions and of fuzzy-numbervalued functions are given. We need the following useful concepts. Let (X, k · k) be a normed space over R. Similar to the case of real-valued functions, the following concepts in the Definitions 4.2.1-4.2.3 can be introduced. Definition 4.2.1. (i) A generalized algebraic polynomial of degree ≤ n, with Pn k coefficients in X will be an expression of the form Pn (x) = k=0 ck x , where ck ∈ X, k = 0, ..., n and x ∈ [a, b]. A generalized trigonometric polynomial of degree ≤ n with coefficients in X will Pn be an expression of the form Tn (x) = a0 + k=1 [ak cos(kx) + bk sin(kx)], where a0 , ak , bk ∈ X, k = 1, ..., n and x ∈ R. (ii) Denote by Pn [a, b], Tn the sets of all generalized algebraic and trigonometric polynomials of degree R b ≤ n with coefficients in X, respectively, kf k∞ = supx {kf (x)k}, kf kp = ( a kf (x)kp dx)1/p , if f : [a, b] → X, kf kp = R 2π ( 0 kf (x)kp dx)1/p , if f : R → X is 2π-periodic, 1 ≤ p < ∞. Also, if kf k∞ < ∞ then we write that f ∈ C([a, b]; X) (or f ∈ C2π (R; X)) and if kf kp < ∞, 1 ≤ p < ∞, we write f ∈ Lp ([a, b]; X) (or f ∈ Lp2π (R; X)), depending if f is defined on [a, b] or 2π-periodic on R, respectively. Definition 4.2.2. f : [a, b] → X will be R b called Riemann integrable on [a, b], if there exists an element I ∈ X denoted by a f (x)dx, with the following property : for any ε > 0, there exists δ > 0, such that for any division of [a, b], d : a = x0 < ... < xn = b with the norm ν(d) < δ and any intermediary points ξi ∈ [xi , xi+1 ], we have Pn−1 kS(f ; d, ξi ) − Ik < ε, where S(f ; d, ξi ) = i=0 f (ξi )(xi+1 − xi ). Also, denoteR by Lp ([a, b]; X) = {f : [a, b] → X; f is pth Bochner-Lebesgue b integrable and a kf (x)kp dx < +∞}, 1 ≤ p < ∞, where the equality between two p functions in L ([a, b]; X) is considered in the almost everywhere sense. For p = +∞ we consider Lp ([a, b]; X) = C([a, b]; X).

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Definition 4.2.3. (i) For f : [a, b] → X, f ∈ Lp ([a, b]; X), the kth Lp -modulus of smoothness of f on [a, b] will be given by !1/p Z b−kh

k∆kh f (x)kp dx

ωk (f ; δ)p = sup{

a

; 0 ≤ h ≤ δ}, if 1 ≤ p < +∞,

and ωk (f ; δ)∞ = sup {sup{k∆kh f (x)k; x, x + kh ∈ [a, b]}}. 0≤h≤δ

For f : R → X, 2π-periodic, f ∈ Lp ([0, 2π]; X) one define Z 2π 1/p ωk (f ; δ)p = sup{ k∆kh f (x)kp dx ; 0 ≤ h ≤ δ}, if 1 ≤ p < +∞, 0

and ωk (f ; δ)∞ = sup {sup{k∆kh f (x)k; x ∈ R}}. 0≤h≤δ

Here ∆kh f (x) = j=0 (−1)k−j kj f (x + jh). (ii) Let f : I → X, f ∈ Lp (I; X), where I is a subinterval of R. The kth Ditzian-Totik Lp -modulus of smoothness will be given by Pk

k

ωφk (f ; δ)Lp := ωφk (f ; δ)p = sup k∆hφ(x) f (x)kLp (I) , 0≤h≤δ

k

1 ≤ p ≤ +∞, where φ(x) is a suitable step-weight attached to I and ∆h f (x) = Pk k j k j=0 (−1) j f (x + kh/2 − jh), if x, x ± kh/2 ∈ I, ∆h f (x) = 0, otherwise. Here R kf kp := kf kLp(I) = ( I kf (x)kp dx)1/p , if 1 ≤ p < +∞ and kf k∞ = sup{kf (x)k; x ∈ I}. In particular, for [a, b] = [0, 1] and p = ∞, we have ωφ2 (f ; δ) = sup{sup{kf (x + hφ(x)) − 2f (x) + f (x − hφ(x))k; x ∈ I2,h },

respectively, where I2,h

h ∈ [0, δ]}, i p 1−h 1−h2 = − 1+h x(1 − x), δ ≤ 1. 2 , 1+h2 , φ(x) = h

2

Remark. In the applications we will encounter the following step-weight functions φ: φ2 (x) = (1 − x2 ), if I = [−1, 1], φ2 (x) = x(1 − x), if I = [0, 1], φ2 (x) = x, φ2 (x) = x2 , or φ2 (x) = x(1 + x), if I = [0, +∞). The main tool used in our proofs is represented by the following well-known result in Functional Analysis.

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Theorem 4.2.4. (see e.g. Muntean [141], p. 183) Let (X, k · k) be a normed space over the real or complex numbers and denote by X ∗ the conjugate space of X. Then, kxk = sup{|x∗ (x)| : x∗ ∈ X ∗ , k|x∗ k| ≤ 1}, for all x ∈ X. Let f : [0, 1] → X be continuous on [0, 1]. The generalized Bernstein and Bernstein-Kantorovich polynomials attached to f can be defined by Bn (f )(x) =

n X k=0

k pn,k (x)f ( ), n

and Kn (f )(x) = (n + 1)

n X

pn,k (x)

k=0

Z

(k+1)/(n+1)

f (t)dt, k/(n+1)

Rb respectively, where pn,k (x) = nk xk (1 − x)n−k and the integral a f (t)dt is defined as the limit for m → ∞ in the norm k · k, of the all (classical defined) Riemann P sums m i=0 (xi+1 − xi )f (ξi ). Also, the generalized trigonometric polynomials of Jackson-type attached to a 2π-periodic continuous function f : R → X, can be defined by Z π 3 Kn (x − t)f (t)dt, Jn (f )(x) = 2πn(2n2 + 1) −π 4 . where Kn (t) = sin(nt/2) sin(t/2) We present Theorem 4.2.5. (Gal [92]) Let f : [0, 1] → X be continuous on [0, 1]. There exist the absolute constants C1 , C2 , such that for all n ∈ N we have : (i) 1 1 C1 ωφ2 (f ; √ )∞ ≤ kBn (f ) − f k∞ ≤ C2 ωφ2 (f ; √ )∞ , n n where kf k∞ = sup{kf (x)k; x ∈ [0, 1]} ; (ii) 1 1 C1 [ωφ2 (f ; √ )∞ + ω1 (f ; )∞ ] ≤ kKn (f ) − f k∞ n n 1 1 ≤ C2 [ωφ2 (f ; √ )∞ + ω1 (f ; )∞ ]; n n (iii)

x(1 − x) kBn (f )(x) − f (x)k ≤ M n

α/2

∀x ∈ [0, 1], if and only if

ω2 (f ; δ)∞ = O(δ α ), where α ≤ 2 ;

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(iv) If, in addition, f : R → X is continuous and 2π-periodic then 1 kJn (f ) − f k∞ ≤ C1 ω2 (f ; )∞ . n Proof. For x∗ ∈ X ∗ , 0 < k|x∗ k| ≤ 1 let us define the function g : [0, 1] → K, g(x) = x∗ (f (x)). Obviously g is continuous on [0, 1]. (i) By Knoop-Zhou [116], Totik [192] we have 1 1 C1 ωφ2 (g; √ )∞ ≤ kBn (g) − gk∞ ≤ C2 ωφ2 (g; √ )∞ , n n where kgk∞ = sup{|g(x)|; x ∈ [0, 1]}. By the linearity of x∗ we get 1 ωφ2 (g; √ )∞ n = sup{sup{|x∗ [f (x + hφ(x)) − 2f (x) + f (x − hφ(x))]|; x ∈ I2,h }, √ h ∈ [0, 1/ n]}

≤ sup{sup{k|x∗ k| · kf (x + hφ(x)) − 2f (x) + f (x − hφ(x))k; x ∈ I2,h }, √ h ∈ [0, 1/ n]} 1 ≤ ωφ2 (f ; √ )∞ . n

Also, from the linearity of x∗ it easily follows Bn (g)(x) − g(x) = x∗ [Bn (f )(x) − f (x)] and from the right-hand side of the inequalities for g, we get for all x ∈ [0, 1] 1 |x∗ [Bn (f )(x) − f (x)]| ≤ C2 ωφ2 (f ; √ )∞ . n Passing to supremum with x∗ and taking into account Theorem 4.2.4, it follows 1 kBn (f )(x) − f (x)k ≤ C2 ωφ2 (f ; √ )∞ , n for all x ∈ [0, 1], i.e.

1 kBn (f ) − f k∞ ≤ C2 ωφ2 (f ; √ )∞ . n

On the other hand, by the left-hand side of the inequalities for g, i.e. 1 C1 ωφ2 (g; √ )∞ ≤ kBn (g) − gk∞ , n we get 1 C1 ωφ2 (g; √ )∞ ≤ sup{|x∗ [Bn (f )(x) − f (x)]|; x ∈ [0, 1]} n ≤ sup{k|x∗ k| · kBn (f )(x) − f (x)k; x ∈ [0, 1]} ≤ kBn (f ) − f k∞ .

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√ Now, for fixed x ∈ I2,h and h ∈ [0, 1/ n], we have

1 C1 |x∗ [f (x + hφ(x)) − 2f (x) + f (x − hφ(x))]| ≤ ωφ2 (g; √ )∞ ≤ kBn (f ) − f k∞ , n wherefrom passing to supremum with x∗ and taking into account Theorem 4.2.4, we get C1 kf (x + hφ(x)) − 2f (x) + f (x − hφ(x))k ≤ kBn (f ) − f k∞ . √ Passing now to supremum with x ∈ I2,h and h ∈ [0, 1/ n] we obtain 1 C1 ωφ2 (f ; √ )∞ ≤ kBn (f ) − f k∞ , n

which proves the point (i). (ii) By Gonska-Zhou [104] we have 1 1 1 1 C1 [ωφ2 (g; √ )∞ + ω1 (g; )∞ ] ≤ kKn (g) − gk∞ ≤ C2 [ωφ2 (g; √ )∞ + ω1 (g; )∞ . n n n n From the linearity and the continuity of x∗ we easily get Kn (g)(x) − g(x) = x∗ [Kn (f )(x) − f (x)]. Also, since ω1 (g;

1 1 )∞ = sup{|x∗ [f (v) − f (w)]|; v, w ∈ [0, 1], |v − w| ≤ } n n

≤ sup{k|x∗ k| · kf (v) − f (w)k; v, w ∈ [0, 1], |v − w| ≤

1 } n

1 )∞ , n reasoning exactly as for the point (i), we immediately get ≤ ω1 (f ;

1 1 kKn (f ) − f k∞ ≤ C2 [ωφ2 (f ; √ )∞ + ω1 (f ; )∞ ]. n n On the other hand, by the left-hand side of the inequalities for g, i.e. 1 1 C1 [ωφ2 (g; √ )∞ + ω1 (g; )∞ ] ≤ kKn (g) − gk∞ , n n we get 1 1 C1 [ωφ2 (g; √ )∞ + ω1 (g; )∞ ] ≤ sup{|x∗ [Kn (f )(x) − f (x)]|; x ∈ [0, 1]} n n ≤ sup{k|x∗ k| · kKn (f )(x) − f (x)k; x ∈ [0, 1]}

≤ kKn (f ) − f k∞ . √ Now, for any fixed x ∈ I2,h , h ∈ [0, 1/ n], v, w ∈ [0, 1] with |v − w| ≤ have

C1 [|x∗ (f (x + hφ(x)) − 2f (x) + f (x − hφ(x)))| + |x∗ (f (v) − f (w))|] 1 1 ≤ C1 [ωφ2 (g; √ )∞ + ω1 (g; )∞ ] ≤ kKn (f ) − f k∞ , n n

1 n,

we

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wherefrom passing to supremum with x∗ we get C1 sup{|x∗ (f (x + hφ(x)) − 2f (x) + f (x − hφ(x)))|; k|x∗ k| ≤ 1}

≤ C1 sup{|x∗ (f (x + hφ(x)) − 2f (x) + f (x − hφ(x)))| + |x∗ (f (v) − f (w))|; k|x∗ k| ≤ 1} ≤ kKn (f ) − f k∞ ,

and C1 sup{|x∗ (f (v) − f (w))|; k|x∗ k| ≤ 1}

≤ C1 sup{|x∗ (f (x + hφ(x)) − 2f (x) + f (x − hφ(x)))| + |x∗ (f (v) − f (w))|; k|x∗ k| ≤ 1} ≤ kKn (f ) − f k∞ .

By Theorem 4.2.4 we obtain C1 kf (x + hφ(x)) − 2f (x) + f (x − hφ(x))k ≤ kKn )f ) − f k∞ and C1 kf (v) − f (w)k ≤ kKn (f ) − f k∞ . √ Passing now to supremum with x ∈ I2,h and h ∈ [0, 1/ n] and with |v − w| ≤ respectively, we obtain 1 C1 ωφ2 (f ; √ )∞ ≤ kKn (f ) − f k∞ , n and C1 ω1 (f ;

1 )∞ ≤ kKn (f ) − f k∞ , n

implying C1 2 1 1 [ω (f ; √ )∞ + ω1 (f ; )∞ ] ≤ kKn (f ) − f k∞ , 2 φ n n which proves the point (ii) too. (iii) By Berens-Lorentz [37] we have α/2 x(1 − x) |Bn (g)(x) − g(x)| ≤ M ∀x ∈ [0, 1], if and only if n ω2 (g; δ)∞ = O(δ α ). First, let us suppose that kBn (f )(x) − f (x)k ≤ M We get

x(1 − x) n

α/2

∀x ∈ [0, 1].

|Bn (g)(x) − g(x)| = |x∗ [Bn (f )(x) − f (x)]|

≤ k|x∗ k| · kBn (f ) − f k α/2 x(1 − x) ≤M ∀x ∈ [0, 1]. n

1 n,

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This implies ω2 (g; δ)∞ ≤ Cδ α , that is for any fixed x and h satisfying x, x + h, x − h ∈ [0, 1], h ∈ [0, δ] we get |x∗ [f (x + h) − 2f (x) + f (x − h)]| ≤ Cδ α . Passing to supremum with x∗ and taking into account Theorem 4.2.4, it follows kf (x + h) − 2f (x) + f (x − h)k ≤ Cδ α . Passing now to supremum with x and h as above, we obtain ω2 (f ; δ)∞ ≤ Cδ α . Conversely, let us suppose that ω2 (f ; δ)∞ ≤ Cδ α . For x and h satisfying x, x + h, x − h ∈ [0, 1], h ∈ [0, δ] we have |g(x + h) − 2g(x) + g(x − h)| = |x∗ [f (x + h) − 2f (x) + f (x − h)]|

≤ k|x∗ k| · kf (x + h) − 2f (x) + f (x − h)k ≤ kf (x + h) − 2f (x) + f (x − h)k

≤ ω2 (f ; δ)∞ ≤ Cδ α .

Passing to supremum with x and h as above, it follows ω2 (g; δ)∞ ≤ Cδ α , which implies α/2 x(1 − x) |Bn (g)(x) − g(x)| ≤ M ∀x ∈ [0, 1]. n This implies α/2 x(1 − x) ∗ ∀x ∈ [0, 1]. |x [Bn (f )(x) − f (x)]| ≤ M n Passing to supremum with x∗ and taking into account Theorem 4.2.4 we immediately obtain α/2 x(1 − x) ∀x ∈ [0, 1], kBn (f )(x) − f (x)k ≤ M n which proves (iii). (iv) By Lorentz [126], p. 56 we have

1 )∞ . n Reasoning exactly as in the first part of the proof of (ii), we immediately get the desired conclusion. kJn (g) − gk∞ ≤ Cω2 (g;

Remark. In the recent paper Anastassiou-Gal [15], results concerning best approximation by generalized polynomials with coefficients in vector spaces over R or C were obtained. In what follows, for f : [0, +∞) → X continuous and bounded on [0, +∞), we can attach the following operators : ∞ X Sn (f )(x) = sn,k (x)f (k/n), k=0

Vn (f )(x) =

∞ X k=0

vn,k (x)f (k/n),

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(n/x)n Pn (f )(x) = (n − 1)!

Z

∞

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e−nu/x un−1 f (u)du,

0

k x (1 − x)−n−k , called the generwhere sn,k (x) = e−nx (nx)k /k!, vn,k (x) = n+k−1 k alized Sz´ asz-Mirakjan, Baskakov and Post-Widder operators, respectively. Denote by Ln (f )(x) any from the above operators (including Kn (f )(x)) and by Ln,r (f )(x) the combination of the form (see Ditzian-Totik [64], p. 116) Ln,r (f )(x) = Pr−1 i=0 Ci (n)Lni (f )(x), where ni and Ci (n) satisfy the relations (9.27), a), b), c) and d) in Ditzian-Totik [64], p. 116. The main result of this section is formally exactly the same as that for realvalued functions (see Ditzian-Totik [64], p. 117, Theorem 9.3.2) and can be stated as follows. Theorem 4.2.6. (Anastassiou-Gal [14]) Let (X, k · k) be a real normed space and f : I → X be continuous and bounded on I, where I ⊂ R is a subinterval of R. Then 1 kLn,r (f ) − f kB ≤ M [ωφ2r (f ; √ )B + n−r kf kB ], n for all n, r ∈ N and for α < 2r kLn,r (f ) − f kB ≤ C1 n−α/2 iff ωφ2r (f ; h)B ≤ C2 hα . Here ωφ2r (f ; h) is given by Definition 4.2.3, (ii) and : 1], φ2 (x) = x(1 − x), B = Lp [0, 1], 1 ≤ p < ∞, kf kB = R 1 1) I =p [0,1/p ( 0 kf (x)k dx) , if Ln (f ) := Kn (f ) ; 2) I = [0, +∞), φ2 (x) = x, B = Cb [0, +∞), ( kf kB = sup{kf (x)k; x ∈ [0, +∞)} < +∞ iff f ∈ Cb [0, +∞)), for Ln (f ) := Sn (f ) ; 3) I = [0, +∞), φ2 (x) = x(1 + x), B = Cb [0, +∞), if Ln (f ) := Vn (f ) ; 4) I = [0, +∞), φ2 (x) = x2 , B = Cb [0, +∞), for Ln (f ) := Pn (f ) . Proof. For f : I → X, X ∗ the conjugate of X and x∗ ∈ B1 = {x∗ ∈ X ∗ ; k|x∗ k| ≤ 1}, let us define g : I → R by g(x) =R x∗ (f (x)). Since x∗ is linear P and continuous, it commutes with and integral and therefore we easily get ∗ ∗ Ln (g)(x) = x [Ln (f )(x)] and Ln,r (g)(x) = x [Ln,r (f )(x)]. We will prove only the cases 1) and 3) in the statement, since the other cases are similar. Case 1). Let 1 ≤ p < ∞. By Ditzian-Totik [64], p. 117, Theorem 9.3.2 we have 1 kKn,r (g) − gkLp[0,1] ≤ M [ωφ2r (g; √ )Lp [0,1] + n−r kgkLp[0,1] ], n for all n, r ∈ N and for α < 2r kKn,r (g) − gkLp[0,1] ≤ C1 n−α/2 iff ωφ2r (g; h)Lp [0,1] ≤ C2 hα .

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But Kn,r (g)(x) − g(x) = x∗ [Kn,r (f )(x) − f (x)] and Z 1 1/p 2r 2r p |∆δφ(x) g(x)| dx k∆δφ(x) g(x)kLp [0,1] = = ≤ ≤

Z

Z

Z

0

1

|x

0

∗

2r (∆δφ(x) g(x))|p dx

1 ∗

[k|x k| ·

0 1

0

1/p

2r k∆δφ(x) f (x)k]p dx

2r [k∆δφ(x) f (x)k]p dx

1/p

1/p

,

which immediately implies ωφ2r (g; √1n )Lp [0,1] ≤ ωφ2r (f ; √1n )Lp [0,1] . R1 Similarly, kgkLp[0,1] = ( 0 |x∗ (f (x))|p dx)1/p ≤ kf kLp[0,1] . Therefore, for all x∗ ∈ B1 we get

1 kKn,r (g) − gkLp[0,1] ≤ M [ωφ2r (f ; √ )Lp [0,1] + n−r kf kLp[0,1] ], n

i.e. Z

1 0

|x∗ (f (x) − Kn,r (f )(x)|p dx

1/p

1 ≤ M [ωφ2r (f ; √ )Lp [0,1] + n−r kf kLp[0,1] ]. n

In what follows, we need the following equality : for any F : [0, 1] → X with kF kLp < +∞, we have Z 1 Z 1 sup{|x∗ (F (x))|p ; x∗ ∈ B1 }dx = sup{ |x∗ (F (x))|p dx; x∗ ∈ B1 }. 0

0

∗

p

∗

p

Indeed, by |x (F (x))| ≤ sup{|x (F (x))| ; x ∈ B1 }, ∀x ∈ [0, 1], integrating and then passing to supremum with x∗ ∈ B1 , we immediately get Z 1 Z 1 sup{|x∗ (F (x))|p ; x∗ ∈ B1 }dx. sup{ |x∗ (F (x))|p dx; x∗ ∈ B1 } ≤ 0

0

To prove the converse inequality, from the definition of supremum, for each ε > 0, there exists y ∗ ∈ B1 (depending on ε), such that sup{|x∗ (F (x))|p ; x∗ ∈ B1 } ≤ |y ∗ (F (x))|p + ε, ∀x ∈ [0, 1]. Integrating, we easily obtain Z 1 Z sup{|x∗ (F (x))|p ; x∗ ∈ B1 }dx ≤ 0

1 0

|y ∗ (F (x))|p dx + ε

≤ sup{

Z

0

1

|x∗ (F (x))|p dx; x∗ ∈ B1 } + ε.

Passing with ε → 0, we get the converse inequality too, which proves the claimed equality.

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Consequently, we obtain Z 1 sup{|x∗ (f (x) − Kn,r (f )(x))|p ; x∗ ∈ B1 }dx 0

= sup{

Z

0

1

|x∗ (f (x) − Kn,r (f )(x))|p dx; x∗ ∈ B1 }

1 ≤ {M [ωφ2r (f ; √ )Lp [0,1] + n−r kf kLp[0,1] ]}p . n Since Theorem 4.2.4 immediately implies kxkp = sup{|x∗ (x)|p ; x∗ ∈ B1 }, 0 < p < ∞, from the previous relations we get 1 kKn,r (f ) − f kLp[0,1] ≤ M [ωφ2r (f ; √ )Lp [0,1] + n−r kf kLp[0,1] ]. n Now, in order to prove the converse result for Kn,r (f )(x), firstly let us suppose that kKn,r (f ) − f kLp [0,1] ≤ C1 n−α/2 . We have 1/p Z 1 |x∗ (Kn,r (f )(x) − f (x))|p dx kKn,r (g) − gkLp[0,1] = ≤

Z

0

1 0

k|x∗ k|p k(Kn,r (f )(x) − f (x)kp dx

≤ kKn,r (f ) − f kLp [0,1] ≤ C1 n−α/2 ,

1/p

for all x∗ ∈ B1 . From Ditzian-Totik [64], p. 117, Theorem 9.3.2, it follows that ωφ2r (g; h)Lp [0,1] ≤ C2 hα ,

where C2 depends only on C1 (i.e. C2 does not depend on x∗ ∈ B1 ). Then, for fixed 0 ≤ δ ≤ h, x ∈ [0, 1], we have Z 1 1/p 2r 2r p ∗ ≤ C 2 hα . |x (∆δφ(x) f (x))| dx k∆δφ(x) g(x)kLp [0,1] = 0

Passing to supremum with x∗ ∈ B1 and taking again into account that the integral commutes with supremum, we get Z 1 1/p 2r p k∆δφ(x) f (x)k dx ≤ C2 hα , ∀δ ∈ (0, h]. 0

Passing now to supremum with δ, we get ωφ2r (f ; h)Lp [0,1] ≤ C2 hα . Conversely, let us suppose that ωφ2r (f ; h)Lp [0,1] ≤ C2 hα . We get ωφ2r (g; h)Lp [0,1] ≤ ωφ2r (f ; h)Lp [0,1] ≤ C2 hα ,

which by Ditzian-Totik [64], p. 117, Theorem 9.3.2 implies the inequality kKn,r (g)− gkLp[0,1] ≤ C1 n−α/2 , i.e. Z 1 1/p |x∗ [Kn,r (f )(x) − f (x)]|p dx ≤ C1 n−α/2 , ∀x∗ ∈ B1 . 0

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Passing to supremum with x∗ ∈ B1 and reasoning as for the above lines, we finally arrive at 1/p Z 1 p kKn,r (f )(x) − f (x)k dx ≤ C1 n−α/2 , 0

which proves the Case 1. Case 3. The proof is a little simpler than in the Case 1. Let B = Cb [0, ∞). First from |x∗ (u)| ≤ k|x∗ k| · kuk ≤ kuk, ∀x∗ ∈ B1 , we immediately get ωφ2r (g; √1n )B ≤ ωφ2r (f ; √1n )B , kgkB ≤ kf kB , where kf kB = sup{kf (x)k; x ∈ [0, +∞)}. From Ditzian-Totik [64], p.117, Theorem 9.3.2 this implies 1 kVn,r (g) − gkB ≤ M [ωφ2r (f ; √ )B + n−r kf kB ]. n ∗ Let x ∈ B1 , x ∈ [0, +∞) be fixed. From the above inequality we obtain 1 |x∗ (Vn,r (f )(x) − f (x))| ≤ M [ωφ2r (f ; √ )B + n−r kf kB ]. n Passing to supremum with x∗ ∈ B1 and taking into account Theorem 4.2.4 we get 1 kVn,r (f )(x) − f (x)k ≤ M [ωφ2r (f ; √ )B + n−r kf kB ], n i.e. passing to supremum with x ∈ [0, ∞) it follows 1 kVn,r (f ) − f kB ≤ M [ωφ2r (f ; √ )B + n−r kf kB ]. n Now, let us suppose kVn,r (f ) − f kB ≤ C1 n−α/2 .

It follows

|Vn,r (g)(x) − g(x)| = |x∗ (Vn,r (f )(x) − f (x))|

≤ k|x∗ k| · kVn,r (f )(x) − f (x)k ≤ C1 n−α/2 ,

i.e. kVn,r (f ) − f kB ≤ C1 n−α/2 . Then, from the same Theorem 9.3.2 in Ditzian-Totik [64] it follows ωφ2r (g; h)B ≤ C2 hα . Let 0 ≤ δ ≤ h and x be fixed. We get 2r

2r

|∆δφ(x) g(x)| = |x∗ (∆δφ(x) f (x))| ≤ ωφ2r (g; h)B ≤ C2 hα ,

for all x∗ ∈ B1 . Passing to supremum with x∗ ∈ B1 , by Theorem 4.2.4 it follows 2r k∆δφ(x) f (x)k ≤ C2 hα , and passing to supremum with δ and x, we get ωφ2r (f ; h)B ≤ C 2 hα . Conversely, let us suppose ωφ2r (f ; h)B ≤ C2 hα . It follows ωφ2r (g; h)B ≤ C2 hα , which from the same Theorem 9.3.2 in Ditzian-Totik [64] implies kVn,r (g) − gkB ≤ C1 n−α/2 . This means that for all x∗ ∈ B1 and x ∈ [0, ∞), we have |x∗ (Vn,r (f )(x) − f (x))| ≤ C1 n−α/2 .

Passing to supremum first with x∗ ∈ B1 , by Theorem 4.2.4 and then passing to supremum with x ∈ [0, +∞), finally we arrive at kVn,r (f ) − f kB ≤ C1 n−α/2 , which proves the theorem.

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Some applications of the above results to the approximation of random functions by random polynomials and of fuzzy-number-valued functions by fuzzy polynomials can be obtained. First, let us recall that if (S, B, P ) is a probability space (P is the probability), then the set of almost sure (a.s.) finite real random variables is denoted by L(S, B, P ) and it is a normed (Banach) space with respect to the norm R kgk = S |g(t)|dP (t). Here, for g1 , g2 ∈ L(S, B, P ) we consider g1 = g2 if g1 (t) = g2 (t), a.s. t ∈ S. A random function defined on [0, 1] is a mapping f : [0, 1] → L(S, B, P ) and we denote f (x)(t) ∈ R by f (x, t). For this kind of f and the Bernstein random Pn polynomials defined by Bn (f )(x, t) = k=0 pn,k (x)f ( nk , t), a direct consequence of Theorem 4.2.5, (i) is the following. Corollary 4.2.7. (Gal [92]) If f : [0, 1] → L(S, B, P ) is continuous on [0, 1], then 1 1 C1 ωφ2 (f ; √ )∞ ≤ kBn (f ) − f k∞ ≤ C2 ωφ2 (f ; √ )∞ , n n R where kf k∞ = sup{kf (x)k; x ∈ [0, 1]} = sup{ S |f (x, t)|dP (t); x ∈ [0, 1]}. Now, for the random function f : I → L(S, B, P ), continuous and bounded on I, where I = [0, +∞), let us consider the random operators given by Ln,r (f )(x, t), where Ln,r are defined by Theorem 4.2.6 and Ln,r (f )(x, t) means that the usual Ln,r is applied to the random function f (x, t) considered as function of x only (t ∈ S is fixed, arbitrary). A direct consequence of Theorem 4.2.6 is the following (keeping the notations there). Corollary 4.2.8. (Anastassiou-Gal [14]) If f : [0, +∞) → L(S, B, P ) is continuous and bounded on [0, +∞), then 1 kLn,r (f ) − f kB ≤ M [ωφ2r (f ; √ )B + n−r kf kB ], n for all n, r ∈ N and for α < 2r

kLn,r (f ) − f kB ≤ C1 n−α/2 iff ωφ2r (f ; h)B ≤ C2 hα .

Here kf kB = sup{kf (x)k; x ∈ [0, +∞)} = sup{

Z

S

|f (x, t)|dP (t); x ∈ [0, +∞)}

and f is called bounded on [0, +∞) if kf kB < +∞. In what follows, we present some applications to the approximation of fuzzynumber-valued functions by fuzzy polynomials. First let us recall a few facts concerning the fuzzy-number valued functions. Given a set X 6= ∅, a fuzzy subset of X is a mapping u : X → [0, 1] and obviously any classical subset A of X can be considered as a fuzzy subset of X defined by χA : X → [0, 1], χA (x) = 1, if x ∈ A, χA (x) = 0 if x ∈ X \ A. (see e.g. Zadeh [204][11]).

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Let us denote by RF the class of fuzzy subsets of real axis R (i.e. u : R → [0, 1]), satisfying the following properties: (i) ∀u ∈ RF , u is normal i.e. ∃xu ∈ R with u (xu ) = 1; (ii) For all u ∈ RF , u is convex fuzzy set, i.e. u (tx + (1 − t) y) ≥ min {u (x) , u (y)} , ∀t ∈ [0, 1] , x, y ∈ R ; (iii) ∀u ∈ RF , u is upper semi-continuous on R; (iv) {x ∈ R : u (x) > 0} is compact, where A denotes the closure of A. Then RF is called the space of fuzzy real numbers (see e.g. Dubois-Prade [66]). Remark. Obviously R ⊂ RF , because any real number x0 ∈ R, can be described as the fuzzy number whose value is 1 for x = x0 and 0 otherwise. r 0 For 0 < r ≤ 1 and u ∈ RF define [u] = {x ∈ R; u (x) ≥ r} and [u] = r {x ∈ R; u (x) > 0}. Then it is well known that for each r ∈ [0, 1], [u] is a bounded closed interval. For u, v ∈ RF and λ ∈ R, we have the sum u ⊕ v and the prodr r r r r uct λ u defined by [u ⊕ v] = [u] + [v] , [λ u] = λ [u] , ∀r ∈ [0, 1], where r r r [u] + [v] means the usual addition of two intervals (as subsets of R) and λ [u] means the usual product between a scalar and a subset of R (see e.g. Dubois-Prade [66], Congxin-Zengtai [201]). Define D : RF × RF → R+ ∪ {0} by r r r , , u+ − v + D (u, v) = sup max ur− − v− r∈[0,1]

r r where [u] = ur− , ur+ , [v]r = v− , v+ . The following properties are known (Dubois-Prade [66]): D (u ⊕ w, v ⊕ w) = D (u, v), ∀u, v, w ∈ RF D (k u, k v) = |k| D (u, v) , ∀u, v ∈ RF , ∀k ∈ R; D (u ⊕ v, w ⊕ e) ≤ D (u, w)+D (v, e) , ∀u, v, w, e ∈ RF and (RF , D) is a complete metric space. Also, we need the following Riemann integral, as particular case of the Henstock integral introduced by Congxin-Zengtai [201]. A function f : [a, b] → RF , [a, b] ⊂ R is called Riemann integrable on [a, b], if there exists I ∈ RF , with the property: ∀ > 0, ∃δ > 0, such that for any division of [a, b], d : a = x0 < ... < xn = b of norm ν (d) < δ, and for any points ξi ∈ [xi , xi+1 ], i = 0, n − 1, we have ! ∗ n−1 X f (ξi ) (xi+1 − xi ) , I < , D r

i=0

Rb where means sum with respect to ⊕. Then we denote I = a f (x) dx. A crucial result for our reasonings will be the following known result. P∗

Theorem 4.2.9. (see e.g. Congxin-Zengtai [201]) RF can be embedded in B = ¯ 1] × C[0, ¯ 1], where C[0, ¯ 1] is the class of all real valued bounded functions f : C[0,

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[0, 1] → R such that f is left continuous for any x ∈ (0, 1], f has right limit for any x ∈ [0, 1) and f is right continuous at 0. With the norm k·k = supx∈[0,1] |f (x)|, ¯ 1] is a Banach space. Denote k·k the usual product norm i.e. k(f, g)k = C[0, B B max {kf k , kgk}. Let us denote the embedding by j : RF → B, j(u) = (u− , u+ ). Then j(RF ) is a closed convex cone in B and j satisfies the following properties: (i) j(s u ⊕ t v) = s · j(u) + t · j(v) for all u, v ∈ RF and s, t ≥ 0 (here “·” and “+” denote the scalar multiplication and addition in B); (ii) D(u, v) = kj(u) − j(v)kB (i.e. j embeds RF in B isometrically.) Now, for f : [0, 1] → RF a fuzzy-number-valued function, the generalized Bernstein and Bernstein-Kantorovich polynomials attached to f can be defined by Bn (f )(x) =

n X k=0

and Kn (f )(x) = (n + 1)

n X k=0

k pn,k (x)f ( ), n

pn,k (x)

Z

(k+1)/(n+1)

f (t)dt, k/(n+1)

Rb respectively, where pn,k (x) = nk xk (1 − x)n−k and the integral a f (t)dt is defined as the limit for m → ∞ in the distance D, of the all (classical defined) Riemann Pm sums i=0 (xi+1 − xi ) f (ξi ). (Here all the sums are with respect to the operation ⊕.) Also, let us define the following moduli of continuity and smoothness of f : ω1 (f ; δ)∞ = sup{D(f (x + h), f (x)); x, x + h ∈ [0, 1], 0 ≤ h ≤ δ}, ωφ2 (f ; δ)∞ = sup{D[f (x + hφ(x)) ⊕ f (x − hφ(x)), 2 f (x)]; x, x + hφ(x), x − hφ(x) ∈ [0, 1], 0 ≤ h ≤ δ}. Here φ2 (x) = x(1 − x). We present Theorem 4.2.10. (Anastassiou-Gal [14]) Let f : [0, 1] → RF be continuous on [0, 1]. There exist the absolute constants C1 , C2 , such that for all n ∈ N we have : (i) 1 1 C1 ωφ2 (f ; √ )∞ ≤ sup{D[Bn (f )(x), f (x)]; x ∈ [0, 1]} ≤ C2 ωφ2 (f ; √ )∞ . n n (ii) 1 1 C1 [ωφ2 (f ; √ )∞ + ω1 (f ; )∞ ] ≤ n n 1 1 sup{D[Kn (f )(x), f (x)]; x ∈ [0, 1]} ≤ C2 [ωφ2 (f ; √ )∞ + ω1 (f ; )∞ ]. n n

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Proof. Define g : [0, 1] → X by g(x) = j[f (x)], x ∈ [0, 1], where j is given by ¯ 1] × C[0, ¯ 1] endowed with the norm in Theorem 4.2.9, Theorem 4.2.9 and X = C[0, denoted by k · k. (i) According to Theorem 4.2.5, there exist the absolute constants C1 , C2 , such that for all n ∈ N we have 1 1 C1 ωφ2 (g; √ )∞ ≤ kBn (g) − gk∞ ≤ C2 ωφ2 (g; √ )∞ , n n where kf k∞ = sup{kf (x)k; x ∈ [0, 1]}. By Theorem 4.2.9, (ii), we notice kg(x + h) − g(x)k = kj[f (x + h)] − j[f (x)]k = D[f (x + h), f (x)], kg(x + hφ(x)) + g(x − hφ(x)) − 2g(x)k

= kj[f (x + hφ(x)) ⊕ f (x − hφ(x))] − j[2 f (x)]k = D[f (x + hφ(x)) ⊕ f (x − hφ(x)), 2 f (x)],

which immediately implies ω1 (g; δ)∞ = ω1 (f ; δ)∞ and ωφ2 (g; δ)∞ = ωφ2 (f ; δ)∞ , for all δ ≥ 0. Since on the other hand, from the linearity of j over the positive scalars we have kg(x) − Bn (g)(x)k = kj[f (x)] − j[Bn (f )(x)]k = D[f (x), Bn (f )(x)],

we easily arrive at the estimates in the statement. (ii) Since j is linear over the positive scalars and j commutes with the integral, we get Kn (g)(x) = j[Kn (f )(x)] and by similar reasoning as for the above point (i), we get the desired estimates. The theorem is proved. 4.2.2

Complex Variable Case

First we recall some known concepts and results. Definition 4.2.11. (see e.g. Hille-Phillips [109], p. 92–93) Let (X, k · k) be a complex Banach space, R > 1 and f : DR → X. We say that f is holomorphic on DR if for any x∗ ∈ B1 = {x∗ : X → C; x∗ linear and continuous, k|x∗ k| ≤ 1}, the function g : DR → C given by g(z) = x∗ [f (z)], is holomorphic on DR . (Here k| · k| represents the usual norm in the dual space X ∗ .) We denote by A(DR ; X) the space of all functions f : DR → X which are continuous on DR and holomorphic on DR . It is a Banach space with respect to the norm kf kR = max{kf (z)k; z ∈ DR } Note that everywhere in this subsection (X, k · k) will be a complex Banach space. Theorem 4.2.12. (see e.g. Hille-Phillips [109], p. 93) If f : DR → X is holomorphic on DR , then f (z) is continuous (as mapping between two metric spaces) and differentiable in the sense that exists f 0 (z) ∈ C given by

f (z + h) − f (z)

0

= 0, lim − f (z)

h→0 h

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uniformly with respect to z in any compact subset of DR . Theorem 4.2.13. (see e.g. Hille-Phillips [109], p. 97) If f : DR → X is holomorphic on DR , then we have the Taylor expansion f (z) =

∞ X f (n) (0) n z , n! n=0

z ∈ DR ,

where the series converges uniformly on any compact subset of DR . For the beginning, to f : DR → X holomorphic on DR with R > 1, let us attach the Bernstein operator of complex variable n X n k Bn (f )(z) = z (1 − z)n−k f (k/n), |z| ≤ R. k k=0

Theorem 4.2.14. Let (X, k · k) be a complex Banach space. Suppose that R > 1 P∞ and f : DR → X is holomorphic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR , with ck ∈ X, for all k. (i) Let 1 ≤ r < R be arbitrary fixed. For all |z| ≤ r and n ∈ N, we have kBn (f )(z) − f (z)k ≤

Kr (f ) , n

P∞ j−2 < ∞. where 0 < Kr (f ) = 3r(1+r) j=2 j(j − 1)kcj kr 2 (ii) If 1 ≤ r < r1 < R are arbitrary fixed, then for all |z| ≤ r and n, p ∈ N, kBn(p) (f )(z) − f (p) (z)k ≤

Kr1 (f )p!r1 , n(r1 − r)p+1

where Kr1 (f ) is given as at the above point (i). (iii) Let r ∈ [1, R). Then for all n ∈ N, |z| ≤ r, we have

2

Bn (f )(z) − f (z) − z(1 − z) f 00 (z) ≤ 5(1 + r) · Mr (f ) ,

2n 2n n P∞ 2 k−2 where Mr (f ) = k=3 kck kk(k − 1)(k − 2) r < ∞.

Proof. Let x∗ ∈ X ∗ be with k|x∗ k| ≤ 1 and define g :: DR → C by g(z) = x∗ [f (z)], for all z ∈ DR . By Definition 4.2.11 g is holomorphic (analytic) in DR . (i) Applying to g Theorem 1.1.2, (i), for all |z| ≤ r and n ∈ N it follows |Bn (g)(z) − g(z)| ≤

Cr (g) , n

P∞ j−2 , and aj are the coefficients in the where Cr (g) = 3r(1+r) j=2 j(j − 1)|aj |r 2 series representation of the analytic function g(z) = x∗ [f (z)]. But the linearity and P ∗ j ∗ ∗ continuity of x∗ implies g(z) = ∞ j=0 x (cj )z and |aj | = |x (cj )| ≤ k|x k| · kcj k ≤ ∗ kcj k, for all j. Therefore, since by the linearity of x we have |Bn (g)(z) − g(z)| = |x∗ [Bn (f )(z) − f (z)]|, it follows |x∗ [Bn (f )(z) − f (z)]| ≤

Kr (f ) , n

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P∞ j−2 with Kr (f ) = 3r(1+r) . j=2 j(j − 1)kcj kr 2 ∗ Passing above to supremum with k|x k ≤ 1 and taking into account Theorem 4.2.4, we obtain kBn (f )(z) − f (z)k ≤

Kr (f ) , n

for all |z| ≤ r and n ∈ N. (ii) Applying now to g Theorem 1.1.2, (ii) and taking into account that g (p) (z) = x∗ [f (p) (z)], similar reasonings with those from the above point (i) prove the desired estimate. (iii) It is immediate by applying Theorem 1.1.3, (ii) to g and by using similar reasoning with those from the above points (i) and (ii). As in the case of complex-valued functions, we can prove that in fact the order of approximation in Theorem 4.2.14 is exactly n1 . In this sense we present Theorem 4.2.15. Let (X, k · k) be a complex Banach space. Suppose that R > 1 P∞ and f : DR → X is holomorphic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR , with ck ∈ X, for all k. If f is not of the form f (z) = c0 + c1 z, with c0 , c1 ∈ X, then for any r ∈ [1, R) we have Kr (f ) , n ∈ N, n where kf kr = max{kf (z)k; |z| ≤ r} and the constant Kr (f ) depends only on f and r. kBn (f ) − f kr ≥

Proof. By the hypothesis on f , there exists at least one cs 6= 0 with s ≥ 2. As a consequence of Hahn-Banach theorem, there exists y ∗ ∈ X ∗ , with k|y ∗ k| = 1 and y ∗ (cs ) = kcs k 6= 0. Define g : DR → C by g(z) = y ∗ [f (z)], for all z ∈ DR . It follows that g(z) cannot be of the form g(z) = az + b with a, b ∈ C. Indeed, supposing the contrary, from P∞ the linearity and continuity of y ∗ we obtain g(z) = j=0 y ∗ (cj )z j = az + b, which implies y ∗ (cj ) = 0, for all j ≥ 2, a contradiction. Applying now Theorem 1.1.4 to g (see better the proof of Theorem 1.5.3), there exists an index n0 such that for all n ≥ n0 and |z| ≤ r we have

1 e1 (1 − e1 )g 00

, |Bn (g)(z) − g(z)|r ≥

n 4 r

e1 (1−e1 )g00 where

> 0. 4 r This is equivalent to

00 1 ∗ e1 (1 − e1 )f ∗

. |y [Bn (f )(z) − f (z)]| ≥ y

n 4 r

But since

|y ∗ [Bn (f )(z) − f (z)]| ≤ k|y ∗ k| · kBn (f )(z) − f (z)k ≤ kBn (f )(z) − f (z)k,

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it follows that for all |z| ≤ r and n ≥ n0 we have

00 1 ∗ e1 (1 − e1 )f

.

kBn (f )(z) − f (z)k ≥ y

n 4 r

For 1 ≤ n ≤ n0 −1 we reason as in the proof of Theorem 1.1.4, so that the conclusion in the statement is immediate. If for f : DR → X holomorphic in DR we consider the Butzer’s linear combination of complex Bernstein polynomials defined by the recurrence [2q−2]

q [2q] q L[0] n (f )(z) = Bn (f )(z), (2 − 1)Ln (f )(z) = 2 L2n

(f )(z) − Ln[2q−2] (f )(z),

where z ∈ C, q = 1, 2, ...,, then taking into account the Remarks 2), 3), 4), and 5) after the proof of Theorem 1.4.1 and Theorem 1.4.2, by similar reasonings as above we obtain the following. Theorem 4.2.16. Let (X, k · k) be a complex Banach space. Suppose that R > 1, q is a given natural number and f : DR → X is holomorphic in DR . If f is not of the form f (z) = c0 + c1 z + ... + cq z q , with c0 , c1 , ..., cq ∈ X, then for any r ∈ [1, R) and n ∈ N we have

1

[2q−2]

(f ) − f ∼ q ,

Ln n r where kf kr = max{kf (z)k; |z| ≤ r}. Remarks. 1) By using the above method, the results in Chapter 1 can be extended in a similar manner to the case of vector-valued Bernstein-type operators of a complex variable. 2) The method in this section can be used for other type too of vector-valued operators of a complex variable. In what follows we illustrate this method for some convolution-kind operators. P∞ For f : DR → X holomorphic on DR , f (z) = k=0 cj z j , cj ∈ X for all j, let us attach the convolution operators of complex variable of de la Vall´ee Poussin kind and of Riesz-Zygmund kind, given by Z π n 2n 1 X 1 iu zj cj f (ze )Kn (u)du = 2n Pn (f )(z) = n + j 2π −π n j=0 =

n X j=0

cj

(n!)2 zj , (n − j)!(n + j)!

and Rn,s (f )(z) =

n−1 X k=0

s k ck 1 − z k , n ∈ N, n

respectively, where s ∈ N. Let x∗ ∈ X ∗ be with k|x∗ k| = 1 and for f ∈ A(DR ; X) define g : DR → C by g(z) = x∗ [f (z)]. Taking into account Theorems 3.2.1, 3.2.2 and 3.2.3 applied to

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Pn (g)(z) and Theorems 3.2.6, 3.2.8 applied to Rn,s (g)(z) and using similar reasonings with those in the proofs of Theorems 4.2.14 and 4.2.15, we immediately obtain the following two results. Theorem 4.2.17. Let (X, k · k) be a complex Banach space. Suppose that R > 1 P∞ and f : DR → X is holomorphic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR , with ck ∈ X, for all k. (i) Let r ∈ [1, R) be arbitrary fixed. For all |z| ≤ r and n ∈ N we have kPn (f )(z) − f (z)k ≤

Mr (f ) , n

P∞ where Mr (f ) = k=1 kck kk 2 rk < ∞. (ii) If 1 ≤ r < r1 < R and p ∈ N then for all |z| ≤ r and n ∈ N we have kPn(p) (f )(z) − f (p) (z)k ≤

r1 p!Mr1 (f ) , (r1 − r)p+1 n

where Mr1 (f ) is given at the point (i). (iii) Let r ∈ [1, R). For all |z| ≤ r and n ∈ N we have

0 00

Pn (f )(z) − f (z) + zf (z) + zf (z) ≤ Ar (f ) ,

n n n2 P∞ where Ar (f ) = k=1 kck kk 4 rk < ∞. (iv) If f is not of the form f (z) = c0 for all z, then for any r ∈ [1, R) we have kPn (f ) − f kr ≥

Kr (f ) , n ∈ N, n

where kf kr = max{kf (z)k; |z| ≤ r} and the constant Kr (f ) depends only on f and r. Theorem 4.2.18. Let (X, k · k) be a complex Banach space. Suppose that R > 1 P∞ and f : DR → X is holomorphic in DR , that is f (z) = k=0 ck z k , for all z ∈ DR , with ck ∈ X, for all k. (i) If 1 ≤ r < r1 < R and s, p ∈ N then for all |z| ≤ r and n ∈ N we have (p) kRn,s (f )(z) − f (p) (z)k ≤

r1 p!Cr1 ,s (f ) , (r1 − r)p+1 ns

P∞ where Cr1 ,s (f ) = k=0 kck kk s rk < ∞. (ii) Let r ∈ [1, R) and s ∈ N. For all |z| ≤ r and n ∈ N we have

s X

Ar,s (f ) 1 j (j)

Rn,s (f )(z) − f (z) +

≤ , α z f (z) j,s

s n j=1 ns+1

P∞ where Ar,s (f ) = k=1 kck kk s+1 rk < ∞ and the real numbers αj,s are defined by Lemma 3.2.7. For other related results let us mention the following two results.

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Theorem 4.2.19. (Anastassiou-Gal [16]) Let (X, k · k) be a complex normed space and f ∈ A(D1 ; X). (i) Define Z π 1 Jn (f )(z) = f (zeiu )Kn (u) du, 2πn0 [2(n0 )2 + 1] −π Rπ where Kn (u) = [sin(n0 u/2)/ sin(u/2)]4 , n0 = [n/2] + 1, and −π is the classical Riemann integral for vector-valued functions. Then Jn (f )(z) is a polynomial in z of degree ≤ n, with coefficients in X, which satisfies the estimate 1 , ∀z ∈ D1 , kf (z) − Jn (f )(z)k ≤ Cω2 f ; n ∂D1 where

ωp (f ; δ)∂D1 = sup k∆pu f (eix )k; |x| ≤ π, |u| ≤ δ ,

p = 2, 3, . . . ;

(ii) Define the polynomials in z (with coefficients in X) Z π p+1 X p+1 (−1)k In,p (f )(z) = − Kn,r (u) f (zeiku ) du, k −π k=1

where Kn,r (u) = λn,r [sin(nu/2)/ sin(u/2)]2r , r is the smallest R π integer which satisfies r ≥ (p + 2)/2 and the constants λn,r are chosen such that −π Kn,r (u) du = 1. Then we have 1 , ∀z ∈ D1 ; kIn,p (f )(z) − f (z)k ≤ Cp ωp+1 f ; n ∂D1 (iii) Define Vn (f )(z) = 2L2n (f )(z) − Ln (f )(z), where Ln (f )(z) =

1 2nπ

Z

π

f (zeiu )Fn (u) du,

−π

Fn (u) = [sin(nu/2)/ sin(u/2)]2 . Then Vn (f )(z) is a polynomial of degree ≤ 2n − 1 in z, with coefficients in X which satisfies the estimate kf (z) − Vn (f )(z)k ≤ 4En (f )∞ (D1 ),

∀z ∈ D1 ,

where En (f )∞ (D1 ) = inf{kf − P kD1 ; P polynomial of degree ≤ n in z, with coefficients in X}, kf kD1 = sup{kf (z)k; z ∈ D1 }. (iv) For f ∈ A(D1 ; X) define Qn,ξ (f )(z) and Wn,ξ (f )(z) by Z π n+1 X 1 n+1 f (zeiku ) k · (−1) Qn,ξ (f )(z) = − 2 du, π 2 2 k −π u + ξ ξ arctg ξ k=1

Wn,ξ (f )(z) = −

Z π n+1 X 2 2 1 n+1 · (−1)k f (zeiku )e−u /ξ du. 2C(ξ) k −π k=1

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Then we have kQn,ξ (f )(z) − f (z)k ≤ K(n, ξ)ωn+1 (f ; ξ)∂D1 ,

∀z ∈ D1 , ξ ∈ (0, 1], n ∈ N

and ∀z ∈ D1 , ξ ∈ (0, 1], n ∈ N,

kWn,ξ (f )(z) − f (z)k ≤ Cn ωn+1 (f ; ξ)∂D1 , where K(n, ξ) =

R π/ξ 0

(u+1)n+1 u2 +1 −1 π tan ξ

du

, Cn =

R∞ 0

2

(1 + u)n+1 e−u du Rπ . −u2 du 0 e

(v) If for f ∈ A(D1 ; X) we consider the operators Z ξ π f (zeiu ) Qξ (f )(z) = du, z ∈ D1 , ξ > 0, π −π u2 + ξ 2 1 Wξ (f )(z) = √ πξ

Z

π

f (zeiu )e−u

2

/ξ

du,

−π

z ∈ D1 , ξ > 0,

then we have kQξ (f )(z) − f (z)k ≤ C

ω2 (f ; ξ)∂D1 , ξ

∀z ∈ D1 , ξ ∈ (0, 1]

and kWξ (f )(z) − f (z)k ≤ C

p ω2 (f ; ξ)∂D1 + C ξk|f k|D1 , ξ

∀z ∈ D1 , ξ ∈ (0, 1],

where k|f k|D1 = sup{kf (z)k; |z| ≤ 1}.

Remark. For the proof of Theorem 4.2.19, as above is used the x∗ -functional method to the corresponding results for complex-valued functions in the Sections 3.1 and 3.2. Theorem 4.2.20. (Aral-Gal [32]) Let f ∈ A(DR ; X) where (X, k · k) is a complex normed space. If for λ > 0, 0 < q < 1, we consider the q-operators Z ∞ f zeit (1 − q) dt, Pλ (f ; q, z) ≡ Pλ (f ; z) := 2 [λ]q ln q −1 −∞ Eq (1−q)|t| [λ] q

Wλ (f ; q, z) ≡ Wλ (f ; z) := then we have

π

q

1 [λ]q q 1/2 ; q

1/2

Z

∞ −∞

f zeit dt, t2 Eq [λ] q

1 kPλ (f ; z) − f (z)k ≤ (R + 1)(1 + )ω1 (f ; [λ]q )DR , q q q −1/2 1/2 (1 − q ) ω1 f ; [λ]q kWλ (f ; z) − f (z)k ≤ (R + 1) 1 + q

, DR

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321

for all z ∈ DR , where ω1 (f ; δ)DR = sup{kf (z1) − f (z2 )k; z1 , z2 ∈ DR , |z1 − z2 | ≤ δ}. Proof. Let x∗ ∈ B1 and define g(z) = x∗ [f (z)], g : DR → C. By Theorem (?, I don’t know yet its number) we have |Pλ (g; z) − g(z)| ≤ 2(1 + q1 )ω1 (g; [λ]q )DR , for all z ∈ DR , where ω1 (g; δ)DR = sup{|x∗ [f (z1 ) − f (z2 )]|; z1 , z2 ∈ DR , |z1 − z2 | ≤ δ}

≤ sup{kf (z1 ) − f (z2 )k; z1 , z2 ∈ DR , |z1 − z2 | ≤ δ} = ω1 (f ; δ)DR .

Therefore, we obtain |x∗ [Pλ (f ; z) − f (z)]| ≤ 2(1 + q1 )ω1 (f ; [λ]q )DR , for all x∗ ∈ B1 , and passing here to supremum, according to Theorem 4.2.4 it follows the required estimate. The proof in the case of Wλ (f ; z) is similar. 4.3

Strong Approximation by Complex Taylor Series

In this section we show that some classical results in the strong approximation by Fourier series can be extended to complex approximation by Taylor series in the unit disk. Let g(x) be a continuous and 2π periodic function and let ∞

a0 X + (ak cos kx + bk sin kx) 2 k=1 Pn be its Fourier series. Denote by sn (f )(x) = a20 + k=1 (ak cos kx + bk sin kx) and En (g) the best approximation of g in the uniform norm by trigonometric polynomials of degree ≤ n. Also, for m ≤ n let us define the de la Vall´ee Poussin sum by Pn 1 [ ] σn,m (g)(x) = m+1 k=n−m sk (g)(x). Concerning σn,m (g)(x), in Stechkin 185 the general result g(x) ∼

kg − σn,m (g)k ≤ C

n X En−m+k (g)

k=0

n+k+1

, 0 ≤ m ≤ n, m = 0, 1, ...,

was obtained, where k · k denotes the uniform norm in the space of all continuous and 2π-periodic functions and C > 0 is an absolute constant independent of g, n and m. Defining the strong de la Vall´ee Poussin means by Vn,m (g)(x) =

1 m+1

n X

k=n−m

|sk (g)(x) − g(x)|,

the above inequality was extended to Vn,m (g) in Leindler [122] as follows kVn,m (g)k ≤ C

n X En−m+k (g) k=0

n+k+1

, 0 ≤ m ≤ n, m = 0, 1, ...,

where again C > 0 is an absolute constant independent of g, n and m.

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Moreover, this last inequality was generalized in Leindler-Meir [123] to the strong de la Vall´ee Poussin means with exponent p ≥ 1, ( )1/p n X 1 (p) p Vn,m (g)(x) = |sk (g)(x) − g(x)| m+1 k=n−m

by the inequality (p) kVn,m (g)k

≤ C(log n)

1−(1/p)

n p X En−m+k (g) k=0

n+k+1

!1/p

, 0 ≤ m ≤ n, n ≥ 2,

where C > 0 is independent of g, m, n and p. For p = 1 this result becomes that in Leindler [122], which clearly implies the inequality in Stechkin [185]. Now, for D1 = {z ∈ C; |z| ≤ 1} denote A(D1 ) = {f : D1 → C; f is analytic in D1 },

endowed with the uniform norm kf kD1 = max{|f (z)|; |z| ≤ 1}. Clearly f ∈ A(D1 ) means that there exists R > 1 such that f is analytic in DR = {z ∈ C; |z| < R}. For f ∈ A(D1 ) and 0 ≤ m ≤ n, let us define the de la Vall´ee Poussin mean n X 1 σn,m (f )(z) = Tk (f )(z), m+1 k=n−m

Pk

f (j) (0) j j=0 j! z

where Tk (f )(z) = represents the kth Taylor partial sum of f . In Gal [95], the proof of Theorem 2.1 (see also the proof of Theorem 3.2.1 in the book Gal [77] ) we pointed out the validity of the following inequality analogous to that in Stechkin [185], given by n X En−m+k (f ) , 0 ≤ m ≤ n, m = 0, 1, ..., kf − σn,m (f )kD1 ≤ C n+k+1 k=0

where En (f ) = inf{kf − P kD1 ; P is complex polynomial of degree ≤ n} and C > 0 is an absolute constant independent of f , n and m. The next theorem gives a positive answer to a problem raised in the book Gal [77] (see there the Remark 3 after the proof of Theorem 3.2.1, (iii), pp. 229-230 ) concerning the complex analogues of the two above mentioned generalizations of the Stechkin’s inequality, obtained for Fourier series in Leindler [122] and Leindler-Meir [123]. Theorem 4.3.1. (Gal [96]) For f ∈ A(D1 ) let us consider the strong de la Vall´ee Poussin means with exponent p ≥ 1, !1/p n X 1 (p) p ∆n,m (f )(z) = |Tk (f )(z) − f (z)| . m+1 k=n−m

For all 0 ≤ m ≤ n and n ≥ 2 we have k∆(p) n,m (f )kD1

≤ C[log(n)]

1−(1/p)

n p X En−m+k (f )

k=0

m+k+1

!1/p

,

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where C > 0 is independent of f , m, n and p. Proof.

P∞ First denoting z = reit ∈ D1 and f (z) = f (zeit ) = k=0 ck rk eikt , we get Z π 1 f (reit )e−ikt dt = ck rk , for all k = 0, 1, ..., . 2π −π

Then, for any k = 1, 2, ..., we observe that it follows Z Z eikθ π e−ikθ π ck rk eikθ = f (reit )e−ikt dt + f (reit )eikt dt 2π −π 2π −π Z π 1 = f (reit )[eik(θ−t) + eik(t−θ) ]dt 2π −π Z 1 π f (reit ) cos[k(t − θ)]dt = π −π Z 1 π = f (rei(t+θ) ) cos(kt)dt. π −π

This immediately implies

1 Tn (f )(re ) = π iθ

=

1 π

Z Z

π

f (re −π π

i(t+θ)

) 1+

n X

k=1

!

cos(kt) dt

f (rei(t+θ) )Dn (t)dt.

−π

By the maximum modulus principle, in all the estimates we can take |z| = 1, that is z = eiθ . Therefore, denoting f (eiθ ) = F [cos(θ), sin(θ)] + iG[cos(θ), sin(θ)], or for the simplicity of notations, f (eiθ ) := F (θ) + iG(θ), we can write Tn (f )(eiθ ) = sn (F )(θ) + isn (G)(θ), for all θ ∈ [−π, π]. By the Minkowski’s inequality and by applying (3), for all θ ∈ [−π, π] we immediately obtain iθ (p) (p) (p) (p) |∆(p) n,m (f )(e )| ≤ |Vn,m (F )(θ)| + |Vn,m (G)(θ)| ≤ kVn,m (F )k + kVn,m (G)k  !1/p !1/p  p p n n X X En−m+k (F ) En−m+k (G) . ≤ C(log n)1−(1/p)  + n+k+1 n+k+1 k=0

k=0

Now, let Ek (f ) = kf − P ∗ kD1 where P ∗ (z) is the polynomial of best approximation of degree ≤ k . By the maximum modulus principle we get Ek (f ) = max {|f (eiθ ) − P ∗ (eiθ )|} = max {|F (θ) − P1∗ (θ) + i[G(θ) − P2∗ (θ)]|} θ∈[−π,π] θ∈[−π,π] q [F (θ) − P1∗ (θ)]2 + [G(θ) − P2∗ (θ)]2 , = max θ∈[−π,π]

where P ∗ (eiθ ) = P1∗ (θ) + iP2∗ (θ) with P1∗ and P2∗ trigonometric polynomials of degree ≤ k.

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Since Ek (F ) ≤ max {|F (θ) − P1∗ (θ)|} θ∈[−π,π]

and Ek (G) ≤ max {|G(θ) − P2∗ (θ)|}, θ∈[−π,π]

by the above inequalities we easily get Ek (F ) ≤ Ek (f ), Ek (G) ≤ Ek (f ) and the estimate in the statement. Remark. For p = 1 it is clear that the estimate in Theorem 4.3.1 also implies the estimate for kf − σn,m (f )k obtained in the proof of Theorem 3.2.1 in the book Gal [77], p. 227 (see also the proof of Theorem 2.1 in Gal [95] ). 4.4

Bibliographical Notes and Open Problems

Theorems 4.1.3, 4.1.4 and Corollary 4.1.5, Theorems 4.2.14, 4.2.15, 4.2.16, 4.2.17 and 4.2.18 appear for the first time here. Open Problem 4.4.1. Let us suppose that f ∈ A(G) (that is f is analytic on G), where G is a continuum in C (that is a compact connected subset of C). If we denote by Fk (z), k = 0, 1, 2, ..., the Faber polynomials attached to G, then from P Theorem 2 in Suetin [186], p. 52, it is known that f (z) = ∞ ak (f )Fk (z), where Pk=0 n the series is uniformly convergent in G. Also, for Pn (z) = k=0 ak (f )Fk (z), let us define the de la Vall´ee Poussin-type means σn,m (f ; G)(z) =

1 m+1

n X

k=n−m

Pk (f )(z), z ∈ G, m ≤ n

and the strong de la Vall´ee Poussin-type means with exponent p ≥ 1, !1/p n X 1 ∆(p) |Pk (f )(z) − f (z)|p , z ∈ G, m ≤ n. n,m (f : G)(z) = m+1 k=n−m

It is an open question if the following three inequalities can hold kf − σn,m (f ; G)kG ≤ C

n X En−m+k (f ; G) k=0

k∆n,m (f ; G)kG ≤ C

k∆(p) n,m (f ; G)kG

≤ C[log(n)]

n+k+1

, m ≤ n, m = 0, 1, ...,

p n X En−m+k (f ; G) , m ≤ n, n ≥ 2, m+k+1 k=0

1−(1/p)

n p X En−m+k (f ; G)

k=0

m+k+1

!1/p

, m ≤ n, n ≥ 2,

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where k · kG denotes the uniform norm on C(G), C > 0 is an absolute constant independent of f , n, m and p but dependent on G and En (f ; G) = inf{kf − P kG ; P is complex polynomial of degree ≤ n}. Open Problem 4.4.2. A famous result of Bernstein [44], [45] states that for any λ > 0, λ not even integer, there exists finite the limit limn→∞ nλ En (|x|λ )∞ > 0, where En (|x|λ )∞ = inf{max{|P (x) − |x|λ |; x ∈ [−1, 1]}; P ∈ Pn (R)}, and Pn (R) denotes the set of all real polynomials (that is with real coefficients) of degree ≤ n (for details and generalizations of this result to weighted approximation see the recent book of Ganzburg [98]). On the other hand, taking into account that |x|λ is convex for λ ≥ 1 and concave for 0 < λ < 1, denoting En(+2) (f )∞ = inf{kf − P k∞ ; P ∈ Pn (R), P 00 (x) ≥ 0, ∀x ∈ [−1, 1]}, and En(−2) (f )∞ = inf{kf − P k∞ ; P ∈ Pn (R), P 00 (x) ≤ 0, ∀x ∈ [−1, 1]}, it has sense to consider here the open problem if there exists finite the lim(−2) (+2) its limn→∞ nλ En (|x|λ )∞ for 0 < λ < 1 and limn→∞ nλ En (|x|λ )∞ for λ≥1? Note that since by Kopotun-Leviatan-Shevchuk [119], for f convex (or concave, respectively) on [−1, 1] and λ > 0 we have En (f )∞ = O(n−λ )

iff En(±2) (f )∞ = O(n−λ ), (−2) (nλ En (|x|λ )∞ )n∈N

n → ∞,

it easily follows that the sequences for 0 < λ < 1 and (+2) (nλ En (|x|λ )∞ )n∈N for λ ≥ 1, are bounded. More general, we can consider the following open question concerning the shape preserving limit results. Thus, taking into account that by Leviatan-Shevchuk [124], (+2) for a convex function f we have En (f )Lp ≤ En−2 (f 00 )Lp , for all n ≥ 2 and 0 ≤ p ≤ ∞ and this estimate cannot be improved, it is an open question if for f convex on [−1, 1] and the sequence (λn )n satisfying the conditions in Ganzburg [98], p. 3, we have 1 − λn 1 − λn , lim En(+2) f ; L∞ − < ∞( or ≤ A(+2) (f, L∞ (R)) < ∞) ? σ n→∞ nσ nσ (+2)

Here Aσ (f, L∞ (R) = inf{kf − gkL∞(R) ; g ∈ Bσ , g convex on R} and Bσ denotes the class of all entire functions of exponential type σ. All the above questions could be considered in complex setting too. Thus, denoting kf kD1 = max{|f (z)|; z ∈ D1 } and for f ∈ C(D1 ), En (f )(D1 ) = inf{kP − f kD1 ; P ∈ Pn (C)},

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with Pn (C) representing the set of all complex polynomials of degree ≤ n and with complex coefficients, it is an open question if for λ > 0 there exists finite the limit lim nλ En (|z|λ )(D1 )

n→∞

?

Moreover, since the continuous non-analytic function fλ (z) = |z|λ , z ∈ D1 is obviously convex (transforming any disk of center origin and radius 0 < r < 1 into a convex set), denoting En(CON V ) (fλ )(D1 ) = inf{kP − fλ kD1 ; P ∈ Pn (C), P (0) = 0, P 0 (0) 6= 0, P convex in D1 }, it is an open question if there exists finite the limit lim nλ En(CON V ) (|z|λ )(D1 )

n→∞

?

00 (z) Here the convexity of the polynomial P means that we have Re zP +1 > 0, P 0 (z) for all z ∈ D1 . Similar problems in the complex case can be posed if we replace |z|λ by f (z)λ , |z| ≤ 1, where f (z) = z, if Re(z) ≥ 0 and f (z) = −z if Re(z) < 0.

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Bibliography

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Index

analytic, 155

holomorphic, 155

Baskakov operator, 124, 156, 172 Bernstein polynomials, 156, 163 Bernstein polynomials of quaternion variable, 295 Bernstein’s inequality, 3 Bernstein-Kantorovich, 96 Bernstein-Stancu polynomials, 67, 80 Bernstein-type operators, 299 Bochner-Lebesgue integrable, 300 Butzer’s linear combination, 42, 317

identity’s theorem, 156 iterates, 26, 60, 78, 83 left Bernstein polynomials, 297 Maximum Modulus Theorem, 2, 156 middle Bernstein polynomials, 297 open polydisk, 155 q-Bernstein polynomials, 50

Cauchy’s formula, 2, 156 complex Banach space, 314 complex convolution polynomials, 181 convex, 16, 66, 87 convolution-type operators, 299

random functions, 300 random polynomials, 311 Riemann integrable, 300 Riesz-Zygmund, 188, 317 right Bernstein polynomials, 297

de la Vall´ee Poussin, 181, 317

simultaneous approximation, 14, 68, 76, 119, 126, 133 spirallike, 17, 66, 87 starlike, 16, 66, 87

Favard-Sz´ asz-Mirakjan operator, 103, 156, 166 Fej´er’s means, 188 fuzzy polynomials, 311 fuzzy-number-valued functions, 300

theorem on the identity, 3 univalent, 16

generalized algebraic polynomial, 300 generalized Bernstein and Bernstein-Kantorovich polynomials, 302, 313 generalized operators, 300 generalized Sz´ asz-Mirakjan, Baskakov and Post-Widder operators, 307 generalized trigonometric polynomial, 300 generalized Voronovskaja’s theorems, 35

vector-valued functions, 299 Vitali’s theorem, 1, 156 Voronovskaja-type formula, 9, 52, 70, 98, 107, 116, 130, 149, 159, 168, 174, 179, 183, 190 Weierstrass’s theorem, 2, 156

337

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