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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
Published Vol. 1
Long Time Behaviour of Classical and Quantum Systems edited by S. Graffi & A. Martinez
Vol. 2
Problems in Probability by T. M. Mills
Vol. 3
Introduction to Matrix Theory: With Applications to Business and Economics by F. Szidarovszky & S. Molnár
Vol. 4
Stochastic Models with Applications to Genetics, Cancers, Aids and Other Biomedical Systems by Tan Wai-Yuan
Vol. 5
Defects of Properties in Mathematics: Quantitative Characterizations by Adrian I. Ban & Sorin G. Gal
Vol. 6
Topics on Stability and Periodicity in Abstract Differential Equations by James H. Liu, Gaston M. N’Guérékata & Nguyen Van Minh
Vol. 7
Probabilistic Inequalities by George A. Anastassiou
Vol. 8
Approximation by Complex Bernstein and Convolution Type Operators by Sorin G. Gal
Vol. 9
Distribution Theory and Applications by Abdellah El Kinani & Mohamed Oudadess
Vol. 11 Advanced Inequalities by George A. Anastassiou
Forthcoming Vol. 10 Theory and Examples of Ordinary Differential Equations by Chin-Yuan Lin
ZhangJi - Advanced Inequalities.pmd
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Series on Concrete and Applicable Mathematics – Vol.11
George A Anastassiou University of Memphis, USA
World Scientific NEW JERSEY
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ADVANCED INEQUALITIES Series on Concrete and Applicable Mathematics — Vol. 11 Copyright © 2011 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.
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ISBN-13 978-981-4317-62-7 ISBN-10 981-4317-62-4
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Dedicated to my daughter Peggy
v
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Preface
In this monograph we present univariate and multivariate classical analysis advanced inequalities. This treatise relies on author’s last thirteen years related research work, more precisely see [15]-[61], and it is a natural outgrowth of it. Chapters are self-contained and several advanced courses can be taught out of this book. Extensive background and motivations are given per chapter. A very extensive list of references is given at the end. The topics covered are very diverse. We present recent advances on Ostrowski type inequalities, see Chapters 2-9; on Opial type inequalities, see Chapters 10-16; Poincare and Sobolev type inequalities, see Chapters 17-20. Also we present HardyOpial type inequalities, as well as Ostrowski like ones, see Chapters 21-22. Then we present some work on ordinary and distributional Taylor formulae with estimates for their remainders and applications, see Chapters 23-24. Finally in Chapters 25-29 we study Chebyshev-Gruss, Gruss, Comparison of Means inequalities. Our book’s results appeared for the first time in our published articles mentioned above, and they are mostly optimal, that is usually our inequalities are sharp and attained. They are expected to find applications in many areas of pure and applied mathematics, such as mathematical analysis, probability, ordinary and partial differential equations, numerical analysis, information theory, etc. As such this monograph is suitable for researchers, graduate students, and seminars of the above subjects, also to be in all science libraries. The preparation of book took place during 2008-2009 in Memphis, Tennessee, USA. I would like to thank my family for their dedication and love to me, which was the strongest support during the writing of this monograph. Also many thanks go to my typist Rodica Gal for an excellent and on time technical job. January 1, 2010 George A. Anastassiou Department of Mathematical Sciences The University of Memphis, TN U.S.A. vii
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Contents
Preface
vii
1. Introduction
1
2. Advanced Univariate Ostrowski Type Inequalities
5
2.1 2.2 2.3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Auxilliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Higher Order Ostrowski Inequalities 3.1 3.2
21
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4. Multidimensional Euler Identity and Optimal Multidimensional Ostrowski Inequalities 4.1 4.2 4.3 4.4 4.5
Introduction . Background . Main Results Applications . Sharpness . .
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Auxilliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6. Ostrowski Inequalities on Euclidean Domains 6.1 6.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
21 22
27
5. More on Multidimensional Ostrowski Type Inequalities 5.1 5.2 5.3
5 6 14
27 30 31 53 76 81 81 82 87 93 93 93
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7. High Order Ostrowski Inequalities on Euclidean Domains 7.1 7.2 7.3
99
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Functions on General Domains . . . . . . . . . . . . . . . . . . . . 106
8. Ostrowski Inequalities on Spherical Shells 8.1 8.2 8.3
109
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Addendum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
9. Ostrowski Inequalities on Balls and Shells Via Taylor–Widder Formula 9.1 9.2 9.3 9.4 9.5
Introduction . . . . . . Background . . . . . . Results on the Shell . Results on the Sphere Addendum . . . . . .
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125 . . . . .
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10. Multivariate Opial Type Inequalities for Functions Vanishing at an Interior Point 10.1 10.2
139
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
11. General Multivariate Weighted Opial Inequalities 11.1 11.2
149
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
12. Opial Inequalities for Widder Derivatives 12.1 12.2 12.3
161
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
13. Opial Inequalities for Linear Differential Operators 13.1 13.2
171
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
14. Opial Inequalities for Vector Valued Functions 14.1 14.2 14.3 14.4
125 126 128 132 138
Introduction Background . Results . . . Applications .
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179 179 180 186
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15. Opial Inequalities for Semigroups 15.1 15.2 15.3
187
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
16. Opial Inequalities for Cosine and Sine Operator Functions 16.1 16.2 16.3 16.4
Introduction . Background . Results . . . Applications .
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197 . . . .
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17. Poincar´e Like Inequalities for Linear Differential Operators 17.1 17.2
209
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
18. Poincar´e and Sobolev Like Inequalities for Widder Derivatives 18.1 18.2
215
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
19. Poincar´e and Sobolev Like Inequalities for Vector Valued Functions 19.1 19.2 19.3 19.4
Introduction . Background . Results . . . Applications .
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20. Poincar´e Type Inequalities for Semigroups, Cosine and Sine Operator Functions 20.1 20.2 20.3 20.4 20.5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . Semigroups Background . . . . . . . . . . . . . . . . . . Poincar´e Type Inequalities for Semigroups . . . . . . . . Cosine and Sine Operator Functions Background . . . . Poincar´e Type Inequalities for Cosine and Sine Operator Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
21. Hardy–Opial Type Inequalities 21.1
Results
229 230 231 240
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243 243 245 249
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22. A Basic Sharp Integral Inequality 22.1 22.2
197 197 199 206
271
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
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xii
23. Estimates of the Remainder in Taylor’s Formula 23.1 23.2 23.3 23.4 23.5
Introduction . . . . . . . . . . . . . . . . Some New Bounds for the Remainder . Some Further Bounds of the Remainder Some Inequalities for Special Cases . . . Taylor-Multivariate Case Estimates . . .
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24. The Distributional Taylor Formula 24.1 24.2 24.3
319
331
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Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
29. Multivariate Fink Type Identity Applied to Multivariate Inequalities 29.1 29.2 29.3
305
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
28. Multivariate Chebyshev–Gr¨ uss and Comparison of Integral Means Inequalities 28.1 28.2
293
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
27. Chebyshev–Gr¨ uss Type Inequalities on Spherical Shells and Balls 27.1 27.2
275 276 280 282 284
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
26. Gr¨ uss Type Multivariate Integral Inequalities 26.1 26.2 26.3
. . . . .
Introduction and Background . . . . . . . . . . . . . . . . . . . . . 293 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302
25. Chebyshev–Gr¨ uss Type Inequalities Using Euler Type and Fink Identities 25.1 25.2
. . . . .
365
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381
Bibliography
395
List of Symbols
407
Index
409
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Chapter 1
Introduction
Here we describe the material contained in this monograph. Our results are mostly optimal, i.e. sharp, attained inequalities. We give also applications. The exposed results are brought for the first time in a book form. CHAPTER 2: Very general univariate Ostrowski type inequalities are given, involving the k · k∞ and k · kp , p ≥ 1 norms of the engaged nth order derivative, n ≥ 1. In establishing them, several important univariate identities of Montgomery type are developed. CHAPTER 3: We generalize Ostrowski inequality for higher order derivatives, by using a generalized Euler-type identity. Some of the produced inequalities are sharp, namely attained by basic functions. The rest of the estimates are tight. We give applications to trapezoidal and midpoint rules. Estimates are given with respect to L∞ norm. CHAPTER 4: We present a general multivariate Euler type identity. Using it we derive general tight multivariate high order Ostrowski type inequalities for the estimate on the error of a multivariate function f evaluated at a point from its average. The estimates are involving only the single partial derivatives of f and are with respect to k · kp , 1 ≤ p ≤ ∞. We give specific applications of our main results to the multivariate trapezoid and midpoint rules for functions f differentiable up to order 6. We prove sharpness of our inequalities for differentiation orders m = 1, 2, 4 and with respect to k · k∞ . CHAPTER 5: Very general multidimensional Ostrowski type inequalities are presented, some of them are proved to be sharp. They involve the k · k∞ and k · kp norms of the engaged mixed partial of nth order n ≥ 1. In establishing them, other important multivariate results of Montgomery type identity are developed. CHAPTER 6: The classical Ostrowski inequality for functions on intervals is extended to functions on general domains in Euclidean space. For radial functions on balls the inequality is sharp. CHAPTER 7: The classical Ostrowski inequality for functions on intervals estimates the value of the function minus its average in terms of the maximum of its first derivative. This result is extended to functions on general domains using the 1
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ADVANCED INEQUALITIES
L∞ norm of its nth partial derivatives. For radial functions on balls the inequality is sharp. CHAPTER 8: Here are presented Ostrowski type inequalities over spherical shells. These regard sharp or close to sharp estimates to the difference of the average of a multivariate function from its value at a point. CHAPTER 9: The classical Ostrowski inequality for functions on intervals estimates the value of the function minus its average in terms of the maximum of its first derivative. This result is extended to higher order over shells and balls of RN , N ≥ 1, with respect to an extended complete Chebyshev system and the generalized radial derivatives of Widder type. We treat radial and non-radial functions. CHAPTER 10: Here we generalize Opial inequalities in the multidimensional case over balls. The inequalities carry weights and are proved to be sharp. The functions under consideration vanish at the center of the ball. CHAPTER 11: Here we give Opial type weighted multidimensional inequalities over balls and arbitrary smooth bounded domains. The inequalities are mostly sharp. The functions under consideration vanish on the boundary. CHAPTER 12: Various Lp form Opial type inequalities are given for Widder derivatives. CHAPTER 13: Various Lp form Opial type inequalities are presented for a linear differential operator L, involving its related initial value problem solution y, Ly, the associated Green’s function H and initial conditions point x0 ∈ R. CHAPTER 14: Various Lp form Opial type inequalities are shown for functions valued in a Banach vector space. We give applications in C. CHAPTER 15: Various Lp form Opial type inequalities are given for semigroups with applications. CHAPTER 16: Various Lp form Opial type inequalities are shown for cosine and sine operator functions with applications. CHAPTER 17: Various Lp form Poincar´e like inequalities, forward and reverse, are given for a linear differential operator L, involving its related initial value problem solution y, Ly, the associated Green’s function H and initial conditions at point x0 ∈ R. CHAPTER 18: Various Lp form Poincar´e and Sobolev like inequalities, forward and reverse, are presented involving Widder derivative. CHAPTER 19: Various Lp form Poincar´e and Sobolev like inequalities are given for functions valued in a Banach vector space. We give applications on C. CHAPTER 20: Here we present Poincar´e type general Lp inequalities regarding semigroups, cosine and sine operator functions. We apply inequalities to specific cases of them. CHAPTER 21: Various Lp form Hardy–Opial type sharp integral inequalities are derived involving two functions. CHAPTER 22: A sharp multidimensional integral type inequality is presented involving nth order (n ∈ N) mixed partial derivatives. This is subject to some basic
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Introduction
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boundary condition satisfied by the involved multivariate function. CHAPTER 23: Important estimates of the remainder in Taylor’s formula are given. We treat both univariate and multivariate cases. CHAPTER 24:We derive a distributional Taylor formula with precise integral remainder. We give applications of it and estimates for the remainder. CHAPTER 25: Here we present Chebyshev–Gr¨ uss type univariate inequalities by using the generalized Euler type and Fink identities. The results involve the functions f, g, f (n) , g (n) , n ∈ N, and are with respect to k · kp , 1 ≤ p ≤ ∞. CHAPTER 26: Here are presented Gr¨ uss type multivariate integral inequalities. CHAPTER 27: Here we give Chebyshev–Gr¨ uss type inequalities on R N over spherical shells and balls by extending some basic univariate results of Pachpatte [206]. CHAPTER 28: We present tight multivariate Chebyshev–Gr¨ uss and comparison of integral means inequalities by using a general multivariate Euler type identity. Our results involve the functions f, g and their high order single partials and are with respect to k · kp , 1 ≤ p ≤ ∞. CHAPTER 29: We describe a general multivariate Fink type identity which is a representation formula for a multivariate function. Using it we derive general tight multivariate high order Ostrowski type, comparison of means and Gr¨ uss type inequalities. The estimates involve Lp norms, any 1 ≤ p ≤ ∞. Our presented results are expected to find applications in many areas of pure and applied mathematics, such as mathematical analysis, probability, ordinary and partial differential equations, numerical analysis, information theory, etc.
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Chapter 2
Advanced Univariate Ostrowski Type Inequalities
Very general univariate Ostrowski type inequalities are presented, involving the k · k∞ and k · kp , p ≥ 1 norms of the engaged nth order derivative, n ≥ 1. In proving them, several important univariate identities of Montgomery type are given. This chapter follows [24]. 2.1
Introduction
In 1938, A. Ostrowski [196] proved the following important inequality: Theorem 2.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e., kf 0 k∞ := supt∈(a,b) |f 0 (t)| < +∞. Then " # 1 Z b 2 ) 1 (x − a+b 2 f (t)dt − f (x) ≤ · (b − a)kf 0 k∞ , (2.1) + b − a a 4 (b − a)2
for any x ∈ [a, b]. The constant
1 4
is the best possible.
Since then there has been a lot of activity around these inequalities with important applications to Numerical Analysis and Probability. This chapter is greatly motivated and inspired also by the following results. Theorem 2.2 (see [16]). Let f ∈ C n+1 ([a, b]), n ∈ N and x ∈ [a, b] be fixed, such that f (k) (x) = 0, k = 1, . . . , n. Then it holds 1 Z b kf (n+1) k (x − a)n+2 + (b − x)n+2 ∞ f (y)dy − f (x) ≤ . (2.2) · b − a a (n + 2)! b−a
Inequality (2.2) is sharp. In particular, when n is odd is attained by f ∗ (y) := (y − x)n+1 · (b − a), while when n is even the optimal function is f˜(y) := |y − x|n+α · (b − a),
α > 1.
Clearly inequality (2.2) generalizes inequality (2.1) for higher order derivatives of f. 5
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One of the goals of this chapter is to give a generalization Theorem 4.17, p. 191 of [85], see here Theorem 2.13 later. We would like to mention that result. Theorem 2.3. Let f : [a, b] → R be continuous on [a, b] and twice differentiable function on (a, b), whose second derivative f 00 : (a, b) → R is bounded on (a, b). Then Z b a + b 1 f (b) − f (a) x− f (t)dt − f (x) − b−a a b−a 2 " #2 2 1 (x − a+b 1 1 2 ) ≤ + + 2 (b − a)2 4 12 ·(b − a)2 · kf 00 k∞ ≤
kf 00 k∞ (b − a)2 , 6
for all x ∈ [a, b].
(2.3)
In this chapter we present very general Ostrowski type inequalities in the one dimensional case. Involved integrals here are also of Riemann–Stieltjes form. Along the way to establishing these results, we show several important side univariate related results, generalizing Montgomery’s identity ([188], Ch. XVII, p. 565), see Section 2.2. 2.2
Auxilliary Results
We present the following results we need, which by themselves are of independent interest. Theorem 2.4. Let f : [a, b] → R be n-times differentiable on [a, b], n ∈ N. The nth derivative f (n) : [a, b] → R is integrable on [a, b]. Let x ∈ [a, b]. Define the kernel s−a , a ≤ s ≤ r, b−a P (r, s) := s − b , r < s ≤ b, b−a where r, s ∈ [a, b]. Then
Z b 1 f (s1 )ds1 θ1,n := f (x) − (b − a) a Z b Z b n−2 X f (k) (b) − f (k) (a) · ··· − b−a k=0 | a {z a}
(k+1)th−integral
· · · dsk+1 = =: θ2,n .
Z
b
a
···
Z
b
P (x, s1 )
a
n−1 Y i=1
P (x, s1 )
k Y
i=1
P (si , si+1 ) · ds1 ds2
P (si , si+1 )f (n) (sn )ds1 ds2 · · · dsn (2.4)
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Advanced Univariate Ostrowski Type Inequalities −1 Q
Here and later we make the conventions that
k=0
• = 0,
0 Q
i=1
7
• = 1.
Proof. Applying integration by parts twice we obtain Montgomery identity ([188], Ch. XVII, p. 565) Z b Z b 1 f (s1 )ds1 + P (x, s1 )f 0 (s1 )ds1 . f (x) = b−a a a Doing the same for the derivative of f we have Z b Z b 1 0 0 f (s1 ) = f (s2 )ds2 + P (s1 , s2 )f 00 (s2 )ds2 . b−a a a That is
f (b) − f (a) f (s1 ) = + b−a 0
Similarly for f 00 we get
f 00 (s2 ) = And in general we obtain f (n−1) (sn−1 ) =
Z
f 0 (b) − f 0 (a) + b−a
b
P (s1 , s2 )f 00 (s2 )ds2 .
a
Z
b
P (s2 , s3 )f 000 (s3 )ds3 .
a
f (n−2) (b) − f (n−2) (a) + b−a
Z
b
P (sn−1 , sn )f (n) (sn )dsn . a
We observe that Z b Z b f (b) − f (a) P (x, s1 )ds1 P (x, s1 )f 0 (s1 )ds1 = b−a a a Z bZ b + P (x, s1 )P (s1 , s2 )f 00 (s2 )ds2 ds1 . a
That is f (x) =
Z b Z b f (b) − f (a) 1 f (s1 )ds1 + P (x, s1 )ds1 b−a a b−a a Z bZ b + P (x, s1 )P (s1 , s2 )f 00 (s2 )ds2 ds1 . a
Next we see that Z bZ b a
a
a
P (x, s1 )P (s1 , s2 )f 00 (s2 )ds2 ds1
a
Z bZ b f 0 (b) − f 0 (a) P (x, s1 )P (s1 , s2 )ds2 ds1 = b−a a a Z bZ bZ b + P (x, s1 )P (s1 , s2 )P (s2 , s3 )f 000 (s3 )ds3 ds2 ds1 .
a
a
a
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Hence we get f (x) =
Z b Z b f (b) − f (a) 1 P (x, s1 )ds1 f (s1 )ds1 + b−a a b−a a 0 Z bZ b f (b) − f 0 (a) + P (x, s1 )P (s1 , s2 )ds2 ds1 b−a a a Z bZ bZ b + P (x, s1 )P (s1 , s2 )P (s2 , s3 )f 000 (s3 )ds3 ds2 ds1 . a
a
a
Therefore in general we derive f (x)
Z b Z b 1 f (b) − f (a) f (s1 )ds1 + P (x, s1 )ds1 b−a a b−a a Z bZ b 0 f (b) − f 0 (a) + P (x, s1 )P (s1 , s2 )ds2 ds1 b−a a a 00 Z Z bZ b b f (b) − f 00 (a) P (x, s1 )P (s1 , s2 )P (s2 , s3 )ds3 ds2 ds1 + b−a a a a 000 Z b Z b Z b Z b f (b) − f 000 (a) P (x, s1 )P (s1 , s2 )P (s2 , s3 )P (s3 , s4 )ds4 · · · ds1 + b−a a a a a (n−2) Z b Z b n−2 Y f (b) − f (n−2) (a) + ···+ P (x, s1 ) P (si , si+1 )dsn−1 · · · ds1 ··· b−a i=1 | a {z a} Z b Z b n−1 Y + ··· P (x, s1 ) P (si , si+1 )f (n) (sn )dsn · · · ds1 . a a i=1 | {z }
=
nth−integral
The last equality is written briefly as follows Z b n−2 X f (k) (b) − f (k) (a) 1 f (x) = f (s1 )ds1 + b−a a b−a k=0 Z b Z b ··· · a | {z a}
P (x, s1 )
(k+1)th−integral
+
Z
b
a
···
Z
b
P (x, s1 )
a
So we have proved (2.4).
n−1 Y i=1
P (si , si+1 )ds1 ds2 · · · dsk+1 i=1 k Y
P (si , si+1 )f (n) (sn )ds1 ds2 · · · dsn .
(2.5)
A special very common case follows. Corollary 2.5. Under the assumptions and notations of Theorem 2.4, additionally assume that f (k) (a) = f (k) (b),
k = 0, 1, . . . , n − 2; when n ≥ 2.
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Then we get θ1,n
Z b 1 = f (x) − f (s1 )ds1 b−a a Z b Z b n−1 Y = ··· P (x, s1 ) · P (si , si+1 ) · f (n) (sn )ds1 · · · dsn = θ2,n , (2.6) a
a
i=1
for n ∈ N, x ∈ [a, b]. Proof.
Directly from (2.4).
Next we generalize Montgomery’s identity for Riemann–Stieltjes integrals (see also in [102] a related but different result). Proposition 2.6. Let f : [a, b] → R be differentiable on [a, b]. The derivative f 0 : [a, b] → R is integrable on [a, b]. Let also g : [a, b] → R be of bounded variation, such that g(a) 6= g(b). Let x ∈ [a, b]. We define g(t) − g(a) , a ≤ t ≤ x, g(b) − g(a) P (g(x), g(t)) = g(t) − g(b) , x < t ≤ b. g(b) − g(a) Then
f (x) =
1 (g(b) − g(a))
Z
b
f (t)dg(t) + a
Z
b
P (g(x), g(t))f 0 (t)dt.
(2.7)
a
Proof. Here f is continuous on [a, b], and g of bounded variation on [a, b]. Then f is Riemann–Stieltjes integrable with respect to g, and it holds Z b Z b f (x)dg(x) + g(x)df (x) = f (b)g(b) − f (a)g(a). a
a
Also it holds that g is Riemann–Stieltjes integrable with respect to f . Here f is differentiable on [a, b]. By the above we obtain Z x Z x (g(t) − g(a))f 0 (t)dt = (g(t) − g(a))df (t) a a x Z x = f (t)(g(t) − g(a)) − f (t)dg(t) a a Z x = f (x)(g(x) − g(a)) − f (t)dg(t). a
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Also we have Z
b x
(g(t) − g(b))f 0 (t)dt =
Z
b x
(g(t) − g(b))df (t)
b Z b f (t)dg(t) = (g(t) − g(b))f (t) − x x Z b = (g(b) − g(x))f (x) − f (t)dg(t). x
Adding the above equalities we get Z b Z x (g(t) − g(b))f 0 (t)dt (g(t) − g(a))f 0 (t)dt + x
a
= f (x)(g(b) − g(a)) −
Z
b
f (t)dg(t). a
Hence we derive that f (x) =
Z b 1 1 f (t)dg(t) + (g(b) − g(a)) a (g(b) − g(a)) "Z # Z b x 0 0 · (g(t) − g(b))f (t)dt . (g(t) − g(a))f (t)dt + x
a
So we have established (2.7).
For twice differentiable functions we obtain the following general Montgomery type identity. Proposition 2.7. Let f : [a, b] → R be twice differentiable on [a, b]. The second derivative f 00 : [a, b] → R is integrable on [a, b]. Let also g : [a, b] → R be of bounded variation, such that g(a) 6= g(b). Let x ∈ [a, b]. The kernel P is defined as in Proposition 2.6. Then Z b 1 1 f (t)dg(t) + f (x) = (g(b) − g(a)) a (g(b) − g(a)) ! ! Z Z b
·
+
Proof.
a
Z
b a
b
P (g(x), g(t)) · dt
Z
b a
·
f 0 (t1 )dg(t1 )
a
P (g(x), g(t)) · P (g(t), g(t1 )) · f 00 (t1 )dt1 dt.
Apply (2.7) for f 0 , we get Z b Z b 1 f 0 (t1 )dg(t1 ) + P (g(t), g(t1 ))f 00 (t1 )dt1 . f 0 (t) = (g(b) − g(a)) a a
(2.8)
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Therefore it holds Z b Z b P (g(x), g(t)) P (g(x), g(t)) · f 0 (t)dt = a a " # Z b Z b 1 · f 0 (t1 )dg(t1 ) + P (g(t), g(t1 ))f 00 (t1 )dt1 dt (g(b) − g(a)) a a ! Z b Z b 1 0 · P (g(x), g(t)) · = f (t1 )dg(t1 ) dt (g(b) − g(a)) a a ! Z b Z b P (g(x), g(t)) · + P (g(t), g(t1 ))f 00 (t1 )dt1 dt a
a
! ! Z b 1 0 = · P (g(x), g(t)) · dt · f (t1 )dg(t1 ) (g(b) − g(a)) a a Z bZ b + P (g(x), g(t)) · P (g(t), g(t1 ))f 00 (t1 )dt1 dt. Z
a
b
a
Using the last equality along with (2.7), we have proved (2.8).
At the greatest generality it holds Theorem 2.8. Let f : [a, b] → R be n-times differentiable on [a, b], n ∈ N. The nth derivative f (n) : [a, b] → R is integrable on [a, b]. Let also g : [a, b] → R be of bounded variation, such that g(a) 6= g(b). Let x ∈ [a, b]. The kernel P is defined as in Proposition 2.6. Then Z b 1 1 · f (s1 )dg(s1 ) + f (x) = (g(b) − g(a)) a (g(b) − g(a)) ! Z n−2 b X · f (k+1) (s1 )dg(s1 ) a
k=0
Z b Z b ··· · a | {z a}
P (g(x), g(s1 ))
(k+1)th−integral
+
Proof.
Z
b
a
···
Z
b
P (g(x), g(s1 ))
a
n−1 Y i=1
P (g(si ), g(si+1 )) · ds1 ds2 · · · dsk+1 i=1 k Y
P (g(si ), g(si+1 ))f (n) (sn )ds1 · · · dsn . (2.9)
Similar to Proposition 2.7.
Another important special case follows. Corollary 2.9. All as in Theorem 2.8. Additionally suppose that Z b f (k+1) (s1 )dg(s1 ) = 0, k = 0, 1, . . . , n − 2; when n ≥ 2. a
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Then M1,n
Z b 1 := f (x) − · f (s1 )dg(s1 ) (g(b) − g(a)) a Z b Z b n−1 Y = ··· P (g(x), g(s1 )) P (g(si ), g(si+1 ))f (n) (sn ) · ds1 · · · dsn a
a
i=1
=: M2,n ,
(2.10)
for n ∈ N, x ∈ [a, b]. Proof.
From Theorem 2.8.
A similar multivariate result comes next. Theorem 2.10. Let Q be a compact convex subset of Rk , k ≥ 2; x := (x1 , . . . , xk ) and O := (0, . . . , 0) ∈ Q. Let f ∈ C 1 (Q) and suppose that all partial derivatives of f of order one are coordinatewise absolutely continuous functions. Then Z 1 Z 1Z 1 k X ∂f (t1 t2 x)dt1 dt2 f (x) = f (t1 x)dt1 + xj t1 ∂xj 0 0 0 j=1 +
k X k X
xi xj
i=1 j=1
Z
Z
1
0
1
t21 t2
0
∂ 2 f (t1 t2 x) dt1 dt2 . ∂xi ∂xj
(2.11)
Note. Parameterized integrals as above are met in Radon Transform Theory and in Computerized Tomography. Proof of Theorem 2.10. Set gx (t) := f (tx), x := (x1 , . . . , xk ), x ∈ Rk , k ≥ 2, 0 ≤ t ≤ 1. Then gx (0) = f (0), gx (1) = f (x). Integrating by parts we obtain that Z 1 Z 1 tgx0 (t)dt. gx (t)dt + gx (1) = 0
0
Here we have
gx0 (t) =
k X
xi
i=1
Therefore f (x) = So that f (x) =
Z
1
f (tx)dt + 0
1
f (t1 x)dt1 + 0
Next apply (2.12) for ∂f (x) = ∂xj
Z
Z
1
t 0
k X i=1
∂f ∂xj
Z
1 0
to obtain
∂f (tx). ∂xi
xi
k X
∂f (tx) xi ∂xi i=1
Z
1
t1 0
!
∂f (t1 x) ∂xi
dt.
dt1 .
Z 1 k X ∂f ∂ 2 f (t2 x) (t2 x)dt2 + xi dt2 . t2 ∂xj ∂xi ∂xj 0 i=1
(2.12)
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Hence t1
∂f (t1 x) = t1 ∂xj
Z
Z 1 k X ∂f ∂ 2 f (t2 t1 x) t21 xi (t2 t1 x)dt2 + dt2 . t2 ∂xj ∂xi ∂xj 0 i=1
1 0
Furthermore it holds Z 1 Z 1 Z 1 ∂f (t1 x) ∂f dt1 = (t2 t1 x)dt2 dt1 t1 t1 ∂xj 0 0 0 ∂xj Z 1 Z 1X k ∂ 2 f (t1 t2 x) t21 xi t2 dt2 dt1 . + ∂xi ∂xj 0 0 i=1
(2.13)
Consequently by (2.13) we obtain Z 1 Z 1Z 1 k k X X ∂f (t1 x) ∂f xj t1 xj dt1 = t1 (t1 t2 x)dt2 dt1 ∂x ∂x j j 0 0 0 j=1 j=1 +
k X
xj
j=1
=
k X
xj
j=1
+
Z
1 0
Z
k X k X
0
k X
t21 xi
i=1
1Z
1
t1 0
xj xi
j=1 i=1
Z
1 0
Z
1
t2 0
∂ 2 f (t1 t2 x) dt2 dt1 ∂xi ∂xj
∂f (t1 t2 x)dt2 dt1 ∂xj Z
1 0
t21 t2
∂ 2 f (t1 t2 x) dt2 dt1 . ∂xi ∂xj
At last combining (2.12) and (2.14) we have established (2.11).
(2.14)
Corollary 2.11. All as in Theorem 2.10. Additionally suppose that Z 1Z 1 ∂f (t1 t2 x)dt1 dt2 = 0, all j = 1, . . . , k. t1 ∂x j 0 0 Then f (x) −
Z
1
f (t1 x)dt1 = 0
k X k X i=1 j=1
xi xj
Z
1 0
Z
1 0
t21 t2
∂ 2 f (t1 t2 x) dt1 dt2 , ∂xi ∂xj
(2.15)
for any x ∈ Q; 0 ∈ Q. We will use the next Lemma 2.12. Let f : [a, b] → R be 3-times differentiable on [a, b]. Assume that f 000 is integrable on [a, b]. Let x ∈ [a, b]. Define s−a , a ≤ s ≤ r, b−a P (r, s) := s−b , r < s ≤ b; r, s ∈ [a, b]. b−a
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Then θ1,3 := f (x) −
1 (b − a)
Z
b a
f (s1 )ds1 −
f (b) − f (a) b−a
Z
b
P (x, s1 )ds1 a
Z bZ b f 0 (b) − f 0 (a) P (x, s1 )P (s1 , s2 )ds1 ds2 − b−a a a Z bZ bZ b = P (x, s1 )P (s1 , s2 )P (s2 , s3 )f 000 (s3 )ds1 ds2 ds3 =: θ2,3 . (2.16)
a
Proof.
2.3
a
a
By Theorem 2.4.
Main Results
We present an Ostrowski type inequality. Theorem 2.13. Let f : [a, b] → R be 3-times differentiable on [a, b]. Assume that f 000 is bounded on [a, b]. Let x ∈ [a, b]. Then Z b a+b 1 f (b) − f (a) x− f (s1 )ds1 − f (x) − (b − a) a b−a 2 0 f (b) − f 0 (a) a2 + b2 + 4ab − · x2 − (a + b)x + 2(b − a) 6 000 kf k∞ · A, (2.17) ≤ (b − a)3 where 1 1 1 A := abx4 − a2 b3 x + a3 bx2 − ab2 x3 − a3 b2 x 3 3 3 1 1 1 3 2 + ab x + a2 b2 x2 − a2 bx3 − ax5 − bx5 3 2 2 1 6 3 2 4 3 2 4 1 2 4 2 3 3 + x + a x + b x + b a − a x 6 4 4 3 3 2 3 3 1 3 3 5 4 2 5 4 2 − b x − b a + a x + b x 3 3 12 12 1 4 2 2 5 2 5 1 5 1 a6 b6 . + b a − ba − ab − a x − b5 x + + 3 15 15 6 6 20 20
(2.18)
Inequality (2.17) is attained by f (x) = (x − a)3 + (b − x)3 , in that case both sides of the inequality equal zero.
(2.19)
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Since kf 000 k∞ < +∞, by (2.16) we obtain Z bZ bZ b |θ1,3 | = |θ2,3 | ≤ kf 000 k∞ · |P (x, s1 )| · |P (s1 , s2 )| · |P (s2 , s3 )|ds1 ds2 ds3 a a a ! ! Z b Z b Z b kf 000 k∞ |p(s2 , s3 )|ds3 ds2 ds1 , |p(s1 , s2 )| · |p(x, s1 )| = (b − a)3 a a a
Proof.
(2.20) where p(r, s) := We see that
Z
s − a, s − b,
a ≤ s ≤ r, r < s ≤ b; r, s ∈ [a, b].
b a
|p(s2 , s3 )|ds3 =
(s2 − a)2 + (b − s2 )2 . 2
Furthermore we find with some calculations that Z b (s2 − a)2 + (b − s2 )2 |p(s1 , s2 )| ds2 2 a # " 4 2 1 a+b a+b 7 2 4 + 3(b − a) s1 − + (b − a) . = 6 s1 − 12 2 2 8 To finish calculating the integral in (2.20) we find " 4 Z b 1 a+b |p(x, s1 )| 6 s1 − 12 a 2 # 2 a+b 7 2 4 +3(b − a) s1 − + (b − a) ds1 2 8 " Z x 4 1 a+b (s1 − a) 6 s1 − = 12 a 2 # 2 7 a+b 4 2 + (b − a) ds1 +3(b − a) s1 − 2 8 " 4 Z b 1 a+b + (b − s1 ) 6 s1 − 12 x 2 # 2 7 a+b 2 4 + (b − a) ds1 +3(b − a) s1 − 2 8 = A. Consequently we have |θ1,3 | ≤
kf 000 k∞ · A. (b − a)3
(2.21)
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Next we find that
Z
b
P (x, s1 )ds1 = a
x−
a+b 2
.
(2.22)
Furthermore we obtain Z bZ b P (x, s1 )P (s1 , s2 )ds1 ds2 a a (5a2 + 5b2 + 8ab) 1 2 2 (x + (a + b) ) − (a + b)x − . = 2 6 Therefore we derive Z b 1 θ1,3 = f (x) − f (s1 )ds1 (b − a) a 0 a+b f (b) − f 0 (a) f (b) − f (a) x− − − b−a 2 2(b − a) 2 2 (5a + 5b + 8ab) · (x2 + (a + b)2 ) − (a + b)x − . 6 So we have established inequality (2.17).
(2.23)
(2.24)
We have an interesting special case. Corollary 2.14. All as in Theorem 2.13. Additionally suppose that f (a) = f (b), and f 0 (a) = f 0 (b). Then Z b 1 kf 000 k∞ · A. (2.25) f (s1 )ds1 − f (x) ≤ (b − a)3 (b − a) a Other Ostrowski type results follow:
Theorem 2.15. All as in Theorem 2.4. Additionally assume that kf (n) k∞ < +∞. Then Z b Z b n−1 Y (n) |θ1,n | ≤ kf k∞ · ··· |P (x, s1 )| · |P (si , si+1 )|ds1 · · · dsn . (2.26) a
a
i=1
Theorem 2.16. All as in Proposition 2.6. Additionally suppose that kf 0 k∞ < +∞. Then Z b Z b 1 0 |P (g(x), g(t))|dt. (2.27) f (t)dg(t) − f (x) ≤ kf k∞ · (g(b) − g(a)) a a
Theorem 2.17. All as in Proposition 2.7. Additionally assume that kf 00 k∞ < +∞. Then Z b 1 1 f (x) − f (t)dg(t) − (g(b) − g(a)) a (g(b) − g(a)) ! ! Z b Z b 0 · P (g(x), g(t))dt · f (t1 )dg(t1 ) a
00
≤ kf k∞ ·
a
Z
a
bZ
b
a
!
|P (g(x), g(t))| · |P (g(t), g(t1 ))| · dt1 dt .
(2.28)
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Theorem 2.18. All as in Theorem 2.8. Additionally suppose that kf (n) k∞ < +∞. Then ! Z b n−2 X Z b 1 1 (k+1) f (s1 )dg(s1 ) f (s1 )dg(s1 ) − · f (x) − (g(b) − g(a)) a (g(b) − g(a)) a k=0 Z b Z b k Y · ··· P (g(x), g(s1 )) · P (g(si ), g(si+1 ))ds1 · · · dsk+1 a a i=1 ! Z b Z b n−1 Y ≤ kf (n) k∞ · |P (g(si ), g(si+1 ))|ds1 · · · dsn . ··· |P (g(x), g(s1 ))| · a
a
i=1
(2.29)
Theorem 2.19. All as 2.10. Additionally assume that
in2Theorem
∂ f
for all i, j = 1, . . . , k. γij :=
∂xi ∂xj < +∞, ∞ Then Z 1 Z 1Z 1 k X ∂f f (x) − (t t x)dt dt f (t x)dt − x t 1 2 1 2 1 1 j 1 ∂xj 0 0 0 j=1 k k 1 XX |xi | |xj | · γij . ≤ 6 i=1 j=1
(2.30)
Next we present Lp , p > 1, Ostrowski type results.
Theorem 2.20. All as in Theorem 2.4. Additionally assume that kf (n) kp < +∞, p > 1. Here p, q : p1 + 1q = 1. Then ! Z b Z b n−2 Y (n) |θ1,n | ≤ kf kp · ··· |P (x, s1 )| · |P (si , si+1 )| a
a
i=1
· kP (sn−1 , •)kq ds1 ds2 · · · dsn−1 .
(2.31)
0
Theorem 2.21. All as in Proposition 2.6. Additionally assume that kf kp < +∞, p > 1. Here p, q : p1 + q1 = 1. Then Z b 1 f (t)dg(t) ≤ kf 0 kp · kP (g(x), g(t))kq,t . (2.32) f (x) − (g(b) − g(a)) a Here k · kq,t means integration with respect to t. Theorem 2.22. All as in Proposition 2.8. Additionally assume that kf 00 kp < +∞, p > 1. Here p, q : 1p + q1 = 1. Then Z b 1 1 f (x) − f (t)dg(t) − (g(b) − g(a)) a (g(b) − g(a)) ! Z ! Z b b 0 · P (g(x), g(t))dt f (t1 )dg(t1 ) a
≤ kf 00 kp ·
a
Z
b
a
|P (g(x), g(t))| · kP (g(t), g(t1 )kq,t1 · dt.
(2.33)
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Theorem 2.23. All as in Theorem 2.8. Additionally suppose that kf (n) kp < +∞, p > 1. Here p, q : p1 + q1 = 1. Then Z b 1 1 f (x) − f (s1 )dg(s1 ) − (g(b) − g(a)) a (g(b) − g(a)) ! Z b Z b n−2 X Z b (k+1) · f (s1 )dg(s1 ) · ··· P (g(x), g(s1 )) k=0
a
a
! k Y · P (g(si ), g(si+1 )) · ds1 · · · dsk+1 i=1
≤ kf (n) kp ·
Z
b
a
···
Z
b
a
|P (g(x), g(s1 ))| ·
a
n−2 Y i=1
|P (g(si ), g(si+1 ))|
· kP (g(sn−1 ), g(sn ))kq,sn · ds1 ds2 · · · dsn−1 .
! (2.34)
Finally we present L1 Ostrowski type results. Theorem 2.24. All as in Theorem 2.4. Additionally suppose that kf (n) k1 < +∞. Then ! Z b Z b n−2 Y (n) |θ1,n | ≤ kf k1 · |P (si , si+1 )| |P (x, s1 )| · ··· a
a
i=1
· kP (sn−1 , •)k∞ · ds1 ds2 · · · dsn−1 .
(2.35)
The interested reader can consult with [10], which is a further study on this basic chapter. Theorem 2.25. All as in Proposition 2.6. Additionally assume that kf 0 k1 < +∞. Then Z b 1 f (t)dg(t) ≤ kf 0 k1 · kP (g(x), g(t))k∞,t . (2.36) f (x) − (g(b) − g(a)) a
Here k · k∞,t is taken with respect to t.
Theorem 2.26. All as in Proposition 2.8. Additionally assume that kf 00 k1 < +∞. Then Z b 1 1 f (x) − f (t)dg(t) − (g(b) − g(a)) a (g(b) − g(a)) ! ! Z b Z b 0 · P (g(x), g(t))dt · f (t1 )dg(t1 ) a
00
≤ kf k1 ·
a
Z
b
a
!
|P (g(x), g(t))| · kP (g(t), g(t1 ))k∞,t1 · dt .
(2.37)
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Finally we give Theorem 2.27. All as in Theorem 2.8. Additionally suppose that kf (n) k1 < +∞. Then Z b 1 1 f (x) − f (s1 )dg(s1 ) − · (g(b) − g(a)) a (g(b) − g(a)) ! Z b Z b n−2 X Z b (k+1) P (g(x), g(s1 )) ··· f (s1 )dg(s1 ) · · k=0
·
k Y
i=1
a
a
! P (g(si ), g(si+1 ))ds1 ds2 · · · dsk+1
≤ kf (n) k1 ·
Z
b
a
···
Z
b
a
|P (g(x), g(s1 ))| ·
a
n−2 Y i=1
·kP (g(sn−1 ), g(sn ))k∞,sn · ds1 · · · dsn−1 ) .
|P (g(si ), g(si+1 ))|
! (2.38)
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Chapter 3
Higher Order Ostrowski Inequalities
We present a generalization of Ostrowski inequality for higher order derivatives, by using a generalized Euler-type identity. Some of the produced inequalities are sharp, namely attained by basic functions. The rest of the estimates are tight. We give applications to trapezoidal and midpoint rules. Estimates are given with respect to L∞ norm. This treatement is based on [35].
3.1
Introduction
We mention as motivation to our work the great Ostrowski inequality [196]: " 2 # Z b x − a+b 1 1 2 (b − a)kf 0 k∞ , (3.1) + f (t)dt ≤ f (x) − b−a a 4 (b − a)2
where f ∈ C 0 ([a, b]), x ∈ [a, b], which is a sharp inequality, see [16]. Other motivations come from [10], [24], [16], [98] and [121]. We use here the sequence {Bk (t), k ≥ 0} of Bernoulli polynomials which is uniquely determined by the following identities: Bk0 (t) = kBk−1 (t),
k ≥ 1,
and
Bk (t + 1) − Bk (t) = ktk−1 ,
B0 (t) = 1 k ≥ 0.
The values Bk = Bk (0), k ≥ 0 are the known Bernoulli numbers. We need to mention 1 1 B0 (t) = 1, B1 (t) = t − , B2 (t) = t2 − t + , 2 6 3 1 1 B3 (t) = t3 − t2 + t, B4 (t) = t4 − 2t3 + t2 − , 2 2 30 5 1 5 t2 1 5 . B5 (t) = t5 − t4 + t3 − t, and B6 (t) = t6 − 3t5 + t4 − + 2 3 6 2 2 42 ∗ Let {Bk (t), k ≥ 0} be a sequence of periodic functions of period 1, related to Bernoulli polynomials as Bk∗ (t) = Bk (t),
Bk∗ (t + 1) = Bk∗ (t),
0 ≤ t < 1, 21
t ∈ R.
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We have that B0∗ (t) = 1, B1∗ is a discontinuous function with a jump of −1 at each integer, while Bk∗ , k ≥ 2 are continuous functions. Notice that Bk (0) = Bk (1) = Bk , k ≥ 2. We use the following result. Theorem 3.1. Let f : [a, b] → R be such that f (n−1) , n ≥ 1, is a continuous function and f (n) (x) exists and is finite for all but a countable set of x in (a, b) and that f (n) ∈ L1 ([a, b]). Then for every x ∈ [a, b] we have Z b n−1 X (b − a)k−1 x − a 1 Bk f (t)dt + [f (k−1) (b) − f (k−1) (a)] f (x) = b−a a k! b−a k=1 " # n−1 Z x−a (b − a) x − t + Bn − Bn∗ f (n) (t)dt. (3.2) n! b−a b−a [a,b]
The sum in (3.2) when n = 1 is zero.
Proof. By using Theorem 2 of [98], Exercise 18.41(d), p. 299 in [158] and Problem 14(c), p. 264 in [224]. And that f (n−1) as implied absolutely continuous it is also of bounded variation. If f (n−1) is just absolutely continuous then (3.2) is valid again. Formula (3.2) is a generalized Euler type identity, see also [171]. We set Z b 1 f (t)dt ∆n (x) := f (x) − b−a a n−1 X (b − a)k−1 x − a − [f (k−1) (b) − f (k−1) (a)], x ∈ [a, b]. (3.3) Bk k! b−a k=1
We have by (3.2) that Z (b − a)n−1 x−a x−t ∆n (x) = Bn − Bn∗ f (n) (t)dt. (3.4) n! b − a b − a [a,b] In this chapter we give sharp, namely attained, upper bounds for |∆4 (x)| and tight upper bounds for |∆n (x)|, n ≥ 5, x ∈ [a, b], with respect to L∞ norm. That is generalizing (3.1) for higher order derivatives. High computational difficulties in this direction prevent us for shoming sharpness for n ≥ 5 cases. 3.2
Main Results
We give Theorem 3.2. Let f : [a, b] → R be such that f (n−1) , n ≥ 1, is a continuous function and f (n) (x) exists and is finite for all but a countable set of x in (a, b) and that f (n) ∈ L∞ ([a, b]). Then for every x ∈ [a, b] we have ! Z b (b − a)n−1 x−a ∗ x−t |∆n (x)| ≤ − Bn Bn dt kf (n) k∞ . (3.5) n! b−a b−a a
Proof.
By Theorem 3.1.
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Performing the change of variable method in the integral of (3.5) we derive Corollary 3.3. All assumptions as in Theorem 3.2. Then for every x ∈ [a, b] we have Z 1 (b − a)n x − a (n) (3.6) |∆n (x)| ≤ Bn (t) − Bn b − a dt kf k∞ , n ≥ 1. n! 0
Note. Inequality (3.6) appeared first as Theorem 7, p. 350, in [98], wrongly under the sole assumption of f (n) ∈ L∞ ([a, b]). Also in the rest of [98] the complete assumptions of our Theorem 3.2 are missing, whenever it applies. Using Cauchy–Schwartz inequality we find that 2 !1/2 Z 1 Z 1 x−a Bn (t) − Bn x − a dt ≤ Bn (t) − Bn dt b−a b−a 0 0 s x−a (n!)2 2 |B2n | + Bn , n ≥ 1, (3.7) = (2n)! b−a the last comes by [98], p. 352. We give Corollary 3.4. All assumptions as in Theorem 3.2. Then for every x ∈ [a, b] we have s ! (n!)2 x − a (b − a)n kf (n) k∞ , n ≥ 1. (3.8) |B2n | + Bn2 |∆n (x)| ≤ n! (2n)! b−a Here we elaborate on (3.6) and (3.8). We introduce the parameter λ := We see that
x−a , b−a
a ≤ x ≤ b.
(3.9)
λ = 0 iff x = a, λ = 1 iff x = b, and λ=
1 2
iff x =
a+b . 2
Consider p4 (t) := B4 (t) − B4 (λ) = t4 − 2t3 + t2 − λ4 + 2λ3 − λ2 .
(3.10)
We need to compute I4 (λ) :=
Z
1 0
|p4 (t)|dt,
0 ≤ λ ≤ 1.
(3.11)
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Lemma 3.5. We find 14 3 1 16λ5 4 2 5 − 7λ + 3 λ − λ + 30 , I4 (λ) = 5 3 − 16λ + 9λ4 − 26λ + 3λ2 − 1 , 5 3 10
0 ≤ λ ≤ 1/2,
(3.12)
1/2 ≤ λ ≤ 1,
which is continuous in λ ∈ [0, 1]. We obtain
1 , I4 (0) = I4 (1) = 30 7 1 = I4 . 2 240
(3.13)
Proof. Here all calculations are done by Mathematica 4. The equation p4 (t) = 0 has four real roots, p 1 r1 = 1 − λ, r2 = λ, r3 = 1 − 1 + 4λ − 4λ2 2 and
r4 = We find the following orders:
p 1 1 + 1 + 4λ − 4λ2 . 2
i) if 0 ≤ λ ≤ 1/2, then r3 ≤ 0 ≤ r 2 ≤ r 1 ≤ 1 ≤ r 4 , and ii) if 1/2 ≤ λ ≤ 1, then r3 ≤ 0 ≤ r 1 ≤ r 2 ≤ 1 ≤ r 4 . So we have p4 (t) = (t − r1 )(t − r2 )(t − r3 )(t − r4 ), t ∈ [0, 1]. We easily derive that when 0 ≤ λ ≤ 1/2, p4 (t) is greater equal zero over [λ, 1 − λ] and smaller equal zero over [0, λ] and [1 − λ, 1]. Similarly when 1/2 ≤ λ ≤ 1 we get that p4 (t) ≥ 0 over [1 − λ, λ] and p4 (t) ≤ 0 over [0, 1 − λ] and [λ, 1]. Therefore when 0 ≤ λ ≤ 1/2 we have Z 0 Z 1−λ Z 1−λ I4 (λ) = p4 (t)dt + p4 (t)dt + p4 (t)dt, λ
λ
while when 1/2 ≤ λ ≤ 1 we have Z Z 0 p4 (t)dt + I4 (λ) = 1−λ
1
λ
p4 (t)dt + 1−λ
proving (3.12) after the computations are done.
Z
λ
p4 (t)dt, 1
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Using basic calculus we obtain Lemma 3.6.
1 3 5 min I4 (λ) = I4 = I4 = = 0.01953125, λ∈[0,1] 4 4 256
(3.14)
and max I4 (λ) = I4 (0) = I4 (1) =
λ∈[0,1]
1 = 0.033333 . 30
(3.15)
Consequently by Lemmas 3.5 and 3.6 we get Theorem 3.7. Assumptions as in Theorem 3.2, case of n = 4. For every x ∈ [a, b] it holds (b − a)4 I4 (λ)kf (4) k∞ , (3.16) |∆4 (x)| ≤ 24 where I4 (λ) is given by (3.12) with λ = x−a b−a . Furthermore we have that |∆4 (x)| ≤
(b − a)4 (4) kf k∞ , 720
∀x ∈ [a, b].
(3.17)
Optimality is achieved in Theorem 3.8. Assumptions as in Theorem 3.2, case of n = 4. Inequality (3.17) is sharp, namely it is attained when x = a, b by the functions (t − a)4 and (t − b)4 . Proof.
We have
∆4 (a) = ∆4 (b) =
Z b (b − a) 0 1 f (a) + f (b) − f (t)dt. (3.18) (f (b)−f 0 (a))− 2 12 b−a a
So by (3.17) we have
|∆4 (a)| = |∆4 (b)| ≤
(b − a)4 (4) kf k∞ . 720
(3.19)
Let f (t) = (t − a)4 or f (t) = (t − b)4 , then
(b − a)4 = R.H.S.(3.19), 30 proving that (3.19) is attained. That is proving (3.17) sharp. |∆4 (a)| = |∆4 (b)| =
The trapezoid and midpoint inequalities follow. Corollary 3.9. Assumptions as in Theorem 3.2, case of n = 4. It holds Z b f (a) + f (b) (b − a) 1 0 0 − (f (b) − f (a)) − f (t)dt 2 12 b−a a
(b − a)4 (4) kf k∞ , 720 the last inequality is attained by (t − a)4 and (t − b)4 , that is sharp. ≤
(3.20)
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Furthermore we obtain Z b a + b (b − a) 1 f (t)dt (f 0 (b) − f 0 (a)) − + f 2 24 b−a a ≤
7 (b − a)4 kf (4) k∞ . 5760
(3.21)
Remark 3.10. We do obtain the trapezoidal formula (f (a) + f (b)) (b − a) 0 − (f (b) − f 0 (a)) 2 12 Z b 1 (b − a)3 000 f (t)dt. (3.22) (f (b) − f 000 (a)) − + 720 b−a a
∆5 (a) = ∆5 (b) = ∆6 (a) = ∆6 (b) =
We also find the midpoint formula a+b (b − a) 0 a+b a+b ∆5 = ∆6 =f + (f (b) − f 0 (a)) 2 2 2 24 Z b 7(b − a)3 000 1 − (f (b) − f 000 (a)) − f (t)dt. 5760 b−a a
(3.23)
Using (3.8) and Mathematica 7 we get
Theorem 3.11. Assumptions as in Theorem 3.2, cases of n = 5, 6. It holds 5 (a)|, |∆ (b − a)5 ≤ √ kf (5) k∞ , (3.24) a + b 144 2310 ∆5 2
and
1 |∆6 (a)| ≤ 1440 with
r
101 (b − a)6 kf (6) k∞ , 30030
r 1 7081 a + b 6 (6) ≤ ∆6 46080 2145 (b − a) kf k∞ . 2
(3.25)
(3.26)
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Chapter 4
Multidimensional Euler Identity and Optimal Multidimensional Ostrowski Inequalities We develop and establish a general multivariate Euler type identity. Using it we derive general tight multivariate high order Ostrowski type inequalities for the estimate on the error of a multivariate function f evaluated at a point from its average. The estimates are involving only the single partial derivatives of f and are with respect to k · kp , 1 ≤ p ≤ ∞. We give specific applications of the main results to the multidimensional trapezoid and midpoint rules for functions f differentiable up to order 6. We show sharpness of the inequalities for differentiation orders m = 1, 2, 4 and with respect to k · k∞ . This treatment relies on [38]. 4.1
Introduction
We mention as motivation to this chapter the great Ostrowski inequality, see [196], [16], [21], [24]. Theorem 4.1. It holds # " Z b a+b 2 x − 1 1 2 f (x) − f (t)dt ≤ (b − a)kf 0 k∞ , + b−a a 4 (b − a)2
where f ∈ C 1 ([a, b]), x ∈ [a, b], which is a sharp inequality. The author proved also in [17], [21] the following result:
Theorem 4.2. Let f ∈ C n+1 ([a, b]), n ∈ N and x ∈ [a, b] be fixed, such that f (k) (x) = 0, k = 1, . . . , n. Then Z b 1 kf (n+1) k∞ (x − a)n+2 + (b − x)n+2 . · f (y)dy − f (x) ≤ b − a (n + 2)! b−a a
The last inequality is sharp. In particular, when n is odd it is attained by f ∗ (y) := (y − x)n+1 · (b − a), while when n is even the optimal function is f˜(y) := |y − x|n+α · (b − a),
Recently the author in [35], proved 27
α > 1.
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Theorem 4.3. Let f : [a, b] → R be such that f (n−1) , n ≥ 1, is a continuous function and f (n) (x) exists and is finite for all but a countable set of x in (a, b) and that f (n) ∈ L∞ ([a, b]). Then for every x ∈ [a, b] we have ! Z b (b − a)n−1 x − t x − a ∗ B n dt kf (n) k∞ , − Bn |∆n (x)| ≤ n! b − a b − a a
where
∆n (x) := f (x) − −
n−1 X k=1
1 b−a
Z
b
f (t)dt a
x−a (b − a)k−1 Bk [f (k−1) (b) − f (k−1) (a)], x ∈ [a, b], k! b−a
with Bk (t) the Bernoulli polynomial of degree k ≥ 1. Here {Bk∗ (t), k ≥ 0} is the sequence of periodic functions of period 1, related to Bernoulli polynomials as Bk∗ (t) = Bk (t), 0 ≤ t < 1, Bk∗ (t + 1) = Bk∗ (t), t ∈ R. We give also Corollary 4.4. All assumptions as in Theorem 4.3. Then for every x ∈ [a, b] we have Z 1 x − a (b − a)n (n) |∆n (x)| ≤ Bn (t) − Bn b − a dt kf k∞ , n ≥ 1. n! 0
Note. The last inequality appeared first as Theorem 7, p. 350, in [98], wrongly under the sole assumption of f (n) ∈ L∞ ([a, b]). Theorem 4.3 and Corollary 4.4 along with [35], [98] are the greatest motivations for the Euler identity method we use in this chapter. Further motivation is given by (see also [17], [21]) n Q 1 Theorem 4.5. Let f ∈ C [ai , bi ] , where ai < bi ; ai , bi ∈ R, i = 1, . . . , k, i=1
k Q − and let → x 0 := (x01 , . . . , x0k ) ∈ [ai , bi ] be fixed. Then i=1
Z b1 Z bi Z bk 1 → − ··· ··· f (z1 , . . . , zk )dz1 · · · dzk − f ( x 0 ) k Q ai ak (bi − ai ) a1 i=1
k X
(x0i − ai )2 + (bi − x0i )2
∂f . ≤
2(bi − ai ) ∂zi ∞ i=1
The last inequality is sharp, here the optimal function is f ∗ (z1 , . . . , zk ) :=
k X i=1
|zi − x0i |αi , αi > 1.
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Theorem 4.5 generalizes Theorem 4.1 to multidimension. We also mention (see [17], [21]) Theorem 4.6. Let Q be a compact and convex subset of Rk , k ≥ 1. Let f ∈ α − C n+1 (Q), n ∈ N and → x 0 ∈ Q be fixed such that all partial derivatives fα := ∂∂zαf , k P where α = (α1 , . . . , αk ), αi ∈ Z+ , i = 1, . . . , k, |α| = αi = j, j = 1, . . . , n fulfill i=1 − fα (→ x 0 ) = 0. ZThen Z 1 Dn+1 (f ) → − → − → − − − − f ( z )d z − f ( x 0 ) ≤ (k→ z −→ x 0 k`1 )n+1 d→ z, Vol(Q) (n + 1)!Vol(Q) Q Q where Dn+1 (f ) := max kfα k∞ α : |α|=n+1
and − − k→ z −→ x 0 k`1 := As a related result we give
k X i=1
|zi − x0i |.
Corollary 4.7. Under the assumptions of Theorem 4.6 we find that Z 1 → − → − → − f ( z )d z − f ( x ) 0 Vol(Q) Q "
n+1 # Z X k
∂ 1 −
|zi − x0i | · f d→ z. ≤
∂zi (n + 1)!Vol(Q) Q ∞ i=1 Furthermore, the last inequality is sharp : when n is odd it is attained by k X f ∗ (z1 , . . . , zk ) := (zi − x0i )n+1 , i=1
while when n is even the optimal function is k X f˜(z1 , . . . , zk ) := |zi − x0i |n+αi , αi > 1. i=1
Theorems 4.5, 4.6 and Corollary 4.7 motivate this chapter in the multivariate setting. Here we develop a very general Multivariate Euler type identity, see Theorems 4.14, 4.15, 4.18, 4.21. Based on this identity we prove some general tight multivariate Ostrowski type inequalities, see Theorems 4.32, 4.33, 4.34, 4.35, 4.60. Then we give lots of specific and important applications of Theorems 4.32, 4.33, 4.34, see Corollaries 4.36–4.59. There are produced many multivariate Ostrowski type inequalities for differentiation orders m = 1, . . . , 6, mostly related to multivariate trapezoid and midpoint rules. When we impose some basic and natural boundary conditions, then inequalities become very simple and elegant, see Corollaries 4.56 – 4.59. The surprising fact there is, that only a very small number of sets of boundary conditions is needed comparely to the higher order of differentiation of the involved functions. At the end we establish sharpness of the inequalities with respect to k · k∞ and for differentiation orders m = 1, 2, 4, see Corollaries 4.61–4.69.
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4.2
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Background
We will use Theorem 4.8 ([171, p. 17]). Let f ∈ C n ([a, b]), n ∈ N, x ∈ [a, b]. Then Z b 1 f (x) = f (t)dt + φn−1 (x) + Rn (x), b−a a where for m ∈ N we call m X (b − a)k−1 x−a [f (k−1) (b) − f (k−1) (a)] φm (x) := Bk k! b−a
(4.1)
k=1
with the convention φ0 (x) = 0, and Z (b − a)n−1 b ∗ x − t x−a Bn Rn (x) := − − Bn f (n) (t)dt. (4.2) n! b − a b − a a Here Bk (x), k ≥ 0, are the Bernoulli polynomials, Bk = Bk (0), k ≥ 0, the Bernoulli numbers, and Bk∗ (x), k ≥ 0, are periodic functions of period one, related to the Bernoulli polynomials as and
Bk∗ (x) = Bk (x), 0 ≤ x < 1,
(4.3)
Bk∗ (x + 1) = Bk∗ (x), x ∈ R.
(4.4)
Some basic properties of Bernoulli polynomials follow (see [1, 23.1]). We have 1 1 B0 (x) = 1, B1 (x) = x − , B2 (x) = x2 − x + , 2 6 and Bk0 (x) = kBk−1 (x), k ∈ N, B0∗
B1∗
Bk (x + 1) − Bk (x) = kx
k−1
(4.5) , k ≥ 0.
(4.6)
Clearly = 1, is a discontinuous function with a jump of −1 at each integer, and Bk∗ , k ≥ 2, is a continuous function. Notice that Bk (0) = Bk (1) = Bk , k ≥ 2. We need also the more general Theorem 4.9 (see [35]). Let f : [a, b] → R be such that f (n−1) , n ≥ 1, is a continuous function and f (n) (x) exists and is finite for all but a countable set of x in (a, b) and that f (n) ∈ L1 ([a, b]). Then for every x ∈ [a, b] we have Z b n−1 X (b − a)k−1 x − a 1 f (x) = [f (k−1) (b) − f (k−1) (a)] Bk f (t)dt + b−a a k! b−a k=1 Z x−a (b − a)n−1 ∗ x−t + Bn − Bn f (n) (t)dt. (4.7) n! b−a b−a [a,b]
The sum in (4.7) when n = 1 is zero. If f (n−1) is just absolutely continuous then (4.7) is valid again. Formula (4.7) is a generalized Euler type identity, see also [98].
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4.3
31
Main Results
We make `
∂ f Assumption 4.10. Let f and the existing ∂x ` , all ` = 1, . . . , m; j = 1, . . . , n, be j n Q continuous real valued functions on [ai , bi ]; m, n ∈ N, ai , bi ∈ R. i=1
We give
Proposition 4.11. All as in Assumption 4.10 for m = n = 2, xi ∈ [ai , bi ], i = 1, 2. Then ! Z b1 Z b 2 1 f (s1 , s2 )ds1 ds2 + T2 (x2 ) + T1 (x1 , x2 ), f (x1 , x2 ) = (b1 − a1 )(b2 − a2 ) a2 a1 (4.8) where ! Z b1 1 x2 − a 2 T2 := T2 (x2 ) := B1 (f (s1 , b2 ) − f (s1 , a2 ))ds1 (b1 − a1 ) b2 − a 2 a1 # ! 2 Z b 1 Z b2 " (b2 − a2 ) x2 − a 2 ∂ f ∗ x2 − s 2 + B2 − B2 (s1 , s2 ) ds1 ds2 , 2(b1 − a1 ) a1 a2 b2 − a 2 b2 − a 2 ∂x22 (4.9) and T1 := T1 (x1 , x2 ) := B1 (b1 − a1 ) + 2
Z
b1 a1
x1 − a 1 (f (b1 , x2 ) − f (a1 , x2 )) b1 − a 1 ! 2 x1 − a 1 x − s ∂ f 1 1 − B2∗ B2 (s1 , x2 )ds1 . b1 − a 1 b1 − a 1 ∂x21 (4.10)
By Theorem 4.8 we have Z b1 x1 − a 1 1 f (s1 , x2 )ds1 + B1 (f (b1 , x2 ) − f (a1 , x2 )) f (x1 , x2 ) = b 1 − a 1 a1 b1 − a 1 ! 2 Z x1 − a 1 (b1 − a1 ) b1 ∂ f ∗ x1 − s 1 + B2 − B2 (s1 , x2 )ds1 2 b1 − a 1 b1 − a 1 ∂x21 a1 Z b1 1 f (s1 , x2 )ds1 + T1 (x1 , x2 ). (4.11) = b 1 − a 1 a1 And also we obtain Z b2 1 x2 − a 2 f (s1 , x2 ) = f (s1 , s2 )ds2 + B1 (f (s1 , b2 ) − f (s1 , a2 )) b 2 − a 2 a2 b2 − a 2 ! 2 Z (b2 − a2 ) b2 x2 − a 2 x − s ∂ f 2 2 + − B2∗ B2 (s1 , s2 )ds2 . 2 b − a b − a ∂x22 2 2 2 2 a2
Proof.
(4.12)
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Thus we derive
Z b1 ( Z b2 1 1 f (s1 , s2 )ds2 f (x1 , x2 ) = b 1 − a 1 a1 b 2 − a 2 a2 x2 − a 2 + B1 (f (s1 , b2 ) − f (s1 , a2 )) b2 − a 2 Z x2 − a 2 (b2 − a2 ) b2 B2 + 2 b2 − a 2 a2 ) ! 2 ∂ f ∗ x2 − s 2 − B2 (s1 , s2 )ds2 ds1 + T1 (x1 , x2 ), b2 − a 2 ∂x22
proving the claim.
(4.13)
We continue with Proposition 4.12. All as in Assumption 4.10, case n = 3 and m ∈ N, xi ∈ [ai , bi ], i = 1, 2, 3. Then Z b1 Z b2 Z b3 1 f s1 , s2 , s3 ds1 ds2 ds3 f (x1 , x2 , x3 ) = 3 Q (bi − ai ) a1 a2 a3 i=1
+ T3 (x3 ) + T2 (x2 , x3 ) + T1 (x1 , x2 , x3 ),
(4.14)
where m−1 X (b3 − a3 )k−1 x3 − a3 1 T3 := T3 (x3 ) := Bk (b1 − a1 )(b2 − a2 ) k! b3 − a 3 k=1 ! Z b1 Z b 2 ∂ k−1 f ∂ k−1 f × (s1 , s2 , b3 ) − s1 , s2 , a 3 ds1 ds2 ∂x3k−1 ∂x3k−1 a2 a1 Z b1 Z b2 Z b3 x3 − a 3 (b3 − a3 )m−1 Bm + m!(b1 − a1 )(b2 − a2 ) a1 a2 a3 b3 − a 3 ! x3 − s 3 ∂mf ∗ −Bm (s1 , s2 , s3 )ds1 ds2 ds3 , (4.15) b3 − a 3 ∂xm 3
T2 :=
m−1 X (b2 − a2 )k−1 x2 − a2 Z b1 ∂ k−1 f 1 Bk (s1 , b2 , x3 ) (b1 − a1 ) k! b2 − a 2 ∂x2k−1 a1 k=1 ! Z Z b2 (b2 − a2 )m−1 b1 ∂ k−1 f x2 − a 2 (s1 , a2 , x3 ) ds1 + − Bm m!(b1 − a1 ) a1 b2 − a 2 ∂x2k−1 a2 ! ! m x2 − s 2 ∂ f ∗ − Bm (4.16) (s1 , s2 , x3 )ds2 ds1 , b2 − a 2 ∂xm 2
T2 (x2 , x3 ) :=
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and x1 − a 1 (b1 − a1 )k−1 T1 := T1 (x1 , x2 , x3 ) := Bk k! b1 − a 1 k=1 ! ∂ k−1 f ∂ k−1 f (b1 , x2 , x3 ) − (a1 , x2 , x3 ) ∂x1k−1 ∂x1k−1 ! m Z x1 − s 1 ∂ f x1 − a 1 (b1 − a1 )m−1 b1 ∗ Bm − Bm (s1 , x2 , x3 )ds1 . + m! b1 − a 1 b1 − a 1 ∂xm a1 1 m−1 X
(4.17) When m = 1 then the sums in (4.15)–(4.17) are zero. Proof.
Applying Theorem 4.8 we have 1 f (x1 , x2 , x3 ) = b1 − a 1
Z
b1
f (s1 , x2 , x3 )ds1 + T1 (x1 , x2 , x3 ).
(4.18)
a1
Furthermore we find Z b2 1 f (s1 , s2 , x3 )ds2 b 2 − a 2 a2 ! m−1 X (b2 − a2 )k−1 x2 − a2 ∂ k−1 f ∂ k−1 f Bk (s1 , b2 , x3 ) − (s1 , a2 , x3 ) + k! b2 − a 2 ∂x2k−1 ∂x2k−1 k=1 ! m Z x − s ∂ f x2 − a 2 (b2 − a2 )m−1 b2 2 2 ∗ Bm − Bm (s1 , s2 , x3 )ds2 + m! b − a b − a ∂xm 2 2 2 2 a2 2 f (s1 , x2 , x3 ) =
(4.19) and Z b3 1 f (s1 , s2 , s3 )ds3 b 3 − a 3 a3 ! m−1 X (b3 − a3 )k−1 x3 − a3 ∂ k−1 f ∂ k−1 f + (s1 , s2 , b3 ) − (s1 , s2 , a3 ) Bk k! b3 − a 3 ∂x3k−1 ∂x3k−1 k=1 ! m Z (b3 − a3 )m−1 b3 x3 − a 3 x − s ∂ f 3 3 ∗ + − Bm Bm (s1 , s2 , s3 )ds3 . m! b − a b − a ∂xm 3 3 3 3 a3 3 f (s1 , s2 , x3 ) =
(4.20)
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Putting the above together we derive Z b1 ( Z b2 1 1 f (s1 , s2 , x3 )ds2 f (x1 , x2 , x3 ) = b 1 − a 1 a1 b 2 − a 2 a2 m−1 X (b2 − a2 )k−1 x2 − a2 ∂ k−1 f ∂ k−1 f + (s1 , b2 , x3 ) − (s1 , a2 , x3 ) Bk k! b2 − a 2 ∂x2k−1 ∂x2k−1 k=1 Z x2 − a 2 (b2 − a2 )m−1 b2 Bm + m! b2 − a 2 a2 ) ! m x − s ∂ f 2 2 ∗ − Bm (4.21) (s1 , s2 , x3 )ds2 ds1 + T1 b2 − a 2 ∂xm 2 Z b 1 Z b2 1 = f (s1 , s2 , x3 )ds2 ds1 (b1 − a1 )(b2 − a2 ) a1 a2 m−1 X (b2 − a2 )k−1 x2 − a2 Z b1 ∂ k−1 f 1 Bk (s , b , x3 )ds1 + k−1 1 2 (b1 − a1 ) k! b2 − a 2 a1 ∂x2 k=1 ! Z Z b2 Z b1 k−1 (b2 − a2 )m−1 b1 x2 − a 2 ∂ f B + (s , a , x )ds − m 3 1 k−1 1 2 m!(b1 − a1 ) a1 b2 − a 2 a2 a1 ∂x2 ! ! m x2 − s 2 ∂ f ∗ − Bm (4.22) (s1 , s2 , x3 )ds2 ds1 + T1 . b2 − a 2 ∂xm 2 So far we have f (x1 , x2 , x3 ) =
1 (b1 − a1 )(b2 − a2 )
Z
b1 a1
Z
b2
f (s1 , s2 , x3 )ds2 ds1 + T2 (x2 , x3 ) + T1 (x1 , x2 , x3 ). a2
(4.23)
Finally we conclude that f (x1 , x2 , x3 )
Z
"
Z b3 1 f (s1 , s2 , s3 )ds3 b 3 − a 3 a3 a1 a2 m−1 X (b3 − a3 )k−1 x3 − a3 ∂ k−1 f ∂ k−1 f Bk (s , s , b ) − (s , s , a ) + 1 2 3 1 2 3 k! b3 − a 3 ∂x3k−1 ∂x3k−1 k=1 Z (b3 − a3 )m−1 b3 x3 − a 3 + Bm m! b3 − a 3 a3 # ! m x3 − s 3 ∂ f ∗ −Bm (4.24) (s1 , s2 , s3 )ds3 ds2 ds1 + T2 + T1 b3 − a 3 ∂xm 3
1 = (b1 − a1 )(b2 − a2 )
b1
Z
b2
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=
1 3 Q
i=1
(bi − ai )
Z
Z
b1 a1
b2 a2
Z
35
b3
f (s1 , s2 , s3 )ds1 ds2 ds3 + T3 + T2 + T1 ,
(4.25)
a3
proving the claim.
We further present Proposition 4.13. All as in Assumption 4.10, case of n = 4 and m ∈ N, xi ∈ [ai , bi ], i = 1, . . . , 4. Then 1
f (x1 , x2 , x3 , x4 ) =
+
4 Q
(bi − ai )
i=1 4 X
Z
4 Q
f (s1 , s2 , s3 , s4 )ds1 ds2 ds3 ds4 [ai ,bi ]
i=1
Tj .
(4.26)
j=1
Here 1
T4 := T4 (x4 ) := Z +
i=1
3 Q
[ai ,bi ]
i=1
(bi − ai )
∗ Bm
k=1
x4 − a 4 (b4 − a4 )k−1 Bk k! b3 − a 4
! ∂ k−1 f ∂ k−1 f (s1 , s2 , s3 , b4 ) − (s1 , s2 , s3 , a4 ) ds1 ds2 ds3 ∂x4k−1 ∂x4k−1
(b4 − a4 )m−1 3 Q (bi − ai )m!
i=1
−
3 Q
m−1 X
x4 − s 4 b4 − a 4
Z
4 Q
i=1
!
Bm [ai ,bi ]
x4 − a 4 b4 − a 4
! ∂ mf (s1 , s2 , s3 , s4 ) ds1 ds2 ds3 ds4 , ∂xm 4
(4.27)
m−1 X (b3 − a3 )k−1 x3 − a3 1 Bk T3 := T3 (x3 , x4 ) := (b1 − a1 )(b2 − a2 ) k! b3 − a 3 k=1 ! Z b1 Z b2 k−1 ∂ f ∂ k−1 f × (s1 , s2 , b3 , x4 ) − (s1 , s2 , a3 , x4 ) ds1 ds2 ∂x3k−1 ∂x3k−1 a1 a2 Z b1 Z b2 Z b3 (b3 − a3 )m−1 x3 − a 3 + Bm m!(b1 − a1 )(b2 − a2 ) a1 a2 a3 b3 − a 3 ! ! ! m x3 − s 3 ∂ f ∗ − Bm (4.28) (s1 , s2 , s3 , x4 ) ds1 ds2 ds3 , b3 − a 3 ∂xm 3
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T2 := T2 (x2 , x3 , x4 )
and
m−1 X (b2 − a2 )k−1 x2 − a2 1 := Bk (b1 − a1 ) k! b2 − a 2 k=1 !! Z b1 k−1 ∂ f ∂ k−1 (s1 , b2 , x3 , x4 ) − f (s1 , a2 , x3 , x4 ) ds1 × ∂x2k−1 ∂x2k−1 a1 Z b1 Z b 2 (b2 − a2 )m−1 x2 − a 2 + Bm m!(b1 − a1 ) b2 − a 2 a1 a2 ! ! ! m ∂ f x2 − s 2 ∗ (4.29) (s1 , s2 , x3 , x4 ) ds1 ds2 , − Bm b2 − a 2 ∂xm 2
T1 := T1 (x1 , x2 , x3 , x4 ) :=
k−1 (b1 − a1 )k−1 ∂ f x1 − a 1 (b , x2 , x3 , x4 ) Bk k−1 1 k! b1 − a 1 ∂x 1 k=1 Z ∂ k−1 f x1 − a 1 (b1 − a1 )m−1 b1 − k−1 (a1 , x2 , x3 , x4 ) + Bm m! b1 − a 1 ∂x1 a1 ! m ∂ f x1 − s 1 ∗ (s1 , x2 , x3 , x4 )ds1 . (4.30) − Bm b1 − a 1 ∂xm 1
m−1 X
When m = 1 then the sums in (4.27)–(4.30) are zero. Proof.
By Theorem 4.8 we have
Z b1 1 f (s1 , x2 , x3 , x4 )ds1 b 1 − a 1 a1 m−1 X (b1 − a1 )k−1 x1 − a1 ∂ k−1 f + Bk (b1 , x2 , x3 , x4 ) k! b1 − a 1 ∂x1k−1 k=1 Z (b1 − a1 )m−1 b1 x1 − a 1 ∂ k−1 f Bm (a1 , x2 , x3 , x4 ) + − m! b1 − a 1 ∂x1k−1 a1 ! x1 − s 1 ∂mf ∗ − Bm (s1 , x2 , x3 , x4 )ds1 b1 − a 1 ∂xm 1 Z b1 1 f (s1 , x2 , x3 , x4 )ds1 + T1 (x1 , x2 , x3 , x4 ). = (b1 − a1 ) a1 Similarly we obtain Z b2 1 f (s1 , s2 , x3 , x4 )ds2 f (s1 , x2 , x3 , x4 ) = b 2 − a 2 a2 m−1 X (b2 − a2 )k−1 x2 − a2 ∂ k−1 f + Bk (s1 , b2 , x3 , x4 ) k! b2 − a 2 ∂x2k−1 k=1 f (x1 , x2 , x3 , x4 ) =
(4.31) (4.32)
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Z x2 − a 2 (b2 − a2 )m−1 b2 ∂ k−1 f (s , a , x , x ) + B 1 2 3 4 m m! b2 − a 2 ∂x2k−1 a2 ! m x2 − s 2 ∂ f ∗ − Bm (s1 , s2 , x3 , x4 )ds2 , b2 − a 2 ∂xm 2 Z b3 1 f (s1 , s2 , s3 , x4 )ds3 f (s1 , s2 , x3 , x4 ) = b 3 − a 3 a3 m−1 X (b3 − a3 )k−1 x3 − a3 ∂ k−1 f + Bk (s1 , s2 , b3 , x4 ) k! b3 − a 3 ∂x3k−1 k=1 Z x3 − a 3 (b3 − a3 )m−1 b3 ∂ k−1 f B (s , s , a , x ) + − m 1 2 3 4 m! b3 − a 3 ∂x3k−1 a3 ! m x3 − s 3 ∂ f ∗ − Bm (s1 , s2 , s3 , x4 )ds3 , b3 − a 3 ∂xm 3
37
−
(4.33)
(4.34)
and finally Z b4 1 f (s1 , s2 , s3 , s4 )ds4 f (s1 , s2 , s3 , x4 ) = b 4 − a 4 a4 m−1 X (b4 − a4 )k−1 x4 − a4 ∂ k−1 f (s1 , s2 , s3 , b4 ) + Bk k! b4 − a 4 ∂x4k−1 k=1 Z ∂ k−1 f (b4 − a4 )m−1 b4 x4 − a 4 − (s , s , s , a ) + B 1 2 3 4 m m! b4 − a 4 ∂x4k−1 a4 ! ∂mf x4 − s 4 ∗ (s1 , s2 , s3 , s4 )ds4 . − Bm b4 − a 4 ∂xm 4
(4.35)
Consequently, by using the above we derive f (x1 , x2 , x3 , x4 ) Z b1 " Z b2 1 1 = f (s1 , s2 , x3 , x4 )ds2 (b1 − a1 ) a1 (b2 − a2 ) a2 m−1 X (b2 − a2 )k−1 x2 − a2 ∂ k−1 f + Bk (s1 , b2 , x3 , x4 ) k! b2 − a 2 ∂x2k−1 k=1 Z (b2 − a2 )m−1 b2 ∂ k−1 f x2 − a 2 − k−1 (s1 , a2 , x3 , x4 ) + Bm m! b2 − a 2 ∂x2 a2 # ! m x2 − s 2 ∂ f ∗ −Bm (s1 , s2 , x3 , x4 )ds2 ds1 + T1 b2 − a 2 ∂xm 2
(4.36)
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1 = (b1 − a1 )(b2 − a2 )
Z
b1 a1
Z
b2
f (s1 , s2 , x3 , x4 )ds1 ds2 a2
m−1 X (b2 − a2 )k−1 x2 − a2 Z b1 ∂ k−1 f 1 + Bk (s1 , b2 , x3 , x4 ) (b1 − a1 ) k! b2 − a 2 ∂x2k−1 a1 k=1 Z Z x2 − a 2 (b2 − a2 )m−1 b1 b2 ∂ k−1 f (s1 , a2 , x3 , x4 )ds1 + Bm − m!(b1 − a1 ) a1 a2 b2 − a 2 ∂x2k−1 ! x2 − s 2 ∂mf ∗ − Bm (s1 , s2 , x3 , x4 )ds1 ds2 + T1 (4.37) b2 − a 2 ∂xm 2 Z b 1 Z b2 ( Z b3 1 1 = f (s1 , s2 , s3 , x4 )ds3 (b1 − a1 )(b2 − a2 ) a1 a2 b3 − a3 a3 m−1 X (b3 − a3 )k−1 x3 − a3 ∂ k−1 f (s1 , s2 , b3 , x4 ) Bk + k! b3 − a 3 ∂x3k−1 k=1 Z b3 x3 − a 3 ∂ k−1 f (b3 − a3 )m−1 − Bm (s1 , s2 , a3 , x3 ) + m! b3 − a 3 ∂x3k−1 a3 ! ! ! ∂mf x3 − s 3 ∗ (s1 , s2 , s3 , x4 )ds3 ds1 ds2 + T2 + T1 . − Bm (4.38) b3 − a 3 ∂xm 3 That is, we found that Z 3 X 1 f (s , s , s , x )ds ds ds + Tj . (4.39) f (x1 , x2 , x3 , x4 ) = 3 1 2 3 4 1 2 3 3 Q Q j=1 (bi − ai ) i=1[ai ,bi ] i=1
At last we observe that
f (x1 , x2 , x3 , x4 ) Z b4 Z 1 1 f (s1 , s2 , s3 , s4 )ds4 = 3 3 Q Q [ai ,bi ] b4 − a4 a4 (bi − ai ) i=1 i=1 m−1 X
k−1 (b4 − a4 )k−1 x4 − a 4 f ∂ ∂ k−1 f + Bk (s1 , s2 , s3 , b4 ) − (s1 , s2 , s3 , a4 ) k! b4 − a 4 ∂x4k−1 ∂x4k−1 k=1 Z x4 − a 4 (b4 − a4 )m−1 b4 Bm + m! b4 − a 4 a4 ! ! m 3 X ∂ f x4 − s 4 ∗ −Bm (s , s , s , s )ds ds ds ds + Ti (4.40) 1 2 3 4 4 1 2 3 b4 − a 4 ∂xm 4 i=1 Z 4 X 1 f (s , s , s , s )ds ds ds ds + Tj , (4.41) = 4 1 2 3 4 1 2 3 4 4 Q Q j=1 (bi − ai ) i=1[ai ,bi ] i=1
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proving the claim.
We finally give the following general Multivariate Euler-type identity in Theorem 4.14. All as in Assumption 4.10 for m, n ∈ N, xi ∈ [ai , bi ], i = 1, 2, . . . , n. Then f (x1 , x2 , . . . , xn ) = Q n
i=1
1 (bi − ai )
Z
n Q
[ai ,bi ]
f (s1 , s2 , . . . , sn )ds1 ds2 · · · dsn +
i=1
n X
Tj ,
j=1
(4.42)
where for j = 1, . . . , n we have
Tj := Tj (xj , xj+1 , . . . , xn ) :=
×
Z
j−1 Q i=1
[ai ,bi ]
1 j−1 Q i=1
(bi − ai )
(m−1 X (bj − aj )k−1 xj − aj Bk k! bj − a j k=1
∂ k−1 f (s1 , s2 , . . . , sj−1 , bj , xj+1 , . . . , xn ) ∂xjk−1
!) ∂ k−1 f − k−1 (s1 , s2 , . . . , sj−1 , aj , xj+1 , . . . , xn ) ds1 · · · dsj−1 ∂xj (Z xj − a j (bj − aj )m−1 B + m j Q j−1 bj − a j Q [ai ,bi ] m! (bi − ai ) i=1 i=1 ! ) ! m ∂ f xj − s j ∗ −Bm (s1 , s2 , . . . , sj , xj+1 , . . . , xn ) ds1 ds2 · · · dsj . bj − a j ∂xm j (4.43) When m = 1 then the sum in (4.43) is zero. Proof. Theorem 4.14 is true for n = 3, 4, see Propositions 4.12, 4.13. Assume that Theorem 4.14, is true for n, see (4.42), (4.43). Then we prove it for n+1, n ∈ N. We have Z 1 f (s1 , . . . , sn , xn+1 )ds1 , . . . dsn f (x1 , x2 , . . . , xn , xn+1 ) = Q n n Q [a ,b ] i i (bi − ai ) i=1 +
i=1 n X j=1
where
Tj , for j = 1, . . . , n,
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Tj := Tj xj , xj+1 , . . . , xn , xn+1 =
Bk
xj − a j bj − a j
! Z
j−1 Q i=1
[ai ,bi ]
1 (j−1) Q i=1
(bi − ai )
(m−1 X (bj − aj )k−1 k! k=1
∂ k−1 f (s1 , . . . , sj−1 , bj , xj+1 , . . . , xn , xn+1 ) ∂xjk−1 !) !!
∂ k−1 f ds1 . . . dsj−1 s1 , . . . , sj−1 , aj , xj+1 , . . . , xn , xn+1 ∂xjk−1 (Z xj − a j (bj − aj )m−1 + Bm j Q j−1 bj − a j Q [ai ,bi ] m! (bi − ai ) i=1 i=1 ) !! ! m ∂ f xj − s j ∗ ds1 . . . dsj . s1 , . . . , sj , xj+1 , . . . , xn , xn+1 − Bm bj − a j ∂xm j −
But it holds 1 f (s1 . . . , sn , xn+1 ) = (bn+1 − an+1 )
Z
bn+1
f (s1 , . . . , sn , sn+1 )dsn+1 an+1
! xn+1 − an+1 (bn+1 − an+1 )k−1 + Bk k! bn+1 − an+1 k=1 # " ∂ k−1 f (s1 , . . . sn , bn+1 ) ∂ k−1 f − s , . . . , s , a 1 n n+1 k−1 k−1 ∂xn+1 ∂xn+1 ! m−1 Z b " n+1 bn+1 − an+1 xn+1 − an+1 Bm + m! bn+1 − an+1 an+1 !# ∂ m f (s1 , . . . , sn , sn+1 ) xn+1 − sn+1 ∗ dsn+1 . − Bm bn+1 − an+1 ∂xm n+1 m−1 X
Thus we get f (x1 , x2 , . . . , xn , xn+1 ) ( Z bn+1 Z 1 1 = Q f (s1 , . . . , sn , sn+1 )dsn+1 n n Q (bn+1 − an+1 ) an+1 (bi − ai ) i=1[ai ,bi ] i=1
+ "
m−1 X k=1
(bn+1 − an+1 )k−1 xn+1 − an+1 Bk k! bn+1 − an+1
∂ k−1 f (s1 , . . . , sn , bn+1 ) ∂ k−1 f (s1 , . . . , sn , an+1 ) − k−1 k−1 ∂xn+1 ∂xn+1
#
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" Z xn+1 − an+1 (bn+1 − an+1 )m−1 bn+1 Bm + m! bn+1 − an+1 an+1 ) # m n X xn+1 − sn+1 ∂ f (s1 , . . . , sn , sn+1 ) ∗ −Bm ds . . . , ds + Tj . ds 1 n n+1 bn+1 − an+1 ∂xm n+1 j=1 Hence f (x1 , . . . , xn+1 ) =
1 n+1 Q i=1
(bi − ai )
Z
n+1 Q
f (s1 , . . . , sn+1 ) ds1 . . . dsn+1 + [ai ,bi ]
n+1 X
Tj ,
j=1
i=1
where Tn+1 (xn+1 ) := Tn+1 := Q n
(Z
n Q
i=1
[ai ,bi ]
"
i=1
∂
1 (bi − ai )
(m−1 X (bn+1 − an+1 )k−1 k!
k=1
Bk
xn+1 − an+1 bn+1 − an+1
)) # f (s1 , . . . , sn , bn+1 ) ∂ k−1 f − k−1 s1 , . . . , sn , an+1 ds1 . . . dsn k−1 ∂xn+1 ∂xn+1 (Z "" (bn+1 − an+1 )m−1 xn+1 − an+1 n+1 n + Bm Q Q bn+1 − an+1 [ai ,bi ] m! bi − a i i=1 i=1 ) # m # xn+1 − sn+1 ∂ f ∗ −Bm s1 , . . . , sn , sn+1 ds1 . . . dsn+1 . bn+1 − an+1 ∂xm n+1
k−1
Thus is proving the claim.
Next we rewrite the last theorem. Theorem 4.15. All as in Assumption 4.10 for m, n ∈ N, xi ∈ [ai , bi ], i = 1, 2, . . . , n. Then
f Em (x1 , x2 , . . . , xn ) := f (x1 , x2 , . . . , xn ) Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ]
−
i=1 n X j=1
Aj =
n X j=1
Bj ,
(4.44)
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where for j = 1, . . . , n we have 1
Aj := Aj (xj , xj+1 , . . . , xn ) =
×
Z
j−1 Q i=1
[ai ,bi ]
j−1 Q i=1
(bi − ai )
(m−1 X (bj − aj )k−1 xj − aj Bk k! bj − a j k=1
∂ k−1 f (s1 , s2 , . . . , sj−1 , bj , xj+1 , . . . , xn ) ∂xjk−1
!) ∂ k−1 f , − (s1 , s2 , . . . , sj−1 , aj , xj+1 , . . . , xn ) ds1 · · · dsj−1 ∂xjk−1
(4.45)
and
(Z (bj − aj )m−1 xj − a j Bj := Bj (xj , xj+1 , . . . , xn ) := B m j Q j−1 bj − a j Q [ai ,bi ] m! (bi − ai ) i=1 i=1 ) ! ! m xj − s j ∂ f ∗ − Bm (s1 , s2 , . . . , sj , xj+1 , . . . , xn ) ds1 ds2 · · · dsj . bj − a j ∂xm j (4.46)
When m = 1 then Aj = 0, and Tj = Bj , j = 1, . . . , n. Remark 4.16. Notice above that Tj = Aj + Bj , j = 1, . . . , n. Also we have that f |Em (x1 , x2 , . . . , xn )| ≤
Also by denoting ∆ := f (x1 , . . . , xn ) − Q n
1
i=1
(bi − ai )
Z
n Q
[ai ,bi ]
n X j=1
|Bj |.
f (s1 , . . . , sn )ds1 · · · dsn
(4.47)
(4.48)
i=1
we get
|∆| ≤
n X j=1
(|Aj | + |Bj |).
(4.49)
Later we will estimate Aj , Bj . A general set of suppositions follow
Assumption 4.17. Here m ∈ N, j = 1, . . . , n.We suppose 1) f :
n Q
i=1
2)
∂`f ∂x`j
[ai , bi ] → R is continuous.
are existing real valued functions for all j = 1, . . . , n; ` = 1, . . . , m − 2.
3) For each j = 1, . . . , n we assume that continuous real valued function.
∂ m−1 f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xjm−1
is a
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m
4) For each j = 1, . . . , n we assume that gj (·) := ∂∂xmf (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) j exists and is real valued with the possibility of being infinite only over an at most countable subset of (aj , bj ). 5) Parts #3, #4 are true for all (x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=1 i6=j
6) The functions for j = 2, . . . , n; ` = 1, . . . , m − 2, j−1 j−1 ` z }| { z }| { ∂ f ·, ·, ·, · · · , ·, xj , xj+1 , . . . , xn qj ·, ·, · · · , · := ∂x`j are continuous on
j−1 Q i=1
[ai , bi ], for each (xj , xj+1 , . . . , xn ) ∈
n Q
[ai , bi ].
i=j
7) The functions for each j = 1, . . . , n, j j m }| { z }| { z ∂ f ·, ·, ·, · · · , ·, xj+1 , . . . , xn ∈ L1 ϕj ·, ·, ·, · · · , ·, ·, · := ∂xm j for any (xj+1 , . . . , xn ) ∈ We present
n Q
j Y
i=1
!
[ai , bi ] ,
[ai , bi ].
i=j+1
Theorem 4.18. All as in Assumption 4.17. Then (4.42) and (4.44) are true again. Proof. Use of Assumption 4.17, Theorem 4.9 and it is similar to the proof of the so far results of this chapter. Remark 4.19. The results of Propositions 4.11, 4.12, 4.13 are still valid under Assumption 4.17 for their cases. Some weaker general suppositions follow. Assumption 4.20. Here m ∈ N, j = 1, . . . , n and only the Parts #1, #2, #6, #7 of Assumption 4.17 remain the same. We further suppose that for each j = 1, . . . , n and over [aj , bj ], the function ∂ m−1 f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xjm−1 is absolutely continuous, and this is true for all (x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈ We give
n Y
i=1 i6=j
[ai , bi ].
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Theorem 4.21. All as in Assumption 4.20. Then (4.42) and (4.44) are still true. Remark 4.22. The results of Propositions 4.11, 4.12, 4.13 are again valid under Assumption 4.20 for their cases. We proceed with the estimates of our remainders in (4.42) and (4.44). We need to make Remark 4.23. We are operating under the Assumptions 4.10 or 4.17 or 4.20. We observe for j = 1, . . . , n that |Bj | ≤ Γj ,
(4.50)
where Γj :=
Z xj − sj (bj − aj )m−1 xj − a j ∗ − B B m j m Q j−1 bj − a j bj − a j Q [ai ,bi ] m! (bi − ai ) i=1 i=1 ! ∂mf × m (s1 , s2 , . . . , sj , xj+1 , . . . , xn ) ds1 · · · dsj . (4.51) ∂xj
Here we assume
j ∂ m f z}|{ · · · , xj+1 , . . . , xn ∈ L∞ ∂xm j
for any (xj+1 , . . . , xn ) ∈ Thus we obtain
n Q
i=j+1
j Y
i=1
[ai , bi ]
!
[ai , bi ], any xj ∈ [aj , bj ].
! Z (bj − aj )m−1 xj − sj xj − a j ∗ Γj ≤ j Bn bj − aj − Bm bj − aj ds1 · · · dsj Q j−1 Q [ai ,bi ] m! (bi − ai ) i=1
i=1
m
j
∂ f z }| {
·, ·, ·, · · · , ·, xj+1 , . . . , xn × (4.52)
∂xm
j
j
Q ∞,
[ai ,bi ]
i=1
! Z bj x − a x − s (bj − aj )m−1 j j j j ∗ Bm dsj − Bm = m! bj − a j bj − a j aj
m
j
∂ f z }| {
× m · · · , · · ·, xj+1 , . . . , xn
j
∂xj
Q ∞,
(4.53)
[ai ,bi ]
i=1
(by letting
λj := λj (xj ) :=
xj − a j bj − a j
(4.54)
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we have) (bj − aj )m = m!
m Z 1 j
∂ f z}|{
|Bm (λj ) − Bm (tj )|dtj m · · · , xj+1 , . . . , xn ·
0
∂xj
45
(4.55) ∞,
j Q
[ai ,bi ]
i=1
(bj − aj )m sm!
m
j Z 1
∂ f z}|{
· · · , xj+1 , . . . , xn (Bm (λj ) − Bm (tj ))2 dtj ·
∂xm
0
j
≤
(4.56) ∞,
(using [98], p. 352 we get) (bj − aj )m = m! s !
m
j 2
z}|{ xj − a j (m!) 2
∂ f · · · , xj+1 , . . . , xn |B2m | + Bm ·
m
(2m)! bj − a j
∂xj
j Q
[ai ,bi ]
i=1
. ∞,
j Q
[ai ,bi ]
i=1
(4.57)
We have established Lemma 4.24. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Bj as in (4.46). In particular suppose that ! j j Y ∂ m f z}|{ · · · , xj+1 , . . . , xn ∈ L∞ [ai , bi ] , ∂xm j i=1
for any (xj+1 , . . . , xn ) (xj , xj+1 , . . . , xn ) ∈
n Q
∈
n Q
[ai , bi ], all j
= 1, . . . , n.
Then for any
i=j+1
[ai , bi ] we have
i=j
(bj − aj )m |Bj | = |Bj (xj , xj+1 , . . . , xn )| ≤ m!
m
j
∂ f z}|{
× m · · · , xj+1 , . . . , xn
j
∂xj
Q ∞,
s
(m!)2 xj − a j 2 |B2m | + Bm (2m)! bj − a j ,
!
(4.58)
[ai ,bi ]
i=1
for all j = 1, . . . , n. We continue with
Remark 4.25. Continuing from Remark 4.23. Let pj , qj > 1: j = 1, . . . , n, with the assumption that ! j Y ∂mf (· · · , xj+1 , . . . , xn ) ∈ Lqj [ai , bi ] , ∂xm j i=1
1 pj
+
1 qj
= 1;
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for any (xj+1 , . . . , xn ) ∈
n Q
i=j+1
[ai , bi ], xj ∈ [aj , bj ]. From (4.50) and (4.51) we get Z
(bj − aj )m−1 |Bj | ≤ j−1 Q (bi − ai ) m! i=1
−
∗ Bm
Z ×
xj − a j j Bm bj − aj Q [ai ,bi ]
i=1
!1/pj p xj − sj j ds1 · · · dsj bj − a j
1/qj m qj ∂ f j m (s1 , . . . , sj , xj+1 , . . . , xn ) ds1 · · · dsj Q [ai ,bi ] ∂xj
(4.59)
i=1
1/pj Z bj j−1 Y (bj − aj )m−1 Bm (λj ) = (bi − ai ) j−1 Q a j m! (bi − ai ) i=1 i=1
∗ − Bm
=
!1/pj
p
∂mf xj − sj j
(4.60) dsj j
∂xm (· · · , xj+1 , . . . , xn ) Q bj − a j j qj , [ai ,bi ] i=1
(bj − aj ) m!
m− q1
j
pj
− Bm (tj )| dtj We have derived
j−1 Y
(bi − ai )
i=1 1/pj
!−1/qj Z
1
0
|Bm (λj )
m
∂ f (· · · , xj+1 , . . . , xn )
∂xm j
qj ,
j Q
.
(4.61)
[ai ,bi ]
i=1
Lemma 4.26. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Bj as in (4.46) and pj , qj > 1 : p1j + q1j = 1; j = 1, . . . , n. In particular we suppose that ! j Y ∂mf (· · · , xj+1 , . . . , xn ) ∈ Lqj [ai , bi ] , ∂xm j i=1 for any (xj+1 , . . . , xn ) ∈ (xj , xj+1 , . . . , xn ) ∈
n Q
n Q
[ai , bi ], for all j = 1, . . . , n.
Then for any
i=j+1
[ai , bi ] we have
i=j
|Bj | = |Bj (xj , xj+1 , . . . , xn )| (bj − aj ) ≤ m!
m− q1
j
pj − Bm (tj ) dtj
j−1 Y
(bi − ai )
i=1 1/pj
!−1/qj
Z 1 B m xj − a j bj − a j 0
m
∂ f (· · · , xj+1 , . . . , xn )
∂xm
j
qj ,
j Q
i=1
[ai ,bi ]
.
(4.62)
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When pj = qj = 2, then we obtain !−1/2 1 j−1 (bj − aj )m− 2 Y |Bj | ≤ (bi − ai ) m! i=1
∂mf
× m (· · · , xj+1 , . . . , xn ) j
∂xj
Q
s
! (m!)2 x − a j j 2 |B2m | + Bm (2m)! bj − a j
.
(4.63)
[ai ,bi ]
2,
i=1
The inequalities (4.62), (4.63) are true for all j = 1, . . . , n. At last we make Remark 4.27. Continuing from Remark 4.25. Assume for j = 1, . . . , n that ! j Y ∂mf [ai , bi ] , (· · · , xj+1 , . . . , xn ) ∈ L1 ∂xm j i=1 Q for any (xj+1 , . . . , xn ) ∈ ni=j+1 [ai , bi ], any xj ∈ [aj , bj ]. Then from (4.50) and (4.51) we obtain |Bj | ≤
(bj − aj )m−1 j−1 Q m! (bi − ai ) i=1
Z
j Q
[ai ,bi ]
i=1
! m ∂ f ∂xm (s1 , . . . , sj , xj+1 , . . . , xn ) ds1 · · · dsj j
xj − a j xj − s j ∗
· B m − Bm bj − a j bj − aj ∞,[aj ,bj ]
(bj − aj )m−1 xj − a j xj − s j
∗ = − Bm
B m Qj−1 bj − a j bj − a j m! i=1 (bi − ai ) ∞,[aj ,bj ]
∂ mf
× m (· · · , xj+1 , . . . , xn ) j
Q
∂xj 1,
(4.64)
(4.65)
[ai ,bi ]
i=1
(bj − aj )m−1
Bm (t) − Bm xj − aj
j−1 bj − aj ∞,[0,1] Q m! (bi − ai ) i=1
∂ mf
. (4.66) × m (· · · , xj+1 , . . . , xn ) j
∂xj
Q
(by [98], p. 347) =
[ai ,bi ]
1,
i=1
From [98], pp. 347–348 we have: i) case m = 2r, r ∈ N, then
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xj − a j
= B2r (t) − B2r bj − aj ∞,[0,1] ∞,[0,1] xj − aj ; (4.67) = (1 − 2−2r )|B2r | + 2−2r B2r − B2r bj − a j
Bm (t) − Bm xj − aj
bj − a j
ii) case m = 2r + 1, r ∈ N, then
Bm (t) − Bm xj − aj
bj − a j
xj − a j
= B2r+1 (t) − B2r+1 bj − aj ∞,[0,1] ∞,[0,1] 2(2r + 1)! B2r+1 xj − aj ; (4.68) + ≤ 2r+1 −2r (2π) (1 − 2 ) bj − a j
iii) special case of m = 1, then
Bm (t) − Bm xj − aj
B1 (t) − B1 xj − aj =
bj − a j bj − aj ∞,[0,1] ∞,[0,1] aj + bj 1 = + xj − . 2 2
(4.69)
We have proved
Lemma 4.28. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Bj as in (4.46), j = 1, . . . , n. In particular we suppose for j = 1, . . . , n that ! j Y ∂mf [ai , bi ] , (. . . , xj+1 , . . . , xn ) ∈ L1 ∂xm j i=1 for any (xj+1 , . . . , xn ) ∈ we have
n Q
i=j+1
[ai , bi ]. Then for any (xj , xj+1 , . . . , xn ) ∈
n Q
[ai , bi ]
i=j
m
(bj − aj )m−1
∂ f (. . . , xj+1 , . . . , xn ) j
m
Q j−1 ∂xj Q 1, [ai ,bi ] m! (bi − ai ) i=1 i=1
xj − a j
. (4.70) × B (t) − B m m
bj − aj ∞,[0,1]
|Bj | = |Bj (xj , xj+1 , . . . , xn )| ≤
The special cases follow: 1) When m = 2r, r ∈ N we have |Bj | ≤
2r
(bj − aj )2r−1
∂ f (. . . , xj+1 , . . . , xn ) j 2r
Q
j−1 ∂xj Q 1, [ai ,bi ] (2r)! (bi − ai ) i=1 i=1 xj − aj . × (1 − 2−2r )|B2r | + 2−2r B2r − B2r bj − a j
(4.71)
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2) When m = 2r + 1, r ∈ N we obtain
∂ 2r+1 f
(bj − aj )2r
|Bj | ≤ (. . . , x , . . . , x )
j+1 n 2r+1 j−1 j
Q Q ∂xj 1, [ai ,bi ] (2r + 1)! (bi − ai ) i=1 i=1 2(2r + 1)! xj − aj + B2r+1 . (4.72) × (2π)2r+1 (1 − 2−2r ) bj − a j
3) When m = 1 we get |Bj | ≤
1 j−1 Q i=1
(bi − ai )
∂f
∂xj (. . . , xj+1 , . . . , xn )
1,
j Q
[ai ,bi ]
"
i=1
# aj + bj 1 + xj − . 2 2
(4.73)
We need θ Q − − [ai , bi ], θ ∈ N, where k→ x k := Definition 4.29. Let → x = (x1 , . . . , xθ ) ∈
p
i=1
x21 + · · · + x2θ . Let F :
lus of continuity of F by
θ Q
i=1
[ai , bi ] → R be continuous. We define the (first) modu-
ω1 (F, δ) :=
sup − − all → x ,→ y∈
θ Q
[ai ,bi ],
− − |F (→ x ) − F (→ y )|,
(4.74)
i=1
− − with k→ x −→ y k≤δ
for all δ > 0. Notice 4.30. Under Assumption 4.10 we have valid that ∂ k−1 f ∂ k−1 f (s , . . . , s , a , x , . . . , x ) k−1 (s1 , . . . , sj−1 , bj , xj+1 , . . . , xn ) − 1 j−1 j j+1 n ∂xj ∂xjk−1 j ∂ k−1 f z}|{ · · · , xj+1 , . . . , xn , bj − aj , all j = 1, . . . , n; k = 1, . . . , m − 1. ≤ ω1 ∂xjk−1
(4.75)
We give Lemma 4.31. Suppose Assumption 4.10. Let Aj as in (4.45), j = 1, . . . , n. Then for any (xj , xj+1 , . . . , xn ) ∈
n Y i=j
[ai , bi ]
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we have |Aj | = |Aj (xj , xj+1 , . . . , xn )| ≤ · ω1
m−1 X k=1
xj − aj (bj − aj )k−1 B k b j − a j k!
∂ f z}|{ · · · , xj+1 , . . . , xn ), bj − aj , for all j = 1, . . . , n. (4.76) ∂xjk−1 j
k−1
Putting together all these above auxilliary results, we derive the following multivariate Ostrowski type inequalities.
f Theorem 4.32. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Em (x1 , x2 , . . . , xn ) as in (4.44) and Aj for j = 1, . . . , n as in (4.45), m ∈ N. In particular we suppose that ! j j Y ∂ m f z}|{ [ai , bi ] , · · · , xj+1 , . . . , xn ∈ L∞ ∂xm j i=1
for any (xj+1 , . . . , xn ) ∈
n Q
[ai , bi ], all j = 1, . . . , n. Then
i=j+1
f |Em (x1 , . . . , xn )| = f (x1 , . . . , xn ) − Q n
1
Z
n X
Aj
f (s1 , . . . , sn )ds1 · · · dsn − n Q j=1 (bi − ai ) i=1[ai ,bi ] i=1 s " ! n 1 X (m!)2 xj − a j m 2 ≤ (bj − aj ) |B2m | + Bm m! j=1 (2m)! bj − a j #
m j
∂ f z}|{
· · · , x , . . . , x ) . (4.77) × j+1 n j
∂xm Q j
∞,
[ai ,bi ]
i=1
Proof.
By Theorems 4.15, 4.18, 4.21 and Lemma 4.24.
Next comes f Theorem 4.33. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Em (x1 , . . . , xn ) 1 1 as in (4.44), m ∈ N. Let pj , qj > 1 : pj + qj = 1; j = 1, . . . , n. In particular we assume that ! j Y ∂mf (. . . , xj+1 , . . . , xn ) ∈ Lqj [ai , bi ] , ∂xm j i=1
for any (xj+1 , . . . , xn ) ∈
n Q
i=j+1
[ai , bi ], for all j = 1, . . . , n. Then
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" j−1 − q1 Z 1 n X Y j xj − a j 1 m− q1 f j (bj − aj ) |Em (x1 , . . . , xn )| ≤ (bi − ai ) B m b j − a j m! j=1 0 i=1 #
pj 1/pj m
∂ f
. −Bm (tj ) dtj j
∂xm (. . . , xj+1 , . . . , xn ) Q j qj , [ai ,bi ] i=1
(4.78)
When pj = qj = 2, all j = 1, . . . , n, then
f |Em (x1 , . . . , xn )|
" −1/2 j−1 n Y 1 1 X (bi − ai ) (bj − aj )m− 2 ≤ m! j=1 i=1 s ! x − a (m!)2 j j 2 × |B2m | + Bm (2m)! bj − a j #
m
∂ f
. × m (. . . , xj+1 , . . . , xn ) Q j ∂xj 2, [ai ,bi ]
(4.79)
i=1
Proof.
By Theorems 4.15, 4.18, 4.21 and Lemma 4.26.
Finally we have f Theorem 4.34. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Em (x1 , . . . , xn ) as in (4.44), m ∈ N. In particular we assume for j = 1, . . . , n that
∂mf (. . . , xj+1 , . . . , xn ) ∈ L1 ∂xm j for any (xj+1 , . . . , xn ) ∈
f |Em (x1 , . . . , xn )|
Qn
i=j+1 [ai , bi ].
j Y
i=1
!
[ai , bi ] ,
Then
" n 1 X (bj − aj )m−1 ≤ m! j=1 j−1 Q (bi − ai ) i=1
#
m
∂ f
xj − a j
. j
∂xm (. . . , xj+1 , . . . , xn ) Q
Bm (t) − Bm bj − aj j 1, [ai ,bi ] ∞,[0,1] i=1
(4.80)
The special cases are calculated and estimated further as follows: 1) When m = 2r, r ∈ N, then
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f |E2r (x1 , . . . , xn )| ( n 1 X (bj − aj )2r−1 ≤ (2r)! j=1 j−1 Q (bi − ai ) i=1
!
2r
∂ f
j
∂x2r (. . . , xj+1 , . . . , xn ) Q j 1, [ai ,bi ] i=1
) −2r xj − aj −2r × (1 − 2 )|B2r | + 2 B2r − B2r . bj − a j
(4.81)
2) When m = 2r + 1, r ∈ N, then
f |E2r+1 (x1 , . . . , xn )| ( n X (bj − aj )2r 1 ≤ j−1 (2r + 1)! j=1 Q (bi − ai ) i=1
!
2r+1
∂
f
j
∂x2r+1 (. . . , xj+1 , . . . , xn ) Q 1, [ai ,bi ] j i=1
) x − a 2(2r + 1)! j j . + B2r+1 × (2π)2r+1 (1 − 2−2r ) bj − a j
(4.82)
And at last 3) When m = 1, then |E1f (x1 , . . . , xn )|
≤
n X j=1
(
j−1 Q i=1
(bi − ai )
"
∂f
(. . . , x , . . . , x ) j+1 n
∂xj
#) 1 aj + bj × + xj − . 2 2
Proof.
1
By Theorems 4.15, 4.18, 4.21 and Lemma 4.28.
1,
j Q
i=1
[ai ,bi ]
#
(4.83)
The final general Ostrowski type estimate follows: Theorem 4.35. All as in Assumption 4.10, m, n ∈ N, xi ∈ [ai , bi ], i = 1, 2, . . . , n. Here Bj is as in (4.46), for j = 1, . . . , n. Then Z 1 f (x1 , . . . , xn ) − f (s , s , . . . , s )ds ds · · · ds 1 2 n 1 2 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 " n n m−1 X X X (bj − aj )k−1 B k ≤ |Bj | + k! j=1 j=1 k=1 k−1 # ∂ f xj − aj ω1 (. . . , xj+1 , . . . , xn ), bj − aj . (4.84) bj − a j ∂xjk−1
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Estimates for Bj are given by Lemmas 4.24, 4.26, 4.28. Proof. (4.49). 4.4
By Theorems 4.14, 4.15 and Lemma 4.31, see also Remark 4.16, (4.48),
Applications
Here we apply Theorems 4.32, 4.33, 4.34. We give Corollary 4.36. Suppose Assumptions 4.10 jor 4.17 or4.20, case m = 1. Q ∂f ([ai , bi ]) , for any (xj+1 , . . . , xn ) ∈ (. . . , xj+1 , . . . , xn ) ∈ L∞ i) Assume ∂x j n Q
i=1
[ai , bi ], all j = 1, . . . , n. Then
i=j+1
Z 1 f (x1 , x2 , . . . , xn ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1 s ! " 2 n X
1 1 xj − a j
(bj − aj ) + − ≤
12 bj − a j 2 j=1 #
∂f
× (. . . , xj+1 , . . . , xn ) Q . j ∂xj ∞, [ai ,bi ]
(4.85)
i=1
ii) Let pj , qj > 1 :
1 pj
+
1 qj
= 1; j = 1, . . . , n. In particular we suppose that ! j Y ∂f [ai , bi ] , (. . . , xj+1 , . . . , xn ) ∈ Lqj ∂xj i=1
for any (xj+1 , . . . , xn ) ∈
n Q
[ai , bi ], for all j = 1, . . . , n. Then
i=j+1
Z 1 f (x1 , x2 , . . . , xn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [a ,b ] i i (bi − ai ) i=1 i=1 " pj 1/pj j−1 − q1 Z 1 n X Y xj − a j j 1− q1 j ≤ (bj − aj ) (bi − ai ) bj − aj − tj dtj 0 j=1 i=1 #
∂f
. (4.86) × j
∂xj (. . . , xj+1 , . . . , xn ) Q qj , [ai ,bi ] i=1
When pj = qj = 2, all j = 1, . . . , n, then
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Z 1 f (x1 , . . . , xn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [a ,b ] i i (bi − ai ) i=1 i=1 v v ! u " 2 ! u n q uj−1 Y X u1 1 x − a j j t + (bj − aj ) t (bi − ai ) − ≤ 12 bj − a j 2 i=1 j=1 #
∂f
(. . . , xj+1 , . . . , xn ) . (4.87) × j
Q ∂xj 2, [ai ,bi ] i=1
iii) Here assume for j = 1, . . . , n that
∂f (. . . , xj+1 , . . . , xn ) ∈ L1 ∂xj for any (xj+1 , . . . , xn ) ∈
n Q
j Y
!
[ai , bi ] ,
i=1
[ai , bi ]. Then
i=j+1
Z 1 f (x1 , . . . , xn ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1 # " (
n X
∂f
1
(. . . , xj+1 , . . . , xn ) ≤ j
Q j−1 ∂xj Q 1, [ai ,bi ] j=1 (bi − ai ) i=1 i=1 ) 1 aj + bj × + xj − . 2 2
(4.88)
Notice 4.37. We have for j = 1, . . . , n:
xj − a j =0 bj − a j = 1 1 = 2
λj :=
iff
xj = aj ,
λj
iff
x j = bj , aj + b j xj = . 2
λj
iff
(4.89)
We continue with Corollaries to Corollary 4.36. Corollary 4.38. Suppose Assumptions 4.10 or 4.17 or 4.20, Case m = 1, all xj = aj , j = 1, . . . , n. i) Assume ! j Y ∂f [ai , bi ] , all j = 1, . . . , n. (. . . , aj+1 , . . . , an ) ∈ L∞ ∂xj i=1
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Then Z 1 f (a1 , a2 , . . . , an ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 ) √ ( n
∂f
3 X
≤ . (bj − aj ) (. . . , a , . . . , a ) j+1 n j
∂xj
Q 3 j=1 [ai ,bi ] ∞,
(4.90)
i=1
ii) Let pj , qj > 1 :
1 pj
+
1 qj
= 1; j = 1, . . . , n. In particular we suppose that ! j Y ∂f [ai , bi ] , (. . . , aj+1 , . . . , an ) ∈ Lqj ∂xj i=1
for all j = 1, . . . , n. Then Z 1 f (a1 , . . . , an ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1
≤
j−1 1 " (bj − aj )1− qj Q (bi − ai )−1/qj n X i=1
(pj + 1)1/pj #
∂f
× . j
∂xj (. . . , aj+1 , . . . , an ) Q qj , [ai ,bi ] j=1
(4.91)
i=1
When pj = qj = 2, all j = 1, . . . , n, then Z 1 f (a1 , . . . , an ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 v " √ u n q uj−1 Y 3 X ≤ (bj − aj ) t (bi − ai ) 3 j=1 i=1 #!
∂f
. × j
∂xj (. . . , aj+1 , . . . , an ) Q [ai ,bi ] 2, i=1
iii) Here assume for j = 1, . . . , n, that
∂f (. . . , aj+1 , . . . , an ) ∈ L1 ∂xj Then
j Y
i=1
!
[ai , bi ] .
(4.92)
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Z 1 f (a1 , . . . , an ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [a ,b ] i i (bi − ai ) i=1 i=1 ) (
n
1 X (bj − aj + 1)
∂f (. . . , aj+1 , . . . , an ) . ≤ j
Q ∂xj 2 j=1 j−1 Q [ai ,bi ] 1, (bi − ai ) i=1
(4.93)
i=1
Corollary 4.39. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 1, all xj = bj , j = 1, . . . , n. i) Assume j Y
∂f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂xj
i=1
!
[ai , bi ] ,
for all j = 1, . . . , n. Then Z 1 f (b1 , . . . , bn ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1 # √ n "
∂f
3X
≤ (bj − aj ) . j
∂xj (. . . , bj+1 , . . . , bn ) Q 3 j=1 ∞, [ai ,bi ]
(4.94)
i=1
ii) Let pj , qj > 1 :
1 pj
+
1 qj
= 1; j = 1, . . . , n. In particular we assume that j Y
∂f (. . . , bj+1 , . . . , bn ) ∈ Lqj ∂xj
i=1
!
[ai , bi ] ,
for all j = 1, . . . , n. Then Z 1 f (b1 , . . . , bn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 " j−1 − q1 n Y X j 1− 1 (bi − ai ) ≤ (pj + 1)−1/pj (bj − aj ) qj i=1
j=1
∂f
(. . . , b , . . . , b ) j+1 n
∂xj
qj ,
j Q
i=1
[ai ,bi ]
#
.
(4.95)
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When pj = qj = 2, all j = 1, . . . , n, then Z 1 f (b1 , . . . , bn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 v # √ n "
u uj−1 Y
∂f
3X p t
(bi − ai ) bj − a j ≤ . j
∂xj (. . . , bj+1 , . . . , bn ) Q 3 j=1 2, [ai ,bi ] i=1 i=1
(4.96)
iii) Here suppose for j = 1, . . . , n that ∂f (. . . , bj+1 , . . . , bn ) ∈ L1 ∂xj Then
j Y
!
[ai , bi ] .
i=1
Z 1 f (b1 , . . . , bn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [a ,b ] i i (bi − ai ) i=1 i=1 ( )
n
1 X (1 + bj − aj )
∂f (. . . , bj+1 , . . . , bn ) . ≤ j
Q ∂xj 2 j=1 j−1 Q [ai ,bi ] 1, (bi − ai ) i=1
(4.97)
i=1
Next come the multivariate midpoint rule inequalities.
Corollary 4.40. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 1, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ! j Y aj+1 + bj+1 ∂f an + b n ..., [ai , bi ] , all j = 1, . . . , n. ∈ L∞ ,..., ∂xj 2 2 i=1
Then Z a1 + b 1 a2 + b 2 1 an + b n f f (s , . . . , s )ds · · · ds − , , . . . , 1 n 1 n n n Q Q 2 2 2 [ai ,bi ] (b − a ) i i i=1 i=1
n X
1 (bj − aj ) ∂f . . . , aj+1 + bj+1 , . . . , an + bn (. 4.98) ≤ √ j
∂xj
Q 2 2 2 3 j=1 ∞, [ai ,bi ] i=1
ii) Let pj , qj > 1 :
1 pj
+
1 qj
= 1; j = 1, . . . , n. In particular we suppose that ! j Y aj+1 + bj+1 an + b n ∂f [ai , bi ] , ..., ,..., ∈ L qj ∂xj 2 2 i=1
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for all j = 1, . . . , n. Then Z a1 + b 1 1 an + b n f − f (s , . . . , s )ds · · · ds , . . . , 1 n 1 n n n Q Q 2 2 [ai ,bi ] (b − a ) i i i=1 i=1
!− q1j j−1 Y 1 − p1 1− q1 ≤ (bi − ai ) (pj+1 ) j (bj − aj ) j 2 j=1 i=1 #
∂f a + b a + b n n j+1 j+1
. × ,..., j
Q
∂xj . . . , 2 2 [ai ,bi ] qj , n X
"
(4.99)
i=1
When pj = qj = 2, all j = 1, . . . , n, then Z a1 + b 1 1 an + b n f − , . . . , f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q 2 2 [a ,b ] i i (bi − ai ) i=1 i=1 v " u n uj−1 Y 1 X p t √ (bi − ai ) bj − a j ≤ 2 3 j=1 i=1 #
∂f
a + b a + b j+1 j+1 n n
,..., × . (4.100) j
∂xj . . . ,
Q 2 2 2, [ai ,bi ] i=1
iii) Here assume for j = 1, . . . , n that ∂f aj+1 + bj+1 an + b n ∈ L1 ..., ,..., ∂xj 2 2
j Y
i=1
!
[ai , bi ] .
Then Z a1 + b 1 an + b n 1 f , . . . , f (s , . . . , s )ds · · · ds − 1 n 1 n n n Q Q 2 2 [ai ,bi ] (b − a ) i i i=1 i=1 ) (
n
∂f aj+1 + bj+1 1X an + b n 1
. . . . , ≤ , . . . , j
Q ∂xj 2 j=1 j−1 2 2 Q [ai ,bi ] 1, (bi − ai ) i=1 i=1
(4.101)
Next we treat the case of m = 2 and only for the norms k · k∞ , k · k2 , and specifically for λj = 0, 1, 12 , j = 1, . . . , n. The multivariate trapezoid rule estimates follow immediately. Corollary 4.41. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 2, all xj = aj , j = 1, . . . , n.
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i) Assume ∂2f (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x2j
j Y
i=1
!
[ai , bi ] ,
all j = 1, . . . , n. Then f (a , a , . . . , a ) + f (b , a , . . . , a ) 1 2 n 1 2 n K2 := 2 Z 1 f (s1 , . . . , sn )ds1 · · · dsn − Q n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( n Z 1 1 X f (s1 , s2 , . . . , sj−1 , bj , aj+1 , . . . , an ) + Q j−1 2 j=2 j−1 Q [ai ,bi ] (bi − ai ) i=1 i=1 ) − f (s1 , . . . , sj−1 , aj , aj+1 , . . . , an ) ds1 · · · dsj−1 ( n
2
) X
1 ∂ f
≤ √ (bi − aj )2 . (4.102) j
∂x2 (. . . , aj+1 , . . . , an ) Q 2 30 j=1 j [ai ,bi ] ∞, i=1
ii) Suppose
∂ 2f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x2j
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then K2 ( n
) j−1 −1/2 2 Y X
∂ f
1 3/2
. (bi − ai ) (bj − aj ) ≤ √ j
∂x2 (. . . , aj+1 , . . . , an ) Q 2 30 j=1 j 2, [ai ,bi ] i=1 i=1
(4.103)
We continue with trapezoid rule estimates.
Corollary 4.42. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 2, all xj = bj , j = 1, . . . , n. i) Assume ∂2f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x2j
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then Λ 2 f (b , . . . , b ) + f (a , b , . . . , b ) 1 n 1 2 n := − Q n 2 −
( n " 1 X
2
j=2
1
j−1 Q i=1
(bi − ai )
(Z
i=1
j−1 Q
1 (bi − ai )
Z
n Q
[ai ,bi ]
f (s1 , . . . , sn )ds1 · · · dsn
i=1
f (s1 , . . . , sj−1 , bj , bj+1 , . . . , bn ) [ai ,bi ]
i=1
#) − f (s1 , . . . , sj−1 , aj , bj+1 , . . . , bn ) ds1 · · · dsj−1
2
n X
∂ f 1 . (bj − aj )2 (. . . , bj+1 , . . . , bn ) ≤ √ j 2
Q ∂xj 2 30 j=1 ∞, [ai ,bi ]
(4.104)
i=1
ii) Assume
! j Y ∂2f [ai , bi ] , (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x2j i=1 all j = 1, . . . , n. Then ( n " j−1 −1/2 X Y 1 3/2 Λ2 ≤ √ (bj − aj ) (bi − ai ) 2 30 j=1 i=1
#)
∂2f
. × 2 (. . . , bj+1 , . . . , bn j
Q
∂xj 2,
(4.105)
[ai ,bi ]
i=1
The multivariate midpoint rule estimates follow.
Corollary 4.43. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 2, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ! j Y ∂2f aj+1 + bj+1 an + b n [ai , bi ] , ∈ L∞ ..., ,..., ∂x2j 2 2 i=1 all j = 1, .. . , n. Then a +b an + b n 1 1 ,..., M2 := f 2 2 Z 1 f (s , . . . , s )ds · · · ds − Q 1 n 1 n n n Q (bi − ai ) i=1[ai ,bi ] i=1
#) ( n "
∂2f
X 1 a + b a + b
j+1 j+1 n n ≤ √ (bj − aj )2 2 . . . , . ,...,
j
∂xj
Q 2 2 8 5 j=1 ∞, [a ,b ] i
i
i=1
(4.106)
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ii) Suppose aj+1 + bj+1 an + b n ∂ 2f ∈ L2 . . . , , . . . , ∂x2j 2 2 all j = 1, . . . , n. Then
j Y
61
!
[ai , bi ] ,
i=1
( n " j−1 −1/2 2 Y X
∂ f 1
(bi − ai ) M2 ≤ √ (bj − aj )3/2
∂x2 8 5 j=1 j i=1 #) an + b n aj+1 + bj+1
,..., . ..., j
Q 2 2 2, [ai ,bi ]
(4.107)
i=1
We continue with trapezoid and midpoint rules inequalities for m = 3. Corollary 4.44. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 3, all xj = aj , j = 1, . . . , n. i) Assume ! j Y ∂3f [ai , bi ] , (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x3j i=1 all j = 1, . . . , n. Then f (a , . . . , a ) + f (b , a , . . . , a ) 1 n 1 2 n K3 := 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( X 2 n X (bj − aj )k−1 1 Bk (0) − j−1 k! Q j=1 k=1 (bi − ai ) with j=k6=1 ×
Z
j−1 Q i=1
[ai ,bi ]
i=1
∂ k−1 f (s1 , . . . , sj−1 , bj , aj+1 , . . . , an ) ∂xjk−1
!) ∂ k−1 f − (s1 , . . . , sj−1 , aj , aj+1 , . . . , an ) ds1 · · · dsj−1 ∂xjk−1
3
n X
∂ f 1
√ (bj − aj )3 . ≤ j
∂x3 (. . . , an+1 , . . . , an ) Q 12 210 j=1 j [ai ,bi ] ∞, i=1
(4.108)
ii) Assume ∂ 3f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x3j
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then
" −1/2 3 j−1 n Y X
∂ f 1 5/2
(bi − ai ) K3 ≤ √ (bj − aj )
∂x3 12 210 j=1 j i=1
. (. . . , aj+1 , . . . , an ) j
Q 2,
(4.109)
[ai ,bi ]
i=1
Corollary 4.45. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 3, all xj = bj , j = 1, . . . , n. i) Assume ! j Y ∂3f [ai , bi ] , (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x3j i=1 all j = 1, . . . , n. Then f (b , . . . , b ) + f (a , b , . . . , b ) 1 n 1 2 n Λ3 := 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( 2 ( n X (bj − aj )k−1 X 1 Bk (1) − j−1 k! Q j=1 k=1 (bi − ai ) with j=k6=1 ×
Z
j−1 Q i=1
[ai ,bi ]
i=1
∂ k−1 f (s1 , s2 , . . . , sj−1 , bj , bj+1 , . . . , bn ) ∂xjk−1
!)) ∂ k−1 f − (s , s , . . . , s , a , b , . . . , b ) ds · · · ds 1 2 j−1 j j+1 n 1 j−1 ∂xjk−1
3
n X
1 . (4.110) (bj − aj )3 ∂ f (. . . , bj+1 , . . . , bn ) √ ≤ j 3
Q ∂xj 12 210 j=1 ∞, [ai ,bi ] i=1
ii) Suppose
∂3f (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x3j all j = 1, . . . , n. Then
j Y
!
[ai , bi ] ,
i=1
" j−1 −1/2 3 n Y X
∂ f 1 5/2
(bi − ai ) Λ3 ≤ √ (bj − aj )
∂x3 12 210 j=1 j i=1
. (. . . , bj+1 , . . . , bn ) j
Q 2,
[ai ,bi ]
i=1
(4.111)
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Corollary 4.46. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 3, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ! j Y ∂3f aj+1 + bj+1 an + b n ..., ,..., ∈ L∞ [ai , bi ] , ∂x3j 2 2 i=1
all j = 1, . . . , n. Then a +b an + b n 1 1 ,..., M3 := f 2 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( Z n (bj − aj ) 1 X ∂f + Q j−1 24 j=1 j−1 ∂x Q j [ai ,bi ] (bi − ai ) i=1 i=1 aj+1 + bj+1 an + b n (s1 , s2 , . . . , sj−1 , bj , ,..., 2 2 ) ! aj+1 + bj+1 an + b n ∂f s1 , . . . , sj−1 , aj , ,..., ds1 · · · dsj−1 − ∂xj 2 2 # "
3 n X
1 aj+1 + bj+1 an + b n 3 ∂ f
√ (bj − aj ) 3 . . . , . ≤ ,..., j
Q ∂xj 2 2 12 210 j=1 [ai ,bi ] ∞, i=1
(4.112)
ii) Suppose ∂3f ∂x3j
aj+1 + bj+1 an + b n ..., ,..., 2 2
all j = 1, . . . , n. Then
∈ L2
j Y
!
[ai , bi ] ,
i=1
" j−1 −1/2 n X Y 1 5/2 M3 ≤ √ (bj − aj ) (bi − ai ) 12 210 j=1 i=1
∂3f aj+1 + bj+1 an + b n
,..., × 3 ...,
j
Q
∂xj 2 2 2,
[ai ,bi ]
#
.
(4.113)
i=1
Next we present trapezoid and midpoint rules inequalities for m = 4. Corollary 4.47. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 4, all xj = aj , j = 1, . . . , n. i) Assume ! j Y ∂4f [ai , bi ] , (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x4j i=1
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all j = 1, . . . , n. Then f (a , . . . , a ) + f (b , a , . . . , a ) 1 n 1 2 n K3 = 2 Z 1 −Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( 2 n X X (bj − aj )k−1 1 − j−1 k! Q j=1 (bi − ai ) k=1 (with j=k6=1) i=1
∂ k−1 f × Bk (0) j−1 (s1 , . . . , sj−1 , bj , aj+1 , . . . , an k−1 Q [ai ,bi ] ∂xj i=1 !)) ! ∂ k−1 f (s , s , . . . , s , a , a , . . . , a ) ds · · · ds − 1 2 j−1 j j+1 n 1 j−1 ∂xjk−1
" # n
∂4f
X 1
≤ √ (bj − aj )4 4 (. . . , aj+1 , . . . , an ) . (4.114) j
∂xj
Q 24 630 Z
j=1
∞,
[ai ,bi ]
i=1
ii) Assume j Y
∂ 4f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x4j
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then " −1/2 4 j−1 n X Y
∂ f 1 7/2
K3 ≤ √ (bj − aj ) (bi − ai )
∂x4 24 630 j=1 j i=1 #
(. . . , aj+1 , . . . , an ) . j
Q 2,
(4.115)
[ai ,bi ]
i=1
Corollary 4.48. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 4, all xj = bj , j = 1, . . . , n. i) Assume ∂4f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x4j
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then f (b , . . . , b ) + f (a , b , . . . , b ) 1 n 1 2 n Λ3 = 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( 2 n X (bj − aj )k−1 X 1 − j−1 k! Q j=1 (bi − ai ) k=1 (with j=k6=1) i=1
Z
∂ k−1 f (s1 , s2 , . . . , sj−1 , bj , bj+1 , . . . , bn ) Qj−1 ∂xjk−1 i=1 [ai ,bi ] !)) ∂ k−1 f − (s , s , . . . , s , a , b , . . . , b ) ds · · · ds 1 2 j−1 j j+1 n 1 j−1 ∂xjk−1 ( n " #)
4
X
∂ f
1
(bj − aj )4 . ≤ √ j
4 (. . . , bj+1 , . . . , bn ) Q ∂x 24 630 j=1 j ∞, [ai ,bi ] × Bk (1)
(4.116)
i=1
ii) Suppose j Y
∂4f (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x4j
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then " −1/2 4 j−1 n Y X
∂ f 1
Λ3 ≤ √ (bj − aj )7/2 (bi − ai )
∂x4 24 630 j=1 j i=1 #
(. . . , bj+1 , . . . , bn ) . j
Q 2,
(4.117)
[ai ,bi ]
i=1
Corollary 4.49. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 4, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ∂4f ∂x4j
aj+1 + bj+1 an + b n ..., ,..., 2 2
∈ L∞
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then a +b an + b n 1 1 M3 = f ,..., 2 2 Z 1 −Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( n ( Z (bj − aj ) 1 X ∂f s1 , s2 , . . . , sj−1 , + Q j−1 24 j=1 j−1 Q [ai ,bi ] ∂xj (bi − ai ) i=1 i=1 aj+1 + bj+1 an + b n bj , ,..., 2 2 !)) ! aj+1 + bj+1 an + b n ∂f s1 , s2 , . . . , sj−1 , aj , ds1 · · · dsj−1 ,..., − ∂xj 2 2 ( n " r
∂4f 1 107 X ≤ (bj − aj )4
∂x4 152 35 j=1 j #) aj+1 + bj+1 an + b n
. (4.118) ..., ,..., j
Q 2 2 ∞, [ai ,bi ] i=1
ii) Assume
∂4f ∂x4j
aj+1 + bj+1 an + b n ..., ,..., 2 2
∈ L2
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then !−1/2 ( n " r j−1 Y 1 107 X 7/2 (bi − ai ) (bj − aj ) M3 ≤ 1152 35 j=1 i=1
#)
∂4f aj+1 + bj+1 an + b n
× 4 ..., ,..., .
j
∂xj
Q 2 2 2,
(4.119)
[ai ,bi ]
i=1
Also we present trapezoid and midpoint rules inequalities for m = 5. Corollary 4.50. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 5, all xj = aj , j = 1, . . . , n. i) Assume ! j Y ∂5f [ai , bi ] , (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x5j i=1
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67
all j = 1, . . . , n. Then f (a , . . . , a ) + f (b , a , . . . , a ) 1 n 1 2 n K5 := 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( n X X (bj − aj )k−1 1 − j−1 k! Q j=1 (bi − ai ) k∈{1,2,4} (with j=k6=1) Z
i=1
∂ k−1 f (s , . . . , sj−1 , bj , aj+1 , . . . , an ) j−1 k−1 1 Q [ai ,bi ] ∂xj i=1 ! !)) k−1 ∂ f − (s , . . . , sj−1 , aj , aj+1 , . . . , an ) ds1 · · · dsj−1 k−1 1 ∂xj ( n " #) r
5
5 X 1 5 ∂ f
. (bj − aj ) 5 (. . . , aj+1 , . . . , an ) Q ≤ j 720 462 j=1 ∂xj [ai ,bi ] ∞, × Bk (0)
i=1
(4.120)
ii) Suppose ∂ 5f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x5j
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r −1/2 j−1 Y 5 X 1 (bi − ai ) (bj − aj )9/2 K5 ≤ 720 462 j=1 i=1
#)
∂5f
× 5 (. . . , aj+1 , . . . , an ) . j
∂xj
Q 2,
(4.121)
[ai ,bi ]
i=1
Corollary 4.51. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 5, all xj = bj , j = 1, . . . , n. i) Assume ∂5f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x5j
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then f (b , . . . , b ) + f (a , b , . . . , b ) 1 n 1 2 n Λ5 := 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( n X X (bj − aj )k−1 1 − j−1 k! Q j=1 (bi − ai ) k∈{1,2,4} (with j=k6=1) Z
i=1
∂ k−1 f (s , . . . , sj−1 , bj , bj+1 , . . . , bn ) j−1 k−1 1 Q [ai ,bi ] ∂xj i=1 ! !)) k−1 ∂ f − (s , . . . , sj−1 , aj , bj+1 , . . . , bn ) ds1 · · · dsj−1 k−1 1 ∂xj ( n " #) r
5
5 X 1 5 ∂ f
. (bj − aj ) 5 (. . . , bj+1 , . . . , bn ) Q ≤ j 720 462 j=1 ∂xj [ai ,bi ] ∞, × Bk (1)
i=1
(4.122)
ii) Suppose j Y
∂5f (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x5j
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r −1/2 j−1 Y 5 X 1 (bi − ai ) (bj − aj )9/2 Λ5 ≤ 720 462 j=1 i=1
#)
∂5f
× 5 (. . . , bj+1 , . . . , bn ) . j
∂xj
Q 2,
(4.123)
[ai ,bi ]
i=1
Corollary 4.52. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 5, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ∂5f ∂x5j
aj+1 + bj+1 an + b n ..., ,..., 2 2
∈ L∞
j Y
i=1
!
[ai , bi ] ,
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69
all j = 1, . . . , n. Then a +b an + b n 1 1 ,..., M5 := f 2 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( n X (bj − aj )k−1 X 1 − j−1 k! Q j=1 (bi − ai ) k∈{2,4} i=1
Z an + b n aj+1 + bj+1 1 ∂ k−1 f ,..., s1 , . . . , sj−1 , bj , × Bk j−1 k−1 Q 2 2 2 [ai ,bi ] ∂xj i=1 !)) ! ∂ k−1 f an + b n aj+1 + bj+1 ds · · · ds − , . . . , s , . . . , s , a , 1 j−1 1 j−1 j 2 2 ∂xjk−1 ( n " r
∂5f 1 5 X ≤ (bj − aj )5
∂x5 . . . , 720 462 j=1 j #) aj+1 + bj+1 an + b n
. (4.124) ,..., j
Q 2 2 ∞, [ai ,bi ] i=1
ii) Assume
∂5f ∂x5j
aj+1 + bj+1 an + b n ..., ,..., 2 2
∈ L2
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r −1/2 j−1 Y 1 5 X 9/2 (bi − ai ) M5 ≤ (bj − aj ) 720 462 j=1 i=1
∂5f aj+1 + bj+1 an + b n
× 5 ..., ,...,
j
∂xj
Q 2 2 2,
[ai ,bi ]
#)
.
(4.125)
i=1
Finally we present trapezoid and midpoint rules inequalities for m = 6.
Corollary 4.53. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all xj = aj , j = 1, . . . , n. i) Assume ! j Y ∂6f [ai , bi ] , (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x6j i=1
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all j = 1, . . . , n. Then f (a , . . . , a ) + f (b , a , . . . , a ) 1 n 1 2 n K5 = 2 Z 1 f (s1 , . . . , sn )ds1 · · · dsn − Q n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( n X X (bj − aj )k−1 1 − j−1 k! Q j=1 k∈{1,2,4} (b − a ) i i (with j=k6=1) Z
i=1
∂ k−1 f (s , . . . , sj−1 , bj , aj+1 , . . . , an ) k−1 1 [ai ,bi ] ∂xj i=1 !)) ! k−1 ∂ f − (s , . . . , s , a , a , . . . , a ) ds · · · ds 1 j−1 j j+1 n 1 j−1 k−1 ∂xj #) ( n " r
6
1 101 X 6 ∂ f . (bj − aj ) 6 (. . . , aj+1 , . . . , an ) ≤ j
Q 1440 30030 j=1 ∂xj [ai ,bi ] ∞, × Bk (0)
j−1 Q
i=1
(4.126)
ii) Suppose ∂ 6f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x6j
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r −1/2 j−1 Y 1 101 X (bi − ai ) K5 ≤ (bj − aj )11/2 1440 30030 j=1 i=1
#)
∂6f
× 6 (. . . , aj+1 , . . . , an ) . j
∂xj
Q 2,
(4.127)
[ai ,bi ]
i=1
Corollary 4.54. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all xj = bj , j = 1, . . . , n. i) Assume ∂6f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x6j all j = 1, . . . , n. Then
j Y
i=1
!
[ai , bi ] ,
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71
f (b , . . . , b ) + f (a , b , . . . , b ) 1 n 1 2 n Λ5 = 2 Z 1 − Q f (s1 , . . . , sn )ds1 · · · dsn n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( X X (bj − aj )k−1 1 − j−1 k! Q j=1 k∈{1,2,4} (b − a ) i i (with j=k6=1) Z
i=1
∂ k−1 f (s , . . . , sj−1 , bj , bj+1 , . . . , bn ) k−1 1 [ai ,bi ] ∂xj i=1 !)) ! k−1 ∂ f (s , . . . , s , a , b , . . . , b ) ds · · · ds − 1 j−1 j j+1 n 1 j−1 k−1 ∂xj ( n " #) r
6
101 X 1 6 ∂ f
(bj − aj ) 6 (. . . , bj+1 , . . . , bn ) Q . ≤ j 1440 30030 j=1 ∂xj ∞, [ai ,bi ] × Bk (1)
j−1 Q
i=1
(4.128)
ii) Assume j Y
∂6f (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x6j
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r j−1 −1/2 Y 1 101 X Λ5 ≤ (bi − ai ) (bj − aj )11/2 1440 30030 j=1 i=1
#)
∂6f
. × 6 (. . . , bj+1 , . . . , bn ) j
∂xj
Q 2,
(4.129)
[ai ,bi ]
i=1
Corollary 4.55. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all a +b xj = j 2 j , j = 1, . . . , n. i) Assume ∂6f ∂x6j
aj+1 + bj+1 an + b n ..., ,..., 2 2
∈ L∞
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then a +b an + b n 1 1 M 5 = f ,..., 2 2 Z 1 f (s1 , . . . , sn )ds1 · · · dsn − Q n n Q (bi − ai ) i=1[ai ,bi ] i=1 ( ( n X X (bj − aj )k−1 1 − j−1 k! Q j=1 (bi − ai ) k∈{2,4} i=1
Z an + b n aj+1 + bj+1 1 ∂ k−1 f , . . . , s , . . . , s , b , × Bk 1 j−1 j j−1 k−1 Q 2 2 2 [ai ,bi ] ∂xj i=1 !)) ! aj+1 + bj+1 an + b n ∂ k−1 f ds · · · ds s , . . . , s , a , , . . . , − 1 j−1 1 j−1 j k−1 2 2 ∂xj ( n " r 1 7081 X ≤ (bj − aj )6 46080 2145 j=1 #)
6
∂ f
a + b a + b j+1 j+1 n n
. (4.130) ,..., × j
∂x6 . . . ,
Q 2 2 j ∞, [ai ,bi ] i=1
ii) Suppose
aj+1 + bj+1 an + b n ∂ 6f ..., ,..., ∈ L2 ∂x6j 2 2
j Y
!
[ai , bi ] ,
i=1
all j = 1, . . . , n. Then ( n " r j−1 −1/2 Y 1 7081 X 11/2 M5 ≤ (bj − aj ) (bi − ai ) 46080 2145 j=1 i=1
#)
∂6f aj+1 + bj+1 an + b n
. × 6 ..., ,...,
j
Q
∂xj 2 2 2,
(4.131)
[ai ,bi ]
i=1
At the end we give a simplified special case of Theorems 4.32 and 4.33. Corollary 4.56. Suppose Assumptions 4.10 or 4.17 or 4.20. We further assume that ∂ `f ∂`f (. . . , bj , . . .) = (. . . , aj , . . .), ` ∂xj ∂x`j
(4.132)
for all j = 1, . . . , n and all ` = 0, 1, . . . , m − 2. Here m, n ∈ N, xi ∈ [ai , bi ], i = 1, 2, . . . , n.
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i) In particular we assume that ∂ mf (. . . , xj+1 , . . . , xn ) ∈ L∞ ∂xm j
j Y
i=1
!
[ai , bi ] ,
n Q for any (xj+1 , . . . , xn ) ∈ [ai , bi ], all j = 1, . . . , n. Then i=j+1 Z 1 f (x1 , . . . , xn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 s ! ( n " xj − a j 1 X (m!)2 2 (bj − aj )m ≤ |B2m | + Bm m! j=1 (2m)! bj − a j
#)
∂mf
, × m (. . . , xj+1 , . . . , xn ) j
Q
∂xj ∞,
73
(4.133)
[ai ,bi ]
i=1
true ∀(x1 , . . . , xn ) ∈ ii) Suppose
n Q
[ai , bi ].
i=1
∂mf (. . . , xj+1 , . . . , xn ) ∈ L2 ∂xm j
j Y
i=1
!
[ai , bi ] ,
n Q for any (xj+1 , . . . , xn ) ∈ [ai , bi ], for all j = 1, . . . , n. Then i=j+1 Z 1 f (x1 , . . . , xn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [a ,b ] i i (bi − ai ) i=1 i=1 s ( n " −1/2 j−1 ! Y 1 X (m!)2 xj − a j m− 12 2 ≤ (bi − ai ) (bj − aj ) |B2m | + Bm m! j=1 (2m)! bj − a j i=1
#)
∂ mf
× m (. . . , xj+1 , . . . , xn ) , (4.134) j
∂xj
Q 2,
[ai ,bi ]
i=1
true ∀(x1 , . . . , xn ) ∈ Proof.
n Q
[ai , bi ].
i=1
Clearly here Aj = 0, j = 1, . . . , n. Then proof is obvious.
Similarly as in Corollary 4.56 we obtain Corollary 4.57. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all xj = aj , j = 1, . . . , n. Also assume ∂`f ∂`f (. . . , b , a , . . . , a ) = (. . . , aj , aj+1 , . . . , an ), (4.135) j j+1 n ∂x`j ∂x`j
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for all j = 1, . . . , n, and all ` = 0, 1, 3. i) Assume ∂6f (. . . , aj+1 , . . . , an ) ∈ L∞ ∂x6j
j Y
!
[ai , bi ] , all j = 1, . . . , n.
i=1
Then Z 1 f (a1 , . . . , an ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 " # r
6 n
101 X ∂ f 1
(. . . , a , . . . , a ) (bj − aj )6 . ≤ j+1 n j
Q
∂x6 1440 30030 j=1 j ∞, [ai ,bi ] i=1
(4.136)
ii) Suppose ∂6f (. . . , aj+1 , . . . , an ) ∈ L2 ∂x6j Then
j Y
!
[ai , bi ] , all j = 1, . . . , n.
i=1
Z 1 f (a1 , . . . , an ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1 ( n r j−1 −1/2 Y 1 101 X ≤ (bi − ai ) (bj − aj )11/2 1440 30030 j=1 i=1
#)
∂6f
. × 6 (. . . , aj+1 , . . . , an ) j
∂xj
Q 2,
(4.137)
[ai ,bi ]
i=1
Proof.
By Corollary 4.53.
Corollary 4.58. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all xj = bj , j = 1, . . . , n. Also assume ∂ `f ∂`f (. . . , b , b , . . . , b ) = (. . . , aj , bj+1 , . . . , bn ), j j+1 n ∂x`j ∂x`j for all j = 1, . . . , n and all ` = 0, 1, 3. i) Assume ∂6f (. . . , bj+1 , . . . , bn ) ∈ L∞ ∂x6j
j Y
i=1
!
[ai , bi ] , all j = 1, . . . , n.
(4.138)
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75
Then Z 1 f (b1 , . . . , bn ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1 # " r
6 n X
101 ∂ f 1 (. . . , bj+1 , . . . , bn ) . (bj − aj )6 ≤ j 6
Q 1440 30030 j=1 ∂xj ∞, [ai ,bi ] i=1
(4.139)
ii) Suppose ∂ 6f (. . . , bj+1 , . . . , bn ) ∈ L2 ∂x6j
j Y
!
[ai , bi ] , all j = 1, . . . , n.
i=1
Then Z 1 f (b1 , . . . , bn ) − f (s , . . . , s )ds · · · ds 1 n 1 n n n Q Q [ai ,bi ] (b − a ) i i i=1 i=1 ( n " r j−1 −1/2 Y 101 X 1 (bj − aj )11/2 (bi − ai ) ≤ 1440 30030 j=1 i=1
#)
∂6f
. × 6 (. . . , bj+1 , . . . , bn ) j
Q
∂xj 2,
(4.140)
[ai ,bi ]
i=1
Proof.
By Corollary 4.54.
Corollary 4.59. Suppose Assumptions 4.10 or 4.17 or 4.20, case m = 6, all a +b xj = j 2 j , j = 1, . . . , n. Also assume ∂`f an + b n an + b n aj+1 + bj+1 aj+1 + bj+1 ∂`f , . . . , , . . . , . . . , b , = . . . , a , , j j 2 2 2 2 ∂x`j ∂x`j (4.141) for all j = 1, . . . , n and ` = 1, 3. i) Assume aj+1 + bj+1 an + b n ∂6f . . . , , . . . , ∈ L∞ ∂x6j 2 2
j Y
i=1
!
[ai , bi ] ,
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all j = 1, . . . , n. Then a +b an + b n 1 1 − Q ,..., f n 2 2 ≤
(
r
i=1
"
1 (bi − ai )
Z
f (s , . . . , s )ds · · · ds 1 n 1 n n Q [ai ,bi ]
i=1
n 1 7081 X (bj − aj )6 46080 2145 j=1
∂6f aj+1 + bj+1 an + b n
,..., × 6 ...,
∂xj 2 2
∞,
j Q
[ai ,bi ]
#)
.
(4.142)
i=1
ii) Suppose
∂6f ∂x6j
an + b n aj+1 + bj+1 ,..., ..., 2 2
all j = 1, . . . , n. Then a +b an + b n 1 1 − Q ,..., f n 2 2 ≤
r
(
"
i=1
1 (bi − ai )
Z
∈ L2
j Y
!
[ai , bi ] ,
i=1
f (s , . . . , s )ds · · · ds 1 n 1 n n Q [ai ,bi ]
i=1
j−1 −1/2 n Y 7081 X 1 (bj − aj )11/2 (bi − ai ) 46080 2145 j=1 i=1
#)
∂6f aj+1 + bj+1 aj + b n
,..., . × 6 ...,
j
Q
∂xj 2 2 2,
(4.143)
[ai ,bi ]
i=1
Proof.
By Corollary 4.55.
One can apply similar conditions to (4.132) for the cases of m = 2, 3, 4, 5 and simplify a lot the results of Corollaries 4.41, 4.42 and of Corollaries 4.44 – 4.52, exactly as we did in Corollaries 4.56–4.59 for general m ∈ N and m = 6, etc. 4.5
Sharpness
We need to include f Theorem 4.60. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Em (x1 , . . . , xn ) as in (4.44), m ∈ N. In particular we assume that ! j j Y ∂ m f z}|{ [ai , bi ] , · · · , xj+1 , . . . , xn ∈ L∞ ∂xm j i=1
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Multidimensional Euler Identity and Optimal Multidimensional Ostrowski Inequalities
for any (xj+1 , . . . , xn ) ∈
n Q
77
[ai , bi ], all j = 1, . . . , n. Then
i=j+1
" Z 1 n X xj − a j 1 f m |Em (x1 , . . . , xn )| ≤ (bj − aj ) Bm bj − aj − Bm (tj ) dtj m! j=1 0
#
∂mf
. (4.144) × m (. . . , xj+1 , . . . , xn ) j
Q
∂xj ∞,
[ai ,bi ]
i=1
Proof.
By Remark 4.23, see (4.55).
We give the important f Corollary 4.61. Suppose Assumptions 4.10 or 4.17 or 4.20. Let Em (x1 , . . . , xn ) as in (4.44), m ∈ N. In particular we assume that ! j j Y ∂ m f z}|{ [ai , bi ] , · · · , xj+1 , . . . , xn ∈ L∞ ∂xm j i=1
for any (xj+1 , . . . , xn ) ∈ L∞
n Q
i=1
n Q
[ai , bi ], all j = 1, . . . , n. And also suppose
i=j+1
∂mf ∂xm j
∈
[ai , bi ] , j = 1, . . . , n − 1. Call
(
)
∂mf
. Dm (f ) := max
1≤j≤n ∂xm j ∞
Then f |Em (x1 , . . . , xn )|
(4.145)
" !# Z 1 n x − a Dm (f ) X j j m Bm . − Bm (tj ) dtj (bj − aj ) ≤ m! j=1 bj − a j 0
(4.146)
Comment 4.62. We observe that (see also [98]) Z 1 a +b 2 xj − j 2 j 1 |B1 (λj ) − B1 (tj )|dtj = + I1 (λj ) := , 4 (bj − aj )2 0
where λj =
xj −aj bj −aj ,
(4.147)
j = 1, . . . , n. Notice that
max I1 (λj ) = I1 (0) = I1 (1) =
λj ∈[0,1]
1 , 2
i.e. when xj = aj or bj . Thus we have Corollary 4.63. All here assumed as in Corollary 4.61 when m = 1. Then Z 1 f (x1 , . . . , xn ) − f (s1 , . . . , sn )ds1 · · · dsn n n Q Q (bi − ai ) i=1[ai ,bi ] i=1
(4.148)
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≤
n D1 (f ) X (bj − aj ) . 2 j=1
(4.149)
Inequality (4.149) is sharp, that is attained by f1 (s1 , . . . , sn ) := xj = aj , j = 1, . . . , n, and by f2 (s1 , . . . , sn ) :=
n P
j=1
1, . . . , n.
Proof. i) Case of xj = aj , j = 1, . . . , n. Then
∂f1
∂xj ∞ = 1 and D1 (f1 ) = 1. Clearly then we have L.H.S.(4.149) = R.H.S.(4.149) =
proving sharpness. ii) Case of xj = bj , j = 1, . . . , n. Then and D1 (f2 ) = 1. Clearly we have
∂f2 ∂xj
n P
j=1
(sj − aj ) when
(bj − sj ) when xj = bj , j = ∂f1 ∂xj
= 1, j = 1, . . . , n, i.e.
n
1X (bj − aj ), 2 j=1
∂f2
=1 = −1, j = 1, . . . , n, i.e. ∂x ∞ j n
L.H.S.(4.149) = R.H.S.(4.149) = proving again sharpness.
1X (bj − aj ), 2 j=1
Comment 4.64. We observe that ([98]) Z 1 8 1 I2 (λj ) := |B2 (λj ) − B2 (tj )|dtj = δj3 (x) − δj2 (x) + , j = 1, . . . , n, (4.150) 3 12 0 where
xj − aj +bj 2 δj (xj ) := , bj − a j
xj ∈ [aj , bj ].
Also from [98] we have that
max I2 (λj ) = I2 (0) = I2 (1) =
0≤λj ≤1
1 , 6
(4.151)
i.e. when xj = aj or bj . We continue with Corollary 4.65. All here assumed as in Corollary 4.61 when m = 2. Then n
|E2f (x1 , . . . , xn )| ≤
D2 (f ) X (bj − aj )2 . 12 j=1
(4.152)
n P Inequality (4.152) is sharp, that is attained by f1 (s1 , . . . , sn ) := (sj − aj )2 when j=1 Pn 2 xj = aj , j = 1, . . . , n and by f2 (s1 , . . . , sn ) := j=1 (sj − bj ) when xj = bj , j = 1, . . . , n.
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Multidimensional Euler Identity and Optimal Multidimensional Ostrowski Inequalities
Proof.
∂2f
21 ∂xj
∞
i) Case of xj = aj , j = 1, . . . , n. Then
∂f1 ∂xj
∂ 2 f1 ∂x2j
= 2(sj − aj ),
79
= 2, and
= 2, with D2 (f ) = 2. Clearly then we have n
L.H.S.(4.152) = R.H.S.(4.152) =
1X (bi − aj )2 , 6 j=1
proving sharpness. ∂f2 ii) Case of xj = bj , j = 1, . . . , n. Then ∂x = 2(sj − bj ), j
∂2f
22 = 2, with D2 (f ) = 2. Clearly again we have ∂x ∞
∂ 2 f2 ∂x2j
= 2, and
j
n
L.H.S.(4.152) = R.H.S.(4.152) =
1X (bi − aj )2 , 6 j=1
proving again sharpness.
Comment 4.66. By [10] we have that √ ! 3− 3 = I3 max I3 (λj ) = I3 0≤λj ≤1 6 where I3 (λj ) :=
Z
√ ! √ 3+ 3 3 = , 6 36
(4.153)
j = 1, . . . , n.
(4.154)
1 0
|B3 (λj ) − B3 (tj )|dtj ,
Consequently we have Corollary 4.67. All here assumed as in Corollary 4.61 when m = 3. Then √ n 3D3 (f ) X f |E3 (x1 , . . . , xn )| ≤ (bj − aj )3 . (4.155) 216 j=1 Comment 4.68. We call Im (λj ) := where λj :=
xj −aj bj −aj ,
Z
1 0
|Bm (λj ) − Bm (tj )|dtj ,
(4.156)
j = 1, . . . , n, m ∈ N. In [35] we found that max I4 (λj ) = I4 (0) = I4 (1) =
λj ∈[0,1]
1 . 30
(4.157)
So we give Corollary 4.69. All here supposed as in Corollary 4.61 when m = 4. Then n D4 (f ) X f (bj − aj )4 . |E4 (x1 , . . . , xn )| ≤ (4.158) 720 j=1
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Inequality (4.158) is sharp, that is attained by f1 (s1 , . . . , sn ) := xj = aj , j = 1, . . . , n and by f2 (s1 , . . . , sn ) :=
n P
j=1
1, . . . , n. Proof.
n P
j=1 4
(sj − aj )4 when
(sj − bj ) when xj = bj , j =
Case of xj = aj , j = 1, . . . , n. Then
∂f1 ∂ 2 f1 ∂ 3 f1 ∂ 4 f1 = 4(sj − aj )3 , = 12(sj − aj )2 , = 24(sj − aj ), = 24, 2 3 ∂xj ∂xj ∂xj ∂x4j
4 with ∂∂xf41 ∞ = 24 and D4 (f ) = 24. Clearly then we have j n 1 X L.H.S.(4.158) = R.H.S.(4.158) = (bj − aj )4 , 30 j=1
proving sharpness. ii) Case xj = bj , j = 1, . . . , n. Then
∂ 2 f2 ∂ 3 f2 ∂ 4 f2 ∂f2 2 = 4(sj − bj )3 , = 12(s − b ) , = 24(s − b ), = 24, j j j j ∂xj ∂x2j ∂x3j ∂x4j with D4 (f2 ) = 24. Clearly again we have
n 1 X L.H.S.(4.158) = R.H.S.(4.158) = (bj − aj )4 , 30 j=1
proving again sharpness.
Comment 4.70. Inequality (4.144) is sharper than (4.77), however the integral Im (λj ) (see (4.156)) in its right hand side, is difficult to compute and find its maximum value for m ≥ 5. That is why (4.77) is more practical, also less restrictive, and we used it extensively here in the applications.
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Chapter 5
More on Multidimensional Ostrowski Type Inequalities
Very general multidimensional Ostrowski type inequalities are established, some of them prove to be sharp. They involve the k · k∞ and k · kp norms of the engaged mixed partial of nth order n ≥ 1. In establishing them, other important multivariate results of Montgomery type identity are developed and presented. This chapter relies on [23].
5.1
Introduction
In 1938, A. Ostrowski [196] proved the following inequality: Theorem 5.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e., kf 0 k∞ = sup |f 0 (t)| < t∈(a,b)
+∞. Then " 2 # 1 Z b x − a+b) 1 2 + f (t)dt − f (x) ≤ · (b − a)kf 0 k∞ , b − a a 4 (b − a)2
(5.1)
for all x ∈ [a, b]. The constant 14 is the best possible. Since then there has been a lot of activity around these inequalities with important applications to Numerical Analysis. This chapter is greatly motivated by the following results: k Q [ai , bi ] , where ai < bi ; ai , bi ∈ R, Theorem 5.2 (see [17]). Let f ∈ C 1 i=1
k Q − i = 1, . . . , k, and let → x 0 := (x01 , . . . , x0k ) ∈ [ai , bi ] be fixed. Then i=1 Z bk Z bi Z b1 1 − f (z1 , . . . , zk )dz1 · · · dzk − f (→ x 0 ) ··· ··· k Q ak ai (bi − ai ) a1 i=1
k X
(x0i − ai )2 + (bi − x0i )2
∂f . ≤
2(bi − ai ) ∂zi ∞ i=1
81
(5.2)
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Inequality (5.2) is sharp, here the optimal function is f ∗ (z1 , . . . , zk ) :=
k X i=1
|zi − x0i |αi ,
αi > 1.
Clearly inequality (5.2) generalizes inequality (5.1) to multidimension. We also would like to mention Theorem 5.3 (see [122]). Let f : [a, b] × [c, d] → R be a continuous mapping on ∂2f 00 exists on (a, b) × (c, d) and is in Lp ((a, b) × (c, d)), i.e., [a, b] × [c, d], fx,y := ∂x∂y !1/p Z bZ d 2 ∂ f (x, y) p 00 < +∞, p > 1. kfs,t kp := ∂x∂y dxdy a c
Then
"
− (b − a)
Z
d c
Z Z b d f (s, t)dsdt a c
f (x, t)dt + (d − c)
(x − a)q+1 + (b − x)q+1 ≤ q+1
Z
b
a
# f (s, y)ds − (d − c)(b − a)f (x, y)
1/q 1/q (y − c)q+1 + (d − y)q+1 00 · · kfs,t kp , q+1
(5.3)
for all (x, y) ∈ [a, b] × [c, d], where p1 + 1q = 1. In this chapter we develop more Ostrowski type inequalities in the multidimension, some of them are sharp. Along the way to establishing them we produce important side related results. Among others, one of the purposes of the chapter is to generalize Theorem 5.3, as far as possible, over ×ni=n [ai , bi ], n ∈ N, where we have n f (x1 ,...,xn ) . The main results are Ostrowski type inequalities involving involved ∂ ∂x 1 ···∂xn the norms k · k∞ and k · kp . Reference [62] provided us with important analytical tools. 5.2
Auxilliary Results
The following results are also by themselves of interest. We give Theorem 5.4. Let f : [a, A] × [b, B] × [c, C] → R be a continuous mapping on ∂3f 000 [a, A] × [b, B] × [c, C], and fx,y,z := ∂x∂y∂z exists on [a, A] × [b, B] × [c, C] and is integrable. Let also (x, y, z) ∈ [a, A] × [b, B] × [c, C] be fixed. We define the kernels p : [a, A]2 → R, q : [b, B]2 → R, and θ : [c, C]2 → R: s − a, s ∈ [a, x] p(x, s) := s − A, s ∈ (x, A],
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More on Multidimensional Ostrowski Type Inequalities
q(y, t) :=
θ(z, r) :=
and
Then
Z
A
Z
B
Z
83
t − b, t ∈ [b, y], t − B, t ∈ (y, B] r − c, r ∈ [c, z], r − C, r ∈ (z, C].
C
000 p(x, s)q(y, t)θ(z, r)fs,t,r (s, t, r)dsdtdr Z = {(A − a)(B − b)(C − c) · f (x, y, z)} − (B − b)(C − c)
θ1,3 :=
a
b
c
+ (A − a)(C − c)
+ (C − c) + (A − a) −
Z
A a
Z
B b
Z
Z
A
a B
b
Z
C
Z
Z
Z
B
f (x, t, z)dt + (A − a)(B − b)
b B
f (s, t, z)dsdt + (B − b)
b C
f (x, t, r)dtdr c
Z
Z
A a
Z
A
f (s, y, z)ds a
C
f (x, y, r)dr c
C
f (s, y, r)dsdr c
f (s, t, r)dsdtdr =: θ2,3 .
(5.4)
c
Proof. Z Integrating by parts repeatedly we obtain the following eight equalities: xZ yZ z 000 (s − a)(t − b)(r − c)fs,t,r (s, t, r)drdtds a b c Z x f (s, y, z)ds = (x − a)(y − b)(z − c)f (x, y, z) − (y − b)(z − c) Z y Za z − (x − a)(z − c) f (x, t, z)dt − (x − a)(y − b) f (x, y, r)dr Z xZ z c Z xZ y b f (s, y, r)drds f (s, t, z)dtds + (y − b) + (z − c) Za y Zb z Z x Z y Za z c + (x − a) f (x, t, r)drdt − f (s, t, r)drdtds. (5.5) Similarly, we have Z AZ BZ C x
y
z
b
c
a
b
c
000 (s − A)(t − B)(r − C)fs,t,r (s, t, r)drdtds
Z
= (A − x)(B − y)(C − z)f (x, y, z) − (B − y)(C − z) − (A − x)(C − z) + (C − z) + (A − x)
Z
A
x
Z
B y
Z
Z
y
f (x, t, z)dt − (A − x)(B − y)
f (s, t, z)dtds + (B − y)
y
Z
C z
f (x, t, r)drdt −
Z
A x
Z
Z
B y
A x
Z
Z
C
f (s, y, z)ds
x
B
B
A
Z
C
f (x, y, r)dr z
f (s, y, r)drds z
C
f (s, t, r)drdtds. z
(5.6)
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Furthermore, Z
x a
Z
y b
Z
C z
000 (s − a)(t − b)(r − C)fs,t,r (s, t, r)drdtds
Z
= (x − a)(y − b)(C − z)f (x, y, z) − (y − b)(C − z) − (x − a)(C − z) + (C − z) + (x − a)
Z
x
a Z y
Z
Z
b
y
Z
f (x, t, z)dt − (x − a)(y − b)
f (s, t, z)dtds + (y − b)
b C
f (x, t, r)drdt −
z
Z
x a
Z
Z
y b
a
Z
xZ C
f (s, y, z)ds a
Z
y b
x
C
f (x, y, r)dr z
f (s, y, r)drds
z
C
f (s, t, r)drdtds.
(5.7)
z
Also, Z
x A
Z
y B
Z
z c
000 (s − A)(t − B)(r − c)fs,t,r (s, t, r)drdtds
= (A − x)(B − y)(z − c)f (x, y, z) − (B − y)(z − c) − (A − x)(z − c) + (z − c)
Z
+ (A − x)
A x
Z
Z
B y
Z
B
f (x, t, z)dt − (A − x)(B − y)
y
B
f (s, t, z)dtds + (B − y)
y
Z
z
f (x, t, r)drdt −
c
Z
Z
A x
Z
Z
B y
A x
Z
Z
z
A
f (s, y, z)ds x
Z
z
f (x, y, r)dr c
f (s, y, r)drds c
z
f (s, t, r)drdtds.
(5.8)
c
Next we find, Z
a
xZ B y
Z
C z
000 (s − a)(t − B)(r − C)fs,t,r (s, t, r)drdtds
= (x − a)(B − y)(C − z)f (x, y, z) − (B − y)(C − z) − (x − a)(C − z) + (C − z) + (x − a)
Z
xZ B
a Z B y
Z
Z
y
y
B
f (x, t, z)dt − (x − a)(B − y)
f (s, t, z)dtds + (B − y)
C
z
f (x, t, r)drdt −
Z
x a
Z
B y
Z
x a
Z
C
Z
C
Z
Z
x
f (s, y, z)ds
a C
f (x, y, r)dr z
f (s, y, r)drds z
f (s, t, r)drdtds. z
(5.9)
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More on Multidimensional Ostrowski Type Inequalities
85
Also, Z
A x
Z
y b
Z
z c
000 (s − A)(t − b)(r − c)fs,t,r (s, t, r)drdtds
Z A = (A − x)(y − b)(z − c)f (x, y, z) − (y − b)(z − c) f (s, y, z)ds x Z z Z y f (x, y, r)dr f (x, t, z)dt − (A − x)(y − b) − (A − x)(z − c) + (y − b)
Z
+ (A − x)
A
x
Z
y b
Z
b
Z
z
f (s, y, r)drds + (z − c)
c
Z
z
f (x, t, r)drdt −
c
Z
A x
Z
y b
A
x
Z
z
Z
c
y
f (s, t, z)dtds b
f (s, t, r)drdtds.
(5.10)
c
And, Z
x a
Z
B y
Z
z c
000 (s − a)(t − B)(r − c)fs,t,r (s, t, r)drdtds
= (x − a)(B − y)(z − c)f (x, y, z) − (B − y)(z − c) − (x − a)(z − c) + (z − c) + (x − a)
Z
a
Z
Z
xZ B y
B
y
Z
B
f (x, t, z)dt − (x − a)(B − y)
y
f (s, t, z)dtds + (B − y)
z
f (x, t, r)drdt −
c
Z
Z
x a
Z
B y
Z
x a
Z
z
Z
z
x
f (s, y, z)ds a
Z
z
f (x, y, r)dr c
f (s, y, r)drds c
f (s, t, r)drdtds.
(5.11)
c
Finally we have, Z
A x
Z
y b
Z
C z
000 (s − A)(t − b)(r − C)fs,t,r (s, t, r)drdtds
= (A − x)(y − b)(C − z)f (x, y, z) − (y − b)(C − z) − (A − x)(C − z) + (C − z) + (A − x)
Z
A x
Z
b
Z
y
b yZ C z
Z
y b
Z
f (x, t, z)dt − (A − x)(y − b)
f (s, t, z)dtds + (y − b) f (x, t, r)drdt −
Z
A x
Z
Z
y b
AZ C
x
Z
A
f (s, y, z)ds x
Z
C
f (x, y, r)dr z
f (s, y, r)drds
z
C
f (s, t, r)drdtds. z
(5.12)
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Adding all the right-hand sides of (5.5) – (5.12), we derive:
Big R.H.S. = (A − a)(B − b)(C − c)f (x, y, z) − (B − b)(C − c) − (A − a)(C − c) + (C − c) + (A − a)
Z
A a
Z
B b
Z
Z
B
f (x, t, z)dt − (A − a)(B − b)
b
B
f (s, t, z)dtds + (B − b)
b
Z
C c
Z
f (x, t, r)dtdr −
Z
A a
Z
Z
B b
A
a
Z
Z
C
A
f (s, y, z)ds a
Z
C
f (x, y, r)dr c
f (s, y, r)dsdr c
C
f (s, t, r)drdtds. c
(5.13)
Adding all the left-hand sides of (5.5) – (5.12), we find
Big L.H.S. =
Z
A a
Z
B b
Z
C c
000 p(x, s)q(y, t)θ(z, r)fs,t,r (s, t, r)drdtds.
(5.14)
Clearly we have Big L.H.S. = Big R.H.S. We have established (5.4).
(5.15)
In general we state Theorem 5.5. Let f : ×ni=1 [ai , bi ] → R be a continuous mapping on ×ni=1 [ai , bi ], n f (x1 ,...,xn ) and ∂ ∂x exists on ×ni=1 [ai , bi ] and is integrable. Let also (x1 , . . . , xn ) ∈ 1 ···∂xn ×ni=1 [ai , bi ] be fixed. We define the kernels pi : [ai , bi ]2 → R: pi (xi , si ) :=
si − ai , si ∈ [ai , xi ] si − bi , si ∈ (xi , bi ],
for all i = 1, . . . , n. Then θ1,n := =
Z
n Y
×n i=1 [ai ,bi ] i=1
(
n Y
i=1
pi (xi , si )
(bi − ai )
!
∂ n f (s1 , . . . , sn ) ds1 · · · dsn ∂s1 · · · ∂sn )
· f (x1 , . . . , xn )
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n (X Z 1) n Y − (bj − aj )
i=1
j=1 j6=i
bi ai
k=1 k6=i,j
f (x1 , . . . , si , . . . , xn )dsi
(n2 ) Z n X Y + (bk − ak ) `=1
87
bi
ai
Z
bj
f (x1 , . . . , si , . . . , sj , . . . , xn )dsi dsj aj
!
(`)
− + · · · − + · · · + (−1)n−1 n (X Z n−1) cj · · · dsn (bj − aj ) f (s1 , . . . , xj , . . . , sn )ds1 · · · ds · + (−1)
Z
×n [a ,b ] i=1 i i i6=j
j=1
n
×n i=1 [ai ,bi ]
f (s1 , . . . , sn )ds1 · · · dsn =: θ2,n .
(5.16)
cj means dsj The above ` counts all the (i, j)’s, i < j and i, j = 1, . . . , n. Also ds is missing. Proof.
5.3
Similar to Theorem 5.4.
Main Results
Here we present the consequences of Section 5.2. We have Theorem 5.6. Under the notations and assumptions of Theorem 5.4, additionally 000 we suppose that kfs,t,r k∞ < +∞. Then 000 kfs,t,r k∞ · ((x − a)2 + (A − x)2 ) · ((y − b)2 + (B − y)2 ) 8 ·((z − c)2 + (C − z)2 ) , for all (x, y, z) ∈ [a, A] × [b, B] × [c, C].
|θ2,3 | ≤ Proof.
Notice that
000 |θ2,3 | = |θ1,3 | ≤ kfs,t,r k∞
Also observe that
etc.
Z
Z
A a
|p(x, s)|ds
A a
|p(x, s)|ds =
! Z
B b
|q(y, t)|dt
! Z
(5.17)
C c
|θ(z, r)|dr
!
.
1 {(x − a)2 + (A − x)2 }, 2
The counterpart of the last theorem is Theorem 5.7. Under the notations and assumptions of Theorem 5.5, additionally we suppose that
n
∂ f (s1 , . . . , sn )
∂s1 · · · ∂sn < +∞. ∞
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Then
∂ n f (s1 ,...,sn )
∂s1 ···∂sn 2n
|θ2,n | ≤ Proof.
∞
·
(
n Y
i=1
2
2
)
[(xi − ai ) + (bi − xi ) ] ,
for all (x1 , . . . , xn ) ∈ ×ni=1 [ai , bi ].
(5.18)
As in Theorem 5.6.
The Lp -analogues follow. Theorem 5.8. Under the notations and assumptions of Theorem 5.4, additionally, 000 we assume that fs,t,r ∈ Lp ([a, A] × [b, B] × [c, C]), i.e., !1/p Z AZ BZ C 3 ∂ f (x, y, z) p 000 kfs,t,r kp := < +∞, ∂x∂y∂z dxdydz a b c
where p > 1. Then 000 kfs,t,r kp · [((x − a)q+1 + (A − x)q+1 ) · ((y − b)q+1 + (B − y)q+1 ) |θ2,3 | ≤ (q + 1)3/q · ((z − c)q+1 + (C − z)q+1 )]1/q , for all (x, y, z) ∈ [a, A] × [b, B] × [c, C],
(5.19)
1 p
where q is such that Proof.
1 q
+
= 1.
Notice that
Z Z Z A B C 000 |θ2,3 | = |θ1,3 | = p(x, s)q(y, t)θ(z, r)fs,t,r (s, t, r)dsdtdr a b c Z AZ BZ C 000 ≤ |p(x, s)| |q(y, t)| |θ(z, r)| |fs,t,r (s, t, r)|dsdtdr a
a
c
(by H¨ older’s inequality)
Z
≤ ·
A a
Z
Z
A a
B b
Z
B b
000 = kfs,t,r kp ·
·
Z
Z
C c
Z
000 |fs,t,r (s, t, r)|p dsdtdr
C
q
(|p(x, s)| |q(y, t)| |θ(z, r)|) dsdtdr
c
Z
A a
|p(x, s)|q ds
C q
c
!1/p
|θ(z, r)| dr
!
!
·
Z
B b
!1/q
|q(y, t)|q dt
!
1/q
(x − a)q+1 + (A − x)q+1 = · q+1 1/q q+1 (y − b) + (B − y)q+1 (z − c)q+1 + (C − z)q+1 . · · q+1 q+1 000 kfs,t,r kp
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89
The corresponding general Lp -case follows. Theorem 5.9. Under the notations and assumptions of Theorem 5.5, additionally ∂nf n we assume that ∂x1 ···∂xn ∈ Lp ×i=1 [ai , bi ] , i.e.
n
∂ f (x1 , . . . , xn )
where p > 1.
∂x1 · · · ∂xn < +∞, p Then
|θ2,n | ≤
∂ n f (x1 ,...,xn )
∂x1 ···∂xn (q + 1)n/q
p
·
(
n Y
i=1
[(xi − ai )
for any (x1 , . . . , xn ) ∈ ×ni=1 [ai , bi ], and q : Proof.
1 p
+
q+1
1 q
+ (bi − xi )
q+1
]
)1/q
,
(5.20)
= 1.
Same as in Theorem 5.8.
Remark 5.10. Equalities (5.4) and (5.16) can simplify dramatically, if for instance we suppose in Theorem 5.4 that there exists an (x0 , y0 , z0 ) ∈ [a, A] × [b, B] × [c, C] such that f (x0 , ·, ·) = f (·, y0 , ·) = f (·, ·, z0 ) = 0.
Also in Theorem 5.5 we may assume that there exists an (x01 , x02 , . . . , x0n ) ∈ ×ni=1 [ai , bi ] such that f (x01 , x2 , . . . , xn ) = f (x1 , x02 , x3 , . . . , xn ) = · · · = f (x1 , . . . , xn−1 , x0n ) = 0,
for any (x1 , . . . , xn ) ∈ ×ni=1 [ai , bi ]. So in these particular cases we derive that Z AZ BZ C θ2,3 (x0 , y0 , z0 ) = − f (s, t, r)dsdtdr, (5.21) a
and
θ2,n (x01 , x02 , . . . , x0n )
= (−1)
n
Z
b
×n i=1 [ai ,bi ]
c
f (s1 , . . . , sn )ds1 · · · dsn .
(5.22)
Hence in these cases we have
and
Z Z Z A B C |θ2,3 (x0 , y0 , z0 )| = f (s, t, r)dsdtdr , a b c
(5.23)
Z f (s1 , . . . , sn )ds1 · · · dsn . |θ2,n (x01 , x02 , . . . , x0n )| = ×ni=1 [ai ,bi ]
So according to Theorems 5.6 – 5.9 we get respectively: Z Z Z kf 000 k A B C s,t,r ∞ f (s, t, r)dsdtdr ≤ a b 8 c · ((x0 − a)2 + (A − x0 )2 ) · ((y0 − b)2 + (B − y0 )2 ) · ((z0 − c)2 + (C − z0 )2 ) ;
(5.24)
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and
Z ∂ n f (s ,...,s ) 1 n ∂s1 ···∂sn ∞ f (s1 , . . . , sn )ds1 · · · dsn ≤ ×ni=1 [ai ,bi ] 2n (n ) Y 0 2 0 2 · [(xi − ai ) + (bi − xi ) ] .
(5.25)
i=1
Also it holds Z Z Z A B C f (s, t, r)dsdtdr a b c ≤
and
000 kfs,t,r kp · [((x0 − a)q+1 + (A − x0 )q+1 ) · ((y0 − b)q+1 + (B − y0 )q+1 ) (q + 1)3/q · ((z0 − c)q+1 + (C − z0 )q+1 )]1/q ; (5.26)
∂ n f (x1 ,...,xn ) Z
∂x1 ···∂xn p f (s1 , . . . , sn )ds1 · · · dsn ≤ ×ni=1 [ai ,bi ] (q + 1)n/q )1/q (n Y 0 q+1 0 q+1 [(xi − ai ) + (bi − xi ) ] · .
(5.27)
i=1
Finally we present the optimality of inequalities in Theorems 5.6 and 5.7. Theorem 5.11. Inequalities (5.17) and (5.18) are sharp. Proof. be
It is enough to prove that (5.18) is sharp. Here the optimal function will
f ∗ (s1 , . . . , sn ) :=
n Y
i=1
|si − x0i |α (bi − αi ),
α > 1,
(5.28)
where (x01 , x02 , . . . , x0n ) is fixed in ×ni=1 [ai , bi ]. Notice here that f ∗ (s1 , . . . , x0j , . . . , sn ) = 0,
for all j = 1, . . . , n, and any (s1 , . . . , sn ) ∈ ×ni=1 [ai , bi ].
Therefore by Remark 5.10, inequality (5.18) collapses to inequality (5.25). We see that ! ! n n Y Y ∂ n f ∗ (s1 , . . . , sn ) = αn · (bi − ai ) · |si − x0i |α−1 sign(si − x0i ) , (5.29) ∂s1 · · · ∂sn i=1 i=1 and n ∗ ∂ f (s1 , . . . , sn ) n ∂s1 · · · ∂sn = α ·
n Y
i=1
(bi − ai )
!
·
n Y
i=1
|si −
x0i |α−1
!
.
(5.30)
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Consequently we find that
n ∗
∂ f (s1 , . . . , sn )
∂s1 · · · ∂sn
=α
n Y
n
∞
i=1 n Y
·
i=1
(bi − ai )
(max(bi −
91
! x0i , x0i
− ai ))
α−1
!
.
(5.31)
First we calculate the left-hand side of corresponding inequality (5.25). We have Z f ∗ (s1 , . . . , sn )ds1 · · · dsn ×ni=1 [ai ,bi ] ! ! Z n n Y Y 0 α ds1 · · · dsn (bi − ai ) · |si − xi | = ×n i=1 [ai ,bi ]
= = =
n Y
i=1 n Y
i=1 n Y i=1
That is,
L.H.S.(5.25) =
i=1
(bi − ai ) (bi − ai ) (bi − ai ) n Q
i=1
!Z !
!
R.H.S.(5.25) =
·
(
n Q
i=1
n Y
Z n Y i=1
·
n Y
lim R.H.S.(5.25) =
α→1
n Q
i=1
|si −
i=1
x0i |α dsi
ds1 · · · dsn
!!
− ai )α+1 + (bi − x0i )α+1 α+1
((x0i
− ai )
α+1
+ (bi −
!
.
!
.
x0i )α+1 )
(5.32)
i=1
2n
2
− ai ) + (bi −
Now let α → 1. We find that n Q (bi − ai ) lim L.H.S.(5.25) = i=1 n · α→1 2 and
bi
|si − x0i |α
!
n Q (bi − ai ) · (max(bi − x0i , x0i − ai ))α−1
((x0i
i=1
i=1
ai i=1 n Y (x0i
·
(α + 1)n αn ·
×n i=1 [ai ,bi ]
·
(bi − ai )
And next we observe that
i=1
n Y
(bi − ai ) 2n
That is,
·
n Y
x0i )2 )
((x0i
i=1
n Y
i=1
)
.
(5.33)
2
− ai ) + (bi −
hence proving the sharpness of (5.18).
α→1
!
,
!
((x0i − ai )2 + (bi − x0i )2 ) .
lim L.H.S.(5.25) = lim R.H.S.(5.25),
α→1
x0i )2 )
(5.34)
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Remark 5.12. Another interesting case for (5.4) and (5.16) is to suppose that for specific (x, y, z) ((x1 , . . . , xn ), respectively) all the marginal integrals of f are equal to zero. Then we find Z AZ BZ C f (s, t, r)dsdtdr, (5.35) θ2,3 = (A − a)(B − b)(C − c) · f (x, y, z) − a
and
θ2,n =
n Y
i=1
(bi − ai )
!
· f (x1 , . . . , xn ) + (−1)
n
Z
a
c
×n i=1 [ai ,bi ]
f (s1 , . . . , sn )ds1 · · · dsn .
(5.36) Hence inequalities (5.17), (5.18), (5.19), and (5.20) become again much simpler.
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Chapter 6
Ostrowski Inequalities on Euclidean Domains
The classical Ostrowski inequality for functions on intervals is extended to functions on general domains in Euclidean space. For radial functions on balls the inequality is sharp. This treatment relies on [56].
6.1
Introduction
The classical Ostrowski inequality (of 1938) [196] is 2 ! Z b 1 x − a+b 1 2 (b − a)kf 0 k∞ , + f (y)dy − f (x) ≤ b − a 4 (b − a)2 a
for f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. This was extended from intervals to rectangles in RN , N ≥ 1, see [21], p. 507. For other recent results related to Ostrowski’s inequality, see [110], [126] and [199], [214], [231]. The extension to general domains in RN is presented here. We deduce Ostrowski type inequalities on general bounded domains in RN , and the inequalities are shown to be sharp on balls.
6.2
Main Results
Let N > 1, B(0, R) := {x ∈ RN : |x| < R} be the ball in RN centered at the origin and of radius R > 0. Let S N −1 := {x ∈ RN : |x| = 1} be the unit sphereR in RN . Let dω be the element of surface measure on S N −1 and let ωN = S N −1 dω = N/2 2π . For x ∈ RN −{0} we can write x = rω, where r = |x| > 0 and ω = xr ∈ S N −1 . Γ
N 2
R
N
dy = ωNNR is the Lebesgue measure of the ball. For f ∈ C B(0, R) let Z Z 1 − f (y)dy := f (y)dy, Vol(B(0, R)) B(0,R)
Note that
B(0,R)
B(0,R)
93
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and Z Z 1 − f (rω)dω = f (rω)dω ωN S N −1
S N −1
be the averages of f over the ball and the sphere, respectively. Here f can be real or complex valued. Let Z ˜ := − f (rω)dω f(r) S N −1
be the average of f (x) as x ranges over {y ∈ RN : |y| = r}. Then N (f ) :=
sup x∈B(0,R)
˜∞ |f (x) − f˜(r)| = kf − fk
(6.1)
measures howfar f is from being a radial function. More precisely, N is a seminorm on C B(0, R) , and N (f ) = 0 if and only if f is a radial function, i.e. f (x) = g(r) for some function g ∈ C([0, R]). We view how close f is to being radial by measuring N (f ); the closer f is to being radial, the smaller N (f ) is and conversely. Let Ω be a domain in RN and let Lip(Ω) = f ∈ C(Ω) : |f (x) − f (y)| ≤ K|x − y| for some K > 0 and all x, y ∈ Ω .
(6.2)
The Lipschitz constant of f ∈ Lip(Ω) is
kf kLip = inf{K : K as in (6.2)}. Then X := Lip(Ω) is a Banach space with norm f → kf k∞ + kf kLip =: kf kX . Equivalently, X is the Sobolev space W 1,∞ (Ω) (cf. [139]). We present Theorem 6.1. Let f ∈ Lip(B(0, R)) = W 1,∞ (B(0, R)). Then for x = rω as above Z N +1 |x| R N f (x) − − f (y)dy ≤ N (f ) + N k∇f k∞ 2|x| + R − . RN N (N + 1) N +1 N B(0,R)
(6.3) The constants in (6.3) are best possible, equality can be attained for nontrivial radial functions at any r ∈ [0, R].
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Proof. Let f ∈ Lip(B(0, R)). Then Z f (x) − − f (y)dy B(0,R) Z Z Z R N 0 N −1 0 ˜ ≤ |f (x) − f(r)| + − f (rω 0 )dω 0 − f (sω )s dsdω N ωN R S N −1 0 N −1 S # Z "Z R N |f (rω 0 ) − f (sω 0 )|S N −1 ds dω 0 ≤ N (f ) + N − R 0 S N −1
N ≤ N (f ) + N R
Z Z −
S N −1
0
∂f (ω 0 )
∂r
R
L∞ ((0,R))
|s − r|sN −1 dsdω 0
!
Z R
∂f N N −1
≤ N (f ) + |s − r|s ds
∂r ∞ N 0 L (B(0,R)) R
∂f 2rN +1 R r N N
. + R − = N (f ) +
∂r ∞ N N (N + 1) N +1 N L (B(0,R)) R
Then (6.3) follows since
(6.4)
∂f
≤ k∇f k∞ .
∂r ∞
In particular, a stronger form of (6.3) actually holds in all cases, replacing k∇f k ∞
. Let r ∈ [0, R] and g ∗ (z) = |z − r|. We can view g ∗ as a radial function by ∂f ∂r ∞ on B(0, R). Then sign(z − r), z 6= r, 0 < r < R g ∗0 (z) = 1, r=0 −1 r = R.
Thus kg ∗0 k∞ = 1. Therefore Z R Z R N N ∗ ∗ N −1 N −1 L.H.S.(6.3) = g (z) − N g (s)s ds = |z − r| − N |s − r|s ds R R 0 0 N r 2rN +1 R . = |z − r| − N + RN − R N (N + 1) N +1 N Also N r 2rN +1 R R.H.S.(6.3) = N + RN − . R N (N + 1) N +1 N Hence equality holds in (6.3) at z = r. Note that the function g ∗ (z) = |z − r|, is in C 1 ([0, R]) only for r = 0 and r = R; for 0 < r < R, g ∗ ∈ Lip([0, R]) − C 1 ([0, R]). Of course for 0 < r < R, g ∗ can be 1 approximated by C 1 functions, namely gn (z) = |z − r|1+ n .
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Remark 6.2. A key step in the proof is the fact that we can evaluate exactly Z R Q(r) = |s − r|p(s)ds 0
RR for 0 ≤ r ≤ R, where p is a nonnegative continuous function satisfying 0 p(s)ds = 1 1. In the Ostrowski case (N = 1), p(s) = R ; in our N -dimensional case, p(s) = N −1 Ns RN . This works for many other cases including linear combinations of p(s) = m P aj sqj , where aj > 0, qj ≥ 0 (not necessarily an integer) and j=1
m X
aj
j=1
p(s) =
m P1
j=1
Rqj +1 = 1, qj + 1
aj eλj s , where aj > 0, λj ∈ R − {0}, m1 X j=1
aj
e λj R − 1 λj
= 1,
p(s) = sums of the form aj sin(bj s + cj ) + dj cos(ej s + fj ), RR where the coefficients are such that p(s) ≥ 0 and 0 p(s)ds = 1. The space Lip(Ω) ∩ C0 (Ω) consists of all Lipschitz continuous functions on Ω vanishing on the boundary ∂Ω of Ω. Note that Lip(Ω) ∩ C0 (Ω) = f ∈ W 1,∞ (Ω) ∩ C(Ω) : f = 0 on ∂Ω (cf. [139]). Next comes a more general result when we consider functions over general domains.
Theorem 6.3. Let f ∈ Lip(Ω) ∩ C0 (Ω), where Ω is a bounded domain in RN . Extend f by zero to F on B(0, R), the smallest ball centered at the origin and containing Ω. Then for all x ∈ Ω, Z Z Vol(Ω) − f (y)dy f (x) − − f (y)dy ≤ N (F ) + 1 − Vol(B(0, R)) Ω Ω N 2|x|N +1 |x| R + N k∇f kL∞(Ω) . (6.5) + RN − R N (N + 1) N +1 N Proof.
Let R := inf{R0 > 0 : Ω ⊂ B(0, R0 )}. Then f (x), x ∈ Ω, F (x) := 0, x ∈ B(0, R) − Ω,
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satisfies F ∈ Lip(B(0, R)) ∩ C0 (B(0, R)). Then for x ∈ Ω Z Z Z Z f (x) − − f (y)dy ≤ F (x) − − F (y)dy + − F (y)dy − − f (y)dy Ω
B(0,R)
B(0,R)
Ω
= J1 + J2 ,
where
Z J1 := F (x) − − F (y)dy B(0,R)
and
By Theorem 6.1,
Z 1 1 J2 := − f (y)dy . Vol(B(0, R)) Vol(Ω) Ω
J1 ≤ N (F ) +
2|x|N +1 R |x| N N . k∇f k + R − ∞ RN N (N + 1) N +1 N
and
Vol(Ω) J2 = 1 − Vol(B(0, R)) This completes the proof of Theorem 6.3.
Z − f (y)dy . Ω
Remark 6.4. Note that N (F ) appears in (6.5). In this context, N (f ) does not make sense. Also, N (F ) need not be small (of course, it is small if f is approximately spherically symmetric). Here is a simple example to illustrate that N (F ) can be large. Let x0 ∈ Ω and choose ε > 0, small enough so that B(x0 , ε) ⊂ Ω. Let f ∈ C ∞ (Ω) with f have support in B(x0 , ε) and satisfy f (y) = g(ρ) where ρ = |y − x0 |, for 0 ≤ ρ ≤ ε. Assume further that g is nonincreasing and g(0) = f (x0 ) = M > 0. Fix M , then Z 0 < f (x0 ) − − f (y)dy → M = kf k∞ as ε → 0 + . Ω
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Book˙Adv˙Ineq
Chapter 7
High Order Ostrowski Inequalities on Euclidean Domains
The original Ostrowski the function minus its This result is extended nth partial derivatives. chapter relies on [57].
7.1
inequality for functions on intervals estimates the value of average in terms of the maximum of its first derivative. to functions on general domains using the L∞ norm of its For radial functions on balls the inequality is sharp. This
Introduction
The classical Ostrowski inequality (of 1938) [196] is 1 Z b f (y)dy − f (x) ≤ b − a a
2 ! x − a+b 1 2 (b − a)kf 0 k∞ , + 4 (b − a)2
(7.1)
for f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. Further related work was done in [16]. Inequality (7.1) was extended from intervals to rectangles in R N , N ≥ 1, see [21], p. 507. The extension to general domains in RN was done [56]. The resulting inequality was sharp, and equality was shown to hold for certain radial functions on balls. The right hand side included the factor k∇f k∞ . The extension of (7.1) to high order derivative bounds on an interval was obtained in [21], p. 502. The purpose here is to extend the results of [56] and [21], p. 502, to the higher order case when k∇f k∞ is replaced by max kDα f k∞ for W n,∞ functions on balls |α|=n
and more general domains. The obtained inequalities are sharp on balls.
7.2
Main Results
Let N > 1, B(0, R) := {x ∈ RN : |x| < R} be the ball in RN centered at the origin and of radius R > 0. Let S N −1 := {x ∈ RN : |x| = 1} be the unit sphere in RN . 99
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Let dω be the element of surface measure on S N −1 and let Z 2π N/2 ωN = . dω = Γ(N/2) S N −1
For x ∈ RN − {0} we can write x = rω, where r = |x| > 0 and ω = R N Note that B(0,R) dy = ωNNR is the Lebesgue measure of the ball. For f ∈ C B(0, R) let Z Z 1 f (y)dy, − f (y)dy := Vol(B(0, R)) B(0,R)
x r
∈ S N −1 .
B(0,R)
and
Z Z 1 − f (rω)dω = f (rω)dω ωN S N −1
S N −1
be the averages of f over the ball and the sphere, respectively. Here f can be real or complex valued. Let Z ˜ := − f (rω)dω f(r) S N −1
be the average of f (x) as x ranges over {y ∈ RN : |y| = r}. Therefore N (f ) :=
sup x∈B(0,R)
˜∞ |f (x) − f˜(r)| = kf − fk
(7.2)
measures howfar f is from being a radial function. More precisely, N is a seminorm on C B(0, R) , and N (f ) = 0 if and only if f is a radial function, i.e. f (x) = g(r) for some function g ∈ C([0, R]). We view how close f is to being radial by measuring N (f ); the closer f is to being radial, the smaller N (f ) is and conversely. Let Ω be a domain in RN and let Lip(Ω) = f ∈ C(Ω) : |f (x) − f (y)| ≤ K|x − y| for some K > 0 and all x, y ∈ Ω . (7.3)
The Lipschitz constant of f ∈ Lip(Ω) is
kf kLip = inf{K : K as in (7.3)}.
(7.4)
n,∞
Let n ∈ Z+ = {0, 1, 2, . . .}. Define the Sobolev space W (Ω) in the usual way: W n,∞ (Ω) := u : Ω → C : the distributional derivative D α u exists and is in L∞ (Ω) for 0 ≤ |α| ≤ n = u ∈ C n−1 (Ω) : Dα u ∈ Lip(Ω) for |α| = n . (7.5)
This is a Banach space with norm
kuk := max kDα ukL∞ (Ω) , 0≤|α|≤n
(7.6)
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(see [139]). Clearly W n,∞ (Ω) ⊃ C n (Ω). The first main result is the following. Theorem 7.1. Let f ∈ W n,∞ (B(0, R)). Then for x = rω as above Z f (x) − − f (y)dy B(0,R) ( n−1 Z N X 1 ∂ k f (rω 0 ) 0 ≤ N (f ) + N dω − R k! ∂rk k=1
S N −1
N +k−m ! k X k R × (−1)m rm m N + k − m m=0
∂nf n
n ∞ X RN +n−m n ∂r L (B(0,R)) + (−1)m rm n! N +n−m m m=0 !) n X rN +n n − βn , (−1)m N +n−m m m=0
(7.7)
where n ∈ N, βn = 0 or 2, according as n is even or odd. The constants in (7.7) are best possible, equality can be attained for nontrivial radial functions at any r ∈ [0, R]. We remark that
n
∂ f n
max kDα f k∞ ,
∂rn ≤ kD f k∞ := |α|=n ∞
n n and the larger bound kD f k∞ replacing ∂∂rnf ∞ gives perhaps a simpler version of the inequality (7.7).
Proof. Let f ∈ W n,∞ (B(0, R)). Hence Z Z f (x) − − f (y)dy ≤ |f (x) − f˜(r)| + − f (rω 0 )dω 0 B(0,R) S N −1 Z Z R N 0 N −1 0 − f (sω )s dsdω (7.8) ωN RN S N −1 0 Z Z R N 0 0 N −1 ≤ N (f ) + N − [f (rω ) − f (sω )]s ds dω 0 = N (f ) + (∗). R 0 S N −1
(7.9)
We see by Taylor’s formula that f (sω 0 ) − f (ρω 0 ) =
n−1 X k=1
∂ k f (ρω 0 ) (s − ρ)k + Rn−1 (ρ, s), k!∂rk
(7.10)
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where
Z s
∂ n−1 f (tω 0 ) ∂ n−1 f (ρω 0 ) (s − t)n−2 Rn−1 (ρ, s) := − dt, (7.11) ∂ρn−1 ∂ρn−1 (n − 2)! ρ for fixed ω 0 ∈ S N −1 and all s, ρ ∈ (0, R]. As in p. 500 of [21] we derive n
∂ nf |Rn−1 (ρ, s)| ≤ ∂r ∞ |s − ρ|n , ∀s, ρ ∈ [0, R]. (7.12) n! Consequently it holds " # Z R n−1 Z X ∂ k f (rω 0 ) N k N −1 0 (s − r) + R (r, s) s ds dω (∗) = N − (7.13) n−1 R k!∂rk 0 k=1 N −1 S ( n−1 Z Z R k N X ∂ f (rω 0 ) k N −1 ≤ N − (s − r) s ds dω 0 k R k!∂r 0 k=1 N −1 S ) Z Z R
+
−
S N −1
0
|Rn−1 (r, s)|sN −1 dsdω 0
( n−1 Z Z R ∂ k f (rω 0 ) 0 N X 1 dω − (s − r)k sN −1 ds ≤ N k R k! ∂r 0 k=1 S N −1
∂ nf ) Z R
n ∞ ∂r L (B(0,R)) n N −1 |s − r| s ds =: A. + n! 0
(7.14)
(7.15)
To evaluate the integrals in (7.15) we use an elementary calculation as follows. Lemma 7.2. (i) Z R k X k RN +k−m , k ∈ N. sN −1 (s − r)k ds = (−1)m rm N +k−m m 0 m=0 (ii) Z R n X n RN +n−m N −1 n s |s − r| ds = (−1)m rm N +n−m m 0 m=0 n X n rN +n (−1)m − βn , m N +n−m m=0 where n ∈ N, βn = 0 or 2, according as n is even or odd. Proof.
(i) We see that Z R Z sN −1 (s − r)k ds = 0
! k X k m k−m m (−1) s r ds s m 0 m=0 Z R k X k m m = (−1) r sN +k−m−1 ds m 0 m=0 k X k RN +k−m . (−1)m rm = N +k−m m m=0 R
(7.16)
(7.17)
N −1
(7.18)
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(ii) Using (i) twice, Z
R
s 0
N −1
n
|s − r| ds =
Z
r
s
N −1
0
n
(r − s) ds +
Z
R r
sN −1 (s − r)n ds
n X rN +n n (−1)m = (−1) m N +n−m m=0 n X n RN +n−m − rN +n−m + (−1)m rm N +n−m m m=0 n X n RN +n−m = (−1)m rm N +n−m m m=0 n X rN +n n + . (−1)m [(−1)n − 1] N +n−m m m=0 n
(7.19)
Part (ii) now follows. Continuing the calculation in (7.15) N A= N R
( n−1 Z X 1 ∂ k f (rω 0 ) 0 dω − k! ∂rk k=1
S N −1
N +k−m ! k R m m × (−1) r N + k − m m m=0
∂ nf n
n ∞ X n RN +n−m ∂r L (B(0,R)) (−1)m rm + m n! N +n−m m=0 !) n X rN +n n (−1)m − βn , m N +n−m m=0 k X
(7.20)
by Lemma 7.2. Hence proving (7.7). Let n be even. Let f be a radial function so that f (x) = g(r). Then equality in (7.7) is attained by g(r) = (r − z)n for any fixed z satisfying 0 ≤ z ≤ R. We observe that g (j) (z) = 0, j = 0, 1, . . . , n − 1 and g (n) (z) = n!, kg (n) k∞ = n!. We look at inequality (7.7) evaluating the function at z: L.H.S.(7.7) =
N RN
Z
R 0
(s − z)n sN −1 ds
n N +n−m N X n m m R = (−1) z RN m=0 m N +n−m ! n X n Rn−m m m . = N (−1) z N +n−m m m=0
(7.16)
(7.21) (7.22)
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Because N (f ) = 0 and βn = 0 the right hand side of (7.7) collapses to ! n N +n−m X N n m m R (−1) z R.H.S.(7.7) = N R m N +n−m m=0 ! n X Rn−m n m m (−1) z . (7.23) =N m N +n−m m=0 That is equality holds in (7.7). Note that g ∈ C n B(0, R) if z > 0, but g ∈ W n,∞ (B(0, R)) − C n B(0, R) if z = 0. The optimal function in the case of n odd is g(r) := |z − r|n , 0 ≤ r ≤ R, z is fixed in [0, R].
(7.24)
Direct calculations show, for n odd, g (k) (r) = n(n − 1) · · · (n − k + 1)|r − z|n−k αk (r),
(7.25)
where αk (r) = 1 αk (r) = sign(r − z)
if k is even, if k is odd and r 6= z.
(7.26)
It follows that g ∈ C ∞ ([0, R] − {z}) ∩ C n−1 ([0, R]) and g (n) ∈ L∞ ((0, R)) with a jump discontinuity at r = z, whence g ∈ W n,∞ ((0, R)). It follows that f ∈ W n,∞ (B(0, R)). Moreover, g (k) (z) = 0, for k = 0, 1, . . . , n − 1 and |g (k) (z + ε)| → n! as ε → 0 (with ε 6= 0). Thus kg (n) k∞ = n!. The left-hand side of (7.7) is Z R N |z − s|n sN −1 ds L.H.S.(7.7) = RN 0 n N +n−m X n (7.17) N m m R = (−1) z RN m=0 m N +n−m ! n X n z N +n m −2 (−1) . (7.27) m N +n−m m=0
Since N (f ) = 0 and n is odd, using the calculation on g given above leads to the conclusion that ( n RN +n−m N n! X n (−1)m z m R.H.S.(7.7) = N R n! m=0 m N +n−m !) n N +n X n z −2 (−1)m . (7.28) m N +n−m m=0
Hence equality holds in (7.7). This completes the proof of the sharpness of inequality (7.7). The proof of the theorem is now complete.
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Note that the extremalfunctions which give equality in (7.7) in the case of n odd are not in C n B(0, R) . To find a sequence {fν }ν∈N of approximate extremal functions in C n B(0, R) , simply take fν (x) = gν (r), where 1
gν (r) = |z − r|n+ ν , ν ∈ N.
(7.29)
Next we treat the case of r = R. k
Theorem 7.3. Let f ∈ W n,∞ (B(0, R)) such that ∂∂rfk , k = 1, . . . , n − 1 vanish on ∂B(0, R). Then for all w ∈ S N −1 ,
n Z Rn ∂∂rnf L∞ (B(0,R)) . (7.30) f (Rω) − − f (y)dy ≤ N (f ) + (N + 1) · · · (N + n) B(0,R)
The constants in (7.30) are best possible, equality can be attained for nontrivial radial functions. Proof.
Here from (7.7) we obtain Z f (Rω) − − f (y)dy ≤ N (f ) B(0,R)
n n m n X N Rn n m (−1)
∂ f (−1) + = (∗),
n n! ∂r L∞ (B(0,R)) N +n−m m=0
since 1 − βn = (−1)n . Next, n X n (−1)m tN +n−m−1 = tN −1 (t − 1)n , m m=0
by the Binomial Theorem. Integrate (7.32) over t ∈ [0, 1], the result is Z 1 n X n (−1)m (1 − t)n tN −1 dt = (−1)n m N + n − m 0 m=0 = (−1)n B(n + 1, N ) n!(N − 1)! = (−1)n . (N + n)!
By (7.34) we derive
n
n Rn ∂∂rnf L∞ (B(0,R))
(N − 1)! ∂ f
(∗) = N Rn = ,
∂rn ∞ (N + 1) · · · (N + n) L (B(0,R)) (N + n)!
(7.31)
(7.32)
(7.33)
(7.34)
(7.35)
that is proving (7.30). The function f (x) = g(r) = (R − r)n gives equality in (7.30), whether n is even or odd.
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7.3
Book˙Adv˙Ineq
Functions on General Domains
The next result gives high order Ostrowski type inequalities functions on general bounded domains in RN . Theorem 7.4. Let f ∈ W n,∞ (Ω) ∩ C0n−1 (Ω), where Ω is a bounded domain in RN and C0n−1 (Ω) = {u ∈ C n−1 (Ω) : Dα u = 0 on ∂Ω for |α| ≤ n − 1}. Extend f by zero to F on B(0, R), the smallest ball centered at the origin and containing Ω. Then F ∈ W n,∞ (B(0, R)), and for all x = rω ∈ Ω, Z Z Vol(Ω) − f (y)dy f (x) − − f (y)dy ≤ N (F ) + 1 − Vol(B(0, R)) Ω Ω ( n−1 Z N +k−m ! X k R k ∂ k F (rω 0 ) 0 N X 1 m m (−1) r dω − + N k m R k! ∂r N +k−m k=1
S n−1 n X
m=0
kDn f kL∞(Ω) n RN +n−m (−1)m rm n! N +n−m m m=0 !) n X rN +n n (−1)m − βn , m N +n−m m=0 +
(7.36)
where n ∈ N, βn = 0 or 2, according as n is even or odd. Here kDn f kL∞(Ω) = max kDα f kL∞(Ω) . |α|=n
Proof.
satisfies
Let R := inf{R0 > 0 : Ω ⊂ B(0, R0 )}. Then f (x), x ∈ Ω F (x) := 0, x ∈ B(0, R) − Ω,
(7.37)
(7.38)
F ∈ W n,∞ (B(0, R)) ∩ C0n−1 (B(0, R)). Then for x ∈ Ω Z f (x) − − f (y)dy ≤ Ω
where
Z F (x) − − F (y)dy B(0,R) Z Z + − F (y)dy − − f (y)dy = J1 + J2 , B(0,R) Ω
Z J1 := F (x) − − F (y)dy B(0,R)
(7.39)
(7.40)
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and
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Z 1 1 − J2 := f (y)dy Vol(B(0, R)) Vol(Ω) Ω Z Vol(Ω) (7.41) − f (y)dy . = 1− Vol(B(0, R)) Ω
n Theorem 7.1 applies and we may replace ∂∂rFn L∞ (B(0,R)) by its majorant kDn f kL∞(Ω) . The theorem then follows.
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Chapter 8
Ostrowski Inequalities on Spherical Shells
Here are presented Ostrowski type inequalities over spherical shells. These regard sharp or close to sharp estimates to the difference of the average of a multivariate function from its value at a point. This chapter relies on [47]. 8.1
Introduction
The famous Ostrowski’s inequality (1938), see [196], is ! 1 Z b 2 1 (x − a+b 2 ) (b − a)kf 0 k∞ , f (y)dy − f (x) ≤ + b − a a 4 (b − a)2
for f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. This was generalized from intervals to boxes in RN , N ≥ 1 , see [17], [21], p. 507. Here we establish Ostrowski type inequalities over spherical shells. We present first the sharp results for the radial functions, then we move to the non-radial case. We use the polar method. The estimates in both cases involve radial derivatives of arbitrary order of the engaged function. At the end we give the connection of radial derivative to the ordinary partial derivative of the function. 8.2
Main Results
We make Remark 8.1. Let A be a spherical shell ⊆ RN , N ≥ 1, i.e. A := B(0, R2 ) − B(0, R1 ), 0 < R1 < R2 . Here the ball B(0, R) := {x ∈ RN : |x| < R}, R > 0 , where | · | is the Euclidean norm , also S N −1 := {x ∈ RN : |x| = 1} is the unit sphere in RN with surface 2π N/2 area ωN := Γ(N/2) . For x ∈ RN − {0} one can write uniquely x = rω, where r > 0, ω ∈ S N −1 . 109
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¯ We suppose first that f is radial i.e f (x) = g(r), where r = Let f ∈ C 1 (A). |x|, R1 ≤ r ≤ R2 . Clearly here g ∈ C 1 ([R1 , R2 ]). In general it holds k ∂f ∂r k∞ ≤ k 5 f k∞ , with equality in the radial case. ¯ we have For F ∈ C(A) ! Z R2 Z Z N −1 F (rω)r dr dω. F (x)dx = S N −1
A
We notice that
Z
N N R2 − R1N
and
R1
R2
sN −1 ds = 1,
ωN (R2N − R1N ) . N Let x ∈ A. Then by using the polar method we derive R R R2 R N −1 N f (sω)s ds dω N −1 S R1 f (x) − A f (y)dy = f (x) − V ol(A) ωN (R2N − R1N ) V ol(A) =
R R R N S N −1 R12 g(s)sN −1 ds dω = g(r) − ωN (R2N − R1N ) Z R2 N N −1 = g(r) − g(s)s ds N N R2 − R 1 R1
! Z R2 Z R2 N N −1 N −1 g(s)s ds = g(r)s ds − R2N − R1N R1 R1 =
N N R2 − R1N
≤
N R2N − R1N
R2
R1
N R2N − R1 =
R1
N N R2 − R1N
(8.2)
(8.3)
(8.4)
(8.5)
(8.6)
|g(r) − g(s)| sN −1 ds ≤
Z R2 N 0 kg k∞ |r − s| sN −1 ds R2N − R1N R1 # "Z Z R2 r N −1 0 N −1 (s − r)s ds kg k∞ (r − s)s ds + N
=
Z R2 (g(r) − g(s)) sN −1 ds R1
Z
(8.1)
R1
r
N 2r − (R1N + R2N ) kg 0 k∞ r N
(8.7)
(8.8)
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R1N +1 + R2N +1 − 2rN +1 N +1
+
!#
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.
(8.9)
So we have established the first main result. ¯ be radial, i.e. f (x) = g(r), R1 ≤ r ≤ R2 , x ∈ A. ¯ Theorem 8.2. Let f ∈ C 1 (A) Then R Z R2 f (y)dy N N −1 A f (x) − = g(r) − g(s)s ds ≤ V ol(A) RN − RN 2
N N R2 − R1N
N 2r − (R1N + R2N ) kg k∞ r N 0
R1N +1 + R2N +1 − 2rN +1 N +1
+
=
R1
1
N R2N − R1N
(8.10)
2|x|N − (R1N + R2N ) k5f k∞ |x| N
R1N +1 + R2N +1 − 2|x|N +1 N +1
+
!#
!#
.
(8.11)
Optimality comes next Theorem 8.3. Inequality (8.10) is sharp. More precisely (i) it is asymptotically attained by g ∗ (z) := |z−r|α , 1 < α ≤ k, when 0 < r < R. (ii) It is attained by g ∗ (z) = (z − R1 ), when r = R1 . (iii) It is attained by g ∗ (z) = (z − R2 ), when r = R2 . Proof.
(i) We observe that 0
g ∗ (z) = α|z − r|α−1 sign(z − r) and 0
kg ∗ k∞ = α(max{R2 − r, r − R1 })α−1 , along with g ∗ (r) = 0. We see that L.H.S(8.10) =
α→1
−→
N R2N − R1N
N N R2 − R1N
Z
Z R2
R1
R2 R1
α
|s − r| sN −1 ds
|s − r| sN −1 ds
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=
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N R2N − R1N
" N 2r − (R1N + R2N ) r + N
We also have R.H.S(8.10) =
N N R2 − R1N
" 2rN − (R1N + R2N ) + r N
N R2N − R1N
R1N +1 + R2N +1 − 2rN +1 N +1
!#
. (8.12)
α(max{R2 − r, r − R1 })α−1
R1N +1 + R2N +1 − 2rN +1 N +1
" N 2r − (R1N + R2N ) + r N
!#
α→1
−→
R1N +1 + R2N +1 − 2rN +1 N +1
!#
.
(8.13)
That is lim L.H.S(8.10) = lim R.H.S(8.10),
α→1
α→1
proving sharpness for the case. 0 (ii) We have g ∗ (R1 ) = 0 and kg ∗ k∞ = 1. Thus Z R2 N L.H.S(8.10) = (s − R1 )sN −1 ds R2N − R1N R1 ! # " N R2N − R1N R2N +1 − R1N +1 = − R1 = R.H.S (8.10), (8.14) N +1 N R2N − R1N proving the attainability for the case. (iii) We have g ∗ (R2 ) = 0, and kg ∗0 k∞ = 1. Hence Z R2 N L.H.S(8.10) = (R2 − s)sN −1 ds = R2N − R1N R1 !# " N R2 − R1N R2N +1 − R1N +1 N R2 − = R.H.S (8.10), N N +1 R2N − R1N proving attainability of last case.
(8.15)
We would line to rewrite Theorem 8.3 for the equivalent inequality (8.10) in terms of f . We have ¯ Inequality (8.10) is sharp as follows: Theorem 8.4. Let x ∈ A. (i) Let 0 < |x| < R, then it is asymptotically attained by f ∗ (w) :=| |w| − |x| |α , 1 < α ≤ k.
(ii) Let |x| = R1 , then it is asymptotically attained by f ∗ (w) = |w| − R1 .
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(iii) Let |x| = R2 , then it is asymptotically attained by f ∗ (w) = |w| − R2 . We continue from Remark 8.1 into ¯ n ∈ N , again radial such that f (x) = g(r), where Remark 8.5. Now f ∈ C n (A), ¯ r = |x|, x ∈ A, R1 ≤ r ≤ R2 . Hence g ∈ C n ([R1 , R2 ]). Using the polar method we get again R Z R2 f (y)dy N N −1 A = E := f (x) − (g(r) − g(s))s ds (8.16) . V ol(A) R2N − R1N R1 Let s, r ∈ [R1 , R2 ], then by Taylor’s formula we obtain g(s) − g(r) = where Rn−1 (r, s) :=
Z
n−1 X k=1
s r
As in [21], p. 500, we find
g (k) (r) (s − r)k + Rn−1 (r, s), k!
g (n−1) (t) − g (n−1) (r)
|Rn−1 (r, s)| ≤ ∀ s, r ∈ [R1 , R2 ]. Therefore Z R2 N E= R2N − R1N R1 ≤
N R2N − R1N
"n−1 X k=1
≤
n−1 X k=1
|g (k) (r)| k!
N N R2 − R1N
But one finds that
kg (n) k∞,[R1 ,R2 ] |s − r|n , n!
! g (k) (r) k N −1 (s − r) + Rn−1 (r, s) s ds k!
(8.18)
(8.19)
(8.20)
Z Z # R2 R2 N −1 N −1 k s |Rn−1 (r, s)|ds s (s − r) ds + R1 R1
"n−1 X
R2N +k−m − R1N +k−m N +k−m
(s − t)n−2 dt. (n − 2)!
(8.17)
k=1
|g k (r)| k!
k X k (−1)m rm m m=0
! # kg (n) k Z R2 ∞ N −1 n s |s − r| ds . + n! R1 Z
R2
R1
sn−1 |s − r|n ds
(8.21)
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" n X n m rn−m (−1) = m m=0
rm+N − R1m+N m+N
!
+r
m
R2N +n−m − rN +n−m N +n−m
!#
.
(8.22)
Putting all the above together we have derived ¯ n ∈ N, be radial, i.e. f (x) = g(r), R1 ≤ r ≤ Theorem 8.6. Let f ∈ C n (A), ¯ R2 , x ∈ A. Then R Z R2 f (y)dy N N −1 A f (x) − = g(r) − g(s)s ds V ol(A) R2N − R1N R1 ≤
N R2N − R1N
"n−1 X k=1
+ g (n)
∞
"
k X (−1)m rm |g (k) (r)| m!(k − m)! m=0
" (−1)m rn−m m!(n − m)! m=0 n X
+rm
=
N N R2 − R1N
R2N +k−m − R1N +k−m N +k−m
rm+N − R1m+N m+N
R2N +n−m − rN +n−m N +n−m
!###
k (n−1 X ∂ k f (x) X (−1)m |x|m ∂rk m!(k − m)! m=0 k=1
" " n
n X (−1)m
∂ f
|x|n−m · + n ∂r ∞,A¯ m=0 m!(n − m)! + |x|
m
!
!
R2N +n−m − |x|N +n−m N +n−m
(8.23)
R2N +k−m − R1N +k−m N +k−m |x|m+N − R1m+N m+N !##)
.
!
!
(8.24)
We give ¯ n ∈ N, be radial i.e. f (x) = f (r), R1 ≤ r ≤ Corollary 8.7. Let f ∈ C n (A), i ∂ f (x ) 0 ¯ Suppose R2 , x ∈ A. = g (i) (r0 ) = 0, i = 1, . . . , n − 1, for r0 ∈ [R1 , R2 ], x0 = ∂ i ri N −1 ¯ r0 ω ∈ A, ω ∈ S . Then R Z R2 f (y)dy N N −1 A = g(r0 ) − f (x0 ) − g(s)s ds N N V ol(A) R2 − R 1 R1 N kg (n) k∞ ≤ N R2 − R1N
"
n X
(−1)m m!(n − m)! m=0
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Book˙Adv˙Ineq
Ostrowski Inequalities on Spherical Shells
"
!
r0m+N − R1m+N m+N
r0n−m
+
n
= "
|x0 |n−m
N k ∂∂rnf k∞,A R2N
|x0 |m+N − R1m+N m+N
−
!
R2N +n−m − r0N +n−m N +n−m
r0m
R1N
"
115
!##
n X
(−1)m m!(n − m)! m=0
+ |x0 |m
R2N +n−m − |x0 |N +n−m N +n−m
(8.25)
!##
.
(8.26)
We also have the extreme cases. ¯ n ∈ N, be radial ; f (x) = g(r), r ∈ [R1 , R2 ], x ∈ Corollary 8.8. Let f ∈ C n (A), i ∂ f (x) ¯ x = rω. Suppose that A, ∂r i , i = 1, . . . , n − 1, are zero on ∂B(0, R1 ), i.e. g (i) (R1 ) = 0, i = 1, . . . , n − 1. Then for x ∈ ∂B(0, R1 ) we have R Z R2 f (y)dy N N −1 A = g(R1 ) − f (x) − g(s)s ds V ol(A) R2N − R1N R1 ≤
=
N N R2 − R1N
N R2N − R1N
kg
(n)
k∞
"
n X
(−1)m R1m m!(n − m)! m=0
R2N +n−m − R1N +n−m N +n−m
" n n X (−1)m
∂ f
R1m
∂rn ¯ m!(n − m)! ∞,A m=0
!#
(8.27)
R2N +n−m − R1N +n−m N +n−m
!#
.
(8.28)
Another extreme case follows.
¯ n ∈ N, be radial; f (x) = g(r), r ∈ [R1 , R2 ], x ∈ Corollary 8.9. Let f ∈ C n (A), i f (x) barA; x = rω. Assume that ∂ ∂r i , i = 1, . . . , n − 1, are zero on ∂B(0, R2 ), i.e., (i) g (R2 ) = 0, i = 1, . . . , n − 1. Then for x ∈ ∂B(0, R2 ) we have R Z R2 f (y)dy N N −1 f (x) − A = g(R2 ) − g(s)s ds V ol(A) R2N − R1N R1
=
≤
N N R2 − R1N
N R2N − R1N
kg (n) k∞
"
n X
(−1)m R2n−m m!(n − m)! m=0
" n n X (−1)m
∂ f
R2n−m
∂rn ¯ ∞,A m=0 m!(n − m)!
R2m+N − R1m+N m+N
!#
R2m+N − R1m+N m+N
Optimality follows.
Proposition 8.10. Inequality (8.25) is sharp, namely it is attained by g ∗ (y) := (y − r0 )n ,
y,
r0 ∈ [R1 , R2 ], when n is even.
!#
(8.29)
. (8.30)
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Proof.
Notice (j)
g ∗ (r0 ) = 0,
j = 0, 1, . . . ,
We observe L.H.S(8.25) =
=
N R2N − R1N
Next we see that
"X n n (−1)m r0m m m=0
R.H.S (8.25) = " =
N R2N − R1N
=
!
+
R2 R1
(n)
k∞ = n!.
(s − r0 )n sN −1 ds
R2N +n−m − R1N +n−m N +n−m
!#
.
"X n n (−1)m m m=0
r0m R2N +n−m − r0N +n N +n−m
!##
n X n 1 1 m − (−1) N +m N +n−m m m=0
N R2N − R1N
=
N R2N − R1N
=
N N R2 − R1N
(8.33)
(8.34)
(X n rm RN +n−m n (−1)m 0 2 N +n−m m m=0
) n N +n−m X n m m R1 − (−1) r0 m N +n−m m=0
(8.32)
(X n n rm RN +n−m (−1)m 0 2 N +n−m m m=0
) n n−m m+N X n R1 m r0 − (−1) m+N m m=0
(8.31)
" # (X n m N +n−m n r0n−m R1m+N m r0 R2 (−1) − m N +n−m m+N m=0
+r0n+N
Z
N R2N − R1N
r0n+N − r0n−m R1m+N m+N
N R2N − R1N
n − 1 and kg ∗
(8.35)
) (X n N +n−m n − R1N +n−m ) m m (R2 (−1) r0 N +n−m m m=0 = L.H.S
That is proving the claim.
(8.25).
(8.36)
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The other optimal case follows. Proposition 8.11. Inequality (8.25) is sharp, namely it is asymptotically attained by g ∗ (y) := |y − r0 |n−1+α , y, r0 ∈ [R1 , R2 ], 1 < α ≤ T , in the case of n is odd. g∗
Proof. It holds Also we have g∗ That is
(n)
(
k)
(r0 ) = 0,
k = 0, 1, . . . , n − 1.
(y) = (n − 1 + α)(n − 2 + α) · · · (α + 1)α|y − r0 |α−1 sign(y − r0 ). n (n) ∗ Y (y) = (n − j + α) |y − r0 |α−1 , g j=1
and
(n)
∗
g
∞
Consequently,
=
n Y
j=1
(n − j + α) (max{R2 − r0 , r0 − R1 })α−1 .
Z R2 N α→1 |s − r0 |n−1+α sN −1 ds −→ R2N − R1N R1 Z R2 N |s − r0 |n sN −1 ds R2N − R1N R1 ! " (X n N n r0m+N − R1m+N n−m m = (−1) r0 m+N m R2N − R1N m=0 !#) R2N +n−m − r0N +n−m +r0m =: µ. N +n−m L.H.S(8.25) =
Next we derive
R.H.S(8.25) = (
n X n (−1)m m m=0
(
N N R2 − R1N ( r0n−m
(8.38) (8.39)
(8.40)
Qn (n − j + α) (max{R − r , r − R })α−1 2 0 0 1 j=1 n!
r0m+N − R1m+N m+N
!
+ r0m
R2N +n−m − r0N +n−m N +n−m
(X n n N α→1 (−1)m −→ N N m R2 − R 1 m=0 ! !)) r0m+N − R1m+N R2N +n−m − r0N +n−m m + r0 = µ. m+N N +n−m
r0n−m
(8.37)
!)) (8.41)
(8.42)
That is, L.H.S (8.25), R.H.S(8.25) → µ, proving asymptotic attainability and sharpness of (8.25).
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Optimality of external inequality (8.25) follows. Proposition 8.12. External inequality (8.25) is asymptotically attained, that is sharp as follows: ¯ (i) when n is even, then optimal function is f ∗ (w) := (|w| − |x0 |)n , w ∈ A. ∗ n−1+α (ii) when n is odd, then optional function is f (w) :=| |w| − |x0 | | ,1< ¯ α ≤ T, w ∈ A. Similarly we obtain
Proposition 8.13. Inequalities (8.23), (8.27) and (8.29) are asymptotically attained, therefore sharp, as inequality (8.25). A simple but general result follows. Theorem 8.14. Let ∅ 6= R be a convex bounded region of RN , N ≥ 1. Let f ∈ C 1 (R). Then R Z f (y)dy k∇f k∞ f (x) − R ≤ |x − y|dy, x ∈ R. (8.43) V ol(R) V ol(R) R
Proof.
We observe that R Z 1 f (x) − R f (y)dy = f (x)V ol(R) − f (y)dy V ol(R) V ol(R) R Z 1 (f (x) − f (y))dy = V ol(R) R Z 1 ≤ |f (x) − f (y)|dy V ol(R) R = (z belongs to the line segment from x to y) Z 1 |∇f (z) · (x − y)|dy V ol(R) R ≤
1 V ol(R)
Z
R
k∇f k∞ V ol(R)
|∇f (z)| · |x − y|dy ≤
Specializing on the shell and sphere we have
Z
R
|x − y|dy.
N Γ( N2 ) k∇f k∞ 2π N/2 (R2N − R1N )
Z
A
|x − y| dy,
(8.45)
(8.46)
¯ or f ∈ C 1 (B(0, R)), R > 0. Then Proposition 8.15. Let f ∈ C 1 (A), (i) Z N Γ( N2 ) f (y)dy f (x) − N/2 N 2π (R2 − R1N ) A ≤
(8.44)
¯ x ∈ A;
(8.47)
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also it holds (ii) Z Z N Γ( N2 ) N Γ( N2 ) f (y)dy ≤ |x − y| dy, k∇f k∞ f (x) − N/2 N 2π N/2 RN 2π R B(0,R) B(0,R)
x ∈ B(0, R). More precise Ostrowski type inequalities for general, not necessarily radial functions, follow. ¯ x ∈ A¯ , x = rω, r > 0. Then Theorem 8.16. Let f ∈ C 1 (A), R R f (y)dy A ≤ f (x) − S N −1 f (rω)dω f (x) − V ol(A) ωN +
N R2N − R1N +
" Proof.
|x|
∂f
∂r
∞
|x|
2|x|N − (R1N + R2N ) N
R1N +1 + R2N +1 − 2|x|N +1 N +1
!#
(8.48)
R Γ( N2 ) S N −1 f (rω)dω N k∇f k∞ + ≤ f (x) − 2π N/2 R2N − R1N
2|x|N − (R1N + R2N ) N
+
R1N +1 + R2N +1 − 2|x|N +1 N +1
!#
.
(8.49)
Applying internal (8.10) to f (rω) we get Z R2 N N −1 f (sω)s ds f (rω) − R2N − R1N R1
≤
N R2N − R1N
N
∂f (rω) 2r − (R1N + R2N )
r
∂r N ∞,r∈[R1 ,R2 ] +
≤
N N R2 − R1N +
R1N +1 + R2N +1 − 2rN +1 N +1
!#
(8.50)
∂f 2|x|N − (R1N + R2N )
∂r ¯ |x| N ∞,A R1N +1 + R2N +1 − 2|x|N +1 N +1
!#
.
(8.51)
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Hence it holds R Z N S N −1 f (rω)dω − ωN ωN (R2N − R1N ) S n−1 ≤
N R2n − R1N
Z
R2
f (sω)s R1
ds dω !
∂f 2|x|N − (R1N + R2N )
|x|
∂r N ∞
R1N +1 + R2N +1 − 2|x|N +1 N +1
+
N −1
!#
,
(8.52)
proving the claim.
We continue with ¯ ¯ x = rω, r > 0. Then Theorem 8.17. Let f ∈ C n (A), n ∈ N, x ∈ A, R R N f (y)dy Γ( f (rω)dω ) N −1 S 2 f (x) − A ≤ f (x) − V ol(A) 2π N/2 +
N R2N − R1N
(
Γ( N2 ) 2π N/2
k X (−1)m |x|m m!(k − m)! m=0
(n−1 Z X k=1
S
k ∂ f (rω) ∂rk dω N −1
R2N +k−m − R1N +k−m N +k−m
"
n "X n
∂ f (−1)m
|x|n−m + n ∂r ∞ m=0 m!(n − m)! +|x|
Proof.
m
! )
|x|m+N − R1m+N m+N
R2N +n−m − |x|N +n−m N +n−m
!##)
!
.
Applying internal (8.23) to f (rω) we obtain Z R2 N N −1 f (sω)s ds f (rω) − R2N − R1N R1 ≤
N R2N − R1N
(n−1 X ∂ k f (rω) ∂rk k=1
R2N +k−m − R1N +k−m N +k−m
k X (−1)m |x|m m! (k − m)! m=0
! " n ∂nf X (−1)m
+ n
∂r ∞,A¯ m=0 m!(n − m)!
(8.53)
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Ostrowski Inequalities on Spherical Shells
"
|x|
n−m
|x|m+N − R1m+N m+N
Therefore it holds
≤
!
+ |x|
m
121
R2N +n−m − |x|N +n−m N +n−m
!##)
.
(8.54)
Z Γ( N ) Z 1 2 f (y)dy f (rω)dω − N/2 2π V ol(A) A S N −1
N R2 − R1N
(
Γ(N/2) 2π N/2
k X (−1)m |x|m m!(k − m)! m=0
(n−1 Z X k=1
S
k ∂ f (rω) ∂rk dω N −1
R2N +k−m − R1N +k−m N +k−m
" n "
n X (−1)m
∂ f
+ n |x|n−m ∂r ∞,A¯ m=0 m!(n − m)! +|x|
m
! )
|x|m+N − R1m+N m+N
R2N +n−m − |x|N +n−m N +n−m
!##)
!
,
(8.55)
proving the claim.
We also give i
¯ n ∈ N, such that ∂ fi , i = 1, . . . , n − 1, are zero Proposition 8.18. Let f ∈ C n (A), ∂r on ∂B(0, r0 ), r0 ∈ (R1 , R2 ). Then for x0 ∈ ∂B(0, r0 ) we have R Z f (x0 ) − A f (y)dy ≤ f (x0 ) − Γ(N/2) f (r ω)dω 0 N/2 V ol(A) 2π S N −1 +
N R2N − R1N
" n "X n m
∂ f (−1)
|x0 |n−m
∂rn m!(n − m)! ∞ m=0 +|x0 |
Proof.
m
R2N +n−m − |x0 |N +n−m N +n−m
|x0 |m+N − R1m+N m+N
!##
.
Applying internal (8.25) to f (rω) we find Z R2 N N −1 f (sω)s ds f (r0 ω) − R2N − R1N R1 ≤
N R2N − R1N
" n n X (−1)m
∂ f
∂rn ¯ ∞,A m=0 m!(n − m)!
!
(8.56)
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122
"
|x0 |
n−m
|x0 |m+N − R1m+N m+N
Therefore
≤
N N R2 − R1N
!
+ |x0 |
m
R2N +n−m − |x0 |N +n−m N +n−m
!##
.
Z Γ( N ) Z 1 2 f (r0 ω)dω − f (y)dy N/2 2π V ol(A) A S N −1
" n "X n
∂ f (−1)m
|x0 |n−m
∂rn ∞ m=0 m!(n − m)! +|x0 |
m
R2N +n−m − |x0 |N +n−m N +n−m
|x0 |m+N − R1m+N m+N
!##
.
(8.57)
! (8.58)
Claim is clear.
We present the extreme cases. i ¯ Proposition 8.19. Let f ∈ C n (A), n ∈ N such that ∂∂rfi , i = 1, . . . , n − 1, are zero on ∂B(0, R1 ). Then for x0 ∈ ∂B(0, R1 ) it holds R N Z f (x0 ) − A f (y)dy ≤ f (x0 ) − Γ( 2 ) f (R ω)dω 1 V ol(A) 2π N/2 S N −1
+
N N R2 − R1N
Proof.
n "X n
∂ f (−1)m
R1m
∂rn ∞ m=0 m!(n − m)!
R2N +n−m − R1N +n−m N +n−m
!#
By internal (8.27).
(8.59)
We finish the main results with i ¯ Proposition 8.20. Let f ∈ C n (A), n ∈ N, such that ∂∂rfi , i = 1, . . . , n − 1, are zero on ∂B(0, R2 ). Then for x0 ∈ ∂B(0, R2 ) we find R (N ) Z f (x0 ) − A f (y)dy ≤ f (x0 ) − Γ 2 f (R ω)dω 2 V ol(A) 2π N/2 S N −1
+
N N R2 − R1N
Proof.
n "X n
∂ f (−1)m
R2n−m
∂rn ∞ m=0 m!(n − m)!
By internal (8.29).
R2m+N − R1m+N m+N
!#
.
(8.60)
The radial derivatives appearing in the right hand sides of the inequalities can be expressed and estimated by regular partial derivatives in terms of x1 , . . . , xN . See Addendum next.
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8.3
Book˙Adv˙Ineq
123
Addendum
Let u ∈ C n (B(0, R)), the open ball B(0, R) ⊆ RN , n, N ∈ N. (x1 , · · · , xN ) ∈ B(0, R) and the radial derivative of u is given also by x ∂u(x) = ∇u(x) · , ∂r |x|
That is,
∂u(x) 1 = ∂r |x|
N X ∂u(x)
∂xi
i=1
Here x =
x 6= 0.
xi
!
,
x 6= 0.
(8.61)
In general for 1 ≤ l ≤ n it holds (by induction) that X 1 l! ∂ l u(x) ∂ l u(x) k = · N k ΠN x j , x 6= 0. N l l ∂r |x| Πj=1 kj ! Πj=1 ∂ j xj j=1 j PN k1 ;...;kN ,
j=1
kj =l,kj ∈Z+
(8.62)
For example, when n = N = 2 we obtain 1 ∂ 2 u(x) 2 2∂ 2 u(x) ∂ 2 u(x) 2 ∂ 2 u(x) = x + 2 x x + x , x 6= 0. 1 2 ∂r2 |x|2 ∂x21 1 ∂x1 ∂x2 ∂x22 2 Thus we find l ∂ u(x) ∂rl ≤
or better in brief,
k1 ;...;kN ,
X
PN
j=1 kj =l,kj ∈Z+
l ∂ u(x) ∂rl ≤
∂ l u(x) · N k , x 6= 0, N j Πj=1 kj ! Πj=1 ∂ xj l!
N X ∂ ∂xi i=1
!l
(u(x)) ,
(8.63)
(8.64)
(8.65)
all x ∈ B(0, R) − {0}, all 1 ≤ l ≤ n. So if all the u partial derivatives vanish then the corresponding radial derivative is zero. Consequently, from (8.65) it holds for the essential supreme k · k∞ that !l
l N X
∂
∂ u
≤ (u) < +∞. (8.66)
∂xi
∂rl ∞,B(0,R) ∞,B(0,R) i=1
II. Continuing and specializing in the study of higher order radial derivatives of radial functions. Let now u be radial i.e. u(x) = g(|x|) = g(r); x = rω, r ∈ R+ , ω ∈ S N −1 . Here we will suppose x 6= 0 and N = 2. Let further x1 , x2 6= 0, then by chain rule one has q ∂u r ∂u r g 0 (r) = = , where r = x21 + x22 . (8.67) ∂x1 x1 ∂x2 x2
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That is,
∂u |x| 1 ∂u |x| + . g (r) = 2 ∂x1 x1 ∂x2 x2 0
So one has
Book˙Adv˙Ineq
∂u(x) , |∇u(x)| = |g (r)| = ∂r 0
(8.68)
(8.69)
for any x ∈ B(0, R). ∂u But if u radial, not necessarily ∂x is radial. Here we put x1 = r cos θ, x2 = j r sin θ. Again for x1 , x2 6= 0 and via chain rule we derive ∂ 2 u(x) ∂ 2 u(x) ∂ 2 u(x) ∂ 2 u(x) g 00 (r) = + tan θ = + cot θ. (8.70) ∂x21 ∂x2 ∂x1 ∂x22 ∂x1 ∂x2 That is ∂ 2 u(x) 1 ∂ 2 u(x) ∂ 2 u(x) + + (tan θ + cot θ) . (8.71) g 00 (r) = 2 ∂x21 ∂x22 ∂x1 ∂x2 Or better, by using the Laplacian ∆ we find ∂ 2 u(x) 1 g 00 (r) = ∆u(x) + csc(2 θ), (8.72) 2 ∂x1 ∂x2 or 1 ∂ 2 u(x) |x|2 00 g (r) = ∆u(x) + , (8.73) 2 ∂x1 ∂x2 x1 x2 x1 , x2 6= 0. Similarly, one has that ∂ 3 u(x) ∂ 3 u(x) sin2 θ ∂ 3 u(x) cos θ + 2 sin θ + , x1 , x2 6= 0. (8.74) g 000 (r) = ∂x31 ∂x2 ∂x21 ∂x22 ∂x1 cos θ Also we get ∂ 3 u(x) ∂ 3 u(x) ∂ 3 u(x) cos2 θ g 000 (r) = 2 , x1 , x2 6= 0. (8.75) cos θ + sin θ + ∂x1 ∂x22 ∂x32 ∂x21 ∂x2 sin θ That is, by (8.74) and (8.75) we have ∂ 3 u(x) 1 ∂ 3 u(x) 000 cos θ + sin θ+ g (r) = 2 ∂x31 ∂x32 3 ∂ 3 u(x) cos2 θ ∂ 3 u(x) ∂ 3 u(x) sin2 θ ∂ u(x) + sin θ + cos θ + , x1 , x2 6= 0. 2 ∂x2 ∂x21 ∂x1 ∂x22 ∂x22 ∂x1 cos θ ∂x21 ∂x2 sin θ (8.76) Clearly, it holds 1 ∂ 3 u(x) x1 ∂ 3 u(x) x2 g 000 (r) = + + 2 ∂x31 |x| ∂x32 |x| 2 2 x2 x1 x1 ∂ 3 u(x) x2 ∂ 3 u(x) + 2 + 2 + , x1 , x2 6= 0. (8.77) ∂x21 ∂x2 x2 |x| |x| ∂x22 ∂x1 x1 |x| |x| Not general formula, as in (8.62), can be derived for g (l) (r), l ∈ N, in the radial case. Of course (8.62) is valid for both non-radial and radial cases. Notice in (8.67), (8.70), (8.74) and (8.75) are used less number of terms than in the corresponding non-radial cases.
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Chapter 9
Ostrowski Inequalities on Balls and Shells Via Taylor Widder Formula
The classical Ostrowski inequality for functions on intervals estimates the value of the function minus its average in terms of the maximum of its first derivative. This result is extended to higher order over shells and balls of RN , N ≥ 1, with respect to an extended complete Tschebyshev system and the generalized radial derivatives of Widder type. We treat radial and non-radial functions. This chapter relies on [42].
9.1
Introduction
The classical Ostrowski inequality (of 1938, see [196]) is 2 a+b x− 1 1 Z b 2 f (y)dy − f (x) ≤ + (b − a)kf 0 k∞ , 4 b − a a (b − a)2 for f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. This was extended to RN , N ≥ 1, over balls and shells in [56], [57], [47]. Also this extension was done over boxes and rectangles, see [21], pp. 507-520, and [17], and [16]. The produced Ostrowski inequalities, in the above mentioned references, were mostly sharp and they involved the first and higher order derivatives of the engaged function f . Here we derive a set of very general higher order Ostrowski type inequalities over shells and balls with respect to an extended complete Tschebyshev system (see [163])and generalized derivatives if Widder type (see [243]). The proofs are based on the polar method and the general Taylor-Widder formula (see [243], 1928). These results generalize the higher order Ostrowski type inequalities established in the above mentioned references. 125
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9.2
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Background
The following are taken from [243]. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]),n ≥ 0, and the Wronskians u0 (x) u1 (x) . . . ui (x) u0 (x) u0 (x) . . . u0 (x) 0 1 i , i = 0, 1, . . . , n. Wi (x) := W [u0 (x), u1 (x), . . . , ui (x)] := .. . (i) (i) u0 (x) u(i) 1 (x) . . . ui (x) Assume Wi (x) > 0 over [a, b]. Clearly then φ0 (x) := W0 (x) = u0 (x), φ1 (x) :=
Wi (x)Wi−2 (x) W1 (x) , . . . , φi (x) := , (W0 (x))2 (Wi−1 (x))2
i = 2, 3, . . . , n, are positive on [a, b]. For i ≥ 0, the linear differentiable operator of order i: Li f (x) :=
W [u0 (x), u1 (x), . . . , ui−1 (x), f (x)] , Wi−1 (x)
i = 1, . . . , n + 1;
L0 f (x) := f (x), ∀ x ∈ [a, b]. Then for i = 1, . . . , n + 1 we have Li f (x) = φ0 (x)φ1 (x) . . . φi−1 (x) Consider also
d d 1 d f (x) 1 d 1 d ... . dx φi−1 (x) dx φi−2 (x) dx dx φ1 (x) dx φ0 (x)
u0 (t) u1 (t) u0 (t) u01 (t) 1 0 ... gi (x, t) := ... Wi (t) (i−1) (i−1) u0 (t) u (t) 1 u0 (x) u1 (x)
. . . ui (t) . . . u0i (t) ... . . . , (i−1) . . . ui (t) . . . ui (x)
u0 (x) , ∀ x, t ∈ [a, b]. u0 (t) Note that gi (x, t) as a function of x is a linear combination of u0 (x), u1 (x), . . . , ui (x) and it holds
i = 1, 2, . . . , n; g0 (x, t) :=
gi (x, t)
Z xi−1 Z xi−2 Z x Z x1 φ0 (x) φi (xi )dxi dxi−1 . . . dx1 φi−1 (xi−1 ) ... φ1 (x) = φ0 (t) . . . φi (t) t t t t Z x 1 = φ0 (s) . . . φi (s)gi−1 (x, s)ds, i = 1, 2, . . . , n. φ0 (t) . . . φi (t) t
Example 9.1. ([243]). The sets {1, x, x2 , . . . , xn }, {1, sin x, − cos x, − sin 2x, cos 2x, . . . , (−1)n−1 sin nx, (−1)n cos nx}
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127
fulfill the above theory. We mention Theorem 9.2. (Karlin and Studded (1966), see p. 376, [163]). Let u0 , u1 , . . . , un ∈ C n ([a, b]), n ≥ 0. Then {ui }ni=0 is an extended complete Tschebyshev system on [a, b] iff Wi (x) > 0 on [a, b], i = 0, 1, . . . , n. We also mention Theorem 9.3. (D. Widder, p. 138, [243]). Let the functions f (x), u0 (x), u1 (x), . . . , un (x) ∈ C n+1 ([a, b]), and the Wronskians W0 (x), W1 (x), . . . , Wn (x) > 0 on [a, b], x ∈ [a, b]. Then for t ∈ [a, b] we have f (x) = f (t)
u0 (x) + L1 f (t)g1 (x, t) + · · · + Ln f (t)gn (x, t) + Rn (x), u0 (t)
where Rn (x) :=
Z
x
gn (x, s)Ln+1 f (s)ds. t
For example one could take u0 (x) = c > 0. If ui (x) = xi , i = 0, 1, . . . , n, defined on [a, b], then (x − t)i , t ∈ [a, b]. i! So under the assumptions of Theorem 9.3 we get Z x n u0 (x) X f (x) = f (y) + Li f (y)gi (x, y) + gn (x, t)Ln+1 f (t)dt, ∀ x, y ∈ [a, b]. u0 (y) i=1 y (9.1) If u0 (x) = c > 0, then Z x n X f (x) = f (y) + Li f (y)gi (x, y) + gn (x, t)Ln+1 f (t)dt, ∀ x, y ∈ [a, b]. (9.2) Li f (t) = f (i) (t)
i=1
and gi (x, t) =
y
We call Li the generalized Widder-type derivative. We need
Notation 9.4. Let A be a spherical shell ⊆ RN , N ≥ 1, i.e. A := B(0, R2 ) − B(0, R1 ), 0 < R1 < R2 . Here the ball B(0, R) := {x ∈ RN : |x| < R}, R > 0, where | · | is the Euclidean norm, also S N −1 := {x ∈ RN : |x| = 1} is the unit sphere in RN with surface area 2π N/2 . For x ∈ RN − {0} one can write uniquely x = rω, where r > 0, ωN := Γ(N/2) ω ∈ S N −1 . Let f ∈ C n+1 (A), n ≥ 0. If f is radial i.e. f (x) = g(r), where r = |x|, R1 ≤ r ≤ R2 , then g ∈ C n+1 ([R1 , R2 ]).
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For radial f define θi f (x) := Li g(r),
all i = 1, . . . , n + 1, ∀ x ∈ A.
(9.3)
Here g (i) (r) =
∂ i f (x) , ∂ri
For F ∈ C(A) we have Z Z F (x)dx = A
S N −1
We notice that N N R2 − R1N
and
V ol(A) =
9.3
Z Z
i = 1, . . . , n + 1.
R2
F (rω)r
N −1
R1
R2
dr
!
dω.
sN −1 ds = 1,
(9.4)
(9.5)
R1
ωN (R2N − R1N ) . N
Results on the Shell
We make Remark 9.5. Let here u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]), and W0 , W1 , . . . , Wn > 0 on [R1 , R2 ], 0 < R1 < R2 , n ≥ 0 integer, with u0 (r) = c > 0. Let also f ∈ C n+1 (A). We suppose first that f is radial, i.e. there exists g such that f (x) = g(r), r = |x|, R1 ≤ r ≤ R2 . Clearly g ∈ C n+1 ([R1 , R2 ]). Let x ∈ A. Then by using the polar method (9.4) we obtain ! Z R2 Z Z N −1 f (sω)s ds dω N f (y)dy N −1 R S 1 A (9.6) E(x) := f (x) − = f (x) − N N V ol(A) ωN (R2 − R1 ) ! Z R2 Z g(s)sN −1 ds dω N R1 S N −1 = g(r) − N − RN ) ω (R N 2 1 Z R2 N N −1 = g(r) − g(s)s ds N N R2 − R 1 R1
! Z R2 Z R2 N N −1 N −1 g(s)s ds = g(r)s ds − N N R2 − R 1 R1 R1
(9.7)
(9.8)
(9.9)
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Ostrowski Inequalities on Balls and Shells Via Taylor–WidderFormula
=
N R2N − R1N
Z R2 N −1 (g(r) − g(s))s ds =: (∗). R1
129
(9.10)
Let s, r ∈ [R1 , R2 ], then by generalized Taylor’s formula (9.2) we have n X
g(s) − g(r) = where
Rn (r, s) := But it holds
By calling
we find
Z |Rn (r, s)| ≤ Z Nn (r, s) :=
s r
s
r
Li g(r)gi (s, r) + Rn (r, s),
Z
s
gn (s, t)Ln+1 g(t)dt.
(9.12)
r
|gn (s, t)|dt kLn+1 gk∞,[R1 ,R2 ] .
|gn (s, t)|dt , ∀ s, r ∈ [R1 , R2 ],
|Rn (r, s)| ≤ Nn (r, s)kLn+1 gk∞,[R1 ,R2 ] , ∀ s, r ∈ [R1 , R2 ]. Therefore by (9.10), (9.11), we have # Z R2 "X n N N −1 L g(r)g (s, r) + R (r, s) s ds (∗) = i i n R2N − R1N R1 i=1 ≤
N R2N − R1N
(9.11)
i=1
(9.13)
(9.14)
(9.15)
(9.16)
Z # "X n Z R2 R2 N −1 N −1 |Rn (r, s)|s ds Li g(r)gi (s, r)s ds + R1 R1 i=1
by (9.15) ≤
N N R2 − R1N
"X n i=1
+(kLn+1 gk∞,[R1 ,R2 ] )
Z R2 N −1 |Li g(r)| gi (s, r)s ds R1
Z
R2
Nn (r, s)s
R1
N −1
#
ds .
(9.17)
We have established the following result. Theorem 9.6. Let u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]), W0 , W1 , . . . , Wn > 0 on [R1 , R2 ], 0 < R1 < R2 , n ≥ 0 integer, with u0 (r) = c > 0. Let f ∈ C n+1 (A) be radial, i.e. there exists g such that f (x) = g(r), r = |x|, R1 ≤ r ≤ R2 , x ∈ A. Then Z Z R2 f (y)dy N N −1 A = g(r) − E(x) := f (x) − g(s)s ds N − RN V ol(A) R R 2 1 1
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≤
N R2N − R1N
"X n i=1
Z R2 N −1 gi (s, r)s ds |Li g(r)| R1
+(kLn+1 gk∞,[R1 ,R2 ] )
=
N R2N − R1N
"X n i=1
Z
R2
Nn (r, s)s
N −1
ds
R1
#
(9.18)
Z R2 N −1 gi (s, |x|)s ds |θi f (x)| R1
+(kθn+1 f k∞,A )
Z
R2
#
Nn (|x|, s)sN −1 ds .
R1
We give Corollary 9.7. Same terms and assumptions as in Theorem 9.6. Suppose further that Li g(r0 ) = 0, i = 1, . . . , n, for a fixed r0 ∈ [R1 , R2 ]; considered all x0 = r0 ω ∈ A, for any ω ∈ S N −1 . Then Z Z R2 f (y)dy N N −1 A = g(r0 ) − g(s)s ds E(x0 ) = f (x0 ) − N − RN V ol(A) R R 2 1 1 ≤
=
N R2N − R1N
N R2N − R1N
Z
(kLn+1 gk∞,[R1 ,R2 ] )
(kθn+1 f k∞,A )
Z
R2
Nn (r0 , s)s
ds
R1
R2 R1
N −1
Nn (|x0 |, s)s
N −1
!
!
ds .
(9.19)
Interesting cases also arise when r0 = R1 or R2 . We continue Remark 9.5 with Remark 9.8. Let now f ∈ C n+1 (A), n ≥ 0, x ∈ A, x = rω, r > 0. Clearly for fixed ω ∈ S N −1 , the function f (rω), r ∈ [R1 , R2 ] is radial, it also belongs to C n+1 ([R1 , R2 ]). By applying internal inequality (9.18) we obtain Z R2 N N −1 f (sω)s ds f (rω) − R2N − R1N R1 ≤
N R2N − R1N
"X n i=1
Z R2 N −1 gi (s, r)s ds |(Li f (·ω))(r)| R1
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+(kLn+1 f (·ω)k∞,[R1 ,R2 ] )
Z
R2
Nn (r, s)s
N −1
R1
#
ds .
131
(9.20)
For non-radial f we define again all i = 1, . . . , n + 1, ∀ x ∈ A.
θi f (x) = θi f (rω) := (Li f (·ω))(r),
(9.21)
Here the involved ∂ i f (x) ∂ i f (rω) = , i = 1, . . . , n + 1, i ∂r ∂ri are the radial derivatives. In a sense θi is a generalized radial derivative of Widdertype. Hence "X n N R.H.S.(9.20) ≤ |θi f (rω)| R2N − R1N i=1 Z # Z R2 R2 N −1 N −1 Nn (|x|, s)s ds . gi (s, r)s ds + kθn+1 f k∞,A R1 R1
(9.22)
Therefore, by (9.20) and (9.22) we have N Γ Z Z 1 2 f (y)dy f (rω)dω − 2π N/2 V ol(A) A S N −1
≤
N R2N − R1N
"
Γ(N/2) 2π N/2
(
n Z X i=1
+kθn+1 f k∞,A
Z
) Z R2 gi (s, r)sN −1 ds |θi f (rω)|dω N −1 R1 S R2
R1
#
Nn (|x|, s)sN −1 ds .
(9.23)
We have established Theorem 9.9. Let u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]); W0 , W1 , . . . , Wn > 0 on [R1 , R2 ], 0 < R1 < R2 , n ≥ 0 integer, with u0 (r) = c > 0. Let f ∈ C n+1 (A), x ∈ A; x = rω, r ∈ [R1 , R2 ], ω ∈ S N −1 . Then Z N R Γ f (rω)dω f (y)dy N −1 S 2 ≤ f (x) − E(x) = f (x) − A N/2 V ol(A) 2π N ( ) Γ Z R2 n Z X N 2 N −1 + |θ f (rω)|dω g (s, r)s ds i i R1 R2N − R1N 2π N/2 N −1 S i=1
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Z
+kθn+1 f k∞,A
R2
Nn (|x|, s)s
N −1
R1
#
ds , ∀ x ∈ A.
(9.24)
We give Corollary 9.10. Same terms and assumptions as in Theorem 9.9. Suppose that θi f (r0 ω) = 0, i = 1, . . . , n, for a fixed r0 ∈ [R1 , R2 ], ∀ ω ∈ S N −1 ; also consider all x0 = r0 ω ∈ A for any ω ∈ S N −1 . Then Z Z N f (r ω)dω ) Γ f (y)dy 0 2 N −1 S A E(x0 ) = f (x0 ) − ≤ f (x0 ) − N/2 V ol(A) 2π +
N N R2 − R1N
kθn+1 f k∞,A
Z
R2
R1
Nn (|x0 |, s)s
N −1
!
ds .
(9.25)
When r0 = R1 or R2 is of special interest. 9.4
Results on the Sphere
Notation 9.11. Let N ≥ 1, B(0, R) := {x ∈ RN : |x| < R} be the ball in RN ωN R N . centered at the origin and of radius R > 0. Note that V ol(B(0, R)) = N Let f from B(0, R) into R and consider f be radial, i.e. f (x) = g(r), where r = |x|, 0 ≤ r ≤ R, we assume g ∈ C n+1 ([0, R]), n ≥ 0. Clearly then f ∈ C(B(0, R)). For F ∈ C(B(0, R)) we have ! Z Z Z R
F (x)dx =
S N −1
B(0,R)
F (rω)rN −1 dr
dω.
(9.26)
0
We notice that N RN
Z
R
sN −1 ds = 1.
(9.27)
0
The operator θi in the radial case, is defined as in (9.3), now ∀ x ∈ B(0, R). In the non-radial case, for f ∈ C n+1 (B(0, R)), θi is defined as in (9.21), ∀ x ∈ B(0, R)−{0}. We make Remark 9.12. Let here u0 , u1 , . . . , un ∈ C n+1 ([0, R]) and W0 , W1 , . . . , Wn > 0 on [0, R], R > 0, n ≥ 0 integer, with u0 (r) = c > 0.
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133
We again first suppose that f is radial on B(0, R), i.e. there exists g such that f (x) = g(r), r = |x|, 0 ≤ r ≤ R. Assume further that g ∈ C n+1 ([0, R]). Let x ∈ B(0, R). Then by using the polar method (9.26) we obtain ! Z R Z Z n−1 N g(s)s ds dω f (y)dy 0 S N −1 B(0,R) = g(r) − E(x) := f (x) − N V ol(B(0, R)) ωN R (9.28) Z R (by 9.27) N Z R N N −1 (g(r) − g(s))s ds = g(r) − N g(s)sn−1 ds = =: (∗). R RN 0 0 (9.29) Let s, r ∈ [0, R], then by generalized Taylor’s formula (9.2) we have n X
g(s) − g(r) = where
Rn (r, s) := But it holds
Z |Rn (r, s)| ≤
By calling
Z Nn (r, s) :=
we find
Li g(r)gi (s, r) + Rn (r, s),
(9.30)
i=1
s r
r
s
Z
s
gn (s, t)Ln+1 g(t)dt.
(9.31)
r
|gn (s, t)|dt kLn+1 gk∞,[0,R] .
|gn (s, t)|dt , ∀ s, r ∈ [0, R],
|Rn (r, s)| ≤ Nn (r, s)kLn+1 gk∞,[0,R] , ∀ s, r ∈ [0, R]. Therefore by (9.29) and (9.30), we have Z " n # N R X (∗) = N Li g(r)gi (s, r) + Rn (r, s) sN −1 ds R 0
(9.32)
(9.33)
(9.34)
i=1
≤
N RN
"
Z # n Z R R X N −1 N −1 Li g(r)gi (s, r)s ds + |Rn (r, s)|s ds 0 0
(9.35)
i=1
by
(9.34)
≤
Z " n R N X N −1 |L g(r)| g (s, r)s ds i i 0 RN i=1
+(kLn+1 gk∞,[0,R] )
Z
R
Nn (r, s)s
0
N −1
#
ds .
(9.36)
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We have established the next result. Theorem 9.13. Let u0 , u1 , . . . , un ∈ C n+1 ([0, R]), W0 , W1 , . . . , Wn > 0 on [0, R], R > 0, n ≥ 0 integer, with u0 (r) = c > 0. Let f from B(0, R) into R be radial, i.e. there exists g such that f (x) = g(r), r = |x|, 0 ≤ r ≤ R, ∀ x ∈ B(0, R); further suppose that g ∈ C n+1 ([0, R]). Then Z f (y)dy Z R N B(0,R) n−1 = g(r) − g(s)s ds E(x) := f (x) − N V ol(B(0, R)) R 0 Z # " n Z R R N X N −1 N −1 |Li g(r)| gi (s, r)s ds + (kLn+1 gk∞,[0,R] ) Nn (r, s)s ds ≤ N 0 R 0 i=1 Z " n R N X N −1 |θi f (x)| = N gi (s, |x|)s ds 0 R i=1
+(kθn+1 f k∞,B(0,R) )
Z
R
Nn (|x|, s)s
N −1
#
ds .
0
(9.37)
We give Corollary 9.14. Same terms and assumptions as in Theorem 9.13. Assume further that Li g(r0 ) = 0, i = 1, . . . , n, for a fixed r0 ∈ [0, R]; consider all x0 = r0 ω ∈ B(0, R), any ω ∈ S N −1 . Then Z f (y)dy Z R N B(0,R) n−1 E(x0 ) := f (x0 ) − = g(r ) − g(s)s ds 0 V ol(B(0, R)) RN 0 N ≤ N (kLn+1 gk∞,[0,R] ) R
Z
N = N (kθn+1 f k∞,B(0,R) ) R
Z
R
Nn (r0 , s)s
ds
0 R
0
Nn (|x0 |, s)s
Interesting cases especially arise when r0 = 0 or R. We continue Remark 9.12 with Remark 9.15. Let f be non-radial.
N −1
N −1
!
!
ds .
(9.38)
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Here assume that f ∈ C n+1 (B(0, R)). Consider x ∈ B(0, R) − {0}, i.e. uniquely x = rω, r ∈ (0, R], ω ∈ S N −1 . Then by using again the polar method (9.26) we get Z Z f (y)dy S N −1 f (rω)dω B(0,R) (9.39) − ω V ol(B(0, R)) N Z Z N N −1 f (rω)dω S N −1 = S − ωN by
(9.27)
=
N ωN R N
N − ωN R N N = ωN R N
Z S N −1
Z
Z
S N −1
Z
S N −1
Z
R 0
Z
Z
R
f (sω)s
0
ωN R N
f (rω)s
N −1
f (sω)s
ds dω
ds dω
0
N −1
!
!
R
R 0
N −1
(9.40)
!
!
! ds dω
(f (rω) − f (sω))sN −1 ds dω =: (∗). !
(9.41)
Clearly here f (·ω) ∈ C n+1 ((0, R]). Let ρ ∈ (0, R], then by (9.2) we obtain f (ρω) − f (rω) = where
n X
Rn (r, ρ) := That is
Z
(9.42)
ρ
gn (ρ, t)(Ln+1 (f (·ω)))(t)dt.
(9.43)
r
f (ρω) − f (rω) = with Rn (r, ρ) =
n X (θi f (rω))gi (ρ, r) + Rn (r, ρ),
(9.44)
i=1
Z
ρ
gn (ρ, t)θn+1 f (tω)dt.
(9.45)
θ := kθn+1 f k∞,B(0,R)−{0} < +∞.
(9.46)
We further assume that
Therefore we derive
((Li (f (·ω)))(r))gi (ρ, r) + Rn (r, ρ),
i=1
r
Z |Rn (r, ρ)| ≤
ρ r
|gn (ρ, t)|dt θ.
(9.47)
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By calling
Z Nn (r, ρ) :=
we get
|gn (ρ, t)|dt , ∀ ρ, r ∈ [0, R],
ρ r
(9.48)
|Rn (r, ρ)| ≤ Nn (r, ρ)θ.
(9.49)
Hence by (9.44) we obtain |f (ρω) − f (rω)| ≤ ≤
n X i=1
n X
|θi f (rω)||gi (ρ, r)| + |Rn (r, ρ)|
i=1
|θi f (rω)||gi (ρ, r)| + Nn (r, ρ)θ.
(9.50)
By continuity of f and gi , i = 1, . . . , n,and by taking the limit as ρ −→ 0 in the external inequality (9.50), we find |f (0) − f (rω)| ≤
n X
|θi f (rω)||gi (0, r)| + Nn (r, 0)θ.
i=1
(9.51)
Notice here gn (ρ, t) is jointly continuous in (ρ, t) ∈ [0, R]2 , hence Nn (r, ρ) is continuous in ρ ∈ [0, R]. That is, ∀ s ∈ [0, R] we get |f (sω) − f (rω)| ≤
n X i=1
|θi f (rω)||gi (s, r)| + Nn (r, s)θ.
Consequently by (9.41) and (9.52) we find ! Z R Z N N −1 |f (sω) − f (rω)|s ds dω (∗) ≤ ωN RN S N −1 0 " n Z X N ≤ N ωN R S N −1 i=1 +θ
=
Z n X i=1
Z
Z
S N −1
|θi f (rω)||gi (s, r)|s
0
Z
S N −1
Z
R
R
!
Nn (r, s)sN −1 ds dω
0
R
|θi f (rω)||gi (s, r)|s
0
N −1
N −1
!
ds dω
!
(9.52)
(9.53)
(9.54)
# !
ds dω
!
V ol(B(0, R)) θN +
Z
R
Nn (r, s)sN −1 ds
0
RN
!
(9.55)
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Ostrowski Inequalities on Balls and Shells Via Taylor–WidderFormula
Z
N X = i=1
S N −1
|θi f (rω)|dω
R
|gi (s, r)|s
0
N −1
V ol(B(0, R))
θN + Consequently,
Z
Z
R
Nn (r, s)s
N −1
ds
0
!
RN
n X ≤ i=1
S N −1
|θi f (rω)|dω
Z
R
|gi (s, r)|s
0
V ol(V (0, R))
θN +
Z
R
Nn (r, s)sN −1 ds 0
RN
!
!
ds
.
(9.56)
Z Z f (y)dy S N −1 f (rω)dω B(0,R) ∆ := − ω V ol(B(0, R)) N
Z
137
N −1
!
ds
.
(9.57)
We have proved Theorem 9.16. Let u0 , u1 , . . . , un ∈ C n+1 ([0, R]), W0 , W1 , . . . , Wn > 0 on [0, R], R > 0, n ≥ 0 integer, with u0 (r) = c > 0. Let f ∈ C n+1 (B(0, R)). Suppose that θ := kθn+1 f k∞,B(0,R)−{0} < +∞.
(9.58)
Let x ∈ B(0, R) − {0}, i.e. uniquely x = rω, r ∈ (0, R], ω ∈ S N −1 . Then Z f (y)dy Z B(0,R) 0 0 ≤ f (x) − Γ(N/2) E(x) := f (x) − f (rω )dω N/2 V ol(B(0, R)) 2π S N −1 Z
n X + i=1
S N −1
Z
R
!
|gi (s, r)|sN −1 ds V ol(B(0, R))
|θi f (rω 0 )dω 0
0
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θN +
Z
R
Nn (r, s)s 0
N −1
ds
!
RN
.
(9.59)
We finally give Corollary 9.17. Same terms and assumptions as in Theorem 9.16. Assume further that θi f (r0 ω 0 ) = 0, ∀ ω 0 ∈ S N −1 , for some r0 ∈ (0, R], for all i = 1, . . . , n. Consider all x0 = r0 ω ∈ B(0, R) − {0}, any ω ∈ S N −1 . Then Z f (y)dy Z Γ(N/2) B(0,r) 0 0 E(x0 ) := f (x0 ) − f (|x0 |ω )dω ≤ f (x0 ) − 2π N/2 V ol(B(0, R)) N −1 S ! Z R N −1 θN Nn (|x0 |, s)s ds +
0
RN
.
(9.60)
An interesting case is when r0 = R.
9.5
Addendum
We give Proposition 9.18. Let f ∈ C 1 (B(0, R)) such that f is radial, i.e. f (x) = g(r), r = |x|, 0 ≤ r ≤ R, ∀ x ∈ B(0, R). Then (i) ∃ g 0 ∈ C((0, R]), (9.61) 0 (ii) ∃ g (0) = 0. (9.62) Proof. (i) is obvious. (ii) Let u be a unit vector, h > 0, then f (hu) − f (0) g(h) − g(0) = h h
∇f (0) · hu + o(h) h o(h) ∇f (0) · hu o(h) + = ∇f (0) · u + = h h h o(h) h→0 −→ ∇f (0) · u, by → 0. h The last is true for any unit vector u. Thus ∇f (0) = 0, proving the claim. (by first order multivariate Taylor’s formula) =
Comment 9.19. By Proposition 9.18 we see that, it may well be that g 0 is discontinuous at zero, if only f ∈ C 1 (B(0, R)). Therefore the assumption that g ∈ C n+1 ([0, R]) in Theorem 9.13 seems to be the best.
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Chapter 10
Multivariate Opial Type Inequalities for Functions Vanishing at an Interior Point
In this chapter we generalize Opial inequalities in the multivariate case on the balls. The inequalities carry weights and are proved to be sharp. The functions we study vanish at the center of the ball. This treatment relies on [58]. 10.1
Introduction
Z. Opial [195] and C. Olech [194] in 1960 proved the following famous inequality. Theorem 10.1. Let c > 0, and y(x) be real, continuously differentiable on [0, c], with y(0) = y(c) = 0. Then Z Z c c c 0 (y (x))2 dx. |y(x)y 0 (x)| dx ≤ 4 0 0 c Equality holds for the function y(x) = x on 0, 2 , and y(x) = c − x on 2c , c . In 1962 P. Beesack [75] gave the following improvement.
Theorem 10.2. Let b > 0. If y(x) is real, continuously differentiable on [0, b], and y(0) = 0, then Z b Z b b 0 |y(x)y 0 (x)| dx ≤ (y (x))2 dx. 2 0 0 Equality holds only for y(x) = mx, where m is a constant. Since then many people have worked on this type of inequalities in many directions; for an account see the important monograph of 1995 by R. Agarwal and P. Pang [6]. One motivation for this chapter is the interesting article of W. Troy [237] of 2001. His relevant result follows. Theorem 10.3. Let p > −1. Let a, b ∈ R with 0 ≤ a < b. If y(x) is continuously differentiable on [a, b], and y(a) = 0, then Z b Z b 1 tp |y(t)y 0 (t)| dt ≤ √ (bp+1 − atp )(y 0 (t))2 dt. 2 p + 1 a a 139
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Let |·| denote the Euclidean norm in RN , N ≥ 1. Let B(0, R) := {x ∈ RN : |x| < R} be the open ball of radius R with center 0 in RN . Let S N −1 := {x ∈ RN : |x| = 1} N/2 be the unit sphere in RN , centered at zero. Let ωN = NNπ N (see p. 220, [142]) 2
Γ
2
be its surface area. In this chapter we estimate the integral Z |x|p |u(x)| |∇u(x)| dx, I= B(0,R)
for p ∈ R and u ∈ C 1 (B(0, R)). Using polar coordinates we obtain ! Z Z R p N −1 I= r |u(rω)| |∇u(rω)|r dr dω, S N −1
0
where 0 6= x = rω with r := |x| and ω := xr . For radial (spherically symmetric) functions u, this reduces to Z R ∂u p+N −1 r |u(r)| (r) dr, I = ωN ∂r 0 ∂u ∂u(x) x where here |∇u| = ∂r , with ∂r = ∇u(x) · |x| the radial derivative of u. Spherically symmetric function means that u(x) = u(rω) = u(r).
In general one has ∂u(x) ∂r ≤ |∇u(x)|, We will prove that Z
B(0,R)
for any u ∈ C 1 (B(0, R)).
|x|p |u(x)| |∇u(x)| dx ≤ C
Z
B(0,R)
|x|p |∇u(x)|2 dx
for some constant C and functions vanishing at the origin. The idea is to use ! Z R Z Z N −1 F (rω)r dr dω, F (x) dx = B(0,R)
S N −1
0
and to do first a one-dimensional analysis on the inner integral with ω fixed. The interior constraint (u(0) = 0) becomes a boundary condition (F (0) = 0).
10.2
Main Results
We present a basic result. Theorem 10.4. Let R > 0, N ≥ 1, B(0, R) the ball centered at 0 of radius R in RN . Let p ∈ R such that p + N ∈ (0, 2). Consider u ∈ C 1 (B(0, R)) such that u(0) = 0. Then ! Z Z R p 2 p √ |x| |∇u(x)| dx , |x| |u(x)| |∇u(x)| dx ≤ 2 2−p−N B(0,R) B(0,R) (10.1)
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where ∇ is the gradient operator. Proposition 10.5. Inequality (10.1) is sharp, more precisely it is attained by u(x) = |x|, for all x ∈ B(0, R), when p + N = 1. Proof.
Call r := |x|, 0 ≤ r ≤ R. Clearly u(0) = 0. Notice that ∂u = 1 and |∇u(x)| = 1. ∂r
We observe that L.H.S.(10.1) = ωN
Rp+N +1 , p+N +1
and R.H.S(10.1) = ωN Thus
Rp+N +1 p . 2(p + N ) 2 − (p + N )
L.H.S.(10.1) = R.H.S.(10.1) iff (p + N ) + 1 = 2(p + N ) (calling y := p + N , y ∈ (0, 2)) y + 1 = 2y
p 2 − (p + N ) iff
p 2−y
iff
g(y) := 4y 3 − 7y 2 + 2y + 1 = 0,
y ∈ (0, 2).
See that g(1) = 0, g(0) = 1, g(2) = 9, and 1 g 0 (y) = 12(y − 1) y − . 6
Thus g has critical numbers 1, 61 with local maximum g 16 = 1.1574078, and minimum g(1) = 0. So y = p + N = 1 is the only optimal value making inequality (10.1) attained. Proof of Theorem 10.4. The integral in the R.H.S(10.1) is finite since 2 > p + N > 0. We can rewrite inequality (10.1) by the use of polar coordinates as follows, cf. p. 217, [142]: ! Z Z R p N −1 r |u(rω)| |∇u(rω)|r dr dω S N −1
0
R ≤ √ 2 2−p−N
Z
S N −1
Z
R 0
p
2 N −1
r |∇u(rω)| r
dr
!
!
dω .
(10.2)
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Here 0 = 6 x ∈ B(0, R) is written as x := rω with r := |x|, 0 < r ≤ R, and x ω = |x| ∈ S N −1 . So it is enough to prove that ! Z R Z R R p+N −1 2 p+N −1 √ r |∇u(rω)| dr . r |u(rω)| |∇u(rω)| dr ≤ 2 2−p−N 0 0 (10.3) We set Z r sp+N −1 |∇u(sω)|2 ds, 0 ≤ r ≤ R. z(r) := 0
Here z(r) ≥ 0 and z(0) = 0. Therefore z 0 (r) = rp+N −1 |∇u(rω)|2 ≥ 0,
0 < r ≤ R.
Whence r
p+N −1 2
|∇u(rω)| = (z 0 (r))1/2 ,
0 < r ≤ R.
(10.4)
By the fundamental theorem of calculus we have Z r ∂u(sω) ds. u(rω) = ∂s 0
Consequently it holds Z r Z r p+N −1 ∂u −1 ∂u − p+N 2 2 s |u(rω)| ≤ s ∂s (sω) ds = ∂s (sω) ds 0 0 (by the Cauchy–Schwarz inequality) 2 !1/2 Z r 1/2 Z r −(p+N −1) (p+N −1) ∂u ≤ s ds s ∂s (sω) ds 0 0 2−p−N 1/2 Z r 1/2 r ≤ s(p+N −1) |∇u(sω)|2 ds 2−p−N 0 2−p−N 1/2 r = (z(r))1/2 . 2−p−N So we have proved that |u(rω)| ≤ Furthermore, r
p+N −1 2
r2−p−N 2−p−N
|u(rω)| ≤ √
1/2 √
(z(r))1/2 ,
r (z(r))1/2 , 2−p−N
all 0 ≤ r ≤ R.
all 0 ≤ r ≤ R.
(10.5)
Consequently by (10.4) and (10.5) we derive √ r r(p+N −1) |u(rω)| |∇u(rω)| ≤ √ (z(r))1/2 (z 0 (r))1/2 , all 0 < r ≤ R. (10.6) 2−p−N
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Next we integrate (10.6) and use the Cauchy–Schwarz inequality to get Z R Z R √ 1 r(z(r))1/2 (z 0 (r))1/2 dr rp+N −1 |u(rω)| |∇u(rω)| dr ≤ √ 2 − p − N 0 0 !1/2 Z R R z(r) dz(r) ≤ √ √ 2 2−p−N 0 ! Z R R R p+N −1 2 = √ z(R) = √ r |∇u(rω)| dr , 2 2−p−N 2 2−p−N 0 establishing (10.3).
A similar result follows. Theorem 10.6. Let R > 0, N ≥ 1, B(0, R) the ball centered at 0 of radius R in RN . Let p ∈ R such that p + N > 0. Consider u ∈ C 1 (B(0, R)) such that u(0) = 0. Then Z Z Rp+N p √ |x|1−N |∇u(x)|2 dx. (10.7) |x| |u(x)| |∇u(x)| dx ≤ 2 p + N B(0,R) B(0,R) Proposition 10.7. Inequality (10.7) is sharp, namely it is attained by u(x) = |x|, for all x ∈ B(0, R), when p + N = 1. Proof. Call r := |x|, 0 ≤ r ≤ R. Clearly u(0) = 0. Note that |∇u(x)| = 1. We have that L.H.S.(10.7) = ωN
∂u ∂r
= 1 and
Rp+N +1 , p+N +1
and Rp+N +1 R.H.S.(10.7) = ωN √ . 2 p+N Thus L.H.S.(10.7) = R.H.S.(10.7) iff p (p + N ) + 1 = 2 p + N iff √ y + 1 = 2 y, where y := p + N > 0, y = 1.
iff
Proof of Theorem 10.6. Clearly the R.H.S.(10.7) is finite. As in the proof of Theorem 10.4 it is enough to prove ! Z R Z R Rp+N p N −1 2 r |u(rω)| |∇u(rω)|r dr ≤ √ |∇u(rω)| dr . (10.8) 2 p+N 0 0
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Therefore we begin with Z
R 0
p
Z
N −1
R
r |u(rω)| |∇u(rω)|r dr = rp+N −1 |u(rω)| |∇u(rω)| dr 0 Z R p+N −2 p+N 2 2 = |∇u(rω)| r |u(rω)| dr r 0
Z
≤
0
(by the Cauchy–Schwarz inequality) !1/2 Z !1/2 R R p+N 2 (p+N −1) −1 2 r |∇u(rω)| dr =: (∗). r r (u(rω)) dr 0
We have again u(rω) =
Z
r 0
∂u (sω) ds, ∂s
0 ≤ r ≤ R.
Hence ∂u 1 · (sω) ds ∂s 0 (by the Cauchy–Schwarz inequality) 2 !1/2 1/2 Z r Z r ∂u √ √ 2 (sω) ds ≤ r |∇u(sω)| ds r . ≤ ∂s 0 0
|u(rω)| ≤
That is
Z
r
2
(u(rω)) ≤ r
Z
r 0
|∇u(sω)|2 ds.
(10.9)
Thus by (10.9) we derive Z
(∗) ≤
Z
= −
R
r
p+N
0 R
r
|∇u(rω)| dr
1 p+N
!1/2
·
Z
R
r
(p+N −1)
0
Z
r 0
2
|∇u(sω)| ds dr
!1/2 ! Z R Rp+N 2 · |∇u(rω)| dr |∇u(rω)| dr p+N 0 1/2 Z R p+N 2 =: (∗∗). r |∇u(rω)| dr
p+N
0
2
!1/2
2
0
If A ≥ 0, B ≥ 0 and ε > 0 we have (AB)1/2 ≤
ε 1 A + B. 2 2ε
(10.10)
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Therefore
145
! Z R 1 Rp+N 2 r |∇u(rω)| dr + |∇u(rω)| dr 2ε p + N 0 0 ! Z R 1 p+N 2 − r |∇u(rω)| dr p+N 0 ! Z R 1 1 p+N 2 r |∇u(rω)| dr = ε− 2 ε(p + N ) 0 ! Z R Rp+N 2 + |∇u(rω)| dr 2ε(p + N ) 0 ! Z R 1 Rp+N choosing ε = √ = √ |∇u(rω)|2 dr . p+N 2 p+N 0
ε (∗∗) ≤ 2
Z
R
p+N
!
2
We have established (10.8).
Finally we present a generalization and extension of Theorem 10.4. Theorem 10.8. Let R > 0, N ≥ 1, ρ > 1, α, β > 0, ρ ≥ α + β and p ∈ R such that 0 < p + N < ρ. Consider u ∈ C 1 (B(0, R)) such that u(0) = 0. Then Z α+β |x| (p+N −1) ρ +(1−N ) |u(x)|β |∇u(x)|α dx B(0,R)
ρ−α−β ρ
≤ LωN where L :=
ρ−1 ρ−p−N
β
ρ−1 ρ
Z
B(0,R)
ρ−a β(ρ − 1) + (ρ − α)
|x|p |∇u(x)|ρ dx
ρ−α ρ
α α+β
!
αρ
α+β ρ
R
,
(10.11)
β(ρ−1)+(ρ−α) ρ
.
(10.12)
Proposition 10.9. Inequality (10.11) is sharp, namely it is attained by u(x) = |x|, α for all x ∈ B(0, R), when p + N = 1, ρ = α + β and α α+β = α+β 1+β . Proof. Call r := |x|, 0 ≤ r ≤ R. Clearly u(0) = 0. Note that |∇u(x)| = 1. We have L.H.S.(10.11) = ωN
∂u ∂r
= 1 and
Rβ+1 , β+1
and R.H.S.(10.11) = ωN
αα/ρ β+1 R . ρ
It is obvious now that (10.11) holds as equality.
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Proof of Theorem 10.8. The integral in the R.H.S.(10.11) is finite by p + N > 0. x Here for 0 6= x ∈ B(0, R) we set ω := |x| = xr , where r := |x|, 0 < r ≤ R. That is x = rω. By the fundamental theorem of calculus we have Z r ∂u (sω) ds, 0 ≤ r ≤ R. u(rω) = 0 ∂s We see that
Z r p+N −1 ∂u −1 − p+N (sω) ds = ∂u (sω) ds ρ ρ s s ∂s ∂s 0 0 ρ (using H¨ older’s inequality with indices ρ and ρ−1 ) ρ−1 Z r ρ Z r 1/ρ ρ ∂u (p+N −1) ≤ s− ρ−1 ds s(p+N −1) (sω) ds ∂s 0 0 ρ−1 Z 1/ρ r ρ (ρ−p−N ) ρ−1 s(p+N −1) |∇u(sω)|ρ ds ≤ r ρ . ρ−p−N 0
|u(rω)| ≤
That is,
Z
r
ρ−1 ρ−p−N
+∞ > z(r) :=
Z
|u(rω)| ≤
Here we call
ρ−1 ρ
r 0
ρ−p−N ρ
r
z(r)1/ρ ,
0 ≤ r ≤ R.
s(p+N −1) |∇u(sω)|ρ ds ≥ 0,
with z(0) = 0. We have z 0 (r) = rp+N −1 |∇u(rω)|ρ ≥ 0, and p+N −1 ρ
r We observe that
|u(rω)|β ≤ and r
p+N −1 ρ
β
β
|u(rω)| ≤
α
0 < r ≤ R.
|∇u(rω)|α = (z 0 (r))α/ρ ,
ρ−1 ρ−p−N
ρ−1 ρ−p−N
β
β ρ−1 ρ
ρ−1 ρ
·r
β
rβ
0 < r ≤ R.
(ρ−p−N ) ρ
ρ−1 ρ
(10.13)
(z(r))β/ρ ,
z(r)β/ρ ,
By multiplying (10.13) and (10.14) we get α+β r(p+N −1) ρ |u(rω)|β |∇u(rω)|α β ρ−1 ρ ρ−1 ρ−1 ≤ rβ ρ (z(r))β/ρ (z 0 (r))α/ρ , ρ−p−N
0 ≤ r ≤ R. (10.14)
0 < r ≤ R. (10.15)
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ρ Integrate (10.15) and use H¨ older’s inequality with indices αρ , ρ−α to find Z R α+β r(p+N −1) ρ |u(rω)|β |∇u(rω)|α dr 0 β ρ−1 Z R ρ ρ−1 β ρ−1 ρ ≤ (z(r))β/ρ (z 0 (r))α/ρ dr r ρ−p−N 0 ! ρ−α !α/ρ β ρ−1 Z R Z R ρ ρ ρ−1 ρ−1 β ρ−α β/α 0 ≤ dr · r (z(r)) z (r) dr ρ−p−N 0 0 ! α+β Z R ρ α+β (10.12) . rp+N −1 |∇u(rω)|ρ dr = L · z(R) ρ = L 0
We have established Z R r (p+N −1)
α+β ρ
0
Z
≤ L
+(1−N )
α N −1
|u(rω)| |∇u(rω)| r
R
rp |∇u(rω)|ρ rN −1 dr
0
β
!
α+β ρ
dr
!
.
(10.16)
Integrating (10.16) over S N −1 we find ! Z R Z (p+N −1) α+β +(1−N ) β α N −1 ρ r |u(rω)| |∇u(rω)| r dr dω S N −1
≤ L·
0
Z
S N −1
Z
R 0
rp |∇u(rω)|ρ rN −1 dr
!
α+β ρ
dω =: (∗ ∗ ∗).
If ρ > α + β then apply again H¨ older’s inequality with indices to obtain (∗ ∗ ∗) ≤L
Z
ρ
1 ρ−α−β dω S N −1
ρ−α−β
= LωN
ρ
·
Z
S N −1
ρ−α−β ρ
Z
R 0
·
Z
S N −1
Z
R 0
rp |∇u(rω)|ρ rN −1 dr
From (10.17) and (10.18) we conclude (10.11).
ρ ρ−α−β
rp |∇u(rω)|ρ rN −1 dr !
dω
!
α+β ρ
(10.17)
.
!
ρ α+β
abd
dω
!
α+β ρ
(10.18)
The work of Neˇceav [193] in 1973 is related to this chapter, see also [6, p. 275]. Theorem 10.10 [193]. Let u ∈ C 1 (B(0, R)) be such that u(0) = 0, and N + p < 2, N ≥ 1. Then Z Z R2−N −p 1−N |x| |u(x)| |∇u(x)| dx ≤ |x|p |∇u(x)|2 dx, 2(2 − N − p) B(0,R) B(0,R)
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with equality holding when u(x) = c|x|2−N −p for a real constant c. This result can be compared to Theorem 10.6 if and only if p = 1 − N , in which case the conclusions are the same, with the common constant being R/2.
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Chapter 11
General Multivariate Weighted Opial Inequalities
In this chapter we expose Opial type weighted multivariate inequalities on balls and arbitrary smooth bounded domains. The inequalities are mostly sharp. The functions we consider vanish on the boundary. This chapter relies on [59]. 11.1
Introduction
Z. Opial [195] and C. Olech [194] in 1960 proved the following well-known inequality. Theorem 11.1. Let c > 0, and y(x) be real, continuously differentiable on [0, c], with y(0) = y(c) = 0. Then Z Z c c c 0 (y (x))2 dx. |y(x)y 0 (x)| dx ≤ 4 0 0 c Equality holds for the function y(x) = x on 0, 2 , and y(x) = c − x on 2c , c . In 1962 P. Beesack [75] gave the following variant.
Theorem 11.2. Let b > 0. If y(x) is real, continuously differentiable on [0, b], and y(0) = 0, then Z b Z b b 0 |y(x)y 0 (x)| dx ≤ (y (x))2 dx. 2 0 0 Equality holds only for y = mx, where m is a constant. Since then many researchers have been extending this type of inequalities in many directions, for a thorough account see the important monograph of 1995 by R. Agarwal and P. Pang [6]. An inspiration for this chapter is the interesting article by W. Troy [237] of 2001. His relevant result is as follows. Theorem 11.3. Let p > −1. Let b and c be real with 0 ≤ b < c. If y(x) is continuously differentiable on [b, c], and y(c) = 0, then Z c Z c 1 tp |y(t)y 0 (t)| dt ≤ √ (ctp − bp+1) (y 0 (t))2 dt. 2 p + 1 b b 149
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R. Agarwal in 1981, see [4] and p. 208 of [6], proved the following two dimensional result. Theorem 11.4. If u(t, s) ∈ C (1,1) ([a, T ] × [c, S]), u(a, s) = u(t, c) = 0, then Z TZ S Z Z (T − a)(S − c) T S √ |uts (t, s)|2 dtds. |u(t, s)uts (t, s)| dtds ≤ 2 2 a a c c In 1982, G.S. Yang [247] proved the following Opial-type inequality in two variables: Theorem 11.5. If f (s, t), f1 (s, t), and f12 (s, t) are continuous functions on [a, b]× [c, d], and if f (a, t) = f (b, t) = f1 (s, c) = f1 (s, d) = 0, for a ≤ s ≤ b, c ≤ t ≤ d, then Z Z Z bZ d (b − a)(d − c) b d (f12 (s, t))2 dtds. |f (s, t)| |f12 (s, t)| dtds ≤ 8 a c a c In 1983, C.T. Lin and G.S. Yang [179] generalized Theorem 11.5 in the following form: Theorem 11.6. If f (s, t), f1 (s, t), and f12 (s, t) are continuous functions on [a, b]× [c, d], and if f (a, t) = f (b, t) = f1 (s, c) = f1 (s, d) = 0, for a ≤ s ≤ b, c ≤ t ≤ d, then Z bZ d |f (s, t)|m |f12 (s, t)|n dtds a
≤
n m+n
c
(b − a)(d − c) 4
m Z
b a
Z
d c
|f12 (s, t)|m+n dtds.
The above Theorems 11.4 – 11.6 motivate this chapter. Here we prove weighted Opial type inequalities over a ball (or other domain) in RN , N ≥ 1, where the functions under consideration vanish on the boundary. Let | · | denote the Euclidean distance for vectors in RN . Let B(0, R) := {x ∈ N R : |x| < R} be the open ball of radius R with center 0 in RN . Let S N −1 := {x ∈ RN : |x| = 1} be the unit sphere in RN centered at zero, with N/2 ωN = NNπ N to be its surface area (cf. [142], p. 220). 2
Γ
2
The basic results here are extended to Sobolev spaces of order 1. We mention
Definition 11.7 (cf. [139]). The Sobolev space of order 1 denoted by H01 (Ω) is the completion of Cc1 (Ω). Here Cc1 (Ω) is the space of one time continuously differentiable functions with compact support on the bounded domain Ω ⊂ RN , N ≥ 1. The norm in H01 (Ω) is given by Z 1/2 Z kuk∗ := |u(x)|2 dx + |∇u(x)|2 dx , for u ∈ H01 (Ω). Ω
Ω
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In this chapter we estimate the integral Z I= |x|p |u(x)| |∇u(x)| dx, B(0,R)
for p ∈ R and u ∈
Cc1 (B(0, R)).
I=
Z
S N −1
Using polar coordinates we have ! Z R p N −1 r |u(rω)| |∇u(rω)|r dr dω, 0
where x = rω with r := |x| and ω := xr . For radial (spherically symmetric) functions u, this reduces to Z R ∂u p+N −1 r |u(r)| (r) dr, I = ωN ∂r 0 ∂u x where here |∇u| = ∂r , with ∂u(x) ∂r = ∇u(x) · |x| the radial derivative of u. Spherically symmetric function means that u(x) = u(rω) = u(r).
In general one has ∂u(x) ∂r ≤ |∇u(x)|, We will prove that Z
B(0,R)
for any u ∈ Cc1 (B(0, R)).
|x|p |u(x)| |∇u(x)| dx ≤ C
Z
B(0,R)
|x|p |∇u(x)|2 dx
for some constant C and functions u as above. The idea is to use ! Z R Z Z F (rω)rN −1 dr dω, F (x) dx = B(0,R)
S N −1
0
and to do first a one-dimensional analysis on the inner integral with ω fixed. The boundary condition u = 0 on ∂B(0, R) becomes the boundary condition F (R) = 0. 11.2
Main Results
We present a basic result. Theorem 11.8. Let R > 0, N ≥ 1, B(0, R) the ball centered at 0 of radius R in R N . Let p ∈ R such that p + N ≥ 1. Let u ∈ C 1 (B(0, R)) be such that u(∂B(0, R)) = 0. Then Z Z R |x|p |u(x)| |∇u(x)| dx ≤ |x|p |∇u(x)|2 dx, (11.1) 2 B(0,R) B(0,R) where ∇ is the gradient operator.
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Note that Theorem 11.8 holds for more general choices of u. By a simple approximation argument, the condition u ∈ C 1 (B(0, R)) can be weakened to 1,2 B(0, R) − {0} ∩ C B(0, R) u ∈ Wloc
such that u = 0 on ∂B(0, R) and the right hand side of (11.1) is finite. The extremal in the following proposition satisfies this weaker hypothesis. Similar remarks apply to the Theorems 11.13 and 11.17 below. Proposition 11.9. Inequality (11.1) is sharp, namely it is attained by u(x) = R − |x|, for all x ∈ B(0, R), when p + N = 1. Proof. Call r := |x|, 0 ≤ r ≤ R. Notice u(∂B(0, R)) = 0. Since u(x) = R − r we find ∂u ∂r = −1 and |∇u(x)| = 1. Clearly we get L.H.S.(11.1) = ωN
Rp+N +1 , (p + N )(p + N + 1)
and R.H.S.(11.1) = ωN
Rp+N +1 , 2(p + N )
where ωN is the surface area of the unit sphere S N −1 in RN . Thus L.H.S.(11.1) = R.H.S.(11.1) iff p + N = 1. Proof of Theorem 11.8. The integral in the R.H.S.(11.1) is finite since p+N > 0. We can rewrite inequality (11.1) by the use of polar coordinates as follows ! Z Z R rp |u(rω)| |∇u(rω)|rN −1 dr dω S N −1
0
R ≤ 2
Z
S N −1
Z
R p
0
2 N −1
r |∇u(rω)| r
dr
!
dω.
(11.2)
Here x ∈ B(0, R) is given in polar coordinates by x = rω with r := |x|, 0 ≤ r ≤ R. x Clearly ω = |x| if r > 0. So it is enough to prove that, for each ω, Z R Z R R p p N −1 r |∇u(rω)|2 rN −1 dr. (11.3) r |u(rω)| |∇u(rω)|r dr ≤ 2 0 0
We set
z(r) :=
Z
R r
sp+N −1 |∇u(sω)|2 ds,
0 ≤ r ≤ R.
Here z(r) ≥ 0 and z(R) = 0. Then
−z 0 (r) = rp+N −1 |∇u(rω)|2 ≥ 0,
0 ≤ r ≤ R.
Thus r
p+N −1 2
|∇u(rω)| = (−z 0 (r))1/2 .
(11.4)
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Since u = 0 on ∂B(0, R) we find u(rω) = − We observe that r
p+N −1 2
Z
R r
153
∂u (sω) ds. ∂s
∂u (sω) ds ∂s r Z R ∂u p+N −1 2 ≤ 1·s ∂s (sω) ds
|u(rω)| ≤
Z
R
p+N −1 2
r
r
(by the Cauchy–Schwarz inequality) 2 !1/2 Z R √ p+N −1 ∂u ≤ R−r s ∂s (sω) ds r !1/2 Z R √ √ ≤ R−r sp+N −1 |∇u(sω)|2 ds = R − r(z(r))1/2 . r
That is, we have
r
p+N −1 2
|u(rω)| ≤
√
R − r(z(r))1/2 ,
(11.5)
all 0 ≤ r ≤ R. Multiplying (11.4) by (11.5) we obtain √ rp+N −1 |u(rω)| |∇u(rω)| ≤ R − r(z(r))1/2 (−z 0 (r))1/2 ,
(11.6)
for all 0 ≤ r ≤ R. Therefore by integrating (11.6) and using the Cauchy–Schwarz inequality we get Z R rp+N −1 |u(rω)| |∇u(rω)| dr 0
≤ ≤
Z
R
√
0
Z
R − r(z(r))1/2 (−z 0 (r))1/2 dr
R 0
(R − r) dr
!1/2
−
Z
R
0
z(r)z (r) dr 0
Z R R R p+N −1 r |∇u(rω)|2 dr. = z(0) = 2 2 0 Clearly we have established (11.3), and (11.1) follows. We have
!1/2
Corollary 11.10. Let p ∈ R be such that p + N ≥ 1 and let u ∈ H01 (B(0, R)). Then Z Z R |x|p |u(x)| |∇u(x)| dx ≤ |x|p |∇u(x)|2 dx. (11.7) 2 B(0,R) B(0,R) Proof. By (11.1), inequality (11.7) is valid for u ∈ C 1 (B(0, R)) with u = 0 on ∂B(0, R). If u ∈ H01 (B(0, R)) then there exists a sequence {um } from C 1 (B(0, R)) k·k∗
such that um = 0 on ∂B(0, R) and um −→ u. Then the corollary follows by using (11.1) for um and letting m → ∞.
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Next we are ready to give Theorem 11.11. Let Ω be a smooth bounded domain in RN , N ≥ 1. Let p ∈ R such that p + N ≥ 1. Then there exists a constant K = K(Ω) such that Z Z p |x|p |∇u(x)|2 dx, (11.8) |x| |u(x)| |∇u(x)| dx ≤ K(Ω) Ω
Ω
for all u ∈
H01 (Ω).
Here
R , 2 where R is the radius of the smallest ball centered at the origin and containing Ω. K(Ω) :=
Proof. Since ∂Ω is smooth any function u in H01 (Ω) can be viewed as a member of H01 (B(0, R)) by defining it to be zero on B(0, R) − Ω. For the last step, see p. 245 of [139]. Now Corollary 11.10 applies. Remark 11.12. (i) When u is spherically symmetric then (11.7) reduces to Z R Z R R p+N −1 0 rp+N −1 |u(r)| |u0 (r)| dr ≤ r (u (r))2 dr. (11.9) 2 0 0 (ii) When p = 0 then (11.8) collapses into Z Z |u(x)| |∇u(x)| dx ≤ K(Ω) |∇u(x)|2 dx, Ω
(11.10)
Ω
for all u ∈ H01 (Ω). The counterpart of Theorem 11.8 follows
Theorem 11.13. Let R > 0, N ≥ 1, B(0, R) the ball centered at 0 of radius R in RN . Let p ∈ R be such that p + N > 1. Let u ∈ C 1 (B(0, R)) such that u(∂B(0, R)) = 0. Then √ Z Z R p+N p |x| |u(x)| |∇u(x)| dx ≤ |x|p |∇u(x)|2 dx. (11.11) 2(p + N − 1) B(0,R) B(0,R) Proof. The integral in the R.H.S.(11.11) is finite by p + N > 0. We estimate Z R Z R p+N −2 p+N rp+N −1 |u(rω)| |∇u(rω)|dr = r 2 |∇u(rω)| r 2 |u(rω)| dr 0
0
≤
Z
R
(by the Cauchy–Schwarz inequality) !1/2 Z R
r 0
p+N
2
·
|∇u(rω)| dr
r
p+N −2
0
Since u = 0 on ∂B(0, R) we find u(rω) = −
Z
R r
∂u (sω) ds. ∂s
2
|u(rω)| dr
!1/2
=: (∗∗).
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Therefore |u(rω)| ≤ ≤
Z
R r
√
∂u 1 · (sω) ds ≤ ∂s
R−r
Z
R
r
Z
R
2
1 ds r
|∇u(sω)|2 ds
!1/2
!1/2
Z
.
That is Z
2
|u(rω)| ≤ (R − r)
R
2
R r
2 !1/2 ∂u (sω) ds ∂s
!
|∇u(sω)| ds .
r
Consequently we have Z
(∗∗) ≤ ·
Z
Z
=
=
Z
r
p+N
0
2
|∇u(rω)| ds
R
r 0 R
0 R 0
R
p+N −2
(R − r)
rp+N |∇u(rω)|2 dr rp+N |∇u(rω)|2 dr
Z
!1/2
!
R 2
r
!1/2
!1/2
|∇u(sω)| ds dr · ·
Z
Z
R 0 R 0
Z
s 0
!1/2
rp+N −2 (R − r) dr |∇u(sω)|2 ds
sp+N Rsp+N −1 − p+N −1 p+N
|∇u(sω)|2 ds
!1/2
!1/2
ε 1 using the inequality (AB)1/2 ≤ A + B, for any A ≥ 0, B ≥ 0 and ε > 0 2 2ε ! " ! Z R Z R 1 ε R p+N 2 p+N −1 2 r |∇u(rω)| dr + r |∇u(rω)| dr ≤ 2 2ε p+N −1 0 0 !# Z R 1 − rp+N |∇u(rω)|2 dr p+N 0 ! Z R 1 p+N 2 r |∇u(rω)| dr ε− ε(p + N ) 0 ! Z R R p+N −1 2 r |∇u(rω)| dr + 2ε(p + N − 1) 0 1 by choosing ε = √ p+N ! √ Z R R p+N p+N −1 2 = r |∇u(rω)| dr . 2(p + N − 1) 0
1 = 2
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We have proved that Z R rp+N −1 |u(rω)| |∇u(rω)| dr ≤ 0
Finally we obtain Z Z S N −1
R
r
p+N −1
0
√ R p+N 2(p + N − 1)
Z
!
dw
|u(rω)| |∇u(rω)| dr
√ Z R p+N ≤ 2(p + N − 1) S N −1
Z
R
r
p+N −1
0
2
|∇u(rω)| dr .
R
r
p+N −1
0
!
2
|∇u(rω)| dr
proving (11.11).
!
dω,
Corollary 11.14. Let p ∈ R be such that p + N > 1 and let u ∈ H01 (B(0, R)). Then √ Z Z R p+N |x|p |∇u(x)|2 dx. (11.12) |x|p |u(x)| |∇u(x)| dx ≤ 2(p + N − 1) B(0,R) B(0,R) Proof.
Similar to Corollary 11.10.
Next we are ready to give Theorem 11.15. Let Ω be a smooth bounded domain in RN , N ≥ 1. Let p ∈ R be such that p + N > 1. Then there exists a constant K = K(Ω, p, N ) such that Z Z |x|p |u(x)| |∇u(x)| dx ≤ K |x|p |∇u(x)|2 dx, (11.13) Ω
Ω
for all u ∈ H01 (Ω). Here
√ R p+N K := , 2(p + N − 1)
where R is the radius of the smallest ball centered at the origin and containing Ω. Proof.
Similar to Theorem 11.11 by applying Corollary 11.14.
Remark 11.16. (i) When u is spherically symmetric then (11.12) reduces to √ Z R Z R R p+N p+N −1 0 r |u(r)| |u (r)| dr ≤ rp+N −1 (u0 (r))2 dr. (11.14) 2(p + N − 1) 0 0 (ii) When p = 0 then (11.13) collapses into Z Z ˜ |∇u(x)|2 dx, |u(x)| |∇u(x)| dx ≤ K Ω
Ω
for all u ∈ H01 (Ω), where
˜ := K
√ R N , 2(N − 1)
Theorem 11.8 is generalized as follows.
N ≥ 2.
(11.15)
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Theorem 11.17 Let R > 0, N ≥ 1, ρ > 1, α, β > 0, ρ ≥ α + β, p ≥ 1 − N , and u ∈ C 1 (B(0, R)) with u(∂B(0, R)) = 0. Then Z α+β |x| (p+N −1) ρ +(1−N ) |u(x)|β |∇u(x)|α dx B(0,R)
ρ−α−β ρ
≤ M ωN where M :=
ρ−α (ρ − 1)β + (ρ − α)
ρ−α ρ
Z
·
B(0,R)
|x|p |∇u(x)|ρ dx
α α+β
α/ρ
·R
!
α+β ρ
(ρ−1)β+(ρ−α) ρ
,
(11.16)
.
(11.17)
Proposition 11.18. Inequality (11.16) is sharp, namely it is attained by u(x) := R − |x|, for x ∈ B(0, R), when p + N = 1, ρ = α + β and α = 1. Proof.
Notice that L.H.S.(11.16) = ωN
Rβ+1 , β+1
and R.H.S.(11.16) = ωN
αα/ρ β+1 R . ρ
Equality holds in (11.16) if α
α α+β =
α+β . 1+β
But this implies α = 1. α Here is a proof that if α > 0 and β > 0, then α α+β =
α+β 1+β
implies that α = 1. α
Put = t and = s. Then 0 < t < 1 and s > 0. The condition α α+β = α+β 1+β yields the equation s − st = 1 − t. Put g(x) = xt − xt, for x > 0. Note that g 0 (x) = txt−1 − t = t(xt−1 − 1). It follows that g has a unique global maximum at x = 1. Since g(s) = g(1), we infer that s = 1, and hence α = 1s = 1. α α+β
1 α t
Proof of Theorem 11.17. The integral in the R.H.S.(11.16) is finite by p+N > 0. We have Z R ∂u (sω) ds, all 0 ≤ r ≤ R. u(rω) = − ∂s r ρ Next we apply H¨ older’s inequality with indices ρ and ρ−1 to derive Z R Z R p+N −1 p+N −1 ∂u(sω) p+N −1 ∂u ρ ρ ρ r |u(rω)| ≤ r 1·s ∂s ds ≤ ∂s (sω) ds r r ρ !1/ρ Z R Z R ! ρ−1 ρ p+N −1 ∂u s ≤ ds ∂s (sω) ds r r !1/ρ Z ≤ (R − r)
R
ρ−1 ρ
r
sp+N −1 |∇u(sω)|ρ ds
.
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So by calling
Z
z(r) := we got p+N −1 ρ
r and
r(p+N −1)
β ρ
R
sp+N −1 |∇u(sω)|ρ ds
r
|u(rω)| ≤ (R − r)
ρ−1 ρ
(11.18) 1
(z(r)) ρ ,
|u(rω)|β ≤ (R − r)(ρ−1)
β ρ
(11.19) β
(z(r)) ρ ,
(11.20)
all 0 ≤ r ≤ R. Here notice that z(r) ≥ 0 and z(R) = 0. Also we have Z r sp+N −1 |∇u(sω)|ρ ds −z(r) = R
and
−z 0 (r) = rp+N −1 |∇u(rω)|ρ ≥ 0.
Consequently
p+N −1 ρ
r and
1
|∇u(rω)| = (−z 0 (r)) ρ
α α (11.21) r(p+N −1) ρ |∇u(rω)|α = (−z 0 (r)) ρ , for all 0 ≤ r ≤ R. Multiplying (11.20) by (11.21) we get α+β β β α r(p+N −1) ρ |u(rω)|β |∇u(rω)|α ≤ (R − r)(ρ−1) ρ (z(r)) ρ (−z 0 (r)) ρ , (11.22) for all 0 ≤ r ≤ R. ρ Next we integrate (11.22) and apply H¨ older’s inequality with indices αρ and (ρ−α) to find Z R (p+N −1) α+β ρ r |u(rω)|β |∇u(rω)|α dr 0
≤
Z
R
0
≤
Z
=
Z
= =
(R − r)(ρ−1) R
(R − r)
0 R
(R − r)
0
β ρ
(ρ−1)
(ρ−1)β (ρ−α)
dr
ρ−α (ρ − 1)β + (ρ − α) ρ−α (ρ − 1)β + (ρ − α) Z
R
r 0
p+N −1
β
α
(z(r)) ρ (−z 0 (r)) ρ dr ! ρ−α ρ ρ β ρ−α ρ dr · ! ρ−α ρ
ρ−α ρ
ρ−α ρ
ρ
·
|∇u(rω)| dr
R
!
−
Z
R
Z
R
β ρ
(z(r)) (−z (r))
0
(z(r)) z (r) dr 0
α α+β α+β
αρ
ρ
.
R
α ρ
0
β α
(ρ−1)β+(ρ−α) ρ
0
α α+β
! αρ
αρ
(ρ−1)β+(ρ−α) ρ
(z(0))
α+β ρ
αρ
dr
! αρ
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Book˙Adv˙Ineq
General Multivariate Weighted Opial Inequalities
We have proved that Z R
r(p+N −1)
α+β ρ
0
Z
≤ M Finally we derive Z Z S N −1
R
r 0
(p+N −1)
≤ M
Z
|u(rω)|β |∇u(rω)|α dr R
rp+N −1 |∇u(rω)|ρ dr
0
α+β ρ
Z
S N −1
R
+(1−N )
!
α+β ρ
.
(11.23) !
|u(rω)|β |∇u(rω)|α rN −1 dr dω
p
ρ N −1
r |∇u(rω)| r
0
159
dr
! (α+β) ρ
dω ρ ρ−α−β
(if ρ > α + β then we apply again H¨ older’s inequality with indices ρ α+β to find) Z ρ−α−β ρ ρ ≤M 1 ρ−α−β dω S N −1 ! ! α+β Z Z R ρ · rp |∇u(rω)|ρ rN −1 dr dω S N −1
0
ρ−α−β ρ
= M ωN
Z
Z
S N −1
R
rp |∇u(rω)|ρ rN −1 dr
0
!
dω
!
α+β ρ
We have established inequality (11.16).
and
.
Corollary 11.19. Let all the conditions of Theorem 11.17 hold, except now consider u ∈ H01 (B(0, R)). Then Z α+β |x| (p+N −1) ρ +(1−N ) |u(x)|β |∇u(x)|α dx B(0,R)
ρ−α−β ρ
≤ M ωN Proof.
Similar to Corollary 11.10.
Z
B(0,R)
|x|p |∇u(x)|ρ dx
!
α+β ρ
.
(11.24)
Next we give the final main result of this chapter. Theorem 11.20. Let Ω be a smooth bounded domain in RN , N ≥ 1. Let ρ > 1, α, β > 0, ρ ≥ α + β, p ≥ 1 − N . Then there exists a constant K = K(Ω, α, β, ρ, N ) such that Z Ω
|x|
(p+N −1)
α+β ρ
+(1−N )
|u(x)|β |∇u(x)|α dx Z α+β ρ ≤ K |x|p |∇u(x)|ρ dx , Ω
(11.25)
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for all u ∈ H01 (Ω). Here ρ−α−β ρ
K = M ωN
(11.26)
is the constant of inequality (11.16). The M in (11.26) is given by (11.17), where R is the radius of the smallest ball centered at the origin and containing Ω. Proof.
Similar to Theorem 11.11 by applying Corollary 11.19.
Remark 11.21. (i) When u is spherically symmetric then (11.24) reduces to ! α+β Z R Z R ρ α+β , r(p+N −1) ρ |u(r)|β |u0 (r)|α dr ≤ M rp+N −1 |u0 (r)|ρ dr 0
0
(11.27)
where M is given by (11.17). (ii) When p = 0 then (11.25) collapses into Z
Ω
|x|(N −1)
α+β−ρ ρ
for all u ∈ H01 (Ω).
|u(x)|β |∇u(x)|α dx ≤ K
Z
Ω
|∇u(x)|ρ dx
α+β ρ
,
(11.28)
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Book˙Adv˙Ineq
Chapter 12
Opial Inequalities for Widder Derivatives
Various Lp form Opial type inequalities are given for Widder derivatives. This treatment relies on [36]. 12.1
Introduction
This chapter is greatly motivated by the article of Z. Opial [195]. Theorem 12.1. (Opial [195]) Let x(t) ∈ C 1 ([0, h]) be such that x(0) = x(h) = 0, and x(t) > 0 in (0, h). Then, Z h Z h h 0 0 (x (t))2 dt. (12.1) |x(t)x (t)|dt ≤ 4 0 0 In the last inequality the constant h/4 is the best possible. Opial type inequalities have applications in establishing uniqueness of solution to initial value problems in differential equations, see [244]. 12.2
Background
The following are taken from [243]. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]), n ≥ 0, and the Wronskians u0 (x) u1 (x) . . . ui (x) u0 (x) u0 (x) . . . u0 (x) 0 1 i , Wi (x) := W [u0 (x), u1 (x), . . . , ui (x)] := .. . (i) (i) u0 (x) u(i) 1 (x) . . . ui (x)
(12.2)
i = 0, 1, . . . , n. Here W0 (x) = u0 (x). Assume Wi (x) > 0 over [a, b], i = 0, 1, . . . , n. For i ≥ 0, the differential operator of order i (Widder derivative): Li f (x) :=
W [u0 (x), u1 (x), . . . , ui−1 (x), f (x)] , Wi−1 (x) 161
(12.3)
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i = 1, . . . , n + 1; L0 f (x) := f (x), ∀ x ∈ [a, b]. Consider also u0 (t) u1 (t) u0 (t) u01 (t) 0 1 .. gi (x, t) := . Wi (t) (i−1) (i−1) u0 (t) u1 (t) u0 (x) u1 (x)
i = 1, 2, . . . , n; g0 (x, t) :=
u0 (x) , ∀ x, t ∈ [a, b]. u0 (t)
, (i−1) . . . ui (t) . . . ui (x)
... ...
ui (t) u0i (t)
(12.4)
Example ([243]). Sets of the form {u0 , u1 , . . . , un } are {1, x, x2 , . . . , xn }, {1, sin x, − cos x, − sin 2x, cos 2x, . . . , (−1)n−1 sin nx, (−1)n cos nx}, etc. We also mention the generalized Widder–Taylor’s formula, see [243]. Theorem 12.2. Let the functions f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]), and the Wronskians W0 (x), W1 (x), . . . , Wn (x) > 0 on [a, b], x ∈ [a, b]. Then for t ∈ [a, b] we have u0 (x) f (x) = f (t) + L1 f (t)g1 (x, t) + · · · + Ln f (t)gn (x, t) + Rn (x), (12.5) u0 (t) where Rn (x) :=
Z
x
gn (x, s)Ln+1 f (s)ds.
(12.6)
t
For example ([243]) one could take u0 (x) = c > 0. If ui (x) = xi , i = 0, 1, . . . , n, defined on [a, b], then Li f (t) = f (i) (t)
and gi (x, t) =
(x − t)i , i!
t ∈ [a, b].
We need Corollary 12.3 (on Theorem 12.2). By additionally assuming for fixed x0 ∈ [a, b] that Li f (x0 ) = 0, i = 0, 1, . . . , n, we get that Z x f (x) = gn (x, t)Ln+1 f (t)dt, ∀ x ∈ [a, b]. (12.7) x0
12.3
Results
From now on we are working under the terms and assumptions of Theorem 12.2 and Corollary 12.3. We first present the following
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1 1 + = 1. Then p q 1/p Z x Z w Z x (gn (w, t))p dt dw |f (w)||Ln+1 f (w)|dw ≤ 2−1/q
Theorem 12.4. Let x ≥ x0 , x0 , x ∈ [a, b] and p, q > 1:
x0
x0
x0
×
Z
x x0
|Ln+1 f (w)|q dw
By H¨ older’s inequality we have Z x 1/p Z p |f (x)| ≤ |gn (x, t)| dt
Proof.
x0
Call z(w) := Thus
Z
2/q
.
x x0
(12.8)
q
|Ln+1 f (t)| dt
1/q
.
(12.9)
w
|Ln+1 f (t)|q dt,
x0
x0 ≤ w ≤ x
(z(x0 ) = 0).
z 0 (w) = |Ln+1 f (w)|q and |Ln+1 f (w)| = (z 0 (w))1/q . From (12.9) we obtain |f (w)||Ln+1 f (w)| ≤
Z
w x0
p
|gn (w, t)| dt
1/p
(z(w)z 0 (w))1/q .
Integrating the last inequality over [x0 , x] we obtain 1/p Z x Z x Z w |f (w)||Ln+1 f (w)|dw ≤ (z(w)z 0 (w))1/q dw |gn (w, t)|p dt x0
x0
≤
Z
x x0
Z
=
w x0
Z
1/p Z |gn (w, t)|p dt dw
x x0
x0
Z
w x0
p
|gn (w, t)| dt dw
x
z(w)z 0 (w)dw
x0
1/p
1/q
(z(x))2/q , 21/q
proving the claim of Theorem 12.4. See also Remark 22.11 next. The counter part of the previous result follows. 1 1 + = 1. Then is p q 1/p Z x Z w Z x0 |gn (w, t)|p dt dw |f (w)||Ln+1 f (w)|dw ≤ 2−1/q
Theorem 12.5. Let x ≤ x0 , x0 ∈ [a, b] and p, q > 1:
x
x0
x0
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Z
×
Proof.
x0
|Ln+1 f (w)|q dw
x
2/q
Here take x ≤ x0 and p, q > 1 such that
inequality we have
Z
≤
Z |f (x)| =
x0 x
z(x) :=
Z
x0 x
x
Z
|gn (x, t)||Ln+1 f (t)|dt ≤
Call
x0
x0
p
1/p Z
|Ln+1 f (t)|q dt ≥ 0,
That is −z(x) =
Z
(12.10)
1 1 + = 1. From H¨ older’s p q
gn (x, t)Ln+1 f (t)dt |gn (x, t)| dt
x
.
x0 x
q
|Ln+1 f (t)| dt
1/q
.
(12.11)
z(x0 ) = 0.
x x0
|Ln+1 f (t)|q dt ≤ 0,
and −z 0 (x) = |Ln+1 f (x)|q ≥ 0, and |Ln+1 f (x)| = (−z 0 (x))1/q ,
x ∈ [a, x0 ].
Hence by (12.11) (x ≤ w ≤ x0 ) Z x0 1/p |f (w)||Ln+1 f (w)| ≤ |gn (w, t)|p dt (z(w)(−z 0 (w))1/q . w
Integrating the last inequality over [x, x0 ] we obtain 1/p Z x0 Z x0 Z x0 p (z(w)(−z 0 (w)))1/q dw |gn (w, t)| dt |f (w)||Ln+1 f (w)|dw ≤ w
x
x
≤
Z
x0 x
Z
x0 w
= 2−1/q proving the claim.
1/p Z |gn (w, t)|p dt dw Z
x x0
Z
w x0
x0
z(w)(−z 0 (w))dw
x
1/q
1/p (z(x))2/q , |gn (w, t)|p dt dw
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Next we study the extreme cases. Theorem 12.6. Here p = 1, q = ∞ and x ≥ x0 . Then Z x Z w Z x |f (w)||Ln+1 f (w)|dw ≤ gn (w, t)dt dw kLn+1 f k2∞ . x0
x0
By
Proof.
Z
f (w) = we get |f (w)| ≤ and
Z
(12.12)
x0
w
gn (w, t)Ln+1 f (t)dt x0
w x0
|f (w)||Ln+1 f (w)| ≤
|gn (w, t)|dt kLn+1 f k∞ , Z
w x0
|gn (w, t)|dt kLn+1 f k2∞ .
Integrating the last inequality we find (12.12). See also Remark 12.11 next. Theorem 12.7. Again p = 1, q = ∞, but x ≤ x0 . Then Z x Z w Z x0 |f (w)||Ln+1 f (w)|dw ≤ gn (w, t)|dt dw (kLn+1 f k2∞ ). x
x0
Here x ≤ w ≤ x0 , and
Proof.
Z =
Thus
and
Z
Z |f (w)| =
gn (w, t)Ln+1 f (t)dt x0 Z x0 x0 gn (w, t)Ln+1 f (t)dt ≤ |gn (w, t)|dt kLn+1 f k∞ . w
w
w
|f (w)||Ln+1 f (w)| ≤
x0 x
(12.13)
x0
|f (w)||Ln+1 f (w)|dw ≤
proving the claim.
Z
Z
x0
w
x0 x
|gn (w, t)|dt kLn+1 f k2∞ ,
Z
x0 w
|gn (w, t)|dt dw kLn+1 f k2∞ ,
1 1 + = 1. Then p q Z x Z x −1/q |f (w)||Ln+1 f (w)|dw ≤ 2
Corollary 12.8. Let p, q > 1: Z
Proof.
x0
x0
Z ×
x
x0
By Theorems 12.4, 12.5.
w x0
p
|gn (w, t)| dt dw
2/q . |Ln+1 f (w)|q dw
1/p (12.14)
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In particular when p = q = 2 we have Corollary 12.9. It holds Z x Z −1/2 |f (w)||Ln+1 f (w)|dw ≤ 2 x0
x Z w
x0
2
(gn (w, t)) dt dw
x0
Z x × (Ln+1 f (w))2 dw .
1/2 (12.15)
x0
Furthermore we get
Corollary 12.10. Let p = 1, q = ∞. Then Z x Z x Z w ≤ |f (w)||L f (w)|dw |g (w, t)|dt dw kLn+1 f k2∞ . n+1 n x0
x0
By Theorems 12.6, 12.7.
Proof.
(12.16)
x0
We need to make Remark 12.11. We define (see [243]) φ0 (x) := W0 (x), φ1 (x) :=
W1 (x) ,..., (W0 (x))2
in general φk (x) :=
Wk (x)Wk−2 (x) , (Wk−1 (x))2
k = 2, 3, . . . , n.
The functions φi (x) are positive on [a, b]. According to [243] we get, for x, x0 not fixed, that Z x Z xn−2 Z x1 φ0 (x) gn (x, x0 ) = φn−1 (xn−1 ) ... φ1 (x1 ) φ0 (x0 ) . . . φn (x0 ) x0 x0 x0 Z xn−1 φn (xn )dx1 dx2 . . . dxn × x0
1 = φ0 (x0 ) . . . φn (x0 )
Z
x
φ0 (s) . . . φn (s)gn−1 (x, s)ds.
(12.17)
x0
We get that gn (x, x) = 0, all x ∈ [a, b], and gn (x, x0 ) > 0, x > x0 , x, x0 ∈ [a, b], ∀ n ≥ 1. Also g0 (x, x0 ) > 0 for any x, x0 ∈ [a, b]. To complete the chapter we present 1 1 + = 1, and x > x0 , p q x, x0 ∈ [a, b]. Assume that Ln+1 f is of fixed sign and nowhere zero. Then 1/p Z x Z w Z x (gn (w, t))p dt dw |f (w)||Ln+1 f (w)|dw ≥ 2−1/q
Theorem 12.12. Let 0 < p < 1 and q < 0 be such that
x0
x0
x0
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Opial Inequalities for Widder Derivatives
Z
×
Proof.
x x0
|Ln+1 f (w)|q dw
2/q
167
.
(12.18)
Here for x0 ≤ w ≤ x we have Z w |f (w)| = gn (w, t)|Ln+1 f (t)|dt,
(12.19)
x0
by Remark 22.11. From (12.19) by H¨ older’s inequality we have Z x 1/p Z w 1/q p q |f (w)| ≥ (gn (w, t)) dt , |Ln+1 f (t)| dt x0
(12.20)
x0
for w > x0 . Consider z(w) :=
Z
w x0
|Ln+1 f (t)|q dt > 0,
z(x0 ) = 0.
So that z 0 (w) = |Ln+1 f (w)|q > 0 and|Ln+1 f (w)| = (z 0 (w))1/q , all x0 ≤ w ≤ x. Thus by (12.20) we get Z w 1/p |f (w)||Ln+1 f (w)| ≥ (z(w)z 0 (w))1/q , (gn (w, t))p dt x0
all x0 < w ≤ x. Let x0 < θ ≤ w ≤ x and θ ↓ x0 , then by integration of the last inequality we obtain Z x Z x |f (w)||Ln+1 f (w)|dw |f (w)||Ln+1 f (w)|dw = lim θ↓x0
x0
Z
≥ lim
θ↓x0
≥ lim
θ↓x0
Z
=2
x θ
−1/q
Z Z
=2
x θ
Z
w
p
(gn (w, t)) dt x0
w
p
(gn (w, t)) dt dw x0
x x0
−1/q
Z
w
p
θ
1/p
1/p
Z
x Z w
x0
That is establishing (12.18).
x0
p
(z(w)z (w))
lim
θ↓x0
(gn (w, t)) dt dw x0
1/p
0
Z
x
1/q
dw
!
0
z(w)z (w)dw θ
1/q !
lim (z 2 (x) − z 2 (θ))1/q
θ↓x0
(gn (w, t)) dt dw
1/p
(z(x))2/q .
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We make Remark 12.13. We notice by (12.17) that gn (x, t) < 0,
x < t,
n odd,
gn (x, t) > 0,
x < t,
n even,
where x, t ∈ [a, b]. We give 1 1 + = 1, and x < x0 , p q x, x0 ∈ [a, b]. Let n be odd. Also assume that Ln+1 f is of fixed sign and nowhere zero. Then 1/p Z x Z w Z x0 p −1/q (−gn (w, t)) dt dw |f (w)||Ln+1 f (w)|dw ≥ 2 Theorem 12.14. Let 0 < p < 1 and q < 0 such that
× Proof.
x0
x0
x
Z
x0 x
|(Ln+1 f )(w)|q dw
2/q
.
(12.21)
Here for x ≤ w ≤ x0 we have Z w |f (w)| = gn (w, t)Ln+1 f (t)dt x0
Z =
x0
w
Z gn (w, t)Ln+1 f (t)dt = =
Z
x0
w
x0
(−gn (w, t))Ln+1 f (t)dt
(−gn (w, t))|Ln+1 f (t)|dt.
From (12.22) by H¨ older’s inequality we get Z x0 1/p Z |f (w)| ≥ (−gn (w, t))p dt w
for w < x0 . That is Z |f (w)| ≥
x0 w
|gn (w, t)|p dt
for w < x0 . Consider
z(w) := So that
Z
(12.22)
w
x0 w
1/p Z
x0 w
x0 w
|Ln+1 f (t)|q dt
|Ln+1 f (t)|q dt > 0,
−z(w) =
Z
w x0
|Ln+1 f (t)|q dt
1/q
z(x0 ) = 0.
|Ln+1 f (t)|q dt
1/q
,
(12.23)
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and −z 0 (w) = |Ln+1 f (w)|q > 0, i.e. z 0 (w) < 0. Hence |Ln+1 f (w)| = (−z 0 (w))1/q > 0, all x ≤ w ≤ x0 . Therefore by (12.23) we derive Z |f (w)||Ln+1 f (w)| ≥
x0
p
|gn (w, t)| dt
w
1/p
(z(w)(−z 0 (w))1/q ,
all x ≤ w < x0 . Let x ≤ w ≤ θ < x0 and θ ↑ x0 , then by integration of the last inequality we obtain Z x0 Z θ |f (w)||Ln+1 f (w)|dw = lim |f (w)||Ln+1 f (w)|dw θ↑x0
x
≥ lim
θ↑x0
≥ lim
θ↑x0
=
Z
Z
=2
Z
θ x
x0 x
−1/q
Z
x
Z
x0 w
|gn (w, t)|p dt
x0
p
x0 w
x0 x
= 2−1/q
p
!1/p
1/p
|gn (w, t)| dt dw Z Z
x0
p
x0
Z
x0 w
(z(w)(−z 0 (w)))1/q dw
Z
lim
θ↑x0
θ
1/p
!
0
z(w)(−z (w))dw x
−
lim
θ↑x0
|gn (w, t)| dt dw
w
x
1/p
|gn (w, t)| dt dw
w
Z
Z
θ
x
Z
θ
z(w)dz(w) x
!1/q
!1/q
lim (z 2 (x) − z 2 (θ))1/q
θ↑x0
1/p |gn (w, t)|p dt dw (z(x))2/q .
Hence Z
x0 x
|f (w)||Ln+1 f (w)|dw ≥ 2−1/q
× that is proving (12.21).
Z
x0 x
Z
x x0
Z
w x0
|Ln+1 f (w)|q dw
1/p |gn (w, t)|p dt dw
2/q
,
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Putting things together we have 1 1 + = 1, and x0 , x ∈ [a, b] p q such that x = 6 x0 . Let n be odd. Assume that Ln+1 f is of fixed sign and nowhere zero. Then Z x Z x Z w 1/p −1/q p |f (w)||Ln+1 f (w)|dw ≥ 2 |gn (w, t)| dt dw Corollary 12.15. Let 0 < p < 1 and q < 0 be such that
x0
Proof.
x0
Z ×
x
x0
x0
2/q . |Ln+1 f (w)| dw q
By Theorems 12.12, 12.14.
(12.24)
We give without proof the similar Theorem 12.14, see Remark 12.13. 1 1 + = 1, and x < x0 , p q x, x0 ∈ [a, b]. Let n be even. Also assume that Ln+1 f is of fixed sign and nowhere zero. Then 1/p Z x Z w Z x0 (gn (w, t))p dt dw |f (w)||Ln+1 f (w)|dw ≥ 2−1/q Theorem 12.16. Let 0 < p < 1 and q < 0 such that
x
x0
×
Z
x0 x
x0
|Ln+1 f (w)|q dw
2/q
.
(12.25)
We finish here with 1 1 + = 1, and x0 , x ∈ [a, b] p q such that x = 6 x0 . Let n be even. Assume that Ln+1 f is of fixed sign and nowhere zero. Then Z x 1/p Z x Z w p −1/q (gn (w, t)) dt dw |f (w)||Ln+1 f (w)|dw ≥ 2 Corollary 12.17. Let 0 < p < 1 and q < 0 be such that
x0
x0
Proof.
Z ×
x
x0
x0
2/q . |Ln+1 f (w)| dw
By Theorems 12.12, 12.16.
q
(12.26)
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Chapter 13
Opial Inequalities for Linear Differential Operators
Various Lp form Opial inequalities [195] are given for a Linear Differential Operator L, involving its related initial value problem solution y, Ly, the associated Green’s function H and initial conditions point x0 ∈ R. This chapter follows [18]. 13.1
Background
Here we follow [169], pp. 145–154. Let I be a closed interval of R. Let ai (x), i = 0, 1, . . . , n − 1 (n ∈ N), h(x) be continuous functions on I and let L = D n + an−1 (x)Dn−1 + · · · + a0 (x) be a fixed linear differential operator on C n (I). Let y1 (x), . . . , yn (x) be a set of linear independent solutions to Ly = 0. Here the associated Green’s function for L is y1 (t) · · · yn (t) y1 (t) · · · yn (t) 0 0 0 y1 (t) . . . yn0 (t) , y1 (t) · · · yn (t) .. .. H(x, t) := . . , (n−2) (n−2) y (n−2) (t) · · · y (n−2) (t) y (t) · · · yn (t) n 1 1 (n−1) (n−1) y1 (x) · · · yn (x) y (t) · · · yn (t) 1 which is a continuous function on I 2 . Consider fixed x0 ∈ I, then Z x y(x) = H(x, t)h(t)dt, all x ∈ I x0
is the unique solution to the initial value problem Ly = h; y (i) (x0 ) = 0, i = 0, 1, . . . , n − 1. 13.2
Results
We first give 171
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Proposition 13.1. Let x ≥ x0 ; x0 , x ∈ I and p, q > 1: p1 + 1q = 1. Then 1/p Z x Z w Z x |H(w, t)|p dt dw |y(w)||(Ly)(w)|dw ≤ 2−1/q · x0
·
Z
x
x0
x0
q
2/q
|(Ly)(w)| dw
x0
.
(13.1)
older’s Proof. Here take x ≥ x0 and p, q > 1 such that p1 + 1q = 1. From H¨ inequality we have Z x 1/p Z x 1/q . (13.2) |y(x)| ≤ |H(x, t)|p dt |h(t)|q dt x0
Set
z(w) := Thus
Z
w x0
x0
|h(t)|q dt,
x0 ≤ w ≤ x (z(x0 ) = 0).
z 0 (w) = |h(w)|q and |h(w)| = (z 0 (w))1/q . From (13.2) we obtain |y(w)| |h(w)| ≤
Z
w x0
|H(w, t)|p dt
1/p
· (z(w) · z 0 (w))1/q .
Integrating the last inequality over [x0 , x] we derive 1/p Z x Z x Z w p |y(w)| |h(w)|dw ≤ · (z(w) · z 0 (w))1/q dw |H(w, t)| dt x0
≤ =
Z
Z
x
x0 x x0
Z
Z
x0
w
x0 w x0
x0
1/p Z · |H(w, t)| dt dw
p
x
x0
0
z(w) · z (w)dw
1/p (z(x))2/q |H(w, t)|p dt dw · , 21/q
1/q
proving the claim of Proposition 13.1.
The counterpart of the previous result follows. Proposition 13.2. Let x ≤ x0 ; x0 , x ∈ I and p, q > 1: p1 + 1q = 1. Then 1/p Z x Z w Z x0 p −1/q |H(w, t)| dt dw |y(w)| |(Ly)(w)|dw ≤ 2 · x
x0
·
Z
x0 x
|(Ly)(w)|q dw
x0
2/q
.
(13.3)
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older’s Proof. Here take x ≤ x0 and p, q > 1 such that p1 + 1q = 1. From H¨ inequality we have Z x 0 Z x0 |H(x, t)| |h(t)|dt |y(x)| = H(x, t)h(t)dt ≤ x
x
≤
Set
Z
x0
p
|H(x, t)| dt
x
Z
z(x) := That is
x0 x
x0
Z
x x0
q
|h(t)| dt
x
|h(t)|q dt ≥ 0,
−z(x) = and
1/p Z
1/q
.
(13.4)
z(x0 ) = 0.
|h(t)|q dt ≤ 0,
−z 0 (x) = |h(x)|q ≥ 0, and |h(x)| = (−z 0 (x))1/q , Hence by (13.4) (x ≤ w ≤ x0 ) Z |y(w)| · |h(w)| ≤
x0 w
|H(w, t)|p dt
x ∈ I.
1/p
· (z(w) · (−z 0 (w)))1/q .
Integrating the last inequality over [x, x0 ] we find 1/p Z x0 Z x0 Z x0 p |H(w, t)| dt |y(w)| |h(w)|dw ≤ w
x
x
·(z(w) · (−z 0 (w)))1/q · dw Z x0 Z x0 1/p Z ≤ |H(w, t)|p dt · dw x
= 2−1/q ·
w
Z
x
x0
Z
w
x0
|H(w, t)|p dt · dw
x0 x
1/p
(−z 0 (w) · z(w)) · dw
1/q
· (z(x))2/q ,
proving the claim of Proposition 13.2.
Extreme cases come next. Proposition 13.3. Here p = 1, q = ∞ and x ≥ x0 . Then Z x |y(w)||(Ly)(w)|dw x0
≤
Z
x x0
Z
w x0
|H(w, t)|dt dw · kLyk2∞ .
(13.5)
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From
Proof.
Z
y(w) = we obtain |y(w)| ≤
Z
and
w
H(w, t)h(t)dt x0
|H(w, t)|dt khk∞ ,
x0
Z
|y(w)| |(Ly)(w)| ≤
w
w x0
|H(w, t)|dt · (kLyk∞ )2 .
Integrating the last inequality we obtain (13.5).
Proposition 13.4. Again p = 1, q = ∞, but x ≤ x0 . Then Z x0 |y(w)| |(Ly)(w)|dw x
≤
Z
x x0
Z
w x0
|H(w, t)|dt dw · (k(Ly)k2∞ ).
(13.6)
Here x ≤ w ≤ x0 , and Z x 0 Z w H(w, t)h(t)dt H(w, t)h(t)dt = |y(w)| = w x Z 0 x0 ≤ |H(w, t)|dt khk∞ .
Proof.
w
Therefore |y(w)| |(Ly)(w)| ≤ and
Z
x0
|y(w)| |(Ly)(w)|dw ≤
x
Corollary 13.5. Let p, q > 1: Z
Proof.
x x0
1 p
Z
Z +
x0 w
x0 x
1 q
|H(w, t)|dt (kLyk∞ )2 ,
Z
w
x
x0
|H(w, t)|dt dw
x Z w
x0
x0
· (kLyk∞ )2 .
p
|H(w, t)| dt dw
2/q . |(Ly)(w)|q dw
By Propositions 13.1, 13.2.
= 1. Then
Z −1/q |y(w)| |(Ly)(w)|dw ≤ 2 · Z ·
x0
1/p (13.7)
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In particular when p = q = 2 we obtain Corollary 13.6. It holds Z x Z ≤ 2−1/2 · |y(w)| |(Ly)(w)|dw x0
x x0
Z
w x0
1/2 (H(w, t))2 dt dw
Z x ((Ly)(w))2 dw . ·
(13.8)
x0
Furthermore we have
Corollary 13.7 Let p = 1, q = ∞. Then Z x |y(w)| |(Ly)(w)|dw x0
≤
Proof.
Z
Z
x
x0
w
x0
|H(w, t)|dt dw · kLyk2∞.
(13.9)
By Propositions 13.3 and 13.4.
To complete the chapter we present Proposition 13.8. Let 0 < p < 1 and q be such that x0 , x ∈ I. Suppose that
1 p
+
1 q
= 1, and x > x0 ,
H(w, t) ≥ 0 for x0 ≤ t ≤ w, w ∈ I. Also assume that Ly = h is of fixed sign and nowhere zero. Then Z x Z w 1/p Z x −1/q p |y(w)| |(Ly)(w)|dw ≥ 2 (H(w, t)) dt dw x0
x0
· Proof.
Z
x x0
x0
|(Ly)(w)|q dw
2/q
.
(13.10)
Here for x0 ≤ w ≤ x we have Z w H(w, t)|h(t)|dt. |y(w)| =
(13.11)
x0
From (13.11) by H¨ older’s inequality we derive 1/p Z Z w p (H(w, t)) dt · |y(w)| ≥ x0
for w > x0 . Consider z(w) :=
Z
w x0
|h(t)|q dt,
w x0
q
|h(t)| dt
z(x0 ) = 0.
1/q
,
(13.12)
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So that z 0 (w) = |h(w)|q and |h(w)| = (z 0 (w))1/q , all x0 ≤ w ≤ x. Hence by (13.12) we get 1/p Z w · (z(w) · z 0 (w))1/q , (H(w, t))p dt |y(w)| |h(w)| ≥ x0
all x0 < w ≤ x. Let x0 < θ ≤ w ≤ x and θ ↓ x0 , then by integration of the last inequality we obtain Z x Z x |y(w)| |h(w)|dw = lim |y(w)| |h(w)|dw θ↓x0 θ x0 ! Z Z x
≥ lim
θ↓x0
≥ lim
θ↓x0
= 2
w
θ
Z
−1/q
= 2−1/q
·
1/p
· (z(w) · z 0 (w))1/q · dw
(H(w, t))p dt
x0
x θ
Z
Z
Z
x x0
w p
(H(w, t)) dt dw x0
Z
x Z
x0
w
p
1/p
(H(w, t)) dt dw x0
w
(H(w, t))p dt dw x0
Z
· lim
θ↓x0
1/p
1/p
x 0
z(w)z (w)dw θ
1/q !
· lim (z 2 (x) − z 2 (θ))1/q θ↓x0
· (z(x))2/q .
That is establishing (13.10).
Proposition 13.9. Let 0 < p < 1 and q be such that x0 , x ∈ I. Suppose that
1 p
+
1 q
= 1, and x < x0 ,
H(w, t) ≤ 0 for w ≤ t ≤ x0 , w ∈ I. Also assume that Ly = h is of fixed sign and nowhere zero. Then Z x Z w 1/p Z x0 |y(w)| |(Ly)(w)|dw ≥ 2−1/q · |H(w, t)|p dt dw x
· Proof.
Z
x0 x
x0
x0
q
2/q
|(Ly)(w)| dw
.
(13.13)
Here for x ≤ w ≤ x0 we have Z x0 Z w H(w, t)h(t)dt H(w, t)h(t)dt = |y(w)| = Zw x0 Z xx0 0 (−H(w, t))|h(t)|dt. (−H(w, t))h(t)dt = =
(13.14)
w
w
From (13.14) by H¨ older’s inequality we find Z x0 1/p Z |y(w)| ≥ (−H(w, t))p dt w
x0 w
|h(t)|q dt
1/q
,
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Opial Inequalities for Linear Differential Operators
for w < x0 . That is |y(w)| ≥ for w < x0 . Consider
Z
x0
p
|H(w, t)| dt
w
z(w) := So that
Z
x0
1/p Z
|h(t)|q dt,
w
−z(w) = and
Z
w x0
x0
q
|h(t)| dt
w
177
1/q
,
(13.15)
z(x0 ) = 0.
|h(t)|q dt
−z 0 (w) = |h(w)|q ,
with
|h(w)| = (−z 0 (w))1/q ,
Therefore by (13.15) we find Z |y(w)| |h(w)| ≥
x0 w
all x ≤ w ≤ x0 .
|H(w, t)|p dt
1/p
· (z(w) · (−z 0 (w)))1/q ,
all x ≤ w < x0 . Let x ≤ w ≤ θ < x0 and θ ↑ x0 , then by integration of the last inequality we obtain Z θ Z x0 |y(w)| |h(w)|dw = lim |y(w)| |h(w)|dw θ↑x0 x x ! Z Z θ
≥ lim
θ↑x0
x
Z
≥ lim
θ↑x0
=
Z
= 2
x0
x0 x
−1/q
= 2−1/q · Therefore Z x0 x
θ x
Z ·
w
Z
1/p
|H(w, t)|p dt
x0 w
p
p
|H(w, t)| dt dw
w
Z
|H(w, t)| dt dw
x0
Z
· (z(w) · (−z 0 (w)))1/q · dw
x0 x
Z
x0 Z
x
x0 w x0 w
1/p
p
|H(w, t)|p dt dw
that is proving (13.13).
x0 x
Z
|h(w)|q dw
θ↑x0
−
θ↑x0
|H(w, t)| dt dw
Z
· lim
· lim
|y(w)| |h(w)|dw ≥ 2−1/q · ·
!1/p
1/p
1/p
0
x
(−z(w))(z (w)) · dw
θ
z(w)dz(w) x
!1/q
!1/q
θ↑x0
· (z(x))2/q .
x0
x0
θ
· lim (z 2 (x) − z 2 (θ))1/q
x Z w
2/q
Z
Z
1/p |H(w, t)|p dt dw
,
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Putting things together we have Corollary 13.10. Let 0 < p < 1 and q be such that x 6= x0 . Suppose that
1 p
+
1 q
= 1, and x0 , x ∈ I that
(w − x0 ) · H(w, t) ≥ 0, for all t between x0 , w ∈ I. Also assume that Ly = h is of fixed sign and nowhere zero. Then Z x Z x Z w 1/p p ≥ 2−1/q · |y(w)| |(Ly)(w)|dw |H(w, t)| dt dw x0
Proof.
x0
Z ·
x
2/q . |(Ly)(w)| dw q
x0
x0
By Propositions 13.8 and 13.9.
(13.16)
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Chapter 14
Opial Inequalities for Vector Valued Functions
Various Lp form Opial type inequalities are given for functions valued in a Banach vector space with applications in C. This chapter relies on [52].
14.1
Introduction
This chapter is greatly motivated by the article of Z. Opial [195]. Theorem 14.1. (Opial [195]) Let x(t) ∈ C 1 ([0, h]) be such that x(0) = x(h) = 0, and x(t) > 0 in (0, h). Then Z Z h h h 0 0 (x (t))2 dt. (14.1) |x(t)x (t)|dt ≤ 4 0 0 In the last inequality the constant h/4 is the best possible. Equality holds for the function x(t) = t on [0, h/2] and x(t) = h − t on [h/2, h]. Opial type inequalities have applications in establishing uniqueness of solution to initial value problems in differential equations, see [244]. We are also motivated by [16], [21]. We need
14.2
Background
(see [230], pp. 83–94) Let f (t) be a function defined on [a, b] ⊆ R taking values in a real or complex normed linear space (X, k · k). Then f (t) is said to be differentiable at a point t0 ∈ [a, b] if the limit f (t0 + h) − f (t0 ) (14.2) f 0 (t0 ) := lim h→0 h 179
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exists in X, the convergence is in k · k. This is called the derivative of f (t) at t = t 0 . We call f (t) differentiable on [a,b], iff there exists f 0 (t) ∈ X for all t ∈ [a, b]. Similarly and inductively are defined higher order derivatives of f , denoted 00 f , f (3) , ..., f (k) , k ∈ N, just as for numerical functions. For all the properties of derivatives see [230], pp. 83–86. Let now (X, k · k) be a Banach space, and f : [a, R b b] → X. We define the vector valued Riemann integral a f (t)dt ∈ X as the limit of the vector valued Riemann sums in X, convergence is in k · k. The definition is as for the numerical valued functions. Rb If a f (t)dt ∈ X we call f integrable on [a,b]. For the properties of vector valued Riemann integrals see [230], pp. 86–91. We define the space C n ([a, b], X), n ∈ N, of n-times continuously differentiable functions from [a,b] into X; here continuity is with respect to k · k and defined in the usual way as for numerical functions. Let (X, k · k) be a Banach space and f ∈ C n ([a, b], X), then we have the vector valued Taylor’s formula, see [230], pp. 93–94, and also [227], (IV, 9; 47): It holds En (x, y) := f (y) − f (x) − f 0 (x)(y − x) 1 1 f (n−1) (x)(y − x)n−1 − f 00 (x)(y − x)2 − · · · − 2 (n − 1)! =
1 (n − 1)!
Z
y x
(y − t)n−1 f (n) (t)dt,
∀ x, y ∈ [a, b].
(14.3)
In particular (14.3) is true when X = Rm , Cm , m ∈ N, etc. In case of some x0 ∈ [a, b] such that f (k) (x0 ) = 0, k = 0, 1, ..., n − 1, then Z y 1 f (y) = (y − t)n−1 f (n) (t)dt, ∀ y ∈ [a, b]. (14.4) (n − 1)! x0 In that case En (x0 , y) = f (y). 14.3
Results
Here we consider always X to be a Banach space, n ∈ N, and f ∈ C n ([a, b], X), [a, b] ⊆ R. We fix x0 ∈ [a, b]. We present the first result of the chapter. Theorem 14.2. Let p, q > 1 : Then I1 :=
Z
1 p
y x0
+
1 q
= 1, y ≥ x0 ; y, x0 ∈ [a, b].
kEn (x0 , w)k kf (n) (w)kdw
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Opial Inequalities for Vector Valued Functions 2
≤
(y − x0 )n−1+ p 21/q (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)]1/p
Z
y x0
181
kf (n) (w)kq dw
2/q
.
(14.5)
We give Corollary 14.3. For y ∈ [x0 , b] we have I1 ≤
Z
(y − x0 )n p 2(n − 1)! n(2n − 1)
y x0
kf (n) (w)k2 dw .
Proof of Theorem 14.2. We have by (14.3) that Z y 1 (y − t)n−1 f (n) (t)dt, ∀ y ≥ x0 ; x0 , y ∈ [a, b]. En (x0 , y) = (n − 1)! x0 By H¨ older’s inequality we derive 1 kEn (x0 , y)k ≤ (n − 1)! 1 ≤ (n − 1)!
Z
Z
x0
(y − t)
(14.7)
y x0
y p(n−1)
(14.6)
dt
(y − t)n−1 kf (n) (t)kdt
1/p Z
y x0
kf
(n)
q
(t)k dt
1/q
1
= where z(y) :=
(y − x0 )n−1+ p 1 (z(y))1/q , (n − 1)! (p(n − 1) + 1)1/p
(14.8)
Z
(14.9)
y x0
kf (n) (t)kq dt, x0 ≤ y ≤ b (z(x0 ) = 0).
Thus z 0 (y) = kf (n) (y)kq and kf (n) (y)k = (z 0 (y))1/q .
(14.10)
Therefore we obtain 1
kEn (x0 , y)k kf
(n)
1 (y − x0 )n−1+ p (y)k ≤ (z(y)z 0 (y))1/q . (n − 1)! (p(n − 1) + 1)1/p
Integrating the last inequality we have Z y kEn (x0 , w)k kf (n) (w)kdw x0
≤
1 (n − 1)!(p(n − 1) + 1)1/p
Z
y x0
1
(w − x0 )n−1+ p (z(w)z 0 (w))1/q dw
(14.11)
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Z
1 (n − 1)!(p(n − 1) + 1)1/p
≤
y x0
(w − x0 )p(n−1)+1 dw
1/p Z
y
z(w)z 0 (w)dw
x0
1/q
n−1+ p2
=
21/q (n
(y − x0 ) (z(y))2/q , − 1)![(p(n − 1) + 1)(p(n − 1) + 2)]1/p
(14.12)
proving the claim of the theorem. We continue with
Theorem 14.4. Let p, q > 1 : p1 + q1 = 1, y ∈ [a, x0 ]. Then Z x0 kEn (x0 , w)k kf (n) (w)kdw I2 := y
2
(x0 − y)n−1+ p ≤ 1/q 2 (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)]1/p
Z
x0
kf
y
(n)
q
(w)k dw
2/q
. (14.13)
We give Corollary 14.5. For y ∈ [a, x0 ] we have
(x0 − y)n p I2 ≤ 2(n − 1)! n(2n − 1)
Z
x0 y
kf
(n)
2
(w)k dw .
(14.14)
Proof of Theorem 14.4. We have by (14.3) that
Z x 0
1 n−1 (n)
kEn (x0 , y)k = (y − t) f (t)dt
(n − 1)! y
1 (n − 1)!
≤
≤
1 (n − 1)!
Z
x0 y
Z
x0
y
(t − y)n−1 kf (n) (t)kdt
(t − y)p(n−1) dt
1/p Z
x0 y
kf (n) (t)kq dt
1/q
1
= Here
(x0 − y)n−1+ p (z(y))1/q . (n − 1)!(p(n − 1) + 1)1/p
z(y) :=
Z
x0 y
kf (n) (t)kq dt,
and −z(y) =
Z
y x0
(z(x0 ) = 0),
kf (n) (t)kq dt ≤ 0,
−z 0 (y) = kf (n) (y)kq ≥ 0,
(14.15)
(14.16)
(14.17) (14.18)
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and kf (n) (y)k = (−z 0 (y))1/q ,
a ≤ y ≤ x0 .
(14.19)
Therefore kEn (x0 , y)k kf (n) (y)k 1
(x0 − y)n−1+ p (z(y)(−z 0 (y)))1/q , ≤ (n − 1)!(p(n − 1) + 1)1/p
Hence we find
≤
Z
x0
a ≤ y ≤ x0 .
(14.20)
kEn (x0 , w)k kf (n) (w)kdw
y
1 (n − 1)!(p(n − 1) + 1)1/p
Z
x0 y
1 ≤ (n − 1)!(p(n − 1) + 1)1/p Z
x0
1
(x0 − w)n−1+ p (z(w)(−z 0 (w)))1/q dw Z
x0 y
(x0 − w)
z(w)(−z 0 (w))dw
y
p(n−1)+1
dw
(14.21)
1/p
1/q
(14.22)
2
(x0 − y)n−1+ p (z(y))2/q , ∀ y ∈ [a, x0 ], = 1/q 2 (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)]1/p
proving the claim of the theorem. Combining Theorems 14.2 and 14.4 we obtain
(14.23)
Theorem 14.6. Let p, q > 1 : p1 + q1 = 1 and y, x0 ∈ [a, b]. Then Z y kEn (x0 , w)k kf (n) (w)kdw I := x0
Z y 2/q 2 |y − x0 |n−1+ p (n) q . ≤ 1/q kf (w)k dw 2 (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)]1/p x0
(14.24)
Combining Corollaries 14.3 and 14.5 we derive
Corollary 14.7. For y, x0 ∈ [a, b] it holds I≤
Z y |y − x0 |n (n) 2 . p kf (w)k dw 2(n − 1)! n(2n − 1) x0
A special but important case follows
(14.25)
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Theorem 14.8. Let p, q > 1 : p1 + 1q = 1 and y, x0 ∈ [a, b]. Suppose further that f (k) (x0 ) = 0, k = 0, 1, ..., n − 1. Then Z y (n) kf (w)k kf (w)kdw x0
≤ min
Proof.
Z y 2/q 2 |y − x0 |n−1+ p (n) q , kf (w)k dw 1/q 1/p 2 (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)] x0 Z y ! |y − x0 |n p kf (n) (w)k2 dw . 2(n − 1)! n(2n − 1) x0
By Theorem 14.6 and Corollary 14.7.
(14.26)
We continue with Theorem 14.9. Let p = 1, q = ∞ and y ∈ [x0 , b]. Then Z y (y − x0 )n+1 (n) 2 kf k∞,[x0 ,b] . kEn (x0 , w)k kf (n) (w)kdw ≤ (n + 1)! x0
(14.27)
We have by (14.3) that Z y (y − x0 )n 1 (y − t)n−1 kf (n) (t)kdt ≤ kf (n) k∞,[x0 ,b] kEn (x0 , y)k ≤ . (n − 1)! x0 n! (14.28) Consequently it holds Proof.
kEn (x0 , y)k kf (n)(y)k ≤ kf (n) k2∞,[x0 ,b]
(y − x0 )n , n!
(14.29)
all x0 ≤ y ≤ b. Hence we derive Z y kf (n) k2∞,[x0 ,b] Z y (n) kEn (x0 , w)k kf (w)kdw ≤ (w − x0 )n dw n! x0 x0 =
(y − x0 )n+1 (n) 2 kf k∞,[x0 ,b] . (n + 1)!
The counterpart of Theorem 14.9 follows. Theorem 14.10. Let p = 1, q = ∞ and y ∈ [a, x0 ]. Then
(14.30)
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Z Proof.
x0 y
kEn (x0 , w)k kf (n) (w)kdw ≤
185
(x0 − y)n+1 (n) 2 kf k∞,[a,x0 ] . (n + 1)!
(14.31)
We have by (14.3) that
Z y
1 n−1 (n)
kEn (x0 , y)k = (y − t) f (t)dt
(n − 1)! x0
=
Z x 0
Z x0
1 1 n−1 (n)
≤ (y − t) f (t)dt (t−y)n−1 kf (n) (t)kdt (14.32)
(n − 1)! (n − 1)! y y kf (n) k∞,[a,x0 ] ≤ (n − 1)!
Z
x0
y
(t − y)n−1 dt = kf (n) k∞,[a,x0 ]
(x0 − y)n . n!
(14.33)
Hence it holds kEn (x0 , y)k kf (n) (y)k ≤ kf (n) k2∞,[a,x0 ]
(x0 − y)n , n!
(14.34)
all a ≤ y ≤ x0 . Consequently Z
x0 y
kEn (x0 , w)kf
(n)
(w)kdw ≤
kf (n) k2∞,[a,x0 ] Z
= kf (n) k2∞,[a,x0 ]
n!
x0 y
(x0 − w)n dw
(x0 − y)n+1 . (n + 1)!
(14.35)
Combining Theorems 14.9 and 14.10 we obtain Proposition 14.11. Let p = 1, q = ∞, y, x0 ∈ [a, b]. Then Z y |y − x0 |n+1 (n) 2 (n) ≤ kf k∞ . kE (x , w)k kf (w)kdw n 0 (n + 1)! x0
(14.36)
In particular we get
Proposition 14.12. Let p = 1, q = ∞, y, x0 ∈ [a, b]. Further suppose that f (k) (x0 ) = 0, k = 0, 1, ..., n − 1. Then Z y |y − x0 |n+1 (n) 2 (n) ≤ kf k∞ . (14.37) kf (w)k kf (w)kdw (n + 1)! x0
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14.4
Applications
Here X = C, f ∈ C n ([a, b], C), n ∈ N, x0 , y ∈ [a, b]. Furthermore suppose that f (k) (x0 ) = 0, k = 0, 1, ..., n − 1. Applying Theorem 14.8 we obtain Theorem 14.13. Let p, q > 1 : Then Z ≤ min
1 p
+
1 q
= 1.
y x0
|f (w)| |f
(n)
(w)|dw
Z y 2/q 2 |y − x0 |n−1+ p (n) q |f (w)| dw , 1/q 1/p 2 (n − 1)![(p(n − 1) + 1)(p(n − 1) + 2)] x0 Z y ! |y − x0 |n p |f (n) (w)|2 dw . 2(n − 1)! n(2n − 1) x0
(14.38)
By Proposition 14.12 we derive
Proposition 14.14. Let p = 1, q = ∞. Then Z y |y − x0 |n+1 (n) 2 (n) ≤ kf k∞ . |f (w)| |f (w)|dw (n + 1)! x0
(14.39)
Let now f ∈ C 1 ([a, b], C), x0 , y ∈ [a, b] with f (x0 ) = 0. Applying Theorem 14.13 we get
Theorem 14.15. Let p, q > 1 : p1 + 1q = 1. Then Z y 0 |f (w)| |f (w)|dw x0
Z y 2/q Z y ! 1 |f 0 (w)|q dw , |y − x0 | |f 0 (w)|2 dw . ≤ min |y − x0 |2/p 2 x0 x0
(14.40)
Finally by applying Proposition 14.14 we find
Proposition 14.16. Let p = 1, q = ∞. Then Z y (y − x0 )2 0 2 0 ≤ |f (w)| |f (w)|dw kf k∞ . 2 x0
(14.41)
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Chapter 15
Opial Inequalities for Semigroups
Various Lp form Opial type inequalities are given for semigroups with applications. This chapter relies on [31].
15.1
Introduction
The chapter is also greatly motivated by the article of Z. Opial ([195]): Theorem 15.1 (Opial [195]) Let x(t) ∈ C 1 ([0, h]) be such that x(0) = x(h) = 0, and x(t) > 0 in (0, h). The following inequality holds Z
h 0
|x(t)x0 (t)|dt ≤ (h/4)
Z
h
(x0 (t))2 dt.
(15.1)
0
In the last inequality the constant h/4 is the best possible. Opial type inequalities have applications in establishing uniqueness of solution to initial value problems in differential equations, see [244]. We are also motivated by [16], [21]. The generalized Opial type inequalities we prove here are in Theorems 15.8, 15.10 and 15.15.
15.2
Background
All this background comes from [79] (in general see also [147]). Let X a real or complex Banach space with elements f, g, · · · having norm kf k, kgk, · · · and let E(X) be the Banach algebra of endomorphisms of X. If T ∈ E(X), kT k denotes the norm of T . Definition 15.2. If T (t) is an operator function on the non- negative real axis 0 ≤ t < ∞ to the Banach algebra E(X) satisfying the following conditions: 187
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(i) T (t1 + t2 ) = T (t1 )T (t2 ), (t1 , t2 ≥ 0), (ii) T (0) = I (I = identity operator),
(15.2)
then {T (t); 0 ≤ t < ∞} is called a one-parameter semi-group of operators in E(X). The semi-group {T (t); 0 ≤ t < ∞} is said to be of class C0 if it satisfies the further property (iii)s − lim T (t)f = f, (f ∈ X) t→0+
(15.3)
referred to as the strong continuity of T (t) at the origin. In this chapter we shall assume that the family of bounded linear operators {T (t); 0 ≤ t < ∞} mapping X to itself is a semi-group of class C0 , thus that all three conditions of the above definition are satisfied. Proposition 15.3. (a) kT (t)k is bounded on every finite subinterval of [0, ∞). (b) For each f ∈ X, the vector-valued function T (t)f on [0, ∞) is strongly continuous. Definition 15.4. The infinitesimal generator A of the semi-group {T (t); 0 ≤ t < ∞} is defined by Af = s − lim Aτ f, Aτ f = τ →0+
1 [T (τ ) − I] τ
(15.4)
whenever the limit exists; the domain of A, in symbols D(A), is the set of elements f for which the limit exists. Proposition 15.5. (a) D(A) is a linear manifold in X and A is a linear operator. (b) If f ∈ D(A), then T (t)f ∈ D(A) for each t ≥ 0 and d T (t)f = AT (t)f = T (t)Af dt
(t ≥ 0);
(15.5)
furthermore,
T (t)f − f =
Z
t
T (u)Af du
(t > 0).
(15.6)
0
(c) D(A) is dense in X, i.e. D(A) = X, and A is a closed operator. Definition 15.6. For r = 0, 1, 2, . . . the operator Ar is defined inductively by the relations A0 = I, A1 = A, and D(Ar ) = {f ; f ∈ D(Ar−1 ) and Ar−1 f ∈ D(A)} Ar f = A(Ar−1 f ) = s − lim Aτ (Ar−1 f ) (f ∈ D(Ar )). τ →0+
(15.7)
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For the operator Ar and its domain D(Ar ) we have the following Proposition 15.7. (a) D(Ar ) is a linear subspace in X and Ar is a linear operator. (b) If f ∈ D(Ar ), so does T (t)f for each t ≥ 0 and dr T (t)f = Ar T (t)f = T (t)Ar f. dtr
(15.8)
Furthermore T (t)f −
r−1 k X t
k=0
k!
Ak f =
1 (r − 1)!
Z
t 0
(t − u)r−1 T (u)Ar f du,
(15.9)
the Taylor’s formula for semigroups. Additionally it holds Z tZ t Z t [T (t) − I]r = ··· T (u1 + u2 + · + ur )Ar f du1 du2 . . . dur . 0
0
(15.10)
0
r (c) D(Ar ) is dense in X for r = 1, 2 . . .; moreover, ∩∞ r=1 D(A ) is dense in X. A is a closed operator. Integrals in (15.9) and (15.10) are vector valued Riemann integrals, see [79], [164]. r
15.3
Results
Here always we consider T (t), A as in the Background of this chapter and f ∈ D(Ar ), r ∈ N. We present the first main result. Theorem 15.8. Let p, q > 1 : p1 + q1 = 1. One has Z t k∆(w)f k kT (w)Ar f kdw ≤ 0
2
tr−1+ p 21/q (r − 1)! [(p(r − 1) + 1)(p(r − 1) + 2)]1/p
for all t ∈ R+ = [0, ∞), where
∆(t)f := T (t)f −
Z
t 0
r−1 k X t
k=0
k!
kT (w)Ar f kq dw
Ak f.
We give Corollary 15.9. Let p = q = 2. Then Z
t 0
k∆(w)f k kT (w)Ar f kdw ≤
2/q
,
(15.11)
(15.12)
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2(r − 1)!
tr p
r(2r − 1)
Z
t r
0
2
kT (w)A f k dw , for all t ∈ R+ .
Proof of Theorem 15.8. By Taylor’s formula (15.9) we have Z t 1 (t − u)r−1 T (u)Ar f du. ∆(t)f = (r − 1)! 0 Therefore
k∆(t)f k = ≤ 1 (r − 1)!
≤
Z
1 k (r − 1)!
1 (r − 1)! t 0
Z
t 0
t
(15.14)
(t − u)r−1 T (u)Ar f duk
0
(t − u)r−1 kT (u)Ar f kdu
(t − u)p(r−1) du 1
=
Z
(15.13)
1/p Z
tr−1+ p (r − 1)! (p(r − 1) + 1)1/p
So far we have found that
Z
t 0
t 0
kT (u)Ar f kq du
kT (u)Ar f kq du
1/q
1/q
.
1
tr−1+ p 1/q (z(t)) , k∆(t)f k ≤ (r − 1)! (p(r − 1) + 1)1/p
where
z(t) := Clearly
Z
(15.15)
t 0
kT (u)Ar f kq du, for all t ∈ R+ .
z 0 (w) := kT (w)Ar f kq , 0 ≤ w ≤ t, z(0) = 0.
(15.16)
(15.17)
In particular kT (w)Ar f k = (z 0 (w))
1/q
.
Hence we have that 1
wr−1+ p 1/q k∆(w)f k kT (w)A f k ≤ (z(w)z 0 (w)) . (r − 1)! (p(r − 1) + 1)1/p r
Consequently it follows Z
Z
t 0 t
0
k∆(w)f k kT (w)Ar f kdw ≤ 1
wr−1+ p (z(w)z 0 (w))
1/q
1 (r − 1)! (p(r − 1) + 1)1/p
dw ≤
1 (r − 1)! (p(r − 1) + 1)1/p
(15.18)
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191
Z t 1/q p 1/p Z t r−1+ p1 0 dw · z(w)z (w)dw w 0
0
2 r−1+ p
t 1 · = (r − 1)! (p(r − 1) + 1)1/p (p(r − 1) + 2)1/p 2
1/q z 2 (w) t 2 0
1 tr−1+ p 2/q = · (z(t)) (r − 1)! (p(r − 1) + 1)1/p 21/q (p(r − 1) + 2)1/p Z t 2/q 2 tr−1+ p 1 r q · = kT (u)A f k du , (r − 1)! (p(r − 1) + 1)1/p 21/q (p(r − 1) + 2)1/p 0 (15.19) proving the claim. We give Theorem 15.10. Here p = 1, q = ∞ and suppose w w w w w kT (u)Ar f k w := sup kT (u)Ar f k < ∞. ∞
ThenZ
t
k∆(w)f k kT (w)Ar f kdw ≤
0
where
w2 tr+1 w w w w ||T (u)Ar f k w , for all t ∈ R+ , (r + 1)! ∞
∆(t)f := T (t)f − We have
Proof.
(15.20)
u∈R+
r−1 k X t
k=0
k!
Ak f.
(15.21)
Z t 1 (t − u)r−1 kT (u)Ar f kdu (r − 1)! 0 Z t w w w w (t − u)r−1 du w kT (u)Ar f k w
k∆(t)f k ≤ ≤
That is
1 (r − 1)! 0 w tr w w w = w kT (u)Ar f k w , for all t ∈ R+ . r! ∞
k∆(t)f k ≤ Therefore Hence
w tr w w w w kT (u)Ar f k w , for all t ∈ R+ . r! ∞
k∆(w)f k kT (w)Ar f k ≤ Z
t 0
w2 wr w w w w kT (u)Ar f k w , all 0 ≤ w ≤ t. r! ∞
k∆(w)f k kT (w)Ar f kdw ≤
proving the claim.
∞
Z
t
0
wr dw
(15.22)
w w2 w w w kT (u)Ar f k w
∞
r!
w2 tr+1 w w w = w kT (u)Ar f k w , for all t ∈ R+ , (r + 1)! ∞
(15.23)
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We give Application 15.11 (see also [80]). It is known that the classical diffusion equation ∂2W ∂W = , −∞ < x < ∞, t > 0 ∂t ∂x2
(15.24)
lim W (x, t) = f (x),
(15.25)
with initial condition t→0+
has under general conditions its solution given by Z ∞ 2 1 f (x + u)e−u /4t du, W (x, t, f ) = [T (t)f ](x) = √ 2 πt −∞
(15.26)
the so called Gauss–Weierstrass singular integral. The infinitesimal generator of the semigroup {T (t); 0 ≤ t < ∞} is A = ∂ 2 /∂x2 ([159], p. 578). Here we suppose that f, f (2k) , k = 1, . . . , r, all belong to the Banach space U CB(R), the space of bounded and uniformly continuous functions from R into itself, with norm kf kC := sup |f (x)|.
(15.27)
x∈R
Here we define ∆(t)f (x) := W (x, t, f ) −
r−1 k X t
k=0
k!
f (2k) (x), for all x ∈ R.
(15.28)
Therefore by Theorem 15.8 we derive
Proposition 15.12. Let p, q > 1 : p1 + 1q = 1. Then Z t k∆(t)f kC kW (·, w, f (2r) )kC dw ≤ It := 0
2
tr−1+ p 21/q (r − 1)! [(p(r − 1) + 1)(p(r − 1) + 2)]1/p ·
Z
t 0
kW (·, w, f (2r) )kqC dw
2/q
, for all t ∈ R+ .
We make Remark 15.13. We notice here that 1 |W (x, t, f (2r) )| = √ 2 πt
Z
∞ −∞
f (2r) (x + u)e−u
2
/4t
du
(15.29)
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1 ≤ √ 2 πt
Z
∞ −∞
|f (2r) (x + u)|e−u
1 ≤ kf (2r)kC √ 2 πt
Z
∞
e−u
2
2
193
/4t
/4t
du
du
−∞
= kf (2r) kC · 1 = kf (2r) kC < ∞, for all t ∈ R.
(15.30)
kW (·, t, f (2r) )kC ≤ kf (2r) kC .
(15.31)
That is
And thus we obtain
w w w w Φ := wkW (·, t, f (2r) )kC w . ∞
:= sup kW (·, t, f (2r) )kC ≤ kf (2r) kC < ∞.
(15.32)
t∈R+
We give Proposition 15.14. Here p = 1, q = ∞. Then tr+1 It ≤ Φ2 , for all t ∈ R+ . (r + 1)!
Proof.
(15.33)
By Theorem 15.10.
We finally get Theorem 15.15. Let p, q > 1 : Choose x∗t ∈ [0, rt] such that
1 p
+
1 q
= 1, t ≥ 0.
ρt := kT (x∗t )Ar f k = sup kT (x)Ar f k.
(15.34)
x∈[0,rt]
Then
Z
t 0
k(T (w) − I)r f k kT (rw)Ar f kdw
ρt tr+(1/p) ≤ (rp + 1)1/p Proof.
Z
t 0
r
q
kT (rw)A f k dw
Call Γ(t) := T (t) − I, t ≥ 0.
Then by (15.10) we have
1/q
.
(15.35)
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r
Γ (t)f =
Z tZ 0
t
···
0
Z
t
T (u1 + u2 + · + ur )Ar f du1 du2 . . . dur , t ≥ 0.
0
(15.36)
Therefore it follows that Z tZ t Z t kΓr (t)f k ≤ ··· kT (u1 + u2 + · + ur )Ar f kdu1 du2 . . . dur 0
0
Z t Z
≤ ·
Z t Z
=t
t
0
0
r/p
Z
···
0
0
Z
t
···
0 t
···
0
Z
t
1p du1 du2 . . . dur 0
t
r
q
kT (u1 + u2 + · + ur )A f k du1 du2 . . . dur
0
Z
1/p
t
r
q
1/q
kT (u1 + u2 + · + ur )A f k du1 du2 . . . dur
0
1/q
.
That is we derive
kΓr (t)f k ≤ tr/p t ≥ 0.
Z t Z 0
Set G(t1 , t2 , . . . , tr ) :=
Z
t
···
0
t1 0
Z
Z
t 0
1/q
,
(15.37)
t2
···
0
kT (u1 + u2 + · + ur )Ar f kq du1 du2 . . . dur
Z
tr 0
where
kT (u1 + u2 + · + ur )Ar f kq du1 du2 . . . dur ,
t1 , t2 , . . . , tr ≥ 0.
(15.38)
Thus kΓr (t)f k ≤ tr/p (G(t, t, . . . , t))
1/q
, t ≥ 0.
(15.39)
Clearly we find ∂ r G(t1 , t2 , . . . , tr ) = kT (t1 + t2 + . . . + tr )Ar f kq , all ti ≥ 0, i = 1, r, ∂t1 ∂t2 . . . ∂tr
(15.40)
and in particular we get ∂ r G(t, t, . . . , t) = kT (rt)Ar f kq , ∂t1 ∂t2 . . . ∂tr
(15.41)
and r
kT (rt)A f k = Consequently we have
∂ r G(t, t, . . . , t) ∂t1 ∂t2 . . . ∂tr
1/q
, for all t ≥ 0.
(15.42)
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kΓr (t)f k kT (rt)Ar f k ≤ 1/q ∂ r G(t, t, . . . , t) r/p G(t, t, . . . , t) t , for all t ≥ 0. ∂t1 ∂t2 . . . ∂tr
Thus
Z
t 0
Z
t
kΓr (w)f k kT (rw)Ar f kdw ≤
0
wr/p G(w, w, . . . , w)
∂ r G(w, w, . . . , w) ∂t1 ∂t2 . . . ∂tr
Notice here that
G(w, w, . . . , w) =
Z
w 0
≤w
r
Z
(15.43)
w
···
0
Z
w 0
1/q
dw =: J
(15.44)
kT (u1 + u2 + · + ur )Ar f kq du1 du2 . . . dur
sup kT (x)Ar f kq = wr kT (x∗t )Ar f kq ,
(15.45)
x∈[0,rt]
for some x∗t ∈ [0, rt], 0 ≤ w ≤ t. That is G(w, w, . . . , w) ≤ wr kT (x∗t )Ar f kq .
Hence we derive
(15.46)
1/q ∂ r G(w, w, . . . , w) dw ∂t1 ∂t2 . . . ∂tr 0 r 1/q Z t ∂ G(w, w, . . . , w) = kT (x∗t )Ar f k wr dw ∂t1 ∂t2 . . . ∂tr 0
J ≤
Z
≤ kT (x∗t )Ar f k
t
wr kT (x∗t )Ar f k
Z
t
wrp dw
0
1/p Z t 0
= kT (x∗t )Ar f k
tr+(1/p) (rp + 1)1/p
We have established that Z
t 0
Z
∂ r G(w, w, . . . , w) ∂t1 ∂t2 . . . ∂tr
t
kT (rw)Ar f kq dw
0
dw
1/q
1/q
.
(15.47)
1/q
,
(15.48)
kΓr (w)f k kT (rw)Ar f kdw
tr+(1/p) ≤ kT (x∗t )Ar f k (rp + 1)1/p that is proving the claim.
Z
t 0
r
q
kT (rw)A f k dw
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Chapter 16
Opial Inequalities for Cosine and Sine Operator Functions
Various Lp form Opial type inequalities are given for Cosine and Sine Operator functions with applications. This chapter relies on [45]. 16.1
Introduction
This chapter is also greatly motivated by the article of Z. Opial [195]. Theorem 16.1. (Opial [195]) Let x(t) ∈ C 1 ([0, h]) be such that x(0) = x(h) = 0, and x(t) > 0 in (0, h). Then, Z Z h h h 0 0 (x (t))2 dt. (16.1) |x(t)x (t)|dt ≤ 4 0 0 In the last inequality the constant h/4 is the best possible. Opial type inequalities have applications in establishing uniqueness of solution to initial value problems in differential equations, see [244]. We are also motivated by [16], [21]. We need 16.2
Background
(see [93], [147], [191], [192]) Let (X, k·k) be a real or complex Banach space. By definition, a Cosine operator function is a family {C(t); t ∈ R} of bounded linear operators from X into itself, satisfying (i) C(0) = I, I the identity operator; (ii) C(t + s) + C(t − s) = 2C(t)C(s), ∀ t, s ∈ R; (the last product is composition ) (iii) C(·)f is continuous an R, ∀ f ∈ X. 197
(16.2)
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Notice that C(t) = C(−t) ∀ t ∈ R. The associated Sine operator function S(·) is defined by S(t)f :=
Z
t
∀ t ∈ R,
C(s)f ds, 0
∀ f ∈ X.
(16.3)
The Cosine operator function C(·) is such that kC(t)k ≤ M eωt , for some M ≥ 1, ω ≥ 0, ∀ t ∈ R+ , here k · k is the norm of the operator. The infinitesimal generator A of C(·) is the operator from X into itself defined as Af := lim
t→0+
2 (C(t) − I) f t2
(16.4)
with domain D(A). Operator A is closed and D(A) is dense in X, i.e. D(A) = X, and satisfies: Z t Z t S(s)f ds = C(t)f − f, ∀f ∈ X. (16.5) S(s)f ds ∈ D(A) and A 0
0
Also it holds A = C 00 (0), and D(A) is the set of f ∈ X : C(t)f is twice differentiable at t = 0 ; equivalently, D(A) = {f ∈ X : C(·)f ∈ C 2 (R, X)}.
(16.6)
If f ∈ D(A), then C(t)f ∈ D(A), and C 00 (t)f = C(t)Af = AC(t)f, ∀t ∈ R; C 0 (0)f = 0, see [140], [232]. We define A0 = I, A2 = A ◦ A, ..., An = A ◦ An−1 , n ∈ N. Let f ∈ D(An ), then C(t)f ∈ C 2n (R, X), and C (2n) (t)f = C(t)An f = An C(t)f, ∀t ∈ R, and C (2k−1) (0)f = 0, 1 ≤ k ≤ n, see [191]. For f ∈ D(An ), t ∈ R, we have the Cosine operator function’s Taylor formula ([191], [192]) saying that Tn (t)f := C(t)f −
n−1 X k=0
t2k k A f= (2k)!
Z
t 0
(t − s)2n−1 C(s)An f ds. (2n − 1)!
(16.7)
By integrating (16.7) we obtain the Sine operator function’s Taylor formula Mn (t)f := S(t)f − f t − Z
t 0
t2n−1 t3 Af − · · · − An−1 f = 3! (2n − 1)!
(t − s)2n C(s)An f ds, ∀t ∈ R, (2n)!
(16.8)
all f ∈ D(An ). Integrals in (16.7) and (16.8) are vector valued Riemann integrals, see [79], [164].
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16.3
199
Results
Here, always, we consider C(t), S(t), A as in the Background and f ∈ D(An ), n ∈ N. Theorem 16.2. Let p, q > 1 : p1 + q1 = 1, t ∈ R+ . Then i) Z t I := kTn (w)f k kC(w)An f kdw ≤ 0
2
t2n−1+ p 1/q 2 (2n − 1)![(p(2n − 1) + 1)(p(2n − 1) + 2)]1/p
Z
t 0
kC(w)An f kq dw
2/q
,
(16.9)
and ii) J :=
Z
t 0
kMn (w)f k kC(w)An f kdw ≤ Z
1
t2(n+ p ) 2(2n)![(2pn + 1)(pn + 1)]1/p
t n
0
q
kC(w)A f k dw
2/q
.
(16.10)
We give Corollary 16.3. It holds i) I≤ and
Z
t2n p 2(2n − 1)! 2n(4n − 1)
t 0
kC(w)An f k2 dw, ∀t ∈ R+ ,
(16.11)
ii) t2n+1 p J≤ 2(2n)! (2n + 1)(4n + 1)
Z
t 0
kC(w)An f k2 dw, ∀t ∈ R+ .
(16.12)
Proof of Theorem 16.2. For convenience we call m := 2n − 1. By (16.7) we have Z t (t − s)m Tn (t)f = C(s)An f ds, t ∈ R+ . (16.13) m! 0 Thus
kTn (t)f k ≤ 1 m!
Z
t 0
1 m!
Z
(t − s)pm ds
t 0
(t − s)m kC(s)An f kds ≤
1/p Z
t 0
kC(s)An f kq ds
1/q
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1
tm+ p = m!(pm + 1)1/p
Z
t 0
n
q
kC(s)A f k ds
1/q
.
(16.14)
So far we have found that 1
tm+ p (z(t))1/q , kTn (t)f k ≤ m!(pm + 1)1/p where z(t) :=
Z
(16.15)
t 0
kC(s)An f kq ds, ∀t ∈ R+ .
(16.16)
Clearly z 0 (w) = kC(w)An f kq , 0 ≤ w ≤ t, z(0) = 0.
(16.17)
kC(w)An f k = (z 0 (w))1/q .
(16.18)
Also
Therefore we have that 1
wm+ p (z(w)z 0 (w))1/q . m!(pm + 1)1/p
kTn (w)f k kC(w)An f k ≤ Consequently it holds Z t kTn (w)f k kC(w)An f kdw ≤ 0
≤
1 m!(pm + 1)1/p
Z
t
1 m!(pm + 1)1/p
wpm+1 dw
0
1/p Z 2
=
tm+ p 1 m!(pm + 1)1/p (pm + 2)1/p 2
tm+ p 1 = m!(pm + 1)1/p 21/q (pm + 2)1/p
Z
Z
t 0
t
(16.19)
1
wm+ p (z(w)z 0 (w))1/q dw
0
t
(16.20) z(w)z 0 (w)dw
0
z 2 (w) t 0 2 n
1/q
1/q
q
kC(w)A f k dw
2/q
(16.21)
,
proving (16.9). Setting now m := 2n and working similarly with (16.8) we get (16.10). The proof of the theorem is now complete. We present
(16.22)
Theorem 16.4. Here p = 1, q = ∞, and k kC(s)An f k k∞,+ := sup kC(s)An f k < ∞. s∈R+
Then
(16.23)
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201
i) t2n+1 2 k kC(s)An f k k∞,+ , ∀t ∈ R+ , (2n + 1)!
(16.24)
t2(n+1) 2 k kC(s)An f k k∞,+ , ∀t ∈ R+ . (2(n + 1))!
(16.25)
I≤ and ii) J≤
We have from (16.7) that Z t t2n (t − s)2n−1 kC(s)An f kds ≤ k kC(s)An f k k∞,+ , ∀t ∈ R+ . kTn (t)f k ≤ (2n − 1)! (2n)! 0 (16.26) Thus Proof.
kTn (w)f k kC(w)An f k ≤
w2n k kC(s)An f k k2∞,+ (2n)!
(16.27)
all 0 ≤ w ≤ t. So that I= Z
t
w2n dw 0
Z
t 0
kTn (w)f k kC(w)An f kdw ≤
k kC(s)An f k k2∞ , + t2n+1 = k kC(s)An f k k2∞,+, (2n)! (2n + 1)!
(16.28)
proving (16.24). Similarly from (16.8) we obtain kMn (t)f k ≤
t2n+1 k kC(s)An f k k∞,+, ∀t ∈ R+ . (2n + 1)!
(16.29)
Furthermore it holds kMn (w)f k kC(w)An f k ≤ all 0 ≤ w ≤ t. Therefore Z t J= kMn (w)f k kC(w)An f kdw ≤
w2n+1 k kC(s)An f k k2∞,+, (2n + 1)!
(16.30)
t2(n+1) k kC(s)An f k k2∞,+, ∀t ∈ R+ . (2(n + 1))! 0 (16.31) That is establishing (16.25). The theorem is proved.
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We continue with Theorem 16.5. Let p, q > 1 : p1 + q1 = 1, t ∈ R− . Then i) Z 0 ∗ I := kTn (w)f k kC(w)An f kdw t
2
(−t)2n−1+ p ≤ 1/q 2 (2n − 1)![(p(2n − 1) + 1)(p(2n − 1) + 2)]1/p
Z
0 n
t
q
kC(w)A f k dw
2/q
,
(16.32)
and ii) J ∗ :=
Z
0 t
kMn (w)f k kC(w)An f kdw
1
≤
(−t)2(n+ p ) 2(2n)![(pn + 1)(2pn + 1)]1/p
Z
0
kC(w)An f kq dw
2/q
.
(16.33)
kC(w)An f k2 dw, ∀t ∈ R− ,
(16.34)
t
We give Corollary 16.6. It holds i) t2n p I ≤ 2(2n − 1)! 2n(4n − 1) ∗
and
Z
0 t
ii) J∗ ≤
(−t)2n+1 p 2(2n)! (2n + 1)(4n + 1)
Z
0 t
kC(w)An f k2 dw, ∀t ∈ R− .
Proof of Theorem 16.5. For convenience call m := 2n − 1. By (16.7) we have Z t (t − s)m C(s)An f ds, t ∈ R− . Tn (t)f = m! 0 Thus
Z t
1 m n
(t − s) C(s)A f ds kTn (t)f k =
m!
(16.35)
(16.36)
0
Z 0
Z 0
1 1 m n
= (t − s) C(s)A f ds ≤ (s − t)m kC(s)An f kds m! t m! t ≤
1 m!
Z
0
t
(s − t)pm ds
1/p Z
0
t
kC(s)An f kq ds
1/q
(16.37)
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Opial Inequalities for Cosine and Sine Operator Functions
Z
1
1 (−t)m+ p = m! (pm + 1)1/p So far we have found that
0 t
n
q
kC(s)A f k ds
1/q
203
.
(16.38)
1
(−t)m+ p (z(t))1/q , kTn (t)f k ≤ m!(pm + 1)1/p where z(t) := That is
Z
0 t
−z(t) = and
(16.39)
kC(s)An f kq ds, z(0) = 0.
(16.40)
Z
(16.41)
t 0
kC(s)An f kq ds ≥ 0,
−z 0 (t) = kC(t)An f kq ≥ 0,
(16.42)
kC(t)An f k = (−z 0 (t))1/q , t ∈ R−
(16.43)
and
Hence we have 1
kTn (w)f k kC(w)An f k ≤
(−w)m+ p (z(w)(−z 0 (w)))1/q . m!(pm + 1)1/p
(16.44)
Integrating the last inequality we derive Z 0 kTn (w)f k kC(w)An f kdw t
≤ ≤
1 m!(pm + 1)1/p
1 m!(pm + 1)1/p
Z
0
Z
0
1
(−w)m+ p (z(w)(−z 0 (w))1/q dw
(16.45)
t
(−w)pm+1 dw
t
1/p Z
0
z(w)(−z 0 (w))dw
t
1/q
(16.46)
2
1 (−t)m+ p (z(t))2/q = m!(pm + 1)1/p (pm + 2)1/p 21/q 2
=
(−t)m+ p 21/q m![(pm + 1)(pm + 2)]1/p
Z
0 t
kC(w)An f kq dw
(16.47) 2/q
,
proving (16.32). Setting now m := 2n and working similarly with (16.8) we get (16.33). The proof of the theorem is now complete. Combining Theorems 16.2 and 16.5 we derive Proposition 16.7. Let p, q > 1 :
1 p
+
1 q
= 1, t ∈ R. Then
(16.48)
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204
i) Z t kTn (w)f k kC(w)An f kdw I¯ := 0
and
Z t 2/q 2 |t|2n−1+ p n q , ≤ 1/q kC(w)A f k dw 2 (2n − 1)![(p(2n − 1) + 1)(p(2n − 1) + 2)]1/p 0 (16.49) ii) Z t n ¯ J := kMn (w)f k kC(w)A f kdw 0
≤
Z t 2/q 1 |t|2(n+ p ) n q . kC(w)A f k dw 1/p 2(2n)![(pn + 1)(2pn + 1)] 0
(16.50)
Combining Corollaries 16.3 and 16.6 we obtain Corollary 16.8. It holds i) I¯ ≤ and ii) J¯ ≤
Z t t2n n 2 p kC(w)A f k dw , ∀t ∈ R, 2(2n − 1)! 2n(4n − 1) 0 Z t |t|2n+1 n 2 p kC(w)A f k dw , ∀t ∈ R. 2(2n)! (2n + 1)(4n + 1) 0
(16.51)
(16.52)
Next we present
Theorem 16.9. Here p = 1, q = ∞, and k kC(s)An f k k∞,− := sup kC(s)An f k < ∞. s∈R−
Then i) I∗ ≤
(−t)2n+1 k kC(s)An f k k2∞,− , ∀t ∈ R− , (2n + 1)!
(16.53)
t2(n+1) k kC(s)An f k k2∞,− , ∀t ∈ R− . (2(n + 1))!
(16.54)
and ii) J∗ ≤
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Proof.
Thus
For convenience call m := 2n − 1. By (16.7) we have Z t (t − s)m C(s)An f ds, t ∈ R− . Tn (t)f = m! 0
Z 0
1
(t − s)m C(s)An f ds
m! t Z 0 1 ≤ (s − t)m kC(s)An f kds m! t Z 0 (−t)m+1 m . (s − t) ds = k kC(s)An f k k∞,− (m + 1)! t
Book˙Adv˙Ineq
205
(16.55)
kTn (t)f k =
k kC(s)An f k k∞,− m! Hence ≤
kTn (t)f k ≤ k kC(s)An f k k∞,−
(−t)m+1 , ∀t ∈ R− . (m + 1)!
(16.56) (16.57)
(16.58)
So that it holds kTn (w)f k kC(w)An f k ≤ k kC(s)An f k k2∞,−
(−t)m+1 , t≤w≤0 (m + 1)!
(16.59)
Consequently we get Z Z 0 k kC(s)An f k k2∞,− 0 n ∗ (−w)m+1 dw (16.60) kTn (w)f k kC(w)A f kdw ≤ I = (m + 1)! t t =
(−t)m+2 k kC(s)An f k k2∞,− (m + 2)!
proving (16.53). Similarly, from (16.8), for m = 2n we obtain (16.54).
(16.61)
Finally we give Proposition 16.10. Here p = 1, q = ∞, and
k kC(s)An f k k∞ := sup kC(s)An f k < ∞.
(16.62)
s∈R
Then i) Z t 2n+1 n n 2 ≤ |t| kT (w)f k kC(w)A f kdw n (2n + 1)! k kC(s)A f k k∞ , ∀t ∈ R, 0
(16.63)
and
ii) Z t t2(n+1) n k kC(s)An f k k2∞ , ∀t ∈ R. kMn (w)f k kC(w)A f kdw ≤ (2(n + 1))! 0
Proof.
By Theorems 16.4, 16.9.
(16.64)
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16.4
Book˙Adv˙Ineq
Applications
(see [147], p. 121) Let X be the Banach space of odd, 2π-periodic real functions in the space of d2 bounded uniformly continuous functions from R into itself : BU C(R). Let A := dx 2 with D(An ) = {f ∈ X : f (2k) ∈ X , k = 1, ..., n}, n ∈ N. A generates a Cosine functions C ∗ given by 1 [f (x + t) + f (x − t)] , ∀x, t ∈ R. 2 The corresponding Sine function S ∗ is given by Z t Z t 1 S ∗ (t)f (x) = f (x + s)ds + f (x − s)ds , ∀x, t ∈ R. 2 0 0 C ∗ (t)f (x) =
(16.65)
(16.66)
Here we consider f ∈ D(An ), n ∈ N, as above. By (16.7) we obtain n−1
Tn∗ (t)f :=
X t2k 1 [f (· + t) + f (· − t)] − f (2k) 2 (2k)! k=0
=
Z
t 0
i (t − s)2n−1 h (2n) f (· + s) + f (2n) (· − s) ds, ∀t ∈ R. 2(2n − 1)!
(16.67)
By (16.8) we get Mn∗ (t)f
1 := 2
=
Z
Z t
0
t
f (· + s)ds + 0
Z
t 0
f (· − s)ds −
n X
k=1
t2k−1 f (2(k−1)) (2k − 1)!
i (t − s)2n h (2n) f (· + s) + f (2n) (· − s) ds, ∀t ∈ R. 2(2n)!
(16.68)
Let g ∈ BU C(R), we define kgk = kgk∞ := sup |g(x)| < ∞. x∈R
Notice also that k kC ∗ (s)An f k∞ k∞ = k kC ∗ (s)f (2n) k∞ k∞ = ≤ where
1 k k(f (2n) (· + s) + f (2n) (· − s))k∞ k∞ 2
i 1h k kf (2n)(· + s)k∞ k∞ + k kf (2n) (· − s)k∞ k∞ ≤ Θ < ∞, 2 Θ := kf (2n) k∞ .
We have the following applications. From Proposition 16.7 we derive Proposition 16.11. Let p, q > 1 :
1 p
+
1 q
= 1, t ∈ R. Then
(16.69)
(16.70)
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i) Z t
∗ (2n) ∗ ε1 := kTn (w)f k∞ C (w)f
dw ∞ 0
≤
2/q Z t 2
q |t|2n−1+ p
∗ (2n) , dw (w)f
C
1/q 1/p ∞ 2 (2n − 1)![(p(2n − 1) + 1)(p(2n − 1) + 2)] 0 ∀t ∈ R,
(16.71)
and ii) Z t
∗ ∗ (2n) ε2 := kMn (w)f k∞ C (w)f
dw ∞ 0
Z t 2/q 1
q |t|2(n+ p )
∗ (2n) ≤
C (w)f
dw , ∀t ∈ R. 1/p 2(2n)![(pn + 1)(2pn + 1)] ∞ 0
(16.72)
From Corollary 16.8 we get Corollary 16.12. It holds i)
and ii)
Z t
2 t2n
∗ (2n) p ε1 ≤
C (w)f
dw , ∀t ∈ R, ∞ 2(2n − 1)! 2n(4n − 1) 0
Z t
2 |t|2n+1
∗ (2n) p ε2 ≤
C (w)f
dw , ∀t ∈ R. ∞ 2(2n)! (2n + 1)(4n + 1) 0
(16.73)
(16.74)
Finally from Proposition 16.10 we find
Proposition 16.13. Here p = 1, q = ∞. Then i)
|t|2n+1
(2n) 2 ε1 ≤
f
, ∀t ∈ R, (2n + 1)! ∞
(16.75)
and
ii)
ε2 ≤
t2(n+1)
(2n) 2
f
, ∀t ∈ R. (2(n + 1))! ∞
(16.76)
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Chapter 17
Poincar´ e Like Inequalities for Linear Differential Operators
Various Lp form Poincar´e like inequalities [2], forward and reverse, are given for a linear differential operator L, involving its related initial value problem solution y, Ly, the associated Green’s function H and initial conditions at point x0 ∈ R. This chapter follows [48].
17.1
Background
Here we use [169], pp. 145–154. Let [a, b] ⊂ R, ai (x) , i = 0, 1, . . . , n − 1 (n ∈ N) , h (x) be continuous functions on [a, b] and let L = D n + an−1 (x) Dn−1 + . . . + a0 (x) be a fixed linear differential operator on C n ([a, b]) . Let y1 (x) , . . . , yn (x) be a set of linear independent solutions to Ly = 0. Here the associated Green’s functions for L is y1 (t) . . . yn (t) y 0 (t) . . . y 0 (t) 1 n ... ... ... (n−2) (n−2) y1 (t) . . . yn (t) y1 (x) . . . yn (x) , (17.1) H (x, t) := y1 (t) . . . yn (t) y 0 (t) . . . y 0 (t) 1 n ... ... ... (n−2) (n−2) y1 (t) . . . yn (t) (n−1) (n−1) y (t) . . . yn (t) 1
2
which is a continuous function on [a, b] . Consider a fixed x0 ∈ [a, b] , then Z x y (x) = H (x, t) h (t) dt, ∀x ∈ [a, b]
(17.2)
x0
is the unique solution to the initial value problem
Ly = h; y (i) (x0 ) = 0, i = 0, 1, . . . , n − 1. 209
(17.3)
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Next we suppose all of the above.
17.2
Results
We present the following Poincar´e like inequalities. Theorem 17.1. Let x0 < b and x ∈ [x0 , b] , and p, q > 1 : Then !1/ν Z Z b
1) kykLν (x0 ,b) ≤
x
x0
1 q
+
= 1; ν > 0.
ν/p
p
|H (x, t)| dt
x0
1 p
dx
kLykLq (x0 ,b) .
(17.4)
kLykLq (x0 ,b) .
(17.5)
When ν = q we have 2) kykLq (x0 ,b) ≤
Z
Z
b x0
x
p
|H (x, t)| dt
x0
q/p
dx
!1/q
When ν = p = q = 2 we obtain Z
3) kykL2 (x0 ,b) ≤ Proof.
b x0
Z
x
2
H (x, t) dt dx x0
!1/2
kLykL2 (x0 ,b) .
(17.6)
From (17.2) we have |y (x)| ≤
≤
Z
≤
Z
x x0
x x0
|H (x, t)| |h (t)| dt
|H (x, t)|p dt
x x0
Z
p
|H (x, t)| dt
1/p Z
1/p
Z
x x0
|h (t)|q dt
b x0
q
|h (t)| dt
1/q
!1/q
.
(17.7)
kLykLq (x0 ,b) ,
(17.8)
That is ν
|y (x)| ≤ Hence Z b x0
ν
|y (x)| dx ≤
proving the claim.
Z
Z b
x0
x x0
Z
p
|H (x, t)| dt
x x0
p
ν/p
|H (x, t)| dt
ν/p
ν
!
ν
dx kLykLq (x0 ,b) ,
(17.9)
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211
We continue with Theorem 17.2. Let x0 > a and x ∈ [a, x0 ] , and p, q > 1 : p1 + 1q = 1; ν > 0. Then ν/p !1/ν Z x0 Z x0 p kLykLq (a,x0 ) . (17.10) dx |H (x, t)| dt 1) kykLν (a,x0 ) ≤ x
a
When ν = q we obtain Z x0 Z 2) kykLq (a,x0 ) ≤ a
x0
p
|H (x, t)| dt
x
q/p
dx
!1/q
kLykLq (a,x0 ) .
When ν = p = q = 2 we obtain 1/2 Z x0 Z x0 2 H (x, t) dt dx 3) kykL2 (a,x0 ) ≤ kLykL2 (a,x0 ) . From (17.2) we have
|y (x)| ≤ ≤ ≤ That is
Z
Z
x0
Z
x0
p
ν
|y (x)| ≤
a
Z
|H (x, t)| |h (t)| dt
x
|H (x, t)| dt
x
Therefore Z x0 |y (x)|ν dx ≤
x0
|H (x, t)|p dt
x
(17.12)
x
a
Proof.
(17.11)
Z
x0 a
x0
1/p Z p
|H (x, t)| dt
x
Z
1/p Z
x0
p
x0 x x0
a
q
|h (t)| dt
ν/p
|H (x, t)| dt
x
|h (t)|q dt
1/q
1/q
.
(17.13)
kLykLq (a,x0 ) .
(17.14)
ν
ν/p
!
dx kLykνLq (a,x0 ) ,
proving the claim.
(17.15)
Extreme cases follow Proposition 17.3. Here x0 < b, x ∈ [x0 , b] , and p = 1, q = ∞. Then ν !1/ν Z b Z x 1) kykLν (x0 ,b) ≤ kLykL∞ (x0 ,b) . |H (x, t)| dt dx x0
(17.16)
x0
When ν = 1 we have
2) kykL1 (x0 ,b) ≤
Z
b x0
Z
x x0
!
|H (x, t)| dt dx kLykL∞ (x0 ,b) .
(17.17)
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From (17.2) we have
Proof.
|y (x)| ≤ Z
≤ That is
x x0
Z
ν
|y (x)| ≤
Z
x x0
|H (x, t)| |h (t)| dt
|H (x, t)| dt khkL∞ (x0 ,b) . x
|H (x, t)| dt
x0
ν
ν
kLykL∞ (x0 ,b) ,
and Z
b x0
Z
ν
|y (x)| dx ≤
Z
b x0
x x0
|H (x, t)| dt
ν
(17.18)
!
ν
dx kLykL∞ (x0 ,b) ,
proving the claim.
(17.19)
(17.20)
We continue with Proposition 17.4. Here x0 > a, x ∈ [a, x0 ] , and p = 1, q = ∞. Then ν 1/ν Z x0 Z x0 |H (x, t)| dt dx 1) kykLν (a,x0 ) ≤ kLykL∞ (a,x0 ) .
(17.21)
x
a
When ν = 1 we derive
2) kykL1 (a,x0 ) ≤
Z
Z
x0 a
x0 x
|y (x)| ≤ ≤
Z
That is ν
|y (x)| ≤ Z
(17.22)
From (17.2) we have
Proof.
and
|H (x, t)| dt dx kLykL∞ (a,x0 ) .
x0 a
ν
|y (x)| dx ≤
proving the claim.
Z
x
Z
x0 x
|H (x, t)| |h (t)| dt
x0
x0 a
Z
|H (x, t)| dt khkL∞ (a,x0 ) . x0
|H (x, t)| dt
x
Z
ν
x0 x
|H (x, t)| dt
ν
kLykL∞ (a,x0 ) ,
ν
ν dx kLykL∞ (a,x0 ) ,
(17.23)
(17.24)
(17.25)
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Poincar´ e Like Inequalities for Linear Differential Operators
213
Next we give reverse Poincar´e like inequalities. Theorem 17.5. Let x0 < b, x ∈ [x0 , b] , and 0 < p < 1, q < 0 : p1 + q1 = 1, ν > 0. Suppose H (x, t) ≥ 0 for x0 ≤ t ≤ x, and Ly = h is of fixed sign and nowhere zero. Then !1/ν Z Z x
b
1) kykLν (x0 ,b) ≥
ν/p
p
kLykLq (x0 ,b) .
dx
(H (x, t)) dt
x0
x0
(17.26)
When ν = p we obtain Z
2) kykLp (x0 ,b) ≥
b x0
Z
(H (x, t)) dt dx
!1/p
1/p
!
x
p
x0
kLykLq (x0 ,b) .
(17.27)
dx kLykLq (x0 ,b) .
(17.28)
When ν = 1 we get Z
3) kykL1 (x0 ,b) ≥ Proof.
b x0
Z
x
p
(H (x, t)) dt x0
By (17.2) we have Z x |y (x)| = H (x, t) |h (t)| dt, for all x0 ≤ x ≤ b.
(17.29)
x0
From (17.29) by reverse H¨ older’s inequality we find |y (x)| ≥
≥
Z
Z
x
p
(H (x, t)) dt x0
x
p
(H (x, t)) dt x0
1/p Z
1/p
Z
x x0
b x0
q
|h (t)| dt
q
|h (t)| dt
1/q
!1/q
,
(17.30)
khkνLq (x0 ,b) ,
(17.31)
for all x0 < x ≤ b. That is it holds |y (x)|ν ≥ for all x0 ≤ x ≤ b, and Z b ν |y (x)| dx ≥ x0
proving the claim.
Z
b x0
Z
x
(H (x, t))p dt
x0
Z
x
p
ν/p
(H (x, t)) dt x0
ν/p
!
ν
dx khkLq (x0 ,b) ,
(17.32)
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We continue with Theorem 17.6. Let x0 > a, x ∈ [a, x0 ] , and 0 < p < 1, q < 0 : p1 + q1 = 1, ν > 0. Suppose H (x, t) ≤ 0 for x ≤ t ≤ x0 , and Ly = h is of fixed sign and nowhere zero. Then ν/p !1/ν Z x0 Z x0 p kLykLq (a,x0 ) . (17.33) dx (−H (x, t)) dt 1) kykLν (a,x0 ) ≥ x
a
When ν = p we obtain Z x0 Z 2) kykLp (a,x0 ) ≥ a
When ν = 1 we have Z 3) kykL1 (a,x0 ) ≥ Proof.
x0 a
Z
x0 x
1/p p kLykLq (a,x0 ) . (−H (x, t)) dt dx
x0
p
(−H (x, t)) dt x
From (17.2) we have
Z |y (x)| =
x x0
1/p
!
dx kLykLq (a,x0 ) .
(17.34)
(17.35)
H (x, t) h (t) dt
Z x0 H (x, t) h (t) dt = x Z x 0 = (−H (x, t)) h (t) dt x
=
Z
x0
(−H (x, t)) |h (t)| dt.
x
(17.36)
From (17.36) by reverse H¨ older’s inequality we find Z x0 1/p Z x0 1/q p q |y (x)| ≥ (−H (x, t)) dt |h (t)| dt x
≥ for all a ≤ x < x0 . That is it holds
Z
x0
(−H (x, t))p dt
x
ν
|y (x)| ≥ for all a ≤ x ≤ x0 , and Z x0 ν |y (x)| dx ≥ a
proving the claim.
x
Z
Z
x0 a
x0
1/p Z
p
(−H (x, t)) dt x
Z
x0
p
x0 a
ν/p
(−H (x, t)) dt x
|h (t)|q dt
1/q
ν
kLykLq (a,x0 ) ,
ν/p
!
ν
dx kLykLq (a,x0 ) ,
(17.37)
(17.38)
(17.39)
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Book˙Adv˙Ineq
Chapter 18
Poincar´ e and Sobolev Like Inequalities for Widder Derivatives
Various Lp form Poincar´e and Sobolev like inequalities, forward and reverse, are given involving Widder derivative ([243]). This chapter follows [54]. 18.1
Background
The following come from [243]. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , n ≥ 0, and the Wronskians Wi (x) := W [u0 (x) , u1 (x) , . . . , ui (x)] := u0 (x) u1 (x) . . . ui (x) u0 (x) u0 (x) . . . u0 (x) 0 1 i , .. . (i) (i) u0 (x) u(i) (x) . . . u (x) 1 i
(18.1)
i = 0, 1, . . . , n. Here W0 (x) = u0 (x) . Assume Wi (x) > 0 over [a, b] , i = 0, 1, . . . , n. For i ≥ 0, the differential operator of order i (Widder derivative): Li f (x) :=
W [u0 (x) , u1 (x) , . . . , ui−1 (x) , f (x)] , Wi−1 (x)
i = 1, . . . , n + 1; L0 f (x) := f (x) , ∀x ∈ [a, b] . Consider also u0 (t) u1 (t) u0 (t) u01 (t) 0 1 .. gi (x, t) := . Wi (t) (i−1) (i−1) u0 (t) u1 (t) u0 (x) u1 (x) i = 1, 2, . . . , n; g0 (x, t) :=
u0 (x) u0 (t) ,
∀x, t ∈ [a, b] .
, (i−1) . . . ui (t) . . . ui (x)
... ...
Example ([243]). Sets of the form {u0 , u1 , . . . , un } are 215
(18.2)
ui (t) u0i (t)
(18.3)
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216
n
Book˙Adv˙Ineq
1, x, x2 , . . . , xn ,
1, sin x, − cos x, − sin 2x, cos 2x, . . . , (−1)
n−1
o n sin nx, (−1) cos nx , etc.
We also mention the generalized Widder–Taylor’s formula, see [243].
Theorem 18.1. Let the functions f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , and the Wronskians W0 (x) , W1 (x) , . . . , Wn (x) > 0 on [a, b] , x ∈ [a, b] . Then for t ∈ [a, b] we have f (x) = f (t)
u0 (x) + L1 f (t) g1 (x, t) + . . . + Ln f (t) gn (x, t) + Rn (x) , u0 (t)
where Rn (x) :=
Z
(18.4)
x
gn (x, s) Ln+1 f (s) ds.
(18.5)
t
For example ([243]) one could take u0 (x) = c > 0. If ui (x) = xi , i = 0, 1, . . . , n, defined on [a, b] , then Li f (t) = f (i) (t) and gi (x, t) =
(x − t)i , t ∈ [a, b] . i!
(18.6)
We need Corollary 18.2. (to Theorem 18.1) Additionally suppose that for fixed x 0 ∈ [a, b] we have Li f (x0 ) = 0, i = 0, 1, . . . , n. Then Z x f (x) = gn (x, t) Ln+1 f (t) dt, ∀x ∈ [a, b] . (18.7) x0
Corollary 18.3. (to Theorem 18.1) Let f, u0 ∈ C 1 ([a, b]) , u0 (x) > 0, for all x ∈ [a, b] . Then Z x u0 (x) L1 f (s) + u0 (x) ds, ∀x, t ∈ [a, b] , (18.8) f (x) = f (t) u0 (t) u0 (s) t where L1 f (s) =
W [u0 (s) , f (s)] u0 (s) f 0 (s) − u00 (s) f (s) = . u0 (s) u0 (s)
(18.9)
We need to make Remark 18.4. We define (see [243]) φ0 (x) := W0 (x) , φ1 (x) :=
W1 (x) (W0 (x))
2,...,
in general φk (x) :=
Wk (x) Wk−2 (x) (Wk−1 (x))2
, k = 2, 3, . . . , n.
(18.10)
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217
The functions φi (x) are positive on [a, b] . According to [243] we get, for any x, x0 ∈ [a, b] that Z x Z x1 φ0 (x) φ1 (x1 ) gn (x, x0 ) = ... φ0 (x0 ) . . . φn (x0 ) x0 x0 Z xn−2 Z xn−1 φn−1 (xn−1 ) φn (xn ) dx1 dx2 . . . dxn x0
=
x0
1 φ0 (x0 ) . . . φn (x0 )
Z
x
φ0 (s) . . . φn (s) gn−1 (x, s) ds.
(18.11)
x0
We get that gn (x, x) = 0, all x ∈ [a, b] , and gn (x, x0 ) > 0, x > x0 , x, x0 ∈ [a, b] , any n ∈ N. Also g0 (x, x0 ) > 0 for any x, x0 ∈ [a, b] . By (18.11) we notice that gn (x, t) < 0, x < t, n odd, gn (x, t) > 0, x < t, n even,
(18.12)
where x, t ∈ [a, b] . In the next we work under the terms and assumptions of Theorem 18.1 and Corollary 18.2, and the rest of the above conclusions. 18.2
Results
We give the following weighted Neumann–Poincar´e like inequality. Theorem 18.5. Let p, q > 1 : p1 + u0 > 0 on [a, b] . Then
Z b
f 1 f (t)
1) − dt
u0 b − a a u0 (t)
1 q
Lν (a,b)
When ν = q we obtain
Z b
f f (t) 1
2) − dt
u0 b − a a u0 (t)
When ν = p = q = 2 we have
Z b
f 1 f (t)
3) − dt
u0 b − a a u0 (t)
= 1, ν > 0. Consider f, u0 ∈ C 1 ([a, b]) ,
≤ (b − a)( p
Lq (a,b)
L2 (a,b)
Equivalently,
Z
b a
1
f (x) 1 − u0 (x) b − a
+ ν1 )
L1 f
.
u0 Lq (a,b)
L1 f
≤ (b − a) .
u0 Lq (a,b)
L1 f
. ≤ (b − a)
u0 L2 (a,b)
Z
b a
f (t) dt u0 (t)
!2
dx
(18.13)
(18.14)
(18.15)
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≤ (b − a) Proof.
1 b−a
Let A = f (c) u0 (c)
Rb
f (t) dt. a u0 (t)
Z
2
b a
L1 f (x) u0 (x)
f (c) f (x) − = u0 (x) u0 (c) i.e.
Z
f (x) −A = u0 (x) Z f (x) ≤ − A u0 (x) ≤ (b − a) Therefore
Z
1/p
b
a
dx.
(18.15*)
By the intermediate value theorem there exists
c ∈ [a, b] with = A. Thus by (18.8) we get
Hence
2
x c
x c
Z
x
L1 f (s) ds, u0 (s)
c
(18.16)
L1 f (s) ds, ∀x ∈ [a, b] . u0 (s)
(18.17)
Z b |L1 f (s)| |L1 f (s)| ds ≤ ds u0 (s) u0 (s) a
|L1 f (s)| u0 (s)
ν f (x) ν p u0 (x) − A ≤ (b − a)
q
ds
!1/q
, ∀x ∈ [a, b] .
L1 f ν
, ∀x ∈ [a, b] .
u0 Lq (a,b)
Consequently we obtain
ν
ν Z b
f (x) ν dx ≤ (b − a)( p +1) L1 f , − A
u0 u0 (x) a Lq (a,b)
proving the claim.
Corollary 18.6. (to Theorem 18.5) Let
f
1)
u0
Lν (a,b)
f
2)
u0
equivalently, Z
b a
f (x) u0 (x)
L2 (a,b)
2
It follows the related result.
f (t) dt a u0 (t)
L1 f
≤ (b − a) u0
dx ≤ (b − a)
Z
b
a
(18.20)
,
(18.21)
Lq (a,b)
L1 f
≤ (b − a) ,
u0 Lq (a,b)
2
(18.19)
= 0. Then
1 1 L1 f ( p+ν )
≤ (b − a)
u0
Lq (a,b)
f
3)
u0
Rb
(18.18)
(18.22)
,
(18.23)
L2 (a,b)
L1 f (x) u0 (x)
2
dx.
(18.23*)
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219
Proposition 18.7. Let ν > 0 and consider f, u0 ∈ C 1 ([a, b]) , u0 > 0 on [a, b] . Then
Z b
f f (t) 1 L1 f
1/ν
1) − dt ≤ (b − a) . (18.24)
u0 b − a a u0 (t) u0 L1 (a,b) Lν (a,b)
When ν = 1 we have
Z b
f 1 f (t)
dt 2) −
u0 b − a a u0 (t)
Proof.
L1 (a,b)
L1 f
≤ (b − a) . u0 L1 (a,b)
By (18.18) we have ν Z b f (x) 1 f (t) − dt u0 (x) b − a a u0 (t)
L 1 f ν
≤ , ∀x ∈ [a, b] . u0 L1 (a,b)
(18.25)
(18.26)
So that ν
Z b Z b
L 1 f ν 1 f (t) f (x)
− dt dx ≤ (b − a) , b − a a u0 (t) u0 L1 (a,b) a u0 (x)
proving the claim.
(18.27)
We continue with generalized Dirichlet–Poincar´e like inequalities. Theorem 18.8. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , n ∈ Z+ ; W0 , W1 , . . . , Wn > 0 on [a, b] , and for fixed x0 ∈ [a, b] suppose that Li f (x0 ) = 0, i = 0, 1, . . . , n. Let p, q > 1 : p1 + 1q = 1, ν > 0. Then ν/p !1/ν Z b Z x p 1) kf kLν (x0 ,b) ≤ kLn+1 f kLq (x0 ,b) . (18.28) dx (gn (x, t)) dt x0
2) kf kLq (x0 ,b) ≤
Z
b x0
x0
Z
x
p
(gn (x, t)) dt x0
q/p
dx
!1/q
kLn+1 f kLq (x0 ,b) .
(18.29)
Also, 3) kf kL2 (x0 ,b) ≤ Proof.
Z
b x0
Z
x
2
(gn (x, t)) dt dx x0
!1/2
kLn+1 f kL2 (x0 ,b) .
By (18.7) we have Z |f (x)| =
x x0
gn (x, t) Ln+1 f (t) dt
(18.30)
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220
Z
≤ Z
≤ ≤
Z
x
x
p
(gn (x, t)) dt x0
|f (x)| ≤ and
x0
gn (x, t) |Ln+1 f (t)| dt
x0
ν
b
x0
(gn (x, t))p dt
That is
Z
x
Z
ν
|f (x)| dx ≤
b x0
Z
1/p
1/p Z
x0
|Ln+1 f (t)|q dt
1/q
kLn+1 f kLq (x0 ,b) , ∀x ∈ [x0 , b] .
x
p
(gn (x, t)) dt x0
Z
x
x
ν/p
p
(gn (x, t)) dt x0
ν/p
ν
kLn+1 f kLq (x0 ,b) , !
ν
dx kLn+1 f kLq (x0 ,b) ,
proving the claim.
(18.31)
(18.32)
(18.33)
We give Theorem 18.9. Same assumptions as in Theorem 18.8. Then ν/p !1/ν Z x0 Z x0 p 1) kf kLν (a,x0 ) ≤ |gn (x, t)| dt dx kLn+1 f kLq (a,x0 ) . (18.34) a
2) kf kLq (a,x0 ) ≤
Z
x
x0 a
Z
x0
p
|gn (x, t)| dt
x
q/p
dx
!1/q
kLn+1 f kLq (a,x0 ) . (18.35)
Also, 3) kf kL2 (a,x0 ) ≤ Proof.
Z
x0 a
Z
1/2 2 (gn (x, t)) dt dx kLn+1 f kL2 (a,x0 ) .
x0 x
By (18.7) we have
Z |f (x)| = Z =
≤ ≤
Z
x0 x
Z
x0
x
x0
x
x x0
gn (x, t) Ln+1 f (t) dt
gn (x, t) Ln+1 f (t) dt
|gn (x, t)| |Ln+1 f (t)| dt
|gn (x, t)|p dt
1/p Z
x0 x
|Ln+1 f (t)|q dt
1/q
(18.36)
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Poincar´ e and Sobolev Like Inequalities for Widder Derivatives
≤
Z
x0
p
|gn (x, t)| dt
x
1/p
kLn+1 f kLq (a,x0 ) .
221
(18.37)
That is Z
ν
|f (x)| ≤ and Z
x0
ν
|f (x)| dx ≤
a
Z
x0 a
x0
p
|gn (x, t)| dt
x
Z
x0
ν/p
p
|gn (x, t)| dt
x
ν
kLn+1 f kLq (a,x0 ) ,
ν/p
!
ν
dx kLn+1 f kLq (a,x0 ) ,
proving the claim.
(18.38)
(18.39)
We continue with Proposition 18.10. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , n ∈ Z+ ; W0 , W1 , . . . , Wn > 0 on [a, b] , and for fixed x0 ∈ [a, b] suppose that Li f (x0 ) = 0, i = 0, 1, . . . , n; ν > 0. Then !1/ν Z Z b
1) kf kLν (x0 ,b) ≤
ν
x
gn (x, t) dt
x0
Z
2) kf kL1 (x0 ,b) ≤
kLn+1 f kL∞ (x0 ,b) .
dx
x0
b x0
Z
x x0
!
gn (x, t) dt dx kLn+1 f kL∞ (x0 ,b) .
(18.40)
(18.41)
By (18.7) we have
Proof.
|f (x)| ≤
≤
Z
Z
x0
gn (x, t) |Ln+1 f (t)| dt
x x0
x
gn (x, t) dt kLn+1 f k∞,[x0 ,b] .
(18.42)
That is ν
|f (x)| ≤
Z
x
gn (x, t) dt x0
ν
ν
kLn+1 f k∞,[x0 ,b] ,
(18.43)
and Z
b x0
ν
|f (x)| dx ≤
proving the claim.
Z
b x0
Z
x
gn (x, t) dt x0
ν
!
ν
dx kLn+1 f k∞,[x0 ,b] ,
(18.44)
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We give Proposition 18.11. All as in Proposition 18.10. Then ν 1/ν Z x0 Z x0 kLn+1 f kL∞ (a,x0 ) . |gn (x, t)| dt dx 1) kf kLν (a,x0 ) ≤ 2) kf kL1 (a,x0 ) ≤ Proof.
Z
Z
x0 a
x
By (18.7) we have
Z
≤ That is
ν
|f (x)| ≤ Z
x0 a
ν
|f (x)| dx ≤
proving the claim.
Z
Z
|gn (x, t)| |Ln+1 f (t)| dt
x
|gn (x, t)| dt kLn+1 f k∞,[a,x0 ] .
x
Z
x0
|gn (x, t)| dt
x
Z
ν
x0 x
(18.46)
x0
x0
x0 a
|gn (x, t)| dt dx kLn+1 f kL∞ (a,x0 ) .
x0
|f (x)| ≤
and
(18.45)
x
a
|gn (x, t)| dt
(18.47)
ν
kLn+1 f k∞,[a,x0 ] ,
ν
ν
dx kLn+1 f k∞,[a,x0 ] ,
(18.48)
(18.49)
We continue with reverse Dirichlet–Poincar´e like inequalities. Theorem 18.12. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , n ∈ Z+ ; W0 , W1 , . . . , Wn > 0 on [a, b] , and for a fixed a ≤ x0 < b assume that Li f (x0 ) = 0, i = 0, 1, . . . , n. Let 0 < p < 1, q < 0 : 1p + 1q = 1, ν > 0. Further suppose that Ln+1 f is of fixed sign and nowhere zero on [x0 , b] . Then ν/p !1/ν Z b Z x p (gn (x, t)) dt dx kLn+1 f kLq (x0 ,b) . (18.50) 1) kf kLν (x0 ,b) ≥ x0
x0
2) kf kLp (x0 ,b) ≥
Z
3) kf kL1 (x0 ,b) ≥
Z
Proof.
b x0 b
x0
Z Z
(gn (x, t)) dt dx
!1/p
1/p
!
x
p
x0 x
(gn (x, t))p dt
x0
kLn+1 f kLq (x0 ,b) .
(18.51)
dx kLn+1 f kLq (x0 ,b) .
(18.52)
Here we have by (18.7) and assumption that Z x gn (x, t) |Ln+1 f (t)| dt, |f (x)| = x0
(18.53)
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∀x ∈ [x0 , b] . Hence by reverse H¨ older’s inequality we derive Z x 1/p Z x 1/q p q |f (x)| ≥ (gn (x, t)) dt |Ln+1 f (t)| dt x0
≥
Z
x0
x
p
(gn (x, t)) dt x0
1/p
Z
b x0
q
|Ln+1 f (t)| dt
!1/q
,
(18.54)
kLn+1 f kνLq (x0 ,b) ,
(18.55)
true for all x ∈ (x0 , b] . That is Z
ν
|f (x)| ≥ true for all x ∈ [x0 , b] , and Z b Z ν |f (x)| dx ≥ x0
b x0
x
p
(gn (x, t)) dt x0
Z
x
p
ν/p
(gn (x, t)) dt x0
ν/p
!
ν
dx kLn+1 f kLq (x0 ,b) ,
proving the claim.
(18.56)
We give Theorem 18.13. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]) , n is odd; W0 , W1 , . . . , Wn > 0 on [a, b] , and for a fixed a < x0 ≤ b assume that Li f (x0 ) = 0, i = 0, 1, . . . , n. Let 0 < p < 1, q < 0 : 1p + 1q = 1, ν > 0. Further suppose that Ln+1 f is of fixed sign and nowhere zero on [a, x0 ] . Then ν/p !1/ν Z x0 Z x0 p kLn+1 f kLq (a,x0 ) . dx (−gn (x, t)) dt 1) kf kLν (a,x0 ) ≥ x
a
(18.57)
2) kf kLp (a,x0 ) ≥
Z
3) kf kL1 (a,x0 ) ≥
Z
Proof.
a
x0 Z
x0 a
Z
(−gn (x, t)) dt dx
1/p
1/p
!
x0
p
x x0
p
(−gn (x, t)) dt x
kLn+1 f kLq (a,x0 ) . (18.58)
dx kLn+1 f kLq (a,x0 ) . (18.59)
Here by (18.7) and assumption we have Z x |f (x)| = gn (x, t) Ln+1 f (t) dt x0
Z =
x0
x
gn (x, t) Ln+1 f (t) dt
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Z
(18.12)
=
=
Z
x0 x
x0
x
(−gn (x, t)) Ln+1 f (t) dt
(−gn (x, t)) |Ln+1 f (t)| dt.
So by reverse H¨ older’s inequality we find 1/p Z Z x0 p (−gn (x, t)) dt |f (x)| ≥ x
≥
Z
x0 x
true for all x ∈ [a, x0 ) , and ν
|f (x)| ≥ true for all x ∈ [a, x0 ] . Thus Z Z x0 ν |f (x)| dx ≥ a
(−gn (x, t))p dt
Z
x0 a
x0
1/p
p
(−gn (x, t)) dt x
Z
x0
x0
1/q
kLn+1 f kLq (a,x0 ) ,
(−gn (x, t)) dt x
q
|Ln+1 f (t)| dt
x
ν/p
p
(18.60)
(18.61)
ν
kLn+1 f kLq (a,x0 ) ,
ν/p
!
(18.62)
ν
dx kLn+1 f kLq (a,x0 ) , (18.63)
proving the claim.
We add Theorem 18.14. Now n is even, the rest as in Theorem 18.13. Then ν/p !1/ν Z x0 Z x0 p kLn+1 f kLq (a,x0 ) . (18.64) dx (gn (x, t)) dt 1) kf kLν (a,x0 ) ≥ x
a
2) kf kLp (a,x0 ) ≥
Z
3) kf kL1 (a,x0 ) ≥
Z
Proof.
x0 a x0
a
Z Z
x0
p
1/p
kLn+1 f kLq (a,x0 ) .
(18.65)
1/p
dx kLn+1 f kLq (a,x0 ) .
(18.66)
(gn (x, t)) dt dx x x0
p
(gn (x, t)) dt x
!
Similar to Theorem 18.13.
We continue with Sobolev like inequalities. Theorem 18.15. Same assumptions as in Theorem 18.8. Call ν/p !1/ν Z b Z x p (gj (x, t)) dt dx , Mν,1 := max 0≤j≤n x0 x0
(18.67)
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ν > 0. Then 1) kf kLν (x0 ,b) ≤
Mν,1 n+1
2) kf kLq (x0 ,b) ≤
Mq,1 n+1
n X j=0
225
kLj+1 f kLq (x0 ,b) ,
(18.68)
,
(18.69)
n X kLj+1 f k
Lq (x0 ,b)
j=0
and when ν = p = q = 2 we obtain n M2,1 X 3) kf kL2 (x0 ,b) ≤ kLj+1 f kL2 (x0 ,b) , n+1 j=0
where
M2,1 := max
Z
0≤j≤n
b x0
Z
x
2
(gj (x, t)) dt dx x0
!1/2
(18.70)
.
(18.71)
Proof. The assumptions of Theorem 18.8 are fulfilled for all j = 0, 1, . . . , n. Thus by (18.28) we derive !1/ν Z Z b
kf kLν (x0 ,b) ≤
x
ν/p
p
(gj (x, t)) dt
x0
dx
x0
kLj+1 f kLq (x0 ,b)
≤ Mν,1 kLj+1 f kLq (x0 ,b) , for all j = 0, 1, . . . , n. From (18.72) by addition we get
(n + 1) kf kLν (x0 ,b) ≤ Mν,1
n X j=0
proving the claim.
(18.72)
kLj+1 f kLq (x0 ,b) ,
(18.73)
We continue with Theorem 18.16. Same assumptions as in Theorem 18.8. Call ν/p !1/ν Z x0 Z x0 p , dx |gj (x, t)| dt Mν,2 := max 0≤j≤n x a
ν > 0. Then
1) kf kLν (a,x0 ) ≤
Mν,2 n+1
n X j=0
kLj+1 f kLq (a,x0 ) ,
(18.74)
(18.75)
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2) kf kLq (a,x0 ) ≤
Mq,2 n+1
n X kLj+1 f k
Lq (a,x0 )
j=0
,
(18.76)
and when ν = p = q = 2 we obtain n M2,2 X kLj+1 f kL2 (a,x0 ) , 3) kf kL2 (a,x0 ) ≤ n+1 j=0 where
M2,2 := max
0≤j≤n
Proof.
(Z
x0 a
Z
x0
2
(gj (x, t)) dt dx x
1/2 )
(18.77)
.
Similar to Theorem 18.15, based on Theorem 18.9.
(18.78)
We continue with L∞ results. Proposition 18.17. All as in Proposition 18.10. Call ν !1/ν Z b Z x , gj (x, t) dt dx Kν,1 := max 0≤j≤n x0 x0 ν > 0. Then
1) kf kLν (x0 ,b) ≤
Kν,1 n+1
2) kf kL1 (x0 ,b) ≤
K1,1 n+1
and
Proof.
n X kLj+1 f k
L∞ (x0 ,b)
j=0
n X kLj+1 f k
L∞ (x0 ,b)
j=0
,
(18.79)
.
(18.80)
Similar to Theorem 18.15, based on Proposition 18.10.
Proposition 18.18. All as in Proposition 18.10. Call (Z ν 1/ν ) x0 Z x0 Kν,2 := max |gn (x, t)| dt dx , 0≤j≤n
a
1) kf kLν (a,x0 ) ≤
Kν,2 n+1
2) kf kL1 (a,x0 ) ≤
K1,2 n+1
and
Based on Proposition 18.11.
(18.81)
x
ν > 0. Then
Proof.
n X kLj+1 f k
L∞ (a,x0 )
j=0
n X kLj+1 f k
L∞ (a,x0 )
j=0
,
(18.82)
.
(18.83)
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We continue with reverse Sobolev like inequalities. Theorem 18.19. Assume here that Lj+1 f is of fixed sign and nowhere zero on [x0 , b] , for j = 0, 1, . . . , n. The rest are supposed as in Theorem 18.12. Call ν/p !1/ν Z b Z x p Sν,1 := min , (18.84) dx (gj (x, t)) dt 0≤j≤n x0 x0
ν > 0. Then
1) kf kLν (x0 ,b) ≥
Sν,1 n+1
2) kf kLp (x0 ,b) ≥
Sp,1 n+1
3) kf kL1 (x0 ,b) ≥
S1,1 n+1
and
n X j=0
kLj+1 f kLq (x0 ,b) ,
(18.85)
,
(18.86)
n X kLj+1 f k
Lq (x0 ,b)
j=0
n X j=0
kLj+1 f kLq (x0 ,b) .
(18.87)
Proof. The assumptions of Theorem 18.12 are fulfilled for all j = 0, 1, . . . , n. Thus by (18.50) we obtain !1/ν Z Z b
kf kLν (x0 ,b) ≥
x
ν/p
p
(gj (x, t)) dt
x0
kLj+1 f kLq (x0 ,b)
dx
x0
≥ Sν,1 kLj+1 f kLq (x0 ,b) ,
(18.88)
for all j = 0, 1, . . . , n. From (18.88) by addition we obtain (n + 1) kf kLν (x0 ,b) proving the claim.
n X kLj+1 f k ≥ Sν,1
Lq (x0 ,b)
j=0
,
(18.89)
We continue with Theorem 18.20. Assume all as in Theorem 18.13. Here n = 2k + 1, k ∈ Z + ; ν > 0. Further suppose that Lj+1 f is of fixed sign and nowhere zero on [a, x0 ] for all odds j ∈ [1, n] . Call ν/p !1/ν Z x0 Z x0 p Sν,2 := min (−gj (x, t)) dt dx . (18.90) odd j∈[1,n] a x
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Then 1) kf kLν (a,x0 ) ≥
2) kf kLp (a,x0 ) ≥
3) kf kL1 (a,x0 ) ≥
Sν,2 k+1
Sp,2 k+1
S1,2 k+1
and
(18.91)
(18.92)
(18.93)
X kLj+1 f kLq (a,x0 ) , j∈[1,n] odd
X kLj+1 f kLq (a,x0 ) , j∈[1,n] odd
X kLj+1 f kLq (a,x0 ) . j∈[1,n] odd
Proof. As in Theorem 18.19, based on Theorem 18.13. Since n = 2k + 1, k ∈ Z + , there are (k + 1) odd numbers in [1, n] , so we apply (18.57) (k + 1) times. We finish with Theorem 18.21. Assume all as in Theorem 18.14. Here n = 2k, k ∈ Z+ ; ν > 0. Further suppose that Lj+1 f is of fixed sign and nowhere zero on [a, x0 ] for all evens j ∈ [0, n] . Call ν/p !1/ν Z x0 Z x0 . (18.94) dx (gj (x, t))p dt Sν,3 := min even j∈[0,n] x a Then
1) kf kLν (a,x0 ) ≥
2) kf kLp (a,x0 ) ≥
3) kf kL1 (a,x0 ) ≥
Sν,3 k+1
Sp,3 k+1
S1,3 k+1
and
(18.95)
(18.96)
(18.97)
X kLj+1 f kLq (a,x0 ) , j∈[0,n] even
X kLj+1 f kLq (a,x0 ) , j∈[0,n] even
X kLj+1 f kLq (a,x0 ) . j∈[0,n] even
Proof. As in Theorem 18.19, based on Theorem 18.14. Since n = 2k, k ∈ Z+ , there are (k + 1) even numbers in [0, n] , so we apply (18.64) (k + 1) times.
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Chapter 19
Poincar´ e and Sobolev Like Inequalities for Vector Valued Functions
Various Lp form Poincar´e and Sobolev like inequalities are given for functions valued in a Banach vector space with applications on C. This chapter relies on [44]. 19.1
Introduction
This chapter is motivated by the famous Poincar´e inequality, see [2]: Given a bounded domain Ω ⊂ Rn , n ∈ N, it holds kukLp (Ω) ≤ C k∇ukLp (Ω) for real valued functions u with vanishing mean value over Ω, 1 ≤ p ≤ ∞, under very general assumptions on Ω, where k∇ukLp (Ω) is defined as the Lp −norm of the euclidian norm of ∇u. Especially in [2] is proved for a convex domain Ω ⊂ Rn with diameter d that
d k∇ukL1 (Ω) 2 for any u with zero mean value on Ω, also the constant 1/2 is optimal. We are also motivated by [92], where the authors prove the following: Let 1 < p < ∞, −∞ < a < b < ∞. The best constant C (independent of a, b) for which the 1−dimensional Poincar´e inequality
Rb
f (t) dt
2− 1 a ≤ C (b − a) p kf 0 kLp ([a,b])
f −
b−a 1 kukL1 (Ω) ≤
L ([a,b])
0
holds for all real valued Lipschitz continuous functions f, is C = 12 (1 + p0 )−1/p , where p0 > 1 : p1 + p10 = 1. This chapter is also motivated by the famous Sobolev inequality of the following form, see [139], p. 263: (the Gagliardo–Nirenberg–Sobolev inequality). Assume 1 ≤ p ≤ n. Then exists a constant C, depending only on p and n, such that kukLp∗ (Rn ) ≤ C kDukLp (Rn ) , 229
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for all real valued u ∈ Cc1 (Rn ) . np Here p∗ := n−p , p∗ > p, and Du is the gradient of u. Also we are motivated by the following result, p. 265, [139]: estimates for Sobolev space W 1,p , 1 ≤ p < n. Let U be a bounded, open subset of Rn , and suppose the boundary ∂U is C 1 . Assume 1 ≤ p < n, and real valued u ∈ W 1,p (U ) . ∗ Then u ∈ Lp (U ) , with the estimate kukLp∗ (U ) ≤ C kukW 1,p (U ) , the constant C depending only on p, n and U . So here we derive Poincar´e and Sobolev like inequalities of Lp form for Banach vector space valued functions. Most of the Poincar´e and Sobolev like inequalities here are of Dirichlet type. We assume in this case initial conditions prescribed equal to zero. But we give also a Neumann–Poincar´e like inequality involving the average of the engaged function, see Theorem 19.3. At the end we apply the results to C-valued functions. We need
19.2
Background
(see [230], pp. 83–94) Let f (t) be a function defined on [a, b] ⊆ R taking values in a real or complex normed linear space (X, k·k) . Then f (t) is said to be differentiable at a point t0 ∈ [a, b] if the limit
f (t0 + h) − f (t0 ) (19.1) h→0 h exists in X, the convergence is in k·k . This is called the derivative of f (t) at t = t 0 . We call f (t) differentiable on [a, b] , iff there exists f 0 (t) ∈ X for all t ∈ [a, b] . Similarly and inductively are defined higher order derivatives of f, denoted f 00 , (3) f , . . . , f (k) , k ∈ N, just as for numerical functions. For all the properties of derivatives see [230], pp. 83–86. Let now (X, k·k) be a Banach space, and f : [a, R bb] → X. We define the vector valued Riemann integral a f (t) dt ∈ X as the limit of the vector valued Riemann sums in X, convergence is in k·k . The definition is as for the numerical valued functions. Rb If a f (t) dt ∈ X we call f integrable on [a, b] . If f ∈ C ([a, b] , X), then f is integrable, [230], p. 87. For all the properties of vector valued Riemann integrals see [230], pp. 86–91. We define the space C n ([a, b] , X) , n ∈ N, of n-times continuously differentiable functions from [a, b] into X; here continuity is with respect to k·k and defined in the usual way as for numerical functions. f 0 (t0 ) := lim
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231
Let (X, k·k) be a Banach space and f ∈ C n ([a, b] , X) , then we have the vector valued Taylor’s formula, see [230], pp. 93–94, and also [227], (IV, 9; 47). It holds En (x, y) := f (y) − f (x) − f 0 (x) (y − x) − 1 1 00 2 n−1 f (x) (y − x) − . . . − f (n−1) (x) (y − x) 2 (n − 1)! Z y 1 n−1 (n) = (y − t) f (t) dt, ∀x, y ∈ [a, b] . (n − 1)! x
(19.2)
In particular (19.2) is true when X = Rm , Cm , m ∈ N, etc. In case of some x0 ∈ [a, b] such that f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1, then Z y 1 n−1 (n) (y − t) f (t) dt, ∀y ∈ [a, b] , (19.3) f (y) = (n − 1)! x0 see also [52]. In that case En (x0 , y) = f (y) .
19.3
Results
Here we consider always X to be a Banach space, n ∈ N, and f ∈ C n ([a, b] , X) , [a, b] ⊆ R. We fix x0 ∈ [a, b] ; p, q > 1 : p1 + q1 = 1, ν > 0. We present the first results which are of Dirichlet–Poincar´e like inequalities. Theorem 19.1. Let y ∈ [x0 , b] , x0 < b. Assume f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then
1
1 (b − x0 )(n−1+ p + ν )
f (n)
L (x ,b) q 0 1) kkf kkLν (x0 ,b) ≤ 1/ν . (19.4) 1/p 1 ((n − 1)!) (p (n − 1) + 1) ν n−1+ p +1 When ν = q we obtain
2) kkf kkLq (x0 ,b) ≤
((n − 1)!) (p (n − 1) + 1)1/p (qn)1/q
And for ν = q = p = 2 we have 3) kkf kkL2 (x0 ,b) Proof.
By (19.3) we have
(b − x0 )n
f (n)
Lq (x0 ,b)
n
(b − x0 )
f (n)
L2 (x0 ,b) √ ≤ . √ ((n − 1)!) 2n − 1 2n
Z y
1 n−1 (n)
(y − t) f (t) dt
(n − 1)! x0 Z y
1
n−1 (n) ≤ (y − t)
f (t) dt (n − 1)! x0
kf (y)k =
.
(19.5)
(19.6)
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≤
1 (n − 1)!
Z
y x0
(y − t)
p(n−1)
dt
1/p Z
y x0
)
(y − x0 )( 1
(n)
.
f
(n − 1)! (p (n − 1) + 1)1/p Lq (x0 ,b) 1 n−1+ p
≤ Thus
1/q
(n) q
f (t) dt (19.7)
1 ν n−1+ p
)
1 (y − x0 ) (
(n)
ν kf (y)k ≤ ,
f
ν ((n − 1)!) (p (n − 1) + 1)ν/p Lq (x0 ,b)
ν
and Z
(19.8)
1 ν (n−1+ p )+1
f (n)
ν (b − x ) 0 1 Lq (x0 ,b) ν , kf (y)k dy ≤ ν ((n − 1)!) ν n − 1 + 1 + 1 (p (n − 1) + 1)ν/p x0 p b
(19.9)
proving the claim.
For the next result we need Theorem 19.2. (see [230], p. 90) Let f (t) be an integrable function on an interval [a, b] , with values in a Banach space X. Then the mean value of f (t) on [a, b] belongs to the closed convex hull of the set of values of f (t) on [a, b] . That is we have that Z b n−1 X 1 1 lim f (t) dt = f (τk ) ∆tk , (19.10) b−a a b − a d(Π)→0 k=0
where
Π := {a = t0 ≤ τ0 ≤ t1 ≤ τ1 ≤ t2 ≤ . . . ≤ tn−1 ≤ τn−1 ≤ tn = b} ,
(19.11)
with d (Π) := max {∆t0 , ∆t1 , . . . , ∆tn−1 } ,
(19.12)
∆tk := tk+1 − tk . Next we give Neumann–Poincar´e like inequalities. Theorem 19.3. Here f ∈ C 1 ([a, b] , X) , X a Banach space, p, q > 1 : ν > 0. One has
Z b
1 1 1
≤ (b − a)( p + ν ) kkf 0 kkLq (a,b) . 1)
f − f (t) dt
b−a a
1 p
+
1 q
= 1,
(19.13)
Lν (a,b)
When ν = q we obtain
Z b
1
f (t) dt
2)
f −
b−a a
Lq (a,b)
≤ (b − a) kkf 0 kkLq (a,b) .
(19.14)
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When ν = p = q = 2 we have
Z b
1
3)
f − f (t) dt
b−a a
L2 (a,b)
≤ (b − a) kkf 0 kkL2 (a,b) ,
equivalently,
2 Z b Z b
1
2 f (t) dt ds ≤ (b − a)
f (s) −
b − a a a
Proof.
Z
b a
0
2
!
kf (s)k ds .
Since f ∈ C 1 ([a, b] , X) we have by (19.2) that Z y f (y) = f (x) + f 0 (t) dt, ∀x, y ∈ [a, b] .
233
(19.15)
(19.15*)
(19.16)
x
So by (19.16) we have f (s) = f (τk ) +
Z
s
f 0 (r) dr,
(19.17)
τk
∀s ∈ [a, b] and k = 0, 1, . . . , n − 1. Hence Z s ∆tk ∆tk ∆tk = f (τk ) + f 0 (r) dr, f (s) b−a b − a b − a τk
(19.18)
and f (s)
n−1 1 1 X ∆tk = b−a b−a k=0
·
n−1 X
f (τk ) ∆tk +
k=0
− and
1 b−a
(19.19)
k=0
Therefore we observe that f (s) −
Z s n−1 1 X f 0 (r) dr. ∆tk b−a τk
1 b−a Z
Z
b
f (t) dt = a
b
f (t) dt + a
n−1 1 X f (τk ) ∆tk b−a k=0
Z s n−1 1 X ∆tk f 0 (r) dr, b−a τk k=0
Z b
1
f (t) dt
f (s) −
b−a a
Z b
1 n−1
X 1
≤ f (τk ) ∆tk − f (t) dt
b − a
b−a a k=0
(19.20)
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+
Z s
n−1
1 X 0
∆tk f (r) dr
b−a τk k=0
Z b
1 n−1 X 1
f (τk ) ∆tk − f (t) dt ≤
b − a b−a a k=0
Z s n−1 1 X 0 + ∆tk kf (r)k dr b−a τk k=0
Z
Z b
1 n−1 b X 1
f (τk ) ∆tk − f (t) dt + kf 0 (r)k dr. ≤
b − a b−a a a
(19.21)
k=0
That is we proved
Z b
1
f (t) dt
f (s) −
b−a a
Z Z b
1 n−1
b X 1
≤ kf 0 (r)k dr, ∀s ∈ [a, b] , (19.22) f (t) dt + f (τk ) ∆tk −
b − a
b−a a a k=0
for any partition Π. Taking the limit in both sides of (19.22) as d (Π) → 0 we obtain by (19.10) that
Z Z b
b 1
f (t) dt ≤ kf 0 (r)k dr
f (s) −
b−a a a ≤ (b − a)
1/p
Z
b
a
0
q
kf (r)k dr
Consequently we find
ν Z b
1
ν/p f (t) dt ≤ (b − a)
f (s) −
b−a a
!1/q
Z
b a
q
0
kf (r)k dr
Finally we have
ν Z b Z b
ν 1
+1 f (t) dt ds ≤ (b − a) p
f (s) −
b − a a a
proving the claim.
, ∀s ∈ [a, b] .
!ν/q
(19.23)
, ∀s ∈ [a, b] . (19.24)
Z
b a
0
q
kf (r)k dr
!ν/q
,
(19.25)
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Corollary 19.4. (to Theorem 19.3). Additionally assume that One has 1
1
Rb a
235
f (t) dt = 0.
1) kkf kkLν (a,b) ≤ (b − a) p + ν kkf 0 kkLq (a,b) .
(19.26)
When ν = q we obtain 2) kkf kkLq (a,b) ≤ (b − a) kkf 0 kkLq (a,b) .
(19.27)
When ν = p = q = 2 we have 3) kkf kkL2 (a,b) ≤ (b − a) kkf 0 kkL2 (a,b) , equivalently Z
b a
2
kf (s)k ds ≤ (b − a)
2
Z
b a
2
0
(19.28) !
kf (s)k ds .
(19.28*)
We continue with Theorem 19.5. Let y ∈ [a, x0 ] , x0 > a. Assume f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then
n−1+ p1 + ν1 )
f (n)
(x0 − a)( Lq (a,x0 ) 1) kkf kkLν (a,x0 ) ≤ 1/ν . (19.29) 1/p ((n − 1)!) (p (n − 1) + 1) ν n − 1 + p1 + 1 When ν = q we derive
2) kkf kkLq (a,x0 ) ≤
n
(x0 − a)
f (n)
Lq (a,x0 )
((n − 1)!) (p (n − 1) + 1)
And for ν = q = p = 2 we have 3) kkf kkL2 (a,x0 ) Proof.
By (19.3) we have kf (y)k =
1 (n − 1)!
Z
Z
1/q
.
n
(x0 − a)
f (n)
L2 (a,x0 ) √ ≤ . √ ((n − 1)!) 2n − 1 2n
x0
y
x0 y
(qn)
(19.30)
(19.31)
Z x 0
1 n−1 (n)
(t − y) f (t) dt
(n − 1)! y
1 ≤ (n − 1)! ≤
1/p
(t − y)
(t − y)
p(n−1)
dt
n−1
(n)
f (t) dt
1/p Z
x0
y
1/q
(n) q
f (t) dt
1 n−1+ p
)
(x0 − y)( 1
(n)
. ≤
f
(n − 1)! (p (n − 1) + 1)1/p Lq (a,x0 )
(19.32)
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Hence ν n−1+ p1 )
1 (x0 − y) (
(n)
ν kf (y)k ≤ ,
f
ν ((n − 1)!) (p (n − 1) + 1)ν/p Lq (a,x0 ) ν
and
Z
x0
ν
kf (y)k dy ≤
a
1 ν ((n − 1)!)
1 ν n−1+ p )+1)
f (n)
ν (x0 − a)( ( Lq (a,x0 ) , · ν/p 1 ν n−1+ p +1 (p (n − 1) + 1)
proving the claim.
(19.33)
(19.34)
We continue with L1 results. Proposition 19.6. Let y ∈ [x0 , b] , x0 < b. Suppose f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then
n−1+ ν1 )
f (n)
(b − x0 )( L1 (x0 ,b) 1) kkf kkLν (x0 ,b) ≤ . (19.35) 1/ν (n − 1)! (ν (n − 1) + 1) When ν = 1 we get 2) kkf kkL1 (x0 ,b) ≤ Proof.
By (19.3) , (19.7) we have kf (y)k ≤
≤
(y − x0 )n−1 (n − 1)!
That is
Z
1 (n − 1)!
ν
and b x0
proving the claim.
n!
Z
y x0
.
(19.36)
(y − t)n−1 f (n) (t) dt
(y − x0 )n−1
(n)
(n)
.
f (t) dt =
f
(n − 1)! L1 (x0 ,b) x0 b
kf (y)k ≤ Z
n
(b − x0 )
f (n)
L1 (x0 ,b)
ν
(y − x0 )ν(n−1)
(n)
ν ,
f
ν ((n − 1)!) L1 (x0 ,b)
kf (y)k dy ≤
(b − x0 )
ν(n−1)+1
(n)
ν
f
L1 (x0 ,b)
ν
((n − 1)!) (ν (n − 1) + 1)
(19.37)
(19.38)
,
(19.39)
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The countercase of the last result follows. Proposition 19.7. Let y ∈ [a, x0 ] , x0 > a. Suppose f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then
n−1+ ν1 )
f (n)
(x0 − a)( L1 (a,x0 ) 1) kkf kkLν (a,x0 ) ≤ . (19.40) ((n − 1)!) (ν (n − 1) + 1)1/ν When ν = 1 we get
n
(x0 − a)
f (n)
L1 (a,x0 ) . (19.41) 2) kkf kkL1 (a,x0 ) ≤ n! Proof. By (19.3) and (19.32) we have Z x0
1
n−1 (n) kf (y)k ≤ (t − y)
f (t) dt (n − 1)! y n−1
≤
(x0 − y) (n − 1)!
n−1
= That is ν
(x0 − y) (n − 1)!
kf (y)k ≤ and Z
x0
ν
x0
a
(n)
f (t) dt
(n)
f
ν(n−1)
(x0 − y) ν ((n − 1)!)
kf (y)k dy ≤
a
Z
(x0 − a)
proving the claim.
L1 (a,x0 )
.
(19.42)
(n)
ν
f
(ν(n−1)+1)
L1 (a,x0 )
,
(19.43)
(n)
ν
f
L1 (a,x0 )
((n − 1)!)ν (ν (n − 1) + 1)
,
(19.44)
We continue with Sobolev like inequalities Theorem 19.8. Let y ∈ [x0 , b] , x0 < b. Assume f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then ! n
Mν X
(i)
, (19.45) 1) kkf kkLν (x0 ,b) ≤
f
n Lq (x0 ,b) i=1 where
1 i−1+ p + ν1 ) ( (b − x0 ) Mν := max 1/ν . 1≤i≤n ((i − 1)!) (p (i − 1) + 1)1/p ν i − 1 + 1 + 1 p
When ν = q we derive
2) kkf kkLq (x0 ,b)
Mq ≤ n
n
X
(i)
f
i=1
Lq (x0 ,b)
!
,
(19.46)
(19.47)
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where Mq := max
1≤i≤n
(
For ν = q = p = 2 we have
(b − x0 )
i
((i − 1)!) (p (i − 1) + 1)
3) kkf kkL2 (x0 ,b) where M2 := max
1≤i≤n
M2 ≤ n (
1/p
n
X
(i)
f
i=1
(qi)
1/q
L2 (x0 ,b)
(b − x0 )i √ √ ((i − 1)!) 2i − 1 2i
)
)
!
.
,
.
(19.48)
(19.49)
(19.50)
Proof. By assumption of Theorem 19.1 we have that f ∈ C i ([a, b] , X) , for all i = 1, . . . , n, and that f (k) (x0 ) = 0, k = 0, 1, . . . , i − 1; for all i = 1, . . . , n, n ∈ N. Thus by (19.4) we find
1 + ν1 ) i−1+ p
f (i)
(b − x0 )( Lq (x0 ,b) kkf kkLν (x0 ,b) ≤ 1/ν 1/p ν i − 1 + p1 + 1 ((i − 1)!) (p (i − 1) + 1)
for all i = 1, . . . , n. So that
≤ Mν
f (i)
n kkf kkLν (x0 ,b) ≤ Mν proving the claim.
(19.51)
Lq (x0 ,b)
n
X
(i)
f
i=1
Lq (x0 ,b)
!
,
(19.52)
We continue with Theorem 19.9. Let y ∈ [a, x0 ] , x0 > a. Assume f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1. Then ! n
∆ν X
(i)
1) kkf kkLν (a,x0 ) ≤ , (19.53)
f
n Lq (a,x0 ) i=1 where
i−1+ p1 + ν1 ) ( (x0 − a) ∆ν := max . 1/ν 1≤i≤n ((i − 1)!) (p (i − 1) + 1)1/p ν i − 1 + 1 + 1 p
When ν = q we obtain
2) kkf kkLq (a,x0 )
∆q ≤ n
n
X
(i)
f
i=1
Lq (a,x0 )
!
,
(19.54)
(19.55)
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where ∆q := max
1≤i≤n
(
For ν = q = p = 2 we have
(x0 − a)
((i − 1)!) (p (i − 1) + 1)1/p (qi)1/q
3) kkf kkL2 (a,x0 ) where ∆2 := max
1≤i≤n
Proof.
i
n
X
(i)
f
∆2 ≤ n (
i=1
L2 (a,x0 )
i
(x0 − a) √ √ ((i − 1)!) 2i − 1 2i
)
)
!
.
,
.
239
(19.56)
(19.57)
(19.58)
Based on Theorem 19.5, similar to the proof of Theorem 19.8.
We continue with L1 results. Proposition 19.10. Let y ∈ [x0 , b] , x0 < b. Suppose f (k) (x0 ) = 0, k = 0, 1, . . . , n− 1. Then ! n
θν X
(i)
, (19.59) 1) kkf kkLν (x0 ,b) ≤
f
n i=1 L1 (x0 ,b)
where
θν := max
1≤i≤n
When ν = 1 we get
(
(b − x0 )(
i−1+ ν1 )
((i − 1)!) (ν (i − 1) + 1)
2) kkf kkL1 (x0 ,b)
θ1 ≤ n
where θ1 := max
1≤i≤n
n
X
(i)
f
i=1
(
(b − x0 ) i!
i
)
1/ν
)
L1 (x0 ,b)
.
!
(19.60)
,
.
(19.61)
(19.62)
Based on Proposition 19.6. We have from (19.35) that
1 (b − x0 )(i−1+ ν )
f (i)
L (x ,b) 1 0 kkf kkLν (x0 ,b) ≤ 1/ν ((i − 1)!) (ν (i − 1) + 1) for all i = 1, . . . , n. Consequently we see that
, kkf kkLν (x0 ,b) ≤ θν
f (i)
Proof.
L1 (x0 ,b)
all i = 1, . . . , n. Thus
n kkf kkLν (x0 ,b) ≤ θν proving the claim.
n
X
(i)
f
i=1
L1 (x0 ,b)
!
,
(19.63)
(19.64)
(19.65)
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Similarly we give Proposition 19.11. Let y ∈ [a, x0 ] , x0 > a. Assume f (k) (x0 ) = 0, k = 0, 1, . . . , n− 1. Then ! n
Tν X
(i)
, (19.66) 1) kkf kkLν (a,x0 ) ≤
f
n i=1 L1 (a,x0 )
where
Tν := max
1≤i≤n
(
(x0 − a)(
i−1+ ν1 )
((i − 1)!) (ν (i − 1) + 1)1/ν
)
.
(19.67)
When ν = 1 we get 2) kkf kkL1 (a,x0 ) where T1 := max
1≤i≤n
Proof.
19.4
n
X
(i)
f
T1 ≤ n
i=1
(
(x0 − a)i i!
)
L1 (a,x0 )
!
,
.
(19.68)
(19.69)
Based on Proposition 19.7 as in Proposition 19.10.
Applications
Here X = C, f ∈ C n ([a, b] , C) , n ∈ N, x0 , y ∈ [a, b] . Furthermore suppose that f (k) (x0 ) = 0, k = 0, 1, . . . , n − 1; p, q > 1 : p1 + q1 = 1, ν > 0. First we give the following Poincar´e like inequalities results. Corollary 19.12. Let x0 < b. One has kf kLν (x0 ,b) Proof.
(n)
f Lq (x0 ,b) ≤ 1/ν . 1/p ν n − 1 + 1p + 1 ((n − 1)!) (p (n − 1) + 1) (b − x0 )(
n−1+ p1 + ν1 )
Use of (19.4) .
(19.70)
We continue with Corollary 19.13. One has
Z b
1
f (t) dt
f −
b−a a
Proof.
Use of (19.13) .
≤ (b − a)( p 1
Lν (a,b)
+ ν1 )
kf 0 kLq (a,b) .
(19.71)
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Corollary 19.14. Let x0 > a. One has kf kLν (a,x0 ) Proof.
1 1 (x0 − a)(n−1+ p + ν ) f (n) L (a,x ) q 0 ≤ 1/ν . 1/p 1 ((n − 1)!) (p (n − 1) + 1) ν n−1+ p +1
Use of (19.29).
(19.72)
We continue with L1 Poincar´e like results on C. Corollary 19.15. Let x0 < b. Then (b − x0 )(
kf kLν (x0 ,b) ≤ Proof.
n−1+ ν1 )
(n)
f
L1 (x0 ,b) . 1/ν
((n − 1)!) (ν (n − 1) + 1)
Use of (19.35) .
(19.73)
Corollary 19.16. Let x0 > a. Then kf kLν (a,x0 ) ≤ Proof.
(x0 − a)(
Use of (19.40) .
n−1+ ν1 )
(n)
f
L1 (a,x0 ) . 1/ν
((n − 1)!) (ν (n − 1) + 1)
(19.74)
We finish with Sobolev like inequalities results on C. Corollary 19.17. Let x0 < b. Then kf kLν (x0 ,b)
Mν ≤ n
where Mν as in (19.46). Proof.
n
X
(i)
f i=1
Lq (x0 ,b)
,
Use of (19.45).
kf kLν (a,x0 )
(19.75)
Corollary 19.18. Let x0 > a. Then ∆ν ≤ n
where ∆ν as in (19.54) . Proof.
!
n
X
(i)
f i=1
Lq (a,x0 )
!
Use of (19.53).
,
(19.76)
At last we give L1 Sobolev like results on C. Corollary 19.19. Let x0 < b. Then kf kLν (x0 ,b) where θν as in (19.60) . Proof.
Use of (19.59).
θν ≤ n
n
X
(i)
f i=1
L1 (x0 ,b)
!
,
(19.77)
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Corollary 19.20. Let x0 > a. Then kf kLν (a,x0 ) where Tν as in (19.67) . Proof.
Book˙Adv˙Ineq
Use of (19.66).
Tν ≤ n
n
X
(i)
f i=1
L1 (a,x0 )
!
,
(19.78)
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Chapter 20
Poincar´ e Type Inequalities for Semigroups, Cosine and Sine Operator Functions Here we present Poincar´e type general Lp inequalities regarding Semigroups, Cosine and Sine Operator functions. We give applications. This chapter relies on [50].
20.1
Introduction
This chapter is motivated by the famous Poincar´e inequality, see [2]: Given a bounded domain Ω ⊂ Rn , n ∈ N, it holds kukLp (Ω) ≤ C k∇ukLp (Ω)
for functions u with vanishing mean value over Ω, 1 ≤ p ≤ ∞, under very general assumptions on Ω, where k∇ukLp (Ω) is defined as the Lp −norm of the euclidean norm of ∇u. Especially in [2] is proved for a convex domain Ω ⊂ Rn with diameter d that d kukL1 (Ω) ≤ k∇ukL1 (Ω) 2 for any u with zero mean value on Ω, also the constant 1/2 is optimal. We are also motivated by [92], where the authors prove the following: Let 1 < p < ∞, −∞ < a < b < ∞. The best constant C (independent of a, b) for which the 1−dimensional Poincar´e inequality
Rb
f (t) dt
2− 1 a ≤ C (b − a) p kf 0 kLp ([a,b])
f −
b−a 1 L ([a,b])
−1/p0
, where p0 > 1 : holds for all Lipschitz continuous functions f, is C = 12 (1 + p0 ) 1 1 p + p0 = 1. So here we present Poincar´e type inequalities for Semigroups and for Cosine and Sine Operator functions.
20.2
Semigroups Background
All this background comes from [79] (in general see also [147]). 243
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Let X a real or complex Banach space with elements f, g, . . . having norm kf k , kgk , . . . and let ε (X) be the Banach algebra of endomorphisms of X. If T ∈ ε (X) , kT k denotes the norm of T. Definition 20.1. If T (t) is an operator function on the non-negative real axis 0 ≤ t < ∞ to the Banach algebra ε (X) satisfying the following conditions: (i) T (t1 + t2 ) = T (t1 ) T (t2 ) , (t1 , t2 ≥ 0) , (20.1) (ii) T (0) = I (I = identity operator) , then {T (t) ; 0 ≤ t < ∞} is called a one-parameter semi-group of operators in ε (X) . The semi-group {T (t) ; 0 ≤ t < ∞} is said to be of class C0 if it satisfies the further property (iii) s − lim T (t) f = f, (f ∈ X) t→0+
(20.2)
referred to as the strong continuity of T (t) at the origin. In this chapter we shall assume that the family of bounded linear operators {T (t) ; 0 ≤ t < ∞} mapping X to itself is a semi-group of class C0 , thus that all three conditions of the above definition are satisfied. Proposition 20.2. (a) kT (t)k is bounded on every finite subinterval of [0, ∞) . (b) For each f ∈ X, the vector-valued function T (t) f on [0, ∞) is strongly continuous. Definition 20.3. The infinitesimal generator A of the semi-group {T (t); 0 ≤ t < ∞} is defined by 1 (20.3) Af = s − lim Aτ f, Aτ f = [T (τ ) − I] τ →0+ τ whenever the limit exists; the domain of A, in symbols D (A) , is the set of elements f for which the limit exists. Proposition 20.4. (a) D (A) is a linear manifold in X and A is a linear operator. (b) If f ∈ D (A) , then T (t) f ∈ D (A) for each t ≥ 0 and d T (t) f = AT (t) f = T (t) Af (t ≥ 0) ; dt
furthermore, T (t) f − f =
Z
(20.4)
t
T (u) Af du (t > 0) .
(20.5)
0
(c) D (A) is dense in X, i.e. D (A) = X, and A is a closed operator. Definition 20.5. For r = 0, 1, 2, . . . the operator Ar is defined inductively by the relations A0 = I, A1 = A, and D (Ar ) = f ; f ∈ D Ar−1 and Ar−1 f ∈ D (A) Ar f = A Ar−1 f = s − lim Aτ Ar−1 f (f ∈ D (Ar )) . (20.6) τ →0+
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For the operator Ar and its domain D (Ar ) we have the following Proposition 20.6. (a) D (Ar ) is a linear subspace in X and Ar is a linear operator. (b) If f ∈ D (Ar ) , so does T (t) f for each t ≥ 0 and dr T (t) f = Ar T (t) f = T (t) Ar f. dtr
Moreover r−1 k X t
∆r (t) f := T (t) f −
k=0
k!
Ak f =
1 (r − 1)!
Z
t 0
(20.7)
(t − u)r−1 T (u) Ar f du,
the Taylor’s formula for semigroups. (c) D (Ar ) is dense in X for r = 1, 2, . . . ; furthermore ,
∞ T
(20.8)
D (Ar ) is dense in
r=1
X. Ar is a closed operator. The integral in (20.8) is a vector valued Riemann integral, see [79], [164].
20.3
Poincar´ e Type Inequalities for Semigroups
Here always we consider T (t) , A as in the Semigroups Background and f ∈ D (Ar ) , r ∈ N. Also we consider p, q > 1 : p1 + 1q = 1, along with a, ν ∈ R : a, ν > 0. We present the first result. Theorem 20.7. One has 1) kk∆r (t) f kkLν (0,a) 1
≤
1
ar−1+ p + ν r 1/ν kkT (t) A f kkLq (0,a) . (20.9) 1/p 1 (r − 1)! (p (r − 1) + 1) ν r−1+ p +1
When q = ν we have
2) kk∆r (t) f kkLq (0,a) ≤ Proof.
ar (r − 1)! (p (r − 1) + 1)
1/p
(qr)
1/q
kkT (t) Ar f kkLq (0,a) .
(20.10)
By (20.8) we have
Z t
1 r−1 r
k∆r (t) f k = (t − u) T (u) A f du
(r − 1)! 0 Z t 1 r−1 ≤ (t − u) kT (u) Ar f k du (r − 1)! 0
≤
1 (r − 1)!
Z
t 0
(t − u)p(r−1) du
1/p Z
t 0
kT (u) Ar f kq du
1/q
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1
=
tr−1+ p (r − 1)! (p (r − 1) + 1)
1/p
Z
t 0
q
r
kT (u) A f k du
1/q
.
(20.11)
That is we get
1
tr−1+ p
k∆r (t) f k ≤
(r − 1)! (p (r − 1) + 1)
Thus
1/p
Z
a 0
tν (r−1+ p )
q
kT (u) Ar f k du
1/q
.
(20.12)
1
ν
k∆r (t) f k ≤ and Z
a 0
ν
ν
((r − 1)!) (p (r − 1) + 1)
ν/p
kkT (t) Ar f kkLq (0,a) ,
aν (r−1+ p )+1 kkT (t) Ar f kkLq (0,a) , k∆r (t) f k dt ≤ ν ν/p ((r − 1)!) (p (r − 1) + 1) ν r − 1 + p1 + 1 1
(20.13)
ν
ν
proving the claim.
(20.14)
We continue with Corollary 20.8. (to Theorem 20.7) Case of p = q = 2. We have 1) kk∆r (t) f kkLν (0,a) 1
1
≤
(r − 1)!
p
ar− 2 + ν
(2r − 1) ν r −
1 2
When ν = 2 we derive
2) kk∆r (t) f kkL2 (0,a) ≤
(r − 1)!
We treat next the L1 case.
p
+1
r 1/ν kkT (t) A f kkL2 (0,a) .
ar √ kkT (t) Ar f kkL2 (0,a) . (2r − 1) 2r
(20.15)
(20.16)
Theorem 20.9. One has a(r−1+ ν ) kkT (t) Ar f kkL1 (0,a) 1
1) kk∆r (t) f kkLν (0,a) ≤ For ν = 1 we get
1/ν
(r − 1)! (ν (r − 1) + 1)
2) kk∆r (t) f kkL1 (0,a) ≤ Proof.
We observe that k∆r (t) f k ≤ ≤
1 (r − 1)! tr−1 (r − 1)!
Z
Z
ar kkT (t) Ar f kkL1 (0,a)
t 0 a
0
(t − u)
r!
r−1
.
(20.17)
.
(20.18)
kT (u) Ar f k du
kT (u) Ar f k du.
(20.19)
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So that ν
k∆r (t) f k ≤ and Z
(20.20)
ν
a 0
tν(r−1) ν r ν kkT (u) A f kkL1 (0,a) ((r − 1)!)
ν
k∆r (t) f k dt ≤
aν(r−1)+1 kkT (u) Ar f kkL1 (0,a)
proving the claim.
ν
((r − 1)!) (ν (r − 1) + 1)
,
(20.21)
We finish this section with Application 20.10. (see also [80]) It is known the classical diffusion equation ∂W ∂2W = , ∂t ∂x2
− ∞ < x < ∞, t > 0
(20.22)
with initial condition lim W (x, t) = f (x) ,
t→0+
has under general conditions its solution given by Z ∞ 2 1 W (x, t, f ) = [T (t) f ] (x) = √ f (x + u) e−u /4t du, 2 πt −∞
(20.23)
(20.24)
the so called Gauss–Weierstrass singular integral. The infinitesimal generator of the semigroup {T (t) ; 0 ≤ t < ∞} is A = ∂ 2 /∂x2 ([159], p. 578). Here we suppose that f, f (2k) , k = 1, . . . , r, all belong to the Banach space U CB (R) , the space of bounded and uniformly continuous functions from R into itself, with norm kf kC := sup |f (x)| .
(20.25)
x∈R
Here we define ∆r (t) f (x) := W (x, t, f ) − So by (20.9) we find
r−1 k X t k=0
k!
f (2k) (x) , for all x ∈ R.
(20.26)
∆r (t) f
C Lν (0,a)
1 1
ar−1+ p + ν
(2r) ≤ .
1/ν
W ·, t, f C Lq (0,a) (r − 1)! (p (r − 1) + 1)1/p ν r − 1 + p1 + 1
(20.27)
We notice here that Z 1 ∞ (2r) (2r) −u2 /4t f (x + u) e du W x, t, f = √ 2 πt −∞
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1 ≤ √ 2 πt
Z
∞ −∞
≤ f (2r)
C
2 (2r) (x + u) e−u /4t du f 1 √ 2 πt
Z
∞
e−u
2
/4t
du
(20.28)
−∞
= f (2r) · 1 = f (2r) < ∞, for all t ∈ R. C
That is
C
W ·, t, f (2r) ≤ f (2r) . C
(20.29)
C
Also by (20.10) we obtain
∆r (t) f
C Lq (0,a)
≤
ar (r − 1)! (p (r − 1) + 1)
1/p
(qr)
1/q
W ·, t, f (2r)
C Lq (0,a)
By (20.15) we have (p = q = 2)
∆r (t) f C L 1
≤
1
ar− 2 + ν
1/ν √ (r − 1)! 2r − 1 ν r − 21 + 1
.
(20.30)
ν (0,a)
W ·, t, f (2r)
C L2 (0,a)
.
(20.31)
And by (20.16) (ν = 2) we see that
∆r (t) f
C L2 (0,a)
≤
ar
√
W ·, t, f (2r) . √ C L2 (0,a) (r − 1)! 2r − 1 2r
We finish application with L1 results. By (20.17) we find
∆r (t) f C L a(r−1+ ν ) 1
≤
(r − 1)! (ν (r − 1) + 1)1/ν
and by (20.18) (ν = 1) we obtain
∆r (t) f
C L1 (0,a)
≤
(20.32)
ν (0,a)
W ·, t, f (2r)
C L1 (0,a)
,
ar
.
W ·, t, f (2r) r! C L1 (0,a)
(20.33)
(20.34)
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Cosine and Sine Operator Functions Background
(see [93], [147], [192], [191]) Let (X, k·k) be a real or complex Banach space. By definition, a cosine operator function is a family {C (t) ; t ∈ R} of bounded linear operators from X into itself, satisfying (i) C (0) = I, I the identity operator; (ii) C (t + s) + C (t − s) = 2C (t) C (s) , for all t, s ∈ R;
(20.35)
(the last product is composition) Notice that
(iii) C (·) f is continuous an R, for all f ∈ X. C (t) = C (−t) , for all t ∈ R.
The associated sine operator function S (·) is defined by Z t S (t) f := C (s) f ds, for all t ∈ R, for all f ∈ X.
(20.36)
0
The cosine operator function C (·) is such that kC (t)k ≤ M eω|t| , for some M ≥ 1, ω ≥ 0, for all t ∈ R, here k·k is the norm of the operator. The infinitesimal generator A of C (·) is the operator from X into itself defined as 2 (20.37) Af := lim 2 (C (t) − I) f t→0+ t with domain D (A) . The operator A is closed and D (A) is dense in X, i.e. D (A) = X, and one has Z t Z t S (s) f ds ∈ D (A) and A S (s) f ds = C (t) f − f, for all f ∈ X. (20.38) 0
0
Also one has A = C 00 (0) , and D (A) is the set of f ∈ X such that C (t) f is twice differentiable at t = 0; equivalently, D (A) = f ∈ X : C (·) f ∈ C 2 (R, X) . (20.39)
If f ∈ D (A) , then C (t) f ∈ D (A) , and C 00 (t) f = C (t) Af = AC (t) f, for all t ∈ R; C 0 (0) f = 0, see [140], [232]. We define A0 = I, A2 = A ◦ A, . . . , An = A ◦ An−1 , n ∈ N. Let f ∈ D (An ) , then C (t) f ∈ C 2n (R, X) , and C (2n) (t) f = C (t) An f = An C (t) f, for all t ∈ R, and C (2k−1) (0) f = 0, 1 ≤ k ≤ n, see [191]. For f ∈ D (An ) , t ∈ R, we have the cosine operator function’s Taylor formula ([191], [192]) saying that Z t n−1 2n−1 X t2k (t − s) Tn (t) f := C (t) f − Ak f = C (s) An f ds. (20.40) (2k)! (2n − 1)! 0 k=0
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By integrating (20.40) we obtain the sine operator function’s Taylor formula (see [45]) Mn (t) f := S (t) f − f t − =
Z
t 0
t2n−1 t3 Af − . . . − An−1 f 3! (2n − 1)!
2n
(t − s) C (s) An f ds, for all t ∈ R, (2n)!
(20.41)
all f ∈ D (An ) . The integrals in (20.40) and (20.41) are vector valued Riemann integrals, see [79], [164]. 20.5
Poincar´ e Type Inequalities for Cosine and Sine Operator Functions
Here always we consider C (t) , S (t) , A as in the Cosine and Sine operator functions Background and f ∈ D (An ) , n ∈ N. Also we consider p, q > 1 : 1p + 1q = 1, along with a, ν ∈ R : ν > 0. We present the following result Theorem 20.11. Let a > 0, t ∈ [0, a] . One has 1) kkTn (t) f kkLν (0,a) ≤
When ν = q we have
a(2n−1+ p + ν ) 1
1
1/p
(2n − 1)! (p (2n − 1) + 1)
kkC (t) An f kkLq (0,a) · 1/ν . ν 2n − 1 + p1 + 1
2) kkTn (t) f kkLq (0,a) ≤ ·
(20.42)
a2n (2n − 1)! (p (2n − 1) + 1)1/p
kkC (t) An f kkLq (0,a) (2qn)1/q
.
(20.43)
When ν = q = p = 2 we derive 3) kkTn (t) f kkL2 (0,a) ≤ Proof.
a2n · kkC (t) An f kkL2 (0,a) . √ √ 2 (2n − 1)! n 4n − 1
By (20.40) we have
Z t
1 2n−1 n
(t − s) C (s) A f ds
(2n − 1)! 0 Z t 1 ≤ (t − s)2n−1 kC (s) An f k ds (2n − 1)! 0
kTn (t) f k =
(20.44)
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1 ≤ (2n − 1)!
Z
t
(t − s)
0
p(2n−1)
ds
1/p Z
t 0
n
q
kC (s) A f k ds
251
1/q
1
t2n−1+ p 1 kkC (s) An f kkLq (0,t) . (2n − 1)! (p (2n − 1) + 1)1/p
= That is
(20.45)
1
kTn (t) f k ≤ Consequently ν
kTn (t) f k ≤ and thus
1 t2n−1+ p · kkC (s) An f kkLq (0,a) . (2n − 1)! (p (2n − 1) + 1)1/p
1 tν (2n−1+ p ) 1 ν · kkC (s) An f kkLq (0,a) , ((2n − 1)!)ν (p (2n − 1) + 1)ν/p
Z
a 0
kTn (t) f kν dt ≤
(20.46)
(20.47)
1 ν ((2n − 1)!)
kkC (s) An f kkνLq (0,a) . · (p (2n − 1) + 1)ν/p ν 2n − 1 + p1 + 1 aν (2n−1+ p )+1 1
Therefore
Z
a 0
ν
kTn (t) f k dt
≤
1 (2n − 1)!
a(2n−1+ p + ν ) kkC (s) An f kkLq (0,a) · 1/ν , 1/p ν 2n − 1 + 1p + 1 (p (2n − 1) + 1) 1
proving the claim.
1/ν
(20.48)
1
(20.49)
Next we give the counterpart of the last theorem. Theorem 20.12. Let a < 0, t ∈ [a, 0] . One has 1) kkTn (t) f kkLν (a,0) ≤
(−a)(
1 + ν1 ) 2n−1+ p
kkC (t) An f kkLq (a,0) 1/ν . 1/p (2n − 1)! (p (2n − 1) + 1) ν 2n − 1 + p1 + 1
(20.50)
When ν = q we have
2) kkTn (t) f kkLq (a,0) ≤
a2n kkC (t) An f kkLq (a,0)
(2n − 1)! (p (2n − 1) + 1)1/p (2qn)1/q
.
(20.51)
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When ν = q = p = 2 we obtain 3) kkTn (t) f kkL2 (a,0) ≤ Proof.
By (20.40) we have
a2n kkC (t) An f kkL2 (a,0) . √ √ 2 (2n − 1)! n 4n − 1
(20.52)
Z t
1 2n−1 n
(t − s) C (s) A f ds kTn (t) f k =
(2n − 1)! 0
Z 0
1 2n−1 n
= (t − s) C (s) A f ds
(2n − 1)! t 1 ≤ (2n − 1)!
≤
1 (2n − 1)!
Z
Z
0
(s − t)
t
0
(s − t)
t
p(2n−1)
2n−1
ds
kC (s) An f k ds
1/p Z
0 t
q
kC (s) An f k ds
1/q
) (−t)( 1 kkC (s) An f kkLq (t,0) . (2n − 1)! (p (2n − 1) + 1)1/p 1 2n−1+ p
= That is
kTn (t) f k ≤ Consequently
(−t)(
(2n − 1)! (p (2n − 1) + 1)
and hence
1/p
kkC (t) An f kkLq (a,0) .
1 ν 2n−1+ p ) (−t) (
ν
kTn (t) f k ≤
1 2n−1+ p )
ν
((2n − 1)!)ν (p (2n − 1) + 1)ν/p Z
0 a
kkC (t) An f kkLq (a,0)
(20.53)
(20.54)
(20.55)
1 ν 2n−1+ p )+1 (−a) ( ν kTn (t) f k dt ≤ ν 2n − 1 + p1 + 1
ν
·
kkC (t) An f kkLq (a,0)
((2n − 1)!)ν (p (2n − 1) + 1)ν/p
Consequently it holds Z
0 a
ν
kTn (t) f k dt
≤
(20.56)
1 (2n − 1)!
(−a)(2n−1+ p + ν ) kkC (t) An f kkLq (a,0) · 1/ν , 1/p 1 ν 2n − 1 + p + 1 (p (2n − 1) + 1) 1
proving the claim.
1/ν
.
1
(20.57)
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Next we treat the L1 case Theorem 20.13. Let a > 0, t ∈ [0, a] . One has 1 a(2n−1+ ν ) kkC (t) An f kkL1 (0,a) . 1) kkTn (t) f kkLν (0,a) ≤ 1/ν ((2n − 1)!) (ν (2n − 1) + 1)
(20.58)
When ν = 1 we obtain
2) kkTn (t) f kkL1 (0,a) ≤ Proof.
≤
a2n kkC (t) An f kkL1 (0,a) (2n)!
.
(20.59)
By (20.40) we have
Z t
1 2n−1 n
kTn (t) f k = (t − s) C (s) A f ds
(2n − 1)! 0 Z
1 (2n − 1)!
t
0
(t − s)2n−1 kC (s) An f k ds ≤
t2n−1 (2n − 1)!
Z
a
0
kC (s) An f k ds. (20.60)
That is kTn (t) f k ≤ and kTn (t) f kν ≤ Therefore Z
a 0
t2n−1 kkC (s) An f kkL1 (0,a) . (2n − 1)!
(20.61)
tν(2n−1) ν n ν kkC (s) A f kkL1 (0,a) . ((2n − 1)!)
(20.62)
ν
kTn (t) f kν dt ≤
a(ν(2n−1)+1) kkC (t) An f kkL1 (0,a) ν
((2n − 1)!) (ν (2n − 1) + 1)
.
Finally we find 1 Z a 1/ν a(2n−1+ ν ) kkC (t) An f kkL1 (0,a) ν ≤ kTn (t) f k dt , 1/ν 0 ((2n − 1)!) (ν (2n − 1) + 1)
proving the claim.
(20.63)
(20.64)
Next we give the countercase of the last result. Theorem 20.14. Let a < 0, t ∈ [a, 0] . One has 2n−1+ ν1 ) (−a)( kkC (t) An f kkL1 (a,0) 1) kkTn (t) f kkLν (a,0) ≤ . 1/ν ((2n − 1)!) (ν (2n − 1) + 1)
(20.65)
When ν = 1 we derive
2) kkTn (t) f kkL1 (a,0) ≤
a2n kkC (t) An f kkL1 (a,0) (2n)!
.
(20.66)
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By (20.40) we have
Proof.
Z 0
1 2n−1 n
kTn (t) f k = (t − s) C (s) A f ds
(2n − 1)! t 1 (2n − 1)!
≤
Z
0
t
(s − t)
2n−1
≤ That is we see that
(−t) (2n − 1)!
Z
0 a
2n−1
kC (s) An f k ds
kC (s) An f k ds.
(20.67)
2n−1
kTn (t) f k ≤ and
Consequently we obtain 0 a
ν
kTn (t) f k dt ≤
(−t) ν n ν kkC (t) A f kkL1 (a,0) . ((2n − 1)!)
(−a)(ν(2n−1)+1) kkC (t) An f kkνL1 (a,0) ν
((2n − 1)!) (ν (2n − 1) + 1)
Finally we observe that Z
0 a
ν
kTn (t) f k dt
(20.68)
ν(2n−1)
ν
kTn (t) f k ≤
Z
(−t) kkC (t) An f kkL1 (a,0) (2n − 1)!
1/ν
≤
(−a)(
2n−1+ ν1 )
.
(20.70)
kkC (t) An f kkL1 (a,0)
((2n − 1)!) (ν (2n − 1) + 1)
proving the claim.
(20.69)
1/ν
,
(20.71)
Next we continue with Poincar´e like inequalities involving the Sine operator function. Theorem 20.15. Let a > 0, t ∈ [0, a] . One has 1 1 a(2n+ p + ν ) kkC (t) An f kkLq (0,a) 1) kkMn (t) f kkLν (0,a) ≤ 1/ν . 1/p 1 ((2n)!) (2pn + 1) ν 2n + p + 1
(20.72)
When ν = q we have
2) kkMn (t) f kkLq (0,a) ≤
a2n kkC (t) An f kkLq (0,a)
((2n)!) (2pn + 1)1/p (q (2n + 1))1/q
.
(20.73)
When ν = q = p = 2 we obtain 3) kkMn (t) f kkL2 (0,a) ≤
a2n kkC (t) An f kkL2 (0,a) p . √ ((2n)!) 4n + 1 2 (2n + 1)
(20.74)
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Proof.
255
By (20.41) we have
Z t
1 2n n
(t − s) C (s) A f ds kMn (t) f k =
(2n)! 0 1 ≤ (2n)!
1 (2n)!
≤
Z
Z
t
(t − s)
0
t
(t − s)
0
2pn
ds
2n
kC (s) An f k ds
1/p
kkC (t) An f kkLq (0,a)
1
=
t2n+ p 1/p
(2n)! (2pn + 1)
kkC (t) An f kkLq (0,a) .
(20.75)
Consequently tν (2n+ p ) 1
ν
kMn (t) f k ≤
ν
ν
((2n)!) (2pn + 1)
ν/p
kkC (t) An f kkLq (0,a) ,
(20.76)
and Z
a 0
ν
kkC (t) A f kkLq (0,a) a(ν (2n+ p )+1) , · kMn (t) f k dt ≤ ν ν/p ((2n)!) ν 2n + p1 + 1 (2pn + 1) n
1
ν
proving the claim.
(20.77)
Next we give the counterpart of the last theorem. Theorem 20.16. Let a < 0, t ∈ [a, 0] . One has 1 1 (−a)(2n+ p + ν ) kkC (t) An f kkLq (a,0) 1) kkMn (t) f kkLν (a,0) ≤ 1/ν . 1/p 1 ((2n)!) (2pn + 1) ν 2n + p + 1
(20.78)
When ν = q we have
2) kkMn (t) f kkLq (a,0) ≤
a2n kkC (t) An f kkLq (a,0)
((2n)!) (2pn + 1)1/p (q (2n + 1))1/q
.
(20.79)
When ν = q = p = 2 we get 3) kkMn (t) f kkL2 (a,0) ≤ Proof.
By (20.41) we have kMn (t) f k = ≤
1 (2n)!
a2n kkC (t) An f kkL2 (a,0) p . √ ((2n)!) 4n + 1 2 (2n + 1)
Z 0
1 2n n
(s − t) C (s) A f ds
(2n)! t
Z
0
t
(s − t)2n kC (s) An f k ds
(20.80)
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256
1 ≤ (2n)!
Z
0
(s − t)
t
2pn
ds
1/p
kkC (t) An f kkLq (a,0)
2n+ 1
=
p (−t) 1 kkC (t) An f kkLq (a,0) . 1/p (2n)! (2pn + 1)
(20.81)
Therefore ν
kMn (t) f k ≤
ν 2n+ p1 ) (−t) ( ν
ν
((2n)!) (2pn + 1)
ν/p
kkC (t) An f kkLq (a,0) ,
(20.82)
and Z
0 a
1 ν n (−a)(ν (2n+ p )+1) kkC (t) A f kkLq (a,0) · kMn (t) f k dt ≤ , ν ν/p ((2n)!) (2pn + 1) ν 2n + p1 + 1
ν
proving the claim.
(20.83)
It follows the corresponding L1 treatment Theorem 20.17. Let a > 0, t ∈ [0, a] . One has a(2n+ ν ) kkC (t) An f kkL1 (0,a) 1
1) kkMn (t) f kkLν (0,a) ≤
((2n)!) (2νn + 1)1/ν
.
(20.84)
When ν = 1 we find 2) kkMn (t) f kkL1 (0,a) ≤ Proof.
a(2n+1) kkC (t) An f kkL1 (0,a) (2n + 1)!
.
(20.85)
By (20.41) and (20.75) we have Z t 1 2n kMn (t) f k ≤ (t − s) kC (s) An f k ds (2n)! 0 ≤
t2n kkC (t) An f kkL1 (0,a) . (2n)!
(20.86)
Hence ν
kMn (t) f k ≤
t2νn ν n ν kkC (t) A f kkL1 (0,a) ((2n)!)
(20.87)
and Z
a 0
proving the claim.
ν
ν
kMn (t) f k dt ≤
a(2νn+1) kkC (t) An f kkL1 (0,a) ν
((2n)!) (2νn + 1)
,
(20.88)
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257
The counterpart of last theorem follows. Theorem 20.18. Let a < 0, t ∈ [a, 0] . One has 2n+ ν1 ) (−a)( kkC (t) An f kkL1 (a,0) 1) kkMn (t) f kkLν (a,0) ≤ . 1/ν ((2n)!) (2νn + 1)
(20.89)
When ν = 1 we derive 2) kkMn (t) f kkL1 (a,0) ≤ Proof.
(−a)(2n+1) kkC (t) An f kkL1 (a,0) (2n + 1)!
.
(20.90)
By (20.41) and (20.81) we have Z 0 1 2n kMn (t) f k ≤ (s − t) kC (s) An f k ds (2n)! t 2n
≤
(−t) kkC (t) An f kkL1 (a,0) . (2n)!
(20.91)
Thus ν
kMn (t) f k ≤
(−t)2νn ν kkC (t) An f kkL1 (a,0) ((2n)!)ν
(20.92)
and Z
2νn+1
0 a
ν
kMn (t) f k dt ≤
(−a)
ν
kkC (t) An f kkL1 (a,0)
((2n)!)ν (2νn + 1)
,
proving the claim.
(20.93)
Application 20.19. (see [147], p. 121) Let X be the Banach space of odd, 2π−periodic real functions in the space of d2 bounded uniformly continuous functions from R into itself: BU C (R) . Let A := dx 2 with D (An ) = f ∈ X : f (2k) ∈ X, k = 1, . . . , n , n ∈ N. A generates a Cosine function C ∗ given by 1 C ∗ (t) f (x) = [f (x + t) + f (x − t)] , ∀x, t ∈ R. (20.94) 2 The corresponding Sine function S ∗ is given by Z t Z t 1 ∗ S (t) f (x) = f (x + s) ds + f (x − s) ds , ∀x, t ∈ R. (20.95) 2 0 0 Here we consider f ∈ D (An ) , n ∈ N, as above. By (20.40) we obtain n−1
Tn∗ (t) f :=
X t2k 1 [f (· + t) + f (· − t)] − f (2k) 2 (2k)! k=0
=
Z
t 0
i (t − s)2n−1 h (2n) f (· + s) + f (2n) (· − s) ds, ∀t ∈ R. 2 (2n − 1)!
(20.96)
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By (20.41) we get X Z t Z t n 1 t2k−1 f (· + s) ds + f (· − s) ds − Mn∗ (t) f := f (2(k−1)) 2 0 (2k − 1)! 0 k=1
=
Z
t 0
2n i (t − s) h (2n) f (· + s) + f (2n) (· − s) ds, ∀t ∈ R. 2 (2n)!
(20.97)
Let g ∈ BU C (R) , we define kgk = kgk∞ := sup |g (x)| < ∞. x∈R
Notice also that
kkC ∗ (s) An f k∞ k∞ =
C ∗ (s) f (2n)
∞ ∞
1
(2n)
(· + s) + f (2n) (· − s)
f 2 ∞ ∞
1
(2n)
(· + s) +
f (2n) (· − s) ≤ ≤ Θ < ∞,
f 2 ∞ ∞ ∞ ∞ =
where
Θ := f (2n)
∞
.
(20.98)
(20.99)
Let a > 0, t ∈ [0, a] . By (20.42) we derive
a(2n−1+ p + ν ) 1
kkTn∗
(t) f kkLν (0,a) ≤
Also by (20.72) we get
(2n − 1)! (p (2n − 1) + 1)
1/p
∗
C (t) f (2n)
Lq (0,a) · 1/ν . ν 2n − 1 + p1 + 1
(20.100)
a(2n+ p + ν ) 1
kkMn∗ (t) f kkLν (0,a) ≤
By (20.44) we find
1
1
((2n)!) (2pn + 1)
1/p
∗
C (t) f (2n)
Lq (0,a) · 1/ν . ν 2n + 1p + 1
kkTn∗ (t) f kkL2 (0,a) ≤
(20.101)
a2n √ √ 2 (2n − 1)! n 4n − 1
·
C ∗ (t) f (2n)
L2 (0,a)
.
(20.102)
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By (20.74) we have kkMn∗ (t) f kkL2 (0,a) ≤
a2n √ ((2n)!) 4n + 1
∗
C (t) f (2n)
L2 (0,a) p . · 2 (2n + 1)
(20.103)
Also by (20.58) we obtain kkTn∗ (t) f kkLν (0,a)
≤
1 a(2n−1+ ν )
C ∗ (t) f (2n)
L1 (0,a) ((2n − 1)!) (ν (2n − 1) + 1)
and by (20.84) we observe that kkMn∗
(t) f kkLν (0,a) ≤
1/ν
1
a(2n+ ν )
C ∗ (t) f (2n)
L1 (0,a) ((2n)!) (2νn + 1)1/ν
,
.
(20.104)
(20.105)
Let now a < 0, t ∈ [a, 0] . Then one by (20.50) has kkTn∗
(t) f kkLν (a,0) ≤
and by (20.78) we get
(−a)(
(2n − 1)! (p (2n − 1) + 1)
1/p
∗
C (t) f (2n)
Lq (a,0) · 1/ν . ν 2n − 1 + p1 + 1
kkMn∗ (t) f kkLν (a,0)
By (20.66) we find
2n−1+ p1 + ν1 )
≤
(20.106)
1 1 (−a)(2n+ p + ν )
((2n)!) (2pn + 1)
1/p
∗
C (t) f (2n)
Lq (a,0) · 1/ν . ν 2n + 1p + 1
kkTn∗ (t) f kkL1 (a,0)
≤
and by (20.90) we see that kkMn∗ (t) f kkL1 (a,0) ≤
(−a)
(20.107)
a2n
C ∗ (t) f (2n)
L
(2n+1)
1 (a,0)
(2n)!
∗
C (t) f (2n)
(2n + 1)!
L1 (a,0)
,
(20.108)
.
(20.109)
Similarly one can apply the rest of the results here on Cosine and Sine Operator functions.
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Chapter 21
Hardy Opial Type Inequalities
Various Lp form Hardy–Opial type sharp integral inequalities are presented involving two functions. This chapter follows [32].
21.1
Results
Let f ∈ Lp ([a, b]) and g ∈ Lq ([a, b]), with p, q be such that 1p + 1q = 1. We consider the generalized Hardy type operators (for the basic Hardy operator see [204], p. 306), Tg f (t) =
Z
t
g(s)f (s)ds, a
and Tg∗ f (t) =
Z
b
g(s)f (s)ds. t
We present here integral Hardy–Opial type sharp Lp -inequalities involving or related R R to Tg , TgR∗ . Here L stands for Lebesque integral, R for Riemann integral and (R − S) stands for Riemann–Stieltjes integral. The first result follows. Theorem 21.1. Let p, q > 1 such that x0 ∈ [a, b] be fixed. Then Z s Z w |g(t)f (t)|dt |f (w)|dw L L x0
≤ 2
x0
−1/p
Z s Z L R x0
w x0
1 p
+ q1 = 1, with f ∈ Lp ([a, b]), g ∈ Lq ([a, b]),
!1/q Z L |g(t)|q dt dw
s x0
p
|f (t)| dt
2/p
,
all x0 ≤ s ≤ b. (21.1)
Inequality (21.1) is sharp, that is attained when f (t) = g(t) = c > 0 a constant, for all t ∈ [x0 , b]. 261
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1) For x0 ≤ w ≤ b, by H¨ older’s inequality we have Z w 1/q Z w 1/p Z w L |g(t)f (t)|dt ≤ L |g(t)|q dt L |f (t)|p dt
Proof.
x0
Z = L
where
z(w) := L
Z
w x0
x0
w
x0
|g(t)|q dt
1/q
x0
(z(w))1/p ,
|f (t)|p dt ≥ 0,
z(x0 ) = 0.
(21.2)
(21.3)
Here z is an absolutely continuous function. Hence we have a.e. that z 0 (w) = |f (w)|p ≥ 0,
(21.4)
|f (w)| = (z 0 (w))1/p , a.e.
(21.5)
and
Therefore it holds Z w Z L |g(t)f (t)|dt |f (w)| ≤ L x0
w x0
q
|g(t)| dt
1/q
(z(w)z 0 (w))1/p ,
(21.6)
a.e. on [x0 , b]. Hence by integrating (21.6) over [x0 , s], where s ∈ [x0 , b], we obtain Z s Z w L L |g(t)f (t)|dt |f (w)|dw x0
x0
Z s Z ≤ L L x0
≤ =
w
|g(t)|q dt
x0
1/q
(z(w)z 0 (w))1/p dw
(21.7)
(using again H¨ older’s inequality) 1/q Z s 1/p Z s Z w q 0 R L |g(t)| dt dw L z(w)z (w)dw
x0
x0
Z s Z R L x0
w
x0
q
|g(t)| dt dw
(21.8)
x0
1/q
(21.9)
1/q 2 1/p z (s) |g(t)|q dt dw 2 x0 x0 x0 (21.10) 2/p 1/q Z s Z s Z w |f (t)|p dt |g(t)|q dt dw L . (21.11) L = 2−1/p R
(R − S)
Z
s
x0
z(w)dz(w)
1/p
=
R
Z s Z L
x0
w
x0
That is proving inequality (21.1). 2) The sharpness follows. We see that
L.H.S.(21.1) = R.H.S.(21.1) = establishing attainability of (21.1).
c3 (s − x0 )2 , 2
(21.12)
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We also give Theorem 21.2. Let p, q > 1 such that p1 + q1 = 1, with f ∈ Lp ([a, b]), g ∈ Lq ([a, b]), x0 ∈ [a, b] be fixed. Then Z x0 Z x0 |g(t)f (t)|dt |f (w)|dw L L w
s
≤2
−1/p
R
Z
s
x0
L
Z
x0 w
!1/q Z |g(t)| dt dw L q
x0 s
|f (t)|p dt
all a ≤ s ≤ x0 .
2/p
,
(21.13)
Inequality (21.13) is sharp, that is attained when f (t) = g(t) = c > 0 a constant, for all t ∈ [a, x0 ]. 1) For a ≤ w ≤ x0 , by H¨ older’s inequality we have
Proof.
L
Z
Z L
x0 w
|g(t)f (t)|dt ≤
= where z(w) := L
Z
That is z(w) = L
L
x0 w
Z
w Z x0 w
|g(t)|q dt q
|g(t)| dt
1/q Z L 1/q
|f (t)|p dt − L
Z
x0 w
|f (t)|p dt
1/p
(z(w))1/p ,
|f (t)|p dt ≥ 0,
x0 a
x0
w a
(21.14) (21.15)
z(x0 ) = 0.
(21.16)
|f (t)|p dt.
(21.17)
Here z is an absolutely continuous function. Hence we have a.e. that z 0 (w) = −|f (w)|p ≤ 0,
(21.18)
−z 0 (w) = |f (w)|p ≥ 0,
(21.19)
|f (w)| = (−z 0 (w))1/p .
(21.20)
and a.e. that
and a.e. that
Therefore it holds Z x0 Z L |g(t)f (t)|dt |f (w)| ≤ L w
x0 w
|g(t)|q dt
1/q
(z(w)(−z 0 (w))1/p ,
(21.21)
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a.e. on [a, x0 ]. Integrating inequality (21.21) over [s, x0 ], where s ∈ [a, x0 ], we derive Z x0 Z x0 L |g(t)f (t)|dt |f (w)|dw L s
≤L
Z
s
x0
Z ≤ L =
R
s
Z
Z
w
L
Z
x0
|g(t)|q dt
w
1/q
z(w)(−z 0 (w))
1/p
R
(21.22)
(again using H¨ older’s inequality) 1/p 1/q Z x0 Z x0 x0 z(w)(−z 0 (w))dw |g(t)|q dt dw L L
s
x0
x0
s
w
L
Z
Z
x0
w x0
1/q Z |g(t)|q dt dw −(R − S)
1/q
1/p
2
z (s) 2 w s 1/q Z Z x0 Z x0 |g(t)|q dt dw L L = 2−1/p R =
dw
L
|g(t)|q dt dw
w
s
x0
z(w)dz(w) s
establishing attainability of (21.13).
1/p (21.24) (21.25)
x0
|f (t)|p dt
s
That is proving (21.13). 2) The sharpness follows. We see that
L.H.S.(21.13) = R.H.S.(21.13) =
(21.23)
2/p
.
(21.26)
c3 (x0 − s)2 , 2
(21.27)
We have Corollary 21.3. Let p, q > 1 such that Then " kf gk1 kf k1 ≤ 2−1/p kf k2p Z b Z + R L a
b w
1 1 p+q
= 1, with f ∈ Lp ([a, b]), g ∈ Lq ([a, b]).
Z b Z R L a
|g(t)|q dt dw
!1/q #
w a
|g(t)|q dt dw
!1/q
≤ (b − a)1/q kf k2p kgkq .
(21.28)
Inequality is sharp, attained when f = g = c > 0. Proof.
We apply Theorem 21.1 for x0 = a and s = b. We have Z b Z w L L |g(t)f (t)|dt |f (w)|dw a
≤ 2
a
−1/p
Z b Z R L a
w a
!1/q Z |g(t)|q dt dw L
b a
p
|f (t)| dt
We also apply Theorem 21.2 for x0 = b, s = a. We find Z b Z b L L |g(t)f (t)|dt |f (w)|dw a
≤ 2
−1/p
w
Z b Z R L a
b w
!1/q Z |g(t)| dt dw L q
b a
|f (t)|p dt
2/p
2/p
.
(21.29)
.
(21.30)
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We add (21.29) and (21.30) to observe that Z b Z w Z b L L |g(t)f (t)|dt + L |g(t)f (t)|dt |f (w)|dw a
≤ 2 +
a
−1/p
kf k2p
w
"
a
Z b Z R L a
Z b Z R L
b w
w
|g(t)| dt dw
a
q
q
|g(t)| dt dw
!1/q #
!1/q
.
(21.31)
But it holds L.H.S.(21.31) = kf gk1 kf k1 , proving first inequality in (21.28). Since 1/q < 1 we get R.H.S.(21.31) ≤ 2
−1/p
1 kf k2p 21− q
Z b Z +R L a
= kf k2p
"
b w
"
R
Z b Z L a
|g(t)|q dt dw
Z b Z L R a
b a
(21.32)
w a
#1/q
|g(t)|q dt dw
|g(t)|q dt dw
(21.33) #1/q
= kf k2p kgkq (b − a)1/q ,
(21.34) (21.35)
proving second part of (21.28). Sharpness is obvious, by all three parts being equal to c3 (b − a)2 , when f = g = c > 0. Next comes Corollary 21.4. Let f, g ∈ L2 ([0, 1]). Then Z s Z w L L |g(t)f (t)|dt |f (w)|dw 0
0
√ !1/2 Z s Z s Z w 2 2 ≤ R L g (t)dt dw L f 2 (t)dt , all 0 ≤ s ≤ 1.(21.36) 2 0 0 0
Inequality (21.36) is sharp, that is attained when f (t) = g(t) = 1, all t ∈ [0, 1]. Proof.
Apply Theorem 21.1 for p = q = 2 on [0, 1], and x0 = 0.
We obtain the following Opial type (see [75], [6], p. 8) basic inequality which implies Opial’s inequality [195]. Corollary 21.5. Let f ∈ C 1 ([0, 1]) such that f (0) = 0. Then Z Z s s s 0 (f (t))2 dt, ∀s ∈ [0, 1]. |f (t)f 0 (t)|dt ≤ 2 0 0
(21.37)
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Inequality (21.37) is sharp, namely it is attained by f (t) = t. Proof. We apply (21.36) for g(t) = 1, ∀t ∈ [0, 1], and in the place of f we plug in f 0 . We then have by (21.36) that Z s |f (t)f 0 (t)|dt 0 Z s Z w Z s Z w 0 0 0 |f (w)|dw ≤ = f (t)dt |f (t)|dt |f 0 (w)|dw (21.38) 0
0
0
0
√ Z s 1/2 Z s 2 w dw (f 0 (t))2 dt ≤ 2 0 0 Z s s 0 (f (t))2 dt. = 2 0
(21.39)
(21.40)
That is proving (21.37). Sharpness now is obvious.
The extreme case follows. Proposition 21.6. Let f ∈ L∞ ([a, b]), g ∈ L1 ([a, b]), x0 ∈ [a, b] be fixed. Then i) ! Z s Z w Z s Z w 2 L L |g(t)f (t)|dt |f (w)|dw ≤ kf k∞ R L |g(t)|dt dw , x0
x0
x0
x0
(21.41) for all s ∈ [x0 , b]. Furthermore ii) Z x0 Z x0 Z 2 L L |g(t)f (t)|dt |f (w)|dw ≤ kf k∞ R s
w
s
x0
L
Z
x0 w
!
|g(t)|dt dw , (21.42)
for all s ∈ [a, x0 ]. Both (21.41), (21.42) are sharp, attained by f (t) = g(t) = c > 0. Proof.
Obvious.
To complete the chapter we present Theorem 21.7. Let 0 < q < 1 and p < 0 such that 1p + 1q = 1, a fixed x0 ∈ [a, b] and s ∈ (x0 , b]. Assume that g ∈ Lq ([a, b]), f ∈ Lp ([a, b]), f is nowhere zero and |f | ≤ k a.e., k > 0. Then Z s Z w |g(t)f (t)|dt |f (w)|dw L L x0
≥ 2
−1/p
x0
Z s Z L R x0
w x0
!1/q Z |g(t)| dt dw L q
s x0
|f (t)|p dt
2/p
.
(21.43)
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Inequality (21.43) is sharp, namely it is attained when f (t) = g(t) = c > 0, for all t ∈ [x0 , b]. Proof. Let x0 < w ≤ s. Here |f | > 0 and |f | ≤ k a.e., k > 0. Hence −p > 0 and k p ≤ |f |p a.e. Therefore Z w z(w) := L |f (x)|p dx ≥ k p (w − x0 ) > 0, z(x0 ) = 0,
1 k
≤
1 |f |
a.e.,
(21.44)
x0
and when w2 > w1 we derive Z
z(w2 ) − z(w1 ) = L
w2 w1
|f |p dx ≥ k p (w2 − w1 ) > 0.
(21.45)
That is proving that z(w) is strictly increasing on [x0 , s], and z also is an absolutely continuous function there. Next, by reverse H¨ older’s inequality we obtain L
Z
w x0
Z |g(t)f (t)|dt ≥ L
Z = L
w x0 w x0
1/q Z |g(t)| dt L q
q
|g(t)| dt
1/q
w x0
p
|f (t)| dt
1/p
(21.46)
(z(w))1/p , for x0 < w ≤ s. (21.47)
It holds a.e. on [x0 , s] that z 0 (w) = |f (w)|p > 0 and |f (w)| = (z 0 (w))1/p , a.e. on [x0 , s].
(21.48)
Therefore Z L
w x0
Z ≥ L
|g(t)f (t)|dt |f (w)| w
x0
|g(t)|q dt
1/q
(z(w))1/p (z 0 (w))1/p ,
a.e. on (x0 , s].
(21.49)
Take now x0 < θ ≤ w ≤ s and θ < s. We observe that Z s Z w Z s Z w |g(t)f (t)|dt |f (w)|dw |g(t)f (t)|dt |f (w)|dw = lim L L L L θ↓x0 θ x0 x x0 # " Z0 Z w
s
≥ lim L θ↓x0
θ
L
x0
1/q
|g(t)|q dt
(z(w)z 0 (w))1/p dw
(again by reverse H¨ older’s inequality)
(21.50)
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Z s Z L ≥ lim R θ↓x0
R
=
R
=
= 2
θ
Z s Z L x0
Z s Z L x0
−1/p
w x0 w x0
w x0
= 2
|g(t)| dt dw
|g(t)|q dt dw
|g(t)|q dt dw
Z s Z R L x0
−1/p
q
Z s Z R L x0
w x0 w x0
q
!1/q
!1/q
!1/q
Z lim L
θ↓x0
s
z(w)z 0 (w)dw
θ
Z lim (R − S)
θ↓x0
lim
θ↓x0
|g(t)| dt dw
!1/q
1/p (21.51)
s
z(w)dz(w) θ
z 2 (s) − z 2 (θ) 2
1/p
1/p (21.52)
(z(s))2/p
!1/q Z |g(t)|q dt dw L
s x0
p
|f (t)| dt
2/p
,
proving (21.43). Sharpness same as in Theorem 21.1.
(21.53)
The counterpart of Theorem 21.7 follows. Theorem 21.8. Let 0 < q < 1 and p < 0 such that p1 + 1q = 1, a fixed x0 ∈ [a, b] and s ∈ [a, x0 ). Assume that g ∈ Lq ([a, b]), f ∈ Lp ([a, b]), f is nowhere zero and |f | ≤ k a.e., k > 0. Then Z x0 Z x0 L L |g(t)f (t)|dt |f (w)|dw s
≥ 2
w
−1/p
R
Z
s
x0
L
Z
x0 w
!1/q Z L |g(t)| dt dw q
x0 s
|f (t)|p dt
2/p
. (21.54)
Inequality (21.54) is sharp, that is attained when f (t) = g(t) = c > 0 a constant, for all t ∈ [a, x0 ]. Proof.
Let s ≤ w < x0 . Here again k p ≤ |f |p a.e. Therefore Z x0 z(w) := L |f (t)|p dt ≥ k p (x0 − w) > 0, z(x0 ) = 0,
(21.55)
w
and when w2 < w1 we find
z(w2 ) − z(w1 ) = L
Z
w1 w2
|f (t)|p dt ≥ k p (w1 − w2 ) > 0.
(21.56)
That is proving that z(w) is strictly decreasing on [s, x0 ], and z also is an absolutely continuous function there. Next, by reverse H¨ older’s inequality we obtain Z x0 1/q Z x0 1/p Z x0 q p L |g(t)f (t)|dt ≥ L |g(t)| dt L |f (t)| dt w
Z = L
w
x0
w
|g(t)|q dt
1/q
w
(z(w))1/p , for s ≤ w < x0 .
(21.57)
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z 0 (w) = −|f (w)|p < 0,
(21.58)
−z 0 (w) = |f (w)|p > 0,
(21.59)
|f (w)| = (−z 0 (w))1/p ,
(21.60)
It holds a.e. on [s, x0 ] that
i.e.
and
a.e. on [s, x0 ]. Therefore Z Z x0 |g(t)f (t)|dt |f (w)| ≥ L L w
x0 w
|g(t)|q dt
a.e. on [s, x0 ).
1/q
z(w)(−z 0 (w))
x0
θ
θ↑x0
L
s
" Z Z θ L ≥ lim L θ↑x0
s
|g(t)f (t)|dt |f (w)|dw
w
x0
q
|g(t)| dt
w
1/q
0
z(w)(−z (w)
1/p
(again by reverse H¨ older’s inequality) Z !1/q Z θ Z x0 q lim L |g(t)| dt dw ≥ lim R L θ↑x0
=
=
R
s
x0 Z L
Z x0 Z L R s
= 2−1/p R =2
−1/p
R
Z Z
s
s
x0
q
x0
x0
q
|g(t)| dt dw
w
x0
|g(t)| dt dw
w
L L
that is proving (21.54). Sharpness is obvious. We give
θ↑x0
w
s
Z
,
(21.61)
Take now s ≤ w ≤ θ < x0 and s < θ. We see that Z x0 Z x0 L |g(t)f (t)|dt |f (w)|dw L w s ! Z Z = lim L
1/p
Z Z
x0 w x0 w
!1/q
!1/q
0
s
Z
θ s
z 2 (s) − z 2 (θ) lim θ↑x0 2 !1/q
1/p z(w)dz(w) (21.64)
1/p
(21.65)
(z(s))2/p
!1/q Z |g(t)| dt dw L
(21.62)
1/p z(w) −z (w) dw (21.63)
θ
lim −(R − S)
θ↑x0
|g(t)|q dt dw q
dw
#
x0 s
|f (t)|p dt
(21.66) 2/p
,
(21.67)
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Corollary 21.9. Let p, q > 1 such that p1 + be fixed. Then Z s Z w t e |f (t)|dt |f (w)|dw L L
1 q
= 1, with f ∈ Lp ([a, b]), x0 ∈ [a, b]
x0
x0
1/q Z s 2/p −1/p 1 1 qs qx0 qx0 p ≤2 (e − e ) − e (s − x0 ) L |f (t)| dt , q q x0
(21.68)
for all x0 ≤ s ≤ b. Proof.
By (21.1) we have for all x0 ≤ s ≤ b that Z s Z w L et |f (t)|dt |f (w)|dw L x0
≤ 2−1/p =2
−1/p
proving (21.68).
x0
Z s Z x0
w
x0
1/q Z eqt dt dw L
s
x0
|f (t)|p dt
2/p
(21.69)
1/q Z s 2/p 1 1 qs qx0 qx0 p (e − e ) − e (s − x0 ) L |f (t)| dt ,(21.70) q q x0
We finish the chapter with Corollary 21.10. Let p, q > 1 such that p1 + 1q = 1, with g ∈ Lq ([a, b]), here a > 0, and let r 6= 0. Then Z w Z s 1 |g(t)| L dt dw L r tr a w a 1−rp 2/p Z s Z w 1/q s − a1−rp ≤ 2−1/p R L |g(t)|q dt dw , 1 − rp a a (21.71) for all a ≤ s ≤ b. Proof.
From (21.1) for f (t) = t1r , we obtain Z w Z s 1 |g(t)| dt dw L L r tr a w a !1/q Z Z s Z w ≤ 2−1/p R |g(t)|q dt dw L a
a
all a ≤ s ≤ b. But
proving (21.71).
s a
1 trp
dt
2/p
, (21.72)
Z
s a
1 t
dt = rp
s1−rp − a1−rp , 1 − rp
(21.73)
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Chapter 22
A Basic Sharp Integral Inequality
A sharp multidimensional integral type inequality is presented involving n-th order (n ∈ N) mixed partial derivatives. This is subject to some basic boundary condition satisfied by the involved multivariate function. This chapter is based on [60].
22.1
Introduction
This chapter is motivated and inspired by [16] and [123] about Ostrowski type inequalities; see also Ostrowski’s paper [196], 1938. Though the presented results are quite different from Ostrowski type inequalities, the working spirit is the same. T. Apostol’s book [62] was used as a reference for some basic facts related to integration and differentiation.
22.2
Results
We present Theorem 22.1. Let f ∈ C n (B), n ∈ N, where B = [a1 , b1 ] × · · · × [an , bn ], aj , bj ∈ R, with aj < bj , j = 1, ..., n. Denote by ∂B the boundary of the box B. Assume that f (x) = 0, for all x = (x1 , ..., xn ) ∈ ∂B (in other words we suppose that f (· · ·, aj , · · ·) = f (· · ·, bj , · · ·) = 0, for all j = 1, ..., n). Then Z
m (B) |f (x1 , ..., xn )| dx1 · · · dxn ≤ 2n B
where m(B) = measure) of B.
Qn
j=1 (bj
Z n ∂ f (x1 , ..., xn ) ∂x1 · · · ∂xn dx1 · · · dxn ,
(22.1)
B
− aj ) is the n-th dimensional volume (i.e. the Lebesgue
Theorem 22.2. Inequality (22.1) is sharp (in the sense that equality can be asymptotically attained). Proof of Theorem 22.1. Let (x1 , ..., xn ) ∈ B, i.e. aj ≤ xj ≤ bj , for all j = 1, ..., n. 271
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The assumptions give f (x1 , ..., xn ) = and
Z
x1
···
a1
f (x1 , ..., xn ) = (−1)n
Z
Z
xn an
b1 x1
∂ n f (s1 , ..., sn ) ds1 · · · dsn , ∂s1 · · · ∂sn
···
Z
bn xn
More generally, if we introduce the intervals Ij,0 = [aj , xj ]
and
we have f (x1 , ..., xn ) = (−1)ε1 +···+εn
Z
∂ n f (s1 , ..., sn ) ds1 · · · dsn . ∂s1 · · · ∂sn
Ij,1 = [xj , bj ],
I1,ε1
···
Z
In,εn
j = 1, ..., n,
∂ n f (s1 , ..., sn ) ds1 · · · dsn , ∂s1 · · · ∂sn
(22.2)
where each εj can be either 0 or 1. Adding up (22.2) for all 2n choices of (ε1 , ..., εn ) we derive Z Z X ∂ n f (s1 , ..., sn ) 2n f (x1 , ..., xn ) = ds1 · · · dsn . (−1)ε1 +···+εn ··· ∂s1 · · · ∂sn I1,ε1 In,εn ε ,...,ε 1
n
(22.3) Next by taking absolute values in (22.3) and using basic properties of integrals (noting that the 2n “sub-boxes” I1,ε1 × · · · × In,εn form a partition of B) and the subadditivity property of the absolute value we find that Z n ∂ f (x1 , ..., xn ) 1 dx1 · · · dxn , (22.4) |f (x1 , ..., xn )| ≤ n 2 B ∂x1 · · · ∂xn
true for all (x1 , ..., xn ) ∈ B. Inequality (22.4) has by itself its merits. Finally by integrating (22.4) over B we establish the result.
Proof of Theorem 22.2. Without loss of generality, to establish optimality of (22.1) it is enough to prove sharpness of the following inequality Z Z 1 1 1 0 |f (x)| dx (22.5) |f (x)| dx ≤ 2 0 0 (this is obtained from (22.1) by taking n = 1 and [a1 , b1 ] = [0, 1]). Clearly here it is assumed that f ∈ C 1 [0, 1] with f (0) = f (1) = 0. Let 0 < ε < 1. Consider the function 2 2εx−x , 0 ≤ x ≤ ε, ε2 ε < x ≤ 1 − ε, fε (x) = 1, 2ε(1−x)−(1−x)2 , 1 − ε < x ≤ 1. ε2
Clearly fε (x) ≥ 0 on [0, 1], and fε (0) = fε (1) = 0 (in fact fε (x) = fε (1 − x), that is fε (x) is symmetric about x = 1/2). Furthermore 2(ε−x) 0 ≤ x ≤ ε, ε2 , 0 ε < x ≤ 1 − ε, fε (x) = 0, 2(1−x−ε) , 1 − ε < x ≤ 1, ε2
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hence fε0 is continuous on [0, 1]. Next we calculate and find for fε that the right hand side of (22.5) is 1, while the left hand side of (22.5) is 1 − (2/3)ε, i.e. (22.5) becomes
2 1 − ε ≤ 1. 3 Therefore, by letting ε & 0, we see that equality in (22.5) is asymptotically attained. Remark 22.3. Notice that limε&0 fε (x) = 1 (pointwise) for all x ∈ (0, 1). In this sense equality in (22.4) is also asymptotically attainable. Remark 22.4. There is no function other than the trivial f (x1 , ..., xn ) ≡ 0 that makes (22.1) an equality. Indeed, if there were such an f , then, by continuity (22.4) should be an equality for all (x1 , ..., xn ) ∈ B, thus f should be constant which is impossible due to the boundary condition, unless, of course f ≡ 0. Remark 22.5. The function f of Theorem 22.1 can be complex valued. Remark 22.6. The case n = 2 is quite interesting: Let f ∈ C 2 ([a, b]×[c, d]), a < b, c < d, with f (a, ·) = f (b, ·) = f (·, c) = f (·, d) = 0. Then (22.1) becomes Z Z Z dZ b (b − a)(d − c) d b |∂xy [f (x, y)]| dxdy. |f (x, y)| dxdy ≤ 4 c a c a
Notice that the operator 2∂xy becomes the wave operator ∂yy − ∂xx after a 45◦ rotation of the axes x and y. Remark 22.7. It is a curious fact that inequality (22.1) fails badly in a variety of domains. For example, given a, b > 0, let Dk ⊂ R2 , k = 1, 2, 3, ... , be the domain x2k y 2k Dk = (x, y) ∈ R2 : 2k + 2k ≤ 1 . a b
Notice that Dk ⊂ B = [−a, a] × [−b, b] and we can make Dk as close to B as we wish, by taking k sufficiently large. However, if fk is the polynomial y 2k x2k + − 1, a2k b2k then, of course, fk ∈ C 2 (Dk ) and fk = 0 on ∂Dk , but fk (x, y) =
∂ 2 fk (x, y) ≡ 0, ∂x∂y
hence (22.1) fails completely if Dk is the domain of integration! We finish chapter with Theorem 22.8. Let f ∈ C n+2 (B), n ∈ N, B = [a1 , b1 ] × . . . × [an , bn ], aj , bj ∈ R, with aj < bj , j = 1, . . . , n. Assume that ∂ 2f ∂2f (· · · , a , · · · ) = (· · · , bi , · · · ) = 0, i ∂x2j ∂x2j
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for all i, j = 1, · · · , n. Then Z 2 ∇ f (x1 , . . . , xn ) dx1 , dx2 . . . dxn B
(Πni=1 (bi − ai )) ≤ 2n
Proof.
! n Z n+2 X ∂ f (x1 , . . . , xn ) . dx1 . . . dxn 3 B ∂x1 . . . ∂ xi . . . ∂xn i=1
(22.6)
We observe the following n n 2 X ∂ 2 f (x1 , . . . , xn ) X ∂ 2 f (x1 , . . . , xn ) ∇ f (x1 , . . . , xn ) = ≤ (22.7) ∂x2i ∂x2i i=1 i=1 2 ≤ (by (22.4), set instead of f the function ∂∂xf2 i 2 n ∂ f Z n (x1 , . . . , xn ) X 1 ∂ ∂x2i dx1 . . . dxn n 2 ∂x . . . ∂x 1 n B i=1 n Z ∂ n+2 f (x1 , . . . , xn ) 1 X dx1 . . . dxn . = n (22.8) 3 2 i=1 B ∂x1 . . . ∂xi . . . ∂xn
So we get true that
n Z n+2 X 2 ∂ f (x1 , . . . , xn ) ∇ f (x1 , . . . , xn ) ≤ 1 dx . . . dx , 1 n 3 2n i=1 B ∂x1 . . . ∂xi . . . ∂xn
∀ (x1 , . . . , xn ) ∈ B. Integrating (22.9) over B we get (22.6).
(22.9)
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Chapter 23
Estimates of the Remainder in Taylor’s Formula
Estimates of the remainder in Taylor’s formula are given. This chapter relies on [55].
23.1
Introduction
The following theorem is well known as Taylor’s formula or Taylor’s theorem with the integral remainder. Theorem 23.1. Let f : [a, b] → R and n a positive integer. If f is such that f (n) is absolutely continuous on [a, b], x0 ∈ (a, b), then for all x ∈ (a, b) we have f (x) = Tn (f ; x0 , x) + Rn (f ; x0 , x) ,
(23.1)
where Tn (f ; x0 , ·) is Taylor’s polynomial of degree n, i.e., Tn (f ; x0 , x) =
n k X f (k) (x0 ) (x − x0 )
k!
k=0
(note that f (0) = f and 0! = 1) and the remainder can be given by Z 1 x n (x − t) f (n+1) (t) dt. Rn (f ; x0 , x) = n! x0 For a mapping g : [a, b] → R and two arbitrary points x0 , x ∈ (a, b), define
and
Z kgk[x0 ,x];p :=
x x0
p1 |g (t)|p dt , p ≥ 1
kgk[x0 ,x];∞ := ess
sup t∈[x0 ,x] (t∈[x,x0 ])
|g (t)| .
Using H¨ older’s inequality, we may state the following corollary. 275
(23.2)
(23.3)
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Corollary 23.2.With the above assumptions, we have |Rn (f ; x0 , x)|
|x−x0 |n (n+1) f if f (n+1) ∈ L1 [a, b] ; n! [x0 ,x],1 n+ 1
q |x−x0 |
f (n+1) if f (n+1) ∈ Lq [a, b] , ≤ 1 [x0 ,x],p q n!(nq+1) p > 1, 1p + q1 = 1;
n+1 |x−x | 0
f (n+1) if f (n+1) ∈ L∞ [a, b] . (n+1)! [x0 ,x],∞
(23.4)
For some applications of (23.4) for particular functions, see [101].
23.2
Some New Bounds for the Remainder
The following simple result comes from G.A. Anastassiou in [16]. Lemma 23.3. Suppose that the mapping f : [a, b] → R is such that f (n−1) is absolutely continuous on [a, b] and x0 ∈ (a, b). Then for all x ∈ (a, b) the remainder Rn (f ; x0 , x) in (23.1) can be represented by Rn (f ; x0 , x) =
Proof.
1 (n − 1)!
Z
x x0
h
i f (n) (t) − f (n) (x0 ) (x − t)n−1 dt, n ≥ 1.
(23.5)
We apply Taylor’s formula with the integral remainder for n − 1 obtaining
1 f (x) = Tn−1 (f ; x0 , x) + (n − 1)!
Z
x x0
(x − t)
n−1
f (n) (t) dt
Z x (x − x0 )n (n) 1 n−1 (n) = Tn (f ; x0 , x) − (x − t) f (t) dt f (x0 ) + n! (n − 1)! x0 Z xh i 1 = Tn (f ; x0 , x) + f (n) (t) − f (n) (x0 ) (x − t)n−1 dt, (n − 1)! x0
which produces the representation (23.5).
The following theorem holds. Theorem 23.4. Suppose that f , x0 and x are as in Lemma 23.3. Then we have
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the bounds |Rn (f ; x0 , x)|
|x−x0 |n−1 (n) f − f (n) (x0 ) [x0 ,x],1 if f (n) ∈ L1 [a, b] ; n! n−1+ 1
(n)
q |x−x0 |
f − f (n) (x0 ) ≤ if f (n) ∈ Lq [a, b] , 1 [x0 ,x],p q (n−1)![(n−1)q+1] p > 1, p1 + q1 = 1;
n |x−x | 0 (n) (n)
f − f (x0 ) if f (n) ∈ L∞ [a, b] . n! [x ,x],∞ 0
Proof.
(23.6)
We have
Z x h i 1 n−1 (n) (n) |Rn (f ; x0 , x)| ≤ f (t) − f (x0 ) (x − t) dt (n − 1)! x0 Z x 1 (n) n−1 (n) ≤ dt := M (x0 , x) . f (t) − f (x0 ) |x − t| (n − 1)! x0
If f
(n)
∈ L1 [a, b], then M (x0 , x) ≤
1 (n − 1)!
sup t∈[x0 ,x] (t∈[x,x0 ])
|x − t|
n−1
Z
x
x0
(n) f (t) − f (n) (x0 ) dt
1
n−1 (n) |x − x0 |
f − f (n) (x0 ) n! [x0 ,x],1 and the first inequality in (23.6) is proved. Using H¨ older’s integral inequality, we have, for f (n) ∈ Lp [a, b], that q1 Z x p p1 Z x 1 (n) (n−1)q (n) |x − t| dt (t) − f (x ) dt M (x0 , x) ≤ f 0 (n − 1)! x0 x0 " #1 (n−1)q+1 q
|x − x0 | 1
(n)
(n) =
f − f (x0 ) (n − 1)! (n − 1) q + 1 [x0 ,x],p
1
n−1+ 1q (n) =
f − f (n) (x0 ) 1 |x − x0 | [x0 ,x],p q (n − 1)! [(n − 1) q + 1] and the second inequality in (23.6) is proved. Finally, we have for f (n) ∈ L∞ [a, b] , that Z x 1 n−1 M (x0 , x) ≤ ess sup f (n) (t) − f (n) (x0 ) |x − t| dt (n − 1)! =
t∈[x0 ,x] (t∈[x,x0 ])
1
|x − x0 |n f (n) − f (n) (x0 ) n! [x0 ,x],∞ and the theorem is proved.
x0
=
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The following result for H¨ older type mappings also holds. Theorem 23.5. Suppose that the mapping f : [a, b] → R is such that f (n) is of H − r−H¨ older type. That is, (n) r (23.7) f (t) − f (n) (s) ≤ H |t − s| for all t, s ∈ (a, b) and H > 0 is given. Then we have the inequality: |Rn (f ; x0 , x)| ≤
HB (r + 1, n) r+n , |x − x0 | (n − 1)!
(23.8)
where B (·, ·) is Euler’s beta function. Proof.
As f (n) is of H − r−H¨ older type, we may write Z x 1 (n) n−1 (n) |Rn (f ; x0 , x)| ≤ dt f (t) − f (x0 ) |x − t| (n − 1)! x0 Z x H r n−1 |t − x0 | |x − t| dt =: N (x0 , x) . ≤ (n − 1)! x0
Assume that x0 ≤ x. Then Z x Z r n−1 r+n−1+1 N (x0 , x) = (t − x0 ) (x − t) dt = (x − x0 ) x0
= (x − x0 )
r+n
1 0
(23.9)
tr (1 − t)n−1 dt
B (r + 1, n) .
A similar equality can be obtained if x < x0 . Consequently, in general N (x0 , x) = |x − x0 |
r+n
B (r + 1, n)
and then, by (23.9), we deduce (23.8).
Corollary 23.6. Suppose that the mapping f : [a, b] → R is such that f (n) is L−Lipschitzian on [a, b], i.e., (n) (23.10) f (t) − f (n) (s) ≤ L |t − s| for all t, s ∈ (a, b) ,
where L > 0 is given. Then we have the inequality
n+1
|Rn (f ; x0 , x)| ≤ Proof.
L |x − x0 | (n + 1)!
.
(23.11)
For r = 1 we have B (2, n) =
Z
Using (23.8), we deduce (23.11).
1 0
tn−1 (1 − t) dt =
1 . n (n + 1)
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We can now state the following result as well. Theorem 23.7. Let f, x0 and x be as in Theorem 23.1. Then the remainder Rn (f ; x0 , x) satisfies the bound |Rn (f ; x0 , x)| R
(n+1) x n−1 1
f
|x − t| |t − x | dt if 0 (n−1)! x0 [x0 ,t],∞ R
1 x n−1 1 |t − x0 | q f (n+1) [x0 ,t],p dt if ≤ (n−1)! x0 |x − t|
1 R x n−1
f (n+1) |x − t| dt if (n−1)! x0 [x0 ,t],1 Proof.
f (n+1) ∈ L∞ [a, b] ; f (n+1) ∈ Lq [a, b] , p > 1,
f
(n+1)
1 p
+
1 q
= 1;
∈ L1 [a, b] . (23.12)
As f (n) is absolutely continuous on [a, b] we may write that Z t (n) (n) f (t) − f (x0 ) = f (n+1) (u) du x0
and then, by (23.5), we have the representation: Z x Z t 1 n−1 Rn (f ; x0 , x) = f (n+1) (u) du (x − t) dt. (n − 1)! x0 x0 By (23.13) we may write: Z x Z t 1 (n+1) |x − t|n−1 dt . f (u) du |Rn (f ; x0 , x)| ≤ (n − 1)! x0 x0 Now, if f (n+1) ∈ L∞ [a, b], then Z t
(n+1) . f (u) du ≤ |t − x0 | f (n+1)
(23.13)
(23.14)
[x0 ,t],∞ 1, 1p + q1 =
x0
Also, by H¨ older’s integral inequality we have (for p > 1) that p1 Z t Z t 1 (n+1) p (u) du f (n+1) (u) du ≤ |t − x0 | q f x x0
0
1
= |t − x0 | q f (n+1) [x0 ,t],p
and
Z
t
x0
Z f (n+1) (u) du ≤
Consequently, we have
t
x0
(n+1)
(u) du = f (n+1) f
[x0 ,t],1
|t − x0 | f (n+1) [x0 ,t],∞ Z t
1 (n+1) q (n+1) f (u) du ≤ |t − x0 | f [x0 ,t],p x0
f (n+1) .
.
(23.15)
[x0 ,t],1
Using (23.14) and (23.15), we easily deduce (23.12).
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23.3
Some Further Bounds of the Remainder
Let us consider the Chebychev functional defined by Z b Z b Z b 1 1 h (x) g (x) dx − T (g, h) := h (x) dx g (x) dx, b−a a (b − a)2 a a
(23.16)
where h, g : [a, b] → R are measurable on [a, b] and the involved integrals exist on [a, b]. The following identity which can be proved by direct computation is well known as Korkine’s identity: Z bZ b 1 (h (x) − h (y)) (g (x) − g (y)) dxdy. (23.17) T (g, h) = 2 2 (b − a) a a The following lemma holds. Lemma 23.8. Suppose that the mapping f : [a, b] → R and x0 , x are as in Lemma 23.3. Then we have the representation Rn (f ; x0 , x) hh i i (x − x )n 1 0 + = f (n−1) ; x0 , x − f (n) (x0 ) n! 2 (n − 1)! (x − x0 ) Z xZ x n−1 n−1 × dtds. f (n) (t) − f (n) (s) (x − t) − (x − s) x0
x0
(23.18)
Applying Korkine’s identity, we may write: Z x 1 n−1 f (n) (t) − f (n) (x0 ) (x − t) dt x − x 0 x0 Z x Z x 1 1 n−1 (n) (n) − (x − t) dt f (t) − f (x0 ) dt · x − x 0 x0 (x − x0 )2 x0 Z xZ x 1 n−1 n−1 (n) (n) = f (t) − f (s) (x − t) − (x − s) dtds 2 2 (x − x0 ) x0 x0
Proof.
which is clearly equivalent to: Z x n−1 f (n) (t) − f (n) (x0 ) (x − t) dt x0
h
i (x − x )n−1 0 = f (n−1) (x) − f (n) (x0 ) − (x − x0 ) f (n) (x0 ) n Z xZ x 1 n−1 n−1 dtds f (n) (t) − f (n) (s) (x − t) − (x − s) + 2 (x − x0 ) x0 x0
from where we get (23.18).
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The following theorem holds. Theorem 23.9. Assume that the mapping f : [a, b] → R has the property that f (n) ∈ L2 [a, b] and x0 , x ∈ (a, b). Then we have the inequality: |Rn (f ; x0 , x)| |x − x |n h i 0 ≤ f (n−1) ; x0 , x − f (n) (x0 ) n! n
h i2 12 1 (n − 1) |x − x0 |
(n) 2 (n−1) √ , + − f ; x0 , x
f x − x0 [x0 ,x];2 n! 2n − 1 where
Proof.
(n)
f
[x0 ,x];2
We have, by (23.18), that
Z =
x x0
1 (n) 2 2 f (t) dt .
|Rn (f ; x0 , x)| h |x − x |n i 1 0 ≤ f (n−1) ; x0 , x − f (n) (x0 ) + n! 2 (n − 1)! |x − x0 | Z x Z x i h n−1 n−1 (n) (n) × dtds . f (t) − f (s) (x − t) − (x − s) x0
x0
Using the Cauchy-Buniakowski-Schwartz inequality, we have Z x Z x h i n−1 n−1 (n) (n) f (t) − f (s) (x − t) − (x − s) dtds x0 x0 Z x Z x 21 2 ≤ f (n) (t) − f (n) (s) dtds x0 x0 Z x Z x h 21 i2 n−1 n−1 (x − t) − (x − s) dtds x0 x0 " Z x 2 # 12 Z xh i2 (n) (n) = 2 (x − x0 ) f (t) dt f (t) dt − x0
"
× (x − x0 )
Z
x0
x x0
(x − t)
2
= 2 (x − x0 ) f (n)
2(n−1)
[x0 ,x];2
"
(23.19)
dt −
− f
(n−1)
2(n−1)+1
(x − x0 ) × (x − x0 ) 2 (n − 1) + 1
Z
−
x x0
(x − t)
(x) − f
(x − x0 ) n
n−1
(n−1) n 2
dt
(x0 )
# 12
2 # 21
2 12
(23.20)
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h i2 12 1
(n) 2 (n−1) − f ; x , x
f 0 x − x0 [x0 ,x];2 " # 21 (x − x0 )2n (x − x0 )2n × − 2n − 1 n2
h i2 12 1
(n) 2 (n−1) = 2 |x − x0 | ; x0 , x − f
f x − x0 [x0 ,x];2 12 2 n n − 2n + 1 × |x − x0 | n2 (2n − 1)
h i2 12 n − 1 1
(n) 2 n+1 √ − f (n−1) ; x0 , x = 2 |x − x0 |
f x − x0 [x0 ,x];2 n 2n − 1 n+1
h i2 12 2 (n − 1) |x − x0 | 1
(n) 2 √ = − f (n−1) ; x0 , x
f x − x0 [x0 ,x];2 n 2n − 1 = 2 |x − x0 |
and then
Z x Z x h i 1 n−1 n−1 (n) (n) f (t) − f (s) (x − t) − (x − s) dtds 2 (n − 1)! x0 x0 n
h i2 12 (n − 1) |x − x0 | 1
(n) 2 (n−1) √ . ≤ − f ; x0 , x
f x − x0 [x0 ,x];2 n! 2n − 1 Using (23.20), we deduce (23.19).
23.4
Some Inequalities for Special Cases
In this section we assume that x0 , x ∈ (a, b) and x ≥ x0 . The following theorem holds. Theorem 23.10. Let f : [a, b] → R be such that f (n) is monotonic nondecreasing (nonincreasing) on [x0 , x]. Then we have the inequality: hh i i (x − x )n 0 (23.21) f (x) ≤ (≥) Tn (f ; x0 , x) + f (n−1) ; x0 , x − f (n) (x0 ) n! or, equivalently, n i (x − x0 ) h (n−1) f (x) ≤ (≥) Tn−1 (f ; x0 , x) + f ; x0 , x . (23.22) n! Proof.
We use the following Chebychev inequality T (g, h) ≥ 0 (≤ 0)
(23.23)
provided that (g, h) are synchronous (asynchronous), i.e., we recall that the mappings (g, h) are synchronous (asynchronous) if (g (x) − g (y)) (h (x) − h (y)) ≥ 0 (≤ 0) for all x, y ∈ [a, b] .
(23.24)
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n−1
As the mapping h (t) := (x − t) is monotonic nonincreasing on [x0 , x], then we (n) have, for f nondecreasing, n−1 ≤0 T f (n) , (x − ·)
and then, by (23.18) we conclude that hh i i (x − x )n 0 Rn (f ; x0 , x) ≤ f (n−1) ; x0 , x − f (n) (x0 ) . (23.25) n! (n) The case when f is monotonic nonincreasing goes likewise and we omit the details. The following refinement of Chebychev’s inequality is known (see for example [129]) T (g, h) ≥ max {|T (g, |h|)| , |T (|g| , h)| , |T (|g| , |h|)|} ≥ 0.
(23.26)
Using (23.9), we may improve (23.21) as follows.
Theorem 23.11. Let f : [a, b] → R be such that f (n) is monotonic nonincreasing on [x0 , x]. Then we have the inequality: h i (x − x )n 0 f (x) − Tn−1 (f ; x0 , x) − f (n−1) ; x0 , x n! Z x Z x 1 1 (n) (n) n−1 n−1 (x − t) dt − (t) (t) ≥ (x − x ) dt ≥ 0. f f 0 (n − 1)! x0 n x0 (23.27) Proof.
where
n−1
Apply inequality (23.26) for g = f (n) , h = (x − ·) to derive n−1 (n) T f , (x − ·) ≥ max {|A1 | , |B1 | , |C1 |} ,
(23.28)
A1 = T f (n) , (x − ·)n−1 = T f (n) , (x − ·)n−1 Z x 1 (n) (n) n−1 n−1 = B1 = T f , (x − ·) dt f (t) (x − t) x − x 0 x0 Z x Z x 1 (n) n−1 − (t) dt · (x − t) dt f 2 (x − x0 ) x0 x0 Z x Z x n 1 (x − x0 ) 1 (n) (n) n−1 (t) dt − dt · = f f (t) (x − t) 2 x − x 0 x0 n (x − x0 ) x0 and n−1 n−1 C1 = T f (n) , (x − ·) = B1 . = T f (n) , (x − ·) Now, using the fact that (see Lemma 23.8) hh i i (x − x ) 0 f (x) − Tn (f ; x0 , x) − f (n−1) ; x0 , x − f (n) (x0 ) n! n (x − x0 ) n−1 = T f (n) , (x − ·) (n − 1)! then by the inequality (23.28), we may deduce (23.27).
(23.29)
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The following theorem also holds. Theorem 23.12. Let f : [a, b] → R be such that f (n) is convex (concave) on [x0 , x]. Then we have the inequality: f (x) − Tn (f ; x0 , x) Proof.
1 ≥ n+1 f (n−1) (x0 ) (x − x0 ) . (≤) (n + 1)!
(23.30)
As f (n) is convex (concave) on [x0 , x], we may write that f (n) (t) − f (n) (x0 )
≥ (n−1) f (x0 ) (t − x0 ) , t ∈ [x0 , x] (≤)
which implies that h i n−1 ≥ n−1 f (n) (t) − f (n) (x0 ) (x − t) f (n−1) (x0 ) (t − x0 ) (x − t) , t ∈ [x0 , x] . (≤) Integrating over t on [x0 , x] and using the representation (23.1), we may get Z x 1 n−1 f (n−1) (x0 ) (t − x0 ) (x − t) dt Rn (f ; x0 , x) ≥ (≤) (n − 1)! x0 Z x 1 (t − x0 ) (x − t)n−1 dt f (n−1) (x0 ) = (n − 1)! x0 1 n+1 (n−1) = f (x0 ) (x − x0 ) B (2, n) (n − 1)! 1 n+1 f (n−1) (x0 ) (x − x0 ) = (n + 1)!
and the inequality (23.30) is proved. 23.5
Taylor-Multivariate Case Estimates
Let Q be a compact convex subset of Rk , k ≥ 2; z := (z1 , . . . , zk ) , x0 := (x01 , . . . , x0k ) ∈ Q. Let f : Q → R be such that all partial derivatives of order (n − 1) are coordinatewise absolutely continuous functions, n ∈ N. Also, f ∈ C n−1 (Q). Each nth order α partial derivative is denoted by fα := ∂∂xαf , where α := (α1 , . . . , αk ), αi ∈ Z+ , Pk i = 1, . . . , k and |α| := i=1 αi = n. Consider gz (t) := f (x0 + t (z − x0 )), t ≥ 0. Then gz(j) (t) !j k X ∂ f (x01 + t (z1 − x01 ) , . . . , x0k + t (zk − x0k )) , = (zi − x0i ) ∂xi i=1
(23.31)
for all j = 0, 1, 2, . . . , n − 1.
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(n)
Note that gz (t) is given in a similar way. Example 23.13. Let n = k = 2. Then gz (t) = f (x01 + t (z1 − x01 ) , x02 + t (z2 − x02 )) , t ∈ R and gz0 (t) = (z1 − x01 )
∂f ∂f (x0 + t (z − x0 )) + (z2 − x02 ) (x0 + t (z − x0 )) . ∂x1 ∂x2
In addition, gz00 (t) = (z1 − x01 )
∂f (x0 + t (z − x0 )) ∂x1
0
+ (z2 − x02 ) ·
∂f (x0 + t (z − x0 )) ∂x2
0
∂f 2 ∂f 2 = (z1 − x01 ) (z1 − x01 ) 2 (·) + (z2 − x02 ) (·) ∂x1 ∂x2 ∂x1 ∂f 2 ∂f 2 (·) + (z2 − x02 ) 2 (·) . + (z2 − x02 ) (z1 − x01 ) ∂x1 ∂x2 ∂x2 Hence, ∂f 2 ∂f 2 (·) + (z1 − x01 ) (z2 − x02 ) (·) 2 ∂x1 ∂x2 ∂x1 ∂f 2 ∂f 2 + (z1 − x01 ) (z2 − x02 ) (·) + (z2 − x02 )2 (·) . ∂x1 ∂x2 ∂x22
gz00 (t) = (z1 − x01 )2
(n)
Similarly, we derive the case for n, k ∈ N for gz (t).
(n) Notice that, if kfα k[x0 ,z] exists for all α such that |α| = n, then gz
[0,1]
also
exists; k·k is any type of p−norm (p ∈ [1, ∞]). Therefore, we derive the multivariate Taylor Theorem: Theorem 23.14. With the above assumptions, we have f (z1 , . . . , zk ) = gz (1) =
n (j) X gz (0) j=0
where Rn (z, 0) := or
Z
1 0
Z
t1 0
···
1 Rn (z, 0) = (n − 1)!
A simpler form is
Z Z
tn−1 0
+ Rn (z, 0) ,
(n) (n) gz (tn ) − gz (0) dtn . . . dt1 ,
1 0
j!
(1 − θ)
n−1
f (z1 , . . . , zk ) = gz (1) =
gz(n) (θ) − gz(n) (0) dθ.
n−1 X j=0
(23.32)
(23.33)
(23.34)
(j)
gz (0) ˜ n (z, 0) , +R j!
(23.35)
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where ˜ n (z, 0) := R or
Z
1 0
Z
t1 0
˜ n (z, 0) = R
···
Z
tn−1 0
1 (n − 1)!
Z
1 0
gz(n) (tn ) dtn . . . dt1 ,
(23.36)
(1 − θ)n−1 gz(n) (θ) dθ.
(23.37)
Notice that gz (0) = f (x0 ). For a mapping f : Q → R, z, x0 ∈ Q, Q ⊂ Rk compact and convex, we define p1 Z p |f (y)| dy , p ≥ 1. kf k[x0 ,z];p([z,x0 ]) = (23.38) [x0 ,z]([z,x0 ]) R Here [x0 ,z]([z,x0 ]) is a k th multiple integral. Also, kf k[x0 ,z];∞ := ess
sup
y∈[x0 ,z](y∈[z,x0 ])
|f (y)| ,
(23.39)
where [x0 , z] ≡ [z, x0 ] are line segments in Q. ˜ n (z, 0) as in (23.37). We first find estimates for R Remark 23.15. Let k·k be any norm on the functions from Q to R. Let kfα∗ k[x0 ,z] = max kfα k[x0 ,z] . Then |α|=n
(n)
gz (t) [0,1]
" k
!n #
X
∂
= (zi − x0i ) f (x01 + t (z1 − x01 ) , . . . , x0k + t (zk − x0k ))
∂xi i=1 [0,1] !n k X ≤ |zi − x0i | · kfα∗ k[x0 ,z] , i=1
that is,
(n)
gz (t)
[0,1]
≤ kz − x0 kl1
n
· kfα∗ k[x0 ,z] .
Here k·k[0,1] , k·k[x0 ,z] may be any kind of p−norm (p ∈ [1, ∞]). ˜ n (z, 0). We now state the first result in estimating the remainder R Theorem 23.16. With the above assumptions, we have
(n) (n) 1 if gz ∈ L1 [0, 1] ;
g
z (n−1)! L1 [0,1]
(n) (n) 1 ˜ if gz ∈ Lq [0, 1] ,
g z 1 Rn (z, 0) ≤ Lq [0,1] (n−1)!(pn−p+1) p 1 1 p + q = 1, p > 1;
(n) (n) 1 if gz ∈ L∞ [0, 1] . n! gz [0,1];∞
(23.40)
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Proof.
We have ˜ Rn (z, 0) ≤
≤
1 (n − 1)! 1 (n − 1)!
That is
(n)
Z Z
1 0 1 0
287
(1 − θ)n−1 gz(n) (θ) dθ (n) gz (θ) dθ =
˜ Rn (z, 0) ≤
(n)
gz
1
(n) .
gz (t) (n − 1)! L1 [0,1]
L1 [0,1]
(n − 1)!
,
(23.41)
given that gz ∈ L1 [0, 1], the last is implied by all fα ∈ L1 [x0 , z]. Again we observe that Z 1 1 ˜ n−1 (n) (1 − θ) Rn (z, 0) ≤ gz (θ) dθ (n − 1)! 0 Z 1 q1 p p1 Z 1 1 (n) q n−1 ≤ (1 − θ) dθ gz (θ) dθ (n − 1)! 0 0 p1 Z 1
1
(n) pn−p (1 − θ) dθ =
g z (n − 1)! Lq [0,1] 0
1
(n) . =
1 gz Lq [0,1] (n − 1)! (pn − p + 1) p That is,
˜ Rn (z, 0) ≤
(n)
gz
Lq [0,1] 1
(n − 1)! (pn − p + 1) p
,
(23.42)
(n)
where p, q > 1, 1p + 1q = 1; gz ∈ Lq [0, 1], the last is implied by all fα ∈ Lq [x0 , z]. Also it holds Z 1 1 ˜ n−1 (n) (1 − θ) Rn (z, 0) ≤ gz (θ) dθ (n − 1)! 0 Z 1
1
n−1 (1 − θ) dθ gz(n) ≤ (n − 1)! [0,1];∞ 0
(n)
gz [0,1];∞ , = n! that is
(n) gz [0,1];∞ ˜ , if gz(n) ∈ L∞ [0, 1] , (23.43) Rn (z, 0) ≤ n! the last is implied by all fα ∈ L∞ [x0 , z].
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Remark 23.17. Observe that " gz(n)
(0) =
where x0 := (x01 , . . . , x0k ). Similarly, we find
k X
∂ (zi − x0i ) ∂xi i=1
|Rn (z, 0)| ≤ (n)
given that gz Also
!n #
f (x0 ) ,
(n)
(n)
gz (t) − gz (0)
L1 [0,1]
(23.44)
(n − 1)!
∈ L1 [0, 1] which holds when all fα ∈ L1 [x0 , z].
|Rn (z, 0)| ≤ (n)
(n)
(n)
gz (t) − gz (0)
Lq [0,1] 1
(n − 1)! (pn − p + 1) p
,
(23.45)
where p, q > 1, 1p + 1q = 1; gz ∈ Lq [0, 1], when all fα ∈ Lq [x0 , z]. Furthermore we obtain
(n)
(n)
gz (t) − gz (0) [0,1];∞ |Rn (z, 0)| ≤ , if gz(n) ∈ L∞ [0, 1] , n! when all fα ∈ L∞ [x0 , z]. Suppose now that (for all α such that |α| = n)
(23.46)
|fα (x) − fα (y)| ≤ L · kx − ykβl1 , 0 < β ≤ 1,
(23.47)
for all x, y ∈ Q, L > 0, where k·kl1 is the l1 norm in Rk . Here fα is any partial derivative of order n. Then clearly (for all α such that |α| = n) β
|fα (x0 + t (z − x0 )) − fα (x0 )| ≤ L · tβ · kz − x0 kl1 , (23.48) Pk where kz − x0 kl1 = i=1 |zi − x0i |. Thus, if z 6= x0 , then for at least one i ∈ {1, . . . , k}, we have zi 6= x0i , i.e., kz − x0 kl1 6= 0. So, without loss of generality assume that z 6= x0 , which implies that kz − x0 kl1 6= 0. Hence by (23.33) |Rn (z, 0)| Z 1 Z t1 Z ≤ ··· 0
=
0
X n!
|α|=n
=L
k X i=1
Qk
tn−1 0
X n! Qk |zi − x0i |αi β i=1 L kz − x0 kl1 tβn dtn . . . dt1 α1 ! · · · α k !
|zi − x0i | α1 ! · · · α k ! !n
i=1
|zi − x0i |
|α|=n
αi
· L · kz −
β x0 kl1
·
Z
1
0
Z
t1
0
···
Z
tn−1
0
kz − x0 kβ+n 1 l1 = L Qn . (β + j) (β + j) j=1 j=1
β kz − x0 kl1 Qn
tβn dtn
...
dt1
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Consequently, we state the following result. Theorem 23.18. Let fα satisfy (23.47), then β+n
kz − x0 kl1 , for all z ∈ Q, |α| = n. |Rn (z, 0)| ≤ L · Qn j=1 (β + j)
(23.49)
Another matter to discuss: We have
f (z1 , . . . , zk ) =
n−1 X j=0
where 1 Rn−1 (z, 0) = (n − 2)! =
1 (n − 2)!
Z Z
(j)
gz (0) + Rn−1 (z, 0) , j!
1 0
(1 − θ)
n−2
(1 − θ)
n−2
1 0
(23.50)
gz(n−1) (θ) − gz(n−1) (0) dθ ! Z θ
gz(n) (u) du dθ.
0
(23.51) Furthermore 1 |Rn−1 (z, 0)| ≤ (n − 2)!
Z
1 0
(1 − θ)
n−2
Z
! (n) gz (u) du dθ.
θ 0
(n)
Now, if all fα ∈ L∞ [x0 , z], then gz ∈ L∞ [0, 1], and thus Z θ
(n)
. gz (u) du ≤ θ gz(n) L∞ [0,θ]
0
Let p, q > 1 such that p1 + q1 = 1, then Z ! q1 Z θ θ (n) 1q du g (u) du ≤ 0 z 0
1
= θ q gz(n)
Z
θ 0
Lp [0,θ]
(n)
where gz Also,
(n)
p (n) gz (u) du
∈ Lp [0, 1], when all fα ∈ Lp [x0 , z]. Z
θ 0
(n)
gz (u) du ≤ gz(n)
L1 [0,θ]
where gz ∈ L1 [0, 1], when all fα ∈ L1 [x0 , z]. Thus we state the following result.
Theorem 23.19. With the above assumptions, we have |Rn−1 (z, 0)|
,
! p1
(23.52)
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R1
(n) (n) n−2 1 (1 − θ) · θ · dθ; if gz ∈ L∞ [0, 1] ,
g
z (n−2)! 0 L∞ [0,θ] all fα ∈ L∞ [x0 , z] ;
(n) (n) 1 R 1 (1 − θ)n−2 · θ q1 · dθ; if gz ∈ Lp [0, 1] , 1p + q1 = 1,
gz Lp [0,θ] ≤ (n−2)! 0 p, q > 1, all fα ∈ Lp [x0 , z] ;
1 R1 (n) n−2 (n) (1 − θ) dθ; if g ∈ L1 [0, 1] ,
gz z (n−2)! 0 L1 [0,θ] all fα ∈ L1 [x0 , z] .
(23.53)
Remark 23.20. a) Using Lemma 23.8, we have the representation i 1 h Rn (z, 0) = gz(n−1) (1) − gz(n−1) (0) − gz(n) (0) n! Z 1Z 1 1 n−1 n−1 gz(n) (t) − gz(n) (s) (1 − t) − (1 − s) dtds. + 2 (n − 1)! 0 0 (23.54) (n)
∈ L2 [−1, 2]. By Theorem 23.9 we get (0, 1 ∈ (−1, 2)): (n−1) (n−1) (n) (1) − gz (0) − gz (0) gz |Rn (z, 0)| ≤ n!
(n − 1)
(n) 2 + √
gz [0,1];2 n! 2n − 1 2 12 (n−1) (n−1) − gz (1) − gz (0) , (23.55)
Next, suppose that gz
where
(n)
gz
[0,1];2
:=
Z
1 0
(n)
b) By Theorem 23.10, assuming that gz [0, 1], we have f (z1 , . . . , zk ) ≤ (≥) That is
n (j) X gz (0) j=0
j!
+
h
(n−1)
gz
21 (n) 2 . gz (t) dt
is nondecreasing (nonincreasing) over
(n−1)
(1) − gz
n!
i (n) (0) − gz (0)
.
(23.56)
f (z1 , . . . , zk ) ≤ (≥)
n−1 X j=0
(n−1) (n−1) (j) gz (1) − gz (0) gz (0) + . j! n!
(23.57)
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c) By Theorem 23.11, assuming that gz find that
291
is monotonic increasing on [0, 1], we
(n−1) (n−1) (j) gz (1) − gz (0) gz (0) f (z1 , . . . , zk ) − − j! n! j=0 Z 1 Z 1 1 1 (n) (n) n−1 ≥ (t) (1 − t) dt − (t) dt g g . (n − 1)! 0 z n 0 z n−1 X
(n)
d) At last, by Theorem 23.12, for gz f (z1 , . . . , zk ) −
convex (concave) on [0, 1], we derive
n (j) X gz (0) j=0
(23.58)
j!
(n−1)
≥ (≤)
gz (0) . (n + 1)!
(23.59)
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Chapter 24
The Distributional Taylor Formula
We present a distributional Taylor formula with precise integral remainder. We give applications using it and estimates for the remainder. This chapter relies on [51]. 24.1
Introduction and Background
This chapter is motivated by the important works of R. Estrada-R. P. Kanwal (1990) [135]; (1992), [136]; (1993), [137]; and A. Dur´ an-R. Estrada-R. P. Kanwal (1996), [134]; B. Stankovic (1996), [234], and R. Estrada-R. P. Kanwal (2002), Chapter 3, section 3.2, [138]; I. M. Gel’fand-G. E. Shilov (1964), Vol I, pp. 146-151, and pp. 331-345, [146]. It is also motivated by the seminal works of B. Ziemian (1988), [251]; (1988), [253]; (1989), [252]. All the above authors either give distributional Taylor type asymptotic expansions, or distributional Taylor formulae where the remainder is not precisely specified and it is rather vague. Other times Taylor expansions are only for the delta function distribution. Author’s work with S. S. Dragomir (2001), [55], was also another inspiration for this chapter. Other stimulating works are the ones by V. I. Burenkov (1974), [77] and (1998), [78], see Chapter 3, both for Sobolev’s integral representation. But this chapter would have been impossible without the existence of the excellent monograph “Analysis” by E. H. Lieb-M. Loss, 2001, [178]. We are based mainly on the following result from last. Theorem 24.1. ([178], p. 143, Fundamental theorem of Calculus for distributions) 1,1 Let f ∈ Wloc (Rn ) . Then, for each y ∈ Rn and almost every x ∈ Rn , Z 1 y · ∇f (x + ty) dt. (24.1) f (x + y) − f (x) = 0
Before using (24.1) we would like to give an equivalent result to Theorem 24.1 and some useful definitions, however for full details on basic distribution theory we use we refer the reader to [178], Ch. 6 pp. 135–169. 293
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So we give the equivalent to Theorem 24.1 result. 1,1 Theorem 24.2. Let f ∈ Wloc (Rn ) . Then, almost every x, z ∈ Rn we have, Z 1 f (z) − f (x) = (z − x) · ∇f (x + t (z − x)) dt. (24.2) 0
Proof.
Easily we see that (24.1) is equivalent to (24.2).
Remark 24.3. Notice that (24.1) is true for any y ∈ A ⊆ Rn and almost all x ∈ A such that x + y ∈ A, where A is a compact and convex subset of Rn . The null set of x’s (24.1) does not hold depends on y, f. Similarly (24.2) holds for almost every x, z ∈ A, where A is as above. 1,1 In both cases f ∈ Wloc (Rn ) . The following terms and definitions are taken from [178], pp. 136–144. Let Ω be an open, nonempty subset of Rn , n ≥ 2, Cc∞ (Ω) denotes the space of all infinitely differentiable, complex-valued functions whose support is compact and in Ω. The space of test functions is denoted by D (Ω) . The dual space of D (Ω) is denoted by D0 (Ω) which is all continuous linear functionals T : D (Ω) → C, we call T ’s distributions. By Lploc (Ω) we denote the space of locally pth −power summable functions, 1 ≤ p ≤ ∞. Such functions are Borel measurable functions defined on all of Ω and values in C and with the property kf kLp (K) < ∞, for every compact set K ⊂ Ω. Clearly Lploc (Ω) ⊃ Lp (Ω) and, if r > p, we have Lploc (Ω) ⊃ Lrloc (Ω) . We must mention Theorem 24.4. (Functions are uniquely determined by distributions, see [178], p. 138) Let Ω ⊂ Rn be open and let f, g ∈ L1loc (Ω) . Assume that the distributions defined by f and g are equal, i.e. Z
fΦ = Ω
Z
gΦ
(24.3)
Ω
for all Φ ∈ D (Ω) . Then f (x) = g (x) for almost every x ∈ Ω. By ∇f we denote the distributional gradient of f, that is the n−tuple (∂1 f, ∂2 f, . . . , ∂n f ) , where f ∈ L1loc (Ω) −functions are an important class of distributions. 1,1 We denote by Wloc (Ω) the class of functions from L1loc (Ω) whose distributional (weak) first derivatives are also in L1loc (Ω) . We further define, 1 ≤ p ≤ ∞, 1,p Wloc (Ω) := {f : Ω → C : f ∈ Lploc (Ω) and∂i f, as a distribution in D 0 (Ω) , is a
We have W 1,p (Ω)
1,p Wloc (Ω) ⊃ 1,p ⊂ Wloc (Ω) :
Lploc (Ω) − function for i = 1, . . . , n} .
1,r Wloc
(24.4)
(Ω) if r > p. We can also define the Sobolev space
W 1,p (Ω) := {f : Ω → C : f and
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m,p Wloc
∂i f ∈ Lp (Ω) , i = 1, . . . , n} .
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(24.5)
Similarly we define (Ω) , and W m,p (Ω) , m > 1. In these the Borel measurable functions f : Ω → C and all of its partial distributional derivatives up to order m m,p belong to Lploc (Ω) , and Lp (Ω) , respectively. Clearly W m,p (Ω) ⊂ Wloc (Ω) . We end this section with the known basic and useful results, usually given as exercises. We put them together. m,1 Theorem 24.5. Let f ∈ Wloc (Ω) , Ω is open subset in Rn . Then the distributional partial derivative Dr1 ,...,rk f (x) , x ∈ Ω, remains unchanged almost everywhere, when the indices r1 , . . . , rk are permuted, each ri is a positive integer ≤ m. k+n−1 There are distinct distributional partial derivatives of order k in n k dimensions, however there is a total of nk distributional partial derivatives of order k ≤ m. We demonstrate the first part of the last theorem. 3,1 Example 24.6. Let f ∈ Wloc (Ω) . We see that Z Z Z ∂f ∂ 2 Φ ∂ 2 f ∂Φ ∂3Φ =− = f Ω ∂x ∂y∂z Ω ∂y∂x ∂z Ω ∂x∂y∂z
=− Similarly we obtain
= However
Z
Ω
Z
Z
f Ω
Ω
∂3f Φ, ∀Φ ∈ D (Ω) . ∂z∂y∂x
∂3Φ =− ∂x∂z∂y
∂ 2 f ∂Φ =− ∂z∂x ∂y
Z
Ω
Z
Ω
(24.6)
∂f ∂ 2 Φ ∂x ∂z∂y
∂3f Φ, ∀Φ ∈ D (Ω) . ∂y∂z∂x
(24.7)
∂3Φ ∂ 3Φ = , ∂x∂y∂z ∂x∂z∂y thus
hence
Z Z
Ω
∂3Φ f = Ω ∂x∂y∂z
∂3f Φ= ∂z∂y∂x
Z
Ω
Z
∂3Φ , ∂x∂z∂y
(24.8)
∂ 3f Φ, ∀Φ ∈ D (Ω) . ∂y∂z∂x
(24.9)
f Ω
Therefore by Theorem 24.4 we derive ∂3f ∂3f = , a.e. on Ω, ∂z∂y∂x ∂y∂z∂x etc.
(24.10)
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24.2
Main Results
We present the main results. We start with m,1 Theorem 24.7. Let f ∈ Wloc (Rn ) ; m ∈ N, n ∈ N− {1} . Then for every y ∈ Rn n and almost every x ∈ R we have
f (x + y) − f (x) = m Y
n X
+
y ji
i=1
j1 ,...,jm =1
!Z
m−1 X
n X
∂jl1 ...jl f (x)
j1 ,j2 ,...,jl =1
l=1
[0,1]m
∂jm1 ...jm f
l Y
y ji
i=1
x+
m Y
!
!
tr y dt1 . . . dtm ,
r=1
(24.11)
where y := (y1 , . . . , yn ) , x := (x1 , . . . , xn ) . Equality (24.11) is true for any y ∈ A and almost every x ∈ A such that x+y ∈ A, where A is a compact and convex subset of Rn . The above theorem is equivalent to m,1 Theorem 24.8. Let f ∈ Wloc (Rn ) ; m ∈ N, n ∈ N− {1} . Then for almost every n x, z ∈ R we have
f (z) − f (x) =
+
n X
j1 ,...,jm =1
m Y
i=1
m−1 X l=1
(zji − xji )
n X
∂jl1 ...jl f (x)
j1 ,...,jl =1
!Z
[0,1]m
l Y i=1
∂jm1 ...jm f
x+
m Y
r=1
(zji − xji ) !
!
tr (z − x) dt1 . . . dtm ,
(24.12) where z := (z1 , . . . , zn ) . Equality (24.12) is true for almost every x, z ∈ A, where A is a compact and convex subset of Rn . Another useful equivalent form of Theorem 24.7 follows. m,1 Theorem 24.9. Let f ∈ Wloc (Rn ) ; m ∈ N, n ∈ N− {1} . Then for every y ∈ Rn and almost every x ∈ Rn we have
f (x + y) − f (x) =
+
n X
j1 ,...,jm =1
m Y
i=1
y ji
!Z
[0,1]
m
m X l=1
n X
∂jl1 ...jl f (x)
i=1
j1 ,...,jl =1
∂jm1 ...jm f
x+
l Y
m Y
r=1
tr y
!
y ji
!
− ∂jm1 ...jm f (x) dt1 . . . dtm .
(24.13)
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Above (24.13) is also true for any y ∈ A, almost every x ∈ A : x + y ∈ A, where A is compact, convex subset of Rn . Similarly, Theorem 24.8 is equivalent to m,1 Theorem 24.10. Let f ∈ Wloc (Rn ) ; m ∈ N, n ∈ N− {1} . Then for almost every n x, z ∈ R we have
m X
f (z) − f (x) =
+
n X
j1 ,...,jm =1
m Y
i=1
l=1
n X
∂jl1 ...jl f (x)
j1 ,...,jl =1
(zji − xji )
!Z
l Y i=1
∂jm1 ...jm f
[0,1]m
(zji − xji )
x+
m Y
r=1
− ∂jm1 ...jm f (x) dt1 . . . dtm .
!
tr (z − x)
! (24.14)
The last (24.14) is true for almost every x, z ∈ A, where A is a compact, convex subset of Rn . We demonstrate the validity of Theorem 24.7 by giving the following. 2,1 Proposition 24.11. Let f ∈ Wloc (Rn ) , n ≥ 2. Then for every y ∈ Rn and almost every x ∈ Rn we have
f (x + y) − f (x) = y · ∇f (x) Z n X n X + y j1 y j2 j1 =1 j2 =1
Proof.
1 0
Z
1
∂j2 ∂j1 f (x + t2 t1 y) dt2 dt1 . (24.15) 0
By Theorem 24.1 we have
f (x + y) − f (x) = Let Φ ∈ D (Rn ) , then
Z
Z
Rn
Φ (x)
y j1
j1 =1
Z
1
∂j1 f (x + t1 y) dt1 .
(24.16)
0
Φ (x) [f (x + y) − f (x)] dx =
n X
Rn
n X
y j1
j1 =1
Z
1
∂j1 f (x + t1 y) dt1 0
dx.
(24.17)
Next we apply (24.17) for each ∂j1 f, j1 = 1, . . . , n, to obtain Z Φ (x) [∂j1 f (x + t1 y) − ∂j1 f (x)] dx = Rn
Z
Φ (x) Rn
n X
j2 =1
y j2
Z
1
∂j2 ∂j1 f (x + t2 t1 y) dt2 0
dx.
(24.18)
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That is
Z Z
Rn
Rn
Φ (x) ∂j1 f (x + t1 y) dx =
n X
Φ (x) ∂j1 f (x)
Z
y j2
j2 =1
1 0
∂j22 j1 f (x + t2 t1 y) dt2 dx.
(24.19)
Here all integrands are integrable and we can apply Fubini’s theorem. Therefore we derive Z 1 Z Φ (x) ∂j1 f (x + t1 y) dt1 dx Rn
=
Z
Rn
0
Φ (x) ∂j1 f (x) +
n X
y j2
j2 =1
Z
1 0
Z
1 0
∂j22 j1 f (x + t2 t1 y) dt2 dt1 dx.
(24.20)
Next we apply (24.20) into (24.17) to find Z Φ (x) [f (x + y) − f (x)] dx Rn
(24.17)
=
n X
y j1
j1 =1
Z
n X
(24.20)
=
j1 =1
Z
1 0
Z
1 0
Rn
=
Rn
Rn
Z y j1
So we find from (24.21) that Z Z
Φ (x)
∂j22 j1 f
1 0
Z
1 0
1 0
∂j1 f (x + t1 y) dt1 dx
Φ (x) ∂j1 f (x) +
n X
y j2
j2 =1
(x + t2 t1 y) dt2 dt1 dx .
(24.21)
Φ (x) [f (x + y) − f (x)] dx
Φ (x)
Z
Rn
Z
n X
j1 =1
yj1 ∂j1 f (x) +
n X n X
y j1 y j2
j1 =1 j2 =1
∂j22 j1 f (x + t2 t1 y) dt2 dt1 dx,
for every Φ ∈ D (Rn ) . Using now Theorem 24.4 on (24.22) we get (24.15).
(24.22)
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We continue with the next step of the above established procedure of iteration. 3,1 Proposition 24.12. Let f ∈ Wloc (Rn ) , n ≥ 2. Then for every y ∈ Rn and almost every x ∈ Rn , we have n n X X y j1 y j2 y j3 f (x + y) − f (x) = y · ∇f (x) + yj1 yj2 ∂j22 j1 f (x) + j1 ,j2 =1
n X
+
y j1 y j2 y j3
j1 ,j2 ,j3 =1
Z
[0,1]3
j1 ,j2 ,j3 =1
∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dt2 dt1 .
(24.23)
Here we plug into (24.17) ∂j2 ∂j1 f, for all j1 , j2 ∈ {1, . . . , n} , to obtain Z Φ (x) [∂j2 ∂j1 f (x + t2 t1 y) − ∂j2 ∂j1 f (x)] dx Rn Z 1 Z n X (24.24) y j3 ∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dx. = Φ (x) 0 Rn
Proof.
j3 =1
That is
=
Z
Rn
Z
Φ (x) ∂j22 j1 f (x) +
n X
Rn
Rn
Φ (x) ∂j22 j1 f (x) +
Z
y j3
j3 =1
By Fubini theorem we derive Z Z Φ (x) Z
Φ (x) ∂j22 j1 f (x + t2 t1 y) dx
Rn
n X
[0,1]2
y j3
j3 =1
Z
1 0
∂j22 j1 f
[0,1]
3
(x + t2 t1 y) dt2 dt1
∂j2 j f (x) + 2 1 = n X
j1 ,j2 ,j3 =1
Z
Rn
n X
y j3
j3 =1
Z
[0,1]3
Rn
y j1 y j2 y j3
[0,1]3
dx =
Φ (x) y · ∇f (x) +
n X
y j1 y j2
j1 ,j2 =1
∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dt2 dt1 dx
Φ (x) y · ∇f (x) + Z
!
(24.25)
∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dt2 dt1 dx. (24.26)
Next we put (24.26) into (24.22) to get Z Z Φ (x) [f (x + y) − f (x)] dx = Rn
∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dx.
n X
yj1 yj2 ∂j22 j1 f (x) +
j1 ,j2 =1
∂j33 j2 j1 f (x + t3 t2 t1 y) dt3 dt2 dt1 dx,
true for every Φ ∈ D (Rn ) . By Theorem 24.4 we derive (24.23) .
(24.27)
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4,1 We continue recursively as above but now on Wloc (Rn ) to establish 4,1 Proposition 24.13. Let f ∈ Wloc (Rn ) , n ≥ 2. Then for every y ∈ Rn and almost n every x ∈ R we derive n n X X yj1 yj2 yj3 ∂j31 j2 j3 f (x) + f (x + y)−f (x) = y·∇f (x)+ yj1 yj2 ∂j21 j2 f (x)+ j1 ,j2 =1
n X
y j1 y j2 y j3 y j4
j1 ,j2 ,j3 ,j4 =1
Proof.
=
Z
Φ (x) Rn
Rn
[0,1]
n X
y j4
j4 =1
That is we have
=
4
j1 ,j2 ,j3 =1
∂j41 j2 j3 j4 f (x + t1 t2 t3 t4 y) dt1 dt2 dt3 dt4 .
(24.28)
Now we plug into (24.17) ∂j33 j2 j1 f, for all j1 , j2 , j3 ∈ {1, . . . , n} , to obtain Z Φ (x) ∂j33 j2 j1 f (x + t3 t2 t1 y) − ∂j33 j2 j1 f (x) dx Rn
Z
Z
Z
Rn
Z
1
0
∂j44 j3 j2 j1 f (x + t4 t3 t2 t1 y) dt4
Φ (x) ∂j33 j2 j1 f (x + t3 t2 t1 y) dx
Φ (x) ∂j33 j2 j1 f (x) +
n X
y j4
j4 =1
Z
1 0
dx.
(24.29)
∂j44 j3 j2 j1 f (x + t4 t3 t2 t1 y) dt4 dx. (24.30)
Using again Fubini’s theorem we find ! Z Z 3 Φ (x) ∂j3 j2 j1 f (x + t3 t2 t1 y) dt3 dt2 dt1 dx [0,1]3
Rn
=
Z
Rn
Φ (x) ∂j33 j2 j1 f (x) +
n X
y j4
j4 =1
Z
[0,1]4
∂j44 j3 j2 j1 f (x + t4 t3 t2 t1 y) dt4 dt3 dt2 dt1 dx.
(24.31) Finally we put (24.31) into (24.27) to derive Z Z n X yj1 yj2 ∂j22 j1 f (x) Φ (x) y · ∇f (x) + Φ (x) [f (x + y) − f (x)] dx = Rn
Rn
+
n X
j1 ,j2 ,j3 =1
+
n X
j4 =1
y j4
Z
j1 ,j2 =1
yj1 yj2 yj3 ∂j33 j2 j1 f (x)
∂j44 j3 j2 j1 f (x + t4 t3 t2 t1 y) dt4 dt3 dt2 dt1 dx, [0,1]4
(24.32)
for every Φ ∈ D (Rn ) . Using Theorems 24.4 and 24.5 and Fubini’s theorem we establish (24.28) .
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301
Clearly by mathematical induction we establish m,1 Theorem 24.14. Let f ∈ Wloc (Rn ) , where m ∈ N, n ∈ N− {1} . Then for n n every y ∈ R and almost every x ∈ R we have f (x + y) − f (x) = y · ∇f (x) +
+
n X
yj1 yj2 . . . yjm−1 ∂jm−1 f (x) 1 j2 ...jm−1
j1 ,j2 ,...,jm−1 =1 n X
+
yj1 yj2 ∂j21 j2 f (x)
j1 ,j2 =1
n X
yj1 yj2 yj3 ∂j31 j2 j3 f (x) + . . . +
j1 ,j2 ,j3 =1
n X
y j1 y j2 . . . y jm
j1 ,j2 ,j3 ,...,jm =1
Z
[0,1]m
∂jm1 j2 ...jm f (x + t1 . . . tm y) dt1 dt2 . . . dtm . (24.33)
Writing (24.33) into a compact form we have (24.11) . Remark 24.15. (on Theorem 24.14) 3,1 3 1) Let f ∈ Wloc R , for every y ∈ R3 and almost every x ∈ R3 we get f (x + y) − f (x) = y · ∇f (x) +
+
3 X
y j1 y j2 y j3
j1 ,j2 ,j3 =1
Z
[0,1]3
3 X
yj1 yj2 ∂j21 j2 f (x)
j1 ,j2 =1
∂j31 j2 j3 f (x + t1 t2 t3 y) dt1 dt2 dt3 .
(24.34)
4,1 R4 , for every y ∈ R4 and almost every x ∈ R4 it holds 2) Let f ∈ Wloc f (x + y) − f (x) = y · ∇f (x) +
+
4 X
Z
[0,1]4
yj1 yj2 ∂j21 j2 f (x)
j1 ,j2 =1 4 X
yj1 yj2 yj3 ∂j31 j2 j3 f (x) +
y j1 y j2 y j3 y j4
j1 ,j2 ,j3 ,j4 =1
j1 ,j2 ,j3 =1
·
4 X
∂j41 j2 j3 j4 f (x + t1 t2 t3 t4 y) dt1 dt2 dt3 dt4 .
(24.35)
5,1 3) Let f ∈ Wloc R4 , for every y ∈ R4 and almost every x ∈ R4 we derive f (x + y) − f (x) = y · ∇f (x) +
+
4 X
j1 ,j2 ,j3 =1
yj1 yj2 yj3 ∂j31 j2 j3 f (x) +
4 X
yj1 yj2 ∂j21 j2 f (x)
j1 ,j2 =1
4 X
j1 ,j2 ,j3 ,j4 =1
yj1 yj2 yj3 yj4 ∂j41 j2 j3 j4 f (x)
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4 X
+
y j1 y j2 y j3 y j4 y j5
j1 ,j2 ,j3 ,j4 ,j5 =1
·
Z
[0,1]5
∂j51 j2 j3 j4 j5 f (x + t1 t2 t3 t4 t5 y) dt1 dt2 dt3 dt4 dt5 .
(24.36)
6,1 R4 , for every y ∈ R4 and almost every x ∈ R4 it holds 4) Let f ∈ Wloc f (x + y) − f (x) = y · ∇f (x) +
4 X
yj1 yj2 ∂j21 j2 f (x) +
j1 ,j2 =1
4 X
yj1 yj2 yj3 ∂j31 j2 j3 f (x) +
j1 ,j2 ,j3 =1
4 X
yj1 yj2 yj3 yj4 ∂j41 j2 j3 j4 f (x)
j1 ,j2 ,j3 ,j4 =1 4 X
+
yj1 yj2 yj3 yj4 yj5 ∂j51 j2 j3 j4 j5 f (x)
j1 ,j2 ,j3 ,j4 ,j5 =1 4 X
+
y j1 y j2 y j3 y j4 y j5 y j6
j1 ,j2 ,j3 ,j4 ,j5 ,j6 =1
·
Z
[0,1]6
∂j61 j2 j3 j4 j5 j6 f
x+
24.3
!
tl y dt1 . . . dt6 .
l=1
Notice here that for example 5 ∂32234 f=
6 Y
(24.37)
∂5f ∂5f 5 , ∂44444 f= , etc. 2 ∂x3 ∂x2 ∂x3 ∂x4 ∂x54
Applications
1) Let f ∈ W m,1 (Rn ) ; m ∈ N, n ∈ N− {1} , y ∈ Rn . Consider the convolution operator Z (Lf ) (y) := f (x + y) dx. (24.38) Rn
Using (24.11) and Fubini’s theorem we obtain Z Z (Lf ) (y) − (Lf ) (0) = f (x + y) dx − Rn
=
m−1 X
n X
l=1 j1 ,...,jl =1
l Y
i=1
y ji
!Z
Rn
f (x) dx Rn
∂jl1 ...jl f (x) dx
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The Distributional Taylor Formula n X
+
j1 ,...,jm =1
m Y
y ji
i=1
!Z
[0,1]m
Using (24.13) we find (Lf ) (y)−(Lf ) (0) =
m X l=1
·
Z
[0,1]m
"
L
Z
∂jm1 ...jm f
Rn
∂jm1 ...jm f
n X
j1 ,...,jl =1 m Y
!
l Y
y ji
i=1
!
!
− L
L
∂jl1 ...jl f
∂jm1 ...jm f
!
tr y dx dt1 . . . dtm . (24.39)
r=1
tr y
r=1
x+
m Y
303
(0)+
n X
j1 ,...,jm =1
#
(0) dt1 . . . dtm .
m Y
i=1
y ji
!
(24.40)
2) Let nonempty B measurable subset of Rn , with V ol (B) 6= 0. Let f ∈ (Rn ) ; m ∈ N, n ≥ 2, y ∈ Rn , then it holds !R R R m−1 n l l X X Y f (x) dx f (x + y) dx B ∂j1 ...jl f (x) dx B B − = y ji V ol (B) V ol (B) V ol (B) j ,...,j =1 i=1
m,1 Wloc
l=1
+
n X
j1 ,...,jm =1
m Y
y ji
i=1
!Z
[0,1]
R
m
1
l
tr y dx r=1 dt1 . . . dtm . (24.41) V ol (B)
m B ∂j1 ...jm f
x+
m Q
m,1 3) Let f ∈ Wloc (Rn ) , m ∈ N, n ≥ 2. Then for y ∈ Rn and almost all x ∈ Rn , by (24.11) we derive ! m−1 n l X X Y ∂jl1 ...jl f (x) T := f (x + y) − f (x) − y ji j1 ,...,jl =1 i=1 l=1 ! !Z m m n Y Y X m tr y dt1 . . . dtm < ∞, (24.42) |yji | ≤ ∂j1 ...jm f x + m [0,1] r=1 j1 ,...,jm =1 i=1 m Q m by ∂jm1 ...jm f x + tr y being in L1 ([0, 1] ) , as a function of t1 , . . . , tm . r=1
If additionally we suppose that ∂jm1 ...jm f
x+
m Y
tr y
r=1
then by (24.42) we derive T ≤
n X
j1 ,...,jm =1
m Y
i=1
!
!
m
∈ L∞ ([0, 1] ) ,
|yji | ∂jm1 ...jm f (x + . . . y) ∞,[0,1]m .
(24.43)
4) Let a, b ∈ Rn , we say a ≤ b iff ai ≤ bi , for all i = 1, . . . , n. Let h ∈ Rn , h ≥ 0, f ∈ W m,1 (Rn ) , m ∈ N, n ∈ N− {1} . We define the first L1 modulus of continuity of f as follows: Z ω1 (f, δ)1 := sup |f (x + h) − f (x)| dx, (24.44) 0≤h≤δ
Rn
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0 ≤ δ ∈ Rn , i.e. hi ≤ δi , i = 1, . . . , n. By (24.13) we obtain (for almost every x ∈ Rn ) ! l n m X X l Y ∂j ...j f (x) δ ji |f (x + h) − f (x)| ≤ 1 l m Y
n X
+
δ ji
i=1
j1 ,...,jm =1
!Z
Consequently we have Z
Rn
m ∂j1 ...jm f [0,1]m
|f (x + h) − f (x)| dx ≤
m X l=1
x+
n X
n X
·
[0,1]m
Z
m ∂j1 ...jm f n R
x+
l=1
+
j1 ,...,jm =1
m Y
δ ji
i=1
tr h
r=1
We have derived the conclusion m X ω1 (f, δ)1 ≤ n X
m Y
!Z
n X
!
ω1
tr h
!
∂jm1 ...jm f
−
(x) dt1 . . . dtm .
(24.45)
Rn
δ ji
i=1
l ∂
j1 ...jl
!
f (x) dx
Y l i=1
− ∂jm1 ...jm f (x) dxdt1 . . . dtm .
l
∂
j1 ...jl
j1 ,...,jl =1
[0,1]m
Z
m Y
j1 ,...,jm =1
Z
m Y
r=1
j1 ,...,jl =1
+
i=1
j1 ,...,jl =1
l=1
f
∂jm1 ...jm f,
l Y
L1 (Rn )
m Y
r=1
i=1
tr δ
!
δ ji
δ ji
!
(24.46)
!
dt1 . . . dtm .
(24.47)
1
We finish the chapter with m,1 5) Let f ∈ Wloc (Rn ) ; m ∈ N, n ≥ 2, y0 ∈ Rn be fixed. Then for almost every n x0 ∈ R with the property ∂jl1 ...jl f (x0 ) = 0, for all l = 0, 1, . . . , m − 1, and all j1 , . . . , jl ∈ {1, . . . , n} , by (24.11) it holds f (x0 + y0 ) = n X
j1 ,...,jm =1
m Y
i=1
y0ji
!Z
[0,1]m
∂jm1 ...jm f
x0 +
m Y
r=1
tr y 0
!
dt1 . . . dtm .
(24.48)
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Book˙Adv˙Ineq
Chapter 25
uss Type Inequalities Using Chebyshev Gr¨ Euler Type and Fink Identities
In this chapter we present Chebyshev–Gr¨ uss type univariate inequalities by using the generalized Euler type and Fink identities. The results involve functions f , g, f (n) , g (n) , n ∈ N, and are with respect to k · kp , 1 ≤ p ≤ ∞. This chapter relies on [41]. 25.1
Background
Here we mention the following inspiring and motivating results. ˇ Theorem 25.1 (Cebyˇ sev, 1882, [88]). Let f, g : [a, b] → R absolutely continuous 0 0 functions. If f , g ∈ L∞ ([a, b]), then Z b Z b 1 Z b 1 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a b−a a b−a a ≤
1 (b − a)2 kf 0 k∞ kg 0 k∞ . 12
(25.1)
Also we mention Theorem 25.1∗ (Gr¨ uss, 1935, [150]). Let f, g integrable functions from [a, b] into R, such that m ≤ f (x) ≤ M , ρ ≤ g(x) ≤ σ, for all x ∈ [a, b], where m, M , ρ, σ ∈ R. Then Z b Z b 1 Z b 1 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a b−a a b−a a ≤
1 (M − m)(σ − ρ). 4
(25.1)∗
Let Bk (x), k ≥ 0, the Bernoulli polynomials, Bk := Bk (0), k ≥ 0, the Bernoulli numbers, and Bk∗ (x), k ≥ 0, are the periodic functions of period one, related to the Bernoulli polynomials as Bk∗ (x) = Bk (x), 305
0 ≤ x < 1,
(25.2)
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and Bk∗ (x + 1) = Bk∗ (x),
x ∈ R.
(25.3)
Some basic properties of Bernoulli polynomials follow (see [1, 23.1]). We have B0 (x) = 1,
1 B1 (x) = x − , 2
1 B2 (x) = x2 − x + , 6
and Bk0 (x) = kBk−1 (x),
k∈N
Bk (x + 1) − Bk (x) = kx
k−1
(25.4) ,
k ≥ 0.
(25.5)
Clearly B0∗ = 1, B1∗ is a discontinuous function with a jump of −1 at each integer, and Bk∗ , k ≥ 2, is a continuous function. Notice that Bk (0) = Bk (1) = Bk , k ≥ 2. We need the general Theorem 25.2 (see [35]). Let f : [a, b] → R be such that f (n−1) , n ≥ 1, is a continuous function and f (n) (x) exists and is finite for all but a countable set of x in (a, b) and that f (n) ∈ L1 ([a, b]). Then for every x ∈ [a, b] we have Z b n−1 X (b − a)k−1 x − a 1 [f (k−1) (b) − f (k−1) (a)] f (t)dt + f (x) = Bk b−a a k! b−a k=1 Z x−a (b − a)n−1 x−t + Bn − Bn∗ f (n) (t)dt. (25.6) n! b−a b−a [a,b] The sum in (25.6) when n = 1 is zero. If f (n−1) is just absolutely continuous then (25.6) is valid again. Formula (25.6) is a generalized Euler type identity, see also [98], [171]. We need also Fink’s identity. Theorem 25.3 (Fink, [141]). Let a, b ∈ R, f : [a, b] → R, n ≥ 1, f (n−1) is absolutely continuous on [a, b]. Then Z b n−1 X n − k f (k−1) (a)(x − a)k − f (k−1) (b)(x − b)k n f (t)dt − f (x) = b−a a k! b−a k=1 Z b 1 + (x − t)n−1 k(t, x)f (n) (t)dt, (25.7) (n − 1)!(b − a) a where
k(t, x) := When n = 1 the sum
n−1 P k=1
t − a, t − b,
a ≤ t ≤ x ≤ b, a ≤ x < t ≤ b.
(25.8)
in (25.7) is zero.
ˇ In this chapter based on Theorems 25.2 and 25.3 we present Cebyˇ sev–Gr¨ uss type ∗ (n) (n) inequalities, see (25.1), (25.1) , involving f , g, f , g , n ∈ N, with respect to k · kp , 1 ≤ p ≤ ∞, see Theorems 25.4 and 25.6 and Corollaries 25.5 and 25.7.
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Chebyshev–Gr¨ uss Type Inequalities Using Euler Type andFink Identities
25.2
307
Main Results
We present the first main result based on generalized Euler type identity. Theorem 25.4. Let f, g : [a, b] → R be such that f (n−1) , g (n−1) , n ∈ N, be continuous functions and f (n) , g (n) exist and are finite for all but a countable set of x ∈ (a, b) and that f (n) , g (n) ∈ L1 ([a, b]). Denote n−1 X (b − a)k−1 x − a f f (k−1) (b) − f (k−1) (a) , (25.9) Bk Tn−1 (x) := k! b−a k=1
(T0f (x) = 0),
g Tn−1 (x)
:=
n−1 X k=1
and ∆(f,g)
(b − a)k−1 x−a Bk g (k−1) (b) − g (k−1) (a) , k! b−a (T0g (x) = 0),
Z b Z b 1 f (x)dx f (x)g(x)dx − := g(x)dx b−a a a a Z 1 b g f f (x)Tn−1 (x) + g(x)Tn−1 (x) dx. − 2 a Z
b
1) If f (n) , g (n) ∈ L∞ ([a, b]), then ! ! Z b s x−a (n!)2 (b − a)n 2 |B2n | + Bn dx |∆(f,g) | ≤ 2n! (2n)! b−a a × kf k∞ kg (n) k∞ + kgk∞ kf (n) k∞ . (n)
(25.10)
(n)
1 p
(25.11)
(25.12)
1 q
∈ Lp ([a, b]), where p, q > 1 such that + = 1, then "Z q ! #1/q 1 b Z 1 (b − a)n− p x − a |∆(f,g) | ≤ Bn (t) − Bn b − a dt dx 2n! a 0 × kf kp kg (n) kp + kgkp kf (n) kp . (25.13)
2) If f
,g
When p = q = 2, it holds
"Z ! #1/2 1 b (b − a)n− 2 (n!)2 2 x−a |∆(f,g) | ≤ dx |B2n | + Bn 2n! (2n)! b−a a × kf k2 kg (n) k2 + kgk2 kf (n) k2 .
3) With respect to k · k1 it holds
s (b − a)n x−a
(n!)2 2 |∆(f,g) | ≤ |B2n | + Bn
2n! (2n)! b−a ∞ (n) (n) × kf k1 kg k∞ + kgk1kf k∞ .
(25.14)
(25.15)
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Proof.
where
We have by Theorem 25.2 that Z b 1 f f (t)dt + Tn−1 (x) + Rfn (x), f (x) = b−a a
f Tn−1 (x)
:=
n−1 X k=1
(T0f (x) = 0), and Rfn (x) := −
(b − a)k−1 x−a Bk f (k−1) (b) − f (k−1) (a) , k! b−a Z
(b − a)n−1 n!
b a
Bn∗
x−t b−a
− Bn
x−a b−a
(25.16)
(25.17)
f (n) (t)dt, ∀x ∈ [a, b]. (25.18)
Similarly we derive g(x) = where g Tn−1 (x)
:=
n−1 X k=1
(T0g (x) = 0), and Rgn (x)
(b − a)n−1 := − n!
Z b"
g(x) f (x)g(x) = b−a g(x)f (x) =
So that
Z
f (x) b−a
Z
b a
a
Z
b
b a
Bn∗
b a
− Bn
x−a b−a
#
(25.20)
g (n) (t)dt, ∀x ∈ [a, b]. (25.21)
g g(t)dt + f (x)Tn−1 (x) + f (x)Rgn (x), ∀x ∈ [a, b].
(25.23)
f (x)g(x)dx =
Z
x−t b−a
(25.19)
(25.22)
a
Z b Z b 1 g(x)dx f (x)dx b−a a a Z b Z b f + g(x)Tn−1 (x)dx + g(x)Rfn (x)dx, a
and
g g(t)dt + Tn−1 (x) + Rgn (x),
f f (t)dt + g(x)Tn−1 (x) + g(x)Rfn (x),
b a
Z
x−a (b − a)k−1 Bk g (k−1) (b) − g (k−1) (a) , k! b−a
Then
and
1 b−a
Z b Z b 1 f (x)dx g(x)dx f (x)g(x)dx = b−a a a Z b Z b g + f (t)Tn−1 (x)dx + f (x)Rgn (x)dx. a
(25.24)
a
a
(25.25)
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Chebyshev–Gr¨ uss Type Inequalities Using Euler Type andFink Identities
That is we get
Z
b a
f (x)g(x)dx − = =
Z
Z
b a b a
1 b−a
Z
b
f (x)dx a
f g(x)Tn−1 (x)dx + g f (x)Tn−1 (x)dx +
Z
Z
b
g(x)dx a
b
b a
g(x)Rfn (x)dx
a
Z
309
f (x)Rgn (x)dx.
Therefore we have Z b Z b Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b−a a a a (Z b 1 g f f (x)Tn−1 (x) + g(x)Tn−1 (x) dx = 2 a ) Z b + f (x)Rgn (x) + g(x)Rfn (x) dx ,
(25.26)
(25.27)
a
and
∆(f,g) :=
Z
b a
f (x)g(x)dx −
1 b−a
Z
b
f (x)dx a
Z
b
g(x)dx a
Z 1 b g f − f (x)Tn−1 (x) + g(x)Tn−1 (x) dx 2 a Z 1 b f (x)Rgn (x) + g(x)Rfn (x) dx. = 2 a 1) We estimate ∆(f,g) with respect to k · k∞ . We have # " Z b Z b 1 f g |∆(f,g) | ≤ |Rn (x)|dx . |Rn (x)|dx + kgk∞ kf k∞ 2 a a
But we see that
|Rfn (x)|
That is
(25.28)
(25.29)
Z b x − a (n) ∗ x−t − Bn |f (t)|dt B n b−a b−a a Z b (b − a)n−1 (n) x − a ∗ x−t ≤ − Bn kf k∞ B n dt n! b−a b−a a Z 1 kf (n) k∞ x − a n (b − a) = Bn (t) − Bn dt n! b−a 0 2 !1/2 Z 1 kf (n) k∞ x−a n ≤ (b − a) Bn (t) − Bn dt . (25.30) n! b−a 0
(b − a)n−1 ≤ n!
1/2 !2 Z 1 (n) kf k x − a ∞ (b − a)n Bn (t) − Bn dt . |Rfn (x)| ≤ n! b−a 0
(25.31)
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Using (25.31) into (25.29) we get (b − a)n |∆(f,g) | ≤ kf k∞ kg (n) k∞ + kgk∞ kf (n) k∞ 2n! 2 !1/2 Z b Z 1 x − a dx . dt × Bn (t) − Bn b−a a 0
(25.32)
Finally by [98], p. 352, we find |∆(f,g) | ≤
(b − a)n kf k∞ kg (n) k∞ + kgk∞kf (n) k∞ 2n! s ! Z b (n!)2 x−a 2 × dx , |B2n | + Bn (2n)! b−a a
proving the claim. 2) We estimate ∆(f,g) with respect to k · kp . Let p, q > 1: |∆(f,g) | ≤ Notice that
1 p
+
1 q
(25.33)
= 1. Then
1 kf kpkRgn kq + kgkp kRfn kq . 2
(25.34)
!q Z b q(n−1) x − t x − a (b − a) (n) ∗ f q − Bn |Rn (x)| ≤ B n |f (t)|dt (n!)q b−a b−a a q ! Z b x − a (b − a)q(n−1) ∗ x−t ≤ − Bn B n dt kf (n) kqp . (25.35) (n!)q b − a b − a a
Furthermore it holds that |Rfn (x)|q
kf (n) kqp (b − a)q(n−1)+1 ≤ (n!)q
Also it holds kRfn (x)kq
1 kf (n) kp (b − a)n− p ≤ n!
Z
b a
q ! Z 1 x − a Bn (t) − Bn dt . b−a 0
(25.36)
q ! !1/q Z 1 Bn (t) − Bn x − a dt dx . b−a 0
(25.37) Hence by using (25.37) into (25.34) we get "Z q ! #1/q 1 b Z 1 (b − a)n− p x − a dt dx Bn (t) − Bn |∆(f,g) | ≤ 2n! b−a a 0 × kf kp kg (n) kp + kgkp kf (n) kp , (25.38)
proving the claim. When p = q = 2, then
"Z 2 ! #1/2 1 b Z 1 (b − a)n− 2 x−a |∆(f,g) | ≤ Bn (t) − Bn dt dx 2n! b−a a 0 × kf k2kg (n) k2 + kgk2 kf (n) k2 . (25.39)
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Consequently by [98], p. 352 we derive "Z ! #1/2 1 b (b − a)n− 2 (n!)2 2 x−a |B2n | + Bn dx |∆(f,g) | ≤ 2n! (2n)! b−a a × kf k2 kg (n) k2 + kgk2 kf (n) k2 .
3) We estimate ∆(f,g) with respect to k · k1 . We observe that 1 |∆(f,g) | ≤ 2 ≤
Z
b a
|f (x)| |Rgn (x)|dx
+
Z
b a
|g(x)| |Rfn (x)|dx
!
(25.40)
(25.41)
1 kf k1 kRgn (x)k∞ + kgk1kRfn (x)k∞ . 2
(25.42)
Next we observe that kRfn (x)k∞
Z " #
b x−a (b − a)n−1
(n) ∗ x−t − Bn f (t)dt . Bn =
a n! b−a b−a ∞
But we see that
|Rfn (x)|
That is
(25.43)
Z b x − a (n) ∗ x−t − Bn |f (t)|dt B n b−a b−a a Z 1 (b − a)n (n) Bn (t) − Bn x − a dt . ≤ kf k∞ n! b−a 0
(b − a)n−1 ≤ n!
(25.44)
Z 1
(b − a)n (n) Bn (t) − Bn x − a dt (25.45) kf k∞
n! b − a ∞ 0 (by [98], p. 352)
s
(n!)2 (b − a)n (n) x−a
2 ≤ kf k∞ |B2n | + Bn
. (25.46)
(2n)! n! b−a
kRfn (x)k∞ ≤
∞
Using (25.46) into (25.42) we find
s x−a (b − a)n
(n!)2 2 |B2n | + Bn |∆(f,g) | ≤
2n! (2n)! b−a ∞ (n) (n) × kf k1 kg k∞ + kgk1kf k∞ ,
proving the claim.
(25.47)
We give Corollary 25.5. Let f, g ∈ C([a, b]) and f 0 , g 0 exist and are finite for all but a countable set of x ∈ (a, b) and that f 0 , g 0 ∈ L1 ([a, b]).
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1) If f 0 , g 0 ∈ L∞ ([a, b]), then Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a ≤
√ (b − a) √ 4 3 − ln[3a + 2 3(b − a) − 3b] 48 √ + ln[−3a + 2 3(b − a) + 3b] kf k∞kg 0 k∞ + kgk∞ kf 0 k∞ . (25.48)
2) If f 0 , g 0 ∈ Lp ([a, b]), where p, q > 1 such that p1 + q1 = 1, then Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a ! # "Z 1/q q b Z 1 t − x − a dt dx ≤ 2−1 (b − a)−1/p b−a a 0 × kf kp kg 0 kp + kgkp kf 0 kp .
When p = q = 2, it holds Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a 1 ≤ √ kf k2 kg 0 k2 + kgk2 kf 0 k2 . 2 6 3) With respect to k · k1 it holds Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a 1 ≤ √ kf k1 kg 0 k∞ + kgk1 kf 0 k∞ . 2 3
Proof.
By Theorem 25.4 for n = 1, etc.
(25.49)
(25.50)
(25.51)
Next we present the second main result here based on Fink’s identity. Theorem 25.6. Let f, g : [a, b] → R, n ∈ N, f (n−1) , g (n−1) are absolutely continuous on [a, b]. Denote f Fn−1 (x)
:=
n−1 X
n−k k!
f (k−1) (b)(x − b)k − f (k−1) (a)(x − a)k b−a
,
(25.52)
n−1 X
n−k k!
g (k−1) (b)(x − b)k − g (k−1) (a)(x − a)k b−a
,
(25.53)
k=1
(F0f (x) = 0), g Fn−1 (x) :=
k=1
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(F0g (x) = 0), and ∆(f,g) :=
313
Z b Z b n g(x)dx f (x)dx f (x)g(x)dx − b−a a a a # "Z b 1 f g − g(x)Fn−1 (x) + f (x)Fn−1 (x) dx . 2 a Z
b
(25.54)
1) If f (n) , g (n) ∈ L∞ ([a, b]), then
(b − a)n+1 kf k∞ kg (n) k∞ + kgk∞ kf (n) k∞ . (n + 2)!
|∆(f,g) | ≤
2) If f (n) , g (n) ∈ Lp ([a, b]), where p, q > 1 such that
1 p
+
1 q
(25.55)
= 1, then 2
1/q (b − a)n−1+ q |∆(f,g) | ≤ 2 (qn + 2) B(q(n − 1) + 1, q + 1) (n − 1)! (n) (n) × kf kp kg kp + kgkp kf kp . (25.56) −1/p
−1/q
When p = q = 2, it holds |∆(f,g) | ≤
(b − a)n p kf k2 kg (n) k2 + kgk2 kf (n) k2 . 2 (n − 1)!2 n(n + 1)(4n − 1)
(25.57)
3) With respect to k · k1 it holds |∆(f,g) | ≤
(b − a)n kf k1kg (n) k∞ + kgk1 kf (n) k∞ . 2(n + 1)!
(25.58)
Proof. Since f : [a, b] → R has f (n−1) absolutely continuous over [a, b], by Theorem 25.3 we have that Z b n f f (x) = f (t)dt + Fn−1 (x) + Rfn (x), (25.59) b−a a where
f Fn−1 (x)
:=
n−1 X k=1
and
n−k k!
Rfn (x) := with
f (k−1) (b)(x − b)k − f (k−1) (a)(x − a)k b−a
1 (n − 1)!(b − a) k(t, x) :=
Z
b a
t − a, t − b,
(x − t)n−1 k(t, x)f (n) (t)dt, a ≤ t ≤ x ≤ b, a ≤ x < t ≤ b.
,
(25.60)
(25.61)
(25.62)
Let f, g as above, i.e. g(x) =
n b−a
Z
b a
g g(t)dt + Fn−1 (x) + Rgn (x).
(25.63)
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Then f (x)g(x) =
n g(x) b−a
Z
b a
f f (t)dt + g(x)Fn−1 (x) + g(x)Rfn (x),
Z b n g g(t)dt + f (x)Fn−1 (x) + f (x)Rgn (x). f (x) b−a a Then by integrating we obtain Z b Z b Z b n f (x)g(x)dx = g(x)dx f (x)dx b−a a a a Z b Z b f + g(x)Fn−1 (x)dx + g(x)Rfn (x), f (x)g(x) =
Z
That is
b
f (x)g(x)dx = a
Z
a
a
a
a
Z b Z b n f (x)dx g(x)dx b−a a a Z b Z b g + f (x)Fn−1 (x)dx + f (x)Rgn (x)dx.
b a
f (x)g(x)dx − = =
Z
Z
b a b a
n b−a
Z
b
f (x)dx a
f g(x)Fn−1 (x)dx
Z
+
g f (x)Fn−1 (x)dx +
Z
b
g(x)dx a
b
b a
f (x)Rgn (x)dx.
(25.67)
Next we estimate ∆(f,g) . 1) Estimate with respect to k · k∞ . We have that # " Z b Z b 1 f g |Rn (x)|dx . |∆(f,g) | ≤ |Rn (x)|dx + kgk∞ kf k∞ 2 a a
Z
a
b
a
|Rfn (x)|dx
kf (n) k∞ ≤ (n − 1)!(b − a)
|Rgn (x)|dx ≤
kg (n) k∞ (n − 1)!(b − a)
Z b Z a
Z b Z a
b
|x − t|
a b a
(25.66)
Adding the last we derive Z b Z b Z b n g(x)dx f (x)dx ∆(f,g) := f (x)g(x)dx − b−a a a a "Z # b 1 f g − g(x)Fn−1 (x) + f (x)Fn−1 (x) dx 2 a # "Z b 1 g f f (x)Rn (x) + g(x)Rn (x) dx . = 2 a
But it holds Z b
(25.65)
g(x)Rfn (x)dx
a
Z
(25.64)
n−1
(25.68)
(25.69)
|k(t, x)|dt dx,
|x − t|n−1 |k(t, x)|dt dx. (25.70)
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Consequently we find
! Z b Z b 1 n−1 |x − t| |k(t, x)|dt dx |∆(f,g) | ≤ 2(n − 1)!(b − a) a a × kf k∞ kg (n) k∞ + kgk∞ kf (n) k∞ .
We have that Z b |x − t|n−1 |k(t, x)|dt = a
and
Z
Z
b a
b a
1 (x − a)n+1 + (b − x)n+1 , n(n + 1) !
|x − t|n−1 |k(t, x)|dt dx =
(25.71)
(25.72)
2(b − a)n+2 , n(n + 1)(n + 2)
(25.73)
giving us (b − a)n+1 kf k∞ kg (n) k∞ + kgk∞ kf (n) k∞ , (n + 2)! that is proving the claim. 2) Case of k · kp : Let p, q > 1 such that p1 + q1 = 1. Then |∆(f,g) | ≤
|∆(f,g) | ≤ But it holds 1 kRfn kq = (n − 1)!(b − a) 1 ≤ (n − 1)!(b − a) 1 ≤ (n − 1)!(b − a)
1 kf kpkRgn kq + kgkp kRfn kq . 2
(25.75)
q !1/q Z b Z b (x − t)n−1 k(t, x)f (n) (t)dt dx a a Z b Z a
b
a
Z b "Z a
|x − t|n−1 |k(t, x)| |f (n) (t)|dt
b a
|x − t|
q(n−1)
q
|k(t, x)| dt
q
1/q
kf (n) kp ≤ (n − 1)!(b − a)
Z b Z a
b a
|x − t|
q(n−1)
q
dx
kf
That is kRfn kq
(25.74)
|k(t, x)| dt dx
!1/q
(n)
kp
!1/q
#q
.
(25.76)
dx
!1/q
.
(25.77)
We find that Z b
|x − t|q(n−1) |k(t, x)|q dt = B q(n − 1) + 1, q + 1 (x − a)qn+1 + (b − x)qn+1 , ∀x ∈ [a, b], (25.78) a
and
Z b Z a
=
b a
|x − t|q(n−1) |k(t, x)|q dt dx
B(q(n − 1) + 1, q + 1)
1/q
!1/q
2
21/q (qn + 2)−1/q (b − a)n+ q .
(25.79)
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Consequently by (25.79) and (25.77) it holds kRfn kq ≤
1/q 2 kf (n) kp 2(qn + 2)−1 B(q(n − 1) + 1, q + 1) (b − a)n−1+ q . (n − 1)!
(25.80)
We conclude by (25.80) and (25.75) that
|∆(f,g) | ≤ 2−1/p (qn + 2)−1/q B(q(n − 1) + 1, q + 1) 2
1/q
(b − a)n−1+ q kf kp kg (n) kp + kgkp kf (n) kp , (n − 1)!
×
(25.81)
proving the claim. When p = q = 2 we find
−1/2 |∆(f,g) | ≤ 2−1 n(n + 1)(2n − 1)(2n + 1) (b − a)n × kf k2 kg (n) k2 + kgk2 kf (n) k2 . (n − 1)!
3) Case of k · k1 . We have
|∆(f,g) | ≤ It holds |Rfn (x)| ≤ and kRfn k∞
1 kf k1 kRgn k∞ + kgk1 kRfn k∞ . 2
kf (n) k∞ (n − 1)!(b − a)
Z
b a
(25.83)
|x − t|n−1 |k(t, x)|dt,
Z
b
kf (n) k∞
≤ |x − t|n−1 |k(t, x)|dt
(n − 1)!(b − a) a
(25.82)
(25.84)
∞
kf (n) k∞ k(x − a)n+1 + (b − x)n+1 k∞ = (n + 1)!(b − a) =
kf (n) k∞ (b − a)n . (n + 1)!
(25.85)
That is kRfn k∞ ≤ kf (n) k∞
(b − a)n . (n + 1)!
(25.86)
Finally by (25.86) and (25.83) we obtain |∆(f,g) | ≤ proving the claim.
(b − a)n kf k1kg (n) k∞ + kgk1 kf (n) k∞ , 2(n + 1)!
We end chapter with Corollary 25.7. Let f, g : [a, b] → R be absolutely continuous.
(25.87)
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1) If f 0 , g 0 ∈ L∞ ([a, b]), then Z b Z b 1 Z b 1 g(x)dx f (x)g(x)dx − f (x)dx b − a a (b − a)2 a a ≤
(b − a) kf k∞ kg 0 k∞ + kgk∞ kf 0 k∞ . 6
Book˙Adv˙Ineq
317
(25.88)
2) If f 0 , g 0 ∈ Lp ([a, b]), where p, q > 1 such that p1 + q1 = 1, then Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a 1 1 2 1 ≤ 2− p (q + 2)− q (q + 1)− q (b − a) q −1 kf kp kg 0 kp + kgkp kf 0 kp . (25.89)
When p = q = 2, it holds Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a 1 ≤ √ kf k2kg 0 k2 + kgk2 kf 0 k2 . 2 6 3) With respect to k · k1 it holds Z b Z b 1 Z b 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a (b − a)2 a a 1 kf k1 kg 0 k∞ + kgk1 kf 0 k∞ . ≤ 4 Proof. By Theorem 25.6 for n = 1, etc.
(25.90)
(25.91)
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Chapter 26
Gr¨ uss Type Multivariate Integral Inequalities
Gr¨ uss type multidimensional integral inequalities are presented involving two functions of any number of independent variables. This chapter is based on [27].
26.1
Introduction
One of the most famous integral inequalities was given by Gr¨ uss [150] in 1935 and it can be stated as follows (see [187, p. 296]), ! ! Z b Z b 1 Z b 1 1 f (x)g(x) dx − f (x) dx g(x) dx b − a a b−a a b−a a ≤
1 (M − m)(N − n), 4
where f and g are integrable functions on [a, b] and satisfy the conditions m ≤ f (x) ≤ M,
n ≤ g(x) ≤ N,
for all x ∈ [a, b], where m, M , n, N are given real numbers. A great deal of attention has been given to the above inequality and many articles related to various generalizations, extensions, and variants of it have appeared in the literature; see Chapter X of the book [187] by Mitrinovi´c, Peˇcari´c, and Fink, where more references are given. Here we give the multivariate analog of Gr¨ uss inequality for as many as possible independent variables, given in two variations. Gr¨ uss inequalities for functions of two variables were first given by B. Pachpatte in [200]. His paper is one of our motivations. Next we mention one of his results. We follow the notations of [200] exactly. Here R denotes the set of real numbers and ∆ = [a, b] × [c, d], a, b, c, d ∈ R. We denote by G(∆) the set of continuous functions z : ∆ → R for which D2 D1 z(x, y) = ∂ 2 z(x,y) exists and is continuous on ∆ and belong to L∞ (∆). For any function ∂y∂x z(x, y) ∈ L∞ (∆), we define kzk∞ = sup(x,y)∈∆ |z(x, y)|. 319
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We need the following notation k = (b − a)(d − c), " 2 # a+b 1 2 , H1 (x) = (b − a) + x − 4 2 " 2 # 1 c + d H2 (y) = , (d − c)2 + y − 4 2 " Z b Z F (x, y) = (d − c) f (t, y) dt + (b − a) a
"
G(x, y) = (d − c)
Z
b a
g(t, y) dt + (b − a)
Z
#
d
f (x, s) ds , c d
#
g(x, s) ds , c
M (x, y) = |g(x, y)| kD2 D1 f k∞ + |f (x, y)| kD2 D1 gk∞ , for f, g ∈ G(∆). Then we have Theorem 26.1 ([200]). Let f, g ∈ G(∆). It holds Z Z 1 b d f (x, y)g(x, y) dy dx k a c ! ! Z Z Z Z 1 b d 1 b d f (x, y) dy dx g(x, y) dy dx + k a c k a c Z bZ d 1 (g(x, y)F (x, y) + f (x, y)G(x, y)) dy dx − 2 2k a c Z bZ d 1 ≤ M (x, y)H1 (x)H2 (y) dy dx. 2k 2 a c Another motivation for this work is the approach in [60]. 26.2
Auxiliary Result
We need the following generalized Montgomery type identity. n
n
Theorem 26.2 ([23]). Let f : × [ai , bi ] → R be a continuous mapping on × [ai , bi ], and n
∂ n f (x1 ,...,xn ) ∂x1 ···∂xn
i=1 n
i=1
exists on × [ai , bi ] and is integrable. Let also (x1 , . . ., xn ) ∈ i=1
× [ai , bi ] be fixed. We define the kernels pi : [ai , bi ]2 → R:
i=1
pi (xi , si ) := for all i = 1, . . . , n.
si − ai , si ∈ [ai , xi ], si − bi , si ∈ (xi , bi ],
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Then
θ1,n :=
=
(
Z
n Y
n
× [ai ,bi ] i=1
i=1
n Y
i=1
pi (xi , si )
(bi − ai )
!
· f (x1 , . . . , xn )
n ( ) Z 1 n X Y − (bj − aj ) i=1
j=1 j6=i
bi ai
(n2 ) Z n X Y + (bk − ak ) `=1
k=1 k6=i,j
∂ n f (s1 , . . . , sn ) ds1 · · · dsn ∂s1 · · · ∂sn )
f (x1 , . . . , si , . . . , xn )dsi
bi Z
ai
bj aj
f (x1 , . . . , si , . . . , sj , . . . , xn )dsi dsj
(`)
− + · · · − + · · · + (−1)n−1 n (X ) Z n−1 cj · · · dsn f (s , . . . , x , . . . , s )ds · · · ds (b − a ) · 1 j n 1 j j n Z
× [ai ,bi ] i=1 i6=j
j=1
+ (−1)n
n
× [ai ,bi ]
f (s1 , . . . , sn )ds1 · · · dsn =: θ2,n .
(26.1)
i=1
cj means dsj The above ` counts all the (i, j)’s, i < j and i, j = 1, . . . , n. Also ds is missing.
26.3
Main Results
We present n
Theorem 26.3. Let f, g ∈ C n (B), n ∈ N, where B := × [ai , bi ], ai , bi ∈ R, with i=1
ai < bi , i = 1, . . . , n. Denote by ∂B the boundary of the box B. Suppose that f (x) = g(x) = 0, for all x = (x1 , . . . , xn ) ∈ ∂B (in other words we assume that f (. . . , ai , . . .) = g(. . . , ai , . . .) = f (. . . , bi , . . .) = g(. . . , bi , . . .) = 0,
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for all i = 1, . . . , n). Call Vn := 1 Vn
Z
n Q
i=1
(bi − ai ). Then
|f (x1 , . . . , xn )| |g(x1 , . . . , xn )| dx1 · · · dxn
Z
∂nf 1
|g(x1 , . . . , xn )| ≤ n+1 2 ∂s1 · · · ∂sn ∞ B
∂ng
dx1 · · · dxn .
+ |f (x1 , . . . , xn )| ∂s1 · · · ∂sn ∞ B
(26.2)
Proof. Let (x1 , . . . , xn ) ∈ B, i.e., ai ≤ xi ≤ bi , for all i = 1, . . . , n. By the assumptions we obtain Z xn n Z x1 ∂ f (s1 , . . . , sn ) ··· f (x1 , . . . , xn ) = ds1 · · · dsn , ∂s1 · · · ∂sn an a1 and f (x1 , . . . , xn ) = (−1)
n
Z
b1
···
x1
Z
bn xn
∂ n f (s1 , . . . , sn ) ds1 · · · dsn . ∂s1 · · · ∂sn
In general we introduce the subintervals Ii,0 = [ai , xi ] Then we find that f (x1 , . . . , xn ) = (−1)ε1 +···+εn
and Z
Ii,1 = [xi , bi ],
I1,ε1
···
Z
In,εn
i = 1, . . . , n.
∂ n f (s1 , . . . , sn ) ds1 · · · dsn , ∂s1 · · · ∂sn
(26.3)
where each εi can be either 0 or 1. Adding up (26.3) for all 2n choices for (ε1 , . . . , εn ) we derive Z X (−1)ε1 +···+εn 2n f (x1 , . . . , xn ) = I1,ε1
ε1 ,...,εn
···
Z
n
In,εn
∂ f (s1 , . . . , sn ) ds1 · · · dsn . ∂s1 · · · ∂sn
(26.4)
Next by taking absolute values in (26.4) and using basic properties of integrals (noticing that the 2n “sub-boxes” I1,ε1 × · · · × In,εn form a partition of B) and the subadditivity property of the absolute value we find that Z n ∂ f (x1 , . . . , xn ) 1 dx1 · · · dxn |f (x1 , . . . , xn )| ≤ n 2 B ∂x1 · · · ∂xn
Vn ∂nf
. ≤ n
2 ∂x1 · · · ∂xn ∞
That is we have that
V ∂nf
, |f (x1 , . . . , xn )| ≤ n 2 ∂x1 · · · ∂xn ∞
(26.5)
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is true for all (x1 , . . . , xn ) ∈ B. Similarly it holds
Vn ∂ng
|g(x1 , . . . , xn )| ≤ n 2 ∂x1 · · · ∂xn
323
,
(26.6)
∞
true for all (x1 , . . . , xn ) ∈ B. Therefore
∂nf Vn
, |f (x1 , . . . , xn )| |g(x1 , . . . , xn )| ≤ n |g(x1 , . . . , xn )| (26.7)
2 ∂x1 · · · ∂xn ∞ and
∂ng Vn
. (26.8) |f (x1 , . . . , xn )| |g(x1 , . . . , xn )| ≤ n |f (x1 , . . . , xn )|
2 ∂x1 · · · ∂xn ∞ Hence by adding (26.7) and (26.8) we see that |f (x1 , . . . , xn )| |g(x1 , . . . , xn |
∂nf Vn
≤ n+1 |g(x1 , . . . , xn )| 2 ∂x1 · · · ∂xn ∞
∂ng
+ |f (x1 , . . . , xn )| , (26.9)
∂x1 · · · ∂xn ∞ is true for all (x1 , . . . , xn ) ∈ B. Integrating (26.9) over B we obtain (26.2). Remark 26.4. Inequality (26.9) has by itself its own merits. We also give n Theorem 26.5. Consider the class of functions G := f : × [ai , bi ] → R continui=1 n n n f exists on × [ai , bi ] and belongs to L∞ × [ai , bi ] ous, n ∈ N : the partial ∂x1∂···∂x n i=1 i=1 n with norm k · k∞ . Here (x1 , . . . , xn ) ∈ × [ai , bi ]. Let f, g ∈ G. Denote Vn :=
n Q
i=1
Set
i=1
(bi − ai ), and
(xj − aj )2 + (bj − xj )2 , j = 1, . . . , n. 2 n (X Z n 1) bi Y (bj − aj ) f (x1 , . . . , si , . . . , xn )dsi F1 (x1 , . . . , xn ) := , Hj (xj ) :=
i=1
ai
j=1 j6=i
n (X Z n 1) Y G1 (x1 , . . . , xn ) := (b − a ) j j i=1
F2 (x1 , . . . , xn ) := −
j=1 j6=i
" (n2 ) X `=1
n Y
k=1 k6=i,j
bi ai
(bk − ak )
g(x1 , . . . , si , . . . , xn )dsi ;
Z
bi
ai
Z
bj
aj
· f (x1 , . . . , si , . . . , sj , . . . , xn )dsi dsj
!
(`)
#
,
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324
where ` counts all (i, j)’s, i < j, and i, j = 1, . . . , n, also G2 (x1 , . . . , xn ) := −
" (n2 ) X `=1
n Y
k=1 k6=i,j
(bk − ak )
Z
bi ai
Z
bj aj
· g(x1 , . . . , si , . . . , sj , . . . , xn )dsi dsj ·················· ; n "(n−1 Z X) n (bj − aj ) Fn−1 (x1 , . . . , xn ) := (−1) j=1
!
(`)
#
;
n
× [ai ,bi ] i=1 i6=j
#
cj · · · dsn , · f (s1 , . . . , xj , . . . , sn )ds1 · · · ds
cj means dsj is missing, also where ds Gn−1 (x1 , . . . , xn ) := (−1)
n
n "(n−1 X)
j=1
Also define
Then
(bj − aj )
Z
n
× [ai ,bi ] i=1 i6=j
# c · g(s1 , . . . , xj , . . . , sn )ds1 · · · dsj · · · dsn .
∂nf
Mn (x1 , . . . , xn ) := |g(x1 , . . . , xn )| ∂s1 · · · ∂sn ∞
∂ng
.
+|f (x1 , . . . , xn )| ∂s1 · · · ∂sn ∞
1 Z Γn := f (x1 , . . . , xn )g(x1 , . . . , xn )dx1 · · · dxn n Vn × [ai ,bi ] i=1 Z 1 f (x1 , . . . , xn )dx1 · · · dxn + (−1)n n Vn × [ai ,bi ] i=1 Z 1 · g(x1 , . . . , xn )dx1 · · · dxn n Vn × [ai ,bi ] i=1 " Z n−1 X 1 − 2 g(x , . . . , x ) Fj (x1 , . . . , xn ) 1 n n 2Vn × [ai ,bi ] j=1 i=1
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325
! +f (x1 , . . . , xn ) Gj (x1 , . . . , xn ) dx1 · · · dxn j=1 Z n Y 1 M (x , . . . , x ) Hj (xj ) dx1 · · · dxn . ≤ n 1 n n 2Vn2 × [ai ,bi ] j=1 #
n−1 X
(26.10)
i=1
Proof.
We define the kernels pi : [ai , bi ]2 → R: si − ai , si ∈ [ai , xi ], pi (xi , si ) := si − bi , si ∈ (xi , bi ],
for all i = 1, . . . , n. We also have that Z
[aj ,bj ]
|pj (xj , sj )|dsj = Hj (xj ),
j = 1, . . . , n.
From Theorem 26.5 ([23]) we get Vn f (x1 , . . . , xn ) =
n−1 X
Z
Fj (x1 , . . . , xn ) + (−1)n+1
j=1
+
Z
n
× [ai ,bi ]
f (s1 , . . . , sn )ds1 · · · dsn
i=1
n Y
∂ n f (s1 , . . . , sn ) ds1 · · · dsn . p (x , s ) i i i n ∂s1 · · · ∂sn × [ai ,bi ] i=1
i=1
(26.11)
And also Vn g(x1 , . . . , xn ) =
n−1 X
Gj (x1 , . . . , xn ) + (−1)n+1
j=1
+
Z
Z
n
× [ai ,bi ]
g(s1 , . . . , sn )ds1 · · · dsn
i=1
n Y
∂ n g(s1 , . . . , sn ) pi (xi , si ) ds1 · · · dsn . n ∂s1 · · · ∂sn × [ai ,bi ] i=1
i=1
(26.12)
Next we multiply (26.11) by g(x1 , . . . , xn ) and (26.12) by f (x1 , . . . , xn ) and we add the resulting identities, to find 2Vn f (x1 , . . . , xn )g(x1 , . . . , xn ) " # n−1 n−1 X X = g(x1 , . . . , xn ) Fj (x1 , . . . , xn ) + f (x1 , . . . , xn ) Gj (x1 , . . . , xn ) j=1
j=1
+ An (x1 , . . . , xn ) + Bn (x1 , . . . , xn ). Here we have An (x1 , . . . , xn ) := (−1)
n+1
"
g(x1 , . . . , xn ) ·
+ f (x1 , . . . , xn )
Z
n
× [ai ,bi ]
i=1
(26.13)
Z
n
× [ai ,bi ]
i=1
f (s1 , . . . , sn )ds1 · · · dsn #
g(s1 , . . . , sn )ds1 · · · dsn ,
(26.14)
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and
"
Bn (x1 , . . . , xn ) := g(x1 , . . . , xn )
Z
n Y
n
× [ai ,bi ] i=1
i=1
pi (xi , si ) ·
∂ n f (s1 , . . . , sn ) ds1 · · · dsn ∂x1 · · · ∂sn
# ∂ n g(s1 , . . . , sn ) +f (x1 , . . . , xn ) n ds1 · · · dsn . pi (xi , si ) · ∂s1 · · · ∂sn × [ai ,bi ] i=1 Z
i=1
n Y
We see that
(26.15)
|Bn (x1 , . . . , xn )| ≤ Mn (x1 , . . . , xn )
n Y
j=1
n
Hj (xj ) .
(26.16)
Next we integrate (26.13) over × [ai , bi ] and we derive i=1 Z 1 f (x1 , . . . , xn )g(x1 , . . . , xn )dx1 · · · dxn n Vn × [ai ,bi ] i=1 " Z n−1 X 1 g(x , . . . , x ) Fj (x1 , . . . , xn ) = 1 n n 2 2Vn × [ai ,bi ] j=1 i=1 ! # n−1 X Gj (x1 , . . . , xn ) dx1 · · · dxn + f (x1 , . . . , xn ) + (−1)
·
1 Vn
n+1
Z
1 + 2Vn2
j=1
"
1 Vn
n
× [ai ,bi ]
i=1
Z
Z
n
× [ai ,bi ]
f (x1 , . . . , xn )dx1 · · · dxn
i=1
#
!
g(x1 , . . . , xn )dx1 · · · dxn
n
× [ai ,bi ]
Bn (x1 , . . . , xn )dx1 · · · dxn .
(26.17)
i=1
Consequently we obtain 1 Γn ≤ 2Vn2
Z
n
× [ai ,bi ]
|Bn (x1 , . . . , xn )|dx1 · · · dxn .
(26.18)
i=1
At last using (26.18) along with (26.16) we have established (26.10). Remark 26.6. From (26.13), (26.14), (26.16) we get 2Vn f (x1 , . . . , xn )g(x1 , . . . , xn ) − g(x1 , . . . , xn ) ·
n−1 X j=1
Fj (x1 , . . . , xn ) − f (x1 , . . . , xn )
n−1 X j=1
Gj (x1 , . . . , xn ) − An (x1 , . . . , xn )
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Gr¨ uss Type Multivariate Integral Inequalities
≤ Mn (x1 , . . . , xn )
n Y
j=1
327
n
Hj (xj ) , for (x1 , . . . , xn ) ∈ × [ai , bi ]. i=1
(26.19)
We also give Corollary 26.7 (to Theorem 26.3). Consider the class of functions F := n
n
f ∈
C (B), where n ∈ N, B := × [ai , bi ] such that f (x) = 0, for all x = (x1 , . . . , xn ) ∈ i=1 n Q ∂B (the boundary of B) . Let f ∈ F. Let also g ∈ C n (B). Put Vn := (bi − ai ). i=1
Then
1 Vn
Z
|f (x1 , . . . , xn )| |g(x1 , . . . , xn )| dx1 · · · dxn Z ∂ n (f g) 1 (x , . . . , x ) ≤ n 1 n dx1 · · · dxn . 2 B ∂x1 · · · ∂xn B
(26.20)
Totally the same way as in the proof of Theorem 26.3 we find
Proof.
1 |f (x1 , . . . , xn )| ≤ n 2
Z n ∂ f (x1 , . . . , xn ) ∂x1 · · · ∂xn dx1 · · · dxn , ∀f ∈ F. B
Integrating over B the last one we obtain Z
B
|f (x1 , . . . , xn )| dx1 · · · dxn ≤
Vn 2n
Z n ∂ f (x1 , . . . , xn ) ∂x1 · · · ∂xn dx1 · · · dxn , (26.21) B
true for any f ∈ F. The inequality (26.21) was also proved in [60]. Clearly here f · g ∈ F. Finally applying (26.21) for f · g we establish (26.20).
Finally we have Corollary 26.8 (to Theorem 26.5). Case of n = 3. Here we consider the class 3 3 3 of functions G∗ := f : × [ai , bi ] → R continuous: ∂x1 ∂∂xf2 ∂x3 exists on × [ai , bi ] i=1 i=1 3 3 and belongs to L∞ × [ai , bi ] with norm k · k∞ . Let (x1 , x2 , x3 ) ∈ × [ai , bi ] and i=1
f, g ∈ G∗ . Denote V3 :=
i=1
3 Q
i=1
Hj (xj ) :=
(bi − ai ), and
(xj − aj )2 + (bj − xj )2 , 2
j = 1, . . . , n.
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Put F1 (x1 , x2 , x3 ) :=
3 X i=1
G1 (x1 , x2 , x3 ) :=
3 X i=1
"
3 Y
j=1 j6=i 3 Y
j=1 j6=i
(bj − aj )
Z
(bj − aj )
Z
F2 (x1 , x2 , x3 ) := − (b3 − a3 ) + (b2 − a2 ) + (b1 − a1 ) "
Z
Z
G2 (x1 , x2 , x3 ) := − (b3 − a3 ) + (b2 − a2 ) + (b1 − a1 )
Z
Z
a1 b1
a1 b2 a2
Z
Z
b1
Z
Z
b1 a1 b1
a1 b2 a2
Z
Z
ai
bi ai
b2
f (x1 , si , x3 )dsi ,
g(x1 , si , x3 )dsi ;
f (s1 , s2 , x3 )ds1 ds2 a2 b3
f (s1 , x2 , s3 )ds1 ds3 a3
#
b3
f (x1 , s2 , s3 )ds2 ds3 , a3
Z
Z
bi
b2
g(s1 , s2 , x3 )ds1 ds2 a2 b3
g(s1 , x2 , s3 )ds1 ds3 a3 b3
#
g(x1 , s2 , s3 )ds2 ds3 ; a3
∂ 3f
M3 (x1 , x2 , x3 ) := |g(x1 , x2 , x3 )| ∂s1 ∂s2 ∂s3 ∞
∂3g
+ |f (x1 , x2 , x3 )|
∂s1 ∂s2 ∂s3 . ∞
Then Z Z Z 1 b3 b2 b1 f (x1 , x2 , x3 )g(x1 , x2 , x3 )dx1 dx2 dx3 V3 a 1 a 2 a 3 ! Z b1 Z b2 Z b3 1 − f (x1 , x2 , x3 )dx1 dx2 dx3 V3 a 1 a 2 a 3 ! Z b 1 Z b2 Z b 3 1 g(x1 , x2 , x3 )dx1 dx2 dx3 V3 a 1 a 2 a 3 Z b1 Z b2 Z b3 1 g(x1 , x2 , x3 )(F1 (x1 , x2 , x3 ) − 2V32 a1 a2 a3
+ F2 (x1 , x2 , x3 )) + f (x1 , x2 , x3 )(G1 (x1 , x2 , x3 ) + G2 (x1 , x2 , x3 )) dx1 dx2 dx3
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Gr¨ uss Type Multivariate Integral Inequalities
1 ≤ 2V32
Z
b1 a1
Z
b2 a2
Z
b3 a3
M3 (x1 , x2 , x3 )
3 Y
j=1
329
Hj (xj ) dx1 dx2 dx3 . (26.22)
Comment 26.9. Theorem 26.5 clearly generalizes Theorem 26.1 of [200], that is for n = 2 the corresponding inequalities coincide.
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Chapter 27
uss Type Inequalities on Chebyshev Gr¨ Spherical Shells and Balls
We present Chebyshev–Gr¨ uss type inequalities over spherical shells and balls by extending some basic univariate results of Pachpatte. This chapter is based on [46].
27.1
Introduction
For two absolutely continuous functions f, g : [a, b] → R consider the functional Z b Z b Z b 1 1 1 T (f, g) = f (x)g(x)dx − f (x)dx g(x)dx , (27.1) b−a a b−a a b−a a where the involved integrals exist. In 1882, Chebyshev [88] proved that if f 0 , g 0 ∈ L∞ [a, b], then 1 |T (f, g)| ≤ (b − a)2 kf 0 k∞ kg 0 k∞ . (27.2) 12 In 1935, Gr¨ uss [150] showed that 1 (27.3) |T (f, g)| ≤ (M − θ)(Γ − ∆), 4 provided θ, M , ∆, Γ are real numbers satisfying the condition −∞ < θ ≤ M < ∞, −∞ < ∆ ≤ Γ < ∞ for x ∈ [a, b]. The purpose of this chapter is to extend above fundamental results over spherical shells and balls in RN , N ≥ 1. For that we need the following machinery from Pachpatte [206] which also motivates this chapter. Let f : [a, b] → R be differentiable on [a, b] and f 0 : [a, b] → R is integrable on [a, b]. Let w : [a, b] → [0, ∞) R b be some probability density R t function, that is, an integrable function satisfying a w(t)dt = 1, and W (t) = a w(x)dx for t ∈ [a, b], W (t) = 0 for t < a, and W (t) = 1 for t > b. In [213] Peˇcariˇc has given the following weighted extension of the Montgomery identity: Z b Z b f (x) = w(t)f (t)dt + Pw (x, t)f 0 (t)dt, (27.4) a
a
where Pw (x, t) is the weighted Peano kernel defined by W (t), a ≤ t ≤ x, Pw (x, t) = W (t) − 1, x < t ≤ b. 331
(27.5)
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For some suitable functions w, f, g : [a, b] → R, we define Z Z b Z b w(x)f (x)dx T (w, f, g) = w(x)f (x)g(x)dx − a
a
b a
w(x)g(x)dx , (27.6)
and k · k∞ the usual Lebesgue norm on L∞ [a, b] that is, khk∞ := ess supt∈[a,b] |h(t)| for h ∈ L∞ [a, b]. Pachpatte Theorems 2.1, 2.2 ([206]) follow. Theorem 27.1. Let f, g : [a, b] → R be differentiable on [a, b] and f 0 , g 0 : [a, b] → R integrable on [a, b]. Let w : [a, b] → [0, ∞) be an integrable function satisfying Rare b w(t)dt = 1. Then a Z b |T (w, f, g)| ≤ kf 0 k∞ kg 0 k∞ w(x)H 2 (x)dx, (27.7) a
where
H(x) =
Z
b a
|Pw (x, t)|dt
(27.8)
for x ∈ [a, b] and Pw (x, t) is the weighted Peano kernel given by (27.5). Theorem 27.2. Let f, g, f 0 , g 0 , w be as in Theorem 27.1. Then Z 1 b w(x)[|g(x)| kf 0 k∞ + |f (x)| kg 0 k∞ ]H(x)dx, |T (w, f, g)| ≤ 2 a
(27.9)
where H(x) is defined by (27.8). 27.2
Main Results
We make Remark 27.3. Let A be a spherical shell ⊆ RN , N ≥ 1, i.e. A := B(0, R2 ) − B(0, R1 ), 0 < R1 < R2 . Here the ball B(0, R) := {x ∈ RN : |x| < R}, R > 0, where | · | is the Euclidean norm, also S N −1 := {x ∈ RN : |x| = 1} is the unit sphere in RN N/2 with surface area ωN := 2π N . For x ∈ RN − {0} one can write uniquely x = rω, Γ
2
N −1
where r > 0, ω ∈ S . Let F, G ∈ C 1 (A). First we suppose that F , G are radial, i.e. F (x) = f (r), G(x) = g(r), where r = |x|, R1 ≤ r ≤ R2 . Of course f, g ∈ C 1 ([R1 , R2 ]). We notice that Z R2 N sN −1 ds = 1, R2N − R1N R1
i.e. w(s) := Thus
N sN −1 N RN 2 −R1
, R1 ≤ s ≤ R2 , is a probability density function.
W (s) :=
Z
s
w(τ )dτ = R1
sN − R1N , for s ∈ [R1 , R2 ], R2N − R1N
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also W (s) = 1, for s > R2 ; with W (s) = 0, for s < R1 . That is W is the associate distribution function here. We introduce W (s), R1 ≤ s ≤ r, (27.10) Pw (r, s) = W (s) − 1, r < s ≤ R2 . Peˇcariˇc in [213], proved a weighted extension of Montgomery identity, see (27.4), which in our case is Z R2 Z R2 N sN −1 f (s)ds + f (r) = Pw (r, s)f 0 (s)ds, (27.11) R2N − R1N R1 R1
and
g(r) =
Z
R2
R1
Here we observe that
Z R2 N sN −1 g(s)ds + Pw (r, s)g 0 (s)ds. R2N − R1N R1 Vol(A) =
We denote by
R
ωN (R2N − R1N ) . N
(27.12)
(27.13)
Z Z F (x)G(x)dx 1 F (x)dx − G(x)dx Vol(A) (Vol(A))2 A A Z N F (x)G(x)dx = ωN (R2N − R1N ) A 2 Z Z N − F (x)dx G(x)dx , (27.14) ωN (R2N − R1N ) A A
T˜(F, G) :=
A
the Chebyshev functional in this setting. We notice that Z R2 Z R2 Z Z 1 1 f (r)rN −1 dr. F (x)dx = f (r)rN −1 dr dω = ωN A ωN S N −1 R1 R1 Similarly we find
1 ωN and 1 ωN
Z
Z
G(x)dx = A
F (x)G(x)dx = A
Z
R2
(27.15)
g(r)rN −1 dr,
(27.16)
f (r)g(r)rN −1 dr.
(27.17)
R1
Z
R2 R1
Consequently we have Z R2 N ˜ T (F, G) = f (r)g(r)rN −1 dr R2N − R1N R1 2 Z R2 Z R2 N N −1 N −1 − f (r)r dr g(r)r dr =: T (w, f, g), R2N − R1N R1 R1 (27.18)
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as in Pachpatte [206], see (27.6). By Theorem 2.1 of Pachpatte [206], see Theorem 27.1, we obtain that Z R2 N rN −1 H 2 (r)dr, |T˜(F, G)| ≤ kf 0 k∞ kg 0 k∞ R2N − R1N R1
where
H(r) :=
Z
R2 R1
That is H(r) =
1 N R2 − R1N
r ∈ [R1 , R2 ]. In general it holds
|Pw (r, s)|ds = "
Z
r
W (s)ds + R1
Z
(27.19)
R2
(1 − W (s))ds.
r
2rN +1 + N (R1N +1 + R2N +1 ) N +1
−
r(R1N
+
R2N )
#
, (27.20)
∂F
∂r ≤ k∇F k∞ ,
∞
∂G
∂r ≤ k∇Gk∞ ,
(27.21)
∞
with equality in the radial case. So we derive that
∂F ∂G NI
˜
, |T (F, G)| ≤ ∂r ∞ ∂r ∞ (R2N − R1N )3 (N + 1)2
where
I :=
Z
R2
R1
(27.22)
2 rN −1 2rN +1 + N (R1N +1 + R2N +1 ) − r(N + 1)(R1N + R2N ) dr. (27.23)
One calculates the above integral to find 3N +2 − R13N +2 R2 I=4 + N (R1N +1 + R2N +1 )2 (R2N − R1N ) 3N + 2 (N + 1)2 N (R + R2N )2 (R2N +2 − R1N +2 ) + (N + 2) 1 4N + (R1N +1 + R2N +1 )(R22N +1 − R12N +1 ) 2N + 1 − 2(N + 1)(R1N + R2N )(R22N +2 − R12N +2 ).
(27.24)
We have established the first main result. Theorem 27.4. Let F, G ∈ C 1 (A) that are radial functions. Then |T˜(F, G)| ≤ k∇F k∞ k∇Gk∞
NI
(R2N
−
R1N )3 (N
+ 1)2
.
(27.25)
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We continue from Remark 27.3. Remark 27.5. By Theorem 2.2 of Pachpatte [206], see Theorem 27.2, under the same terms and assumptions. We get Z R2 N 1 ˜ rN −1 |g(r)| kf 0 k∞ + |f (r)| kg 0 k∞ H(r)dr. |T (F, G)| ≤ N N 2 R2 − R 1 R1 (27.26) We have derived the following result. Theorem 27.6. Let F, G ∈ C 1 (A) that are radial functions. Then Z 1 T˜(F, G) ≤ |G(x)| k∇F k∞ + |F (x)| k∇Gk∞ H(|x|)dx, 2 Vol(A) A where 2|x|N +1 + N (R1N +1 + R2N +1 ) 1 H(|x|) = N +1 R2N − R1N − |x|(R1N + R2N ) , x ∈ A.
(27.27)
(27.28)
We continue from Remark 27.5 to transfer our results over the ball.
Remark 27.7. Here we set R := R2 and let R > R1 → 0. We suppose F, G ∈ C 1 (B(0, R)) that are radial. Inequality (27.25) is clearly true when k∇F k∞ , k∇Gk∞ are taken over B(0, R) which is a larger set containing A. We consider here this modified form of (27.25). Let now 0 < R1,n ↓ 0, as n → ∞, i.e. B(0, R1,1 ) ⊃ B(0, R1,2 ) ⊃ · · · ⊃ ∞ T B(0, R1,n ) . . ., with B(0, R1,n ) = {0}. That is, as R1 → 0 then B(0, R1 ) ↓ {0}. Hence
n=1
R →0
1 χB(0,R1 ) −→ χ{0} .
Let h be Lebesque integrable on B(0, R). I.e. |hχB(0,R1 ) | ≤ |h|. Consequently R →0
1 hχ{0} , pointwise over B(0, R). hχB(0,R1 ) −→
Thus, by Dominated convergence Theorem we get Z Z R1 →0 L hχB(0,R1 ) dx −→ L B(0,R)
That is
hχ{0} dx.
B(0,R)
Z
R →0
B(0,R1 )
1 hdx −→ 0.
(27.29)
Applying (27.29) we obtain R B(0,R) F (x)G(x)dx ˜ lim T (F, G) = R1 →0 Vol(B(0, R)) Z Z 1 − F (x)dx G(x)dx . (Vol(B(0, R)))2 B(0,R) B(0,R)
(27.30)
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We also notice that
N R2 4 (N + 1)2 lim (R.H.S.(27.25)) = k∇F k∞ k∇Gk∞ + R1 →0 (N + 1)2 (3N + 2) (N + 2) 4N + − (N + 2) . (27.31) (2N + 1)
So based on (27.25) and the above we derive
Theorem 27.8. Let F, G ∈ C 1 (B(0, R)) that are radial functions. Then R Z Z 1 B(0,R) F (x)G(x)dx F (x)dx G(x)dx − Vol(B(0, R)) (Vol(B(0, R)))2 B(0,R) B(0,R) 4 (N + 1)2 4N N R2 + + − (N + 2) . ≤ k∇F k∞ k∇Gk∞ 2 (N + 1) (3N + 2) (N + 2) (2N + 1) (27.32) Similarly from (27.27) we obtain Theorem 27.9. Let F, G ∈ C 1 (B(0, R)) that are radial functions. Then R Z Z 1 B(0,R) F (x)G(x)dx G(x)dx F (x)dx − Vol(B(0, R)) (Vol(B(0, R)))2 B(0,R) B(0,R) Z 1 ≤ |G(x)| k∇F k∞ + |F (x)| k∇Gk∞ H ∗ (|x|)dx, (27.33) 2 Vol(B(0, R)) B(0,R) where
1 H (|x|) = N R ∗
2|x|N +1 + N RN +1 N +1
− |x|R
N
,
x ∈ B(0, R).
(27.34)
Next we treat not necessarily radial functions in our setting. We give Theorem 27.10. Let F, G ∈ C 1 (A). Then Z "Z Z R2 1 N N −1 F (rω)r dr F (x)G(x)dx − N Vol(A) A (R2 − R1N ) S N −1 R1 # Z R2 G(rω)rN −1 dr dω (27.35) × R1
∂F ∂G NI
≤
∂r ∂r (RN − RN )3 (N + 1)2 2 1 ∞ ∞ NI , (27.36) ≤ k∇F k∞ k∇Gk∞ N (R2 − R1N )3 (N + 1)2 where I is given by (27.24).
Proof. The functions F (rω), G(rω) are considered radial in r, ∀ω ∈ S N −1 . Also there exist ∂F (rω) ∂G(rω) , ∈ C([R1 , R2 ]), ∀ω ∈ S N −1 . ∂r ∂r
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As in (27.18), (27.19) and (27.22), (27.25) we get Z R2 N F (rω)G(rω)rN −1 dr R2N − R1N R1 2 Z R2 Z R2 N N −1 N −1 F (rω)r dr G(rω)r dr (27.37) − R2N − R1N R1 R1
∂G(rω)
∂F (rω) NI
≤
∂r
∂r N N 3 2 ∞,[R1 ,R2 ] ∞,[R1 ,R2 ] (R2 − R1 ) (N + 1)
∂F ∂G NI
≤ (27.38)
∂r ∂r (RN − RN )3 (N + 1)2 . 2 1 ∞ ∞ Consequently we have Z R2 Z N N −1 F (rω)G(rω)r dr dω ωN (R2N − R1N ) S N −1 R1 "Z 2 Z Z R2 # R2 N 1 N −1 N −1 F (rω)r dr − G(rω)r dr dω ωN R2N − R1N R1 S N −1 R1
∂F ∂G NI
(27.39) ≤
∂r ∂r (RN − RN )3 (N + 1)2 . 2 1 ∞ ∞
But it holds Z
S N −1
Z
R2
F (rω)G(rω)r
N −1
R1
dr dω =
Z
F (x)G(x)dx.
(27.40)
A
That is completing the proof.
Next we generalize Theorem 27.6. Theorem 27.11. Let F, G ∈ C 1 (A). Then Z " Z Z R2 N N −1 F (rω)r dr F (x)G(x)dx − A R2N − R1N R1 S N −1 # Z R2 N −1 G(rω)r dr dω × R1 "
# Z
∂G
∂F 1
H(|x|)dx ≤ + |F (x)| |G(x)| 2 A ∂r ∞ ∂r ∞ Z 1 ≤ |G(x)| k∇F k∞ + |F (x)| k∇Gk∞ H(|x|)dx, 2 A where H is given by (27.20).
(27.41)
(27.42)
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Proof. Acting similarly as in the proof of Theorem 27.10 and by (27.26) we have Z R2 N F (rω)G(rω)rN −1 dr R2N − R1N R1 2 Z R2 Z R2 N N −1 N −1 − F (rω)r dr G(rω)r dr R2N − R1N R1 R1 "
#
"Z R 2
∂F
∂G N 1 N −1
H(r)dr.
r |G(rω)| + |F (rω)| ≤ 2 R2N − R1N ∂r ∞ ∂r ∞ R1
(27.43)
Therefore we derive Z R2 Z N N −1 F (rω)G(rω)r dr dω ωN (R2N − R1N ) S N −1 R1 "Z # Z R2 2 Z R2 N 1 G(rω)rN −1 dr dω F (rω)rN −1 dr − ωN R2N − R1N R1 R1 S N −1 " "
Z Z R2
∂F 1 N N −1
r |G(rω)| ≤
∂r 2ωN R2N − R1N S N −1 R1 ∞ #
#
∂G
+ |F (ω)| (27.44)
∂r H(r)dr dω, ∞ which proves the claim.
Next we give general results over the ball. Theorem 27.12. Let F, G ∈ C 1 (B(0, R)). Then Z 1 F (x)G(x)dx Vol(B(0, R)) B(0,R) "Z # Z R Z R N N −1 N −1 G(rω)r dr dω (27.45) F (rω)r dr − N R 0 0 S N −1
∂F ∂G (N + 1)2 4N N R2 4
≤ + + − (N + 2) ∂r ∞ ∂r ∞ (N + 1)2 (3N + 2) (N + 2) (2N + 1) N R2 (N + 1)2 4N 4 ≤ k∇F k∞ k∇Gk∞ + + − (N + 2) . (N + 1)2 (3N + 2) (N + 2) (2N + 1) (27.46) Proof. Here put R := R2 . The functions F (rω), G(rω) are considered radial in r ∈ [0, R], ∀ω ∈ S N −1 . Also there exist ∂F (rω) ∂G(rω) , ∈ C((0, R]), ∂r ∂r
∀ω ∈ S N −1 .
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Then as in (27.37) and (27.38) we obtain Z R N F (rω)G(rω)rN −1 dr RN − R1N R1 Z R 2 Z R N N −1 N −1 (27.47) G(rω)r dr F (rω)r dr − RN − R1N R1 R1
∂F (rω)
∂G(rω) NI
≤
N N 3 2 ∂r ∂r ∞,[0,R] ∞,[0,R] (R − R1 ) (N + 1)
∂F ∂G NI
≤ (27.48)
∂r ∂r (RN − RN )3 (N + 1)2 . 1 ∞ ∞ Taking the limit as R1 ↓ 0 to both sides of (27.14), (27.48) we find Z R 2 Z R N Z R N N −1 N −1 N −1 G(rω)r dr F (rω)r dr F (rω)G(rω)r dr− N R RN 0 0 0
∂F ∂G N R2 4 (N + 1)2 4N
≤ + + − (N + 2) .
∂r ∂r (N + 1)2 (3N + 2) (N + 2) (2N + 1) ∞ ∞ (27.49) Consequently one obtains that Z R N Z N −1 F (rω)G(rω)r dr dω ωN RN S N −1 0 "Z Z R # 2 Z R N 1 N −1 N −1 F (rω)r dr G(rω)r dr dω − ωN R N N −1 0 0 S
∂F ∂G N R2 (N + 1)2 4N 4
≤ + + − (N + 2) .
∂r ∂r (N + 1)2 (3N + 2) (N + 2) (2N + 1) ∞ ∞ (27.50) But it holds Z Z S N −1
R
F (rω)G(rω)r 0
That completes the proof.
N −1
dr dω =
Z
F (x)G(x)dx.
(27.51)
B(0,R)
We end chapter with Theorem 27.13. Let F, G ∈ C 1 (B(0, R)). Then Z "Z Z R # Z R N N −1 N −1 F (x)G(x)dx − N F (rω)r dr G(rω)r dr dω B(0,R) R 0 0 S N −1 # "
Z
∂F
1
+ |F (x)| ∂G H ∗ (|x|)dx ≤ (27.52) |G(x)|
2 B(0,R) ∂r ∞ ∂r ∞ Z 1 ≤ [|G(x)| k∇F k∞ + |F (x)| k∇Gk∞ ]H ∗ (|x|)dx, (27.53) 2 B(0,R)
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where H ∗ is given by (27.34). Proof. Here put R := R2 . The functions F (rω), G(rω) are considered radial in r ∈ [0, R], ∀ω ∈ S N −1 . Also there exist
∂F (rω) ∂G(rω) , ∈ C((0, R]), ∀ω ∈ S N −1 . ∂r ∂r Then as in (27.43) we derive Z R N F (rω)G(rω)rN −1 dr RN − R1N R1 Z R 2 Z R N N −1 N −1 G(rω)r dr F (rω)r dr − RN − R1N R1 R1
"Z R
∂F 1 N
≤ rN −1 |G(rω)|
∂r 2 RN − R1N R1 ∞,B(0,R) #
∂G
+ |F (rω)| H(r)dr.
∂r ∞,B(0,R)
(27.54)
Taking the limit as R1 ↓ 0 to both sides of (27.54) and after simplification, we find Z Z R Z R R N F (rω)G(rω)rN −1 dr − F (rω)rN −1 dr G(rω)rN −1 dr 0 RN 0 0 "Z
R
∂F
1
+ |F (rω)| ∂G H ∗ (r)dr, rN −1 |G(rω)| (27.55) ≤
∂r
∂r 2 0 ∞ ∞ where H ∗ is given by (27.34). Consequently one obtains Z Z R N −1 F (rω)G(rω)r dr dω S N −1 0 "Z Z R # Z R N N −1 N −1 − F (rω)r dr G(rω)r dr dω RN S N −1 0 0 "Z #
Z R
∂G
∂F 1 N −1 ∗
≤ r |G(rω)|
∂r + |F (rω)| ∂r H (r)dr dω, 2 S N −1 0 ∞ ∞
(27.56)
which proves the claim.
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Chapter 28
uss and Multivariate Chebyshev Gr¨ Comparison of Integral Means Inequalities Using a Multivariate Euler Type Identity In this chapter we present tight multivariate Chebyshev–Gr¨ uss and Comparison of Integral means inequalities by using a general multivariate Euler type identity. The results involve functions f, g and their high order single partials and are with respect to k · kp , 1 ≤ p ≤ ∞. This chapter relies on [34]. 28.1
Background
Here we mention the following motivating results. Theorem 28.1. (Chebyshev, 1882, [88]) Let f, g : [a, b] → R absolutely continuous functions. If f 0 , g 0 ∈ L∞ ([a, b]), then ! ! Z b Z b 1 Z b 1 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a b−a a b−a a ≤
1 (b − a)2 kf 0 k∞ kg 0 k∞ . 12
(28.1)
Also we mention Theorem 28.2. (Gr¨ uss, 1935, [150]) Let f, g integrable functions from [a, b] into R, such that m ≤ f (x) ≤ M , ρ ≤ g(x) ≤ σ, for all x ∈ [a, b], where m, M, ρ, σ ∈ R. Then ! ! Z b Z b 1 Z b 1 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a b−a a b−a a ≤
1 (M − m)(σ − ρ). 4
(28.2)
Here Bk (x), k ≥ 0, are the Bernoulli polynomials, Bk = Bk (0), k ≥ 0, the Bernoulli numbers, and Bk∗ (x), k ≥ 0, are periodic functions of period one, related to the Bernoulli polynomials as Bk∗ (x) = Bk (x), 341
0 ≤ x < 1,
(28.3)
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and Bk∗ (x + 1) = Bk∗ (x),
x ∈ R.
(28.4)
Some basic properties of Bernoulli polynomials follow (see [1, 23.1]). We have 1 B1 (x) = x − , 2
B0 (x) = 1,
1 B2 (x) = x2 − x + , 6
(28.5)
and Bk0 (x) = kBk−1 (x),
k ∈ N,
Bk (x + 1) − Bk (x) = kxk−1 ,
k ≥ 0.
(28.6) (28.7)
Clearly B0∗ = 1, B1∗ is a discontinuous function with a jump of −1 at each integer, and Bk∗ , k ≥ 2, is a continuous function. Notice that Bk (0) = Bk (1) = Bk , k ≥ 2. We make Assumption 28.3. Let f and the existing continuous real valued functions on
n Y
i=1
∂lf , all l = 1, . . . , m; j = 1, . . . , n, be ∂xlj
[ai , bi ]; m, n ∈ N, ai , bi ∈ R.
In [29] we proved a general Multivariate Euler-type identity, see next. Theorem 28.4. All as in Assumption 28.3 for m, n ∈ N, xi ∈ [ai , bi ], i = 1, 2, . . . , n. Then Z 1 f (x1 , x2 , . . . , xn ) − n f (s1 , . . . , sn )ds1 . . . dsn n Q Y [ai ,bi ] (bi − ai ) i=1 i=1
−
n X
Afj =
j=1
n X
Bjf ,
(28.8)
j=1
where for j = 1, . . . , n we have Afj
:=
× −
Afj (xj , xj+1 , . . . , xn )
Z
=
1 j−1 Y i=1
k−1
j−1 Q i=1
[ai ,bi ]
(bi − ai )
( m−1 X (bj − aj )k−1 xj − aj Bk ! k! bj − a j k=1
∂ f (s , s , . . . , sj−1 , bj , xj+1 , . . . , xn ) k−1 1 2 ∂xj !) !
∂ k−1 f (s1 , s2 , . . . , sj−1 , aj , xj+1 , . . . , xn ) ds1 . . . dsj−1 ∂xjk−1
(28.9)
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and Bjf
:=
−
Bjf (xj , xj+1 , . . . , xn )
∗ Bm
xj − s j bj − a j
!
(bj − aj )m−1 := ! j−1 Y m! (bi − ai )
(Z
j Q
Bm [ai ,bi ]
i=1
i=1
xj − a j bj − a j
343
) ! ∂ f (28.10) (s1 , s2 , . . . , sj , xj+1 , . . . , xn ) ds1 ds2 . . . dsj ∂xm j m
When m = 1 then Afj = 0, j = 1, . . . , n. A general set of suppositions follow. Assumption 28.5. Here m ∈ N, j = 1, . . . , n. We assume n Y [ai , bi ] → R is continuous. 1) f : i=1
∂lf 2) are existing real valued functions for all j = 1, . . . , n; l = 1, . . . , m − 2. ∂xlj 3) For each j = 1, . . . , n we suppose that ∂ m−1 f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xjm−1
is a continuous real valued function. 4) For each j = 1, . . . , n we suppose that gj (·) :=
∂mf (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xm j
exists and is real valued with the possibility of being infinite only over an at most countable subset of (aj , bj ). 5) Parts #3, #4 are true for all (x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=1 i6=j
6) The functions for j = 2, . . . , n; l = 1, . . . , m − 2, j−1
j−1
z }| { ∂ l f z }| { qj (·, ·, . . . , ·) := (·, ·, . . . , ·, xj , xj+1 , . . . , xn ) ∂xlj are continuous on
j−1 Y i=1
[ai , bi ], for each (xj , xj+1 , . . . , xn ) ∈
7) The functions for each j = 1, . . . , n, j
j
n Y
[ai , bi ].
i=j
z }| { ∂ m f z }| { (·, ·, . . . , ·, xj+1 , . . . , xn ) ∈ L1 ϕj (·, ·, . . . , ·) := ∂xm j
j Y
i=1
!
[ai , bi ] ,
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for any (xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=j+1
We give (see [29])
Theorem 28.6. All as in Assumption 28.5. Then (28.8) is valid again. Some weaker general suppositions follow. Assumption 28.7. Here m ∈ N, j = 1, . . . , n and only the Parts #1, #2, #6, #7 of Assumption 28.5 remain the same. We further assume that for each j = 1, . . . , n and over [aj , bj ], the function ∂ m−1 f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xjm−1 is absolutely continuous, and this is true for all (x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=1 i6=j
We give (see [29]) Theorem 28.8. All as in Assumption 28.7. Then (28.8) is valid again. In this chapter based on Theorems 28.4, 28.6, 28.8 we present Chebyshev–Gr¨ uss type and comparison of integral means multivariate inequalities involving f, g and their high order single partials with respect to k · kp , 1 ≤ p ≤ ∞. 28.2
Main Results
We present the first main result regarding Chebyshev–Gr¨ uss type multivariate inequalities, see also [27], [201]. Theorem 28.9. Let f, g as in Assumptions 28.3 or 28.5 or 28.7. Here Afj , Agj as in (28.9), j = 1, . . . , n. Denote by Z − − − f (→ r )g(→ x )d→ x ∆(f,g) := Q n [ai ,bi ]
i=1
−
Z
1
n Y
i=1
1 − 2
(bi − ai )
"Z
n Q
[ai ,bi ]
i=1
− + g(→ x)
n X j=1
"
n Q
[ai ,bi ]
− − f (→ x )d→ x
i=1
− f (→ x)
n X
! Z
n Q
Agj (xj , . . . , xn )
j=1
Afj (xj , . . . , xn )
!#
# → − dx ,
[ai ,bi ]
i=1
− − g(→ x )d→ x
!
! (28.11)
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→ − x := (x1 , . . . , xn ) ∈
n Y
Book˙Adv˙Ineq
345
[ai , bi ].
! n Y ∂mg ∂mf , m ∈ L∞ [ai , bi ] , for j = Then we have: 1) By assuming that ∂xm ∂xj j i=1 1, . . . , n, it holds n Y (bi − ai ) ( n X i=1 (bj − aj )m−1 |∆(f,g) | ≤ 2m! j=1 ! ! Z bj s xj − a j (m!)2 2 dxj |B2m | + Bm (2m)! bj − a j aj
#) "
∂mg
∂mf
kf k∞ m + kgk∞ m . (28.12)
∂xj
∂xj i=1
∞
∞
1 1 2)Let pj , qj > 1: + = 1; j = 1, . . . , n. We further suppose that pj qj ! n Y ∂mg ∂mf , m ∈ L qj [ai , bi ] . ∂xm ∂xj j i=1 It holds
n
1 X |∆(f,g) | ≤ 2m! j=1
((
(bj − aj )
m− q1
j
!qj /pj !1/qj ) pj Bm xj − aj − Bm (tj ) dtj dxj bj − a j aj 0
#)
"
∂ mf
∂ mg
. (28.13) kf kpj m + kgkpj m
∂xj
∂xj Z
bj
Z
1
qj
qj
When pj = qj = 2, all j = 1, . . . , n, it holds (( n 1 1 X (bj − aj )m− 2 |∆(f,g) | ≤ 2m! j=1
!1/2 ) (m!)2 xj − a j 2 dxj |B2m | + Bm (2m)! bj − a j aj
#) "
∂mg
∂ mf
kf k2 m + kgk2 m .
∂xj
∂xj Z
bj
2
2
3) By assuming again that
∂mg ∂mf , m ∈ L∞ ∂xm ∂xj j
n Y
i=1
!
[ai , bi ] ,
j = 1, . . . , n,
(28.14)
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it holds: i) Case of m = 2r, r ∈ N, then ( n 1 X |∆(f,g) | ≤ (bj − aj )2r 2(2r)! j=1 # "
x − a j j −2r
(1 − 2−2r )|B2r | +
2 B2r − B2r bj − aj ∞,[aj ,bj ]
#) "
∂ 2r g
∂ 2r f
kf k1 2r + kgk1 2r (28.15)
∂xj
∂xj ∞
∞
ii) Case of m = 2r + 1, r ∈ N, then ( n X 1 (bj − aj )2r+1 |∆(f,g) | ≤ 2(2r + 1)! j=1 " #
2(2r + 1)! xj − a j
+ B2r+1 (2π)2r+1 (1 − 2−2r ) bj − aj ∞,[aj ,bj ]
#) "
∂ 2r+1 g
∂ 2r+1 f
kf k1 2r+1 + kgk1 2r+1 .
∂xj
∂xj
∞
(28.16)
∞
iii) Case of m = 1, then ( "
#)
n
∂f
∂g 1X
|∆(f,g) | ≤ . (bj − aj )(1 + bj − aj ) kf k1
∂xj + kgk1 ∂xj 4 j=1 ∞ ∞
(28.17)
When m = 1 then Afj = Agj = 0, j = 1, . . . , n. Since f, g are as in Assumptions 28.3 or 28.5 or 28.7 by (28.8) we obtain Z 1 f (x1 , x2 , . . . , xn ) − n f (s1 , . . . , sn )ds1 . . . dsn n Q Y [ai ,bi ] (bi − ai ) i=1
Proof.
i=1
−
n X
Afj (xj , xj+1 , . . . , xn ) =
j=1
−
1 n Y
(bi − ai )
i=1 n X j=1
Bjf (xj , . . . , xn ),
(28.18)
j=1
and g(x1 , . . . , xn ) −
n X
Z
n Q
g(s1 , . . . , sn )ds1 . . . dsn [ai ,bi ]
i=1
Agj (xj , xj+1 , . . . , xn ) =
n X j=1
Bjg (xj , . . . , xn ).
(28.19)
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347
Consequently it holds f (x1 , . . . , xn )g(x1 , . . . , xn ) Z 1 f (s1 , . . . , sn )ds1 . . . dsn − n g(x1 , . . . , xn ) Q n Y [ai ,bi ] i=1 (bi − ai ) i=1
n X − g(x1 , . . . , xn ) Afj (xj , . . . , xn ) j=1
= g(x1 , . . . , xn )
and
n X j=1
Bjf (xj , . . . , xn ) ,
(28.20)
f (x1 , . . . , xn )g(x1 , . . . , xn ) Z 1 g(s1 , . . . , sn )ds1 . . . dsn f (x1 , . . . , xn ) Q − n n Y [ai ,bi ] i=1 (bi − ai ) i=1
n X g − f (x1 , . . . , xn ) Aj (xj , . . . , xn ) j=1
= f (x1 , . . . , xn )
n X j=1
Bjg (xj , . . . , xn ) .
By integrating the last identities we derive Z f (x1 , . . . , xn )g(x1 , . . . , xn )dx1 . . . dxn n Q
(28.21)
[ai ,bi ]
i=1
−
1
n Y
i=1
Z − =
n Q
(bi − ai )
g(x1 , . . . , xn )dx1 . . . dxn [ai ,bi ]
i=1
f (x1 , . . . , xn )dx1 . . . dxn
n Q
[ai ,bi ]
i=1
Z
n Q
[ai ,bi ]
i=1
Z
Z
n Q
i=1
[ai ,bi ]
g(x1 , . . . , xn )
n X j=1
!
!
Afj (xj , . . . , xn ) dx1 . . . dxn
n X g(x1 , . . . , xn ) Bjf (xj , . . . , xn ) dx1 . . . dxn , j=1
(28.22)
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and
Z
f (x1 , . . . , xn )g(x1 , . . . , xn )dx1 . . . dxn
n Q
[ai ,bi ]
i=1
−
n Y
i=1
Z − =
Z
1
n Q
(bi − ai )
n Q
f (x1 , . . . , xn )dx1 . . . dxn [ai ,bi ]
i=1
g(x1 , . . . , xn )dx1 . . . dxn [ai ,bi ]
i=1
Z
n Q
f (x1 , . . . , xn )
[ai ,bi ]
i=1
Z
n Q
[ai ,bi ]
i=1
n X j=1
f (x1 , . . . , xn )
n X j=1
!
!
Agj (xj , . . . , xn ) dx1
. . . dxn
Bjg (xj , . . . , xn ) dx1 . . . dxn .
(28.23)
− Adding and dividing by 2 the latter two identities we get (→ x := (x1 , . . . , xn )) Z − − − ∆(f,g) := Q f (→ x )g(→ x )d→ x n [ai ,bi ]
i=1
−
Z
1
n Y
i=1
1 − 2
(bi − ai )
"Z
n Q
i=1
− + g(→ x) 1 = 2
[ai ,bi ]
"Z
n X
n Q
[ai ,bi ]
− + g(→ x)
n X
[ai ,bi ]
i=1
− f (→ x)
n X
"
j=1
− f (→ x)
n X
!#
n Q
[ai ,bi ]
i=1
Bjf (xj , . . . , xn )
− − g(→ x )d→ x
!
!
− d→ x
Bjg (xj , . . . , xn )
j=1
j=1
! Z
Agj (xj , . . . , xn )
Afj (xj , . . . , xn )
j=1
i=1
"
n Q
− − f (→ x )d→ x
!#
# → − d x := Γ.
! (28.24)
That is ∆(f,g) = Γ. Next we estimate Γ. 1) Estimate with respect to k · k∞ .
(28.25)
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We have
" Z n 1X − |Bjg (xj , . . . , xn )|d→ x |Γ| ≤ kf k∞ Q n 2 [a ,b ] i i j=1 i=1 # Z f → − |B (x , . . . , x )|d x + kgk ∞
1 = 2
(
n Q
j
[ai ,bi ]
j
i=1
n X
i=1
j=1
+ kgk∞
Z
j−1 Y
n Q
(bi − ai )
[ai ,bi ]
n
!"
kf k∞
Z
n Q
[ai ,bi ]
|Bjg (xj , . . . , xn )|dxj . . . dxn
i=j
|Bjf (xj , . . . , xn )|dxj . . . dxn
i=j
#)
=: (∗)
(28.26)
By Lemma 1 of [29], see there (58), we find v !! u m u 2 (bj − aj ) x − a (m!) j j f 2 t |Bj (xj , . . . , xn )| ≤ |B2m | + Bm m! (2m)! bj − a j
∂mf j
, (28.27) × m (z}|{ . . . , xj+1 , . . . , xn ) j
∂xj
Q ∞,
[ai ,bi ]
i=1
for all j = 1, . . . , n. Similarly for g. Using (28.27) into (28.26) we derive ( n X 1 |Γ| ≤ (∗) ≤ 2m! j=1 Z "
kf k∞
+kgk∞
Z
n Q
i=j+1
Z
bj aj
j−1 Y i=1
!
(bi − ai ) (bj − aj )m
v ! !! u u (m!)2 xj − a j 2 t |B2m | + Bm dxj (2m)! bj − a j
∂mg j
z}|{
m ( . . . , xj+1 , . . . , xn )
∂x [ai ,bi ] j
n Q
i=j+1
∞,
∂mf j
z}|{
m ( . . . , xj+1 , . . . , xn )
∂x [ai ,bi ] j ≤
n Y
i=1
(bi − ai ) ( 2m!
n X j=1
j Q
dxj+1 . . . dxn [ai ,bi ]
!
(28.28)
i=1
∞,
j Q
dxj+1 . . . dxn [ai ,bi ]
i=1
(bj − aj )m−1
!#)
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Z
v ! !! u u (m!)2 x − a j j 2 t dxj |B2m | + Bm (2m)! bj − a j
bj aj
"
∂ mg
kf k∞ m
∂xj
∞
proving the claim (28.12).
#)
∂mf
+ kgk∞ m ,
∂xj
(28.29)
∞
2) Estimate with respect to k · kp , 1 < p < ∞. Here pj , qj > 1: j = 1, . . . , n. We have
" # n 1X g f kf kpj kBj kqj + kgkpj kBj kqj . |Γ| ≤ 2 j=1
The assumption implies
∂mg (. . . , xj+1 , . . . , xn ) ∈ Lqj ∂xm j for almost all (xj+1 , . . . , xn ) ∈
n Y
(bj − aj ) ≤ m!
m− q1
j
j−1 Y i=1
(bi −ai )
!−1/qj
true for almost all (xj+1 , . . . , xn ) ∈
n Q
[ai ,bi ]
− |Bjg |qj d→ x
i=1
Z
n Q
i=1
[ai ,bi ]
!1/qj
( Z
1 0
!
[ai , bi ] ,
i=1
[ai , bi ], j = 1, . . . , n. Then as in Lemma 2 of
Z
1 0
!1/pj pj x − a j j Bm − Bm (tj ) dtj bj − a j
∂ mg
m (. . . , xj+1 , . . . , xn )
∂xj
Thus we derive Z
j Y
(28.30)
i=j+1
[29], see (62) there, we obtain |Bjg |
1 1 + = 1; pj qj
n Y
qj ,
j Q
,
(28.31)
[ai ,bi ]
i=1
[ai , bi ], j = 1, . . . , n.
i=j+1
≤
(bj − aj ) m!
m− q1
j
j−1 Y i=1
(bi − ai )
!−1/qj !
pj ! ! pqj j xj − a j − Bm (tj ) dtj B m bj − a j
q j
∂mg
m (. . . , xj+1 , . . . , xn )
∂xj
qj ,
j Q
)
− d→ x
[ai ,bi ]
i=1
!1/qj
(28.32)
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(bj − aj ) m!
= Z Z
n Q
i=j+1
Z
bj aj
m− q1
j
j−1 Y i=1
!− q1 !
j−1 Y
j
(bi − ai )
i=1
(bi − ai )
351
!1/qj
pj !qj /pj !1/qj ! xj − a j dxj − Bm (tj ) dtj Bm bj − a j
1 0
qj
∂mg
m (. . . , xj+1 , . . . , xn )
∂x [ai ,bi ] j
qj ,
That is we find
j Q
dxj+1 . . . dxn [ai ,bi ]
!1/qj
.
(28.33)
i=1
m−
1
(bj − aj ) qj ≤ m!
pj !qj /pj !1/qj !
∂mg xj − a j
− Bm (tj ) dtj dxj
m ,
∂xj bj − a j kBjg kqj
Z
bj aj
Z
1 0
B m
qj
j = 1, . . . , n. Similarly for f . Using (28.34) into (28.30) we find
n
1 X |Γ| ≤ 2m! j=1 Z
Z
bj aj
(28.34)
1 0
((
(bj − aj )
m− q1
j
pj !qj /pj !1/qj ) ! xj − a j dxj − Bm (tj ) dtj B m bj − a j
"
∂mg
kf kpj m
∂xj
qj
#)
∂mf
+ kgkpj m ,
∂xj
(28.35)
qj
proving the claim (28.13). When pj = qj = 2, all j = 1, . . . , n, then (( n 1 1 X |Γ| ≤ (bj − aj )m− 2 2m! j=1 Z
bj aj
Z
1
Bm 0
"
xj − a j bj − a j
!
− Bm (tj )
!2
!
dtj dxj
#)
∂ mf
∂mg
kf k2 m + kgk2 m
∂xj
∂xj 2
(by [98], p. 352)
2
!1/2 ) (28.36)
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n
1 X = 2m! j=1
((
(bj − aj )
Z
m− 12
bj
xj − a j (m!)2 2 |B2m | + Bm (2m)! bj − a j
aj
!!
dxj
#)
∂mg
∂ mf
kf k2 m + kgk2 m
∂xj
∂xj
"
2
!1/2 ) (28.37)
2
proving the claim (28.14). 3) Estimate with respect to k · k1 . We see that "Z n 1X − − |f (→ x )||Bjg (xj , . . . , xn )|d→ x |Γ| ≤ n Q 2 j=1 [ai ,bi ] i=1
+
Z
n Q
[ai ,bi ]
− − |g(→ x )||Bjf (xj , . . . , xn )|d→ x
i=1
#
(28.38)
# n 1X f g n n . + kgk1 kBj (xj , . . . , xn )k Q ≤ kf k1 kBj (xj , . . . , xn )k Q [ai ,bi ] [ai ,bj ] ∞, ∞, 2 j=1 "
i=j
i=j
(28.39)
By Lemma 3 of [29], (70) there, we have |Bjf (xj , xj+1 , . . . , xn )| ≤
(bj − aj )m−1 ! j−1 Y (bi − ai ) m! i=1
∂mf
m (. . . , xj+1 , . . . , xn )
∂xj
1,
j Q
[ai ,bi ]
i=1
!
xj − a j
Bm (t) − Bm
bj − a j
.
(28.40)
∞,[0,1]
The special cases follow: 1)0 When m = 2r, r ∈ N we have (see again Lemma 3 of [29], (71) there) |Bjf (xj , . . . , xn )| ≤
(bj − aj )2r−1 ! j−1 Y (bi − ai ) (2r)! i=1
∂ 2r f
2r (. . . , xj+1 , . . . , xn )
∂xj
1,
j Q
"
[ai ,bi ]
(1 − 2
−2r
i=1
! # xj − aj −2r )|B2r | + 2 B2r − B2r . bj − a j
(28.41)
Consequently, when m = 2r, r ∈ N, we have kBjf (xj , . . . , xn )k
∞,
n Q
i=j
[ai ,bi ]
(bj − aj )2r
∂ 2r f ≤
2r (2r)! ∂xj
∞
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"
(1 − 2
−2r
0
!
xj − a j
−2r )|B2r | + 2 B2r − B2r
bj − a j
∞,[aj ,bj ]
#
.
353
(28.42)
2) When m = 2r + 1, r ∈ N (see again Lemma 3 of [29], (72) there) |Bjf (xj , . . . , xn )| ≤
(bj − aj )2r ! j−1 Y (bi − ai ) (2r + 1)! i=1
∂ 2r+1
2r+1 f (. . . , xj+1 , . . . , xn )
∂xj
1,
j Q
[ai ,bi ]
"
i=1
! # 2(2r + 1)! xj − aj + B2r+1 . (2π)2r+1 (1 − 2−2r ) bj − a j
(28.43)
Consequently, when m = 2r + 1, r ∈ N we have kBjf (xj , . . . , xn )k "
∞,
n Q
[ai ,bi ]
i=j
(bj − aj )2r+1
∂ 2r+1 f ≤
2r+1
(2r + 1)! ∂xj
!
2(2r + 1)! xj − a j
+ B2r+1
2r+1 −2r
(2π) (1 − 2 ) bj − a j
∞,[aj ,bj ]
∞
#
.
(28.44)
3)0 When m = 1, we get again from Lemma 3 of [29], (73) there, that
∂f
1
(. . . , xj+1 , . . . , xn ) |Bjf (xj , . . . , xn )| ≤ j−1
j
Q ∂xj Y 1, [ai ,bi ] (bi − ai ) i=1 i=1
"
Thus it holds kBjf (xj , . . . , xn )k
∞,
n Q
i=j
1 + xj − 2
[ai ,bi ]
aj + b j 2
! # .
(28.45)
(bj − aj )
∂f ≤
[1 + bj − aj ].
∂xj 2
(28.46)
∞
Similar conclusions follow for g. Using (28.42), (28.44) and (28.46) into (28.38)–(28.39) we derive: 1)00 Case of m = 2r, r ∈ N, then ( n 1 X |Γ| ≤ (bj − aj )2r 2(2r)! j=1 "
!
xj − a j
−2r
−2r (1 − 2 )|B2r | + 2 B2r − B2r
bj − a j
∞,[aj ,bj ]
#
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"
∂ 2r g
kf k1 2r
∂xj
∞
#)
∂ 2r f
+ kgk1 2r .
∂xj
(28.47)
∞
2)00 Case of m = 2r + 1, r ∈ N, then
"
( n X 1 (bj − aj )2r+1 |Γ| ≤ 2(2r + 1)! j=1
!
2(2r + 1)! xj − a j
+ B2r+1
(2π)2r+1 (1 − 2−2r ) bj − a j "
∂ 2r+1 g
kf k1 2r+1
∂xj
∞
∞,[aj ,bj ]
#
#)
∂ 2r+1 f
+ kgk1 2r+1 .
∂xj
(28.48)
∞
3)00 Case of m = 1, then
( " n
∂g 1X
(bj − aj )(1 + bj − aj ) kf k1 |Γ| ≤
∂xj 4 j=1
∞
That is proving all the claims.
#)
∂f
+ kgk1 .
∂xj
(28.49)
∞
When m = 1 our basic assumptions are simplified to Assumption 28.10. Let f and the existing valued functions on
n Y
∂f , j = 1, . . . , n, be continuous real ∂xj
[ai , bi ].
i=1
Assumption 28.11. Here j = 1, . . . , n. We assume n Y 1) f : [ai , bi ] → R is continuous. i=1
2) For each j = 1, . . . , n, we suppose that gj (·) :=
∂f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn ) ∂xj
exists and is real valued with the possibility of being infinite only over an at most countable subset of (aj , bj ). 3) Part #2 is true for all (x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=1 i6=j
4) The functions for each j = 1, . . . , n, j
j ∂f z}|{ ( ·, ·, ·, xj+1 , . . . , xn ) ∈ L1 ϕj (z}|{ . . . ) := ∂xj
j Y
i=1
!
[ai , bi ] ,
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for any (xj+1 , . . . , xn ) ∈
n Y
355
[ai , bi ].
i=j+1
Assumption 28.12. Parts #1, #4 of Assumption 28.11 remain the same. We further suppose that for each j = 1, . . . , n and over [aj , bj ], the function f (x1 , . . . , xj−1 , ·, xj+1 , . . . , xn )
is absolutely continuous, and this is true for all
(x1 , . . . , xj−1 , xj+1 , . . . , xn ) ∈
n Y
[ai , bi ].
i=1 i6=j
We give Corollary 28.13. Let f, g as in Assumptions 28.10 or 28.11 or 28.12. Denote by Z 1 − − K(f,g) := n f (→ x )g(→ x )dx n Q Y [ai ,bi ] (bi − ai ) i=1 i=1
Z Z 1 1 → − → − → − → − f ( x )d x n g( x )d x − n n n . (28.50) Q Q Y Y [ai ,bi ] [ai ,bi ] (bi − ai ) i=1 (bi − ai ) i=1 i=1
i=1
Then we have:
! n Y ∂g ∂f 1) By assuming that , ∈ L∞ [ai , bi ] , for j = 1, . . . , n, it holds ∂xj ∂xj i=1 ( n √ √ 1 X |K(f,g) | ≤ (bj − aj ){4 3 − ln[3aj + 2 3(bj − aj ) − 3bj ] 48 j=1 √ + ln[−3aj + 2 3(bj − aj ) + 3bj ]}
#) "
∂f
∂g
. kf k∞
+ kgk∞
∂xj
∂xj ∞
2)Let pj , qj > 1:
It holds
(28.51)
∞
1 1 + = 1; j = 1, . . . , n. Suppose also that pj qj ! n Y ∂g ∂f , ∈ L qj [ai , bi ] . ∂xj ∂xj i=1
|K(f,g) | ≤
1 2
n Y
i=1
(bi − ai )
!
n X j=1
((
(bj − aj )
1− q1
j
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Z
Z
bj aj
"
1 0
!1/qj ) !qj /pj pj xj − a j dxj bj − aj − tj dtj
∂g
kf kpj
∂xj
qj
#)
∂f
+ kgkpj .
∂xj
When pj = qj = 2, all j = 1, . . . , n, it holds |K(f,g) | ≤
1 48
n Y
i=1
(28.52)
qj
(bi − ai )
!
n X j=1
(
(bj − aj )3/2
√ √ √ {4 3 − ln[3aj + 2 3(bj − aj ) − 3bj ] + ln[−3aj + 2 3(bj − aj ) + 3bj ]} "
3) Assume again
#)
∂f
∂g
. kf k2
+ kgk2
∂xj
∂xj
∂g ∂f , ∈ L∞ ∂xj ∂xj It holds n
1X |K(f,g) | ≤ 4 j=1
(
2
n Y
i=1
(28.53)
2
!
[ai , bi ] ,
j = 1, . . . , n.
#)
"
∂g
∂f (1 + bj − aj )
kf k1 .
+ kgk1
n Y
∂xj
∂xj ∞ ∞ [ai , bi ]
(28.54)
i=1 i6=j
Proof.
By Theorem 28.9 and s 2 ! Z bj (bj − aj ) 1 1 xj − a j dxj = + − 12 b − a 2 24 j j aj
(28.55)
√ √ √ {4 3 − ln[3aj + 2 3(bj − aj ) − 3bj ] + ln[−3aj + 2 3(bj − aj ) + 3bj ]}, j = 1, . . . , n.
Remark 28.14. In the Assumptions 28.3 or 28.5 or 28.7. We further suppose that ∂lf ∂lf (. . . , b , . . . ) = (. . . , aj , . . . ), j ∂xlj ∂xlj
(28.56)
for all j = 1, . . . , n and all l = 0, 1, . . . , m − 2. In that case Afj = 0, j = 1, . . . , n, see (28.9). So if we consider f, g as above, we also find Agj = 0, j = 1, . . . , n.
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In that case, see (28.11), it collapses to Z − − − ∆(f,g) = Q f (→ x )g(→ x )d→ x n [ai ,bi ]
i=1
−
1 n Y
i=1
(bi − ai )
Z
n Q
[ai ,bi ]
− − f (→ x )d→ x
i=1
! Z
n Q
[ai ,bi ]
i=1
! → − → − g( x )d x ,
(28.57)
that is simplified a lot, etc. Remark 28.15. Let [ci , di ] !! ⊆ [ai , bi ], i = 1, . . . , n. Let µ be a probability measure n n Y Y [ci , di ] , P stands for the powerset. [ci , di ], P on i=1
i=1
Let f as in Assumptions 28.3 or 28.5 or 28.7. Then by Theorem 28.4, see (28.8), we have Z Z 1 − − → − → − f (→ x )d→ x M := Q f ( x )µ(d x ) − n n Q Y n [ci ,di ] [ai ,bi ] i=1 (bi − ai ) i=1 i=1
−
n Z X j=1
n Q
[ci ,di ]
i=1
Afj dµ
n Z n Z X X f Bj dµ ≤ |Bjf |dµ. = n n Q Q [ci ,di ] [ci ,di ] j=1
j=1
i=1
(28.58)
(28.59)
i=1
When conditions (28.56) are valid then Z Z 1 → − → − → − → − f ( x )dµ( x ) − n M = Q f ( x )d x . n n Q Y [ci ,di ] [ai ,bi ] i=1 (bi − ai ) i=1
(28.60)
i=1
When m = 1 then again M is given by (28.60). We present the second main result, now is regarding comparison of integral means, see also [30]. Theorem 28.16. Let f as in Assumptions 28.3 or 28.5 or 28.7. Let µ and M as in Remark 28.15, see (28.58). Additionally assume that ! n Y ∂mf [ai , bi ] , j = 1, . . . , n; m ∈ N. ∈ L∞ ∂xm j i=1 Then
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1)
Z
n Q
s
[ci ,di ]
( n
m 1 X m ∂ f M≤ (bj − aj ) m
∂xj m! j=1
(m!)2 2 |B2m | + Bm (2m)!
i=1
xj − a j bj − a j
!
∞
− dµ(→ x)
!)
=: L1 .
(28.61)
1 1 + = 1, j = 1, . . . , n, it holds pj qj !−1/qj ( j−1 n Y 1 X m− q1 j (bi − ai ) (bj − aj ) M≤ m! j=1 i=1
2) Let pj , qj > 1:
Z
×
Z
n Q
[ci ,di ]
i=1
Z
1 0
! !1/pj pj − Bm xj − aj − Bm (tj ) dtj dµ(→ x) bj − a j
q j
∂mf
(. . . , x , . . . , x )
j+1 n n m Q
[ci ,di ] ∂xj
j Q
qj ,
i=1
− dµ(→ x) [ai ,bi ]
×
Z
n Q
[ci ,di ]
(m!)2 2 |B2m | + Bm (2m)!
i=1
2
∂mf
(. . . , x , . . . , x )
j+1 n n m Q
[ci ,di ] ∂xj
2,
i=1
3) (i) Case m = 2r, r ∈ N, it holds n
1 X M≤ (2r)! j=1
(
=: L2 .
j Q
j−1 Y i=1
(bi − ai )
xj − a j bj − a j
!−1/2
!1/2 ! → − dµ( x )
− dµ(→ x) [ai ,bi ]
!1/2 )
= L3 .
i=1
(bj − aj )2r−1 ! j−1 Y (bi − ai ) i=1
Z
∂ 2r f
2r (. . . , xj+1 , . . . , xn ) n Q
∂x [ci ,di ] j i=1
(28.62)
i=1
When pj = qj = 2, j = 1, . . . , n, it holds ( n 1 X 1 (bj − aj )m− 2 M≤ m! j=1 Z
!1/qj )
1,
j Q
i=1
− dµ(→ x) [ai ,bi ]
!
(28.63)
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"
(1 − 2
−2r
#)
xj − a j
−2r =: L4 . )|B2r | + 2 B2r − B2r
bj − a j
359
(28.64)
∞
(ii) Case m = 2r + 1, r ∈ N, it holds
n
X 1 M≤ (2r + 1)! j=1
(
(bj − aj )2r ! j−1 Y (bi − ai ) i=1
Z
∂ 2r+1 f
(. . . , x , . . . , x )
j+1 n n 2r+1 Q
[ci ,di ] ∂xj i=1
1,
j Q
− dµ(→ x) [ai ,bi ]
!
i=1
#)
xj − a j 2(2r + 1)!
=: L5 . + B2r+1 ×
(2π)2r+1 (1 − 2−2r ) bj − a j "
(28.65)
∞
(iii) When m = 1, it holds
n
1X M≤ 2 j=1
(
(1 + bj − aj ) j−1 Y (bi − ai ) i=1
Z
∂f
(. . . , x , . . . , x )
j+1 n n Q
∂x j [ci ,di ] i=1
1,
j Q
− dµ(→ x) [ai ,bi ]
!)
=: L6 .
(28.66)
i=1
Proof. We rely on (28.59). 1) Estimate with respect to k · k∞ . From (28.27) we obtain s
!
(bj − aj )m (m!)2 xj − a j
∂mf f 2 |Bj (xj , . . . , xn )| ≤ , |B2m | + Bm
m
∂xj m! (2m)! bj − a j ∞ (28.67) n Y all j = 1, . . . , n, any (xj , . . . , xn ) ∈ [ai , bi ]. i=j
Thus
n Z X j=1
Z
n Q
[ci ,di ]
n Q
[ci ,di ]
i=1
s
|Bjf |dµ
( n
m 1 X m ∂ f ≤ (bj − aj ) m
∂xj m! j=1
(m!)2 2 |B2m | + Bm (2m)!
i=1
proving the claim (28.61).
xj − a j bj − a j
∞
!) ! → − dµ( x ) ,
(28.68)
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2) Estimate with respect to k·kqj . Here pj > 1, qj > 1:
1 1 + = 1; j = 1, . . . , n. p j qj
By Lemma 2 of [29], see (28.62) there, we derive |Bjf |
(bj − aj ) ≤ m!
m− q1
j
j−1 Y i=1
(bi −ai )
−1/qj
Z
1 0
!1/pj pj x − a j j B m − Bm (tj ) dtj bj − a j
∂mf
m (. . . , xj+1 , . . . , xn )
∂xj true for all (xj , xj+1 , . . . , xn ) ∈ Hence we have n Z X j=1
n Q
[ci ,di ]
|Bjf |dµ
i=1
Z
n Q
[ci ,di ]
i=1
n Y
1 0
×
Z
n Q
i=1
n Q
[ci ,di ]
i=1
[ci ,di ]
0
(28.69)
i=1
[ai , bi ], j = 1, . . . , n.
j−1 Y i=1
(bi − ai )
qj ,
j Q
[ai ,bi ]
#
− dµ(→ x)
)
(28.70)
i=1
j−1 Y i=1
(bi − ai )
!−1/qj
! !1/pj pj x − a j j → − Bm − Bm (tj ) dtj dµ( x ) bj − a j
∂mf
m (. . . , xj+1 , . . . , xn )
∂xj
qj ,
j Q
[ai ,bi ]
! qj
− dµ(→ x)
i=1
proving the claim (28.62). When pj = qj = 2, j = 1, . . . , n, it holds (see p.352 of [98]) ( n Z n X 1 X 1 f |Bj |dµ ≤ (bj − aj )m− 2 n Q m! [ci ,di ] j=1 j=1 i=1
!−1/qj
!1/pj pj x − a j j B m − Bm (tj ) dtj bj − a j
∂ mf
m (. . . , xj+1 , . . . , xn )
∂xj
1
, [ai ,bi ]
( n 1 X m− 1 ≤ (bj − aj ) qj m! j=1
" Z
Z
j Q
i=j
( n 1 X m− 1 (bj − aj ) qj ≤ m! j=1 Z
qj ,
!1/qj )
,
(28.71)
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j−1 Y i=1
(bi − ai )
×
Z
!−1/2
Z
n Q
[ci ,di ]
(m!)2 2 |B2m | + Bm (2m)!
i=1
2
∂mf
m (. . . , xj+1 , . . . , xn ) n Q
∂x [ci ,di ] j
2,
i=1
j Q
xj − a j bj − a j
!1/2 ! → − dµ( x )
− dµ(→ x) [ai ,bi ]
361
!1/2 )
,
i=1
proving the claim (28.63). 3) Estimate with respect to k · k1 . The special cases follow: (i) When m = 2r, r ∈ N, by (28.4) we obtain ( n Z n X (bj − aj )2r−1 1 X f |Bj |dµ ≤ ! n Q j−1 (2r)! j=1 Y j=1 i=1[ci ,di ] (bi − ai ) i=1
Z
∂ 2r f
2r (. . . , xj+1 , . . . , xn ) n Q
∂x [ci ,di ] j
i=1
1,
j Q
− dµ(→ x) [ai ,bi ]
!"
(1 − 2−2r )|B2r |
i=1
#)
xj − a j
−2r
. + 2 B2r − B2r
bj − a j
(28.72)
∞
(ii) When m = 2r + 1, r ∈ N, then by (28.43) it holds ( n Z n X X 1 (bj − aj )2r f |B |dµ ≤ ! n j Q j−1 (2r + 1)! j=1 Y [ci ,di ] j=1 i=1 (bi − ai ) i=1
Z
∂ 2r+1 f
(. . . , x , . . . , x )
j+1 n n 2r+1 Q
[ci ,di ] ∂xj i=1
1,
j Q
− dµ(→ x) [ai ,bi ]
i=1
#)
2(2r + 1)! xj − a j
× . + B2r+1
(2π)2r+1 (1 − 2−2r ) bj − a j "
!
(28.73)
∞
(iii) When m = 1, from (28.45) it holds ( n Z m X 1 X (1 + bj − aj ) f |Bj |dµ ≤ n Q 2 j=1 j−1 Y [ci ,di ] j=1 i=1 (bi − ai ) i=1
Z
∂f
(. . . , xj+1 , . . . , xn )
n Q
∂x j [ci ,di ]
i=1
proving all claims.
1,
j Q
− dµ(→ x) [ai ,bi ]
!)
,
(28.74)
i=1
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We give Corollary 28.17. All as in Theorem 28.16. 1) Let m = 2r, r ∈ N. Then M ≤ min{L1 , L2 , L3 , L4 }.
(28.75)
2) Let m = 2r + 1, r ∈ N. Then M ≤ min{L1 , L2 , L3 , L5 }.
(28.76)
Next we compare means when m = 1. Corollary 28.18. Let f as in Assumptions 28.10 or 28.11 or 28.12. Let µ as in Remark 28.15. Additionally suppose that ! n Y ∂f [ai , bi ] , j = 1, . . . , n. ∈ L∞ ∂xj i=1 Then
Z Z 1 → − → − → − → − M = Q f ( x )dµ( x ) − f ( x )d x n n Q Y n [ci ,di ] [ai ,bi ] i=1 (bi − ai ) i=1 i=1
n X
≤ Z
n Q
[ci ,di ]
s
1 + 12
i=1
j=1
(
∂f
(bj − aj )
∂xj
1 xj − a j − bj − a j 2
2 !
∞
− dµ(→ x)
!)
=: L01 .
(28.77)
1 1 + = 1, j = 1 . . . , n. It holds pj qj ( !−1/qj j−1 n X Y 1− q1 M≤ (bj − aj ) j (bi − ai )
2) Let pj , qj > 1:
j=1
Z
×
Z
n Q
i=1
[ci ,di ]
i=1
Z
1 0
! !1/pj pj xj − a j → − bj − aj − tj dtj dµ( x )
q j
∂f
(. . . , xj+1 , . . . , xn )
n Q
∂x j [ci ,di ]
i=1
qj ,
j Q
i=1
− dµ(→ x) [ai ,bi ]
!1/qj )
=: L02 .
(28.78)
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When pj = qj = 2, j = 1, . . . , n, it holds ( !−1/2 j−1 n Y X p bj − a j (bi − ai ) M≤ i=1
j=1
Z
×
Proof.
Z
n Q
1 + 12
[ci ,di ]
i=1
xj − a j 1 − bj − a j 2
2
∂f
(. . . , xj+1 , . . . , xn )
n Q
∂x j [ci ,di ]
2,
i=1
j Q
!1/2 2 ! → − dµ( x ) − dµ(→ x)
[ai ,bi ]
!1/2 )
=: L03 .
(28.79)
i=1
By Theorem 28.16.
We end chapter with Corollary 28.19. All as in Corollary 28.18. Then M ≤ min{L01 , L02 , L03 , L6 }. Proof.
By Corollary 28.18 and (28.66).
(28.80)
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Chapter 29
Multivariate Fink Type Identity Applied to Multivariate Ostrowski, Comparison of Means and Gr¨ uss Inequalities We present a general multivariate Fink type identity which is a representation formula for a multivariate function. Using it we derive general tight multivariate high order Ostrowski type, comparison of means and Gr¨ uss type inequalities. The estimates involve Lp norms, any 1 ≤ p ≤ ∞. This chapter is based on [39]. 29.1
Introduction
The article that motivates the most this chapter is of A. Fink (1992), see [141], where he presents among others a representation formulae for a univariate function. We generalize this to multi-dimension, see Theorem 29.6. Using this multivariate representation formula we derive multivariate Ostrowski type inequalities, see Theorems 29.9–29.12. We continue the comparison of general multivariate integral mean to Riemann integral average, see Theorem 29.15. Then we present multivariate Gr¨ uss type inequalities, see Theorems 29.17, 29.19, 29.20. At the end we apply the general results to the cases: of a two dimensional function involving partials of order one, see Corollaries 29.23, 29.25–29.33, and of a three-dimensional functions involving second order partials, see Corollaries 29.36, 29.38–29.46. The results here are also motivated by the works of Ostrowski [196], Gr¨ uss [150], and of the author [16], [17], [24], [38]. We would like to mention the following inspiring results. Theorem 29.1 (Ostrowski, 1938, [196]). Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e., kf 0 k∞ = sup |f 0 (t)| < +∞. Then t∈(a,b)
Z 1 b − a
b a
2 # " x − a+b 1 2 f (t)dt − f (x) ≤ (b − a)kf 0 k∞ + 4 (b − a)2
for any x ∈ [a, b]. The constant 1/4 is the best possible.
Theorem 29.2 (Gr¨ uss, 1935, [150]). Let f, g integrable functions from [a, b] into R, such that m ≤ f (x) ≤ M , ρ ≤ g(x) ≤ σ, for all x ∈ [a, b], where m, M , ρ, σ ∈ R. 365
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Then
Z b Z b 1 Z b 1 1 f (x)g(x)dx − f (x)dx g(x)dx b − a a b−a a b−a a 1 (M − m)(σ − ρ). 4
≤
29.2 Here
Main Results m Q
i=1
[ai , bi ] ⊆ Rm , m, n ∈ N.
We mention General Assumptions 29.3. Let f :
m Q
i=1
1) for j = 1, . . . , m we have that
[ai , bi ] → R. We assume
∂ n−1 f (x1 , x2 , . . . , xj−1 , sj , xj+1 , . . . , xm ) ∂xjn−1 is absolutely continuous in sj on [aj , bj ], for every (x1 , x2 , . . . , xj−1 , xj+1 , . . . , xm ) ∈
m Y
[ai , bi ],
i=1 i6=j
2) for j = 1, . . . , m we have that ∂ nf (s1 , . . . , sj , xj+1 , . . . , xm ) ∂xnj is continuous on
j Q
[ai , bi ] for every
i=1
(xj+1 , . . . , xm ) ∈
m Y
[ai , bi ],
i=j+1
3) for each j = 1, . . . , m, and for every ` = 1, . . . , n − 2, we have that ∂`f (s1 , s2 , . . . , sj−1 , xj , . . . , xm ) ∂x`j
is continuous on
j−1 Q i=1
[ai , bi ], for every (xj , . . . , xm ) ∈
4) f is continuous on We mention
m Q
[ai , bi ].
m Q
[aj , bj ].
i=j
i=1
Brief Assumptions 29.4. Let f : i = 1, . . . , m, are continuous on
m Q
i=1
m Q
i=1
[ai , bi ] → R with
[ai , bi ].
∂`f ∂x`i
for ` = 0, 1, . . . , n;
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We mention Definition 29.5. We put si − a i , k(si , xi ) := si − b i ,
ai ≤ s i ≤ xi ≤ b i , ai ≤ xi < s i ≤ b i ,
i = 1, . . . , m.
We present the representation result of Fink type.
Theorem 29.6. Let f as in General Assumptions 29.3 or Brief Assumptions 29.4, m Y (x1 , . . . , xm ) ∈ [ai , bi ]. j=1
Then
f (x1 , . . . , xm ) = Q m
nm
j=1
(bj − aj )
Z
m Q
f (s1 , . . . , sm )ds1 · · · dsm +
[aj ,bj ]
j=1
where for i = 1, . . . , m we set
m X
Ti ,
(29.1)
i=1
n−1 ni−1 X n−k Ti := Ti (xi , . . . , xm ) := i k! Q (bj − aj ) k=1 ×
(Z
j=1
i−1 Q
[aj ,bj ]
j=1
k−1
∂ f (s , s , . . . , si−1 , bi , xi+1 , . . . , xm )(xi − bi )k k−1 1 2 ∂xi
) ∂ k−1 f − (s1 , s2 , . . . , si−1 , ai , xi+1 , . . . , xm )(xi − ai )k ds1 ds2 · · · dsi−1 ∂xik−1 +
ni−1 i Q (n − 1)! (bj − aj ) j=1
Z
i Q
[aj ,bj ]
(xi − si )n−1 k(si , xi )
j=1
! ∂nf × (s1 , s2 , . . . , si , xi+1 , . . . , xm )ds1 ds2 · · · dsi . ∂xni
When n = 1 the sum
n−1 P
(29.2)
in (29.2) is zero.
k=1
For the proof of Theorem 29.6 we need Fink’s identity. Theorem 29.7 (Fink, [141]). Let any a, b ∈ R, f : [a, b] → R, n ≥ 1, f (n−1) is absolutely continuous on [a, b]. Then Z b n−1 X n − k f (k−1) (a)(x − a)k − f (k−1) (b)(x − b)k n f (x) = f (t)dt − b−a a k! b−a k=1 Z b 1 + (x − t)n−1 k(t, x)f (n) (t)dt, (29.3) (n − 1)!(b − a) a
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where k(t, x) :=
When n = 1 the sum
n−1 P
t − a, t − b,
a ≤ t ≤ x ≤ b, a ≤ x < t ≤ b.
in (29.3) is zero.
k=1
Proof of Theorem 29.6. Here we apply (29.3) repeatedly. We have n f (x1 , . . . , xm ) = b1 − a 1
Z
b1
f (s1 , x2 , . . . , xm )ds1 + T1 (x1 , . . . , xm ),
(29.4)
a1
where n−1 X n − k ∂ k−1 f (b1 , x2 , . . . , xm ) 1 (x1 − b1 )k T1 (x1 , . . . , xm ) := k−1 (b1 − a1 ) k! ∂x 1 k=1 ! k−1 ∂ f − k−1 (a1 , x2 , . . . , xm )(x1 − a1 )k ∂x1 Z b1 ∂nf 1 (29.5) (x1 − s1 )n−1 k(s1 , x1 ) n (s1 , x2 , . . . , xm )ds1 . + (n − 1)!(b1 − a1 ) a1 ∂x1
But it holds Z b2 n f (s1 , s2 , x3 , . . . , xm )ds2 b 2 − a 2 a2 n−1 X n − k ∂ k−1 f 1 + (s , b , x3 , . . . , xm )(x2 − b2 )k k−1 1 2 (b2 − a2 ) k! ∂x 2 k=1 ! k−1 ∂ k f (s1 , a2 , x3 , . . . , xm )(x2 − a2 ) − ∂x2k−1 Z b2 1 + (x2 − s2 )n−1 k(s2 , x2 ) (n − 1)!(b2 − a2 ) a2 ∂ nf × n (s1 , s2 , x3 , . . . , xm )ds2 . ∂x2 f (s1 , x2 , . . . , xm ) =
(29.6)
Combining (29.4) and (29.6) we obtain Z b1 Z b 2 n2 f (s1 , s2 , x3 , . . . , xm )ds1 ds2 (b1 − a1 )(b2 − a2 ) a1 a2 + T2 (x2 , x3 , . . . , xm ) + T1 (x1 , . . . , xm ), (29.7)
f (x1 , x2 , . . . , xm ) =
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where
× − + ×
n−1 X n − k n T2 (x2 , x3 , . . . , xm ) := (b1 − a1 )(b2 − a2 ) k! k=1 Z b1 k−1 ∂ f (s , b , x3 , . . . , xm )(x2 − b2 )k k−1 1 2 ∂x a1 2 ! k−1 ∂ f k (s1 , a2 , x3 , . . . , xm )(x2 − a2 ) ds1 ∂x2k−1 Z b1 Z b2 n (x2 − s2 )n−1 k(s2 , x2 ) (n − 1)!(b1 − a1 )(b2 − a2 ) a1 a2 ∂nf (s1 , s2 , x3 , . . . , xm )ds1 ds2 . ∂xn2
(29.8)
Next we observe that
Z b3 n f (s1 , s2 , s3 , x4 , . . . , xm )ds3 b 3 − a 3 a3 n−1 X n − k ∂ k−1 f 1 + (s , s , b , x4 , . . . , xm )(x3 − b3 )k k−1 1 2 3 (b3 − a3 ) k! ∂x 3 k=1 ! k−1 ∂ f k − k−1 (s1 , s2 , a3 , x4 , . . . , xm )(x3 − a3 ) ∂x3 Z b3 1 + (x3 − s3 )n−1 k(s3 , x3 ) (n − 1)!(b3 − a3 ) a3 ∂nf × n (s1 , s2 , s3 , x4 , . . . , xm )ds3 . (29.9) ∂x3 f (s1 , s2 , x3 , . . . , xm ) =
Combining (29.7) and (29.9) we derive Z b1 Z b2 Z b3 n3 f (s1 , s2 , s3 , x4 , . . . , xm )ds1 ds2 ds3 f (x1 , x2 , . . . , xm ) = 3 Q (bi − ai ) a1 a2 a3 i=1
+ T3 (x3 , x4 , . . . , xm ) + T1 + T2 ,
where T3 (x3 , x4 , . . . , xm ) :=
n2 3 Q
j=1
Z
b1
Z
b2
(bj − aj )
!
(29.10)
n−1 X k=1
n−k k!
∂ k−1 f (s1 , s2 , b3 , x4 , . . . , xm )(x3 − b3 )k ∂x3k−1 a1 a2 !! ∂ k−1 f k − (s1 , s2 , a3 , x4 , . . . , xm )(x3 − a3 ) ds1 ds2 ∂x3k−1 ×
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+ (n − 1)!
n2 3 Q
j=1
(bj − aj )
! Z
3 Q
[aj ,bj ]
(x3 − s3 )n−1 k(s3 , x3 )
j=1
! ∂nf × (s1 , s2 , s3 , x4 , . . . , xm )ds1 ds2 ds3 . ∂xn3
(29.11)
We also observe that
Z b4 n f (s1 , s2 , s3 , s4 , x5 , . . . , xm )ds4 f (s1 , s2 , s3 , x4 , . . . , xm ) = b 4 − a 4 a4 n−1 X n − k ∂ k−1 f 1 + (s , s , s , b , x , . . . , xm )(x4 − b4 )k k−1 1 2 3 4 5 (b4 − a4 ) k! ∂x 4 k=1 ∂ k−1 f k (s1 , s2 , s3 , a4 , x5 , . . . , xm )(x4 − a4 ) − ∂x4k−1 Z b4 1 (x4 − s4 )n−1 k(s4 , x4 ) + (n − 1)!(b4 − a4 ) a4 ∂nf × (s1 , s2 , s3 , s4 , x5 , . . . , xm )ds4 . (29.12) ∂xn4 Combining (29.10) and (29.12) we obtain f (x1 , x2 , . . . , xm ) Z n4 f (s1 , s2 , s3 , s4 , x5 , . . . , xm )ds1 ds2 ds3 ds4 = 4 4 Q Q (bj − aj ) j=1[aj ,bj ] j=1
+
4 X
Tj ,
(29.13)
j=1
where
T4 (x4 , x5 , . . . , xm ) :=
n3 4 Q
j=1
(bj − aj )
!(n−1 " X n − k Z k=1
k!
3 Q
[aj ,bj ]
j=1
∂ k−1 f (s1 , s2 , s3 , b4 , x5 , . . . , xm )(x4 − b4 )k ∂x4k−1
#) ∂ k−1 f k − (s1 , s2 , s3 , a4 , x5 , . . . , xm )(x4 − a4 ) ds1 ds2 ds3 ∂x4k−1 ! Z n3 (x4 − s4 )n−1 k(s4 , x4 ) + 4 Q 4 Q [aj ,bj ] (n − 1)! (bj − aj ) j=1 j=1
! ∂nf × (s1 , s2 , s3 , s4 , x5 , . . . , xm )ds1 ds2 ds3 ds4 . ∂xn4
(29.14)
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etc. proving the claim. We make
Remark 29.8. For i = 1, . . . , m denote Ai (xi , xi+1 , . . . , xm ) :=
×
(Z
ni−1 i Q
j=1
i−1 Q
j=1
[aj ,bj ]
(bj − aj )
! n−1 X n − k k!
k=1
∂ k−1 f (s1 , s2 , . . . , si−1 , bi , xi+1 , . . . , xm )(xi − bi )k ∂xik−1
) ∂ k−1 f k − k−1 (s1 , s2 , . . . , si−1 , ai , xi+1 , . . . , xm )(xi − ai ) ds1 ds2 . . . dsi−1 . ∂xi (29.15) When n = 1 then Ai = 0, i = 1, . . . , m. Also we denote Bi (xi , xi+1 , . . . , xm ) :=
ni−1 i Q (n − 1)! (bj − aj ) j=1
×
! Z
!
i Q
[aj ,bj ]
(xi − si )n−1 k(si , xi )
j=1
∂nf (s1 , s2 , . . . , si , xi+1 , . . . , xm )ds1 · · · dsi . ∂xni
(29.16)
That is Ti = A i + B i ,
i = 1, . . . , m.
(29.17)
Thus by (29.1) we derive Ef (x1 , . . . , xm ) := f (x1 , . . . , xm ) Z m m X X nm −Q f (s , . . . , s )ds · · · ds − A = Bi . (29.18) 1 m 1 m i m m Q i=1 i=1 (bj − aj ) j=1[ai ,bj ] j=1
Thus |Ef (x1 , . . . , xm )| ≤
m X i=1
|Bi |.
(29.19)
Next we estimate Ef via some Ostrowski type inequalities, see Theorems 29.9 – 29.12.
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Theorem 29.9. Assume all as in Theorem 29.6. Then m P (
−i−
∂ n f z}|{ ni−1 i=1
· · · , x , . . . , x ) |Ef (x1 , . . . , xm )| ≤ i+1 m
(bi − ai ) (n − 1)! ∂xni ∞ " n n bi (bi − xi ) − ai (xi − ai ) × n ( n−1 X n − 1 (n + 3)xn+1 n−λ−1 i (−1) + (λ + 2)(n − λ + 1) λ λ=0 )#) λ n−λ+1 m λ+2 Y xn−λ−1 b xi a i i + i [aj , bj ]. (29.20) , ∀(x1 , . . . , xm ) ∈ − (n − λ + 1) (λ + 2) j=1
Proof.
We observe that Z ni−1 |Bi | ≤ |xi − si |n−1 |k(si , xi )| i Q i Q [aj ,bj ] (n − 1)! (bj − aj ) j=1 j=1
! n ∂ f × n (s1 , s2 , . . . , si , xi+1 , . . . , xm ) ds1 · · · dsi ∂xi
−i−
∂ n f z}|{ ni−1
≤ · · · , xi+1 , . . . , xm
(n − 1)!(bi − ai ) ∂xni ∞ Z bi |xi − si |n−1 |k(si , xi )|dsi =: (∗). ×
(29.21)
ai
We find that Z bi ai
= =
|xi − si |n−1 |k(si , xi )|dsi Z
Z
xi ai xi ai
"
|xi − si |n−1 |k(si , xi )|dsi + (xi − si )n−1 (si − ai )dsi +
Z
Z
bi xi bi
xi
|xi − si |n−1 |k(si , xi )|dsi
(si − xi )n−1 (bi − si )dsi
bi (bi − xi )n − ai (xi − ai )n = n ( n−1 X n − 1 (n + 3)xn+1 n−λ−1 i + (−1) λ (λ + 2)(n − λ + 1) λ=0 λ n−λ+1 )# xn−λ−1 bλ+2 xi a i i i − =: γi . + (n − λ + 1) (λ + 2) Therefore we get
n −i−
∂ f z}|{ ni−1
(∗) ≤ · · · , xi+1 , . . . , xm
γi . (n − 1)!(bi − ai ) ∂xn i
∞
(29.22)
(29.23)
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Finally we find |Ef (x1 , . . . , xm )| ≤
m X i=1
proving the claim.
n −i−
∂ f z}|{
ni−1
· · · , x , . . . , x ) i+1 m γi , n
(n − 1)!(bi − ai ) ∂xi ∞
(29.24)
We continue with Theorem 29.10. Assume all as in Theorem 29.6, and n
m Y
∂ f ∈ L∞ [aj , bj ] , ∂xni j=1
i = 1, . . . , m.
Then m P
n
∂ f
|Ef (x1 , x2 , . . . , xm )| ≤ (n − 1)! ∂xni ∞ " ( bi (bi − xi )n − ai (xi − ai )n ni−1 × (bi − ai ) n ( λ n−λ+1 n−1 X n−1 xi a i (n + 3)xn+1 i − + (−1)n−λ−1 (λ + 2)(n − λ + 1) (n − λ + 1) λ λ=0 )#) m Y bλ+2 xn−λ−1 i i , ∀(x1 , . . . , xm ) ∈ [ai , bj ]. + (λ + 2) j=1 i=1
Proof.
From Theorem 29.9.
(29.25)
Next we give Theorem 29.11. Suppose all as in Theorem 29.6, and let p, q > 1 such that 1 1 p + q = 1. Then ni−1 i=1 |Ef (x1 , . . . , xm )| ≤ i−1 Q (n − 1)! 1/p (bj − aj ) (bi − ai ) m P
j=1
1/q (xi − ai )nq+1 + (bi − xi ) B((n − 1)q + 1, q + 1)
n −i−
m Y
∂ f z}|{
× · · · , x , . . . , x ) , ∀(x , . . . , x ) ∈ [aj , bj ]. i+1 m 1 m
∂xn
i p j=1
×
nq+1
(29.26)
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Proof.
We notice that
|Bi | ≤
n −i−
∂ f z}|{ ni−1
∂xn · · · , xi+1 , . . . , xm i Q i p (n − 1)! (bj − aj ) j=1
Z
×
=
bi
ai
1/q !1/q i−1 Y (bj − aj ) |xi − si |q(n−1) |k(si , xi )|q dsi j=1
(by [242], p. 256)
ni−1 i−1 1/p Q (bj − aj ) (n − 1)!(bi − ai ) j=1
nq+1
× ((xi − ai ) + (bi − xi )nq+1 )B((n − 1)q + 1, q + 1)
n −i−
∂ f z}|{
× · · · , x , . . . , x i+1 m ,
∂xn
i
1/q
(29.27)
p
proving the claim.
We also present Theorem 29.12. Suppose all as in Theorem 29.6. Then
|Ef (x1 , . . . , xm )| ≤
m P
i=1
(n − 1)!
(
ni−1 i Q
j=1
Proof.
(bj − aj )
max(xi − ai , bi − xi )
n
)
n −i− m Y
∂ f z}|{
, ∀(x , . . . , x ) ∈ [aj , bj ]. × n · · · , xi+1 , . . . , xm 1 m
∂xi 1 j=1
We observe that
|Bi | ≤
ni−1 i Q (n − 1)! (bj − aj )
sup si ∈[ai ,bi ]
j=1
n −i−
∂ f z}|{
× n · · · , xi+1 , . . . , xm
∂xi
1
|xi − si |n−1 |k(si , xi )|
(29.28)
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375
n ni−1 max(xi − ai , bi − xi ) i Q (n − 1)! (bj − aj ) j=1
n −i−
∂ f z}|{
× n · · · , xi+1 , . . . , xm
,
∂xi
(29.29)
1
proving the claim.
We give Corollary 29.13. Assume all as in Theorem 29.6. Then |Ef (x1 , . . . , xm )| ≤ min R.H.S.(29.20), R.H.S.(29.26), R.H.S.(29.28)},
where p, q > 1 : Proof.
1 p
+
1 q
= 1, ∀(x1 , . . . , xm ) ∈
By (29.20), (29.26) and (29.28).
m Q
[ai , bi ].
i=1
We proceed with the comparison of integral means. For that we make Remark 29.14. Let [ci , di ] ⊆ [ai , bi ], i = 1, . . . , m. Let µ be a probability measure on !! m m Y Y [ci , di ] , [ci , di ], P i=1
i=1
P stands for the powerset. Then Z Z nm f (s1 , . . . , sm )ds1 · · · dsm f (x1 , . . . , xm )dµ − Q Q m m Q m [ci ,di ] (bj − aj ) j=1[aj ,bj ] i=1 j=1
− A (x , . . . , x )dµ i i m m Q [ci ,di ] i=1 i=1 Z Z = Q E (x , . . . , x )dµ |Ef (x1 , . . . , xm )|dµ. ≤ Q f 1 m m m [ci ,di ] [ci ,di ] m Z X
i=1
(29.30)
i=1
We conclude
Theorem 29.15. Suppose all as in Theorem 29.6, and
∂nf ∂xn i
∈ L∞
m Q
j=1
[aj , bj ] ,
i = 1, . . . , m. Let [ci , di ] ⊆ [ai , bi ], i = 1, . . . , m and µ be a probability measure on !! m m Y Y [ci , di ] . [ci , di ], P i=1
i=1
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Then Z Z nm f (x1 , . . . , xm )dµ − Q f (s1 , . . . , sm )ds1 · · · dsm Q m m Q m [cj ,dj ] [aj ,bj ] (b − a ) j j j=1 j=1 j=1
−
m Z X i=1
m P
A (x , . . . , x )dµ i i m m Q [ci ,di ]
i=1
"
n ( Z
∂ f ni−1 1
b (bi − xi )n dµ ≤ i m Q (n − 1)! ∂xni ∞ (bi − ai ) n [cj ,dj ] j=1 ! Z i=1
− ai
m Q
[cj ,dj ]
(xi − ai )n dµ
j=1
R m xn+1 dµ ( (n + 3) Q [cj ,dj ] i n−1 j=1 + (−1)n−λ−1 (λ + 2)(n − λ + 1) λ λ=0 R n−λ+1 R m m ai xλi dµ bλ+2 xn−λ−1 dµ Q Q i i [cj ,dj ] [cj ,dj ] j=1 j=1 − + . (n − λ + 1) (λ + 2) n−1 X
Proof.
We use (29.25) and (29.30).
(29.31)
Similar results to (29.31) can be derived by integrating against µ the inequalities (29.26) and (29.28). We continue with Gr¨ uss type inequalities m Q
Remark 29.16. Let f, g :
i=1
obtain
f (x1 , . . . , xm ) = Q m
nm
j=1
+
[ai , bi ] → R as in Theorem 29.6. Then by (29.1) we
(bj − aj )
m X
Z
m Q
[aj ,bj ]
f (s1 , . . . , sm )ds1 · · · dsm
j=1
Tif (xi , xi+1 , . . . , xm ),
(29.32)
i=1
where Tif as in (29.2), and g(x1 , . . . , xm ) = Q m +
j=1 m X i=1
nm (bj − aj )
Z
m Q
[aj ,bj ]
g(s1 , . . . , sm )ds1 · · · dsm
j=1
Tig (xi , xi+1 , . . . , xm ), ∀(x1 , . . . , xm ) ∈
m Y
i=1
[aj , bj ].
(29.33)
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Therefore f (x1 , . . . , xm )g(x1 , . . . , xm ) Z nm = Q g(x , . . . , x ) f (s1 , . . . , sm )ds1 · · · dsm 1 m m m Q [a ,b ] j j (bj − aj ) j=1 j=1
m X
+ g(x1 , . . . , xm )
!
Tif (xi , . . . , xm ) ,
i=1
and
(29.34)
f (x1 , . . . , xm )g(x1 , . . . , xm ) Z nm f (x , . . . , x ) g(s1 , . . . , sm )ds1 · · · dsm = Q 1 m m m Q [aj ,bj ] (bj − aj ) j=1 j=1
m X
+ f (x1 , . . . , xm )
Tig (xi , . . . , xm )
i=1
!
.
(29.35)
Consequently after integration we get Z f (x1 , . . . , xm )g(x1 , . . . , xm )dx1 · · · dxm m Q [aj ,bj ]
j=1
= Q m
nm
j=1
(bj − aj )
Z × m
Q
[aj ,bj ]
j=1
+
Z
m Q
[aj ,bj ]
j=1
"
Z
m Q
[aj ,bj ]
j=1
g(x1 , . . . , xm )dx1 · · · dxm
f (x1 , . . . , xm )dx1 · · · dxm
g(x1 , . . . , xm )
m X
Tif (xi , . . . , xm )
i=1
!#
Similarly Z we have f (x1 , . . . , xm )g(x1 , . . . , xm )dx1 · · · dxm m Q
dx1 · · · dxm . (29.36)
[aj ,bj ]
j=1
nm
= Q m
j=1
(bj − aj )
Z × m
Q
[aj ,bj ]
j=1
+
Z
m Q
j=1
[aj ,bj ]
"
Z
m Q
[aj ,bj ]
j=1
f (x1 , . . . , xm )dx1 · · · dxm
g(x1 , . . . , xm )dx1 · · · dxm
f (x1 , . . . , xm )
m X i=1
Tig (xi , . . . , xm )
!#
dx1 · · · dxm .
(29.37)
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By (29.36) and (29.37) we obtain Z
m Q
[aj ,bj ]
f (x1 , . . . , xm )g(x1 , . . . , xm )dx1 · · · dxm
j=1
−Q m
j=1
n
m
(bj − aj )
Z × m
Q
[aj ,bj ]
j=1
=
=
Z
Z
m Q
[aj ,bj ]
"
[aj ,bj ]
"
j=1
m Q
j=1
Z
m Q
[aj ,bj ]
j=1
f (x1 , . . . , xm )dx1 · · · dxm
g(x1 , . . . , xm )dx1 · · · dxm
g(x1 , . . . , xm )
X m
Tif (xi , . . . , xm )
i=1
f (x1 , . . . , xm )
X m
#
Tig (xi , . . . , xm )
i=1
dx1 · · · dxm
#
dx1 · · · dxm .
(29.38)
We conclude that Z
m Q
[aj ,bj ]
j=1
Z × m
− f gd→ x − Q m
j=1
nm (bj − aj )
1 − − g(→ x )d→ x= Q 2 [aj ,bj ]
j=1
− + f (→ x)
X m i=1
Tig (xi , . . . , xm )
"Z
#
Z
m Q
m Q
[aj ,bj ]
j=1
[aj ,bj ]
j=1
#
"
− − f (→ x )d→ x
− g(→ x)
X m
Tif (xi , . . . , xm )
i=1
− d→ x .
(29.39)
Remember by (29.17) for i = 1, . . . , m that Tif (xi , . . . , xm ) = Afi (xi , . . . , xm ) + Bif (xi , . . . , xm ),
(29.40)
Tig (xi , . . . , xm ) = Agi (xi , . . . , xm ) + Big (xi , . . . , xm ).
(29.41)
and
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We call and observe that ∆(f,g)
Z nm → − → − m fd x := Q f gd x − Q m m Q [a ,b ] [aj ,bj ] j j (bj − aj ) j=1 j=1 Z
j=1
Z × m
1 − gd→ x− Q 2 [aj ,bj ]
j=1
− + f (→ x) 1 = 2
"Z
X m
[aj ,bj ]
j=1
− + f (→ x)
X m
m Q
[aj ,bj ]
j=1
#
− x Agi (xi , . . . , xm ) d→
i=1
m Q
"Z
"
− g(→ x)
X m
Big (xi , . . . , xm )
#
×
m Q
[aj ,bj ]
j=i
+kf k∞
(
m i−1 X Y i=1
j=1
(bj − aj )
Z
m Q
· · · dxm
[aj ,bj ]
# → − d x =: Γ(f, g).
Clearly we obtain that ( m i−1 " X Y 1 (bj − aj ) |∆(f,g) | ≤ kgk∞ 2 i=1 j=1 |Bif (xi , . . . , xm )|dxi
i=1
Bif (xi , . . . , xm )
i=1
i=1
Z
#
X m f → − g( x ) Ai (xi , . . . , xm )
(29.42)
!)
|Big (xi , . . . , xm )|dxi
· · · dxm
j=i
!)#
(29.43)
We state the established Gr¨ uss type result. Theorem 29.17. Let f, g :
m Q
i=1
[ai , bi ] → R as in Theorem 29.6. Then
Z Z m n − − m f gd→ x − Q f d→ x Q m Q m [aj ,bj ] [a ,b ] j j (bj − aj ) j=1 j=1
Z × m Q
j=1
− + f (→ x)
j=1
1 − gd→ x− 2 [aj ,bj ]
X m i=1
"Z
m Q
j=1
Agi (xi , . . . , xm )
.
[aj ,bj ]
#
"
# → − dx
− g(→ x)
X m i=1
Afi (xi , . . . , xm )
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!) ( m i−1 " Z X Y 1 f (bj − aj ) Q |Bi (xi , . . . , xm )|dxi · · · dxm kgk∞ ≤ m 2 [aj ,bj ] i=1 j=1 + kf k∞
( m i−1 X Y i=1
j=1
j=i
(bj − aj )
Z
We make Remark 29.18. Let p, q > 1 :
1 p
+
m Q
[aj ,bj ]
|Big (xi , . . . , xm )|dxi
· · · dxm
j=i
!)#
.(29.44)
1 q
= 1. Suppose m Y Bif , Biq ∈ Lq [aj , bj ] , i = 1, . . . , m. j=i
From (29.42) and H¨ older’s inequality we obtain
|Γ(f, g)| "m Z 1 X − − |g(→ x )||Bif (xi , . . . , xm )|d→ x ≤ m Q 2 i=1 [aj ,bj ] j=1 !# Z g → − → − + |f ( x )||B (x , . . . , x )|d x m Q
i
i
[aj ,bj ]
m
j=1
(m" Z 1/q 1 X f − q → |B (x , . . . , x )| d x kgkp ≤ i m m i Q 2 i=1 [aj ,bj ] j=1 Z 1/q #) g − q → |B (x , . . . , x )| d x +kf k p
m Q
[aj ,bj ]
i
i
j=1
=
( m " i−1 1 X Y
2 i=1 Z × m Q
j=1
[aj ,bj ]
j=i
+kf kp
Z
m Q
(bj − aj )
1/q (
kgkp 1/q
|Bif (xi , . . . , xm )|q dxi · · · dxm
[aj ,bj ]
|Big (xi , . . . , xm )|q dxi
j=i
(29.45)
m
· · · dxm
1/q )#)
.
(29.46)
We have established the following Gr¨ uss type result. Qm Theorem 29.19. Let f, g : i=1 [ai , bi ] → R as in Theorem 29.6. Let p, q > 1 : p1 + 1q = 1. Assume m Y f g Bi , Bi ∈ Lq [aj , bj ] , i = 1, . . . , m. j=i
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Then
Z Z m n − − m f gd→ x − Q f d→ x Q m Q m [aj ,bj ] [a ,b ] j j (bj − aj ) j=1 j=1
×
Z
j=1
1 − gd→ x− m Q 2 [aj ,bj ]
j=1
− + f (→ x)
X m
"Z
×
Z
m Q
Agi (xi , . . . , xm )
[aj ,bj ]
j=i
+ kf kp
Z
m Q
m Q
[aj ,bj ]
j=1
i=1
381
#
"
− g(→ x)
X m
Afi (xi , . . . , xm )
i=1
( m " i−1 # 1/q ( 1 X Y − (bj − aj ) kgkp d→ x ≤ 2 j=1 i=1 1/q
|Bif (xi , . . . , xm )|q dxi · · · dxm
[aj ,bj ]
|Big (xi , . . . , xm )|q dxi
j=i
1/q )#) . · · · dxm
(29.47)
We derive the last Gr¨ uss type result. Q Theorem 29.20. Let f, g : m i=1 [ai , bi ] → R as in Theorem 29.6. Assume ∂ ng , ∂xn i
i = 1, . . . , m are continuous. Then
Z Z m n − − m f gd→ x − Q f d→ x Q m Q m [aj ,bj ] [aj ,bj ] (b − a ) j j j=1 j=1
Z × Q m
j=1
Proof. 29.3
∂nf , ∂xn i
j=1
1 − gd→ x− 2 [aj ,bj ]
"Z
m Q
j=1
!
[aj ,bj ]
"
− g(→ x)
X m
Afi (xi , . . . , xm )
i=1
# X m g → − → − + f( x ) Ai (xi , . . . , xm ) d x i=1 (m
n #)
n X
1 ∂ f
+ kf k1 ∂ g ≤ ni−1 (bi − ai )n kgk1 . (29.48) n
n
2(n − 1)! i=1 ∂xi ∞ ∂xi ∞ Use of (29.45).
Applications
I) We treat the case of n = 1, m = 2. Let [a1 , b1 ] × [a2 , b2 ] ⊆ R2 . General Assumptions 29.21. Let f : [a1 , b1 ] × [a2 , b2 ] → R. We assume
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1) for j = 1, 2 we have that f (s1 , x2 ), f (x1 , s2 ) are absolutely continuous in s1 ∈ [a1 , b1 ], s2 ∈ [a2 , b2 ], for every x2 ∈ [a2 , b2 ], x1 ∈ [a1 , b1 ], respectively. ∂f ∂f 2) ∂x (s1 , x2 ), ∂x (s1 , s2 ) are continuous on [a1 , b1 ], [a1 , b1 ] × [a2 , b2 ], respectively, 1 2 for every x2 ∈ [a2 , b2 ]. 3) f is continuous on [a1 , b1 ] × [a2 , b2 ]. We mention Brief Assumptions 29.22. Let f : [a1 , b1 ] × [a2 , b2 ] → R with f , continuous on [a1 , b1 ] × [a2 , b2 ]. We give
∂f ∂f ∂x1 , ∂x2
being
Corollary 29.23. Let f as in General Assumptions 29.21 or Brief Assumptions 29.22, (x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. Then Z 1 f (x1 , x2 ) = f (s1 , s2 )ds1 ds2 + T1 + T2 , (29.49) (b1 − a1 )(b2 − a2 ) [a1 ,b1 ]×[a2 ,b2 ] where Z b1 1 ∂f T1 = T1 (x1 , x2 ) = K(s1 , x1 ) (s1 , x2 )ds1 (b1 − a1 ) a1 ∂x1 = B1 (x1 , x2 ), (29.50) Z b1 Z b 2 1 ∂f T2 = T2 (x2 ) = (s1 , s2 )ds1 ds2 K(s2 , x2 ) (b1 − a1 )(b2 − a2 ) a1 a2 ∂x2 = B2 (x2 ). (29.51) Proof. By Theorem 29.6. We need Remark 29.24. Denote here Ef (x1 , x2 ) = f (x1 , x2 ) − = B1 + B2 .
1 (b1 − a1 )(b2 − a2 )
Z
b1 a1
Z
b2
f (s1 , s2 )ds1 ds2 a2
(29.52)
That is |Ef (x1 , x2 )| ≤ |B1 | + |B2 |. We give the following special Ostrowski type inequalities.
(29.53)
Corollary 29.25. Suppose all as in Corollary 29.23. Then Z b1 Z b2 1 |Ef (x1 , x2 )| = f (x1 , x2 ) − f (s1 , s2 )ds1 ds2 (b1 − a1 )(b2 − a2 ) a1 a2
∂f [a21 + b21 − x1 (a1 + b1 )]
≤
∂x1 (·, x2 ) (b1 − a1 ) ∞,[a1 ,b1 ]
∂f
[a22 + b22 − x2 (a2 + b2 )]
+ ,
∂x2 (·, ·) (b2 − a2 ) ∞,[a1 ,b1 ]×[a2 ,b2 ]
Proof.
∀(x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. By Theorem 29.9.
(29.54)
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We continue with ∂f ∂f , ∂x ∈ L∞ ([a1 , b1 ] × Corollary 29.26. Suppose all as in Corollary 29.23, and ∂x 1 2 [a2 , b2 ]). Then (
)( 2
∂f ∂f a1 + b21 − x1 (a1 + b1 )
,
|Ef (x1 , x2 )| ≤ max
∂x1 ∂x2 b1 − a 1 ∞ ∞ ) 2 a2 + b22 − x2 (a2 + b2 ) , ∀(x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. (29.55) + b2 − a 2
Proof.
By (29.54).
Next we have Corollary 29.27. Suppose all as in Corollary 29.23. Then Z b1 Z b2 1 f (s1 , s2 )ds1 ds2 f (x1 , x2 ) − (b1 − a1 )(b2 − a2 ) a1 a2 r
1 (x1 − a1 )3 + (b1 − x1 )3
∂f (·, x2 ) ≤
(b1 − a1 ) 3 ∂x1 2,[a1 ,b1 ] r
3 3
1 (x2 − a2 ) + (b2 − x2 ) ∂f
√ (·, ·) + ,
3 ∂x (b2 − a2 ) b1 − a1 2 2,[a1 ,b1 ]×[a2 ,b2 ]
(29.56)
for all (x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. Proof.
By (29.26).
We present Corollary 29.28. Assume all as in Corollary 29.23. Then Z b1 Z b 2 1 f (s1 , s2 )ds1 ds2 f (x1 , x2 ) − (b1 − a1 )(b2 − a2 ) a1 a2
∂f
1
≤ max(x1 − a1 , b1 − x1 )
∂x1 (·, x2 ) b1 − a 1 1,[a1 ,b1 ]
∂f 1
max(x2 − a2 , b2 − x2 ) (·, ·) +
(b1 − a1 )(b2 − a2 ) ∂x2
,
1,[a1 ,b1 ]×[a2 ,b2 ]
(29.57)
for all (x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. Proof.
By (29.28).
We finish Ostrowski type inequality applications for n = 1, m = 2 with
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Corollary 29.29. Assume all as in Corollary 29.23. Then Z b1 Z b2 1 f (s1 , s2 )ds1 ds2 f (x1 , x2 ) − (b1 − a1 )(b2 − a2 ) a1 a2 ≤ min R.H.S.(29.54), R.H.S.(29.56), R.H.S.(29.57) ,
(29.58)
for all (x1 , x2 ) ∈ [a1 , b1 ] × [a2 , b2 ]. We proceed with the related comparison of means.
∂f ∂f Corollary 29.30. Suppose all as in Corollary 29.23, and ∂x , ∂x ∈ L∞ [a1 , b1 ] × 1 2 [a2 , b2 ]. Let [ci , di ] ⊆ [ai , bi], i = 1, 2 and µ be a probability measure on [c1 , d1 ] × [c2 , d2 ] , P [c1 , d1 ] × [c2 , d2 ] . Then Z Z 1 f (s , s )ds ds f (x1 , x2 )dµ(x1 , x2 ) − Q 1 2 1 2 2 2 Q (b − a )(b − a ) 1 1 2 2 [aj ,bj ] [ci ,di ] j=1 j=1 "
(
Z #) Z
∂f 1
≤ b1 b1 − Q x1 dµ − a1 x1 dµ − a1 2 2
∂x1 Q [cj ,dj ] [cj ,dj ] ∞ (b1 − a1 ) j=1 j=1 "
(
#) Z Z
∂f 1
. x2 dµ − a2 x2 dµ − a2 b2 b2 − Q + 2 2 Q ∂x2 ∞ (b2 − a2 ) [cj ,dj ] [cj ,dj ] j=1
j=1
(29.59)
Proof.
By (29.31).
We present the Gr¨ uss type inequalities for n = 1, m = 2. Corollary 29.31. Let f, g : [a1 , b1 ] × [a2 , b2 ] × R as in Corollary 29.23. Then Z Z 1 − − 2 f gd→ x − 2 f d→ x Q Q Q 2 [aj ,bj ] [aj ,bj ] (bj − aj ) j=1 j=1 Z ×
j=1
" (Z 1 → − gd x ≤ kgk∞ |B1f (x1 , x2 )|dx1 dx2 2 2 Q Q 2 [aj ,bj ] [aj ,bj ] j=1 j=1 (Z ) Z b2
+ (b1 − a1 )
+ (b1 − a1 ) Proof.
a2
Z
b2 a2
By (29.44).
We continue with
|B2f (x2 )|dx2 |B2g (x2 )|dx2
+ kf k∞
)#
2 Q
[aj ,bj ]
|B1g (x1 , x2 )|dx1 dx2
j=1
.
(29.60)
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Corollary 29.32. Let f, g : [a1 , b1 ] × [a2 , b2 ] → R as in Corollary 29.23. Suppose 2 Y f g Bi , Bi ∈ L2 [aj , bj ] , i = 1, 2. j=i
Then Z − f gd→ x − 2 Q Q 2 [aj ,bj ] j=1
Z ×
j=1
[aj ,bj ]
j=1
− f d→ x
1/2
2 Q
[aj ,bj ]
j=1
(
(B1g (x1 , x2 ))2 dx1 dx2 Z
p
b1 − a1 kgk2
Proof.
By (29.47).
+
(bj − aj )
2 Q
(" Z 1/2 1 f → − 2 kgk2 (B1 (x1 , x2 )) dx1 dx2 gd x ≤ 2 2 Q Q 2 [aj ,bj ] [aj ,bj ] j=1 j=1 # Z
+kf k2 "
1
Z
b2 a2
(B2f (x2 ))2 dx2
1/2
+ kf k2
Z
b2 a2
(B2g (x2 ))2 dx2
1/2 )#)
.
(29.61)
We also give Corollary 29.33. Let f, g : [a1 , b1 ] × [a2 , b2 ] → R as in Corollary 29.23. Suppose ∂f ∂g ∂xi , ∂xi , i = 1, 2 are continuous. Then Z Z 1 − − 2 f d→ x f gd→ x − 2 Q Q Q 2 [aj ,bj ] [aj ,bj ] (bj − aj ) j=1 j=1 Z ×
Proof.
j=1
( "
# 1
∂f
∂g → −
(b1 − a1 ) kgk1 + kf k1 gd x ≤ 2 Q 2 ∂x1 ∞ ∂x1 ∞ [aj ,bj ] j=1 "
#)
∂g
∂f
+ (b2 − a2 ) kgk1 . (29.62)
∂x2 + kf k1 ∂x2 ∞ ∞ By (29.48).
II) Here we treat the case of n = 2, m = 3. Let
3 Q
j=1
General Assumptions 29.34. Let f :
3 Q
j=1
We assume
[aj , bj ] ⊆ R3 .
[aj , bj ] → R.
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1) that
∂f ∂f ∂f ∂x1 (s1 , x2 , x3 ), ∂x2 (x1 , s2 , x3 ), ∂x3 (x1 , x2 , s3 ) 3 Q
s1 , s2 , s3 , respectively, for every (x1 , x2 , x3 ) ∈ 2) We have that
2 ∂2f (s , x2 , x3 ), ∂∂xf2 (s1 , s2 , x3 ), ∂x21 1 2 3 Q
[a1 , b1 ], [a1 , b1 ] × [a2 , b2 ], 3) f is continuous on
3 Q
j=1
are absolutely continuous in
[aj , bj ]. j=1 ∂2f (s , s , s ) ∂x23 1 2 3
are continuous on 3 Q [aj , bj ]. [aj , bj ], respectively, for any (x2 , x3 ) ∈ j=2
[aj , bj ].
j=1
We mention Brief Assumptions 29.35. Let f :
3 Q
j=1
j = 1, 2, 3, are continuous on
3 Q
[aj , bj ].
[aj , bj ] → R with
∂`f ∂x`j
for ` = 0, 1, 2;
j=1
We present
Corollary 29.36. Let f as in General Assumptions 29.34 or Brief Assumptions 3 Q 29.35, (x1 , x2 , x3 ) ∈ [aj , bj ]. Then j=1
f (x1 , x2 , x3 ) =
8 3 Q
j=1
(bj − aj )
Z
3 Q
f (s1 , s2 , s3 )ds1 ds2 ds3 + [aj ,bj ]
3 X
Ti .
(29.63)
i=1
j=1
Here for i = 1, 2, 3 we put 1 T1 = T1 (x1 , x2 , x3 ) = (b1 − a1 ) (
f (b1 , x2 , x3 )(x1 − b1 ) − f (a1 , x2 , x3 )(x1 − a1 )
×
) ∂2f (x1 − s1 )k(s1 , x1 ) 2 (s1 , x2 , x3 )ds1 , + ∂x1 a1 (Z b1 2 T2 = T2 (x2 , x3 ) = f (s1 , b2 , x3 )(x2 − b2 ) (b1 − a1 )(b2 − a2 ) a1 − f (s1 , a2 , x3 )(x2 − a2 ) ds1 ) Z ∂2f + Q (x2 − s2 )k(s2 , x2 ) 2 (s1 , s2 , x3 )ds1 ds2 , 2 ∂x2 [aj ,bj ] Z
b1
j=1
(29.64)
(29.65)
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T3 = T3 (x3 ) =
4 3 Q
j=1
(bj − aj )
(Z
2 Q
[aj ,bj ]
387
f (s1 , s2 , b3 )(x3 − b3 )
j=1
− f (s1 , s2 , a3 )(x3 − a3 ) ds1 ds2 ) Z ∂2f (x3 − s3 )k(s3 , x3 ) 2 (s1 , s2 , s3 )ds1 ds2 ds3 . + Q 3 ∂x3 [aj ,bj ]
(29.66)
j=1
We need
Remark 29.37. We denote A1 = A1 (x1 , x2 , x3 ) =
also
f (b1 , x2 , x3 )(x1 − b1 ) − f (a1 , x2 , x3 )(x1 − a1 ) , (29.67) b1 − a 1
B1 = B1 (x1 , x2 , x3 ) Z b1 1 ∂2f = (x1 − s1 )k(s1 , x1 ) 2 (s1 , x2 , x3 )ds1 , (29.68) (b1 − a1 ) a1 ∂x1 Z b1 2 A2 = A2 (x2 , x3 ) = f (s1 , b2 , x3 )(x2 − b2 ) (b1 − a1 )(b2 − a2 ) a1 − f (s1 , a2 , x3 )(x2 − a2 ) ds1 , (29.69) Z 2 B2 = B2 (x2 , x3 ) = (x2 − s2 )k(s2 , x2 ) 2 (b1 − a1 )(b2 − a2 ) Q [aj ,bj ] j=1
2
∂ f (s1 , s2 , x3 )ds1 ds2 , ∂x22 Z 4 f (s1 , s2 , b3 )(x3 − b3 ) A3 = A3 (x3 ) = 3 2 Q Q (bj − aj ) j=1[aj ,bj ] ×
and
(29.70)
j=1
− f (s1 , s2 , a3 )(x3 − a3 ) ds1 ds2 , (29.71) Z 4 B3 = B3 (x3 ) = 3 (x3 − s3 )K(s3 , x3 ) 3 Q Q [aj ,bj ] (bj − aj ) j=1 j=1
Clearly here
∂2f × (s1 , s2 , s3 )ds1 ds2 ds3 . ∂x23
Ti = Ai + Bi , i = 1, 2, 3. We also denote 8 Ef (x1 , x2 , x3 ) := f (x1 , x2 , x3 ) − 3 Q (bj − aj )
(29.72) (29.73)
j=1
×
Z
3 Q
j=1
[aj ,bj ]
f (s1 , s2 , s3 )ds1 ds2 ds3 −
3 X i=1
Ai =
3 X i=1
Bi .
(29.74)
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Therefore |Ef (x1 , x2 , x3 )| ≤ We give
3 X i=1
|Bi |.
(29.75)
Corollary 29.38. Assume all as in Corollary 29.36. Then
( −i− 3
∂ 2 f z}|{ X 2i−1
|Ef (x1 , x2 , x3 )| ≤
2 · · · , x3
∂xi
(bi − ai ) i=1 ∞ ( " X 1 2 2 5x3i bi (bi − xi ) − ai (xi − ai ) 1−λ + (−1) × 2 (λ + 2)(3 − λ) λ=0 !)#) 3 Y x1−λ bλ+2 xλi ai3−λ i + i , ∀(x1 , x2 , x3 ) ∈ [aj , bj ]. (29.76) − 3−λ λ+2 j=1
Proof.
By (29.20).
We continue with Corollary 29.39. Suppose all as in Corollary 29.36, and 3 Y ∂2f ∈ L∞ [aj , bj ] , i = 1, 2, 3. ∂x2i j=1
Then
Proof.
( 3 2 X
∂ f 2i−1
|Ef (x1 , x2 , x3 )| ≤
∂x2 i ∞ (bi − ai ) i=1 " ( X 1 bi (bi − xi )2 − ai (xi − ai )2 5x3i × (−1)1−λ + 2 (λ + 2)(3 − λ) λ=0 !)#) 3 Y xλi ai3−λ x1−λ bλ+2 i − , ∀(x1 , x2 , x3 ) ∈ [aj bj ]. (29.77) + i 3−λ λ+2 j=1 By (29.25).
Next we give Corollary 29.40. Assume all as in Corollary 29.36. Then ( ! 3 X 2i−1 |Ef (x1 , x2 , x3 )| ≤ i−1 1/2 Q i=1 (bi − ai ) (bj − aj ) × for all (x1 , x2 , x3 ) ∈ Proof.
Q3
By (29.26).
r
j=1
(xi − ai )5 + (bi − xi )5 30
j=1 [aj , bj ].
) ! −i−
∂ 2 f z}|{
2 · · · , x3 ,
∂xi
(29.78)
2
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We also present Corollary 29.41. Suppose all as ( in Corollary 29.36. Then 3 X 2 2i−1 max(xi − ai , bi − xi ) |Ef (x1 , x2 , x3 )| ≤ i Q i=1 (bj − aj ) j=1
Proof.
2 −i−
∂ f z}|{
× 2 · · · , x3
,
∂xi
1
By (29.28).
We give
∀(x1 , x2 , x3 ) ∈
3 Y
[aj , bj ].
(29.79)
j=1
Corollary 29.42. Assume all as in Corollary 29.36. Then |Ef (x1 , x2 , x3 )| ≤ min R.H.S.(29.76), R.H.S.(29.78), R.H.S.(29.79)}, (29.80) Q3 for all (x1 , x2 , x3 ) ∈ j=1 [aj , bj ]. Proof.
By Corollary 29.13.
Next we do the comparison of means. Corollary 29.43. Assume all asin Corollary 29.36, and 3 2 Y ∂ f ∈ L∞ [aj , bj ] , i = 1, 2, 3. ∂x2i j=1
Let [ci , di ] ⊆ [ai , bi ], i = 1, 2, 3 and µ be a probability!! measure on 3 3 Y Y [ci , di ], P [ci , di ] . i=1
Then Z f (x1 , x2 , x3 )dµ − 3 Q Q 3 [cj ,dj ] j=1
j=1
3 Z X − A (x , ·, x )dµ i i 3 3 Q [cj ,dj ] i=1
i=1
8 (bj − aj )
Z
3 Q
f (s1 , s2 , s3 )ds1 ds2 ds3 [aj ,bj ]
j=1
j=1
( " Z 3 2 i−1 X
∂ f 2 1
≤ bi Q (bi − xi )2 dµ 3
∂x2 (b − a ) 2 i i [c ,d ] i ∞ j j i=1 j=1 R 5 Q x3i dµ 3 ( X Z 1 [cj ,dj ] j=1 (xi − ai )2 dµ + (−1)1−λ −ai Q 3 (λ + 2)(3 − λ) [cj ,dj ] j=1
ai3−λ
− Proof.
λ=0
R
3 Q
[cj ,dj ]
j=1
(3 − λ) By (29.31).
bλ+2 i
xλi dµ
+
R
3 Q
j=1
[cj ,dj ]
xi1−λ dµ !)#)
(λ + 2)
.
(29.81)
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We continue with the Gr¨ uss type inequalities for the case of n = 2, m = 3. 3 Q [aj , bj ] → R. Then Corollary 29.44. Suppose all as in Corollary 29.36 for f, g : j=1 Z Z 8 − − 3 f gd→ x − 3 f d→ x Q Q Q 3 [aj ,bj ] [aj ,bj ] (bj − aj ) j=1 j=1 Z ×
j=1
1 − gd→ x− 3 Q 2 [aj ,bj ]
j=1
"Z
3 Q
[aj ,bj ]
j=1
− g(→ x)
X 3 i=1
Afi (xi , ·, x3 )
# g → − → − + f( x ) Ai (xi , ·, x3 ) d x i=1 !) ( 3 i−1 " Z X Y 1 f |Bi (xi , ·, x3 )|dxi · · · dx3 (bj − aj ) Q kgk∞ ≤ 3 2 [aj ,bj ] i=1 j=1 X 3
+kf k∞ Proof.
(
#
"
j=i
3 X i=1
By (29.44).
i−1 Y
j=1
(bj − aj )
It follows
Z
3 Q
[aj ,bj ]
|Big (xi , ·, x3 )|dxi
· · · dx3
j=i
!)#
. (29.82)
Q3
→ R as in Corollary 29.36. Suppose 3 Y f g Bi , Bi ∈ L2 [aj , bj ] , i = 1, 2, 3.
Corollary 29.45. Let f, g :
j=1 [aj , bj ]
j=i
Then
Z − f gd→ x − 3 Q Q 3 [aj ,bj ] j=1
Z ×
j=1
(bj − aj )
1 − gd→ x− 3 Q 2 [aj ,bj ]
j=1
− + f (→ x)
3 Q
#
3 Q
[aj ,bj ]
j=1
[aj ,bj ]
j=1
"
− g(→ x)
[aj ,bj ]
(Bif (xi , ·, x3 ))2 dxi · · · dx3
Z
3 Q
j=i
[aj ,bj ]
− f d→ x
X 3 i=1
Afi (xi , ·, x3 )
# ( 3 " i−1 1/2 ( 1 X Y → − dx ≤ (bj − aj ) kgk2 2 i=1 j=1 1/2
Agi (xi , ·, x3
j=i
+ kf k2
"Z
3 Q
X 3 i=1
Z ×
8
Z
(Big (xi , ·, x3 ))2 dxi
· · · dx3
!1/2 )#)
.
(29.83)
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Multivariate Fink Type Identity Applied to Multivariate Inequalities
Proof.
391
By (29.47).
We also have Corollary 29.46. Let f, g :
3 Q
j=1 ∂2g , ∂x2i
[aj , bj ] → R as in Corollary 29.36. Suppose
i = 1, 2, 3 are continuous. Then Z Z 8 − − 3 f d→ x f gd→ x − 3 Q Q Q 3 [aj ,bj ] [aj ,bj ] (bj − aj ) j=1 j=1 Z ×
Proof.
∂2f , ∂x2i
j=1
1 − gd→ x− 3 Q 2 [aj ,bj ]
j=1
"Z
3 Q
[aj ,bj ]
j=1
#
"
− g(→ x)
X 3 i=1
Afi (xi , ·, x3 )
# X 3 g → − → − Ai (xi , ·, x3 ) d x + f( x ) i=1 " ( 3
2
2 #)
∂ f
∂ f 1 X i−1
2 (bi − ai )2 kgk1 ≤ .
∂x2 + kf k1 ∂x2 2 i=1 i ∞ i ∞
(29.84)
By (29.48).
Finally we give a second inductive proof of Theorem 29.6. Another Proof of Theorem 29.6. We have that f (x1 , . . . , xm , xm+1 ) Z m X nm = Q f (s , . . . , s , x )ds · · · ds + Ti , (29.85) 1 m m+1 1 m m m Q i=1 (bj − aj ) j=1[aj ,bj ] j=1
where for i, . . . , m, we put
Ti := Ti (xi , . . . , xm , xm+1 ) := (Z
ni−1 i Q
j=1
i−1 Q
j=1
[aj ,bj ]
(bj − aj )
n−1 X k=1
(n − k) k!
∂ k−1 f (s1 , . . . , si−1 , bi , xi+1 , . . . , xm , xm+1 )(xi − bi )k ∂xik−1
) ∂ k−1 f k − (s1 , . . . , si−1 , ai , xi+1 , . . . , xm , xm+1 )(xi − ai ) ds1 · · · dsi−1 ∂xik−1 ! Z ni−1 + (xi − si )n−1 k(si , xi ) i Q i Q [aj ,bj ] (n − 1)! (bj − aj ) j=1 j=1
! ∂nf (s1 , . . . , si , xi+1 , . . . , xm , xm+1 )ds1 . . . dsi . ∂xni
(29.86)
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ADVANCED INEQUALITIES
392
But by Fink’s identity, see (29.3),we have f (s1 , sm , xm+1 ) = +
n (bm+1 − am+1 )
Z
bm+1
f (s1 , . . . , sm , sm+1 )dsm+1 am+1
n−1 X n − k ∂ k−1 f 1 (s1 , . . . , sm , bm+1 )(xm+1 − bm+1 )k k−1 (bm+1 − am+1 ) k! ∂x m+1 k=1 k−1 ∂ f − k−1 (s1 , . . . , sm , am+1 )(xm+1 − am+1 )k ∂xm+1 Z bm+1 1 (xm+1 − sm+1 )n−1 + (n − 1)!(bm+1 − am+1 ) am+1 n ∂ f (s1 , . . . , sm , sm+1 )dsm+1 . (29.87) k sm+1 , xm+1 ∂xnm+1
Next we put (29.87) into (20.85). We obtain f (x1 , . . . , xm+1 ) ( Z Z bm+1 nm n = Q f (s1 , . . . , sm , sm+1 )dsm+1 m m Q (bm+1 − am+1 ) am+1 (bj − aj ) j=1[aj ,bj ] j=1
n−1 X n − k ∂ k−1 f 1 + (s1 , . . . , sm , bm+1 )(xm+1 − bm+1 )k k−1 (bm+1 − am+1 ) k! ∂x m+1 k=1 ∂ k−1 f k − k−1 (s1 , . . . , sm , am+1 )(xm+1 − am+1 ) ∂xm+1 Z bm+1 1 + (xm+1 − sm+1 )n−1 (n − 1)!(bm+1 − am+1 ) am+1 ) n m X ∂ f k sm+1 , xm+1 Ti . (s1 , . . . , sm , sm+1 )dsm+1 ds1 . . . dsm + ∂xnm+1 i=1
Therefore
f (x1 , . . . , xm+1 ) =
nm+1 m+1 Q j=1
where
(bj − aj )
Z
m+1 Q
f (s1 , . . . , sm+1 )ds1 . . . dsm+1 + [aj ,bj ]
i=1
j=1
Tm+1 := Tm+1 (xm+1 ) :=
m+1 X
nm m+1 Q j=1
(bj − aj )
n−1 X k=1
n−k k!
Ti ,
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Multivariate Fink Type Identity Applied to Multivariate Inequalities
Z
Book˙Adv˙Ineq
393
∂ k−1 f (s1 , . . . , sm , bm+1 )(xm+1 − bm+1 )k k−1 [aj ,bj ] ∂xm+1 j=1 ∂ k−1 f k − k−1 (s1 , . . . , sm , am+1 )(xm+1 − am+1 ) ds1 . . . dsm ∂xm+1 Z nm + (xm+1 − sm+1 )n−1 m+1 Q m+1 Q [aj ,bj ] (n − 1)! (bj − aj ) j=1 m Q
j=1
k sm+1 , xm+1 That is proving the claim.
∂ nf (s1 , . . . , sm , sm+1 )ds1 . . . dsm+1 . ∂xnm+1
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Bibliography
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List of Symbols
k · kp , Lp norm, 17 k · k∞ , L∞ norm, 5 C n ([a, b]), n times continuously differentiable functions, 5 k · k1 , L1 norm, 47 Bk , Bernoulli polynomial, 21 Bk∗ , Bernoulli related periodic functions, 21 ω1 , modulus of continuity, 49 k · kq , Lq norm, 17 B(0, R), ball in RN , 99 S N −1 , unit sphere in RN , 99 kf kLip , Lipschitz constant, 100 W 1,∞ (Ω), Sobolev space, 94 ∇, gradient, 110 Vol, volume, 110 Dα , distributional derivative, 100 W n,∞ (Ω), 100 T W n,∞ (Ω) C0n−1 (Ω), 106 ωN , 93
| · |, 150 B(0, R), 141 T Lip(Ω) C0 (Ω), 96 RN , 140 C, 241 BU C(R), 206 D(Ar ), 188 C(t), 249 S(t), R 249 L R , 261 R , 261R (R − S) , 261 m(B), 271 T (g, h), 280 1,1 Wloc (Rn ), 294 p Lloc (Ω), 294 D0 (Ω), 294 D(Ω), 294 W m,p (Ω), 295 PW (x, t), 331
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Index
Absolutely continuous, 12 asymptotically attained, 111 attained inequality, 143 ball in RN , 132 Banach algebra, 187 Banach space, 187 Bernoulli polynomials, 305 Bernoulli numbers, 305 best constant, 229 boundary conditions, 29 boundary, 271 Cauchy–Schwarz inequality, 23 Chebychev functional, 333 Chebychev’s inequality, 283 Chebyshev–Gr¨ uss type inequality, 331 compact and convex, 29 Cosine and Sine operator functions, 197 diffusion equation, 192 Dirichlet–Poincar´e like inequality, 219 distribution, 293 distributional derivative, 295 distributional Taylor formula, 293 Euclidean domain, 99 Euclidean norm, 109 Euler-type identity, 342 extended complete Tschebyshev system, 125
409
Fink identity, 305 general Taylor–Widder formula, 127 generalized Euler type identity, 306 generalized radial derivatives of Widder type, 125 Green’s function, 171 Gr¨ uss inequality, 319 Hardy–Opial type inequality, 261 H¨ older’s inequality, 173 infinitesimal generator, 188 initial value problem, 171 integral means inequality, 341 Korkine’s identity, 280 Linear Differential operator, 171 Lipschitz constant, 94 mixed partial, 271 modulus of continuity, 49 Montgomery identity, 331 multivariate Euler type identity, 39 multivariate Fink identity, 367 multivariate Gr¨ uss inequality, 365 multivariate means inequality, 357, 358 multivariate Ostrowski inequality, 365 multivariate Taylor formula, 285 multivariate trapezoid and midpoint rules, 27
Book˙Adv˙Ineq
17th August 2010
410
15:56
World Scientific Book - 9.75in x 6.5in
ADVANCED INEQUALITIES
Neumann–Poincar´e like inequality, 232 non-radial, 109 Opial inequality, 149 Opial type inequality, 150 Ostrowski inequality, 93 Ostrowski type inequality, 383 Poincar´e inequality, 229 Poincar´e like inequality, 209 polar method, 109 radial function, 109 Riemann–Stieltjes integral, 261 remainder, 275 reverse inequality, 209
semigroup, 244 seminorm, 100 sharp inequality, 125 Sobolev like inequality, 215 Sobolev space, 94 spherical shell, 332 surface area, 332 Taylor formula, 198 test function, 294 unit sphere, 332 vector valued function, 188, 229 vector valued Riemann integral, 230 weighted Peano kernel, 331
Book˙Adv˙Ineq
