Numer Algor DOI 10.1007/s11075-007-9099-x ORIGINAL PAPER
Convergence of polyharmonic splines on semi-regular grids Z×aZn for a → 0 O. Kounchev · H. Render
Received: 15 November 2006 / Accepted: 19 April 2007 © Springer Science + Business Media B.V. 2007
Abstract Let p, n ∈ N with 2 p ≥ n + 2, and let Ia be a polyharmonic spline of order p on the grid Z × aZn which satisfies the interpolating conditions Ia ( j, am) = d j (am) for j ∈ Z, m ∈ Zn where the functions d j : Rn → R and the parameter a > 0 are given. Let Bs (Rn ) be the set of all integrable functions f : Rn → C such that the integral f s := f (ξ ) 1 + |ξ |s dξ Rn
is finite. The main result states that for given σ ≥ 0 there exists aconstant c > 0 such that whenever d j ∈ B2 p (Rn ) ∩ C (Rn ) , j ∈ Z, satisfy d j2 p ≤ D · (1 + | j|σ ) for all j ∈ Z there exists a polyspline S : Rn+1 → C of order p on strips such that |S (t, y) − Ia (t, y)| ≤ a2 p−1 c · D · (1 + |t|σ ) for all y ∈ Rn , t ∈ R and all 0 < a ≤ 1. Keywords Radial basis functions · Interpolation · Polysplines · Polyharmonic splines Mathematics Subject Classifications (2000) Primary 41A05 · 65D10 · Secondary 41A15
O. Kounchev (B) Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Acad. G. Bonchev St. 8, Sofia, Bulgaria e-mail:
[email protected] H. Render Departamento de Matématicas y Computación, Universidad de La Rioja, Edificio Vives, Luis de Ulloa s/n., 26004 Logroño, Spain
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1 Introduction Let be a discrete subset of Rn+1 and let p be a natural number satisfying 2 p ≥ n + 2. We define SH p Rn+1 , to be the set of all tempered distributions u on Rn+1 which are 2 p − n − 2 continuously differentiable and such that p u (x) = 0 for all x ∈ Rn+1 \ . Here p is the pth iterate of the Laplace operator =
n+1
∂2 j=1 ∂ x2j
for a natural
one calls the distribution u ∈ number the discrete subset = Z p. For SH p Rn+1 , polyharmonic cardinal spline of order p. Fundamental results about polyharmonic cardinal splines and interpolation properties have been proved In particular, it is proven in Proposition 2 in [15] that u ∈ in [15]. SH p Rn+1 , satisfies the equation p u (x) = aγ δ (x − γ ) for all x ∈ Rn+1 , n+1
γ ∈
where the aγ s are constants and δ is the Dirac distribution for the point 0. In the present paper we shall be concerned with interpolation problems on semi-regular grids a , i.e. grids of the form a := Z × aZn
(1)
where aZn is the set {am : m ∈ Zn } and a > 0 is a positive real number. This kind of interpolation problems typically occur in practice, when the measurements of data are taken in one direction with a small step size a > 0 while in the second direction they are distributed with a relatively large step size. Let us illustrate this kind of problems by an instructive example from Geophysics, elaborated in [8, chapter 6]: data of the magnetic field over the Cobb Offset (California) are collected by air planes on 13 parallel lines in R2 with approximately 200 data points which are relatively uniformly distributed on every line and we may assume that the step size is approximately a; the distance between each two neighbouring lines is very large compared with a and thus we are close to the setting of our problem where the grid is given by a . Due to the reversals of the magnetic field the neighboring layers of the magma have opposite signs and thus the data oscillate wildly. This kind of data provide a good test for every smoothing method. Interpolation for this kind of data can be subsumed under the general theory of interpolation with Radial Basis Functions for scattered data (see [4, 5, 15, 17, 20]). Indeed, the general theory allows the data even to be arbitrarily scattered, and from a theoretical point of view there seems to be no big advantage to know that the data have been taken in a very regular way. However, we can use this information in a completely different way: using univariate interpolation methods applied to the discrete data given on each line j × R one can find a spline function d j : R → C interpolating the data on the line. Thus we may assume that the data are not anymore given in a discrete manner but by means of functions d j
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defined on the whole line j × R ; thus the data d j may have smoothness up to a certain prescribed order. From the general theory of polysplines, developed by the first author in [8], one can find a polyspline S interpolating the data function d j on each line; let us recall that in this case the polyspline S is a function polyharmonic of order p in the open strips ( j, j + 1) × R, which is continuously differentiable up to order 2 p − 2 ; note that the smoothness is independent of the dimension. We refer to [8, chapter 6] where polysplines are defined as piecewise solutions of elliptic equations of order 2 p which are smooth of order 2 p − 2. In [8, chapter 6] the performance of the polysplines is compared with other smoothing methods in the above-mentioned example of magnetic data; in [13] one can find applications of polysplines to Medical Imaging. It is an essential feature of the polyspline theory that the data are given on hypersurfaces and not in a discrete way. This fact implies a remarkable advantage of the polysplines: The polysplines may be defined for any order p (independent of the dimension n) while the polyharmonic splines require the condition 2 p ≥ n + 1 for Rn+1 . In particular, the biharmonic case p = 2 already gives a C2 interpolation result. The smoothness order 2 p − n − 2 for the polyharmonic splines on Rn+1 is related to the smoothness order of the fundamental solution R p of the polyharmonic operator p which satisfies p R p (x) = δ (x) ; for the explicit form of R p see [1, p. 2] or [8, Theorem 10.37]. Polysplines have smoothness of order 2 p − 2. Roughly speaking, polysplines may be considered by means of the Green formula as simple layer polyharmonic potentials, i.e. as convolutions of the fundamental solution R p with some continuous measures which are supported on the data hypersurfaces; such convolutions increase the smoothness of R p , cf. [7, p. 346], [8]. In this paper we want to compare the interpolation methods for polyharmonic splines and polysplines. We shall idealize and simplify the above-described framework of practical by assuming that the mea examples surements have provided data values dγ γ ∈Z×aZn for the grid a defined in (1). This assumption has the advantage that the interpolation polyharmonic splines Ia (satisfying I (γ ) = dγ for all γ ∈ a = Z × aZn ) can be defined in an analytical way using Fourier methods. Similarly, interpolation with polysplines can be performed in an explicit way since the geometry of the data sets is very regular, see [2, 8–10, 12] . In the present paper we shall need the definition of a polyspline only in a very special setting: a function u : Rn+1 → C is a cardinal polyspline of order p on strips if u is 2 p − 2 times continuously differentiable and p u (x) = 0 for all x in the open strips ( j, j + 1) × Rn with j ∈ Z. The interpolation polyharmonic spline Ia depends on the step size a > 0 in the direction of the second variable. The main problem which we address is the convergence of the polyharmonic interpolants Ia on the grid a for given data and for a −→ 0. In order that the problem is well-posed we need an assumption that the data depend on the parameter a > 0 in a reasonable way. This is achieved by assuming that functions d j : Rn → R, j ∈ Z, are given such that the data for step size a > 0 are provided by the values d j(am) for
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Then there exists ( j, m) ∈ Zn+1 , and that these data are ofpolynomial growth. a unique polyharmonic spline Ia ∈ SH p Rn+1 , Z × aZn such that Ia ( j, am) = d j(am) for all m ∈ Zn , j ∈ Z,
(2)
see [15] or [11]. Later we shall assume that the data functions d j : R → R, j ∈ Z for the interpolation problem are sufficiently smooth. In [11] we have obtained a qualitative solution of the convergence problem: the polyharmonic splines Ia (with respect to the grid a ) of order p interpolating given data functions d j : Rn → R, j ∈ Z, on the grid a , converge pointwise to a limit function S, and this limit function S is a polyspline of order p on strips. In this sense we see that polysplines are a continuous version of polyharmonic splines. However, the method used in [11] permitted only data satisfying the restrictive assumption n
dj = 0
for | j| > N,
(3)
where N is some positive integer. In the present paper we shall give a quantitative answer to the convergence problem by replacing (3) by the weaker assumption of a sufficient decay. Roughly speaking, we shall prove that the polyharmonic splines Ia of order p (on the grid a ) interpolating given data functions d j : Rn → R, j ∈ Z, on the grid a , converge to a limit function S with a rate of convergence a2 p−1 for a → 0, see Theorem 1 below. Thus in the present paper we establish the link between the Schoenberg type interpolation for polyharmonic splines proved in [15] and the Schoenberg type interpolation for polysplines proved in [2] and [3]. In order to formulate the main result precisely, let us introduce now some notations. By Bs (Rn ) we denote the set of all integrable functions f : Rn → C such that the integral f s := f (ξ ) 1 + |ξ |s dξ (4) Rn
is finite. Here f denotes as usual the Fourier transform of a function
f (ω) := Rn e−ix,ω f (x) dx. For an extended study of f : Rn → C, defined by similar function spaces see Definition 10.1.6 in Hörmander [6]. Further Ck (Rn ) denotes the set of all k-times continuously differentiable functions f : Rn → C for k ∈ N0 ∪ {∞} . We shall write C (Rn ) for C0 (Rn ) . The main result of the present paper is the following: Theorem 1 Let σ ≥ 0 and p ∈ N with 2 p ≥ n + 2. Then there exists a constant c > 0 such that whenever d j ∈ B2 p (Rn ) ∩ C (Rn ) , j ∈ Z, satisfy d j ≤ D · (1 + | j|σ ) (5) for all j ∈ Z, 2p there exists a polyspline S : Rn+1 → C of order p on strips such that |S (t, y) − Ia (t, y)| ≤ a2 p−1 c · D · (1 + |t|σ ) for all y ∈ Rn , t ∈ R and all 0 < a ≤ 1. Here Ia ∈ SH p Rn+1 , Z × aZn−1 is the polyharmonic spline satisfying (2).
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The result of Theorem 1 may be compared with the error estimates available in [14, Theorem 2] for cardinal polyharmonic splines: if f is in the Sobolev 2p space Ls Rn+1 (namely, the space of order 2 p, see [14] for the precise definition) and Sσ f its polyharmonic interpolant on the grid σ −1 · Zn+1 , then there exists a constant C such that for all σ > 0 1s 1s 2p s s −2 p | f (x) − (Sσ f ) (x)| dx ≤ σ f (x) dx . C Our result is of different nature since we consider a limit process for the grid Z × aZn , a → 0. General results concerning the error estimates of f and its interpolant Sf can also be found in [16]. The paper is organized in the following way: in Section 1 we recall and improve some facts from [11] concerning the representation of polyharmonic splines L p,a, f which interpolate data at the points (0, am) , m ∈ Zn , given by the values f (am) , m ∈ Zn , of a function f, and zero data on ( j, am) , m ∈ Zn , j ∈ Z \ {0} , see (8) and (9). The main result in Section 2 is the following: if f ∈ B2 p (Rn ) ∩ C (Rn ) then there exist constants η > 0 and C > 0 (independent of f ) and there exists a cardinal polyspline L p, f on strips such that the following estimate holds p L p,a, f (t, y) − L p, f (t, y) ≤ a2 p−1 Ce−ηt · f (ξ ) 1 + |ξ |2 dξ (6) for all t ∈ R and y ∈ Rn and for all 0 < a ≤ 1. Due to the exponential decay in estimate (6) we can give in Section 3 an estimate for the case of infinitely many data functions d j as described in Theorem 1. In the Appendix we provide the somewhat technical proofs of Theorems 5, 6 and 7 which contain the basic estimates.
2 Basic facts for interpolation with polyharmonic splines Let us remind that throughout the paper we assume 2 p ≥ n + 2. polyharmonic spline of order p for the grid A function L p,a is a fundamental a whenever L p,a is in SH p Rn+1 , a and L p,a (0) = 1, and L p,a (γ ) = 0
for all γ ∈ a \ {0} .
Define the function ϕ by putting ϕ (ω) := |ω|−2 p for ω = (s, ξ ) with s ∈ R and n ξ ∈ R , and put Sϕ,a (ω) :=
k∈Z m∈Zn
Then L p,a (x) =
an−1 (2π )
n+1
1 2 p . (s + 2π k)2 + ξ − 2πa−1 m
Rn+1
eix,ω
ϕ (ω) dω Sϕ,a (ω)
for x ∈ Rn+1 .
(7)
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Assume that f : Rn → C is a function of polynomial growth: we define L p,a, f as the polyharmonic spline of order p for the grid a such that L p,a, f ((0, am)) = f (am) and La, p, f (( j, am)) = 0 for j ∈ Z \ {0}
(8)
and for all m ∈ Zn . Indeed, L p,a, f is given by
L p,a, f (t, y) :=
f (am) L p,a ((t, y − am)) .
(9)
m∈Zn
The following result was proven in [11]: Theorem 2 Let a > 0 and suppose that for the continuous function f : Rn → C there exist constants A > 0 and δ > 0 such that for all x ∈ Rn and ω ∈ Rn | f (x)| ≤ A (1 + |x|)−n−δ and f (ω) ≤ A (1 + |ω|)−n−δ . (10) Then the polyharmonic spline L p,a, f of order p for the grid a , defined in (9), is given by Ba, p (s, ξ, y) 1 eits eiy,ξ L p,a, f (t, y) = f (ξ ) dsdξ. (11) 2π Rn R Sϕ,a (s, ξ ) Zn by Here the function Ba, p is defined for all (s, ξ ) ∈ Rn+1 \ {0} × 2π a Ba, p (s, ξ, y) =
m∈Zn
1 − 2π iy,m .
2 p e a s2 + ξ − 2πa−1 m
(12)
Clearly, for each s = 0 the function ξ → Ba, p (s, ξ, y) is well-defined Zn . Since since 2 p ≥ n + 2. Note that Ba, p (s, ξ, y) has poles at {0} × 2π a n Ba, p (s, ξ, y) ≤ Ba, p (s, ξ, 0) ≤ Sϕ,a (s, ξ ) we have for all ξ ∈ R and s = 0 Ba, p (s, ξ, y) ≤ Ba, p (s, ξ, 0) ≤ 1. 0≤ (13) Sϕ,a (s, ξ ) Sϕ,a (s, ξ ) Further, we have shown in [11, Theorem 6] that there exists a constant C > 0 such that for all |s| > 0 and for all ξ ∈ Rn and for all 0 < a ≤ 1 Ba, p (s, ξ, y) ≤
m∈Zn
1 C n 2 p ≤ s2 p (1 + |s|) . s2 + ξ − 2πa−1 m
(14)
We now extend Theorem 2 in the following way: Theorem 3 Let a > 0 and suppose that f ∈ B2 p−n−2 (Rn ) ∩ C (Rn ) . Then Ba, p (s, ξ, y) 1 L p,a, f (t, y) = eits eiy,ξ f (ξ ) dsdξ. (15) 2π Rn R Sϕ,a (s, ξ )
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Proof Let us denote the expression on the right hand side in (15) by S p,a, f . The assumption f ∈ B2 p−n−2 (Rn ) implies that f is integrable. Formula (13) and (14) show that Ba, p (s, ξ, y) f (ξ ) g y (s, ξ ) := Sϕ,a (s, ξ ) is an integrable function. Hence S p,a, f is well-defined, and it is a continuous and bounded function, in particular it is a tempered distribution. Obviously, the integrand is differentiable with respect to the variable t and using (14) is easy to see that t → S p,a, f (t, y) is differentiable up to the order 2 p − n − 2. Moreover it can be shown by using similar estimates as in (14) that the function y → Ba, p (s, ξ, y) is 2 p − n − 2 differentiable with 2π |β| β m −a dβ 2π Ba, p (s, ξ, y) =
p e− a iy,m . β 2 dy 2 −1 m∈Zn s + ξ − 2πa m It follows that S p,a, f ∈ C2 p−n−2 Rn+1 . Now we want to show that S p,a, f is polyharmonic of order p on Rn+1 \ a . Since f is integrable there ex∞ n ∈ C R with compact support such that ists a sequence of functions ϕ ( ) k
f (ξ ) − ϕk (ξ ) dξ → 0 for k → ∞. Let us define fk by Fourier inversion of ϕk , i.e. by putting fk (x) =
1 (2π )n
eix,ξ ϕk (ξ ) dξ.
(16)
Clearly, then fk = ϕk . It follows that for x = (t, y) S p,a, f (x) − S p,a, f (x) ≤ 1 k 2π
Rn
Ba, p (s, ξ, 0) f (ξ ) − ϕk (ξ ) dsdξ Sϕ,a (s, ξ ) R
(17)
and by (13) and (14) it follows that the right hand side tends to zero: thus we have uniform convergence on the left hand side. Since ϕk ∈ C∞ (Rn ) has compact support it follows that each ϕk satisfies (10). Hence S p,a, fk ∈ SH p Rn+1 , a by Theorem 2. In particular, S p,a, fk is polyharmonic of order p on Rn+1 \ a . As S p,a, f is the uniform limit of S p,a, fk we conclude that S p,a, f is polyharmonic of order p on Rn \ a . As pointed out in [15, p. 147], S p,a, f with the proven properties is an element of SH p Rn+1 , a .
Since f (ξ ) − ϕk (ξ ) dξ → 0 we conclude from (16) that fk (x) converges
f (ξ ) dξ. Since f is continuous the Fourier inversion formula to (2π1 )n eix,ξ shows that fk (x) converges to f (x) . By Theorem 2 S p,a, fk interpolates fk at the lattice points γ ∈ a , hence we conclude by (17) that S p,a, f (γ ) = limk−→∞ fk (γ ) = f (γ ) . The proof is complete.
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3 The main result First, let us recall Theorem 6 from [11]. Theorem 4 Suppose that f ∈ B2 p (Rn ) ∩ C (Rn ) satisfies the decay condition (10). Then the polyharmonic spline L p,a, f of order p for the grid a defined in (9) converges pointwise for a → 0 to the function defined by 1 1 dsdξ, (18) eiy,ξ eits f (ξ ) L p, f (t, y) = p 2 2 2π Rn R S p (s, ξ ) s + |ξ | where S p (s, ξ ) :=
k∈Z
1
p . (s + 2π k)2 + |ξ |2
(19)
Moreover the function L p, f is a polyspline. Madych and Nelson have shown in [15] the fundamental fact that the function (s, ξ ) −→ 1/Sϕ,a (s, ξ ) defined in (7) has an analytic extension to some strip in Cn+1 . We need an analogous result for the function (19). Since this function is not periodic in ξ the argument is more involved. The proof of the following three theorems will be given in the Appendix. Theorem 5 There exists ε > 0 such that for each ξ ∈ Rn the function s → 1/S p (s, ξ ) can be extended analytically to the strip {z ∈ C : |Imz| < ε } . Theorem 6 There exists ε > 0 such that for all z = s+iη with |s| ≤ π and |η| ≤ ε and for all ξ ∈ Rn 1 p , 3 · S p (z, ξ ) ≥ 2 2 k∈Z (z + 2π k) + |ξ |
(20)
where the function S p is defined in (19). Theorem 7 There exists ε > 0 such that for all z = s+iη with |s| ≤ π and |η| ≤ ε and for all ξ ∈ Rn and for all a with 0 < a ≤ 1 1 3 Sϕ,a (z, ξ ) ≥ p , 2 −1 2 k∈Z m∈Zn (z + 2π k) + ξ − 2πa m
(21)
where the function Sϕ,a is defined in (7). In particular, for all 0 < a ≤ 1 and |s| ≤ π and |η| ≤ ε and all ξ ∈ Rn p 1 ≤ 3 (s + iη)2 + |ξ |2 . Sϕ,a (s + iη, ξ )
(22)
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Now we are able to state our second result, containing the rate of the convergence established in Theorem 4: Theorem 8 There exists a constant η > 0 such that for every function f ∈ B2 p (Rn ) ∩ C (Rn ) the following estimate holds for all x = (t, y) ∈ R × Rn L p,a, f (x)− L p, f (x) ≤ 6e−ηt
Rn
f (ξ )
S p s + iη, ξ +2π ma−1 dsdξ. Sϕ,a (s+iη, ξ ) n
π
−π m∈Z m =0
Proof It is easy to see from (11) and (18) that
L p,a, f
1 (t, y) − L f (t, y) = 2π
where A (ξ, y, t) :=
∞
−∞
Rn
eiξ,y f (ξ ) A (ξ, y, t) dsdξ,
eits R (s, ξ ) ds and
R (s, ξ ) :=
Ba, p (s, ξ, y) 1 . − p 2 2 Sϕ,a (s, ξ ) S p (s, ξ ) s + |ξ |
By Theorem 5, for each ξ ∈ Rn the function s −→ R (s, ξ ) has an analytic extension for z ∈ C with |Imz| < ε. Hence, by Cauchy’s theorem, A (ξ, y, t) =
∞
−∞
eit(s+iη) R (s + iη, ξ ) ds.
In order to estimate this integral we define
× Ba, p (s, ξ,
y) :=
m∈Zn ,m =0
e− a iy,m 2π
2 p . s2 + ξ + 2πa−1 m
(23)
− p × Clearly, Ba, p (s, ξ, y) = s2 + |ξ |2 + Ba, p (s, ξ, y) and a simple calculation shows that R (s, ξ ) = R1 (s, ξ ) + R2 (s, ξ ) where
R1 (s, ξ ) = R2 (s, ξ ) =
× Ba, p (s, ξ, y)
Sϕ,a (s, ξ )
,
S p (s, ξ ) − Sϕ,a (s, ξ ) . p Sϕ,a (s, ξ ) s2 + |ξ |2 S p (s, ξ )
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π
∞ Using the simple identity −∞ g (s) ds = ∞ k=−∞ −π g (s + 2π k) ds and the periodicity of Sϕ,a (s, ξ ) with respect to s we obtain ∞ ∞ π × Ba, p (s+2π k + iη, ξ, y) it(s+iη) −ηt e R1 (s + iη, ξ ) ds = e ei(s+2π k)t ds. Sϕ,a (s+iη, ξ ) −∞ −π k=−∞
× Ba, p
in (23) and apply the triangle inequality and Recall now the definition of use then the non-trivial inequality (20 ) valid for |s| ≤ π : ∞ × Ba, p (s + 2π k + iη, ξ, y) k=−∞ ∞
1 2 p 2 −1 m∈Zn m =0 k=−∞ (s + 2π k + iη) + ξ + 2πa m S p s + iη, ξ + 2πa−1 m . ≤3 ≤
m∈Zn m =0
∞ Thus −∞ eit(s+iη) R1 (s + iη, ξ ) ds is bounded by the constant π S p s + iη, ξ + 2πa−1 m −ηt ds. M := 3e Sϕ,a (s + iη, ξ ) m∈Zn m =0 −π
(24)
Similarly, if we define the function d (s, ξ ) :=
S p (s, ξ ) − Sϕ,a (s, ξ ) , Sϕ,a (s, ξ ) S p (s, ξ )
which is periodic in s, we obtain ∞ ∞ it(s+iη) −ηt e R2 (s + iη, ξ ) ds = e −∞
π
k=−∞ −π
ei(s+2π k)t d (s + iη, ξ )
(s + 2π k + iη)2 + |ξ |2
p ds.
Using again (20) we obtain ∞ π it(s+iη) −ηt |d (s + iη, ξ )| · S p (s + iη, ξ ) ds ≤ 3e e R + iη, ξ ds (s ) 2 −∞ −π π S p (s + iη, ξ ) − Sϕ,a (s + iη, ξ ) ds. = 3e−ηt Sϕ,a (s + iη, ξ ) −π The estimate S p (s + iη, ξ ) − Sϕ,a (s + iη, ξ ) ≤
S p s + iη, ξ + 2πa−1 m
m∈Zn m =0
shows that
∞ −∞
eit(s+iη) R2 (s + iη, ξ ) ds ≤ M
where M is defined in (24). The proof is complete.
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We need the following Proposition 9 For all s ∈ R and σ > 0 the following inequality holds A :=
∞
1 1+σ p ≤ 2p . 2 σ 2 k=−∞ (s + 2π k) + σ
Proof From the proof of Theorem 6 in [11] applied to the case n = 1 and a = 1 one obtains that A≤
∞ 1 2 1 +
p . σ 2p σ 2p π 2 k2 k=1 1 + σ 2 σ π
∞
1 dy. By an obvious estimate the (1+y2 ) p last integral attains maximum for p = 1; for p = 1 it is a table integral equal to π . This ends the proof. 2
The latter sum can be estimated by
0
In the next Theorem we refine the estimate in Theorem 8 by estimating the right–hand side. Theorem 10 There exist constants η > 0 and C > 0 such that for every function f ∈ B2 p (Rn ) ∩ C (Rn ) the estimate p L p,a, f (t, y) − L f (t, y) ≤ a2 p−1 Ce−ηt · f (ξ ) 1 + |ξ |2 dξ holds for all t ∈ R and y ∈ Rn and for all 0 < a ≤ 1. Proof Define the set Ia := ξ ∈ Rn : |ξ | ≤ πa−1 . 1. We are going to estimate at first the error π S p s + iη, ξ + 2π ma−1 1 f (ξ ) dsdξ. E1 (Ia ) := Sϕ,a (s + iη, ξ ) 2π Ia −π m∈Zn m =0 Since |z| ≥ |Re z| and Re (s + 2π k + iη)2 = (s + 2π k)2 − η2 we have 2 2 (s + iη + 2π k)2 + ξ − 2πa−1 m ≥ (s + 2π k)2 + ξ − 2πa−1 m − η2 . Thus we can estimate ∞ 1 S p s + iη, ξ + 2π ma−1 ≤ p . 2 −1 2 2 k=−∞ (s + 2π k) + ξ − 2πa m − η
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For ξ ∈ Ia we have ξ −2π ma−1 ≥ 2π ma−1 −|ξ | ≥ πma−1 for every m = 2 0. Further η ≤ 1 and a ≤ 1 imply 1 πma−1 ≥ η2 for m = 0, and therefore 2
2 2 2 1 σ 2 := ξ − 2πa−1 m − η2 ≥ πma−1 − η2 ≥ πma−1 . 2
(25)
Hence Proposition 9 and (25) yield ∞
1+ 1 1 + |σ | p ≤ ≤ 2p 2 p 2 σ 2 k=−∞ (s + 2π k) + σ
ξ − 2πa−1 m2 − η2 . πma−1 2 p
Since 1 ≤ a−1 and |ξ | ≤ πa−1 for ξ ∈ Ia we obtain 2 1 + ξ − 2πa−1 m − η2 ≤ 1 + ξ − 2πa−1 m ≤ a−1 (1 + π + 2π |m|) . Thus we proved S p s+iη, ξ +2π ma−1 ≤ 2 p a2 p−1 (1+π +2π |m|)/|πm|2 p . Further we know by (22) p p 1 ≤ 3 (s + iη)2 + |ξ |2 ≤ 3 2π 2 + |ξ |2 . Sϕ,a (s + iη, ξ ) Thus we obtain the upper bound p 2 E1 (Ia ) ≤ 3a2 p−1 2 p f (ξ ) 2π + |ξ |2 dξ Ia
1 + π + 2π |m| . |πm|2 p m∈Zn m =0
By enlarging the domain of integration to Rn we obtain the desired estimate. 2. Now let us discuss the integral π S p s + iη, ξ + 2π ma−1 1 f (ξ ) dsdξ. E2 (a) := Sϕ,a (s + iη, ξ ) 2π Rn \Ia −π m∈Zn m =0 From (21) and the definition of S p and Sϕ,a follows immediately S p s + iη, ξ + 2π ma−1 ≤ 3 Sϕ,a (s + iη, ξ ) m∈Zn m =0
f (ξ ) dξ. For for all s ∈ R, ξ ∈ Rn and 0 < a ≤ 1. Hence, E2 (a) ≤ 3 Rn \Ia ξ∈ / Ia we have the trivial estimate 1 + |ξ |2 ≥ a−2 , hence p p f (ξ ) 1+|ξ |2 2p f (ξ ) dξ = f (ξ ) 1 + |ξ |2 dξ. p dξ ≤ a 2 1 + |ξ | Rn \Ia Rn \Ia Rn \Ia The proof is complete.
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4 Proof of the main Theorem 1 Proof of Theorem 1 Let η > 0 and C > 0 as in Theorem 10, and let σ > 0 as in the assumption of Theorem 1. By a classical result, see e.g. Schoenberg [19], or [18], or [8, p. 297, Lemma 15.3], there exists a constant D (η, σ ) > 0 and a constant R > 0 such that for all t ∈ R with |t| ≥ R the following inequality holds: ∞
| j|σ e−η|t− j| ≤ D (η, σ ) |t|σ .
j=−∞
σ −η|t− j| Since the function F (t) := ∞ is bounded on [−R, R] we can j=−∞ | j| e find a constant D2 = D2 (η, σ ) such that ∞
| j|σ e−η|t− j| ≤ D2 (1 + |t|σ )
(26)
j=−∞
for all t ∈ R. The constant c > 0 in Theorem 1 will be defined as c := C · D2 . Let now d j ∈ B2 p (Rn ) ∩ C (Rn ) , j ∈ Z, satisfying d j2 p ≤ D (1 + | j|σ ) for all we substitute f by j ∈ Z. For each j ∈ Z let L p,d j be defined by (18) where d j. We define S by the Lagrange scheme S (t, y) = ∞ j=−∞ L p,d j (t − j, y) and recall from [2] (for the case p = 2) or from [3] for general p ∈ N, that S is indeed a polyspline on strips of order p. Similarly, we can write Ia (t, y) = ∞ j=−∞ L p,a,d j (t − j, y) . By Theorem 10 it follows that |S (t, y) − Ia (t, y)| ≤
∞ L p,d (t − j, y) − L p,a,d (t − j, y) j j j=−∞
≤
∞
a2 p−1 Ce−η|t− j| ·
p d j (ξ ) 1 + |ξ |2 dξ.
j=−∞
Using (5) and (26) we arrive at
|S (t, y) − Ia (t, y)| ≤ a2 p−1 CD
∞
e−η|t− j| (1 + | j|σ )
j=−∞
≤a This ends the proof.
2 p−1
CDD2 (1 + |t|σ ) .
Acknowledgements Both authors acknowledge the support of the Institutes Partnership Project with the Alexander von Humboldt Foundation, Bonn. The second author is supported in part by Grant MTM2006-13000-C03-03 of the D.G.I. of Spain.
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Appendix Proof of the Theorems 5, 6 and 7 The function S p (s, ξ ) in (19) is defined for all pairs (s, ξ ) ∈ Rn+1 \ where the set is given by := {(s, ξ ) : s ∈ 2π Z, |ξ | = 0} . Proof of Theorem 6 Define a function qξ by qξ (z) := z2 + |ξ |2 . For z = s + iη we have qξ (s + 2π k + iη) = (s + 2π k)2 − η2 + |ξ |2 + 2iη (s + 2π k) .
(27)
Thus Re qξ (z+2π k) = (s+2π k) −η +|ξ | which implies Re qξ (z+2π k) ≥ 4π 2 k2 − 2π 2 |k| − π 2 ≥ π 2 k2 for |η| ≤ π and |s| ≤ π and k ∈ Z, k = 0. Hence, we have proved for k ∈ Z, k = 0 qξ (z + 2π k) ≥ Re qξ (z + 2π k) ≥ π 2 k2 . (28) 2
2
2
It follows that qξ (z + 2π k) = 0 for k = 0, |s| ≤ π and |η| ≤ π, hence z → −p qξ (z + 2π k) is well defined and analytic for k = 0, |s| < π and |η| < π. For each ξ ∈ Rn define 1 S×p (z, ξ ) := . (29) p q + 2π k) (z k∈Z\{0} ξ For |s| ≤ π and |η| ≤ π, we obtain by (28) the inequality 1 1 × ≤ p =: M < ∞. S p (z, ξ ) ≤ 2 p q (z + 2π k) π k2 p ξ k∈Z\{0}
(30)
k∈Z\{0}
We want to show (20). By multiplying (20) with qξ (z) p it suffices to show the following statement: there exists ε > 0 with ε < π such that for z = s + iη with |s| ≤ π and |η| ≤ ε, and for all ξ ∈ Rn holds qξ (z) p p × . (31) 3 1 + qξ (z) S p (z, ξ ) ≥ 1 + qξ (z + 2π k) p k∈Z\{0} The constant ε > 0 for the estimate (31) will be chosen inthe following way: p 1 , where we first take δ with 1 > δ > 0 so small that 2δ 2 + δ ≤ min 12 , 2M have defined M in (30). Then choose ε > 0 so small that π 2επ δ ≤ , and arctan . (32) ε≤ 2 2π δ 4p Now let us prove the estimate. We consider two cases. 1. In the first case we s2 + |ξ |2 ≤ δ 2 . Since |η| ≤ ε, formula (27) assume 2 2 2 for 0 implies qξ (z) ≤ s + |ξ | + ε + 2π ε. Using (32) we arrive at k= qξ (z) ≤ 2δ 2 + δ, hence qξ (z) p ≤ 2δ 2 + δ p ≤ 1 . 2M
(33)
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Then (30) and (33) yield p 1 p p 1 + qξ (z) S×p (z, ξ ) ≥ 1 − qξ (z) S×p (z, ξ ) ≥ 1 − qξ (z) M ≥ . (34) 2 Hence, for s2 + |ξ |2 ≤ δ 2 holds 3 p 3 1 + qξ (z) S×p (z, ξ ) ≥ . 2
(35)
On the other hand, (33) and the definition of M in (30) give qξ (z) p 1 1 3 p ≤ 1 + p ≤ . 1+ qξ (z + 2π k) 2M 2 qξ (z + 2π k) k∈Z\{0} k∈Z\{0}
(36)
Clearly (35) and (36) imply (31) for s2 + |ξ |2 ≤ δ 2 . 2. Assume now as the second (and more complicated) case that s2 + |ξ |2 ≥ δ 2 . The inequality |w| ≥ Rew yields p Re 1 + qξ (z) S×p (z, ξ ) ≥ 1 + k∈Z\{0}
p
qξ (z) p
qξ (z + 2π k)
.
(37)
Roughly speaking, we shall show that each factor in the summands on the right-hand side of (37) has an argument which is small (using that η is small) so each summand will have an argument which will be less than π/4. More precisely, we want to show that for z = s + iη with |s| ≤ π, |η| ≤ ε, and k ∈ Z, the argument ϕ (z, ξ, k) in (38) qξ (z + 2π k) = qξ (z + 2π k) · eiϕ(z,ξ,k) p
can be estimated by |ϕ (z, ξ, k)| ≤ π/4 p. Hence, qξ (z + 2π k) has an argument with an absolute value lower or equal to π/4. Thus p p q (z) qξ (z) ξ cos σ, Re p = p q (z + 2π k) q (z + 2π k) ξ
√
ξ
where |σ | ≤ π/4. Since cos σ ≥ 1/ 2 for |σ | ≤ π/4, we obtain p p q (z) qξ (z) 1 ξ . Re p ≥√ p q (z + 2π k) 2 q (z + 2π k) ξ
(39)
ξ
Now (37) and (39) lead to inequality (31). We still have to prove (38). For k = 0 we shall need our assumption s2 + 2 2 |ξ | ≥ δ : the estimate qξ (s + iη) ≥ Re qξ (s + iη) ≥ s2 + |ξ |2 − ε2 ≥ 12 δ 2 1 holds since by (32) follows ε ≤ 2π δ < 12 δ. Further Imqξ (s + iη) = |ηs| ≤ επ implies Im qξ (s + iη) ε2π ≤ 2 . Re qξ (s + iη) δ Our choice of ε > 0 in (32) implies that arg qξ (s + iη) ≤ arctan ε2π ≤ δ2 π/4 p, where arg z denotes the argument of the complex number z.
Numer Algor
For k = 0 the inequality (28) yields Re qξ (s + 2π k + iη) ≥ π 2 k2 , and clearly Im qξ (s + 2π k + iη) = η (s + 2π k) . Hence, for k ∈ Z \ {0} Im qξ (s + 2π k + iη) |k| + 1 ε (π + 2π |k|) =ε ≤ 2ε. ≤ 2 2 Re qξ (s + 2π k + iη) π k k2 we have arctan 2ε ≤ π/4 p. Hence arg qξ (s + 2π k + iη) ≤ Since 2ε ≤ ε2π δ2 π/4 p for all k ∈ Z, k = 0. This ends the proof. Proof of Theorem 5 It is easy to see that qξ (z + 2π k) = 0 for all z = s + iη with |s| ≤ 32 π and |η| ≤ π. Thus z −→ S×p (z, ξ ) defined in (29) is an analytic function for z = s + iη with |s| < 32 π and |η| < π . It follows from equation (21) p that 1 + qξ (z) S×p,0 (z, ξ ) = 0 for all z = s + iη with |s| ≤ π and |η| ≤ ε. Hence p by continuity there exists δξ such that 1 + qξ (z) S×p,0 (z, ξ ) = 0 for all z = s + iη with |s| < π + δξ and |η| ≤ ε. Hence p
qξ (z) 1 = p S p (z, ξ ) 1 + qξ (z) S×p (z, ξ ) is an analytic extension of the function 1/S p (s, ξ ) for z = s + iη with |s| < π + δξ and |η| < ε. Since S p (z, ξ ) = S p (z + 2π k, ξ ) for all k ∈ Z it follows that 1/S p (z, ξ ) is analytic for |Imz| < ε. The proof is complete. Proof of Theorem 7 Let us write ξ = (ξ1 , ..., ξn ) and m = (m1 , ..., mn ) . Since ξ j ≤ may and do assume that ξ −→ Sϕ,a (s + iη, ξ ) is 2πa−1 Zn -periodic we πa−1 for j = 1, ..., n. It is easy to see that ξ j − 2πa−1 m j ≥ πa−1 m j for j = 1, ..., n and all m ∈ Zn . It follows that ξ − 2πa−1 m2 ≥ π 2 a−2 |m|2 ≥ π 2 |m|2 . (40) Now we claim that 2 π2 |m|2 . (41) (s + iη + 2π k)2 + ξ − 2πa−1 m ≥ π 2 k2 + 2 2 2 Indeed, (s +iη +2π k)2 + ξ − 2πa−1 m ≥ Re (s+iη+2π k)2 + ξ − 2πa−1 m and in case of k = 0 we have Re (s + iη + 2π k)2 ≥ π 2 k2 by (28). Hence (41) is proven for k√ = 0. If k = 0 we use that |z| ≥ Re z and (40) in order to obtain for |η| ≤ π/ 2 2 1 (s + iη)2 + ξ − 2πa−1 m ≥ π 2 |m|2 + s2 − η2 ≥ π 2 |m|2 − π 2 . 2 Hence (41) is proven for the case k = 0 as well. We conclude that S× ϕ,a (s + iη, ξ ) :=
(k,m)∈Zn+1 (k,m) =0
1 2 p 2 (s + iη + 2π k) + ξ − 2πa−1 m
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is bounded for all |s| ≤ π, |η| ≤ π, 0 < a ≤ 1, and ξ ∈ Rn since × 1 S (s + iη, ξ ) ≤
p =: M. ϕ,a 2 π2 2 2 (k,m)∈Zn+1 ,(k,m) =0 π k + 2 |m| Now it suffices to prove: p 3 1 + qξ (z) S× ϕ,a (z, ξ ) ≥ 1 +
(k,m)∈Zn+1 (k,m) =0
(42)
p p q (z) ξ 2 p . (43) 2 (z + 2π k) + ξ − 2πa−1 m
Choose 1 > δ > 0 and ε > 0 as in the proof of Theorem 6 but with M now defined in (42), and conclude in the same fashion that (43) holds for the first case s2 + |ξ |2 ≤ δ 2 . In the second case s2 + |ξ |2 ≥ δ 2 we notice that this implies that s2 + ξ − 2πa−1 m2 ≥ δ 2 for all m ∈ Zn since trivially ξ − 2πa−1 m ≥ π |m| ≥ δ for all m ∈ Zn , m = 0. Note that 2 (z + 2π k)2 + ξ − 2πa−1 m = qξ −2πa−1 m (z + 2π k) . In the proof of Theorem 6 we have seen that −1 qξ −2πa−1 m (z + 2π k) = qξ −2πa−1 m (z + 2π k) eiϕ (z,ξ −2πa m,k) where ϕ z, ξ − 2πa−1 m, k ≤ π/4 p. Now we argument in a way analogous to the proof of Theorem 6: p p q (z) ξ 1 + q p (z) S× (z, ξ ) ≥ 1 + Re 2 p ϕ,a ξ 2 (z + 2π k) + ξ − 2πa−1 m (k,m)∈Zn+1 (k,m) =0
≥1+
(k,m)∈Zn+1 (k,m) =0
The proof is complete.
p p q (z) 1 ξ √ p 2 (z + 2π k)2 + ξ − 2πa−1 m2
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