Bulletin T.CXXXIII de l’Académie serbe des sciences et des arts − 2006 Classe des Sciences mathématiques et naturelles Sciences mathématiques, N\underline{o} 31

ON POSITIVITY PROPERTIES OF FUNDAMENTAL CARDINAL POLYSPLINES^{1}

H. RENDER^{2}

(Presented at the 1st Meeting, held on February 24, 2006)

A b s t r a c t. Polysplines on strips of order p are natural generalizations of univariate splines. In [3] and [4] interpolation results for cardinal polysplines on strips have been proven. In this paper the following problems will be addressed: (i) positivity of the fundamental polyspline on the strip [−1, 1] × ℝ^{n}, and (ii) uniqueness of interpolation for polynomially bounded cardinal polysplines.

AMS Mathematics Subject Classification (2000): 41A82

Key Words: Cardinal splines, L-splines, fundamental spline, polyharmonic functions, polysplines

1. Introduction

A function f : U → ℂ defined on an open subset U of the euclidean space ℝ^{n+1} is polyharmonic of order p if it is 2p times continuously differentiable

1This paper was presented at the Conference GENERALIZED FUNCTIONS 2004, Topics in PDE, Harmonic Analysis and Mathematical Physics, Novi Sad, September 22– 28, 2004

2The research of the author is partially supported by Dirección General de Investigación (Spain) under grant BFM2003–06335–C03–03.

and Δ^{p}f (x) = 0 for all x ∈ U, where Δ^{p} is the p-th iterate of the Laplace operator Δ = \frac{∂^{2}}{∂x_{1}^{2}} +...+ \frac{∂^{2}}{∂x_{n+1}^{2}} . A famous example in the area of interpolation with polyharmonic functions are the so-called thin–plate splines (and more generally, polyharmonic splines) which are linear combinations of translates of the function φ defined by

φ (x) = |x|^{2} log |x|;  (1)

it is well known that (1) is the fundamental solution of the biharmonic operator Δ^{2} in ℝ^{2}. Since the appearance of the fundamental work of Duchon [8] such ”splines” have been used by numerous authors for interpolation purposes in the multivariate case, see, for example, the papers of W. Madych and S. Nelson [18], K. Jetter [10], and the recent monograph [5]. In all these examples one interpolates data prescribed on a (finite or countable) set of discrete points.

An alternative and completely different ”data concept” is provided by the notion of polyspline, introduced by O. Kounchev in [11], and extensively discussed in [12]. Polysplines distinguish from the widely spread data principle and allow to interpolate functions prescribed on surfaces of codimension 1; for a concrete application see [17]. As in [3],[4] and [15] we consider here the case that data functions are prescribed on parallel equidistant hyperplanes. Let us recall that a function S : ℝ^{n+1} → ℂ is a cardinal polyspline of order p on strips, when S is a 2p − 2 times continuously differentiable function on ℝ^{n+1} which is polyharmonic of order p on the strips (j, j + 1) × ℝ^{n}, j ∈ ℤ, where as usually (a, b) denotes the open interval in ℝ with endpoints a, b, and ℤ is the set of all integers. Note that for n = 0 (with the identification ℝ^{0} = {0} and ℝ × {0} = ℝ) a cardinal polyspline of order p on strips is just a cardinal spline on the real line ℝ of degree 2p − 1 (hence of order 2p), as discussed by I. Schoenberg in his celebrated monograph [21] (or [22]). In passing, let us remark that in the recent paper [15] it has been proved that the cardinal polysplines on strips occur as a natural limit of polyharmonic splines considered on the lattice ℤ × aℤ^{n} when the positive number a ⟶ 0, and an estimate of the rate of convergence has been given in [16]. A discussion of wavelet analysis of cardinal polysplines can be found in [12] and [13].

In the first section we recall briefly the main results about interpolation with polysplines presented by A. Bejancu, O. Kounchev and the author in [4] (for the case p = 2 see [3]). An important tool are so-called fundamental cardinal polysplines which can be seen as the multivariate analog of the fundamental cardinal spline L^{0} : ℝ → ℝ which is by definition the unique cardinal spline which has exponential decay and the interpolation property

L^{0} (0) = 1 and L^{0} (j) = 0 for j ∈ ℤ, j ≠ 0.  (2)

We call a polyspline L_{f} a fundamental cardinal polyspline with respect to the data function f : ℝ^{n} → ℂ if

L_{f} (0, y) = f (y) and L_{f} (j, y) = 0 for j ∈ ℤ \ {0} , y ∈ ℝ^{n}  (3)

and if there exists C > 0 and ε > 0 such that |L_{f} (t, y)| ≤ Ce^{−ε|t|} for all y ∈ ℝ^{n}, t ∈ ℝ. The existence of fundamental cardinal polysplines is guaranteed by Theorem 2, and the reader may take formula (9) as a defining formula.

It is a well-known fact that the fundamental cardinal spline L^{0} defined in (2) is non-negative on the unit interval [−1, 1] , see [7]. One aim of this paper is to discuss the question whether the fundamental cardinal polyspline L_{f} : ℝ^{n+1} → ℂ is non-negative on the strip [−1, 1]×ℝ^{n} for any non-negative integrable function f : ℝ^{n} → [0, ∞) . Unfortunately, we have not been able to give a positive answer to this question, although numerical experiments support this conjecture. However, in the second section we shall prove that the non-negativity of L_{f} on [−1, 1] × ℝ^{n} for any non-negative integrable function f : ℝ^{n} → [0, ∞) is equivalent to the positive definiteness of a certain family of functions ξ ⟼ L^{ξ} (t) where t ranges over [−1, 1] . Here L^{ξ} is the fundamental cardinal L-spline L^{ξ} : ℝ → ℝ (cf. [19] and [3] for definition and details) which can be written as

L^{ξ} (t) = \frac{1}{2π} ∫_{−∞}^{∞} e^{its} \frac{1}{(s^{2} + |ξ|^{2})^{p} S_{p} (s, ξ)}ds,  (4)

where

S_{p} (s, ξ) := ∑_{k∈ℤ} \frac{1}{((s + 2πk)^{2} + |ξ|^{2})^{p}} .  (5)

In the third section we shall show that for the special, and much simpler, case p = 1 the fundamental cardinal polyspline L_{f} is non-negative on the strip [−1, 1]×ℝ^{n} for any non-negative integrable function f : ℝ^{n} → [0, ∞). Moreover we give a simplified formula for the fundamental cardinal polyspline L_{f} in the case p = 1.

The last section is devoted to the question under which conditions interpolation with cardinal polysplines on strips is unique. A simple example shows that even for the case p = 1 there is no uniqueness if we do not impose some growth conditions. The author believes that for polynomially bounded polysplines interpolation is unique; in the last section it is proved that this is true for the case p = 1. It is hoped that the results presented here motivate further research on the subject.

Let us recall some terminology and notation: the Fourier transform of an integrable function f : ℝ^{n} → ℂ is defined by

f̂(ξ) := ∫_{ℝ^{n}} e^{−i⟨y,ξ⟩}f (y) dy.

By B_{s} (ℝ^{n}) we denote the set of all measurable functions f : ℝ^{n} → ℂ such that the integral

∥f∥_{s} := ∫_{ℝ^{n}} |f̂(ξ)| (1 + |ξ|^{s}) dξ  (6)

is finite (see Definition 10.1.6 in Hörmander [9], vol. 2). By S (ℝ^{n}) we denote the Schwartz space of rapidly decreasing functions on ℝ^{n}, see [25, p. 19]. A function f : ℝ^{n} → ℝ is radially symmetric if f (x) depends only on the Euclidean norm |x| = \sqrt{x_{1}^{2} + ... + x_{n}^{2}}.

2. Interpolation with Polysplines

In this section we recall the interpolation theorem for cardinal polysplines of order p proved by A. Bejancu, O. Kounchev and the present author. As mentioned above, this result formally includes the theorem of I. Schoenberg about cardinal spline interpolation by setting n = 0. But it should be emphasized that the proof of Theorem 1 relys on results of Ch. Micchelli in [19] about cardinal interpolation with so-called L-splines which itself is a generalization of Schoenberg’s theorem.

Theorem 1. Let γ ≥ 0 be fixed. Let integrable functions f_{j} : ℝ^{n} → ℂ be given such that f_{j} ∈ B_{2p−2} (ℝ^{n}) ∩ L^{1} (ℝ^{n}) , and assume that the following growth condition holds

∥f_{j}∥_{2p−2} ≤ C (1 + |j|^{γ})  for all j ∈ ℤ.  (7)

Then there exists a polyspline S of order p on strips satisfying

S (j, y) = f_{j} (y)  for y ∈ ℝ^{n}, j ∈ ℤ,  (8)

as well as the growth estimate

|S (t, y)| ≤ D (1 + |t|^{γ}) for all y ∈ ℝ^{n}.

An important step in the proof of the last theorem is the following:

Theorem 2. Let f ∈ L_{1} (ℝ^{n})∩B_{2p−2} (ℝ^{n}) and define L^{ξ} as in (4). Then the function L_{f} defined by

L_{f} (t, y) := \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}f̂(ξ) L^{ξ} (t) dξ  (9)

is a polyspline of order p such that

\{ \begin{matrix}L_{f} (0, y) = f (y) & for y ∈ ℝ^{n}, \\ L_{f} (j, y) = 0 & for y ∈ ℝ^{n}, for all j ≠ 0. \end{matrix}

There exists a constant C > 0 and η > 0 such that for every multi-index α ∈ ℕ_{0}^{n+1} with |α| ≤ 2 (p − 1) , the decay estimate

|\frac{∂^{α}}{∂x^{α}} L_{f} (x)| ≤ Ce^{−η|t|} ∥f∥_{|α|}  (10)

holds for all x = (t, y) ∈ ℝ^{n+1}.

Theorem 1 can be deduced from Theorem 2 by considering the Lagrangetype representation

S (t, y) = ∑_{j=−∞}^{∞} L_{f_{j}} (t − j, y) .  (11)

Details and proofs can be found in [4] and [3]. In this paper we shall make use only of formula (9) which can be taken as a definition for L_{f}. What we need in this paper is the following fact which also shows that (9) is well-defined.

Theorem 3. There exist constants C > 0 and η > 0, such that for all t ∈ ℝ, ξ ∈ ℝ^{n} the following estimate holds:

|L^{ξ} (t)| ≤ Ce^{−η|t|}.  (12)

A proof for p = 2 can be found in [3], and for arbitrary p in [4].

3. A conjecture about positivity of the fundamental spline

Recall that a function g : ℝ^{n} → ℂ is positive definite if for all y_{1}, ..., y_{N} ∈ ℝ^{n} and for all complex numbers c_{1}, ..., c_{N} the inequality

∑_{k,l=1}^{N} c_{l}\overline{c_{k}}g (y_{k} − y_{l}) ≥ 0

holds; for properties of positive definite functions we refer to [23], cf. also the nice introduction [24]. It is well-known that the product of two positive definite functions is positive definite. Moreover it is elementary to see that the Fourier transform ĝ of a non-negative function g ∈ L_{1} (ℝ^{n}) is positive definite. Conversely, if g ∈ L_{1} (ℝ^{n}) is positive definite then the Fourier transform is a non-negative function on ℝ^{n} (Theorem of Mathias).

Properties of the fundamental cardinal spline L^{0} : ℝ → ℝ have been investigated by de Boor and Schoenberg in [7]. One particularly nice property is that L^{0} has an alternating sign on the intervals (k, k + 1) for k ∈ ℕ_{0}, i.e., that

(−1)^{k} L^{0} (x + k) ≥ 0

for all k ∈ ℕ_{0}, x ∈ (0, 1). Numerical experiments have lead us to formulate the following conjecture:

Conjecture 4. Let f ∈ L_{1} (ℝ^{n}) ∩ B_{2p−2} (ℝ^{n}) . If f is non-negative then the fundamental polyspline L_{f} has an alternating sign on the strips (k, k + 1)× ℝ^{n} for k ∈ ℕ_{0}, i.e., that

(−1)^{k} L_{f} (t + k, y) ≥ 0

for all k ∈ ℕ_{0}, t ∈ (0, 1) and y ∈ ℝ^{n}.

Note that for k = 0 the conjecture implies that L_{f} (t, y) ≥ 0 for all (t, y) ∈ [−1, 1] × ℝ^{n}. The following result shows that the latter property is equivalent to the positive definiteness of the function ξ ⟼ L^{ξ} (t) for each t ∈ [−1, 1] . Note that this formulation is independent of the data function f.

Theorem 5. Let t ∈ ℝ be fixed. Then the following statements are equivalent

(i) The function ξ ⟼ L^{ξ} (t) is positive definite.

(ii) For each non-negative f ∈ L_{1} (ℝ^{n}) such that f̂ ∈ L_{1} (ℝ^{n}) the fundamental cardinal polyspline L_{f} is non-negative on {t} × ℝ^{n}.

(iii) For each non-negative, radially symmetric function f ∈ S (ℝ^{n}) the function L_{f} is non-negative on {t} × ℝ^{n}.

P r o o f. For (i) → (ii) let f ∈ L_{1} (ℝ^{n}) be non-negative, clearly then f̂ is positive definite. By assumption, ξ ⟼ L^{ξ} (t) is positive definite. By the above remarks the function ξ ⟼ f̂(ξ) L^{ξ} (t) is positive definite. Since by Theorem 3 the function ξ ⟼ L^{ξ} (t) is bounded, we know that ξ ⟼ f̂(ξ) L^{ξ} (t) is integrable. By the theorem of Mathias (see [24, p. 412]) the (inverse) Fourier transform is non-negative, i.e., that for all y ∈ ℝ^{n}

L_{f} (t, y) = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}f̂(ξ) L^{ξ} (t) dξ ≥ 0.

The implication (ii) → (iii) is trivial.

Let us show that (iii) → (i). We use arguments from the proof of Bochner’s theorem in [1, p. 196]: Let us define f_{δ} (y) := e^{− \frac{1}{2} δ|y|^{2}} which is radially symmetric and in the Schwartz class. By Theorem 3 the function ξ ⟼ L^{ξ} (t) is bounded. Hence g_{ε} defined by g_{ε} (ξ) := L^{ξ} (t) e^{−ε|ξ|^{2}} is integrable for any ε > 0. Parseval’s identity yields

∫_{ℝ^{n}} f_{δ} (y) ĝ_{ε} (y) dx = ∫_{ℝ^{n}} f̂_{δ} (ξ) g_{ε} (ξ) dξ.  (13)

On the other hand, assumption (iii) implies that

L_{f_{ε}} (t, y) = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}e^{−ε|ξ|^{2}}L^{ξ} (t) dξ ≥ 0.

Thus ĝ_{ε} (y) = (2π)^{n} L_{f_{ε}} (t, −y) ≥ 0 for all y ∈ ℝ^{n}. So we obtain from (13) that

∫_{ℝ^{n}} e^{− \frac{1}{2} δ|y|^{2}} ĝ_{ε} (y) dy = |∫_{ℝ^{n}} e^{− \frac{1}{2} δ|y|^{2}} ĝ_{ε} (y) dy| ≤ M ∫_{ℝ^{n}} f̂_{δ} (ξ) dξ,  (14)

where M is a constant such that |g_{ε} (ξ)| ≤ M for all ξ ∈ ℝ^{n} and for all 0 < ε ≤ 1. Since

∫_{ℝ^{n}} f̂_{δ} (ξ) dξ = (2π)^{n} f_{δ} (0) ≤ (2π)^{n}

we conclude from (14) and Fatou’s lemma that ĝ_{ε} is integrable. Now the inversion formula

g_{ε} (ξ) = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}ĝ_{ε} (y) dy.  (15)

shows that g_{ε} (ξ) := L^{ξ} (t) e^{−ε|ξ|^{2}} is positive definite. Then L^{ξ} (t) = lim_{ε→0}g_{ε} (ξ) for each ξ ∈ ℝ^{n} (and fixed t), and since the pointwise limit of positive definite functions is again positive definite, it follows that ξ ⟼ L^{ξ} (t) is positive definite. □

4. Positivity of fundamental cardinal polysplines on [−1, 1] × ℝ^{n} for p = 1.

Recall that a function g : ℝ^{n} → ℂ vanishes at infinity if for each ε > 0 there exists a compact subset K of ℝ^{n} such that |f (x)| < ε for all x ∈ ℝ^{n}\K. Now we want to prove

Theorem 6. Let p = 1. Let f ∈ L_{1} (ℝ^{n}) such that f̂ ∈ L_{1} (ℝ^{n}). If f is non-negative then L_{f} defined in (9) is a non-negative function on ℝ^{n+1}.

P r o o f. From the definition of L_{f} and L^{ξ} it follows that

L_{f} (t, y) = \frac{1}{(2π)^{n+1}} ∫_{ℝ^{n}} ∫_{−∞}^{∞} e^{i⟨y,ξ⟩}e^{its} \frac{f̂(ξ)}{(s^{2} + |ξ|^{2}) S_{1} (s, ξ)}dsdξ.

Further it can be shown that (ξ, s) ⟼ f̂(ξ) / (s^{2} + |ξ|^{2}) S_{1} (s, ξ) is integrable. The Lemma of Riemann-Lebesgue (see [25, p. 2]) shows that L_{f} : ℝ^{n+1} → ℂ vanishes at infinity. Now the next theorem applied to L_{f} and j ∈ ℤ, shows that L_{f} is a non-negative function. □

Theorem 7. Let S : ℝ^{n+1} → ℂ be a cardinal polyspline of order 1 on strips which vanishes at infinity and let j ∈ ℤ. If

S (j, y) ≥ 0 and S (j + 1, y) ≥ 0 for all y ∈ ℝ^{n}

then S is non-negative on [j, j + 1] × ℝ^{n}.

P r o o f. Let ε > 0 be arbitrary. Since S vanishes at infinity we can find R > 0 such that |S (t, y)| < ε if |t| > R or |y| > R. Define G_{R} = [j, j + 1] × {y ∈ ℝ^{n} : |y| ≤ R + 1} . Then S (t, y) ≥ −ε for (t, y) in the boundary of G_{R}. Since S is a harmonic function in the interior of G_{R} and continuous on G_{R} the minimum principle yields that S (t, y) ≥ −ε for all (t, y) ∈ G_{R}. Hence S (t, y) ≥ −ε for given (t, y) ∈ G_{R}. Since ε > 0 is arbitrary we obtain S (t, y) ≥ 0 and the proof is accomplished. □

In the rest of this section we want to give an explicit formula for L^{ξ} in the case that p = 1 (see (16)) which clearly leads to a simpler formula for fundamental cardinal polysplines, see formula (17). From formula (16) one can see that ξ ⟼ L^{ξ} (t) is positive definite for each t ∈ [−1, 1] , so one obtains with Theorem 5 a second proof that L_{f} is non-negative on [−1, 1] × ℝ^{n} for a non-negative data function f ∈ L^{1} (ℝ^{n}) such that f̂ ∈ L^{1} (ℝ^{n}) . Unfortunately, for p ≥ 2 we do not have simple formulas for L^{ξ}.

Let us compute S_{p} (s, ξ) defined in (5) for p = 1. An application of Poisson’s summation formula (see [6, p. 204]) shows that

2 ∑_{k∈ℤ} \frac{y}{y^{2} + (x + 2πk)^{2}} = ∑_{k∈ℤ} e^{−|k|y}e^{ikx} = \frac{1 − e^{−2y}}{1 − 2e^{−y} cos x + e^{−2y}} .

We apply this to x := s and y := |ξ| > 0 and obtain for S_{p} defined in (5) with p = 1

S_{1} (s, ξ) = \frac{1 − e^{−2|ξ|}}{2 |ξ| (1 − 2e^{−|ξ|} cos s + e^{−2|ξ|})}.

Hence we obtain

L^{ξ} (t) = \frac{1}{π} \frac{|ξ|}{1 − e^{−2|ξ|}} ∫_{−∞}^{∞} e^{its} \frac{1 − 2e^{−|ξ|} cos s + e^{−2|ξ|}}{s^{2} + |ξ|^{2}} ds.

Since 2e^{its} cos s = e^{its} (e^{is} + e^{−is}) = e^{i(t+1)s} + e^{is(t−1)} we see that L^{ξ} (t) is equal to

\frac{|ξ|}{π} \frac{1 + e^{−2|ξ|}}{1 − e^{−2|ξ|}} ∫_{−∞}^{∞} \frac{e^{its}}{s^{2} + |ξ|^{2}} ds − \frac{|ξ|}{π} \frac{e^{−|ξ|}}{1 − e^{−2|ξ|}} ∫_{−∞}^{∞} \frac{e^{i(t+1)s} + e^{is(t−1)}}{s^{2} + |ξ|^{2}} ds.

Since ∫_{−∞}^{∞} e^{its} \frac{1}{s^{2}+|ξ|^{2}} ds = \frac{π}{|ξ|}e^{−|t|·|ξ|} a straightforward computation shows that

L^{ξ} (t) = \frac{1}{e^{|ξ|} − e^{|−ξ|}} [(e^{|ξ|} + e^{−|ξ|}) e^{−|t|·|ξ|} − e^{−|t+1|·|ξ|} − e^{−|t−1|·|ξ|}] .

If t ≥ 1 one obtains easily L^{ξ} (t) = 0. For 0 ≤ t ≤ 1 one has

L^{ξ} (t) = \frac{e^{|ξ|(1−t)} − e^{−(1−t)·|ξ|}}{e^{|ξ|} − e^{−|ξ|}} = sinh (|ξ| (1 − t)) ; sinh |ξ|.

We now summarize the result:

Corollary 8. Let p = 1. For |t| ≥ 1 the function L^{ξ} vanishes and for 0 ≤ t ≤ 1

L^{ξ} (t) = sinh (|ξ| (1 − t)) ; sinh |ξ|.  (16)

In case ξ = 0 the function t ⟼ L^{0} (t) is a linear spline and L^{0} (t) = 1 − t for 0 ≤ t ≤ 1.

Now Theorem 2 for p = 1 can be read as follows:

Theorem 9. Let f ∈ L_{1} (ℝ^{n}) such that f̂ ∈ L_{1} (ℝ^{n}) . Then there exists a continuous function L_{f} : ℝ^{n+1} → ℂ which is harmonic in (−1, 0)× ℝ^{n} and (0, 1) × ℝ^{n} such that

L_{f} (0, y) = f (y)

for y ∈ ℝ^{n}, and it vanishes for all (t, y) ∈ ℝ^{n+1} with |t| ≥ 1. Further for 0 ≤ t ≤ 1

L_{f} (t, y) = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}f̂(ξ) sinh (|ξ| (1 − t)) ; sinh |ξ|dξ.  (17)

The fundamental linear interpolation spline has nice symmetry properties around x = \frac{1}{2}. In the following we want to formulate a symmetry property for cardinal polysplines of order 1. Formula (17) suggests that we have to use the addition theorem for sinh x = \frac{1}{2} (e^{x} − e^{−x}):

sinh x − sinh y = 2 cosh \frac{x + y}{2} sinh \frac{x − y}{2} .  (18)

Proposition 10. For 0 ≤ s ≤ t ≤ 1 the following relation holds

L^{ξ} (s) = L^{ξ} (t) + 2L^{ξ} (1 − \frac{t − s}{2}) cosh \frac{(2 − s − t) |ξ|}{2}.  (19)

P r o o f. Put x = (1 − s) |ξ| and y = (1 − t) |ξ| in (18): then x + y = (2 − s − t) |ξ| and x − y = (t − s) |ξ| and we have

sinh [(1 − s) |ξ|] − sinh [(1 − t) |ξ|] = 2 cosh \frac{(2 − s − t) |ξ|}{2} sinh \frac{(t − s) |ξ|}{2}.

(20)

Now divide (20) by sinh |ξ| and use formula (16). □

As an illustration put s = \frac{1}{2} − δ and t = \frac{1}{2} + δ in (19). Then

L^{ξ} (\frac{1}{2} − δ) − L^{ξ} (\frac{1}{2} + δ) = 2 cosh (\frac{1}{2} |ξ|) · L^{ξ} (1 − δ)

Multiply (19) with f̂(ξ) e^{i⟨y,ξ⟩} and integrate with respect to dξ. Then (17) implies that for an integrable function f the following formula holds:

L_{f}(\frac{1}{2} − δ, y) − L_{f}(\frac{1}{2} + δ, y) = \frac{2}{(2π)^{n}} ∫_{ℝ^{n}} e^{i⟨y,ξ⟩}f̂(ξ) cosh(\frac{1}{2} |ξ|)L^{ξ} (1 − δ) dξ.

5. Uniqueness of interpolation for polynomially bounded polysplines for p = 1

In this section we want to prove uniqueness results for interpolation: suppose that S_{1} and S_{2} are two polysplines interpolating the same data. It is clear that S_{2} − S_{1} vanishes on {j} × ℝ^{n} for all j ∈ ℤ. We would like to conclude that S_{2} − S_{1} = 0. The following simple example shows that we have to impose some conditions on the interpolation polysplines even in the case p = 1 in order to obtain uniqueness:

Example 11. There exists a harmonic function f on ℝ^{2} which vanishes on all hyperplanes {j} × ℝ, j ∈ ℤ without being identically zero, namely

f (t, y) = sin πt · e^{πy}.

As mentioned in the introduction we believe that interpolation is unique if we assume that S is polynomially bounded, i.e., that there exists a polynomial p (x) such that

|S (x)| ≤ |p (x)|

for all x ∈ ℝ^{n+1}.

In the following we shall prove this for p = 1. In the case that S_{1} and S_{2} vanish at infinity we could use Theorem 7 applied to S_{2} − S_{1} and S_{1} − S_{2}: then S_{2} − S_{1} and S_{1} − S_{2} are non-negative functions on the whole space, hence S_{2} − S_{1} = 0.

Instead of the minimum principle we will use the Schwarz reflection principle for harmonic functions (see e.g., [2, p. 66]) in order to prove uniqueness. Reflection principles for polyharmonic functions have been investigated by several authors and we refer to [20] for a nice introduction. However, it seems that the latter results can not be used for a proof of uniqueness of interpolation for polysplines of order p > 1.

Proposition 12. Suppose that S : ℝ^{n+1} → ℂ is a cardinal polyspline of order 1 on strips with S (j, y) = 0 for all j ∈ ℤ and y ∈ ℝ^{n}. Then there exists a harmonic function h : ℝ^{n+1} → ℂ such that

h (t, y) = S (t, y) for t ∈ (0, 1) and y ∈ ℝ^{n},  (21)
h (j, y) = 0 for j ∈ ℤ and y ∈ ℝ^{n},  (22)

and for each natural number N

max_{|y|≤N,t∈ℝ} |h (t, y)| ≤ max_{|y|≤N,0≤t≤1} |S (t, y)| .  (23)

P r o o f. Clearly S is a harmonic function on the strip (0, 1)×ℝ^{n}, and it is continuous on the closure of the strip. By the Schwarz reflection principle, S can be extended to a continuous function S_{1} on [−1, 1] × ℝ^{n} by defining

S_{1} (−t, y) = −S (t, −y) for t ∈ [−1, 0]

which is harmonic on (−1, 1) × ℝ^{n}. Further S_{1} (−1, y) = −S (1, −y) = 0 for all y ∈ ℝ^{n}, so S_{1} vanishes on the boundary of the new strip [−1, 0] × ℝ^{n} and clearly the maximum of |h| on {(t, y) : |y| ≤ N, −1 ≤ t ≤ 0} can be estimated by

max_{|y|≤N,−1≤t≤0} |S_{1} (t, y)| ≤ max_{|y|≤N,0≤t≤1} |S (t, y)| .

Now apply the same procedure to S_{1} : [−1, 0]×ℝ^{n} at the hyperplane {−1}× ℝ^{n}, obtaining an extension S_{2} on [−2, 0] × ℝ^{n} of S_{1} with

max_{|y|≤N,−2≤t≤−1} |S_{2} (t, y)| ≤ max_{|y|≤N,−1≤t≤0} |S_{1} (t, y)| ≤ max_{|y|≤N,0≤t≤1} |S (t, y)| .

Proceed in this way for negative j ∈ ℤ, then for positive j ∈ ℤ and we arrive at a harmonic function h : ℝ^{n+1} → ℂ with the desired properties. □

Theorem 13. Let S : ℝ^{n+1} → ℂ be a cardinal polyspline of order 1 on strips vanishing on the affine hyperplanes {j} × ℝ^{n}, j ∈ ℤ. If S is polynomially bounded then S is identically zero.

P r o o f. By Proposition 12 there exists a harmonic function h : ℝ^{n+1} → ℂ with (21), (22) and (23). Since S is polynomially bounded, (23) implies that h is polynomially bounded. It follows that h is a harmonic polynomial, see [2, p. 41]. A polynomial h (t, y) which vanishes on the hyperplanes {j} × ℝ^{n+1} for all j ∈ ℤ is identically zero: the equation h (0, y) = 0 for all y ∈ ℝ^{n} implies that the (finite) Taylor expansion of h (t, y) contains only nontrivial summands where the variable t occurs. Hence h (t, y) = t · h_{1} (t, y) with a polynomial h_{1}. Similarly, h_{1} (1, y) = 0 for all y ∈ ℝ^{n} implies that h_{1} (t, y) = (t − 1) h_{2} (t, y) . Hence we can write

h (t, y) = t (t − 1) .... (t − m) h_{m} (t, y) .

If m is bigger than the total degree of h we obtain a contradiction, showing that h must be zero. By (21) we conclude that S must be zero on (0, 1)×ℝ^{n}. In order to show that S is zero on ℝ^{n+1} consider the polyspline S_{j} defined by S_{j} (t, y) = S (t − j, y) for (t, y) ∈ ℝ^{n+1}, j ∈ ℤ. By the above, S_{j} is zero in (0, 1) × ℝ^{n}. Hence S must be zero on (j, j + 1) × ℝ^{n}. □

Corollary 14. Interpolation with polynomially bounded cardinal polysplines of order 1 on strips is unique.

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