MINIMAL DEGREE UNIVARIATE PIECEWISE POLYNOMIALS WITH PRESCRIBED SOBOLEV REGULARITY Amal Al-Rashdan & Michael J. Johnson* Department of Mathematics Kuwait University P.O. Box: 5969 Safat 136 Kuwait yohnson1963@hotmail.com* Abstract. For k {1, 2, 3,... }, we construct an even compactly supported piecewise polynomial ψ k whose Fourier transform satisfies A k (1 + ω 2 ) k b ψ k (ω) B k (1 + ω 2 ) k, ω R, for some constants B k A k >. The degree of ψ k is shown to be minimal, and is strictly less than that of Wendland s function φ 1,k 1 when k > 2. This shows that, for k > 2, Wendland s piecewise polynomial φ 1,k 1 is not of minimal degree if one places no restrictions on the number of pieces. 1. Introduction A function Φ L 1 (R d ) is said to have Sobolev regularity k > if its Fourier transform Φ(ω) := (2π) d/2 R d Φ(x)e ıx ω dx satisfies A(1 + ω 2 ) k Φ(ω) B(1 + ω 2 ) k, ω R d, for some constants B A >. Such functions are useful in radial basis function methods since the generated native space will equal the Sobolev space W2 k(rd ). The reader is referred to Schaback [3] for a description of the current state of the art in the construction of compactly supported functions Φ having prescribed Sobolev regularity. Wendland (see [4] and [5]) has constructed radial { functions Φ d,l (x) = φ d,l ( x ), where φ d,l is a piecewise polynomial of the form φ d,l (t) =, p being a polynomial. For d {1, 2, 3,...} p( t ), t 1, t > 1 and l {, 1, 2,...}, with the case d = 1, l = excluded, Φ d,l has Sobolev regularity k = l+(d+1)/2 and the degree of the piecewise polynomial φ d,l, namely d/2 +3l+1, is minimal with respect to this property. A natural question to ask is whether the degree of φ d,l would still be minimal if we considered functions of the form Φ(x) = φ( x ) where φ is a piecewise polynomial having bounded support. In this note, we answer this question in the univariate case d = 1. Specifically, we construct a compactly supported even piecewise Key words and phrases. positive definite function, compactly supported, B-spline. 1 Typeset by AMS-TEX
2 PIECEWISE POLYNOMIALS WITH PRESCRIBED SOBOLEV REGULARITY polynomial ψ k, with Sobolev regularity k (see Theorem 2.8), and we show that the degree of ψ k, namely 2k, is minimal (see Theorem 2.1). In comparison with Wendland s function Φ 1,k 1 (which has Sobolev regularity k when k > 1), we see that deg ψ k = deg φ 1,k 1, if k = 2, while deg ψ k = 2k < 3k 2 = deg φ 1,k 1 when k > 2. 2. Results Wendland s piecewise polynomial φ d,l can be identified as a constant multiple of the B-spline having l + 1 knots at the nodes 1 and 1 and d/2 + l + 1 knots at. This can be verified simply by observing that φ d,l and the above-mentioned B-spline have the same degree, d/2 + 3l + 1, and satisfy the same number of continuity conditions across each of the nodes 1,, 1, namely d/2 + 2l + 1 at 1, 1 and 2l + 1 at. It is well understood in the theory of B-splines that multiple knots are to be avoided if one wishes to keep the degree low, and with this in mind, we define ψ k as follows. For k = 1, 2, 3,..., let ψ k be the B-spline having knots k,..., 2, 1, ;, 1, 2,..., k (note that is the only double knot). For easy reference, we display ψ k (t) (normalized) for t [, k] and k = 1, 2, 3: { 8 24t ψ 1 (t) = (1 t) 2 2 + 24t 3 7t 4, t [, 1], ψ 2 (t) = (2 t) 4, t (1, 2] 198 27t 2 + 27t 4 18t 5 + 37t 6, t [, 1] ψ 3 (t) = 153 + 27t 945t 2 + 9t 3 45t 4 + 9t 5 8t 6, t (1, 2] (3 t) 6, t (2, 3] We begin by mentioning several salient facts about the B-spline ψ k which can be found in [1, pp. 18 131]. The piecewise polynomial ψ k is supported on [ k, k], positive on ( k, k), even and of degree 2k. Furthermore, it is 2k 1 times continuously differentiable on R\{} and 2k 2 times continuously differentiable on all of R. It follows from this that the 2k 1 order derivative, D 2k 1 ψ k, is a piecewise linear function which is supported on [ k, k] and is continuous except at the origin where it has a jump discontinuity. Consequently, the 2k order derivative has the form D 2k ψ k = 2πa δ + 2πaj (χ [j 1,j) + χ [ j,1 j) ), for some constants a, a 1, a 2,..., a k and where δ is the Dirac δ-distribution defined by δ (f) = f(). We can thus express the Fourier transform of D 2k ψ k as ( D 2k ψ k ) (ω) = a + 2 a j sin(jω) sin((j 1)ω) ω = a + 2(a j a j+1 ) sin(jω), ω with a k+1 :=, whence it follows that (2.1) ψk (ω) = (ıω) ( 2k D 2k ( 1) ψ k ) (ω) k = a ω + 2(a j a j+1 ) sin(jω).
A. AL-RASHDAN & M.J. JOHNSON 3 Lemma 2.2. Let β R. Then there exist unique scalars c 1, c 2,..., c k R such that (2.3) β + c j cos(jω) = O( ω 2k ) as ω. Proof. Define g(w) = β + k i=1 c i cos(iω). Since g C (R) is even, (2.3) holds if and only if D 2l g() = for l =, 1, 2,..., k 1. These conditions form the system of linear equations [c 1, c 2,..., c k ]A = [ β,,,..., ], where A is the k k matrix given by A(i, j) = ( 1) j 1 i 2j 2. Writing A(i, j) = ( i 2 ) j 1, we recognize A as a nonsingular Vandermonde matrix, and therefore, (2.3) holds if and only if [c 1, c 2,..., c k ] = [ β,,,..., ]A 1. Theorem 2.4. Let β, c 1, c 2,..., c k R be such that (2.3) holds. Then (2.5) β + c j cos(jω) = βα k (1 cos ω) k, ω R, where α k > is defined by 1 α k = 1 π (1 cos ω)k dω. Proof. Since cos j ω span{1, cosω, cos2ω,..., coskω} for j =, 1,..., k, it follows that there exist b j R such that (1 cos ω) k = b + k b j cos(jω). Note that < 1 = 1 α k π (1 cos ω) k dω = 1 π b dω + b j 1 π cos(jω) dω = b, and hence βα k (1 cos ω) k = β + k βα kb j cos(jω). Since βαk (1 cos ω) k = O( ω 2k ) as ω, it follows from the lemma that c j = βα k b j for j = 1, 2,..., k, and therefore (2.5) holds. Corollary 2.6. Let a be as in (2.1). Then ( 1) k a > and (2.7) ψk (ω) = ( 1)k a α k ω (1 cos t) k dt, ω. Proof. It follows from (2.1) that ψ k (ω) = ( 1)k f(ω), where f(ω) := a ω + k 2(a j a j+1 ) sin(jω). Since ψ k is supported on [ k, k] and positive on ( k, k), it follows that ψ k is continuous (in fact entire) and ψ k () >. Consequently, f(ω) = O( ω 2k+1 ) as ω. Since f is infinitely differentiable, it follows that f (ω) = a + k 2j(a j a j+1 ) cos(jω) = O( ω 2k ) as ω, and so by Theorem 2.4, f (ω) = a α k (1 cos ω) k. Since f() =, we can write f(ω) = ω f ω (t) dt = a α k (1 cos t)k dt, and hence obtain (2.7). That ( 1) k a > is now evident since < ψ k () = lim ψ ω + k (w). Remark. At this point, it is also easy to show that ψ k (ω) = ( 1)k a (ω + b j sin(jω)), ω, where the scalars {b j } are determined by the fact that ψ k is continuous at.
4 PIECEWISE POLYNOMIALS WITH PRESCRIBED SOBOLEV REGULARITY Theorem 2.8. For k {1, 2, 3,...}, ψ k has Sobolev regularity k; that is, there exist constants B k A k > such that (2.9) A k (1 + ω 2 ) k ψ k (ω) B k (1 + ω 2 ) k, ω R. Proof. As in the proof of Corollary 2.6, let us write ψ k (ω) = ( 1)k f(ω), where f(ω) := a ω+ k 2(a f(ω) j a j+1 ) sin(jω). Since lim ω ω = a, it follows that lim w w 2k ψ k (w) = ( 1) k a. Since ( 1) k a > (by Corollary 2.6), it follows that there exists N > such that (2.9) holds for ω N. That ψ k (ω) > for all ω > follows easily from Corollary 2.6, and since ψ k is continuous and ψ k () >, we see that (2.9) holds for ω N. We finally conclude that (2.9) holds for all ω R since ψ k is an even function. We now show that the degree of ψ k is minimal. Theorem 2.1. If ψ is an even, compactly supported, piecewise polynomial satisfying (2.9), then the degree of ψ is at least 2k. Proof. Let ψ be an even, compactly supported piecewise polynomial satisfying (2.9) and let the l-th derivative of ψ be the first discontinuous derivative of ψ (if ψ is itself discontinuous then l = ). Then D l+1 ψ can be written as (2.11) D l+1 ψ = g + n 2πcj δ xj, where g L 1 (R) and c j is the height (possibly ) of the jump discontinuity at x j. We can then express the Fourier transform of ψ as ψ(ω) = (ıω) l 1 ( D l+1 ψ ) (ω) = (ıω) l 1 (ĝ(ω) + Θ(ω)), where Θ(ω) = n c je ıxjω. Since Θ is bounded and ĝ(ω) has limit as ω (by the Riemann-Lebesgue lemma), it follows that ψ(ω) = O( ω l 1 ) as ω. With the left side of (2.9) in view, we conclude that l + 1 2k. Since Θ is a non-trivial almost periodic function (see [2, pp.9 14]), it follows that Θ(ω) o(1) as ω, and with the right side of (2.9) in view, we see that l + 1 2k. Therefore, l + 1 = 2k and we conclude that ψ k is 2k 2 times continuously differentiable. Since ψ k is compactly supported (ie. ψ k is not a polynomial), it follows that ψ k has degree at least 2k 1 (see [1, pp. 96-12]). In order to show that the degree of ψ k is at least 2k, let us assume to the contrary that the degree equals 2k 1. In this case the l = 2k 1 derivative of ψ k is piecewise constant and hence g = and ψ(ω) = ( 1) k ω 2k Θ(ω). Since ψ is continuous at, it follows that Θ() =. Since Θ is an almost periodic function, there exist values ω 1 < ω 2 < such that lim n ω n = and lim n Θ(ω n ) = ; but this contradicts the left side of (2.9). Therefore, ψ has degree at least 2k.
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