MINIMAL DEGREE UNIVARIATE PIECEWISE POLYNOMIALS WITH PRESCRIBED SOBOLEV REGULARITY

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1 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 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 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 > 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) t 3 7t 4, t [, 1], ψ 2 (t) = (2 t) 4, t (1, 2] t t 4 18t t 6, t [, 1] ψ 3 (t) = t 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 ]. 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).

3 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 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 ]). 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.

5 A. AL-RASHDAN & M.J. JOHNSON 5 References 1. C. de Boor, A practical guide to splines, Applied Mathematical Sciences 27, Springer-Verlag, New York, C. Corduneanu, Almost periodic functions, Chelsea Publishing Company, New York, R. Schaback, The missing Wendland functions, Adv. Comp. Math. 34 (211), H. Wendland, Piecewise polynomial, positive definite and compactly supported radial functions of minimal degree, Adv. Comp. Math. 4 (1995), H. Wendland, Scattered Data Approximation, Cambridge Monographs on Applied and Computational Mathematics, Cambridge University Press, Cambridge, 25.