DAMTP 000/NA04 On the semi-norm of raial basis function interpolants H.-M. Gutmann Abstract: Raial basis function interpolation has attracte a lot of

Transcription

1 UNIVERSITY OF CAMBRIDGE Numerical Analysis Reports On the semi-norm of raial basis function interpolants H.-M. Gutmann DAMTP 000/NA04 May, 000 Department of Applie Mathematics an Theoretical Physics Silver Street Cambrige CB3 9EW Englan

2 DAMTP 000/NA04 On the semi-norm of raial basis function interpolants H.-M. Gutmann Abstract: Raial basis function interpolation has attracte a lot of interest in recent years. For popular choices, for example thin plate splines, this problem has a variational formulation, i.e. the interpolant minimizes a semi-norm on a certain space of raial functions. This gives rise to a function space, calle the native space. Every function in this space has the property that the semi-norm of an arbitrary interpolant to this function is uniformly boune. In applications it is of interest whether a suciently smooth function belongs to the native space. In this paper we give sucient conitions on the ierentiability of a function with compact support, in the case of cubic, linear an thin plate splines. In the case of multiquarics an Gaussian functions, it is shown that the only compactly supporte function that satises these conitions is ientically zero. Department of Applie Mathematics an Theoretical Physics, University of Cambrige, Silver Street, Cambrige CB3 9EW, Englan. May, 000.

3 On the semi-norm of raial basis function interpolants 3 Introuction Let n pairwise ierent points x ; : : : ; x n IR be given, where n an are any positive integers. Further, let ; : : : ; n be real numbers, p be in m, the space of polynomials of egree at most m, an enote the Eucliean norm by k:k. A raial basis function is of the form s(x) = n In this paper we consier the following choices of : i (kx? x i k) + p(x); x IR : (.) (r) = r (linear); (r) = r 3 (cubic); (r) = r log r (thin plate spline); (r) = p r + c (multiquaric); (r) = e?r (Gaussian); Let the matrix IR nn be ene by 9 >= >; r 0: (.) () i := (kx i? x k); i; = ; : : : ; n: (.3) In all the cases in (.), is conitionally enite. Specically, letting m 0 be in the cubic an thin plate spline case, 0 in the linear an multiquaric case an? in the Gaussian case, we obtain (see e.g. [6]) if IR n is any nonzero vector that satises n (?) m 0+ T > 0; (.4) i q(x i ) = 0 8 q m ; (.5) where? is the space that contains only the zero function. We choose m to be an integer that is not less than m 0. Now x n IN an the centres x ; : : : ; x n IR. The interpolation problem can be pose in the following way. Fin a raial basis function s of the form (.) that satises n s(x i ) = f i ; i = ; : : : ; n i q(x i ) = 0; q m 9 >= >; : (.6) Let ^m be the imension of m. Further, let p ; : : : ; p ^m be a basis of this linear space. The matrix P is ene by P := 0 p (x ) p ^m (x ).. p (x n ) p ^m (x n ) C A :

4 4 H.-M. Gutmann In the Gaussian case with m =?, P is omitte. Further, let F IR n be the vector whose entries are the ata values f ; : : : ; f n. Therefore the system (.6) can be written as P F P T = ; (.7) 0 c 0 ^m where IR n has the components i, where c IR ^m an where 0 ^m is the zero in IR ^m. The components of c are the coecients of the polynomial p with respect to the basis p ; : : : ; p ^m. Powell [6] shows that the interpolation matrix A = P P T 0 IR (n+ ^m)(n+ ^m) (.8) is nonsingular for any prescribe m m 0, if an only if x ; : : : ; x n satisfy q m an q(x i ) = 0; i = ; : : : ; n =) q 0: (.9) An interesting an useful property is that the raial basis function that solves the interpolation problem (.6) is the solution of a minimization problem. For any choice of in (.) an m m 0, let us ene the linear space A ;m as the space of functions of the form N i (kx? y i k) + p(x); x IR ; where N is any positive integer, where y ; : : : ; y N are any pairwise ierent points in IR, where p is any polynomial in m an where = ( ; : : : ; N ) T satises N i q(y i ) = 0 8 q m : (.0) On this space a semi-inner prouct an a semi-norm can be ene. Let s an u be any functions in A ;m, i.e. s(x) = N(s) i (kx? y i k) + p(x) an u(x) = The semi-inner prouct is the expression hs; ui := (?) m 0+ N(s) N(u) = Clearly, it is bilinear. To show symmetry, by employing N(s) i q(y i ) = 0 an N(u) = (kx? z k) + q(x): i u(y i ): (.) p(z ) = 0;

5 On the semi-norm of raial basis function interpolants 5 we euce hs; ui = (?) m 0+ By (.4), the expression hs; si = (?) m 0+ = (?) m 0+ = (?) m 0+ = (?) m 0+ N(s) N(s) N(s) N(u) = N(u) = 0 N(u) N(u) = (ky i? z k) + q(y i ) i (ky i? z k) = N(s) i (kz? y i k) + p(z ) s(z ) = hu; si : i s(y i ) = (?) m 0+ N(s) i;= A A i (ky i? y k) (.) is strictly positive, if satises (.5) for n = N(s) an any m m 0. Thus (.) is inee a semi-inner prouct. Then (hs; si) = is a semi-norm on the space A ;m with null space m. Schaback [8] shows that the solution of (.6) can be characterize as follows. Theorem Let be any raial basis function from (.), an let m be chosen such that m m 0. Given are points x ; : : : ; x n in IR that satisfy (.9) an values f ; : : : ; f n in IR. Let s be the raial function of the form (.) that solves the system (.6). Then s minimizes the semi-norm hg; gi = on the set of functions g A ;m that satisfy g(x i ) = f i ; i = ; : : : ; n: (.3) Proof: P n(g) Let g(x) := = (kx? y k) + q(x) A ;m satisfy (.3). We consier the semi-norm of g? s. Since s(x i ) = g(x i ); i = ; : : : ; n, it has the value hg? s; g? si = hg; gi? hg; si + hs; si = hg; gi + hs; si? (?) m 0+ = hg; gi + hs; si? (?) m 0+ = hg; gi + hs; si? hs; si n n i g(x i ) i s(x i ) = hg; gi? hs; si : (.4)

6 6 H.-M. Gutmann Thus, we euce the require conition hs; si = hg; gi?hg? s; g? si hg; gi : One can associate with a particular in (.) an m m 0 an interesting function space, the so-calle native space N ;m (see Schaback [9]). Schaback an Wenlan [0] show that it can be characterize as the space of all functions F : IR! IR for which the following conition hols. Conition There exists a real number C, that only epens on F, such that, for any choice of interpolation points x ; : : : ; x n IR for which (.9) hols, the interpolant s n to F at these points satises hs n ; s n i C: For many types of raial basis functions, it is unknown what functions F belong to N ;m. It is the subect of this paper to prove Conition for suciently ierentiable functions F with boune support. A useful application of the results presente here is given by a metho for global optimization (Gutmann [3]). Here the global minimum of a continuous function f : D! IR is sought, where D IR is compact. If x ; : : : ; x n D an their function values have been calculate, the next point x n+ is etermine as follows. Choose an estimate of the global minimum, f say. For each y D n fx ; : : : ; x n g there exists a raial basis function s y A ;m that interpolates (x ; f(x )); : : : ; (x n ; f(x n )) an (y; f ). We take the view that the \least bumpy" of these interpolants yiels the most promising location for evaluating the obective function. The \bumpiness" of s y is measure by its semi-norm. This means that x n+ minimizes hs y ; s y i ; y D n fx ; : : : ; x n g. A crucial part of the proof of convergence of this metho is to show that suitable functions with boune support satisfy Conition. Another use is that the assumptions that guarantee convergence can be weakene if the obective function f itself satises Conition. We propose a new technique that relates the semi-norm introuce above to a semi-norm ene for raial basis function interpolants on an innite regular gri. For that purpose we generalize the semi-norm by allowing innite sums in (.), an show in Section that a semi-inner prouct an a semi-norm are well-ene. In Section 3, the extene semi-norm is applie to interpolants s h on the innite regular gri h with the mesh-size h > 0, where we restrict our attention to interpolation of functions with boune support. Proposition 8 shows that, if the interpolants on h have uniformly boune semi-norms, as h! 0, then Conition hols. A conition on f that ensures the uniform bouneness of these semi-norms is given in Theorem.

7 On the semi-norm of raial basis function interpolants 7 Multi-inex notation will be use in the following sections. For a multi-inex = ( ; : : : ; ) T IN, the orer an the factorial! are ene as := i an! := The power x ; x IR, an the erivative D u(x) for a suciently smooth function u : IR! IR are respectively. x := Y Y i!: x i i an D u(x) u(x) ; The semi-norm for raial basis functions with innitely many centres We let the linear space ~A ;m be the space of functions of the form where s(x) = i (kx? x i k) + p(x); x IR ; (.) i q(x i ) = 0 8 q m : (.) Here is a raial basis function from (.), (x i ) iin is a sequence in IR, the x i being pairwise ierent, ( i ) iin is a sequence in IR an p is a polynomial in m where m m 0, m 0 being ene in Section. We require ( i ) iin an (x i ) iin to satisfy an i < ; Every choice of in (.) has the property i (kx i k) < ; (.3) i q(x i ) < ; q m : (.4) (kx? yk) [ + (kxk) + (kyk)]; x; y IR ; (.5) where is a constant. Thus (.3) implies that the sum in expression (.) is absolutely convergent for each x IR. Note that, if m is increase, then fewer functions of the form (.) satisfy (.).

8 8 H.-M. Gutmann Let s(x) = i (kx? x i k) + p(x) an u(x) = = (kx? y k) + q(x) be elements of ~A ;m, an let P = fz k : k INg be the union of fx ; x ; : : :g an fy ; y ; : : :g, where any point that is in both of the last sets is inclue only once in P. New coecients ~ k an ~ k are ene by an ~ k := i ; if z k = x i for an i IN 0; otherwise ~ k := which gives the expressions s(x) = k= Then the sum s + u is ; if z k = y for a IN 0; otherwise ; ~ k (kx? z k k) + p(x) an u(x) = (s + u)(x) = k= an the prouct of s with a scalar is (s)(x) = k= ~ k (kx? z k k) + q(x): ( ~ k + ~ k )(kx? z k k) + (p + q)(x); x IR ; ( i )(kx? x i k) + p(x): Therefore ~ A ;m is a linear space. It contains A ;m as ene in Section. Next, by analogy with expression (.), we introuce the bilinear form hs; ui := (?) m 0+ i u(x i ); (.6) where s an u are as above. We will show that it is a well-ene semi-inner prouct. We consier the formula hs; ui = (?) m 0+ = (?) m 0+ i u(x i ) " # i (kx i? y k) + q(x i ) = :

9 On the semi-norm of raial basis function interpolants 9 P Equation (.) shows that i iq(x i ) vanishes. (.5) provie the absolute convergence conition = Further, inequalities (.3) an i (kx i? y k) < ; (.7) the functions s an u being in ~A ;m. Therefore hs; ui an hs; si are well-ene. Symmetry can be erive easily now, as the sums in the ouble series can be exchange. Thus using (.) again, hs; ui = (?) m 0+ = (?) m 0+ " = = i (ky? x i k) + p(y ) s(y ) = hu; si : It remains to prove that hs; si is nonnegative. Denition (.6) gives the value hs; si = (?) m 0+ i;= i (kx i? x k): If only nitely many i are nonzero, then s is of the form (.) an the i satisfy (.0). Hence hs; si is a nite sum, an the assertion follows from (.4). However, if innitely many i are nonzero, we have to argue in a ierent way. As in the previous section, let ^m be the imension of m, an let p ; : : : ; p ^m be a basis of m. There exist ^m pairwise ierent points y ; : : : ; y ^m anywhere in IR that satisfy (.9) for n = ^m, if x i = y i ; i = ; : : : ; ^m. Thus the matrix P := 0 p (y ) p (y ^m ).. p ^m (y ) p ^m (y ^m ) is nonsingular. For large n, let (n) be the solution of the system where the right-han sie vector has the components b (n) k =? n C A P (n) = b (n) ; (.8) i p k (x i ); k = ; : : : ; ^m: When n tens to innity, each component of b (n) tens to zero by (.). Therefore, (n) = P? b (n) also tens to zero. #

10 0 H.-M. Gutmann The choices of b (n) an (n) provie the equations n Therefore the function n i q(x i ) + i (kx? x i k) + ^m = (n) q(y ) = 0; q m : ^m = (n) (kx? y k); x IR ; is in the linear space A ;m of Section. Thus the conitional positive or negative eniteness of yiels the inequality (?) m 0+ n i;= i (kx i? x k) + (?) m 0+ + (?) m 0+ ^m = ^m i;= (n) n i (kx i? y k) (n) i (n) (ky i? y k) 0: The rst term in the sum on the left-han sie tens to hs; si, as n tens to innity, whereas the other two terms ten to zero, since (n) tens to zero an since the enition (.) gives the limit lim n! n = (ky i? x k) = s(y i )? p(y i ); i = ; : : : ; ^m: Thus the left-han sie, as n!, provies a sequence of nonnegative numbers that ten to hs; si. It follows that hs; si is nonnegative as require. We state the result formally. Theorem 3 hs; ui ene in (.6) is a semi-inner prouct on ~A ;m, an therefore hs; si = is a semi-norm on this space. Now one can exten the variational principle in Theorem to ~A ;m. Theorem 4 Let be any raial basis function from (.), let m be chosen such that m m 0, where m 0 is ene in the previous section, an let one of the two following statements hol: (i) The ata are nitely many points x ; : : : ; x n in IR, that satisfy (.9), an the values f ; : : : ; f n in IR. Let s be the raial function of the form (.) that solves the system (.6).

11 On the semi-norm of raial basis function interpolants (ii) The ata are a sequence of pairwise ierent points (x i ) iin in IR an a sequence of values (f i ) iin in IR. Assume there is a raial basis function s A ~ ;m of the form (.) that interpolates (x i ; f i ); i IN. In both cases s minimizes the semi-norm hg; gi = on the set of functions g ~ A ;m that satisfy g(x i ) = f i ; i = ; : : : ; n; (.9) or respectively. g(x i ) = f i ; i IN; (.0) Proof: P Let g(x) := = (kx? y k) + q(x) ~A ;m satisfy (.9) or (.0). We consier the semi-norm of g? s. Equation (.4) is vali in case (i), an is the ientity hg? s; g? si = hg; gi? hs; si : For case (ii), the nite sums of equation (.4) are replace by the innite series i g(x i ) an i s(x i ); an the same argument applies, since s(x i ) = g(x i ); i IN. Therefore, in both cases, hs; si = hg; gi? hg? s; g? si hg; gi ; which completes the proof. 3 Application to interpolation on the innite regular gri An important special case of interpolation at innitely many points is interpolation on an innite regular gri h, where h > 0 is the mesh-size. It is known [6] that an interpolant to a function F : IR! IR can be written in the form u h (x) = k F (hk) x h? k ; x IR ; (3.) provie that the sum (3.) is absolutely convergent for all x IR, where is the unique carinal spline (x) = (kx? k); x IR ; (3.)

12 H.-M. Gutmann that satises (k) = k0 ; k. Because of the rapi ecrease of (x) when kxk tens to innity, the absolute convergence conition is achieve for many functions F, incluing all boune functions. In this section we restrict our attention to continuous functions F that have boune support, i.e. there exists a boune subset U IR such that F (x) = 0 for all x IR nu. It follows that every F is boune. First, we want to show that the interpolant to F on h is in A ~ ;m. Then we apply the results of the previous section to show that, if the semi-norm of this interpolant is uniformly boune, as h! 0, then the semi-norm of an interpolant at nitely many arbitrary points is also uniformly boune. Conitions on the ierentiability of F that are sucient for the uniform bouneness of the semi-norm of the interpolant on h will be investigate in the next section. Since the argument of in (3.) is x=h? k, an not x? hk as is neee for the form (.), we ene h (r) := (hr); r 0, an we let the coecients ; ; have the values that cause the function h (x) = h (kx? k); x IR ; (3.3) to satisfy the Lagrange conitions h (k) = k0 ; k. Then, instea of u h in (3.), we consier the interpolant s h (x) = ; x IR : (3.4) k F (hk) h x h? k The coecients of the carinal functions (3.) an (3.3) can be written explicitly as (see [6]) i e T t = t; P` ; (3.5) ^(kt + `k) [0;] an = [0;] i e T t P` ^h (kt + `k) t; ; (3.6) where ^(k:k) an ^ h (k:k) are the Fourier transforms of (k:k) an h (k:k), respectively. The Fourier transform of an absolutely integrable function : IR! IR is the integral ^(t) = IR e?i xt t (x) x; t IR : The generalize transforms ^ are known for all choices of in (.) an can be

13 On the semi-norm of raial basis function interpolants 3 foun in [6]. They take the values ^(r) = C r?? (linear); ^(r) = C r??3 (cubic); ^(r) = C r?? (thin plate spline); ^(r) = C (c=r) (+)= K (+)= (cr) (multiquaric); ^(r) = C e?r =4 (Gaussian); 9 >= >; r > 0: (3.7) In each case C is a constant that oes not epen on t an c, an r = ktk; t IR. K (+)= is a moie Bessel function that is positive for r > 0, an that ecays exponentially as r tens to innity (Abramowitz an Stegun []). The enition h (r) = (hr); r 0, implies ^ h (ktk) = h? ^ ktk h ; t IR n f0g: (3.8) The coecients ecrease rapily (cf. [6]). Specically, they have the property C h kk?`; kk! ; (3.9) where C h epens only on h, an ` is + in the linear an the multiquaric cases, + in the thin plate spline case an + 3 in the cubic case. In the Gaussian case, any integer ` is amissible. There is a ierence between even an o imensions, i.e. the ecay is faster, often exponential, in either even or o imensions, but the boun (3.9) is sucient for our purposes. For any h > 0, let N h be the nite set fk : hk Ug. Thus (3.4) can be expresse in the form s h (x) = kn h F (hk) h x h? k = F (hk) kn h = F (hk) kn h = kn h F (hk) h x h? k? (kx? hk? hk)?k (kx? hk); x IR : P By (.5) an (3.9) the sum?k (kx? hk) is absolutely convergent for each k N h an each x IR. It follows from the niteness of N h that we can write s h (x) = " kn h F (hk)?k # (kx? hk) = (kx? hk);

14 4 H.-M. Gutmann where Because of (3.9), the coecients have the property = := F (hk)?k ; : (3.0) kn h F (hk)?k kn h ^Ch kk?`; kk! ; (3.) with the values of ` specie after (3.9) an ^C h epening on h an F. From (3.0) we can euce an (khk) = F (hk)?k kn h kf k < ; kf k kn h kn h?k?k (khk) < ; so the satisfy the require conitions (.3). The remaining question for eciing if s h is in ~A ;m is whether they also satisfy equation (.). If the coecients of the carinal function (3.3) have the properties q() < an q() = 0 8 q m ; (3.) then, because q(x+k); x IR ; is in m for every k, we euce the ientities q() =?k q( + k) = 0 8 q m ; 8 k : (3.3) Thus the enition (3.0) implies q() = F (hk) q()?k kn h = kn h F (hk)?k q() = 0; q m; which is the require conition (.). Therefore the interpolant (3.4) is in the linear space ~ A ;m if the conitions (3.) are achieve for the choice of m. The following remark shows that one cannot expect any m m 0 to be amissible.

15 On the semi-norm of raial basis function interpolants 5 Remark 5 We consier linear splines in one imension, i.e. (r) = r. Then the carinal function (3.3) takes the form h (x) = h x + h? h x + x? h; h x IR; so ust three of the coecients f : g are nonzero. Thus the rst of the conitions P (3.) is achieve for every m. Further, for q(x) an q(x) = x, the sum q() takes the values h? h + h = 0 an h (?)? h 0 + h = 0: On the other han, for q(x) = x the sum takes the nonzero value h? h 0 + h = h : Thus the coecients fail to satisfy the constraints (3.) for m =. Proposition 6 gives values of m that satisfy the constraints (3.), but the Gaussian case is exclue. Proposition 6 Let be any of the raial basis functions in (.), except the Gaussian function. Let m take the value ; + ; + an in the linear, cubic, thin plate spline an multiquaric case respectively. Then the coecients of the carinal function (3.3) satisfy an this sum is absolutely convergent. q() = 0 8 q m ; Proof: See Buhmann [], chapter 5. We summarize the results obtaine so far in the following theorem. Theorem 7 Let be any of the raial basis functions in (.). Let the polynomial space m be chosen such that m takes an integer value in [0; ], [; +], [; +] an [0; ] in the linear, cubic, thin plate spline an multiquaric case respectively. In the Gaussian case, the polynomial space is omitte, i.e. m = m 0 =?. Let F : IR! IR be a continuous function with boune support. Then the interpolant

16 6 H.-M. Gutmann to F on h of the form (3.4) lies in the space ~A ;m, an its semi-norm is the square root of hs h ; s h i = (?) m 0+ F : (3.4) Now we turn to the secon problem pose in the secon paragraph of this section. We use the results of the previous section to stuy the behaviour of the semi-norm of a function (.) that interpolates a continuous function F with boune support at nitely many points. It turns out that this semi-norm can be boune by a constant that oes not epen on the particular choice of the interpolation points, if the semi-norm of the interpolant on h is uniformly boune, as h! 0. Proposition 8 Let be any of the raial basis functions in (.), an let m take one of the values state in Theorem 7. F : IR! IR is a continuous function that has boune support. For h > 0 let s h be the interpolant (3.4) to F on h. Further, let pairwise ierent points x ; : : : ; x n IR satisfy (.9), an let s n A ;m be the unique function (.) that interpolates F at x ; : : : ; x n. If hs h ; s h i is uniformly boune by a constant K > 0, as h! 0, then hs n ; s n i K: Proof: We regar the ata points x ; : : : ; x n as xe. For a suciently small h > 0 there exist pairwise ierent points x ; : : : ; x n h such that kx i? x i k < h; i = ; : : : ; n; (3.5) where k:k enotes the maximum norm in IR. The entries of the interpolation matrix (.8) are continuous with respect to the ata points. Thus there exists h 0 > 0 such that, for all h < h 0, (3.5) hols an the nonsingularity of the interpolation matrix is preserve, if each x i is replace P by x i. For any h < h 0, let s n n be the raial basis function interpolant i (kx? x i k) + p (x) that satises s n (x i ) = F (x i ); i = ; : : : ; n: The continuity of F yiels the continuity of the right-han sie vector of the system (.7), an the nonsingularity of (.8) for all h < h 0 implies the continuity of the solution vector (.7), consiere as a function of the ata points, in

17 On the semi-norm of raial basis function interpolants 7 (x ; : : : ; x n ). Formula (.) then yiels that the semi-norm as a function of the ata points is continuous in (x ; : : : ; x n ). Thus, as lim x i = x i ; i = ; : : : ; n, h!0 hs n ; s n i = lim h!0 s n ; s n Since x i h ; i = ; : : : ; n; Theorem 4 implies s n ; s n hsh ; s h i : : (3.6) By assumption, the right-han sie is boune above by a constant K that oes not epen on h. Therefore (3.6) proves the assertion. 4 Sucient conitions for a uniform boun on the semi-norm Proposition 8 shows that F is in N ;m if the semi-norm of the interpolant s h on h is uniformly boune. In this section we investigate conitions on F that guarantee this property. We will n that it is sucient if F is in the function space ~N ene below. The enition requires the remark that (?) m 0+ ^h (ktk) is nonnegative for h > 0, which is prove as follows. Using (3.6) an (3.0), we write the nonnegative expression (3.4) as hs h ; s h i = (?) m 0+ [0;] P` ^ h (kt + `k) t: F (hk)e?ikt t k (4.) In all the cases in (3.7), ^ oes not change sign. Therefore formula (3.8) implies that (?) m 0+ ^h (ktk) is nonnegative. Denition 9 The linear space ~N is the space of continuous functions f : IR! IR that are absolutely integrable an whose Fourier transform ^f satises IR ^f(t) It is enowe with the inner prouct (f; g) := ^(ktk) t < : ^f(t)^g(t) IR ^(ktk) t

18 8 H.-M. Gutmann that inuces the norm (f; f) = = 4 IR ^f(t) ^(ktk) t 3 5 = : Remark 0 The Fourier transform of a function f ~N is absolutely integrable which can be shown as follows. If we enote the close unit ball by B an observe that ^(k:k) is in L (IR n B) in all the cases in (3.7), we n IR ^f(t) t = B B ^f(t) t + ^f(t) t + IR nb IR nb ^f(t) r r ^(ktk) t ^(ktk) ^f(t) ^(ktk) t A = IRnB Thus the inverse Fourier transform theorem hols, i.e. f(x) = ^f(t)e itx t; x IR : IR = ^(ktk) t < : The following lemma shows that the raial basis function interpolant s h to a function F with boune support is in ~N. Lemma Let F : IR! IR be a continuous function with boune support, let s h be the interpolant (3.4) to F on h, an let g be any function in ~ N. Then (i) s h is in ~N, an hs h ; s h i = (s h ; s h ). (ii) (g; s h ) = (?) m 0+ g, where has the value (3.0). Proof of (i): First we obtain the Fourier transform of s h from expression (3.4). Because of the rapi ecrease of the carinal spline h, s h is absolutely integrable. For each k, the Fourier transform of h (:=h? k) is \ : h (t) h? k = h e?ihkt t ^ h (ht); t IR :

19 On the semi-norm of raial basis function interpolants 9 Moreover, ^ h is relate to ^ h (see [6]) by the equation ^ h (ht) = ^ h (hktk) P` ^ h (hkt + `=hk) ; t IR : an (3.8) provies ^ h (hktk) = h? ^(ktk); t IR n f0g. Thus the Fourier transform of s h becomes ^s h (t) = h F (hk)e?ihkt t ^(ktk) ; t IR k P` : (4.) ^(kt + `=hk) Further, for every, ^s h t + = h F (hk)e?ihkt t ^(kt + =hk) h k P` ^(kt + `=hk) ; t IR ; (4.3) which implies ^s h t + = h F (hk)e?ihkt t ; t IR : (4.4) h k Therefore expression (4.) an application of (4.3) an (4.4) yiel h [0;=h] hs h ; s h i = F (hk)e?ihkt t t P k ` [0;=h] = ^s h t + h ^(kt + =hk) t: ^(kt + `=hk) All the terms in the sum are nonnegative, an hs h ; s h i exists. Therefore the monotone convergence theorem provies ^s h hs h ; s h i = (?) m 0+ = (?) m 0+ [0;=h] t + h IR ^s h (t) ^(ktk) t: If follows that s h is in ~ N, an hs h ; s h i = (s h ; s h ). ^(kt + =hk) t

20 0 H.-M. Gutmann Proof of (ii): For, (3.6) an (3.0) show that [0;] = F (hk)e?ikt t k Thus can be expresse as! e it t P` ^ h (kt + `k) t: ; ; are the Fourier coecients of the perioic function k F (hk)e?ikt t! P` ^h (kt + `k) ; t IR ; an (3.) shows a ecay of the coecients that provies! e?it t = P` ^h (kt + `k) ; t IR : (4.5) k F (hk)e?ikt t Expressions (4.) an (4.5) imply e?iht t = ^s h(t) ^(ktk) ; t IR : Thus from Denition 9 an ^ = (?) m 0+ ^ we obtain (g; s h ) = (?) m 0+ IR ^g(t) e iht t t: (4.6) P For n IN, ene S n : IR! IR by S n (t) := kk n e iht t ; t IR. Then the function ^g(:)s n (:) converges to the integran of (4.6) as n!, an satises the boun ^g(t)s n (t) A ^g(t); t IR : The function on the right han sie is integrable, because g ominate convergence theorem (g; s h ) = (?) m 0+ IR ^g(t)e iht t t: ~N, so by the Thus the inverse Fourier transform formula gives the require property (g; s h ) = (?) m 0+ g: Now we are reay to state an prove the theorem.

21 On the semi-norm of raial basis function interpolants Theorem Let be any of the functions in (.), an choose m accoring to Theorem 7. Let F : IR! IR be in ~N an have boune support. For h > 0 enote the interpolant (3.4) to F on h by s h. Then hs h ; s h i is boune above by a number that oes not epen on h. Proof: Lemma (i) shows that s h ~N an (s h ; s h ) = hs h ; s h i. By analogy to the proof of Theorem 4, using Lemma (ii), one obtains Thus 0 (F? s h ; F? s h ) = (F; F )? (F; s h ) + (s h ; s h ) = (F; F )? (?) m 0+ F + (s h ; s h ) = (F; F )? (s h ; s h ): hs h ; s h i = (s h ; s h ) (F; F ): Because (F; F ) oes not epen on h, it serves as the require number. In the linear, cubic an thin plate spline cases, a boun on hs h ; s h i can be obtaine for a class of functions F. Specically, let = in the linear case, = in the thin plate spline case an = 3 in the cubic case. Then the following result hols. Corollary 3 Let (r) = r, (r) = r log r or (r) = r 3, an let m be as in Theorem 7. Let F C (+)= (IR ) if + is even, or F C (++)= (IR ) if + is o, where is ene above, an let F have boune support. For h > 0 let s h be the interpolant (3.4) on h. Then hs h ; s h i is uniformly boune above, so F satises Conition. Proof: It will be prove that F is in the linear space ~N. Because F has boune support, it is absolutely integrable. Further, let be (+)= if + is even, or (++)= if + is o. Integration by parts shows that the Fourier transform ^F satises [D F (t) = (it) ^F (t); t IR ; (4.7) for any multi-inex with orer =. Also, as D F has boune support an thus is in L (IR ), the Plancherel formula gives [ D F (t) t = D F (x) x < ; = : (4.8) IR IR Moreover, expression (3.7) states that ^ has the form ^(ktk) = C ktk + ; t IR n f0g;

22 H.-M. Gutmann for a constant C, an a generalization of the binomial formula provies! ktk =! t ; t IR : = Thus, if + is even, the term (F; F ) of Denition 9 has the value (F; F ) = C? ^F (t) ktk t IR! = C? ^F (t) t t! IR =! = C? [ D! F (t) t < : = Analogously, if + is o, one obtains (F; F ) = C? = IR! [! IR D F (t) ktk t: For each multi-inex with orer, (4.7) shows that lim t!0 [ D F (t) ktk? = 0: It follows from (4.8) that (F; F ) is nite. Thus, in both cases, F ~N. Therefore Theorem provies a uniform boun on the semi-norm of interpolants to F on h. Proposition 8 then completes the proof that F satises Conition. This corollary inclues some well-known special cases incluing the following one. Remark 4 Consier natural cubic splines in one imension, i.e. =, (r) = r 3 an m =. The variational principle (e.g. Powell [5]) states that for F C (IR) with boune support, the cubic spline interpolant s at nitely many points satises IR s 00 (x) x F 00 (x) x: IR It can be shown easily that the left han sie equals hs; si. Thus the constant (=) R F 00 (x) x is a uniform boun on hs; si. The existence of such a boun is inclue in Corollary 3, an the corollary also hols for m = an m = 3. In the multiquaric an Gaussian cases, however, there is no useful class of functions with boune support. In these cases, ^(ktk) ecays exponentially, as ktk!. One can show that the only function in ~N with boune support is the zero function. This follows from a Paley-Wiener theorem (see Katznelson [4] for a one-imensional version).

23 On the semi-norm of raial basis function interpolants 3 Theorem 5 Let F : IR! IR be continuous an compactly supporte, an let e k:k ^F L (IR ) for some > 0. Then F 0. Proof: The assumption e k:k ^F L (IR ) shows that ^F is absolutely integrable. Therefore the inverse Fourier transform formula F (x) = ^F (t)e itx t; x IR ; (4.9) IR hols. Let D := fz C : Im z < g. For any arbitrary v IR with kvk = ene g : D! C as g(z) = ^F (t)e i(zv) T t t; z D: We show that g is well-ene an continuous on D. For z D ^F (t)e iz(vt t) t = ^F (t)e?im z(vt t) t IR IR = ^F (t)e ktk t IR?ktk e e?im z(v T t) IR ^F e k:k L (IR ) IR = e?(im z+)ktk t : The right han sie is nite, so g is well-ene on D. The continuity of g can be euce as follows. Fix z D. Then there is IR such that Im z < <. For w D suciently close to z, Im w <, which provies the boun ^F (t)e iw(vt t) = ^F (t) e?(vt t)im w ^F (t) e ktk ; t IR : The function on the right han sie is integrable, because a similar argument as above gives IR ^F (t) e ktk t = ^F (t)e ktk e?(?)ktk t IR ^F e k:k L (IR ) IR = e?(?)ktk t < : Therefore, as w tens to z, the ominate convergence theorem implies g(z) = lim ^F (t)e iw(vt t) t w!z = lim w!z IR IR ^F (t)e iw(vt t) t = lim w!z g(w):

24 4 H.-M. Gutmann Thus g is continuous on D. Further, Morera's Theorem (see e.g. Ruin [7]) shows that g is analytic on D. Let T be any triangle in the interior of D, an enote its bounary Then, by Fubini's Theorem, g(z)z = ^F (t)e i(zv) T t IR IR = ^F (t) e i(vt t)z z t = 0; because e i(vt t)z is analytic for every t IR. As T is arbitrary, g is analytic. Now choose IR such that v is outsie the support of F. Because g an F (: v) coincie on the real line, g an all its erivatives vanish at. Therefore the Taylor expansion of g aroun yiels that g 0 on D. Thus F ( v) = 0 for all IR. Because v has been chosen arbitrarily, this proves the theorem. It follows that in the multiquaric an Gaussian cases ~N contains no nonzero function with compact support. Corollary 6 Let (r) = p r + c or (r) = e?r. If F ~N has boune support, then F 0. Proof: In both cases consiere here, ^(k:k) ecays exponentially, i.e. there exist > 0 an K > 0 such ^(ktk) Ke?ktk ; t IR n B; (4.0) where B is the close unit ball. Now F ~ N provies IR nb ^F (t) e ktk t K IR nb ^F (t) ^(ktk) t < : It follows that e k:k ^F L (IR ). Theorem 5 now implies the require result. 5 Conclusions We have introuce a new technique to prove Conition for suciently smooth functions with boune support. It provies an extension of well-known results in the linear, cubic an thin plate spline cases. In particular, the choice of m is less restricte than before.

25 On the semi-norm of raial basis function interpolants 5 The situation in the multiquaric an Gaussian cases, however, is isappointing. The exponential ecay of ^ prevents a uniform boun on the semi-norms when F has boune support. As mentione in the introuction, the application to global optimization we have in min requires Conition for such functions. Thus the presente approach is not useful in these cases. In the other cases the question arises whether there exist functions with boune support that are not in ~N but still satisfy Conition. In particular, it is interesting to investigate how the semi-norm behaves when F is less ierentiable than require by Corollary 3. Acknowlegements. I am very grateful to my supervisor M.J.D. Powell. He ha the iea of extening the semi-norm to interpolants on the innite regular gri, an he was always prepare to help over iculties. Also I woul like to thank Bra Baxter an Michael Johnson for their avice. Further, this research has been supporte by the German Acaemic Exchange Service through a octoral scholarship (HSP III) an by the Engineering an Physical Sciences Research Council. References [] M. Abramowitz an I.A. Stegun. Hanbook of Mathematical Functions, volume 55 of Applie Mathematics Series (National Bureau of Stanars). Unite States Department of Commerce, 964. [] M.D. Buhmann. Multivariable Interpolation using Raial Basis Functions. Ph.D. Dissertation, University of Cambrige, 989. [3] H.-M. Gutmann. A raial basis function metho for global optimization. Report DAMTP 999/NA, University of Cambrige, 999. [4] Y. Katznelson. An Introuction to Harmonic Analysis. Dover Publications, New York, 976. [5] M.J.D. Powell. Approximation Theory an Methos. Cambrige University Press, 98. [6] M.J.D. Powell. The Theory of Raial Basis Function Approximation in 990. In W.A. Light, eitor, Avances in Numerical Analysis, Volume : Wavelets, Subivision Algorithms an Raial Basis Functions, pages 05{0. Oxfor University Press, 99. [7] W. Ruin. Real an Complex Analysis. Mc Graw-Hill, Singapore, 987.

26 6 H.-M. Gutmann [8] R. Schaback. Comparison of raial basis function interpolants. In K. Jetter an F. Utreras, eitors, Multivariate Approximations: From CAGD to Wavelets, pages 93{305. Worl Scientic, Singapore, 993. [9] R. Schaback. Native Hilbert Spaces for Raial Basis Functions I. In M.W. Muller, M.D. Buhmann, D.H. Mache, an M. Felten, eitors, New Developments in Approximation Theory, International Series of Numerical Mathematics, Vol. 3, pages 55{8. Birkhauser Verlag, Basel, 999. [0] R. Schaback an H. Wenlan. Inverse an Saturation Theorems for Raial Basis Function Interpolation. Preprint, Universitat Gottingen, 998.