g(ω)

Rd

√

= g ∈ L 2 (Rd ).

is well de¬ned, continuous, an element of L 2 (Rd ), and satis¬es f

Furthermore we have

| f (x) ’ f n (x)| ¤ (2π)’d/2 (ω) ’ f n (ω) dω

g(ω)

Rd

¤ (2π)’d/2 g ’ f n L 1 (Rd ) ,

L 2 (Rd )

which means that f is also real-valued and hence that f ∈ G. Finally,

f fn fn

= (2π)’d/4 √ = (2π )’d/4 g ’ √

f ’ fn ’√ ’0

G

(Rd ) (Rd )

L2 L2

for n ’ ∞. Thus G is complete. It remains to show that (· ’ ·) is the reproducing kernel

of G. First of all, is bounded by (0) and in L 1 (Rd ), so that it is also in L 2 (Rd ). Moreover,

(· ’ y)§ = L 1 (Rd ) ,

L 2 (Rd )

10.3 Native spaces for conditionally positive de¬nite kernels 141

(· ’ y) ∈ G for every y ∈ G. The reproduction follows immediately from

so that

f (ω) (ω)e’iωT y

( f, (· ’ y))G = (2π)’d/2 dω

(ω)

Rd

= (2π )’d/2

T

f (ω)eiω y dω

Rd

= f (y).

Because of the previous result, native spaces are in one way a generalization of Sobolev

spaces. Let us clarify this. Remember that for s > d/2 the Sobolev space of order s is

de¬ned as

H s (Rd ) = f ∈ L 2 (Rd ) © C(Rd ) : f (·)(1 + · ∈ L 2 (Rd ) .

2 s/2

2)

Hence, if has a Fourier transform that decays only algebraically then its native space is

a Sobolev space.

∈ L 1 (Rd ) © C(Rd ) satis¬es

Corollary 10.13 Suppose that

c1 (1 + ω 2 )’s ¤ (ω) ¤ c2 (1 + ω 2 )’s , ω ∈ Rd

2 2

with s > d/2 and two positive constants c1 ¤ c2 . Then the native space N (Rd ) corre-

sponding to coincides with the Sobolev space H s (Rd ), and the native space norm and

the Sobolev norm are equivalent.

10.3 Native spaces for conditionally positive de¬nite kernels

As in the case of positive de¬nite functions we generalize the notion of a conditionally

positive de¬nite function to a conditionally positive de¬nite kernel.

De¬nition 10.14 Suppose that P is a ¬nite-dimensional subspace of C( ), ⊆ Rd . A

continuous symmetric kernel : — ’ R is said to be conditionally positive de¬nite

with respect to P if, for any N pairwise distinct centers x1 , . . . , x N ∈ and all

on

± ∈ R N \ {0} with

N

± j p(x j ) = 0 (10.2)

j=1

for all p ∈ P, the quadratic form

N N

± j ±k (x j , xk ) (10.3)

j=1 k=1

is positive.

142 Native spaces

The domain ⊆ Rd can still be quite arbitrary, except that now it should contain at

least one P-unisolvent subset. In contrast with the de¬nition of a conditionally positive

de¬nite function we will allow a more general space P. A conditionally positive de¬nite

function of order m is therefore a conditionally positive de¬nite kernel with respect to

P = πm’1 (Rd ). Again, we restrict ourselves here to real-valued kernels. Everything said

about complex-valued kernels in the last section remains true.

We can proceed as in the case of a positive de¬nite kernel to construct the native space

of a conditionally positive kernel. Hence, we start with the linear space

N

± j (·, x j ) : N ∈ N, ± ∈ R N , x1 , . . . , x N ∈ ,

F ( ) :=

j=1

N

± j p(x j ) = 0 for all p ∈ P ,

with

j=1

which becomes a pre-Hilbert space by introduction of the inner product

N M N M

± j (·, x j ), βk (·, yk ) ± j βk (x j , yk ).

:=

j=1 k=1 j=1 k=1

Note that the additional conditions in the de¬nition of F ( ) ensure the de¬niteness of the

inner product.

Continuing as in the case of positive de¬nite kernels, we can form the Hilbert-space

completion F ( ) of F ( ) with respect to (·, ·) . But since (·, x) is in general not

included in F ( ), we cannot use f (x) = ( f, (·, x)) as in the case of positive de¬nite

kernels to derive that point evaluation functionals are continuous on F ( ) and thus on

F ( ). Hence we cannot interpret the abstract elements of F ( ) as functions in this way.

To arrive at an interpretation of the elements of F ( ) as functions we have to make a

detour.

To this end we choose a P-unisolvent subset = {ξ1 , . . . , ξ Q } ⊆ with | | = dim P =

Q elements and a Lagrange basis p j , 1 ¤ j ¤ Q, of P with respect to this set of centers.

This choice will be ¬xed for the rest of this section. Next, we de¬ne a functional

Q

δ(x) := δx ’ pk (x)δξk , x∈ ,

k=1

and a function

Q

y

δ(x) (·, y) = (·, x) ’ pk (x) (·, ξk ), x∈ .

G(·, x) := (10.4)

k=1

The functional obviously satis¬es δ(x) ( p) = 0 for all p ∈ P; therefore it lies in F ( )— and

so has a continuous extension to F ( ). The function is obviously an element of F ( ).

Furthermore, for x ∈ we have the representation

δ(x) ( f ) = ( f, G(·, x)) , f ∈ F ( ),

10.3 Native spaces for conditionally positive de¬nite kernels 143

which extends to F ( ) by continuity. Hence, if δ(x) is now to play the role of δx we have

to de¬ne the mapping

R : F ( ) ’ C( ), R( f )(x) := ( f, G(·, x)) .

The range of R is indeed a subset of C( ). By Cauchy“Schwarz we have again

|R f (x) ’ R f (y)| ¤ f G(·, x) ’ G(·, y) ,