√

g ∈ L loc (Rd √{0}) shows that the generalized

\

This, together with the obvious fact

√ 2

= g ∈ L 2 (Rd ). Thus we have

Fourier transform of f is indeed g. Obviously f

shown that f ∈ G0 and hence that G0 is complete. The given representation formula follows

from the reproducing-kernel property and the Fourier transform of κ(·, y).

10.4 Further characterizations of native spaces

With Theorems 10.12 and 10.21 we have already had two examples of equivalent repre-

sentations of the native Hilbert space. In this section we want to give several other charac-

terizations. We start with two characterizations for conditionally positive de¬nite kernels,

continuing to use the notation of the previous section. After that we will give a characteri-

zation valid only for positive de¬nite kernels.

The ¬rst equivalent formulation we want to give is based on ¬nitely supported linear

functionals on C( ) that vanish on P. To be more precise we form the set

N

L P ( ) := » N ,±,X = ± j δx j : N ∈ N, ± ∈ R N , x1 , . . . , x N ∈ ,

j=1

with » N ,±,X ( p) = 0 for all p ∈ P

and equip it with the inner product

N M

(» N ,±,X , » M,β,Y ) := ± j βk (x j , yk ).

j=1 k=1

10.4 Further characterizations of native spaces 151

Obviously there is a one-to-one relation between L P ( ) and F ( ), given simply by

L P ( ) ’ F ( ), » ’ »x (·, x),

where »x means action with respect to the variable x. This shows in particular that »( f ) =

(»x (·, x), f ) for all » ∈ L P ( ) and all f ∈ F ( ), so that the norm de¬ned on L P gives

the operator norm if L P ( ) is interpreted as a subspace of the dual space of F ( ).

However, we do not want to exploit this relation explicitly any further; we shall use it

whenever it is appropriate. Instead, we will look at the space of all functions on which these

functionals are continuous.

with respect to P.

Theorem 10.22 Suppose that is conditionally positive de¬nite on

De¬ne

G = { f ∈ C( ) : |»( f )| ¤ C f » for all » ∈ L P ( )}.

This space carries the semi-norm

|»( f )|

| f |G = sup .

»

»∈L P ( )

»=0

Then we have N ( ) = G and both semi-norms are equal.

Proof Suppose that f ∈ N ( ). To show that f ∈ G we ¬rst remark that for

Q

G(·, x) = (·, x) ’ (·, ξk ) pk (x)

k=1

and a general » ∈ L P ( ) we have

Q

» (G(·, x)) = » (·, x) ’ (·, ξk )»( pk ) = »x (·, x).

x x

k=1

Using the reproduction formula f (x) = f (x) + ( f, G(·, x))N given in Theorem

P ()

10.17 we ¬nd that

»( f ) = »( f ) + »x ( f, G(·, x))N

P ()

= ( f, » x

(·, x))N ()

¤ | f |N » ,

()

and thus f is an element of G. Furthermore we have established that | f |G ¤ | f |N ( ) .

Now let us assume that f is an element of G and that we want to prove that f belongs to

the native space. This f allows us to de¬ne a linear functional

F f : F ( ) ’ R, »x (·, x) ’ »( f ),

which is continuous because of the de¬nition of G. Hence F f has a continuous extension

152 Native spaces

to F ( ) and we can use Riesz™ representation theorem to represent this extension by

F f (g) = (g, S f ) for all g ∈ F ( ).

Here S f is the Riesz representer for this functional F f . For proving f ∈ N ( ) it suf¬ces

to show that f and R(S f ) differ only by an element of P. To see this, we use the de¬nition

R(S f )(x) = (S f, G(·, x)) and that for every μ ∈ L P ( ) we have μx G(·, x) = μx (·, x).

This gives

μ( f ’ R S f ) = μ( f ) ’ (S f, μx G(·, x))

= μ( f ) ’ F f (μx G(·, x))

= μ( f ) ’ F f (μx (·, x))

= μ( f ) ’ μ( f )

= 0.

If we specify that μ = δ(x) then we end up with

Q

f (x) = R(S f )(x) + f (ξk ) pk (x).

k=1

Finally, since S f ∈ F ( ) we can choose a sequence » j ∈ L P ( ) such that »x (·, x) ’

j

S f in F ( ). Hence » j ( f ) = (S f, »x (·, x)) ’ S f 2 and » j ’ Sf for j ’

j

∞. This allows us to make the bound

|» j ( f )| 2

Sf

| f |G ≥ lim = = | f |N ( ).

j’∞ » j Sf

The previous result is not only interesting in its own right; it has also the following

consequence.

Corollary 10.23 The space N ( ) is independent of the particular choice of =

{ξ1 , . . . , ξ Q }. The semi-norms for any two different choices are equal.

Our next characterization allows us to determine whether a function belongs to the native

space, simply by looking at sequences of interpolants. Thus it is numerically applicable

because it is based only on function values. For its proof we need a result that will also play

an important role in a later chapter.

Lemma 10.24 Suppose that X = {x1 , . . . , x N } ⊆ is P-unisolvent. Denote the unique

interpolant, based on a conditionally positive de¬nite kernel and the set X , of a function

f ∈ N ( ) by s f,X . Then we have

( f ’ s f,X , s)N =0

()

for every s ∈ span{ (·, x j ) : x j ∈ X } © F ( ) + P. In particular, this gives

( f ’ s f,X , s f,X )N = 0.

()

10.4 Further characterizations of native spaces 153

Proof Any such function s can be written in the form s = »x (·, x) + P (s) with a certain

linear functional » = N ± j δx j ∈ L P ( ). The reproduction formula of Theorem 10.17

j=1

gives