(·, x) ’ = (x, x) + (y, y) ’ 2 (x, y).

2

(·, y)

Furthermore, we have R f (x) = f (x) for all x ∈ and all f ∈ F ( ).

Lemma 10.8 The linear mapping R : F ( ) ’ C( ) is injective.

Proof R f = 0 for an f ∈ F ( ) would mean that ( f, (·, x)) = 0 for all x ∈ or

f ⊥F ( ). But F ( ) is the completion of F ( ). Hence, the only element from F ( )

perpendicular to F ( ) is f = 0.

After this technical interlude we are able to de¬ne the native Hilbert space of a positive

de¬nite kernel .

De¬nition 10.9 The native Hilbert function space corresponding to the symmetric positive

de¬nite kernel : — ’ R is de¬ned by

N ( ) := R(F ( )).

It carries the inner product

:= (R ’1 f, R ’1 g) .

( f, g)N ()

Indeed, the space so de¬ned is a Hilbert space of continuous functions on with repro-

ducing kernel . Since (·, x) is an element of F ( ) for x ∈ it is unchanged under R

and hence

f (x) = (R ’1 f, (·, x)) = ( f, (·, x))N (10.1)

()

for all f ∈ N ( ) and x ∈ .

Theorem 10.10 Suppose that : — ’ R is a symmetric positive de¬nite kernel. Then

its associated native space N ( ) is a Hilbert function space with reproducing kernel .

Hence, positive (semi-)de¬nite kernels and reproducing kernels of Hilbert function spaces

are the same thing. In the rest of this section we will discuss the uniqueness of the native

space and will give a special characterization that is handier than the abstract de¬nition

given so far.

We already know by construction that F ( ) ⊆ N ( ) is dense in N ( ) and that

=f for all f ∈ F ( ).

f N()

These last two properties make the native space unique in the following sense.

Theorem 10.11 Suppose that is a symmetric positive de¬nite kernel. Suppose further

that G is a Hilbert space of functions f : ’ R with reproducing kernel . Then G is the

native space N ( ) and the inner products are the same.

Proof From the second remark after the proof of Theorem 10.4 we know that F ( ) ⊆

G and f G = f N ( ) for all f ∈ F ( ). Now take f ∈ N ( ). Then there exists a

10.2 Native spaces for positive de¬nite kernels 139

Cauchy sequence { f n } ⊆ F ( ) converging to f in N ( ) (remember that R ’1 f n = f n ).

By Theorem 10.3 we know that f is given pointwise by f (y) = limn’∞ f n (y). However,

f n is also a Cauchy sequence in G that converges to a g ∈ G. But now the reproducing-kernel

property of G gives g(x) = limn’∞ f n (x). Thus f = g ∈ G, which means that N ( ) ⊆ G

with f N ( ) = f G for all f ∈ N ( ). Next, suppose that N ( ) does not equal G.

Since N ( ) is closed we can then ¬nd an element g ∈ G \ {0} orthogonal to N ( ). But

this means that g(y) = (g, (·, y))G = 0 for all y ∈ , which contradicts g = 0. Finally,

because the norms are the same so are the inner products, by polarization.

This uniqueness result allows us to give another characterization of the native space in

the case where = Rd and is translation invariant. This result uses Fourier transforms

and shows that the native space actually consists of smooth functions.

∈ C(Rd ) © L 1 (Rd ) is a real-valued positive de¬nite func-

Theorem 10.12 Suppose that

tion. De¬ne

G := f ∈ L 2 (Rd ) © C(Rd ) : f ∈ L 2 (Rd )

and equip this space with the bilinear form

f (ω)g(ω)

( f, g)G := (2π)’d/2 f = (2π )’d/2

,g dω.

(ω)

L 2 (Rd ) Rd

Then G is a real Hilbert space with inner product (·, ·)G and reproducing kernel (· ’ ·).

Hence G is the native space of on Rd , i.e. G = N (Rd ), and both inner products coincide.

In particular, every f ∈ N (Rd ) can be recovered from its Fourier transform f ∈ L 1 (Rd ) ©

L 2 (Rd ).

Note that we use the sloppy notation N (Rd ) instead of the precise notation N d

(·’·) (R ),

for simplicity. This should cause no misunderstandings.

∈ L 1 (Rd ). For f ∈ G this means in particular

Proof From Corollary 6.12 we know that

that f ∈ L 1 (Rd ) because

1/2 1/2

| f (ω)|2

| f (ω)|dω ¤ .

dω (ω)dω

(ω)

Rd Rd Rd

Hence Plancherel™s theorem and the continuity of f and ( f )∨ allow us to recover f pointwise

from its Fourier transform via

f (x) = (2π)’d/2 f (ω)ei x ω dω.

T

Rd

Since (·, ·)G is obviously R-bilinear, it is an inner product if it is real-valued and posi-

tive de¬nite. For a real-valued f the L 2 -Fourier transform satis¬es f (x) = f (’x) almost

140 Native spaces

everywhere, so that

f (ω)g(ω) f (ω)g(ω) f (’ω)g(’ω)

dω = + dω

(ω) (ω) (’ω)

Rd ω1 >0

f (ω)g(ω) + f (ω)g(ω)

= dω

(ω)

ω1 >0

[ f (ω)g(ω)]

=2 dω ∈ R,

(ω)

ω1 >0

proving that the bilinear form is indeed real-valued. As a weighted L 2 -inner product, (·, ·)G

is also positive de¬nite.

Thus far we know that G is a pre-Hilbert space, and our next step is to prove that it

is complete. For this reason suppose that { f n } is a Cauchy sequence in G. This means

√

is a Cauchy sequence in L 2 (Rd ). Thus there exists a function g ∈ L 2 (Rd )

that f n

√ √

’ g in L 2 (Rd ). From our initial assumptions we can conclude that g ∈

with f n

L 1 (R ) © L 2 (R ). Namely,

d d

1/2 1/2

(ω) dω ¤ |g(ω)| dω 2

g(ω) (ω)dω

Rd Rd Rd

and

2

(ω) dω ¤ L 2 (Rd ) .

2

g(ω) g

L ∞ (Rd )

Rd

Thus

f (x) := (2π )’d/2 (ω)ei x ω dω,

T