The previous theorem allows us to characterize native spaces for conditionally positive

de¬nite kernels in a way similar to the positive de¬nite result in Theorem 10.11. As a

consequence, we can derive a characterization based on (generalized) Fourier transforms

comparable to the one in Theorem 10.12.

Proposition 10.19 Suppose that G ⊆ C( ) carries a semi-inner product (·, ·)G with null

space P ⊆ G such that G0 := {g ∈ G : g(ξk ) = 0, 1 ¤ k ¤ Q} is a Hilbert space with re-

producing kernel κ. Then G is the native space corresponding to on .

Proof Since obviously G = G0 • P and N ( ) = R(F ( )) • P it suf¬ces to show that

R(F ( )) = G0 . This can be done in a way similar to that in the proof of Theorem 10.11.

146 Native spaces

N

First of all, for f = ± j (·, x j ) ∈ F ( ) we have

j=1

N N

R f (x) = f (x) ’ f (x) = ± j G(x, x j ) = ± j κ(x, x j ),

P

j=1 j=1

showing that R(F ( )) ⊆ G0 . Moreover, for f 1 , f 2 ∈ R(F ( )) it is true that

( f 1 , f 2 )N ( ) = ( f 1 , f 2 )G .

For every f ∈ F ( ) there exists a Cauchy sequence ( f n ) ⊆ R(F ( )) satisfying in

particular f (x) = limn’∞ f n (x) for all x ∈ . However, since f n is also a Cauchy sequence

in G0 there exists a limit g ∈ G0 . From the continuity of the inner product we can derive again

that g(x) = limn’∞ f n (x). Hence, f ≡ g. Finally, if g ∈ G0 is perpendicular to R(F ( ))

then all inner products (g, κ(·, x))G , x ∈ , have to vanish. But this means that g ≡ 0.

The native space carries only a semi-inner product. This semi-inner product has the null

space P. The following simple trick de¬nes an inner product on N ( ) and makes the

native space a reproducing-kernel Hilbert space.

Theorem 10.20 The native space N ( ) corresponding to a conditionally positive de¬nite

kernel carries the inner product

Q

+

( f, g) := ( f, g)N f (ξk )g(ξk ).

()

k=1

With this inner product N ( ) becomes a reproducing-kernel Hilbert space with reproduc-

ing kernel

Q

K (x, y) = κ(x, y) + pk (x) pk (y),

k=1

where κ is the kernel from Theorem 10.18.

Proof Obviously, the new inner product is symmetric, bilinear, and nonnegative. If

Q

0 = ( f, f ) = ( f, f )N )+ | f (ξk )|2

(

k=1

for an f ∈ N ( ) then each summand has to be zero. But ( f, f )N ( ) = 0 means that

f ∈ P, and the additional information f (ξk ) = 0 for 1 ¤ k ¤ Q coupled with the choice of

the set leads to f ≡ 0. Hence, (·, ·) is positive de¬nite. Let us come to the reproducing-

kernel property. Since p is a Lagrangian basis for and since κ(ξk , ·) = 0, we have

Q Q Q Q

f (ξk )K (ξk , x) = f (ξk )κ(ξk , x) + f (ξk ) p (x) p (ξk )

=1

k=1 k=1 k=1

Q

= f (ξk ) pk (x).

k=1

10.3 Native spaces for conditionally positive de¬nite kernels 147

Moreover, since G(·, x), κ(·, x), and K (·, x) differ only by a polynomial, Theorem 10.17

gives the representation

f (x) = f (x) + ( f, κ(·, x))N

P ()

Q

= f (ξk )K (ξk , x) + ( f, K (·, x))N ()

k=1

= ( f, K (·, x)).

In the case of positive de¬nite functions, we have seen that the native space on Rd can

be characterized by using Fourier transforms. Something similar is possible in the case of

conditionally positive de¬nite functions if we use generalized Fourier transforms instead.

Theorem 10.21 Suppose that ∈ C(Rd ) is an even conditionally positive de¬nite function

of order m ∈ N0 . Suppose further that has a generalized Fourier transform of order

m that is continuous on Rd \ {0}. Let G be the real vector space consisting of all functions

f ∈ C(Rd ) that are slowly increasing and have a generalized Fourier transform f of order

√

∈ L 2 (Rd ). Equip G with the symmetric bilinear form

m/2 that satis¬es f

f (ω)g(ω)

( f, g)G = (2π)’d/2 dω.

(ω)

Rd

Then G is the native space corresponding to , i.e. G = N (Rd ), and the semi-inner product

(·, ·)N (Rd ) coincides with the semi-inner product (·, ·)G . Furthermore, every f ∈ N ( ) has

the representation

Q

f (x) + (2π)’d/2 ω

pk (x)eiξk ω dω.

T

T

f (x) = ’

f (ω) ei x

P

Rd k=1

Proof Obviously (·, ·)G is R-bilinear. It is also symmetric if it is real-valued. This can be

shown in the same way as in the proof of Theorem 10.12 as long as we know that the

generalized Fourier transform f satis¬es f (ω) = f (’ω) almost everywhere. Since this

relation is satis¬ed for all real-valued test functions γ ∈ Sm , the de¬nition of a generalized

Fourier transform and γ (’·)(ω) = γ (’ω) for such test functions ensures that

f (ω)γ (ω)dω = =

f (ω)γ (ω)dω f (ω)γ (’ω)dω

Rd Rd Rd

= f (ω)γ (’ω)dω = f (’ω)γ (ω)dω.

Rd Rd

The uniqueness of the generalized Fourier transform gives the stated relation.

Note that πm’1 (Rd ) ⊆ G is the null space of (·, ·)G by Proposition 8.10. With regard to

Proposition 10.19 it remains to show that the space G0 = {g ∈ G : g(ξk ) = 0, 1 ¤ k ¤ Q}

is a Hilbert space of functions with reproducing kernel κ. This is done in three steps.

148 Native spaces

Step 1 κ(·, y) is an element of G0 for all y ∈ Rd .

Obviously κ(ξk , y) = 0 for all 1 ¤ k ¤ Q and y ∈ Rd . Next, κ(·, y) has the generalized

Fourier transform

Q

§ ’i y T ω

pk (y)e’iξk ω

T

κ(·, y) (ω) = ’

(ω) e

k=1

of order m/2. To see this we have to show that

κ(·, y)§ (ω)γ (ω)dω

κ(ω, y)γ (ω)dω =

Rd Rd

for every γ ∈ Sm . By setting ξ0 = y, ±0 = 1, and ±k = ’ pk (y), 1 ¤ k ¤ Q, Lemma 8.11