Q Q

g(ω) := γ (ω) e’i y ω

pk (y)e’iξk ω ±k e’iξk ω ∈ S2m

T T

T

’ = γ (ω)

k=1 k=0

for all γ ∈ Sm . Moreover,

Q

g(ω) = γ (ω + y) ’ pk (y)γ (ω + ξk ).

k=1

Since κ(·, y) = G(·, y) ’ P G(·, y) this gives

Q

κ(ω, y)γ (ω)dω = (ω ’ y) ’ pk (y) (ω ’ ξk ) γ (ω)dω

Rd Rd k=1

= (ω)g(ω)dω = (ω)g(ω)dω

Rd Rd

Q

(ω) e’i y ω

pk (y)e’iξk ω γ (ω)dω.

T

T

= ’

Rd k=1

√

Finally, κ(·, y)§ / ∈ L 2 (Rd ) because

2

|κ(·, y)§ (ω)|2 Q

’i y T ω ’iξkT ω

dω = ’

(ω) e pk (y)e dω

(ω)

Rd Rd k=1

2

Q

’iξkT ω

= ±k e

(ω) dω

Rd k=0

Q

= (2π) ±k ± (ξk ’ ξ ) < ∞

d/2

,k=0

by Corollary 8.13.

Step 2 κ is the reproducing kernel of G0 .

10.3 Native spaces for conditionally positive de¬nite kernels 149

To show this we use the test functions g (x) = ( /π)d/2 e’

2

x

from Theorem 5.20 again.

2

From our previous observations we know that the function

Q

ω

pk (ω)eiξk ω g (ω)

T

T

γ (ω) := ei y ’

k=1

belongs to Sm . Since |g (ω)| ¤ (2π)’d/2 for all ω ∈ Rd and g (ω) ’ (2π)’d/2 for ’ ∞

we can conclude that

f (ω)κ(·, y)§ (ω)

( f, κ(·, y))G = lim g (ω)dω

’∞ Rd (ω)

Q

i yT ω

pk (y)eiξk ω g (ω)dω

T

= lim ’

f (ω) e

’∞ Rd

k=1

= lim f (ω)γ (ω)dω

’∞ Rd

= lim f (ω)γ (ω)dω

’∞ Rd

Q

= lim f (ω) g (ω ’ y) ’ pk (y)g (ω ’ ξk ) dω

’∞ Rd

k=1

Q

= f (y) ’ pk (y) f (ξk )

k=1

= f (y)

for all f ∈ G0 .

Step 3 G0 is complete.

Suppose that { f n } ⊆ G0 is a Cauchy sequence. Because

| f n (x) ’ f m (x)| ¤ f n ’ f m G κ(x, x) ,

1/2

(10.6)

the sequence { f n (x)} is a Cauchy sequence in R and we can de¬ne a function f by f (x) :=

limn’∞ f n (x). On account of (10.6) the sequence f n converges uniformly on compact

subsets of Rd and hence the function f is continuous. Next, note that f n G is bounded

since f n is a Cauchy sequence in G0 . From the fact that is slowly increasing and from

| f n (x)|2 ¤ G κ(x, x)

2

fn

Q Q

¤C (0) ’ 2 pk (x) (x ’ ξk ) + pk (x) p j (x) (ξ j ’ ξk ) ,

k=1 j,k=1

we can deduce that f is also slowly increasing. Obviously f satis¬es f (ξk ) = 0 for 1 ¤ k ¤

√

Q. Finally, since { f n } is a Cauchy sequence in G0 , { f n / } must be a Cauchy sequence in

√

L 2 (R ) and must converge to a function g ∈ L 2 (R ). It remains to show that g

d d

is the

150 Native spaces

generalized Fourier transform of f . To this end, choose γ ∈ Sm . From

f n (ω) ’ (ω)g(ω) γ (ω) dω

Rd

f n (ω)

= ’ g(ω) (ω)γ (ω) dω

Rd (ω)

1/2

fn

¤ √ ’g (ω)|γ (ω)|2 dω

L 2 (Rd ) Rd

’0 for n ’ ∞,

we see that

f (ω)γ (ω)dω = lim f n (ω)γ (ω)dω

n’∞ Rd

Rd

= lim f n (ω)γ (ω)dω

n’∞ Rd

= (ω)g(ω)γ (ω)dω.