but this time with

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

2

Q

’2 [ p j (x) ’ p j (y)] (x, ξ j ) ’ (y, ξ j )

j=1

Q

+ [ pk (x) ’ pk (y)]{ p j (x) ’ p j (y)} (ξ j , ξk ).

j,k=1

Following the lines of the previous section, the next step is to show that R is injective.

Lemma 10.15 The linear mapping R : F ( ) ’ C( ) is injective.

Proof R f = 0 for f ∈ F ( ) means that ( f, G(·, x)) = 0 for all x ∈ . Now, we choose

an arbitrary h = N ± j (·, x j ) from F ( ). For the coef¬cients of this element we

j=1

have

Q

N N N

± j G(·, x j ) = ± j (·, x j ) ’ ±j (·, ξk ) pk (x j )

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

Q N

=h’ (·, ξk ) ± j pk (x j )

k=1 j=1

= h,

showing that ( f, h) = 0 for all h ∈ F ( ). But because F ( ) is the completion of F ( )

this means that f = 0.

This allows us to interpret F ( ) as a space of functions. But since G(·, x) and δ(x)

have the dispensable property G(·, ξk ) = δ(ξk ) ≡ 0, 1 ¤ k ¤ Q, we ¬nd that R f (ξk ) = 0,

1 ¤ k ¤ Q, for all f ∈ F ( ). However, f (ξk ) does not vanish for all f ∈ F ( ), so that

it is not reasonable to take R(F ( )) as the de¬nition of the native space. The right choice

comes from a closer look at how R acts on F ( ). Here, we ¬nd for f = β j (·, x j )

that

Q

R f (x) = ( f, G(·, x)) = f (x) ’ pk (x) f (ξk ) = f (x) ’ f (x),

P

k=1

144 Native spaces

introducing the projection operator

Q

: C( ) ’ P, P( f ) = f (ξk ) pk . (10.5)

P

k=1

Note that, with this operator, G(·, x) can also be written as G(·, x) = (·, x) ’ x

(·, x).

P

De¬nition 10.16 The native space corresponding to a symmetric kernel that is condi-

tionally positive de¬nite on with respect to P is de¬ned by

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

The space is equipped with a semi-inner product via

= (R ’1 ( f ’ f ), R ’1 (g ’ .

( f, g)N P g))

P

()

Let us investigate this de¬nition in more detail. First of all, the sum R(F ( )) + P is direct

because any element of R(F ( )) vanishes on and this set is P-unisolvent, which means

that zero is the only element from P that vanishes there. From this fact we can also deduce

that a general element f = R(g) + p satis¬es f (ξk ) = p(ξk ), 1 ¤ k ¤ Q, and hence can

be written as f = R(g) + P ( f ). This, again, means that f ’ P ( f ) is an element of

R(F ( )) and that the semi-inner product is well de¬ned. Obviously its null space is given

by P. Moreover, if f ∈ F ( ) then the de¬nition ensures that | f |N ( ) = f .

Finally, we have the following Taylor expansion of f ∈ N ( ), which can also be seen

as the generalization of the reproducing-kernel property of positive de¬nite kernels.

Theorem 10.17 Suppose that : — ’ R is a symmetric kernel that is conditionally

positive de¬nite on with respect to P ⊆ C( ). Every f ∈ N ( ) can be written as

f (x) = f (x) + ( f, G(·, x))N

P ()

with the function G from (10.4) and the projector from (10.5).

P

Proof Since G(·, x) ∈ F ( ) for every x ∈ we have, as remarked earlier, G(·, x) =

R ’1 (G(·, x) ’ P G(·, x)). From the de¬nitions we can derive that

f (x) + (R ’1 ( f ’

f (x) = f ), G(·, x))

P P

f (x) + (R ’1 ( f ’ f ), R ’1 (G(·, x) ’

= P G(·, x)))

P P

= f (x) + ( f, G(·, x))N ( ).

P

Note that in the case of a positive de¬nite kernel both the de¬nition of the native space in

De¬nition 10.16 and the representation formula from Theorem 10.17 reduce to the de¬nition

and the reproducing-kernel formula given in De¬nition 10.9 and formula (10.1) since in

this case G(·, x) reduces to and N ( ) to R(F ( )).

10.3 Native spaces for conditionally positive de¬nite kernels 145

The next theorem gives some more information on the connection with reproducing

kernels. Even if N ( ) does not have a reproducing kernel, the main part R(F ( )) will

do so.

Theorem 10.18 The bilinear form (·, ·)N is, on the space

()

X ( ) = { f ∈ N ( ) : f (ξk ) = 0, 1 ¤ k ¤ Q} = R(F ( )),

an inner product, which makes this space a Hilbert space. Moreover, this space has the

reproducing kernel

Q Q

κ(x, y) := (x, y) ’ pk (x) (ξk , y) ’ p (y) (x, ξ )

=1

k=1

Q Q

+ pk (x) p (y) (ξk , ξ ),

k=1 =1

i.e. every f ∈ X ( ) has the representation

f (x) = ( f, κ(·, x))N ( ).

Proof We know that the linear mapping R : F ( ) ’ X ( ) is isometric and bijective.

Since F ( ) is a Hilbert space, so is X ( ). In particular the bilinear form (·, ·)N ( )

becomes an inner product on X ( ).

The kernel κ is a symmetric function and satis¬es κ(·, y) = RG(·, y) for all y ∈ . This

shows that κ(·, y) is indeed an element of X ( ) for every y ∈ . Finally, for f ∈ X ( )

and x ∈ we obtain

= (R ’1 f, G(·, x))

( f, κ(·, x))N ()

f (x) + (R ’1 ( f ’ f ), R ’1 (G(·, x) ’

= P G(·, x)))

P P

= f (x)