( f ’ s f,X )(x) = f ’ s f,X )(x) + ( f ’ s f,X , G(·, x))N ( ), x∈ .

P(

If we use »x G(·, x) = »x (·, x), the fact that » vanishes on elements of P, and the fact that

s f,X interpolates f in x j , yielding »( f ’ s f,X ) = 0, we can conclude that

0 = »( f ’ s f,X )

= »( f ’ s f,X )) + ( f ’ s f,X , »x (·, x))N

P( ()

= ( f ’ s f,X , s)N ( ).

Finally, s f,X satis¬es the conditions imposed on s.

For later reasons we state an immediate consequence.

Corollary 10.25 In the situation of Lemma 10.24 we have the estimates |s f,X |N ¤

()

| f |N ( ) and | f ’ s f,X |N ( ) ¤ | f |N ( ) .

Proof According to Lemma 10.24 the functions f ’ s f,X and s f,X are mutually orthogonal.

As a consequence we can deduce the Pythagorean law

| f ’ s f,X |2 + |s f,X |2 = | f |2 ( ),

N N N

() ()

which gives immediately both bounds.

Theorem 10.26 Let be a conditionally positive de¬nite kernel on with respect to P.

Denote by s f,X the interpolant to a function f ∈ C( ) based on a P-unisolvent X using

. Then f belongs to the native space N ( ) if and only if there exists a constant c f such

that |s f,X |N ( ) ¤ c f for all P-unisolvent X ⊆ . Moreover, in the case f ∈ N ( ) the

smallest possible constant c f is given by | f |N ( ) .

Proof If f ∈ N ( ), Corollary 10.25 shows that |s f,X |N ( ) ¤ | f |N ( ) , which gives for

such an f the upper bound c f ¤ | f |N ( ) if c f is the minimal choice. Next, let us assume

|s f,X |N ( ) ¤ c f for all P-unisolvent X ⊆ . For an arbitrary

N

» N ,±,X = ± j δx j ∈ L P ( ),

j=1

we choose a P-unisolvent set Y ⊇ X and let s f,Y be the interpolant on this set Y to f . Then

s f,Y belongs to N ( ) and we have » N ,±,X ( f ’ s f,Y ) = 0. Thus we can make the estimate

|» N ,±,X ( f )| ¤ |» N ,±,X ( f ’ s f,Y )| + |» N ,±,X (s f,Y )|

¤ » N ,±,X |s f,Y |N ()

¤ c f » N ,±,X .

As this holds for all » N ,±,X we have f ∈ N ( ) and | f |N ¤ c f by Theorem10.22.

()

154 Native spaces

Our next characterization is only for positive de¬nite and not for conditionally positive

de¬nite kernels. Moreover, we assume the set ⊆ Rd to be compact. We need some

preparatory results on embeddings.

Lemma 10.27 Suppose that ⊆ Rd is compact and is a symmetric positive de¬nite

kernel on . Then the native space N ( ) has a continuous linear embedding into L 2 ( ).

Proof Since is the reproducing kernel of its native space we have

| f (x)|2 = |( f, (·, x))|2 ¤f N ( )=

2 2 2

(·, x) f (x, x)

N N() N()

()

for every f ∈ N ( ) and x ∈ . This implies that f L 2 ( ) ¤ C f N ( ) with C 2 =

(x, x)d x. The latter integral is ¬nite because is continuous and is compact.

Now we introduce the integral operator T : L 2 ( ) ’ L 2 ( ), de¬ned by

T v(x) := v ∈ L 2 ( ), x∈ .

(x, y)v(y)dy, (10.7)

Obviously T v is continuous. But it is also an element of the native space.

Proposition 10.28 Suppose that is a symmetric positive de¬nite kernel of the compact

set ⊆ Rd . Then the integral operator T maps L 2 ( ) continuously into the native space

N ( ). It is the adjoint of the embedding operator of the native space N ( ) into L 2 ( ),

i.e. it satis¬es

( f, v) L 2 ( = ( f, T v)N ( ), f ∈ N ( ), v ∈ L 2 ( ). (10.8)

)

The range of T is dense in N ( ).

Proof We use the characterization of Theorem 10.22 to show that T v ∈ N ( ). Hence

we will pick an arbitrary » ∈ L( ) (recall that we do not have any side conditions), and we

see that

|»(T v)| ¤ v »x (·, x) L2( ), ¤ Cv » ,

L2( ) L2( )

by Lemma 10.27. This gives T v N ( ) ¤ C v L 2 ( ) . To prove (10.8) we start with an

f ∈ F ( ) and ¬nd, by the reproducing-kernel property, that

N N

( f, v) L 2 ( = ±j (x, x j )v(x)d x = ± j T v(x j )

)

j=1 j=1

N

= ± j (T v, (·, x j ))N = ( f, T v)N

() ()

j=1

for all v ∈ L 2 ( ). Since F ( ) is dense in N ( ) and N ( ) is continuously embedded

into L 2 ( ) the general case follows by continuous extension. The ¬nal statement is a

consequence of the general properties of adjoint mappings. The closure of the range of the

operator T is the orthogonal complement of the kernel of its adjoint operator. But this is the

whole space.

10.4 Further characterizations of native spaces 155

It is well known that in our situation the operator T : L 2 ( ) ’ L 2 ( ) is a compact

operator. Moreover, it satis¬es

(T v, v) L 2 ( = (T v, T v)N ≥0

) ()

for all v ∈ L 2 ( ). For such an operator, Mercer™s theorem (see Pogorzelski [154] for exam-

ple) guarantees the existence of a countable set of positive eigenvalues ρ1 ≥ ρ2 ≥ · · · > 0

and continuous eigenfunctions {•n }n∈N ⊆ L 2 ( ) such that T •n = ρn •n . Furthermore, {φn }

is an orthonormal basis for L 2 ( ) and the kernel possesses the absolutely and uniformly

convergent representation

∞

(x, y) = ρn •n (x)•n (y).

n=1

This allows us to derive our ¬nal characterization for native spaces.

Theorem 10.29 Suppose is a symmetric positive de¬nite kernel on a compact set

⊆ R . Then its native space is given by

d

∞

1

N ( )= f ∈ L 2( ) : |( f, •n ) L 2 ( ) |2 < ∞ (10.9)

ρn

n=1

and the inner product has the representation

∞

1

( f, g)N = ( f, •n ) L 2 ( ) (g, •n ) L 2 ( ) , f, g ∈ N ( ). (10.10)

ρn

n=1

Proof Denote the set on the right-hand side of (10.9) by G and the inner product on the right-

hand side of (10.10) by (·, ·)G . We start by showing that G ⊆ N ( ) and f N ( ) ¤ f G

for all f ∈ G. If f ∈ G is given, f is continuous because

1/2 1/2

∞ ∞ ∞

|( f, •n ) L 2 ( ) |2

|( f, •n ) L 2 ( ) •n (x)| ¤ ρn |•n (x)| 2

ρn

n=1 n=1 n=1

= f (x, x).

G