Moreover, for » = ± j δx j ∈ L( ) we ¬nd that

j=1

1/2 1/2

∞ ∞

1

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

ρn

n=1 n=1

1/2

∞

N

= ± j ±k ρn •n (x j )•n (xk )

f G

j,k=1 n=1

1/2

N

= ± j ±k (x j , xk )

f G

j,k=1

= » ,

f G

which leads to the desired result by Theorem 10.22.

156 Native spaces

It remains to show that N ( ) ⊆ G and that f G ¤ f N ( ) for all f ∈ N ( ).

To achieve this, we start by looking at the dense subset T (L 2 ( )) ⊆ N ( ). For an

element f = T v, v ∈ L 2 ( ), we can conclude from the L 2 expansion of v that f =

∞

n=1 ρn (v, •n ) L 2 ( ) •n , so that ( f, •n ) L 2 ( ) = ρn (v, •n ) L 2 ( ) . This allows us to calculate

its native space norm:

∞

= (v, T v) L 2 ( = (v, •n ) L 2 ( ) ( f, •n ) L 2 (

2

f N() ) )

n=1

∞

1

= |( f, •n ) L 2 ( ) |2 ,

ρn

n=1

so that · N ( ) and · G are the same on T (L 2 ( )). For an arbitrary f ∈ N ( ) we

choose a sequence { f j } ⊆ T (L 2 ( )) with f ’ f j N ( ) ’ 0 for j ’ ∞. For N , j ∈ N

we have the bound

N

1

|( f j , •n ) L 2 ( ) |2 ¤ f j N ( ).

ρn

n=1

Since f j converges to f the sum is uniformly bounded in j and N . Using the fact that the

native space is continuously embedded in L 2 ( ) and letting j tend to in¬nity gives therefore

N

1

|( f, •n ) L 2 ( ) |2 ¤ f N ( ).

ρn

n=1

Hence, we can let N tend to in¬nity also, which shows that f ∈ G and f ¤f N ( );

G

this actually establishes norm equality.

Picard™s theorem on the range of a compact integral operator gives also

Corollary 10.30 Suppose that is a symmetric positive de¬nite kernel on a compact set

⊆ Rd . Then the range of the integral operator (10.7) is given by

∞ |( f, •n )|2 2 (

L )

T (L 2 ( )) = f ∈ L 2( ) : <∞ .

ρn

2

n=1

This space will play a particular role in the context of improved error estimates for radial

basis function interpolants in Section 11.5.

10.5 Special cases of native spaces

In this section we want to take a closer look at the native spaces of two instances of basis

functions. In the ¬rst instance we investigate the compactly supported functions of Section

9.4. The other class of functions is provided by certain thin-plate splines and powers. It will

turn out that the native spaces are Sobolev and Beppo Levi spaces, respectively.

Let us start with the compactly supported functions d,k = φd,k ( · 2 ) from Chapter 9.

We know that such a function has a classical radial Fourier transform d,k = Fd φd,k ( · 2 ).

10.5 Special cases of native spaces 157

It is our goal to show that this Fourier transform decays as (1 + · 2 )’d’2k’1 , which

means by Corollary 10.13 that the associated native space is the Sobolev space H s (Rd )

with s = k + d/2 + 1/2.

We start our investigation by restricting ourselves to the odd-dimensional case. To this

end let us set d = 2n + 1 and m = n + k. With this we derive

Fd φd,k (r ) = Fd+2k φ d/2 +k+1 (r ) = F2m+1 φm+1 (r )

r

r ’3m’2

= (r ’ t)m+1 t m+1/2 Jm’1/2 (t)dt.

0

Thus we can use the representation (6.9) for bounding the Fourier transform.

Lemma 10.31 For every m ∈ N0 there exists a constant Cm such that

F2m+1 φm+1 (r ) ¤ Cm r ’2m’2

for all r > 0.

Proof By Lemma 6.19 we know that F2m+1 φm+1 (r ) = Bm f m (r )r ’3m’2 with a certain

constant Bm and a nonnegative function f m de¬ned in that lemma. Thus, it suf¬ces to show

that f m (r ) ¤ Cr m . We now show by induction that, more precisely, f m (r ) ¤ 2m+1r m /m!.

In the case m = 0 we have f 0 (r ) = 1 ’ cos r , which obviously satis¬es f 0 (r ) ¤ 2. Now

suppose that everything is settled for m ≥ 0. Then

r r

2m+1 m 2m+2

f m+1 (r ) = f m (t) f 0 (r ’ t)dt ¤ t 2dt = r m+1 ,

(m + 1)!

m!

0 0

which completes our proof.

We want to point out that the constant Cm of the last lemma is given explicitly. Moreover,

even if this bound is valid for r > 0 it is only of interest for large r . For r close to zero we

know that F2m+1 φm+1 is bounded by a constant.

Next we turn to the lower bounds on F2m+1 φm+1 . Since F1 φ1 coincides, up to a constant,

with (1 ’ cos r )/r 2 there is no chance of getting a lower bound for this function. But in all

other cases, i.e. m ≥ 1, it is possible.

Lemma 10.32 For every m ∈ N there exist constants rm , cm > 0 such that

F2m+1 φm+1 (r ) ≥ cm r ’2m’2

for all r ≥ rm . Moreover, F2m+1 φm+1 is strictly positive on [0, ∞).

Proof The proof is by induction on m. Again we use the representation F2m+1 φm+1 (r ) =

Bm f m (r )r ’3m’2 of Lemma 6.19 and concentrate on showing that f m (r ) ≥ cm r m for r ≥ rm .

If m = 1 it is easy to compute that

r

f 1 (r ) = f 0 (t) f 0 (r ’ t)dt = r + 1 r cos r ’ 3 sin r ≥ 1 r ’ ≥ 1r

3

2 2 2 2 4

0

158 Native spaces

if r ≥ 6. Hence we have found r1 and c1 . Now suppose that our statement is true for m ≥ 1.

Then for m + 1 we have

r

f m+1 (r ) = f m (t)[1 ’ cos(r ’ t)] dt

0

r

≥ cm t m [1 ’ cos(r ’ t)] dt