r

cm m+1 cm m+1

= ’ ’ cm t m cos(r ’ t) dt

r r

m+1 m+1 m rm

for r ≥ rm . We use integration by parts to bound the last integral via

r r

t cos(r ’ t) dt ¤ |rm sin(r ’ rm )| + m t m’1 sin(r ’ t) dt

m m

rm rm

r

¤ rm + m

m

t m’1 dt

rm

= + r ’ rm = r m .

m m m

rm

Hence

cm m+1 cm m+1 cm

f m+1 (r ) ≥ ’ cm r m ’ rm ≥ r m+1

r

m+1 m+1 2(m + 1)

for suf¬ciently large r ≥ rm+1 .

Since φm+1 is nonnegative and nonvanishing, F2m+1 φm+1 is positive de¬nite on R2m+1

according to Corollary 6.9. This means in particular that F2m+1 φm+1 (0) > 0 by Theorem

6.2. Furthermore, the function f 1 is the (Laplace-)convolution of two nonnegative func-

tions with isolated zeros. Thus it has to be positive on (0, ∞). Finally f m , m ≥ 2, is the

(Laplace-)convolution of a positive function with 1 ’ cos r and, therefore, also positive on

(0, ∞).

This ¬nishes our investigations in the odd-dimensional case. Next we turn to even space

dimensions. To this end we set d = 2n, m = n + k, to get

r

Fd φd,k (r ) = F2m φm+1 (r ) = r ’3m’1 (r ’ t)m+1 t m Jm’1 (t) dt.

0

In the odd-dimensional case the functions f n de¬ned by Laplace-transform conform-

ing convolution were an appropriate tool for determining the Fourier transform of φm+1 .

Something similar is true in the even-dimensional case. Let us de¬ne

r

g0 (r ) = J0 (t)dt,

0

r

gm (r ) = f m’1 (r ’ t)g0 (t)dt, m ≥ 1,

0

with f m from Lemma 6.19. Then the result analogous to Lemma 6.19 becomes

10.5 Special cases of native spaces 159

√

Lemma 10.33 Let Am = 2m (m + 1)! (m + 1/2)/ π for m ∈ N. Then

F2m φm+1 (r ) = Am r ’3m’1 gm (r ).

Proof The proof is similar to the proof of Lemma 6.19. On the one hand, the Fourier

transform can be written as F2m φm+1 (r ) = r ’3m’1 h(r ) with h(r ) = 0 h 1 (r ’ t)h 2 (t) dt,

r

h 1 (t) = t m+1 and h 2 (t) = t m Jm’1 (t). The Laplace transform of h 1 is given by Lh 1 (r ) =

(m + 1)! r ’m’2 . Setting ν = m ’ 1 ≥ 0 in Lemma 5.7 yields

2m (m + 1/2)r

Lh 2 (r ) = √ ,

π(1 + r 2 )m+1/2

so that the Laplace transform of h becomes

Am

Lh(r ) = .

+ r 2 )m+1/2

r m+1 (1

On the other hand, we know by Lemma 5.8 that

∞

1

J0 (t)e’r t dt =

(1 + r 2 )1/2

0

so that

∞ ∞

t

1 1

J0 (s)ds e’r t dt = J0 (s)e’r s ds =

Lg0 (r ) = .

r (1 + r 2 )1/2

r

0 0 0

Moreover, f m’1 has Laplace transform L f m’1 (r ) = r ’m (1 + r 2 )’m , as shown in the proof

of Lemma 6.19. This gives

1

Lgm (r ) =

+ r 2 )m+1/2

r m+1 (1

and h = Am gm by uniqueness, which completes the proof.

This iterative representation of the Fourier transform F2m φm+1 together with the bounds

on f m allow us to ¬nd upper and lower bounds for F2m φm+1 .

Lemma 10.34 For every m ∈ N there exists a constant Cm > 0 such that

F2m φm+1 (r ) ¤ Cm r ’2m’1

for all r > 0. Moreover, for m ≥ 2 there exist an rm > 0 and a cm > 0 such that

F2m φm+1 (r ) ≥ cm r ’2m’1

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

Proof By Lemma 5.9 we know that limt’∞ g0 (t) = 1. Thus there exists a t0 > 0 such that

1/2 ¤ g0 (t) ¤ 3/2 for all t ≥ t0 . Since g0 is also continuous this means in particular that

g0 is bounded. Moreover, from the proof of Lemma 10.31 we know that f m’1 (t) ¤ Ct m’1 ,

160 Native spaces

which gives

r r

|gm (r )| ¤ | f m’1 (t)||g0 (r ’ t)|dt ¤ C t m’1 dt = Cr m .

0 0

Since F2m φm+1 (r ) = Am r ’3m’1 gm (r ) this proves the upper bound.

For the lower bound we ¬rst choose an r0 > 0 such that g0 (r ) ≥ 1/2 and f m’1 (r ) ≥

Cr m’1 for all r ≥ r0 /2. The latter was done in the proof of Lemma 10.32. Since both g0

and f m’1 are nonnegative we can obtain the estimate

r/2 r/2

gm (r ) ≥ f m’1 (t)g0 (r ’ t)dt ≥ C t m’1 dt ≥ Cr m

r0 /2 r0 /2

for suf¬ciently large r ≥ r0 . The use of F2m φm+1 (r ) = Am r ’3m’1 gm (r ) ¬nishes the proof

for this part.

Finally, since g0 (r ) > 0 for all r > 0 and f m’1 (r ) > 0 or f m’1 (r ) = 1 ’ cos r if m ≥ 2 or

m = 1 respectively, gm has to be positive on (0, ∞). Since φm+1 is nonnegative, F2m φm+1

is positive de¬nite on R2m , showing that F2m φm+1 has to be positive at zero and hence

everywhere.

Now that we have complete control over the Fourier transform of φd,k we can state and

proof our main result.

Theorem 10.35 Let d,k = φd,k ( · 2 ) denote the compactly supported radial basis func-

tion of minimal degree that is positive de¬nite on Rd and in C 2k . Let d ≥ 3 if k = 0. Then

there exist constants c1 , c2 > 0 depending only on d and k such that

c1 (1 + ω 2 )’d’2k’1 ¤ ¤ c2 (1 + ω 2 )’d’2k’1

d,k (ω)

for all ω ∈ Rd . This means in particular that

N (Rd ) = H d/2+k+1/2 (Rd ),

d,k

i.e. the native space for these basis functions is a classical Sobolev space.

Proof The preceding results show that both upper and lower bounds are valid for suf¬-

ciently large arguments r = ω 2 , say ω 2 ≥ r0 . But as d,k is a continuous and positive

function on Rd the bounds have to hold with possibly worse constants on the whole of Rd .