= (1 ’ )k (x)g (x)d x,

Rd

since possesses 2k continuous derivatives around zero and the integrals are actually only

integrals over {x : x 2 ¤ 1/ }. The last integral converges for ’ ∞ to (1 ’ )k (0).

Hence, by Fatou™s lemma and g (ω) ’ (2π)’d/2 we get

(1 + ω 2 )k dμ(ω) ¤ (2π)d/2 (1 ’ )k (0).

2

Rd

Thus we can differentiate up to 2k times under the integral in (6.7), which means that is

in C 2k everywhere.

6.3 Radial functions

Even though we have used it already in several places we will now recall the de¬nition of

a radial function.

De¬nition 6.15 A function : Rd ’ R is said to be radial if there exists a function φ :

[0, ∞) ’ R such that (x) = φ( x 2 ) for all x ∈ Rd .

In the Gaussians and the multiquadrics we have already found examples of radial and

positive de¬nite functions without using the fact that they are radial. Now we want to exploit

radiality in more detail.

A radial function has the advantage of a very simple structure. This is motivation enough

to investigate whether such a univariate function is positive de¬nite in the following sense.

De¬nition 6.16 We will call a univariate function φ : [0, ∞) ’ R positive de¬nite on Rd

if the corresponding multivariate function (x) := φ( x 2 ), x ∈ Rd , is positive de¬nite.

The smoothness of the multivariate function is determined by the smoothness of the

even extension of the univariate function φ. This is the reason why we always assume φ to

be an even function de¬ned on all of R by even extension.

Any radial function is obviously even. From Theorem 6.3 we know that we can restrict

ourselves to real coef¬cients in the quadratic form.

Lemma 6.17 Suppose that a univariate function φ is positive de¬nite on Rd ; then it is also

positive de¬nite on Rk with k ¤ d.

Proof The proof of this lemma is obvious because Rk is a subspace of Rd .

Theorem 5.26 tells us how to compute the Fourier transform of a radial function. The

Fourier transform is again radial and thus can be expressed as a univariate function. Hence,

6.3 Radial functions 79

Bochner™s characterization can be reduced to a univariate setting. For example, the radial

version of Theorem 6.11 is given by

Theorem 6.18 Suppose that φ ∈ C[0, ∞) satis¬es r ’ r d’1 φ(r ) ∈ L 1 [0, ∞). Then φ is

positive de¬nite on Rd if and only if φ is bounded and

∞

Fd φ(r ) := r ’(d’2)/2 φ(t)t d/2 J(d’2)/2 (r t)dt

0

is nonnegative and nonvanishing.

Proof From r ’ r d’1 φ(r ) ∈ L 1 [0, ∞) we know that := φ( · is in L 1 (Rd ). Fur-

2)

thermore, we have (x) = Fd φ( x 2 ).

Note that the operator Fd de¬ned by Fd φ = acts on univariate functions. Hence, if

working with radial functions we are in a situation where we can do most of the analysis in

a univariate setting, which often makes things easier.

We will demonstrate this concept by giving another example of a positive de¬nite func-

tion. It suf¬ces to show that the univariate function Fd φ is nonnegative in contrast with

having to show that the multivariate function is nonnegative.

The function that we are now going to investigate differs from all positive de¬nite func-

tions we have encountered so far in two ways. First, it has a compact support. Second, it is

not positive de¬nite on every Rd as it was the case for Gaussians and inverse multiquadrics

(see the remarks after the proof of Theorem 6.13). We will see that these two features do

not appear together accidentally. On the contrary, the ¬rst implies the second.

Lemma 6.19 De¬ne the functions f 0 (r ) = 1 ’ cos r and

r

f n (r ) = f 0 (t) f n’1 (r ’ t)dt

0

for n ≥ 1. Let

2n+1/2 n!(n + 1)!

Bn = .

√

π

Then f n has the representation

r

(r ’ t)n+1 t n+1/2 Jn’1/2 (t)dt = Bn f n (r ). (6.9)

0

Proof De¬ne the integral on the left-hand side of (6.9) to be g(r ). Thus g is the convolution

r

g(r ) = 0 g1 (r ’ s)g2 (s)ds of the functions g1 (s) := s n+1 and g2 (s) := s n+1/2 Jn’1/2 (s).

This form of convolution differs slightly from the kind about which we have learnt in

the context of Fourier transforms. This new form of convolution is compatible with the

∞

Laplace transform L in the sense that Lg(r ) = 0 g(t)e’r t dt is the product of the Laplace

transforms of g1 and g2 . These transforms can be computed for r > 0 in the following

manner. The ¬rst function g1 can be handled by using the integral representation for the

80 Positive de¬nite functions

-function:

∞ ∞

(n + 2)

s n+1 e’r s ds = r ’n’2 t n+1 e’t dt =

Lg1 (r ) =

r n+2

0 0

(n + 1)!

= .

r n+2

The Laplace transform of g2 was computed in Lemma 5.7. If we set ν = n ’ 1/2 > ’1,

Lemma 5.7 yields

∞

n! 2n+1/2r

s n+1/2 Jn’1/2 (s)e’r s ds = √

Lg2 (r ) =

π (1 + r 2 )n+1

0

for r > 0. Together these two expressions give

2n+1/2 n!(n + 1)! 1

Lg(r ) = .

√ n+1 (1 + r 2 )n+1

π r

It is easily shown, however, that the function f 0 (r ) = 1 ’ cos r has Laplace transform

1/[r (1 + r 2 )]. Thus, since f n is the n-fold convolution of this function with itself, we get

1

L f n (r ) = .

r n+1 (1 + r 2 )n+1

By the uniqueness of the Laplace transform this leads to

2n+1/2 n!(n + 1)!

g(r ) = √ f n (r )

π

as stated.

The reason for proving this lemma is provided by the integral in (6.9). Obviously it

represents the Fourier transform of a radial function. And we now know that this Fourier

transform is nonnegative and nonvanishing. To give the associated function φ itself, let us

introduce the cutoff function (·)+ , which is de¬ned by

for x ≥ 0,

x

(x)+ =

for x < 0,

0

and the notation x , which denotes the largest integer less than or equal to x.

Theorem 6.20 The truncated power function

φ (r ) = (1 ’ r )+

is positive de¬nite on Rd provided that ∈ N satis¬es ≥ d/2 + 1.

Proof Let us start with the case of an odd space dimension d = 2n + 1 and = d/2 +

1 = n + 1. We have to check whether the function F2n+1 φn+1 is nonnegative and nonvan-

ishing. In this special situation it takes the form