is at least continuous by Lemma 9.8. Furthermore, ψ is positive de¬nite on Rd+2k . Thus

˜

ψ must possess at least d/2 + k continuous derivatives at 1, which means that ψ(r ) =

˜ ˜

d/2 +k+1

(1 ’ r )+ q(r ) with a polynomial q. But, because of the minimal degree, q must be a

constant. Thus ψ = I k ψ and φd,k can differ only by a constant factor.

˜

For convenience, we list the simplest cases in Table 9.1, where = denotes equality up to

™

a positive constant factor. We use this notation also in the next corollary, in which we give

the explicit form for the most important cases.

Corollary 9.14 The functions φd,k , k = 0, 1, 2, 3 have the following form:

d/2 +1

φd,0 (r ) = (1 ’ r )+ ,

+1

φd,1 (r ) = (1 ’ r )+ [( + 1)r + 1],

™

+2

φd,2 (r ) = (1 ’ r )+ [( + 4 + 3)r 2 + (3 + 6)r + 3],

2

™

+3

φd,3 (r ) = (1 ’ r )+ [( +9 + 23 + 15)r 3

3 2

™

+ (6 + 36 + 45)r 2 + (15 + 45)r + 15],

2

where we have used := d/2 + k + 1.

Proof The form for k = 0 is obvious. For k = 1 we have to apply I once. A simple

computation shows for r ∈ [0, 1] that

(1 ’ r ) +1

1

φd,1 (r ) = I φ (r ) = t(1 ’ t) dt = [( + 1)r + 1] .

( + 1)( + 2)

r

The other cases are dealt with in the same spirit.

130 Compactly supported functions

Fig. 9.1 The C 0 -, C 2 -, and C 4 -function for R3 (on the left) and the C 2 -, function in R2 (on the right).

The left-hand part of Figure 9.1 shows the univariate functions φ3,0 , φ3,1 , and φ3,2 , which

are positive de¬nite on Rd for d ¤ 3; the right-hand part shows the C 2 -function φ3,1 (r ) =

(1 ’ r )4 (4r + 1) in R2 .

9.5 Generalizations

In the example following Lemma 9.8 we have already seen another class of compactly

supported radial basis functions. While these basis functions are still of the form (9.1),

one can also consider basis functions that do not have a polynomial representation within

their support. The reason for choosing univariate polynomials was to get a simple evalu-

ation. But obviously a simple evaluation can be achieved by other (univariate) functions

also.

While it took rather a long time for the ¬rst compactly supported radial basis functions

to be found, it is now quite simple to construct a variety of them by different tools. Besides

the operators I and D we want to discuss two further techniques here. Both use existing

positive de¬nite functions to construct new ones.

The ¬rst idea is to apply an operator T to the basis function . This operator has to be

nonnegative in the Fourier domain, i.e. it has to satisfy T ≥ 0 if ≥ 0. Then the resulting

function is also positive de¬nite. If we are interested in compactly supported functions then

we have to make sure that T respects the compact support of a function. The most important

example is the classical Laplace operator.

Lemma 9.15 Suppose that ∈ C 2 (Rd ) © L 1 (Rd ) is positive de¬nite. Denote by the

Laplace operator = d ‚ 2 /‚x2 . If ’ is integrable then it is also positive de¬nite.

j=1 j

If is radial, so is . If has compact support, so has .

Proof Since all involved functions are integrable, the Fourier transform of is given

= ’ · 2 . This proves the positive de¬niteness of ’

2

by . The Laplace operator

9.5 Generalizations 131

= φ( ·

applied to a radial function is a radial operator. If 2) then

d ’1

(x) = φ ( x 2 ) + φ ( x 2 ).

x2

The statement about the support is obvious.

The second possible construction technique we want to mention is a specialization of the

¬rst one and is crucial to Bochner™s and Schoenberg™s characterizations. It uses the fact that

if a positive de¬nite function is integrated against a nonnegative measure then the result is

also positive de¬nite.

Proposition 9.16 Suppose that f ∈ L 1 [0, ∞) is nonnegative and positive on a set U of

positive Lebesgue measure. Suppose further that a bounded kernel K : [0, ∞) — [0, ∞) ’

R is given satisfying

(1) K (·, r ) is measurable for all r ≥ 0,

(2) K (t, ·) is positive semi-de¬nite on Rd for all t > 0 and positive de¬nite on Rd for t ∈ U .

Then

∞

φ(r ) := K (t, r ) f (t)dt (9.6)

0

is positive de¬nite on Rd .

Proof Since K (t, ·) is positive semi-de¬nite for every t > 0, it is in particular continuous.

Standard arguments yield that φ is also continuous. Finally, if pairwise distinct x1 , . . . , x N ∈

Rd and ± ∈ R N \ {0} are given, we can see at once that

∞

N N

± j ±k φ( x j ’ xk 2 ) = ± j ±k K (t, x j ’ xk 2 ) f (t)dt ≥ 0.

0

j,k=1 j,k=1

Actually, the quadratic form must be positive because otherwise the set of points where the

integrand does not vanish must have measure zero. This is impossible since it contains the

set U .

A typical choice for K is K (t, r ) = ψ(r/t) with a compactly supported function ψ.

The compact support makes K well de¬ned. For example, one could use the compactly

supported functions φd,k .

Proposition 9.16 does not work only for kernels K such that K (t, ·) is positive de¬nite.

It is also possible to relax this condition if the choices for the function f are further

restricted. For example, in [34, 35] Buhmann makes the choices K (t, r ) = (1 ’ r 2 /t)» and

+

δρ

±

f (t) = t (1 ’ t )+ . He discusses thoroughly all choices of the parameters ±, δ, », and ρ

such that φ from (9.6), which now takes the form

∞

ρ

(1 ’ r 2 /t)» t ± (1 ’ t δ )+ dt,

φ(r ) = (9.7)

+

0

becomes positive de¬nite. The result is

132 Compactly supported functions

Theorem 9.17 Let 0 < δ ¤ 1 , ρ ≥ 1 be real numbers, and suppose that » = 0 and ± are

2