we see that

1 k!

= (’1)k+1 .

lim

(’β/2)(β ’ 2k)

β’2k 2

Now we can apply the dominated convergence theorem to derive

γ (ω)

k!

γ (x)d x = 22k+d/2 (k + d/2)(’1)k+1

2k

x log x dω

2

2

ω d+2k

2

Rd Rd 2

for all γ ∈ S2m , which gives the stated Fourier transform.

Now it is easy to decide whether the functions just investigated are conditionally positive

de¬nite. As mentioned before we state the minimal m. The case of inverse multiquadrics

was treated in Theorem 7.15.

: Rd ’ R are conditionally positive de¬nite of

Corollary 8.18 The following functions

order m:

(x) = (’1) β (c2 + x 2 )β , β > 0, β ∈ N, m= β ,

(1) 2

β

β/2

(x) = β > 0, β ∈ 2N, m = β/2 ,

(2) (’1) x 2,

(x) = k ∈ N, m = k + 1.

(’1)k+1 x 2k log

(3) x 2,

2

8.4 Radial conditionally positive de¬nite functions

As in case of (unconditionally) positive semi-de¬nite functions it is possible to derive a

characterization of conditionally positive semi-de¬nite and radial functions from Theorem

8.14. We omit the details here and refer the interested reader to the article [77] by Guo

et al. Instead, we turn to univariate functions that are conditionally positive (semi-)de¬nite

on every Rd and derive a result along the lines of the characterization of Schoenberg. In

particular, the univariate function φ acts again as a multivariate function via φ( · 2 ). 2

Theorem 8.19 (Micchelli) Suppose that φ ∈ C[0, ∞) © C ∞ (0, ∞) is given. Then the func-

tion = φ( · 2 ) is conditionally positive semi-de¬nite of order m ∈ N0 on every Rd if

2

and only if (’1)m φ (m) is completely monotone on (0, ∞).

114 Conditionally positive de¬nite functions

Proof Suppose that (’1)m φ (m) is completely monotone on (0, ∞). We know from Theorem

7.11 that it can be represented by

∞

e’r t dμ(t)

(’1)m φ (m) (r ) =

0

with a nonnegative Borel measure μ on [0, ∞). As we do not want to assume φ (m) to be

continuous in zero, the measure μ will not be ¬nite. Hence, we de¬ne φ (r ) = φ(r + ) for

> 0. Using Taylor™s formula gives us

m’1

φ ( ) (0) r

1

φ (r ) = r+ (r ’ t)m’1 φ (m) (t)dt

(m ’ 1)!

! 0

=0

∞

m’1

φ ( ) (0) r

(’1)m

(r ’ t)m’1 e’(t+ )s dμ(s)dt

= r+

(m ’ 1)!

! 0 0

=0

∞

m’1

φ ( ) (0) r

(’1)m

e’ (r ’ t)m’1 e’ts dtdμ(s),

= r+ s

(m ’ 1)!

! 0 0

=0

where we have applied Fubini™s theorem in the last step. A further application of Taylor™s

formula to the function r ’ e’r s shows that

m’1 r

(’1) j (’s)m

e’r s = (r ’ t)m’1 e’st dt.

(r s) j +

(m ’ 1)!

j! 0

j=0

Inserting this representation for the inner integral into the representation of φ leads ¬nally

to

∞

m’1

φ ( ) (0) m’1

(’1) j j j ’ s dμ(s)

e’r s ’

φ (r ) = r+ .

rs e

sm

! j!

0

=0 j=0

Now suppose pairwise distinct x1 , . . . , x N ∈ Rd and ± ∈ R N satisfying (8.1) for all p ∈

πm’1 (Rd ) are given. Then we can conclude from Proposition 8.4 that

∞

N N

dμ(s)

± j ±k e’s x j ’xk

e’

2

± j ±k φ ( x j ’ xk 2 ) = ≥0

s

2

2

sm

0

j,k=1 j,k=1

for every > 0, since the Gaussian is positive de¬nite. Note that the last integral is well

de¬ned by an argument similar to that in the proof of Lemma 8.11. But as the function on

the left-hand side is a continuous function in we can let tend to zero, showing that the

quadratic form is nonnegative for = 0 also.

For the necessary part let us assume that φ( · 2 ) is conditionally positive de¬nite of order

2

m. We want to show by induction on m that in this case (’1)m φ (m) is completely monotone.

For m = 0 this is Schoenberg™s result given in Theorem 7.13. For the induction step we

assume that the result is true for m and we want to conclude it for m + 1. Suppose φ( · 2 ) 2

is conditionally positive semi-de¬nite of order m + 1 on every Rd . Fix a dimension d. For

8.4 Radial conditionally positive de¬nite functions 115

h > 0 we consider the (d + 1)-variate function