since γ ∈ Sm .

∞

For the second part we choose a ¬xed test function χ ∈ C0 (Rd ), which is identically

equal to one in a neighborhood of the origin. Then we de¬ne for an arbitrary g ∈ S the

function

D β g(0) β

γ (x) = g(x) ’ x χ(x), x ∈ Rd ,

β!

|β|<m

which is clearly in Sm and has Fourier transform

D β g(0) |β| β

γ (ω) = g(ω) ’ i D χ (ω).

β!

|β|<m

Hence (8.6) yields

0= (x)γ (x)d x

Rd

D β g(0) |β|

(x)D β γ (x)d x

= (x)g(x)d x ’ i

β!

Rd Rd

|β|<m

i |β|

cβ D β g(0),

= (x)g(x)d x ’

β!

Rd |β|<m

8.2 An analogue of Bochner™s characterization 105

(x)D β χ (x)d x. If we ¬nally use D β g(0) =

the constants cβ being de¬ned by cβ =

(2π )’d/2 i |β| g(x)x β d x we derive

cβ (’1)|β| β

(x) ’ (2π)’d/2 x g(x)d x = 0

β!

Rd |β|<m

for all g ∈ S. Approximation by convolution from Theorem 5.20 shows that is indeed a

polynomial of degree less than m.

On our way to deriving a Bochner-type result we need to know how to construct functions

from S2m . A simple trick is to employ centers and coef¬cients, which satisfy the side

condition for a conditionally positive de¬nite function.

Lemma 8.11 Suppose that pairwise distinct x1 , . . . , x N ∈ Rd and ± ∈ C N \ {0} are given

such that (8.1) is satis¬ed for all p ∈ πm’1 (Rd ). Then

N

± j ei x j ω = O( ω

T

m

2)

j=1

holds for ω ’ 0.

2

Proof The expansion of the exponential function leads to

∞

N N

ik

± j ei x j ω =

T

± j (x T ω)k .

j

k!

j=1 k=0 j=1

For ¬xed ω ∈ Rd we have pk (x) := (x T ω)k ∈ πk (Rd ). Thus (8.1) ensures that the ¬rst m

terms vanish:

∞

N N

ik

± j ei x j ω =

T

± j (x T ω)k ,

j

k!

k=m

j=1 j=1

which gives the stated behavior.

Now it is time to state and prove our main theorem. It states that the order of the generalized

Fourier transform, which is nothing other than the order of the singularity of the Fourier

transform at the origin, determines the minimal order of a conditionally positive de¬nite

function.

Theorem 8.12 Suppose : Rd ’ C is continuous, slowly increasing, and possesses a

generalized Fourier transform of order m, which is continuous on Rd \ {0}. Then is

conditionally positive de¬nite of order m if and only if is nonnegative and nonvanishing.

Proof Suppose that is nonnegative and nonvanishing. Suppose further that pairwise

distinct x1 , . . . , x N ∈ R and ± ∈ C N \ {0} satisfy (8.1) for all p ∈ πm’1 (Rd ). De¬ne

d

N

± j ±k (x + (x j ’ xk ))

f (x) :=

j,k=1

106 Conditionally positive de¬nite functions

and

2

N N

(x j ’xk )

ixT x j T

γ (x) := ±je g (x) = ± j ±k e i x g (x)

j=1 j,k=1

where g (x) = ( /π)d/2 e’ x 2 is the test function from Theorem 5.20. On account of

2

γ ∈ S and Lemma 8.11 we have γ ∈ S2m . Furthermore, we can compute the Fourier trans-

form:

N

γ (x) = (2π )’d/2 (x j ’xk )

g (ω)e’i x ω dω

T T

± j ±k eiω

Rd j,k=1

N

± j ±k (2π)’d/2 g (ω)e’iω (x’(x j ’xk ))

T

= dω

Rd

j,k=1

N

= ± j ±k g (x ’ (x j ’ xk )),

j,k=1

because g = g . Collecting these facts gives, together with De¬nition 8.9,

N

f (x)g (x)d x = ± j ±k g (x ’ (x j ’ xk ))d x

(x)

Rd Rd j,k=1

= (x)γ (x)d x

Rd