k=1

As is positive de¬nite we have

1

N N

±T A ±= ± ±j ’ xj) ’ xj)

1 (x 2 (x

2 ,X

1

=1 j=1

N N N

= ± ±j ’ xj) s k s jk »k

1 (x

=1 j=1 k=1

N N N

= »k ± s k ± j s jk ’ xj)

1 (x

=1 j=1

k=1

N N N

≥ »1 ± ±j ’ xj)

1 (x s k s jk

=1 j=1 k=1

N

= »1 |± |2 1 (0).

=1

The last expression is nonnegative for all ± ∈ C N and positive unless ± = 0.

We now come back to the question of real-valued positive de¬nite functions and their

characterization. First of all, it is now clear that a positive semi-de¬nite function is real-

valued if and only if it is even. But we can also restrict ourselves to real coef¬cient vectors

± ∈ R N in a quadratic form.

Theorem 6.3 Suppose that : Rd ’ R is continuous. Then is positive de¬nite if and

only if is even and we have, for all N ∈ N, for all ± ∈ R N \ {0}, and for all pairwise

distinct x1 , . . . , x N ,

N N

± j ±k (x j ’ xk ) > 0.

j=1 k=1

Proof If is positive de¬nite and real-valued then it is even by Theorem 6.2; it obviously

satis¬es the second condition.

6.2 Bochner™s characterization 67

satis¬es the given conditions then we have with ± j = a j + ib j

If, however,

N N

± j ±k (x j ’ xk ) = (a j ak + b j bk ) (x j ’ xk )

j,k=1 j,k=1

N

+i ak b j [ (x j ’ xk ) ’ (xk ’ x j )].

j,k=1

As is even the second sum on the right-hand side vanishes. The ¬rst sum is nonnegative

because of the assumption and vanishes only if all the a j and b j vanish.

6.2 Bochner™s characterization

One of the most celebrated results on positive semi-de¬nite functions is their character-

ization in terms of Fourier transforms, which was established by Bochner in 1932 for

d = 1, [28], and 1933 for general d, [29]. This result is one of the reasons for introducing

complex-valued positive de¬nite functions. The idea behind it is easily described. Suppose

∈ C(Rd ) © L 1 (Rd ) has an integrable Fourier transform ∈ L 1 (Rd ). Then by the Fourier

inversion formula we can recover from its Fourier transform:

(x) = (2π)’d/2 (ω)ei x ω dω.

T

Rd

This means that a quadratic form involving can be expressed as follows:

N N N

± j ±k (x j ’ xk ) = (2π)’d/2 (x j ’xk )

T

± j ±k (ω)eiω dω

Rd

j=1 k=1 j,k=1

2

N

ixT ω

’d/2

= (2π) ±je

(ω) dω.

j

Rd j=1

Hence, if is nonnegative then the function is positive semi-de¬nite. The analysis also

shows that it is unimportant whether we recover from its Fourier or its inverse Fourier

transform.

Even though we will see that every positive de¬nite and integrable function has an

integrable Fourier transform, we cannot use this approach in the case of a nonintegrable

function. For a complete characterization we have to replace the measure with Lebesgue

density by a more general Borel measure μ. We will see that this suf¬ces to characterize

every positive semi-de¬nite function as the Fourier transform of such a measure.

To prove this characterization we need two auxiliary results. The ¬rst is a characterization

of positive semi-de¬nite functions as integrally positive semi-de¬nite functions.

: Rd ’ C is positive semi-de¬nite if and only if

Proposition 6.4 A continuous function

is bounded and satis¬es

(x ’ y)γ (x)γ (y)d xd y ≥ 0 (6.3)

Rd Rd

for all test functions γ from the Schwartz space S.

68 Positive de¬nite functions

Proof Suppose that is a positive semi-de¬nite function. Since is bounded and γ ∈ S

decays rapidly, the integral (6.3) is well de¬ned. Moreover, for every > 0 there exists a

closed cube W ⊆ Rd such that

(x ’ y)γ (x)γ (y)d xd y ’ (x ’ y)γ (x)γ (y)d xd y < .

2

Rd Rd W W

But the double integral over the cubes is the limit of Riemannian sums. Hence, we can ¬nd

x1 , . . . , x N ∈ Rd and weights w1 , . . . , w N such that

N

(x ’ y)γ (x)γ (y)d xd y ’ (x j ’ xk )γ (x j )w j γ (xk )wk < .

2

W W j,k=1

This means that

N

(x ’ y)γ (x)γ (y)d xd y > (x j ’ xk )γ (x j )w j γ (xk )wk ’ .

Rd Rd j,k=1

Letting tend to zero and using that is positive semi-de¬nite shows indeed that (6.3) is

true for all γ ∈ S. As a positive semi-de¬nite function is also bounded.

Conversely, let us assume that is bounded and satis¬es (6.3). Because of the assumption

imposed on and γ we can rewrite the double integral as

(x ’ y)γ (x)γ (y)d xd y = (x)γ — γ (x)d x

Rd Rd Rd

using the notation γ (x) = γ (’x). Next, suppose that x1 , . . . , x N ∈ Rd and ±1 , . . . , ± N ∈ C

are given. We choose

N

γ = γm = ± j g2m (· ’ x j ),

j=1

where gm (x) = (m/π )d/2 e’m ω 2 is the function from Theorem 5.20. Then, using standard

2

techniques to compute the Fourier transform, we see that

N N

’iω T x j ’d/2

± j e’iω xj ’ x 2 /(8m)

2

T