If γ ∈ S is of the form γ = |ψ|2 with ψ ∈ S then we ¬nd with f := ψ ∨ and f (x) = f (’x)

that

∨

»(γ ) = (x) | f |2 (x)d x

Rd

= (2π )’d/2 (x) f — f (x)d x

Rd

= (2π )’d/2 (x ’ y) f (x) f (y)d xd y

Rd Rd

≥ 0,

where the last inequality follows from Proposition 6.4.

Hence » is nonnegative on the set of all functions γ from the Schwartz space S of the

form γ = |ψ|2 with ψ ∈ S.

∞

We wish to extend this relation to all γ ∈ C0 (Rd ) with γ ≥ 0. To this end we form

γ + µ2 G, where G is the Gaussian G(x) = e’ x 2 /2 and µ > 0. This function is in S and

2

positive on Rd . Thus it possesses a square root, say γ1 := γ + µ 2 G, which is clearly in

C ∞ . It is also in S because it coincides with µG(·/2) for suf¬ciently large x. Altogether,

this leads us to

0 ¤ »(|γ1 |2 ) = »(γ ) + µ 2 »(G),

which establishes the desired result with µ ’ 0. But we know that » is a nonnegative

∞

linear functional de¬ned on C0 (Rd ) and this allows us to use Proposition 6.5 to obtain a

nonnegative Borel measure μ on Rd such that

»(γ ) = γ (x)dμ(x), γ ∈ C0 (Rd ). (6.5)

Rd

Next we apply approximation by convolution, as provided in Theorem 5.22, to show that μ

is ¬nite and that » can be used to represent . Approximation by convolution uses a family

{gm } with gm = m d g(m·) where g ∈ C(Rd ) © L 1 (Rd ) has L 1 -norm one and is nonnegative.

Here, we need in addition that g also is nonnegative and in C0 (Rd ). This can be achieved

∞

by choosing a nonnegative function g0 ∈ C0 (Rd ) and de¬ning g via g = (g0 — g0 )∨ with

∞

possible renormalization. Note that this construction even leads to g ∈ S and g ∈ C0 (Rd ).

Let us show that the measure μ has a ¬nite total mass. First of all we have

(x)gm (x)d x = gm (x)dμ(x) = g(x/m)dμ(x).

Rd Rd Rd

72 Positive de¬nite functions

Since g(x/m) tends to (2π )’d/2 and is nonnegative we can apply Fatou™s lemma to

derive

(2π )’d/2 dμ(x) = lim g(x/m)dμ(x)

Rd m’∞

Rd

¤ lim g(x/m)dμ(x)

m’∞ Rd

= lim (x)gm (x)d x

m’∞ Rd

= (0),

since is bounded. This shows that the total mass of μ is indeed bounded by (2π)d/2 (0).

Finally, Theorem 5.22 gives

(x) = lim (ω)gm (x ’ ω)dω

m’∞ Rd

e’iω x gm (’ω)dμ(ω)

T

= lim

m’∞ Rd

= (2π )’d/2 e’i x ω dμ(ω),

T

Rd

since μ is ¬nite. This is what we intended to show.

The advanced reader will certainly have noticed that the functional » in the proof of

Theorem 6.6 is nothing other than the distributional Fourier transform of . Moreover,

Proposition 6.5 just states that every nonnegative distribution can be interpreted as a Borel

measure.

With the interpolation problem in mind, we are more interested in positive de¬nite than

in positive semi-de¬nite functions. Unfortunately, a complete characterization of positive

de¬nite functions leads to a very technical discussion of the Borel measures involved. The

interested reader might have a look at the paper [39] by Chang.

But all the important functions that we will encounter in this book have either a discrete

measure or a measure with Lebesgue density. In such a situation it is far easier to decide

whether a positive semi-de¬nite function is also positive de¬nite. In fact, it will turn out that

all positive semi-de¬nite functions having a measure with a continuous Lebesgue density

that is not identically zero are positive de¬nite. We will now derive suf¬cient conditions on

the measure μ to ensure that is positive de¬nite.

Lemma 6.7 Suppose that U ⊆ Rd is open. Suppose, further, that x1 , . . . , x N ∈ Rd are

pairwise distinct and that c ∈ C N . If N c j e’i x j ω = 0 for all ω ∈ U then c ≡ 0.

T

j=1

Proof By successive analytic continuation in each coordinate (or by the identity theorem, to

be more precise) we can derive that the assumption actually means that N c j e’i x j ω = 0

T

j=1

6.2 Bochner™s characterization 73

for all ω ∈ Rd . Now take a test function f ∈ S. Then

§

N N

’i x T ω

0= f (ω) = c j f (· ’ x j )

cje (ω)

j

j=1 j=1

for all ω ∈ Rd implies that

N

c j f (x ’ x j ) = 0, x ∈ Rd .

j=1

Now take f to be compactly supported with f (0) = 1 and support contained in the ball

around zero with radius < min j=k x j ’ xk 2 . This gives

N

0= c j f (xk ’ x j ) = ck f (0) = ck

j=1

for 1 ¤ k ¤ N .

Knowing when the exponentials are linearly independent, we can give a suf¬cient con-

dition for a function to be positive de¬nite.

Theorem 6.8 A positive semi-de¬nite function is positive de¬nite if the carrier of the

measure μ in the representation (6.4) contains an open subset.

Proof Denote the open subset by U . Then μ(U ) = 0 and thus we can conclude from the

proof of Theorem 6.6 that for any pairwise distinct x1 , . . . , x N ∈ Rd and any ± ∈ C N we

must have

N

± j e’i x j x = 0,

T