e

j=1 j=1

which leads to

(γm — γm )§ (ω) = (2π)d/2 |γm |2 (ω)

2

N

’d/2 ’iω T x j

e’ ω 2 /(4m)

= (2π) ±je 2

j=1

N

± j ±k e’iω (x j ’xk )

T

= gm (ω)

j,k=1

§

N

= ± j ±k gm (· ’ (x j ’ xk )) (ω).

j=1

6.2 Bochner™s characterization 69

This together with the uniqueness of the Fourier transform on S and Theorem 5.20 allows

us conclude that

N N

± j ±k (x j ’ xk )= ± j ±k lim (x)gm (x ’ (x j ’ xk ))d x

m’∞ Rd

j,k=1 j,k=1

= lim (x)γm — γm (x)d x,

m’∞ Rd

and the last expression is nonnegative by our assumption.

The second auxiliary result that we need on our way to proving Bochner™s theorem is a

generalization of Riesz™ representation theorem, Theorem 5.33. For applying this theorem

to Rd , it is not necessary to have a linear functional that is bounded. It suf¬ces that the

functional is bounded only in a local sense. Moreover, the functional needs only to be

∞

de¬ned on functions from C0 (Rd ).

∞

Proposition 6.5 Suppose that » : C0 (Rd ) ’ C is a linear functional which is nonnegative,

∞

meaning that »( f ) ≥ 0 for all f ∈ C0 (Rd ) with f ≥ 0. Then » has an extension to C0 (Rd )

and there exists a nonnegative Borel measure μ such that

»( f ) = f (x)dμ(x)

Rd

for all f ∈ C0 (Rd ).

Proof The proof is divided into three steps. In the ¬rst step we show that a nonnegative

linear functional is locally bounded. Hence, let K be a compact subset of Rd . Denote by

∞ ∞

C0 (K ) the subset of functions from C0 (Rd ) having support in K . We use the notation

∞

C0 (K ) similarly. We want to show that |»( f )| ¤ C K f L ∞ (K ) for all f ∈ C0 (K ) with a

∞

constant C K depending only on K . To this end we choose a function ψ ∈ C0 (Rd ) with

∞

ψ|K ≡ 1 and ψ ≥ 0. If f ∈ C0 (K ) is real-valued then

L ∞ (K ) ψ ± f ≥0

f

obviously implies that »( f ) ∈ R and

|»( f )| ¤ »(ψ) f L ∞ (K ) .

If f is complex-valued then we choose a θ ∈ R such that eiθ »( f ) ∈ R. Using »( ( f )) =

(»( f )), which we have just proven, a simple computation shows that »( (eiθ f )) =

eiθ »( f ). Applying the previous result to (eiθ f ) gives therefore

|»( f )| = |eiθ »( f )| = |»( (eiθ f ))| ¤ »(ψ) (eiθ f ) L ∞ (K )

¤ »(ψ) f L ∞ (K ) .

Hence we can de¬ne C K to be »(ψ).

∞

The second step deals with the local extension of ». Being restricted to C0 (K ), the

functional » is continuous and thus has a unique extension to C0 (K ). The well-known

70 Positive de¬nite functions

∞

construction starts for f ∈ C0 (K ) with a sequence { f j } ⊆ C0 (K ) that converges uniformly

to f and de¬nes »( f ) as the limit of »( f j ). Since { f j } can be chosen as the convolution of

f with a family of nonnegative and compactly supported functions, the extension of » to

C0 (K ) is also nonnegative.

By Theorem 5.33 we can ¬nd a unique Borel measure μ K de¬ned on the Borel sets B(K )

such that

»( f ) = for all f ∈ C0 (K ).

f (x)dμ K (x)

K

Since » when restricted to C0 (K ) is continuous, the measure μ K is ¬nite.

In the ¬nal step we use this result to de¬ne a pre-measure on the ring F d , which is the

collection of ¬nite unions of semi-open cubes. See the start of Section 5.3 for the notation.

To this end we let K j := {x ∈ Rd : x 2 ¤ j}. The pre-measure μ is now de¬ned

as follows. For A ∈ F d we choose a j ∈ N suf¬ciently large that A ⊆ K j and de¬ne

μ(A) := μ K j (A). The function μ is well de¬ned and indeed a pre-measure if its de¬ni-

tion is independent of the choice of j. But this is the case since μ K j+1 |B(K j ) = μ K j by

uniqueness.

Hence, by Proposition 5.31 there exists a Borel measure μ on Rd such that

»( f ) = f (x)dμ(x)

Rd

for all f ∈ C0 (Rd ).

After these preparatory steps we are now in a position to formulate and prove Bochner™s

result.

Theorem 6.6 (Bochner) A continuous function : Rd ’ C is positive semi-de¬nite if

and only if it is the Fourier transform of a ¬nite nonnegative Borel measure μ on Rd , i.e.

(x) = μ(x) = (2π)’d/2 e’i x ω dμ(ω),

T

x ∈ Rd . (6.4)

Rd

Proof We ¬rst assume that is the Fourier transform of a ¬nite nonnegative Borel measure

and show that is positive semi-de¬nite. With the usual x1 , . . . , x N ∈ Rd and ± ∈ C N we

get, as in the introductory part of this section,

N N

± j ±k (x j ’ xk ) = (2π)’d/2 e’i(x j ’xk ) ω dμ(ω)

T

± j ±k

Rd

j,k=1 j,k=1

2

N

’i x T ω

’d/2

= (2π ) ±je dμ(ω) ≥ 0.

j

Rd j=1

The last inequality holds because of the conditions imposed on the measure μ. As it is the

Fourier transform of the ¬nite measure μ, the function must be continuous.

6.2 Bochner™s characterization 71

is positive semi-de¬nite. Then we de¬ne a functional » on the

Next let us suppose that

Schwartz space S by

(x)γ ∨ (x)d x,

»(γ ) := γ ∈ S.