59.21. POSITIVE DEFINITE FUNCTIONS, BOCHNER’S THEOREM 1943

for all f ∈Cc (Rn). Thus from the dominated convergence theorem, the above holds for allf ∈S also. Hence for all f ∈S and considering µ as an element of S∗,

F−1µ (F f ) = µ ( f ) =

∫Rn

f (y)dµ (y)

=1

(2π)n/2

∫Rn

ψ (x)F ( f )(x)dx

=1

(2π)n/2 F (ψ)( f )≡ 1

(2π)n/2 ψ (F f ) .

It follows that in S∗,

ψ = (2π)n/2 F−1µ

Thusψ (t) =

∫Rn

eit·xdµ

in L1. Since the right side is continuous and the left is given continuous at t = 0 and equalto 1 there, it follows

1 = ψ (0) =∫Rn

ei0·xdµ = µ (Rn)

and so µ is a probability measure as claimed. This proves the lemma.The following is Bochner’s theorem.

Theorem 59.21.7 Let ψ be positive definite, continuous at 0, and ψ (0) = 1. Then thereexists a unique Radon probability measure µ such that ψ = φ µ .

Proof: If ψ ∈ L1 (Rn,mn) , then the result follows from Lemma 59.21.6. By Lemma59.21.3 ψ is bounded. Consider

ψ t (x)≡ ψ (x)1

(2πt)n/2 e−12t |x|

2.

Then ψ t (0) = 1, x→ψ t (x) is continuous at 0, and ψ t ∈ L1 (Rn,mn) . Therefore, by Lemma59.21.6 there exists a unique Radon probability measure µ t such that

ψ t (x) =∫Rn

eix·ydµ t (y) = φ µt(x)

Now letting t→ ∞,limt→∞

ψ t (x) = limt→∞

φ µt(x) = ψ (x) .

By Levy’s theorem, Theorem 59.19.7 it follows there exists µ, a probability measure onB (Rn) such that ψ (x) = φ µ (x) . The measure is unique because the characteristic func-tions are uniquely determined by the measure. This proves the theorem.