61.5. BOCHNER’S THEOREM IN INFINITE DIMENSIONS 2005

Next suppose the conditions hold. Define for t ∈ Rn and {y1, · · · ,yn} ⊆ Λ

ψ{y1,··· ,yn} (t)≡ ψ

(n

∑j=1

t jy j

).

Then ψ{y1,··· ,yn} is continuous at 0, equals 1 there, and is positive definite. It follows fromBochner’s theorem, Theorem 59.21.7 on Page 1943 there exists a measure µ{y1,··· ,yn} de-fined on the Borel sets of Rn such that

ψ

(n

∑j=1

t jy j

)= ψ{y1,··· ,yn} (t) =

∫Rn

eit·xdµ{y1,··· ,yn} (x) . (61.5.16)

Thus if{

y1, · · · ,yn,yn+1, · · · ,yp}⊆ Λ,

ψ

(n

∑j=1

t jy j +p−n

∑j=1

s jy j+n

)= ψ{y1,··· ,yp} (t,s)

=∫Rp−n

eis·x∫Rn

eit·xdµ{y1,··· ,yp} (x)

I need to verify the measures are consistent to use Kolmogorov’s theorem. Specifically, Ineed to show

µ{y1,··· ,yp}(A×Rp−n)= µ{y1,··· ,yn} (A) .

Lettingλ (A) = µ{y1,··· ,yp}

(A×Rp−n)

it follows ∫Rn

eit·xdλ =∫Rp−n

∫Rn

eit·xdµ{y1,··· ,yp} (x)

=∫Rp−n

ei0·x∫Rn

eit·xdµ{y1,··· ,yp} (x)

= ψ

(n

∑j=1

t jy j +p−n

∑j=1

0y j+n

)

= ψ

(n

∑j=1

t jy j

)=

∫Rn

eit·xdµ{y1,··· ,yn} (x)

and so, by uniqueness of characteristic functions,

λ = µ{y1,··· ,yn}

which verifies the necessary consistency condition for Kolmogorov’s theorem.