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In mathematics, Bochner's theorem characterizes the Fourier transform of a positive finite Borel measure on the real line.

Background edit

Given a positive finite Borel measure μ on the real line R, the Fourier transform Q of μ is the continuous function

Q(t)=\int _{\mathbb {R} }e^{-itx}d\mu (x).

Q is continuous since for a fixed x, the function e^-itx is continuous and periodic. The function Q is a positive definite function, i.e. the kernel K(x, y) = Q(y - x) is positive definite, which can be shown via a direct calculation.

The theorem edit

Bochner's theorem says the converse is true, i.e. every positive definite function Q is the Fourier transform of a positive finite Borel measure. A proof can be sketched as follows.

Let F₀(R) be the family of complex valued functions on R with finite support, i.e. f(x) = 0 for all but finitely many x. The positive definite kernel K(x, y) induces a sesquilinear form on F₀(R). This in turn results in a Hilbert space

({\mathcal {H}},\langle \;,\;\rangle )

whose typical element is an equivalence class [g]. For a fixed t in R, the "shift operator" U_t defined by (U_tg)(x) = g(x - t), for a representative of [g] is unitary. ^[1]

In fact the map

t\;{\stackrel {\Phi }{\mapsto }}\;U_{t}

is a strongly continuous representation of the additive group R. ^[2]

By the Stone-von Neumann theorem, there exists a (possibly unbounded) self-adjoint operator A such that

U_{-t}=e^{-iAt}.\;

This implies there exists a finite positive Borel measure μ on R where

\langle U_{-t}[e_{0}],[e_{0}]\rangle =\int e^{-iAt}d\mu (x),

where e₀ is the element in F₀(R) defned by e₀(m) = 1 if m = 0 and 0 otherwise. Because

\langle U_{-t}[e_{0}],[e_{0}]\rangle =K(-t,0)=Q(t),

the theorem holds.

Bochner's theorem can be generalized. Instead of positive definite function Q, one can consider distributions of positive type. Bochner-Schwarz theorem then states that a distribution is of positive type if and only if it is a tempered distribution and the Fourier transform of a positive measure of at most polynomial growth.

Notes edit

^ It is sufficient to verify the unitarity of [f] for f in F₀(R), since such functions are dense in ${\mathcal {H}}$ . If e_n is the element in F₀(R) defned by e_n(m) = δ_nm, then
$\langle U_{t}[e_{n}],U_{t}[e_{m}]\rangle =\langle [e_{n+t}],[e_{m+t}]\rangle =K(m+t,n+t)=K(m,n)=Q_{n-m}.\;$
^ We have
$\langle U_{t}[e_{x}],[e_{x}]\rangle =\langle U_{t}[e_{x}],[e_{x}]\rangle =K(x-t,x)=Q(t).$
Since Q is continuous, Φ is weakly continuous. Since the unitaries lies in the unit ball, Φ is also strongly continuous.

Reference edit

M. Reed and B. Simon, Methods of Modern Mathematical Physics, vol. II, Academic Press, 1975.

[1] It is sufficient to verify the unitarity of [f] for f in F₀(R), since such functions are dense in ${\mathcal {H}}$ . If e_n is the element in F₀(R) defned by e_n(m) = δ_nm, then
$\langle U_{t}[e_{n}],U_{t}[e_{m}]\rangle =\langle [e_{n+t}],[e_{m+t}]\rangle =K(m+t,n+t)=K(m,n)=Q_{n-m}.\;$

[2] We have
$\langle U_{t}[e_{x}],[e_{x}]\rangle =\langle U_{t}[e_{x}],[e_{x}]\rangle =K(x-t,x)=Q(t).$
Since Q is continuous, Φ is weakly continuous. Since the unitaries lies in the unit ball, Φ is also strongly continuous.

[1]

[2]