Gaussian measures and Bochner’s theorem

1 Fourier transforms of measures

Let $m_n$ be normalized Lebesgue measure on ${ℝ}^n$: $d{m}_n(x)={(2\pi )}^{-n/2}dx$. If $\mu$ is a finite positive Borel measure on ${ℝ}^n$, the Fourier transform of $\mu$ is the function $\widehat{\mu }:{ℝ}^n\to ℂ$ defined by

$$\widehat{\mu }(\xi )={\int }_{{ℝ}^n}{e}^{-i\xi \cdot x}𝑑\mu (x),\xi \in {ℝ}^n.$$

One proves using the dominated convergence theorem that $\widehat{\mu }$ is continuous. If $f\in L^1({ℝ}^n)$, the Fourier transform of $f$ is the function $\widehat{f}:{ℝ}^n\to ℂ$ defined by

$$\widehat{f}(\xi )={\int }_{{ℝ}^n}{e}^{-i\xi \cdot x}f(x)𝑑m_n(x),\xi \in {ℝ}^n.$$

Likewise, using the dominated convergence theorem, $\widehat{f}$ is continuous. One proves that if $f\in L^1({ℝ}^n)$ and $\widehat{f}\in L^1({ℝ}^n)$ then, for almost all $x\in {ℝ}^n$,

$$f(x)={\int }_{{ℝ}^n}{e}^{ix\cdot \xi }\widehat{f}(\xi )𝑑m_n(\xi ).$$

As

$$\widehat{\mu }(0)={\int }_{{ℝ}^n}𝑑\mu (x)=\mu ({ℝ}^n),$$

$\mu$ is a probability measure if and only if $\widehat{\mu }(0)=1$. (By a probability measure we mean a positive measure with mass $1$.)

If $\varphi \in L^1({ℝ}^n)$ and $\widehat{\varphi }\in L^1({ℝ}^n)$, then, inverting the Fourier transform,

$⟨\varphi ,\mu ⟩$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}\varphi (x)𝑑\mu (x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}\widehat{\varphi }(\xi )e^{ix\cdot \xi }𝑑m_n(\xi )\right)𝑑\mu (x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\widehat{\varphi }(\xi ){\displaystyle {\int }_{{ℝ}^n}}{e}^{i\xi \cdot x}𝑑\mu (x)𝑑m_n(\xi )$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\widehat{\varphi }(\xi )\widehat{\mu }(-\xi )𝑑m_n(\xi )$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\widehat{\varphi }(-\xi )\widehat{\mu }(\xi )𝑑m_n(\xi ).$

Theorem 1.

If $\mu$ and $\nu$ are finite Borel measures on ${ℝ}^n$ and $\widehat{\mu }=\widehat{\nu }$, then $\mu =\nu$.

Proof.

To prove that $\mu =\nu$ it suffices to prove that for any ball $B$ in ${ℝ}^n$ we have $\mu (B)=\nu (B)$. Let ${\varphi }_n\in C_c^{\mathrm{\infty }}({ℝ}^n)\to {\chi }_B$ pointwise. On the one hand, by the dominated convergence theorem, $⟨{\varphi }_n,\mu ⟩\to \mu (B)$ and $⟨{\varphi }_n,\nu ⟩\to \nu (B)$ as $n\to \mathrm{\infty }$. On the other hand, because $\widehat{\mu }=\widehat{\nu }$ we have

$$⟨{\varphi }_n,\mu ⟩={\int }_{{ℝ}^n}{\widehat{\varphi }}_n(-\xi )\widehat{\mu }(\xi )𝑑m_n(\xi )={\int }_{{ℝ}^n}{\widehat{\varphi }}_n(-\xi )\widehat{\nu }(\xi )𝑑m_n(\xi )=⟨{\varphi }_n,\nu ⟩.$$

Therefore $\mu (B)=\nu (B)$, and it follows that $\mu =\nu$. ∎

2 Gaussian measures

Let ${\lambda }_1,\mathrm{\dots },{\lambda }_n>0$, and let $\mathrm{\Lambda }:{ℝ}^n\to {ℝ}^n$ be the linear map defined by $\mathrm{\Lambda }{e}_i={\lambda }_i{e}_i$. Define

$$d\mu (x)=\sqrt{det\mathrm{\Lambda }}\exp \left(-\frac{1}{2}x\cdot \mathrm{\Lambda }x\right)d{m}_n(x),$$

called a Gaussian measure.

Theorem 2.

$$\widehat{\mu }(\xi )=\exp \left(-\frac{1}{2}\xi \cdot {\mathrm{\Lambda }}^{-1}\xi \right),\xi \in {ℝ}^n.$$

Proof.

We have

$\widehat{\mu }(\xi )$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}{e}^{-i\xi \cdot x}\sqrt{det\mathrm{\Lambda }}\exp \left(-{\displaystyle \frac{1}{2}}x\cdot \mathrm{\Lambda }x\right)𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^d}}{e}^{-i{\xi }_1{x}_1-\mathrm{\cdots }-i{\xi }_n{x}_n}\sqrt{{\lambda }_1\mathrm{\cdots }{\lambda }_n}\exp \left(-{\displaystyle \frac{1}{2}}{\lambda }_1{x}_1^2-\mathrm{\cdots }-{\displaystyle \frac{1}{2}}{\lambda }_n{x}_n^2\right)𝑑m_n(x)$

$=$ ${\displaystyle \prod _{j=1}^n}{I}_j,$

where

$$I_j={\int }_{ℝ}{e}^{-i{\xi }_j{x}_j}\sqrt{{\lambda }_j}\exp \left(-\frac{1}{2}{\lambda }_j{x}_j^2\right)𝑑m_1(x_j).$$

Using

$$-i{\xi }_j{x}_j-\frac{1}{2}{\lambda }_j{x}_j^2=-\frac{{\lambda }_j}{2}\left({\left(x_j+\frac{i{\xi }_j}{{\lambda }_j}\right)}^2+\frac{{\xi }_j^2}{{\lambda }_j^2}\right)=-\frac{{\lambda }_j}{2}{\left(x_j+\frac{i{\xi }_j}{{\lambda }_j}\right)}^2-\frac{{\xi }_j^2}{2{\lambda }_j},$$

we get, doing contour integration,

$I_j$ $=$ ${\displaystyle {\int }_{ℝ}}\sqrt{{\lambda }_j}\exp \left(-{\displaystyle \frac{{\lambda }_j}{2}}{\left(x_j+{\displaystyle \frac{i{\xi }_j}{{\lambda }_j}}\right)}^2\right)\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right)𝑑m_1(x_j)$

$=$ ${\displaystyle {\int }_{ℝ}}\sqrt{{\lambda }_j}\exp \left(-{\displaystyle \frac{{\lambda }_j{x}_j^2}{2}}\right)\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right)𝑑m_1(x_j)$

$=$ ${\displaystyle {\int }_{ℝ}}\sqrt{{\lambda }_j}\exp (-y_j^2)\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right)\sqrt{{\displaystyle \frac{2}{{\lambda }_j}}}𝑑m_1(y_j)$

$=$ $\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right){\displaystyle {\int }_{ℝ}}\sqrt{2}\exp (-y_j^2)𝑑m_1(y_j)$

$=$ $\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right){\displaystyle {\int }_{ℝ}}{\displaystyle \frac{1}{\sqrt{\pi }}}\exp (-y_j^2)𝑑y_j$

$=$ $\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right).$

Therefore, as ${\mathrm{\Lambda }}^{-1}\xi ={\sum }_{j=1}^n\frac{{\xi }_j}{{\lambda }_j}{e}_j$ and $\xi \cdot {\mathrm{\Lambda }}^{-1}\xi ={\sum }_{j=1}^n\frac{{\xi }_j^2}{{\lambda }_j}$,

$\widehat{\mu }(\xi )$ $=$ ${\displaystyle \prod _{j=1}^n}\exp \left(-{\displaystyle \frac{{\xi }_j^2}{2{\lambda }_j}}\right)$

$=$ $\exp \left(-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{j=1}^n}{\displaystyle \frac{{\xi }_j^2}{{\lambda }_j}}\right)$

$=$ $\exp \left(-{\displaystyle \frac{1}{2}}\xi \cdot {\mathrm{\Lambda }}^{-1}\xi \right).$

∎

From the above theorem we get

$$\widehat{\mu }(0)=1,$$

and hence a Gaussian measure is a probability measure.

For $h\in {ℝ}^n$, define $T_h:{ℝ}^n\to {ℝ}^n$ by $T_h(x)=x-h$. If $E$ is a Borel subset of ${ℝ}^n$, because ${\chi }_{T_{-h}(E)}={\chi }_E\circ T_h$,

$$({(T_h)}_{*}\mu )(E)=\mu (T_h^{-1}(E))=\mu (T_{-h}(E))={\int }_{{ℝ}^n}{\chi }_{T_{-h}(E)}𝑑\mu ={\int }_{{ℝ}^n}{\chi }_E\circ T_h𝑑\mu .$$

Then, because $T_h\circ T_{-h}={\mathrm{id}}_{{ℝ}^n}$,

${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E\circ T_h𝑑\mu$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E\circ T_h(x)\sqrt{det\mathrm{\Lambda }}\exp \left(-{\displaystyle \frac{1}{2}}x\cdot \mathrm{\Lambda }x\right)𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E(x)\sqrt{det\mathrm{\Lambda }}\exp \left(-{\displaystyle \frac{1}{2}}(T_{-h}x)\cdot (\mathrm{\Lambda }{T}_{-h}x)\right)d({(T_{-h})}_{*}{m}_n)(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E(x)\sqrt{det\mathrm{\Lambda }}\exp \left(-{\displaystyle \frac{1}{2}}(T_{-h}x)\cdot (\mathrm{\Lambda }{T}_{-h}x)\right)𝑑m_n(x).$

As $\mathrm{\Lambda }$ is self-adjoint $\mathrm{\Lambda }x\cdot h=x\cdot \mathrm{\Lambda }h$,

$(T_{-h}x)\cdot (\mathrm{\Lambda }{T}_{-h}x)$ $=$ $(x+h)\cdot (\mathrm{\Lambda }(x+h))$

$=$ $(x+h)\cdot (\mathrm{\Lambda }x+\mathrm{\Lambda }h)$

$=$ $x\cdot \mathrm{\Lambda }x+x\cdot \mathrm{\Lambda }h+h\cdot \mathrm{\Lambda }x+h\cdot \mathrm{\Lambda }h$

$=$ $x\cdot \mathrm{\Lambda }x+2x\cdot \mathrm{\Lambda }h+h\cdot \mathrm{\Lambda }h.$

Therefore,

$({(T_h)}_{*}\mu )(E)$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E(x)\exp \left(-{\displaystyle \frac{1}{2}}\left(2x\cdot \mathrm{\Lambda }h+h\cdot \mathrm{\Lambda }h\right)\right)𝑑\mu (x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\chi }_E(x)\exp \left(-x\cdot \mathrm{\Lambda }h-{\displaystyle \frac{1}{2}}h\cdot \mathrm{\Lambda }h\right)𝑑\mu (x).$

This shows that the Radon-Nikodym derivative of ${(T_h)}_{*}\mu$ with respect to $\mu$ is

$$\frac{d{(T_h)}_{*}\mu }{d\mu }(x)=\exp \left(-x\cdot \mathrm{\Lambda }h-\frac{1}{2}h\cdot \mathrm{\Lambda }h\right).$$

3 Positive-definite functions

We say that a function $\varphi :{ℝ}^n\to ℂ$ is positive-definite if $x_1,\mathrm{\dots },x_r\in {ℝ}^n$ and $c_1,\mathrm{\dots },c_r\in ℂ$ imply that

$$\sum _{i,j=1}^r{c}_i\overline{c_j}\varphi (x_i-x_j)\ge 0;$$

in particular, the left-hand side is real.

Using $r=1$, $c_1=1$, we have for any $x_1\in {ℝ}^n$ that $\varphi (x_1-x_1)\ge 0$, i.e. $\varphi (0)\ge 0$. For $x\in {ℝ}^n$, using $r=2$, $x_1=x,x_2=0$ and choosing fitting $c_1,c_2\in ℂ$ gives

$$\varphi (-x)=\overline{\varphi (x)},$$

and using this with $c_2=1$ and for appropriate $c_1$ gives

$$|\varphi (x)|\le \varphi (0).$$

For $f,g\in L^1({ℝ}^n)$, the convolution of $f$ and $g$ is the function $f*g:{ℝ}^n\to ℂ$ defined by

$$(f*g)(x)={\int }_{{ℝ}^n}f(y)g(x-y)𝑑m_n(y),x\in {ℝ}^n,$$

and ${\parallel f*g\parallel }_{L^1}\le {\parallel f\parallel }_{L^1}{\parallel g\parallel }_{L^1}$, a case of Young’s inequality. For $f:{ℝ}^n\to ℂ$, we denote by $\mathrm{supp}f$ the essential support of $f$; if $f$ is continuous, then $\mathrm{supp}f$ is the closure of the set $\{x\in {ℝ}^n:f(x)\ne 0\}$. A fact that we will use later is

$$\mathrm{supp}(f*g)\subseteq \overline{\mathrm{supp}f+\mathrm{supp}g}.$$

We denote by $f^{*}$ the function defined by $f^{*}(x)=\overline{f(-x)}$.

$C_c({ℝ}^n)$ is the set of all $f\in C({ℝ}^n)$ for which $\mathrm{supp}f$ is a compact set. The set $C_c({ℝ}^n)$ is dense in the Banach space $C_0({ℝ}^n)$ and also in the Banach space $L^1({ℝ}^n)$; $C_c({ℝ}^n)$ is not a Banach space or even a Fréchet space, and thus does not have a robust structure itself, but is used because it is easier to prove things for it which one then extends in some way to spaces in which the set is dense. The proof of the following theorem follows Folland.

Theorem 3.

If $\varphi :{ℝ}^n\to ℂ$ is positive-definite and continuous and $f\in C_c({ℝ}^n)$, then

$$\int (f^{*}*f)\varphi \ge 0.$$

Proof.

Write $K=\mathrm{supp}f$, and define $F:{ℝ}^n\times {ℝ}^n\to ℂ$ by

$$F(x,y)=f(x)\overline{f(y)}\varphi (x-y).$$

$F$ is continuous, and $\mathrm{supp}F\subseteq K\times K$, hence $\mathrm{supp}F$ is compact. Thus $F\in C_c({ℝ}^n\times {ℝ}^n)$; in particular $F$ is uniformly continuous on $K\times K$, and it follows that for each $ϵ>0$ there is some $\delta >0$ such that if and then . The collection $\{B_{\delta }(x):x\in K\}$ covers $K$ and hence there are finitely many distinct $x_i\in K$ such that the collection $\{B_{\delta }(x_i):i\}$ covers $K$. Then $\{B_{\delta }(x_i)\times B_{\delta }(x_j):i,j\}$ covers $K\times K$. Let $E_i$ be pairwise disjoint, measurable, and satisfy $x_i\in E_i\subseteq B_{\delta }(x_i)$. The collection $\{E_i:i,\}$ covers $K$, so the collection $\{E_i\times E_j:i,j\}$ covers $K\times K$.

Define

$$R=\sum _{i,j}{\int }_{E_i\times E_j}(F(x,y)-F(x_i,x_j))𝑑m_n(x)𝑑m_n(y).$$

$R$ satisfies

$|R|$ $\le$ ${\displaystyle \sum _{i,j}}{\displaystyle {\int }_{E_i\times E_j}}|F(x,y)-F(x_i,x_j)|𝑑m_n(x)𝑑m_n(y)$

$\le$ ${\displaystyle \sum _{i,j}}{\displaystyle {\int }_{E_i\times E_j}}ϵ𝑑m_n(x)𝑑m_n(y)$

$=$ $ϵ{\displaystyle \sum _{i,j}}{m}_n(E_i)m_n(E_j)$

$=$ $ϵm_n{(K)}^2.$

We obtain

${\displaystyle {\int }_{K\times K}}F(x,y)𝑑m_n(x)𝑑m_n(y)$ $=$ ${\displaystyle \sum _{i,j}}{\displaystyle {\int }_{E_i\times E_j}}F(x,y)𝑑m_n(x)𝑑m_n(y)$

$=$ ${\displaystyle \sum _{i,j}}F(x_i,x_j)m_n(E_i)m_n(E_j)+R$

$=$ ${\displaystyle \sum _{i,j}}f(x_i)\overline{f(x_j)}\varphi (x_i-x_j)m_n(E_i)m_n(E_j)+R.$

Using $c_i=f(x_i)m_n(E_i)$, the fact that $\varphi$ is positive-definite means that the sum is $\ge 0$. Therefore

$${\int }_{K\times K}F(x,y)𝑑m_n(x)𝑑m_n(y)\ge -|R|\ge -ϵm_n{(K)}^2.$$

This is true for all $ϵ>0$, hence

$${\int }_{{ℝ}^n}{\int }_{{ℝ}^n}f(x)\overline{f(y)}\varphi (x-y)𝑑m_n(x)𝑑m_n(y)={\int }_{K\times K}F(x,y)𝑑m_n(x)𝑑m_n(y)\ge 0.$$

But

${\displaystyle {\int }_{{ℝ}^n}}(f^{*}*f)(x)\varphi (x)𝑑m_n(x)$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}{f}^{*}(y)f(x-y)𝑑m_n(y)\right)\varphi (x)𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}\overline{f(-y)}f(x-y)\varphi (x)𝑑m_n(x)𝑑m_n(y)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}\overline{f(-y)}f(x)\varphi (x+y)𝑑m_n(x)𝑑m_n(y)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}\overline{f(y)}f(x)\varphi (x-y)𝑑m_n(x)𝑑m_n(y).$

∎

Corollary 4.

If $\varphi :{ℝ}^n\to ℂ$ is positive-definite and continuous and $f\in L^1({ℝ}^n)$, then

$$\int (f^{*}*f)\varphi \ge 0.$$

Proof.

Let $f_n\in C_c({ℝ}^n)$ converge to $f$ in $L^1({ℝ}^n)$ as $n\to \mathrm{\infty }$; that there is such a sequence is given to us by the fact that $C_c({ℝ}^n)$ is a dense subset of $L^1({ℝ}^n)$. Using

$f_n^{*}*f_n-f^{*}*f$ $=$ $f_n^{*}*f_n-f_n^{*}*f+f_n^{*}*f-f^{*}*f$

$=$ $f_n^{*}*(f_n-f)+(f_n^{*}-f^{*})*f$

$=$ $f_n^{*}*(f_n-f)+{(f_n-f)}^{*}*f,$

and ${\parallel g^{*}\parallel }_{L^1}={\parallel g\parallel }_{L^1}$, we get

${\parallel f_n^{*}*f_n-f^{*}*f\parallel }_{L^1}$ $\le$ ${\parallel f_n^{*}*(f_n-f)\parallel }_{L^1}+{\parallel {(f_n-f)}^{*}*f\parallel }_{L^1}$

$\le$ ${\parallel f_n^{*}\parallel }_{L^1}{\parallel f_n-f\parallel }_{L^1}+{\parallel {(f_n-f)}^{*}\parallel }_{L^1}{\parallel f\parallel }_{L^1}$

$=$ ${\parallel f_n\parallel }_{L^1}{\parallel f_n-f\parallel }_{L^1}+{\parallel f_n-f\parallel }_{L^1}{\parallel f\parallel }_{L^1},$

which converges to $0$ because ${\parallel f_n-f\parallel }_{L^1}\to 0$. Therefore, because $\varphi$ is bounded,

$${\int }_{{ℝ}^n}(f_n^{*}*f_n)\varphi 𝑑m_n\to {\int }_{{ℝ}^n}(f^{*}*f)\varphi 𝑑m_n.$$

As ${\int }_{{ℝ}^n}(f_n^{*}*f_n)\varphi 𝑑m_n\ge 0$ for each $n$, this implies that ${\int }_{{ℝ}^n}(f^{*}*f)\varphi 𝑑m_n\ge 0$. ∎

It is straightforward to prove that the Fourier transform of a finite positive Borel measure is a positive-definite function; one ends up with the expression

$${\int }_{{ℝ}^n}{\left|\sum _{j=1}^n{c}_j{e}^{i{\xi }_j\cdot x}\right|}^2𝑑\mu (x),$$

which is finite and nonnegative because $\mu$ is finite and positive respectively. We have established already that the Fourier transform of a finite positive Borel measure $\mu$ on ${ℝ}^n$ is continuous and satisfies $\widehat{\mu }(0)=1$. Bochner’s theorem is the statement that a function with these three properties is indeed the Fourier transform of a finite positive Borel measure. Our proof of the following theorem follows Folland.

Theorem 5 (Bochner).

If $\varphi :{ℝ}^n\to ℂ$ is positive-definite, continuous, and satisfies $\varphi (0)=1$, then there is some Borel probability measure $\mu$ on ${ℝ}^n$ such that $\varphi =\widehat{\mu }$.

Proof.

Let $\{{\psi }_U\}$ be an approximate identity. That is, for each neighborhood $U$ of $0$, ${\psi }_U$ is a function such that $\mathrm{supp}{\psi }_U$ is compact and contained in $U$, $\psi \ge 0$, ${\psi }_U(-x)={\psi }_U(x)$, and ${\int }_{{ℝ}^n}{\psi }_U𝑑m_n=1$. For every $f\in L^1({ℝ}^n)$, an approximate identity satisfies ${\parallel f*{\psi }_U-f\parallel }_{L^1}\to 0$ as $U\to \{0\}$.

We have ${\psi }_U^{*}={\psi }_{-U}$, so

$$\mathrm{supp}({\psi }_U^{*}*{\psi }_U)\subseteq \overline{\mathrm{supp}{\psi }_{-U}+\mathrm{supp}{\psi }_U}=\mathrm{supp}{\psi }_{-U}+\mathrm{supp}{\psi }_U\subseteq -U+U,$$

and as always, ${\int }_{{ℝ}^n}f*g𝑑m_n={\int }_{{ℝ}^n}f𝑑m_n{\int }_{{ℝ}^n}g𝑑m_n$. Therefore $\{{\psi }_U^{*}*{\psi }_U\}$ is an approximate identity:

For $f,g\in L^1({ℝ}^n)$, define

$${⟨f,g⟩}_{\varphi }={\int }_{{ℝ}^n}(g^{*}*f)\varphi 𝑑m_n.$$

One checks that this is a positive Hermitian form; positive means that ${⟨f,f⟩}_{\varphi }\ge 0$ for all $f\in L^1({ℝ}^n)$, and this is given to us by Corollary 4. Using the Cauchy-Schwarz inequality,

$${|{⟨f,g⟩}_{\varphi }|}^2\le {⟨f,f⟩}_{\varphi }{⟨g,g⟩}_{\varphi }.$$

We have laid out the tools that we will use. Let $f\in L^1({ℝ}^n)$. ${\psi }_U*f\to f$ in $L^1$ as $U\to \{0\}$, and as $\varphi$ is bounded this gives ${\int }_{{ℝ}^n}({\psi }_U^{*}*f)\varphi 𝑑m_n\to {\int }_{{ℝ}^n}f\varphi 𝑑m_n$ as $U\to \{0\}$. Because $\{{\psi }_U^{*}*{\psi }_U\}$ is an approximate identity, ${\int }_{{ℝ}^n}({\psi }_U^{*}*{\psi }_U)\varphi 𝑑m_n\to \varphi (0)$ as $U\to \{0\}$. That is, we have ${⟨f,{\psi }_U⟩}_{\varphi }\to {\int }_{{ℝ}^n}f\varphi 𝑑m_n$ and ${⟨{\psi }_U,{\psi }_U⟩}_{\varphi }\to \varphi (0$ as $U\to \{0\}$, and as $\varphi (0)=1$, the above statemtn of the Cauchy-Schwarz inequality produces

$${\left|{\int }_{{ℝ}^n}f\varphi 𝑑m_n\right|}^2\le {\int }_{{ℝ}^n}(f^{*}*f)\varphi 𝑑m_n.$$ (1)

With $h=f^{*}*f$, the inequality (1) reads

$${\left|{\int }_{{ℝ}^n}f\varphi 𝑑m_n\right|}^2\le {\int }_{{ℝ}^n}h\varphi 𝑑m_n.$$

Defining $h^{(1)}=h$, $h^{(2)}=h*h$, $h^{(3)}=h*h*h$, etc., applying (1) to $h$ gives, because $h^{*}=h$,

$${\left|{\int }_{{ℝ}^n}h\varphi 𝑑m_n\right|}^2\le {\int }_{{ℝ}^n}{h}^{(2)}\varphi 𝑑m_n.$$

Then applying (1) to $h^{(2)}$, which satisfies ${(h^{(2)})}^{*}=h^{(2)}$,

$${\left|{\int }_{{ℝ}^n}{h}^{(2)}\varphi 𝑑m_n\right|}^2\le {\int }_{{ℝ}^n}{h}^{(4)}\varphi 𝑑m_n.$$

Thus, for any $m\ge 0$ we have

$\left|{\displaystyle {\int }_{{ℝ}^n}}f\varphi 𝑑m_n\right|$ $\le$ ${\left|{\displaystyle {\int }_{{ℝ}^n}}{h}^{\left(2^m\right)}\varphi 𝑑m_n\right|}^{2^{-(m+1)}}$

$\le$ ${\parallel h^{\left(2^m\right)}\parallel }_{L^1}^{2^{-(m+1)}}$

$=$ ${\left({\parallel h^{\left(2^m\right)}\parallel }_{L^1}^{2^{-m}}\right)}^{1/2},$

since ${\parallel \varphi \parallel }_{\mathrm{\infty }}=\varphi (0)=1$.

With convolution as multiplication, $L^1({ℝ}^n)$ is a commutative Banach algebra, and the Gelfand transform is an algebra homomorphism $L^1({ℝ}^n)\to C_0({ℝ}^n)$ that satisfies

$${\parallel \widehat{g}\parallel }_{\mathrm{\infty }}=\underset{k\to \mathrm{\infty }}{lim}{\parallel g^{(k)}\parallel }_{L^1}^{1/k},g\in L^1({ℝ}^n);$$

for $L^1({ℝ}^n)$, the Gelfand transform is the Fourier transform. Write the Fourier transform as $ℱ:L^1({ℝ}^n)\to C_0({ℝ}^n)$. Stating that the Gelfand transform is a homomorphism means that $ℱ(g_1*g_2)=ℱ(g_1)ℱ(g_2)$, because multiplication in the Banach algebra $C_0({ℝ}^n)$ is pointwise multiplication. Then, since a subsequence of a convergent sequence converges to the same limit,

$$\underset{m\to \mathrm{\infty }}{lim}{\left({\parallel h^{\left(2^m\right)}\parallel }_{L^1}^{2^{-m}}\right)}^{1/2}={\left({\parallel \widehat{h}\parallel }_{\mathrm{\infty }}\right)}^{1/2}.$$

But

$$\widehat{h}=ℱ(f^{*}*f)=ℱ(f^{*})ℱ(f)=\overline{ℱ(f)}ℱ(f)={\left|ℱ(f)\right|}^2,$$

so

$${\left({\parallel \widehat{h}\parallel }_{\mathrm{\infty }}\right)}^{1/2}={\left({\parallel {|\widehat{f}|}^2\parallel }_{\mathrm{\infty }}\right)}^{1/2}={\parallel \widehat{f}\parallel }_{\mathrm{\infty }}.$$

Putting things together, we have that for any $f\in L^1({ℝ}^n)$,

$$\left|{\int }_{{ℝ}^n}f\varphi 𝑑m_n\right|\le {\parallel \widehat{f}\parallel }_{\mathrm{\infty }}.$$

Therefore $\widehat{f}\mapsto {\int }_{{ℝ}^n}f\varphi 𝑑m_n$ is a bounded linear functional $ℱ(L^1({ℝ}^n))\to ℂ$, of norm $\le 1$. Using $\varphi (0)=1$, one proves that this functional has norm $1$. (If we could apply this inequality to $ℱ(\delta )$ the two sides would be equal, thus to prove that the operator norm is 1, one applies the inequality to a sequence of functions that converge weakly to $\delta$.) We take as known that $ℱ(L^1({ℝ}^n))$ is dense in the Banach space $C_0({ℝ}^n)$, so there is a bounded linear functional $\mathrm{\Phi }:C_0({ℝ}^n)\to ℂ$ whose restriction to $ℱ(L^1({ℝ}^n))$ is equal to $\widehat{f}\mapsto {\int }_{{ℝ}^n}f\varphi 𝑑m_n$, and $\parallel \mathrm{\Phi }\parallel =1$.

Using the Riesz-Markov theorem, there is a regular complex Borel measure $\mu$ on ${ℝ}^n$ such that

$$\mathrm{\Phi }(g)={\int }_{{ℝ}^n}g𝑑\mu ,g\in C_0({ℝ}^n),$$

and $\parallel \mu \parallel =\parallel \mathrm{\Phi }\parallel$; $\parallel \mu \parallel$ is the total variation norm of $\mu$, $\parallel \mu \parallel =|\mu |({ℝ}^n)$. Then for $f\in L^1({ℝ}^n)$ we have

${\displaystyle {\int }_{{ℝ}^n}}f\varphi 𝑑m_n$ $=$ $\mathrm{\Phi }(\widehat{f})$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\widehat{f}𝑑\mu$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}{e}^{-i\xi \cdot x}f(x)𝑑m_n(x)\right)𝑑\mu (\xi )$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}f(x)\left({\displaystyle {\int }_{{ℝ}^n}}{e}^{-ix\cdot \xi }𝑑\mu (\xi )\right)𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}f(x)\widehat{\mu }(x)𝑑m_n(x).$

That this is true for all $f\in L^1({ℝ}^n)$ implies that $\varphi =\widehat{\mu }$. As $\mu ({ℝ}^n)=\widehat{\mu }(0)=\varphi (0)=1$ and $\parallel \mu \parallel =\parallel \mathrm{\Phi }\parallel =1$ we have $\mu ({ℝ}^n)=\parallel \mu \parallel$, and this implies that $\mu$ is positive measure, hence, as $\mu ({ℝ}^n)=1$, a probability measure.

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