The Gelfand transform, positive linear functionals, and positive-definite functions

1 Introduction

In this note, unless we say otherwise every vector space or algebra we speak about is over $ℂ$.

If $A$ is a Banach algebra and $e\in A$ satisfies $xe=x$ and $ex=x$ for all $x\in A$, and also $\parallel e\parallel =1$, we say that $e$ is unity and that $A$ is unital.

If $A$ is a unital Banach algebra and $x\in A$, the spectrum of $x$ is the set $\sigma (x)$ of those $\lambda \in ℂ$ for which $\lambda e-x$ is not invertible. It is a fact that if $A$ is a unital Banach algebra and $x\in A$, then $\sigma (x)\ne \mathrm{\varnothing }$.

If $A$ and $B$ are Banach algebras and $T:A\to B$ is a map, we say that $T$ is an isomorphism of Banach algebras if $T$ is an algebra isomorphism and an isometry.

Theorem 1 (Gelfand-Mazur).

If $A$ is a Banach algebra and every nonzero element of $A$ is invertible, then there is an isomorphism of Banach algebras $A\to ℂ$.

Proof.

Let $x\in A$. $\sigma (x)\ne \mathrm{\varnothing }$. If ${\lambda }_1,{\lambda }_2\in \sigma (x)$, then neither ${\lambda }_{1}e-x$ nor ${\lambda }_{2}e-x$ is invertible, so they are both $0$: $x={\lambda }_{1}e$ and $x={\lambda }_{2}e$, whence ${\lambda }_1={\lambda }_2$. Therefore $\sigma (x)$ has precisely one element, which we denote by $\lambda (x)$, and which satisfies

$$x=\lambda (x)e.$$

If $x,y\in A$, then $x+y=\lambda (x)e+\lambda (y)e=(\lambda (x)+\lambda (y))e$ and also $x+y=\lambda (x+y)e$, so $\lambda (x+y)=\lambda (x)+\lambda (y)$. If $x\in A$ and $\alpha \in ℂ$, then $\alpha x=\alpha \lambda (x)e$ and also $\alpha x=\lambda (\alpha x)e$, so $\lambda (\alpha x)=\alpha \lambda (x)$. Hence $x\mapsto \lambda (x)$ is linear. If ${\lambda }_0\in ℂ$, then $\lambda ({\lambda }_{0}e)={\lambda }_0$, showing that $x\mapsto \lambda (x)$ is onto. If $\lambda (x)=\lambda (y)$ then $x=\lambda (x)e=\lambda (y)e=y$, showing that $x\mapsto \lambda (x)$ is one-to-one. Therefore $x\mapsto \lambda (x)$ is a linear isomorphism $A\to ℂ$.

If $x\in A$, then $x=\lambda (x)e$ gives

$$\parallel x\parallel =\parallel \lambda (x)e\parallel =|\lambda (x)|\parallel e\parallel =|\lambda (x)|,$$

showing that the map $x\mapsto \lambda (x)$ is an isometry $A\to ℂ$. ∎

2 Complex homomorphisms

An ideal $J$ of an algebra $A$ is said to be proper if $J\ne A$. An ideal is called maximal if it is a maximal element in the collection of proper ideals of $A$ ordered by set inclusion.

The following theorem, which is proved using the fact that a maximal ideal is closed, the fact that a quotient of a Banach algebra with a closed ideal is a Banach algebra, and the Gelfand-Mazur theorem, states some basic facts about algebra homomorphisms from a Banach algebra to $ℂ$.

Theorem 2.

If $A$ is a commutative unital Banach algebra and $\mathrm{\Delta }$ is the set of all nonzero algebra homomorphisms $A\to ℂ$, then:

1. If $M$ is a maximal ideal of $A$ then there is some $h\in \mathrm{\Delta }$ for which $M=\ker h$.

2. If $h\in \mathrm{\Delta }$ then $\ker h$ is a maximal ideal of $A$.

3. $x\in A$ is invertible if and only if $h(x)\ne 0$ for all $h\in \mathrm{\Delta }$.

4. $x\in A$ is invertible if and only if $x$ does not belong to any proper ideal of $A$.

5. $\lambda \in \sigma (x)$ if and only if there is some $h\in \mathrm{\Delta }$ for which $h(x)=\lambda$.

3 The Gelfand transform and maximal ideals

Suppose that $A$ is a commutative unital Banach algebra and that $\mathrm{\Delta }$ is the set of all nonzero algebra homomorphisms $A\to ℂ$. For each $x\in A$, we define $\widehat{x}:\mathrm{\Delta }\to ℂ$ by

$$\widehat{x}(h)=h(x),h\in \mathrm{\Delta }.$$

We call $\widehat{x}$ the Gelfand transform of $x$, and we call the map $\mathrm{\Gamma }:A\to {ℂ}^{\mathrm{\Delta }}$ defined by $\mathrm{\Gamma }(x)=\widehat{x}$ the Gelfand transform.

We define $\widehat{A}=\{\widehat{x}:x\in A\}$, and we call the set $\mathrm{\Delta }$ with the initial topology for $\widehat{A}$ the maximal ideal space of $A$. That is, the topology of $\mathrm{\Delta }$ is the coarsest topology on $\mathrm{\Delta }$ such that each $\widehat{x}:\mathrm{\Delta }\to ℂ$ is continuous. If $X$ is a topological space, we denote by $C(X)$ the set of all continuous functions $X\to ℂ$. $C(X)$ is a commutative unital algebra, although it need not be a Banach algebra.

The radical of $A$, denoted $\mathrm{rad}A$, is the intersection of all maximal ideals of $A$. If $\mathrm{rad}A=\{0\}$, we say that $A$ is semisimple.

The following theorem establishes some basic facts about the Gelfand transform and the maximal ideal space.

Theorem 3.

If $A$ is a commutative unital Banach algebra and $\mathrm{\Delta }$ is the maximal ideal space of $A$, then:

1. $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an algebra homomorphism with $\ker \mathrm{\Gamma }=\mathrm{rad}A$.

2. If $x\in A$, then $\mathrm{im}\widehat{x}=\sigma (x)$.

3. $\mathrm{\Delta }$ is a compact Hausdorff space.

Proof.

Let $x,y\in A$ and $\alpha \in ℂ$. For $h\in \mathrm{\Delta }$,

$$\mathrm{\Gamma }(\alpha x+y)(h)=h(\alpha x+y)=\alpha h(x)+h(y)=\alpha \mathrm{\Gamma }(x)(h)+\mathrm{\Gamma }(y)(h)=(\mathrm{\Gamma }(x)+\mathrm{\Gamma }(y)(h),$$

showing that $\mathrm{\Gamma }(\alpha x+y)=\alpha \mathrm{\Gamma }(x)+\mathrm{\Gamma }(y)$, and

$$\mathrm{\Gamma }(xy)(h)=h(xy)=h(x)h(y)=\mathrm{\Gamma }(x)(h)\mathrm{\Gamma }(y)(h)=(\mathrm{\Gamma }(x)\mathrm{\Gamma }(y))(h),$$

showing that $\mathrm{\Gamma }(xy)=\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)$. Therefore $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an algebra homomorphism. $x\in \ker \mathrm{\Gamma }$ is equivalent to $h(x)=0$ for all $h\in \mathrm{\Delta }$, which is equivalent to $x\in \ker h$ for all $h\in \mathrm{\Delta }$. But by Theorem 2, $\{\ker h:h\in \mathrm{\Delta }\}$ is equal to the set of all maximal ideals of $A$, so $x\in \ker \mathrm{\Gamma }$ is equivalent to $x\in \mathrm{rad}A$, i.e. $\ker \mathrm{\Gamma }=\mathrm{rad}A$.

Let $x\in A$. If $\lambda \in \mathrm{im}\widehat{x}$ then there is some $h\in \mathrm{\Delta }$ for which $\widehat{x}(h)=\lambda$, and by Theorem 2, this yields $\lambda \in \sigma (x)$. Hence $\mathrm{im}\widehat{x}\subseteq \sigma (x)$. If $\lambda \in \sigma (x)$, then by Theorem 2 there is some $h\in \mathrm{\Delta }$ for which $h(x)=\lambda$, i.e. there is some $h\in \mathrm{\Delta }$ for which $\widehat{x}(h)=\lambda$, i.e. $\lambda \in \mathrm{im}\widehat{x}$. Hence $\sigma (x)\subseteq \mathrm{im}\widehat{x}$. Therefore, $\mathrm{im}\widehat{x}=\sigma (x)$.

It is straightforward to check that the topology of $\mathrm{\Delta }$ is the subspace topology inherited from $A^{*}$ with the weak-* topology; in particular, the topology of $\mathrm{\Delta }$ is Hausdorff. Therefore, to prove that $\mathrm{\Delta }$ is compact it suffices to prove that $\mathrm{\Delta }$ is a weak-* compact subset of $A^{*}$. Let

$$K=\{\lambda \in A^{*}:\parallel \lambda \parallel \le 1\}.$$

By the Banach-Alaoglu theorem, $K$ is a weak-* compact subset of $A^{*}$. If $h\in \mathrm{\Delta }$, then because $h$ is an algebra homomorphism $A\to ℂ$ it follows that $\parallel h\parallel \le 1$. Thus, $\mathrm{\Delta }\subset K$. Therefore, to prove that $\mathrm{\Delta }$ is compact it suffices to prove that $\mathrm{\Delta }$ is a weak-* closed subset of $A^{*}$.

Suppose that $h_i\in \mathrm{\Delta }$ is a net that weak-* converges to $\lambda \in A^{*}$. Then $h_i(e)\to \lambda (e)$, i.e. $1\to \lambda (e)$, so $\lambda (e)=1$. Thus $\lambda \ne 0$. Let $x,y\in A$. On the one hand, $h_i(xy)\to \lambda (xy)$, and on the other hand, $h_i(x)\to \lambda (x)$ and $h_i(y)\to \lambda (y)$, so $h_i(x)h_i(y)\to \lambda (x)\lambda (y)$ and hence $h_i(xy)=h_i(x)h_i(y)\to \lambda (x)\lambda (y)$. Therefore, $\lambda (xy)=\lambda (x)\lambda (y)$, and because $\lambda \in A^{*}$ is linear, this shows that $\lambda :A\to ℂ$ is an algebra homomorphism, and hence that $\lambda \in \mathrm{\Delta }$. Therefore $\mathrm{\Delta }$ is a weak-* closed subset of $A^{*}$. ∎

If $A$ is a commutative unital Banach algebra, the above theorem shows that $\mathrm{\Gamma }:A\to \widehat{A}$ is an algebra isomorphism if and only if $\mathrm{rad}A=\{0\}$, i.e., $\mathrm{\Gamma }$ is an algebra isomorphism if and only if $A$ is semisimple.

The above theorem tells us that if $A$ is a commutative unital Banach algebra and $x\in A$, then $\mathrm{im}\widehat{x}=\sigma (x)$. This gives us

$${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}=\rho (x),$$ (1)

where $\rho (x)$ is the spectral radius of $x$, defined by

$$\rho (x)=sup\{|\lambda |:\lambda \in \sigma (x)\}.$$

Therefore, $\widehat{x}=0$ is equivalent to $\rho (x)=0$, and so by the above theorem, $x\in \mathrm{rad}A$ is equivalent to $\rho (x)=0$. Moreover, it is a fact that $\rho (x)\le \parallel x\parallel$. Therefore,

$${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}\le \parallel x\parallel .$$ (2)

In the proof of Theorem 3 we used the fact that the norm of any algebra homomorphism from a Banach algebra to $ℂ$ is $\le 1$. In particular, this means that any algebra homomorphism from a Banach algebra to $ℂ$ is continuous. The following theorem shows that any algebra homomorphism from a Banach algebra to a commutative unital semisimple Banach algebra is continuous.

Theorem 4.

Suppose that $A$ is a Banach algebra and that $B$ is a commutative unital semisimple Banach algebra. If $\psi :A\to B$ is an algebra homomorphism, then $\psi$ is continuous.

Proof.

Because $\psi :A\to B$ is linear, to prove that $\psi$ is continuous, by the closed graph theorem it suffices to prove that

$$G=\{(x,\psi (x)):x\in A\}$$

is closed in $A\times B$. To prove that $G$ is closed in $A\times B$, it suffices to prove that if $(x_n,y_n)\in G$ converges to $(x,y)\in A\times B$ then $(x,y)\in G$.

Let $h\in {\mathrm{\Delta }}_B$. Then $\varphi =h\circ \psi :A\to ℂ$ is an algebra homomorphism. Because $h:B\to ℂ$ and $\varphi :A\to ℂ$ are algebra homomorphisms with codomain $ℂ$, they are both continuous. Therefore, $h(y_n)\to h(y)$ and $\varphi (x_n)\to \varphi (x)$. Therefore,

$$h(y)=limh(y_n)=limh(\psi (x_n))=lim(h\circ \psi )(x_n)=lim\varphi (x_n)=\varphi (x)=h(\psi (x)),$$

so $h(y-\psi (x))=0$. This is true for all $h\in {\mathrm{\Delta }}_B$, hence $y-\psi (x)\in \mathrm{rad}B$. But $B$ is semisimple, so $y-\psi (x)=0$, i.e. $y=\psi (x)$, so $(x,y)\in G$. ∎

If $A$ is a commutative unital Banach algebra and $x\in A$, we recorded in (2) that ${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}\le \parallel x\parallel$. The following lemma shows that if $\parallel x^2\parallel ={\parallel x\parallel }^2$ and $x\ne 0$, then $inf\frac{{\parallel \widehat{x}\parallel }_{\mathrm{\infty }}}{\parallel x\parallel }\ge 1$, hence that ${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}=\parallel x\parallel$. Therefore, if $\parallel x^2\parallel ={\parallel x\parallel }^2$ for all $x\in A$, then $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an isometry.

Lemma 5.

Let $A$ be a commutative unital Banach algebra. If

$$r=\underset{x\ne 0}{inf}\frac{\parallel x^2\parallel }{{\parallel x\parallel }^2},s=\underset{x\ne 0}{inf}\frac{{\parallel \widehat{x}\parallel }_{\mathrm{\infty }}}{\parallel x\parallel },$$

then $s^2\le r\le s$.

Theorem 3 shows that if $A$ is a commutative unital Banach algebra, then $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an algebra homomorphism. Therefore $\mathrm{\Gamma }(A)=\widehat{A}$ is a subalgebra of $C(\mathrm{\Delta })$. Moreover, Theorem 3 also shows that $\mathrm{\Delta }$ is a compact Hausdorff space. Therefore, $C(\mathrm{\Delta })$ is a unital Banach algebra with the supremum norm. (If $X$ is a topological space then $C(X)$ is an algebra, but need not be a Banach algebra.) For $\widehat{A}$ to be a Banach subalgebra of $C(\mathrm{\Delta })$ it is necessary and sufficient that $\widehat{A}$ be a closed subset of the Banach algebra $C(\mathrm{\Delta })$. The following theorem gives conditions under which this occurs.

Theorem 6.

If $A$ is a commutative unital Banach algebra, then $A$ is semisimple and $\widehat{A}$ is a closed subset of $C(\mathrm{\Delta })$ if and only if there exists some such that ${\parallel x\parallel }^2\le K\parallel x^2\parallel$ for all $x\in A$.

Proof.

Suppose that there is some such that $x\in A$ implies that ${\parallel x\parallel }^2\le K\parallel x^2\parallel$. Then

$$r=\underset{x\ne 0}{inf}\frac{\parallel x^2\parallel }{{\parallel x\parallel }^2}\ge \underset{x\ne 0}{inf}\frac{\parallel x^2\parallel }{K\parallel x^2\parallel }=\frac{1}{K}.$$

By Lemma 5, with $s={inf}_{x\ne 0}\frac{{\parallel \widehat{x}\parallel }_{\mathrm{\infty }}}{\parallel x\parallel }$ we have

$$\frac{1}{K}\le s,$$

hence ${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}\ge \frac{1}{K}\parallel x\parallel$. Thus, if $x\in A$ then ${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}\ge \frac{1}{K}\parallel x\parallel$, from which it follows that $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is one-to-one. Since $\mathrm{\Gamma }$ is one-to-one, by Theorem 3 we get that $A$ is semisimple. Suppose that ${\widehat{x}}_n\in \widehat{A}$ converges to $\widehat{x}\in \widehat{A}$, i.e. ${\parallel {\widehat{x}}_n-\widehat{x}\parallel }_{\mathrm{\infty }}\to 0$, i.e. ${\parallel \mathrm{\Gamma }(x_n-x)\parallel }_{\mathrm{\infty }}\to 0$. But ${\parallel \mathrm{\Gamma }(x_n-x)\parallel }_{\mathrm{\infty }}\ge \frac{1}{K}\parallel x_n-x\parallel$, so $\parallel x_n-x\parallel \to 0$, showing that ${\mathrm{\Gamma }}^{-1}:\widehat{A}\to A$ is bounded. Therefore $\mathrm{\Gamma }:A\to \widehat{A}$ is bilipschitz, and so $\widehat{A}$ is a complete metric space, from which it follows that $\widehat{A}$ is a closed subset of $C(\mathrm{\Delta })$.

Suppose that $A$ is semisimple and that $\widehat{A}$ is a closed subset of $C(\mathrm{\Delta })$. The fact that $A$ is semisimple gives us by Theorem 3 that $\mathrm{\Gamma }:A\to \widehat{A}$ is a bijection. The fact that $\widehat{A}$ is closed means that $\widehat{A}$ is a Banach algebra. Because $\mathrm{\Gamma }:A\to \widehat{A}$ is continuous, linear, and a bijection, by the open mapping theorem it follows that there are positive real numbers $a,b$ such that if $x\in A$ then

$$a\parallel x\parallel \le {\parallel \mathrm{\Gamma }x\parallel }_{\mathrm{\infty }}\le b\parallel x\parallel .$$

Then ${inf}_{x\ne 0}\frac{{\parallel \widehat{x}\parallel }_{\mathrm{\infty }}}{\parallel x\parallel }\ge a$. By Lemma 5, it follows that ${inf}_{x\ne 0}\frac{\parallel x^2\parallel }{{\parallel x\parallel }^2}\ge a^2$. Hence, for all $x\ne 0$ we have ${\parallel x\parallel }^2\le K\parallel x^2\parallel$, with $K=\frac{1}{a^2}$. ∎

4 L¹

Let $M({ℝ}^n)$ denote the set of all complex Borel measures on ${ℝ}^n$, and let $S:{ℝ}^n\times {ℝ}^n\to {ℝ}^n$ be defined by $S(x,y)=x+y$. For ${\mu }_1,{\mu }_2\in M({ℝ}^n)$, we denote by ${\mu }_1\times {\mu }_2$ the product measure on ${ℝ}^n\times {ℝ}^n$, and we define the convolution of ${\mu }_{\mathrm{1}}$ and ${\mu }_{\mathrm{2}}$ to be ${\mu }_1*{\mu }_2=S_{*}({\mu }_1\times {\mu }_2)$, the pushforward of ${\mu }_1\times {\mu }_2$ with respect to $S$. That is, if $E$ is a Borel subset of ${ℝ}^n$, then

$({\mu }_1*{\mu }_2)(E)$ $=$ $(S_{*}({\mu }_1\times {\mu }_2))(E)$

$=$ $({\mu }_1\times {\mu }_2)(S^{-1}(E))$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}{\chi }_E(x+y)𝑑{\mu }_1(x)𝑑{\mu }_2(y).$

With convolution as multiplication, $M({ℝ}^n)$ is an algebra.

If $\mu \in M({ℝ}^n)$, the variation of $\mu$ is the measure $|\mu |\in M({ℝ}^n)$, where for a Borel subset $E$ of ${ℝ}^n$, we define $|\mu |(E)$ to be the supremum of ${\sum }_{A\in \pi }|\mu (A)|$ over all partitions $\pi$ of $E$ into finitely many disjoint Borel subsets. The total variation of $\mu$ is $\parallel \mu \parallel =|\mu |({ℝ}^n)$. One proves that $\parallel \cdot \parallel$ is a norm on $M({ℝ}^n)$ and that with this norm, $M({ℝ}^n)$ is a Banach algebra.

Let $m_n$ be Lebesgue measure on ${ℝ}^n$, let $\delta$ be the Dirac measure on ${ℝ}^n$, and let $A$ be the set of those $\mu \in M({ℝ}^n)$ for which there is some $f\in L^1({ℝ}^n)$ and some $\alpha \in ℂ$ with which

$$d\mu =fd{m}_n+\alpha d\delta .$$

One proves that $A$ is a Banach subalgebra of $M({ℝ}^n)$. $A$ is a unital Banach algebra, with unity $\delta$. In particular, $A$ is a unital Banach algebra that contains the Banach algebra $L^1({ℝ}^n)$.

If $f+\alpha \delta ,g+\beta \delta \in A$ (identifying $f\in L^1({ℝ}^n)$ with the complex Borel measure whose Radon-Nikodym derivative with respect to $m_n$ is $f$), then

$$(f+\alpha \delta )*(g+\beta \delta )=(f*g+\beta f+\alpha g)+\alpha \beta \delta ,$$ (3)

where

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

If $t\in {ℝ}^n$, let $e_t(x)=\exp (it\cdot x)$, and if $f\in L^1({ℝ}^n)$, define $\widehat{f}:{ℝ}^n\to ℂ$, the Fourier transform of $f$, by

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

If $t\in {ℝ}^n$, define $h_t:A\to ℂ$ by

$$h_t(f+\alpha \delta )=\widehat{f}(t)+\alpha ,f+\alpha \delta \in A,$$

and define $h_{\mathrm{\infty }}:A\to ℂ$ by

$$h_{\mathrm{\infty }}(f+\alpha \delta )=\alpha ,f+\alpha \delta \in A.$$

By (3) it is apparent that for each $t\in {ℝ}^n\cup \{\mathrm{\infty }\}$, the map $h_t$ is a homomorphism of algebras. It can be proved that $\mathrm{\Delta }=\{h_t:t\in {ℝ}^n\}\cup \{h_{\mathrm{\infty }}\}$. Let ${ℝ}^n\cup \{\mathrm{\infty }\}$ be the one-point compactification of ${ℝ}^n$, and define $T:{ℝ}^n\cup \{\mathrm{\infty }\}\to \mathrm{\Delta }$ by $T(t)=h_t$, which is a bijection.

Suppose that $t_k\to t$ in ${ℝ}^n$. If $f+\alpha \delta \in A$, then because $\widehat{f}:{ℝ}^n\to ℂ$ is continuous, we have

$$T(t_k)(f+\alpha \delta )=h_{t_k}(f+\alpha \delta )=\widehat{f}(t_k)+\alpha \to \widehat{f}(t)+\alpha =T(t)(f+\alpha \delta ).$$

Suppose that $t_k\to \mathrm{\infty }$. If $f+\alpha \delta \in A$, then by the Riemann-Lebesgue lemma we have $\widehat{f}(t_k)\to 0$, and hence

$$T(t_k)(f+\alpha \delta )=\widehat{f}(t_k)+\alpha \to \alpha =h_{\mathrm{\infty }}(f+\alpha \delta )=T(\mathrm{\infty })(f+\alpha \delta ).$$

Therefore, $T:{ℝ}^n\cup \{\mathrm{\infty }\}\to \mathrm{\Delta }$ is continuous.

Suppose that $h_{t_k}\to h_t$ in $\mathrm{\Delta }$, $t_k,t\in {ℝ}^n$. If $f+\alpha \delta \in A$, then $h_{t_k}(f+\alpha \delta )\to h_t(f+\alpha \delta )$. But $h_{t_k}(f+\alpha \delta )=\widehat{f}(t_k)+\alpha$ and $h_t(f+\alpha \delta )=\widehat{f}(t)+\alpha$, so $\widehat{f}(t_k)\to \widehat{f}(t)$. Because this is true for all $f\in L^1({ℝ}^n)$, it follows that $t_k\to t$. Suppose that $h_{t_k}\to h_{\mathrm{\infty }}$ in $\mathrm{\Delta }$, $t_k\in {ℝ}^n$. If $f+\alpha \delta \in A$, then $h_{t_k}(f+\alpha \delta )\to h_{\mathrm{\infty }}(f+\alpha \delta )$, i.e. $\widehat{f}(t_k)+\alpha \to \alpha$, i.e. $\widehat{f}(t_k)\to 0$. Because this is true for all $f\in L^1({ℝ}^n)$, it follows that $t_k\to \mathrm{\infty }$. Therefore, $T^{-1}:\mathrm{\Delta }\to {ℝ}^n\cup \{\mathrm{\infty }\}$ is continuous, and so $\mathrm{\Delta }$ is homeomorphic to the one-point compactification of ${ℝ}^n$.

5 Involutions

If $A$ is an algebra, an involution of $A$ is a map ${}^{*}:A\to A$ satisfying

1. ${(x+y)}^{*}=x^{*}+y^{*}$

2. ${(\alpha x)}^{*}=\overline{\alpha }{x}^{*}$

3. ${(xy)}^{*}=y^{*}{x}^{*}$

4. $x^{**}=x$.

We say that $x$ is self-adjoint if $x^{*}=x$.

Following Rudin, if $A$ is a Banach algebra with an involution ${}^{*}:A\to A$ satisfying

$$\parallel x{x}^{*}\parallel ={\parallel x\parallel }^2,x\in A,$$

we say that $A$ is a $B^{*}$-algebra.

The following theorem shows that a commutative unital $B^{*}$-algebra with maximal ideal space $\mathrm{\Delta }$ is isomorphic as a $B^{*}$-algebra to $C(\mathrm{\Delta })$. (An isomorphism of $B^{\mathrm{*}}$-algebras is an isomorphism of Banach algebras that preserves the involution; the involution on $C(\mathrm{\Delta })$ is $(x\mapsto f(x))\mapsto (x\mapsto \overline{f(x)})$.)

Theorem 7 (Gelfand-Naimark).

If $A$ is a commutative unital $B^{*}$-algebra, then $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an isomorphism of Banach algebras, and if $x\in A$ then $\mathrm{\Gamma }(x^{*})=\overline{\mathrm{\Gamma }(x)}$.

Proof.

Let $u\in A$ be self-adjoint, let $h\in \mathrm{\Delta }$, and let $h(u)=\alpha +i\beta$. For $t\in ℝ$, put $z=u+ite$. We have

$$h(z)=h(u)+h(ite)=\alpha +i\beta +it=\alpha +i(\beta +t),$$

and

$$z{z}^{*}=(u+ite)(u-ite)=u^2+t^{2}e,$$

hence

$${\alpha }^2+{(\beta +t)}^2={|h(z)|}^2\le {\parallel z\parallel }^2=\parallel z{z}^{*}\parallel =\parallel u^2+t^{2}e\parallel \le {\parallel u\parallel }^2+t^2,$$

i.e.

$${\alpha }^2+{\beta }^2+2\beta t\le {\parallel u\parallel }^2.$$

Because this is true for all $t\in ℝ$, it follows that $\beta =0$. Therefore, if $u\in A$ is self-adjoint then $h(u)\in ℝ$.

Furthermore, if $x\in A$ then with $2u=x+x^{*}$ and $2v=i(x^{*}-x)$ we have $x=u+iv$ with $u$ and $v$ self-adjoint. Then $x^{*}=u-iv$, and so

$$h(x^{*})=h(u-iv)=h(u)-ih(v)=\overline{h(x)}.$$

This shows that if $x\in A$ then $\mathrm{\Gamma }(x^{*})=\overline{\mathrm{\Gamma }(x)}$. In particular, $\widehat{A}$ is closed under complex conjugation. If $h_1\ne h_2$, then there is some $x\in A$ for which $h_1(x)\ne h_2(x)$, i.e. $\widehat{x}(h_1)\ne \widehat{x}(h_2)$, so $\widehat{A}$ separates points in $\mathrm{\Delta }$. Because $\widehat{A}$ is a unital Banach algebra, it follows from the Stone-Weierstrass theorem that $\widehat{A}$ is dense in $C(\mathrm{\Delta })$.

Let $x\in A$. With $y=x{x}^{*}$, we have $y^{*}={(x{x}^{*})}^{*}=x^{**}{x}^{*}=x{x}^{*}=y$, from which it follows that $\parallel y^2\parallel ={\parallel y\parallel }^2$. Assume by induction that $\parallel y^m\parallel ={\parallel y\parallel }^m$, for $m=2^n$. Then, as ${(y^m)}^{*}=y^m$,

$$\parallel y^{2m}\parallel =\parallel y^m{y}^m\parallel =\parallel y^m{(y^m)}^{*}\parallel ={\parallel y^m\parallel }^2={({\parallel y\parallel }^m)}^2={\parallel y\parallel }^{2m}.$$

The spectral radius formula gives

$$\rho (y)=lim{\parallel y^n\parallel }^{1/n},$$

and because $\parallel y^m\parallel ={\parallel y\parallel }^m$ for $m=2^n$, we have $lim{\parallel y^m\parallel }^{1/m}=\parallel y\parallel$. Because the limit of this subsequence is $\parallel y\parallel$, the limit of ${\parallel y^n\parallel }^{1/n}$ is also $\parallel y\parallel$, so we obtain

$$\rho (y)=\parallel y\parallel .$$

But (1) tells us ${\parallel \widehat{y}\parallel }_{\mathrm{\infty }}=\rho (y)$, so we have ${\parallel \widehat{y}\parallel }_{\mathrm{\infty }}=\parallel y\parallel$. Because $y=x{x}^{*}$, using $\mathrm{\Gamma }(x^{*})=\overline{\mathrm{\Gamma }(x)}$ and the fact that $\mathrm{\Gamma }$ is an algebra homomorphism, we get

$$\mathrm{\Gamma }(y)=\mathrm{\Gamma }(x{x}^{*})=\mathrm{\Gamma }(x)\mathrm{\Gamma }(x^{*})=\mathrm{\Gamma }(x)\overline{\mathrm{\Gamma }(x)}={|\mathrm{\Gamma }(x)|}^2.$$

That is, $\widehat{y}={|\widehat{x}|}^2$ and with ${\parallel \widehat{y}\parallel }_{\mathrm{\infty }}=\parallel y\parallel$ we obtain

$${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}^2={\parallel \widehat{y}\parallel }_{\mathrm{\infty }}=\parallel y\parallel =\parallel x{x}^{*}\parallel ={\parallel x\parallel }^2,$$

i.e.

$${\parallel \widehat{x}\parallel }_{\mathrm{\infty }}=\parallel x\parallel .$$

This shows that $\mathrm{\Gamma }:A\to C(\mathrm{\Delta })$ is an isometry. In particular, $\mathrm{\Gamma }$ maps closed sets to closed sets, so $\widehat{A}=\mathrm{\Gamma }(A)$ is a closed subset of $C(\mathrm{\Delta })$. We have already established that $\widehat{A}$ is dense in $C(\mathrm{\Delta })$, so $\widehat{A}=C(\mathrm{\Delta })$. The fact that $\mathrm{\Gamma }$ is an isometry yields that $\mathrm{\Gamma }$ is one-to-one, and the fact that $\widehat{A}=C(\mathrm{\Delta })$ means that that $\mathrm{\Gamma }$ is onto, hence $\mathrm{\Gamma }$ is a bijection, and therefore it is an isomorphism of algebras. Because $\mathrm{\Gamma }$ is an isometry, it is an isomorphism of Banach algebras. ∎

The following theorem states conditions under which a self-adjoint element of a unital Banach algebra with an involution has a square root.

Theorem 8.

Let $A$ be a unital Banach algebra with an involution ${}^{*}:A\to A$. If $x\in A$ is self-adjoint and $\sigma (x)$ contains no real $\lambda$ with $\lambda \le 0$, then there is some self-adjoint $y\in A$ satisfying $y^2=x$.

If $A$ is a Banach algebra and $x\in A$, we say that $x\in A$ is normal if $x{x}^{*}=x^{*}x$. If $A$ is a Banach algebra with involution ${}^{*}:A\to A$, by $x\ge 0$ we mean that $x$ is self-adjoint and $\sigma (x)\subseteq [0,\mathrm{\infty })$, and we say that $x$ is positive. The following theorem states basic facts about the spectrum of elements of a unital $B^{*}$-algebra.

Theorem 9.

If $A$ is a unital $B^{*}$-algebra, then:

1. If $x$ is self-adjoint, then $\sigma (x)\subseteq ℝ$.

2. If $x$ is normal, then $\rho (x)=\parallel x\parallel$.

3. If $x\in A$, then $\rho (x{x}^{*})={\parallel x\parallel }^2$.

4. If $x\ge 0$ and $y\ge 0$, then $x+y\ge 0$.

5. If $x\in A$, then $x{x}^{*}\ge 0$.

6. If $x\in A$, then $e+x{x}^{*}$ is invertible.

6 Positive linear functionals

Suppose that $A$ is a Banach algebra with an involution ${}^{*}:A\to A$. If $F:A\to ℂ$ is a linear map such that $F(x{x}^{*})$ is real and $\ge 0$ for all $x\in A$, we say that $F$ is a positive linear functional. In particular, if $h\in \mathrm{\Delta }$ and $x\in A$, then from Theorem 7 we have $h(x^{*})=\overline{h(x)}$, and so $h(x{x}^{*})=h(x)h(x^{*})=h(x)\overline{h(x)}={|h(x)|}^2\ge 0$. Thus, the elements of $\mathrm{\Delta }$ are positive linear functionals.

We shall use the following theorem to prove the theorem after it.

Theorem 10.

If $X$ is a real or complex Banach space, $X_1$ and $X_2$ are closed subspaces of $X$, and $X=X_1+X_2$, then there is some such that for every $x\in X$ there are $x_1\in X_1,x_2\in X_2$ satisfying $x=x_1+x_2$ and

$$\parallel x_1\parallel +\parallel x_2\parallel \le \gamma \parallel x\parallel .$$

The following theorem establishes some basic properties of positive linear functionals on a unital Banach algebra with an involution.

Theorem 11.

Suppose that $A$ is a unital Banach algebra with an involution ${}^{*}:A\to A$. If $F:A\to ℂ$ is a positive linear functional, then:

1. $F(x^{*})=\overline{F(x)}$.

2. ${|F(x{y}^{*})|}^2\le F(x{x}^{*})F(y{y}^{*})$.

3. ${|F(x)|}^2\le F(e)F(x{x}^{*})\le F{(e)}^2\rho (x{x}^{*})$.

4. If $x$ is normal, then $|F(x)|\le F(e)\rho (x)$.

5. If $A$ is commutative, then $\parallel F\parallel =F(e)$.

6. If there is some $\beta$ such that $\parallel x^{*}\parallel \le \beta \parallel x\parallel$ for all $x\in A$, then $\parallel F\parallel \le {\beta }^{1/2}F(e)$.

7. $F$ is a bounded linear map.

Proof.

Suppose that $x,y\in A$. For any $\alpha \in ℂ$, we have on the one hand $F((x+\alpha y){(x+\alpha y)}^{*})\ge 0$, and on the other hand

$$F((x+\alpha y){(x+\alpha y)}^{*})=F((x+\alpha y)(x^{*}+\overline{\alpha }{y}^{*}))=F(x{x}^{*}+\overline{\alpha }x{y}^{*}+\alpha y{x}^{*}+{|\alpha |}^{2}y{y}^{*}).$$

Therefore,

$$F(x{x}^{*})+\overline{\alpha }F(x{y}^{*})+\alpha F(y{x}^{*})+{|\alpha |}^{2}F(y{y}^{*})\ge 0.$$ (4)

Applying (4) with $\alpha =1$ gives

$$F(x{x}^{*})+F(x{y}^{*})+F(y{x}^{*})+F(y{y}^{*})\ge 0.$$

In particular, this expression is real, and because $F(x{x}^{*})$ and $F(y{y}^{*})$ are real we get that $F(x{y}^{*})+F(y{x}^{*})$ is real, so $\mathrm{Im}F(y{x}^{*})=-\mathrm{Im}F(x{y}^{*})$. Applying (4) with $\alpha =i$ gives

$$F(x{x}^{*})-iF(x{y}^{*})+iF(y{x}^{*})+F(y{y}^{*})\ge 0.$$

In particular, this expression is real, and so $-iF(x{y}^{*})+iF(y{x}^{*})$ is real, i.e. $F(x{y}^{*})-F(y{x}^{*})$ is imaginary, so $\mathrm{Re}F(y{x}^{*})=\mathrm{Re}F(x{y}^{*})$. Therefore $F(y{x}^{*})=\overline{F(x{y}^{*})}$. Using $y=e$ yields

$$F(x^{*})=\overline{F(x)}.$$

Suppose that $x,y\in A$ and that $F(x{y}^{*})\ne 0$. For any $t\in ℝ$, using (4) with $\alpha =\frac{t}{|F(x{y}^{*})|}F(x{y}^{*})$ gives

$$F(x{x}^{*})+\frac{t}{|F(x{y}^{*})|}\overline{F(x{y}^{*})}F(x{y}^{*})+\frac{t}{|F(x{y}^{*})|}F(x{y}^{*})F(y{x}^{*})+t^{2}F(y{y}^{*})\ge 0,$$

i.e.

$$F(x{x}^{*})+t|F(x{y}^{*})|+\frac{t}{|F(x{y}^{*})|}F(x{y}^{*})F(y{x}^{*})+t^{2}F(y{y}^{*})\ge 0,$$

and as $F(y{x}^{*})=F({(x{y}^{*})}^{*})=\overline{F(x{y}^{*})}$, we have

$$F(x{x}^{*})+2t|F(x{y}^{*})|+t^{2}F(y{y}^{*})\ge 0.$$

For $t=-\frac{|F(x{y}^{*})|}{F(y{y}^{*})}$ this is

$$F(x{x}^{*})-2\frac{{|F(x{y}^{*})|}^2}{F(y{y}^{*})}+\frac{{|F(x{y}^{*})|}^2}{F(y{y}^{*})}\ge 0,$$

i.e.

$${|F(x{y}^{*})|}^2\le F(x{x}^{*})F(y{y}^{*}).$$

Suppose that $x\in A$. Because $x{e}^{*}=x$ and $e{e}^{*}=e$, we have

$${|F(x)|}^2\le F(e)F(x{x}^{*}).$$

We shall prove that $F(x{x}^{*})\le F(e)\rho (x{x}^{*})$. Let $t>\rho (x{x}^{*})$. It then follows that $\sigma (te-x{x}^{*})$ is contained in the open right half-plane, and thus by Theorem 8 there is some self-adjoint $u\in A$ satisfying $u^2=te-x{x}^{*}$. Then

$$F(te-x{x}^{*})=F(u^2)=F(u{u}^{*})\ge 0,$$

so

$$F(x{x}^{*})\le tF(e).$$

Because this is true for all $t>\rho (x{x}^{*})$, we obtain

$$F(x{x}^{*})\le F(e)\rho (x{x}^{*}).$$

Suppose that $x$ is normal. It is a fact that if $x$ and $y$ belong to a unital Banach algebra and $xy=yx$, then $\sigma (xy)\subseteq \sigma (x)\sigma (y)$. Thus $\sigma (x{x}^{*})\subseteq \sigma (x)\sigma (x^{*})$, from which we get

$$\rho (x{x}^{*})\le \rho (x)\rho (x^{*}).$$

It is a fact that $\sigma (x^{*})=\overline{\sigma (x)}$,, so we have $\rho (x)=\rho (x^{*})$, and thus

$$\rho (x{x}^{*})\le \rho {(x)}^2.$$

But ${|F(x)|}^2\le F{(e)}^2\rho (x{x}^{*})$, so we have ${|F(x)|}^2\le F{(e)}^2\rho {(x)}^2$, i.e.

$$|F(x)|\le F(e)\rho (x).$$

Suppose that $A$ is commutative, and let $x\in A$. Since $A$ is commutative, $x$ is normal and hence we have $|F(x)|\le F(e)\rho (x)$, and as always we have $\rho (x)\le \parallel x\parallel$. Therefore, for every $x\in A$ we have

$$|F(x)|\le F(e)\parallel x\parallel .$$

This implies that $\parallel F\parallel \le F(e)$, and because the above inequality is an equality for $x=e$, we have $\parallel F\parallel =F(e)$.

Suppose that there is some $\beta$ such that $\parallel x^{*}\parallel \le \beta \parallel x\parallel$ for all $x\in A$. We have $\rho (x{x}^{*})\le \parallel x{x}^{*}\parallel \le \parallel x\parallel \parallel x^{*}\parallel \le \beta {\parallel x\parallel }^2$. (We merely stipulated that $A$ is a unital Banach algebra with an involution; if we had demanded that $A$ be a $B^{*}$-algebra, then we would have $\parallel x{x}^{*}\parallel =\parallel x\parallel \parallel x^{*}\parallel ={\parallel x\parallel }^2$.) Using ${|F(x)|}^2\le F{(e)}^2\rho (x{x}^{*})$ then gives us ${|F(x)|}^2\le \beta F{(e)}^2{\parallel x\parallel }^2$, hence

$$|F(x)|\le {\beta }^{1/2}F(e)\parallel x\parallel .$$

If $F(e)=0$, then ${|F(x)|}^2\le 0$ for all $x\in A$, and hence $F=0$, which indeed is bounded. Otherwise, $F(e)>0$, and $F$ is bounded if and only if $\frac{1}{F(e)}F$ is bounded. Therefore, to prove that $F$ is bounded it suffices to prove that $F$ is bounded in the case where $F(e)=1$.

Let $H$ be the set of all self-adjoint elements of $A$. $H$ and $iH$ are real vector spaces. For any $x\in A$, defining $2u=x+x^{*}$ and $2v=i(x^{*}-x)$, we have $x=u+iv$, and $u,v$ are self-adjoint. It follows that

$$A=H+iH.$$

Because the elements of $H$ are self-adjoint, the restriction of $F$ to $H$ is a real-linear map $H\to ℝ$. For $u\in H$, because $u$ is self-adjoint it is in particular normal, and so $|F(x)|\le F(e)\rho (x)\le F(e)\parallel x\parallel =\parallel x\parallel$, because $F(e)=1$. Hence the restriction of $F$ to $H$ is a real-linear map $H\to ℝ$ with norm $1$, and therefore there is a unique bounded real-linear map $\mathrm{\Phi }:\overline{H}\to ℝ$ whose restriction to $H$ is equal to the restriction of $F$ to $H$, and $\parallel \mathrm{\Phi }\parallel =1$.

Suppose that $y\in \overline{H}\cap i\overline{H}$. There are $u_n\in H$ with $u_n\to y$ and there are $v_n\in H$ with $i{v}_n\to y$. Then $u_n^2\to y^2$ and $-v_n^2\to y$, or $v_n^2\to -y^2$. Because ${|F(u_n)|}^2\le F(e)F(u_n{u}_n^{*})=F(u_n^2)$, we have

$${|F(u_n)|}^2\le F(u_n^2)\le F(u_n^2+v_n^2).$$

Because $u_n$ and $v_n$ are self-adjoint, $u_n^2+v_n^2$ is normal and hence

$$|F(u_n^2+v_n^2)|\le F(e)\rho (u_n^2+v_n^2)=\rho (u_n^2+v_n^2)\le \parallel u_n^2+v_n^2\parallel ,$$

and so we have

$${|F(u_n)|}^2\le \parallel u_n^2+v_n^2\parallel .$$

But $u_n^2\to y$ and $v_n^2\to -y$, so $\parallel u_n^2+v_n^2\parallel \to \parallel y-y\parallel =0$. Therefore, $F(u_n)\to 0$, and so

$$\mathrm{\Phi }(y)=limF(u_n)\to 0.$$

That is, if $y\in \overline{H}\cap i\overline{H}$, then $F(y)=0$.

Because $A=H+iH$, certainly $A=\overline{H}+i\overline{H}$, so by Theorem 10 there is some such that for all $x\in A$, there are $x_1\in \overline{H}$ and $x_2\in \overline{H}$ satisfying

$$x=x_1+i{x}_2,\parallel x_1\parallel +\parallel x_2\parallel \le \gamma \parallel x\parallel .$$

Let $x\in A$ and let $x=x_1+i{x}_2$, where $x_1,x_2$ satisfy the above, and let $x=u+iv$ with $u,v\in H$, namely $2u=x+x^{*}$ and $2v=i(x^{*}-x)$. Supposing that $x_1-u$ and $x_2-v\in \overline{H}\cap i\overline{H}$, which Rudin asserts but whose truth is not apparent to me, we obtain $F(x_1-u)=0$ and $F(x_2-v)=0$, or $F(x_1)=F(u)$ and $F(x_2)=F(v)$. Then,

$$F(x)=F(u+iv)=F(u)+iF(v)=F(x_1)+iF(x_2)=\mathrm{\Phi }(x_1)+i\mathrm{\Phi }(x_2),$$

and therefore, because $\parallel \mathrm{\Phi }\parallel =1$ and because $\parallel x_1\parallel +\parallel x_2\parallel \le \gamma \parallel x\parallel$,

$$|F(x)|\le |\mathrm{\Phi }(x_1)+i\mathrm{\Phi }(x_2)|\le |\mathrm{\Phi }(x_1)|+|\mathrm{\Phi }(x_2)|\le \parallel x_1\parallel +\parallel x_2\parallel \le \gamma \parallel x\parallel ,$$

showing that $\parallel F\parallel \le \gamma$, and in particular that $F$ is bounded. ∎

7 The Riesz-Markov theorem and extreme points

We say that a positive Borel measure $\mu$ on a compact Hausdorff space $X$ is regular if for every Borel subset $E$ of $X$ we have

$$\mu (E)=sup\{\mu (F):F\text{ is compact and }F\subseteq E\}$$

and

$$\mu (E)=inf\{\mu (G):G\text{ is open and }E\subseteq G\}.$$

We say that a complex Borel measure $\mu$ on a compact Hausdorff space is regular if the positive Borel measure $|\mu |$ is regular, and we write $\parallel \mu \parallel =|\mu |(X)$. The following is the Riesz-Markov theorem, stated for complex Borel measures on a compact Hausdorff space.

Theorem 12 (Riesz-Markov).

Suppose that $X$ is a compact Hausdorff space. If $\mathrm{\Lambda }$ is a bounded linear functional on $C(X)$, then there is one and only one regular complex Borel measure $\mu$ on $X$ satisfying

$$\mathrm{\Lambda }f={\int }_{X}f𝑑\mu ,f\in C(X).$$

This measure $\mu$ satisfies $\parallel \mu \parallel =\parallel \mathrm{\Lambda }\parallel$.

The following theorem uses the Riesz-Markov theorem to define a correspondence between positive linear functionals on a commutative unital Banach algebra with a symmetric involution and regular positive Borel measures on its maximal ideal space.

Theorem 13.

Suppose that $A$ is a commutative unital Banach algebra with an involution ${}^{*}:A\to A$ satisfying

$$h(x^{*})=\overline{h(x)},x\in A,h\in \mathrm{\Delta }.$$ (5)

Let $K$ be the set of all positive linear functionals $F:A\to ℂ$ satisfying $F(e)\le 1$, and let $M$ be the set of all regular positive Borel measures $\mu$ on $\mathrm{\Delta }$ satisfying $\mu (\mathrm{\Delta })\le 1$. $K$ and $M$ are convex sets. If $\mu \in M$, then $F:A\to ℂ$ defined by

$$F_{\mu }(x)={\int }_{\mathrm{\Delta }}\widehat{x}𝑑\mu ,x\in A,$$

belongs to $K$, and this map $\mu \mapsto F_{\mu }$ is an isometric bijection $M\to K$.

Proof.

If $F_1,F_2\in K$ and $0\le t\le 1$, then $(1-t)F_1+t{F}_2$ is linear, and it is straightforward to check that it is positive. Moreover, $((1-t)F_1+t{F}_2)(e)=(1-t)F_1(e)+t{F}_2(e)\le (1-t)+t=1$, so $(1-t)F_1+t{F}_2\in K$. Therefore $K$ is a convex set.

Suppose that ${\mu }_1,{\mu }_2\in M$, that $a_1,a_2$ are nonnegative real numbers, and let $\mu =a_1{\mu }_1+a_2{\mu }_2$. If $E$ is a Borel subset of $\mathrm{\Delta }$, then for any $ϵ>0$ there are compact subsets $F_1,F_2$ of $\mathrm{\Delta }$ such that and . With $F=F_1\cup F_2$, we have

$\mu (F)$ $=$ $a_1{\mu }_1(F)+a_2{\mu }_2(F)$

$\ge$ $a_1{\mu }_1(F_1)+a_2{\mu }_2(F_2)$

$\ge$ $a_1({\mu }_1(E)+ϵ)+a_2({\mu }_2(E)+ϵ)$

$=$ $\mu (E)+(a_1+a_2)ϵ.$

It follows that $\mu (E)=sup\{\mu (F):F\text{ is compact and }F\subseteq E\}$. If $E$ is a Borel subset of $\mathrm{\Delta }$, then for any $ϵ>0$ there are open subsets $G_1,G_2$ of $\mathrm{\Delta }$ such that ${\mu }_1(E)>{\mu }_1(G_1)-ϵ$ and ${\mu }_2(E)>{\mu }_2(G_2)-ϵ$. With $G=G_1\cap G_2$, we have

$\mu (G)$ $=$ $a_1{\mu }_1(G)+a_2{\mu }_2(G)$

$\le$ $a_1{\mu }_1(G_1)+a_2{\mu }_2(G_2)$

$a_1({\mu }_1(E)+ϵ)+a_2({\mu }_2(E)+ϵ)$

$=$ $\mu (E)+(a_1+a_2)ϵ.$

It follows that $\mu (E)=inf\{\mu (G):G\text{ is open and }E\subseteq G\}$. Therefore, $\mu =a_1{\mu }_1+a_2{\mu }_2$ is a regular positive Borel measure. In particular, if $0\le t\le 1$ and $a_1=1-t$, $a_2=t$, then $\mu$ is a regular positive Borel measure. Finally, for $\mu =(1-t){\mu }_1+t{\mu }_2$, $0\le t\le 1$, we have, because ${\mu }_1(\mathrm{\Delta })\le 1$ and ${\mu }_2(\mathrm{\Delta })\le 1$,

$$\mu (\mathrm{\Delta })=(1-t){\mu }_1(\mathrm{\Delta })+t{\mu }_2(\mathrm{\Delta })\le (1-t)+t=1,$$

so $\mu \in M$, showing that $M$ is a convex set.

Let $\mu \in M$. It is apparent that $F_{\mu }:A\to ℂ$ is linear. For $x\in A$, we have $\mathrm{\Gamma }(x{x}^{*})=\mathrm{\Gamma }(x)\mathrm{\Gamma }(x^{*})$, and as $\mathrm{\Gamma }(x^{*})=\overline{\mathrm{\Gamma }(x)}$ by (5), we get $\mathrm{\Gamma }(x{x}^{*})={|\mathrm{\Gamma }(x)|}^2$. As ${|\mathrm{\Gamma }(x)|}^2(h)\ge 0$ for all $h\in \mathrm{\Delta }$, we have

$$F_{\mu }(x{x}^{*})={\int }_{\mathrm{\Delta }}\mathrm{\Gamma }(x{x}^{*})𝑑\mu ={\int }_{\mathrm{\Delta }}{|\mathrm{\Gamma }(x)|}^2𝑑\mu \ge 0,$$

showing that $F_{\mu }$ is a positive linear functional. Furthermore, $\widehat{e}(h)=h(e)=1$ for all $h\in \mathrm{\Delta }$, so

$$F_{\mu }(e)=\mu (\mathrm{\Delta })\le 1,$$

showing that $F_{\mu }\in K$.

If $x\in \mathrm{rad}A$, then $\rho (x)=0$ by (1), and so $F(x)=0$ by Theorem 11. We define $\widehat{F}:\widehat{A}\to ℂ$

$$\widehat{F}(\widehat{x})=F(x);$$

this makes sense because if $\widehat{x}=\widehat{y}$ then $\mathrm{\Gamma }(x-y)=0$, and so by Theorem 3 we have $x-y\in \mathrm{rad}A$ and hence $F(x-y)=0$, i.e. so $F(x)=F(y)$. For $x\in A$, $x$ is normal because $A$ is commutative so we have by Theorem 11 that

$$|\widehat{F}(\widehat{x})|=|F(x)|\le F(e)\rho (x)$$

and by (1) we have $\rho (x)={\parallel \widehat{x}\parallel }_{\mathrm{\infty }}$, so

$$|\widehat{F}(\widehat{x})|\le F(e){\parallel \widehat{x}\parallel }_{\mathrm{\infty }}.$$

As $\widehat{F}(\widehat{e})=F(e)$, it follows that $\parallel \widehat{F}\parallel =F(e)$. By (5) and because $\widehat{A}$ separates points in $\mathrm{\Delta }$, applying the Stone-Weierstrass we obtain that $\widehat{A}$ is dense in $C(\mathrm{\Delta })$. Because $\widehat{F}$ is a continuous linear functional on the dense subspace $\widehat{A}$ of $C(\mathrm{\Delta })$, there is a unique continuous linear functional $\mathrm{\Lambda }$ on $C(\mathrm{\Delta })$ such that $\mathrm{\Lambda }=\widehat{F}$ on $\widehat{A}$, and $\parallel \mathrm{\Lambda }\parallel =\parallel \widehat{F}\parallel$. Applying Theorem 12, there is one and only one regular complex Borel measure $\mu$ on $X$ that satisfies

$$\mathrm{\Lambda }f={\int }_{\mathrm{\Delta }}f𝑑\mu ,f\in C(\mathrm{\Delta }),$$ (6)

and $\parallel \mu \parallel =\parallel \mathrm{\Lambda }\parallel =\parallel \widehat{F}\parallel =F(e)$. It follows that $\mu \mapsto F_{\mu }$ is one-to-one. Because $\widehat{e}(h)=1$ for all $h\in \mathrm{\Delta }$,

$$\mu (\mathrm{\Delta })={\int }_{\mathrm{\Delta }}{\chi }_{\mathrm{\Delta }}𝑑\mu ={\int }_{\mathrm{\Delta }}\widehat{e}𝑑\mu =\mathrm{\Lambda }\widehat{e}=\widehat{F}(\widehat{e})=F(e)=\parallel \mu \parallel =|\mu |(\mathrm{\Delta }).$$

The fact that $\mu (\mathrm{\Delta })=|\mu |(\mathrm{\Delta })$ implies that $\mu$ is a positive measure. The above equalities also state $\mu (\mathrm{\Delta })=F(e)$, and since $F\in K$ we have $F(e)\le 1$, hence $\mu (\mathrm{\Delta })\le 1$. Therefore, $\mu \in M$. For $x\in A$, as $\widehat{x}\in C(\mathrm{\Delta })$ we have by (6) that

$$F_{\mu }(x)={\int }_{\mathrm{\Delta }}\widehat{x}𝑑\mu =\mathrm{\Lambda }\widehat{x}=\widehat{F}(\widehat{x})=F(x),$$

showing that $F=F_{\mu }$. This shows that $\mu \mapsto F_{\mu }$ is onto, and therefore $\mu \mapsto F_{\mu }$ is a bijection $M\to K$. ∎

Because the map $\mu \mapsto F_{\mu }$ in the above theorem is an isometric bijection $M\to K$, it follows that that $\mu$ is an extreme point of $M$ if and only if $F_{\mu }$ is an extreme point of $K$.

It is a fact that the set of extreme points of the set of regular Borel probability measures on a compact Hausdorff space $X$ is $\{{\delta }_x:x\in X\}$. Given this, one proves that the set of extreme points of $M$ is $\{0\}\cup \{{\delta }_h:h\in \mathrm{\Delta }\}$. For $x\in A$, $F_0(x)=0$, i.e. $F_0=0$. For $h\in \mathrm{\Delta }$ and $x\in A$,

$$F_{{\delta }_h}(x)={\int }_{\mathrm{\Delta }}\widehat{x}𝑑{\delta }_h=\widehat{x}(h)=h(x),$$

so $F_{{\delta }_h}=h$. Therefore, the extreme points of $K$ are $\{0\}\cup \mathrm{\Delta }$, that is, the set of algebra homomorphisms $A\to ℂ$.

Corollary 14.

Suppose that $A$ is a commutative unital Banach algebra with involution ${}^{*}:A\to A$ satisfying

$$h(x^{*})=\overline{h(x)},x\in A,h\in \mathrm{\Delta }.$$

If $K$ is the set of all positive linear functionals $F:A\to ℂ$ satisfying $F(e)\le 1$, then

$$\mathrm{ext}K=\{0\}\cup \mathrm{\Delta }.$$

Moreover, it is straightforward to check that the set $K$ in the above corollary is a weak-* closed subset of $A^{*}$: if $F_i\in K$ is a net that weak-* converges to $\mathrm{\Lambda }\in A^{*}$, one checks that $\mathrm{\Lambda }(x{x}^{*})\ge 0$ for all $x\in A$ and that $\mathrm{\Lambda }e\le 1$. By the Banach-Alaoglu theorem, the set $B=\{\mathrm{\Lambda }\in A^{*}:\parallel \mathrm{\Lambda }\parallel \le 1\}$ is weak-* compact, and if $F\in K$ then $\parallel F\parallel =F(e)$ by Theorem 11 and $F(e)\le 1$, so $K\subseteq B$. Hence, $K$ is a weak-* compact subset of $A^{*}$. Therefore, the Krein-Milman theorem tells us that $K$ is equal to the weak-* closure of the convex hull of the set of its extreme points, and by the above corollary this means that $K$ is equal to the weak-* closure of the convex hull of $\{0\}\cup \mathrm{\Delta }$.

8 Positive definite functions

A function $\varphi :{ℝ}^n\to ℂ$ is said to be positive-definite if $r\ge 1$, $x_1,\mathrm{\dots },x_r\in {ℝ}^n,c_1,\mathrm{\dots },c_r\in ℂ$ imply that

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

in particular, for $\varphi$ to be positive-definite demands that the left-hand side of this inequality is real.

A positive-definite function need not be measurable. For example, $ℝ$ is a vector space over $\mathbb{Q}$, and if $\psi :ℝ\to ℝ$ is a vector space automorphism of $ℝ$ over $\mathbb{Q}$, one proves that $x\mapsto e^{i\psi (x)}$ is a positive-definite function $ℝ\to ℂ$, and that there are $\psi$ for which $x\mapsto e^{i\psi (x)}$ is not measurable.

The following theorem states some basic facts about positive-definite functions. More material on positive-definite functions is presented in Bogachev; for example, if $\varphi :{ℝ}^n\to ℂ$ is a measurable positive-definite function, then there is a continuous positive-definite function ${ℝ}^n\to ℂ$ that is equal to $\varphi$ almost everywhere.

Theorem 15.

If $\varphi :{ℝ}^n\to ℂ$ is positive-definite, then

$$\varphi (0)\ge 0,$$ (7)

and for all $x\in {ℝ}^n$ we have

$$\overline{\varphi (x)}=\varphi (-x)$$ (8)

and

$$|\varphi (x)|\le \varphi (0).$$ (9)

Proof.

Using $r=1$ and $c_1=1$, we have for all $x_1\in {ℝ}^n$ that $\varphi (x_1-x_1)\ge 0$, i.e. $\varphi (0)\ge 0$.

Using $r=2$, for $x_1,x_2\in {ℝ}^n$ and $c_1,c_2\in ℂ$ we have

$$c_1\overline{c_1}\varphi (x_1-x_1)+c_1\overline{c_2}\varphi (x_1-x_2)+c_2\overline{c_1}\varphi (x_2-x_1)+c_2\overline{c_2}\varphi (x_2-x_2)\ge 0.$$

Take $x_1=x$ and $x_2=0$, with which

$${|c_1|}^2\varphi (0)+c_1\overline{c_2}\varphi (x)+c_2\overline{c_1}\varphi (-x)+{|c_2|}^2\varphi (0)\ge 0;$$

in particular, the left-hand side is real, and because $\varphi (0)$ is real by (7), this implies that $c_1\overline{c_2}\varphi (x)+c_2\overline{c_1}\varphi (-x)$ is real. That is, it is equal to its complex conjugate:

$$c_1\overline{c_2}\varphi (x)+c_2\overline{c_1}\varphi (-x)=\overline{c_1}{c}_2\overline{\varphi (x)}+\overline{c_2}{c}_1\overline{\varphi (-x)}.$$

The fact that this holds every $c_1,c_2\in ℂ$ implies that $\overline{\varphi (x)}=\varphi (-x)$.

Again using that

$$c_1\overline{c_1}\varphi (x_1-x_1)+c_1\overline{c_2}\varphi (x_1-x_2)+c_2\overline{c_1}\varphi (x_2-x_1)+c_2\overline{c_2}\varphi (x_2-x_2)\ge 0,$$

with $x_1=x,x_2=0$ and $c_2=1$ we get

$${|c_1|}^2\varphi (0)+c_1\varphi (x)+\overline{c_1}\varphi (-x)+\varphi (0)\ge 0.$$

Applying (8) gives

$${|c_1|}^2\varphi (0)+c_1\varphi (x)+\overline{c_1\varphi (x)}+\varphi (0)\ge 0.$$

For $c_1\in ℂ$ such that $|c_1|=1$,

$$2\varphi (0)+2\mathrm{R}\mathrm{e}\left(c_1\varphi (x)\right)\ge 0,$$

or

$$-\mathrm{Re}\left(c_1\varphi (x)\right)\le \varphi (0).$$

Thus, taking $c_1\in ℂ$ such that $|c_1|=1$ and for which $-\mathrm{Re}\left(c_1\varphi (x)\right)=|\varphi (x)|$, we get $|\varphi (x)|\le \varphi (0)$. ∎

The following lemma about positive-definite functions follows a proof in Bogachev.

Lemma 16.

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

$${\int }_{{ℝ}^n}{\int }_{{ℝ}^n}\varphi (x-y)f(x)f(y)𝑑m_n(x)𝑑m_n(y)\ge 0.$$

Proof.

For $r\ge 2$ and for any $x_1,\mathrm{\dots },x_r$ and $c_1=1,\mathrm{\dots },c_r=1$, we have

$$\sum _{j,k=1}^r\varphi (x_j-x_k)\ge 0,$$

or

$$r\varphi (0)+\sum _{j\ne k}\varphi (x_j-x_k)\ge 0.$$

By (9), $\varphi$ is bounded. It follows that we can integrate both sides of the above inequality over ${({ℝ}^n)}^r$ with respect to the positive measure

$$f(x_1)\mathrm{\cdots }f(x_r)d{m}_n(x_1)\mathrm{\cdots }d{m}_n(x_r).$$

Writing

$$I={\int }_{{ℝ}^n}f(x)𝑑m_n(x),$$

we obtain

$$r\varphi (0)I^r+\sum _{j\ne k}{\int }_{{ℝ}^n}\mathrm{\cdots }{\int }_{{ℝ}^n}\varphi (x_j-x_k)f(x_1)\mathrm{\cdots }f(x_r)𝑑m_n(x_1)\mathrm{\cdots }𝑑m_n(x_r)\ge 0,$$

and so, writing

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

we have

$$r\varphi (0)I^r+\sum _{j\ne k}J{I}^{r-2}\ge 0,$$

or

$$r\varphi (0)I^r+r(r-1)J{I}^{r-2}\ge 0.$$

If $I=0$, then because $f$ is nonnegative it follows that $f$ is $0$ almost everywhere, in which case $J=0$, so the claim is true. If $I>0$, then dividing by $r(r-1)I^{r-2}$ we obtain

$$\frac{1}{r-1}\varphi (0)I^2+J\ge 0.$$

This inequality holds for all $r\ge 2$, so taking $r\to \mathrm{\infty }$ yields

$$J\ge 0,$$

which is the claim. ∎

For $f,g\in L^1({ℝ}^n)$,

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

The support of a function $f:{ℝ}^n\to ℂ$, denoted $\mathrm{supp}f$, is the closure of the set $\{x\in {ℝ}^n:f(x)\ne 0\}$. We denote by $C_c({ℝ}^n)$ the set of all continuous functions $f:{ℝ}^n\to ℂ$ such that $\mathrm{supp}f$ is a compact set. It is straightforward to check that an element of $C_c({ℝ}^n)$ is uniformly continuous on ${ℝ}^n$. The following theorem is similar to the previous lemma, but applies to functions that need not be nonnegative.

Theorem 17.

For $f:{ℝ}^n\to ℂ$, define $\tilde{f}:{ℝ}^n\to ℂ$ by $\tilde{f}(x)=\overline{f(-x)}$. If $\varphi :{ℝ}^n\to ℂ$ is a continuous positive-definite function, then for all $f\in C_c({ℝ}^n)$, we have

$${\int }_{{ℝ}^n}(f*\tilde{f})\psi \ge 0.$$

If $\mu$ is a complex Borel measure on ${ℝ}^n$, the Fourier transform of $\mu$ is the function $\widehat{\mu }:{ℝ}^n\to ℂ$ defined by

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

One proves using the dominated convergence theorem that $\widehat{\mu }$ is continuous.

Theorem 18.

If $\mu$ is a finite positive Borel measure on ${ℝ}^n$, then $\widehat{\mu }:{ℝ}^n\to ℂ$ is positive-definite.

Proof.

For ${\xi }_1,\mathrm{\dots },{\xi }_r\in {ℝ}^n$ and $c_1,\mathrm{\dots },c_r\in ℂ$, we have

${\displaystyle \sum _{j,k=1}^r}{c}_j\overline{c_k}\widehat{\mu }({\xi }_j-{\xi }_k)$ $=$ ${\displaystyle \sum _{j,k=1}^r}{c}_j\overline{c_k}{\displaystyle {\int }_{{ℝ}^n}}{e}^{-i({\xi }_j-{\xi }_k)\cdot x}𝑑\mu (x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \sum _{j,k=1}^r}{c}_j{e}^{-i{\xi }_j\cdot x}\overline{c_k{e}^{-i{\xi }_k\cdot x}}d\mu (x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle \sum _{j=1}^r}{c}_j{e}^{-i{\xi }_j\cdot x}\right)\overline{\left({\displaystyle \sum _{k=1}^r}{c}_k{e}^{-i{\xi }_k\cdot x}\right)}𝑑\mu (x)$

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

$\ge$ $0.$

∎

The following proof of Bochner’s theorem follows an exercise in Rudin.

Theorem 19 (Bochner).

If $\varphi :{ℝ}^n\to ℂ$ is continuous and positive-definite, then there is some finite positive Borel measure $\nu$ on ${ℝ}^n$ for which $\varphi =\widehat{\nu }$.

Proof.

Let $A$ be the Banach algebra defined in §4, whose elements are those complex Borel measures $\mu$ on ${ℝ}^n$ for which there is some $f\in L^1({ℝ}^n)$ and some $\alpha \in ℂ$ such that

$$d\mu =fd{m}_n+\alpha d\delta ,$$

where $m_n$ is Lebesgue measure on ${ℝ}^n$. For $f+\alpha \delta ,g+\beta \delta \in A$, we have

$$(f+\alpha \delta )*(g+\beta \delta )=(f*g+\beta f+\alpha g)+\alpha \beta \delta ;$$

we are identifying $f\in L^1({ℝ}^n)$ with the complex Borel measure whose Radon-Nikodym derivative with respect to $m_n$ is $f$. The norm on $A$ is the total variation norm of a complex measure; one checks that for $f+\alpha \delta$ this is

$$\parallel f+\alpha \delta \parallel =\parallel f\parallel +|\alpha |,$$

where $\parallel f\parallel ={\int }_{{ℝ}^n}|f(x)|𝑑m_n(x)$.

For $f\in L^1({ℝ}^n)$, we define $\tilde{f}\in L^1({ℝ}^n)$ by $\tilde{f}(x)=\overline{f(-x)}$, and we define ${}^{*}:A\to A$ by

$${(f+\alpha \delta )}^{*}=\tilde{f}+\overline{\alpha }\delta ,f+\alpha \delta \in A.$$

On the one hand,

${((f+\alpha \delta )*(g+\beta \delta ))}^{*}$ $=$ ${((f*g+\beta f+\alpha g)+\alpha \beta \delta )}^{*}$

$=$ $\widetilde{f*g}+\overline{\beta }\tilde{f}+\overline{\alpha }\tilde{g}+\overline{\alpha \beta }\delta$

$=$ $\widetilde{f*g}+\overline{\beta }\tilde{f}+\overline{\alpha }\tilde{g}+\overline{\alpha \beta }\delta ,$

and

$$\widetilde{f*g}(x)=\overline{{\int }_{{ℝ}^n}f(y)g(-x-y)𝑑m_n(y)}.$$

On the other hand,

${(g+\beta \delta )}^{*}*{(f+\alpha \delta )}^{*}$ $=$ $(\tilde{g}+\overline{\beta }\delta )*(\tilde{f}+\overline{\alpha }\delta )$

$=$ $(\tilde{g}*\tilde{f}+\overline{\alpha }\tilde{g}+\overline{\beta }\tilde{f})+\overline{\beta \alpha }\delta ,$

and

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

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

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

Therefore we have

$${((f+\alpha \delta )*(g+\beta \delta ))}^{*}={(g+\beta \delta )}^{*}*{(f+\alpha \delta )}^{*}.$$

Thus ${}^{*}:A\to A$ is an involution (the other properties demanded of an involution are immediate).

We define $F:A\to ℂ$ by

$$F(f+\alpha \delta )={\int }_{{ℝ}^n}f\varphi 𝑑m_n+\alpha \varphi (0),f+\alpha \delta \in A.$$

It is apparent that $F$ is linear, and because $|\varphi (x)|\le \varphi (0)$ for all $x$,

$|F(f+\alpha \delta )|$ $\le$ $\left|{\displaystyle {\int }_{{ℝ}^n}}f\varphi 𝑑m_n\right|+|\alpha |\varphi (0)$

$\le$ ${\displaystyle {\int }_{{ℝ}^n}}|f||\varphi |𝑑m_n+|\alpha |\varphi (0)$

$\le$ $\varphi (0){\displaystyle {\int }_{{ℝ}^n}}|f|𝑑m_n+|\alpha |\varphi (0)$

$=$ $\varphi (0)\parallel f+\alpha \delta \parallel ,$

from which it follows that $\parallel F\parallel =\varphi (0)$, and in particular that $F$ is bounded. Let $A_0=\{f+\alpha \delta \in A:f\in C_c({ℝ}^n)\}$. Because $F:A\to ℂ$ is bounded and $A_0$ is a dense subset of $A$, to prove that $F$ is a positive linear functional it suffices to prove that for all $f+\alpha \delta \in A_0$ we have $F((f+\alpha \delta )*{(f+\alpha \delta )}^{*})\ge 0$.

For $g\in C_c({ℝ}^n)$, by Theorem 17 we obtain

$$F(g*g^{*})=F(g*\tilde{g})={\int }_{{ℝ}^n}(g*\tilde{g})\varphi 𝑑m_n\ge 0.$$ (10)

Define $\eta :{ℝ}^n\to ℝ$ by

and for $ϵ>0$, define ${\eta }_{ϵ}:{ℝ}^n\to ℝ$ by ${\eta }_{ϵ}(x)={ϵ}^{-n}\eta \left(\frac{x}{ϵ}\right)$. Let $f+\alpha \delta \in A_0$ and define $g_{ϵ}=f+\alpha {\eta }_{ϵ}\in C_c({ℝ}^n)$. From (10) we have $F(g_{ϵ}*g_{ϵ}^{*})\ge 0$ for any $ϵ>0$. On the other hand,

$F(g_{ϵ}*g_{ϵ}^{*})$ $=$ $F((f+\alpha {\eta }_{ϵ})*(\tilde{f}+\overline{\alpha }{\eta }_{ϵ}))$

$=$ $F(f*\tilde{f}+\overline{\alpha }f*{\eta }_{ϵ}+\alpha {\eta }_{ϵ}*\tilde{f}+{|\alpha |}^2{\eta }_{ϵ}*{\eta }_{ϵ})$

$=$ $F(f*\tilde{f})+\overline{\alpha }{\displaystyle {\int }_{{ℝ}^n}}(f*{\eta }_{ϵ})\varphi 𝑑m_n+\alpha {\displaystyle {\int }_{{ℝ}^n}}({\eta }_{ϵ}*\tilde{f})\varphi 𝑑m_n$

$+{|\alpha |}^2{\displaystyle {\int }_{{ℝ}^n}}({\eta }_{ϵ}*{\eta }_{ϵ})\varphi 𝑑m_n.$

We take as granted that

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

as $ϵ\to 0$, that

$${\int }_{{ℝ}^n}({\eta }_{ϵ}*\tilde{f})\varphi 𝑑m_n\to {\int }_{{ℝ}^n}\tilde{f}\varphi 𝑑m_n$$

as $ϵ\to 0$, and that

$${\int }_{{ℝ}^n}({\eta }_{ϵ}*{\eta }_{ϵ})\varphi 𝑑m_n\to \varphi (0)$$

as $ϵ\to 0$. Furthermore,

$F((f+\alpha \delta )*{(f+\alpha \delta )}^{*})$ $=$ $F((f+\alpha \delta )*(\tilde{f}+\overline{\alpha }\delta ))$

$=$ $F(f*\tilde{f}+\overline{\alpha }f+\alpha \tilde{f}+{|\alpha |}^2)$

Thus

$$F(g_{ϵ}*g_{ϵ}^{*})\to F((f+\alpha \delta )*{(f+\alpha \delta )}^{*})$$

as $ϵ\to 0$. Since $F(g_{ϵ}*g_{ϵ}^{*})\ge 0$ for all $ϵ>0$, it follows that

$$F((f+\alpha \delta )*{(f+\alpha \delta )}^{*})\ge 0.$$

Therefore, $F:A\to ℂ$ is a positive linear functional.

Because $F$ is a positive linear functional and $F(e)=F(\delta )=1$, we can apply Theorem 12, according to which there is a regular positive Borel measure $\mu$ on $\mathrm{\Delta }$ satisfying

$$F(f+\alpha \delta )={\int }_{\mathrm{\Delta }}\mathrm{\Gamma }(f+\alpha \delta )𝑑\mu ,f+\alpha \delta \in A,$$

and hence, from the definition of $F$,

$${\int }_{{ℝ}^n}f\varphi 𝑑m_n+\alpha \varphi (0)={\int }_{\mathrm{\Delta }}\mathrm{\Gamma }(f+\alpha \delta )𝑑\mu ,f+\alpha \in A.$$

We state the following again from §4 for easy access. If $t\in {ℝ}^n$, define $h_t:A\to ℂ$ by

$$h_t(f+\alpha \delta )=\widehat{f}(t)+\alpha ,f+\alpha \delta \in A,$$

and also define $h_{\mathrm{\infty }}:A\to ℂ$ by

$$h_{\mathrm{\infty }}(f+\alpha \delta )=\alpha ,f+\alpha \delta \in A.$$

Let ${ℝ}^n\cup \{\mathrm{\infty }\}$ be the one-point compactification of ${ℝ}^n$. We proved in §4 that the map $T:{ℝ}^n\cup \{\mathrm{\infty }\}\to \mathrm{\Delta }$ defined by $T(t)=h_t$ is a homeomorphism. With $\nu ={(T^{-1})}_{*}\mu$ we have $\mu =T_{*}\nu$, and then

${\displaystyle {\int }_{\mathrm{\Delta }}}\mathrm{\Gamma }(f+\alpha \delta )𝑑\mu$ $=$ ${\displaystyle {\int }_{\mathrm{\Delta }}}\mathrm{\Gamma }(f)𝑑\mu +\alpha {\displaystyle {\int }_{\mathrm{\Delta }}}\mathrm{\Gamma }(\delta )𝑑\mu$

$=$ ${\displaystyle {\int }_{\mathrm{\Delta }}}\mathrm{\Gamma }(f)d(T_{*}\nu )+\alpha {\displaystyle {\int }_{\mathrm{\Delta }}}{\chi }_{\mathrm{\Delta }}𝑑\mu$

$=$ ${\displaystyle {\int }_{{ℝ}^n\cup \{\mathrm{\infty }\}}}\mathrm{\Gamma }(f)\circ T𝑑\nu +\alpha \mu (\mathrm{\Delta })$

$=$ ${\displaystyle {\int }_{{ℝ}^n\cup \{\mathrm{\infty }\}}}\mathrm{\Gamma }(f)\circ T(t)𝑑\nu (t)+\alpha F(\delta )$

$=$ ${\displaystyle {\int }_{{ℝ}^n\cup \{\mathrm{\infty }\}}}{h}_t(f)𝑑\nu (t)+\alpha \varphi (0)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}\widehat{f}(t)𝑑\nu (t)+\alpha \varphi (0).$

Therefore

$${\int }_{{ℝ}^n}f\varphi 𝑑m_n+\alpha \varphi (0)={\int }_{{ℝ}^n}\widehat{f}(t)𝑑\nu (t)+\alpha \varphi (0),$$

i.e.

$${\int }_{{ℝ}^n}f\varphi 𝑑m_n={\int }_{{ℝ}^n}\widehat{f}(t)𝑑\nu (t).$$

As

${\displaystyle {\int }_{{ℝ}^n}}\widehat{f}(t)𝑑\nu (t)$ $=$ ${\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}{e}^{-it\cdot x}f(x)𝑑m_n(x)\right)𝑑\nu (t)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}f(x){\displaystyle {\int }_{{ℝ}^n}}{e}^{-ix\cdot t}𝑑\nu (t)𝑑m_n(x)$

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

we have

$${\int }_{{ℝ}^n}f\varphi 𝑑m_n={\int }_{{ℝ}^n}f\widehat{\nu }𝑑m_n.$$

This is true for all $f\in L^1({ℝ}^n)$, from which it follows that $\varphi =\widehat{\nu }$. ∎