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

Jordan Bell
June 24, 2015

## 1 Introduction

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

If $A$ is a Banach algebra and $e\in A$ satisfies $xe=x$ and $ex=x$ for all $x\in A$, and also $\left\|e\right\|=1$, 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 $\lambda\in\mathbb{C}$ 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 $\sigma(x)\neq\emptyset$.11 1 Walter Rudin, Functional Analysis, second ed., p. 253, Theorem 10.13.

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 $A\to\mathbb{C}$.

###### Proof.

Let $x\in A$. $\sigma(x)\neq\emptyset$. 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 $\alpha\in\mathbb{C}$, 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 $\lambda_{0}\in\mathbb{C}$, 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 $A\to\mathbb{C}$.

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

 $\left\|x\right\|=\left\|\lambda(x)e\right\|=|\lambda(x)|\left\|e\right\|=|% \lambda(x)|,$

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

## 2 Complex homomorphisms

An ideal $J$ of an algebra $A$ is said to be proper if $J\neq A$. 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 $\mathbb{C}$.22 2 Walter Rudin, Functional Analysis, second ed., p. 277, Theorem 11.5.

###### Theorem 2.

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

1. 1.

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

2. 2.

If $h\in\Delta$ then $\ker h$ is a maximal ideal of $A$.

3. 3.

$x\in A$ is invertible if and only if $h(x)\neq 0$ for all $h\in\Delta$.

4. 4.

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

5. 5.

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

## 3 The Gelfand transform and maximal ideals

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

 $\hat{x}(h)=h(x),\qquad h\in\Delta.$

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

We define $\widehat{A}=\{\hat{x}:x\in A\}$, and we call the set $\Delta$ with the initial topology for $\widehat{A}$ the maximal ideal space of $A$. That is, the topology of $\Delta$ is the coarsest topology on $\Delta$ such that each $\hat{x}:\Delta\to\mathbb{C}$ is continuous. If $X$ is a topological space, we denote by $C(X)$ the set of all continuous functions $X\to\mathbb{C}$. $C(X)$ 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.33 3 Walter Rudin, Functional Analysis, second ed., p. 280, Theorem 11.9.

###### Theorem 3.

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

1. 1.

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

2. 2.

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

3. 3.

$\Delta$ is a compact Hausdorff space.

###### Proof.

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

 $\Gamma(\alpha x+y)(h)=h(\alpha x+y)=\alpha h(x)+h(y)=\alpha\Gamma(x)(h)+\Gamma% (y)(h)=(\Gamma(x)+\Gamma(y)(h),$

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

 $\Gamma(xy)(h)=h(xy)=h(x)h(y)=\Gamma(x)(h)\Gamma(y)(h)=(\Gamma(x)\Gamma(y))(h),$

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

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

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

 $K=\{\lambda\in A^{*}:\left\|\lambda\right\|\leq 1\}.$

By the Banach-Alaoglu theorem, $K$ is a weak-* compact subset of $A^{*}$. If $h\in\Delta$, then because $h$ is an algebra homomorphism $A\to\mathbb{C}$ it follows that $\left\|h\right\|\leq 1$.44 4 Walter Rudin, Functional Analysis, second ed., p. 249, Theorem 10.7. Thus, $\Delta\subset K$. Therefore, to prove that $\Delta$ is compact it suffices to prove that $\Delta$ is a weak-* closed subset of $A^{*}$.

Suppose that $h_{i}\in\Delta$ 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 $\lambda\neq 0$. 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 $\lambda:A\to\mathbb{C}$ is an algebra homomorphism, and hence that $\lambda\in\Delta$. Therefore $\Delta$ is a weak-* closed subset of $A^{*}$. ∎

If $A$ is a commutative unital Banach algebra, the above theorem shows that $\Gamma:A\to\widehat{A}$ is an algebra isomorphism if and only if $\mathrm{rad}\,A=\{0\}$, i.e., $\Gamma$ 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 $\mathrm{im}\,\hat{x}=\sigma(x)$. This gives us

 $\left\|\hat{x}\right\|_{\infty}=\rho(x),$ (1)

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

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

Therefore, $\hat{x}=0$ 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 $\rho(x)\leq\left\|x\right\|$.55 5 Walter Rudin, Functional Analysis, second ed., p. 253, Theorem 10.13. Therefore,

 $\left\|\hat{x}\right\|_{\infty}\leq\left\|x\right\|.$ (2)

In the proof of Theorem 3 we used the fact66 6 Walter Rudin, Functional Analysis, second ed., p. 249, Theorem 10.7. that the norm of any algebra homomorphism from a Banach algebra to $\mathbb{C}$ is $\leq 1$. In particular, this means that any algebra homomorphism from a Banach algebra to $\mathbb{C}$ is continuous. The following theorem shows that any algebra homomorphism from a Banach algebra to a commutative unital semisimple Banach algebra is continuous.77 7 Walter Rudin, Functional Analysis, second ed., p. 281, Theorem 11.10.

###### 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 theorem88 8 Walter Rudin, Functional Analysis, second ed., p. 51, Theorem 2.15. 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 $h\in\Delta_{B}$. Then $\phi=h\circ\psi:A\to\mathbb{C}$ is an algebra homomorphism. Because $h:B\to\mathbb{C}$ and $\phi:A\to\mathbb{C}$ are algebra homomorphisms with codomain $\mathbb{C}$, they are both continuous. Therefore, $h(y_{n})\to h(y)$ and $\phi(x_{n})\to\phi(x)$. Therefore,

 $h(y)=\lim h(y_{n})=\lim h(\psi(x_{n}))=\lim(h\circ\psi)(x_{n})=\lim\phi(x_{n})% =\phi(x)=h(\psi(x)),$

so $h(y-\psi(x))=0$. This is true for all $h\in\Delta_{B}$, 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 $\left\|\hat{x}\right\|_{\infty}\leq\left\|x\right\|$. The following lemma99 9 Walter Rudin, Functional Analysis, second ed., p. 282, Lemma 11.11. shows that if $\left\|x^{2}\right\|=\left\|x\right\|^{2}$ and $x\neq 0$, then $\inf\frac{\left\|\hat{x}\right\|_{\infty}}{\left\|x\right\|}\geq 1$, hence that $\left\|\hat{x}\right\|_{\infty}=\left\|x\right\|$. Therefore, if $\left\|x^{2}\right\|=\left\|x\right\|^{2}$ for all $x\in A$, then $\Gamma:A\to C(\Delta)$ is an isometry.

###### Lemma 5.

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

 $r=\inf_{x\neq 0}\frac{\left\|x^{2}\right\|}{\left\|x\right\|^{2}},\qquad s=% \inf_{x\neq 0}\frac{\left\|\hat{x}\right\|_{\infty}}{\left\|x\right\|},$

then $s^{2}\leq r\leq s$.

Theorem 3 shows that if $A$ is a commutative unital Banach algebra, then $\Gamma:A\to C(\Delta)$ is an algebra homomorphism. Therefore $\Gamma(A)=\widehat{A}$ is a subalgebra of $C(\Delta)$. Moreover, Theorem 3 also shows that $\Delta$ is a compact Hausdorff space. Therefore, $C(\Delta)$ 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 $C(\Delta)$ it is necessary and sufficient that $\widehat{A}$ be a closed subset of the Banach algebra $C(\Delta)$. The following theorem gives conditions under which this occurs.1010 10 Walter Rudin, Functional Analysis, second ed., p. 282, Theorem 11.12.

###### Theorem 6.

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

###### Proof.

Suppose that there is some $0 such that $x\in A$ implies that $\left\|x\right\|^{2}\leq K\left\|x^{2}\right\|$. Then

 $r=\inf_{x\neq 0}\frac{\left\|x^{2}\right\|}{\left\|x\right\|^{2}}\geq\inf_{x% \neq 0}\frac{\left\|x^{2}\right\|}{K\left\|x^{2}\right\|}=\frac{1}{K}.$

By Lemma 5, with $s=\inf_{x\neq 0}\frac{\left\|\hat{x}\right\|_{\infty}}{\left\|x\right\|}$ we have

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

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

Suppose that $A$ is semisimple and that $\widehat{A}$ is a closed subset of $C(\Delta)$. The fact that $A$ is semisimple gives us by Theorem 3 that $\Gamma:A\to\widehat{A}$ is a bijection. The fact that $\widehat{A}$ is closed means that $\widehat{A}$ is a Banach algebra. Because $\Gamma:A\to\widehat{A}$ is continuous, linear, and a bijection, by the open mapping theorem1111 11 Walter Rudin, Functional Analysis, second ed., p. 49, Corollary 2.12. it follows that there are positive real numbers $a,b$ such that if $x\in A$ then

 $a\left\|x\right\|\leq\left\|\Gamma x\right\|_{\infty}\leq b\left\|x\right\|.$

Then $\inf_{x\neq 0}\frac{\left\|\hat{x}\right\|_{\infty}}{\left\|x\right\|}\geq a$. By Lemma 5, it follows that $\inf_{x\neq 0}\frac{\left\|x^{2}\right\|}{\left\|x\right\|^{2}}\geq a^{2}$. Hence, for all $x\neq 0$ we have $\left\|x\right\|^{2}\leq K\left\|x^{2}\right\|$, with $K=\frac{1}{a^{2}}$. ∎

## 4 L1

Let $M(\mathbb{R}^{n})$ denote the set of all complex Borel measures on $\mathbb{R}^{n}$, and let $S:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}^{n}$ be defined by $S(x,y)=x+y$. For $\mu_{1},\mu_{2}\in M(\mathbb{R}^{n})$, we denote by $\mu_{1}\times\mu_{2}$ the product measure on $\mathbb{R}^{n}\times\mathbb{R}^{n}$, and we define the convolution of $\mu_{1}$ and $\mu_{2}$ 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 $\mathbb{R}^{n}$, then

 $\displaystyle(\mu_{1}*\mu_{2})(E)$ $\displaystyle=$ $\displaystyle(S_{*}(\mu_{1}\times\mu_{2}))(E)$ $\displaystyle=$ $\displaystyle(\mu_{1}\times\mu_{2})(S^{-1}(E))$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\chi_{E}(x+y)d\mu_{1}(x% )d\mu_{2}(y).$

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

If $\mu\in M(\mathbb{R}^{n})$, the variation of $\mu$ is the measure $|\mu|\in M(\mathbb{R}^{n})$, where for a Borel subset $E$ of $\mathbb{R}^{n}$, 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 $\left\|\mu\right\|=|\mu|(\mathbb{R}^{n})$. One proves that $\left\|\cdot\right\|$ is a norm on $M(\mathbb{R}^{n})$ and that with this norm, $M(\mathbb{R}^{n})$ is a Banach algebra.1212 12 See Walter Rudin, Real and Complex Analysis, third ed., chapter 6.

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

 $d\mu=fdm_{n}+\alpha d\delta.$

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

If $f+\alpha\delta,g+\beta\delta\in A$ (identifying $f\in L^{1}(\mathbb{R}^{n})$ 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

 $(f*g)(x)=\int_{\mathbb{R}^{n}}f(y)g(x-y)dm_{n}(y).$

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

 $\hat{f}(t)=\int_{\mathbb{R}^{n}}fe_{-t}dm_{n},\qquad t\in\mathbb{R}^{n}.$

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

 $h_{t}(f+\alpha\delta)=\hat{f}(t)+\alpha,\qquad f+\alpha\delta\in A,$

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

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

By (3) it is apparent that for each $t\in\mathbb{R}^{n}\cup\{\infty\}$, the map $h_{t}$ is a homomorphism of algebras. It can be proved that $\Delta=\{h_{t}:t\in\mathbb{R}^{n}\}\cup\{h_{\infty}\}$.1313 13 Walter Rudin, Functional Analysis, second ed., p. 285. Let $\mathbb{R}^{n}\cup\{\infty\}$ be the one-point compactification of $\mathbb{R}^{n}$, and define $T:\mathbb{R}^{n}\cup\{\infty\}\to\Delta$ by $T(t)=h_{t}$, which is a bijection.

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

 $T(t_{k})(f+\alpha\delta)=h_{t_{k}}(f+\alpha\delta)=\hat{f}(t_{k})+\alpha\to% \hat{f}(t)+\alpha=T(t)(f+\alpha\delta).$

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

 $T(t_{k})(f+\alpha\delta)=\hat{f}(t_{k})+\alpha\to\alpha=h_{\infty}(f+\alpha% \delta)=T(\infty)(f+\alpha\delta).$

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

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

## 5 Involutions

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

1. 1.

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

2. 2.

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

3. 3.

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

4. 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

 $\left\|xx^{*}\right\|=\left\|x\right\|^{2},\qquad x\in A,$

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

The following theorem shows that a commutative unital $B^{*}$-algebra with maximal ideal space $\Delta$ is isomorphic as a $B^{*}$-algebra to $C(\Delta)$.1414 14 Walter Rudin, Functional Analysis, second ed., p. 289, Theorem 11.18. (An isomorphism of $B^{*}$-algebras is an isomorphism of Banach algebras that preserves the involution; the involution on $C(\Delta)$ is $(x\mapsto f(x))\mapsto(x\mapsto\overline{f(x)})$.)

###### Theorem 7 (Gelfand-Naimark).

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

###### Proof.

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

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

and

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

hence

 $\alpha^{2}+(\beta+t)^{2}=|h(z)|^{2}\leq\left\|z\right\|^{2}=\left\|zz^{*}% \right\|=\left\|u^{2}+t^{2}e\right\|\leq\left\|u\right\|^{2}+t^{2},$

i.e.

 $\alpha^{2}+\beta^{2}+2\beta t\leq\left\|u\right\|^{2}.$

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

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 $\Gamma(x^{*})=\overline{\Gamma(x)}$. In particular, $\widehat{A}$ is closed under complex conjugation. If $h_{1}\neq h_{2}$, then there is some $x\in A$ for which $h_{1}(x)\neq h_{2}(x)$, i.e. $\hat{x}(h_{1})\neq\hat{x}(h_{2})$, so $\widehat{A}$ separates points in $\Delta$. Because $\widehat{A}$ is a unital Banach algebra, it follows from the Stone-Weierstrass theorem that $\widehat{A}$ is dense in $C(\Delta)$.

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

 $\left\|y^{2m}\right\|=\left\|y^{m}y^{m}\right\|=\left\|y^{m}(y^{m})^{*}\right% \|=\left\|y^{m}\right\|^{2}=(\left\|y\right\|^{m})^{2}=\left\|y\right\|^{2m}.$

The spectral radius formula1515 15 Walter Rudin, Functional Analysis, second ed., p. 253, Theorem 10.13. gives

 $\rho(y)=\lim\left\|y^{n}\right\|^{1/n},$

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

 $\rho(y)=\left\|y\right\|.$

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

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

That is, $\hat{y}=|\hat{x}|^{2}$ and with $\left\|\hat{y}\right\|_{\infty}=\left\|y\right\|$ we obtain

 $\left\|\hat{x}\right\|_{\infty}^{2}=\left\|\hat{y}\right\|_{\infty}=\left\|y% \right\|=\left\|xx^{*}\right\|=\left\|x\right\|^{2},$

i.e.

 $\left\|\hat{x}\right\|_{\infty}=\left\|x\right\|.$

This shows that $\Gamma:A\to C(\Delta)$ is an isometry. In particular, $\Gamma$ maps closed sets to closed sets, so $\widehat{A}=\Gamma(A)$ is a closed subset of $C(\Delta)$. We have already established that $\widehat{A}$ is dense in $C(\Delta)$, so $\widehat{A}=C(\Delta)$. The fact that $\Gamma$ is an isometry yields that $\Gamma$ is one-to-one, and the fact that $\widehat{A}=C(\Delta)$ means that that $\Gamma$ is onto, hence $\Gamma$ is a bijection, and therefore it is an isomorphism of algebras. Because $\Gamma$ 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.1616 16 Walter Rudin, Functional Analysis, second ed., p. 294, Theorem 11.26.

###### 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 $\lambda\leq 0$, 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 $xx^{*}=x^{*}x$. If $A$ is a Banach algebra with involution ${}^{*}:A\to A$, by $x\geq 0$ we mean that $x$ is self-adjoint and $\sigma(x)\subseteq[0,\infty)$, and we say that $x$ is positive. The following theorem states basic facts about the spectrum of elements of a unital $B^{*}$-algebra.1717 17 Walter Rudin, Functional Analysis, second ed., p. 294, Theorem 11.28.

###### Theorem 9.

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

1. 1.

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

2. 2.

If $x$ is normal, then $\rho(x)=\left\|x\right\|$.

3. 3.

If $x\in A$, then $\rho(xx^{*})=\left\|x\right\|^{2}$.

4. 4.

If $x\geq 0$ and $y\geq 0$, then $x+y\geq 0$.

5. 5.

If $x\in A$, then $xx^{*}\geq 0$.

6. 6.

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

## 6 Positive linear functionals

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

We shall use the following theorem to prove the theorem after it.1818 18 Walter Rudin, Functional Analysis, second ed., p. 137, Theorem 5.20.

###### 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 $\gamma<\infty$ 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

 $\left\|x_{1}\right\|+\left\|x_{2}\right\|\leq\gamma\left\|x\right\|.$

The following theorem establishes some basic properties of positive linear functionals on a unital Banach algebra with an involution.1919 19 Walter Rudin, Functional Analysis, second ed., p. 296, Theorem 11.31.

###### Theorem 11.

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

1. 1.

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

2. 2.

$|F(xy^{*})|^{2}\leq F(xx^{*})F(yy^{*})$.

3. 3.

$|F(x)|^{2}\leq F(e)F(xx^{*})\leq F(e)^{2}\rho(xx^{*})$.

4. 4.

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

5. 5.

If $A$ is commutative, then $\left\|F\right\|=F(e)$.

6. 6.

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

7. 7.

$F$ is a bounded linear map.

###### Proof.

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

 $F((x+\alpha y)(x+\alpha y)^{*})=F((x+\alpha y)(x^{*}+\overline{\alpha}y^{*}))=% F(xx^{*}+\overline{\alpha}xy^{*}+\alpha yx^{*}+|\alpha|^{2}yy^{*}).$

Therefore,

 $F(xx^{*})+\overline{\alpha}F(xy^{*})+\alpha F(yx^{*})+|\alpha|^{2}F(yy^{*})% \geq 0.$ (4)

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

 $F(xx^{*})+F(xy^{*})+F(yx^{*})+F(yy^{*})\geq 0.$

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

 $F(xx^{*})-iF(xy^{*})+iF(yx^{*})+F(yy^{*})\geq 0.$

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

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

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

 $F(xx^{*})+\frac{t}{|F(xy^{*})|}\overline{F(xy^{*})}F(xy^{*})+\frac{t}{|F(xy^{*% })|}F(xy^{*})F(yx^{*})+t^{2}F(yy^{*})\geq 0,$

i.e.

 $F(xx^{*})+t|F(xy^{*})|+\frac{t}{|F(xy^{*})|}F(xy^{*})F(yx^{*})+t^{2}F(yy^{*})% \geq 0,$

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

 $F(xx^{*})+2t|F(xy^{*})|+t^{2}F(yy^{*})\geq 0.$

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

 $F(xx^{*})-2\frac{|F(xy^{*})|^{2}}{F(yy^{*})}+\frac{|F(xy^{*})|^{2}}{F(yy^{*})}% \geq 0,$

i.e.

 $|F(xy^{*})|^{2}\leq F(xx^{*})F(yy^{*}).$

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

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

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

 $F(te-xx^{*})=F(u^{2})=F(uu^{*})\geq 0,$

so

 $F(xx^{*})\leq tF(e).$

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

 $F(xx^{*})\leq F(e)\rho(xx^{*}).$

Suppose that $x$ is normal. It is a fact that if $x$ and $y$ belong to a unital Banach algebra and $xy=yx$, then $\sigma(xy)\subseteq\sigma(x)\sigma(y)$.2020 20 Walter Rudin, Functional Analysis, second ed., p. 293, Theorem 11.23. Thus $\sigma(xx^{*})\subseteq\sigma(x)\sigma(x^{*})$, from which we get

 $\rho(xx^{*})\leq\rho(x)\rho(x^{*}).$

It is a fact that $\sigma(x^{*})=\overline{\sigma(x)}$,2121 21 Walter Rudin, Functional Analysis, second ed., p. 288, Theorem 11.15., so we have $\rho(x)=\rho(x^{*})$, and thus

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

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

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

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

 $|F(x)|\leq F(e)\left\|x\right\|.$

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

Suppose that there is some $\beta$ such that $\left\|x^{*}\right\|\leq\beta\left\|x\right\|$ for all $x\in A$. We have $\rho(xx^{*})\leq\left\|xx^{*}\right\|\leq\left\|x\right\|\left\|x^{*}\right\|% \leq\beta\left\|x\right\|^{2}$. (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 $\left\|xx^{*}\right\|=\left\|x\right\|\left\|x^{*}\right\|=\left\|x\right\|^{2}$.) Using $|F(x)|^{2}\leq F(e)^{2}\rho(xx^{*})$ then gives us $|F(x)|^{2}\leq\beta F(e)^{2}\left\|x\right\|^{2}$, hence

 $|F(x)|\leq\beta^{1/2}F(e)\left\|x\right\|.$

If $F(e)=0$, then $|F(x)|^{2}\leq 0$ 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 $H\to\mathbb{R}$. For $u\in H$, because $u$ is self-adjoint it is in particular normal, and so $|F(x)|\leq F(e)\rho(x)\leq F(e)\left\|x\right\|=\left\|x\right\|$, because $F(e)=1$. Hence the restriction of $F$ to $H$ is a real-linear map $H\to\mathbb{R}$ with norm $1$, and therefore there is a unique bounded real-linear map $\Phi:\overline{H}\to\mathbb{R}$ whose restriction to $H$ is equal to the restriction of $F$ to $H$, and $\left\|\Phi\right\|=1$.

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 $iv_{n}\to y$. Then $u_{n}^{2}\to y^{2}$ and $-v_{n}^{2}\to y$, or $v_{n}^{2}\to-y^{2}$. Because $|F(u_{n})|^{2}\leq F(e)F(u_{n}u_{n}^{*})=F(u_{n}^{2})$, we have

 $|F(u_{n})|^{2}\leq F(u_{n}^{2})\leq F(u_{n}^{2}+v_{n}^{2}).$

Because $u_{n}$ and $v_{n}$ are self-adjoint, $u_{n}^{2}+v_{n}^{2}$ is normal and hence

 $|F(u_{n}^{2}+v_{n}^{2})|\leq F(e)\rho(u_{n}^{2}+v_{n}^{2})=\rho(u_{n}^{2}+v_{n% }^{2})\leq\left\|u_{n}^{2}+v_{n}^{2}\right\|,$

and so we have

 $|F(u_{n})|^{2}\leq\left\|u_{n}^{2}+v_{n}^{2}\right\|.$

But $u_{n}^{2}\to y$ and $v_{n}^{2}\to-y$, so $\left\|u_{n}^{2}+v_{n}^{2}\right\|\to\left\|y-y\right\|=0$. Therefore, $F(u_{n})\to 0$, and so

 $\Phi(y)=\lim F(u_{n})\to 0.$

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 $\gamma<\infty$ such that for all $x\in A$, there are $x_{1}\in\overline{H}$ and $x_{2}\in\overline{H}$ satisfying

 $x=x_{1}+ix_{2},\qquad\left\|x_{1}\right\|+\left\|x_{2}\right\|\leq\gamma\left% \|x\right\|.$

Let $x\in A$ and let $x=x_{1}+ix_{2}$, 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,

 $F(x)=F(u+iv)=F(u)+iF(v)=F(x_{1})+iF(x_{2})=\Phi(x_{1})+i\Phi(x_{2}),$

and therefore, because $\left\|\Phi\right\|=1$ and because $\left\|x_{1}\right\|+\left\|x_{2}\right\|\leq\gamma\left\|x\right\|$,

 $|F(x)|\leq|\Phi(x_{1})+i\Phi(x_{2})|\leq|\Phi(x_{1})|+|\Phi(x_{2})|\leq\left\|% x_{1}\right\|+\left\|x_{2}\right\|\leq\gamma\left\|x\right\|,$

showing that $\left\|F\right\|\leq\gamma$, 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

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

and

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

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 $\left\|\mu\right\|=|\mu|(X)$. The following is the Riesz-Markov theorem, stated for complex Borel measures on a compact Hausdorff space.2222 22 Walter Rudin, Real and Complex Analysis, third ed., p. 130, Theorem 6.19.

###### Theorem 12 (Riesz-Markov).

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

 $\Lambda f=\int_{X}fd\mu,\qquad f\in C(X).$

This measure $\mu$ satisfies $\left\|\mu\right\|=\left\|\Lambda\right\|$.

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.2323 23 Walter Rudin, Functional Analysis, second ed., p. 299, Theorem 11.33.

###### Theorem 13.

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

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

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

 $F_{\mu}(x)=\int_{\Delta}\hat{x}d\mu,\qquad x\in A,$

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 $0\leq t\leq 1$, then $(1-t)F_{1}+tF_{2}$ is linear, and it is straightforward to check that it is positive. Moreover, $((1-t)F_{1}+tF_{2})(e)=(1-t)F_{1}(e)+tF_{2}(e)\leq(1-t)+t=1$, so $(1-t)F_{1}+tF_{2}\in K$. 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 $\Delta$, then for any $\epsilon>0$ there are compact subsets $F_{1},F_{2}$ of $\Delta$ such that $\mu_{1}(E)<\mu_{1}(F_{1})-\epsilon$ and $\mu_{2}(E)<\mu_{2}(F_{2})-\epsilon$. With $F=F_{1}\cup F_{2}$, we have

 $\displaystyle\mu(F)$ $\displaystyle=$ $\displaystyle a_{1}\mu_{1}(F)+a_{2}\mu_{2}(F)$ $\displaystyle\geq$ $\displaystyle a_{1}\mu_{1}(F_{1})+a_{2}\mu_{2}(F_{2})$ $\displaystyle\geq$ $\displaystyle a_{1}(\mu_{1}(E)+\epsilon)+a_{2}(\mu_{2}(E)+\epsilon)$ $\displaystyle=$ $\displaystyle\mu(E)+(a_{1}+a_{2})\epsilon.$

It follows that $\mu(E)=\sup\{\mu(F):\textrm{F is compact and F\subseteq E}\}$. If $E$ is a Borel subset of $\Delta$, then for any $\epsilon>0$ there are open subsets $G_{1},G_{2}$ of $\Delta$ such that $\mu_{1}(E)>\mu_{1}(G_{1})-\epsilon$ and $\mu_{2}(E)>\mu_{2}(G_{2})-\epsilon$. With $G=G_{1}\cap G_{2}$, we have

 $\displaystyle\mu(G)$ $\displaystyle=$ $\displaystyle a_{1}\mu_{1}(G)+a_{2}\mu_{2}(G)$ $\displaystyle\leq$ $\displaystyle a_{1}\mu_{1}(G_{1})+a_{2}\mu_{2}(G_{2})$ $\displaystyle<$ $\displaystyle a_{1}(\mu_{1}(E)+\epsilon)+a_{2}(\mu_{2}(E)+\epsilon)$ $\displaystyle=$ $\displaystyle\mu(E)+(a_{1}+a_{2})\epsilon.$

It follows that $\mu(E)=\inf\{\mu(G):\textrm{G is open and E\subseteq G}\}$. Therefore, $\mu=a_{1}\mu_{1}+a_{2}\mu_{2}$ is a regular positive Borel measure. In particular, if $0\leq t\leq 1$ and $a_{1}=1-t$, $a_{2}=t$, then $\mu$ is a regular positive Borel measure. Finally, for $\mu=(1-t)\mu_{1}+t\mu_{2}$, $0\leq t\leq 1$, we have, because $\mu_{1}(\Delta)\leq 1$ and $\mu_{2}(\Delta)\leq 1$,

 $\mu(\Delta)=(1-t)\mu_{1}(\Delta)+t\mu_{2}(\Delta)\leq(1-t)+t=1,$

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

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

 $F_{\mu}(xx^{*})=\int_{\Delta}\Gamma(xx^{*})d\mu=\int_{\Delta}|\Gamma(x)|^{2}d% \mu\geq 0,$

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

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

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 $\widehat{F}:\widehat{A}\to\mathbb{C}$

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

this makes sense because if $\hat{x}=\hat{y}$ then $\Gamma(x-y)=0$, 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

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

and by (1) we have $\rho(x)=\left\|\hat{x}\right\|_{\infty}$, so

 $|\widehat{F}(\hat{x})|\leq F(e)\left\|\hat{x}\right\|_{\infty}.$

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

 $\Lambda f=\int_{\Delta}fd\mu,\qquad f\in C(\Delta),$ (6)

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

 $\mu(\Delta)=\int_{\Delta}\chi_{\Delta}d\mu=\int_{\Delta}\hat{e}d\mu=\Lambda% \hat{e}=\widehat{F}(\hat{e})=F(e)=\left\|\mu\right\|=|\mu|(\Delta).$

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

 $F_{\mu}(x)=\int_{\Delta}\hat{x}d\mu=\Lambda\hat{x}=\widehat{F}(\hat{x})=F(x),$

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 $\{\delta_{x}:x\in X\}$.2424 24 Barry Simon, Convexity: An Analytic Viewpoint, p. 128, Example 8.16. Given this, one proves that the set of extreme points of $M$ is $\{0\}\cup\{\delta_{h}:h\in\Delta\}$. For $x\in A$, $F_{0}(x)=0$, i.e. $F_{0}=0$. For $h\in\Delta$ and $x\in A$,

 $F_{\delta_{h}}(x)=\int_{\Delta}\hat{x}d\delta_{h}=\hat{x}(h)=h(x),$

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

###### Corollary 14.

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

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

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

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

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 $\Lambda\in A^{*}$, one checks that $\Lambda(xx^{*})\geq 0$ for all $x\in A$ and that $\Lambda e\leq 1$. By the Banach-Alaoglu theorem, the set $B=\{\Lambda\in A^{*}:\left\|\Lambda\right\|\leq 1\}$ is weak-* compact, and if $F\in K$ then $\left\|F\right\|=F(e)$ by Theorem 11 and $F(e)\leq 1$, so $K\subseteq B$. Hence, $K$ is a weak-* compact subset of $A^{*}$. Therefore, the Krein-Milman theorem2525 25 Walter Rudin, Functional Analysis, second ed., p. 75, Theorem 3.23. 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 $\{0\}\cup\Delta$.

## 8 Positive definite functions

A function $\phi:\mathbb{R}^{n}\to\mathbb{C}$ is said to be positive-definite if $r\geq 1$, $x_{1},\ldots,x_{r}\in\mathbb{R}^{n},c_{1},\ldots,c_{r}\in\mathbb{C}$ imply that

 $\sum_{i,j=1}^{r}c_{i}\overline{c_{j}}\phi(x_{i}-x_{j})\geq 0;$

in particular, for $\phi$ 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, $\mathbb{R}$ is a vector space over $\mathbb{Q}$, and if $\psi:\mathbb{R}\to\mathbb{R}$ is a vector space automorphism of $\mathbb{R}$ over $\mathbb{Q}$, one proves that $x\mapsto e^{i\psi(x)}$ is a positive-definite function $\mathbb{R}\to\mathbb{C}$, 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 $\phi:\mathbb{R}^{n}\to\mathbb{C}$ is a measurable positive-definite function, then there is a continuous positive-definite function $\mathbb{R}^{n}\to\mathbb{C}$ that is equal to $\phi$ almost everywhere.2626 26 Vladimir I. Bogachev, Measure Theory, vol. 1, p. 221, Theorem 3.10.20. See also Anthony W. Knapp, Basic Real Analysis, p. 406.

###### Theorem 15.

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

 $\phi(0)\geq 0,$ (7)

and for all $x\in\mathbb{R}^{n}$ we have

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

and

 $|\phi(x)|\leq\phi(0).$ (9)
###### Proof.

Using $r=1$ and $c_{1}=1$, we have for all $x_{1}\in\mathbb{R}^{n}$ that $\phi(x_{1}-x_{1})\geq 0$, i.e. $\phi(0)\geq 0$.

Using $r=2$, for $x_{1},x_{2}\in\mathbb{R}^{n}$ and $c_{1},c_{2}\in\mathbb{C}$ we have

 $c_{1}\overline{c_{1}}\phi(x_{1}-x_{1})+c_{1}\overline{c_{2}}\phi(x_{1}-x_{2})+% c_{2}\overline{c_{1}}\phi(x_{2}-x_{1})+c_{2}\overline{c_{2}}\phi(x_{2}-x_{2})% \geq 0.$

Take $x_{1}=x$ and $x_{2}=0$, with which

 $|c_{1}|^{2}\phi(0)+c_{1}\overline{c_{2}}\phi(x)+c_{2}\overline{c_{1}}\phi(-x)+% |c_{2}|^{2}\phi(0)\geq 0;$

in particular, the left-hand side is real, and because $\phi(0)$ is real by (7), this implies that $c_{1}\overline{c_{2}}\phi(x)+c_{2}\overline{c_{1}}\phi(-x)$ is real. That is, it is equal to its complex conjugate:

 $c_{1}\overline{c_{2}}\phi(x)+c_{2}\overline{c_{1}}\phi(-x)=\overline{c_{1}}c_{% 2}\overline{\phi(x)}+\overline{c_{2}}c_{1}\overline{\phi(-x)}.$

The fact that this holds every $c_{1},c_{2}\in\mathbb{C}$ implies that $\overline{\phi(x)}=\phi(-x)$.

Again using that

 $c_{1}\overline{c_{1}}\phi(x_{1}-x_{1})+c_{1}\overline{c_{2}}\phi(x_{1}-x_{2})+% c_{2}\overline{c_{1}}\phi(x_{2}-x_{1})+c_{2}\overline{c_{2}}\phi(x_{2}-x_{2})% \geq 0,$

with $x_{1}=x,x_{2}=0$ and $c_{2}=1$ we get

 $|c_{1}|^{2}\phi(0)+c_{1}\phi(x)+\overline{c_{1}}\phi(-x)+\phi(0)\geq 0.$

Applying (8) gives

 $|c_{1}|^{2}\phi(0)+c_{1}\phi(x)+\overline{c_{1}\phi(x)}+\phi(0)\geq 0.$

For $c_{1}\in\mathbb{C}$ such that $|c_{1}|=1$,

 $2\phi(0)+2\mathrm{Re}\,\left(c_{1}\phi(x)\right)\geq 0,$

or

 $-\mathrm{Re}\,\left(c_{1}\phi(x)\right)\leq\phi(0).$

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

The following lemma about positive-definite functions follows a proof in Bogachev.2727 27 Vladimir I. Bogachev, Measure Theory, vol. 1, p. 221, Lemma 3.10.19.

###### Lemma 16.

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

 $\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\phi(x-y)f(x)f(y)dm_{n}(x)dm_{n}(y)% \geq 0.$
###### Proof.

For $r\geq 2$ and for any $x_{1},\ldots,x_{r}$ and $c_{1}=1,\ldots,c_{r}=1$, we have

 $\sum_{j,k=1}^{r}\phi(x_{j}-x_{k})\geq 0,$

or

 $r\phi(0)+\sum_{j\neq k}\phi(x_{j}-x_{k})\geq 0.$

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

 $f(x_{1})\cdots f(x_{r})dm_{n}(x_{1})\cdots dm_{n}(x_{r}).$

Writing

 $I=\int_{\mathbb{R}^{n}}f(x)dm_{n}(x),$

we obtain

 $r\phi(0)I^{r}+\sum_{j\neq k}\int_{\mathbb{R}^{n}}\cdots\int_{\mathbb{R}^{n}}% \phi(x_{j}-x_{k})f(x_{1})\cdots f(x_{r})dm_{n}(x_{1})\cdots dm_{n}(x_{r})\geq 0,$

and so, writing

 $J=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\phi(x-y)f(x)f(y)dm_{n}(x)dm_{n}(y),$

we have

 $r\phi(0)I^{r}+\sum_{j\neq k}JI^{r-2}\geq 0,$

or

 $r\phi(0)I^{r}+r(r-1)JI^{r-2}\geq 0.$

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

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

This inequality holds for all $r\geq 2$, so taking $r\to\infty$ yields

 $J\geq 0,$

which is the claim. ∎

For $f,g\in L^{1}(\mathbb{R}^{n})$,

 $(f*g)(x)=\int_{\mathbb{R}^{n}}f(y)g(x-y)dm_{n}(y).$

The support of a function $f:\mathbb{R}^{n}\to\mathbb{C}$, denoted $\mathrm{supp}f$, is the closure of the set $\{x\in\mathbb{R}^{n}:f(x)\neq 0\}$. We denote by $C_{c}(\mathbb{R}^{n})$ the set of all continuous functions $f:\mathbb{R}^{n}\to\mathbb{C}$ such that $\mathrm{supp}f$ is a compact set. It is straightforward to check that an element of $C_{c}(\mathbb{R}^{n})$ is uniformly continuous on $\mathbb{R}^{n}$. The following theorem is similar to the previous lemma, but applies to functions that need not be nonnegative.2828 28 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 85, Proposition 3.35.

###### Theorem 17.

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

 $\int_{\mathbb{R}^{n}}(f*\widetilde{f})\psi\geq 0.$

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

 $\hat{\mu}(\xi)=\int_{\mathbb{R}^{n}}e_{-\xi}d\mu,\qquad\xi\in\mathbb{R}^{n}.$

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

###### Theorem 18.

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

###### Proof.

For $\xi_{1},\ldots,\xi_{r}\in\mathbb{R}^{n}$ and $c_{1},\ldots,c_{r}\in\mathbb{C}$, we have

 $\displaystyle\sum_{j,k=1}^{r}c_{j}\overline{c_{k}}\hat{\mu}(\xi_{j}-\xi_{k})$ $\displaystyle=$ $\displaystyle\sum_{j,k=1}^{r}c_{j}\overline{c_{k}}\int_{\mathbb{R}^{n}}e^{-i(% \xi_{j}-\xi_{k})\cdot x}d\mu(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\sum_{j,k=1}^{r}c_{j}e^{-i\xi_{j}\cdot x}% \overline{c_{k}e^{-i\xi_{k}\cdot x}}d\mu(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\left(\sum_{j=1}^{r}c_{j}e^{-i\xi_{j}\cdot x% }\right)\overline{\left(\sum_{k=1}^{r}c_{k}e^{-i\xi_{k}\cdot x}\right)}d\mu(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\left|\sum_{j=1}^{r}c_{j}e^{-i\xi_{j}\cdot x% }\right|^{2}d\mu(x)$ $\displaystyle\geq$ $\displaystyle 0.$

The following proof of Bochner’s theorem follows an exercise in Rudin.2929 29 Walter Rudin, Functional Analysis, second ed., p. 303, Exercise 14. Other references on Bochner’s theorem are the following: Barry Simon, Convexity: An Analytic Viewpoint, p. 153, Theorem 9.17; Edwin Hewitt and Kenneth A. Ross, Abstract Harmonic Analysis, vol. II, p. 293, Theorem 33.3; Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 220, Theorem 3.9.16; Walter Rudin, Fourier Analysis on Groups, p. 19, Theorem 1.4.3; Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 170; Vladimir I. Bogachev, Measure Theory, vol. II, p. 121, Theorem 7.13.1; Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 95, Theorem 4.18.

###### Theorem 19 (Bochner).

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

###### Proof.

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

 $d\mu=fdm_{n}+\alpha d\delta,$

where $m_{n}$ is Lebesgue measure on $\mathbb{R}^{n}$. 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 $f\in L^{1}(\mathbb{R}^{n})$ 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

 $\left\|f+\alpha\delta\right\|=\left\|f\right\|+|\alpha|,$

where $\left\|f\right\|=\int_{\mathbb{R}^{n}}|f(x)|dm_{n}(x)$.

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

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

On the one hand,

 $\displaystyle((f+\alpha\delta)*(g+\beta\delta))^{*}$ $\displaystyle=$ $\displaystyle((f*g+\beta f+\alpha g)+\alpha\beta\delta)^{*}$ $\displaystyle=$ $\displaystyle\widetilde{f*g}+\overline{\beta}\widetilde{f}+\overline{\alpha}% \widetilde{g}+\overline{\alpha\beta}\delta$ $\displaystyle=$ $\displaystyle\widetilde{f*g}+\overline{\beta}\widetilde{f}+\overline{\alpha}% \widetilde{g}+\overline{\alpha\beta}\delta,$

and

 $\widetilde{f*g}(x)=\overline{\int_{\mathbb{R}^{n}}f(y)g(-x-y)dm_{n}(y)}.$

On the other hand,

 $\displaystyle(g+\beta\delta)^{*}*(f+\alpha\delta)^{*}$ $\displaystyle=$ $\displaystyle(\widetilde{g}+\overline{\beta}\delta)*(\widetilde{f}+\overline{% \alpha}\delta)$ $\displaystyle=$ $\displaystyle(\widetilde{g}*\widetilde{f}+\overline{\alpha}\widetilde{g}+% \overline{\beta}\widetilde{f})+\overline{\beta\alpha}\delta,$

and

 $\displaystyle(\widetilde{g}*\widetilde{f})(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\widetilde{g}(y)\widetilde{f}(x-y)dm_{n}(y)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\overline{g(-y)}\overline{f(-x+y)}dm_{n}(y)$ $\displaystyle=$ $\displaystyle\overline{\int_{\mathbb{R}^{n}}g(-y-x)f(y)dm_{n}(y)}.$

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 $F:A\to\mathbb{C}$ by

 $F(f+\alpha\delta)=\int_{\mathbb{R}^{n}}f\phi dm_{n}+\alpha\phi(0),\qquad f+% \alpha\delta\in A.$

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

 $\displaystyle|F(f+\alpha\delta)|$ $\displaystyle\leq$ $\displaystyle\left|\int_{\mathbb{R}^{n}}f\phi dm_{n}\right|+|\alpha|\phi(0)$ $\displaystyle\leq$ $\displaystyle\int_{\mathbb{R}^{n}}|f||\phi|dm_{n}+|\alpha|\phi(0)$ $\displaystyle\leq$ $\displaystyle\phi(0)\int_{\mathbb{R}^{n}}|f|dm_{n}+|\alpha|\phi(0)$ $\displaystyle=$ $\displaystyle\phi(0)\left\|f+\alpha\delta\right\|,$

from which it follows that $\left\|F\right\|=\phi(0)$, and in particular that $F$ is bounded. Let $A_{0}=\{f+\alpha\delta\in A:f\in C_{c}(\mathbb{R}^{n})\}$. Because $F:A\to\mathbb{C}$ 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 $F((f+\alpha\delta)*(f+\alpha\delta)^{*})\geq 0$.

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

 $F(g*g^{*})=F(g*\widetilde{g})=\int_{\mathbb{R}^{n}}(g*\widetilde{g})\phi dm_{n% }\geq 0.$ (10)

Define $\eta:\mathbb{R}^{n}\to\mathbb{R}$ by

 $\eta(x)=\begin{cases}\exp\left(-\frac{1}{1-|x|^{2}}\right)&|x|<1\\ 0&|x|\geq 1,\end{cases}$

and for $\epsilon>0$, define $\eta_{\epsilon}:\mathbb{R}^{n}\to\mathbb{R}$ by $\eta_{\epsilon}(x)=\epsilon^{-n}\eta\left(\frac{x}{\epsilon}\right)$. Let $f+\alpha\delta\in A_{0}$ and define $g_{\epsilon}=f+\alpha\eta_{\epsilon}\in C_{c}(\mathbb{R}^{n})$. From (10) we have $F(g_{\epsilon}*g_{\epsilon}^{*})\geq 0$ for any $\epsilon>0$. On the other hand,

 $\displaystyle F(g_{\epsilon}*g_{\epsilon}^{*})$ $\displaystyle=$ $\displaystyle F((f+\alpha\eta_{\epsilon})*(\widetilde{f}+\overline{\alpha}\eta% _{\epsilon}))$ $\displaystyle=$ $\displaystyle F(f*\widetilde{f}+\overline{\alpha}f*\eta_{\epsilon}+\alpha\eta_% {\epsilon}*\widetilde{f}+|\alpha|^{2}\eta_{\epsilon}*\eta_{\epsilon})$ $\displaystyle=$ $\displaystyle F(f*\widetilde{f})+\overline{\alpha}\int_{\mathbb{R}^{n}}(f*\eta% _{\epsilon})\phi dm_{n}+\alpha\int_{\mathbb{R}^{n}}(\eta_{\epsilon}*\widetilde% {f})\phi dm_{n}$ $\displaystyle+|\alpha|^{2}\int_{\mathbb{R}^{n}}(\eta_{\epsilon}*\eta_{\epsilon% })\phi dm_{n}.$

We take as granted that

 $\int_{\mathbb{R}^{n}}(f*\eta_{\epsilon})\phi dm_{n}\to\int_{\mathbb{R}^{n}}f% \phi dm_{n}$

as $\epsilon\to 0$, that

 $\int_{\mathbb{R}^{n}}(\eta_{\epsilon}*\widetilde{f})\phi dm_{n}\to\int_{% \mathbb{R}^{n}}\widetilde{f}\phi dm_{n}$

as $\epsilon\to 0$, and that

 $\int_{\mathbb{R}^{n}}(\eta_{\epsilon}*\eta_{\epsilon})\phi dm_{n}\to\phi(0)$

as $\epsilon\to 0$. Furthermore,

 $\displaystyle F((f+\alpha\delta)*(f+\alpha\delta)^{*})$ $\displaystyle=$ $\displaystyle F((f+\alpha\delta)*(\widetilde{f}+\overline{\alpha}\delta))$ $\displaystyle=$ $\displaystyle F(f*\widetilde{f}+\overline{\alpha}f+\alpha\widetilde{f}+|\alpha% |^{2})$

Thus

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

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

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

Therefore, $F:A\to\mathbb{C}$ 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 $\Delta$ satisfying

 $F(f+\alpha\delta)=\int_{\Delta}\Gamma(f+\alpha\delta)d\mu,\qquad f+\alpha% \delta\in A,$

and hence, from the definition of $F$,

 $\int_{\mathbb{R}^{n}}f\phi dm_{n}+\alpha\phi(0)=\int_{\Delta}\Gamma(f+\alpha% \delta)d\mu,\qquad f+\alpha\in A.$

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

 $h_{t}(f+\alpha\delta)=\hat{f}(t)+\alpha,\qquad f+\alpha\delta\in A,$

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

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

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

 $\displaystyle\int_{\Delta}\Gamma(f+\alpha\delta)d\mu$ $\displaystyle=$ $\displaystyle\int_{\Delta}\Gamma(f)d\mu+\alpha\int_{\Delta}\Gamma(\delta)d\mu$ $\displaystyle=$ $\displaystyle\int_{\Delta}\Gamma(f)d(T_{*}\nu)+\alpha\int_{\Delta}\chi_{\Delta% }d\mu$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}\cup\{\infty\}}\Gamma(f)\circ Td\nu+\alpha\mu% (\Delta)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}\cup\{\infty\}}\Gamma(f)\circ T(t)d\nu(t)+% \alpha F(\delta)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}\cup\{\infty\}}h_{t}(f)d\nu(t)+\alpha\phi(0)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\hat{f}(t)d\nu(t)+\alpha\phi(0).$

Therefore

 $\int_{\mathbb{R}^{n}}f\phi dm_{n}+\alpha\phi(0)=\int_{\mathbb{R}^{n}}\hat{f}(t% )d\nu(t)+\alpha\phi(0),$

i.e.

 $\int_{\mathbb{R}^{n}}f\phi dm_{n}=\int_{\mathbb{R}^{n}}\hat{f}(t)d\nu(t).$

As

 $\displaystyle\int_{\mathbb{R}^{n}}\hat{f}(t)d\nu(t)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}e^{-it\cdot x}f(x% )dm_{n}(x)\right)d\nu(t)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}f(x)\int_{\mathbb{R}^{n}}e^{-ix\cdot t}d\nu(% t)dm_{n}(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}f(x)\hat{\nu}(x)dm_{n}(x),$

we have

 $\int_{\mathbb{R}^{n}}f\phi dm_{n}=\int_{\mathbb{R}^{n}}f\hat{\nu}dm_{n}.$

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