\classheader{2012-11-30}

Last time, we discussed Banach algebras, and now we'll add a new piece of structure to them.

\begin{definition}
A $\ast$-algebra is a Banach algebra $A$ equipped with an anti-involution $\ast$, sending $a$ to
$a^*$, such that $(ab)^*=b^*a^*$, $(a^*)^*=a$, $(\lambda a)^*=\overline{\lambda} a^*$, and
$\abs{a^*}
= \abs{a}$ for all $a,b\in A$ and $\lambda\in\C$.
\end{definition}

%\begin{definition}
%A $\ast$-algebra is a Banach algebra $A$ equipped with an anti-involution $\ast$, sending $a$ to
%$a^*$, such that $(ab)^*=b^*a^*$, $(a^*)^*=a$, $(\lambda a)^*=\overline{\lambda} a^*$, and
%$\abs{a^*a}=\abs{a}^2$ for all $a,b\in A$ and $\lambda\in\C$. The last condition implies $\abs{a^*}
%= \abs{a}$. %jw-- Ginzburg only required this last part, but I think you need the |a^*a| = |a|^2 for what
%%we do; Wikipedia agrees?
%\end{definition}
\begin{examples}
$\text{}$

\begin{itemize}
\item Let $X$ be a compact Hausdorff space, and let $A=C(X)$, with norm $|f|=\max_{x\in X}|f(x)|$. Then the map $f\mapsto f^*$ defined by $f^*(x)=\overline{f(x)}$ makes $A$ a $\ast$-algebra.
\item Let $V$ be a finite-dimensional hermitian vector space, and let $A=\End(V)$, with norm $|a|=\max_{v\in V\setminus \{0\}}\frac{|a(v)|}{|v|}$. Then the map $a\mapsto a^*$ (the hermitian adjoint) makes $A$ a $\ast$-algebra.
\end{itemize}
\end{examples}
\begin{remark}
Let $\widehat{A}=\Hom_\mathrm{alg}(A,\C)$, which is a compact Hausdorff space. There is a natural evaluation map $\ev:A\to C(\widehat{A})$. If $A=C(X)$, then $\widehat{A}=X$, using our identification of the points of $X$ with maximal ideals. Specifically, the map
\[\ev:C(X)\to C(\widehat{C(X)})=C(X)\]
is the identity.
\end{remark}
\begin{theorem}
Let $A$ be a \textbf{commutative} $\ast$-algebra. Then
\begin{enumerate}
\item The evaluation map $\ev:A\to C(\widehat{A})$ commutes with anti-involution, i.e. for all $\chi\in\widehat{A}$,
\[\overline{\ev(a)(\chi)}=\ev(a^*)(\chi).\]
\item $\im(\ev)$ is dense in $C(\widehat{A})$.
\end{enumerate}
\end{theorem}
\begin{proof}
First, we'll show that 1 implies 2. We use the Stone-Weierstrass theorem:\vspace{-0.2in}
\begin{quotation}
\begin{theorem}[Stone-Weierstrass]
Let $X$ be a compact Hausdorff space, and let $B\subset C(X)$ be a subalgebra that separates points, and that is stable under the involution of complex conjugation. Then $B$ is dense (with respect to the uniform convergence norm) in $C(X)$.
\end{theorem}
\end{quotation}
In our setting, statement 1 implies that $\im(\ev)$ is stable under conjugation, and $\im(\ev)$ separates points essentially by definition. Thus, statement 2 follows from statement 1.

Now we will prove statement 1. First, we will need the following fact: if $u\in A$ is invertible, then $\lambda\in\spec(u)\iff\lambda^{-1}\in\spec(u^{-1})$. This is clear: $u-\lambda$ is invertible $\iff$ $u^{-1}-\lambda^{-1}$ is invertible.

Recall that for a Banach algebra $A$, we can define $\exp:A\to A$ by
\[\exp(a)=1+a+\frac{a^2}{2}+\cdots,\]
and the fact that $|a^n|\leq|a|^n$ ensures that this is absolutely convergent. We claim that $\lambda\in\spec(a)\implies e^\lambda\in\spec(\exp(a))$. This follows from the fact that
\[\frac{\exp(a)-e^\lambda}{a-\lambda}=1+\cdots\]
as a formal power series absolutely convergent in $a$ (take the Taylor series of the holomorphic function $\frac{e^z - e^\lambda}{z-\lambda}$). Therefore if $\exp(a) - e^\lambda$ is
invertible, we can define $(a-\lambda)^{-1}$. 

Let $A$ be a $\ast$-algebra. We say that $a$ is hermitian (respectively, skew-hermitian) if $a^*=a$ (respectively, if $a^*=-a$). We say that $u$ is unitary if $u^*u=uu^*=1$. Clearly $a$ is hermitian $\iff$ $\sqrt{-1}\, a$ is skew-hermitian. Let $H(A)$ be the set of hermitian elements of $A$ (which is a sub-$\R$-vector space of $A$), and let $U(A)$ be the set of unitary elements (which is a subgroup of $A^\times$). It is easy to see that $u\in U(A)$ implies $\abs{u}=1$. Observe that 
\[A=H(A)\oplus \sqrt{-1}\,H(A),\]
and specifically, for $a\in A$, we have $a=x+iy$ where $x=\frac{a+a^*}{2},y=\frac{a-a^*}{2i}\in H(A)$.

We claim that $a\in H(A)$ $\iff$ $\exp(ita)\in U(A)$ for any $t\in\R$. This holds because
\[(e^{ita})^*=1^*+(ita)^*+(i^2t^2a^2/2)^*+\cdots=e^{-ita},\]
proving the $\implies$ direction, and the $\impliedby$ direction follows from considering the derivative
\[a=\frac{1}{i}\left(\frac{d}{dt}\,e^{ita}\right)\bigg|_{t=0}.\]
Because $|u|=1$ for any $u\in U(A)$, for any $\lambda\in\spec(u)$ we have $|\lambda|\leq \abs u = 1$. But $u^*=u^{-1}$ is also unitary, so for any $\lambda^{-1}\in\spec(u^{-1})$, we have $|\lambda^{-1}|\leq|u^{-1}|=1$. Therefore $|\lambda|=1$, so that $\spec(u)\subseteq\S^1\subset\C$ for any unitary $u$.

Now, we observe that for any $a\in H(A)$, we have $\spec(a)\subseteq\R$. We can see this as follows. For any $t\in \R$, we know that $e^{ita}\in U(A)$, and therefore for any $\mu\in\spec(e^{ita})$, we must have $|\mu|=1$. But $\lambda\in\spec(a)$ implies that $e^{it\lambda}\in\spec(e^{ita})$, so that $|e^{it\lambda}|=1$, and therefore $\lambda\in\R$.

We claim that for any $\chi:A\to\C$ in $\widehat{A}$, we have $\chi(a^*)=\overline{\chi(a)}$ for any $a\in A$. We write $a=x+iy$ where $x,y\in H(A)$, and recall that we proved last time $\spec(b)=\{\chi(b)\mid \chi\in\widehat{A}\}$ (this is where we need commutativity of $A$). Then $\chi(x) \in \spec(x) \subset \R$, $\chi(y) \in \spec(y) \subset \R$, and
\[\chi(a)=\chi(x+iy)=\chi(x)+i\chi(y)\]
implies that
\[\chi(a^*)=\chi(x-iy)=\chi(x)-i\chi(y)=\overline{\chi(x)+i\chi(y)}.\]
Thus, we have proven statement 1 in our theorem.
\end{proof}

Note that if $a \in H(A)$, then $\abs{\ev(a)} = \limsup \abs{a^n}^{\frac 1 n} = \abs a$ since
$\abs{a^2} = \abs{a}^2$. For general $a \in A$, we have using part (1) of the Theorem that 
$\abs{\ev(a)}^2 = \abs{\ev(a^* a)} = \abs{a^* a} = \abs{a}^2$. It follows that $\ev$ is an isometry.

%% a web reference: http://qchu.wordpress.com/2012/07/17/banach-algebras-the-gelfand-representation-and-the-commutative-gelfand-naimark-theorem/

\begin{example}
Let $G$ be a locally compact group, and $\int$ be a left invariant integral. We can consider $L^1(G)$, which is the closure of the space of continuous functions with compact support. We claim that $L^1(G)$ together with the convolution product 
%\[(f_1\star f_2)(g)=\int f_1(gh^{-1})f_2(h)\,dh\]
\[(f_1\star f_2)(g)=\int f_1(h)f_2(h^{-1} g)\,dh\]
(we can't use $\ast$ for convolution, it'd be too confusing) and the anti-involution $f^*(g)=\overline{f(g^{-1})}$ makes $L^1(G)$ into a $\ast$-algebra.

Suppose that $G$ is abelian. Then $L^1(G)$ is commutative under $\star$, so all of our previous work applies (sort of). We can consider the evaluation map $\ev:L^1(G)\to C(\widehat{L^1(G)})$. We first suggest that $\widehat{L^1(G)}=\widehat{G}$. Here $\widehat{G}$ denotes the Pontryagin dual of $G$, i.e. the continuous group homomorphisms $G \to
\mathbb{S}^1$. We can see that there may be an issue because the Gelfand representations we have considered are always compact\footnote{In fact one can define $\widehat A$ for a commutative Banach algebra $A$ without unit, in which case $\widehat A$ is not guaranteed to be compact.}, while $\widehat{G}$ may only be locally compact in general. The issue arises because $L^1(G)$ is not quite a [unital] algebra, since it need not have a unit (the unit under convolution is not the function 1, but rather the Dirac
delta, which is not in $L^1$ unless $G$ is discrete, that is, unless $\widehat{G}$ is compact). We fix this by formally adjoining a unit (even if there was one already in it) by defining
\[L^1(G)^{\thicksim}=\C 1\oplus L^1(G)\]
such that $L^1(G)$ is an ideal in $L^1(G)^\thicksim$ and $\abs{c \oplus a} = \abs c + \abs a$. 
This makes $L^1(G)^\thicksim$ a unital Banach $\ast$-algebra. 

\begin{proposition} The Fourier transform defines a group homeomorphism 
\[ \widehat G \cup \{\infty\} \cong \widehat{L^1(G)^\thicksim} \]
where the LHS is a one-point compactification of $\widehat G$. 
\end{proposition}
\begin{proof} %jw
Let $\zeta: L^1(G)^\thicksim \to \C$ be an algebra homomorphism. We proved that $\ev$ is an isometry, so $\zeta$ restricted to $L^1(G)$ is a bounded linear functional with $\lVert\zeta\rVert_1 = 1$. The dual of $L^1(G)$ is $L^\infty(G)$ so $\zeta$ may be considered as an almost everywhere bounded function on $G$. Pick $f \in L^1(G)$ such that $\zeta(f) = \int_G \zeta(h)f(h)dh \ne 0$. Then define $\chi(g) = \frac{\int \zeta(gh)f(h)dh}{\int \zeta(h)f(h)dh}$. The fact that $\zeta$ is an algebra homomorphism and has norm $1$ implies that $\chi : G \to \mathbb{S}^1$ is a unitary character of $G$, i.e. $\chi \in \widehat G$. Then one sees that $\zeta$ as a functional corresponds to the Fourier transform of $\chi$. See Bourbaki, Vol IX, \emph{Th\'eories spectrales}, Ch 2, \textsection 1, Proposition 1.1 for details.
\end{proof}


\end{example}
\begin{theorem}[N.W.]
If $f\in C(\S^1)$ and $f$ nowhere vanishes, then $\text{Fourier}(f)$ is absolutely convergent implies that $\text{Fourier}(\frac{1}{f})$ is absolutely convergent.
\end{theorem}
\begin{proof}
We let $G=\Z$, so that $\widehat{G}=\S^1$. Consider $\ell^1(\Z)$, and the evaluation map $\ev:\ell^1(\Z)\to C(\S^1)$ sending $\varphi$ to $\text{Fourier}(\varphi)\in C(\S^1)$. There exists some $\varphi\in \ell^1(\Z)$ such that $\text{Fourier}(\varphi)=\ev(\varphi)=f$, and \[0\notin \{\text{values of }f\}=\{\text{values of }\ev(\varphi)\}.\] Therefore $0\notin \spec(\varphi)$, and therefore there exists a $\varphi^{-1}\in\ell^1(\Z)$ (inverse with respect to convolution). We have $\text{Fourier}(\varphi^{-1})=\frac{1}{f}$.
\end{proof}




