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\begin{document}
\begin{center}
 { \bfseries
Math 123 Homework Assignment \#1}\\
Due Friday, April 8th.
\end{center}

\noindent
{\bf Part I:}
\begin{ques}
  Suppose that $X$ is a normed vector space.  Then $X$ is a Banach
  space (that is, $X$ is complete) if and only if every absolutely
  convergent series in $X$ is convergent.
\end{ques}

\begin{ques}
  Let $X$ be a normed vector space and suppose that $S$ and $T$ are
  bounded linear operators on $X$.  Show that $\|ST\|\le\|S\|\|T\|$.
\end{ques}

\begin{ques}
  Let $X$ be a locally compact Hausdorff space.
  Show that $C_{0}(X)$ is a closed subalgebra of $C^{b}(X)$.
\end{ques}

\begin{ques}
  Let $A$ be a unital Banach algebra.  Show that $x\mapsto x^{-1}$ is
  continuous from $G(A)$ to $G(A)$.  (Hint:
  $(a-h)^{-1}-a^{-1}=((1-a^{-1}h)^{-1}-1)a^{-1}$.) 
\end{ques}
\begin{ans}
  Using the ``hint'', suppose that 
$a\in G(A)$ and let $b=a-h$ for some $h\in A$ with
  \begin{equation}\label{eq:3}
    \|h\|\le \frac{\|a^{-1}\|^{-1}}2.
  \end{equation}
Then standard arguments show that $b\in G(A)$.  (Recall that if
$\|c\|<1$, then $1-c\in G(A)$ and $(1-c)^{-1}=1+c+c^{2}+\cdots $.)
Now we observe that
\begin{align*}
  \|b^{-1}-a^{-1}\|&=\|(a-h)^{-1}-a^{-1}\| \\
&= \bigl\| \bigl((1-a^{-1}h)-1\bigr)a^{-1}\bigr\|\\
&\le \Bigl\|\sum_{n=1}^{\infty} (a^{-1}h)^{n}\Bigr\|\|a^{-1}\| \\
&\le \sum_{n=1}^{\infty}\|a^{-1}h\|^{n}\|a^{-1}\| \\
&= \frac{\|a^{-1}h\|\|a^{-1}\|}{1-\|a^{-1}h\|}\\
\intertext{which, in view of \eqref{eq:3} and $\|a^{-1}h\|\le
  \|a^{-1}\|\|h\|$, is}
&\le 2\|a^{-1}\|^{2}\|h\|.
\end{align*}
Since the latter term tends to $0$ with $h$, we're done.

However, Peng Peng Yu came up with a cleaner argument that does not
require $A$ to be a Banach algebra.  Here it is enough that $A$ be a
normed algebra (and therefore with a submultiplicative norm).

Fix $a\in A$ and $\epsilon>0$.  Let
  $\delta=\frac12\min\bigl(\epsilon\|a^{-1}\|^{-2},
  \|a^{-1}\|^{-1}\bigr)$. We just need to verify that if $b\in
  B_{\delta}(a)\cap G(A)$, then $\|a^{-1}-b^{-1}\|<\epsilon$.  However, we first
  notice that by the ``reverse triangle inequality'',
  \begin{align*}
\|b^{-1}\|-\|a^{-1}\|&\le \|b^{-1}-a^{-1}\| \\
&=\|b^{-1}(a-b)a^{-1}\| \\
&\le \|b^{-1}\|\|a-b\|\|a^{-1}\| \\
\intertext{which, since $\|a-b\|<\frac12\|a^{-1}\|^{-1}$, is}
&<\frac12\|b^{-1}\|.
  \end{align*}
In particular,  this implies that
\begin{equation}
  \label{eq:41}
  \|b^{-1}\|<2\|a^{-1}\| \quad\text{if $b\in B_{\delta}(a)\cap G(A)$. }
\end{equation}
But then if $b\in B_{\delta}(a)\cap G(A)$, we have \eqref{eq:41} and
$\|a-b\|<\frac\epsilon2\|a^{-1}\|^{-2}$.  Therefore
\begin{align*}
  \|b^{-1}-a^{-1}\|&=\|b^{-1}(a-b)a^{-1}\|\\
&\le \|b^{-1}\|\|a-b\|\|a^{-1}\|\\
&<\epsilon.
\end{align*}
This is what we wanted to show.
\end{ans}


\bigbreak
\vfill
\hrule
\vfill

\noindent{\bf Part II:}

\begin{ques}\label{ques:one}
  Suppose that $X$ is a compact Hausdorff space.  If $E$ is a
closed subset of $X$, define $I(E)$ to be the ideal in $C(X)$ of
functions which vanish on $E$.
\begin{enumerate}
\item 
Let $J$ be a closed ideal in $C(X)$ and let $E=\set{x\in X: 
\hbox{$f(x)=0$ for all $f\in J$}}$.  Prove that if $U$ is an open
\nbhd{} of $E$ in $X$, then there is a $f\in J$ such that $0\le
f(x)\le 1$ for all $x$ and such that $f(x)=1$
for all $x$ in the compact set $X\setminus U$.
\item 
Conclude that $J=I(E)$ in part~(a), and hence, conclude
that {\it every closed\/} ideal in $C(X)$ has the form $I(E)$ for
some closed subset $E$ of $X$.
\end{enumerate}
\end{ques}
\begin{ans}
  Fix ${x_0}\in X\setminus U$.
By definition of $E$, there is a $f_{x_0}\in J$ with $f_{x_0}(x_0)\not=0$.
Since $|f|^2=\bar f f\in J$ if $f\in J$,
we may as well assume that $f_{x_0}(x)\ge0$
for all $x\in X$, and since $J$ is a subalgebra, we may also
assume that $f_{x_0}(x_0)>1$.  Since $X\setminus U$ is compact,
there are $x_1,\dots x_n\in X$ so that $f=\sum_k f_{x_k}$
satisfies $f\in J$ and $f(x)>1$ for all $x\in X\setminus U$.
Observe that $g=\min(1,1/f)$ is in $C(X)$\footnote{If $a,b\in
C(X)$, then so are $\min(a,b)=(a+b)/2-|a-b|/2$ and $\max(a,b) =
(a+b)/2 + |a-b|/2$.  In the above, we can replace $f$ by $\max(f,1/2)$
without altering $g$.}.  Since $fg\in J$, we are done with
part~(a).

Notice that we have proved that there is a $f\in J$ such that $0\le
f(x)\le1$ for all $x\in X$ and $f(x)=1$ for all $x\notin U$.
Thus if $h$ is any function in $I(E)$ and $\epsilon>0$, then $U =
\set{x\in X:|h(x)|>\epsilon}$ is a \nbhd{} of $E$ in $X$.  Then
we can choose $f\in J$ as above and $\|fh-h\|_\infty<\epsilon$.
Thus $h\in \overline J = J$.  This suffices as we have $J\subseteq
I(E)$ by definition. (Notice that if $E=\emptyset$, then we can take
$U=\emptyset$ in the above and then $J=C(X)$.)

\noindent{\bf Remark}:  Notice that we have established a 1-1
correspondence between the closed subsets $E$ of $X$ and the
closed ideals $J$ of $C(X)$: it follows immediately from 
Urysohn's Lemma\footnote{For a reference, see
Pedersen's {\it Analysis Now}: Theorems 1.5.6~and 1.6.6 or, more
generally, Proposition~1.7.5.}
that if $E$ is closed and $x\notin E$, then there
is a $f\in I(E)$ with $f(x)\not=0$.
Thus $I(E)\not=I(F)$ if $E$ and
$F$ are distinct closed sets.
\end{ans}
\ifshowanswers\else
\vfill\goodbreak\fi

\begin{ques}\label{ques:two}
  Suppose that $X$ is a (non-compact) locally compact Hausdorff space.
  Let $\xp$ be the {\it one-point compactification\/} of $X$ (also
  called the Alexandroff compactification: see [Kelly; Theorem 5.21]
  or [Folland, Proposition~4.36]).  Recall that $\xp=X\cup\set \infty$
  with $U\subseteq \xp$ open if and only if either $U$ is an open
  subset of $X$ or $\xp\setminus U$ is a {\it compact\/} subset of
  $X$.
\begin{enumerate}
\item 
Show that $f\in C(X)$ belongs to $C_0(X)$ if and only if the
extension
$$\tilde f(\tilde x)=
\begin{cases}
  f(\tilde x)&\text{if $\tilde x\in X$, and}\\
0&\text{if $\tilde x=\infty$.}
\end{cases}$$
is continuous on $\xp$.
\item 
Conclude that $C_0(X)$ can be identified with the maximal ideal
of $C(\xp)$ consisting of functions which `vanish at $\infty$.'
\end{enumerate}
\end{ques}
\begin{ans}
  \def\tf{\tilde f}\def\tx{\tilde x}%
Suppose $\tilde f$ is continuous at $x=\infty$, and that $\epsilon>0$.
Then $U=\set{\tx\in\xp:|\tf(\tx)|<\epsilon}$ is an open \nbhd{}
of $\infty$ in $\xp$.  But then $X\setminus U$ is compact; but
that means $\set{x\in X:|f(x)|\ge\epsilon}$ is compact.  That is,
$f\in C_0(X)$ as required.

For the converse, suppose that $f\in C_0(X)$, and that
$V$ is open in $\C$.  If $0\notin
V$, then $\tf^{-1}(V)=f^{-1}(V)$ is open in $X$, and therefore,
open in $\xp$.  On the other hand, if $0\in V$, then there is a
$\epsilon>0$ so that $\set{z\in\C:|z|<\epsilon}\subseteq V$.
Thus, $\xp\setminus \tf^{-1}(V)=\set{x\in X:f(x)\notin V}\bigcap
\set{x\in X:|f(x)|\ge\epsilon}$.  Since the first set is closed
and the second compact, $\xp\setminus \tf^{-1}(V)$ is  a 
compact subset of $X$, and
$\tf^{-1}(V)$ is a open \nbhd{} of $\infty$ in $\xp$.  This
proves part~(a).

Part~(b) is immediate: each $f\in C_0(X)$ has a (unique)
extension to a function in $C(\xp)$ and this identifies $C_0(X)$
with the ideal $I\bigl(\set{\infty}\bigr)$ in $C(\xp)$.  In view of
question~\ref{ques:one} above, $I\bigl(\set{\infty}\bigr)$ is maximal among closed
ideals in $C(\xp)$, and, as maximal ideals are automatically
closed, maximal among all proper ideals.
\end{ans}

\begin{ques}
  Use the above to establish the following ideal theorem for $C_0(X)$.
\medbreak
\noindent{\bf Theorem}: Suppose that $X$ is a locally compact
Hausdorff space.  Then every closed ideal $J$ in $C_0(X)$ is of
the form
$$J=\set{f\in C_0(X):\hbox{$f(x)=0$ for all $x\in E$}}$$
for some closed subset $E$ of $X$.
\end{ques}
\begin{ans}
  Suppose that $J$ is a closed ideal in $C_0(X)$.  Then $J$ is, in
view of question~\ref{ques:two}(b) above, a closed subalgebra of $C(\xp)$.
I claim the result will follow once it is observed that $J$ is actually
an ideal in $C(\xp)$.   In that case, $J=I(E\cup\set\infty)$,
where $E\subseteq X$ is such that $E\cup\set\infty$ is closed in
$\xp$.  Thus $\xp\setminus(E\cup\set\infty)=X\setminus E$ is open
in $X$, and $E $ is closed in $X$.

The easy way to verify the claim, is to observe that, in view of
the fact that $C_0(X)$ is a maximal ideal in $C(\xp)$,
$C(\xp)=\set{f
+ \lambda: \hbox{$f\in C_0(X)$ and $\lambda\in \C$ }}$.  (Here $\lambda
\in \C$ is identified with the constant function on $\xp$.)
Then, since $J$ is an algebra, $f(g+\lambda)=fg+\lambda f$
belongs to $J$ whenever $f$ does.
\end{ans}

\bigbreak
\vfill
\hrule
\vfill

\noindent{\bf Part III:}

\begin{ques}
  Assume you remember enough measure theory to show that if $f,g\in
L^1\bigl([0,1]\bigr)$, then
\begin{equation}
  \label{eq:1}
  f*g(t)=\int_0^t f(t-s)g(s)\,ds
\end{equation}
exists for almost all $t\in[0,1]$, and defines an element of $L^1\bigl([0,
1]\bigr)$.  Let $A$ be the algebra consisting of the Banach space $L^1\bigl([0,1]\bigr)$
with multiplication defined by \eqref{eq:1}.
\begin{enumerate}
\item Conclude that $A$ is a
commutative Banach algebra: that is, show that
$f*g=g*f$, and that $\|f*g\|_1\le\|f\|_1\,\|g\|_1$.
\item Let $f_0$ be the constant function $f_0(t)=1$ for all $t\in[0,1]$.
Show that
\begin{equation}
  \label{eq:2}
  f_0^n(t):= f_0*\cdots* f_0(t) = t^{n-1}/(n-1)!,
\end{equation}
and hence,
\begin{equation}
  \label{eq:3}
  \|f_0^n\|_1={1\over n!}.
\end{equation}
\item Show that \eqref{eq:2} implies that $f_0$ generates $A$ as a
  Banach algebra: that is, $\operatorname{alg}(f)$ is norm dense.
  Conclude from \eqref{eq:3} that the spectral radius $\rho(f)$ is
  zero for all $f\in A$.
\item Conclude that $A$ {\it has no nonzero complex homomorphisms}.
\end{enumerate}
\end{ques}
\begin{ans}
  First compute that\footnote{For a reference for Tonelli's Theorem (the `uselful'
version of Fubini's Theorem), see
[{\it Analysis Now}, Corollary~6.6.8], or much
better, see Royden's {\it Real Analysis}.
On the other hand, if you are worried about the calculus style
manipulation of limits, consider the integrand
$$F(s,t)=
\begin{cases}
  |f(t-s)g(s)|&\text{if $0\le s\le t\le1$, and}\\
0&\text{otherwise.}
\end{cases}$$}
  \begin{align*}
    \|f*g\|_1&=\int_0^1 |f*g(t)|\,dt\\
&\le \int_0^1\int_0^t|f(t-s)g(s)|\,ds\,dt\\
\intertext{which, using Tonelli's 
Theorem, is}
&=\int_0^1|g(s)|\Bigl(\int_s^1|f(t-s)|\,dt
\Bigr)\,ds\\
&=\int_0^1|g(s)|\Bigl(\int_0^{1-s}|f(u)|\,du\Bigr) \,ds\\
&\le\|f\|_1\|g\|_1.
\end{align*}


To show that $f*g=g*f$ it suffices, in view of the above, to
consider continuous functions.  Thus, the usual calculus
techniques apply.  In particular,
\begin{align*}
  f*g(t)&=\int_0^tf(t-s)g(s)\,ds\\
&=-\int_t^0 f(u)g(t-u)\,du=g*f(t).
\end{align*}
This proves (a).  However, (b) is a simple
induction argument.

Now for (c): the calculation \eqref{eq:2} shows that
$\operatorname{alg}(f_0)$ contains all polynomials.  Since the polynomials
are uniformly dense in $C[0,1]$, and the later is dense in
$L^1$, we can conclude that $\operatorname{alg}(f_0)$ is norm dense.

Next, observe that \eqref{eq:3} not only implies that
$\rho(f_0)=0$, but that $\rho(f_0^k)=0$ as well for any
positive integer $k$.  However, it is not immediately
clear that every element of $\operatorname{alg}(f_0)$ has spectral radius zero.
However, there is an easy way to see this.  Let
\def\tA{\widetilde A}%
$\tA$ be the unitalization of $A$ (i.e., $\tA:=A\oplus\C$), and
recall that $a\in A$ has spectral radius zero ($a$ is called
{\it quasi-nilpotent}) if and only if \def\th{\tilde h}%
$\th(a)=0$ for all $\th\in\widetilde\Delta=\Delta(\tA)$.
Since each $\th$ is a continuous algebra homomorphism,
$\ker(\th)$ is a closed ideal in $\tA$, and it follows that the
collection of quasi-nilpotent elements is actually a {\it
closed ideal\/} of
$A$ given by\footnote{This result is of interest in its own
right.  Note that $A$ is always a maximal ideal in $\tA$, and
so $\mathop{\rm rad}(A)$ is always contained in $A$ itself.}
$$\mathop{\rm
rad}(A)=\bigcap\nolimits_{\th\in\widetilde\Delta}
\ker(\th).$$
Since each $f_0^k$ is in $\mathop{\rm rad}(A)$, so is the
{\it closed\/} algebra (in fact, the closed ideal) generated by
$f_0$.  Thus,
$\mathop{\rm rad}(A)=A$ in this case, which is what was to be
shown.

Of course, (d) is an immediate consequence of
(c): if $\rho\in\Delta(A)$, then by defintion there
is a $f\in A$ such that $\rho(f)\not=0$.  But then
$\rho(f)\ge|h(f)|>0$, which contradicts the fact that
$\mathop{\rm rad}(A)=A$.
\end{ans}

\begin{ques}
  Here we want to give an example of a unital commutative Banach
  algebra $A$ where the Gelfand transform induces and injective
  isometric map of $A$ onto a proper subalgebra of $C(\Delta)$.  For
  $A$, we want to take the \emph{disk algebra}.  There are a couple of
  ways that the disk algebra arises in the standard texts, but the
  most convenient for us is to proceed as follows.  Let $D=\set{z\in
    \C:|z|<1}$ be the open unit disk.  We'll naturally write
  $\overline D$ for its closure $\set{z\in\C:|z|\le 1}$, and $\T$ for
  its boundary.  Then $A$ will be the subalgebra of $C(\overline D)$
  consisting of functions which are holomorphic on $D$.  Using
  Morera's Theorem, it is not hard to see that $A$ is closed in
  $C(\overline D)$, and therefore a unital commutative Banach
  algebra.\footnote{The maximum modulus principal implies that the map
    sending $f\in C(\overline D)$ to its restriction to $\T$ is an
    isometric isomorphism of $A$ onto a closed subalgebra $A(D)$ in
    $C(\T)$.  Of course, our analysis applies equally well to $A(D)$.}
  Notice that for each $z\in \overline D$, we obtain $\phi_{z}\in
  \Delta$ by $\phi_{z}(f):=f(z)$.  We'll get the example we want by
  showing that $z\mapsto \phi_{z}$ is a homeomorphism $\Psi$ of
  $\overline D$ onto $\Delta$.  For convenience, let $p_{n}\in A$ be
  given by $p_{n}(z)=z^{n}$ for $n=0,1,2,\dots$, and let $\mathcal{P}$
  be the subalgebra of polynomials spanned by the $p_{n}$.
  \begin{enumerate}
  \item First observe that $\Psi$ is injective.  (Consider $p_{1}$.)
  \item If $f\in A$ and $0<r<1$, then let $f_{r}(z):=f(rz)$.  Show
    that $f_{r}\to f$ in $A$ as $r\to 1$.
  \item Conclude that $\mathcal{P}$ is dense in $A$.  (Hint: show that
    $f_{r}\in \overline{\mathcal{P}}$ for all $0<r<1$.)
  \item Now show that $\Psi$ is surjective.  (Hint: suppose that $h\in
    \Delta$.  Then show that $h=\phi_{z}$ where $z=h(p_{1})$.)
  \item Show that $\Psi$ is a homeomorphism.  (Hint: $\Psi$ is clearly
    continuous and both $\overline D$ and $\Delta$ are compact and
    Hausdorff.)
  \item Observe that if we use the above to identify $\Delta$ and
    $\overline D$, then the Gelfand transform is the identify on $A$,
    and $A$ is a proper subalgebra of $C(\overline D)$.
  \end{enumerate}
\end{ques}

\begin{ques}
  In this problem, we want to prove an old result to due Wiener about
  functions with absolutely converent Fourier series using the
  machinery of Gelfand theory.  
Recall that if $\phi\in C(\T)$, then the \emph{Fourier coefficients}
 of $\phi$ are given by\footnote{I've used $\check \phi$ in place of
   the traditional $\hat \phi$ to 
distinguish it from the (other) Gelfand transform to be used in the problem.}
 \begin{equation*}
   \check \phi(n):=\frac1{2\pi}\int_{-\pi}^{\pi} \phi(e^{it})e^{-int}\,dt.
 \end{equation*}
 In some cases --- for example if $\phi$ has two continuous
 derivatives --- the Fourier coefficients are \emph{absolutely
   convergent} in the sense that $n\mapsto \check\phi(n)$ defines an
 element of $\ell^{1}(\Z)$.\footnote{Recall that
   $\ell^{1}(\Z)=L^{1}(\Z,\nu)$, where $\nu$ is counting measure, is
   the set of functions $f:\Z\to\C$ such that
   $\lim_{N\to\infty}\sum_{n=-N}^{n=N}|f(n)|<\infty$.}  

We aim to prove the following:
\medbreak
  \noindent{\bf Theorem:} (Wiener) \emph{
    Suppose the $\phi\in C(\T)$ never vanishes and has an absolutely
    convergent Fourier series.  Then $\psi:=1/\phi$ also has an
    absolutely convergent Fourier series.}
\smallbreak
I suggest the following strategy.
  \begin{enumerate}
  \item If $f$ and $g$ are in $\ell^{1}(\Z)$, then their \emph{convolution},
    $f*g$ is defined by
    \begin{equation*}
      f*g(n)=\sumz{m}f(m)g(n-m).
    \end{equation*}
    Show that $f*g\in \ell^{1}(\Z)$ (so that in particular, $f*g(n)$
    is defined for all $n$), and that convolution makes $\ell^{1}(\Z)$
    into a unital, commutative Banach algebra.  (Here,
    $1_{\ell^{1}(\Z)}=\charfcn{\sset0}$.\footnote{If $S$ is a subset of $X$, I
      use $\charfcn{S}$ for the characteristic function of $S$, which
      takes the value $1$ on $S$, and $0$ otherwise.})
  \item Let $\Delta=\Delta(\ell^{1}(\Z))$ be the maximal ideal space
    of $\ell^{1}(\Z)$ equipped with its compact, Hausdorff Gelfand
    topology.  If $z\in \T$, then define $h_{z}:A\to \C$ by
  \begin{equation*}
    h_{z}(f)=\sumz{n} f(n)z^{n}.
  \end{equation*}
Show that $h_{z}\in \Delta$.
\item Let $w=\charfcn{1}\in \ell^{1}(\Z)$.  If $h\in \Delta$, then
  show that $h=h_{z}$ where $z=h(w)$.  (Hint: If $f\in \ell^{1}(\Z)$,
  then $f=\sumz n f(n)w^{n}$ in norm in $\ell^{1}(\Z)$, where for
  example, $w^{2}=w*w=\charfcn 2$ and $w^{-1}=\charfcn {-1}$.)
\item Show that $z\mapsto h_{z}$ is a homeomorphism $\Phi$ of $\T$ onto
  $\Delta$.  (Hint: Since both $\T$ and $\Delta$ are compact
  Hausdorff sets, it suffices to see that $\Phi$ is a continuous
  bijection.  To show that $\Phi$ is continuous, observe that
  functions of the form $\sum_{n=-N}^{n=N} f(n)w^{n}$ are dense in $\ell^{1}(\Z)$.)
\item Since we can identify $\T$ with $\Delta$, if $f\in \ell^{1}(\Z)$, we will
  view the Gelfand transform of $f$ as a continuous function on $\T$.
  (So that we write $\hat f(z)$ in place of $\hat f(h_{z})$.)  
Show that if $\phi=\hat f$ for some $f\in A$, then $\check \phi=f$.
\item Conclude that the image $\mathfrak A$ of $\ell^{1}(\Z)$ in $C(\T)$ under the Gelfand
  transform is exactly the set of $\phi$ in $C(\T)$ whose Fourier
  coefficients are absolutely convergent.  (That is, $\mathfrak A$ is 
the collection of
  $\phi\in C(\T)$ such that $n\mapsto \check\phi(n)$ is in
  $\ell^{1}(\Z)$.)
\item Now prove Wiener's Theorem as stated above.
(Hint: More or less by assumption, $\phi=\hat f$ for some $f$ in $\ell^{1}(\Z)$.
Show that $f$ must be invertible in $\ell^{1}(\Z)$ and consider the Gelfand transform of
the inverse of $f$.)
\begin{ans}
  By Theorem~J from lecture, $\sigma(f)=\set{\phi(z):z\in\T}$.  Since
  $\phi$ never vanishes, $0\notin\sigma(f)$ and $f$ is invertible.
  Let $g=f^{-1}$, and consider $\hat g$.  Since $g*f=1_{A}$, we have
  $\hat g \hat f=\hat g \phi=1_{C(\T)}$.  It follows that $\hat
  g=\psi$, and hence $\psi$ has an absolutely convergent Fourier
  series --- namely $g$.
\end{ans}
  \end{enumerate}

\end{ques}
\end{document}

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