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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