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\begin{document}
\noindent{MATH5002: Ongoing corrections and comments}
\section{Theorem 3.26}
At the top of page~124, why is it ``elementary to see'' that $f(A)$ is open?
By a ``topological vector space'' I mean a vector space which has a topology making
the vector space operations continuous. Any normed space, or locally convex space,
is a topological vector space.
\begin{lemma}
Let $E$ be a topological vector space, and let $f:E\rightarrow\mathbb R$ be a
continuous linear functional. Then either $f=0$ or $f$ is an open mapping.
\end{lemma}
\begin{proof}
Suppose $f\not=0$, so there is $x_0\in E$ with $f(x_0)\not=0$.
Let $A\subseteq E$ be open, and let $a\in A$. As $f(x_0)\not=0$ we can
find some scalar $\lambda$ such that $f(\lambda x_0) = \lambda f(x_0)$ does
not equal $f(a)$. Set $x_1=\lambda x_0$. Consider the function
$\varphi:\mathbb R\rightarrow E; t\mapsto tx_1 +(1-t)a$. This is continuous
(as it only involves the vector space operations) and $\varphi(0)=a\in A$.
As $A$ is open, there is $\epsilon>0$ such that $\varphi(t)\in A$ if
$-\epsilon < t <\epsilon$. Then consider the set $\{ f(\varphi(t)) :
-\epsilon < t <\epsilon\}$, which is the image of a line segment under a linear
map, and is hence the open interval between $-\epsilon f(x_1) + (1+\epsilon)
f(a)$ and $\epsilon f(x_1) + (1-\epsilon) f(a)$. The actual values are
irrelevant-- the point is that $f(x_1)\not=f(a)$, so this is a proper open
interval containing $f(a)$. Thus every point in $f(A)$ has an open
neighbourhood, and we conclude that $f(A)$ is open.
\end{proof}
In the proof of part (iii) (also on page~124) it's written ``since $A$
is compact, there is a convex open neighbourhood $V$ of $0_E$ with $(A+V)
\cap B=\emptyset$''. Why is this? We first need a lemma.
\begin{lemma}
Let $E$ be a real locally convex space. For each $x\in E$ and each
open set $U$ containing $x$, there is a convex open neighbourhood $W$ of
$0_E$ with $x+W+W \subseteq U$, where $x+W+W=\{ x+w+v : w,v\in W \}$.
\end{lemma}
\begin{proof}
As the topology is translation invariant, $U=x+V$ for some open set $V$
containing $0$. As addition is continuous and $0_E+0_E=0_E$, there are
neighbourhoods $V_1,V_2$ of $0_E$ with $V_1+V_2 \subseteq V$. As the
topology is locally convex, we can find a convex open neighbourhood $W$ of
$0_E$ with $W\subseteq V_1\cap V_2$. Thus $W+W\subseteq V$, or equivalently,
$x+W+W \subseteq U$.
\end{proof}
We now show the claimed result. For each $a\in A$, as $B$ is closed
and $a\not\in B$, by the lemma, there is a convex open neighbourhood $W_a$
of $0_E$ with $a+W_a+W_a \cap B = \emptyset$. Then the family
$\{ a+W_a : a\in A \}$ is an open cover for $A$, so as $A$ is
compact, there is a finite subcover, say $\{ a_i+W_{a_i} : 1\leq i\leq n\}$.
Set $V=W_{a_1} \cap\cdots\cap W_{a_n}$, which is a convex open neighbourhood
of $0_E$. Then
\[ A+V \subseteq \bigcup_{i=1}^n a_i + W_{a_i} + V
\subseteq \bigcup_{i=1}^n a_i + W_{a_i} + W_{a_i}, \]
which is disjoint from $B$, as required.
\section{Lemma~3.38}
Let $T:E\rightarrow F$ be a (bounded) linear map.
The first part of this lemma suggests that it is obvious to see that if
\[ \text{for all } y\in F \text{ there is }x\in E \text{ with }
T(x)=y, \|x\|\leq K\|y\|, \]
then $T$ is open. Why is this?
Well, let $U\subseteq E$ be open. Then let $y_0\in T(U)$, so $y_0=T(x_0)$ for some
$x_0\in U$. As $U$ is open, $B(x_0,\epsilon)\subseteq U$ for some $\epsilon>0$.
Let $z\in F$ with $\|z\| < \epsilon/K$, so by assumption, there is $x'\in E$ with
$T(x')=z$ and $\|x'\|\leq K\|z\|<\epsilon$. Then $x_0+x' \in B(x_0,\epsilon)\subseteq U$
and $T(x_0+x') = y_0+z$. As $z$ was arbitrary, we've shown that $B(y_0,\epsilon/K)
\subseteq T(U)$. As $y_0$ are arbitrary, we've shown that $T(U)$ is open.
If you think about it for a moment, we have actually just proved the following lemma:
\begin{lemma}
Let $E,F$ be normed space and $T:E\rightarrow F$ be linear. Let $B=\{x\in E:
\|x\|<1\}$ be the open unit ball of $E$, and suppose that $T(B)$ contains an open
neighbourhood of $0$. Then $T$ is open.
\end{lemma}
\section{Theorem~3.40}
What is the ``elementary argument'' in the proof. I think it is the following.
We know that $\overline{T(B_N)}$ has non-empty interior, which means we can find
$y\in \overline{T(B_N)}$ and $\epsilon>0$ with $B(y,\epsilon) \subseteq
\overline{T(B_N)}$. Thus we can find $(x_n)\subseteq E$ with $\|x_n\|< N$ for all
$n$, and with
$T(x_n)\rightarrow y$. Let $w\in F$ with $\|w\| < \epsilon/2$. Then
$y+2w \in B(y,\epsilon) \subseteq \overline{T(B_N)}$ and so we can find
$(x_n')\subseteq E$ with $\|x_n'\|0$ let $U=\{ k\in K : |g(k)|<\epsilon\}$
so $U$ is an open set containing $F$. For each $x\not\in U$, as $x\not\in K$,
we can find $f_x\in J$ with $f_x(x)=1$ say (by definition of $F$ we can find
$f_x\in J$ with $f_x(x)\not=0$, and then rescale). Then $U_x=\{ k\in K:
|f_x(k)|>1/2 \}$ is open and contains $x$. As $K\setminus U$ is
closed, hence compact, we can find $x_1,\cdots,x_n$ with $U_{x_1}\cup
\cdots\cup U_{x_n} \supseteq K\setminus U$.
Given $f\in J$, notice that $|f|^2 = f \overline{f} \in J$ as $J$ is an ideal.
Thus $h = |f_{x_1}|^2 + \cdots + |f_{x_n}|^2\in J$. Then $h(k)=0$ for each
$k\in F$, while for each $x\not\in U$, there is $i$ with $x\in U_{x_i}$, and so
$h(x)>(1/2)^2 = 1/4$.
Now consider\footnote{Thanks to George Berkley for point this trick out.}
$g_n\in C(K)$ defined by
\[ g_n(x) = g(x) \frac{n h(x)}{1+nh(x)}. \]
Notice that $nh(x)/(1+nh(x)) \in [0,1)$ for all $x$ and $n$.
If $x\not\in U$ then $h(x)>1/4$ and so $nh(x)/(1+nh(x)) \rightarrow 1$ as
$n\rightarrow\infty$, \emph{uniformly} for $x\not\in U$. In particular, if
$n$ is large, then $|g_n(x)-g(x)| < \epsilon$ for all $x\not\in U$.
If $x\in U$ then $|g(x)|<\epsilon$ and so also $|g_n(x)|<\epsilon$, and so
$|g_n(x)-g(x)| < 2\epsilon$. We conclude that for $n$ large, $g_n$ approximates
$g$ in the supremum norm. However, notice that
\[ g_n = \frac{ng}{1+nh} h, \]
and so as $J$ is an ideal, $g_n\in J$.
As $J$ is closed, we conclude that $g\in J$, as required.
\subsection{An example}
Consider $B=C^1([0,1])$ the continuous
differentiable functions on $[0,1]$ with the norm $\|f\|=\|f\|_\infty +
\|f'\|_\infty$. This is a natural Banach function algebra on $[0,1]$
(see Exercise~4.12).
Let $J$ be the collection of functions with $f(1/2)=f'(1/2)=0$.
This is a linear subspace, and an ideal, as for any $g$,
\[ (gf)(1/2) =0, \qquad (gf)'(1/2) = g'(1/2)f(1/2) + g(1/2) f'(1/2) =0. \]
It's easy to see that it's closed (thanks to the norm we used). However,
I claim that the associated $F$ must be $\{1/2\}$, and so $I(F)\not=J$.
Indeed, the function $f(x)=(x-1/2)^2$ is in $J$ but vanishes only at $1/2$.
What goes wrong with the above proof? The problem is that while $\|g_n-g\|$
is small, we have no control over $\| g_n' - g'\|$.
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