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\ifsolutions
\title[Math 617 (Winter 2024): Solutions~2]{Math 617 (Winter 2024,
        Phillips): Partial Solutions to Homework~2}
\date{24~January 2024}
\else
\title[Math 617 (Winter 2024): Homework~2]{Math 617 (Winter 2024,
        Phillips): Homework~2}
\date{17~January 2024}
\fi

\begin{document}

% \setcounter{section}{-1}

\maketitle

This homework assignment is due Wednesday 24~January 2024.

\ifsolutions
Solutions have not yet been properly proofread.

Problems and all other items use two independent numbering sequences.
This is annoying, but necessary to preserve
the problem numbers in the solutions files.
\fi


\begin{pbm}[Problem~B; worth two ordinary problems]\label{Pb2406_LEF}
This is a collection of standard facts about bounded linear operators
which are not in Chapter~5 of Rudin,
but are in functional analysis books.
Please try to do them from scratch,
without looking up the proofs in textbooks.

Recall that if $E$ and $F$ are normed vector spaces,
then $L (E, F)$ is the set of all bounded (equivalently, \ct)
linear maps from $E$ to~$F$.
(This space is also often called $B (E, F)$.)
Further, recall that if $a \in L (E, F)$, then
\[
\| a \| = \sup \bigl( \bigl\{ \| a \xi \| \colon
       {\mbox{$\xi \in E$ and $\| \xi \| \leq 1$}} \bigr\} \bigr).
\]
%
\begin{enumerate}
% [label=$\mathrm{(\arabic*)}$]
%
\item\label{I_2406_LEF_Nm}
Let $E$ and $F$ be normed vector spaces.
Prove that $\| \cdot \|$ is a norm on $L (E, F)$.
%
\item\label{I_2406_LEF_Cp}
Let $E$ be a normed vector space and let $F$ be a Banach space.
Prove that $L (E, F)$ is a Banach space.
%
\item\label{I_2406_LEF_SbMlt}
Let $E_1$, $E_2$, and $E_3$ be normed vector spaces.
Let $b \in L (E_1, E_2)$ and let $a \in L (E_2, E_3)$.
Prove that $\| a b \| \leq \| a \| \| b \|$.
%
%
\item\label{I_2406_LEF_SbMfd}
Let $E$ and $F$ be normed vector spaces, with $E$ \fd.
Prove that every linear map $a \colon E \to F$ is bounded.
(This is not as straightforward as one might initially think.)
%
\end{enumerate}
%
\end{pbm}

% The main part is Part~(\ref{I_2406_LEF_Cp}).

\ifsolutions

\begin{proof}[Proof of Part~(\ref{I_2406_LEF_Nm})]
% No proof has been written yet, but this is just calculations.
We mostly use the characterization of $\| a \|$
(done in class, but stated here in two parts for convenience of writing)
%
\begin{equation}\label{Eq_24208_Inf}
\| a \| = \inf \bigl( \bigl\{ M \in [0, \I)
      \colon {\mbox{for all $\xi \in E$ we
      have $\| a \xi \| \leq M \| \xi \|$}} \bigr\} \bigr)
\end{equation}
%
and, for all $\xi \in E$,
%
\begin{equation}\label{Eq_24208_Bound}
\| a \xi \| \leq \| a \| \| \xi \|.
\end{equation}
%
Nonemptiness of the set in~(\ref{Eq_24208_Inf}) is the definition
of boundedness for a linear map,
and~(\ref{Eq_24208_Bound}) says this set contains its infimum.
(It is necessary to specify $M \geq 0$ in~(\ref{Eq_24208_Inf}),
since otherwise if $E = \{ 0 \}$ and $a \in L (E, F)$ is the
zero operator, then $\| a \| = - \I$.)

It is obvious that if $a \in L (E, F)$ then $\| a \| \geq 0$.

Suppose $\| a \| = 0$.
By~(\ref{Eq_24208_Bound}), for any $\xi \in E$ we have
$\| a \xi \| \leq \| a \| \| \xi \| = 0$, so $a \xi = 0$.
Since this holds for all $\xi \in E$, we get $a = 0$.

For subadditivity, let $a, b \in L (E, F)$.
For all $\xi \in E$ we have, using subadditivity of the norm on~$F$
at the second step and~(\ref{Eq_24208_Bound}) at the third step,
\[
\| (a + b) \xi \|
 = \| a \xi + b \xi \|
 \leq \| a \xi \| + \| b \xi \|
 \leq \| a \| \| \xi \| + \| b \| \| \xi \|
 = \bigl( \| a \| + \| b \| \bigr) \| \xi \|.
\]
Since this is true for all $\xi \in E$, by~(\ref{Eq_24208_Inf})
we get $\| a + b \| \leq \| a \| + \| b \|$.

To prove homogeneity, for $\ld \in \C$ and $a \in L (E, F)$ we calculate:
\[
\| \ld a \|
 = \sup_{\| \xi \| \leq 1 \|} \| (\ld a) \xi \|
 = \sup_{\| \xi \| \leq 1 \|} \| \ld \cdot (a \xi) \|
 = \sup_{\| \xi \| \leq 1 \|} | \ld | \| a \xi \|
 = | \ld | \sup_{\| \xi \| \leq 1 \|} \| a \xi \|
 = | \ld | \| a \|.
\]
This completes the solution.
\end{proof}

\begin{proof}[Alternate proof of homogeneity]
For all $\xi \in E$ we have, using homogeneity of the norm on~$F$
at the second step and~(\ref{Eq_24208_Bound}) at the third step,
\[
\| (\ld a) \xi \|
 = \| \ld \cdot (a \xi) \|
 = | \ld | \| a \xi \|
 \leq | \ld | \| a \| \| \xi \|.
\]
Therefore, by~(\ref{Eq_24208_Inf}),
%
\begin{equation}\label{Eq_24208_ScInq}
\| \ld a \| \leq | \ld | \| a \|.
\end{equation}
%
If $\ld = 0$, nonnegativity of the norm on $L (E, F)$ gives
$\| \ld a \| = 0 = | \ld | \| a \|$.
Otherwise, apply (\ref{Eq_24208_ScInq}) with $\ld^{-1}$ in place of $\ld$
and $\ld a$ in place of~$a$, getting
$\| a \| \leq | \ld |^{- 1} \| \ld a \|$.
Combining this with~(\ref{Eq_24208_ScInq})
gives $\| \ld a \| = | \ld | \| a \|$.
\end{proof}

\begin{proof}[Proof of~(\ref{I_2406_LEF_Cp})]
Let $(a_n)_{n \in \N}$ be a Cauchy sequence in $L (E, F)$.
Let $\xi \in E$.
For $m, n \in \N$, we have
\[
\| a_m \xi - a_n \xi \|
 = \| (a_m - a_n) \xi \|
 \leq \| \xi \| \| a_m - a_n\|,
\]
It is now immediate that $(a_n \xi)_{n \in \N}$
is a Cauchy sequence in~$F$.
Since $F$ is complete, $\limi{n} a_n \xi$ exists.
Denote this limit by $a \xi$.

Using continuity of addition and scalar multiplication,
and linearity of $a_n$ for $n \in \N$,
one checks immediately that $a$ is linear.

In any normed vector space, we have
$\bigl| \| \xi \| - \| \et \| \bigr| \leq \| \xi - \et \|$.
Applying this to $L (E, F)$,
it is immediate that $(\| a_n \|)_{n \in \N}$ is a Cauchy sequence
in~$\R$.
Therefore $M = \sup_{n \in \N} \| a_n \|$ is finite.
For $\xi \in E$, we have
\[
\| a \xi \|
 = \limi{n} \| a_n \xi \|
 \leq \limsup_{n \to \I} \| a_n \| \| \xi \|
 \leq M \| \xi \|.
\]
Thus $a$ is bounded.

It remains to show that $\limi{n} \| a_n - a \| = 0$.
(Without doing this, the proof is not complete.)
So let $\ep > 0$.
Choose $N \in \N$ such that
for all $m, n \geq N$ we have $\| a_m - a_n \| < \frac{\ep}{2}$.
Let $n \in \N$ satisfy $n \geq N$.
Then for any $\xi \in E$ we have
\[
\| (a_n - a) \xi \|
  = \| a_n \xi - a \xi \|
  = \limi{m} \| a_n \xi - a_m \xi \|
  \leq \limsup_{n \to I} \| a_n - a_m \| \| \xi \|
  \leq \frac{\ep}{2} \| \xi \|.
\]
This shows that $\| a_n - a \| \leq \frac{\ep}{2} < \ep$.
Thus $\limi{n} \| a_n - a \| = 0$.
\end{proof}


\begin{proof}[Proof of~(\ref{I_2406_LEF_SbMlt})]
No proof has been written yet, but this is just a calculation.
\end{proof}


\begin{proof}[Proof of~(\ref{I_2406_LEF_SbMfd})]
Let $\et_1, \et_2, \ldots, \et_n$ form a basis for~$E$.

Define
\[
S = \bigl\{ ( \af_1, \af_2, \ldots, \af_n ) \in \C^n \colon
  \| ( \af_1, \af_2, \ldots, \af_n ) \|_2 = 1 \bigr\},
\]
which is a compact subset of $\C^n$,
and define a \cfn{} $f \colon S \to [0, \I)$ by
\[
f ( \af_1, \af_2, \ldots, \af_n )
 = \Biggl\| \sum_{j = 1}^n \af_j \et_j \Biggr\|.
\]
Set $c = \inf \bigl( \{ f (\af) \colon \af \in S \bigr\} \bigr)$.
Since $\et_1, \et_2, \ldots, \et_n$ are linearly independent,
$0 \not\in {\operatorname{Ran}} (f)$, so $c > 0$.
Set
\[
M = \frac{1}{c} \sum_{j = 1}^n \| a \et_j \|.
\]
We will show $\| a \| \leq M$.

Let $\xi \in E$ satisfy $\| \xi \| = 1$.
Then there are $\bt_1, \bt_2, \ldots, \bt_n \in \C$ such that
$\xi = \sum_{j = 1}^n \bt_j \et_j$.
Set $\bt = (\bt_1, \bt_2, \ldots, \bt_n)$ and for $j = 1, 2, \ldots, n$
set $\af_j = \| \bt \|_2^{-1} \bt_j$.
Then $( \af_1, \af_2, \ldots, \af_n ) \in S$.
Therefore $\bigl\| \| \bt \|_2^{-1} \xi \bigr\| \geq c$.
Since $\| \xi \| = 1$, we get
\[
\frac{1}{c} \geq \| \bt \|_2 \geq \max_{1 \leq j \leq n} | \bt_j |,
\]
whence
\[
\| a \xi \|
 = \Biggl\| \sum_{j = 1}^n \bt_j a \et_j \Biggr\|
 \leq \left( \max_{1 leq j \leq n} | \bt_j | \right)
       \sum_{j = 1}^n \| a_j \et_j \|
 \leq \frac{1}{c} \sum_{j = 1}^n \| a \et_j \|
 = M.
\]

For general nonzero $\xi \in E$, applying this to $\| \xi \|^{-1} \xi$
gives $\| a \xi \| \leq M \| \xi \|$.
This is also true for $\xi = 0$.
\end{proof}

\begin{rmk}\label{R_4015_basis}
It is common to write,
``let $\{ \et_1, \et_2, \ldots, \et_n \}$ be a basis for~$E$'',
but this means something different.
For example, $\bigl\{ (1, 0), \, (1, 0), \, (0, 1) \bigr\}$
is a basis for $\C^2$,
because
\[
\bigl\{ (1, 0), \, (1, 0), \, (0, 1) \bigr\}
 = \bigl\{ (1, 0), \, (0, 1) \bigr\}.
\]
\end{rmk}

\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~11 in Chapter~5 of Rudin;
 worth two ordinary problems]\label{2016_NP_LipAf}
Let $\af \in (0, 1]$ and let $[a, b] \S \R$ be a compact interval.

For $f \colon [a, b] \to \C$ define
\[
M_{\af, f}
 = \sup_{s \neq t} \frac{ | f (s) - f (t) |}{| s - t |^{\af} },
\]
and
\[
\| f \|_{\af} = | f (a) | + M_{\af, f}
\andeqn
\| f \|'_{\af} = \| f \|_{\I} + M_{\af, f}.
\]
Then define
\[
{\mathrm{Lip}}^{\af} ([a, b])
  = \big\{ f \colon [a, b] \to \C \colon M_{\af, f} < \I \big\}.
\]
We write ${\mathrm{Lip}}^{\af}$ for short when $[a, b]$ is understood.

Prove that ${\mathrm{Lip}}^{\af}$ is vector space,
that $\| \cdot \|_{\af}$ and $\| \cdot \|'_{\af}$ are norms
on ${\mathrm{Lip}}^{\af}$,
and that ${\mathrm{Lip}}^{\af}$ is a Banach space with respect
to each of these norms.
\end{pbm}

The functions $f \in {\mathrm{Lip}}^{\af}$ are said to satisfy
a {\emph{Lipschitz condition of order $\af$}},
and when $\af = 1$ to be {\emph{Lipschtiz functions}}.
The definition makes sense for any metric space
$X$ in place of $[a, b]$,
and the function spaces are Banach spaces (with the same
proofs) whenever in addition $X$ is compact.

\begin{proof}[Hint]
The proofs for both norms are essentially the same.
One can avoid repeating some of the work by showing that there
are constants $c_1, c_2 > 0$
(depending on $a$, $b$, and $\af$)
such that for all $f \in {\mathrm{Lip}}^{\af}$
we have $c_1 \| f \|_{\af} \leq \| f \|'_{\af} \leq c_2 \| f \|_{\af}$.
\phantom\qedhere
\end{proof}

\ifsolutions
The solution is presented as a number of lemmas,
organized so as to minimize the work.

\begin{lem}\label{L_6Z02_Lip_1}
Let $f \colon [a, b] \to \C$ and suppose that $M_{\af, f} < \I$.
Then $f$ is \ct.
\end{lem}

The proof is standard in earlier courses, and is omitted.

\begin{lem}\label{L_6Z02_Lip_2}
Let $f \in {\mathrm{Lip}}^{\af}$.
Then $\| f \|_{\af} \leq \| f \|'_{\af}$.
\end{lem}

\begin{proof}
This is immediate.
\end{proof}

\begin{lem}\label{L_6Z02_Lip_7}
Let $c = 1 + \max (1, \, (b - a)^{\af} )$.
Then $\| f \|'_{\af} \leq c \| f \|_{\af}$ for all
$f \in {\mathrm{Lip}}^{\af}$.
\end{lem}

\begin{proof}
Let $f \in {\mathrm{Lip}}^{\af}$.
For any $t \in [a, b]$, we have $| t - a | \leq b - a$, so
\[
\begin{split}
| f (t) |
 & \leq | f (a) | + | f (t) - f (a) |
   \leq | f (a) | + M_{\af, f} | t - a |^{\af}   \\
 & \leq | f (a) | + M_{\af, f} (b - a )^{\af}
   \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}.
\end{split}
\]
Therefore $\| f \|_{\I} \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af}$.
So
\[
\| f \|'_{\af} = \| f \|_{\I} + M_{\af, f}
  \leq \max (1, \, (b - a)^{\af} ) \| f \|_{\af} + \| f \|_{\af}
  = c \| f \|_{\af}.
\]
This completes the proof.
\end{proof}

\begin{lem}\label{L_6Z02_Lip_3}
The set ${\mathrm{Lip}}^{\af}$ is a vector space,
and $\| \cdot \|_{\af}$ and $\| \cdot \|'_{\af}$
are norms on ${\mathrm{Lip}}^{\af}$.
\end{lem}

\begin{proof}
If $f \in {\mathrm{Lip}}^{\af}$ then $M_{\af, f} < \I$ by definition,
and then clearly $\| f \|_{\I} < \I$.
Therefore $\| f \|_{\af}' < \I$ by Lemma~\ref{L_6Z02_Lip_7}.

We claim that if $f, g \colon [a, b] \to \C$ and $\ld \in \C$, then
\[
M_{\af, f + g} \leq M_{\af, f} + M_{\af, g}
\andeqn M_{\af, \ld f} = | \ld | \cdot M_{\af, f}.
\]
For the first,
\[
\begin{split}
M_{\af, f + g}
& = \sup_{s \neq t}
    \frac{ | f (s) + g (s) - f (t) - g (t) |}{| s - t |^{\af} }
  \leq \sup_{s \neq t} \left(
    \frac{ | f (s) - f (t) |}{| s - t |^{\af} }
        + \frac{ | g (s) - g (t) |}{| s - t |^{\af} } \right)
\\
& \leq \sup_{s \neq t} \frac{ | f (s) - f (t) |}{| s - t |^{\af} }
        + \sup_{s \neq t} \frac{ | g (s) - g (t) |}{| s - t |^{\af} }
  = M_{\af, f} + M_{\af, g}.
\end{split}
\]
For the second,
\[
\begin{split}
M_{\af, \ld f}
& = \sup_{s \neq t} \frac{ | \ld f (s) - \ld f (t) |}{| s - t |^{\af} }
  = \sup_{s \neq t} | \ld |
      \left( \frac{ | f (s) - f (t) |}{| s - t |^{\af} } \right)
\\
& = | \ld | \sup_{s \neq t}
      \left( \frac{ | f (s) - f (t) |}{| s - t |^{\af} } \right)
  = | \ld | \cdot M_{\af, f}.
\end{split}
\]
The claim is proved.

The known properties of $| \cdot |$ and $\| \cdot \|_{\I}$
now immediately imply
\[
\| f + g \|_{\af} \leq \| f \|_{\af} + \| g \|_{\af}
\andeqn \| \ld f \|_{\af} = | \ld | \cdot \| f \|_{\af},
\]
and
\[
\| f + g \|'_{\af} \leq \| f \|'_{\af} + \| g \|'_{\af}
\andeqn \| \ld f \|'_{\af} = | \ld | \cdot \| f \|'_{\af}.
\]
It immediately follows that ${\mathrm{Lip}}^{\af}$ is closed
under addition and scalar multiplication, so is a vector space.
To show that $\| \cdot \|_{\af}$ and $\| \cdot \|'_{\af}$
are norms, it only remains to show that if
$\| f \|_{\af} = 0$ or $\| f \|'_{\af} = 0$ then $f = 0$.
By Lemma~\ref{L_6Z02_Lip_2}
we need only consider the case $\| f \|_{\af} = 0$.
If $\| f \|_{\af} = 0$ then clearly $M_{\af, f} = 0$,
from which it follows that $| f (s) - f (t) | = 0$ whenever $s \neq t$.
That is, $f$ is constant.
But $\| f \|_{\af} = 0$ also implies $| f (a) | = 0$,
whence $f = 0$.
\end{proof}

\begin{lem}\label{L_6Z02_Lip_4}
The space ${\mathrm{Lip}}^{\af}$ is complete in $\| \cdot \|'_{\af}$.
\end{lem}

\begin{proof}
Let $(f_n)_{n \in \N}$ be a Cauchy sequence in ${\mathrm{Lip}}^{\af}$
with respect to $\| \cdot \|'_{\af}$.
Since $\| f \|_{\I} \leq \| f \|'_{\af}$ for any $f$,
it follows from Lemma~\ref{L_6Z02_Lip_1} that
$(f_n)_{n \in \N}$ is a Cauchy sequence in $C ([a, b])$.
Therefore there is $f \in C ([a, b])$ such that
$\| f_n - f \|_{\I} \to 0$, that is, $(f_n)_{n \in \N}$
converges uniformly to $f$.

Set $M = \sup_{n \in \N} \| f_n \|'_{\af}$.
Since $(f_n)_{n \in \N}$ is a Cauchy sequence
with respect to $\| \cdot \|'_{\af}$, we get $M < \I$.
Now let $s, t \in [a, b]$ with $s \neq t$.
Since $f_n (s) \to f (s)$ and $f_n (t) \to f (t)$,
we get
\[
\frac{ | f (s) - f (t) |}{| s - t |^{\af} }
  = \limi{n} \frac{ | f_n (s) - f_n (t) |}{| s - t |^{\af} } \leq M.
\]
Since $s$ and $t$ are arbitrary, we get $M_{\af, f} \leq M < \I$.
Therefore $f \in {\mathrm{Lip}}^{\af}$.

We must still show that $\limi{n} \| f_n - f \|'_{\af} = 0$.
Let $\ep > 0$.
Choose $N$ so large that whenever $m, n \geq N$ then
$\| f_m - f_n \|'_{\af} < {\textstyle{\frac{1}{3}}} \ep$.
We are going to show that $n \geq N$ implies
$\| f_n - f \|'_{\af} < \ep$, and the main part is to
estimate $M_{\af, \, f_n - f}$.
So let $n \geq N$.
Let $s, t \in [a, b]$ with $s \neq t$.
Then
\[
\begin{split}
\frac{ | [f_n (s) - f (s)] - [f_n (t) - f (t)] |}{| s - t |^{\af} }
 & = \limi{m} \frac{
   | [f_n (s) - f_m (s)] - [f_n (t) - f_m (t)] |}{| s - t |^{\af} }
                 \\
 & \leq \sup_{m \geq N} M_{\af, \, f_n - f_m}
   \leq {\textstyle{\frac{1}{3}}} \ep.
\end{split}
\]
Taking the supremum over all $s$ and $t$ such that $s \neq t$,
we get
$M_{\af, \, f_n - f} \leq {\textstyle{\frac{1}{3}}} \ep
 < {\textstyle{\frac{2}{3}}} \ep$.
Therefore
\[
\| f_n - f \|'_{\af} = \| f_n - f \|_{\I} + M_{\af, \, f_n - f}
 < {\textstyle{\frac{1}{3}}} \ep + {\textstyle{\frac{2}{3}}} \ep
 = \ep.
\]
That is, $n \geq N$ implies $\| f_n - f \|'_{\af} < \ep$.
\end{proof}

\begin{rmk}\label{R_24208_No_CX_cpt}
Proving Lemma~\ref{L_6Z02_Lip_4}
without using completeness of $C ([a, b])$
requires only one or two more sentences,
since continuity of the limit function is not used.
This change eliminates the need for Lemma~\ref{L_6Z02_Lip_1}.
\end{rmk}

\begin{rmk}\label{R_6Z02_Lip_5}
It is almost as easy to prove directly that 
${\mathrm{Lip}}^{\af}$ is complete in $\| \cdot \|_{\af}$.
One extra piece of reasoning is required, which is
in any case contained in the proof of Lemma~\ref{L_6Z02_Lip_7} below.
\end{rmk}

\begin{rmk}\label{R_6Z02_Lip_6}
In the proof of Lemma~\ref{L_6Z02_Lip_4}, it is essential to prove that
$\limi{n} \| f_n - f \|'_{\af} = 0$.
Without this, the lemma has not been proved.
\end{rmk}

\begin{prp}\label{P_6Z02_Lip_8}
The space ${\mathrm{Lip}}^{\af}$ is a Banach space in
the norm $\| \cdot \|_{\af}$ and also in the norm $\| \cdot \|'_{\af}$.
\end{prp}

\begin{proof}
For $\| \cdot \|'_{\af}$, combine Lemma~\ref{L_6Z02_Lip_3}
and Lemma~\ref{L_6Z02_Lip_4}.
For $\| \cdot \|_{\af}$, according to Lemma~\ref{L_6Z02_Lip_3}
we need only verify
that ${\mathrm{Lip}}^{\af}$ is complete in this norm.
So let $(f_n)_{n \in \N}$ be a Cauchy sequence in ${\mathrm{Lip}}^{\af}$
with respect to $\| \cdot \|_{\af}$.
It follows from Lemma~\ref{L_6Z02_Lip_7}
that $(f_n)_{n \in \N}$ is a Cauchy sequence
with respect to $\| \cdot \|'_{\af}$.
Lemma~\ref{L_6Z02_Lip_4} provides $f \in {\mathrm{Lip}}^{\af}$ such that
$\limi{n} \| f - f_n \|'_{\af} = 0$.
It is now immediate from Lemma~\ref{L_6Z02_Lip_2} that
$\limi{n} \| f - f_n \|_{\af} = 0$.
\end{proof}

\begin{rmk}\label{R_6Z02_Lip_9}
Let $E$ be a vector space.
Two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on $E$ are
called {\emph{equivalent}} if there are $c_1, c_2 > 0$ such that
$c_1 \| \xi \|_1 \leq \| \xi \|_2 \leq c_2 \| \xi \|_1$ for all
$\xi \in E$.
It is easily checked that this relation is an equivalence relation.
Also, one checks easily that $\| \cdot \|_1$ and $\| \cdot \|_2$
are equivalent \ifo\  both the identity maps
\[
\id_E \colon (E, \| \cdot \|_1) \to (E, \| \cdot \|_2) \andeqn
\id_E \colon (E, \| \cdot \|_2) \to (E, \| \cdot \|_1)
\]
are \ct.

Lemma~\ref{L_6Z02_Lip_2} and Lemma~\ref{L_6Z02_Lip_7}
show that $\| \cdot \|_{\af}$ and $\| \cdot \|'_{\af}$
are equivalent, and the proof of Proposition~\ref{P_6Z02_Lip_8}
shows that
a vector space which is complete in some norm is also complete
in any equivalent norm.
\end{rmk}
\fi

\begin{pbm}[Problem~16 in Chapter~5 of Rudin]\label{Pb2409_ClGr}
Prove the following theorem
(the Closed Graph Theorem).
Let $E$ and $F$ be Banach spaces,
and let $a \colon E \to F$ be a linear map.
Suppose that whenever $(\xi_n)_{n \in \N}$
is a sequence in~$E$, $\xi \in E$, $\et \in F$, and
\[
\limi{n} \xi_n = \xi
\andeqn
\limi{n} a \xi_n = \et,
\]
then $a \xi = \et$.
Prove that $a$ is \ct.

Moreover, prove that, without the linearity hypothesis,
the statement is false, even for $E = F = \R$.
\end{pbm}

\begin{proof}[Hint]
Make $E \oplus F$ into a Banach space with the
standard vector space operations
and the norm $\| (\xi, \et) \| = \| \xi \| + \| \et \|$
for $\xi \in E$ and $\et \in F$.
(You should
check that this formula gives a complete norm on $E \oplus F$,
but this is easy.)
Define $G \S E \oplus F$ by
\[
G = \big\{ (\xi, a \xi) \colon \xi \in E \big\}.
\]
Prove that $G$ s a Banach space,
and that the restriction to $G$ of the first coordinate
projection $E \oplus F \to E$ is bijective and \ct.

(There are other choices for the norm on $E \oplus F$
which work equally well,
such as $\| (\xi, \et) \| = \max ( \| \xi \|, \| \et \| )$.)

For the last part, define $f \colon \R \to \R$ by
\[
f (x) = \begin{cases}
   x^{-1} & x \neq 0
       \\
   0 & x = 0.
\end{cases}
\]
\phantom\qedhere
\end{proof}

\ifsolutions

\begin{proof}[Solution to the first part]
We use the notation of the hint.

It is well known that $E \oplus F$ as described is a vector space.
It is easy to check that the formula
$\| (\xi, \et) \| = \| \xi \| + \| \et \|$
does in fact define a norm on $E \oplus F$; the proof is omitted.
Moreover, one easily checks that, using this norm,
one has $\limi{n} (\xi_n, \et_n) = (\xi, \et)$
\ifo\  $\limi{n} \xi_n = \xi$ and $\limi{n} \et_n = \et$.

Define $p \colon E \oplus F \to E$ and
$q \colon E \oplus F \to F$
by $p (\xi, \et) = \xi$ and $q (\xi, \et) = \et$
for all $\xi \in E$ and $\et \in F$.
These are well known to be linear.
Moreover, clearly $\| p (\xi, \et) \| \leq \| (\xi, \et) \|$
and $\| q (\xi, \et) \| \leq \| (\xi, \et) \|$.
In particular, $p$ and $q$ are \ct.

We prove that $E \oplus F$ is complete.
Consider a Cauchy sequence in $E \oplus F$.
It has the form $((\xi_n, \et_n))_{n \in \N}$
with $\xi_n \in E$ and $\et_n \in F$ for all $n \in \N$.
Since $p$ and $q$ reduce norms, it is immediate that
$(\xi_n)_{n \in \N}$ and $(\et_n)_{n \in \N}$
are Cauchy sequences.
Therefore $\xi = \limi{n} \xi_n$ and $\et = \limi{n} \et_n$ exist.
To complete the proof of completeness,
we show that $\limi{n} (\xi_n, \et_n) = (\xi, \et)$.
But this is immediate from the observation at the end
of the second paragraph.

Let $G$ be as in the hint.
We are assuming that for every sequence $(\xi_n)_{n \in \N}$ in $E$
such that both $\xi = \limi{n} \xi_n$ and $\et = \limi{n} a \xi_n$
exist,
one has $a \xi = \et$.
By the observation at the end of the second paragraph,
this says that if $(\gm_n)_{n \in \N}$ is a sequence in $G$
such that $\limi{n} \gm_n$ exists,
then $\limi{n} \gm_n \in G$.
Thus $G$ is closed.
It is immediate from the linearity of $a$ that $G$ is
a vector subspace of $E \oplus F$.
Thus $G$ is a Banach space.

The map $p |_G \colon G \to E$ is clearly bijective.
Since it is \ct{} and linear,
the Open Mapping Theorem implies that $(p |_G)^{-1}$ is \ct.
Therefore $a = (q |_G) \circ (p |_G)^{-1}$ is \ct.
\end{proof}

It is a standard fact,
which you need not reprove in the future,
that $E \oplus F$ as defined here is a Banach space
and that the projections $p$ and $q$ are linear and norm reducing.

\begin{proof}[Proof for the nonlinear example]
It is obvious that $f$ is not \ct{} at $0$.
To see that the graph $G \S \R^2$ of $f$ is closed,
define $F \colon \R^2 \to \R$ by $f (x, y) = x y$,
and write $G = \{ (0, 0) \} \cup F^{-1} ( \{ 1 \} )$.
This is a union of two closed sets because $F$ is \ct.
\end{proof}

Since many people have used sequences in the past,
here is a proof in terms of sequences.

\begin{proof}[Second proof of the nonlinear example]
It is obvious that $f$ is not \ct{} at $0$.
To see that the graph $G \S \R^2$ of $f$ is closed,
let $((x_n, y_n))_{n \in \N}$ be a sequence in~$G$,
and suppose that $\limi{n} (x_n, y_n) = (x, y)$.
We need to prove that $(x, y) \in G$.

We have, in particular, $\limi{n} x_n = x$.
First suppose that $x \neq 0$.
Then there is $N \in \N$ such that for all $n \geq N$
we have $x_n \neq 0$.
For such values of~$n$, we have $y_n = \frac{1}{x_n}$.
Since the function $h (t) = \frac{1}{t}$ is \ct{} at~$x$,
it follows that $\limi{n} \frac{1}{x_n} = \frac{1}{x}$.
Therefore $y = \frac{1}{x}$, whence $(x, y) \in G$.

Now suppose that $x = 0$.
If $x_n = 0$ for all but finitely many~$n$,
then $y_n = 0$ for all but finitely many~$n$,
so $\limi{n} (x_n, y_n) = (0, 0)$, which is in~$G$.
Otherwise, we show that $\limi{n} y_n$ does not exist,
which is a contradiction.
Let $M \in (0, \I)$;
we show that for all $N \in \N$ there is $n \in \N$
with $n \geq N$ and such that $| y_n | > M$.
Apply the definition of $\limi{n} x_n = 0$
with $\frac{1}{M}$ in place $\ep$,
getting $N_0 \in \N$.
Since there are infinitely many $n \in \N$ such that $x_n \neq 0$,
we can find $n \in \N$ such that $x_n \neq 0$ and $n > \max (N, N_0)$.
Then $n \geq N$ and $| y_n | > \frac{1}{\ep} = M$, as desired.
This shows that $\limi{n} y_n$ does not exist.
\end{proof}

\begin{rmk}\label{R_7223_BallMistake}
The following claim has been made in the past:
If $(x, y) \in \R^2 \SM G$ and
\[
\dt = \min ( \| (x, y) \|, \, \| (x, y) - (x, 1/x) \|,
             \,  \| (x, y) - (1/y, y) \| )
\]
(the minimum of the distances from $(x, y)$ to the three
points $(0, 0)$, $(x, 1/x)$, and $(1/y, y)$),
then the open ball $B_{\dt} (x, y)$ does not intersect $G$.
This is not true.
Take $x = y = \frac{5}{8}$.
Then $(1, 1) \in G$ and
\[
\dt = \min \left( \frac{5 \sqrt{2}}{8}, \, \frac{39}{40} \right)
  > \frac{3 \sqrt{2}}{8} = \| (x, y) - (1, 1) \|,
\]
so $(1, 1) \in B_{\dt} (x, y) \cap G$.
\end{rmk}
\fi

\end{document}
