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\ifsolutions
\title[Math 617 (Winter 2024): Solutions~3]{Math 617 (Winter 2024,
        Phillips): Solutions to Homework~3}
\date{31~January 2024}
\else
\title[Math 617 (Winter 2024): Homework~3]{Math 617 (Winter 2024,
        Phillips): Homework~3}
\date{24~January 2024}
\fi
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\begin{document}

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\ifsolutions
\else
\vspace*{-1ex}
\fi

\maketitle

This homework assignment is due Wednesday 31~January 2024.

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

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


\begin{pbm}[Problem~6 in Chapter~5 of Rudin]\label{Pb2024_4124_HB_Hilb}
Let $H$ be a Hilbert space, let $M \S H$ be a closed subspace,
and let $\om_0 \in M^*$.
Prove that there is a {\emph{unique}} $\om \in H^*$
such that $\om |_M = \om_0$ and $\| \om \| = \| \om_0 \|$.
Moreover, prove that $\om$ vanishes on~$M^{\perp}$.
\end{pbm}

\ifsolutions

\begin{proof}[Solution]
Since $M$ is a Hilbert space, there is $\et_0 \in M$
such that for all $\xi \in M$
we have $\om_0 (\xi) = \langle \xi, \et_0 \rangle$.
Moreover, $\| \et_0 \| = \| \om_0 \|$.
Define $\om \in H^*$ by $\om (\xi) = \langle \xi, \et_0 \rangle$
for $\xi \in H$.
Obviously $\om |_M = \om_0$.
We have $\| \om \| = \| \et_0 \| = \| \om_0 \|$.
Also, if $\xi \in M^{\perp}$,
then, since $\et_0 \in M$, we have
$\om (\xi) = \langle \xi, \et_0 \rangle = 0$.

It remains to prove that if $\rh \in H^*$
satisfies $\rh |_M = \om_0$ and $\| \rh \| = \| \om_0 \|$,
then $\rh = \om$.
Given $\rh$, there is $\et \in H$ such that for all $\xi \in M$
we have $\rh (\xi) = \langle \xi, \et \rangle$.
Since $\rh |_M = \om_0$, for all $\xi \in M$ we have
\[
0 = \rh (\xi) - \om (\xi)
  = \langle \xi, \et \rangle - \langle \xi, \et_0 \rangle
  = \langle \xi, \et - \et_0 \rangle.
\]
In particular, $\langle \et_0, \et - \et_0 \rangle = 0$.
Therefore
\[
\| \et_0 \|^2 + \| \et - \et_0  \|^2
 = \| \et \|^2
 = \| \rh \|^2
 = \| \om_0 \|^2
 = \| \et_0 \|^2.
\]
Thus $\| \et - \et_0  \| = 0$, whence $\et = \et_0$, so $\rh = \om$.
\end{proof}
\fi

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



\begin{pbm}[Problem~18 in Chapter~5 of Rudin]\label{Pb2024_4124_ConvDnSet}
Let $E$ be a normed vector space, let $F$ be a Banach space,
and let $(a_n)_{n \in \N}$ be a bounded sequence in $L (E, F)$.
Suppose that there is a dense set $S \S E$
such that $\limi{n} a_n \xi$ exists for all $\xi \in S$.
Prove that $\limi{n} a_n \xi$ exists for all $\xi \in E$.
\end{pbm}

\ifsolutions

\begin{proof}[Solution]
For $\xi \in E$ set $g (\xi) = \limi{n} a_n \xi$.
Also set $M = 1 + \sup_{n \in \N} \| a_n \|$.

We claim that for $\xi \in E$, the sequence $(a_n \xi)_{n \in \N}$
is Cauchy.
Since $E$ is complete, the result will follow.

To prove the claim, let $\xi \in E$ and let $\ep > 0$.
Choose $\et \in S$ such that $\| \xi - \et \| < \ep / (4 M)$.
Choose $N \in \N$ such that for all $n \geq N$ we have
$\| a_n \et - g (\et) \| < \frac{\ep}{4}$.
Let $m, n \geq N$.
Then
\[
\begin{split}
\| a_m \xi - a_n \xi \|
& \leq \| a_m \| \| \xi - \et \| + \| a_m \et - g (\et) \|
  + \| g (\et) - a_n \et \| + \| a_n \| \| \xi - \et \|
\\
& < M \left( \frac{\ep}{4 M} \right) + \frac{\ep}{4} + \frac{\ep}{4}
     + M \left( \frac{\ep}{4 M} \right)
  = \ep.
\end{split}
\]
The claim is proved, and therefore the solution is complete.
\end{proof}

We took $M = 1 + \sup_{n \in \N} \| a_n \|$
rather than $M = \sup_{n \in \N} \| a_n \|$
to avoid the possibility of dividing by zero.

The following example shows that completeness of $F$ is necessary,
even if $E$ is assumed complete.
(It was not asked for in the problem.)

\begin{exa}\label{E_4131_F_not_cp}
Let $E = l^1 (\N)$ with its usual norm.
Let $F$ be the set of sequences in $l^1 (\N)$ with finite support,
with the $l^1$~norm.
For $n \in \N$ define $a_n \in L (E, F)$ by,
for $\xi = (\xi_1, \xi_2, \ldots ) \in L^1 (\N)$,
\[
a \xi = \bigl( \xi_1, \xi_2, \ldots, \xi_n, 0, 0, \ldots \bigr).
\]
Then $\| a_n \| \leq 1$.
Set $S = F$.
Then for all $\xi \in S$ we have $\limi{n} a_n \xi = \xi$.
However, if $\xi \in E \SM S$,
for example if $\xi = ( n^{-2} )_{n \in \N}$,
then $\limi{n} a_n \xi$ does not exist in~$F$.
\end{exa}

\fi


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


\begin{pbm}[An expansion of Problem~17 in Chapter~5 of
 Rudin]\label{2017_NP_MultOp_p}
(This problem is worth two ordinary problems.)
Let $\mu$ be a n onzero positive measure on a measurable space~$X$.
Let $p \in [1, \I)$.
For $f \in L^{\infty} (X, \mu)$,
let $m (f) \colon  L^p (X, \mu) \to  L^p (X, \mu)$
be defined by $m (f) (\xi) (x) = f (x) \xi (x)$,
that is, $m (f)$ is the multiplication operator by $f$.
%
\begin{enumerate}
% [label=$\mathrm{(\arabic*)}$]
%
\item\label{I_4124_MultOp_Nbd}
Prove that $\| m (f) \| \leq \| f \|_{\I}$
for all $f \in L^{\infty} (X, \mu)$.
%
\item\label{I_4124_MultOp_NmEq}
Prove
that $\| m (f) \| = \| f \|_{\I}$ for all $f \in L^{\infty} (X, \mu)$
\ifo{} $\mu$ is semifinite.
%
\item\label{I_4124_MultOp_WhenSurj}
Assume that $\mu$ is semifinite.
Give, with proof, a characterization in terms of~$f$
of those $f \in L^{\infty} (X, \mu)$ for which the operator
$m (f)$ is surjective.
%
\item\label{I_4124_MultOp_SjInj}
Assume that $\mu$ is semifinite.
Let $f \in L^{\infty} (X, \mu)$, and suppose that $m (f)$ is surjective.
Prove that $m (f)$ is injective.
%
\item\label{I_4124_MultOp_MSj}
Give\DIFaddbegin \DIFadd{, with proof, }\DIFaddend an example of a finite measure $\mu$
on a \mb{} space~$X$ and $f \in L^{\infty} (X, \mu)$
such that that $m (f)$ is injective \DIFdelbegin \DIFdel{.
}\DIFdelend but not surjective.
%
\end{enumerate}
%
\end{pbm}

Recall that
a measure $\mu$ on $X$ is called {\emph{semifinite}}
if for every measurable set $E \S X$ with $\mu (E) > 0$,
there is a measurable set $F \S E$ with $0 < \mu (F) < \I$.

\begin{exa}\label{2017_SemiFinite}
Here are some examples of measures which are semifinite
and some which are not.
(This isn't an exercise.)
\begin{enumerate}
\item\label{2017_SemiFinite_SigmaF}
Every $\sm$-finite measure is semifinite.
\item\label{2017_SemiFinite_Ct}
Counting measure on $\R$ is semifinite but not $\sm$-finite.
\item\label{2017_SemiFinite_InfTriv}
On any set $X$ take the measurable sets to be $\varnothing$
and $X$, and take $\mu (\varnothing) = 0$ and $\mu (X) = \I$.
Then $\mu$ is not semifinite.
\item\label{2017_SemiFinite_CtbleComp}
On $\R$ take the measurable sets to be the countable sets
and their complements.
Take $\mu (E) = 0$ if $E$ is countable and
$\mu (E) = \I$ if $\R \SM E$ is countable.
Then $\mu$ is not semifinite.
\end{enumerate}
\end{exa}

I know of no real use for measures which are not semifinite.

\ifsolutions
\vspace{1ex}

We break the solution into several propositions.


\begin{prp}\label{4124_H6_1}
For every positive measure $\mu$ and every $f \in L^{\I} (\mu)$,
we have $m (f) \in L (L^p (\mu))$ and $\| m (f) \| \leq \| f \|_{\I}$.
\end{prp}

\begin{proof}
We estimate $\| m (f) \xi \|$ for $\xi \in L^p (\mu)$.
We have
\[
\| m (f) \xi \|^p = \int_X | f (x) \xi (x) |^p \, d \mu (x)
  \leq \int_X \| f \|_{\I}^p \| \xi (x) |^p \, d \mu (x)
  = \| f \|_{\I}^p \| \xi \|^p.
\]
It is clear that $m (f)$ is linear, so we conclude that
$m (f)$ is a bounded linear map with $\| m (f) \| \leq \| f \|_{\I}$.
\end{proof}

%  The correct condition for equality has a standard name,
%  as given in the following definition.
%
%  \begin{dfn}\label{4124_H6_2}
%  A measure $\mu$ on $X$ is called {\emph{semifinite}}
%  if for every measurable set $E \S X$ with $\mu (E) > 0$,
%  there is a measurable set $F \S E$ with $0 < \mu (F) < \I$.
%  \end{dfn}

%  \begin{exa}\label{4124_H6_3}
%  \begin{enumerate}
%  \item\label{4124_H6_3_1}
%  Every $\sm$-finite measure is semifinite.
%  \item\label{4124_H6_3_2}
%  Counting measure on $\R$ is semifinite but not $\sm$-finite.
%  \item\label{4124_H6_3_3}
%  On any set $X$ take the measurable sets to be $\varnothing$
%  and $X$, and take $\mu (\varnothing) = 0$ and $\mu (X) = \I$.
%  Then $\mu$ is not semifinite.
%  \item\label{4124_H6_3_4}
%  On $\R$ take the measurable sets to be the countable sets
%  and their complements.
%  Take $\mu (E) = 0$ if $E$ is countable and
%  $\mu (E) = \I$ if $\R \SM E$ is countable.
%  Then $\mu$ is not semifinite.
%  \end{enumerate}
%  \end{exa}

%  \begin{rmk}\label{4124_H6_4}
%  I know of no real use for measures which are not semifinite.
%  \end{rmk}

\begin{prp}\label{4124_H6_5}
For every semifinite
positive measure $\mu$ and every $f \in L^{\I} (\mu)$,
we have $\| m (f) \| = \| f \|_{\I}$.
\end{prp}


\begin{proof}
We need only show $\| m (f) \| \geq \| f \|_{\I}$.
Let $c < \| f \|_{\I}$; we show $\| m (f) \| \geq c$.
Set $E = \{ x \in X \colon | f (x) | \geq c \}$.
Then $\mu (E) > 0$ by the definition of the essential supremum.
By semifiniteness,
there is a measurable set $F \S E$ with $0 < \mu (F) < \I$.
Define $\xi = (\mu (F))^{-1/p} \ch_F$.
Then $\xi \in L^p (\mu)$ and $\| \xi \|_p = 1$.
We have
\[
\| m (f) \xi \|^p = \int_F [(\mu (F))^{-1/2}| f (x) |]^p \, d \mu (x)
  \geq \left( \frac{1}{\mu (F)} \right) \cdot c^p \cdot \mu (F)
  = c^p.
\]
So $\| m (f) \xi \| \geq c$.
Thus $\| m (f) \| \geq c$.
\end{proof}


\begin{prp}\label{4124_H6_6}
For every positive measure $\mu$ which is not semifinite,
there exists $f \in L^{\I} (\mu)$ such that $\| f \|_{\I} = 1$
and $m (f) = 0$.
\end{prp}


\begin{proof}
By definition, there is a measurable set $E \S X$ such that
$\mu (E) > 0$, and with the property that for every measurable
$F \S E$ we have $\mu (F) = 0$ or $\mu (F) = \I$.

We first claim that every $\xi \in L^p (\mu)$ must vanish
almost everywhere on $E$.
The set $\left\{ x \in E \colon | \xi (x) | \geq \frac{1}{n} \right\}$
is a measurable subset of $E$ with finite measure,
hence has measure zero.
Therefore
\[
\mu ( \left\{ x \in E \colon \xi (x) \neq 0 \right\} )
 \leq \DIFdelbegin \DIFdel{\sum_{n = 1}^{\infty}
   }\DIFdelend \DIFaddbegin \DIFadd{\sup_{n \in \N}
   }\DIFaddend \mu \left( \left\{ x \in E \colon
       | \xi (x) | \geq {\textstyle{ \frac{1}{n} }} \right\} \right)
 = 0.
\]
This proves the claim.

Now set $f = \ch_E$.
Since $\mu (E) \neq 0$, we have $\| f \|_{\I} = 1$.
However, for any $\xi \in L^p (\mu)$, the function $f \xi$
vanishes almost everywhere on $E$ (because $\xi$ does),
and vanishes everywhere on $X \SM E$ (because $\ch_E$ does).
Thus $m (f) \xi = 0$.
So $m (f)$ is the zero operator.
\end{proof}


We now consider when $m (f)$ is surjective.
%  I will give a solution only for the case that the measure $\mu$
%  is semifinite.
%  The parts of the next proposition which actually
%  answer the question in the sense intended are the last two.
%  The lemma is given separately for use in an example afterwards.

\begin{lem}\label{4124_H6_7}
Let $\mu$ be a positive measure,
and let $f \in L^{\I} (\mu)$.
Suppose $\mu (  \{ x \in X \colon f (x) = 0 \} ) = 0$.
Then $m (f)$ is injective.
\end{lem}

\begin{proof}
Set $E = \{ x \in X \colon f (x) = 0 \}$.
Let $\xi \in L^p (\mu)$ with $m (f) \xi = 0$.
Then $\xi = 0$ almost everywhere on $X \SM E$.
Since $\mu (E) = 0$,
we get $\xi = 0$ almost everywhere on $X$,
so $\xi$ is the zero element of $L^p (\mu)$.
\end{proof}


\begin{prp}\label{4124_H6_8}
Let $\mu$ be a semifinite positive measure,
and let $f \in L^{\I} (\mu)$.
\Tfae:
\begin{enumerate}
\item\label{4124_H6_8_1}
$m (f)$ is surjective.
\item\label{4124_H6_8_2}
$m (f)$ is bijective.
\item\label{4124_H6_8_3}
$1/f \in L^{\I} (\mu)$.
\item\label{4124_H6_8_4}
There exists $\af > 0$ such that
$\mu ( \{ x \in X \colon | f (x) | < \af \} ) = 0$.
\end{enumerate}
\end{prp}


\begin{proof}
The equivalence of~(\ref{4124_H6_8_3})
and~(\ref{4124_H6_8_4}) is easy and is omitted.
That~(\ref{4124_H6_8_3}) implies~(\ref{4124_H6_8_2}) is clear,
since $m (f)$ has the inverse
operator $m (1/f)$.
That~(\ref{4124_H6_8_2}) implies~(\ref{4124_H6_8_1}) is trivial.
We complete the proof by showing that~(\ref{4124_H6_8_1})
implies~(\ref{4124_H6_8_2})
and that~(\ref{4124_H6_8_2}) implies~(\ref{4124_H6_8_4}).

Assume~(\ref{4124_H6_8_1}).
Set
\[
E = \{ x \in X \colon f (x) = 0 \}.
\]
We claim that $\mu (E) = 0$.
Suppose not.
By semifiniteness, there is $F \S E$ with $0 < \mu (F) < \I$.
Then $\ch_F \in L^p (\mu)$,
but $\ch_F$ is not in the range of $m (f)$ because
$f \xi$ vanishes on $F$ for every $\xi \in L^p (\mu)$.
This proves the claim.
It now follows from Lemma~\ref{4124_H6_7} that $m (f)$ is bijective,
which is~(\ref{4124_H6_8_2}).

Now assume~(\ref{4124_H6_8_2}).
The Open Mapping Theorem provides $c > 0$ such that
$\| m (f) \xi \| \geq c \| \xi \|$ for all $\xi \in L^p (\mu)$.
Set
\[
F = \left\{ x \in X
   \colon | f (x) | \leq {\textstyle{\frac{1}{2}}} c \right\}.
\]
We claim $\mu (F) = 0$, which will prove~(\ref{4124_H6_8_4}).
If not, by semifiniteness, there is $G \S F$ with $0 < \mu (G) < \I$.
Then $\ch_G \in L^p (\mu)$ and satisfies
$\| \ch_G \| = [\mu (g)]^{1/2}$ and
\[
\| m (f) \ch_G \|^p = \int_G | f |^p \, d \mu
 \leq \left( {\textstyle{\frac{1}{2}}} c \right)^p \mu (G),
\]
so that
$\| m (f) \ch_G \| \leq {\textstyle{\frac{1}{2}}} c [\mu (g)]^{1/2}$.
Since $\mu (G) \neq 0$, this contradicts the choice of $c$ above.
\end{proof}

\begin{rmk}\label{4124_H6_9}
It is not really necessary to use the Open Mapping Theorem
in the proof of Proposition~\ref{4124_H6_8}, but it
\DIFaddbegin \DIFadd{significantly }\DIFaddend simplifies the proof.
\DIFaddbegin \DIFadd{We demonstrate with a direct proof of the implication from
(\ref{4124_H6_8_1}) to~(\ref{4124_H6_8_4}).
}\DIFaddend \end{rmk}


\DIFaddbegin \begin{proof}[Alternate proof
 of (\ref{4124_H6_8_1}) imples~(\ref{4124_H6_8_4})
 in Proposition~\ref{4124_H6_8}]
\DIFadd{Assume that~(\ref{4124_H6_8_4}) fails.
Set $E = \bigl\{ x \in X \colon f (x) = 0 \bigr\}$.
There are two cases.
}

\DIFadd{First suppose that $\mu (E) > 0$.
By semifiniteness, there is $F \S E$ with $0 < \mu (F) < \I$.
Then $\ch_F \in L^p (X, \mu)$.
However, $\ch_F \not\in {\operatorname{Ran}} (m (f))$,
because for all $\xi \in L^p (X, \mu)$ we have $m (f) \xi |_F = 0$
almost everywhere.
}

\DIFadd{So suppose that $\mu (E) = 0$.
For $n \in \N$ set
}\[
\DIFadd{S_n = \left\{ x \in X \colon
  \frac{1}{n + 1} \leq | f (x) | < \frac{1}{n} \right\}.
}\]
\DIFadd{If there are only finitely many $n \in \N$ such that $\mu (S_n) > 0$,
then there is $N \in \N$ such that $\mu (S_n) = 0$ for all $n \geq N$.
Therefore
}\[
\DIFadd{\mu \left( \left\{ x \in X \colon
  | f (x) | < \frac{1}{N} \right\} \right)
 = \mu \left( \coprod_{n = N}^{\I} S_n \right)
 = \sum_{n = N}^{\I} \mu (S_n)
 = 0,
}\]
\DIFadd{contradicting the failure of~(\ref{4124_H6_8_4}).
So there is a sequence $n (1) < n (2) < \cdots$ in $\N$
such that for all $k \in \N$ we have $\mu (S_{n (k)}) > 0$.
By semifiniteness, there is $T_k \S S_{n (k)}$ such that $0 < \mu (T_k) < \I$.
}

\DIFadd{For $k \in \N$ set
}\[
\DIFadd{r_k = \frac{1}{k^{1 / (2 p)} n (k) \mu (T_k)^{1 / p} }.
}\]
\DIFadd{The sets $T_1, T_2, \ldots$ are disjoint.
Set $T = \bigcup_{k = 1}^{\I} T_k$.
Then we can define $\et \colon X \to [0, \I)$ by
}\[
\DIFadd{\et (x)
 = \begin{cases}
   r_k & \hspace*{1em} {\mbox{$k \in \N$ and $x \in T_k$}}
       \\
   0 & \hspace*{1em} x \in X \SM T.
\end{cases}
}\]
\DIFadd{We have
}\[
\DIFadd{\int_X | \et |^p \, d \mu
 = \sum_{k = 1}^{\I} r_k^p \mu (T_k)
 = \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} n (k)^p}
 \leq \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} \cdot k^p}
 \leq \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2} \cdot k}
 < \I.
}\]
\DIFadd{so $\et \in L^p (X, \mu)$.
}

\DIFadd{Suppose $\xi \in L^p (X, \mu)$ and $m (f) \xi = \et$.
For $k \in \N$ and almost every $x \in T_k$, we then have
}\[
\DIFadd{| \xi (x) |
  = \left| \frac{\et (x)}{f (x)} \right|
  = \left| \frac{r_k}{f (x)} \right|
  \geq r_k n (k).
}\]
\DIFadd{Therefore, using the Monotone Convergence Theorem,
}\[
\DIFadd{\int_X | \xi |^p \, d \mu
 \geq \sum_{k = 1}^{\I} \int_{T_k} | \xi |^p \, d \mu
 \geq \sum_{k = 1}^{\I} r_k^p n (k)^p \mu (T_k)
 = \sum_{k = 1}^{\I} \frac{1}{k^{1 / 2}}
 = \I.
}\]
\DIFadd{This contradicts $\xi \in L^p (X, \mu)$.
So $m (f)$ is not surjective.
}\end{proof}

\DIFaddend \begin{exa}\label{4124_H6_10}
Proposition~\ref{4124_H6_8} fails whenever $\mu$ is not semifinite.
With $f$ as in the proof of Proposition~\ref{4124_H6_6},
the function $1 - f$
satisfies~(\ref{4124_H6_8_1}) and~(\ref{4124_H6_8_2})
but not~(\ref{4124_H6_8_3}) or~(\ref{4124_H6_8_4}).
\end{exa}

\begin{rmk}\label{4124_H6_11}
The conditions in Proposition~\ref{4124_H6_8} are not equivalent to
injectivity of $m (f)$.
Example: with $\mu$ being Lebesgue measure on $[0, 1]$,
take $f (t) = t$ for all $t$.
Then $m (f)$ is injective by Lemma~\ref{4124_H6_7}.
\DIFaddbegin \DIFadd{However, $m (f)$ is not surjective, by the criterion in
Proposition \ref{4124_H6_8}(\ref{4124_H6_8_4}).
}\DIFaddend \end{rmk}
\fi


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


\begin{pbm}[Problem 8(c) in Chapter~5 of Rudin]\label{Pb_4124_NPDualCt}
Let $E$ be a normed vector space,
and let $(\xi_n)_{n \in \N}$ be a sequence in~$E$.
Suppose that $\limi{n} \om (\xi_n)$ exists for all $\om \in E^*$.
Prove that $(\xi_n)_{n \in \N}$ is bounded.
\end{pbm}

\ifsolutions

\begin{proof}[Solution]
Let $\Ld \colon E \to E^{**}$ be the standard map, that is,
$\Ld (\xi) (\om) = \om (\xi)$ for $\xi \in E$ and $\om \in E^*$.
We have seen that $\| \Ld (\xi) \| = \| \xi \|$ for all $\xi \in E$.

For $\om \in E^*$, we have
$\sup_{n \in \N} | \Ld (\xi_n) (\om) | = \sup_{n \in \N} | \om (\xi_n) |$.
Since $\limi{n} \om (\xi_n)$ exists,
it follows that $\sup_{n \in \N} | \Ld (\xi_n) (\om) | < \infty$.
\DIFdelbegin \DIFdel{So }\DIFdelend \DIFaddbegin \DIFadd{Since $E^*$ is complete, }\DIFaddend the Uniform Boundedness Principle
implies that $\sup_{n \in \N} \| \Ld (\xi_n) \| < \infty$.
\DIFdelbegin \DIFdel{Therefore }\DIFdelend \DIFaddbegin \DIFadd{So }\DIFaddend $\sup_{n \in \N} \| \xi_n \| < \infty$.
\end{proof}
\fi

\end{document}
