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\ifsolutions
\title[Math 617 (Winter 2024): Solutions~6]{Math 617 (Winter 2024,
        Phillips): Solutions to Homework~6}
\date{21~February 2024}
\else
\title[Math 617 (Winter 2024): Homework~6]{Math 617 (Winter 2024,
        Phillips): Homework~6}
\date{14~February 2024}
\fi

\begin{document}

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\maketitle

\ifsolutions
\indent
Essentially no proofreading has been done, and there are slight gaps.
\fi

\begin{pbm}[Problem~5 in Chapter~7 of Rudin]\label{2017_NP_PropOfConv}
This problem counts as five regular problems.

Let $M$ be the Banach space of all complex Borel measures on~$\R$.
Recall that $\| \mu \| = | \mu | (\R)$ for $\mu \in M$.

Let $s \colon \R^2 \to \R$
be the function $s (x, y) = x + y$ for $x, y \in \R$.
For $\mu, \nu \in M$,
define $\mu * \nu$ to be the set function given by
$(\mu * \nu) (E) = (\mu \times \nu) (s^{-1} (E))$
for every Borel set $E \S \R$.
(The function $\mu * \nu$
is called the {\emph{convoluton}} of $\mu$ and~$\nu$.)
%
\begin{enumerate}
%
\item\label{2017_NP_PropOfConv_IsMeas}
Let $\mu, \nu \in M$.
Prove that $\mu * \nu$ is a complex Borel measure on~$\R$
which satisfies $\| \mu * \nu \| \leq \| \mu \| \| \nu \|$.
%
\item\label{2017_NP_PropOfConv_Char}
Let $\mu, \nu \in M$.
Prove that $\mu * \nu$ is the unique measure $\ld \in M$
which satisfies
\[
\int_{\R} f \, d \ld
 = \int_{\R} \left( \int_{\R} f (x + y) \, d \mu (x) \right) \, d \nu (y)
\]
for all $f \in C_0 (\R)$.
%
\item\label{2017_NP_PropOfConv_Alg}
Prove that the operation
$(\mu, \nu) \mapsto \mu * \nu$, from $M \times M$ to~$M$,
is commutative,
associative,
distributes over addition,
and satisfies $\af (\mu * \nu) = \af \mu * \nu = \mu * \af \nu$
for all $\af \in \C$.
(Thus, $M$, with multiplication defined by
$\mu \cdot \nu = \mu * \nu$,
is a commutative algebra over~$\C$.
Since we already know that $M$ is a Banach space,
the inequality in~(\ref{2017_NP_PropOfConv_IsMeas})
now means that $M$ is in fact a complex Banach algebra.)
%
\item\label{2017_NP_PropOfConv_Formula}
Let $\mu, \nu \in M$.
Prove that for every Borel set $E \S \R$,
\[
(\mu * \nu) (E)
 = \int_{\R} \mu \bigl( \bigl\{ x - t \colon x \in E \bigr\} \bigr)
    \, d \nu (t).
\]
%
\item\label{2017_NP_PropOfConv_Discrete}
Say that $\mu \in M$ is {\emph{discrete}}
if there is a countable set $S \S \R$
such that $\R \SM S$ is $\mu$-null,
and say that $\mu \in M$ is {\emph{continuous}}
if $\mu ( \{ x \} ) = 0$ for every $x \in \R$.
Prove that if $\mu, \nu \in M$ are both discrete,
then $\mu * \nu$ is discrete.
Prove that if $\mu, \nu \in M$ and $\mu$ is \ct,
then $\mu * \nu$ is \ct.
%
\item\label{2017_NP_PropOfConv_ACm}
As usual,
let $m$ be Lebesgue measure on~$\R$.
(Note that $m \not\in M$.)
Prove that if $\mu, \nu \in M$ and $\mu \ll m$,
then $\mu * \nu \ll m$.
%
\item\label{2017_NP_PropOfConv_Subalg}
Prove that the discrete measures form a closed subalgebra of~$M$
and that the \ct{} measures form a closed ideal in~$M$.
%
\item\label{2017_NP_PropOfConv_L1}
Recall that if $\ld$ is a (nonnegative) measure on $(X, \cM)$
and $f \colon X \to \C$
is integrable or nonnegative,
then $f \cdot \ld$ is the (complex or nonnegative) measure on~$X$
defined by $(f \cdot \ld) (E) = \int_E f \, d \ld$.
Prove that $f \mapsto f \cdot m$
defines an isometric linear map which
preserves the multiplication given by convolution
from $L^1 (\R)$ to $\big\{ \mu \in M \colon \mu \ll m \big\}$.
Use this fact to prove that
the operation $(f, g) \mapsto f * g$
makes $L^1 (\R)$ a commutative Banach algebra.
Also prove that $\big\{ \mu \in M \colon \mu \ll m \big\}$
is a closed ideal in~$M$.
%
\item\label{2017_NP_PropOfConv_Unital}
Prove that the algebra $M$ is unital,
but that $L^1 (\R)$ is not unital.
\end{enumerate}
%
\end{pbm}

You may use without proof the obvious analog of Fubini's Theorem
for complex measures.
It is proved by writing each complex measure as a linear combination
of four nonnegative measures, but the product then has $16$ terms.
Alternatively

\begin{proof}[Hint for Part~(\ref{2017_NP_PropOfConv_Unital})]
The easiest way to show that $L^1 (\R)$ is not unital is to
see what would happen to an identity element under the
Fourier transform map $f \mapsto {\widehat{f}}$.
Feel free to use that, even if we haven't yet discussed
Fourier transforms.

A more useful method, because it generalizes better,
to to prove (using the definitions in a previous homework problem)
a sufficient special case of the fact
that if $f \in L^1 (\R)$ and $g \in L^{\infty} (\R)$,
then $f * g$ is \ct.
\phantom\qedhere
\end{proof}

Almost all of this works for any locally compact Hausdorff group~$G$
in place of~$\R$,
and with a (left) Haar measure $\mu$ in place of~$m$.
The algebra is then called $M (G)$, the {\emph{measure algebra}} of~$G$.
The algebra $L^1 (G)$ is unital \ifo{} $G$ has the discrete topology;
in this case, the Haar measure can be taken to be counting measure,
and $M (G) = L^1 (G, \mu)$.
If $G$ is not commutative,
one must be a little more careful with the formulas.
The outcome is that commutativity of~$G$,
of $M (G)$, and of $L^1 (G)$, are all equivalent.

\ifsolutions

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_IsMeas})]
We need to know that $\mu * \nu$ is defined,
that is, that $s^{-1} (E)$ is in the product $\sm$-algebra
of two copies of the Borel subsets of~$\R$ when $E$ is Borel.
This is immediate from two facts:
first, the product $\sm$-algebra
of two copies of the Borel subsets of~$\R$
is exactly the Borel subsets of~$\R^2$,
and if $E$ is Borel then $s^{-1} (E)$ is Borel.
The first fact was proved in the proof that Lebesgue measure on~$\R^2$
is the product of two copies of Lebesgue measure on~$\R$.
The second follows from continuity of~$s$.

That $\mu * \nu$ is a complex measure
is now immediate from the fact that inverse images
preserve countable disjoint unions and~$\E$.

For the norm estimate, we first claim that
$| \mu \times \nu | = | \mu | \times | \nu |$
To prove the claim,
write $\mu = h \cdot | \mu |$ and $\mu = k \cdot | \nu$
for Borel functions $h, k \colon \R \to \C$ with $| h (x) | = 1$
and $| k (x) | = 1$ for all $x \in \R$.
Then, using Fubini's Theorem at the first and third steps
(the measurability criterion is immediate),
if $f \colon \R^2 \to \C$ is a bounded Borel function, then
\[
\begin{split}
\int_{\R \times \R} f \, d (\mu \times \nu)
& = \int_{\R} \left( \int_{\R} f (x, y) \, d \mu (x) \right) \, d \nu (y)
\\
& = \int_{\R} \left( \int_{\R} f (x, y) h (x) \, d | \mu | (x) \right)
     k (y) \, d | \nu | (y)
 = \int_{\R \times \R} f l \, d (| \mu | \times | \nu |).
\end{split}
\]
In particular, if $E \S \R \times \R$ is Borel, taking $f = \ch_E$
gives
\[
(\mu \times \nu) (E) = \int_{E} l \, d (| \mu | \times | \nu |).
\]
Since $E$ is an arbitrary Borel set
and $| l (x, y) | = 1$ for all $x, y \in \R$,
this implies the claim.

To prove the norm estimate, we need to show that if $(E_n)_{n \in \N}$
is a family of disjoint Borel sets
such that $\R = \coprod_{n = 1}^{\I} E_n$,
then $\sum_{n = 1}^{\I} | (\mu * \nu) (E_n) | \leq \| \mu \| \| \nu \|$.
We have
\[
\begin{split}
\sum_{n = 1}^{\I} | (\mu * \nu) (E_n) |
& = \sum_{n = 1}^{\I} | (\mu \times \nu) (s^{-1} (E_n)) |
  \leq \sum_{n = 1}^{\I} | \mu \times \nu | (s^{-1} (E_n))
\\
& = \sum_{n = 1}^{\I} \bigl( | \mu | \times | \nu | \bigr) (s^{-1} (E_n))
  = \bigl( | \mu | \times | \nu | \bigr) (\R^2)
\\
& = | \mu | (\R) \cdot | \nu | (\R)
  = \| \mu \| \| \nu \|.
\end{split}
\]
This completes the proof.
\end{proof}

\begin{proof}[Alternate proof of the norm estimate
 in Part~(\ref{2017_NP_PropOfConv_IsMeas})]
We need to show that if $(E_n)_{n \in \N}$
is a family of disjoint Borel sets
such that
%
\begin{equation}\label{Eq_222_Disj}
\R = \coprod_{n = 1}^{\I} E_n,
\end{equation}
%
then $\sum_{n = 1}^{\I} | (\mu * \nu) (E_n) | \leq \| \mu \| \| \nu \|$.
We have,
using the Monotone Convergence Theorem at the sixth step
and (\ref{Eq_222_Disj}) and countable additivity of $| \mu |$
at the seventh step,
\[
\begin{split}
\sum_{n = 1}^{\I} | (\mu * \nu) (E_n) |
& = \sum_{n = 1}^{\I} | (\mu \times \nu) (s^{-1} (E_n)) |
\\
& = \sum_{n = 1}^{\I} \left| \int_{\R}
    \nu \bigl( \bigl\{ y \in \R \colon (x, y) \in s^{-1} (E_n) \bigr\} \bigr)
     \, d \mu (x) \right|
\\
& = \sum_{n = 1}^{\I} \left| \int_{\R}
    \nu \bigl( \bigl\{ y \in \R \colon x + y \in E_n \bigr\} \bigr)
     \, d \mu (x) \right|
\\
& \leq \sum_{n = 1}^{\I} \int_{\R}
    \bigl| \nu \bigl( \bigl\{ y \in \R \colon x + y \in E_n \bigr\}
        \bigr) \bigr|
     \, d | \mu | (x)
\\
& \leq \sum_{n = 1}^{\I} \int_{\R}
    | \nu | \bigl( \bigl\{ y \in \R \colon x + y \in E_n \bigr\} \bigr)
     \, d | \mu | (x)
\\
& = \int_{\R} \sum_{n = 1}^{\I}
    | \nu | \bigl( \bigl\{ y \in \R \colon x + y \in E_n \bigr\} \bigr)
     \, d | \mu | (x)
\\
& = \int_{\R} | \nu | (\R) \, d | \mu | (x)
  = | \mu | (\R) \cdot | \nu | (\R)
  = \| \mu \| \| \nu \|.
\end{split}
\]
This completes the proof.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Char})]
The Riesz Representation Theorem for $C_0 (\R)$ implies the such a
measure $\ld$ is unique.

We must therefore prove that $\ld$ satisfies the condition.
For a Borel set $E \S \R$,
we have $\ch_{s^{-1} (E)} (x, y) = \ch_E (x + y)$.
Therefore, using Fubini's Theorem at the last step,
%
\begin{equation}\label{Eq_4222_CharSt}
\begin{split}
\int_{\R} \ch_E \, d (\mu * \nu)
& = (\mu * \nu) (E)
  = (\mu \times \nu) (s^{-1} (E))
\\
& = \int_{\R \times \R} \ch_{s^{-1} (E)} \, d (\mu \times \nu)
  = \int_{\R}
 \left( \int_{\R} \ch_E (x + y) \, d \mu (x) \right) \, d \nu (y).
\end{split}
\end{equation}
%
Thus, the required formula holds for characteristic functions
of Borel sets.
Therefore it holds for Borel simple functions.
Since every bounded Borel function is a uniform limit of
Borel simple functions and the measures are finite,
the formula holds for all bounded Borel functions,
and in particular for all functions in $C_0 (\R)$.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Alg})]
No solution has been written yet,
but, using Fubini's Theorem multiple times,
everything here is essentially immediate.
(Commutativity uses the fact that the group $\R$ is commutative.
Associativity uses associativity of addition in~$\R$.)
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Formula})]
This follows from~(\ref{Eq_4222_CharSt})
(as written: for the characteristic function of a Borel set~$E$)
together with the fact that
\[
\int_{\R} \ch_E (x + y) \, d \mu (x)
 = \mu \bigl( \bigl\{ x - t \colon x \in E \bigr\} \bigr)
\]
for every Borel set~$E$.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Discrete})]
I didn't find a definition of $\mu$-null in the edition of Rudin's book
I am using.
Here, however, are two equivalence characterizations
(for any complex measure on $(X, \cM)$:
%
\begin{enumerate}
% [label=$\mathrm{(\arabic*)}$]
%
\item\label{I_4222_MNull}
$E \in \cM$ is $\mu$-null if $| \mu | (E) = 0$.
%
\item\label{I_4222_SubQ}
$E \in \cM$ is $\mu$-null if whenever $F \S E$ is in $\cM$,
then $\mu (F) = 0$.
%
\end{enumerate}
That (\ref{I_4222_MNull}) implies~(\ref{I_4222_SubQ})
follows from the fact that $| \mu | (E)$ is the supremum over
all \mb{} partitions $E = \coprod_{n = 1}^{\I} E_n$
of $\sum_{n = 1}^{\I} | \mu (E_n) |$, by takinf $E_1 = F$.
The reverse implication is immediate from the same relation.

For the part about discrete measures,
choose subsets $E, F \S \R$ such that $\R \SM E$ is $\mu$-null
and $\R \SM F$ is $\nu$-null.
Set $S = E + F$, which is also countable.
If $Y \S \R \SM S$,
then $x \in Y$ and $y \in F$ imply $x - y \in \R \SM E$.
Thus, $x \in Y$ implies that
$\bigl\{ x - y \colon x \in Y \bigr\} \S \R \SM F$.
Therefore, by Part~(\ref{2017_NP_PropOfConv_Formula}),
\[
(\mu * \nu) (Y)
 = \int_{\R} \mu \bigl( \bigl\{ x - y \colon x \in Y \bigr\} \bigr)
    \, d \nu (y)
 = \int_{\R} 0 \, d \nu (y)
 = 0.
\]

For the part about \ct{} measures, let $z \in \R$.
Then, by Part~(\ref{2017_NP_PropOfConv_Formula}),
\[
(\mu * \nu) ( \{ z \} )
 = \int_{\R} \mu \bigl( \bigl\{ x - t \colon x \in \{ z \} \bigr\} \bigr)
    \, d \nu (t)
 = \int_{\R} \mu ( \{ z - t \} ) \, d \nu (t)
 = \int_{\R} 0 \, d \nu (t)
 = 0.
\]
This completes the solution.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_ACm})]
Let $E \S \R$ be a Borel set such that $m (E) = 0$.
By translation invariance of~$m$,
for all $y \in \R$, we also have
$m \bigl( \bigl\{ x - y \colon x \in E \bigr\} \bigr) = 0$,
so that also $\mu \bigl( \bigl\{ x - y \colon x \in E \bigr\} \bigr) = 0$.
By Part~(\ref{2017_NP_PropOfConv_Formula}),
\[
(\mu * \nu) (E)
 = \int_{\R} \mu \bigl( \bigl\{ x - y \colon x \in E \bigr\} \bigr)
    \, d \nu (y)
 = \int_{\R} 0 \, d \nu (y)
 = 0.
\]
This completes the solution.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Subalg})]
It is easy to see that both the sets of discrete and \ct{} measues
are vector subspaces of~$M$;
proofs are omitted.
The algebraic statements involving products
in both parts are immediate from Part~(\ref{2017_NP_PropOfConv_Discrete}).

We claim that that the set of discrete measures is closed.
Let $(\mu_n)_{n \in \N}$ be a sequence of discrete measures which
converges (in norm) to a measure~$\mu$.
For $n \in \N$ choose a countable set $S_n \S \R$
such that $\R \SM S_n$ is $\mu_n$-null.
Set $S = \bigcup_{n = 1}^{\I} S_n$, which is a countable set
such that $\R \SM S$ is $\mu_n$-null for all $n \in \N$.
Then for $E \S \R \SM S$, we have
$\mu (E) = \limi{n} \mu_n (E) = \limi{n} 0 = 0$.
Thus $\R \SM S$ is $\mu$-null, so $\mu$ is discrete.
The claim is proved.


We claim that the set of \ct{} measures is closed.
Let $(\mu_n)_{n \in \N}$ be a sequence of \ct{} measures which
converges (in norm) to a measure~$\mu$.
For every $x \in \R$, we have
$\mu ( \{ x \} ) = \limi{n} \mu_n (\{ x \}) = \limi{n} 0 = 0$.
So $\mu$ is \ct.
The claim is proved.
\end{proof}

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_L1}) (sketch)]
It follows from Theorem 6.13 of Rudin's book that
$f \mapsto f \cdot m$ is linear and isometric from
$L^1 (\R)$ to~$M$.
Therefore its range is a closed subspace.

Preservation of convolution is a computation
(which requires Fubini's Theorem); not yet written.
Since we already showed that $M$ is a commutative Banach algebra,
this shows that $L^1 (\R)$ is isometrically isomorphic to a closed
subalgebra of~$M$, and is hence a commutative Banach algebra.

The ideal property follows from Part~(\ref{2017_NP_PropOfConv_ACm}).
\end{proof}

It is also easy to prove directly
that $\big\{ \mu \in M \colon \mu \ll m \big\}$
is closed in~$M$.

\begin{proof}[Solution to Part~(\ref{2017_NP_PropOfConv_Unital})]
For $x \in M$ let $\dt_x$ be the ``point mass measure at~$x$'',
that is,
for a Borel set $E \S \R$,
\[
\dt_x (E)
 = \begin{cases}
   1 & x \in E
        \\
   0 & x \not\in E.
\end{cases}
\]
(The notation,
and its generalizations, is fairly standard.)
Then $\dt_x \in M$ for all $x \in \R$.

We claim that $\dt_0$ is an identity for~$M$.
By commutativity, it is enough to show that $\mu * \dt_0 = \mu$ for
every $\mu \in M$.
Recall from Part~(\ref{2017_NP_PropOfConv_Formula})
that is $\mu, \nu \in M$, $E \S \R$ is a Borel set,
and for $t \in \R$ we define $F_t \S \R$ by
$F_t = \big\{ x - t \colon x \in E \big\}$, then
\[
(\mu * \nu) (E) = \int_{\R} \mu (F_t) \, d \nu (t).
\]
Putting $\nu = \dt_0$ and using $F_0 = E$, we get
$(\mu * \dt_0) (E) = \mu (F_0) = \mu (E)$.

We now show that $L^1 (\R)$ does not have an identity.
The easiest way to do this is to consider
the map $\ph \colon L^1 (\R) \to C_0 (\R)$
given by $\ph (f) = {\widehat{f}}$
for $f \in L^1 (\R)$.
This map is well defined by Theorem 9.6 of Rudin's book,
it is a homomorphism by Theorem 9.2(c) of Rudin's book,
and it is injective by Theorem 9.12 of Rudin's book.
If $L^1 (\R)$ had an identity~$e$,
then $\ph (e) \in C_0 (\R)$ would be a nonzero element
satisfying $\ph (e)^2 = \ph (e)$.
Clearly no such nonzero element exists.
\end{proof}

\begin{proof}[Alternate solution to Part~(\ref{2017_NP_PropOfConv_Unital})]
We first claim that if $f \in L^1 (\R)$ and $a, b \in \R$ satisfy $a < b$,
then $f * \ch_{[a, b]}$ is \ct.
To prove the claim, let $\ep > 0$, and choose $\dt_0 > 0$ so small
that whenever $E \S \R$ is measurable and $m (E) < \dt_0$, then
$\int_E | f | \, d m < \ep$.
(That this can be done is a standard fact for $L^1 (X, \mu)$ for any
positive measure~$\nu$, and should have been done in Math 616.)
Set $\dt = \frac{1}{2} \dt_0$.

For any $t \in \R$, we have
\[
(f * \ch_{[a, b]}) (t)
 = \int_{\R} f (s) \ch_{[a, b]} (t - s) \, d m (s)
 = \int_{t - b}^{t - a} f \, d m.
\]
Therefore, if $t_1, t_2 \in \R$ with $| t_1 - t_2 | < \dt$,
the symmestric difference
\[
E = [t_1 - b, \, t_1 - a ] \triangle [t_2 - b, \, t_2 - a ]
\]
satisfies $m (E) < 2 \dt = \dt_0$, so
\[
\left| (f * \ch_{[a, b]}) (t_1) - (f * \ch_{[a, b]}) (t_2) \right|
  = \left| \int_E f \, d m \right|
  \leq \int_E | f | \, d m
  < \ep.
\]
This proves the claim.

If now $f \in L^1 (\R)$ is an identity for $L^1 (\R)$,
then $f * \ch_{[-1, 1]} = \ch_{[-1, 1]}$ almost everywhere.
The function $f * \ch_{[-1, 1]}$ is \ct.
But there is no \cfn~$g$ such that $g = \ch_{[-1, 1]}$,
since, if there were, the indefinite integral $F$ of $\ch_{[-1, 1]}$
would be differentiable everywhere, which is not the case.
\end{proof}

\fi

\end{document}
