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\ifsolutions
\title[Math 617 (Winter 2024): Solutions~7]{Math 617 (Winter 2024,
        Phillips): Solutions to Homework~7}
\date{28~February 2024}
\else
\title[Math 617 (Winter 2024): Homework~7]{Math 617 (Winter 2024,
        Phillips): Homework~7}
\date{21~February 2024}
\fi

\begin{document}

\maketitle

\indent
Conventions on measures:
$m$ is ordinary Lebesgue measure,
${\overline{m}} = (2 \pi)^{-1/2} m$,
and in expressions of the form $\int_{\R} f (x) \, d x$,
ordinary Lebesgue measure is assumed.

\ifsolutions
Little proofreading has been done.

Some parts of problems have several different solutions
(as many as four).
\fi

\begin{pbm}[Rudin, Chapter~9, Problem~1]\label{Pb_9_1}
Let $f \in L^1 (\R, {\overline{m}} )$,
and suppose that $f \neq 0$ and $f (x) \geq 0$ for all $x \in \R$.
Prove that $\big| {\widehat{f}} (y) \big| < {\widehat{f}} (0)$
for all $y \in \R \SM \{ 0 \}$.
\end{pbm}

The problem in Rudin is not clearly stated.
It is likely to be interpreted as assuming the stronger hypothesis
$f (x) > 0$ for all $x \in \R$.
The stronger assumption doesn't help with the proof.

\ifsolutions
\begin{proof}[Solution]
We have
\[
{\widehat{f}} (y)
=
% \frac{1}{\sqrt{2 \pi}}
   \int_{\R} e^{- i y x} f (x) \, d {\overline{m}} (x)
\]
for $y \in \R$.
When $y = 0$,
the integrand is $f (x) \geq 0$,
so ${\widehat{f}} (0) \geq 0$,
and the desired inequality at least makes sense.
Moreover, for all $y \in \R$,
we have
\[
\big| {\widehat{f}} (y) \big|
 \leq \int_{\R} | e^{- i y x} f (x) | \, d {\overline{m}} (x)
 = \int_{\R}  f (x) \, d {\overline{m}} (x)
 = {\widehat{f}} (0).
\]
We need therefore only prove that the inequality is strict
when $y \neq 0$.

Let $y \in \R \SM \{ 0 \}$.
Choose $\te \in \R$
such that $e^{i \te} {\widehat{f}} (y) = \big| {\widehat{f}} (y) \big|$.
Then
\[
e^{i \te} {\widehat{f}} (y)
  = {\operatorname{Re}} \big( e^{i \te} {\widehat{f}} (y) \big)
  = \int_{\R} {\operatorname{Re}} (e^{ i ( \te -  y x )})
           f (x) \, d {\overline{m}} (x)
   = \int_{\R} \cos (\te - y x) f (x) \, d {\overline{m}} (x).
\]
Therefore
\[
{\widehat{f}} (0) - \big| {\widehat{f}} (y) \big|
 = \int_{\R} [1 - \cos (\te - y x)] f (x) \, d {\overline{m}} (x).
\]

Set
\[
E = \{ x \in \R \colon f (x) > 0 \}
\andeqn
S = \left\{ \frac{2 \pi n + \te}{y} \colon n \in \Z \right\}.
\]
Then ${\overline{m}} (E) > 0$
and $S$ is countable,
so ${\overline{m}} \big( E \cap (\R \SM S) \big) > 0$.
We have $1 - \cos (\te - y x) > 0$
for all $x \in \R \SM S$,
so
\[
[1 - \cos (\te - y x)] f (x) > 0
\]
for all $x \in E \cap (\R \SM S)$.
Since also $[1 - \cos (\te - y x)] f (x) \geq 0$ for all $x \in \R$,
Theorem 1.39(a) of Rudin now implies that
\[
\int_{\R} [1 - \cos (\te - y x)] f (x) \, d {\overline{m}} (x) > 0.
\]
So ${\widehat{f}} (0) - \big| {\widehat{f}} (y) \big| > 0$.
\end{proof}

\begin{proof}[Alternate solution]
We prove $\big| {\widehat{f}} (y) \big| \leq {\widehat{f}} (0)$
as in the first solution.

Now assume that $y \in \R \SM \{ 0 \}$
and $\big| {\widehat{f}} (y) \big| = {\widehat{f}} (0)$.
Define $g (x) = e^{- i y x} f (x)$ for $x \in \R$.
Then for all $x \in \R$,
we have $| g (x) | = f (x)$,
so
\[
\int_{\R} | g | \, d {\overline{m}}
 = \int_{\R} f \, d {\overline{m}}
 = {\widehat{f}} (0)
 = \big| {\widehat{f}} (y) \big|
 = \left| \int_{\R} g \, d {\overline{m}} \right|.
\]
Theorem 1.39(c) of Rudin's book gives
a constant $\af$ such that
$\af e^{- i y x} f (x) = f (x)$ for almost all $x \in \R$.
Since $y \neq 0$,
the set $E = \{ x \in \R \colon \af e^{- i y x} = 1 \}$
is (at most) countable.
So ${\overline{m}} (E) = 0$.
The function $f$ vanishes off~$E$,
so $f$ is the zero element of $L^1 (\R, {\overline{m}} )$.
\end{proof}
\fi

I have restated the next problem in labelled parts for convenience.
It counts as three ordinary problems.

\begin{pbm}[Rudin, Chapter~9, Problem~2]\label{Pb_9_2}
\mbox{}
\begin{enumerate}
\item\label{2017_NP_617_4__1}
Compute the Fourier transform of
the characteristic function of an interval.
\item\label{2017_NP_617_4__2}
For $n \in \N$
let $g_n$ be the characteristic function of $[-n, \, n]$,
and let $h$ be the characteristic function of $[-1, \, 1]$.
Compute $g_n * h$ explicitly.
(It is piecewise linear.)
\item\label{2017_NP_617_4__3}
For $x \in \R \setminus \{ 0 \}$ and $n \in \N$, set
\[
f_n (x) = \frac{ \sin (x) \sin (n x)}{x^2}.
\]
Prove that there is a constant $c$ such that
$g_n * h$ is the Fourier transform of $c f_n$.
\item\label{2017_NP_617_4__4}
Let $f_n$ be as in part~(\ref{2017_NP_617_4__3}).
Prove that $\limi{n} \| f_n \|_1 = \infty$.
\item\label{2017_NP_617_4__5}
Conclude that $\big\{ {\widehat{f}} \colon f \in L^1 (\R) \big\}$
is a {\emph{proper}} subset of $C_0 (\R)$.
\item\label{2017_NP_617_4__6}
Prove that $\big\{ {\widehat{f}} \colon f \in L^1 (\R) \big\}$
is dense $C_0 (\R)$.
\end{enumerate}
\end{pbm}

\ifsolutions
\begin{proof}[Solution to part~(\ref{2017_NP_617_4__1})]
This is just a computation.
(Reminder: we are using
${\overline{m}} = \left( \frac{1}{\sqrt{2 \pi}} \right) m$
in the definition of convolution
as well as in the definition of the Fourier transform.)
The result is
\[
{\widehat{\ch_{[a, b]}}} (t) = \left\{ \begin{array}{ll}
     \frac{i}{\sqrt{2 \pi} \cdot t}
                      \left( e^{- i b t} - e^{- i a t} \right)
                  & \hspace{3em}  t \neq 0 \\
     \left( \frac{1}{\sqrt{2 \pi}} \right) (b - a)
                  & \hspace{3em}  t = 0.
    \end{array} \right.
\]
Of course, ${\widehat{\ch_{[a, b)}}}$, ${\widehat{\ch_{(a, b]}}}$,
and ${\widehat{\ch_{(a, b)}}}$ are the same.
\end{proof}

\begin{proof}[Solution to~(\ref{2017_NP_617_4__2})]
This is also a computation.

Set $p_n (x) = m ( [x - n, \, x + n] \cap [-1, \, 1])$,
which is given by the formula
\[
p_n (x) = \left\{ \begin{array}{ll}
     0     & \hspace{3em}  x \leq - n - 1 \\
 x + n + 1 & \hspace{3em}  - n - 1 \leq x \leq - n + 1 \\
     2     & \hspace{3em}  - n + 1 \leq x \leq n - 1 \\
-x + n + 1 & \hspace{3em}  n - 1 \leq x \leq n + 1 \\
     0     & \hspace{3em}  n + 1 \leq x.
    \end{array} \right.
\]
(For $x \in [- n - 1, \, n + 1]$,
but not for $x$ not in this interval,
the formula can be written as
$\min (1, \, x + n) - \max (-1, \, x - n)$.)

The outcome of  the computation is
$(g_n * h) = \frac{1}{\sqrt{2 \pi}} p_n (x)$.

Here are graphs for $n = 1$ and for $n = 3$:

\vspace{1ex}
\centerline{ \includegraphics[width=16cm]{C1} }
\vspace{1ex}

\vspace{1ex}
\centerline{ \includegraphics[width=16cm]{C3} }
\vspace{1ex}
\noindent
(These are not required as part of the solution.)
\end{proof}

\begin{proof}[Solution to part~(\ref{2017_NP_617_4__3})]
For $t \neq 0$,
Theorem 9.2(c) of Rudin gives the first step in the following
calculation, and part~(a) of this problem gives the second step:
\[
{\widehat{(g_n * h)}} (t)
  = {\widehat{ g_n }} (t) {\widehat{ h }} (t)
  = \left( \frac{i}{\sqrt{2 \pi} \cdot t}
                  \left( e^{- i n t} - e^{ i n t} \right)  \right)
     \left( \frac{i}{\sqrt{2 \pi} \cdot t}
                  \left( e^{- i t} - e^{ i t} \right)  \right).
\]
Using the identity
$\sin (\te) = \left( e^{i \te} - e^{- i \te} \right) / (2 i)$,
one can rewrite the last expression as $\frac{2}{\pi} f_n (t)$.

Unfortunately, this isn't quite what we want.
We will get what we do want using the Fourier inversion theorem
and a bit of trickery.

Clearly $g_n, h \in L^1 (\R)$,
so $g_n * h \in L^1 (\R)$.

We next show that $f_n \in L^1 (\R)$.
Define
\[
b_n (x) = \left\{ \begin{array}{ll}
     x^{-2}   & \hspace{3em}  | x | \geq \frac{1}{\sqrt{n}} \\
     n        & \hspace{3em}  | x | < \frac{1}{\sqrt{n}}.
    \end{array} \right.
\]
It is clear that $| f_n (x) | \leq x^{-2}$ for all $x \neq 0$.
Also, writing
\[
f_n (x) = n \left( \frac{ \sin (x)}{x} \right)
               \left( \frac{\sin (n x)}{n x} \right)
\]
and using $| \sin (y) | \leq | y |$ for all $y \in \R$,
we get $| f_n (x) | \leq n$ for all $x \neq 0$.
We may as well take $f_n (0) = n$.
(This makes $f_n$ \ct{} at $0$.)
Then the inequalities above hold for all $x$,
and imply that $| f_n (x) | \leq b_n (x)$ for all $x$.
Clearly $b_n \in L^1 (\R)$,
so $f_n \in L^1 (\R)$.

The Fourier Inversion Theorem therefore gives the first step
in the following calculation.
At the second step, we use the fact that $f_n$ is an even function,
and at the third step we change variables,
replacing $t$ with $- t$.
We get:
\[
\begin{split}
(g_n * h) (x)
& = \left( \frac{2}{\pi} \right) \left( \frac{1}{\sqrt{2 \pi}} \right)
      \int_{- \I}^{\I} e^{i t x} f_n (t) \, d t
         \\
& = \left( \frac{2}{\pi} \right) \left( \frac{1}{\sqrt{2 \pi}} \right)
      \int_{- \I}^{\I} e^{i t x} f_n (- t) \, d t
         \\
& = \left( \frac{2}{\pi} \right) \left( \frac{1}{\sqrt{2 \pi}} \right)
      \int_{- \I}^{\I} e^{- i t x} f_n (t) \, d t
  = \left( \frac{2}{\pi} \right) {\widehat{f_n}} (x).
\end{split}
\]
This is what is wanted.
\end{proof}

\begin{proof}[Alternate solution
 to part~(\ref{2017_NP_617_4__3}) (sketch)]
Directly compute
\[
\frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} e^{i t x} (g_n * h) (t) \, d t.
\]
One will have to compute three different integrals,
corresponding to three different formulas for $(g_n * h) (t)$
on parts of its domain where it is nonzero.
The most complicated term has the form $\int_r^s t e^{i t x} \, d t$,
which can be done by integration by parts.
Details are omitted.
The result is $\frac{2}{\pi} f_n$.

One checks that $f_n \in L^2 (\R) \cap L^1 (\R)$, as in first solution.
Using Theorem~9.13 of Rudin, it follows that
$\frac{2}{\pi} {\widehat{f_n}} = g_n * h$.
\end{proof}

\begin{proof}[Second alternate solution to
 part~(\ref{2017_NP_617_4__3}) (sketch)]
(This solution is not recommended.)
Imitate the arguments in Chapter~9 of Rudin,
but exchanging $\exp (i t x)$ and $\exp (- i t x)$ everywhere.
Going as far as the Fourier inversion theorem,
one gets the result that if $f \in L^1 (\R)$ and the function
\[
f^{\vee} (t)
 = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} e^{i t x} f (x) \, d x
\]
is also in $L^1 (\R)$,
then for almost all~$x$ we have
\[
f (x)
= \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} e^{i t x} f^{\vee} (x) \, d t.
\]
Either use the analog of Theorem~9.2(c) of Rudin, or directly compute,
to get $(g_n * h)^{\vee} = \tfrac{2}{\pi} f_n$,
and apply the formula above to deduce
that $\tfrac{2}{\pi} {\widehat{f_n}} = g_n * h$.
\end{proof}

\begin{proof}[Third alternate solution to
 part~(\ref{2017_NP_617_4__3}) (sketch)]
Use the same method as in the second alternate solution.
However, instead of repeating all the work in Chapter~9 of Rudin,
deduce the results needed from the ones already there.
For the Fourier inversion theorem, this is done as follows.
For $f \in L^1 (\R)$,
define $R (f) (x) = f (- x)$.
Then $f^{\vee}$ is by definition $R \big( {\widehat{f}} \, \big)$,
so $R (f^{\vee}) = {\widehat{f}}$.
% which by a change of variables is easily seen to equal
% ${\widehat{R (f)}}$.
Since $f \in L^1 (\R)$ \ifo{} $R (f) \in L^1 (\R)$,
we see that if $f$ and $f^{\vee}$ are both in $L^1 (R)$,
then for almost every~$x$ we have,
changing the variable from $t$ to $- t$ at the first step,
\[
\frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} e^{- i t x} f^{\vee} (t) \, d t
  = \frac{1}{\sqrt{2 \pi}}
      \int_{- \I}^{\I} e^{i t x} R (f^{\vee}) (t) \, d t
  = \frac{1}{\sqrt{2 \pi}}
      \int_{- \I}^{\I} e^{i t x} {\widehat{f}} (t) \, d t
  = f (t).
\]
The formula for $(g_n * h)^{\vee}$ can be obtained in a similar way.
\end{proof}

\begin{proof}[Solution to part~(\ref{2017_NP_617_4__4})]
Since $\lim_{x \to 0} x^{-1} \sin (x) = 0$,
there exists $r > 0$ such that $x^{-1} \sin (x) > \frac{1}{2}$
whenever $| x | \leq r$.

Let $k \in \N$ satisfy $k \pi / n \leq r$.
Then for $x \in [(k - 1) \pi / n, \, k \pi / n]$ we have
\[
|f_n (x)|
 \geq \left( \frac{1}{2} \right)
               \left( \frac{| \sin (n x) |}{k \pi / n} \right)
 = \frac{n | \sin ( n x) |}{2 k \pi},
\]
whence
\[
\int_{(k - 1) \pi / n}^{k \pi / n} |f_n (x)| \, d x
   \geq \frac{n}{2 k \pi}
        \int_{(k - 1) \pi / n}^{k \pi / n}| \sin ( n x) | \, d x
   = \frac{1}{k \pi}.
\]
Accordingly,
\[
\| f_n \|_1
 \geq \frac{1}{\pi} \sum_{1 \leq k \leq n r / \pi} \frac{1}{k}.
\]
Since $r > 0$,
\[
\limi{n} \| f_n \|_1
  \geq \frac{1}{\pi} \sum_{k = 1}^{\I} \frac{1}{k}
  = \I,
\]
as desired.
\end{proof}

\begin{proof}[Solution to part~(\ref{2017_NP_617_4__5})]
Let $F \colon L^1 (\R) \to C_0 (\R)$ be the Fourier transform.
Then $F$ is linear, bounded, and injective.
If it were surjective, the Open Mapping Theorem would imply
that its inverse $F^{-1}$ would also be bounded.
But we have seen that $\| g_n * h \|_{\I} = \sqrt{2 / \pi}$ for all $n$,
while $\limi{n} \| F^{-1} (g_n * h) \|_1 = \I$.
So $F^{-1}$ can't be bounded, and $F$ is not surjective.
\end{proof}

\begin{proof}[Solution to part~(\ref{2017_NP_617_4__6})]
We already know that $C_{\mathrm{c}} (\R)$ is dense in $C_0 (\R)$.

Next, let $E_n$ be the set of all \cfn{s} $l$ on $\R$ with compact
support such that $l$ is linear on every interval
$[r/n, \, (r + 1)/n]$ with $r \in \Z$.
Such a function is clearly determined by its values $l (r/n)$
for $r \in \Z$.
Indeed, if we define
\[
l_0 (x) = \left\{ \begin{array}{ll}
     0     & \hspace{3em}  x \leq - 1 \\
 x + 1     & \hspace{3em}  - 1 \leq x \leq 0 \\
-x + 1     & \hspace{3em}  0 \leq x \leq 1 \\
     0     & \hspace{3em}  n + 1 \leq x,
    \end{array} \right.
\]
then
\[
l (x) = \sum_{r \in \Z} l (r/n) l_0 (n x - r).
\]
Note that $l (r/n) \neq 0$ for only finitely many $r \in \Z$.

Using uniform continuity (details omitted), one checks that
$\bigcup_{n \in \N} E_n$ is dense in $C_{\mathrm{c}} (\R)$.
Also, we saw above that $g_1 * h$
is in the range of the Fourier transform.
Using the formula for it derived above,
and Theorems 9.2(a)~and~9.2(e) of Rudin,
one sees that the functions $x \mapsto l_0 (n x - r)$
are all in the range of the Fourier transform.
It follows that $\bigcup_{n \in \N} E_n$
is in the range of the Fourier transform.
Therefore the range of the Fourier transform is dense $C_0 (\R)$.
\end{proof}

\begin{proof}[Alternate solution to part~(\ref{2017_NP_617_4__6})]
Set $A = \{ {\widehat{f}} \colon f \in L^1 (\R) \}$.
Then $A$ is a vector subspace of $C_0 (\R)$.
It is closed under complex conjugation, by Theorem 9.2(d) of Rudin.
It is closed under multiplication, by Theorem 9.2(c) of Rudin
and because $L^1 (\R)$ is closed under convolution.
It separates the points, that is, if $x, y \in \R$,
then there is $g \in A$ such that $g (x) \neq g (y)$.
Indeed, we can take $g = {\widehat{\ch_{[a, b]}}}$ for suitable
$a$ and $b$.
Using the same functions, we see that for every $x \in \R$
there is $g \in A$ such that $g (x) \neq 0$.
The version of the Stone-Weierstrass Theorem for locally
compact spaces now implies that $A$ is dense in $C_0 (\R)$.
(To get this from the statement for compact spaces,
extend elements of $C_0 (\R)$ to be \cfn{s} on the one point
compactification of $\R$, and apply the version for compact spaces
to the linear span of $A$ and the constant functions.)
\end{proof}

\begin{proof}[Second alternate solution to~(\ref{2017_NP_617_4__6})]
(This solution assumes Rudin's Problem 9.7.)

The Schwartz space (the space $S$ of Problem~9.7)
is dense in $C_0 (\R)$,
because the set of $C^{\I}$ functions with compact support
is dense in $C_0 (\R)$.
(This needs proof,
but is a standard result from another part of analysis.)
Every function in $S$ is the Fourier transform of a function in~$S$,
by Problem~9.7, and $S$ is easily seen to be contained in $L^1 (\R)$.
\end{proof}
\fi

\begin{pbm}[Rudin, Chapter~9, Problem~8]\label{Pb_9_8}
Let $p \in [1, \I]$,
and let $q \in [1, \I]$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$.
Prove that if $f \in L^p (\R)$ and $g \in L^q (\R)$,
then $f * g$ is uniformly \ct.
If $1 < p < \I$, prove that $f * g \in C_0 (\R)$.
Show by example that $f * g$ need not be in $C_0 (\R)$ when $p = 1$.
\end{pbm}

\ifsolutions
% {\emph{Solution:}}
In the solution,
it will be convenient to use the notation $\ta_t (f) (x) = f (x - t)$
and $\sm (f) (x) = f (- x)$ for $f$ in the vector space of all measurable
functions from $\R$ to~$\C$
and for $x, t \in \R$.
The maps $\ta$ and $\sm$ are linear,
and the invariance properties of Lebesgue measure imply
that $\| \sm (f) \|_p = \| f \|_p$
for all $p \in [1, \I]$ and all $f \in L^p (\R)$,
and that $\| \ta_t (f) \|_p = \| f \|_p$ for all $p \in [1, \I]$,
all $t \in \R$, and all $f \in L^p (\R)$.

Also, recall that in this problem convolution is supposed to
be defined using Lebesgue measure as normalized for use with
the Fourier transform:
\[
(f * g) (x)
  = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} f (x - y) g (y) \, d y.
\]
However,
exactly the same statements are true without this normalization.

We break the solution into several parts.

\begin{prp}\label{P:P81}
Let $p \in [1, \I]$,
and let $q \in [1, \I]$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$.
If $f \in L^p (\R)$ and $g \in L^q (\R)$, then $f * g$ is uniformly \ct.
\end{prp}

\begin{proof}
We assume $p < \I$.
(Otherwise, exchange $f$ and $g$ and use $f * g = g * f$.)
Let $\om \colon L^q (\R) \to \C$ be the \ct{} linear functional
given by
\[
\om (f) = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} f g \, d m.
\]
Then
\[
(f * g) (x)
  = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} f (x - y) g (y) \, d y
  = \om ( \sm ( \ta_x (f))).
\]
We know that $x \mapsto \ta_x (f)$ is uniformly \ct{}
by Theorem~9.5 of Rudin's book,
and $\sm$ and $\om$ are uniformly \ct{} because they are \ct{} and
linear.
So $f * g$ is the composite of uniformly \cfn{s} and hence
uniformly \ct, as desired.
\end{proof}

\begin{proof}[Alternate proof of Proposition~\ref{P:P81}]
We assume $p < \I$.
(Otherwise, exchange $f$ and $g$ and use $f * g = g * f$.)

We first claim that
for $t \in \R$ we have $\ta_t (f * g) = \ta_t (f) * g$.
To prove this, let $x \in \R$.
Then
\[
\begin{split}
\ta_t (f * g) (x)
& = (f * g) (x - t)
  = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} f (x - t - y) g (y) \, d y
\\
& = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I}
   \ta_t (f) (x - y) g (y) \, d y
  = [\ta_t (f) * g] (x).
\end{split}
\]
The claim is proved.

Next we claim that for every $h \in L^p (\R)$, $k \in L^q (\R)$,
and $x \in \R$,
we have $| (h * k) (x) | \leq \| h \|_p \| k \|_q$.
This is essentially immediate from H\"{o}lder's inequality;
the calculation is carried out in the proof of Lemma~\ref{L:L82}.
(This part of the proof doesn't use Proposition~\ref{P:P81}.)

Now let $\ep > 0$.
Use Theorem~9.5 of Rudin to choose $\dt > 0$
such that for all $t \in \R$ with $| t | < \dt$,
we have $\| \ta_t (f) - f \|_p < \ep / (\| g \|_q + 1)$.
For $| t | < \dt$ and $x \in \R$, we then have,
using the first claim at the second step
and the second claim at the third step,
\[
\begin{split}
| (f * g) (x - t) - (f * g) (x) |
& = | \ta_t (f * g) (x) - (f * g) (x) |
  = | ([\ta_t (f) - f] * g) (x) |
\\
& \leq \| \ta_t (f) - f \|_p \| g \|_q
  < \left( \frac{\ep}{\| g \|_q + 1} \right) \| g \|_q
  < \ep.
\end{split}
\]
This calculation shows that $f * g$ is uniformly \ct.
\end{proof}

The following proof has also been used.
In effect,
it incorporates the proof of Theorem~9.5 of Rudin's book.

\begin{proof}[Second alternate proof of Proposition~\ref{P:P81}]
We assume $p < \I$.
(Otherwise, exchange $f$ and $g$ and use $f * g = g * f$.)

We first claim that for every $h \in L^p (\R)$, $k \in L^q (\R)$,
and $x \in \R$,
we have
$| (h * k) (x) | \leq \| h \|_p \| k \|_q$.
This is essentially immediate from H\"{o}lder's inequality;
the calculation is carried out in the proof of Lemma~\ref{L:L82}.
(This part of the proof doesn't use Proposition~\ref{P:P81}.)

Now let $\ep > 0$.
Choose $f_0 \in C_{\mathrm{c}} (\R)$ such that
\[
\| f_0 - f \|_p < \frac{\ep}{4 (\| g \|_g + 1)}.
\]
Choose $r \geq 0$
such that $\supp (f_0) \S [-r, r]$.
Define
\[
\rh = \frac{\ep}{2 \| g \|_q + 1}
   \left( \frac{\sqrt{2 \pi}}{2 r + 1} \right)^{1/p}.
\]
Since $f_0$ is uniformly \ct,
we can choose $\dt > 0$ such that whenever $x_1, x_2 \in \R$
satisfy $| x_1 - x_2 | < \dt$,
then $| f_0 (x_1) - f_0 (x_2) | < \rh$.

Now suppose that $x_1, x_2 \in \R$
satisfy $| x_1 - x_2 | < \min (1, \dt)$;
we prove that
\[
| (f * g) (x_1) - (f * g) (x_2) | < \ep.
\]
\Wolog{} $x_1 \leq x_2$.
Define functions $h_1, h_2 \c \R \to \C$
by $h_1 (y) = f_0 (x_1 - y)$ and $h_2 (y) = f_0 (x_2 - y)$
for $y \in \R$.
(These appear in the integrands when we consider
$(f * g) (x_1)$ and $(f * g) (x_2)$.)
If $h_j (y) \neq 0$
then $y \in [x_j - r, \, x_j + r]$,
so that (since $x_1 \leq x_2$)
for every $x \not\in [x_1 - r, \, x_2 + r]$
we have $h (x_1) - h (x_2) = 0$.
Meanwhile,
if $x \in [x_1 - r, \, x_2 + r]$
then $| h_1 (x) - h_2 (x) | < \rh$.
Since $x_2 < x_1 + 1$,
it therefore follows that
\[
\| h_1 - h_2 \|_p^p
 \leq \rh^p \left( \frac{1}{\sqrt{2 \pi}} \right)
          m ( [x_1 - r, \, x_2 + r] )
  < \rh^p \left( \frac{1}{\sqrt{2 \pi}} \right) (2 r + 1)
  = \left( \frac{\ep}{2} \right)^p,
\]
whence
\[
\| h_1 - h_2 \|_p < \frac{\ep}{2 \| g \|_q + 1}.
\]
Now,
using the claim at the beginning of the proof several times,
\[
\begin{split}
& | (f * g) (x_1) - (f * g) (x_2) |
\\
& \hspace*{3em} {\mbox{}}
 \leq | ([f - f_0] * g) (x_1) | + | ([f - f_0] * g) (x_2) |
    + | (f_0 * g) (x_1) - (f_0 * g) (x_2) |
\\
& \hspace*{3em} {\mbox{}}
 \leq 2 \| f - f_0 \|_p \| g \|_q
    + \left| \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I}
      [ f_0 (x_1 - y) - f_0 (x_2 - y) ] g (y) \, d y \right|
\\
& \hspace*{3em} {\mbox{}}
 = 2 \| f - f_0 \|_p \| g \|_q
    + \left| \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I}
      [ h_1 (x) - h_2 (x) ] g (y) \, d y \right|
\\
& \hspace*{3em} {\mbox{}}
 \leq 2 \| f - f_0 \|_p \| g \|_q + \| h_1 - h_2 \|_p \| g \|_q
\\
& \hspace*{3em} {\mbox{}}
  < \left( \frac{2 \ep}{4 (\| g \|_g + 1)} + \frac{\ep}{2 \| g \|_q + 1}
             \right) \| g \|_q
  < \ep.
\end{split}
\]
This completes the proof.
\end{proof}

\begin{lem}\label{L:L82}
Let $p \in [1, \I]$,
and let $q \in [1, \I]$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$.
If $f \in L^p (\R)$ and $g \in L^q (\R)$,
then $f * g \in L^{\I} (\R)$
and $\| f * g \|_{\I} \leq \| f \|_p \| g \|_q$.
\end{lem}

\begin{proof}
For every $x \in \R$ we have,
using H\"{o}lder's inequality at the third step,
\[
| (f * g) (x) |
  = \left| \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I}
              f (x - y) g (y) \, d y \right|
 \leq \| (\sm \circ \ta_x ) (f) \|_p \| g \|_q
 = \| f \|_p \| g \|_q.
\]
Therefore $f * g$ is bounded by $\| f \|_p \| g \|_q$.
That $f * g$ is \mb{} follows immediately from Proposition~\ref{P:P81}.
\end{proof}

\begin{prp}\label{P:P83}
Let $p \in (1, \I)$,
and let $q \in (1, \I)$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$.
If $f \in L^p (\R)$ and $g \in L^q (\R)$,
then $f * g \in C_0 (\R)$.
\end{prp}

\begin{proof}
Lemma~\ref{L:L82} implies that $(f, g) \to f * g$ is a (jointly)
\ct{} map from $L^p (\R) \times L^q (\R)$ to $L^{\I} (\R)$.
It follows from Theorem~3.17 of Rudin that $C_0 (\R)$ is a
closed subspace of $L^{\I} (\R)$.
(Complete subsets of metric spaces are necessarily closed.)
Therefore is suffices to find dense subsets
$S \S L^p (\R)$ and $T \S L^q (\R)$ such that $f * g \in C_0 (\R)$
whenever $f \in S$ and $g \in T$.

We take $S = T = C_{\mathrm{c}} (\R)$.
Density follows from Theorem~3.14 of Rudin.
Let $f, g \in C_{\mathrm{c}} (\R)$.
Proposition~\ref{P:P81} implies that $f * g$ is \ct.
Choose $M$ such that $\supp (f), \, \supp (g) \S [-M, \, M]$.
If $| x | > 2 M$, then for every $y \in \R$ at least one of
$y$ and $x - y$ must be in $\R \SM [-M, \, M]$,
so $f (x - y) g (y) = 0$.
It follows that $(f * g) (x) = 0$.
We have shown that $f * g$ has compact support,
so $f * g \in C_{\mathrm{c}} (\R) \S C_0 (\R)$.
\end{proof}

\begin{proof}[Alternate proof]
We prove directly that $f * g$ vanishes at infinity.
Let $\ep > 0$.

Since $| f |^p$ and $| g |^q$ are integrable,
there are $M, N \in [0, \I)$ such that
\[
\frac{1}{\sqrt{2 \pi}} \int_{\R \SM [- M, \, M]} | f (x) |^p \, d x
  < \left( \frac{\ep}{2 \| g \|_q + 1} \right)^p
\]
and
\[
\frac{1}{\sqrt{2 \pi}} \int_{\R \SM [- N, \, N]} | g (x) |^q \, d x
  < \left( \frac{\ep}{2 \| f \|_p + 1} \right)^q.
\]
We claim that $| (f * g) (x) | < \ep$ whenever $| x | > M + N$.

To prove this, 
set
\[
f_0 = \ch_{[- M, \, M]} f
\andeqn
g_0 = \ch_{[- N, \, N]} g.
\]
Then for every $x \in \R$ we have
\[
| (f * g) (x) |
 \leq \big| [ f * (g - g_0) ] (x) \big|
         + \big| [ (f - f_0) * g_0 ] (x) \big|
         + | (f_0 * g_0) (x) |.
\]
Whenever $| x | > M + N$,
we have $(f_0 * g_0) (x) = 0$,
because in the integrand in its definition is always zero.
Using Lemma~\ref{L:L82},
for every $x \in \R$ we have
\[
\begin{split}
\big| [ f * (g - g_0) ] (x) \big|
& \leq \| f \|_p \| g - g_0 \|_q
  \\
& = \| f \|_p
    \left( \frac{1}{\sqrt{2 \pi}} 
           \int_{\R \SM [- N, \, N]} | g (x) |^q \, d x \right)^{1/q}
  \leq \frac{\ep \| f \|_p}{2 \| f \|_p + 1}
  < \frac{\ep}{2}
\end{split}
\]
and (since clearly $\| g_0 \|_q \leq \| g \|_q$)
\[
\begin{split}
\big| [ (f - f_0) * g ] (x) \big|
& \leq \| f - f_0 \|_p \| g_0 \|_q
  \\
& = \| g_0 \|_q
    \left( \frac{1}{\sqrt{2 \pi}} 
            \int_{\R \SM [- M, \, M]} | f (x) |^p \, d x \right)^{1/p}
  \leq \frac{\ep \| g_0 \|_q}{2 \| g \|_q + 1}
  < \frac{\ep}{2}.
\end{split}
\]
So whenever $| x | > M + N$ we have $| (f * g) (x) | < \ep$.
\end{proof}

\begin{prp}\label{P:P84}
There exists $f \in L^1 (\R)$ and $g \in L^{\I} (\R)$
such that $f * g \not\in C_0 (\R)$.
\end{prp}

\begin{proof}
Take $f = \ch_{[- 1, \, 1]}$ and $g = 1$.
Then for every $x \in \R$ we have
\[
(f * g) (x)
  = \frac{1}{\sqrt{2 \pi}} \int_{- \I}^{\I} f (x - y) g (y) \, d y
  = \left( \frac{1}{\sqrt{2 \pi}} \right) m ([x - 1, \, x + 1])
  = \sqrt{\frac{2}{\pi}}.
\]
Thus $f * g$ is a nonzero constant function, so does not vanish at $\I$.
\end{proof}
\fi

\end{document}
