Further question on Rudin Real and Complex Analysis exercise 9.10 on convolutions The other day I asked for and received great help on my question Rudin Real and Complex Analysis exercise 9.10 on convolutions. The exercise in Rudin Real and Complex Analysis goes on to ask you to prove that
$(f * h_\lambda)(x)\to f(x)$ a.e. if $f\in L^1$, as $\lambda\to 0$. Section 9.7 lists the definition of $h_\lambda$:
\begin{equation}
  \begin{split}
    H(t)&=e^{-|t|}\\
    h_\lambda(x)&=\int_{-\infty}^\infty H(\lambda t)e^{itx}\,dm(t)
    \quad(\lambda>0)\\
    &=\sqrt{\frac{2}{\pi}}\frac{\lambda}
    {\lambda^2+x^2}.
  \end{split}
\end{equation}
where $dm(t)$ represents $\frac{1}{\sqrt{2\pi}}dt$. Section 9.7 also gives the
property
\begin{equation}\label{1}\tag{1}
  \int_{-\infty}^\infty h_\lambda(x)\,dm(x)=1,
\end{equation}
Section 9.8 proves that if $f\in L^1$, then
\begin{equation}\tag{2}\label{2}
  (f * h_\lambda)(x)
  =\int_{-\infty}^\infty H(\lambda t)\hat{f}(t)e^{ixt}\,dm(t),
\end{equation}
where $\hat{f}$ is the Fourier transform of $f$ and where convolution (for
chapter 9) is defined for $f,g\in L^1$ as
$$(f * g)(x)=\int_{-\infty}^\infty f(x-y)g(y)\,dm(y).$$
Section 9.9 proves that if
$g\in L^\infty$ and $g$ is continuous at a point $x$, then
\begin{equation}\tag{3}\label{3}
  \lim_{\lambda\to 0}\,(g * h_\lambda)(x)=g(x).
\end{equation}
Section 9.10 proves that if $1\leq p<\infty$ and $f\in L^p$, then
\begin{equation}\tag{4}\label{4}
  \lim_{\lambda\to 0}\,||f * h_\lambda-f||_p=0.
\end{equation}
Finally, section 9.11 proves that if $f\in L^1$ and $\hat{f}\in L^1$, and if
$$g(x)=\int_{-\infty}^\infty\hat{f}(t)e^{ixt}\,dm(t)\quad(x\in R),$$
then $g\in C_0$ and $f(x)=g(x)$ a.e. (where $C_0$ is the space of continuous
functions $f$ on $R$ which "vanish at infinity"; i.e., to every $\epsilon>0$
there exists a compact set $K$ such that $|f(x)|<\epsilon$ for all $x$ not in
$K$.)
If I knew that $\hat{f}\in L^1$, then it would be easy: since
$H(\lambda t)\to 1$ as $\lambda\to 0$, \eqref{2}, the Dominated Convergence
Theorem, and 9.11 would constitute a proof. However, there is no reason to
suppose $\hat{f}\in L^1$.
From equation \eqref{4} with $p=1$, I know that given any sequence
$\{\lambda_n\}$ such that $\lim_{n\to\infty}\lambda_n=0$, there is a
subsequence $\{\lambda_{n_k}\}$ such that
$\lim_{k\to\infty}\,((f * h_{\lambda_{n_k}})(x)-f(x))=0$ for almost every
$x\in R$, but that is not enough to get
$\lim_{n\to\infty}\,((f * h_{\lambda_n})(x)-f(x))=0$ for almost every $x$.
So, I tried a couple of different approaches, none of which worked. In one
approach, I assume that $x$ is a Lebesgue point of $f$ (as are almost all
$x\in R$) and wrote (using \eqref{1})
\begin{equation}
  \begin{split}
    |(f * h_\lambda)&(x)-f(x)|\\
    &=\Biggl|\int_{-\infty}^\infty (f(x-y)-f(x))h_\lambda(y)\,dm(y)\Biggr|\\
    &\leq\int_{-\infty}^\infty |f(x-y)-f(x)|
    \frac{1}{\lambda}h_1\Bigl(\frac{y}{\lambda}\Bigr)\,dm(y)\\
    &=\int_{-\lambda}^\lambda\frac{|f(x-y)-f(x)|}{\lambda}
    h_1\Bigl(\frac{y}{\lambda}\Bigr)\,dm(y)
    +\int_{|y|\geq\lambda}|f(x-y)-f(x)|h_\lambda(y)\,dm(y).
  \end{split}
\end{equation}
The first term goes to $0$ as $\lambda\to 0$ since $x$ is a Lebesgue point
of $f$ and since $||h_1||_\infty=\sqrt{2/\pi}$. However, I had no luck showing that the second term didn't blow up.
I also tried approximating $f$ with a sequence in $C_c$ converging
to $f$ in the $L^1$-norm. Then a subsequence $\{g_n\}$
converges to $f$ pointwise for almost every $x\in R$. Then if I have a sequence $\{\lambda_k\}$
converging to $0$, I wrote for such $x$
\begin{equation}
  \begin{split}
    |(f * h_{\lambda_k})&(x)-f(x)|\\
    &\leq|((f-g_n)* h_{\lambda_k})(x)|+|(g_n * h_{\lambda_k})(x)-g_n(x)|
    +|g_n(x)-f(x)|.
  \end{split}
\end{equation}
Then, assuming I can switch the order of the limits (which I haven't bothered
to justify since it turned out the result was not useful), the third term
on the right vanishes by the choice of $\{g_n\}$ and $x$, while for fixed
$g_n$, the second term on the right vanishes by \eqref{3}. However, I am
unable to show the iterated limit of the first term on the right goes to $0$.
This, despite the fact that I know $||f-g_n||_1\to 0$ as $n\to\infty$.
In another attempt, I tried using \eqref{2}, but it lead me right back to my
first approach above.
At this point, I'm hoping someone out there can help point me in the right
direction.
EDIT
After thinking some more about this, it seems that the second approach gets
me closer than I originally thought. Instead of my last displayed equation
above, let's write
\begin{equation}
  \begin{split}
    \lim_{k\to\infty}\,(f * h_{\lambda_k})(x)
    &=\lim_{k\to\infty}\lim_{n\to\infty}\,(f * h_{\lambda_k})(x)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(5)\\
    &=\lim_{k\to\infty}\lim_{n\to\infty}\,
    (((f-g_n) * h_{\lambda_k})(x)+(g_n * h_{\lambda_k})(x))\\
    &=\lim_{k\to\infty}\lim_{n\to\infty}\,
    \int_{-\infty}^\infty(f-g_n)(x-y)h_{\lambda_k}(y)\,dm(y)
    +\lim_{k\to\infty}\lim_{n\to\infty}\,(g_n * h_{\lambda_k})(x).
  \end{split}
\end{equation}
We have that $\lim_{n\to\infty}\,(f-g_n)(x-y)h_{\lambda_k}(y)=0$ for
almost every $y\in R$ and
\begin{equation}
  |(f-g_n)(x-y)h_{\lambda_k}(y)|
  \leq\sqrt{\frac{2}{\pi}}\frac{1}{\lambda_k}|(f-g_n)(x-y)|\in L^1
\end{equation}
for all $n$, so by the Dominated Convergence Theorem
$$\lim_{n\to\infty}\int_{-\infty}^\infty(f-g_n)(x-y)h_{\lambda_k}(y)\,dm(y)=0.$$
Now, if only I could demonstrate the necessary uniformity, I could switch
the order of the limits in the second term of the last line of (5) and get
\begin{equation}
  \begin{split}
    \lim_{k\to\infty}\,(f * h_{\lambda_k})(x)
    &=\lim_{k\to\infty}\lim_{n\to\infty}\,(g_n * h_{\lambda_k})(x)\\
    &=\lim_{n\to\infty}\lim_{k\to\infty}\,(g_n * h_{\lambda_k})(x)\\
    &=\lim_{n\to\infty}\,g_n(x)\qquad\text{(by \eqref{3})}\\
    &=f(x)
  \end{split}
\end{equation}
for almost every $x\in R$.
So it comes down to needing to show that I can switch the limits in either
the first or second terms of the last line of (5). Right now, I don't see
how to get the necessary uniformity to do that.
 A: Finally, after more extensive searching, I found this post Question about Lebesgue point in Fourier Transform (Big Rudin chapter 9) which asks the same question as I did
and then provides an answer. The answer depends on material from chapter 11 of Rudin Real and Complex Analysis (which is two chapters past the one in which the original exercise appears). So I decided to distill the part of the relevant proof in Rudin (Theorem 11.20)
and the work from the above quoted post into a self-contained answer that, in principle, could have been worked out by anyone who studied up to the point where the exercise appears.
[Equation numbers refer to both this post and my original post.]
Let $f\in L^1$ and let $x$ be a Lebesgue point of $f$. As in Rudin chapter 9,
let $m$ be Lebesgue measure on $R$ divided by $\sqrt{2\pi}$. Define
$$k_x(y)=\chi_{[x-1,x+1]}(y)f(x)\qquad(y\in R).$$
Then $k_x\in L^1$ and by (1)
\begin{equation}
  \begin{split}
    (k_x*h_\lambda)(x)
    &=\int_{-\infty}^\infty k_x(x-y)h_\lambda(y)\,dm(y)
    =\int_{-\infty}^\infty\chi_{[x-1,x+1]}(x-y)f(x)h_\lambda(y)\,dm(y)\\
    &=\int_{-1}^1 f(x)h_\lambda(y)\,dm(y)
    =f(x)\int_{-1}^1\sqrt{\frac{2}{\pi}}\frac{\lambda}{\lambda^2+y^2}\,dm(y)\\
    &=-f(x)\frac{1}{\sqrt{2\pi}}\sqrt{\frac{2}{\pi}}
    \int_{\cot^{-1}(-\lambda^{-1})}^{\cot^{-1}(\lambda^{-1})}dt
    \qquad\text{(having substituted $t=\cot^{-1}(y/\lambda)$)}\\
    &=-\frac{f(x)}{\pi}[\cot^{-1}(\lambda^{-1})-\cot^{-1}(-\lambda^{-1})]
    =-\frac{f(x)}{\pi}[\cot^{-1}(\lambda^{-1})-(\pi-\cot^{-1}(\lambda^{-1}))]\\
    &=f(x)[1-\frac{2}{\pi}\cot^{-1}(\lambda^{-1})]
  \end{split}
\end{equation}
so
\begin{equation}
  \lim_{\lambda\to 0}\,(k_x*h_\lambda)(x)
  =f(x)-\frac{2f(x)}{\pi}\lim_{\lambda\to 0}\cot^{-1}(\lambda^{-1})=f(x).
\end{equation}
Let $f_x=f-k_x\in L^1$. Then
\begin{equation}
  \begin{split}
    \lim_{\lambda\to 0}\,(f*h_\lambda)(x)
    &=\lim_{\lambda\to 0}\,((f_x+k_x)*h_\lambda)(x)
    =\lim_{\lambda\to 0}\,(f_x*h_\lambda)(x)
    +\lim_{\lambda\to 0}\,(k_x*h_\lambda)(x)\\
    &=\lim_{\lambda\to 0}\,(f_x*h_\lambda)(x)+f(x).
  \end{split}
\end{equation}
To complete the proof, we will show that $\lim_{\lambda\to 0}\,(f_x*h_\lambda)(x)=0$.
Let $\epsilon>0$ be given. Since $x$ is a Lebesgue point of $f$,
\begin{equation}
  \begin{split}
    \lim_{\lambda\to 0}\frac{1}{2\lambda}
    \int_{x-\lambda}^{x+\lambda}|f_x(y)|\,dm(y)
    &=\lim_{\lambda\to 0}\frac{1}{2\lambda}
    \int_{x-\lambda}^{x+\lambda}|f(y)-k_x(y)|\,dm(y)\\
    &=\lim_{\substack{\lambda\to 0\\\lambda\in(0,1)}}
    \frac{1}{2\lambda}\int_{x-\lambda}^{x+\lambda}|f(y)-f(x)|\,dm(y)=0,
  \end{split}
\end{equation}
since, in the last integral, restricting $\lambda$ to $(0,1)$ restricts
$y$ to $[x-1,x+1]$ where $k_x(y)=f(x)$. Therefore, there exists $\delta>0$
such that for every $\lambda\in(0,\delta]$
\begin{equation}\tag{6}\label{6}
  \frac{1}{2\lambda}\int_{x-\lambda}^{x+\lambda}|f_x(y)|\,dm(y)
  <\frac{\epsilon}{\sqrt{2\pi}}.
\end{equation}
Let $\{\lambda_n\}\subseteq(0,1)$ be such that $\lim_{n\to\infty}\,\lambda_n=0$
and consider the sequence $\{s_n\}$ where
$$s_n(y)=h_{\lambda_n}(x-y)\chi_{[x-\delta,x+\delta]^c}(y)|f_x(y)|\qquad(y\in R).$$
Since $\lambda_n\in(0,1)$ we have that
$\lambda_n^2+(x-y)^2(1-\lambda_n)>0$ so that
$\lambda_n^2+(x-y)^2>\lambda_n(x-y)^2$ whence
$s_n(y)\leq|f_x(y)|/\delta^2\in L^1$.
Now $\lim_{n\to\infty}s_n(y)=0$ for all $y\in R$,
so by the Dominated Convergence Theorem,
$$\lim_{n\to\infty}\int_{-\infty}^\infty s_n(y)\,dm(y)=0.$$
Therefore
\begin{equation}\tag{7}\label{7}
  \lim_{\lambda\to 0}\int_{[x-\delta,x+\delta]^c}|f_x(y)|h_\lambda(x-y)\,dm(y)=0.
\end{equation}
Our next goal is to prove that
\begin{equation}\tag{8}\label{8}
  \sup_{\lambda>0}\int_{x-\delta}^{x+\delta}|f_x(y)|h_\lambda(x-y)\,dm(y)
  \leq\sup_{\lambda\in(0,\delta]}\frac{\sqrt{2\pi}}{2\lambda}
  \int_{x-\lambda}^{x+\lambda}|f_x(y)|\,dm(y)\equiv M.
\end{equation}
(This will be an adaptation of the second half of Rudin's proof of Theorem 11.20.)
Let $\lambda>0$. Given $n\geq 0$, let $h_{2^n+1}^{(n)}=0$ and for each
$j=1,2,\dots,2^n$ let $I_j^{(n)}=[x-j\delta/2^n,x+j\delta/2^n]$,
$h_j^{(n)}=h_\lambda(j\delta/2^n)$, and define
$$s_n=\sum_{j=1}^{2^n}(h_j^{(n)}-h_{j+1}^{(n)})\chi_{I_j^{(n)}}.$$
Then each $s_j$ is a simple measurable function, and for all $y\in R$
\begin{equation}
  0\leq s_1(y)\leq s_2(y)\leq\cdots\leq\chi_{[x-\delta,x+\delta]}(y)h_\lambda(x-y)
  \text{ and}\\
  \lim_{n\to\infty}\,s_n(y)=\chi_{[x-\delta,x+\delta]}(y)h_\lambda(x-y).
\end{equation}
In words, we are constructing an increasing sequence of step approximations
from below to
$\chi_{[x-\delta,x+\delta]}(y)h_\lambda(x-y)$. We rely on the fact that
$h_\lambda(t)$ is even and strictly decreases from a maximum at $t=0$ as $|t|$
increases.
We then have that for all $y\in R$
\begin{equation}
  0\leq|f_x(y)|s_1(y)\leq|f_x(y)|s_2(y)\leq\cdots\text{ and}\\
  \lim_{n\to\infty}|f_x(y)|s_n(y)
  =|f_x(y)|\chi_{[x-\delta,x+\delta]}(y)h_\lambda(x-y).
\end{equation}
By the Monotone Convergence Theorem applied twice,
\begin{equation}
  \begin{split}
    \int_{x-\delta}^{x+\delta}|f_x(y)|h_\lambda(x-y)\,dm(y)
    &=\lim_{n\to\infty}\int_{-\infty}^\infty|f_x(y)|s_n(y)\,dm(y)\\
    &=\lim_{n\to\infty}\sum_{j=1}^{2^n}(h_j^{(n)}-h_{j+1}^{(n)})
    \int_{x-j\delta/2^n}^{x+j\delta/2^n}|f_x(y)|\,dm(y)\\
    &\leq\lim_{n\to\infty}\sum_{j=1}^{2^n}(h_j^{(n)}-h_{j+1}^{(n)})
    \frac{2j\delta}{2^n\sqrt{2\pi}}M\\
    &=M\lim_{n\to\infty}\int_{-\infty}^\infty s_n(y)\,dm(y)
    =M\int_{x-\delta}^{x+\delta}h_\lambda(x-y)\,dm(y)\\
    &\leq M\int_{-\infty}^\infty h_\lambda(x-y)\,dm(y)=M,
  \end{split}
\end{equation}
from which \eqref{8} follows. By \eqref{6}, $M\leq\epsilon$, so
\begin{equation}
  \limsup_{\lambda\to 0}\int_{x-\delta}^{x+\delta}|f_x(y)|h_\lambda(x-y)\,dm(y)
  \leq\epsilon.
\end{equation}
Since $\epsilon>0$ was arbitrary,
$$\limsup_{\lambda\to 0}\int_{x-\delta}^{x+\delta}|f_x(y)|h_\lambda(x-y)\,dm(y)=0,$$
whence
\begin{equation}\tag{9}\label{9}
 \lim_{\lambda\to 0}\int_{x-\delta}^{x+\delta}|f_x(y)|h_\lambda(x-y)\,dm(y)=0.
\end{equation}
From \eqref{7} and \eqref{9}, we see that
\begin{equation}
  \lim_{\lambda\to 0}\,(|f_x|*h_\lambda)(x)=
  \lim_{\lambda\to 0}\int_{-\infty}^\infty|f_x(y)|h_\lambda(x-y)\,dm(y)=0.
\end{equation}
Finally,
\begin{equation}
  \begin{split}
    |(f_x*h_\lambda)(x)|
    &=\Biggl|\int_{-\infty}^\infty f_x(y)h_\lambda(x-y)\,dm(y)\Biggr|\\
    &\leq\int_{-\infty}^\infty|f_x(y)|h_\lambda(x-y)\,dm(y)
    =(|f_x|*h_\lambda)(x),
  \end{split}
\end{equation}
so $\lim_{\lambda\to 0}\,(f_x*h_\lambda)(x)=0$, as was required to prove.
