Is my proof true? (problem about operator theory and measure theory)

Question

The following is the problem which I wrote a solution. Please look at a glance and leave your comments. Please let me know if there is a flaw in my solution or please make a hint for another solution.

Problem: Let $\mathbb{A}$ be a measurable subset of $\mathbb{R}$. For every $f \in L^{1}(\mathbb{R})$ and for every $ y \in \mathbb{R}.$ Let $$T(f,y)=\int_{\mathbb{A}}f(x-y)dx $$ (a) Prove that for every $y\in \mathbb{R},$ the function $f\mapsto T(f,y)$ is continuous on $ L^{1}(\mathbb{R}). $

(b) Prove that for every $ f\in L^{1}(\mathbb{R}) $ the function $y\mapsto T(f,y)$ is continuous on $\mathbb{R}.$

Here is what I have done:

(a) Let $f_{n}\to f $ on $ L^{1}(\mathbb{R}). $ Then $$\lim\limits_{n\to\infty}\int_{\mathbb{R}}\vert f_{n}- f \vert dx=0 $$ Now, $$\int_{\mathbb{R}}\vert T(f_{n},y)-T(f,y) \vert dy= \int_{\mathbb{R}}\vert \int _{\mathbb{A}}( f_{n}(x-y)-f(x-y))dx\vert dy $$ $$\leq \int_{\mathbb{A}}( \int _{\mathbb{R}}\vert f_{n}(x-y)-f(x-y)\vert dy) dx\to 0 $$ (b)Let $y_{n}\to y,$ we have that $$\lim\limits_{n\to\infty}\vert T(f,y_{n})-T(f,y) \vert= \lim\limits_{n\to\infty}\vert \int_{\mathbb{A}}(f(x-y_{n})-f(x-y)) dx\vert$$ Since $f\in L^{1}(\mathbb{R})$ so there exist $s_{m}\in C_{c}(\mathbb{R})$ such that $s_{m}\to f.$ Therefore, $$ \lim\limits_{n\to\infty} \int_{\mathbb{A}}(f(x-y_{n})-f(x-y))dx=\lim\limits_{n\to\infty}\int_{\mathbb{A}}\lim\limits_{m\to\infty}(s_{m}(x-y_{n})-s_{m}(x-y))dx $$ $$=\lim\limits_{m\to\infty}\int_{\mathbb{A}}\lim\limits_{n\to\infty}(s_{m}(x-y_{n})-s_{m}(x-y))dx=0 $$

Answer 1

A possible proof for 2.

$\forall\varepsilon>0$

1. $\exists\ \delta>0\ s.t. \forall F\subset\mathbb{R} \ \&\ m(F)<\delta$, $\int_F|f(x-y_n)-f(x-y)|dx<\varepsilon/4$ for all $n$.

2. $\exists N$ enough large such that $\int_{\mathbb{R}\backslash[-N, N]}|f(x-y_n)-f(x-y)|dx<\varepsilon/4$ for all $n$.

3. $\exists$ close set $E\subset\mathbb{R}$ such that $m(\mathbb{R}-E)<\delta$ and $f$ is continuous on $E$.

We can easily get these conclusions by a simple inequality $\int_S|f(x-y_n)-f(x-y)|dx\leq\int_S|f(x-y_n)|dx+\int_S|f(x-y)|dx$, where $S$ is any measurable set.

let $x_n=y_n-y\to0$, and we have

\begin{align} &|\int_Af(x-y_n)-f(x-y)dx|\\ & \leq\int_\mathbb{R}|f(x-y_n)-f(x-y)|dx\\ & \leq\varepsilon/4+\int_{[-N, N]}|f(x-y_n)-f(x-y)|dx\\ & =\varepsilon/4+\int_{[-N-y, N-y]}|f(x-x_n)-f(x)|dx \end{align}

Let $G\triangleq E\cap(E+x_n)\cap[-N-y, N-y]$

So $ [-N-y, N-y]\subset G+[\mathbb{R}-E]+[\mathbb{R}-(E+x_n)]$

Due to conclusion 3

original formula $\leq3\varepsilon/4+\int_G|f(x-x_n)-f(x)|dx$

f is uniform continuous on G. So when $|x_n|$ is small enough

original formula $\leq\varepsilon$

Answer 2

Your proof for 1 is incorrect: You do not know if $T(f,y)$ is in $L^1(\Bbb R)$ so $\int_{Bbb R} |T(f,y)-T(f_n,y)|\,dy$ does not make sense. For example if $f=\chi_{[0,1]}$ and $\Bbb A= \Bbb R$ you have $T(f,y)=1$ which is certainly not integrable.

To correct it: \begin{align} |T(f,y)-T(f_n,y)|&=\left|\int_{\Bbb A}f(x-y)-f_n(x-y)\,dx\right|≤\int_{\Bbb A}|f(x-y)-f_n(x-y)|\,dx\\ &≤\int_{\Bbb R}|f(x-y)-f_n(x-y)|\,dx=\|f-f_n\|_{L^1} \end{align} This converges to zero so $T$ is continuous in $f$ for fixed $y$.

Your proof for 2 is also incorrect: It is unclear exactly what you want to do but it looks like you want to pull the limit in $n$ past the integral and swap it with the limit in $m$ in the following expression: $$\lim_n \int_{\Bbb A} \lim_m f(x-y)- s_m(x-y_n)\,dx$$ You cannot do this without comment. I'm not sure about a correct proof at the moment.