Suppose $A$ and $B$ are measurable subsets of $\mathbb{R}$ of finite positive measure. Show that the convolution $\chi_A*\chi_B$ is continuous and not identically $0$. Use this to prove that $A+B$ contains a segment.
We have $$(\chi_A*\chi_B)(x)=\int_{-\infty}^{\infty} \chi_A(x-y)\chi_B(y)~dm(y)$$
For $y\notin B$ we have $\chi_B(y)=0$. So,
$$(\chi_A*\chi_B)(x)=\int_B\chi_A(x-y)~dm(y)$$
For $x-y\notin A$ we have $\chi_A(x-y)=0$
$x-y\in A\Rightarrow y\in x-A$ so, we have $y\in B\cap (x-A)$. So,
$$(\chi_A*\chi_B)(x)=\int_{B\cap(x-A)}~dm(y)=m(B\cap(x-A))$$
So, $$(\chi_A*\chi_B)(x)=m(B\cap(x-A))$$
EDIT 1 : Consider $x_n\rightarrow x $ ( converging sequence) then i see that
$$\cdots (B\cap(x_n-A))\supseteq (B\cap(x_{n+1}-A))\supseteq(B\cap(x_{n+2}-A))\supseteq\cdots (B\cap(x-A))$$
So, $m(B\cap (x_n-A))\rightarrow m(B\cap (x-A))$
(EDIT : i have said this before but then some one said that convergence may not imply that containment but now i tried and i strongly feel it holds if it converges)
I guess this would be sufficient to say that convolution is continuous….
$\epsilon-\delta$ way of proving continuity..
Given $\epsilon >0$ i have to choose $\delta>0$ such that $|x-y|<\delta$ implies $$\big|(\chi_A*\chi_B)(x)-(\chi_A*\chi_B)(y)\big|=\big|m(B\cap(x-A))-m(B\cap(y-A))\big|<\epsilon$$
I know $m(A)-m(B)=m(A\cap B^c)$ if $B\subset A$. I am not so sure if that holds in general or if it holds at least in this case…
Suppose every thing is super perfect then i would get
$$\big|m(B\cap(x-A))-m(B\cap(y-A))\big|=\big|m(B\cap(x-A))\cap (B\cap(y-A))^c\big|=\big| m(B\cap (x-A)\cap (y-A)^c)\big|$$
I do not know where to go from here..
Even if i prove this is continuous i am not very sure how to use this and prove that $A+B$ contains a segment…
Please suggest some path…
Best Answer
Here is another proof for the continuity of $\chi_A*\chi_B$. Let us in fact show, more generally, that if $f$ is integrable on $\mathbb R$ and $g$ is measurable and bounded, then $f*g$ is a continuous function. (Then take $f:=\chi_A$ and $g:=\chi_B$, of course).
The function $f*g(x)=\int_{\mathbb R} f(x-y)g(y)\, dy$ is clearly well-defined at every point $x\in\mathbb R$. Moreover, one may write $$f*g(x)=\int_{\mathbb R} \tau_x f(y) g(y)\, dy\, ,$$ where $\tau_xf(y)=f(x-y)$. In an even more formal way, $$f*g(x)=\Phi_g (\tau_x f)\, ,$$ where $\Phi_g :L^1(\mathbb R)\to \mathbb R$ is the linear functional defined by $\Phi_g(u)=\int_{\mathbb R} u(y)g(y)\, dy$. This makes sense, and $\Phi_g$ is a continuous linear functional on $L^1(\mathbb R)$, because $g$ is bounded.
So, to prove that $f*g$ is a continuous function, you only have to check that the map $x\mapsto \tau_x f$ is continuous from $\mathbb R$ into $L^1(\mathbb R)$.
But now, there is no miracle: you need to use some approximation argument, just like in the other answers already given to your question. For example, you can first check that the map $x\mapsto \tau_x f$ is indeed continuous if $f$ is continuous and compactly supported (this should cause no difficulty); and then, taking a sequence $(f_n)$ of continuous and compactly supported functions tending to $f$ with respect to the $L^1$ norm, observe that $\tau_xf_n\to \tau_x f$ uniformly on $\mathbb R$ as $n\to\infty$, because $\Vert \tau_x f_n-\tau_x f\Vert_{L^1}=\Vert f_n-f\Vert_{L^1}$.
(A proof can be found on Real and Complex Analysis, 3rd Edition, theorem 9.5. In fact, this function is uniformly continuous from $\mathbb{R}$ into $L^p(\mathbb{R})$ where $1 \le p < \infty$)