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,


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,


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...


3 Answers 3


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 $1 \le p < \infty$)

  • $\begingroup$ Got it!! Thank you :) $\endgroup$
    – user87543
    Nov 17, 2014 at 4:57
  • $\begingroup$ You're welcome! $\endgroup$
    – Etienne
    Nov 17, 2014 at 9:28
  • $\begingroup$ Now that I'm 100 years late, I just want to add that if you run through this exact proof but keep $f$ to be a characteristic function, you don't need an approximation argument at the end. You can argue that if $|x-y| < \delta$, then $\int |\chi_A (x-y) - \chi_A(z-y)| dy= 0$. $\endgroup$
    – Dan1618
    Jul 29, 2020 at 17:06
  • $\begingroup$ @Etienne How are the weights $w_A, w_B$ assigned to the individual component distributions $\chi_A, \chi_B$ represented in a convolution, and how can they be extracted? $\endgroup$
    – develarist
    Dec 4, 2020 at 13:52

For the part about the open interval notice that if $m(B\cap(x-A))>0$ then $x\in A+B$, so it's enough to show that $m(B\cap(x-A))>0$ on some interval. But this function is continuous and not identically zero.

Added: Proof of continuity.

So pick a sequence of continuous functions $f_k$ with the followign properties: $\sup_\mathbb{R} f_k \leq 1$, $f_k(x)\to \chi_B(x)$ for a.e. $x\in \mathbb{R}$, and $\int_\mathbb{R} | f_k - \chi_B| dx \to 0$.

First notice that for any integrable function $h$ and bounded $g$ we have, for every $x\in \mathbb{R}$, $$ |h*g(x)|=\left| \int_\mathbb{R} h(x-y)g(y)dy \right| \leq \sup_\mathbb{R} |g| \int_ \mathbb{R} |h(x-y)|dy =\sup_\mathbb{R} |g| \int_\mathbb{R} |h(z)|dz. $$ This is just a change of variables in the last equality.

Now putting $h= \chi_A$ and $g=f_k-\chi_B$ we arrive at $$ |\chi_A*\chi_B(x)-\chi_A*f_k(x)| \leq \int_\mathbb{R} | f_k-\chi_B| dy\to 0, \qquad \text{as } k\to \infty. $$
Therefore $\chi_A*f_k$ converges uniformly on $\mathbb{R}$ to $\chi_A*\chi_B$. Therefore it's enough to prove that $\chi_A*f_k$ is continuous. To see this fix $k$ and take a sequence $x_n\to x$, and define $F_n(y)= f_k(x_n-y)$, and $F(y)=f_k(x-y)$, then since $f_k$ is continuous we get that $F_n(y)\to F(y)$ everywhere. Moreover $|F_n(y)|\leq 1$ (since $|f_k|\leq 1$), and the constant function $1$ is integrable over $A$. Therefore the DCT gives $$ \lim_{n\to \infty}\chi_A *f_k(x_n)=\lim_{n\to \infty}\int_A F_n(y)dy = \int_A F(y)dy = \chi_A*f_k(x). $$ This finishes the proof.

  • $\begingroup$ Sorry, my argument for continuity was wrong (In general you can't conclude that $\chi_A(x_n-y)\to \chi_A(x-y)$). You can get it easily if you know that continuous functions approximate integrable functions. $\endgroup$
    – Jose27
    Nov 7, 2014 at 7:01
  • $\begingroup$ Containing an interval part is fine... :) i know that continuous functions approximate integrable functions... what should be my next step then? $\endgroup$
    – user87543
    Nov 7, 2014 at 7:01
  • $\begingroup$ @PraphullaKoushik: I've now added a proof of continuity using approximations by continuous functions. $\endgroup$
    – Jose27
    Nov 7, 2014 at 22:52
  • $\begingroup$ This seem to be fine but it is a bit confusing.... i would be more happy if you can suggest something which uses only definition of convolution... $\endgroup$
    – user87543
    Nov 8, 2014 at 6:23

I like to think about the continuity of $f(x) = m(A \cap (-B + x))$ in this way: When $A, B$ are finite unions of open intervals, this is true and for every $\epsilon > 0$, there are finite unions of open intervals $U, V$ such that $m(A \Delta U), m(B \Delta V) < \epsilon$.

Also continuity implies that if $f(a) > 0$ then $f(x) > 0$ on some interval $I$ around $a$ so that $I \subseteq A + B$.

  • $\begingroup$ I am sorry, i did not understand your first paragraph.. $\endgroup$
    – user87543
    Nov 7, 2014 at 8:44

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