Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ and $g$ be functions on $\mathbb{R}^n$. Let $x_0$ be a given point in the unit ball $B(0,1)$. I am looking for sufficient conditions for the convolution $$ (f \ast g)(x) = \int_{B(0,1)} f(y)g(x-y) dy $$ to be continuous at $x_0$.

I would appreciate simple proofs or references to proofs that conditions given in an answer are sufficient.

In my specific application, $f$ and $g$ are continuous in $B(0,1) \setminus \{0\}$ and $x_0 \neq 0$, but I would be very interested to see conditions for other (more general) situations as well.

I would also be very interested to see conditions for the situation where $B(0,1)$ is replaced by $\mathbb{R}^n$.

Thanks very much!

share|improve this question
1  
maxpower you have to extend the domain of $g$ a bit more in order to compute $g(x-y)$ for any $x\in B(0,1)$. I guess that you have to let $g$ to be defined at least in $B(0,2)$. –  Leandro Nov 2 '11 at 3:13
1  
in the OP's application $x_0 = 0$, so $h(x_0) = (f \ast g)(x_0)$ does make sense but asking whether $h$ is continuous at $x_0$ does not since the $h$ is only defined at the single point $x_0$. However, if you let $g$ be define on $B(0, 1 + \epsilon)$ for any $\epsilon > 0$, then $h$ is defined on a neighborhood of $x_0$ so the question makes sense. –  user12014 Nov 2 '11 at 4:09
    
    
You guys are right. I have edited the question. –  maxpower Nov 2 '11 at 4:52
add comment

1 Answer

Continuity is not needed in general, but sufficient integrability is helpful: Let $p>1$ and $q$ be conjugate exponents ($1/p + 1/q = 1$) and $f\in L^p(\mathbb{R}^n)$, $g\in L^q(\mathbb{R}^n)$. Then by Hölder's inequality for any $x,h\in\mathbb{R^n}$: $$ |f*g(x+h) - f*g(x)| \leq \|f\|_{L^p} (\int |g(x+h-y) - g(x-y)|^q dy)^{1/q}. $$ The last term vanishes for $h\to 0$ since $L^q$ functions are "continuous in the $q$th mean" (which can be seen by density of the continuous functions with compact support).

share|improve this answer
    
$q\ne\infty$ for sure. –  AD. Nov 3 '11 at 13:07
    
Oh yes, you are right! –  Dirk Nov 3 '11 at 13:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.