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.

This regards measures on $d$-dimensional Euclidean space $\mathbb R^d$ and their associated densities.
A super-level set of a density $f : \mathbb R^d \to \mathbb R^+$ at level $t$ is the set $\{x \in \mathbb R^d \mid f(x) \geq t\}$.
Say a density is super-compact if for every $t>0$ the super-levelset at level $t$ of $f$ is compact.

Given two super-compact densities $f,g$, is the convolution $f*g$ also super-compact?

More specifically, I am interested in the case where one of the two super-compact densities $g$ is a Gaussian kernel, and the other $f$ may have compact support, that is even for $t=0$ the super-levelset of $f$ is compact. But I am hoping there is a more general known result about preserving compactness on convolutions.
If this is not true, then what properties does $f$ need so that $f*g$ is super-compact?

share|improve this question
2  
This doesn't make sense. A measure is not a function from ${\mathbb R}^d \to {\mathbb R}^+$, it is a function on sets. Or are you thinking about the density for a measure that is absolutely continuous with respect to Lebesgue measure? –  Robert Israel Nov 26 '12 at 7:00
    
Thanks Robert - you are right. I have tried to edit the question to correct this issue. –  Jeff P Nov 26 '12 at 7:52
    
How about we skip the super and just say that the density is compact. I've never heard of a notion of compact density before, I don't think it exists... does it? 'Cause the "super" sounds a little weird. –  Patrick Da Silva Nov 26 '12 at 8:45
    
Patrick: Take for example when $g$ is a Gaussian kernel. It has infinite support, and thus is not bounded, and is not compact. This property is passed along under convolution, so $f*g$ is also not compact. I defined super-compact because it is precisely the property I need for $f*g$, in particular where $g$ is the Gaussian kernel. –  Jeff P Nov 26 '12 at 15:51
add comment

1 Answer

up vote 0 down vote accepted

If $f$ and $g$ are both integrable and square-integrable, and $f$ is super-compact, and $f*g$ is continuous, then $f*g$ is super-compact. (This covers your specific case, at least if the compactly supported $f$ is continuous.) Note we don't even need $f$ and $g$ to be nonnegative.

We need to prove that the $t$-level sets are both closed and bounded. First, closed is easy: the $t$-level set is the inverse image of the closed interval $[t,\infty)$ under the map $f*g\colon \mathbb R^d\to\mathbb R$, and hence it is automatically closed (by the topological definition of continuity).

As for bounded: given $\varepsilon>0$, choose $R=R(\varepsilon)$ so large that $|f(x)|<\varepsilon$ for $|x|>R$ (by super-compactness) and that the integral of $|g(x)|^2$ over the complement of the ball of radius $R$ is less than $\varepsilon$ (by square-integrability). Then for $|x|>2R$, $$ (f*g)(x) = \int_{\mathbb R^d} f(y)g(x-y) \,dy = \int_{|y|\le R} f(y)g(x-y) \,dy + \int_{|y|> R} f(y)g(x-y) \,dy. $$ In the first integral, use Cauchy-Schwarz: \begin{align*} \bigg| \int_{|y|\le R} f(y)g(x-y) \,dy \bigg| &\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|y|\le R} |g(x-y)|^2 \,dy \bigg)^{1/2} \\ &\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|z|> R} |g(z)|^2 \,dy \bigg)^{1/2} < \|f\|_2 \varepsilon^{1/2}. \end{align*} (Here we've used the fact that $|x|>2R$ and $|y|\le R$ imply $|z|>R$ in the change of variables $z=x-y$.) For the second integral, $$ \bigg| \int_{|y|> R} f(y)g(x-y) \,dy \bigg| < \varepsilon \int_{|y|> R} |g(x-y)| \,dy \le \|g\|_1 \varepsilon. $$ Thus $|(f*g)(x)|$ is bounded by $\|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$ for all $|x|>2R(\varepsilon)$. In other words, the $t$-level set is bounded for all $t \ge \|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$, since $\varepsilon$ can be taken arbitrarily small, so can $t$.

share|improve this answer
    
Greg: Thanks for the answer. I see how this shows that every super-levelset for $t>0$ of $f*g$ is bounded, but it also needs to be closed to compact. This is the part I am having trouble with and I don't see how this shows that. –  Jeff P Nov 26 '12 at 15:47
    
The $t$-superlevel set of $f*g$ is the inverse image of the closed set $[t,\infty)$, so it is automatically closed when $f*g$ is continuous, for example. –  Greg Martin Nov 26 '12 at 23:57
    
I don't think the argument is so easy. The density $f*g$ could have an infinite number of disjoint points that are all local maximum with value $t$. Then the $t$-superlevel set is not closed since it is an infinite size point set. –  Jeff P Nov 27 '12 at 1:09
    
If so then $f*g$ wouldn't be continuous, which is why I added that clause. –  Greg Martin Nov 27 '12 at 7:09
    
That "a continuous density (i.e. $f*g$) that has bounded support must have finite local maxima at the same function value" sounds like exactly the sort of result I need. My worry was –  Jeff P Nov 27 '12 at 16:40
show 6 more comments

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.