Compactness of Convolutions of Compact Measures

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?

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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

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 $t$.
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