Differentiability of convolution

Question

First let me say that I have used the search bar and looked through all the "differentiability of convolution" questions that I saw, but none of them seem to cover this case. (If one of them did and I missed it, of course please post the link and I will look more closely).

Suppose $h\in L^\infty(\mathbb{R})$. How nice does $g$ have to be in order to guarantee that $h * g \in C^1(\mathbb{R})$?

In my particular situation $g$ is a Schwartz function, infinitely differentiable and all of its derivatives decay faster than any polynomial, and $h$ is continuous and bounded.

I know that if $h \in L^p$ for $p<\infty$ then $f *g \in C^\infty$, and I know it works if $g$ is compactly supported, but in my case neither of these two hold.

All I have is that $h$ is continuous and bounded, and $g$ is Schwartz.

EDIT: $h*g$ being differentiable almost everywhere would actually suffice for my purposes.

EDIT2: More specifically, $g(x) = e^{-x^2}$.

Answer 1

Let us focus on dominating the difference quotient's integrand, if this can be done the result follows by dominated convergence. By the mean value theorem, for any $i$, we have $$ \frac{g(x-y+te_i) - g(x-y)}{t} = \partial_i g(\xi(x-y,t)) $$ where $\xi(u,t)$ lies on the segment between $u$ to $u+te_i$. Without loss we consider $|t|<1$. Then $$ \left|\frac{g(x-y+te_i) - g(x-y)}{t} h(y) \right| \leq | \partial_i g(\xi(x-y,t))|\cdot |h(y)| \leq C |\partial_i g(\xi(x-y,t))| $$ where $|h|\leq C$ since $h$ is bounded, and this is $$ \leq C \sup_{z \in B(x-y,1)} |\partial_i g(z)| $$ so it suffices to show $$\int \sup_{z \in B(x-y,1)} |\partial_i g(z) | dy < \infty.$$

Since $g \in \mathcal{S}(\mathbb{R}^n)$ we may choose $r>0$ such that $|y|^{n+1}\cdot |\partial_i g(y)| \leq K$ for all $|y|>r$. Since $\partial_i g$ is bounded it suffices to show $$\int\limits_{|y|>|x|+r+2} \sup_{z \in B(x-y,1)} |\partial_i g(z) | dy < \infty.$$

Note that for $|y|>|x|+r+2$ we then have $$ \sup_{z \in B(x-y,1)} |\partial_i g(z) | \leq\frac{K}{(|x-y|-1)^{n+1}} $$ and since $$ \int\limits_{|y|>|x|+r+2}\frac{K}{(|x-y|-1)^{n+1}}dy < \infty $$ the result follows.

Answer 2

Let me discuss the case where $g\in \mathcal{S}(\mathbb{R}^{n})$. As $(h * g)(x)=\int\limits_{\mathbb{R}^{n}}g(x-y)h(y)\, \mathrm{d}y$ you may pull any differential inside by means of dominated convergence and conclude that you have a smooth bounded function.