Differentiability of convolution 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,  $g(x) = e^{-x^2}$.
 A: 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.
A: 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.
