How differentiable is the convolution of two continuous functions?

Question

The question is really simple: Given $$ f, g\in C^\alpha_c(\mathcal{R}^d) $$ is $$ f*g\in C^d_c? $$

I came up with a formal argument using the decay of the Fourier transform of continuous functions, but it is really formal and I would appreciate a reference.

What I have though so far, which has a few holes is the following:

1) Because the Fourier transform, by interpolation, goes from $L^p\rightarrow L^q$, for every $p<\infty$, and I am thinking about compactly supported continuous functions, their Fourier transform has to behave roughly like $L^1$ in the far field. Using that if a function is in $L^\infty$ its Fourier transform is uniformly continuous, I claim that $\hat{f}, \hat{g}\sim O(\frac{1}{|\chi|^d})$ when $\chi\to\infty$.

1*) This step is the one I don't trust, though I am looking for a continuous function, for which it's fourier transform does not have this decay. I boiled it down to find a continuous function that has no integrable derivatives at all, any ideas? Because if we take $f,g\in H^{\frac{d}{2}}$ borderline continuous, just by dividing the derivatives we get

$$ \Delta^{\frac{d}{2}}(f*g)=\Delta^{\frac{d}{4}}f*\Delta^{\frac{d}{4}}g\sim L^2*L^2\in C^0 $$

Therefore, $f*g\in C^d$.

2) We use the Convolution Theorem, so $\widehat{f*g}\sim O(\frac{1}{|\chi|^{2d}})$.

3) Finally, if $h\in L^2$ and $\hat{h}\sim O(\frac{1}{|\chi|^{d+\alpha}})$. Then, $h\in C^{\alpha-\epsilon}$ $\forall \epsilon>0$. Therefore, $f*g\in C^{d-\epsilon}$ $\forall \epsilon>0$.

I know this argument is not completely rigorous, though it shows me that dimension should play a role.

Assuming nobody believes in the result (I don't think it's true, but it would be nice, what is in fact true), I have a similar question, which might be easier:

We know that if $\frac{1}{p}+\frac{1}{q}=1$ and $f\in L^p$ and $g\in L^q$ and assume they are compactly supported then, $$ f*g\in C^0_c $$ This can be easily seen by using than $L^q=(L^p)^*$, and the fact that convolution is translation invariant.

But, what happens if $p$ is better than the conjugate of $q$?

Lets say $f\in L^3$ and $g\in L^2$, is it true that $$ f*g\in C^\alpha_c, $$ for some $\alpha>0$?

Even simpler, take $f\in C^\alpha$ and $g\in L^2$, do you get $$ f*g\in C^{\alpha +\beta}_c, $$ for some $\beta>0$?

All of this assuming compact support of both functions.

Answer 1

$C^\alpha_c*C^\beta_c\subset C^{\alpha+\beta}_c$ but not much better than that. If you want it through Fourier analysis, the shortest route to go is to periodize and to use the Bernstein description of periodic $C^\alpha$ with non-integer $\alpha$ as the set of continuous functions $f$ such that for every $n$, there exists a trigonometric polynomial of degree $n$ such that $\|f-p\|_C\le Cn^{-\alpha}$. Then writing $f=p+r$ and $g=q+h$ and noticing that $p*g$ and $r*q$ are polynomials, you get away with the trivial bound $Cn^{-(\alpha+\beta)}$ for the term $r*h$, which you treat as an error term. The case of integer $\alpha$ (or $\beta$) is trivial: just differentiate $f$ all the way you can and only then switch to $g$.