# How differentiable is the convolution of two continuous functions?

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.

• Just to clarify: when you write $C^d_c$ do you mean $C^\alpha_c$? – Yemon Choi Aug 24 '13 at 3:46
• Why don't you just write it by definition and then take derivative under the integral? – Tomas Aug 24 '13 at 6:07
• To clarify, I do mean $C^d_c$, though I think it might be true that it is $C^{d-\epsilon}_c$ $\forall \epsilon>0$. I don't know how to make the argument not in Fourier space, because all the arguments that I can think of are not dependent in the dimension. – Sloth-Meister Aug 24 '13 at 14:30
• It has nothing to do with $d$. Just follow shanlin's advise. – Alexandre Eremenko Aug 24 '13 at 15:25
• I would be very surprised if convolving two functions in $C^1_c(R^8)$ gave me something in $C^4_c(R^8)$. I'm not convinced by your Step 1, for instance – Yemon Choi Aug 24 '13 at 18:39

$C^\alpha_c*C^\beta_c\subset C^{\alpha+\beta}_c$ 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 $g$.
• Thank you, I never thought it to be really true. But, sometimes it is nice to dream! And to learn a characterization of $C^\alpha$, I hope to use it someday. – Sloth-Meister Aug 26 '13 at 3:11
• @MatiasG.Delgadino the gain in differentiability that Fedja points out was never in question. You were claiming something which I think is much stronger, namely that in higher dimensions one can convolve two $C^1_c$ functions and get something $C^d_c$ for $d \gg 2$. (Perhaps I have misread something?) – Yemon Choi Aug 26 '13 at 3:14
• The Bernstein theorem goes both ways and if you consider series with lacunary spectra (say, $\{\pm 2^k\}$ in dimension $1$, but the dimension hardly matters here), it can be easily restated in terms of coefficients directly, so the counterexample is at hand too. – fedja Aug 26 '13 at 3:20