Let $f$ be a compactly supported $C^{\alpha}$ function (that is Holder continuous with exponent $\alpha$) and let $g$ be a compactly supported $C^\beta$ function. What can we say about Holder continuity of their convolution $$ h(x):=\int f(z-x) g (z) dz? $$
It is quite clear that $h$ is in class $C^{\max(\alpha,\beta)}$, but is it possible to say something better? Is it true that $h$ is in class $C^{\alpha+\beta}$? How one can prove something like this?
The statement is actually true. I guess it is possible to prove in many different ways. For the purpose of generality, and because it is the only proof that I came up with by chance, my proof will refer to Hölder-Besov spaces over a torus $\mathbb(T)$.
Let us suppose that $f \in \mathcal{C}^{\alpha} = \mathcal{B}^{\alpha}_{\infty , \infty}$ and $g \in \mathcal{C}^{\beta}$ with $\alpha, \beta \in \mathbb{R}$ (note: the regularity is not supposed to be positive - and you get exactly the effect you are looking for in case of differentiability. The only problems appear if the regularities are integer: $\mathcal{C}^{1}$ is not exactly the space of continuously differentiable functions!).
Now we have a partition of the unity generated by $(\chi, \rho)$ through which we compute the norm of $f$. WLOG (up to a constant in the norm) we use a different partition for the norm of $g$: $(\chi, \psi)$ with $\psi$ supported in an annulus (larger than the one of $\rho$) such that $\psi \cdot \rho = \rho.$
Now we can start the computations: $$ ||f * g||_{\mathcal{B}^{\alpha + \beta}_{\infty , \infty}} = \sup_j 2^{(\alpha + \beta)j}||\Delta_j f * g||_{\infty}$$
We get $$\Delta_j f * g = \mathcal{F}^{-1}(\rho_j \mathcal{F}(f * g)) = \mathcal{F}^{-1}(\rho_j \mathcal{F}f \cdot \mathcal{F}g) =$$ $$ = \mathcal{F}^{-1}((\rho_j \mathcal{F}f) \cdot (\psi_j\mathcal{F}g)) = (\Delta_j f )*( \Delta_j g)$$
Hence we get that:
$$||f * g||_{\mathcal{B}^{\alpha + \beta}_{\infty , \infty}} = \le \sup_j 2^{(\alpha + \beta)j}||(\Delta_j f )*( \Delta_j g) ||_{\infty} \lesssim \sup_j 2^{(\alpha + \beta)j}||\Delta_j f ||_{\infty} \cdot || \Delta_j g||_{\infty} \le$$ $$\le ||f ||_{\mathcal{B}^{\alpha}_{\infty , \infty}} \cdot ||g ||_{\mathcal{B}^{\beta}_{\infty , \infty} }$$
While the proof given by Kore-N is very nice and easy to extend to more general scales of function spaces, I thought someone may find it useful to have a more direct argument. So here it is.
Write $\Delta_h f(x) = f(x + h) - f(x)$. Suppose that $f \in C^\alpha$ and $g \in C^\beta$, and one of them — say, $g$ — is compactly supported. Since $$ \Delta_h^2 (f * g)(x) = (\Delta_h f) * (\Delta_h g) ,$$ we have $$ |\Delta_h^2 (f * g)(x)| \leqslant \|\Delta_h f\|_\infty \|\Delta_h g\|_1 \leqslant C |h|^\alpha |h|^\beta ,$$ where $C$ depends on the Hölder constants of $f$ and $g$ and the size of the support of $g$.
It is now a well-known fact that the above condition implies that $f * g$ is in the Hölder–Zygmund space $\Lambda^{\alpha + \beta}$, which:
coincides with the space of Hölder continuous functions when $\alpha + \beta < 1$;
coincides with the space traditionally denoted by $C^{1,\alpha+\beta-1}$ if $1 < \alpha + \beta < 2$;
is strictly larger than the class of Lipschitz-continuous functions, but it is reasonably close when $\alpha + \beta = 1$.
I am not aware of any simple proof of this fact. A very nice and reasonably elementary argument is given by Stein in his Singular Integrals and Differentiability Properties of Functions book (see Proposition V.8 there).
Remarks:
More generally, we may assume that $p, q, r \in [1, \infty]$ are such that $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ (as in Young's inequality), and $\|\Delta_h f\|_p \leqslant C |h|^\alpha$ and $\|\Delta_h g\|_q \leqslant C |h|^\beta$. Then $\|\Delta_h^2 (f * g)\|_r \leqslant C' |h|^{\alpha + \beta}$.
The above argument clearly carries over to arbitrary locally compact Abelian groups.
Yet another approach uses the fact that Hölder–Zygmund spaces are interpolation spaces between the usual classes $C^k$, and proceeds as in the standard proof of Young's inequality by real interpolation.
