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:

$$||f * g||_{\mathcal{B}^{\alpha + \beta}_{\infty , \infty}} = \le \sup_j 2^{(\alpha + \beta)j}||(\Delta_j f )*( \Delta_j g) ||_{\infty} \lesssim \sup_j 2^{(\alpha + \beta)j}||\Delta_j f ||_{\infty} \cdot || \Delta_j g||_{\infty} \le$$ $$\le ||f ||_{\mathcal{B}^{\alpha}_{\infty , \infty}} \cdot ||g ||_{\mathcal{B}^{\beta}_{\infty , \infty} }$$

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