while solving a completely different problem, I have come to this result:
Let $f,g: \mathbb R\to \mathbb R$ be two functions bounded by $C_f,C_g$ respectively. If at least one, say $f$ is in $L^1$, the convolution $f*g$ is $2C_f\cdot C_g-$Lipschitz continuous
My proof:
Let $\varepsilon>0$, and $x\in \mathbb R$. Let $a,b$ be an interval such that $$\int_{\mathbb R-[a,b]} |f(y)|\ dy\le \varepsilon$$ ($a,b$ exists since $f\in L^1$) then, we have $|f*g(x+z)-f*g(x)|=$ $$\bigg |\int_{\mathbb R} f(y)g(x+z-y)dy- \int_{\mathbb R}f(y)g(x-y)dy\bigg |\le 2C_g\varepsilon + \bigg |\int_a^b f(y)g(x+z-y)dy- \int_a^b f(y)g(x-y)dy\bigg | $$ which can be rewritten as $$2C_g\varepsilon + \bigg |\int_b^{b+z} f(y)g(x-y)dy- \int_a^{a+z} f(y)g(x-y)dy\bigg |\le 2C_g\varepsilon + 2C_fC_f|z|$$
this being valid for every $\varepsilon>0$, we get that the whole function is $2C_fC_g$ Lipschitz.
If this result is correct, Where can I find a reference for this?
and, more important Is this true also in higher dimension?
EDIT:
Unfortunately, the result is not true, a possible counterexample is given by the couple
$$f(x) = \sum_{i=0}^{100} 1_{[i,i+1/100]}(x),\ g(x)=1_{[0,1/100]}(x).$$
As pointed by PhoemueX in the comment, the rewriting at step 4 is incorrect.