# existence of the convolution with a continuously differentiable function

It is well known (using Fubini's theorem) that the convolution of two $$f,g\in L^1(\mathbb{R})$$ functions is again in $$L^1$$, and thus $$f \ast g(x)=\int_\mathbb{R}{f(x-t)g(t)dt}$$ converges for almost all $$x$$. I wanted to ask: in case $$g\in C^1(\mathbb{R})$$ is continuously differentiable with a bounded derivative (or even integrable if necessary), does this necessarily imply that the integral defining $$f \ast g(x)$$ converges for all $$x \in \mathbb{R}$$? I am trying to find a minimal condition (not strong properties like compact support) for the integral to converge everywhere.

• What are your assumptions on $f$ or is it part of your question what assumptions on $f$ guarantee the well-definedness of a convolution for all continuous $g$? – Jonas Lenz Dec 8 '18 at 21:40
• If $g$ is bounded then $\int_\mathbb{R} |h(x)g(x)|dx< \infty$ for any $h\in L^1$. That $g$ is bounded is indeed the minimal condition for $f \ast g$ to be bounded and continuous if you only know that $f \in L^1$. If $g$ isn't bounded you can still have that $f \ast g$ is bounded and/or continuous for many (but not all) $f \in L^1$ – reuns Dec 9 '18 at 2:17
• @JonasLenz I don't think assuming more than "f is integrable" is needed, you can see for example this post: [math.stackexchange.com/questions/3029627/… – pitariver Dec 9 '18 at 7:15
• @reuns indeed if g is bounded the problem is solved (do you have reference as to why $f \ast g$ is continues in this case? what about a reference to the cases where it is valid only for some special $f \in L^1$?). I am quite happy with that, though I think I also came up with the solution in case g is integrable and uniformly continues, not sure if this implies boundness. – pitariver Dec 9 '18 at 7:38
• $f \ast g$ is continuous because $f(.+1/n) \to f$ in $L^1$ (a consequence of that the compactly supported continuous functions are dense in $L^1$) – reuns Dec 9 '18 at 7:59