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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.

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  • $\begingroup$ 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 $g$? $\endgroup$ – Jonas Lenz Dec 8 '18 at 21:40
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    $\begingroup$ 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) $f \in L^1$ $\endgroup$ – reuns Dec 9 '18 at 2:17
  • $\begingroup$ @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/… $\endgroup$ – pitariver Dec 9 '18 at 7:15
  • $\begingroup$ @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. $\endgroup$ – pitariver Dec 9 '18 at 7:38
  • $\begingroup$ $f \ast g$ is continuous because $f(.+1/n) \to f$ in $L^1$ (a consequence of that the compactly supported continuous functions are dense in $L^1$) $\endgroup$ – reuns Dec 9 '18 at 7:59

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