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In Terry's Tao blog, Exercise $1$(ii),

Show that if $f \in C^\infty_c({\bf R}^d)$ and $g: {\bf R}^d \rightarrow {\bf R}$ is absolutely integrable and compactly supported, then the convolution $f*g$ is also in $C^\infty_c({\bf R}^d)$. (Hint: first show that $f*g$ is continuously differentiable with $\nabla(f*g) = (\nabla f)*g$.

By definition of convolution, $(f * g)(x) = \int f(y)g(x-y) dy$. However, I have no idea how to show that it is continuously differentiable.

In the answer given by bruziuz, it is shown that for $f,g \in L^1$, we have $\nabla(f*g) = (\nabla f) * g$. Can we say the same to the above question?

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  • If $f \in C^\infty_c$ and $g \in L^1$ then $|g \ast f(x+h)- g \ast f(x)| = |\int g(y) (f(x+h-y)-f(x-y)) dy| \le |g|_{L^1} \sup_x |f(x+h)-f(x)|$. Since $f'$ (or $|\nabla f|$) is bounded then $\sup_x |f(x+h)-f(x)| \le C|h|$ and $g \ast f$ is uniformly continuous. The same argument applies for showing $g \ast f$ is uniformly $C^k$ for every $k$ – reuns Apr 03 '17 at 03:46
  • If $g$ is only $L^1_{loc}$ then locally the same argument works again, since $f$ is compactly supported. A little more work and you get that $\partial_\alpha(f \ast g) =(\partial_\alpha f) \ast g$ – reuns Apr 03 '17 at 03:49

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