$\newcommand{\Rn}{\mathbb{R}^N}$
$\newcommand{\dxi}[2]{\frac{\partial #1}{\partial x_{#2 } } } $
$\newcommand{\conv}[2]{#1 \star #2}$
$\newcommand{\ball}{B(0, {1 \over n})}$
I would like to show the convolution of two functions is smooth.
$\phi_n \in C^\infty_c( \Rn )$ which is a mollifier or a Dirac sequence.
$f\in L^p(\Rn)\bigcup C(\Rn)$ with $p \in [1,\infty)$
To show this I think I must show $\nabla( \conv{ \phi_n }{ f})(x)=\conv{(\nabla\phi_n)}{ f(x)}$. The rest will follow by induction and the convolution will be as smooth as the mollifier. Taking first the case $f\in C(\Rn)$, it will be sufficient to show that $\dxi{}{i}(\conv{\phi_n}{f})(x)=\conv{\dxi{\phi_n}{i}}{f}(x)$.
By definition
$$(\conv{f}{g})(x)=\int f(x-y)g(y)dy$$
so since $\phi_n$ has compact support on $\ball$
$$\begin{align}
\dxi{}{i}(\conv{\phi_n}{f})(x) &= \lim_{h\to 0}\frac{ (\conv{ \phi_n }{f} )(x+\hat{e}_ih) – (\conv{ \phi_n }{f} )(x)}{h}\\
&= \lim_{h\to 0}\frac{1}{h}\left( \int_{\ball} \left( \phi_n(x-y+\hat{e}_ih)-\phi_n(x-y) \right) f(y) dy \right)
\end{align}$$
Since $\phi_n \in C_c^\infty(\Rn)$ by mean value theorem
$$\phi_n(x-y+\hat{e}_ih)-\phi_n(x-y) = h \dxi{\phi_n}{i}(x-y+c)$$
for some $c$ ( what conditions ? ) and it follows that
$$ \dxi{}{i}(\conv{\phi_n}{f})(x) = \lim_{h\to 0} (\conv{ \dxi{\phi_n}{i}}{f})(x+c) $$
This is where I am stuck, is it valid to only show the derivative in one direction, what are the conditions on $c$ for the mean value theorem, and how am I allowed to take limits here? It would seem to me that the convolution must be continuous and $c \to 0$ as $h \to 0$. Thanks in advance, it would be appreciated if any flaws in the argument so far could be explained and perhaps a clue or starting point for $f \in L^p(\Rn)$.
Best Answer
$c = h^* e_i$ for some $h^* \in (0, h)$, so $\|c\| < h$ and hence $c \to 0$ as $h \to 0$. Note that $C(\mathbb{R}^N)$ is dense in $L^p(\mathbb{R}^N)$, which should do the trick.