Let $1 \leq p \leq \infty$, with $\frac{1}{p} +\frac{1}{q}=1$. Show that if $f \in L^p(\mathbb{R}^n)$ and $g \in L^q(\mathbb{R}^n)$, then
a) The convolution $f\ast g$ is bounded and continuous on $\mathbb{R}^n$.
b) In addition, prove that if either $f \in (L^p \cap C^m)(\mathbb{R}^n)$, or $g \in (L^q \cap C^m)(\mathbb{R}^n)$, then $f \ast g \in C^m$
Proof
a) Using Holder's inequality, it is easy to check that $f \ast g$ is bounded, that is:
$$|f\ast g| \leq \|f\|_p \|g\|_q.$$
For $h \in \mathbb{R}^n$, such that $h = (h_1,...,h_n)$ and $h_i >0$ for all $1 \leq i \leq n$. Define $f_h(x) = f(x+h)$, then it is known that $\|f_n - f\|_p \to 0$ as $|h| \to 0$. To show $f \ast g$ is continuous, it suffice to show that for each $x \in \mathbb{R}^n$ we have $|(f\ast g)(x+h) - f(x)| \to 0$ as $|h| \to 0$.
Fix $\epsilon >0$. Choose $|h|$ small such that $\|f_h -f\|_p \leq \frac{\epsilon}{\|g\|_q}$
\begin{align} |(f*g)(x+h) - (f*g)(x)| \leq \|f_h-f\|_p \|g\|_q< \epsilon. \end{align}
b) Assume $g \in (L^q \cap C^m)(\mathbb{R}^n)$.
Edit For $x \in \mathbb(R)^n$ $$\frac{(f*g)_(x+h)^{(k)}-(f*g)^{(k)}(x)}{|h|} = \int_{\mathbb{R}^n} f(y) \frac{g^{(k)}(x+h-y)-g^{k}(x-y)}{|h|} dy$$
Take the limit as $|h| \to 0$
$(f*g)^{(k+1)}(x)= \int_{\mathbb{R}^n} g^{k+1}(x-y) f(y) dy.$
How can I assure the above integral exists?
I do not know if $g^{(k+1)}$ is compactly supported or not!!
Let $g^{k}$ be a continuous function, $1 \leq k \leq m-1$. It is sufficient to show that $(f*g)^{k+1}$ is continuous in $\mathbb{R}^n$, where $f^{(k)} =\frac{d^k}{dx^k}$. Noting that $\|g_h^{(k+1)}-g^{(k+1)}\|_q \to 0$ as $|h| \to 0$. Fix $\epsilon >0$, choose $|h|$ small such that $\|g_h^{(k+1)}-g^{(k+1)}\|_q < \frac{\epsilon}{\|f\|_p}$, then we have
$$|(f \ast g)^{(k+1)}(x+h)-(f \ast g)^{(k+1)}(x)| \leq \|f\|_p \|g_h^{(k+1)}-g^{(k+1)}\|_q < \epsilon.$$
Is my proof correct?