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We put, $C_{0}(\mathbb R)=$ The space of continuous functions on $\mathbb R$ vanishing at $\infty$; $C^{k}(\mathbb R)=$ The space of all functions $\mathbb R$ whose derivative of order $\leq k$ exist and are continuous; $C_{c}^{\infty}(\mathbb R)=$ The space of $C^{\infty}$ functions on $\mathbb R$ whose support is compact.

It is well-known that, If $f\in L^{1}(\mathbb R), g\in C^{k}(\mathbb R),$ and $\partial^{\alpha}g$ is bounded for $|\alpha| \leq k,$ then $f\ast g\in C^{k}(\mathbb R)$ and $\partial^{\alpha} (f\ast g) = f\ast (\partial^{\alpha} g)$ for $|\alpha| \leq k.$ (Bit roughly speaking, convolution of smooth functions with $L^{1}-$ integrable functions gives us smooth functions)

My Question is: Suppose $f\in L^{p}(\mathbb R)\cap C_{0}(\mathbb R); (1<p<\infty), g\in C^{\infty}_{c}(\mathbb R).$ Then (1) Can we expect $f\ast g \in C^{k}(\mathbb R),$ and $\partial^{\alpha} (f\ast g)= f\ast (\partial^{\alpha}g)$ for $|\alpha| \leq k$ for every $k\in \mathbb N.$ (2) Also, assume, $f$ has a compact support. Then can expect $f\ast g$ also has a compact support ?

Thanks,

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    $\begingroup$ The answer to the question about compact support can be found here $\endgroup$ – Jonas Dahlbæk Jul 21 '14 at 10:30
  • $\begingroup$ If $g$ has compact support, then so does $\partial^\alpha g$. In particular, $\partial^\alpha g$ is bounded for every $\alpha$. $\endgroup$ – Daniel Fischer Jul 21 '14 at 10:31
  • $\begingroup$ @DanielFischer; thanks; but for such $g;$ what can we say about $f\ast g;$ for $f\in L^{p}\cap C_{0} , (p>1);$ can say anything about $f\ast g;$ is it again a smooth ?(I know, for p=1, it is smooth); thanks; $\endgroup$ – Inquisitive Jul 21 '14 at 10:59
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    $\begingroup$ For $g\in C_c^\infty$ and $f\in L^1_\text{loc}$, so in particular for $f\in L^p$, the convolution $f\ast g$ is in $C^\infty$. If $f\in L^p$, then also $f\ast g \in L^p$, and if $f$ has compact support, then $f\ast g \in C_c^\infty$. $\endgroup$ – Daniel Fischer Jul 21 '14 at 11:02
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Assume $g$ has support in $[-M,M]$. Then, for $x\in[-N,N]$, $$ h(x)=\int f(x-y)g(y)dy=\int_{[-M,M]} f(x-y)g(y)dy=\int \tilde f(x-y)g(y)dy $$ where $\tilde f=1_{[-N-M,N+M]}f$, $1_K$ denoting the indicator function of the set $K$. Now show that $\tilde f\in L^1$ and conclude that $h\in C^k$.

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