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Let $f(x) \in L^p(\mathbb{R})$ and $K \in C^m(\mathbb{R})$. Can I then say that $(f \ast K) (x) = \int_{\mathbb{R}} f(t) K(x-t) dt$ is in $C^m$?

I know that this is true if $K$ has compact support, but I was wondering if it is possible to have a stronger result (perhaps $K$ vanishing at $\infty$?).

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2 Answers 2

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No. If $K$ does not have compact support, then $f*K$ need not even be bounded. Consider the function $K = 1 / \log( 3 + |x|)$, and $f(x) = (1 + |x|)^{-1/p - \epsilon}$. For $\epsilon > 0$. Then $f\in L^p$, but the convolution integral does not converge anywhere, even when $K$ vanishes at $\infty$.

What you can do is to require $K$ decays sufficiently rapidly. What you need is also Young's inequality for convolutions, which implies that

$$ \sup |f * K| \leq \|f \|_p \|K\|_q $$

if $1/p + 1/q = 1$. So in particular if $K\in C^m(\mathbb{R}) \cap W^{m,q}(\mathbb{R})$, where $W^{m,q}$ is the Sobolev space of $m$-times weakly differentiable, $q$-integrable functions with $q = p / (p-1)$, you can conclude that the convolution is $C^m$.

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I wrote an article on analysis recently and I included the following relevant result (with proof) in the article; I hope it is helpful:

Theorem Let $f\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ for some $1\leq p \leq \infty$. Also, let $g\in L^1(\mathbb{R}^n)$ be a function all of whose partial derivatives of the first order exist and are such that $\frac{\partial g}{\partial x_i}$ is bounded on $\mathbb{R}^n$ for all $1\leq i\leq n$. We conclude that the partial derivatives of the convolution $f\ast g$ of the first order exist on $\mathbb{R}^n$. In fact, $\frac{\partial (f\ast g)}{\partial x_i}=f\ast (\frac{\partial g}{\partial x_i})$ for all $1\leq i\leq n$.

Proof. First note that the convolution $f\ast g\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ by Minkowski's inquality and is therefore finite (and well-defined) a.e. Let us fix $1\leq i\leq n$. Note that

$\frac{\left(f\ast g\right)\left(x+he_i\right)-\left(f\ast g\right)(x)}{h} - \left(f\ast \left(\frac{\partial g}{\partial x_i}\right)\right)\left(x\right)$

$= \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right) \right]dy$

In particular,

$\frac{\partial \left(f\ast g\right)}{\partial x_i}\left(x\right)$

$=\lim_{h\to 0} \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]dy$

$= \int_{\mathbb{R}^{n}} \left[\lim_{h\to 0} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]\right]dy$

$= \int_{\mathbb{R}^n} f\left(y\right)\frac{\partial g}{\partial x_i}\left(x-y\right) dy$

$= \left(f\ast \frac{\partial g}{\partial x_i}\right)\left(x\right)$

We will justify this computation using the Lebesgue dominated convergence theorem. In particular, we will show that if $x\in \mathbb{R}^n$ is fixed, the expression $\left|\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$ is bounded by an $L^1(f)$ function in $y$ for all $h>0$ sufficiently small. (Let us recall that $L^1(f)$ is the $L^1$ space associated to the complex measure $\mu_f$ defined by $\mu_f(E)=\int_{E} f$ for every measurable $E\subseteq \mathbb{R}^n$. Clearly, every constant function is in $L^1(f)$.) However, this is an easy consequence of the mean value theorem: we know that there exists $\delta>0$ such that $0<h<\delta$ implies

$\left|\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right] - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$

$\leq 2\sup_{c\in\mathbb{R}^n}\left|\frac{\partial g}{\partial x_i}\left(c\right)\right|$

and the result now follows from the hypotheses. Q.E.D.

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How would you show $2\sup_{c\in\mathbb{R}^n}\left|\frac{\partial g}{\partial x_i}\left(c\right)\right| \in L^1$. Since the underlying space is $\mathbb{R}^n$, boundedness does not necessarily lead to integrability. –  newbie May 26 '13 at 21:17
    
@newbie Obviously, it's been a while since I posted this answer (nearly two years!) but the point is that $2\sup_{c\in \mathbb{R}^n} \left|\frac{\partial g}{\partial x_i}(c)\right|\in L^1(f)$. In my answer, I have defined $L^1(f)$ as the $L^1$ space associated to the complex measure $\mu_f$ defined by $\mu_f(E)=\int_{E} f$ for every measurable $E\subseteq \mathbb{R}^n$. Of course, it is easy to check that every bounded function is in $L^1(f)$. (The product of a bounded function and an integrable function (in the usual sense!) is integrable.) –  Amitesh Datta May 29 '13 at 8:53
    
Thanks I get it now. –  newbie May 29 '13 at 18:45

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