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Let $f \in L^p\left(\Bbb R^n\right)$ and $g \in C_c^\infty \left(\Bbb R^n\right)$. Show $f \ast g \in C^\infty\left(\Bbb R^n\right)$ for $1 \le p \le \infty$.

Let $x=(x_1,x_2,\ldots,x_n)$ and $y=(x_1+h,x_2,\ldots,x_n)$, $h \ne 0$.
The first step is to show $$ \partial_{x_1} \left(f \ast g\right)= f \ast \left(\partial_{x_1}g\right), $$ through the dominated convergence theorem.
Is it possible to bound $$ \int_{\Bbb R^n} \left|f(t)\right| \frac{\left|g(x-t)-g(y-t)\right|}{|x-y|}dt $$ with the maximal function of $g$?
Or resorting to the mean value theorem $$ g(x-t)-g(y-t)=\int_0^1 Jg\left(x+t(y-x)\right)dt \cdot (y-x) $$ is the only way, where $Jg$ is the Jacobian matrix of $g$.

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Why do you want a bound with the maximal function of $g$? –  Davide Giraudo Nov 12 '12 at 13:11
    
@DavideGiraudo There is no particular reason. I recently learn about the maximal function, and I was wondering if it was applicable here. –  Nicolas Essis-Breton Nov 12 '12 at 13:58

1 Answer 1

up vote 1 down vote accepted

Here is a proof without using dominated convergence: Let $e_1 = (1,0,\ldots,0) \in \mathbb{R}^n$, $x \in \mathbb{R}^n$, $t \in \mathbb{R}$. Then

$$(f \ast g)(x+t \cdot e_1)-(f \ast g)(x) = \int_{\mathbb{R}^n} (g(x+t \cdot e_1-y)-g(x-y)) \cdot f(y) \, dy \\ = \int_{\mathbb{R}^n} \int_0^t \partial_1 g(x+\tau \cdot e_1-y) \, d\tau f(y) \, dy \\ \stackrel{\text{Fubini}}{=} \int_0^t \int_{\mathbb{R}^n} \partial_1 g(x+t \cdot e_1-y) \cdot f(y) \, dy \, d\tau \\ = \int_0^t \underbrace{((\partial_1 g) \ast f)}_{\text{continuous}} (x+\tau \cdot e_1) \, d\tau$$

Thus (by First Fundamental Theorem of Calculus): $t \mapsto (f \ast g)(x+t \cdot e_1)$ is differentiable and (for $t=0$) we obtain $\partial_1 (f \ast g) = f \ast (\partial_1 g)$.

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