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I am trying to prove that if $f \in C_c^k(\mathbb{R}^n)$ and $g$ is Lebesgue integrable on $\mathbb{R}^n$, then the derivatives of $f * g$ equal $$D^\alpha(f * g)(x) = \int_{\mathbb{R}^n}(D^\alpha f)(x-y)g(y)dy$$ and are continuous for all multi-indexes of order up to $k$.

I know the version for $\mathbb{R}$, however I am interested in this $n$-dimensional generalization.

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1 Answer 1

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Well, since $f\in C_c^k$, you have that $D^{\alpha}f\in L^{\infty}$, and thus, $|D^{\alpha} f(x-y) g(y)|\leq \|D^{\alpha} f\|_{\infty} |g(y)|\in L^1$. Thus, the desired simply follows from the dominated convergence theorem for differentiation under the integral, since for every fixed $y$,

$$ D^{\alpha}( f(x-y)g(y))=D^{\alpha}(f)(x-y)g(y) $$

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  • $\begingroup$ Note that continuity of these derivatives also follows from dominated convergence. $\endgroup$ Commented Jan 26, 2020 at 12:08
  • $\begingroup$ Why is $D^\alpha f \in L^\infty$? I know that $f \in L^\infty$. $\endgroup$
    – Naah
    Commented Jan 26, 2020 at 14:13
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    $\begingroup$ $D^{\alpha} f$ is continuous, since $f\in C^k$ and there exists a compact set $K$ such that $f|_{\mathbb{R}^n\setminus K}\equiv 0$. Clearly, for $x\in \mathbb{R}^n\setminus K,$ we also have $D^{\alpha} f\equiv 0$ so $D^{\alpha} f\in C_c$. Any continuous function with compact support is $L^{\infty}$. $\endgroup$ Commented Jan 26, 2020 at 14:24

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