# Is the Convolution of a Schwartz Function with an $L^{p}$-Function a Smooth $L^{p}$-Function?

Let $n \in \mathbb{N}$ and $p \in \mathbb{R}_{\geq 1}$. If $f \in \mathscr{S}(\mathbb{R}^{n})$ and $g \in {L^{p}}(\mathbb{R}^{n})$, then it is a well-known fact from real analysis that the convolution $f \star g$ is defined almost everywhere on $\mathbb{R}^{n}$ and that $$\| f \star g \|_{p} \leq \| f \|_{1} \| g \|_{p}. \quad (\text{Young's Inequality})$$ This implies that $f \star g \in {L^{p}}(\mathbb{R}^{n})$.

My question: Is it true that $f \star g$ is a smooth $L^{p}$-function?

Thank you very much for your help!

• Yes, the convolution is smooth. To see that this is so, you should convince yourself that you can move the derivative of $f \ast g$ on to $f$ or $g$. Since $f$ is smooth, you can differentiate the convolution by moving the derivatives to $f$ and using the fact that it is differentiable. – Elchanan Solomon Nov 17 '13 at 6:30
• @IsaacSolomon Dear Solomon, a related post is here, you may be interested, math.stackexchange.com/questions/1509579/… – user186225 Nov 2 '15 at 16:12