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We have the Schwartz space $\mathcal{S}$ of $C^\infty(\mathbb{R^n})$ functions $h$ such that $(1+|x|^m)|\partial^\alpha h(x)|$ is bounded for all $m \in \mathbb{N_0}$ and all multi-indices $\alpha$. We are given an $f \in L^p$ with $1\leqslant p < \infty$ and $g \in \mathcal{S}$. We want to show that $f \star g \in \mathcal{S}$, where $\star$ denotes the convolution operator.

I have already shown that $f \star g \in C^\infty$ by proving that $\partial^\alpha (f \star g) = f \star (\partial^\alpha g)$. Now I need to show that $(1+|x|^m)|\partial^\alpha(f \star g)(x) = (1+|x|^m)|f \star (\partial^\alpha g)(x)|$ is bounded. Since $\mathcal{S}$ is closed under differentiation, it suffices to consider $\alpha = 0$. I write $ \int_{\mathbb{R}^n}f(y)g(x-y)(1+|x|^m)dy $ and try to bound it but can't seem to make it work out. Could anyone help me proceed?

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    @MattRosenzweig Yes it does. It removed it.2015-07-15

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This is not true. For an example take $f(x) = (1+\|x\|)^{-\alpha}$. If $p\alpha > n$, then $f \in L^p(\mathbb{R}^n)$. Now take a non-negative $\mathcal{C}^\infty$ bump function $g$ with integral $1$, supported in $\|x\|\le 1$. Then $g \in \mathcal{S}$ and it is easy to show that $|(f\star g)(x)| \ge (2+\|x\|)^{-\alpha}$, since the integral is an average of $f$ over the ball of radius $1$ centered at $x$. This implies $f\star g \notin \mathcal S$.

Roughly speaking, convoluting with a Schwarz function can not radically alter the decay at $\infty$, so the best estimates you can hope for are those for smooth $L^p$ functions.