How to show convolution of an $L^p$ function and a Schwartz function is a Schwartz function 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?
 A: 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.
A: Lukas Geyer's answer is direct, and we can construct counterexamples in a different point of view.
If people had read the proof of the Bernstein lemma, e.g., Lemma 2.1 in Bahouri-Chemin-Danchin's book,
he will get the following example easily.
Take an $L^{2}(R^N)$ function $u(\xi)$ with support in $B(0,1)$. Let $k(\xi)$ be a $C_{0}^{\infty}(R^N)$ function satisfying $k(\xi) = 1$ for $\xi \in B(0,1)$.
Write $U  = F^{-1} u$ and $K = F^{-1}k$, where $F^{-1}$ is the inverse Fourier transform.
Then $K$ is a Schwartz function, and $U$ is a Schwartz function if and only if $u$ is a Schwartz function.
By noticing that
$$
u(\xi) =  k(\xi) u(\xi) \quad\text{for}\quad \xi \in R^{N}. 
$$
We get
$$
U = K \ast U. 
$$
Moreover, by Young's inequality, $U \in L^{p}$ for $p \ge 2$.
So in general, we can not hope that the convolution of an $L^{p}$ function and a Schwartz function is a Schwartz function.
[Reference]
Bahouri, Hajer; Chemin, Jean-Yves; Danchin, Raphaël Fourier analysis
and nonlinear partial differential equations. Grundlehren der
mathematischen Wissenschaften [Fundamental Principles of Mathematical
Sciences], 343. Springer, Heidelberg, 2011.
