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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?

Davide Giraudo
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nullUser
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  • Are you sure that your claim is correct? As far as I know one can only show that the convolution of two Schwartz functions is a again a Schwartz function. – saz Nov 17 '12 at 10:12
  • @saz I don't know if it is true, but the problem says to prove it. None of the books I have seen state this result, I have seen that $f \star g \in \mathcal{S}$ if $f,g \in \mathcal{S}$. If it is not true, a counterexample would be nice. – nullUser Nov 17 '12 at 15:23
  • Saz, How to prove that the convolution of two functions in S(R) is in S(R) – Henrique Tyrrell Dec 12 '12 at 17:35
  • @saz How to prove that the convolution of two schwartz functions is a schwartz funtion? – Henrique Tyrrell Dec 12 '12 at 17:36
  • @HenriqueTyrrell If $f \in \mathcal{S}$, then $\hat{f} \in \mathcal{S}$ where $\hat{f}$ denotes the Fourier transform. Moreover, $\widehat{f \ast g} = \hat{f} \cdot \hat{g}$, thus $\widehat{f \ast g} \in \mathcal{S}$ for $f,g \in \mathcal{S}$. Hence $f \ast g \in \mathcal{S}$. (The Fourier transform $\mathcal{S} \ni f \mapsto \hat{f} \in \mathcal{S}$ is bijective.) – saz Dec 12 '12 at 18:30
  • @MattRosenzweig Yes it does. It removed it. – Davide Giraudo Jul 15 '15 at 12:01

2 Answers2

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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.

Lukas Geyer
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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.

wangtwo
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