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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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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
    
It at least hold when $p=1$, using properties of Fourier transform. –  Davide Giraudo Nov 17 '12 at 13:15
    
@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

1 Answer 1

up vote 7 down vote accepted

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.

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