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Suppose $f\in L^1(R^n)$ and $g\in S(R^n)$, where $S(R^n)$ is Schwartz space. Then, Can I have estimation like following? $$ |[f*g](x)|\leq\frac{1}{(1+|x|)^{s}}, $$ for some $s>n$. If it is correct, how to prove it? If it is not correct, what additional assumptions does it require?

Lin Xuelei
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  • You need some constant factor on the right side possibly depening on $g$. Since $g\in S(\mathbb R^{n})$ implies $kg\in S(\mathbb R^{n})$ for every positive integer $k$ you cannot possibly have any non-zero $f$ satisfying the stated inequality for every $g\in S(\mathbb R^{n})$. – Kavi Rama Murthy Apr 10 '18 at 10:25

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Of course the answer to the question as stated is "of course not"; the sensible version of the question is whether we have $$ |[f*g](x)|\leq\frac{c}{(1+|x|)^{s}}. $$

The answer to that question is still no, although it's not so obvious.

Take $n=1$ just to simplify the notation. Choose $g\in\mathcal S(\Bbb R)$ with $g\ge0$ and $g(x)\ge1$ for all $x\in[-1,1]$. Given a sequence $a_n\to\infty$, let $$f=\sum_{n=1}^\infty\frac1{n^2}\chi_{[a_n,a_n+1]}.$$

Then $f*g(a_n)\ge\frac1{n^2}$, so if $a_n$ blows up fast enough, in particular if $(1+|a_n|)^s/n^2\to\infty$, then $(1+|x|)^sf*g(x)$ is not bounded.

Note The same argument shows that $f*g\in C_0$ is the most that can be said about how fast $f*g$ vanishes at infinity: If $\phi\in C_0$ it is not true that $f*g=O(\phi)$ for every $f\in L^1$ and $g\in\mathcal S$.

David C. Ullrich
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  • Thanks for your answer. But if $f$ is fixed, maybe it is possible to choose a special $g\in S^{R^n}$ to satisfy that estimation? For example, I can start with Fourier transform. If $\hat{f}$ is Fourier transform of $f$, then I can choose a Schwartz function, say, $\hat{g}$ such that $\hat{f}\hat{g}$ is $s$-times differentiable with all of the derivatives belonging to $L^1(R^n)$. So, now that question is "for a uniformly continuous function $\hat{f}$, can I choose a Schwartz function $\hat{g}$ so that $\hat{f}\hat{g}$ is $s$-times differentiable for some $s>n$"? – Lin Xuelei Apr 11 '18 at 05:01
  • It could be that given $f$ there exists such a $g$. But the argument you suggest can't work; if we assume $\hat g\ne0$ to rule out trivialities then $\hat f\hat g$ is not going to be smoother than $\hat f$. – David C. Ullrich Apr 11 '18 at 12:11