2

Let $f(x)$ a Schwartz function and $g(x)$ in $L^2(\mathbb{R})$.

Is possible get

\begin{align} \int_{\mathbb{R}}|x||(f*g)(x)|dx<\infty? \end{align}

My attempt: By Holder's inequality, \begin{align} |(f*g)(x)|\leq \int_{\mathbb{R}}|f(x-y)g(y)|\,dy\leq \left\|y\mapsto f(x-y)\right\|_{L^2(\mathbb{R})}\left\|g\right\|_{L^2(\mathbb{R})} \end{align}

then, with $|x|\leq 1+x^2$,

\begin{align} \int_{\mathbb{R}}|x||(f*g)(x)|\,dx\leq \left\|g\right\|_{L^2(\mathbb{R}}\int_{\mathbb{R}}(1+x^2)\left\|y\mapsto f(x-y)\right\|_{L^2(\mathbb{R})} \end{align} and I would like that $\left\|y\mapsto f(x-y)\right\|_{L^2(\mathbb{R})}$ is a Schwartz function in $x$...

Is there a estimate for I wish?

Thanks!

eraldcoil
  • 3,508

1 Answers1

1

No. If $g(x) = (1+|x|)^{-1}$, and $f$ is non-negative, then the integral equals, by Fubini's theorem, $$ \int_{\mathbb{R}} |x| \int_{\mathbb{R}} \frac{f(y)}{1 + |x-y|} \ \mathrm{d}y\ \mathrm{d}x = \int_{\mathbb{R}} f(y) \int_{\mathbb{R}} \frac{|x|}{1 + |x-y|} \ \mathrm{d}x \ \mathrm{d}y = +\infty $$ because for every $y$, $$ \int_{\mathbb{R}} \frac{|x|}{1 + |x-y|} \ \mathrm{d}x= +\infty $$

fwd
  • 3,290
  • Right. Indeed, probably the sharpest thing in terms of integrability (though smoothness is assured) about a convolution of $L^2$ function and Schwartz, is that the result is in $L^2$. And integration against $|x|$ will not be finite. The smoothness doesn't help. – paul garrett Jan 14 '24 at 23:21