Amy Bright

2022-11-10

If ${\int }_{|x|\le r}|f\left(x\right)|dx\le \left(r+1{\right)}^{a}$, then ${\int }_{\mathbb{R}}|f\left(x\right)|{e}^{-|tx|}dx<\mathrm{\infty }$ for $t\ne 0$
Let f be some function in ${L}_{loc}^{1}\left(\mathbb{R}\right)$ such that, for some $a\in \mathbb{R}$
${\int }_{|x|\le r}|f\left(x\right)|dx\le \left(r+1{\right)}^{a}$
for all $r\ge 0$. Show that $f\left(x\right){e}^{-|tx|}\in {L}^{1}\left(\mathbb{R}\right)$ for all $t\in \mathbb{R}\setminus \left\{0\right\}$
I'm having a hard time finding use of the bound described above. Any help would be appreciated.

tektonikafrs

Assume without loss of generality that $f⩾0$ everywhere and that t>0, then note that, for every x,
${e}^{-t|x|}={\int }_{|x|}^{\mathrm{\infty }}t{e}^{-tr}dr$
hence, applying Tonelli's theorem (aka Fubini for nonnegative functions), one gets
${\int }_{\mathbb{R}}f\left(x\right){e}^{-t|x|}dx=t{\int }_{\mathbb{R}}{\int }_{|x|}^{\mathrm{\infty }}f\left(x\right){e}^{-tr}dr\phantom{\rule{thinmathspace}{0ex}}dx=t{\int }_{0}^{\mathrm{\infty }}{e}^{-tr}\left({\int }_{|x|⩽r}f\left(x\right)dx\right)dr$
Thanks to the hypothesis about the innermost integrals, one gets
${\int }_{\mathbb{R}}f\left(x\right){e}^{-t|x|}dx⩽t{\int }_{0}^{\mathrm{\infty }}{e}^{-tr}\left(r+1{\right)}^{a}dr$

Do you have a similar question?

Recalculate according to your conditions!