Amy Bright

2022-11-10

If ${\int}_{|x|\le r}|f(x)|dx\le (r+1{)}^{a}$, then ${\int}_{\mathbb{R}}|f(x)|{e}^{-|tx|}dx<\mathrm{\infty}$ for $t\ne 0$

Let f be some function in ${L}_{loc}^{1}(\mathbb{R})$ such that, for some $a\in \mathbb{R}$

$${\int}_{|x|\le r}|f(x)|dx\le (r+1{)}^{a}$$

for all $r\ge 0$. Show that $f(x){e}^{-|tx|}\in {L}^{1}(\mathbb{R})$ for all $t\in \mathbb{R}\setminus \{0\}$

I'm having a hard time finding use of the bound described above. Any help would be appreciated.

Let f be some function in ${L}_{loc}^{1}(\mathbb{R})$ such that, for some $a\in \mathbb{R}$

$${\int}_{|x|\le r}|f(x)|dx\le (r+1{)}^{a}$$

for all $r\ge 0$. Show that $f(x){e}^{-|tx|}\in {L}^{1}(\mathbb{R})$ for all $t\in \mathbb{R}\setminus \{0\}$

I'm having a hard time finding use of the bound described above. Any help would be appreciated.

tektonikafrs

Beginner2022-11-11Added 15 answers

Assume without loss of generality that $f\u2a7e0$ 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

$${\int}_{\mathbb{R}}f(x){e}^{-t|x|}dx=t{\int}_{\mathbb{R}}{\int}_{|x|}^{\mathrm{\infty}}f(x){e}^{-tr}dr\phantom{\rule{thinmathspace}{0ex}}dx=t{\int}_{0}^{\mathrm{\infty}}{e}^{-tr}({\int}_{|x|\u2a7dr}f(x)dx)dr$$

Thanks to the hypothesis about the innermost integrals, one gets

$${\int}_{\mathbb{R}}f(x){e}^{-t|x|}dx\u2a7dt{\int}_{0}^{\mathrm{\infty}}{e}^{-tr}(r+1{)}^{a}dr$$

$${e}^{-t|x|}={\int}_{|x|}^{\mathrm{\infty}}t{e}^{-tr}dr$$

hence, applying Tonelli's theorem (aka Fubini for nonnegative functions), one gets

$${\int}_{\mathbb{R}}f(x){e}^{-t|x|}dx=t{\int}_{\mathbb{R}}{\int}_{|x|}^{\mathrm{\infty}}f(x){e}^{-tr}dr\phantom{\rule{thinmathspace}{0ex}}dx=t{\int}_{0}^{\mathrm{\infty}}{e}^{-tr}({\int}_{|x|\u2a7dr}f(x)dx)dr$$

Thanks to the hypothesis about the innermost integrals, one gets

$${\int}_{\mathbb{R}}f(x){e}^{-t|x|}dx\u2a7dt{\int}_{0}^{\mathrm{\infty}}{e}^{-tr}(r+1{)}^{a}dr$$

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