If $\int_{|x| \leq r} |f(x)|dx \leq (r+1)^a$, then $\int_{\mathbb{R}} |f(x)|e^{-|tx|} dx < \infty$ for $t \neq 0$

Question

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

$$\int_{|x| \leq r} |f(x)|dx \leq (r+1)^a$$

for all $r \geq 0$. 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.

Answer 1

Assume without loss of generality that $f\geqslant0$ everywhere and that $t>0$, then note that, for every $x$, $$e^{-t|x|}=\int_{|x|}^\infty te^{-tr}dr$$ hence, applying Tonelli's theorem (aka Fubini for nonnegative functions), one gets $$\int_\mathbb Rf(x)e^{-t|x|}dx=t\int_\mathbb R\int_{|x|}^\infty f(x)e^{-tr}dr\,dx=t\int_0^\infty e^{-tr}\left(\int_{|x|\leqslant r}f(x)dx\right)dr$$ Thanks to the hypothesis about the innermost integrals, one gets $$\int_\mathbb Rf(x)e^{-t|x|}dx\leqslant t\int_0^\infty e^{-tr}(r+1)^adr$$ The last integral obviously converges hence we are done (with an explicit upper bound if need be).

Answer 2

Hint: The given bound is a polynomial growth estimate. Exponential decay always wins over polynomial growth. To get the given estimate in play, integrate by parts.

Added later: To see how integration by parts can work here, assume WLOG $f\ge 0$ and $t>0.$ The integral can be written

$$\int_0^\infty (f(x) + f(-x)) e^{-tx}\, dx.$$

For $y\ge 0,$ define $F(y) = \int_0^y (f(x) + f(-x))\, dx.$ Then

$$\int_0^y (f(x) + f(-x)) e^{-tx}\, dx = F(x)e^{-tx}\big|_0^y + t\int_0^y F(x) e^{-tx}\, dx.$$

Answer 3

A different proof. Assume $t>0$. \begin{align} \int_{\Bbb R}|f(x)|\,e^{-t|x|}\,dx&=\sum_{n=0}^\infty\int_{n\le|x|<n+1}|f(x)|\,e^{-t|x|}\,dx\\ &\le\sum_{n=0}^\infty e^{-t|n|}\int_{|x|<n+1}|f(x)|\,dx\\ &\le\sum_{n=0}^\infty e^{-t|n|}(2+n)^a\\ &<\infty. \end{align}