0
$\begingroup$

If infinite integral $\int_a^{+\infty}|f(x)|dx$ converges, i.e. $\int_a^{+\infty}f(x)dx$ absolutely converges, then $f$ is bounded on $[a,+\infty)$?


Thanks a lot.

$\endgroup$
1
$\begingroup$

Consider $$f(x) = \begin{cases} \frac{1}{2\sqrt{x}} & x \in (0,1) \\ 0 & \text{else} \end{cases}$$

Then $$\int_0^\infty f(x) dx = \int_0^1 \frac{1}{2\sqrt{x}} dx = 1$$ but $$\lim_{x\to 0} f(x) = +\infty$$

so $f$ is not bounded on $[0,\infty)$

$\endgroup$
  • $\begingroup$ If $f$ is continuous on $[a,+\infty)$? $\endgroup$ – James Chan Feb 14 '17 at 17:10
  • $\begingroup$ Then your statement holds due to the necessary condition $\lim_{x\to\infty} |f(x)| = 0$ hence $|f| \le \varepsilon$ for each $\varepsilon > 0$ and an arbitrage large $M>0$. And $f$ is bounded on the compact interval $[a,M]$ because it's continuous, so it holds $$|f| \le \max\{\max_{x\in [a,M]} |f(x)|,\varepsilon\}$$ $\endgroup$ – Gono Feb 15 '17 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.