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.


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 $[0,\infty)$

  • $\begingroup$ 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

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