I'm currently trying to prove it by applying the Fundamental Theorem of Calculus (as per a hint): $$|f'(x)| \le 1/(1+x^2)$$ $$|\int_a^b f'(x)| \le \int_a^b 1/(1+x^2)$$ $$|f(b) - f(a)| \le \int_a^b 1/(1+x^2).$$

I am not sure how I can get from this to $f$ being bounded on $\mathbb{R}.$ Does anyone have any pointers on how to proceed? Should I be trying to find a counterexample instead? Thanks!


Since for any $x\in\Bbb{R}$ $$ f(x)=\int_0^{x}f'(x)\:dx+f(0) $$ We have $$ |f(x)|\leqslant|f(0)|+\int_0^{x}|f'(x)|\:dx<|f(0)|+\int_0^{\infty}\frac1{1+x^2}dx=|f(0)|+\frac{\pi}{2} $$

  • $\begingroup$ Is it possible to show this without using infinity in the integral? $\endgroup$ – mxdg Dec 8 '15 at 2:53
  • $\begingroup$ In my class, we haven't done indefinite integrals yet, so I don't think we can use it. Can we just use the Fundamental Theorem of Calculus and do $\int_0^x 1/(1+x^2) dx = \arctan x - \arctan 0 \le \pi/2 - 0$? (Is $\arctan x \le \pi/2$ considered a common fact?) $\endgroup$ – mxdg Dec 8 '15 at 4:11
  • $\begingroup$ Yes, you can use that. It is basically same. $\endgroup$ – Math Wizard Dec 8 '15 at 5:17

You're nearly there. Note that the antiderivative of $\frac{1}{x^2+1}$ is $\tan^{-1}(x)$.

  • $\begingroup$ I saw that, but I wasn't sure how to use it, since $|f(b) - f(a)|$ being bounded doesn't imply that $|f(x)|$ is bounded in general, right? $\endgroup$ – mxdg Dec 7 '15 at 6:35
  • 1
    $\begingroup$ @mxdg If $|f(b)-f(a)|<M$ for all $a,b$, then $|f(x)|<M+|f(0)|$ forall $x$ $\endgroup$ – Hagen von Eitzen Dec 7 '15 at 7:24

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