# Prove or give a counterexample: If $|f'(x)| \le 1/(1+x^2)$ for all $x \in \mathbb{R}$, then $f$ is bounded on $\mathbb{R}$.

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 $$|f(x)|\leqslant|f(0)|+\int_0^{x}|f'(x)|\:dx<|f(0)|+\int_0^{\infty}\frac1{1+x^2}dx=|f(0)|+\frac{\pi}{2}$$
• 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?) – mxdg Dec 8 '15 at 4:11
You're nearly there. Note that the antiderivative of $\frac{1}{x^2+1}$ is $\tan^{-1}(x)$.
• 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? – mxdg Dec 7 '15 at 6:35
• @mxdg If $|f(b)-f(a)|<M$ for all $a,b$, then $|f(x)|<M+|f(0)|$ forall $x$ – Hagen von Eitzen Dec 7 '15 at 7:24