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I consider a function $f:\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}_0$ which is differentiable.

I wonder which conditions will guarantee that $f'$ is bounded.

I know that boundedness of $f$ is not sufficient condition for boundedness of its derivative $f'$, e.g. $f(x) = \sin(x\cos(x)).$

I wonder if the fact that

  • $\lim_{x\rightarrow +\infty} f(x) = const.$
  • $f$ is bounded

are sufficient conditions for boundedness of $f'$.

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  • $\begingroup$ Controlling $f$ at infinity is clearly not enough. You can have a highly oscillatory function around some finite point, say $f(x)=(x-1)^2\sin(1/(x-1))$ between $0$ and $2$, $f(1)=0$ and $f$ extended in some smooth, bounded way outside $[0,2]$ with $\lim_{x\to\infty}f(x)=c$. Such function is differentiable, bounded with a finite limit at infinity yet the derivative is not bounded, as $f'(x)$ oscillates wildly near $x=1$ ($f'(x)=2(x-1)\sin 1/(x-1)+\cos 1/(x-1)$) $\endgroup$ – GReyes Mar 30 at 16:36
  • $\begingroup$ @GReyes "Controlling ff at infinity is clearly not enough." Good point. It's not clear how relevant it is, since the conditions he gives do not imply that $f'$ is bounded "at infinity""... $\endgroup$ – David C. Ullrich Mar 30 at 17:03
  • $\begingroup$ @DavidC.Ullrich Sure. The derivative can oscillate at infinity as well, as it happens in your example. The first example that came to my mind was one that oscillates around a finite point though.. anyways, as you mention, you need your function to be Lipschitz continuous. $\endgroup$ – GReyes Mar 30 at 17:23
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No, those two conditions do not imply that $f'$ is bounded. Consider for example $$f(x)=\frac{\sin\left(e^x\right)}{x^2+1}.$$

In case it helps, "the" condition is very simple:

Triviality: Suppose that $f:[0,\infty)\to\Bbb R$ is differentiable. Then $f'$ is bounded if and only if there exists $c$ such that $|f(x)-f(y)|\le c|x-y|$ for all $x,y$.

"if": definition of $f'$.

"only if": Mean Value Theorem.

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Take$$\begin{array}{rccc}f\colon&\mathbb R^+&\longrightarrow&\mathbb R_0^+\\&x&\mapsto&\dfrac{x\sin\left(\frac1x\right)}{1+x^2}.\end{array}$$Then $\lim_{x\to\infty}f(x)=0$ and $(\forall x\in\mathbb R):f(x)\in(0,1)$. However, $f'$ is unbounded.

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