# Boundedness of derivative of $f(x)$

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 $$f'$$.

• 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 $x=1$ ($f'(x)=2(x-1)\sin 1/(x-1)+\cos 1/(x-1)$) – GReyes Mar 30 at 16:36
• @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""... – David C. Ullrich Mar 30 at 17:03
• @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. – GReyes Mar 30 at 17:23

## 2 Answers

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.

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.