Two questions:

  1. Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuously differentiable and bounded. Is its derivative $f'$ bounded as well?

  2. What about in the case of $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$, where $f$ is continuously differentiable, and its total derivative $f'$?

Thanks in advance.

  • $\begingroup$ What do you mean by total derivative? In any case, for (2) take a function $\mathbb R^n\to\mathbb R^n$ which is the identity in all coordinates but one, and there make it act like the example from (1). $\endgroup$ Sep 20, 2012 at 21:35

2 Answers 2


No, e.g. $\sin (x^2){}{}{}{}$.

  • $\begingroup$ Thanks (I'll accept as soon as the webpage allows me!). What if $f:\mathbb{R}\rightarrow\mathbb{R}$ is strictly increasing (sorry I know its a different question...)? $\endgroup$
    – jkn
    Sep 20, 2012 at 21:38
  • $\begingroup$ @JuanKuntz I think the answer is still no. Intuitively, the derivative must be very small for most large $x$, but could have sharp peaks. $\endgroup$ Sep 20, 2012 at 21:51

A strictly increasing example: Define $g:\mathbb R_{\geq 0}\to\mathbb R_{\geq 0}$ by $$g(x)=\begin{cases} n^4x-n^5+n &\text{if }n-\frac{1}{n^3}\leq x+1\leq n\\ n+n^5-n^4x &\text{if }n\leq x+1\leq n+\frac{1}{n^3}\\ 0 &\text{otherwise} \end{cases}$$ which intuitively is just a continuous function which is $0$ almost everywhere but has very narrow, tall spikes at each natural number. Note that $g(n-1)=n$, so $g$ is unbounded, yet $$\int_0^\infty g(x)dx = \sum\limits_{n=1}^\infty n\frac{1}{n^3}=\frac{\pi^2}{6}$$ which is finite. Define $f:\mathbb R\to\mathbb R$ by $$f(x)=\begin{cases} \arctan x+\int_0^x g(y)dy &\text{if } x\geq 0\\ \arctan x-\int_0^{-x} g(y)dy &\text{if } x\leq 0\\ \end{cases} $$ which is continuously differentiable, bounded and strictly increasing. Since the derivative of $\arctan x$ is bounded, the derivative of $f$ differs from $g$ by a bounded function, so must be unbounded.

  • $\begingroup$ Thanks for that, nice example (although I have no idea how you went around finding it!). $\endgroup$
    – jkn
    Sep 21, 2012 at 1:26
  • $\begingroup$ @JuanKuntz The intuition is that I construct the derivative first, and try to make it unbounded yet have bounded integral. The natural way to do this is to make it have high, narrow peaks. $\endgroup$ Sep 21, 2012 at 2:19
  • $\begingroup$ got it, thanks again $\endgroup$
    – jkn
    Sep 21, 2012 at 10:03

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