9
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '12 at 21:35
14
$\begingroup$

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

$\endgroup$
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 '12 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 '12 at 21:51
6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thanks for that, nice example (although I have no idea how you went around finding it!). $\endgroup$
    – jkn
    Sep 21 '12 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 '12 at 2:19
  • $\begingroup$ got it, thanks again $\endgroup$
    – jkn
    Sep 21 '12 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.