3

During review of some basic real analysis, I have come across an old qualifying exam question for which in its proof I want to show the following:

Let $f(x)$ be a real-valued function on $[0,1]$ which is differentiable up to and including the endpoints. Then $\sup_{x\in[0,1]}|f'(x)| \leq C$.

I feel like there should be some simple proof of the above, but I am not seeing it right now (the differentiability at the endpoints immediately reminds me of Darboux's theorem).

My idea is to assume it is not bounded. Then there exists a sequence $\{x_n\} \subset [0,1]$ and a $x_0 \in [0,1]$ such that $x_n$ converges to $x_0$, and yet $|f'(x_n)| \geq n$ for every $n \in \mathbb{N}$. But I can't seem to get things to work.

Am I missing something here?

Lost in a Maze
  • 1,634
  • No.Define not continuos not bounded but integrable function on $[0,1]$.Then it's integral will be you function $f(x)$. In your case $f(x)$ must be continuously differentiable namely $C^1$. – Alexander Vigodner Jul 27 '14 at 04:57
  • 2
    See http://math.stackexchange.com/q/257584. – Jonas Meyer Jul 27 '14 at 05:08
  • @AlexanderVigodner: That method of constructing an example doesn't generally work. The integral of an integrable function need not be everywhere differentiable. (E.g., one necessary condition is that there are no jump discontinuities.) – Jonas Meyer Jul 27 '14 at 05:09
  • I agree. I had to be more specific. – Alexander Vigodner Jul 27 '14 at 06:26

1 Answers1

9

Well, no. Take $$ f(x) = x^2 \sin \left( \frac{1}{x^2} \right) $$ for $x \neq 0$ but $$ f(0) = 0. $$

Differentiable at the origin because it is squeezed between $-x^2$ and $x^2.$ Differen tiable everywhere else with $$ f'(x) = 2 x \sin \left( \frac{1}{x^2} \right) - \frac{ 2}{x} \cos \left( \frac{1}{x^2} \right) $$

Will Jagy
  • 146,052