4
$\begingroup$

You have $f \in C^\infty([0,1])$ with $f > 0$. Then $\sqrt{f}$ is easily seen to be differentiable . Prove that there exists a constant $C$ independent of $f$ such that:

\sup_{x\in[0,1]}\left\lvert \left(\sqrt f\right)'(x)\right\rvert \leq C \left(1 + \sup_{x\in[0,1]}\lvert f(x)\rvert + \sup_{x\in[0,1]}\lvert f'(x)\rvert + \sup_{x\in[0,1]} \lvert f''(x)\rvert\right).

I wonder if someone can give a nice proof of this.

  • 2
    Evidently (at least) four people thought that this question was useful, clear, and showed research effort. Therefore the counterexample will be useful to them as well- and therefore not a waste of time:)2012-03-15

1 Answers 1

5

I don't get it. Let's look at $f(x)=x+\varepsilon$ with $\varepsilon>0$. Then \sqrt f '=1/(2\sqrt{x+\varepsilon}), so \sup\sqrt f '=\varepsilon^{-1/2}/2, $\sup f=1+\varepsilon$, \sup f'=1, \sup f''=0. And the question asks us to show that there exists a constant $C$ such that $ \frac1{2\sqrt{\varepsilon}}\le C(3+\varepsilon) $ for all $\varepsilon>0$. That is surely impossible.

  • 0
    Your comment came while I was typing. I just woke up so you didn't waste any of my time. No worries.2012-03-15