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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.

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Is $f \geq 0$ or $f >0$? If $f(x)=x$ I don't think $f$ is differentiable.... –  N. S. Mar 15 '12 at 4:13
Edited the problem. –  Hammerhead Mar 15 '12 at 4:22
Sorry to waste your time guys. Obviously $f(x) = x + \epsilon$ is a family of counterexamples :( –  Hammerhead Mar 15 '12 at 5:23
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:) –  treble Mar 15 '12 at 6:35

1 Answer 1

up vote 5 down vote accepted

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.

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you are right. See my edit. Sorry for wasting your time. –  Hammerhead Mar 15 '12 at 5:30
Your comment came while I was typing. I just woke up so you didn't waste any of my time. No worries. –  Jyrki Lahtonen Mar 15 '12 at 5:32

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