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.
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.