Lipschitizianity of the square root of a positive $C^2$ function

Question

I was trying to solve this exercise. Let $f\in C^2(\mathbb{R})$ a strictly positive function such that $f''$ is bounded. Then prove that $\sqrt{f}$ is Lipschitz. A first idea was to prove that it's first derivative is bounded, but i find the expression $\dfrac{f'}{\sqrt{f}}$ which gave me no informations, also deriving again this expression i'm not able to use the bound on the second derivative.

Answer 1

For any $x$ and $y$ we have

$$0\leq f(y)=f(x)+\int_{x}^y f'(s)ds=f(x)+\int_x^y[ f'(x)+\int_{x}^s f''(t)]dtds\\ \leq f(x)+(y-x)f'(x)+|y-x|^2/2\max_{\mathbb{R}}|f''|$$

Now let $y-x=\pm\sqrt{f(x)}$ and you obtain that

$$0\leq (1+C)f(x)\pm\sqrt{f(x)}f'(x)\\ \Rightarrow \frac{|f'(x)|}{\sqrt{f(x)}}\leq (1+C)$$

This proves the claim for strictly positive $f$, as you noted. But you can also deduct the claim for $f\geq 0$ by approximating it with $f_\varepsilon:=f+\varepsilon$ and then taking limits (this taking limits requires further arguments!)