TITLE: Prove that $\sqrt {f(x)}$ is Lipschitz
QUESTION [6 upvotes]: Let $f(x)\in C^2(\mathbb{R}), f(x)\geq0,f''(x)\leq1,$ prove that $\sqrt{f(x)}$ is a Lipschitz function .
I can prove that $f(x)$ is uniformly continuous by the inequality without the condition of the $f''(x)$, so I want to ask someone for a better answer. 
Thanks in advance.

REPLY [7 votes]: Hint: Using Taylor expansion show that for all real $x,t$:
$$ 0 \leq f(x+t) \leq f(x)+f'(x)t + \frac{t^2}{2} = f(x)-\frac{f'(x)^2}{2} 
+ \frac12 (t+f'(x))^2$$
which implies $f(x)\geq f'(x)^2/2$ and you deduce that $\sqrt{f}$ is $\frac{1}{\sqrt{2}}$-Lipschitz by bounding  its derivative. The result is optimal as shown by the example $f(x)=x^2/2$