Prove that $\sqrt {f(x)}$ is Lipschitz

Question

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.

Answer 1

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 $f(x)=x^2/2$