Can you help me with this exercise?
Let $f \in C^2(\mathbb{R}) $ a function $ f(x) > 0, \forall x \in \mathbb{R} $ and $\|f''\|_\infty < \infty $ , prove that $\sqrt f$ is Lipschitz continuous.
My attempt: i tried assuming that $f'' \ge 0$ , then $f'$ is increasing and the following limits exist:
$$ \lim_{x \to +\infty} f'(x) , \lim_{x \to +\infty} f(x)$$
then i can calculate using L'Hôpital's rule ( $f(x)$ is definitely increasing or decreasing): $$ L=\lim_{x \to +\infty} |(\sqrt f(x))'| = \lim_{x \to +\infty} |\frac{f'(x)}{2\sqrt f}| = \lim_{x \to +\infty} |\frac{f''(x)\sqrt f}{f'(x)}| \leq \frac{\|f''\|_\infty}{2L}$$
and so L must be finite, similar with $-\infty$ limit, so $(\sqrt f(x))'$ is bounded. However I'm not sure that $\lim_{x \to +\infty}|\frac{f''(x)\sqrt f}{f'(x)}|$ always exists. Anyway I can't solve the other cases.