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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.

mrprottolo
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1 Answers1

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Call $\|f''\|_\infty =M$. I claim that $$\left|\frac{f'(x)}{2\sqrt{f(x)}}\right|\leq {\sqrt M}$$ from which the Lipschitz continuity of $\sqrt{f}$ follows.

Suppose by contradiction there is a point in which the opposite inequality holds, wlog $0$. Then $f'(0)^2>4M f(0)$. From the bound $f''(x)\leq M$ it follows that $f$ stays below the second degree polynomial function which has the same value and derivative at $0$: $$f(x)\leq Mx^2+bx+c$$ where $f(0)=c$ and $b=f'(0)$. In particular the discriminant of the polynomial is $$\Delta =b^2-4Mc=f'(0)^2-4Mf(0)>0$$ which implies that the polynomial has roots, therefore $f$ becomes negative, which is a contradiction.

Del
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