The behavior of $\sqrt{f}f''/f'$ for a strictly convex function

Question

Let us consider $f:\mathbb{R}\longrightarrow [0,+\infty)$ is of class $\mathrm{C}^2$ and is strictly convex with $$\min_{x\in\mathbb{R}} f(x) = f(0) = 0$$ If $f''(x) \geq \alpha > 0$ for some $\alpha >0$ then $f$ is called "strongly convex". Then $$ f(x) \geq f(y) + f'(y)(x-y) + \frac{\alpha}{2}(x-y)^2$$ and as a consequence (see https://rkganti.wordpress.com/2015/08/19/lipschitz-functions-and-convexity/) we have $$ f(x) \leq \frac{1}{2\alpha} f'(x)^2$$ for all $x\in \mathbb{R}$, which implies that \begin{equation}\label{xxx} \limsup_{x\longrightarrow 0} \left|\frac{\sqrt{f(x)}}{f'(x)}\right| < \infty.\tag{$\star$} \end{equation}

In fact, all we need to show \eqref{xxx} is $f''(0) > 0$. I am trying to show that \eqref{xxx} is also true if $f''(0) = 0$. To be precise, I am trying to show that

Given the condition above: $f:\mathbb{R}\longrightarrow [0,+\infty)$ is of class $\mathrm{C}^2$ and is strictly convex with $$\min_{x\in\mathbb{R}} f(x) = f(0) = 0$$ then \begin{equation}\label{xxxx} \limsup_{x\longrightarrow 0} \left|\frac{f''(x)}{f'(x)}\sqrt{f(x)}\right| < \infty.\tag{$\star\star$} \end{equation}

Answer 1

This is not true. Let $g$ be a continuous function such that $x^6 \le g(x)\le x^6+x^2$ for all $x\in\mathbb{R}$, and more precisely $g(x)=x^6$ except in the intervals $(2^{-n}, 2^{-n} + 2^{-8n})$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields

1. $|f'(x)|\le |x|^7/7 + \sum_{2^{-n} \le |x|} 2^{-2n}2^{-8n}$. The geometric sum is dominated by its largest term, which is at most $|x|^{10}$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| \le |x|^7$ for all sufficiently small $|x|$.

2. $ f(x) \ge x^8/56$ by comparison of the second derivatives of both sides.

Thus, at the points where $g(x) = x^6+x^2$ we have $$ \left|\frac{f''(x)}{f'(x)}\sqrt{f(x)}\right| \ge \frac{x^2}{x^7}\frac{x^4}{\sqrt{56}} = \frac{1}{x\sqrt{56}} $$ which is unbounded as $x\to 0$.

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Let's generalize the above to get a $C^{2,\alpha}$ counterexample for any $\alpha\in (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define $$ g(x) = |x|^a + \sum_{n=1}^\infty 2^{-bn}\max(0, 1- 2^{cn}|x-2^{-n}|) $$ where the sum is a formal way of saying: add a triangle of height $2^{-bn}$ with the base $[2^{-n}-2^{-cn}, 2^{-n}+2^{-cn}]$.

We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.

Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.

1. Integration yields $|f'(x)| \le |x|^{a+1}/(a+1) + 2\sum_{2^{-n} \le |x|} 2^{-bn}2^{-cn}$. The geometric sum is dominated by its largest term, which is of size $x^{b+c} = x^{a+1}$. Thus, $|f'(x)| = O(|x|^{a+1})$ as $x\to 0$.

2. $f(x) \ge x^{a+2}/((a+1)(a+2))$ by comparison of the second derivatives of both sides.

At the points $x=2^{-n}$ we have $f''(x) \ge x^{b}$, which implies, for some constant $\nu>0$, $$ \left|\frac{f''(x)}{f'(x)}\sqrt{f(x)}\right| \ge \nu \frac{x^b}{x^{a+1}} x^{(a+2)/2} = \nu x^{b - a/2} = \nu x^{-1/2} $$ which is unbounded as $x\to 0$.