A limit associated with a convex function

Question

Let $f:[0,\infty)\to [0,\infty)$ be a $C^1$, strictly increasing, convex function with $f(0)=0$. It follows that $f(x)/x$ is non-decreasing and $xf^\prime (x)/f(x)\ge 1$ on $(0,\infty )$. Suppose there exists some constant $k>1$ such that $\liminf_{x\to 0}f(x)/x^k =0$. Does it necessarily follow that $\liminf_{x\to 0}xf^\prime (x)/f(x)> 1$?

Answer 1

Counterexample: Define $a_n= 1/n!, n=2,\dots$ For these $n$ set $b_n=a_n-a_n^2.$ It's worth drawing a picture, visualizing the intervals $[a_{n+1},a_n],$ with $a_{n+1}<b_n<a_n$ and $b_n$ very close to $a_n.$

Define $g$ to be a continuous "staircase function": For $x\ge 1/2,$ set $g=1/2.$ On $[a_{n+1},a_n],$ set $g=a_{n+1}$ on $[a_{n+1},b_n],$ and on $[b_n,a_n)$ let $g$ be the line through $(b_n,a_{n+1})$ and $(a_n,a_n).$ Finally set $g(0)=0.$ Note that on $[0,\infty),$ $g$ is continuous, $g(x)\le x$, and $g(a_n)=a_n, n=2,3,\dots.$

For $x\ge 0,$ define $f(x)=\int_0^x g(t)\,dt.$ Then $f'=g.$ Because $g>0$ on $(0,\infty),$ $f$ is strictly increasing. And since $g$ is nondecreasing, $f$ is convex. Also note that since $g(t) \le t,$ $f(x)\le x^2/2.$ Thus $f(x)/x^k\to 0$ with $k=3/2>1.$

We have $f'(b_n)=g(b_n) = a_{n+1}.$ Also, $f(b_n) > a_{n+1}(b_n-a_{n+1}).$ Thus

$$\frac{f'(b_n)b_n}{f(b_n)} < \frac{b_n}{b_n-a_{n+1} }\to 1.$$