Does $f'(x)>\frac{f(x)}{x}$ for $x>0$ imply $f$ is convex?

Question

Let $f\in C^\infty([0,\infty))$, $f(0)=0$, $f(x)>0$ for $x>0$, $f'(x)>\frac{f(x)}{x}$ for $x>0$. Is $f$ a convex function?

If we set $h(x)=xf'(x)-f(x)$, then $h(0)=0$, $h(x)>0$ for positive $x$ and $h'(x)=xf''(x)$. Suppose we can find smooth function $h$ satisfying the conditions above and that $h'$ is not always positive, then $f(x)=x(\int \frac{h(x)}{x^2}dx+C)$ is a counterexample. There are many choices of $h$, for example, $h(x)=1-\frac{cos x}{1+x}$.

The question is that this integral $\int_a^x \frac{h(t)}{t^2}dt$ may only be meanful for $a>0$. How can we construct $h$ such that when $a=0$ the improper integral exists? Or is there another way to construct counterexamples?

Remark: This is kind of similar to this question.

Answer 1

Let us consider a simple polynomial function $f(x) = ax^4+bx^3+cx^2$. Obviously we have $f(0)=0$, to satisfy the condition, we need \begin{equation} f'(x) = 4ax^3+3bx^2+2cx > \frac{f(x)}{x} = ax^3+bx^2+cx. \end{equation} Thus $3ax^2+2bx + c > 0$ will be held for any $x$, i.e., \begin{equation} a>0, \quad b^2 - 3ac < 0. \end{equation} To violate the convexity, we need some $x_0 > 0$, such that \begin{equation} f''(x_0) = 12a x_0^2+6b x_0 + 2c < 0. \end{equation} Thus what we need is \begin{equation} \left\{ \begin{array}{c} {a>0,} \\ {b^2 - 3ac < 0,} \\ {\exists x_0, 12ax_0^2 + 6bx_0 + 2c <0.} \end{array} \right. \end{equation}

Here we simply choose $x_0=1$, and we can easily find a feasible solution: \begin{equation} a=0.1, \quad b=-0.4, \quad c=0.55. \end{equation}

And I plot a simple figure in the following:

Answer 2

Observe that your condition $f'(x)>\frac{f(x)}{x}$ is equivalent to $$x\left(\frac{f(x)}{x}\right)'>0.$$Thus, for $x>0$, you ask if a strictly increasing function $g(x)\in C(0,\infty)$ of the form $g(x)=\frac{f(x)}{x}$ implies that $f(x)$ is convex.

This is obviously not true, since it is not hard to function which would violate this (e.g. see the polynomial @Ze-Nan Li answer).