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

A: 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).
