How to prove that a convex function $f(x)$ with $f(0) = 0$ satisfes $f'(x)x \ge f(x)$? How to prove that a convex function $f(x)$ with $f(0) = 0$ satisfes $f'(x)x > f(x)$?
I am not sure how to use the property of convexity to solve this.
 A: Straight-forward approach:
We are interested in the expression $xf'(x) > f(x) \Leftrightarrow xf'(x) - f(x) > 0$.
In such cases, in Calculus, a good way of working our way is naming a function $F(x) = xf'(x) - f(x)$ be the expression that we are interested in and studying that.
Note that $F'(x) = xf''(x) + f'(x) - f'(x) = xf''(x) > 0 $ for $ x>0$. But, that means that $F(x)$ is an increasing function for $x > 0$. Can you conclude now ?
Alternative via MVT:
If $f$ is convex, then $f''(x) > 0$. This means that $f'(x)$ is an increasing function.
Apply the Mean Value Theorem for $f$ now, over an interval $[0,x]$. Then, there exists an element $\xi \in (0,x)$ such that :
$$f'(\xi) = \frac{f(x) - f(0)}{x - 0} \Leftrightarrow f'(\xi) = \frac{f(x)}{x}$$
But, as we said the function $f'(x)$ is increasing, thus for any $x > \xi$ it is $f'(x) > f'(\xi)$. 
That means that $f'(x) > \frac{f(x)}{x} \Leftrightarrow f'(x)x > f(x)$ for $x >0$.
Note that when $x=0$ the strict inequality does not hold, thus : $xf'(x) \geq f(x)$.
A: This proof uses only the definition of a convex function and existence of $f'(x)$. 
If $h>0$ then $x=(t)(0)+(1-t)(x+h)$ where $t =\frac h {x+h}$. Hence $f(x) \leq tf(0)+(1-t) f(x+h)$. I will leave it to you to re-write this inequality as $\frac {f(x+h) -f(x)} h  \geq \frac {f(x)-f(0)} x$. Letting $h \to 0$ gives $f'(x) \geq \frac {f(x)} x$ for  $x>0$. The proof for $x<0$ is similar.
Taking $f\equiv 0$ we see that strict inequality need not hold. 
A: You can easily construct a convex function that is not differentiable and attains $$f(0) = 0$$
The easiest example is probably $f(x) = |x|$ which has derivative everywhere except at $x=0$.
Now to take this one step further, just split line segments (for example already approximating some convex function) into smaller line segments. The derivative will have a jump at every knot between these lines. Now do that systematically, perhaps in a dyadic fashion similar to how discrete wavelets are constructed. You will get a function with as many discontinuities in the derivative as you want.
So your question needs to be attached with the big IF it is differentiable.
