Q:

The right derivative of a convex function $f$ is right continuos $\iff$ $f$ is differentiable

Let be $f$ a convex function defined on an open set. We know from theory that $f'_{+},f'_{-}$ both exist not decreasing.
Claim : The right derivative of a convex function $f$ is right continuos $\iff$ $f$ is differentiable.
I Think $[\Leftarrow]$ following directly from definition because if $f'$ exists, both left and right limits exist finite and has to be coincident $f'_{+} = f'_{-} = l \in \mathbb{R}$ so $f'_{+}$ is continuos.
I'm stuck with $[\Rightarrow]$. I was unable to appoach the problem by any angle. 
Any help,hint or solution would be appreciated.

A:

This is not true. Let $f(x)=0$ for $x<0$ and $f(x)=x$ for $x \geq 0$. Then $f'(x+)$ is right continuous but $f$ is  not differentiable at $0$. 

