Let be $f$ a convex function defined on an open set. We know from theory that $f'_{+},f'_{-}$ both exist not decreasing.
I want to prove or in case it is false to refute this: The right derivative, $f'_{+}$, of a convex function $f$ is continuous $\iff$ $f$ is differentiable.
I'm stuck with $[\Rightarrow]$. But also for $[\Leftarrow]$ I don't have a formal proof.
Any help,hint or solution would be appreciated. (on both arrows)
Best Answer
Suppose $f$ is convex and differentiable. Let $g(x)=f'_{+}(x)=f'(x)$. It is well known that $f(x)=f(0)+\int_0^{x} g(t)\, dt$. If possible suppose the increasing function $g$ has a jump discontinuity at $x_0$. Consider $\frac {f(x_0+\epsilon)+f(x_0-\epsilon)-2f(x_0)} {2\epsilon}$. This tends to $=0$ but $\frac {f(x_0+\epsilon)-f(x_0-\epsilon)-2f(x_0)} {2\epsilon} =\frac 1 {2\epsilon} \int_{x_0} ^{x_0+\epsilon} g(t) \, dt-\frac 1 {2\epsilon} \int_{x_0-\epsilon} ^{x_0} g(t) \, dt\geq c/2$ where $c$ is the jump $g(x_0+)-g(x_0-)$. This proves that $g$ is necessarily continuous. The converse part is true even without convexity and it is proved in the link in my comment above.
Simpler proof using IVP: $f'$ has IVP because all derivatives do. Since it is also increasing it cannot have any jumps. (Just look at the graph to see why this is true).