Right derivative increasing implies convex I want to prove the following: If we have a function $f$ (not necessarily differentiable), but with right and left derivative $$f'_{\pm}(x) \equiv\lim_{y \rightarrow x^{\pm}} \frac{f(y)-f(x)}{y-x}$$ existing at all $x$ in $\mathbb{R}$ with $f'_+(x) \leq f'_+(y)$ $\forall x \leq y$, then $f$ is convex.
This is easily proven if $f$ is differentiable using the mean value theorem, but nothing like the same strategy (as far as I can see) works for this example.  Any help would be massively appreciated!
 A: One can show that
$$
\frac{f(y)-f(x)}{y-x} \le f_+'(y) \le \frac{f(z)-f(y)}{z-y}
$$
for $x < y < z$, which implies that $f$ is convex.
For the right inequality, fix $y$ and consider the function
$$
 h(z) = f(z) - f(y) - f_+'(y) (z-y)
$$
for $z \ge y$. Then $h$ has a right derivative with
$$
 h'_+(z) = f'_+(z) - f'_+(y) \ge 0 \, .
$$
It follows that $h$ is non-decreasing, compare $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing. So $h(z) \ge h(y) = 0$, and that implies the desired estimate.
The proof of the left inequality works similarly.
A: Using mean value inequality
A version of the mean value inequality is:
Let $f : [a, b ] \to \mathbb R$ be a continuous, right (or left) differentiable map with $m \le f_+^\prime(x) \le M$ for all $x \in [a,b]$. Then
$$m(y-x) \le f(y) - f(x) \le M(y -x)$$ for all $x,y \in [a,b]$.
You can see here for proof. See third approach where you'll realize that the proof can be adapted without major change for right differentiable maps only.
Based on this version of the mean value inequality, the result you're looking for follows in a straightforward way. And by the way, the mean value inequality is interesting by itself!
