# Convexity and Proof of one sided Derivative

Working on some real analysis work, I've been able to show that for a function $f$, which is convex on $[a,b]$, for $a\leq x_1< x_2< x_3\leq b$: $$\frac{f(x_2)-f(x_1)}{x_2-x_1} \leq \frac{f(x_3)-f(x_2)}{x_3-x_2}$$ and for $h>0$, for some $x_0 \in [a,b]$: \begin{equation} \frac{f(x_0+h)-f(x_0)}{h} \end{equation} is non-decreasing.

I'm stuck needing to show that:

1). $\exists \; c_0\in\mathbb{R}$ such that $\frac{f(x_0+h)-f(x_0)}{h} > c_0$ for all $h>0$. Intuition tells me $c_0 = f_+'(x_0)$, but i'm having issues with the proof.

2) $\lim\limits_{h\to 0^+} \frac{f(x_0+h)-f(x_0)}{h}= f_+'(x_0)$

Thanks for any help

Consider some $k > 0$ and any $h > 0$ such that

$$a < x_0-k < x_0 < x_0 + h < b.$$

Using your inequality for a convex function, we have

$$\frac{f(x_0) - f(x_0-k)}{k}= \frac{f(x_0) - f(x_0-k)}{x_0 - (x_0-k)} \leqslant \frac{f(x_0+h) - f(x_0)}{x_0+h - x_0} = \frac{f(x_0+h) - f(x_0)}{h}.$$

Thus, $[f(x_0 + h) - f(x_0)]/h$ is bounded below by $[f(x_0) - f(x_0 -k)]/k.$ Hence, there exists a greatest lower bound $c_0 = \inf\{[f(x_0 + h) - f(x_0)]/h: x_0 < x_0 + h \leqslant b\}$

As you have already shown that $[f(x_0 + h) - f(x_0)]/h$ is non-increasing, we have existence of the limit

$$\lim_{h \to 0+} \frac{f(x_0+h) - f(x_0)}{h} = c_0 := f'_+(x_0).$$