TITLE: convex function right left derivatives
QUESTION [14 upvotes]: Hello how to show the followings for a convex function $f(x)$:
Let $f(x_0) \in R$ 
then 
$\frac{f(x_0 + \epsilon) - f(x_0)}{\epsilon}$ 
is nondecreasing in $\epsilon$
Similarly how to show the left and right derivatives of $f(x)$ always exists 
and right derivative is larger than the left derivative.
Thanks a lot!

REPLY [8 votes]: You can prove it using next:
Theorem 1. Function $\varphi:(a,b)\to \mathbb{R}$ is convex iff for every $a<x<y<z<b$ it follows $$\left| \begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
\varphi(x) & \varphi(y) & \varphi(z) \end{array} \right|\ge 0$$ iff 
\begin{equation}\frac{\varphi(y)-\varphi(x)}{y-x} \le \frac{\varphi(z)-\varphi(x)}{z-x} \le \frac{\varphi(z)-\varphi(y)}{z-y} \quad\quad\quad(1)\end{equation}
Theorem 2. For convex function $\varphi:(a,b)\to \mathbb{R}$ and for every $x_0 \in (a,b)$ function $\Delta_{\varphi,x_0}(x)=\dfrac{\varphi(x)-\varphi(x_0)}{x-x_0}$ for $x \neq x_0$ is monotone nondecreasing function on $(a,x_0) \cup (x_0,b)$.
Hint (for Theorem 2): Using relation $(1)$ and setting $x_0 \in \{x,y,z\}$ it "easy" to show the statement of theorem.
Theorem 3. Every convex function $\varphi:(a,b)\to \mathbb{R}$ have  right-hand derivative on $(a,b)$ and $\varphi'_-\le \varphi'_+$ on $(a,b)$.
Proof. Let $a<x<y<z<b$. For monotone nondecreasing function $\Delta_{\varphi,y}$ exist left and right-hand limis, so from $(1)$ when $x\to y-$ and $z \to y+$ we get
$$\begin{align}\varphi'_-(y)&=\lim_{x\to y-} \frac{\varphi(x)-\varphi(y)}{x-y}\\
&=\sup_{a<x<y}\frac{\varphi(y)-\varphi(x)}{y-x}\\
 &\le \inf_{y<z<b}\frac{\varphi(z)-\varphi(y)}{z-y}\\
&=\lim_{z\to y+}\frac{\varphi(z)-\varphi(y)}{z-y}=\varphi'_+(y)\end{align}$$
And we are using here Theorem 2 for part $\lim=\sup$, etc.