convex function right left derivatives 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!
 A: 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.
A: Somewhat schematic, and we drop some details. Complete proof can be found in, e.g., Jan van Tiel, "Convex Analysis: an Introductory Text". 
For simplicity, assume that we consider a convex function $f:\mathbb{R}\rightarrow\mathbb{R}$, i.e., everywhere defined. It is an easy consequence of the definition of convexity (a simple rearrangement), that if you inscribe a triangle $ABC$ in the graph of $f$, with $x_A<x_B<x_C$ (draw this figure), then the slopes of the line segments are ordered as $AB<AC<BC$.
As in the formulation of the  question, let $x_A = x_0$, $x_B = x_0 + \epsilon$, $x_C = x_0 + \epsilon'$, where $\epsilon' > \epsilon$. Note that the slope of $AB$ is simply $(f(x_0+\epsilon)-f(x_0))/\epsilon$ and the slope of $AC$ is $(f(x_0+\epsilon')-f(x_0))/\epsilon'$. It follows that the slope is nondecreasing in $\epsilon$.
To show that the right derivative always exists, we shift our attention to the case $x_A < x_0 = x_B < x_0+\epsilon = x_C$. As $\epsilon\rightarrow 0$, the slope of $BC$ is becomes the right derivative -- if it exists. But the slope of $BC$ is always bounded below by the slope $AB$. Moreover, it is nonincreasing as $\epsilon\rightarrow 0$. It follows that the limit exists.
Similarly, one can show the existence of the left derivative. That the left derivative is not greater than the right derivative follows from simple triangle slope considerations.
A: Hint: Use $f(x_0+\epsilon) = f(\tfrac{\epsilon}{t}(x_0+t) + (1-\tfrac{\epsilon}{t})x_0)$, note that there is a convex combination inside for $t\geq \epsilon$, and rearrange to obtain that
$$ 
\frac{f(x_0+\epsilon) - f(x_0)}{\epsilon}\leq \frac{f(x_0+t) - f(x_0)}{t}.$$
