equivalent definition of convex functions I got the following definition of a convex functions on $\mathbb R$:
Let $I \subseteq \mathbb R$ be an interval and $f \to \mathbb R$. Then $f$ is called convex, if for all $x_1,x_2 \in I$ and all $\lambda \in [0,1]$ holds that
$f((1-\lambda)x_1 + \lambda x_2) \le (1 - \lambda)f(x_1) + \lambda f(x_2)$.
I wonder why this is equivalent to the following:
For all $x_0 \in I$ there exists some number $c(x_0)$ such that $f(x) \ge f(x_0) + c(x - x_0), $ $ x \in I$.
A proof and or a geometrical interpretation would be helpful.
 A: *

*$S=\{(x,y):\ x\in I, y\geq f(x)\}$ is convex.

*Let $I \subseteq \mathbb R$ be an interval and $f \to \mathbb R$. Then $f$ is called convex, if for all $x_1,x_2 \in I$ and all $\lambda \in [0,1]$ holds that $f((1-\lambda)x_1 + \lambda x_2) \le (1 - \lambda)f(x_1) + \lambda f(x_2)$.

*For all $x_0 \in I$ there exists some number $c(x_0)$ such that $f(x) \ge f(x_0) + c(x - x_0), $ $ x \in I$.


If (1) then $(1-\lambda)(x_1,f(x_1))+\lambda(x_2,f(x_2)) = ((1-\lambda)x_1+\lambda x_2,(1 - \lambda)f(x_1) + \lambda f(x_2)) \in S$. This means (2).
If (2), $(x_1,y_1),(x_2,y_2)\in S$ and $\lambda\in[0,1]$. Then $(1-\lambda)y_1+\lambda y_2\geq (1-\lambda)f(x_1)+\lambda f(x_2)\geq f((1-\lambda)x_1+\lambda x_2)$. This means that $((1-\lambda)x_1+\lambda x_2, (1-\lambda)y_1+\lambda y_2)\in S$. This means that (1).
If (3) then (1). Because $S$ turns out to be equal to the intersection of all the half-planes $\{y\geq f(x_0)+c_{x_0}(x-x_0)\}$, which are convex and the rectangle $I\times\mathbb{R}$, which is also convex.
If (1) then proving (3) is this kind of standard business.
A: The claim is WRONG.  
Take $I = [-1,1]$, and   $f(x) =  - \sqrt{1-x^2}$, then clearly $f$ is convex.But for $x_0 = 1 $ there exist No $c \in R$ such that  $$f(x) \ge f(x_0) + c(x - x_0)$$
for all $x \in I.$
The claim is true provided that  $x_0$ is chosen in interior point of $I$, i.e.,  $ x_0 \in \text{int }I = (a,b)$. Then 
one possible argument (but not best) to prove the claim can be an argument like  in @cooper.hat 's answer by following correction 
$$   R(a,x_0)  \leq c_- =   \limsup_{h \uparrow 0}R(x_0,x_0+h) \le \liminf_{h \downarrow 0}R(x_0,x_0+h) = c_+  \leq R(x_0, b ) $$
A: Here is another approach:
Suppose $f$ is convex, $x \neq y$, and $R(x,y) = {f(x)-f(y) \over x-y }$ (see https://en.wikipedia.org/wiki/Convex_function#Properties) .
It is straightforward to show that if $ x<y<z$, then
$R(x,y) \le R(y,z)$. This can be seen graphically by noticing that the slopes of the secants joining points on the graph of $f$ at $x,y,z$ are non decreasing.
Pick $x_0$, then the above shows that
$c_- = \limsup_{h \uparrow 0}R(x_0,x_0+h) \le \liminf_{h \downarrow 0}R(x_0,x_0+h) = c_+$.
Choose $c_0 \in [c_-,c_+]$, then
if $x>x_0$ we have $R(x,x_0) \ge c_0$ and if $x<x_0$ we have
$R(x,x_0) \le c_0$. Combining shows that
$f(x) \ge f(x_0) + c(x-x_0)$ for all $x$.
For the other direction, suppose $\lambda \in (0,1)$ and
$x_0 = (1-\lambda) x_1 + \lambda x_2$. Let $c_0$ be the constant $c(x_0)$, then we have
$f(x_1)-f(x_0) \ge c_0(x_1-x_0)$ and $f(x_2)-f(x_0) \ge c_0(x_2-x_0)$.
Hence $(1-\lambda) (f(x_1)-f(x_0)) + \lambda (f(x_2)-f(x_0)) \ge (1-\lambda )c_0(x_1-x_0) + \lambda c_0(x_2-x_0)$ which reduces to
$(1-\lambda) f(x_1) + \lambda f(x_2) \ge f(x_0)$.
