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.
Best Answer
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
$$ R(a,x_0) \leq c_- = \limsup_{h \uparrow 0}R(x_0,x_0+h) \le \liminf_{h \downarrow 0}R(x_0,x_0+h) = c_+ \leq R(x_0, b ) $$