Characterization of convex functions From now on $I$ will always denote an open interval of $\mathbb{R}$
Let $f \; : \; I \to \mathbb{R}$ be a function, $f$ is said to be convex iff
$$f((1-\gamma)x + \gamma y) \leq (1 - \gamma)f(x) + \gamma f(y) \hspace{0.4cm} \forall x,y \in I \; , \; \forall \gamma \in ]0,1[$$
A well known criterion for convexity is the following
First criterion of convexity
If $f \; I \to \mathbb{R}$ is a function such that
$$(A1)\;\;f'(x) \text{ exists } \;\; \forall x \in I$$
$$(A2)\;\;f'(x) \leq f'(y) \hspace{0.3cm} \forall x,y \in I \; : \; x \leq y$$
Then $f$ is convex
The criterion gives a sufficient condition which is not necessary, the example $f(x) = |x|$ shows that. I tried to generalize the criterion to find a condition which is sufficient and necessary for the convexity, this is what I managed to do.
Let $f$ be a convex function, it can be proved that the function
$$\begin{split} \Phi \; : \; (I \times I) \setminus \Delta &\to \mathbb{R} \\
(x,y) &\mapsto \Phi(x,y) = \frac{f(x) - f(y)}{x - y} \end{split}$$
where $(I \times I) \setminus \Delta = \{ (x,y) \in I \times I \; : \; x \neq y\}$
is increasing both in $x$ and in $y$.
From this property it's pretty easy to show that
$$(B1) \; f'^-(x) \text{ and } f'^+(x) \hspace{0.4cm} \text{ both exists and are finite} \hspace{0.3cm} \forall x \in I$$
$$(B2) \; f'^-(x) \leq f'^+(x) \hspace{0.5cm} \forall x \in I$$
$$(B3) \; f'^+(x) \leq f'^-(y) \hspace{0.5cm} \forall x,y \in I \; : \; x < y$$
where $f'^-(x) := \lim_{y \to x^-}{\frac{f(x) - f(y)}{x-y}}$ , $f'^+(x) := \lim_{y \to x^+}{\frac{f(x) - f(y)}{x-y}}$
now let $D := \{ x \in I \; \text{ such that $f'(x)$ doesn't exist} \}$, in other words $D$ is the set of the point of non differentiability of $f$
from the properties stated above it's easy to see that
$D \subseteq \{ \text{ discontinuity points of $f'^-$} \}$
and because $f'^-$ is increasing ( what I mean with "increasing" is what some people mean with "non-decreasing" ) the discontinuity points of $f'^-$ are at most countable, therefore the point of non derivability of $f$ are at most countable, therefore $f$ is almost everywhere differentiable
So although $f$ isn't necessarily everywhere differentiable it is almost everywhere differentiable
Generalized first criterion of convexity
Let $f \; : \; I \to \mathbb{R}$ be a function, if $f$ is convex if then (B1),(B2) and (B3) holds. furthermore the non differentiable points of $f$ are at most countable.
My question is, are the condition $(B1),(B2)$ and $(B3)$ also sufficient for the convexity of $f$ ?
If so, how can I prove it? If they're not, what is a counterexample?
There is also another criterion of convexity
Second criterion of convexity if $f \; : \; I \to \mathbb{R}$ is a function such that
$$(C1) f''(x) \;\; \text{ exists } \forall x \in I$$
$$(C2) f''(x) \geq 0 \;\; \forall x \in I$$
then $f$ is convex
Can the Second criterion of convexity be generalized in a similar way to the first? Is there a result about the maximum cardinality of the points where $f$ isn't twice differentabile ? (similar to the result "the points where $f$ isn't differentiable are at most countable")
 A: I managed to prove the condition stated by River Li. This proves that the Generalized first criterion of convexity $\implies$ convexity
to prove I use this generalized version of the Fermat's stationary point Theorem
Generalized Fermat's stationary point theorem
Let $g \; [0,1] \; \to \mathbb{R}$ be a function, let $\gamma_0 \in ]0,1[$ then
if $\gamma_0$ is a local minimum point and if both $g'^-(\gamma_0)$ and $g'^+(\gamma_0)$ exists, then $g'^-(\gamma_0) \leq 0 \leq g'^+(\gamma_0)$
if $\gamma_0$ is a local maximum point and if both $g'^-(\gamma_0)$ and $g'^+(\gamma_0)$ exists, then $g'^+(\gamma_0) \leq 0 \leq g'^-(\gamma_0)$
The proof of the Generalized Fermat's stationary point theorem is not very different from the proof of the Standard Fermat's stationary point Theorem
Now to prove the statement I first prove a very useful Lemma
Lemma
Let $f \; : \; I \to \mathbb{R}$ be a function such that (B1),(B2) and (B3) holds, then
$$f'^+(x) \leq \frac{f(y) - f(x)}{y - x} \leq f'^-(y)$$
forall $x < y$ in $I$
Proof of the Lemma
Let $x,y \in I$ such that $x < y$
Let $g \; : \; [0,1] \to \mathbb{R}$ defined in this way
$$g(\gamma) := f(x) + \gamma(f(y) - f(x)) - f((1-\gamma)x + \gamma y)$$
I know that $g(0) = g(1) = 0$, also $g$ is continuos because $f$ is continuos ($f$ is continuos because the existence of $f'^+$ and $f'^-$ implies the continuity).
It's quite easy to show that for all $\gamma \in ]0,1[$ $g'^+(\gamma$) and $g'^-(\gamma)$ both exists and
$$g'^-(\gamma) = f(y) - f(x) - f'^-((1-\gamma)x + \gamma y)(y - x)$$
$$g'^+(\gamma) = f(y) - f(x) - f'^+((1-\gamma)x + \gamma y)(y - x)$$
It's clear $g$ either has an internal maximum point or an internal minimum point.
Maximum point case let's assume that $g$ has an internal maximum point, let's call this point $\gamma_0$, then thanks to the Generalized Fermat's stationary point theorem I know that
$$g'^+(\gamma_0) \leq 0 \leq g'^-(\gamma_0)$$
meaning
$$f(y) - f(x) - f'^+((1-\gamma_0)x + \gamma_0 y)(y - x) \leq 0$$
$$0 \leq f(y) - f(x) - f'^-((1-\gamma_0)x + \gamma_0 y)(y - x)$$
manipulating these inequalities gives
$$f'^-((1-\gamma_0)x + \gamma_0 y) \leq \frac{f(y) - f(x)}{y - x} \leq f'^+((1-\gamma_0)x + \gamma_0 y)$$
and using (B2) I get
$$f'^+(x) \leq  f'^-((1-\gamma_0)x + \gamma_0 y) \leq \frac{f(y) - f(x)}{y - x} \leq f'^+((1-\gamma_0)x + \gamma_0 y)\leq f'^-(y)$$
So in particular
$$\label{disuguaglianza} f'^+(x) \leq \frac{f(y) - f(x)}{y - x} \leq f'^-(y)$$
Minimum point case let's assume that $g$ has an internal minimum point, let's call this point $\gamma_0$, then thanks to the Generalized Fermat's stationary point theorem I know that
$$g'^-(\gamma_0) \leq 0 \leq g'^+(\gamma_0)$$
meaning
$$f(y) - f(x) - f'^-((1-\gamma_0)x + \gamma_0 y)(y - x) \leq 0$$
$$0 \leq f(y) - f(x) - f'^+((1-\gamma_0)x + \gamma_0 y)(y - x)$$
manipulating these inequalities gives
$$f'^+((1-\gamma_0)x + \gamma_0 y) \leq \frac{f(y) - f(x)}{y - x} \leq f'^-((1-\gamma_0)x + \gamma_0 y)$$
but because of $(B2)$ I have that
$$f'^+((1-\gamma_0)x + \gamma_0 y) = \frac{f(y) - f(x)}{y - x} = f'^-((1-\gamma_0)x + \gamma_0 y)$$
and $f$ is differentiable in $(1-\gamma_0)x + \gamma_0y$, in any case using (B3) i have once again
$$f'^+(x) \leq \frac{f(y) - f(x)}{y - x} \leq f'^-(y)$$
which completes the proof of the Lemma
Now let's dive into the real proof.
Let $f \; : \; I \to \mathbb{R}$ be a function such that $(B1),(B2)$ and $(B3)$ holds, then $f$ is convex
Proof*
As I said I will prove the condition stated by River Li, i.e.
$$\frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(b)}{c - b}$$
for all $a < b < c$ in $I$
By using the lemma first with $x = a, y =b$ and then with $x = b , y = c$ and using (B2) and (B3) I get
$$\frac{f(b) - f(a)}{b - a} \leq f'^-(b) \leq f'^+(b) \leq \frac{f(c) - f(b)}{c - b}$$
which proves
$$\frac{f(b) - f(a)}{b - a} \leq  \frac{f(c) - f(b)}{c - b}$$
which proves the convexity of $f$.
Therefore the following the Generalized first criterion of convexity is also a sufficient condition for the convexity, therefore the condition (B1),(B2) and (B3) are a complete characterization of the convexity.
Another question still remains, is there a Generalized second criterion of convexity ?
