I am looking for a reference or a result about the extension of locally convex increasing functions in a neighborhood of $0$ by a strictly convex and increasing functions in $(0, \infty)$.
A function $f : \mathbb{R} → \mathbb{R}$ is said to be locally convex in a neighborhood of $0$ if and only if, $\exists \, \beta > 0$ such that function $g$ is convex where $g=\left\{(x, f(x)): -\beta \leq x \leq +\beta \, \right\}$
Let $R \subseteq \mathbb{R}$, a function $f : R\to \mathbb{R}$ is said to be convex if $\forall x, y, z\in \mathbb{R}$, such that $x < y < z$, then
$$f(y) < (y - x) \dfrac{(f(z) - f(x))}{z - x}.$$
This is impossible in general. Consider $$ f(x) = \max( - \sqrt{1-x}, - \sqrt{1+x}). $$ It is convex on $[-1,1]$, has infinite slope at $x=\pm 1$, so cannot be extended to a convex function $f:\mathbb R \to \mathbb R$.