The existence of continuous and concave/convex extension of discrete function Given a finite $A\subset R$ and a nondecreasing $f:A\rightarrow R$ such that if $c\geq \frac{a+b}{2}$ then $f(c)\geq \frac{f(a)+f(b)}{2}$, is there always exist a continous, nondecreasing, twice differentiable, and concave extension $g$ of $f$ such that $g_{|A}=f$? If the intuition is correct, what formal results can I cite?

You can't always extend to a differentiable function. Intuitively, if you plot the points $(a,f(a))$ in the plane and any three of them are collinear, then any concave (or convex) extension must be coincide with the line segments connecting them.
For example, suppose $\{(a,f(a)) \ | a \in A \} = \{(0,0),(1,1),(2,2),(3,2),(4,2)$}.
Then any concave extension must satisfy:
$$
g(x) = \begin{cases}
x & 0 \le x \le  2 \\
2 & 2 \le x \le 4
\end{cases}
$$
This is not differentiable at $x=2$.
So why is it true that a concave extension must satisfy the above? It's pretty clear if you draw a picture. For example, for $x \in (0,1)$, the concavity between the points $0,x,1$ implies $g(x) \ge x$, but concavity on the points $x,1,2$ implies $g(x) \le x$. So you must have $g(x)=x$ for all  $x \in (0,1)$. Likewise for the other segments, $g$ must be the affine function connecting the graph of $f$.
