Let $x,y\in \mathbb R^{+}$. I have some doubts about a question on the functions thery, in general. If $f$ is a concave (and continue) function and $$\lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$$ can I infer that there exist $M>0$ such that $\forall x,y>M$ it has: $f(x+y)\leq f(x)+f(y)$ and $f(\alpha x)<\alpha f(x)$?
However, for such classes of functions we have that $f(x+y)\leq f(x)+f(y)$, $\forall x,y>M$?
I would say yes (I think that $f$ should be continue almost everywhere). This because $f(x+y) = f(x)+f(y)$ it's a linearity property, from this you can prove also that $f(\alpha x) = \alpha f(x)$ (I will look for an analogous question and I will report in the comments). The limit condition you gave tells that there is a point M such that $\vert f(x) \vert < x\, , \forall x > M$ but since the $f$ is concave absolute value is not necessary.
Thus $f(x)$ is a monotonic increasing function, concave, thus it must growth less than a linear function. This means precisely $f(x+y) \leq f(x) + f(y)$.
Finally, if nothing changes in the function (it remains concave and has no discontinuities) then you can say $\exists M$ such that $f(x+y) \leq f(x) + f(y) \forall x,y > M$.
