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My problem is as follows: Let f(x,y) be a general concave function (i.e. any concave function), and let S be points (x,y) such that f(x,y) greater than or equal to zero.

I understand that this means S is the set of points on or below f(x,y) and >0, and I can show how it is convex in a sketch/graph. I just don't understand how you show it mathematically in a sufficient way using the definition of a convex set.

Any help would be greatly appreciated.

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If $\xi_0=(x_0,y_0)$ and $\xi_1=(x_1,y_1)$ are two points in $S$ then we must have $f(\xi_0)\geq 0$ and $f(\xi_1)\geq 0$. Let $0\leq t\leq 1$ and consider $\xi_t=(1-t) \xi_0+t\xi_1$. By concavity of $f$ we must have: $$ f(\xi_t) \geq (1-t) f(\xi_0) + t f(\xi_1) \geq 0$$ so also $\xi_t\in S$.

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    Thank you! Do you think you could explain the inequalities, with words, if it's not too much of a hassle? It would help my understanding, and I would greatly appreciate it. – 2017-02-01
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    One definition of concavity is that given two points $A$ and $B$ on the graph of a function, the graph should be above the segment between the two points. In the present case $A=(x_0,y_0,f(x_0,y_0))$ and $B=(x_1,y_1,f(x_1,y_1))$ and the first inequality express what I just said above the point $(1-t)\xi_0+t\xi_1$ in the $(x,y)$-plane. – 2017-02-01