Let us take a continuous concave function $f:[0,1]^n \rightarrow \mathbb{R}_+$. The convex hull of its graph is given by, $co(G(f))=\{(x,y)\in[0,1]^n \times \mathbb{R}_+:x=\sum\limits^n_{i=1}\alpha_ix_i,y=\sum\limits^n_{i=1}\alpha_if(x_i),\alpha_i \geq 0, \sum\limits^n_{i=1}\alpha_i=1.\}$ My question is, can any point $(x,y) \in co(G(f))$ be represented as $(x,y)=\alpha(x_1,f(x_1))+(1-\alpha)(x_2,f(x_2))$? I'm aware of the Caratheodory theorem which says we need at most $n$ points for the convex combination, which is why I used $n$ in the definition. The question is whether this can be done by exactly two.
Any point in the convex hull of the graph of a concave function can be represented as a convex combination of two points
convex-geometryconvex-hulls
Best Answer
Here is the counterexample: $$ f(x,y) = x(1-x) + y(1-y). $$ Then $f(x,y)\ge0$ for all $(x,y)\in [0,1]^2$, and $f$ is zero exactly at the four corners of $[0,1]^2$.
Then $[0,1]^2 \times \{0\} \subset conv(G_f)$, but $(0.5, 0.25, 0)$ is not a convex combination of two points of $G_f$.