a property for points in convex hull

Question

Let $A\subset\mathbb{R}^2$ and $b=(b_1,b_2)$ is in the convex hull of $A$. Prove that for any $x=(x_1,x_2)\in\mathbb{R}^2$, there exists $a=(a_1,a_2)\in A$ such that $a_1x_1+a_2x_2\le b_1x_1+b_2x_2$.

I tried to use dot product to reformulate the problem and here is the beginning of my proof:

Assume the statement is false. Then there is an $x$ such that $(a-b)\cdot x>0$ for all $a\in A$.

But I can't find a contradiction from here. Can you finish my proof or provide another proof? I also hope that there is a geometric proof.

Answer 1

By Carathéodory's theorem, $b$ is in a triangle with vertices in $A$.

The linear functional $(u_1,u_2) \mapsto u_1x_1+u_2x_2$ is continuous and so attains a global minimum on the boundary of the triangle. Since the level curves are straight lines, the minimum must be attained at a vertex of the triangle, which is in $A$.

To get a clear picture, you may assume that the functional is $(u_1,u_2) \mapsto u_2$ after applying a rotation to the plane. Then the claim is clear, because one of the vertices must be at the lowest possible height.

Answer 2

More generally, if $T: X \to Y$ is a linear transformation of vector spaces and $C$ is the convex hull of $A \subseteq X$, then $T(C)$ is the convex hull of $T(A)$. In this case, you're taking the linear transformation of $\mathbb R^2$ given by $(s,t) \to x_1 s + x_2 t$.