Convex set: Can a point in a convex set be generated by a convex combination of points outside the set.

Question

I am stuck with the following problem: Consider a set $S$ which is convex. Let points $\pi_1, \pi_2 \notin S$. Can a convex combination of $\pi_1$ and $\pi_2$ $\bar{\pi} = \lambda \pi_1 + (1-\lambda) \pi_2$ be such that $\bar{\pi} \in S$. It does not look possible. Is there any formal proof for this statement?

Answer 1

Yes, of course. Take the $[-1,1]\times [-1,1]$ square. Consider $\pi_1 = (0,2)$ and $\pi_2 = (0,-2)$. Their convex combination is in the square for any $\lambda$ in $[1/4,3/4]$.

Answer 2

Others have already given you concrete examples. Here is a geometric interpretation that will make clear why your claim is false.

A convex combination of two points $\pi_1$ and $\pi_2$ is a point on the line segment $\overline{\pi_1 \pi_2}$. Your question becomes the following: can you pick two points $\pi_1,\pi_2 \notin S$ such that the line segment $\overline{\pi_1 \pi_2}$ intersects $S$?