Whether this set is convex or not?

Question

Consider the closed disk of radius 1 at the origin. Let it be called set S. Now is the set $S'=S\setminus \{(1,0),(0,1)\}$ convex? I feel like it is convex but I am not sure how to prove. It basically boils down to saying than (1,0) can never be written as $\lambda x_1 + (1-\lambda)x_2$ and neither do (0,1) can be written as this for any $x_1, x_2 \in S'$

Answer 1

The key here is that the points you're removing are extremal. So I would suggest the following approach:

1. Show that the two points you're removing are extremal
2. Show that a convex set $K$ is still convex if you remove an extremal point $x_0$

For (1): A point is $x$ is extremal in a set $K$ if whenever we have $x + y \in K$ and $x - y \in K$, it follows that $y = 0$. That any point on the boundary is extremal for $S$ follows like this: Assume $x \in \partial S$ and $x + y$, $x - y \in S$. Then we have $\|x\| = 1$ and $\|x + y\|$, $\|x -y\| \le 1$. Now observe \begin{align*} 4&= (2x,2x)\\ &= ([x+y] + [x-y],[x+y] + [x-y])\\ &= (x+y,x+y) + 2(x+y,x-y) + (x-y,x-y)\\ &= \|x+y\|^2 + \|x-y\|^2 + 2(x+y,x-y)\\ &\le 2 + 2(x+y,x-y) \end{align*} In other words, we have $1 \le (x+y,x-y)$. By the Cauchy-Schwarz inequality, we also have $(x+y,x-y) \le \|x+y\|\|x-y\| \le 1$ and the strict version of Cauchy-Schwarz tells us that this can only happen if $x+y$ and $x-y$ are identical (which means $y = 0$).

For (2): For any two distinct points in $K$, the connecting line segment cannot contain $x_0$ since that point is extremal. Thus it is still contained in $K \setminus \{ x_0 \}$.