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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 $x_1, x_2 \in S'$

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Do you mean closed disk? –  1015 Feb 20 '13 at 4:52
    
@julien yes S is a closed disk. –  Satvik Feb 20 '13 at 4:55
    
The keyword is strict convexity. See here: math.stackexchange.com/questions/232276/… –  1015 Feb 20 '13 at 5:06
    
@julien I am not sure about the use of what you said about space being strictly convex. I can prove that if the above happens then both $x_1$ and $x_2$ must lie on the boundary of the disk. –  Satvik Feb 20 '13 at 5:08
    
You can prove that if $x_1,x_2$ are two distinct points in the closed disk, then any interior points of the segment $[x_1,x_2]$ has norm stricly less than the max of $\|x_1\|$ and $\|x_2\|$. No matter what, you will have to use this at some point. This is called the strict convexity of the unit ball. And that's basically what you are trying to prove. So your answer is in the link I provided above. –  1015 Feb 20 '13 at 5:18

1 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 $K \setminus \{ x_0 \}$.

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