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Feb 2, 2022 at 23:42 comment added afshi7n Thanks! Yes, I understand that. I’m saying your argument seems to be generating a contradiction even for r=2 (as it doesn’t use the fact that r>2). But that would mean the argument is flawed—as you said the statement is trivially correct for r=2
Feb 1, 2022 at 22:53 comment added Robert Israel If $r=2$ there's nothing to prove, as $p$ is already a convex combination of two extreme points of $K$.
Jan 31, 2022 at 22:24 comment added afshi7n where is it used that $r>2$? It seems like the same argument works for $r=2$..
Jan 16, 2020 at 17:31 history edited Robert Israel CC BY-SA 4.0
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Jan 16, 2020 at 17:07 history edited Robert Israel CC BY-SA 4.0
added 376 characters in body
Jan 16, 2020 at 17:01 history answered Robert Israel CC BY-SA 4.0