Talk:Polar body

"If $K$ is a convex set containing the zero element in its interior then $K^\circ$ ... is again a convex neighbourhood of the origin." — Really? Even if $K$ is the whole space $V$? Boris Tsirelson (talk) 22:59, 23 October 2017 (CEST)

...bounded...bounded... Thank you. Richard Pinch (talk) 08:29, 24 October 2017 (CEST)

Another doubt. Indeed, $K^\circ$ is closed and bounded; but "compact"? Do you implicitly assume finite dimension? In fact, if $V$ is complete, then $K^\circ$ is weakly compact. Boris Tsirelson (talk) 10:01, 24 October 2017 (CEST)

Yes, I do. Richard Pinch (talk) 19:13, 26 October 2017 (CEST)

Hm. So, what now? Will you make this assumption explicit? Boris Tsirelson (talk) 20:59, 26 October 2017 (CEST)

Do you have any other comments before I do that? Richard Pinch (talk) 23:00, 26 October 2017 (CEST)

Well, now I've read it to the end. Here are some comments. The support function is defined (in the linked article), but the distance function is not (unless we treat "is given by" as "is defined by", that is define the distance function as just the support function of the polar body). "Given a distance function..., the corresponding closed convex set..." — is it clear enough that this $X$ is the closure of the convex hull of the initial $X$? Also, is it clear enough that the polar body of the polar body of $K$ is the closure of $K$? For now the duality is not emphasized, and the presentation does not look symmetric w.r.t. the (compact convex) body and the dual body. Boris Tsirelson (talk) 23:30, 26 October 2017 (CEST)

Oops, I forgot one more remark. In order to use (here) the definition of the support function (given there) we need first to identify the dual space to $V$ with $V$ itself (according to the given inner product). Boris Tsirelson (talk) 23:53, 26 October 2017 (CEST)