1
$\begingroup$

Consider the following properties of a Banach space:

the intersection of any support hyperplane with the unit sphere is

(S) a singleton (this is the strict convexity);

(SF) finite-dimensional set;

(SC) compact in the norm topology.

It is easy to see that these properties are equivalent to the fact that any closed convex subset of a unit sphere is singleton/finite-dimensional/ compact.

Q1: Are (SF) and (SC) different?

Q2: Were these conditions considered in the literature? Do they have names?

Q3: Are there any necessary or sufficient conditions for them? In particular, are there any dual/predual conditions?

$\endgroup$
2
$\begingroup$

I think that one can answer Question 1 in the positive by studying the following construction: consider $X=\ell_2\oplus \mathbb{R}$. Consider in $\ell_2$ an infinitely dimensional compact ellipsoid $E$ centered at $0$ with all axes of length $<1$. Denote the unit ball of $\ell_2$ by $B$. Consider in $X$ the norm whose unit ball is the closure of the convex hull of the union of three sets: $B\oplus \{0\}$, $E\oplus \{-1\}$, and $E\oplus \{1\}$. I would expect this to be (SC), but it is obviously not (SF).

As for literature, you can check references listed in the book by Day, Normed Linear Spaces, book Deville-Godefroy-Zizler, and the survey of Godefroy in the "Handbook of the Geometry of Banach spaces".

$\endgroup$
  • $\begingroup$ What is an infinitely dimensional compact ellipsoid? When I was thinking about this problem, I came up with a similar example (taking $E$ to be just some infinitely dimensional symmetric compact), but then failed to understand what the "faces" of the obtained "frustum" are. Could you please elaborate? $\endgroup$ – erz Apr 1 '17 at 2:59
  • $\begingroup$ I meant the following: the image of the unit ball of $\ell_2$ under the action of the operator of coordinatewise multiplication onto a positive sequence convergent to $0$. After that I would try to show that all sections of the obtained unit ball which are parallel to $\ell_2$ are strictly convex, and thus all faces, except the top and bottom are one-dimensional. $\endgroup$ – August Cleaner Apr 1 '17 at 3:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.