# A generalization of strict convexity

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?

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).
• 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? – erz Apr 1 '17 at 2:59
• 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. – August Cleaner Apr 1 '17 at 3:41