# An open problem in convex geometry

Is it possible to find four norms $\| \cdot\|_k$ $( 1 \leq k \leq 4)$ on the plane such that a three-dimensional normed space containing four subspaces isometric to these normed planes does not exist?

An equivalent statement:

Is it possible to find four $0$-symmetric convex bodies in the plane such that a three-dimensional $0$-symmetric convex body containing four central sections linearly equivalent to these four bodies does not exist?

This seems to be an old problem, but I'm not sure of its origin. The earliest reference I've found is

Rolewicz, S., Plane sections of centrally symmetric convex bodies. Israel J. Math. 4 1966 135–138.

Does anyone know the status of this problem? Some earlier references?

Just for completeness, let me add that Bessaga proved (C. Bessaga,A note on universal Banach spaces of a finite dimension, Bull. Acad. Polon. Sci.6 (1958), 97–101.) that

Given $n > 2$, it is possible to find $N = N(n)$ norms $\| \cdot\|_k$ $( 1 \leq k \leq N)$ on the plane such that an n-dimensional normed space containing $N$ subspaces isometric to these normed planes does not exist.

Rolewicz mentions that it is not clear whether we can take $N(n) = n+1$.

Added on 08/01/2014. Trying to guess what possible four normed planes cannot be linearly and isometrically embedded in any one three-dimensional normed space: Does there exist a three-dimensional centrally symmetric convex body that contains four central sections that are linearly equivalent to a circle, a square, a regular hexagon, and a regular octagon?

• Do you think that these bodies need to be smooth? – Suvrit Jan 7 '14 at 23:28
• @Suvrit: No, I don't think so. In fact, it is not clear to me how "stable" the problem is. Suppose you find the four normed planes, will other four planes very close (Banach-Mazur distance) to them satisfy the same thing? If so, we can stick to smooth bodies or to polygons and it won't make a difference. – alvarezpaiva Jan 8 '14 at 12:48
• The problem is stable, Alvarez. That follows from the fact that the three dimensional spaces are compact in the Banach-Mazur distance. – Bill Johnson Jan 8 '14 at 20:59
• I'm missing something: why there should be a 3-dimensional normed space containing two linear subspaces isometric say to the Euclidean norm $\|\cdot\|$ and to $10000\|\cdot\|$ ? – Pietro Majer Aug 15 '16 at 16:08
• @PietroMajer: the isometries for different subspaces can be different. The Euclidean space is then a 3-space containing two linear subspaces isometric say to the Euclidean norm on a plane and to 10000 the Euclidean norm on another plane. – alvarezpaiva Aug 21 '16 at 17:23