# Self-dual finite-dimensional complex normed spaces

Suppose $X$ is a complex normed space of dimension 2 or 3 and $X$ is isometrically isomorphic to its dual. Is $X$ a Hilbert space?

Remarks: There are easy counterexamples in the real case, and in higher dimensions one can construct counterexamples from sums of 2-dimensional spaces which are not isometric to their duals. Similarly a 3-dimensional counterexample can be constructed from a 2-dimensional counterexample.

-
Did you try to tensorize your counterexamples i.e. $\C\otimes_\R V$ with the natural extension of the norm and see what happens ? –  Duchamp Gérard H. E. Feb 27 '13 at 18:17
The simplest counterexample is given by the $\ell_1$ and $\ell_\infty$ norms. They are of course dual to each other; they are also isometric to each other only in the real 2-dimensional case. –  Mark Meckes Feb 27 '13 at 18:24
@Bill: Thanks adding the banach-spaces tag. I realized later I should have included that one. I also considered adding the ask-johnson tag, but you had already been here by the time I got to it. –  Mark Meckes Feb 28 '13 at 14:09
I have no idea what the answer is, Mark. Do you know real two dimensional examples other than the obvious ones you mentioned? Is there a classification of them? –  Bill Johnson Feb 28 '13 at 14:47
Any norm whose unit ball is a regular polygon gives a real two dimensional example. I don't know if there are others; in particular I don't know whether there is a continuous family of examples, or whether there exist smooth examples. I discussed the problem with Szarek briefly yesterday and he raised the latter question in particular, but neither of us has given it serious thought. –  Mark Meckes Feb 28 '13 at 14:57