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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.

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  • $\begingroup$ Did you try to tensorize your counterexamples i.e. $\C\otimes_\R V$ with the natural extension of the norm and see what happens ? $\endgroup$ Feb 27, 2013 at 18:17
  • $\begingroup$ 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. $\endgroup$ Feb 27, 2013 at 18:24
  • $\begingroup$ @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. $\endgroup$ Feb 28, 2013 at 14:09
  • $\begingroup$ 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? $\endgroup$ Feb 28, 2013 at 14:47
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    $\begingroup$ To answer one question raised in the comments: there are continuously many examples on $\mathbf{R}^2$, such as the hybrid $\ell_p/\ell_q$ norms (for $p,q$ conjugate) defined as the $\ell_p$ norm on the NW/SE quadrants and the $\ell_q$ norm on the NE/SW quadrants. Variants of the constuction give also smooth examples. $\endgroup$ Jun 1, 2017 at 14:25

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A self-dual norm on $\mathbf{C}^2$ need not be Hilbertian. Here is an example: $$ \|(z_1,z_2)\| = \begin{cases} |z_1|+|z_2| & if \ \ \ |z_1| \leq |z_2| \\ 2|z_1| & if \ \ \ |z_1| \geq |z_2|. \end{cases} $$

The dual norm $\|\cdot\|^*$ satisfies $\|(z_1,z_2)\|^*=\frac{1}{2}\|(z_2,z_1)\|$.

Here a picture for $z_1$, $z_2$ real with $\|\cdot\|=1$ in blue and $\|\cdot\|^*=1$ in red.

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  • $\begingroup$ Nice! $ $ $ $ $ $ $ $ $\endgroup$ Jun 2, 2017 at 12:47

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