TITLE: Self-dual finite-dimensional complex normed spaces
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: 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.