Self-dual normed spaces which are not Hilbert spaces Are there any examples of non-Hilbert normed spaces which are isomorphic (in the norm sense) to their dual spaces? Or, is there any result in Functional Analysis which says that if a space is self-dual it has to be Hilbert space. 
Since, we want isomorphism in the norm sense, examples like $\mathbb{R}^{n}$ are ruled out. The norms of the space and its dual have to be equal and not just equivalent. 
Thank you. 
 A: Another family of infinitely many examples: take $Y$ to be a reflexive Banach space which is not a Hilbert space, then $Y\oplus Y^*$ is isometrically isomorphic to its dual, without being a Hilbert space.
If the isomorphism verifies additional properties, then the result is true. Namely, if a Banach space is isometric to its dual, under certain conditions, it is a Hilbert space. See Theorems 2 and 4 in http://arxiv.org/pdf/0907.1813.pdf and reference therein for similar results. An example of this kind of results is the following:
Theorem: Suppose that $X$ is a Banach space and $\phi:X\to X^*$ is an antilinear isomorphism. If, for all $x\in X$, $x$ is orthogonal (in Birkhoff-James' sense) to $ker(\phi(x))$, then $X$ is a Hilbert space.
A: I have two, and perhaps infinitely many, examples in finite dimension $n$.
n=2. Take $X={\mathbb R}^2$ with $\ell^1$-norm
$$\|x\|_1=|x_1|+|x_2|.$$
Then $X^*={\mathbb R}^2$ has the $\ell^\infty$-norm
$$\|y\|_\infty=\max(|y_1|,|y_2|).$$
I turns out that
$$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$
and thus $X'$ is isometric to $X$, via $x\mapsto(x_1+x_2,x_1-x_2)$.
More generally, suppose that in $\mathbb R^n$, we have a convex polytope $T$ that is self-dual and is symmetric under $x\leftrightarrow-x$. Let $\|\cdot\|_T$ be the gauge associated with $T$. Then $X=(\mathbb R^n, \|\cdot\|_T)$ is isometric to $X'$ because $T$ is the unit ball of $X$ and $T'=T$ is that of $X'$.
For instance, if n=4, the polyoctahedron (= octaplex) has these properties, thus there is an $\mathbb R^4$ that is isometric to its dual, yet is not Hilbert. If $n\ge3$, the  simplex is self-dual but not centro-symmetric.
This raises two questions:

Does there exist other centro-symmetric self dual convex polytopes? Maybe there exist one in any even dimension ...
Is it possible to deform the examples above so as to replace the polygone/-tope by a ball with a smooth boundary?

