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. 



Self-dual normed spaces which are not Hilbert spaces
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?
> 
> 
> 


