# 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.

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Look at t.b.'s answer here: math.stackexchange.com/questions/65609/…. –  Davide Giraudo Mar 15 '12 at 19:37
According to Davide's comment: It would in fact be very interesting to have "natural" example for such a space, i.e. one that is not constructed as $X \oplus X^\ast$ (or in a similar way) with a reflexive space $X$. I have googling for such spaces some time ago and couldn't find any. To me, the question (with this additional condition) is reasearch level and shouldn't b closed. –  Ralph Mar 15 '12 at 19:49
@Bill: That's interesting. So "indecomposable normed space that is isometric to its dual $\Rightarrow$ Hilbert space" would be a reasonable conjecture ? –  Ralph Mar 15 '12 at 20:33
@Ralph: isn't every Hilbert space of dimension at least $2$ decomposable? –  Qiaochu Yuan Mar 15 '12 at 21:05
"Indecomposable" means "not isomorphic to the direct sum of two infinite dimensional subspaces". @Ralph: IMO it is reasonable to conjecture that there is no indecomposable space that is even just isomorphic to its dual. –  Bill Johnson Mar 15 '12 at 21:24

## 2 Answers

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?

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I guess the easy way to visualize this is that the unit balls in both cases are squares, but different sizes and rotated by 45 degrees. –  Nate Eldredge Mar 16 '12 at 0:55
Is a self-dual norm in ${\mathbb R^3}$ necessarily Euclidean? –  alvarezpaiva Mar 16 '12 at 17:39
@alvarezpaiva. This remains a question, apparently open. Even if the unit ball is a polyhedron, which has to be centro-symmetric, why should it be regular? –  Denis Serre Mar 17 '12 at 8:17
Unless I'm doing something silly, $\mathbb{R}^3$ equipped with the norm $\sqrt{(|x_1|+|x_2|)^2+|x_3|^2}$ is isometric to its dual. –  Mark Meckes Feb 27 '13 at 17:54

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.

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Valerio, this is precisely the construction that was linked to in comments above –  Yemon Choi Mar 16 '12 at 9:35
Ah, OK. It's such a short construction that I think it takes more time to find the right link, copy it... than write it down directly. –  Valerio Capraro Mar 16 '12 at 11:33