21

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.

Uday
  • 2,209
  • 6
    Look at t.b.'s answer here: http://math.stackexchange.com/questions/65609/isometric-to-dual-implies-hilbertable. – Davide Giraudo Mar 15 '12 at 19:37
  • 3
    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
  • @David I have got an example. Thanks. @Ralph Are $X \oplus X^{*}$ related to symplectic spaces? – Uday Mar 15 '12 at 19:56
  • Maybe there exists a indecomposable Banach space that is isometric to its dual. I am pretty sure that it is open whether such a space exists. – Bill Johnson Mar 15 '12 at 20:14
  • 1
    @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
  • 2
    @Ralph: isn't every Hilbert space of dimension at least $2$ decomposable? – Qiaochu Yuan Mar 15 '12 at 21:05
  • @Ralph: Your comment didn't assume completeness of the space; was that intentional? – Andreas Blass Mar 15 '12 at 21:10
  • Ralph, the question you suggest is indeed research level, but it is somewhat imprecisely defined, and it isn't the question the OP asked. What counts as a "natural" Banach space anyway? Would you say Tsirelson is unnatural? Gowers-Maurey? – Yemon Choi Mar 15 '12 at 21:13
  • Voting to close is not a statement "no version of this question is interesting". It is, IMHO, a statement "in its present form we do not want to leave the question open for answers" – Yemon Choi Mar 15 '12 at 21:14
  • @Andreas: Yes, because the dual of a normed space is complete (at least over $\mathbb{R}$ or $\mathbb{C}$) and hence a self-dual space a implicitely complete. – Ralph Mar 15 '12 at 21:19
  • 2
    "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
  • @Yemon: Forgive me, I'm just lazy (i.e. if Uday hadn't posted the question I would had done someday in the future. So I saw the opportunity to get "my question" answered along the way). – Ralph Mar 15 '12 at 21:40
  • @Ralph Since the question in your mind is floating around in a half-baked manner. I would prefer(if it is okay for you), you edit the question and mark as CW. – Uday Mar 15 '12 at 21:53
  • It is elementary to show that an indecomposable space that is isomorphic to its dual would necessarily be quasi-reflexive (but not reflexive) and cannot be hereditarily indecomposable. Quasi-reflexive hereditarily indecomposable spaces are known to exist (see the book of Argyros and Todorcevic, Ramsey methods in analysis), but I do not know if anyone has tried and succeeded to construct a quasi-reflexive space that is indecomposable but not hereditarily indecomposable. – Philip Brooker Mar 15 '12 at 21:55
  • Why can it not be reflexive and HI, Phil? Maybe this is elementary, but I do not see a proof. – Bill Johnson Mar 15 '12 at 22:04
  • @Uday: CW is, IMO, not appropriate for an edited version of this question. – Bill Johnson Mar 15 '12 at 22:05
  • 1
    @Bill: sorry, my previous comment was ill-thought-out, and I just raced back from having a shower to my computer to try to correct it before anyone noticed! No luck :-) . Anyway, I think I was confused in my head by a related problem that has been on my mind: does there exist a Banach space $X$ and $n>3$ such that $X$ is isomorphic to the $n$th dual of $X$, but to no previous dual? An affirmative answer implies a negative solution to the Schroeder-Bernstein problem for Banach spaces (which Gowers has of course already achieved in general). Such an $X$ would of course be non-reflexive. – Philip Brooker Mar 15 '12 at 22:25
  • Sorry, in my previous comment I meant $n>2$, and $X$ not isomorphic to any previous dual besides the $0$th dual (namely, itself). – Philip Brooker Mar 15 '12 at 22:28
  • 1
    That is a nice problem, Phil. A related problem is to built a space that is isomorphic to a subspace of codimension $n$ but not one of codimension $n-1$. The Kalton-Peck space may be such an example with $n=2$ but I cannot prove it. I don't recall such an example from the G-M theory but I could have forgotten... – Bill Johnson Mar 15 '12 at 22:32
  • See my edits to my answer, given one more example and suggesting infinitely many. – Denis Serre Mar 16 '12 at 07:17

2 Answers2

23

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?

Community
  • 1
Denis Serre
  • 51,599
  • 8
    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 00:55
  • 2
    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 08:17
  • 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
  • I am sorry to break the code of A&Q in this community but I don't have enough reputation to raise questions in the comment space. Would anyone care to explain to me how the example Denis constructed to be non-Hilbert? In addition, how to see that the dual space is a normed vector space? – dragonxlwang Feb 25 '15 at 04:34
  • 1
    The Denis unit ball is a square, but the unit ball for any two-dimensional Hilbert space is an ellipse. So the Denis space is not isometric to a Hilbert space. (It is, of course, linearly homeomorphic to a Hilbert space.) – Gerald Edgar Feb 25 '15 at 13:52
  • 2
    I recently asked this question (and now came accross this post and your answer). The answer to my post also answers your questions about self-dual polytopes: there are centrally-symmetric self-dual polytopes in all dimensions. Remarkably, the answer refers to a paper in Banach space theory: Reisner, S., "Certain Banach spaces associated with graphs and CL-spaces with 1- unconditional bases". It seems this question has been asked in this context before. – M. Winter Jan 06 '21 at 23:48
6

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.

Valerio Capraro
  • 5,894