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Are there $X,Y$ real Banach spaces, such that $X\subsetneqq Y$ (strictly contained) and $X^\star=Y^\star$, where $\star$ denotes the topological dual?

This property is not true for Hilbert spaces, nor even for $L^p$ spaces, so I was thinking to try some function space: continuous functions or bounded...

Any help is appreciated.

Thank you

Pedro
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Tomás
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  • Just a random thought: if you take $\ell_2$ and $A \subset \ell_2$ which is defined as the subset made of all sequences with first coordinate zero, they are both isometric to $\ell_2$, so their duals are isometric, which is how I interpreted $X^\star = Y^\star$. – Second May 13 '13 at 12:54
  • @Second: I don't think this is what the OP wanted; if $X \subset Y$, then $Y^$ is naturally mapped to $X^$ (a linear form on $Y$ induces a linear form on $X$). I believe the question is whether this map is an isomorphism? – Najib Idrissi May 13 '13 at 13:15
  • Do you mean that the transpose of the inclusion map is a topological isomorphism? Or that there exists a topological isomorphism whatsoever? – egreg May 13 '13 at 13:30
  • What do you mean by $X^=Y^$ exactly? That the duals are isomorphic (via a continuous isomorphism)? Isometric? – Julien May 13 '13 at 16:01
  • Yes @julien , there is an isomorphism between then. – Tomás May 13 '13 at 16:09
  • Should we understand your question like Second and I did, or like nik? Please clarify, thanks. – Julien May 13 '13 at 18:33
  • @julien , I am a little confused here. If $X\subset Y$, can I identify $Y^\star$ with a subspace of $X^\star$? – Tomás May 13 '13 at 18:44
  • If $X\subseteq Y$, you get a canonical surjection $Y^\longrightarrow X^$ by restriction. Hahn-Banach show it is surjective. Then you can mod out by the nullspace, which is the orthogonal of $X$ in $Y^*$. That's isometric, then. – Julien May 13 '13 at 18:47
  • Well ok @julien. For example, we have that $\mathbb{R}\subset \mathbb{R}^2$, but $(\mathbb{R}^N)^\star=\mathbb{R}^N$, so $\mathbb{R}^2\subset \mathbb{R}$. What I am doing wrong? – Tomás May 13 '13 at 18:50
  • I think there is a confusion between the notions of dual and orthogonal here. – Julien May 13 '13 at 18:51
  • @julien, you mean that in the answers down there, they used the notion of orthogonality, insted fo duality? – Tomás May 13 '13 at 18:58
  • No, they use the notion of duality. Me too, actually. But you seem to think that $X\subseteq Y$ implies canonically that $Y^\subseteq X^$. This is false. What the two answers show is that the canonical surjection $Y^\longrightarrow X^$ is not injective when $X$ is a proper closed subspace of $Y$. – Julien May 13 '13 at 19:04
  • Hmm ok @julien, now things clarified for me. But why do you deleted your answer? – Tomás May 13 '13 at 19:10
  • Because I was not sure what you were asking. If it answers your question, I'll undelete. – Julien May 13 '13 at 19:11

2 Answers2

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It seems that you are asking whether $X^*$ and $Y^*$ can be isomorphic (there exists a (bi)continuous isomorphism between them) when $X$ is a proper closed subspace of a Banach space $Y$.

Yes, this can happen. Even with Hilbert spaces. I would say especially with Hilbert spaces. It suffices that $X$ and $Y$ be isomorphic. And that's necessary if they are reflexive, like in the Hilbert case.

E.g.

a) $X=\ell^2(\mathbb{N})\subsetneq \ell^2(\mathbb{Z})=Y$ has $X^*\simeq Y^*\simeq \ell^2(\mathbb{N})$ isometrically.

b) More generally, for an infinite dimensional separable Hilbert space $H$ and any proper closed infinite dimensional subspace $K\subsetneq H$, we have $K^*\simeq H^*\simeq \ell^2(\mathbb{N})$ isometrically.

c) In $\ell^\infty(\mathbb{N})$, we have $c_0\subsetneq c$ and $c_0^*\simeq c^*\simeq \ell^1$ isometrically where $c$ (resp. $c_0$) is the subspace of sequences that converge (resp. to $0$).

d) And here is a somehow silly example: take $X=C([0,1])\subsetneq C([0,2]=Y$ with the obvious embedding which extends a function continuously by a constant on $[1,2]$. Then $X^*\simeq Y^*$ isometrically as follows from Riesz representation theorem. Or much easier, from the fact $X$ and $Y$ are obviously isometric.

Julien
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  • I think that my comment after my question, had mislead the answers of the other guys, nonetheless I haven't say anything specific about the canonical surjection. – Tomás May 13 '13 at 19:21
  • Of course, none of these isometric isomorphisms of dual spaces is in any sense canonical. // The most interesting instance is in c) where $c_0$ and $c$ are not isometrically isomorphic (in the examples in a), b) and d) it is easy to see that $X$ and $Y$ are isometrically isomorphic). Worse, still: it is known that $\ell^1$ has a continuum of non-isomorphic pre-duals. – Martin May 14 '13 at 01:24
  • @Martin Thanks for the note. But you scared me, I thought I had said something wrong... – Julien May 14 '13 at 01:34
  • No, no worries, you didn't -- although I tend to read an equality sign to mean at least that there is a canonical isomorphism if it isn't literal equality, which is probably what triggered the first sentence of my comment. – Martin May 14 '13 at 01:37
  • @Martin Oh, I see. You're right. I pasted the OP's question...I'll edit. – Julien May 14 '13 at 01:38
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Choose a point $p$ in $Y$ outside $X$ and construct a linear functinal $f$ on $X+\mathbb R p$ which vanishes on $X$ whereas $f(p)=1$. By the Hahn-Banach theorem, $f$ extends to a functional on $Y$. Clearly therefore the duals of $X$ and $Y$ are never the same.

Mikhail Katz
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