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Let $X$ be a Banach space. Is it possible that there exists a closed subspace $Y$ of $X^*$ such that $Y^{**}\cong X^*$ but $Y^*\not\cong X$?

What about $Y^{***}\cong X^{**}$ but $Y^{**}\not\cong X^*$?

(Here $\cong$ means isometric isomorphism)

I am almost certain the answer is yes to both questions, so I'm more concerned if anyone has an explicit example. As far as my motivation goes, please see this question.

Miles Gould
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    For the first question consider $X = L_1$, then $X^* = L_\infty$. Let $Y$ be a subspace $X^$ which is isomorphic to $c_0$. On one hand, $Y^{} \cong \ell_\infty \cong L_\infty = X^$. On the other hand, $L_1$ is known not to be isomorphic to a dual space, and thus $Y^* \not\cong X$. – KeeperOfSecrets Jan 30 '25 at 21:06
  • @KeeperOfSecrets I’m not sure I understand your reasoning. – Miles Gould Jan 30 '25 at 21:32
  • @KeeperOfSecrets What are $L_1,L_\infty$ in this case? – Miles Gould Jan 30 '25 at 21:57
  • $L_1$ is the Banach space of integrable functions on the interval $[0,1]$. $L_\infty$ is the Banach space of essentially bounded functions on the interval $[0,1]$ (both with respect to the Lebesgue measure). – KeeperOfSecrets Jan 30 '25 at 22:58
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    I used three classical results: the dual of $L_1$ is $L_\infty$, $L_1$ is not isomorphic to a dual space, and $L_\infty$ is isomorphic to $\ell_\infty$ (this also implies that $L_\infty$ contains a subspace isomorphic to $c_0$). – KeeperOfSecrets Jan 30 '25 at 23:01
  • @KeeperOfSecrets Okay, I was unsure of the last step, I had no idea $L_\infty$ and $\ell^\infty$ were isometrically isomorphic! – Miles Gould Jan 31 '25 at 05:52
  • @KeeperOfSecrets Does the embedding $c_0\hookrightarrow L_\infty$ make it a $*$-subalgebra? – Miles Gould Jan 31 '25 at 06:09
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    They are not isometrically isomorphic, only isomorphic. I thought that by "$\cong$" you mean just an isomorphism relation (my bad, sorry!). – KeeperOfSecrets Jan 31 '25 at 10:19
  • Is $\ell_{\infty}$ isomorphic to $L_{\infty}$? – Jose Gutierrez Feb 17 '25 at 07:42
  • @JoseGutierrez Yes, there is a linear homeomorphism between the two. – Miles Gould Feb 17 '25 at 18:14
  • Looks like this argument depends on axiom of choise :(

    https://math.stackexchange.com/questions/110438/is-there-an-explicit-isomorphism-between-l-infty0-1-and-ell-infty

     – Jose Gutierrez Feb 18 '25 at 09:01
  • @JoseGutierrez Yes, but if you want them not to be isomorphic, you basically need $\ell^1,\ell^\infty$ to be reflexive, which is very non-standard. – Miles Gould Feb 18 '25 at 14:55

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I write some result before:

  • Let $E$, $F$ Banach spaces, then $T:E \rightarrow F$ is continuous iff $T:(E,\sigma(E,E^*)) \rightarrow (F,\sigma(F,F^*))$ is continuos (weak topology).

  • Let $E$ and $F$ normed spaces. $S:F^* \rightarrow E^*$ lineal. Then $S^*$ is continuous in weak topology iff Exist a bounded linear operator $T:E \rightarrow F$ with $T^* = S$

  • $T:E \rightarrow F$ an bounded operator between Banach spaces. Then $T$ is an isomorphism iff $T^*$ is an isomorphism

Then if $X$ and $Y$ are Bananch space and $S:Y^{**} \rightarrow X^* $ is an isomorphism then $S^*$ is an isomorphism, and, in particular, is continuous in weak topology. So exist a $T:X \rightarrow Y^*$ with $T^*=S$ and T must be an isomorphism.

Then $Y^{**} ≅ X^* \Longrightarrow Y^* ≅ X$ (isomorphic)

Maybe there are something bad because @KeepOfSecrets add a counterexample with $X=L_1$ and $c_0 \cong Y \subset X^* = L_\infty$

If someone finds an error, please notify.

Jose Gutierrez
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  • I am so sorry I never saw this! I am asking for an isometric isomorphism, and it seems that @KeepOfSecrets has a counterexample to this, so I'm not sure where an error may be. – Miles Gould Feb 14 '25 at 16:34
  • Ah Right! In my case both subspace are Banach space. The counterexample is using $c_0$. I'm going to update the answer. – Jose Gutierrez Feb 15 '25 at 18:16
  • But $c_0$ is a Banach space. – Miles Gould Feb 15 '25 at 20:35
  • my bad! i though he is talking about $c_{00}$. – Jose Gutierrez Feb 17 '25 at 07:26
  • I think maybe the flaw is that isomorphisms are weak-to-weakly continuous, I'm not sure that is true. – Miles Gould Feb 18 '25 at 21:38
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    I have searched for my references and this is true. (Teorem 3.10 Brezis) This is not the error :/ "Let $E$ and $F$ be two Banach spaces and let $T$ be a linear operator from $E$ into $F$. Assume that $T$ is continuous in the strong topologies. Then T is continuous from E weak $\sigma(E,E^)$ into $F$ weak $\sigma(F,F^)$ and conversely." – Jose Gutierrez Feb 19 '25 at 07:56
  • By conversely, do you mean an operator which is weak-to-weakly continuous is also norm-to-norm continuous? – Miles Gould Feb 19 '25 at 15:28
  • Yes, it is proved using the closed graph theorem – Jose Gutierrez Feb 20 '25 at 13:41