Timeline for A question related to the separable quotient problem
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16 events
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| Jan 8, 2022 at 17:39 | comment | added | Bill Johnson | I see; thanks, Spiros. | |
| Jan 8, 2022 at 8:57 | comment | added | S Argyros | Non separable Banach spaces $X$ with both $X$ and $X^*$ saturated by HI spaces not containing reflexive subspace seems to me that could exist. But this type of construction admits a resolution of the Identity. Hence has a separable quotient. On the other hand I strongly believe that the dual of non separable subspaces of $l^\infty$ contains an unconditional sequence. | |
| Jan 8, 2022 at 8:41 | comment | added | S Argyros | @Bill The general statement is the following. If $X$ is HI and T a bounded linear operator from $X$ to $Y$ then either T is strictly singular or it is isomorphism on a finite codimensional subspace of $X$. The result is included in the Memoirs AMS with Tolias. However the statement if $Z^*$ is HI then Z is separable follows immediately from your result that non separable spaces have decomposable dual. | |
| Jan 8, 2022 at 0:07 | comment | added | Bill Johnson | I do not understand "Since $𝑍^∗$ is HI this operator is an isomorphism on a subspace of $𝑍^∗$ of finite codimension". The operator does not range in $𝑍^∗$. What am I missing? | |
| Jan 7, 2022 at 21:06 | comment | added | S Argyros | Assume that $Z^*$ remains HI . Denote by $Z$ the quotient of $X$ with its dual the space $Z^*$. Choose $(z_n)_n$ a sequence in the unit ball of $Z$ that norms the space $W$. The sequence $(z_n )_n$ defines a w* continuous operator from $Z^*$ to $l^\infty$ which is isometric on the subspace $W$. Since $Z^*$ is HI this operator is an isomorphism on a subspace of $Z^*$ of finite codimension . Assume that it is isomorphism on the whole $Z^*$. then the unit ball of $Z^*$ is w* metrisable which yields that $Z$ is separable. | |
| Jan 7, 2022 at 20:52 | comment | added | S Argyros | Here are some thoughts concerning the general SQP. Assume that the initial space $X$ does not have a separable quotient. Then $X^*$ does not contain neither an unconditional basic sequence nor a reflexive subspace. By Gowers dichotomy $X^*$ is saturated by separable HI subspaces not containing reflexive subspace.Each one of them is either like Gowers Tree space or its predual. (A terrified structure for anyone thinking for a counterexample to SQP). In the case where $X$ fails SQP consider a separable W of $X^*$ which is HI and let $Z^*$ its w* closure. | |
| Jan 6, 2022 at 13:58 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 13:45 | comment | added | S Argyros | @YCor I think that now is clear. Thanks! | |
| Jan 6, 2022 at 13:26 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 13:12 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 13:04 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 12:46 | history | edited | YCor | CC BY-SA 4.0 |
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| Jan 6, 2022 at 12:21 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 12:11 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 12:04 | history | edited | S Argyros | CC BY-SA 4.0 |
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| Jan 6, 2022 at 11:56 | history | asked | S Argyros | CC BY-SA 4.0 |