Complemented subspaces constructed from finite pieces- part II
$begingroup$
This is a follow up to: Complemented subspace constructed from finite pieces
Suppose $Y=overline{cup E_n}$ is a closed subspace of a separable Banach space X, where each $E_n$ is a $n$-dimensional subspace, $K$-complemented in $X$, and for any $n$, $E_nsubseteq E_{n+1}$. Can one conclude that $Y$ is complemented in $X$?
In light of the answer to the previous question, a related question would be the following:
Is $c_0$ complemented in every separable subspace of $l_infty$ that contains it. I suspect the answer is no, but cannot think of a counterexample.
fa.functional-analysis banach-spaces
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
$endgroup$
add a comment |
$begingroup$
This is a follow up to: Complemented subspace constructed from finite pieces
Suppose $Y=overline{cup E_n}$ is a closed subspace of a separable Banach space X, where each $E_n$ is a $n$-dimensional subspace, $K$-complemented in $X$, and for any $n$, $E_nsubseteq E_{n+1}$. Can one conclude that $Y$ is complemented in $X$?
In light of the answer to the previous question, a related question would be the following:
Is $c_0$ complemented in every separable subspace of $l_infty$ that contains it. I suspect the answer is no, but cannot think of a counterexample.
fa.functional-analysis banach-spaces
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
$endgroup$
add a comment |
$begingroup$
This is a follow up to: Complemented subspace constructed from finite pieces
Suppose $Y=overline{cup E_n}$ is a closed subspace of a separable Banach space X, where each $E_n$ is a $n$-dimensional subspace, $K$-complemented in $X$, and for any $n$, $E_nsubseteq E_{n+1}$. Can one conclude that $Y$ is complemented in $X$?
In light of the answer to the previous question, a related question would be the following:
Is $c_0$ complemented in every separable subspace of $l_infty$ that contains it. I suspect the answer is no, but cannot think of a counterexample.
fa.functional-analysis banach-spaces
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
$endgroup$
This is a follow up to: Complemented subspace constructed from finite pieces
Suppose $Y=overline{cup E_n}$ is a closed subspace of a separable Banach space X, where each $E_n$ is a $n$-dimensional subspace, $K$-complemented in $X$, and for any $n$, $E_nsubseteq E_{n+1}$. Can one conclude that $Y$ is complemented in $X$?
In light of the answer to the previous question, a related question would be the following:
Is $c_0$ complemented in every separable subspace of $l_infty$ that contains it. I suspect the answer is no, but cannot think of a counterexample.
fa.functional-analysis banach-spaces
fa.functional-analysis banach-spaces
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
edited Jan 5 at 20:06
Mikhail Ostrovskii
3,2871029
3,2871029
asked Dec 25 '18 at 1:09
user129564user129564
1024
asked Dec 25 '18 at 1:09
user129564user129564
1024
asked Dec 25 '18 at 1:09
user129564user129564
1024
1024
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer to your question in the last paragraph is "Yes", it is a contents of the well-known Sobczyk Theorem, see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I.
The answer to the first question is still negative. You can find an example of this type in W.B. Johnson, J. Lindenstrauss, Examples of L1 spaces, Ark. Mat. 18 (1980), no. 1, 101–106.
The answer is positive if the space is reflexive - you can consider the weak limit of the projections.
Added on January 5, 2018. Many other examples giving a negative answer to the first question can be constructed combining the result of Zippin [Israel J. Math. 26 (1977), no. 3-4, 372–387] stating that any separable infinite-dimensional Banach space which is not isomorphic to $c_0$ can be embedded into a separable Banach space as an uncomplemented subspace, and the well-known fact that a Banach space can represented as a closure of the union of increasing sequence of finite-dimensional subspaces which are uniformly close to $ell_infty^n$ without being isomorphic to $c_0$ (see, e.g., very exotic examples in Bourgain, Pisier [Bol. Soc. Brasil. Mat. 14 (1983), no. 2, 109–123].)
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
$endgroup$
add a comment |
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$begingroup$
The answer to your question in the last paragraph is "Yes", it is a contents of the well-known Sobczyk Theorem, see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I.
The answer to the first question is still negative. You can find an example of this type in W.B. Johnson, J. Lindenstrauss, Examples of L1 spaces, Ark. Mat. 18 (1980), no. 1, 101–106.
The answer is positive if the space is reflexive - you can consider the weak limit of the projections.
Added on January 5, 2018. Many other examples giving a negative answer to the first question can be constructed combining the result of Zippin [Israel J. Math. 26 (1977), no. 3-4, 372–387] stating that any separable infinite-dimensional Banach space which is not isomorphic to $c_0$ can be embedded into a separable Banach space as an uncomplemented subspace, and the well-known fact that a Banach space can represented as a closure of the union of increasing sequence of finite-dimensional subspaces which are uniformly close to $ell_infty^n$ without being isomorphic to $c_0$ (see, e.g., very exotic examples in Bourgain, Pisier [Bol. Soc. Brasil. Mat. 14 (1983), no. 2, 109–123].)
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
$endgroup$
add a comment |
$begingroup$
The answer to your question in the last paragraph is "Yes", it is a contents of the well-known Sobczyk Theorem, see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I.
The answer to the first question is still negative. You can find an example of this type in W.B. Johnson, J. Lindenstrauss, Examples of L1 spaces, Ark. Mat. 18 (1980), no. 1, 101–106.
The answer is positive if the space is reflexive - you can consider the weak limit of the projections.
Added on January 5, 2018. Many other examples giving a negative answer to the first question can be constructed combining the result of Zippin [Israel J. Math. 26 (1977), no. 3-4, 372–387] stating that any separable infinite-dimensional Banach space which is not isomorphic to $c_0$ can be embedded into a separable Banach space as an uncomplemented subspace, and the well-known fact that a Banach space can represented as a closure of the union of increasing sequence of finite-dimensional subspaces which are uniformly close to $ell_infty^n$ without being isomorphic to $c_0$ (see, e.g., very exotic examples in Bourgain, Pisier [Bol. Soc. Brasil. Mat. 14 (1983), no. 2, 109–123].)
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
$endgroup$
add a comment |
$begingroup$
The answer to your question in the last paragraph is "Yes", it is a contents of the well-known Sobczyk Theorem, see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I.
The answer to the first question is still negative. You can find an example of this type in W.B. Johnson, J. Lindenstrauss, Examples of L1 spaces, Ark. Mat. 18 (1980), no. 1, 101–106.
The answer is positive if the space is reflexive - you can consider the weak limit of the projections.
Added on January 5, 2018. Many other examples giving a negative answer to the first question can be constructed combining the result of Zippin [Israel J. Math. 26 (1977), no. 3-4, 372–387] stating that any separable infinite-dimensional Banach space which is not isomorphic to $c_0$ can be embedded into a separable Banach space as an uncomplemented subspace, and the well-known fact that a Banach space can represented as a closure of the union of increasing sequence of finite-dimensional subspaces which are uniformly close to $ell_infty^n$ without being isomorphic to $c_0$ (see, e.g., very exotic examples in Bourgain, Pisier [Bol. Soc. Brasil. Mat. 14 (1983), no. 2, 109–123].)
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
$endgroup$
The answer to your question in the last paragraph is "Yes", it is a contents of the well-known Sobczyk Theorem, see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I.
The answer to the first question is still negative. You can find an example of this type in W.B. Johnson, J. Lindenstrauss, Examples of L1 spaces, Ark. Mat. 18 (1980), no. 1, 101–106.
The answer is positive if the space is reflexive - you can consider the weak limit of the projections.
Added on January 5, 2018. Many other examples giving a negative answer to the first question can be constructed combining the result of Zippin [Israel J. Math. 26 (1977), no. 3-4, 372–387] stating that any separable infinite-dimensional Banach space which is not isomorphic to $c_0$ can be embedded into a separable Banach space as an uncomplemented subspace, and the well-known fact that a Banach space can represented as a closure of the union of increasing sequence of finite-dimensional subspaces which are uniformly close to $ell_infty^n$ without being isomorphic to $c_0$ (see, e.g., very exotic examples in Bourgain, Pisier [Bol. Soc. Brasil. Mat. 14 (1983), no. 2, 109–123].)
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
edited Jan 5 at 20:17
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
answered Dec 25 '18 at 2:08
Mikhail OstrovskiiMikhail Ostrovskii
3,2871029
3,2871029
add a comment |
add a comment |
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