Every finite-dimensional subspace is one-complemented
$begingroup$
Let $X$ be a Banach space. It is known that if every closed subspace of $X$ is one-complemented, then $X$ is isometrically isomorphic to a Hilbert space.
Now if every finite-dimensional subspace of $X$ is one-complemented, is it true that is $X$ isometrically isomorphic to a Hilbert space?
functional-analysis banach-spaces
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space. It is known that if every closed subspace of $X$ is one-complemented, then $X$ is isometrically isomorphic to a Hilbert space.
Now if every finite-dimensional subspace of $X$ is one-complemented, is it true that is $X$ isometrically isomorphic to a Hilbert space?
functional-analysis banach-spaces
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
$endgroup$
2
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06
add a comment |
$begingroup$
Let $X$ be a Banach space. It is known that if every closed subspace of $X$ is one-complemented, then $X$ is isometrically isomorphic to a Hilbert space.
Now if every finite-dimensional subspace of $X$ is one-complemented, is it true that is $X$ isometrically isomorphic to a Hilbert space?
functional-analysis banach-spaces
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
$endgroup$
Let $X$ be a Banach space. It is known that if every closed subspace of $X$ is one-complemented, then $X$ is isometrically isomorphic to a Hilbert space.
Now if every finite-dimensional subspace of $X$ is one-complemented, is it true that is $X$ isometrically isomorphic to a Hilbert space?
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
asked Feb 10 '15 at 9:34
Marten WortelMarten Wortel
1426
1426
2
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06
add a comment |
2
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06
2
2
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.
References:
P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.
S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
$endgroup$
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
add a comment |
$begingroup$
It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.
This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
$endgroup$
add a comment |
$begingroup$
I never heard of result you stated. I take it for granted.
Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
$endgroup$
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1141840%2fevery-finite-dimensional-subspace-is-one-complemented%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.
References:
P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.
S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
$endgroup$
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
add a comment |
$begingroup$
Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.
References:
P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.
S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
$endgroup$
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
add a comment |
$begingroup$
Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.
References:
P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.
S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
$endgroup$
Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case.
References:
P. Bohnenblust, A characterization of complex Hilbert spaces, Portugaliae Math., 3 (1942), 103–109.
S. Kakutani, Some characterizations of Euclidean space, Japan. J. Math. 16 (1939), 93–97.
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
answered Nov 26 '15 at 13:36
Tomek KaniaTomek Kania
12.1k11943
12.1k11943
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
add a comment |
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
1
1
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
$begingroup$
Good references!
$endgroup$
– Norbert
Nov 26 '15 at 16:01
add a comment |
$begingroup$
It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.
This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
$endgroup$
add a comment |
$begingroup$
It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.
This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
$endgroup$
add a comment |
$begingroup$
It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.
This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
$endgroup$
It is known that if every closed subspace of X is one-complemented, then X is isometrically isomorphic to a Hilbert space.
This is true for spaces of dimension at least three, right? So, the argument of Norbert does not work. You can not prove that a two dimensional vector space is a Hilbert space by proving that all the one dimensional subspaces are one-complemented.
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
answered Dec 4 '18 at 16:03
Jorge Tomás RodríguezJorge Tomás Rodríguez
112
112
add a comment |
add a comment |
$begingroup$
I never heard of result you stated. I take it for granted.
Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
$endgroup$
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
add a comment |
$begingroup$
I never heard of result you stated. I take it for granted.
Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
$endgroup$
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
add a comment |
$begingroup$
I never heard of result you stated. I take it for granted.
Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
$endgroup$
I never heard of result you stated. I take it for granted.
Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the beginning of the question we get that $E$ is a Hilbert space.
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
answered Feb 10 '15 at 22:06
NorbertNorbert
45.7k773160
45.7k773160
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
add a comment |
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
$begingroup$
No need to take it for granted :-)
$endgroup$
– Tomek Kania
Nov 26 '15 at 13:36
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1141840%2fevery-finite-dimensional-subspace-is-one-complemented%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Perhaps you should explain (or link to an explanation of) what "one-complemented" means.
$endgroup$
– Omnomnomnom
Feb 10 '15 at 12:19
$begingroup$
@Omnomnomnom I suppose one-complemented means "complemented by projection of norm one".
$endgroup$
– Norbert
Feb 10 '15 at 22:06