Mixing
orthonormal basis of infinite dimensional Hilbert space H is not a basis of H as vector space?
$begingroup$
Apparently the orthonormal basis $(e_n)_{nin mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?
real-analysis functional-analysis vector-spaces hilbert-spaces orthonormal
asked Jan 21 at 12:37
JudithJudith
61
$endgroup$
add a comment |
$begingroup$
Apparently the orthonormal basis $(e_n)_{nin mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?
real-analysis functional-analysis vector-spaces hilbert-spaces orthonormal
asked Jan 21 at 12:37
JudithJudith
61
$endgroup$
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40
add a comment |
$begingroup$
Apparently the orthonormal basis $(e_n)_{nin mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?
real-analysis functional-analysis vector-spaces hilbert-spaces orthonormal
asked Jan 21 at 12:37
JudithJudith
61
$endgroup$
Apparently the orthonormal basis $(e_n)_{nin mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?
real-analysis functional-analysis vector-spaces hilbert-spaces orthonormal
real-analysis functional-analysis vector-spaces hilbert-spaces orthonormal
asked Jan 21 at 12:37
JudithJudith
61
asked Jan 21 at 12:37
JudithJudith
61
asked Jan 21 at 12:37
JudithJudith
61
asked Jan 21 at 12:37
JudithJudith
61
asked Jan 21 at 12:37
JudithJudith
61
61
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40
add a comment |
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
answered Jan 21 at 13:02
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
First show or recall that every $xin H$ has a unique representation $$x=sum a_ne_n,$$with $sum|a_n|^2<infty$. So if $$x=sum_{n=1}^infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
A basis means that for everu $uin H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=sum_{ngeq 0}a_ne_n$ where the sequence $sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
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%2f3081829%2forthonormal-basis-of-infinite-dimensional-hilbert-space-h-is-not-a-basis-of-h-as%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
$begingroup$
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
answered Jan 21 at 13:02
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
answered Jan 21 at 13:02
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
answered Jan 21 at 13:02
KlausKlaus
2,12711
$endgroup$
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
answered Jan 21 at 13:02
KlausKlaus
2,12711
answered Jan 21 at 13:02
KlausKlaus
2,12711
answered Jan 21 at 13:02
KlausKlaus
2,12711
answered Jan 21 at 13:02
KlausKlaus
2,12711
2,12711
add a comment |
add a comment |
$begingroup$
First show or recall that every $xin H$ has a unique representation $$x=sum a_ne_n,$$with $sum|a_n|^2<infty$. So if $$x=sum_{n=1}^infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
First show or recall that every $xin H$ has a unique representation $$x=sum a_ne_n,$$with $sum|a_n|^2<infty$. So if $$x=sum_{n=1}^infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
First show or recall that every $xin H$ has a unique representation $$x=sum a_ne_n,$$with $sum|a_n|^2<infty$. So if $$x=sum_{n=1}^infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
First show or recall that every $xin H$ has a unique representation $$x=sum a_ne_n,$$with $sum|a_n|^2<infty$. So if $$x=sum_{n=1}^infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 21 at 13:34
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
add a comment |
add a comment |
$begingroup$
A basis means that for everu $uin H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=sum_{ngeq 0}a_ne_n$ where the sequence $sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
add a comment |
$begingroup$
A basis means that for everu $uin H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=sum_{ngeq 0}a_ne_n$ where the sequence $sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
add a comment |
$begingroup$
A basis means that for everu $uin H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=sum_{ngeq 0}a_ne_n$ where the sequence $sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
A basis means that for everu $uin H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=sum_{ngeq 0}a_ne_n$ where the sequence $sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 12:40
Tsemo AristideTsemo Aristide
59.1k11445
59.1k11445
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
add a comment |
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
1
1
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
$begingroup$
@Judith think about $v=(1, 1/2^2, ldots, 1/n^2, ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space.
$endgroup$
– Shubham Namdeo
Jan 21 at 12:57
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%2f3081829%2forthonormal-basis-of-infinite-dimensional-hilbert-space-h-is-not-a-basis-of-h-as%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
$begingroup$
You mean Hamel basis?
$endgroup$
– gabriele cassese
Jan 21 at 12:40