Is a linearly independent set whose span is dense a Schauder basis?
$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
$endgroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,43821446
2,43821446
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
$endgroup$
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%2f3171184%2fis-a-linearly-independent-set-whose-span-is-dense-a-schauder-basis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
$endgroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
edited 2 hours ago
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
answered Apr 1 at 21:33
GEdgarGEdgar
63.4k269174
63.4k269174
add a comment |
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%2f3171184%2fis-a-linearly-independent-set-whose-span-is-dense-a-schauder-basis%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
