Is a linearly independent set whose span is dense a Schauder basis? The Next CEO of Stack...
Compensation for working overtime on Saturdays
Salesforce opportunity stages
What does it mean 'exit 1' for a job status after rclone sync
Why do we say “un seul M” and not “une seule M” even though M is a “consonne”?
A hang glider, sudden unexpected lift to 25,000 feet altitude, what could do this?
Horror film about a man brought out of cryogenic suspension without a soul, around 1990
Physiological effects of huge anime eyes
Is it possible to create a QR code using text?
Does int main() need a declaration on C++?
Is it possible to make a 9x9 table fit within the default margins?
Can Sri Krishna be called 'a person'?
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what pi version
Are British MPs missing the point, with these 'Indicative Votes'?
Calculating discount not working
It it possible to avoid kiwi.com's automatic online check-in and instead do it manually by yourself?
Simplify trigonometric expression using trigonometric identities
How to pronounce fünf in 45
Find the majority element, which appears more than half the time
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
Why did the Drakh emissary look so blurred in S04:E11 "Lines of Communication"?
How to implement Comparable so it is consistent with identity-equality
Compilation of a 2d array and a 1d array
Read/write a pipe-delimited file line by line with some simple text manipulation
Planeswalker Ability and Death Timing
Is a linearly independent set whose span is dense a Schauder basis?
The Next CEO of Stack OverflowCoordinate functions of Schauder basisLinearly independentSchauder basis for a separable Banach spaceWhat is the difference between a Hamel basis and a Schauder basis?Hamel basis for subspacesExistence of weak Schauder-basis for concrete example.Isomorphisms with invariant linearly independent dense subset.Linear independence and Schauder basisWhy isn't every Hamel basis a Schauder basis?Schauder basis that is not Hilbert 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 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
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 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
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 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
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 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
2,38621446
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 $C[0,1]$, but is not a Schauder basis of that space. (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, linearly independent set) that have no Schauder basis at all.
answered 54 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
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 $C[0,1]$, but is not a Schauder basis of that space. (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, linearly independent set) that have no Schauder basis at all.
answered 54 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (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, linearly independent set) that have no Schauder basis at all.
answered 54 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (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, linearly independent set) that have no Schauder basis at all.
answered 54 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (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, linearly independent set) that have no Schauder basis at all.
answered 54 mins ago
GEdgarGEdgar
63.3k268172
answered 54 mins ago
GEdgarGEdgar
63.3k268172
answered 54 mins ago
GEdgarGEdgar
63.3k268172
answered 54 mins ago
GEdgarGEdgar
63.3k268172
63.3k268172
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
