Is a linearly independent set whose span is dense a Schauder basis? The Next CEO of Stack...
pgfplots: How to draw a tangent graph below two others?
How can a day be of 24 hours?
Can Sri Krishna be called 'a person'?
Car headlights in a world without electricity
What steps are necessary to read a Modern SSD in Medieval Europe?
Why did early computer designers eschew integers?
Why was Sir Cadogan fired?
It it possible to avoid kiwi.com's automatic online check-in and instead do it manually by yourself?
MT "will strike" & LXX "will watch carefully" (Gen 3:15)?
Can I cast Thunderwave and be at the center of its bottom face, but not be affected by it?
How to find if SQL server backup is encrypted with TDE without restoring the backup
Planeswalker Ability and Death Timing
Is a distribution that is normal, but highly skewed, considered Gaussian?
Why did Batya get tzaraat?
How exploitable/balanced is this homebrew spell: Spell Permanency?
subequations: How to continue numbering within subequation?
Calculating discount not working
What is the difference between 'contrib' and 'non-free' packages repositories?
Does Germany produce more waste than the US?
How seriously should I take size and weight limits of hand luggage?
Incomplete cube
How do I keep Mac Emacs from trapping M-`?
What does it mean 'exit 1' for a job status after rclone sync
Mathematica command that allows it to read my intentions
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
2,39121446
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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour 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
