Uniform convergence of generalised Fourier seriesConvergence of Fourier Series of $L^1$ FunctionsConvergence of Fourier series in L^infty-normFejer's theorem and convergence of Fourier series in measureGeneralized Stone Weierstrass theoremAre weak and strong convergence of sequences not equivalent?On the convergence of the the function series $sum_n=0^infty(-1)^nfracf^(n)(x)n!x^n$In what ways is the standard Fourier basis optimal?Integrating a series expansion of $mboxfrac(x)lfloor xrfloor$ coming from Fourier series of sawtooth functionFourier basis for sub-Gaussian spaces?Convergence of Eigenvalues and Eigenvectors for Uniformly Form-Bounded Operators
Uniform convergence of generalised Fourier series
Convergence of Fourier Series of $L^1$ FunctionsConvergence of Fourier series in L^infty-normFejer's theorem and convergence of Fourier series in measureGeneralized Stone Weierstrass theoremAre weak and strong convergence of sequences not equivalent?On the convergence of the the function series $sum_n=0^infty(-1)^nfracf^(n)(x)n!x^n$In what ways is the standard Fourier basis optimal?Integrating a series expansion of $mboxfrac(x)lfloor xrfloor$ coming from Fourier series of sawtooth functionFourier basis for sub-Gaussian spaces?Convergence of Eigenvalues and Eigenvectors for Uniformly Form-Bounded Operators
$begingroup$
Suppose $u_n$ is an orthonormal basis of smooth functions on $S^1$.
Does there exist a smooth function $u$ such that the generalised Fourier series
$$u=sum_ninmathbbN langle u,u_nrangle u_n $$
does not converge uniformly?
We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.
fa.functional-analysis fourier-analysis
asked May 16 at 3:31
GoonfiendGoonfiend
1304
$endgroup$
add a comment |
$begingroup$
Suppose $u_n$ is an orthonormal basis of smooth functions on $S^1$.
Does there exist a smooth function $u$ such that the generalised Fourier series
$$u=sum_ninmathbbN langle u,u_nrangle u_n $$
does not converge uniformly?
We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.
fa.functional-analysis fourier-analysis
asked May 16 at 3:31
GoonfiendGoonfiend
1304
$endgroup$
1
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52
add a comment |
$begingroup$
Suppose $u_n$ is an orthonormal basis of smooth functions on $S^1$.
Does there exist a smooth function $u$ such that the generalised Fourier series
$$u=sum_ninmathbbN langle u,u_nrangle u_n $$
does not converge uniformly?
We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.
fa.functional-analysis fourier-analysis
asked May 16 at 3:31
GoonfiendGoonfiend
1304
$endgroup$
Suppose $u_n$ is an orthonormal basis of smooth functions on $S^1$.
Does there exist a smooth function $u$ such that the generalised Fourier series
$$u=sum_ninmathbbN langle u,u_nrangle u_n $$
does not converge uniformly?
We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.
fa.functional-analysis fourier-analysis
fa.functional-analysis fourier-analysis
asked May 16 at 3:31
GoonfiendGoonfiend
1304
asked May 16 at 3:31
GoonfiendGoonfiend
1304
asked May 16 at 3:31
GoonfiendGoonfiend
1304
asked May 16 at 3:31
GoonfiendGoonfiend
1304
asked May 16 at 3:31
GoonfiendGoonfiend
1304
1304
1
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52
add a comment |
1
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52
1
1
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The series need not converge uniformly, nor even pointwise everywhere.
For instance, fix $x in S^1$ and consider the space $C^infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, dots in C^infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
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%2fmathoverflow.net%2fquestions%2f331643%2funiform-convergence-of-generalised-fourier-series%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
$begingroup$
The series need not converge uniformly, nor even pointwise everywhere.
For instance, fix $x in S^1$ and consider the space $C^infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, dots in C^infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
$endgroup$
add a comment |
$begingroup$
The series need not converge uniformly, nor even pointwise everywhere.
For instance, fix $x in S^1$ and consider the space $C^infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, dots in C^infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
$endgroup$
add a comment |
$begingroup$
The series need not converge uniformly, nor even pointwise everywhere.
For instance, fix $x in S^1$ and consider the space $C^infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, dots in C^infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
$endgroup$
The series need not converge uniformly, nor even pointwise everywhere.
For instance, fix $x in S^1$ and consider the space $C^infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, dots in C^infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
answered May 16 at 3:56
Nate EldredgeNate Eldredge
20.7k374119
20.7k374119
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f331643%2funiform-convergence-of-generalised-fourier-series%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
1
$begingroup$
I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) ne 0$ then the series will not even converge pointwise.
$endgroup$
– Nate Eldredge
May 16 at 3:40
$begingroup$
Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious?
$endgroup$
– Goonfiend
May 16 at 3:46
$begingroup$
Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$.
$endgroup$
– Nate Eldredge
May 16 at 3:48
$begingroup$
Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it.
$endgroup$
– Goonfiend
May 16 at 3:52