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

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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













3












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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















3












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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













3












3








3





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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












  • 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










1 Answer
1






active

oldest

votes


















6












$begingroup$

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 $u(x)$.






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    1 Answer
    1






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    active

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    6












    $begingroup$

    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 $u(x)$.






    share|cite|improve this answer









    $endgroup$

















      6












      $begingroup$

      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 $u(x)$.






      share|cite|improve this answer









      $endgroup$















        6












        6








        6





        $begingroup$

        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 $u(x)$.






        share|cite|improve this answer









        $endgroup$



        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 $u(x)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 16 at 3:56









        Nate EldredgeNate Eldredge

        20.7k374119




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