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Multi tool use
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 |
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1 Answer
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$begingroup$
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 |
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nl4JAthq,e6
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