Function with uniformly but non absolutely convergent Fourier series The Next CEO of Stack OverflowPointwise but not uniform convergence of a Fourier seriesA function whose derivatives always have a convergent fourier seriesProve that the Fourier series of $dfrac1f$ is absolutely convergent.Absolutely convergent but not convergentAbsolute convergence of fourier seriesAbsolutely but not uniformly convergentconditionally convergent but not absolutely convergent seriesConvergent Fourier series of continuous functionevery absolutely convergence in interval of convergence is uniformlyContinuous periodic function with Fourier series behaving like $1/n$?
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Function with uniformly but non absolutely convergent Fourier series
The Next CEO of Stack OverflowPointwise but not uniform convergence of a Fourier seriesA function whose derivatives always have a convergent fourier seriesProve that the Fourier series of $dfrac1f$ is absolutely convergent.Absolutely convergent but not convergentAbsolute convergence of fourier seriesAbsolutely but not uniformly convergentconditionally convergent but not absolutely convergent seriesConvergent Fourier series of continuous functionevery absolutely convergence in interval of convergence is uniformlyContinuous periodic function with Fourier series behaving like $1/n$?
0
$begingroup$
Is there an example of a periodic continuous function on $mathbbR$ such that its Fourier series is uniformly convergent (to $f$), but it is not absolutely convergent ?
real-analysis sequences-and-series fourier-analysis
asked Mar 28 at 9:19
Phil-WPhil-W
32818
$endgroup$
add a comment |
0
$begingroup$
Is there an example of a periodic continuous function on $mathbbR$ such that its Fourier series is uniformly convergent (to $f$), but it is not absolutely convergent ?
real-analysis sequences-and-series fourier-analysis
asked Mar 28 at 9:19
Phil-WPhil-W
32818
$endgroup$
add a comment |
0
0
0
$begingroup$
Is there an example of a periodic continuous function on $mathbbR$ such that its Fourier series is uniformly convergent (to $f$), but it is not absolutely convergent ?
real-analysis sequences-and-series fourier-analysis
asked Mar 28 at 9:19
Phil-WPhil-W
32818
$endgroup$
Is there an example of a periodic continuous function on $mathbbR$ such that its Fourier series is uniformly convergent (to $f$), but it is not absolutely convergent ?
real-analysis sequences-and-series fourier-analysis
real-analysis sequences-and-series fourier-analysis
asked Mar 28 at 9:19
Phil-WPhil-W
32818
asked Mar 28 at 9:19
Phil-WPhil-W
32818
asked Mar 28 at 9:19
Phil-WPhil-W
32818
asked Mar 28 at 9:19
Phil-WPhil-W
32818
asked Mar 28 at 9:19
Phil-WPhil-W
32818
32818
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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$begingroup$
$sum frac sin (nx) nlog , n$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
|
show 1 more comment
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1 Answer
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1 Answer
1
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oldest
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active
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0
$begingroup$
$sum frac sin (nx) nlog , n$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
|
show 1 more comment
0
$begingroup$
$sum frac sin (nx) nlog , n$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
|
show 1 more comment
0
0
0
$begingroup$
$sum frac sin (nx) nlog , n$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$sum frac sin (nx) nlog , n$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 28 at 9:25
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
71.5k53170
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
|
show 1 more comment
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
If you assume the well-known fact that $sum frac sin (nx) n$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^-a, a > frac12$ and even $a_n$ ~ $n^-frac12(log n)^-1-b, b>0$ which is best we can do since by Plancherel we know that $Sigma^2 < infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $Sigma = infty$ for any $b>0$
$endgroup$
– Conrad
Mar 28 at 12:31
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
Use $Sigmafrac 1 nlog , n = infty$ and $x=fracpi2$ and the absolute sum is the odd part of $Sigmafrac 1 nlog , n$
$endgroup$
– Conrad
Mar 28 at 12:47
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
@Conrad You are right.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 13:00
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive
$endgroup$
– Conrad
Mar 28 at 14:12
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
$begingroup$
@Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $sum a_n sin(nx)$ is uniformly convergent iff $na_n to 0$.
$endgroup$
– Kavi Rama Murthy
Mar 28 at 23:21
|
show 1 more comment
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