Mixing
2$pi$-peridodic and continuous function with non summable Fourier coeffficients
$begingroup$
Can someone tell me an example of 2$pi$-periodic and continuous function, f, with Fourier coefficients $hat{f}(n);forall{nin{mathbb{Z}}}$ such that $sum|hat{f}(n)|>infty$?
Thanks.
real-analysis fourier-analysis fourier-series
asked Jan 1 at 11:47
mathlifemathlife
629
$endgroup$
add a comment |
$begingroup$
Can someone tell me an example of 2$pi$-periodic and continuous function, f, with Fourier coefficients $hat{f}(n);forall{nin{mathbb{Z}}}$ such that $sum|hat{f}(n)|>infty$?
Thanks.
real-analysis fourier-analysis fourier-series
asked Jan 1 at 11:47
mathlifemathlife
629
$endgroup$
$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09
add a comment |
$begingroup$
Can someone tell me an example of 2$pi$-periodic and continuous function, f, with Fourier coefficients $hat{f}(n);forall{nin{mathbb{Z}}}$ such that $sum|hat{f}(n)|>infty$?
Thanks.
real-analysis fourier-analysis fourier-series
asked Jan 1 at 11:47
mathlifemathlife
629
$endgroup$
Can someone tell me an example of 2$pi$-periodic and continuous function, f, with Fourier coefficients $hat{f}(n);forall{nin{mathbb{Z}}}$ such that $sum|hat{f}(n)|>infty$?
Thanks.
real-analysis fourier-analysis fourier-series
real-analysis fourier-analysis fourier-series
asked Jan 1 at 11:47
mathlifemathlife
629
asked Jan 1 at 11:47
mathlifemathlife
629
asked Jan 1 at 11:47
mathlifemathlife
629
asked Jan 1 at 11:47
mathlifemathlife
629
asked Jan 1 at 11:47
mathlifemathlife
629
629
$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09
add a comment |
$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09
$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Convergence of Fourier Series is a delicate business but everything is known. The theorem you are looking for is the following:
(Du Bois-Reymond 1873) There is a continuous function $f: [-pi, pi] rightarrow mathbb{C}$ such that $limsup_{N} sum_{|n| le N} widehat{f}(n) rightarrow infty$.
A standard proof of this result can be found all over the internet but I think the most elementary construction is given here.
Just as a reference, if you know that your function is in $C^1$ then the Fourier series converges uniformly. Furthermore, if you know that your function is in $L^p$ for $p > 1$ then your Fourier series converges pointwise for everything in your compact interval except possibly on a set of measure $0$. This is the celebrated Carleson-Hunt Theorem.
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Convergence of Fourier Series is a delicate business but everything is known. The theorem you are looking for is the following:
(Du Bois-Reymond 1873) There is a continuous function $f: [-pi, pi] rightarrow mathbb{C}$ such that $limsup_{N} sum_{|n| le N} widehat{f}(n) rightarrow infty$.
A standard proof of this result can be found all over the internet but I think the most elementary construction is given here.
Just as a reference, if you know that your function is in $C^1$ then the Fourier series converges uniformly. Furthermore, if you know that your function is in $L^p$ for $p > 1$ then your Fourier series converges pointwise for everything in your compact interval except possibly on a set of measure $0$. This is the celebrated Carleson-Hunt Theorem.
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
$endgroup$
add a comment |
$begingroup$
Convergence of Fourier Series is a delicate business but everything is known. The theorem you are looking for is the following:
(Du Bois-Reymond 1873) There is a continuous function $f: [-pi, pi] rightarrow mathbb{C}$ such that $limsup_{N} sum_{|n| le N} widehat{f}(n) rightarrow infty$.
A standard proof of this result can be found all over the internet but I think the most elementary construction is given here.
Just as a reference, if you know that your function is in $C^1$ then the Fourier series converges uniformly. Furthermore, if you know that your function is in $L^p$ for $p > 1$ then your Fourier series converges pointwise for everything in your compact interval except possibly on a set of measure $0$. This is the celebrated Carleson-Hunt Theorem.
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
$endgroup$
add a comment |
$begingroup$
Convergence of Fourier Series is a delicate business but everything is known. The theorem you are looking for is the following:
(Du Bois-Reymond 1873) There is a continuous function $f: [-pi, pi] rightarrow mathbb{C}$ such that $limsup_{N} sum_{|n| le N} widehat{f}(n) rightarrow infty$.
A standard proof of this result can be found all over the internet but I think the most elementary construction is given here.
Just as a reference, if you know that your function is in $C^1$ then the Fourier series converges uniformly. Furthermore, if you know that your function is in $L^p$ for $p > 1$ then your Fourier series converges pointwise for everything in your compact interval except possibly on a set of measure $0$. This is the celebrated Carleson-Hunt Theorem.
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
$endgroup$
Convergence of Fourier Series is a delicate business but everything is known. The theorem you are looking for is the following:
(Du Bois-Reymond 1873) There is a continuous function $f: [-pi, pi] rightarrow mathbb{C}$ such that $limsup_{N} sum_{|n| le N} widehat{f}(n) rightarrow infty$.
A standard proof of this result can be found all over the internet but I think the most elementary construction is given here.
Just as a reference, if you know that your function is in $C^1$ then the Fourier series converges uniformly. Furthermore, if you know that your function is in $L^p$ for $p > 1$ then your Fourier series converges pointwise for everything in your compact interval except possibly on a set of measure $0$. This is the celebrated Carleson-Hunt Theorem.
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
answered Jan 1 at 15:00
Sandeep SilwalSandeep Silwal
5,84311236
5,84311236
add a comment |
add a comment |
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$begingroup$
try some continuous function that approximates the sign function in $[-pi,pi]$. My idea comes from the fact that the Fourier coefficients of the periodic extension of the sign function in $[-pi,pi]$ are $1/n$ for odd $n$ (an zero for even $n$)
$endgroup$
– Masacroso
Jan 1 at 12:10
$begingroup$
You may also need to prevent your function from being differentiable in order to force the coefficients to be large.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:12
$begingroup$
I have tried a Weirstrass type function: $f(t)=sumfrac{1}{n!}cos((n!)^2t)$. But it doesn't work no?
$endgroup$
– mathlife
Jan 1 at 12:20
$begingroup$
Would $sin(csc x)sin x$ work with the value zero at multiples of $pi$ work? The function is continuous but nondifferentiable at these multiples of $pi$.
$endgroup$
– Oscar Lanzi
Jan 1 at 12:29
$begingroup$
I'm quite sure there is no function whose Fourier coefficients sum to a value strictly greater than infinity.
$endgroup$
– md2perpe
Jan 1 at 14:09