Mixing
Given that $f in L^2(mathbb{T})$ and the sequence of Fourier coefficients $(hat{f_n})in l^1(mathbb{Z})$, must...
3
$begingroup$
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1058
$endgroup$
|
show 5 more comments
3
$begingroup$
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1058
$endgroup$
$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
2
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
2
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
1
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
2
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14
|
show 5 more comments
3
3
3
0
$begingroup$
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1058
$endgroup$
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1058
asked Jan 4 at 1:11
kkckkc
1058
edited Jan 4 at 1:20
kkc
asked Jan 4 at 1:11
kkckkc
1058
asked Jan 4 at 1:11
kkckkc
1058
asked Jan 4 at 1:11
kkckkc
1058
1058
$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
2
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
2
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
1
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
2
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14
|
show 5 more comments
$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
2
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
2
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
1
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
2
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14
$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
2
2
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
2
2
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
1
1
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
2
2
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14
|
show 5 more comments
1 Answer
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If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
add a comment |
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2
$begingroup$
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
add a comment |
2
$begingroup$
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
add a comment |
2
2
2
$begingroup$
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.9k42580
58.9k42580
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
add a comment |
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
$endgroup$
– kkc
Jan 4 at 4:43
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
$begingroup$
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
$endgroup$
– kkc
Jan 4 at 4:45
add a comment |
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$begingroup$
What is $hat{f}_n$?
$endgroup$
– Ben W
Jan 4 at 1:15
2
$begingroup$
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
$endgroup$
– reuns
Jan 4 at 1:26
2
$begingroup$
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
$endgroup$
– Ben W
Jan 4 at 1:32
1
$begingroup$
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
$endgroup$
– reuns
Jan 4 at 3:59
2
$begingroup$
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
$endgroup$
– Ben W
Jan 5 at 14:14