Fourier transform of distributions
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First, I'm sorry is a bit too incoherent and too vague - I've only recently started learning about distributions and am just trying to get a general idea of what's happening.
There's a few things I would like to make sure I understand correctly:
1 We work with functions as distributions by identifying a function $f$ with the map $psi mapsto int f psi$. This is a well defined injection into the space of distributions.
2 We can naturally extend the definition of a fourier transform to tempered distributions by defining $F(T)(psi) = T(F(psi))$. To check that this is really an extension, we would like to show that $F(T_f) = T_{F(f)}$. The crucial observation here is that $int f F(g) = int F(f) g$ - this implies $F(T_f)(psi) = int f F(psi) = int F(f) psi = T_{F(f)} (psi)$.
3 Define $rect_n$ as $rect_n(x)=1$ for $x in langle -n,n rangle$ and $0$ otherwise. Then the pointwise limit is $H=1$. Now apparently the fourier transform is continuous, so we should have that $F(H)(psi) = int H F(psi) = psi(0) = delta (psi)$ is the same thing as $lim_{n rightarrow infty} F(rect_n)(psi) = lim_{n rightarrow infty} int F(rect_n) psi = lim_{n rightarrow infty} int F(rect_n) psi$.
This should mean (I think) that $F(rect_n)$ approaches the dirac delta. But $F(rect_n) (xi)= int_{-n}^{n} e^{2pi ixi x} dx = frac{sin(2pi xi N)}{pi xi}$. But that doesn't appear to be the case - the function doesn't pointwise converge to $0$ for say $xi = 1$.
Perhaps my conclusion that it converges to dirac delta is wrong, despite the fact that $lim_{n rightarrow infty} int F(rect_n) psi = psi (0)$. Or maybe I'm not working with the proper definition of continuity in this space.
fourier-analysis fourier-series fourier-transform
asked Jan 17 at 11:15
John PJohn P
54727
$endgroup$
add a comment |
$begingroup$
First, I'm sorry is a bit too incoherent and too vague - I've only recently started learning about distributions and am just trying to get a general idea of what's happening.
There's a few things I would like to make sure I understand correctly:
1 We work with functions as distributions by identifying a function $f$ with the map $psi mapsto int f psi$. This is a well defined injection into the space of distributions.
2 We can naturally extend the definition of a fourier transform to tempered distributions by defining $F(T)(psi) = T(F(psi))$. To check that this is really an extension, we would like to show that $F(T_f) = T_{F(f)}$. The crucial observation here is that $int f F(g) = int F(f) g$ - this implies $F(T_f)(psi) = int f F(psi) = int F(f) psi = T_{F(f)} (psi)$.
3 Define $rect_n$ as $rect_n(x)=1$ for $x in langle -n,n rangle$ and $0$ otherwise. Then the pointwise limit is $H=1$. Now apparently the fourier transform is continuous, so we should have that $F(H)(psi) = int H F(psi) = psi(0) = delta (psi)$ is the same thing as $lim_{n rightarrow infty} F(rect_n)(psi) = lim_{n rightarrow infty} int F(rect_n) psi = lim_{n rightarrow infty} int F(rect_n) psi$.
This should mean (I think) that $F(rect_n)$ approaches the dirac delta. But $F(rect_n) (xi)= int_{-n}^{n} e^{2pi ixi x} dx = frac{sin(2pi xi N)}{pi xi}$. But that doesn't appear to be the case - the function doesn't pointwise converge to $0$ for say $xi = 1$.
Perhaps my conclusion that it converges to dirac delta is wrong, despite the fact that $lim_{n rightarrow infty} int F(rect_n) psi = psi (0)$. Or maybe I'm not working with the proper definition of continuity in this space.
fourier-analysis fourier-series fourier-transform
asked Jan 17 at 11:15
John PJohn P
54727
$endgroup$
1
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
1
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20
add a comment |
$begingroup$
First, I'm sorry is a bit too incoherent and too vague - I've only recently started learning about distributions and am just trying to get a general idea of what's happening.
There's a few things I would like to make sure I understand correctly:
1 We work with functions as distributions by identifying a function $f$ with the map $psi mapsto int f psi$. This is a well defined injection into the space of distributions.
2 We can naturally extend the definition of a fourier transform to tempered distributions by defining $F(T)(psi) = T(F(psi))$. To check that this is really an extension, we would like to show that $F(T_f) = T_{F(f)}$. The crucial observation here is that $int f F(g) = int F(f) g$ - this implies $F(T_f)(psi) = int f F(psi) = int F(f) psi = T_{F(f)} (psi)$.
3 Define $rect_n$ as $rect_n(x)=1$ for $x in langle -n,n rangle$ and $0$ otherwise. Then the pointwise limit is $H=1$. Now apparently the fourier transform is continuous, so we should have that $F(H)(psi) = int H F(psi) = psi(0) = delta (psi)$ is the same thing as $lim_{n rightarrow infty} F(rect_n)(psi) = lim_{n rightarrow infty} int F(rect_n) psi = lim_{n rightarrow infty} int F(rect_n) psi$.
This should mean (I think) that $F(rect_n)$ approaches the dirac delta. But $F(rect_n) (xi)= int_{-n}^{n} e^{2pi ixi x} dx = frac{sin(2pi xi N)}{pi xi}$. But that doesn't appear to be the case - the function doesn't pointwise converge to $0$ for say $xi = 1$.
Perhaps my conclusion that it converges to dirac delta is wrong, despite the fact that $lim_{n rightarrow infty} int F(rect_n) psi = psi (0)$. Or maybe I'm not working with the proper definition of continuity in this space.
fourier-analysis fourier-series fourier-transform
asked Jan 17 at 11:15
John PJohn P
54727
$endgroup$
First, I'm sorry is a bit too incoherent and too vague - I've only recently started learning about distributions and am just trying to get a general idea of what's happening.
There's a few things I would like to make sure I understand correctly:
1 We work with functions as distributions by identifying a function $f$ with the map $psi mapsto int f psi$. This is a well defined injection into the space of distributions.
2 We can naturally extend the definition of a fourier transform to tempered distributions by defining $F(T)(psi) = T(F(psi))$. To check that this is really an extension, we would like to show that $F(T_f) = T_{F(f)}$. The crucial observation here is that $int f F(g) = int F(f) g$ - this implies $F(T_f)(psi) = int f F(psi) = int F(f) psi = T_{F(f)} (psi)$.
3 Define $rect_n$ as $rect_n(x)=1$ for $x in langle -n,n rangle$ and $0$ otherwise. Then the pointwise limit is $H=1$. Now apparently the fourier transform is continuous, so we should have that $F(H)(psi) = int H F(psi) = psi(0) = delta (psi)$ is the same thing as $lim_{n rightarrow infty} F(rect_n)(psi) = lim_{n rightarrow infty} int F(rect_n) psi = lim_{n rightarrow infty} int F(rect_n) psi$.
This should mean (I think) that $F(rect_n)$ approaches the dirac delta. But $F(rect_n) (xi)= int_{-n}^{n} e^{2pi ixi x} dx = frac{sin(2pi xi N)}{pi xi}$. But that doesn't appear to be the case - the function doesn't pointwise converge to $0$ for say $xi = 1$.
Perhaps my conclusion that it converges to dirac delta is wrong, despite the fact that $lim_{n rightarrow infty} int F(rect_n) psi = psi (0)$. Or maybe I'm not working with the proper definition of continuity in this space.
fourier-analysis fourier-series fourier-transform
fourier-analysis fourier-series fourier-transform
asked Jan 17 at 11:15
John PJohn P
54727
asked Jan 17 at 11:15
John PJohn P
54727
asked Jan 17 at 11:15
John PJohn P
54727
asked Jan 17 at 11:15
John PJohn P
54727
asked Jan 17 at 11:15
John PJohn P
54727
54727
1
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
1
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20
add a comment |
1
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
1
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20
1
1
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
1
1
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20
add a comment |
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1
$begingroup$
$frac{sin(2pi xi N)}{pi xi} = N text{sinc}(Nxi)$, $int frac{sin(2pi xi N)}{pi xi} psi(xi)dxi = int text{sinc}(xi) psi(xi/N)dxi$, it $to int text{sinc}(xi) psi(0)dxi$ at least when $psi' in L^1$
$endgroup$
– reuns
Jan 17 at 12:47
1
$begingroup$
$F(rect_n)$ "should" tend to $delta$ _in the sense of distributions", not pointwise. It does.
$endgroup$
– David C. Ullrich
Jan 17 at 15:20