Mixing
Hypotheses on Plancherel's theorem
Plancherel's theorem is stated as (e.g. in Rudin's Real and Complex Analysis)
If $fin L^1 cap L^2$ then
$$ |f|_2 = |hat f|_2 $$
where $hat f$ is the Fourier transform of $f$. On the other hand, Parseval's formula
$$ int f,overline{g}, d x = int hat{f}~overline{hat{g}}, d x$$
should hold whenever $f,hat f, gin L^1$.
My question is: is the requirement $fin L^2$ in Plancherel's theorem needed just to have the two norms to be finite or is there some (more or less hidden) detail that I'm missing and that makes the statement to actually be false if $fnotin L^2$?
real-analysis fourier-analysis lebesgue-integral parsevals-identity
asked Nov 18 '18 at 11:30
Manlio
902719
add a comment |
Plancherel's theorem is stated as (e.g. in Rudin's Real and Complex Analysis)
If $fin L^1 cap L^2$ then
$$ |f|_2 = |hat f|_2 $$
where $hat f$ is the Fourier transform of $f$. On the other hand, Parseval's formula
$$ int f,overline{g}, d x = int hat{f}~overline{hat{g}}, d x$$
should hold whenever $f,hat f, gin L^1$.
My question is: is the requirement $fin L^2$ in Plancherel's theorem needed just to have the two norms to be finite or is there some (more or less hidden) detail that I'm missing and that makes the statement to actually be false if $fnotin L^2$?
real-analysis fourier-analysis lebesgue-integral parsevals-identity
asked Nov 18 '18 at 11:30
Manlio
902719
If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48
add a comment |
Plancherel's theorem is stated as (e.g. in Rudin's Real and Complex Analysis)
If $fin L^1 cap L^2$ then
$$ |f|_2 = |hat f|_2 $$
where $hat f$ is the Fourier transform of $f$. On the other hand, Parseval's formula
$$ int f,overline{g}, d x = int hat{f}~overline{hat{g}}, d x$$
should hold whenever $f,hat f, gin L^1$.
My question is: is the requirement $fin L^2$ in Plancherel's theorem needed just to have the two norms to be finite or is there some (more or less hidden) detail that I'm missing and that makes the statement to actually be false if $fnotin L^2$?
real-analysis fourier-analysis lebesgue-integral parsevals-identity
asked Nov 18 '18 at 11:30
Manlio
902719
Plancherel's theorem is stated as (e.g. in Rudin's Real and Complex Analysis)
If $fin L^1 cap L^2$ then
$$ |f|_2 = |hat f|_2 $$
where $hat f$ is the Fourier transform of $f$. On the other hand, Parseval's formula
$$ int f,overline{g}, d x = int hat{f}~overline{hat{g}}, d x$$
should hold whenever $f,hat f, gin L^1$.
My question is: is the requirement $fin L^2$ in Plancherel's theorem needed just to have the two norms to be finite or is there some (more or less hidden) detail that I'm missing and that makes the statement to actually be false if $fnotin L^2$?
real-analysis fourier-analysis lebesgue-integral parsevals-identity
real-analysis fourier-analysis lebesgue-integral parsevals-identity
asked Nov 18 '18 at 11:30
Manlio
902719
asked Nov 18 '18 at 11:30
Manlio
902719
asked Nov 18 '18 at 11:30
Manlio
902719
asked Nov 18 '18 at 11:30
Manlio
902719
asked Nov 18 '18 at 11:30
Manlio
902719
902719
If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48
add a comment |
If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48
If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48
If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48
add a comment |
2 Answers
2
active
oldest
votes
If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
add a comment |
If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$
If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$
However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
add a comment |
If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
add a comment |
If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
If $f,g in L^{1}$ there is no reason $int f overline {g} , dx$ exists. Ex: $f(x)=g(x)=x^{-1/2}$ for $0<|x|<1$ and $0$ otherwise.
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
answered Nov 18 '18 at 12:04
Kavi Rama Murthy
50.3k31854
50.3k31854
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
add a comment |
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
Doesn't $int f,bar g dx$ simply diverges in your example? The same happens to the right-hand side of Parseval's equality, so you would still have $infty = infty$. Sorry, I'm probably missing something very basic.
– Manlio
Nov 19 '18 at 9:30
add a comment |
If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$
If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$
However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
add a comment |
If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$
If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$
However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
add a comment |
If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$
If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$
However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
If $f,gin L^1$, then
$$
int f hat{g}dx = int ghat{f}dx.
$$
If $f,hat{f} in L^1$, then $f= (hat{f})^{vee}$, which allows you to swap $f$ and $hat{f}$ in the above in order to obtain the identity that you need. Then you need
$$
f,hat{f},g in L^1.
$$
However, $|f|_2 = |hat{f}|_2$ makes sense for any $fin L^2$ because $hat{f}$ can be defined as the $L^2$ limit of the truncated Fourier transform integral:
$$
hat{f} = L^2-lim_{Rrightarrowinfty}widehat{fchi_{[-R,R]}}
$$
and $fchi_{[-R,R]}in L^1cap L^2$ for $fin L^2$.
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
answered Nov 19 '18 at 22:37
DisintegratingByParts
58.5k42579
58.5k42579
add a comment |
add a comment |
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If $p>2$, the this link says that there exists $fin L^p$ whose Fourier transform, as in distribution sense, is not a proper function. So it seems that we do need certain restrictions on the regularity of $f$.
– Sangchul Lee
Nov 19 '18 at 22:48