Hypotheses on Plancherel's theorem











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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 $fnotin L^2$?










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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 at 22:48















up vote
2
down vote

favorite
1












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 $fnotin L^2$?










share|cite|improve this question






















  • 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 at 22:48













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





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 $fnotin L^2$?










share|cite|improve this question













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






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asked Nov 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 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 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 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 at 22:48










2 Answers
2






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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.






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  • 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 at 9:30


















up vote
0
down vote













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 $fin L^2$.






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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    up vote
    1
    down vote













    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.






    share|cite|improve this answer





















    • 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 at 9:30















    up vote
    1
    down vote













    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.






    share|cite|improve this answer





















    • 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 at 9:30













    up vote
    1
    down vote










    up vote
    1
    down vote









    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 18 at 12:04









    Kavi Rama Murthy

    43.8k31852




    43.8k31852












    • 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 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 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 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 at 9:30










    up vote
    0
    down vote













    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 $fin L^2$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      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 $fin L^2$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        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 $fin L^2$.






        share|cite|improve this answer












        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 $fin L^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 22:37









        DisintegratingByParts

        58k42477




        58k42477






























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