Multiplication formula for Fourier transform












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Citing Stein and Weiss' "Introduction to Fourier Analysis", if $f,gin L^1$ then we have the so-called Multiplication formula:
$$ int hat f g,dx = int f,hat g, dx $$
The proof is just a couple of lines:



$$ int hat f(x) g(x),dx = intleft( int f(t)e^{-2pi i, tx},dtright) g(x), dx = \ intleft( int g(x)e^{-2pi i, tx},dxright) f(t), dt = int f(t),hat g(t), dt$$



My question is: in order to apply Fubini's theorem, shouldn't we require also that $~fgin L^1$? I don't see why the hypotheses of Fubini's theorem are met otherwise.










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    0














    Citing Stein and Weiss' "Introduction to Fourier Analysis", if $f,gin L^1$ then we have the so-called Multiplication formula:
    $$ int hat f g,dx = int f,hat g, dx $$
    The proof is just a couple of lines:



    $$ int hat f(x) g(x),dx = intleft( int f(t)e^{-2pi i, tx},dtright) g(x), dx = \ intleft( int g(x)e^{-2pi i, tx},dxright) f(t), dt = int f(t),hat g(t), dt$$



    My question is: in order to apply Fubini's theorem, shouldn't we require also that $~fgin L^1$? I don't see why the hypotheses of Fubini's theorem are met otherwise.










    share|cite|improve this question

























      0












      0








      0







      Citing Stein and Weiss' "Introduction to Fourier Analysis", if $f,gin L^1$ then we have the so-called Multiplication formula:
      $$ int hat f g,dx = int f,hat g, dx $$
      The proof is just a couple of lines:



      $$ int hat f(x) g(x),dx = intleft( int f(t)e^{-2pi i, tx},dtright) g(x), dx = \ intleft( int g(x)e^{-2pi i, tx},dxright) f(t), dt = int f(t),hat g(t), dt$$



      My question is: in order to apply Fubini's theorem, shouldn't we require also that $~fgin L^1$? I don't see why the hypotheses of Fubini's theorem are met otherwise.










      share|cite|improve this question













      Citing Stein and Weiss' "Introduction to Fourier Analysis", if $f,gin L^1$ then we have the so-called Multiplication formula:
      $$ int hat f g,dx = int f,hat g, dx $$
      The proof is just a couple of lines:



      $$ int hat f(x) g(x),dx = intleft( int f(t)e^{-2pi i, tx},dtright) g(x), dx = \ intleft( int g(x)e^{-2pi i, tx},dxright) f(t), dt = int f(t),hat g(t), dt$$



      My question is: in order to apply Fubini's theorem, shouldn't we require also that $~fgin L^1$? I don't see why the hypotheses of Fubini's theorem are met otherwise.







      integration fourier-analysis lebesgue-integral






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      asked Nov 27 '18 at 23:07









      Manlio

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          $int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.






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            $int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.






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              $int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.






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                $int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.






                share|cite|improve this answer












                $int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.







                share|cite|improve this answer












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                answered Nov 27 '18 at 23:12









                Kavi Rama Murthy

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