Multiplication formula for Fourier transform
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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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
add a comment |
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
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
integration fourier-analysis lebesgue-integral
asked Nov 27 '18 at 23:07
Manlio
902719
902719
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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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1 Answer
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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.
$int int |f(t)g(x)|, dt, dx=(int|f(t)|, dt )(int |g(x)|, dx) <infty$ so Fubini's Theorem is applicable.
answered Nov 27 '18 at 23:12
Kavi Rama Murthy
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