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Problem #7.2, MA598R, Summer 2009, Weigel

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Definition of Fourier Transform:

$ \widehat{f} = \int_{-\infty}^{\infty} f(t) e^{-\imath xt} dt $

Definition of Convolution:

$ f\ast g(x) = \int_{-\infty}^{\infty} f(x-y) g(y) dy $

Now, onto the problem:

$ \widehat{f}(x)\widehat{g}(x) = \int_{-\infty}^{\infty} f(t) e^{-\imath xt} dt \int_{-\infty}^{\infty} g(t') e^{-\imath xt'} dt' $

Now, since f and g are both $ L^{1} $, this integral exists, so by Fubini's Theorem, we may rewrite it as:

$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) g(t') e^{-\imath xt - \imath xt'} dt' dt $

Now, use a change of variables (namely let $ t'-t = t' $)

$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) g(t'-t) e^{-\imath xt'} dt' dt $

Now, apply Fubini's Theorem again (since all of these are equalities, we don't need to check that the integral exists, since it's automatic), to get:

$ = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(t) g(t'-t) dt \right) e^{-\imath xt'} dt' = \widehat{f\ast g} $

"And that's all I have to say about that" -Forrest Gump

(Written by Nicholas Stull)

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