(New page: 7.2 Define the Fourier transform of <math>f \in L^1(\mathbb{R})</math> by <math>\widehat{f}(x) = \int_{-\infty}^{\infty} f(t) e^{-ixt}dt</math> If <math>f</math>, <math>g \in L(\mathbb{R...) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | 7.2 Define the Fourier transform of <math>f \in L^1(\mathbb{R})</math> by | + | [[Category:MA598RSummer2009pweigel]] |
| + | [[Category:MA598]] | ||
| + | [[Category:math]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:real analysis]] | ||
| + | |||
| + | == Problem #7.2, MA598R, Summer 2009, Weigel == | ||
| + | Define the Fourier transform of <math>f \in L^1(\mathbb{R})</math> by | ||
<math>\widehat{f}(x) = \int_{-\infty}^{\infty} f(t) e^{-ixt}dt</math> | <math>\widehat{f}(x) = \int_{-\infty}^{\infty} f(t) e^{-ixt}dt</math> | ||
| Line 6: | Line 13: | ||
<math>\widehat{f * g}(x) = \widehat{f}(x)\widehat{g}(x)</math> | <math>\widehat{f * g}(x) = \widehat{f}(x)\widehat{g}(x)</math> | ||
| − | + | ---- | |
| − | Proof | + | ==Proof== |
| + | Applying the definitions of Fourier transform and convolution, followed by Fubini (since <math>f, g \in L(\mathbb{R})</math>) we have: | ||
<math>\begin{align}\widehat{f * g}(x) &= \int_{\mathbb{R}}(f * g)(t)e^{-ixt}dt\\ | <math>\begin{align}\widehat{f * g}(x) &= \int_{\mathbb{R}}(f * g)(t)e^{-ixt}dt\\ | ||
| Line 16: | Line 24: | ||
&= \widehat{f}(x)\widehat{g}(x) | &= \widehat{f}(x)\widehat{g}(x) | ||
\end{align}</math> | \end{align}</math> | ||
| + | ---- | ||
| + | [[The_Ninja%27s_Solutions|Back to Ninja Solutions]] | ||
| + | |||
| + | [[MA_598R_pweigel_Summer_2009_Lecture_7|Back to Assignment 7]] | ||
| + | |||
| + | [[MA598R_%28WeigelSummer2009%29|Back to MA598R Summer 2009]] | ||
Latest revision as of 05:52, 11 June 2013
Problem #7.2, MA598R, Summer 2009, Weigel
Define the Fourier transform of $ f \in L^1(\mathbb{R}) $ by
$ \widehat{f}(x) = \int_{-\infty}^{\infty} f(t) e^{-ixt}dt $
If $ f $, $ g \in L(\mathbb{R}) $, show
$ \widehat{f * g}(x) = \widehat{f}(x)\widehat{g}(x) $
Proof
Applying the definitions of Fourier transform and convolution, followed by Fubini (since $ f, g \in L(\mathbb{R}) $) we have:
$ \begin{align}\widehat{f * g}(x) &= \int_{\mathbb{R}}(f * g)(t)e^{-ixt}dt\\ &= \int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(t-y)g(y)dy\right)e^{-ixt}dt\\ &= \int_{\mathbb{R}}g(y)\left(\int_{\mathbb{R}}f(t-y)e^{-ixt}dt\right)dy\\ &= \int_{\mathbb{R}}g(y)e^{-ixy}\left(\int_{\mathbb{R}}f(t-y)e^{-ix(t-y)}dt\right)dy\\ &= \int_{\mathbb{R}}g(y)e^{-ixy}\widehat{f}(x)dy\\ &= \widehat{f}(x)\widehat{g}(x) \end{align} $
