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Friday, March 17. 2006

Properties of the Fourier transform, II

I think I found out where the mentioned property comes from: The solution lies in the application of a substitution in the mentioned integral. It is well known that a translation of any (Lebesgue integrable) function leads to a modulation of its Fourier transform, i.e., $\mathscr{F}(\mathrm{T}_a f) = \mathrm{M}_{-a}\mathscr{F}f$ , where $\mathrm{T}_a f(x) := f(x-a)$ and $\mathrm{M}_b f(x) := e^{2\pi\mathrm{i}bx}f(x)$ for $x\in\mathbb{R}$ . But, it is even a common property for any $\mathrm{L}^1$ -function $f$ that $\mathscr{F}f_{1/a}(\xi) = |a|\mathscr{F}f(a\xi)$ , where we again use the symbolism $f_a(x):=f(ax)$ . This becomes clear by using the substitution $x:=ay,\quad dx=|a|dy$ :

$\begin{eqnarray*}\displaystyle\mathscr{F}f_{1/a}(\xi) &=& \int_{\mathbb{R}}f\left(\frac{x}{a}\right)e^{-2\pi\mathrm{i}x\xi}\,dx\\ &=& |a|\int_{\mathbb{R}}f(y)e^{-2\pi\mathrm{i}ay\xi}\,dy\\ &=& |a|\mathscr{F}f(a\xi)\end{eqnarray*}$

So, it’s clear () that this means $\mathscr{F}f_a(\xi) = \frac{1}{|a|}\mathscr{F}f\left(\frac{\xi}{a}\right)\quad\forall f\in\mathrm{L}^1$ , and not only for the Gaussian function $\gamma(x) = e^{-x^2/2}$ . And, for $x\in\mathbb{R}^n$ , the substitution $x:=ay$ leads to $dx = |a|^n dy$ , because if $x_i := ay_i$ , then $dx_i = |a|dy_i$ and therefore $\textstyle\prod_{i=1}^ndx_i = |a|^n\prod_{i=1}^n dy_i$ . I’ll have to recall what the “value of the functional determinant” is.

Regarding the proof that the Gaussian function is an eigenfunction of the Fourier transform, HGFei once handed us one out which seems to be more elegant than the proofs which use the well definedness of solutions for initial value problems of linear common differential equations.

Posted by Stephan Paukner in Master's Thesis at 10:18 | Comments (0) | Trackbacks (0)

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