prb5.tex (3261B)
1 \include{preamble.tex} 2 3 \begin{document} 4 \maketitle 5 \tableofcontents 6 7 \section{Sheet 5} 8 \subsection{Fourier Transform} 9 In this section we prove the linearity of the Fourier Transform $\mathcal{F}$ on 10 $L^1(\mathbb{R}^d)$. For $f, g \in L^1(\mathbb{R}^d)$ and $\lambda, \mu \in 11 \mathbb{R}$ the linearity condition for $\mathcal{F}$ is the following 12 \begin{align} 13 \mathcal{F}(\lambda f + \mu g) = \lambda \mathcal{F}(f) + \mu 14 \mathcal{F}(g). 15 \end{align} 16 We start by using the Fourier transform definition for $x, \xi \in \mathbb{R}^d$ 17 \begin{align} 18 \mathcal{F}(\lambda f + \mu g)(\xi) &= \int_{\mathbb{R}^d} (\lambda f(x)+ 19 \mu g(x)) e^{-2\pi i \langle x, \xi\rangle}\ dx =\\ 20 &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle} + \mu 21 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 22 &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ 23 \int_{\mathbb{R}^d} \mu 24 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 25 &= \lambda \int_{\mathbb{R}^d} f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ 26 \mu \int_{\mathbb{R}^d} 27 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 28 &= \lambda \mathcal{F}(f)(\xi) + \mu \mathcal{F}(g)(\xi) 29 \end{align} 30 \subsection{Identities of the Fourier transform} 31 The following are three identities of the Fourier transform 32 33 \begin{table}[h!] 34 \centering 35 \begin{tabular}{| l | c | c |} 36 \hline 37 & $g(x)$ & $\hat{g}(\xi)$ \\ \hline \hline 38 (1) & $f(x-x_0)$ & $e^{-2\pi ix_0 \xi} \hat{f}(\xi)$ \\ \hline 39 (2) & $e^{2\pi i \xi_0 x} f(x)$ & $f(\xi - \xi_0)$ \\ \hline 40 (3) & $f(ax)$ & $\frac{1}{a} \hat{f}(\frac{\xi}{a})$\\ \hline 41 \end{tabular} 42 \caption{Identities of the Fourier transform for $a > 0, 43 \xi_0, x \in \mathbb{R}$} 44 \end{table} 45 We start with (1) 46 \begin{align} 47 \widehat{f(x-x_0)} 48 &= \int_\mathbb{R} f(x-x_0) e^{-2\pi i x \xi}\ dx= 49 \;\;\;\;\;\; (y = x-x_0)\\ 50 &= \int_\mathbb{R} f(y) e^{-2\pi i (y+x_0) \xi}\ 51 dy=\\ 52 &= e^{-2\pi i x_0 \xi} \int_\mathbb{R}f(y)e^{-2\pi i y 53 \xi}\ dy=\\ 54 &= e^{-2\pi i x_0 \xi} \hat{f}(\xi). 55 \end{align} 56 For (2) we have 57 \begin{align} 58 \widehat{e^{2\pi i x \xi_0} f(x)} 59 &= \int_\mathbb{R} e^{2\pi i x \xi_0} f(x) e^{-2\pi i x \xi}\ dx =\\ 60 &= \int_\mathbb{R} f(x) e^{-2\pi i x (\xi -\xi_0)}\ dx=\\ 61 &= \hat{f}(\xi - \xi_0). 62 \end{align} 63 For (3) we have 64 \begin{align} 65 \widehat{f(ax)} 66 &= \int_\mathbb{R} f(ax) e^{-2\pi i \xi x}\ dx = \qquad \text{sub: 67 $(y=ax)$}\\ 68 &= \int_\mathbb{R} \frac{1}{a}f(y) e^{-2\pi i \frac{\xi}{a} y}\ dy=\\ 69 &= \frac{1}{a} \hat{f}\left(\frac{\xi}{a}\right). 70 \end{align} 71 \subsection{The Box-Function} 72 Consider the following Box-Function 73 \begin{align} 74 \Pi(x) := 75 \begin{cases} 76 1\;\;\;\;\;\; -\frac{3}{2} < x < \frac{1}{2}\\ 77 0\;\;\;\;\; \text{else} 78 \end{cases} 79 \end{align} 80 The Fourier transform of this function is 81 \begin{align} 82 \widehat{\Pi(x)} 83 &= \int_\mathbb{R} \Pi(x) e^{-2\pi i x\xi}\ dx=\\ 84 &= \int_{-\frac{3}{2}}^{\frac{1}{2}} e^{-2\pi i x \xi}\ dx 85 =\frac{-1}{2\pi i \xi} e^{-2\pi i x\xi} 86 \bigg|_{-\frac{3}{2}}^{\frac{1}{2}}=\\ 87 &= \frac{1}{2\pi i \xi} \left(e^{3\pi i \xi} - e^{-\pi i \xi}\right)=\\ 88 &= \frac{e^{\pi i \xi}\sin(2\pi\xi)}{\pi \xi}. 89 \end{align} 90 91 %\printbibliography 92 \end{document}