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1 \documentclass[a4paper]{article} 2 3 4 \usepackage[T1]{fontenc} 5 \usepackage[utf8]{inputenc} 6 \usepackage{mlmodern} 7 8 %\usepackage{ngerman} % Sprachanpassung Deutsch 9 10 \usepackage{graphicx} 11 \usepackage{geometry} 12 \geometry{a4paper, top=15mm} 13 14 \usepackage{subcaption} 15 \usepackage[shortlabels]{enumitem} 16 \usepackage{amssymb} 17 \usepackage{amsthm} 18 \usepackage{mathtools} 19 \usepackage{braket} 20 \usepackage{bbm} 21 \usepackage{graphicx} 22 \usepackage{float} 23 \usepackage{yhmath} 24 \usepackage{tikz} 25 \usetikzlibrary{patterns,decorations.pathmorphing,positioning} 26 \usetikzlibrary{calc,decorations.markings} 27 28 %\usepackage[backend=biber, sorting=none]{biblatex} 29 %\addbibresource{uni.bib} 30 31 \usepackage[framemethod=TikZ]{mdframed} 32 33 \tikzstyle{titlered} = 34 [draw=black, thick, fill=white,% 35 text=black, rectangle, 36 right, minimum height=.7cm] 37 38 39 \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} 40 \usepackage[parfill]{parskip} 41 \usepackage{lipsum} 42 43 44 \usepackage{tcolorbox} 45 \tcbuselibrary{skins,breakable} 46 47 \pagestyle{myheadings} 48 49 \markright{Popović\hfill Applied Analysis\hfill} 50 51 52 \title{University of Vienna\\ Faculty of Mathematics\\ 53 \vspace{1cm}Applied Analysis Problems 54 } 55 \author{Milutin Popovic} 56 57 \begin{document} 58 \maketitle 59 \tableofcontents 60 61 \section{Sheet 5} 62 \subsection{Fourier Transform} 63 In this section we prove the linearity of the Fourier Transform $\mathcal{F}$ on 64 $L^1(\mathbb{R}^d)$. For $f, g \in L^1(\mathbb{R}^d)$ and $\lambda, \mu \in 65 \mathbb{R}$ the linearity condition for $\mathcal{F}$ is the following 66 \begin{align} 67 \mathcal{F}(\lambda f + \mu g) = \lambda \mathcal{F}(f) + \mu 68 \mathcal{F}(g). 69 \end{align} 70 We start by using the Fourier transform definition for $x, \xi \in \mathbb{R}^d$ 71 \begin{align} 72 \mathcal{F}(\lambda f + \mu g)(\xi) &= \int_{\mathbb{R}^d} (\lambda f(x)+ 73 \mu g(x)) e^{-2\pi i \langle x, \xi\rangle}\ dx =\\ 74 &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle} + \mu 75 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 76 &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ 77 \int_{\mathbb{R}^d} \mu 78 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 79 &= \lambda \int_{\mathbb{R}^d} f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+ 80 \mu \int_{\mathbb{R}^d} 81 g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\ 82 &= \lambda \mathcal{F}(f)(\xi) + \mu \mathcal{F}(g)(\xi) 83 \end{align} 84 \subsection{Identities of the Fourier transform} 85 The following are three identities of the Fourier transform 86 87 \begin{table}[h!] 88 \centering 89 \begin{tabular}{| l | c | c |} 90 \hline 91 & $g(x)$ & $\hat{g}(\xi)$ \\ \hline \hline 92 (1) & $f(x-x_0)$ & $e^{-2\pi ix_0 \xi} \hat{f}(\xi)$ \\ \hline 93 (2) & $e^{2\pi i \xi_0 x} f(x)$ & $f(\xi - \xi_0)$ \\ \hline 94 (3) & $f(ax)$ & $\frac{1}{a} \hat{f}(\frac{\xi}{a})$\\ \hline 95 \end{tabular} 96 \caption{Identities of the Fourier transform for $a > 0, 97 \xi_0, x \in \mathbb{R}$} 98 \end{table} 99 We start with (1) 100 \begin{align} 101 \widehat{f(x-x_0)} 102 &= \int_\mathbb{R} f(x-x_0) e^{-2\pi i x \xi}\ dx= 103 \;\;\;\;\;\; (y = x-x_0)\\ 104 &= \int_\mathbb{R} f(y) e^{-2\pi i (y+x_0) \xi}\ 105 dy=\\ 106 &= e^{-2\pi i x_0 \xi} \int_\mathbb{R}f(y)e^{-2\pi i y 107 \xi}\ dy=\\ 108 &= e^{-2\pi i x_0 \xi} \hat{f}(\xi). 109 \end{align} 110 For (2) we have 111 \begin{align} 112 \widehat{e^{2\pi i x \xi_0} f(x)} 113 &= \int_\mathbb{R} e^{2\pi i x \xi_0} f(x) e^{-2\pi i x \xi}\ dx =\\ 114 &= \int_\mathbb{R} f(x) e^{-2\pi i x (\xi -\xi_0)}\ dx=\\ 115 &= \hat{f}(\xi - \xi_0). 116 \end{align} 117 For (3) we have 118 \begin{align} 119 \widehat{f(ax)} 120 &= \int_\mathbb{R} f(ax) e^{-2\pi i \xi x}\ dx = \qquad \text{sub: 121 $(y=ax)$}\\ 122 &= \int_\mathbb{R} \frac{1}{a}f(y) e^{-2\pi i \frac{\xi}{a} y}\ dy=\\ 123 &= \frac{1}{a} \hat{f}\left(\frac{\xi}{a}\right). 124 \end{align} 125 \subsection{The Box-Function} 126 Consider the following Box-Function 127 \begin{align} 128 \Pi(x) := 129 \begin{cases} 130 1\;\;\;\;\;\; -\frac{3}{2} < x < \frac{1}{2}\\ 131 0\;\;\;\;\; \text{else} 132 \end{cases} 133 \end{align} 134 The Fourier transform of this function is 135 \begin{align} 136 \widehat{\Pi(x)} 137 &= \int_\mathbb{R} \Pi(x) e^{-2\pi i x\xi}\ dx=\\ 138 &= \int_{-\frac{3}{2}}^{\frac{1}{2}} e^{-2\pi i x \xi}\ dx 139 =\frac{-1}{2\pi i \xi} e^{-2\pi i x\xi} 140 \bigg|_{-\frac{3}{2}}^{\frac{1}{2}}=\\ 141 &= \frac{1}{2\pi i \xi} \left(e^{3\pi i \xi} - e^{-\pi i \xi}\right)=\\ 142 &= \frac{e^{\pi i \xi}\sin(2\pi\xi)}{\pi \xi}. 143 \end{align} 144 145 %\printbibliography 146 \end{document}