141 lines
5.8 KiB
TeX
141 lines
5.8 KiB
TeX
% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Fourier Transform on $L^2(\realn^d)$}
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\begin{defi}[Hilbert space]
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For this section we introduce the Hilbert space of Lebesgue square-integrable functions
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\[
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L^2(\realn^d) := \set[\norm{f}_{L^2}^2 = \int_{\realn^d} \abs{f(x)}^2 \dd{x} < \infty]{f: \realn^d \rightarrow \cmpln \text{ measurable}}
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\]
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This space is also important in quantum mechanics, as wave functions are elements of $L^2$.
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\end{defi}
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\begin{defi}
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The space $L^2(\realn^d)$ is a Hilbert space, i.e. a complete, normed vector space with an inner product
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\[
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\innerproduct{f}{g} := \int_{\realn^d} \conj{f(x)} g(x) \dd{x}
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\]
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that has the following properties:
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\begin{enumerate}[(i)]
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\item $\innerproduct{f}{f} \ge 0$ and $\innerproduct{f}{f} = 0 \iff f = 0$
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\item $\innerproduct{f}{g} = \conj{\innerproduct{g}{f}}$
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\item $\innerproduct{f}{g + \lambda h} = \innerproduct{f}{g} + \lambda\innerproduct{f}{h}$
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\end{enumerate}
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(ii) and (iii) imply
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\[
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\innerproduct{\lambda f}{g} = \conj{\lambda} \innerproduct{f}{g}
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\]
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The inner product induces a norm
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\[
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\norm{f}_{L^2}^2 = \innerproduct{f}{f} \left( = \int_{\realn} \underbrace{\conj{f(x)}f(x)}_{\abs{f(x)}^2} \dd{x} \right)
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\]
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Since the Fourier transform cannot directly be defined for $L^2(\realn^d)$, we will first consider the space of rapidly decreasing functions,
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the so called \textit{Schwartz space} $\mathcal{S}(\realn^d)$.
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\end{defi}
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\begin{defi}[Schwartz space]
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The Schwartz space $\mathcal{S}(\realn^d)$ is defined as the function space
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\[
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\mathcal{S}(\realn^d) := \set[x \mapsto x^{\beta} \partial^{\alpha} f \text{ bounded,} \quad \forall \alpha, \beta \in \natn_0^d]{f \in C^{\infty}(\realn^d)}
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\]
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\end{defi}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item Smooth functions with compact support $f \in C^{\infty}(\realn^d)$ are also elements of $\mathcal{S}(\realn^d)$, for example
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\[
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f(x) = \begin{cases}
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\exp\left(-\sum_{j=1}^d \rec{1 - \abs{x_j}^2}\right), & \abs{x_j} < 1 \\
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0, & \text{else}
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\end{cases}
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\]
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\item For every polynomial $p(x)$, the function
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\[
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f(x) = p(x)e^{-\abs{x}^2}
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\]
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defines a function in $\mathcal{S}(\realn^d)$.
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\end{enumerate}
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\end{eg}
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\begin{rem}
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Because of the continuity and the rapid decrease towards infinity we can find that
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\[
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\mathcal{S}(\realn^d) \subset L^1(\realn^d) \cap L^2(\realn^d)
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\]
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and one can show that $\mathcal{S}(\realn^d)$ is dense in $L^2(\realn^d)$, i.e.
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\[
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\forall f \in L^2(\realn^d) ~\exists \anyseqdef[f]{\mathcal{S}(\realn^d)}: \quad \norm{f_n - f}_{L^2} \conv{n \rightarrow \infty} 0
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\]
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\end{rem}
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\begin{thm}
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Let $f \in \mathcal{S}(\realn^d)$. Then $\hat{f} \in \mathcal{S}(\realn^d)$ and the restriction of the Fourier transform to $\mathcal{S}(\realn^d)$
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\[
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\mathcal{F}_{\mathcal{S}}: \mathcal{S}(\realn^d) \longrightarrow \mathcal{S}(\realn^d)
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\]
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is an isomorphism. Furthermore
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\[
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\innerproduct{\hat{f}}{\hat{g}} = \innerproduct{f}{g}, \quad \forall f, g \in \mathcal{S}(\realn^d)
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\]
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with the inner product
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\[
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\innerproduct{f}{g} = \int_{\realn^d} \conj{f(x)}g(x) \dd{x}
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\]
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\end{thm}
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\begin{proof}
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To prove that $\hat{f} \in \mathcal{S}$ we use the fact that
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\begin{equation}
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k^{\beta} \partial^{\alpha} \hat{f}(k) = (-i)^{\abs{\alpha} + \abs{\beta}} \mathcal{F}_{\mathcal{S}} \left[ \partial^{\beta} x^{\alpha} f(k) \right], \quad k \in \realn^d, ~\forall \alpha, \beta \in \natn_0^d
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\end{equation}
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Next we want to prove that $\mathcal{F}_{\mathcal{S}}$ is an isomorphism. This is trivial however since
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\begin{equation}
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\forall f \in \mathcal{S}(\realn^d): \quad \inv{\mathcal{F}_{\mathcal{S}}}\mathcal{F}_{\mathcal{S}}(f) = f
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\end{equation}
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To prove the final statement we can explicitly calculate
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\begin{equation}
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\begin{split}
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\innerproduct{\hat{f}}{\hat{g}} &= \int_{\realn^d} \conj{\hat{f}(k)} \hat{g}(k) \dd{k} \\
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&= \int_{\realn^d} \conj{\left( \int_{\realn^d} f(x) e^{-ikx} \dd{x} \frac{\dd{x}}{(2\pi)^{\frac{d}{2}}} \right)} \hat{g}(x) \dd{k} \\
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&= \int_{\realn^d} \left( \int_{\realn^d} \conj{f(x)} e^{ikx} \frac{\dd{x}}{(2\pi)^{\frac{d}{2}}} \right) \hat{g}(k) \dd{k} \\
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&= \int_{\realn^d} \conj{f(x)} \underbrace{\left(\int_{\realn^d} \hat{g}(k) e^{ikx} \frac{\dd{k}}{(2\pi)^{\frac{d}{2}}}\right)}_{\check{\hat{g}}(x) = g(x)} \dd{x} \\
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&= \int_{\realn^d} \conj{f(x)} g(x) \dd{x} = \innerproduct{f}{g}
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\end{split}
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\end{equation}
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\end{proof}
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\begin{rem}
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Since not all functions $f \in L^2(\realn^d)$ are integrable, the limit
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\[
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\lim_{R \rightarrow \infty} \int_{\abs{x} < R} f(x) e^{-ikx} \frac{\dd{x}}{(2\pi)^{\frac{d}{2}}} = \hat{f}(k)
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\]
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doesn't converge for every $k \in \realn^d$, only for almost every.
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\end{rem}
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\begin{thm}
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The Fourier transform $\mathcal{F}_{\mathcal{S}}$ can be uniquely and continuously continued on $L^2(\realn^d)$. The resulting mapping
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\[
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\mathcal{F}: L^2(\realn^d) \longrightarrow L^2(\realn^d)
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\]
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is linear and unitary, i.e. $\forall f, g \in L^2(\realn^d)$ we have
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\[
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\innerproduct{\mathcal{F}(f)}{\mathcal{F}(g)} = \innerproduct{f}{g}
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\]
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which is also known as the Plancherel identity.
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\end{thm}
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\begin{proof}
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\noproof
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\end{proof}
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\begin{rem}
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The continuiuty of $\mathcal{F}$ doesn't imply that $\hat{f}$ is continuous. Secondly, the Plancherel identity also yields
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\[
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\norm{f}_{L^2} = \sqrt{\innerproduct{f}{f}} = \sqrt{\innerproduct{\hat{f}}{\hat{f}}} = \norm{\hat{f}}_{L^2}
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\]
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\end{rem}
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\end{document} |