FACT: 任意週期為 2π 之連續函數 f 必存在一組 trigonometric polynomial P:=N∑−Ncneinx 使得 對任意 x∈R 而言,
|P(x)−f(x)|<ε
但是上述定理無法告訴我們何時週期函數可以被表示成 Fourier Series,故下面的定理尤為重要,通常用此定理判斷某週期函數是否可以表示成 Fourier Series (Pointwise sense)。
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Theorem 0: Pointwise Convergence of Fourier Series
若 對某些 x∈[−π,π] 而言, 存在 δ>0 與 M>0 使得 對任意 t∈(−δ,δ)
|t|<δ⇒|f(x+t)−f(x)|<M|t|則 lim
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接著我們介紹另一個相對於 逐點收斂的重要的結果,稱作 L^2 收斂,亦即 Fourier Series 在 convergence in L^2。
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Theorem 1: Fourier Series Converges in L^2
假設 f, g 為 週期 2 \pi 之 週期連續函數,且
f(x) \sim \sum_{-\infty}^\infty c_n e^{inx}; \;\; g(x) \sim \sum_{-\infty}^\infty \gamma_n e^{inx} 則
\mathop {\lim }\limits_{N \to \infty } \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left| {f\left( x \right) - {S_N}\left( {f;x} \right)} \right|}^2}} dx = \mathop {\lim }\limits_{N \to \infty } \left\| {f - {S_N}} \right\|_2^2 = 0其中 {S_N}\left( {f;x} \right): = \sum\limits_{ - N}^N {{c_n}{e^{inx}}}
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Proof:
令 \varepsilon >0,目標要證明 ||f - S_N(f)||_2 < \varepsilon 。由於 f 為週期 2 \pi 之 週期連續函數,由 FACT 可知必存在一組 trigonometric polynomial P 使得 對任意 x \in \mathbb{R} 而言,
|P(x) - f(x)| < \varepsilon \Rightarrow ||P - f||_2 < \varepsilon 若 P 有階數為 N_0 階,則由於 S_N(f) 為最佳近似 f (請參閱先前BLOG文章 或者 Rudin Theorem8.11) ,對 N \ge N_0,
||f - S_N(f)||_2 \le ||f - P||_2 < \varepsilon \ \ \ \ \square
我們有以下的 Parseval's identity:
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Theorem 2: Parseval's Theorem
假設 f, g 為 週期 2 \pi 之 週期連續函數,且
f(x) \sim \sum_{-\infty}^\infty c_n e^{inx}; \;\; g(x) \sim \sum_{-\infty}^\infty \gamma_n e^{inx} 則
\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right){g^*}\left( x \right)} dx = \sum\limits_{ - \infty }^\infty {{c_n}\gamma _n^*} =================
Proof:
首先觀察
\begin{array}{l} \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{S_N}\left( f \right){g^*}\left( x \right)} dx = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {\sum\limits_{ - N}^N {{c_n}} {e^{inx}}{g^*}\left( x \right)} dx\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \frac{1}{{2\pi }}\sum\limits_{ - N}^N {{c_n}} \underbrace {\int_{ - \pi }^\pi {{e^{inx}}{g^*}\left( x \right)} dx}_{ = \gamma _n^*2\pi } = \sum\limits_{ - N}^N {{c_n}} \gamma _n^* \end{array} 接著我們檢驗
\begin{array}{l} \left| {\int_{ - \pi }^\pi {f\left( x \right){g^*}\left( x \right)} dx - \int_{ - \pi }^\pi {{S_N}\left( f \right){g^*}\left( x \right)} dx} \right| = \left| {\int_{ - \pi }^\pi {\left[ {f\left( x \right) - {S_N}\left( f \right)} \right]{g^*}\left( x \right)} dx} \right|\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} \le \int_{ - \pi }^\pi {\left| {\left[ {f\left( x \right) - {S_N}\left( f \right)} \right]{g^*}\left( x \right)} \right|} dx\\ {\rm{by}}\begin{array}{*{20}{c}} {} \end{array}{\rm{Schwarz}}\begin{array}{*{20}{c}} {} \end{array}{\rm{inequality}}\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} \le {\left( {\int_{ - \pi }^\pi {{{\left| {f\left( x \right) - {S_N}\left( f \right)} \right|}^2}} dx \cdot \int_{ - \pi }^\pi {{{\left| {{g^*}\left( x \right)} \right|}^2}} dx} \right)^{\frac{1}{2}}} \to 0 \end{array}當 N \rightarrow \infty。注意到上式收斂成立是因為我們 \int |g|^2 有界 且 \int |f - S_N| \rightarrow 0 當 N \rightarrow \infty (由前面的 Theorem 1) \square
Comment:
一般而言,Parseval's Thoerem 泛指下式:令 前述 Parseval's Theorem g(x) := f(x),則
\begin{array}{l} \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right){f^*}\left( x \right)} dx = \sum\limits_{n = - \infty }^\infty {{c_n}c_n^*} \\ \Rightarrow \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left| {f\left( x \right)} \right|}^2}} dx = \sum\limits_{n = - \infty }^\infty {{{\left| {{c_n}} \right|}^2}} \end{array}
現在我們看個例子 說明 Parseval Theorem 怎麼使用。
Example:Application of the Parseval Theorem/ Pointwise Convergence Theorem
假設 0 < \delta < \pi,
f\left( x \right): = \left\{ \begin{array}{l} 1,\begin{array}{*{20}{c}} {}&{} \end{array}\left| x \right| < \delta \\ 0,\begin{array}{*{20}{c}} {}&{} \end{array}\delta < \left| x \right| \le \pi \end{array} \right.且對任意 x \in \mathbb{R}, f(x+2 \pi) = f(x)
(a) 試求 Fourier Series Coefficient
(b) 試證 \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} = \frac{{\pi - \delta }}{2}
Solution
在求解之前我們先確認 f(x) 具有 Fourier Series,首先注意到 f 為週期函數 (週期為 2 \pi) 接著我們檢驗其是否滿足我們的 逐點收斂 (Theorem 0) 條件:
給定 x =0 ,我們取題目中給定的 \delta >0 檢驗對任意 t \in (-\delta, \delta) ,觀察
\begin{array}{l} \left| {f\left( {x + t} \right) - f\left( x \right)} \right| = \left| {f\left( t \right) - f\left( 0 \right)} \right|\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \left| {1 - 1} \right| = 0 \le M\left| t \right| \end{array}故我們知道其滿足 Theorem 0,亦即 f 有 Fourier Series 且 f(x) = \sum_{-\infty}^\infty c_n e^{inx} 現在我們可以開始解題:
(a) 首先針對 c_0 可知
{c_0}: = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right)dx} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta {1dx} = \frac{\delta }{\pi }
另外對 n \neq 0
\begin{array}{l} {c_n}: = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right){e^{inx}}dx} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta {1{e^{inx}}dx} \\ \begin{array}{*{20}{c}} {}&{}&{} \end{array} = \frac{1}{{n\pi }}\frac{{{e^{in\delta }} - {e^{ - in\delta }}}}{{2i}} = \frac{1}{{n\pi }}\sin \left( {n\delta } \right) \end{array}注意到上述結果暗示了
f\left( x \right) = \frac{\delta }{\pi } + 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}{e^{inx}}}
(b) 我們要證明 \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} = \frac{{\pi - \delta }}{2} 注意到 part (a) 求出的 Fourier Series coefficient 的平方 出現在等號左方,暗示了我們可使用 Parserval's Theorem 亦即
\begin{array}{l} \sum\limits_{n = - \infty }^\infty {{c_n}^2} = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left| {f\left( x \right)} \right|}^2}} dx\\ \Rightarrow {c_0}^2 + 2\sum\limits_{n = 1}^\infty {{c_n}^2} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta 1 dx\\ \Rightarrow {\left( {\frac{\delta }{\pi }} \right)^2} + 2\sum\limits_{n = 1}^\infty {{{\left( {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}} \right)}^2}} = \frac{\delta }{\pi }\\ \Rightarrow \sum\limits_{n = 1}^\infty {\left( {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} \right)} = \frac{{\pi - \delta }}{2} \end{array}
\begin{array}{l} \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{S_N}\left( f \right){g^*}\left( x \right)} dx = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {\sum\limits_{ - N}^N {{c_n}} {e^{inx}}{g^*}\left( x \right)} dx\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \frac{1}{{2\pi }}\sum\limits_{ - N}^N {{c_n}} \underbrace {\int_{ - \pi }^\pi {{e^{inx}}{g^*}\left( x \right)} dx}_{ = \gamma _n^*2\pi } = \sum\limits_{ - N}^N {{c_n}} \gamma _n^* \end{array} 接著我們檢驗
\begin{array}{l} \left| {\int_{ - \pi }^\pi {f\left( x \right){g^*}\left( x \right)} dx - \int_{ - \pi }^\pi {{S_N}\left( f \right){g^*}\left( x \right)} dx} \right| = \left| {\int_{ - \pi }^\pi {\left[ {f\left( x \right) - {S_N}\left( f \right)} \right]{g^*}\left( x \right)} dx} \right|\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} \le \int_{ - \pi }^\pi {\left| {\left[ {f\left( x \right) - {S_N}\left( f \right)} \right]{g^*}\left( x \right)} \right|} dx\\ {\rm{by}}\begin{array}{*{20}{c}} {} \end{array}{\rm{Schwarz}}\begin{array}{*{20}{c}} {} \end{array}{\rm{inequality}}\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} \le {\left( {\int_{ - \pi }^\pi {{{\left| {f\left( x \right) - {S_N}\left( f \right)} \right|}^2}} dx \cdot \int_{ - \pi }^\pi {{{\left| {{g^*}\left( x \right)} \right|}^2}} dx} \right)^{\frac{1}{2}}} \to 0 \end{array}當 N \rightarrow \infty。注意到上式收斂成立是因為我們 \int |g|^2 有界 且 \int |f - S_N| \rightarrow 0 當 N \rightarrow \infty (由前面的 Theorem 1) \square
Comment:
一般而言,Parseval's Thoerem 泛指下式:令 前述 Parseval's Theorem g(x) := f(x),則
\begin{array}{l} \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right){f^*}\left( x \right)} dx = \sum\limits_{n = - \infty }^\infty {{c_n}c_n^*} \\ \Rightarrow \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left| {f\left( x \right)} \right|}^2}} dx = \sum\limits_{n = - \infty }^\infty {{{\left| {{c_n}} \right|}^2}} \end{array}
現在我們看個例子 說明 Parseval Theorem 怎麼使用。
Example:Application of the Parseval Theorem/ Pointwise Convergence Theorem
假設 0 < \delta < \pi,
f\left( x \right): = \left\{ \begin{array}{l} 1,\begin{array}{*{20}{c}} {}&{} \end{array}\left| x \right| < \delta \\ 0,\begin{array}{*{20}{c}} {}&{} \end{array}\delta < \left| x \right| \le \pi \end{array} \right.且對任意 x \in \mathbb{R}, f(x+2 \pi) = f(x)
(a) 試求 Fourier Series Coefficient
(b) 試證 \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} = \frac{{\pi - \delta }}{2}
Solution
在求解之前我們先確認 f(x) 具有 Fourier Series,首先注意到 f 為週期函數 (週期為 2 \pi) 接著我們檢驗其是否滿足我們的 逐點收斂 (Theorem 0) 條件:
給定 x =0 ,我們取題目中給定的 \delta >0 檢驗對任意 t \in (-\delta, \delta) ,觀察
\begin{array}{l} \left| {f\left( {x + t} \right) - f\left( x \right)} \right| = \left| {f\left( t \right) - f\left( 0 \right)} \right|\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \left| {1 - 1} \right| = 0 \le M\left| t \right| \end{array}故我們知道其滿足 Theorem 0,亦即 f 有 Fourier Series 且 f(x) = \sum_{-\infty}^\infty c_n e^{inx} 現在我們可以開始解題:
(a) 首先針對 c_0 可知
{c_0}: = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right)dx} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta {1dx} = \frac{\delta }{\pi }
另外對 n \neq 0
\begin{array}{l} {c_n}: = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right){e^{inx}}dx} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta {1{e^{inx}}dx} \\ \begin{array}{*{20}{c}} {}&{}&{} \end{array} = \frac{1}{{n\pi }}\frac{{{e^{in\delta }} - {e^{ - in\delta }}}}{{2i}} = \frac{1}{{n\pi }}\sin \left( {n\delta } \right) \end{array}注意到上述結果暗示了
f\left( x \right) = \frac{\delta }{\pi } + 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}{e^{inx}}}
(b) 我們要證明 \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} = \frac{{\pi - \delta }}{2} 注意到 part (a) 求出的 Fourier Series coefficient 的平方 出現在等號左方,暗示了我們可使用 Parserval's Theorem 亦即
\begin{array}{l} \sum\limits_{n = - \infty }^\infty {{c_n}^2} = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left| {f\left( x \right)} \right|}^2}} dx\\ \Rightarrow {c_0}^2 + 2\sum\limits_{n = 1}^\infty {{c_n}^2} = \frac{1}{{2\pi }}\int_{ - \delta }^\delta 1 dx\\ \Rightarrow {\left( {\frac{\delta }{\pi }} \right)^2} + 2\sum\limits_{n = 1}^\infty {{{\left( {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}} \right)}^2}} = \frac{\delta }{\pi }\\ \Rightarrow \sum\limits_{n = 1}^\infty {\left( {\frac{{{{\sin }^2}\left( {n\delta } \right)}}{{{n^2}\delta }}} \right)} = \frac{{\pi - \delta }}{2} \end{array}
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