現在考慮週期連續函數 $f$，並取 $c_n$ 為 $f$ 的 Fourier Series Coefficient，則我們可以定義 $N$ 項 Partial sum $S_N(f;x)$ 如下：
\[
S_N(f;x) :=\sum_{n=-N}^N c_n e^{inx}
\]其中 $c_n = \frac{1}{2 \pi}\int_{-\pi}^\pi f(x) e^{-inx}dx$
為了簡化符號，我們現在定義 Dirichlet kernel $D_N(t) $ 如下
\[
D_N(t) := \sum_{n =-N}^N e^{int}
\]讀者可自行驗證上述 Dirichlet kernel 滿足
\[{D_N}(t): = \sum\limits_{n = - N }^N {{e^{int}}} = \frac{{\sin \left( {\left( {N + 1/2} \right)t} \right)}}{{\sin \left( {t/2} \right)}}\]且 $\int_{ - \pi }^\pi {{D_N}(t)dt = 2\pi } $
現在讓 $n \to \infty$，我們想問何時 上述的 Partial sum 是否收斂到原函數? ;i.e., $$f(x) =?= \sum_n c_n e^{inx} $$ 答案是當 $f$ 為連續函數 或者滿足某程度的連續條件；則 我們前述定義的 Partial sum $S_N(f;x)$ 可以 "逐點收斂" 到原函數 $f$。我們將此重要的結果記錄成以下定理：
================
Theorem 1: Sufficient Condition For Pointwise Convergence of Fourier Series
若 對某些 $x \in [-\pi, \pi]$ 而言， 存在 $\delta >0$ 與 $M>0$ 使得 對任意 $t \in (-\delta, \delta)$
\[
|t| < \delta \Rightarrow |f(x+t) - f(x)| < M\;|t|
\]則 $ \displaystyle \lim_{N \rightarrow \infty}S_N(f;x) = f(x)$
===============
Comments:
1. 上述定理中的條件：存在 $\delta >0$ 與 $M>0$ 使得 $|t| < \delta \Rightarrow |f(x+t) - f(x)| < M\;|t|$ 一般稱為 Lipschitz Condition。
2. 上述定理並 不 保證 均勻收斂!!!
Proof (Theorem 1) : 我們要證 $S_N(f;x) \to f(x)$ 逐點收斂；亦即 $\lim_{N \rightarrow \infty}S_N(f;x) = f(x)$ ；故取 $x$ 滿足假設條件，且給定 $\varepsilon>0$ 我們要證 存在 $N>0$ 使得 $n \ge N \Rightarrow |S_N(f;x) - f(x)| <\varepsilon$ 。現在觀察
\[\begin{array}{*{20}{l}}
{|{S_N}(f;x) - f(x)| = |\sum\limits_{n = - N}^N {{c_n}{e^{inx}}} - f(x)|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = |\frac{1}{{2\pi }}\sum\limits_{n = - N}^N {\left( {\int_{ - \pi }^\pi {f\left( t \right)} {e^{ - int}}dt} \right)} {e^{inx}} - f(x)|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = |\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( t \right)} \sum\limits_{n = - N}^N {{e^{in\left( {x - t} \right)}}dt} - f(x)|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = |\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( t \right)} {D_N}\left( {x - t} \right)dt - f(x)|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = |\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( t \right)} {D_N}\left( {x - t} \right)dt - \underbrace {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( x \right)} {D_N}\left( t \right)dt}_{ = f\left( x \right)}|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}|\int_{x - \pi }^{x + \pi } {f\left( {x - s} \right)} {D_N}\left( s \right)ds - \int_{ - \pi }^\pi {f\left( x \right)} {D_N}\left( t \right)dt|}
\end{array}\]注意到由於 $D_N$ 與 $f$ 為 週期 $2 \pi$函數，故其 $D_N f$ 亦為週期 $2 \pi$函數，故若我們對其積分 其積分範圍可以是 滿足總長為 $2 \pi$ 任意範圍 即可。亦即我們可繼續改寫前式如下：
\[\small \begin{array}{*{20}{l}}
{|{S_N}(f;x) - f(x)| = \frac{1}{{2\pi }}|\int_{x - \pi }^{x + \pi } {f\left( {x - s} \right)} {D_N}\left( s \right)ds - \int_{ - \pi }^\pi {f\left( x \right)} {D_N}\left( t \right)dt|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}|\int_{ - \pi }^\pi {f\left( {x - s} \right)} {D_N}\left( s \right)ds - \int_{ - \pi }^\pi {f\left( x \right)} {D_N}\left( t \right)dt|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}|\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]} {D_N}\left( t \right)dt|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]} \frac{{\sin \left( {\left( {N + 1/2} \right)t} \right)}}{{\sin \left( {t/2} \right)}}dt} \right|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]} \frac{{\sin \left( {Nt} \right)\cos \left( {t/2} \right) + \cos \left( {Nt} \right)\sin \left( {t/2} \right)}}{{\sin \left( {t/2} \right)}}dt} \right|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{2\pi }}\left| \begin{array}{l}
\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]} \frac{{\sin \left( {Nt} \right)\cos \left( {t/2} \right)}}{{\sin \left( {t/2} \right)}}dt\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]} \frac{{\cos \left( {Nt} \right)\sin \left( {t/2} \right)}}{{\sin \left( {t/2} \right)}}dt
\end{array} \right|}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} \le \frac{1}{{2\pi }}\left[ \begin{array}{l}
\left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}\cos \left( {t/2} \right)} \right]} \sin \left( {Nt} \right)dt} \right|\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} + \left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}\sin \left( {t/2} \right)} \right]} \cos \left( {Nt} \right)dt} \right|
\end{array} \right]}
\end{array}
\]注意到若 $|t| \le \delta$ 則下列兩式
\[\left\{ \begin{array}{l}
\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}} \right]\cos \left( {t/2} \right)\\
\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}} \right]\sin \left( {t/2} \right)
\end{array} \right.
\]兩者皆為有界 $(|f(x+t) - f(x)| \le M |t| < M \delta)$。故
\[\small \begin{array}{*{20}{l}}
{|{S_N}(f;x) - f(x)| \le \frac{1}{{2\pi }}\left[ \begin{array}{l}
\left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}\cos \left( {t/2} \right)} \right]} \sin \left( {Nt} \right)dt} \right|\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} + \left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left[ {f\left( {x - t} \right) - f\left( x \right)} \right]}}{{\sin \left( {t/2} \right)}}\sin \left( {t/2} \right)} \right]} \cos \left( {Nt} \right)dt} \right|
\end{array} \right]}\\
{\begin{array}{*{20}{l}}
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} \le \frac{M}{{2\pi }}\left[ {\left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left| t \right|\cos \left( {t/2} \right)}}{{\sin \left( {t/2} \right)}}} \right]} \sin \left( {Nt} \right)dt} \right| + \left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left| t \right|\sin \left( {t/2} \right)}}{{\sin \left( {t/2} \right)}}} \right]} \cos \left( {Nt} \right)dt} \right|} \right]}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} \le \underbrace {\frac{M}{{2\pi }}\left[ {\left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left| t \right|}}{{\sin \left( {t/2} \right)}}} \right]} \sin \left( {Nt} \right)dt} \right| + \left| {\int_{ - \pi }^\pi {\left[ {\frac{{\left| t \right|}}{{\sin \left( {t/2} \right)}}} \right]} \cos \left( {Nt} \right)dt} \right|} \right] \to 0}_{\left( {by\begin{array}{*{20}{c}}
{}
\end{array}\mathop {\lim }\limits_{n \to \infty } {c_n} = 0} \right)}}
\end{array}}
\end{array}\]
以下我們看個例子:
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 \left( {n\delta } \right)}}{{n\pi }}} = \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 point-wise convergence (Theorem 1) 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 1，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 \left( {n\delta } \right)}}{{n\pi }}} = \frac{{\pi - \delta }}{2} $ 故由 part (a) 可知
\[\begin{array}{l}
f\left( x \right) = \frac{\delta }{\pi } + 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}{e^{inx}}} \\
\Rightarrow f\left( 0 \right) = \frac{\delta }{\pi } + 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}} \\
\Rightarrow 1 = \frac{\delta }{\pi } + 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}} \\
\Rightarrow \frac{{\pi - \delta }}{{2\pi }} = \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\delta } \right)}}{{n\pi }}} \ \ \ \ \ \square
\end{array}
\]
==========
Theorem:
令 $f \in C^1([-\pi,\pi])$ 且 periodic ，則 $S_N(f) \to f$ uniformly。
==========
Proof:
要證明 $S_N(f) \to f$ uniformly；首先注意到 $f \in C^1([-\pi,\pi])$ 故自動滿足 Lipschitz condition；由 Theorem 1 亦即我們有 對任意 $x$，$S_N(f;x) \to f(x)$ pointwise。
故我們只需證明 $S_N(f)$ 均勻收斂 無須證明他收斂到 $f$ (why? 因為 limit 的唯一性質保證如果已經有 $S_N(f)$ 逐點收斂到某函數 $f$ 且又知道 $S_N(f)$ 均勻收斂 則 $S_N(f)$ 必定要均勻收斂到 $f$)。
那麼現在問題變成如何證明 $S_N(f)$ 均勻收斂 ? 回憶 $S_N(f)$ 定義：
\[{S_N}\left( {f;x} \right): = \sum\limits_{n = - N}^N {{c_n}{e^{inx}}} \]我們需要額外的工具 幫助我們證明上述 summation 收斂。回憶：( Weierstrass M-test ：若 函數數列 $g_n(x)$ 為連續 且 $|g_n(x)| \le M_n$ 且 $\sum_n M_n < \infty$，則 $\sum_n |g_n(x)|$ 均勻收斂。)
現在觀察 $\left| {{c_n}{e^{inx}}} \right| \le \left| {{c_n}} \right| $；另外由於 $f \in C^1$ 我們可定義 $c_n'$ 為 $f'$ 的 Fourier Series Coefficient，則
\[{c_n}' = in\left( {{c_n}} \right)
\]故由前述結果可推知
\[\left| {{c_n}{e^{inx}}} \right| \le \left| {{c_n}} \right| = \left| {\frac{{{c_n}'}}{{in}}} \right| = \left| {\frac{{{c_n}'}}{n}} \right| \ \ \ \ (**)
\]現在利用 一個不等式工具： 對任意 $a,b \ge 0$，$a \cdot b \le \frac{1}{2} |a^2 + b^2|$；現在選 $a:=c_n'$ 與 $b:=1/n$ 則利用上述不等式可推得
\[\begin{array}{l}
\left| {{c_n}{e^{inx}}} \right| \le \left| {{c_n}} \right| = \left| {\frac{{{c_n}'}}{n}} \right| \le \frac{1}{2}|{\left( {{c_n}'} \right)^2} + {\left( {\frac{1}{n}} \right)^2}|\\
\Rightarrow \left| {{c_n}{e^{inx}}} \right| \le \underbrace {\frac{1}{2}{{\left( {{c_n}'} \right)}^2} + \frac{1}{2}\frac{1}{{{n^2}}}}_{: = {M_n}}
\end{array}\]現在對 $M_n$ 取 summation 可得
\[\begin{array}{l}
\sum\limits_{n = - \infty }^\infty {{M_n}} = \sum\limits_{n = - \infty }^\infty {\left[ {\frac{1}{2}{{\left( {{c_n}'} \right)}^2} + \frac{1}{2}\frac{1}{{{n^2}}}} \right]} \\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left| {{c_0}} \right| + \sum\limits_{n = - \infty ;n \ne 0}^\infty {\left[ {\frac{1}{2}{{\left( {{c_n}'} \right)}^2} + \frac{1}{2}\frac{1}{{{n^2}}}} \right]} \\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left| {{c_0}} \right| + \frac{1}{2}\underbrace {\sum\limits_{n = - \infty ;n \ne 0}^\infty {{{\left( {{c_n}'} \right)}^2}} }_{Term1} + \frac{1}{2}\underbrace {\sum\limits_{n = - \infty ;n \ne 0}^\infty {\left[ {\frac{1}{{{n^2}}}} \right]} }_{ < \infty }
\end{array}\]上式中的 Term 1 可利用 Parseval's Theorem ：($\sum\limits_n | {c_n}{|^2} = \underbrace {\int_{ - \pi }^\pi | f(x){|^2}dx}_{: = \left\| f \right\|_{{L^2}}^2}$)，由於 $f \in C^1([-\pi,\pi])$ 故 \[\sum\limits_n | {c_n}{|^2} = \underbrace {\int_{ - \pi }^\pi | f(x){|^2}dx}_{: = \left\| f \right\|_{{L^2}}^2} < \sup {\left| f \right|^2}2\pi < \infty \]故 Term 1 亦為有界。至此我們證明了
\[
\sum_n M_n <\infty
\]由 Weierstrass M-test 可知 $\sum_n c_n e^{inx}$ 均勻收斂 亦即 $S_N(f)$ 均勻收斂，又由於 $S_N(f) \to f$ 逐點收斂，故 $S_N(f) \to f$ 均勻收斂 $\square$
以下我們看個例子：
Example
令 $f : [-\pi, \pi] \to \mathbb{C}$ 無窮可微 的解析函數 且滿足 $f^{(k)}(-\pi) = f^{(k)}(\pi)$ 對任意 $k \in \mathbb{Z}^+ \cup \{0\}$
(a) 若 $c_n$ 為 $f(x)$ 的 Fourier Series coefficient。試求 $f'(x)$ 的 Fourier Series Coefficient $c_n'$
(b) 試證 $n c_n \to 0$
Proof:
(a) 首先注意到 $f$ 為 解析函數，且 $f^{(k)}(-\pi) = f^{(k)}(\pi)$ 對任意 $k \in \mathbb{Z}^+ \cup \{0\}$ 故此函數滿足 Theorem 1 我們可說 $f$ 具有 Fourier Series 如下
\[f\left( x \right) = \sum\limits_{n = - \infty }^\infty {{c_n}{e^{inx}}} \]故
\[f'\left( x \right) = \sum\limits_{n = - \infty }^\infty {{c_n}in{e^{inx}}} : = \sum\limits_{n = - \infty }^\infty {{c_n}'{e^{inx}}} \Rightarrow {c_n}' = {c_n}in\]
(b) 由於我們知道 $c_n ' = in c_n$ 故
\[\begin{array}{l}
{c_n}' = in{c_n} \Rightarrow \frac{{{c_n}'}}{i} = n{c_n}\\
\Rightarrow \left| {n{c_n} - 0} \right| = \left| {\frac{{{c_n}'}}{i}} \right| \to 0
\end{array}\](利用 Bessel's inequality: $\lim_{n} c_n' \to 0$)
另外若 $f$ 為週期連續函數，則我們可以透過三角多項式逼近，此結果紀錄如下
==========
FACT: 若 $f$ 為 週期 $2 \pi$ 的連續函數 且 若 $\varepsilon >0$ ，則 存在 trigonometric polynomial $P$ 使得 對任意 $x \in \mathbb{R}$
\[
|P(x) - f(x)| < \varepsilon
\]==========
Proof: omitted. (via Weierstrass Approximation Theorem)
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