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《西江月·证明》（佚名）：即得易见平凡，仿照上例显然。留作习题答案略，读者自证不难。反之亦然同理，推论自然成立，略去过程QED，由上可知证毕。
程其襄、张奠宙等《实变函数与泛函分析基础（第三版）》第九章第3节，第255页
设\(e_0(t)\equiv \frac{1}{\sqrt{2}}\), \(e_1(t)=\cos t\), \(e_2(t)=\sin t\), \(e_3(t)=\cos 2t\), \(e_4(t)=\sin 2t\), \(\cdots\), \(e_{2n-1}(t)=\cos nt\), \(e_{2n}(t)=\sin nt\), \(\cdots\). 令\(M=\{e_i\}_{i=0}^\infty\), 我们已经知道, \(M\)是Hilbert空间
\[L^2[-\pi,\pi],\quad \langle f, g\rangle = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)\overline{g(t)} {\rm d} t, \ f,g\in L^2[-\pi,\pi]
\]
中的规范正交系.
我们按以下步骤证明, 三角函数系\(M\)是\(L^2[a,b]\)中的完全规范正交系.
Weierstrauss三角逼近定理: 设\(f\in C[-\pi,\pi]\), 并且\(f(-\pi)=f(\pi)\), 则对任意\(\varepsilon>0\), 存在三角多项式
\[T(t)=a_0+\sum_{k=1}^m \left(a_k \cos kt+b_k \sin kt \right), \quad t\in [-\pi,\pi],
\]
使得
\[\max_{t\in [-\pi,\pi]} \left|f(t)-T(x)\right|<\varepsilon.
\]
Proof. (Fejér核方法)
由于\(f\in C[-\pi,\pi]\), 则\(f\in L^2[-\pi,\pi]\). 对任意\(e_k\in M\), \(f\)关于\(e_k\)的Fourior系数为
\[a_k=\langle f,e_k \rangle =
\begin{cases}
\frac{1}{\pi}\int_{-\pi}^{\pi} \frac{1}{\sqrt{2}} f(t) {\rm d} t, &k=0,\\frac{1}{\pi}\int_{-\pi}^{\pi} f(t) \cos nt {\rm d} t, &k=2n-1,\\frac{1}{\pi}\int_{-\pi}^{\pi} f(t) \sin nt {\rm d} t, &k=2n.
\end{cases}
\]
记\(f\)的Fourior级数的前\(n\)项部分和为
\[S_n=\sum_{k=0}^n \langle f,e_k \rangle e_k =\sum_{k=0}^n a_k e_k,
\]
则\(S_n\in C[-\pi,\pi]\subset L^2[-\pi,\pi]\), 并且\(S_n(-\pi)=S_n(\pi)\). 将Fourior系数代入, 得
\[\begin{equation*}\begin{split}
S_n(t)&=&\frac{a_0}{\sqrt{2}}+\sum_{k=1}^n\left(a_{2k-1}\cos nt +a_{2k}\sin nt \right)\&=&\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(s){\rm d} s \&&+\frac{1}{\pi} \sum_{k=1}^n \left[\left( \int_{-\pi}^\pi f(s)\cos k s{\rm d} s \right)\cos kt + \left( \int_{-\pi}^\pi f(s)\sin k s{\rm d} s \right)\sin kt \right]\&=&\frac{1}{\pi} \int_{-\pi}^\pi f(s) \left[\frac{1}{2} +\sum_{k=1}^n \left( \cos ks \cos kt+ \sin ks\sin kt \right)\right] {\rm d} s\&=&\frac{1}{\pi} \int_{-\pi}^\pi f(s) \left[ \frac{1}{2} +\sum_{k=1}^n \cos k(s-t) \right] {\rm d} s.
\end{split}\end{equation*}
\]
将\(S_n(t)\)和\(f(t)\)延拓成\(\Bbb{R}\)上的以\(2\pi\)为周期的连续函数, 并令\(\tau=s-t\), 则
\[S_n(t)=\frac{1}{\pi} \int_{-\pi -t}^{\pi -t} f(t+\tau) \left[\frac{1}{2} +\sum_{k=1}^n \cos k\tau \right] {\rm d} \tau.
\]
注意到上式右端积分中的被积函数也是以\(2\pi\)为周期的连续函数, 因此在\([-\pi-t,\pi-t]\)上的积分等于\([-\pi,\pi]\)上的积分, 从而
\[S_n(t)=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t+\tau) \left[\frac{1}{2} +\sum_{k=1}^n \cos k\tau \right] {\rm d} \tau,\quad t\in [-\pi,\pi].
\]
注意到(积化和差)
\[\cos kx \sin \frac{x}{2} =\frac{1}{2}\left[\sin \left( k+\frac{1}{2} \right) x -\sin \left(k-\frac{1}{2} \right)x \right],
\]
则
\[\left( \frac{1}{2}+\sum_{k=1}^n \cos kx \right)\sin \frac{x}{2}=\frac{1}{2} \sin \left(n+\frac{1}{2}\right) x,
\]
于是
\[S_n(t)=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t+\tau) \frac{ \sin \left(n+\frac{1}{2}\right) \tau}{2\sin \frac{\tau}{2}} {\rm d} \tau,\quad t\in [-\pi,\pi].
\]
令(称为Cesàro和)
\[\begin{equation*}\begin{split}
\sigma_n(t) &=& \frac{S_0(t)+S_1(t)+\cdots +S_{n-1}(t)}{n}\&=&\frac{1}{2n\pi} \int_{-\pi}^{\pi} \left[\sum_{k=0}^{n-1} \frac{ \sin \left(k+\frac{1}{2}\right) \tau}{\sin \frac{\tau}{2}} \right] f(t+\tau){\rm d} \tau.
\end{split}\end{equation*}
\]
显然, \(\sigma_n\)也是一个三角多项式. 注意到(积化和差)
\[\sin \left(k+\frac{1}{2}\right) x \cdot \sin \frac{x}{2} =\frac{1}{2}\left[\cos k x -\cos (k+1)x \right],
\]
则
\[\sum_{k=0}^{n-1} \sin \left(k+\frac{1}{2}\right) x=\frac{1-\cos nx}{2\sin \frac{x}{2}}=\frac{\sin^2 \frac{nx}{2}}{\sin \frac{x}{2}}.
\]
令(称为Fejér核)
\[\Phi_n(x) =\frac{1}{2n\pi} \left[ \frac{\sin^2 \frac{nx}{2}}{\sin^2 \frac{x}{2}} \right],
\]
于是
\[\sigma_n(t)=\int_{-\pi}^{\pi} \Phi_n(\tau)f(t+\tau){\rm d} \tau.
\]
下证Fejér核\(\Phi_n(t)\)满足以下3条性质:
(i) \(\Phi_n(x)\geq 0\);
(ii) \(\int_{-\pi}^{\pi} \Phi_n(x) {\rm d} x=1\).
(iii) 对任意固定的\(\delta\in (0,\pi)\), 记
\[\eta_n(\delta)=\int_{-\pi}^{-\delta} \Phi_n(x){\rm d} x=\int_{\delta}^{\pi} \Phi_n(x) {\rm d} x,
\]
则\(\lim\limits_{n\to \infty} \eta_n(\delta)=0.\)
性质(i)显然成立.
注意到Fejér核\(\Phi_n(t)\)和函数\(f\)无关. 当\(f(t)\equiv 1\)时, \(f\)关于\(e_k\in M\)的Fourior系数为
\[a_0=\sqrt{2};\quad a_k=0,\ \forall k\in \Bbb{N}_+.
\]
所以
\[S_n(t)\equiv S_0(t)=1,\quad \forall n\in \Bbb{N}_+,
\]
从而
\[\int_{-\pi}^{\pi} \Phi_n(t) {\rm d} t=\sigma_n(t)= \frac{S_0(t)+S_1(t)+\cdots +S_{n-1}(t)}{n}=\frac{n}{n}=1.
\]
性质(ii)得证.
当\(0<\delta\leq x\leq \pi\)时, \(\sin \frac{x}{2}\geq \sin {\frac{\delta}{2}}>0\), 从而
\[\Phi_n(x)=\frac{1}{2n\pi} \left[ \frac{\sin^2 \frac{nx}{2}}{\sin^2 \frac{x}{2}} \right]\leq \frac{1}{2n\pi} \frac{1}{\sin^2 \frac{\delta}{2}},
\]
\[0\leq \eta_n(\delta) \leq \frac{\pi-\delta}{2\pi \sin^2 \frac{\delta}{2}}\cdot \frac{1}{n},
\]
所以\(\lim\limits_{n\to \infty} \eta_n(\delta)=0.\) 性质(iii)得证.
现在\(f\)已经延拓成了\(\Bbb{R}\)上的\(2\pi\)周期函数, 则\(f\)在\(\Bbb{R}\)上有界并且一致连续, 即存在\(M>0\)使得
\[|f(x)|\leq M,\quad \forall x\in \Bbb{R};
\]
对任意\(\varepsilon>0\), 存在\(\delta>0\), 使得
\[\forall x‘,x‘‘\in \Bbb{R}\ :\ |x‘-x‘‘|<\delta,
\]
都有
\[|f(x‘)-f(x‘‘)|<\varepsilon.
\]
利用上述\(\delta>0\)以及Fejér核\(\Phi_n(t)\)的性质(ii), 我们有
\[\begin{equation*}\begin{split}
f(t)-\sigma_n(t)&=&\left( \int_{-\pi}^{\pi}\Phi_n(\tau) {\rm d} \tau \right) f(t)- \int_{-\pi}^{\pi}\Phi_n(\tau) {\rm d} \tau f(t+\tau){\rm d} \tau\&=&\int_{-\pi}^{\pi}\Phi_n(\tau) \left[ f(t)-f(t+\tau)\right] {\rm d} \tau\&=:&J_-+J_0+J_+,
\end{split}\end{equation*}
\]
其中
\[\begin{equation*}\begin{split}
J_-&=& \int_{-\pi}^{-\delta} \Phi_n(\tau) \left[ f(t)-f(t+\tau)\right] {\rm d} \tau,\J_0&=& \int_{-\delta}^{\delta} \Phi_n(\tau) \left[ f(t)-f(t+\tau)\right] {\rm d} \tau,\J_+&=& \int_{\delta}^{\pi} \Phi_n(\tau) \left[ f(t)-f(t+\tau)\right] {\rm d} \tau,
\end{split}\end{equation*}
\]
利用Fejér核\(\Phi_n(t)\)的性质(iii)以及函数\(f\)的有界性和一致连续性, 就有
\[|J_-|\leq 2M\eta_n(\delta),\quad |J_+|\leq 2M\eta_n(\delta),\quad |J_0|\leq \frac{\varepsilon}{2} \int_{-\delta}^{\delta} \Phi_n(\tau){\rm d}\tau <\frac{\varepsilon}{2}.
\]
由于\(\lim\limits_{n\to \infty} \eta_n(\delta)=0\), 则存在正整数\(N\), 使得对任意\(n>N\), 都有
\[2M\eta_n(\delta)<\frac{1}{4}\varepsilon,
\]
从而
\[|f(t)-\sigma_n(t)|<\varepsilon,\quad \forall t\in \Bbb{R}, \quad\forall n>N.
\]
令\(T(t)=\sigma_n(t)\), \(n>N\), 则\(T\)是\([-\pi,\pi]\)上的三角多项式并且
\[\max_{t\in [-\pi,\pi]} |f(t)-T(t)|<\varepsilon.
\]
- Step2. 设\(T\)是\([-\pi,\pi]\)上的一个三角多项式, 则\(T\in C[-\pi,\pi]\), 同时也有\(T\in L^2[-\pi,\pi]\). 证明: \(T\)关于三角函数系\(M\)满足Parseval等式, 即
\[\|T\|^2=\sum_{e\in M}|\langle T,e \rangle |^2.
\]
Proof. 设\(T\)是\([-\pi,\pi]\)上的三角多项式,
\[T(t)=a_0+\sum_{k=1}^m \left(a_k \cos kt+b_k \sin kt\right)=\sum_{i=0}^{2m} c_i e_i(t),\quad t\in [-\pi,\pi],
\]
其中
\[c_i=
\begin{cases}
a_0,& i=0,\a_k,& i=2k-1\b_k,& i=2k.
\end{cases}
\]
由于\(M=\{e_i\}_{i=0}^\infty\)是\(L^2[-\pi,\pi]\)中的规范正交系, 则
\[\|T\|^2=\sum_{i=0}^{2m}\|c_i e_i\|^2=\sum_{i=0}^{2m}|c_i|^2.
\]
另一方面, 对任意\(l\in \Bbb{N}\), 由\(L^2[-\pi,\pi]\)中内积的定义, 就有
\[\begin{equation*}\begin{split}
&&\langle T,e_l\rangle\&=&\frac{1}{\pi} \int_{-\pi}^{\pi} \left[\sum_{i=0}^{2m} c_i e_i(t) \right]\overline{e_l(t)} {\rm d} t\&=&\sum_{i=0}^{2m} c_i\left[\frac{1}{\pi}\int_{-\pi}^{\pi} e_i(t)\overline{e_l(t)} \right]{\rm d} t\&=&\sum_{i=0}^{2m} c_i\langle e_i, e_l\rangle
\end{split}\end{equation*}
\]
由于\(M=\{e_i\}_{i=0}^\infty\)是\(L^2[-\pi,\pi]\)中的规范正交系, 则
\[\begin{cases}
\langle T,e_l\rangle=c_l,& l\leq 2m,\\langle T,e_l\rangle=0,&l>2m.
\end{cases}
\]
综上,
\[\sum_{e\in M}|\langle T,e\rangle|^2=\sum_{i=0}^\infty |\langle T,e_i \rangle|^2=\sum_{i=0}^{2m} |c_i|^2 =\|T\|^2.
\]
- Step3. 利用Steklov定理证明\(M\)是\(L^2[a,b]\)中的完全规范正交系.
Steklov定理: 设\(M\)是Hilbert空间\(X\)中规范正交系. 若Parseval等式在\(X\)的某个稠密子集\(A\)上成立, 即对任意\(x\in A\), 都有
\[x=\sum_{e\in M} \langle x,e \rangle e,
\]
则\(M\)是完全规范正交系.
Proof. 将\([-\pi,\pi]\)上的三角多项式全体集合记为\({\rm Tri}[-\pi,\pi]\), 将\([-\pi,\pi]\)上满足\(f(-\pi)=f(\pi)\)的连续函数全体集合记为\(C(\Bbb{T})\), 显然\({\rm Tri}[-\pi,\pi]\subset C(\Bbb{T})\subset L^2[-\pi,\pi]\), 并且\({\rm Tri}[-\pi,\pi]\)和\(C(\Bbb{T})\)在\(L^2\)-范数
\[\|f\|=\left[ \frac{1}{\pi} \int_{-\pi}^{\pi} |f(t)|^2 {\rm d} t\right]^{\frac{1}{2}}
\]
下都是\(L^2[-\pi,\pi]\)的赋范线性子空间. 由Step1, 对任意\(f\in C(\Bbb{T})\)以及任意\(\varepsilon>0\), 存在\(T\in {\rm Tri}[-\pi,\pi]\)使得
\[\max_{t\in [-\pi,\pi]} |f(t)-T(t)|<\sqrt{\pi}\varepsilon,
\]
从而
\[\begin{equation*}\begin{split}
\|f-T\|&=&\left[\frac{1}{\pi}\int_{-\pi}^{\pi} \left| f(t)-T(t)\right|^2 {\rm d} t \right]^\frac{1}{2}\&\leq &\left[\frac{1}{\pi}\int_{-\pi}^{\pi} \max_{t\in [-\pi,\pi]}\left| f(t)-T(t)\right|^2 {\rm d} t \right]^\frac{1}{2}\&<&\varepsilon,
\end{split}\end{equation*}
\]
所以\({\rm Tri}[-\pi,\pi]\)按照\(L^2\)-范数在\(C(\Bbb{T})\)中稠密.
下证\(C(\Bbb{T})\)按\(L^2\)-范数在\(C[-\pi,\pi]\)中稠密.
对任意\(f\in C[-\pi,\pi]\), 以及任意\(n\in \Bbb{N}_+\), 令
\[\phi_n(t)=
\begin{cases}
f(t),&t\in \left[-\pi,\pi -\frac{2\pi}{n}\right]\k_n(t-\pi)+f(-\pi),& t\in \left( \pi -\frac{2\pi}{n},\pi \right],
\end{cases}
\]
其中
\[k_n=\frac{f(-\pi)-f\left(\pi -\frac{2\pi}{n}\right)}{\pi-\left(\pi-\frac{2\pi}{n}\right)},
\]
显然, \(\phi_n\in C(\Bbb{T})\).
令\(C=\max\limits_{t\in [-\pi,\pi]}|f(t)|\), 则\(\max\limits_{t\in [-\pi,\pi]}|\phi_n(t)|\leq C\), 从而
\[\begin{equation*}\begin{split}
&&\|f-\phi_n\|^2 \&=&\frac{1}{\pi}\int_{-\pi}^{\pi} |f(t)-\phi_n(t)|^2 {\rm d} t\&=&\frac{1}{\pi}\int_{\pi -\frac{2\pi}{n}}^{\pi} \left |f(t)-\phi_n(t)\right|^2 {\rm d} t\&\leq &\frac{1}{\pi}\int_{\pi -\frac{2\pi}{n}}^{\pi} 4C^2 {\rm d} t\&=&\frac{C^2}{\pi} \cdot \frac{2\pi}{n} =\frac{2C^2}{n}\to 0\quad (n\to \infty).
\end{split}\end{equation*}
\]
所以\(C(\Bbb{T})\)按\(L^2\)-范数在\(C[-\pi,\pi]\)中稠密.
将\([-\pi,\pi]\)上的有界可测函数全体集合记为\(M_b[-\pi,\pi]\), 显然在\(L^2\)范数下\(M_b[-\pi,\pi]\)也是\(L^2[-\pi,\pi]\)的赋范线性子空间. 下证\(C[-\pi,\pi]\)按\(L^2\)-范数在\(M_b[-\pi,\pi]\)中稠密.
任取\(f\in M_b[-\pi,\pi]\), 设
\[|f(x)|\leq K,\quad a.e.\ x\in [-\pi,\pi].
\]
对任意\(\varepsilon>0\), 由Lusin定理, 存在\([-\pi,\pi]\)上的连续函数\(g\)以及闭集\(F\subset [-\pi,\pi]\)使得
(i) \(f(t)=g(t)\), \(\forall t\in F\);
(ii) \(m\left([-\pi,\pi]\setminus F\right)<\frac{\varepsilon^2}{4K^2}\);
(iii) \(\max\limits_{t\in [-\pi,\pi]} |g(t)|\leq K\).
于是,
\[\begin{equation*}\begin{split}
&&\|f-g\|^2\&=&\int_{-\pi}^{\pi} |f(t)-g(t)|^2{\rm d} t\&=&\int_{[-\pi,\pi]\setminus F} |f(t)-g(t)|^2{\rm d} t\&\leq & 4K^2 m\left([-\pi,\pi]\setminus F\right)\&<&\varepsilon^2,
\end{split}\end{equation*}
\]
即\(\|f-g\|<\varepsilon\). 所以\(C[-\pi,\pi]\)按\(L^2\)-范数在\(M_b[-\pi,\pi]\)中稠密.
下证\(M_b[-\pi,\pi]\)按\(L^2\)-范数在\(L^2[-\pi,\pi]\)中稠密.
任取\(f\in L^2[-\pi,\pi]\), 对任意\(n\in \Bbb{N}_+\), 令
\[ f_n(t)=
\begin{cases}
f(t),& |f(t)|\leq n,\0,&|f(t)|>n,
\end{cases}
\]
则\(f_n\in M_b[-\pi,\pi]\), 并且
\[\|f_n-f\|^2=\int_{-\pi}^{\pi} |f_n(x)-f(x)|^2{\rm d} t=\int_{\{ t\in [-\pi,\pi]\ |\ |f(t)|>n\} } |f(t)|^2 {\rm d} t,
\]
从而
\[\|f\|^2 \geq \int_{\{ t\in [-\pi,\pi]\ |\ |f(t)|>n\} } |f(t)|^2 {\rm d} t\geq n^2 m \{ t\in [-\pi,\pi]\ |\ |f(t)|>n\},
\]
即
\[m \{ t\in [-\pi,\pi]\ |\ |f(t)|>n\} \leq \frac{1}{n^2}\|f\|^2,\quad \forall n\in \Bbb{N}_+. \tag{1}
\]
由于\(|f|^2\in L^1[-\pi,\pi]\), 由Lebesgue积分的绝对连续性, 对任意\(\varepsilon>0\), 存在\(\delta>0\), 使得对于任意的可测集\(A\subset [-\pi,\pi]\)且\(m(A)<\delta\), 都有
\[\int_A |f(t)|^2 {\rm d} t<\varepsilon^2.
\]
另一方面, 根据(1)式, 对上述\(\delta>0\), 存在正整数\(N\), 使得对任意\(n>N\), 都有
\[m \{ t\in [-\pi,\pi]\ |\ |f(t)|>n\}<\delta,
\]
从而
\[\|f_n-2\|^2=\int_{\{ t\in [-\pi,\pi]\ |\ |f(t)|>n\} } |f(t)|^2 {\rm d} t<\varepsilon^2,
\]
即
\[\|f_n-f\|< \varepsilon, \quad \forall n>N.
\]
所以\(M_b[-\pi,\pi]\)按\(L^2\)-范数在\(L^2[-\pi,\pi]\)中稠密.
综上, 按照稠密性的传递关系, \({\rm Tri}[-\pi,\pi]\)按照\(L^2\)-范数在\(L^2[-\pi,\pi]\)中稠密.
由Step2可知, 对任意\(f\in {\rm Tri}[-\pi,\pi]\), \(f\)关于规范正交系\(M\)成立Parseval等式. 根据Steklov定理, \(M\)是\(L^2[\pi,\pi]\)中的完全规范正交系.
L^2 [?π,π] 中三角函数系的完全性
标签：img block pad var max ber lan ado bbb
原文地址：https://www.cnblogs.com/sunfenglong/p/12877574.html
