实分析作业-第三周

第三周

1. 证明：
   1. $\overline{e^{ix}}=e^{-ix}$
      $$\begin{array}{rl}& \because e^{ix}=cosx+isinx\\ & \therefore \overline{e^{ix}}=cosx-isinx\\ & 又e^{-ix}=cos(-x)+isin(-x)=cosx-isinx\\ & \therefore \overline{e^{ix}}=e^{-ix}\end{array}$$
   2. $e^{ix}{e}^{iy}=e^{i(x+y)}$
      $$\begin{array}{rl}& e^{ix}{e}^{iy}=(cosx+isinx)(cosy+isiny)\\ & =cosxcosy-sinxsiny+i(sinxcosy+cosxsiny)\\ & =cos(x+y)+isin(x+y)\\ & =e^{i(x+y)}\end{array}$$
   3. $e^{i\lambda x}=e^{\lambda ix}$
      $$\begin{array}{rl}& e^{i\lambda x}=cos(\lambda x)+isin(\lambda x)\\ & e^{\lambda ix}=(cosx+isinx{)}^{\lambda }\\ 先证明\lambda \in \mathbb{Z}时& \\ & 设A_n:e^{inx}=e^{nix}\\ & 显然A_0,A_1为真\\ & 则A_n,A_{n-1}为真时，\\ & e^{i(n+1)x}=(cosnxcosx-sinnxsinnx)+i(sinnxcosx+cosnxsinx)\\ & e^{i(n-1)x}=(cosnxcosx+sinnxsinnx)+i(sinnxcosx-cosnxsinx)\\ \therefore & e^{i(n+1)x}+e^{i(n-1)x}\\ & =2cosnxcosx+2isinnxcosx=2cosx\cdot e^{inx}\\ \therefore & e^{i(n+1)x}=2cosx(cosx+isinx{)}^n-(cosx+isinx{)}^{n-1}\\ & =(cosx+isinx{)}^{n-1}[2co{s}^{2}x+2icosxsinx-1]\\ & =(cosx+isinx{)}^{n-1}[2co{s}^{2}x+2icosxsinx+(isinx{)}^2]\\ & =(cosx+isinx{)}^{n-1}(cosx+isinx{)}^2\\ & =(cosx+isinx{)}^{n+1}\\ 证毕& \\ 当\lambda \in \mathbb{R}时& \\ & 令\lambda =n+c,n\in \mathbb{Z},0<c<1\\ & 则e^{i\lambda x}=e^{inx}{e}^{icx}\\ & 由于e^{inx}=(e^{ix}{)}^n\\ & 只需证明e^{icx}=(e^{ix}{)}^c即可\end{array}$$
   4. ${\int }_a^b{e}^{ix}dx=\frac{1}{i}(e^{ib}-e^{ia})$
      $$\begin{array}{rl}{\int }_a^b{e}^{ix}dx& ={\int }_a^{b}cosxdx+i{\int }_a^{b}sinxdx\\ & =sinb-sina-i(cosb-cosa)\\ & =\frac{1}{i}(isinb-isina+cosb-cosa)\\ & =\frac{1}{i}(e^{ib}-e^{ia})\end{array}$$
2. 已知$l^2=\{(\cdots ,a_{-n},\cdots ,a_0,\cdots ,a_n,\cdots ),\forall n\in \mathbb{R},a_n\in \mathbb{C},\sum _{-\infty }^{+\infty }|a_n{|}^2<+\infty \}$，
   证明：
   1. $l^2$为给定加法和数乘下的向量空间
      $$\begin{array}{rl}& 容易看出给定加法满足交换性和结合性\\ & 加法零元为0=(\cdots ,0,\cdots )\\ & \mathbf{a}=(\cdots ,a_{-n},\cdots ,a_0,\cdots ,a_n,\cdots )的加法逆元为(\cdots ,-a_{-n},\cdots ,-a_0,\cdots ,-a_n,\cdots )\\ & 对于给定数乘，容易看出\lambda =1是其乘法单位元\\ & 根据给定加法和数乘，计算(\lambda +\mu )\mathbf{a}和\lambda \mathbf{a}+\mu \mathbf{a}，可以得到给定数乘满足结合律\\ & \therefore 此时l^2是向量空间\end{array}$$
   2. 给定的运算构成$l^2$上的内积
      $$\begin{array}{rl}内积& 符合下面五条性质：\\ & ①：⟨\mathbf{v},\mathbf{v}⟩\ge 0\\ & ②：⟨\mathbf{v},\mathbf{v}⟩=0\leftrightarrows \mathbf{v}=0\\ & ③：⟨\mathbf{u}+\mathbf{w},\mathbf{v}⟩=⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{w},\mathbf{v}⟩\\ & ④：⟨\mathbf{u},\mathbf{v}⟩=\overline{⟨\mathbf{v},\mathbf{u}⟩}\\ & ⑤：⟨\lambda \mathbf{u},\mathbf{v}⟩=\lambda ⟨\mathbf{u},\mathbf{v}⟩\\ & 逐条进行证明\\ & 令\mathbf{v}=(\cdots ,a_{-n},\cdots ,a_0,\cdots ,a_n,\cdots )\\ & \mathbf{u}=(\cdots ,b_{-n},\cdots ,b_0,\cdots ,b_n,\cdots )\\ & \mathbf{w}=(\cdots ,c_{-n},\cdots ,c_0,\cdots ,c_n,\cdots )\\ ①：& \\ & 则⟨\mathbf{v},\mathbf{v}⟩=\sum _{-\infty }^{+\infty }{a}_n\overline{a_n}\\ & 已知a_n\overline{a_n}\ge 0\\ & 故⟨\mathbf{v},\mathbf{v}⟩=\sum _{-\infty }^{+\infty }{a}_n\overline{a_n}\ge 0\\ ②：& \\ & ⟨\mathbf{v},\mathbf{v}⟩=0\\ \leftrightarrows & a_n\overline{a_n}=0,\forall n\in \mathbb{Z}\\ \leftrightarrows & a_n=0,\forall n\in \mathbb{Z}\\ \leftrightarrows & \mathbf{v}=0\\ ③：& \\ & ⟨\mathbf{u}+\mathbf{w},\mathbf{v}⟩\\ =& \sum _{-\infty }^{+\infty }(b_n+c_n)\overline{a_n}\\ =& \sum _{-\infty }^{+\infty }{b}_n\overline{a_n}+c_n\overline{a_n}\\ =& \sum _{-\infty }^{+\infty }{b}_n\overline{a_n}+\sum _{-\infty }^{+\infty }{c}_n\overline{a_n}\\ =& ⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{w},\mathbf{v}⟩\\ ④：& \\ & 易得对c_1,c_2\in \mathbb{C}有\\ & c_1\cdot \overline{c_2}=\overline{c_2}\cdot c_1\\ & c_1\cdot \overline{c_2}与\overline{c_1}\cdot c_2共轭\\ & \overline{c_1+c_2}=\overline{c_1}+\overline{c_2}\\ 故& \overline{⟨\mathbf{v},\mathbf{u}⟩}=\overline{\sum _{-\infty }^{+\infty }{b}_n\overline{a_n}}=\sum _{-\infty }^{+\infty }\overline{b_n\overline{a_n}}=\sum _{-\infty }^{+\infty }\overline{b_n}{a}_n=\sum _{-\infty }^{+\infty }{a}_n\overline{b_n}=⟨\mathbf{u},\mathbf{v}⟩\\ ⑤：& \\ & ⟨\lambda \mathbf{u},\mathbf{v}⟩\\ =& \sum _{-\infty }^{+\infty }\lambda b_n\overline{a_n}\\ =& \lambda \sum _{-\infty }^{+\infty }{b}_n\overline{a_n}\\ =& \lambda ⟨\mathbf{u},\mathbf{v}⟩\\ & 因此给定运算⟨\mathbf{v},\mathbf{u}⟩=\sum _{-\infty }^{+\infty }{a}_n\overline{b_n}可以构成l^2上的内积\end{array}$$
   3. 给定运算是$l^2$上的距离
      $$\begin{array}{rl}距离& 符合下列性质：\\ & ①：d(\mathbf{u},\mathbf{v})\ge 0，当且仅当\mathbf{u}=\mathbf{v}时等号成立\\ & ②：d(\lambda \mathbf{u},\lambda \mathbf{v})=\lambda d(\mathbf{u},\mathbf{v}),\lambda >0\\ & ③：d(\mathbf{u},\mathbf{v})\le d(\mathbf{u},\mathbf{w})+d(\mathbf{w},\mathbf{v})\\ 下面& 逐条证明：\\ ①：& \\ & \because |a_n-b_n{|}^2\ge 0,\forall n\in \mathbb{Z}，当且仅当a_n=b_n时等号成立\\ & \therefore d(\mathbf{u},\mathbf{v})=0\\ \leftrightarrows & \sqrt{\sum _{-\infty }^{+\infty }|a_n-b_n{|}^2}=0\\ \leftrightarrows & \sum _{-\infty }^{+\infty }|a_n-b_n{|}^2=0\\ \leftrightarrows & |a_n-b_n{|}^2=0,\forall n\in \mathbb{Z}\\ \leftrightarrows & a_n=b_n,\forall n\in \mathbb{Z}\\ \leftrightarrows & \mathbf{u}=\mathbf{v}\\ ②：& \\ 首先& |\lambda a_n-\lambda b_n{|}^2\\ =& |\lambda r_a[cos{\theta }_a+isin{\theta }_a]-\lambda r_b[cos{\theta }_b+isin{\theta }_b]{|}^2\\ =& [\lambda r_{a}cos{\theta }_a-\lambda r_{b}cos{\theta }_b{]}^2+[\lambda r_{a}sin{\theta }_a-\lambda r_{b}sin{\theta }_b{]}^2\\ =& {\lambda }^2|a_n-b_n{|}^2\\ & \therefore d(\lambda \mathbf{u},\lambda \mathbf{v})\\ =& \sqrt{\sum _{-\infty }^{+\infty }|\lambda a_n-\lambda b_n{|}^2}\\ =& \sqrt{\sum _{-\infty }^{+\infty }{\lambda }^2|a_n-b_n{|}^2}\\ =& \lambda \sqrt{\sum _{-\infty }^{+\infty }|a_n-b_n{|}^2}\\ =& \lambda d(\mathbf{u},\mathbf{v})\end{array}$$
   4. 借助复数空间的完备性，证明$l^2$在上述定义下是完备的
      $$\begin{array}{rl}反证法：& \\ & 假设存在柯西列\{{\mathbf{a}}_i{\}}_{i=0}^{+\infty }收敛于\mathbf{x}\notin l^2\\ & 则至少有\sum _{-\infty }^{+\infty }|x_n{|}^2=+\infty \cdots \cdots (\ast )\\ 而& 对\forall ϵ>0,\exists M>0,\text{s.t. for all }s>M\\ & d(a_s,x)<ϵ\\ & 即\sqrt{\sum _{-\infty }^{+\infty }|a_{sn}-x_n{|}^2}<ϵ\\ & 不妨设k_n\in \mathbb{R}，使得\sum _{-\infty }^{+\infty }{k}_n^2=1，可知\mathbf{k}=(\cdots ,k_n,\cdots )\in l^2\\ 则& \sqrt{\sum _{-\infty }^{+\infty }|a_{sn}-x_n{|}^2}\sqrt{\sum _{-\infty }^{+\infty }{k}_n^2}<ϵ\\ \Rightarrow & \sqrt{\sum _{-\infty }^{+\infty }|a_{sn}-x_n|k_n}<ϵ（柯西-施瓦茨不等式）\\ \Rightarrow & \sqrt{\sum _{-\infty }^{+\infty }|a_{sn}-x_n|}<ϵ（k_n<1是平凡的）\\ \Rightarrow & \sqrt{\sum _{-\infty }^{+\infty }|x_n|-\sum _{-\infty }^{+\infty }|a_{sn}|}<ϵ（三角不等式）\\ \Rightarrow & \sum _{-\infty }^{+\infty }|x_n|<{ϵ}^2+\sum _{-\infty }^{+\infty }|a_{sn}|<+\infty \\ \Rightarrow & \sum _{-\infty }^{+\infty }|x_n{|}^2<\sum _{-\infty }^{+\infty }|x_n|<+\infty \\ & 与(\ast )矛盾，故\mathbf{x}\in l^2\\ & 故l^2在给定距离定义下是完备的\end{array}$$
3. 定义$d(f,g)=\frac{1}{2\pi }{\int }_0^{2\pi }|f(x)-g(x){|}^{2}dx$，证明此定义下有函数列$\{f_n(x){\}}_{n=1}^{+\infty }$
   $$f_n(x)=\{\begin{array}{cc}0& 0\le x\le \frac{1}{n}\\ -lnx& \frac{1}{n}<x\le 2\pi \end{array}$$是柯西列，但不可积。
   证明:
   $$\begin{array}{rl}给定& 一个任意小的\delta >0，我们取一个足够大的整数N，使得当m>n>N时\\ & d(f_m,f_n)\\ =& \frac{1}{2\pi }{\int }_0^{2\pi }|f_m(x)-f_n(x){|}^{2}dx\\ =& \frac{1}{2\pi }[{\int }_0^{\frac{1}{m}}|0-0{|}^{2}dx+{\int }_{\frac{1}{m}}^{\frac{1}{n}}|-lnx-0|dx+{\int }_{\frac{1}{n}}^{2\pi }|-lnx+lnx|dx]\\ =& \frac{1}{2\pi }{\int }_{\frac{1}{m}}^{\frac{1}{n}}-lnxdx\\ =& \frac{1}{2\pi }[x(lnx-1){]}_{\frac{1}{m}}^{\frac{1}{n}}\\ =& \frac{lnm+1}{m}+\frac{lnn+1}{n}，（不妨m=n+1,n=\lceil e^{{\delta }^2/2-1}\rceil ）\\ \le & 2\frac{lnn+1}{n}\le 2\frac{{\delta }^2/2-1+1}{\delta }=\delta \\ & 故\{f_n\}在该距离定义下是柯西列，\underset{n\to +\infty }{\lim }{f}_n(x)=\{\begin{array}{cc}0& x=0\\ -lnx& 0<x\le 2\pi \end{array}\\ & 易知{\int }_0^{2\pi }-lnxdx不可积\end{array}$$
4. 证明命题2.1，特别地，证明$\hat{f}(n)=0,\forall n\in \mathbb{Z}\Rightarrow {\int }_{-\pi }^{\pi }(ϵ+cosx{)}^{n}f(x)dx=0$
   证明：
   $$\begin{array}{rl}& 反证法：假设f({\theta }_0)>0\\ & 不妨令f是[-\pi ,\pi ]上的周期函数，记|f(x)|\le B\\ & 设p_k(\theta )=[ϵ+cos(\theta -{\theta }_0){]}^k=\sum _{n=0}^k{a}_{kn}co{s}^n(\theta -{\theta }_0)=\sum _{n=0}^k[b_{kn}cosn(\theta -{\theta }_0)+c_{kn}sinn(\theta -{\theta }_0)]\\ & 其中ϵ满足:\\ 1.& p_1(\theta )<1-\frac{ϵ}{2},\forall \theta 满足f(\theta )>\frac{f({\theta }_0)}{2}，取\delta 满足|\theta |<\delta <\frac{\pi }{2}+{\theta }_0\\ 2.& p_1(\theta )\ge 1+\frac{ϵ}{2},\forall \theta 满足|\theta |<\eta <\delta ，\eta 是人为取定的\\ & \end{array}$$
   （粗糙地标注了${\theta }_0=0$时$\eta ,\delta$的可能取值）
   

$$\begin{array}{rl}\because & \hat{f}(n)=0,\forall n\in \mathbb{Z}\\ \therefore & {\int }_{-\pi }^{\pi }f(x)e^{inx}dx=0\\ \therefore & {\int }_{-\pi }^{\pi }f(x)cosxdx+i{\int }_{-\pi }^{\pi }f(x)sinxdx=0\\ \therefore & {\int }_{-\pi }^{\pi }f(x)cosnxdx={\int }_{-\pi }^{\pi }f(x)sinnxdx=0，即f(x)与三角多项式的积分为0\\ \therefore & {\int }_{-\pi }^{\pi }f(\theta )p_k(\theta )d\theta =0\cdots \cdots (\ast )\\ & 又有|{\int }_{-\pi }^{\pi }f(\theta )p_k(\theta )d\theta -{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta +{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta -0|\\ & \ge |{\int }_{-\pi }^{\pi }f(\theta )p_k(\theta )d\theta -{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta |-|{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta |\\ & =|{\int }_{|\theta |<\delta }f(\theta )p_k(\theta )d\theta |-|{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta |\\ & 而|{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta |\\ & \le |{\int }_{|\theta |\ge \delta }B{p}_k(\theta )d\theta |\\ & \le |{\int }_{|\theta |\ge \delta }B[1-\frac{ϵ}{2}{]}^{k}d\theta |\\ & \le |{\int }_{-\pi }^{\pi }B[1-\frac{ϵ}{2}{]}^{k}d\theta |=2\pi B(1-\frac{ϵ}{2}{)}^k\\ 同时& |{\int }_{|\theta |<\delta }f(\theta )p_k(\theta )d\theta |\\ =& |{\int }_{\eta \le |\theta |<\delta }f(\theta )p_k(\theta )d\theta +{\int }_{|\theta |<\eta }f(\theta )p_k(\theta )d\theta |\\ \le & |{\int }_{\eta \le |\theta |<\delta }f(\theta )p_k(\theta )d\theta |+|{\int }_{|\theta |<\eta }f(\theta )p_k(\theta )d\theta |\\ 且& |{\int }_{\eta \le |\theta |<\delta }f(\theta )p_k(\theta )d\theta |\ge 0\\ & |{\int }_{|\theta |<\eta }f(\theta )p_k(\theta )d\theta |\\ \ge & |{\int }_{|\theta |<\eta }\frac{f({\theta }_0)}{2}{p}_k(\theta )d\theta |\\ \ge & |{\int }_{|\theta |<\eta }\frac{f({\theta }_0)}{2}[1+\frac{ϵ}{2}{]}^{k}d\theta |\\ =& \eta f({\theta }_0)(1+\frac{ϵ}{2}{)}^k\\ 故& |{\int }_{-\pi }^{\pi }f(\theta )p_k(\theta )d\theta |\\ \ge & |{\int }_{|\theta |<\eta }f(\theta )p_k(\theta )d\theta |-|{\int }_{|\theta |\ge \delta }f(\theta )p_k(\theta )d\theta |\\ \ge & \eta f({\theta }_0)(1+\frac{ϵ}{2}{)}^k-2\pi B(1-\frac{ϵ}{2}{)}^k\\ \ge & \eta f({\theta }_0)(1+\frac{ϵ}{2}{)}^k=+\infty ,(k\to +\infty )\\ & 与结论(\ast )矛盾\\ & f({\theta }_0)<0时同理\\ & 故f({\theta }_0)=0\end{array}$$