1.空间$S$令$S$为一切实(或复)数列$$\bexx\sex{\xi_1,\xi_2,\cdots,\xi_n,\cdots}\eex$$组成的集合,在$S$中定义距离为$
1.>空间
Saria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">S
class="MathJax_strut" border="0" src="about:blank">>
令 Sid="MathJax-Element-2-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">S
class="MathJax_strut" border="0" src="about:blank">>
为一切实(或复)数列 \bex
x=\sex{\xi_1,\xi_2,\cdots,\xi_n,\cdots} \eex
组成的集合, 在
Sid="MathJax-Element-4-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">S
class="MathJax_strut" border="0" src="about:blank">> 中定义距离为
\bex \rho(x,y) =\sum_{k=1}^\infty
\frac{1}{2^k}\frac{\xi_k-\eta_k}{1+\sev{\xi_k-\eta_k}}, \eex 其中
x=\sex{\xi_1,\xi_2,\cdots,\xi_k,\cdots},\
y=\sex{\eta_1,\eta_2,\cdots,\eta_k,\cdots}id="MathJax-Element-6-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">. 求证
Sid="MathJax-Element-7-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为一个完备的距离空间.
证明: \rhoid="MathJax-Element-8-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
是 Sid="MathJax-Element-9-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
上的一个距离. 这是显然的, 因为 \bex
\frac{\sev{a+b}}{1+\sev{a+b}} &=&\frac{1}{1/\sev{a+b}+1} \leq
\frac{1}{1/\sex{\sev{a}+\sev{b}}+1} =\frac{\sev{a}+\sev{b}}{1+\sev{a}+\sev{b}}\\
&\leq& \frac{\sev{a}}{1+\sev{a}} +\frac{\sev{b}}{1+\sev{b}}. \eex
id="MathJax-Element-10-Frame" class="MathJax">
为证
Sid="MathJax-Element-11-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 完备, 设
\sed{x^i}\subset Sid="MathJax-Element-12-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为
Cauchyid="MathJax-Element-13-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 列, 其中
\bex
x^i=\sex{\xi^i_1,\xi^i_2,\cdots,\xi^i_n,\cdots}. \eexid="MathJax-Element-14-Frame" class="MathJax">
则
\bex \rho(x^i,x^j)=\sum_{k=1}^\infty
\frac{1}{2^k} \frac{\sev{\xi^i_k-\xi^j_k}} {1+\sev{\xi^i_k-\xi^j_k}} \to 0 \quad
(i,j\to \infty). \eexid="MathJax-Element-15-Frame" class="MathJax">
于是对固定的
kid="MathJax-Element-16-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">, 任意的
\ve\in (0,1)id="MathJax-Element-17-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">, 可取
i,jid="MathJax-Element-18-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 充分大使
\bex \frac{1}{2^k}
\frac{\sev{\xi^i_k-\xi^j_k}} {1+\sev{\xi^i_k-\xi^j_k}} \leq \rho(x^i,x^j)
<\frac{\ve}{2^{k+1}}, \eexid="MathJax-Element-19-Frame" class="MathJax">
\bex \sev{\xi^i_k-\xi^j_k} <\frac{\ve}{2-\ve}
<\ve. \eexid="MathJax-Element-20-Frame" class="MathJax">
从而
\sed{\xi^i_k}_{i=1}^\inftyid="MathJax-Element-21-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为实(或复) 的
Cauchyid="MathJax-Element-22-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 列,
\xi^i_k\to \xi_k\ (i\to\infty) id="MathJax-Element-23-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">. 记
\bex x=\sex{\xi_1,\xi_2,\cdots,\xi_n,\cdots},
\eexid="MathJax-Element-24-Frame" class="MathJax">
则由
\bex \rho(x^i,x) &=&\sum_{k=1}^\infty
\frac{1}{2^k} \frac{\sev{\xi^i_k-\xi_k}} {1+\sev{\xi^i_k-\xi_k}}\\
&\leq&\sum_{k=1}^N \frac{1}{2^k}
\frac{\sev{\xi^i_k-\xi_k}}{1+\sev{\xi^i_k-\xi_k}} +\frac{1}{2^N}\quad
(N\mbox{ 待定})\\ &<&\ve\quad (\mbox{当 } N,i\mbox{ 充分大时})
\eexid="MathJax-Element-25-Frame" class="MathJax">
知
x^i\to xid="MathJax-Element-26-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 于
(S,\rho)id="MathJax-Element-27-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 上. 这就证明了
Sid="MathJax-Element-28-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 是完备的.
>2.判断基本列为收敛列的充要条件
在一个度量空间 (\scrX,\rho)id="MathJax-Element-29-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
上, 求证: 基本列是收敛列, 当且仅当其中存在一串收敛子列.
证明: \raid="MathJax-Element-30-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
这是显然的. \laid="MathJax-Element-31-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
设 \sed{x_n}id="MathJax-Element-32-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
为基本列, 且其一子列 \sed{x_{n_k}}id="MathJax-Element-33-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
收敛到 xid="MathJax-Element-34-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, 则由 \bex \rho(x_n,x) &\leq&
\rho(x_n,x_{n_k}) +\rho(x_{n_k},x)\\ &<&\ve\quad (\mbox{当 }
k,n\mbox{ 充分大时}) \eex
id="MathJax-Element-35-Frame" class="MathJax">
知
x_n\to xid="MathJax-Element-36-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">.
>3.空间
C_0aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
设 Fid="MathJax-Element-38-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
是只有有限项不为 0id="MathJax-Element-39-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
的实数列全体, 在 Fid="MathJax-Element-40-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
上引进距离 \bex \rho(x,y)=\sup_{k\geq
1}\sev{\xi_k-\eta_k}, \eex
id="MathJax-Element-41-Frame" class="MathJax">
其中
x=\sed{\xi_k}\in Fid="MathJax-Element-42-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">,
y=\sed{\eta_k}\in Fid="MathJax-Element-43-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">, 求证
\sex{F,\rho}id="MathJax-Element-44-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 不完备, 并指出它的完备化空间.
证明: \sex{F,\rho}id="MathJax-Element-45-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
不完备的, 这是因为对列 \bex
x^n=\sex{1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n},0,\cdots}\in F \eex
id="MathJax-Element-46-Frame" class="MathJax">
而言,
\bex m>n\ra \rho(x^n,x^m)
=\frac{1}{n+1}\to 0 \quad (\mbox{当 } n,m\to \infty), \eexid="MathJax-Element-47-Frame" class="MathJax">
但
\bex x^n\to
x=\sex{1,\frac{1}{2},\frac{1}{3},\cdots, \frac{1}{n},\cdots} \not\in F.
\eexid="MathJax-Element-48-Frame" class="MathJax">
Faria-readOnly="true" class="MathJax MathJax_Processing" role="textbox"> 的完备化空间为
C_0id="MathJax-Element-50-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">, 即所有以
0id="MathJax-Element-51-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为极限的数列全体.
>4.多项式全体在可积函数类中稠密
求证: [0,1]id="MathJax-Element-52-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
上的多项式全体按距离 \bex \rho(p,q)=\int_0^1
\sev{p(x)-q(x)}\rd x\quad (p,q\mbox{ 是多项式}) \eex
id="MathJax-Element-53-Frame" class="MathJax">
是不完备的, 并指出它的完备化空间.
证明: 记 \dps{P_n(x)=\sum_{k=0}^n
\frac{x^k}{k!}}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
, 则由 \bex m>n\ra \rho(P_m,P_n) =\int_0^1
\sum_{k=n+1}^n \frac{x^k}{k!}\rd x =\sum_{k=n+1}^m\frac{1}{(k+1)!} \leq
\sum_{k=n+1}^\infty \frac{1}{(k+1)k} =\frac{1}{n+1} \eex
id="MathJax-Element-55-Frame" class="MathJax">
知
\sed{P_n}id="MathJax-Element-56-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为
Cauchyid="MathJax-Element-57-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 列, 但
\bex \rho(P_n,e^x) =\int_0^1
\sum_{k=n+1}^\infty \frac{x^k}{k!}\rd x =\sum_{k=n+1}^\infty \frac{1}{(k+1)!}
=\frac{1}{n+1}\to 0\quad (\mbox{当 }n\to\infty) \eexid="MathJax-Element-58-Frame" class="MathJax">
知
\rho\to e^xid="MathJax-Element-59-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">. 于是由
e^xid="MathJax-Element-60-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 不是多项式知题中所述空间是不完备的, 且其完备空间为
L^1[0,1] id="MathJax-Element-61-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">.
>5.判断列收敛的一个充分条件
在完备的度量空间 (\scrX,\rho)id="MathJax-Element-62-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
中给定点列 \sed{x_n}id="MathJax-Element-63-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, 如果 \forall\ \ve>0,id="MathJax-Element-64-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
存在基本列 \sed{y_n}id="MathJax-Element-65-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, 使得 \bex \rho(x_n,y_n)<\ve\quad (n\in
\bbN), \eex
id="MathJax-Element-66-Frame" class="MathJax">
求证
\sed{x_n}id="MathJax-Element-67-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 收敛.
证明: 由 \bex \rho(x_n,x_m) &\leq&
\rho(x_n,y_n)+\rho(y_n,y_m)+\rho(y_m,x_m)\\
&<&\frac{2\ve}{3}+\rho(y_n,y_m)\\ &<&\ve\quad
(\mbox{当 }n,m\mbox{ 充分大}) \eex
id="MathJax-Element-68-Frame" class="MathJax">
知
\sed{x_n}id="MathJax-Element-69-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox"> 为基本列, 而收敛.