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[家里蹲大学数学杂志]第041期中山大学数计学院 2008 级数学与应用数学专业《泛函分析》期末考试试题 A

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1 ( 10 分 ) 设 Xbubuko.com,布布扣bubuko.com,布布扣 是 Banach 空间, fbubuko.com,布布扣bubuko.com,布布扣 Xbubuko.com,布布扣bubuko.com,布布扣 上的线性泛函. 求证: fL(X)bubuko.com,布布扣bubuko.com,布布扣 的充分必要条件是

N(f)={xX; f(x)=0}bubuko.com,布布扣bubuko.com,布布扣
Xbubuko.com,布布扣bubuko.com,布布扣 的闭线性子空间.

证明: 必要性. 设 N(f)?xbubuko.com,布布扣nbubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣 , 则

f(x)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣f(xbubuko.com,布布扣nbubuko.com,布布扣)(fXbubuko.com,布布扣?bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣0=0.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
充分性. 用反证法. 若 fbubuko.com,布布扣bubuko.com,布布扣 无界, 则
? nN, ? xbubuko.com,布布扣nbubuko.com,布布扣X, s.t. |f(xbubuko.com,布布扣nbubuko.com,布布扣)|>nxbubuko.com,布布扣nbubuko.com,布布扣,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣xbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣f(xbubuko.com,布布扣nbubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣<1bubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
ybubuko.com,布布扣nbubuko.com,布布扣=xbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣f(xbubuko.com,布布扣nbubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣?xbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣f(xbubuko.com,布布扣1bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
则由 (1)bubuko.com,布布扣bubuko.com,布布扣
N(f)?ybubuko.com,布布扣nbubuko.com,布布扣?xbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣f(xbubuko.com,布布扣1bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣?N(f),bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
这与 N(f)bubuko.com,布布扣bubuko.com,布布扣 的闭性矛盾, 而有结论.

 

 

2 ( 12 分 ) 设 Xbubuko.com,布布扣bubuko.com,布布扣 , Ybubuko.com,布布扣bubuko.com,布布扣 Bbubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣 空间, Dbubuko.com,布布扣bubuko.com,布布扣 Xbubuko.com,布布扣bubuko.com,布布扣 的线性子空间并且 A: DYbubuko.com,布布扣bubuko.com,布布扣 是线性映射. 求证:

(1)如果 Abubuko.com,布布扣bubuko.com,布布扣 连续且是闭算子, 则 Ybubuko.com,布布扣bubuko.com,布布扣 完备蕴含 Dbubuko.com,布布扣bubuko.com,布布扣 闭;

(2)如果 Abubuko.com,布布扣bubuko.com,布布扣 是单射的闭算子, 则 Abubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣 也是闭算子.

证明:

(1)设 D?xbubuko.com,布布扣nbubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣 , 则由 Abubuko.com,布布扣bubuko.com,布布扣 连续知

Axbubuko.com,布布扣nbubuko.com,布布扣?Axbubuko.com,布布扣mbubuko.com,布布扣A?xbubuko.com,布布扣nbubuko.com,布布扣?xbubuko.com,布布扣mbubuko.com,布布扣0(n,m),bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
{Axbubuko.com,布布扣nbubuko.com,布布扣}bubuko.com,布布扣bubuko.com,布布扣 Ybubuko.com,布布扣bubuko.com,布布扣 中 Cauchy 列, 设其收敛到 yYbubuko.com,布布扣bubuko.com,布布扣 , 则由 Abubuko.com,布布扣bubuko.com,布布扣 的闭性知 xDbubuko.com,布布扣bubuko.com,布布扣 , Ax=ybubuko.com,布布扣bubuko.com,布布扣 . 此即证明了 Dbubuko.com,布布扣bubuko.com,布布扣 完备.

(2)设

R(A)=D(Abubuko.com,布布扣?1bubuko.com,布布扣)?ybubuko.com,布布扣nbubuko.com,布布扣y,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
D(A)=R(Abubuko.com,布布扣?1bubuko.com,布布扣)=Abubuko.com,布布扣?1bubuko.com,布布扣ybubuko.com,布布扣nbubuko.com,布布扣x,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
则由 Abubuko.com,布布扣bubuko.com,布布扣 闭知
xD(A)  Ax=y,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
yR(A)=D(Abubuko.com,布布扣?1bubuko.com,布布扣)  x=Abubuko.com,布布扣?1bubuko.com,布布扣y.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

 

 

3 ( 10 分 ) 设 Xbubuko.com,布布扣bubuko.com,布布扣 是 Banach 空间, TL(X)bubuko.com,布布扣bubuko.com,布布扣 . 如果 T<1bubuko.com,布布扣bubuko.com,布布扣 , 则算子 I?Tbubuko.com,布布扣bubuko.com,布布扣 有有界逆算子, 并且

bubuko.com,布布扣bubuko.com,布布扣(I?T)bubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣1?Tbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

证明: 设 Sbubuko.com,布布扣nbubuko.com,布布扣=bubuko.com,布布扣nbubuko.com,布布扣k=0bubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣 , 则对 ? m>nbubuko.com,布布扣bubuko.com,布布扣

Sbubuko.com,布布扣nbubuko.com,布布扣?Sbubuko.com,布布扣mbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣k=n+1bubuko.com,布布扣mbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣k=n+1bubuko.com,布布扣mbubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣Tbubuko.com,布布扣n+1bubuko.com,布布扣bubuko.com,布布扣1?Tbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0(n).bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
于是 {Sbubuko.com,布布扣nbubuko.com,布布扣}bubuko.com,布布扣bubuko.com,布布扣 作为 Banach \scrL(\scrX,\scrX)L(X,X)bubuko.com,布布扣bubuko.com,布布扣 的 Cauchy 列, 收敛到 \bex S=\sum_{k=1}^\infty T^k. \eex
S=bubuko.com,布布扣k=1bubuko.com,布布扣bubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
又由 \bex \sen{(I-T)S_n-I} =\sen{-T^{n+1}}\to 0\quad\sex{n\to\infty} \eex
(I?T)Sbubuko.com,布布扣nbubuko.com,布布扣?I=bubuko.com,布布扣bubuko.com,布布扣?Tbubuko.com,布布扣n+1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0(n)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
(I-T)^{-1}=S(I?T)bubuko.com,布布扣?1bubuko.com,布布扣=Sbubuko.com,布布扣bubuko.com,布布扣 , 而 \bex \sen{(I-T)^{-1}}=\sen{S} =\sen{\sum_{k=0}^\infty T^k} \leq \sum_{k=0}^\infty \sen{T}^k =\frac{1}{1-\sen{T}}. \eex
bubuko.com,布布扣bubuko.com,布布扣(I?T)bubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=S=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣k=0bubuko.com,布布扣bubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣k=0bubuko.com,布布扣bubuko.com,布布扣Tbubuko.com,布布扣kbubuko.com,布布扣=1bubuko.com,布布扣1?Tbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

 

 

4 ( 10 分 ) 设 \mathcal{X}Xbubuko.com,布布扣bubuko.com,布布扣 是赋范线性空间, \{x_1,\cdots,x_n\}{xbubuko.com,布布扣1bubuko.com,布布扣,?,xbubuko.com,布布扣nbubuko.com,布布扣}bubuko.com,布布扣bubuko.com,布布扣 nnbubuko.com,布布扣bubuko.com,布布扣 个线性无关元. 求证: \exists\ \{f_1,\cdots,f_n\}\subset \scrX^*? {fbubuko.com,布布扣1bubuko.com,布布扣,?,fbubuko.com,布布扣nbubuko.com,布布扣}?Xbubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣 使得 \sef{f_i,x_j}=\delta_{ij}\quad(\forall\ i,j=1,\cdots,n).

?fbubuko.com,布布扣ibubuko.com,布布扣,xbubuko.com,布布扣jbubuko.com,布布扣?=δbubuko.com,布布扣ijbubuko.com,布布扣(? i,j=1,?,n).bubuko.com,布布扣bubuko.com,布布扣

证明: 记 \dps{M_i=span_{j\neq i}\sed{x_j}}Mbubuko.com,布布扣ibubuko.com,布布扣=spanbubuko.com,布布扣jibubuko.com,布布扣{xbubuko.com,布布扣jbubuko.com,布布扣}bubuko.com,布布扣bubuko.com,布布扣 , 则 d_i=\rho(x_i,M_i)>0 , 于是由点与线性子空间分离的 Hahn-Banach 定理, \bex \exists\ \tilde f_i\in\scrX^*,\ s.t.\ \tilde f_i(x_i)=d_i,\quad \tilde f_i|_{M_i}=0. \eex

\dps{f_i=\frac{\tilde f_i}{d_i}} , 则 \sed{f_i} 即为所求.

 

 

5 ( 10 分 ) 设 \mathcal{H} 是 Hilbert 空间, T\in \mathcal{L}(\mathcal{H}) ||T||\leq 1 . 证明: 若 Tx=x , 则 T^*x=x .

证明: 由 Tx=x ||T||\leq 1 ||T||=1,

||T^*||=||T||=1.
于是 \bex ||T^*x-x||^2 &=&(T^*x-x,T^*x-x)\\ &=&||T^*x||^2-(T^*x,x)-(x,T^*x)+||x||^2\\ &=&||T^*x||^2-(x,Tx)-(Tx,x)+||x||^2\\ &=&||T^*x||^2-||x||^2\quad(\mbox{由 } Tx=x)\\ &\leq& ||T^*||^2\cdot ||x||^2-||x||^2\\ &=&0\quad(\mbox{由 } ||T^*||=1). \eex
因此 \bex T^*x=x. \eex

 

 

6 ( 16 分 ) 设 \mathcal{H} 是 Hilbert 空间, T:\mathcal{H}\to \mathcal{H} 是线性算子且满足 (Tx,y)=(x,Ty)\quad (\forall\ x,y\in \mathcal{H}).

求证:

(1) T^*=T , 此时称 T 为自共轭算子;

(2) 对任意 x\in \calH , (Tx,x) 是实的;

(3) 算子 T 的本征值是实的;

(4) 对应于 的不同本征值 \lambda_1 , \lambda_2 的本征元 x_1 , x_2 是正交的.

证明:

(1) 往证 T 是闭算子, 而由 D(T)=\mathcal{H} 及闭图像定理知 T\in \mathcal{L}(\mathcal{H}) . 事实上, 设 \mathcal{H}\ni x_n\to x,\ Tx_n\to y , 则于 (Tx_n,z)=(x_n,z)\quad (\forall\ z\in \mathcal{H})

中令 n\to\infty ,有 (y,z)=(x,Tz)=(Tx,z)\quad(\forall\ z\in \mathcal{H}).
于是 y=Tx.

(2) 由 \beex\bea (Tx,x)&=(x,Tx)\quad\sex{T\mbox{ 自共轭}}\\ &=\overline{\sex{Tx,x}}\quad\sex{\mbox{内积定义}} \eea\eeex

即知 (Tx,x)\in \bbR .

(3) 设 \lambda\in \bbC 适合 \bex \exists\ 0\neq x\in \calH,\ s.t.\ Tx=\lambda x, \eex

\bex \lambda \sen{x}^2 =\lambda(x,x) =\sex{\lambda x,x} =\sex{Tx,x}\in \bbR, \eex
于是 \bex \lambda=\frac{\sex{Tx,x}}{\sen{x}^2}\in \bbR. \eex

(4)由 \bex & &(\lambda_1-\lambda_2)(x_1,x_2)\\ & &= (\lambda_1x_1,x_2)-(x_1,\lambda_2x_2)\quad\sex{\lambda_i\in\bbR}\\ & &=(Tx_1,x_2)-(x_1,Tx_2)\quad\sex{x_i \mbox{ 是 } T \mbox{ 的相对于 }\lambda_i\mbox{ 的本征元}}\\ & &=0\quad\sex{T\mbox{ 自共轭}}. \eex

即知结论.

 

 

7 ( 12 分 ) 设 \varphi\in C[0,1] , T:\ L^2[0,1]\to L^2[0,1] 是由 (Tf)(x)=\varphi(x)\int_0^1\varphi(t)f(t)\ dt\quad(\forall\ f\in L^2[0,1])

给出的线性算子. 求证:

(1) T 是自共轭算子 (定义见题 6);

(2) \exists\ \lambda\geq 0 , 使得 T^2=\lambda T , 由此求出 T 的谱半径 r_\sigma(T) .

证明:

(1)对 \forall\ f,\ g\in L^2[0,1] , 由 \bex (Tf,g) &=&\int_0^1 \sez{ \varphi(x)\int_0^1 \varphi(t)f(t)\ dt }\cdot g(x)\ dx\\ &=&\int_0^1 \varphi(t)f(t)\ dt \cdot \int_0^1 \varphi(x)g(x)\ dx\\ &=&\int_0^1 \varphi(x)f(x)\ dx \cdot \int_0^1 \varphi(t)g(t)\ dt\\ &=&\int_0^1 f(x)\cdot \sez{\varphi(x)\int_0^1 \varphi(t)g(t)\ dt}\ dx\\ &=&(f,Tg) \eex

T^*=T , 而 T 为自共轭算子.

(2)由 \bex (T^2f)(x) &=&[T(Tf)](x)\\ &=&\varphi(x)\int_0^1 \varphi(t)(Tf)(t)\ dt\\ &=&\varphi(x) \int_0^1 \sez{ \varphi(t) \cdot \varphi(t) \int_0^1 \varphi(s)f(s)\ ds }\ dt\\ &=&\int_0^1 \varphi^2(t)dt\cdot \varphi(x)\int_0^1 \varphi(s)f(s)\ ds\\ &=&\int_0^1 \varphi^2(t)dt\cdot (Tf)(x)\quad (\forall\ f\in L^2[0,1]) \eex

T^2=\lambda T,
其中 \lambda=\int_0^1 \varphi^2(t)dt.
由数学归纳法易知 T^n=\lambda^{n-1}T\quad(n\geq 1),
T 的谱半径 r_\sigma(T)=\lim_{n\to\infty}||T^n||^\frac{1}{n} =\lim_{n\to\infty} \lambda^\frac{n-1}{n}||T||^\frac{1}{n} =\lambda =\int_0^1 \varphi^2(t)dt.
倒数第二个等号是因为若 \varphi\equiv 0 , 则 \lambda=0 , T=0 ; 若 \varphi\not\equiv 0 , 则 ||T||\neq 0 .

 

8 ( 10 分 ) 设 \scrX 是赋范线性空间, M \scrX 的闭子空间. 证明: 如果 \sed{x_n}\subset M x_n\rhu x_0 , 则 x_0\in M .

证明: 用反证法. 若 x_0\not\in M , 则 d=\rho(x_0,M)>0 , 由点与线性子空间分离的 Hahn-Banach 定理, \bex \exists\ f\in\scrX^*,\ s.t.\ f(x_0)=d>0,\quad f|_M=0. \eex

于是由 x_n\rhu x_0 \bex 0<d=f(x_0)=\lim_{n\to\infty}f(x_n) =\lim_{n\to\infty}0=0. \eex
这是一个矛盾, 故有结论.

 

9 ( 10 分 ) 设 \calH 是 Hilbert 空间, \sed{x_n}\subset \calH , x\in \calH . 证明:

(1) x_n\rhu x 当且仅当对任意 y\in\calH , (x_n,y)\to (x,y) ;

(2) x_n\to x 当且仅当 x_n\rhu x \sen{x_n}\to \sen{x} .

证明:

(1) 这是 Riesz 表示定理与弱收敛定义的直接结论.

(2) 必要性显然. 往证充分性. \bex \sen{x_n-x}^2&=&\sex{x_n-x,x_n-x}\\ &=&\sen{x_n}^2-\sex{x_n,x}-\sex{x,x_n}+\sen{x}^2\\ &\to&\sen{x}^2-\sex{x,x}-\sex{x,x}+\sen{x}^2=0\quad\sex{n\to\infty}. \eex

 

[家里蹲大学数学杂志]第041期中山大学数计学院 2008 级数学与应用数学专业《泛函分析》期末考试试题 A,布布扣,bubuko.com

[家里蹲大学数学杂志]第041期中山大学数计学院 2008 级数学与应用数学专业《泛函分析》期末考试试题 A

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原文地址:http://www.cnblogs.com/zhangzujin/p/3758260.html

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