1
2.6.1 设 X 是 B 空间,求证:
( )XL 中的可逆(有有界逆)算子集
是开的.
证明 设 1A, A� � ( )X ,L
考虑当 � 0> 充分小时,是否
有 ( ) 1A I �+� � ( )X .L 注意到
( )1A I A I A ,�+� = +� 根据引理
2.2.6 , 当 1A 1�� < 时,
( ) 11I A ��+� � ( )XL 故当
1
1
A�� < 时, ( )
11I A
��+� �
( )X ,L 而
( ) ( ) 11 1 1A I I A A�� � �+� = +� �
( )X .L
这里用到, ( )A,B X ,L�
1AB BA I B A.�= =
= 事实上,
2
x ABx
1Bx x B .
= =�
�=� = �
即 B 单
射.
BA I y X,= �
� � Bx y=
有解 x ABx Ay.= = 即 B 满射.
1BA I B B I
1 1B B BA A.
�= =
� �= =
2.6.2 设 A 是闭线性算于,
1 n, ,� �� ( )p A�� 两两互异.又
设 ix 是对应于 i� 的本征元
( i 1,2, ,n= � ) 求证 1 2 nx ,x , ,x�
是线性无关的.
证明 用反证法. 令 mx 为第一个
可由它的前面 m 1� 个向量线
性
出的向量, 即
m 1
m k k
k 1
x x
�
=
= �� (1)
且 1 2 m 1x ,x , ,x �� 线性无关, 对(1)
两边施以 mI A� � 得到
3
( ) ( )
m 1
m m k m k
k 1
0 I A x I A x
�
=
= � � = � � ��
( )
m 1
k m k k
k 1
x
�
=
= � � � ��
1 2 m 1x ,x , ,x �� � 线性无关
( ) ( )
( )
m k
k m k
k
0 k 1,2, ,m 1
0 k 1,2, ,m 1 .
� ��
�� � � � = = �
� = = �
�
�
于是(1) mx 0. = 与 mx 为特征
向量矛盾.
2.6.3 在双边 2l 空间上,考察右推
移算于A:
( n n 1 1 0, 1, n 1 nx , , , , , , , , l� � + � �= � � � � � � �� � � �
( n n 1 1 0, 1, n 1 ny Ax , , , , , , , ,� � + � �= = � � � � � � �� � � �
Ax x A 1,= =
4
( )n n n
n
A x x A 1 r A lim A 1.� ��= = = =
0,� =
( )I A x 0 Ax 0 x 0� � = = =
0,� � ( )I A x 0� � =
( )k k 1 0 k��� � � = �Z
(1)
2
1 1 1
1 0, 2 1 0, ,� � �� = � � = � = � �
n
1
n 0,�� = � 同理 nn 0,�� =� �
由此可见, 如果 0 0,� = 则
n� ( )0 z=
�Z x 0.=
如果 0 0,� �
22 2 2
n 0 n n
n n 1 n 1
,
+� +� +�
�
=�� = =
� < +� � + � + � < +�� � �
即得
5
2 2 22n 2n1 1
0 0 0
n 1 n 1
+� +�
� �
= =
� +� +� � < +� �� �
0� � 矛盾.
因此只能 0 n0� = �
( )0 z=
�Z x 0.= 于是
( )p A ?.� =
再证
( )r A ?� = . ( ) 2R I A .� � =�
即证 ( ) { }R I A .�� � = �
设 ( )z R I A ,� � � 则对
2x ,
��
( )( I A x, z 0� � =
( )k k 1 k
k
z 0
+�
�
=��
�� � � =�
( )
n
n 2x (0,0, ,0,1,0, ) ,=
�����
� � ��
6
( ) ( )
n n 1
n nx (0,0, ,0, ,0, ), Ax (0,0, ,0,1,0, ),
+
� = � =
����� �����
� � � �
( )nx A� �
( )
n 1
nx (0,0, , , 1,0, )
+
= � �
�����
� �
( ) ( )( )n nx Ax , z 0� � =
( )n n 1z z 0 n+� � =
�Z
(2)
(2) 与 (1) 完全类似, 同理可得
z .=�
再证 ( )c A� { }1 .= � = 要证
( )R I A� � � �
( )R I A� � =�
7
先看 1.� =
2x ,
��
( ) ( )k k k 1I A x y y x x k .�� = � = � �Z
特别对
k 0
y ( ,0,0, ,0, 1 ,0, ,0,
=
�
=� � � � )
2�� , 但是
0 1x x 1,�� =
2
2
x
0 1 2 0
0 1x
2 1 1
x x x x 0
x x 0
x x x 0
�
� � �
= = = =
� =
= = =
�
�
�
�
�
�
矛盾.
由此可见, ( )y R I A .� � 即有
( ) 2R I A .� � � 再证
( ) 2R I A .� =� 设
( )n n 1 1 0, 1, n 1 n, , , , , , , , l� � + � �� = � � � � � � �� � � � �
8
取 N, s.t. 2 2n
n N 1
,
�
= +
� <��
令 ( )
( )
j
j
j N
y ,
0 j N 1
� ���
=
> +�!
N 1 0, 1, N jy ( ,0,0, , , , , ,0,0 ) y y .� �= � � � � � � � =� � � �
为了证明 ( )y R I A ,�� 即
证 2x ,
�� s.t.
( ) ( )k k k 1I A x y y x x k�� = � = � �Z
注意到 ( )
( )
j
j
j N
y ,
0 j N 1
� ���
=
> +�!
( )k k 1 kx x k N .�� =� �
9
( )k k 1 k N 1x x 0 k N 1 x x k N 1 .� +� = > + =
令
( )
( )
N 1
j
j k 1
k
k N
x ,
0 k N 1
+
= +
�� � ���
=
� " +!
显然
{ } 2kx x ,= �� 并满足
( )k k k 1y x x k .�= � �Z
从而 ( )I A x y,� = 即
( )y R I A .�� ( )c1 A .� ��
对于一般的 1,� = 可以化归
1,� = 情况.
10
( ) k k k 1
k 1 k k 1
k k k 1 k k k 1
y I A x
y x x y x x
�
� �
� �
= � � � =� � �
=� � � � =� � �
� �
当 1� = 时,
{ }
{ }
2 2
k
2 2
k
x
.
y
� = � �
�= � �
� �
� �
� �
� �
重复上面证明即可.
2.6.4 在 2l 空间上, 考察左推
移算子
( ) (1, 2 n 1 n 2 n 1 nA : x x , ,x ,x , x , ,x ,x , .� �� � � �
求证: ( ) { }p A | 1 ;� = � � <�C
11
( ) { } ( ) ( )c p cA | 1 ; A A A .� = � � = � =� ���C
证明(1) 1� > 时 ,
( )A .� #�
( ) 21, 2 n 1 nx x x , ,x ,x , l�= � � �
( )2 n 1 ny Ax x , ,x ,x , ,�= = � �
即
( ) ( ) ( )2 3 k 11 2 kAx x , Ax x , , Ax x ,+= = =� �
2 22 2
n n
n 2 n 1
Ax x x x A 1.
� �
= =
= � = �� �
又
0x (0,1,0, ,0,= � � )
2 ,��
0Ax (1,0, ,0,= � � )
0
0
Ax
xA 1," = A 1� = 当
12
1 A� > = 时 , ( )A .� #�
(2) ( ) { }p A | 1 .� = � � <�C
记 { }D | 1 .= � � <�C
对于 D,� � 数列 { }n 20 l
�
� �
.
( ) ( ) ( )2 2 2A 1, , , , , 1, , , ,� � = � � =� � �� � � �
( )p A ,�� �� 而 ( )21, , ,� � �
便是相应的特征向量.
反之, 设 Ax x,=� x ,� �
2x l�
. 则
( )n n nn 1 n 2x A x x ,x , 0+ +� = = � � � � �
13
1 D.� < � �
(3)
( ) { }c A | 1 .� $ � � =�C
记 { }
( ) 21, 2 n 1 n
C | 1 .
x x x , ,x ,x , l�
= � � =
= � �
�C
�
先看 1,� =
( ) k k k 1y I A x y x x += � � = �
1 1 2y x x= �
2 2 3y x x= �
3 3 4y x x= �
�
k 1 k 1 ky x x� �= �
k k k 1y x x += �
�
k k
j 1 k 1 k 1 1 j
j 1 j 1
y x x x x y ,+ +
= =
= � = �� �
利用这个公式,我们可以从 y 求
14
出 x, 即 ( ) 1I A .�
�
显然,非零分量个数有限的 y 在
( )R I A� 中. 事实上,设 y 的非
零分量个数为 K, 取
K
1 j
j 1
x y ,
=
=�
( )
( )
k
k 1 1 j 2j 1
k 1
x x y k 1,2, ,K
x l .
x 0 k K .
+
=
+
� = � =��
� =
>!
�
�
注意到非零分量个数有限的 y 在
2l 中稠密,故有 ( ) 2R I A l .� =
( ) 2R I A l .� � 例如
{} 21j j 1y l ,
�
=
= � 但是 ( )y R I A .� �
事实上,按 k 1jk 1 1
j 1
x x ,+
=
= � � 求得
15
的 { }k k 1x x ,
�
=
= 使得 kx .� ��
故 2x l .�
于是 ( )c1 A .��
对于适合 1� = 的一般 � ,可以
化归 1� = 情形.事实上,
( ) k k 1 kI A x y x x y+� � = � � � =
( )k k 1 kk k 1 k 1x x y k 1,2, .++ +� � �� � = = �
令 ( )k kk k 1
def def
x y
k k, k 1,2, ,+� �� = � = = �
则有
( )k k 1 k k 1,2, .+� � � =� = � 此即
化归 1� = 情形. 于是
起来,我们有
( ) { }p A | 1 ,� = � � <�C
( ) { }c A | 1 ,� = � � =�C
( )r A .� =%
16
2.6.5 在 ( )2L 0,� 上,考察微分算
子
( ) ( ) ( )1dxdtA : x t , D A H 0,= �
求证(1) ( ) { }p A Re 0 ;� = � � <�C
(2) ( ) { }c A Re 0 ;� = � � =�C
(3) ( )r A ?.� =
(1) ( )bit tddt e e ,�=�
a ib,� = +
当 Re� = a 0,< 时,
( )t at ibt 2e e e L 0, ,� = & ��
( ) ( )t t 2ddt e e L 0,� �=� ��
( )t 1e H 0, .� ��
( )p A .� �� 即得
( ) { }p A Re 0 .� = � � <�C
17
(2) dxdt bix y� =
( )bit bitddt xe ye� � =
( )bit bit bit0x ce e y e d� '( = + ' '
( )( )bit bit0x e c y e d� '(= + ' ' ,
当 ( )y ' � span { }ne�''
时, ( ) ( )2y L 0, ,' �� 且当
( )c n 1 !=� + 时 , 使得
( ) ( )2t0c y d L 0,(+ ' ' � �
( )x D A ,� 因为 { }ne�'' 构
成 ( )2L 0,� 的完全系, 所以
span { }ne�'' 在 ( )2L 0,� 中稠
密. 即 ( ) ( )2R A bi L 0, .� = � 由此
推出 ( )cb , bi A .
��R � 即得
( ) { }c A Re 0 .� = � � =�C
(3)当 Re� = a 0> 时,
18
( ) ( ) ( )( ) (
?
dx
0 dt x y d 0 x D A y 0.�( � � ' ' =
' =�
( )0x, y C 0, ,
�
��
( ) ( )dx0 dt x y d 0�( � � ' ' =
( ) ( )dx0 0dty d xy d 0� �( (' ' � � ' ' =
( ) ( ) ( )d0 0dt y x d xy d 0� �( ( ' ' ' + � ' ' =
( ) ( )( ) ( )d0 dt y y x d 0�( ' +� ' ' ' =
( ) ( )ddt y y 0 ' +� ' =
( )
Re 0
y Ce C 0.
�>
�' ' = =
( )y 0. Re 0 ' = � >
( ) ( )2R A I L 0, � � = �
( )r A ���
( )rRe 0 A� � ���
( )
( )
r
r
Re 0 A
Re 0 A
� > ��� )
*� � ��� +
( )r A ?. � =
19
计算细节:
( ) ( )t ibn t ib nx x0 0nL x t e e dt t e dt
� ��
( (= =
( ) ( )t ibn 1x0n 1L x t e dt
� �+
(+ =
( ) ( )t ibn 1 x ib nx0x e e n 1 t e dt� �+ � � (=� + +
( ) ( )n 1 x ib nx e e n 1 L x
+ � �=� + +
( ) ( ) ( )t ibn nL x n! e P x
� �= +
( ) ( ) ( ) ( )t ibn 1 n 1L x n 1 ! e P x
� �
+ + = + +
( ) ( )2nL x n! L 0, � ��