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2[1].6

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2[1].6 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�...
2[1].6
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, � ��
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