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BÀI TẬP CHƯƠNG I 1.5/ Giả sử E=C[0,1] và là một hàm cố định, A:E E là một ánh xạ xác định bởi (A(x))(t) = Chứng minh rằng A tuyến tính và liên tục. Xác định chuẩn của A. Giải + Chứng minh A tuyến tính: Ta cần chứng minh : = = = = Vậy A tuyến tính + Chứng minh A liên tục: A bị chặn Kết hợp với A tuyến tính ta có A liên tục + Tính Từ chứng minh bị chặn ta có (1) Chọn , (2) Từ(1), (2) suy ra :

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BI TP CHNG I

BI TP CHNG I

1.5/ Gi s E=C[0,1] v l mt hm c nh, A:E ( E l mt nh x xc nh bi (A(x))(t) = Chng minh rng A tuyn tnh v lin tc. Xc nh chun ca A.Gii

+ Chng minh A tuyn tnh:Ta cn chng minh

:

=

=

=

=

Vy

( A tuyn tnh+ Chng minh A lin tc:

(

( A b chn

Kt hp vi A tuyn tnh ta c A lin tc

+ Tnh

T chng minh b chn ta c

(1)

Chn ,

(

(

(2)

T(1), (2) suy ra :

1.6/ Gi s X= v nh x c xc nh bi

a/ Chng t X l khng gian con Banach ca C[0,1]

b/ CMR: l nh x tuyn tnh lin tc. Tm

c/ Hi c l n nh, ton nh khng?

Gii

a/(

( khi

Ta c:

(

(1)

khi

Ta c:

(

(2)

T (1) v (2) suy ra X l khng gian con(*)

Vi C[0,1] l khng gian Banach nn chng minh X l khng gian Banach ta chng minh X ng trong [0,1]

V nn

Vy ( X l khng gian Banach con

b/ + Chng minh tuyn tnh:

Ta cn chng minh

:

=

=

=

=

=

Vy

( tuyn tnh

+ Chng minh lin tc:

=

( ( b chnKt hp vi tuyn tnh ta c lin tc

+ Tnh

T chng minh b chn ta c (1)

Chn

(

(

Cho , ta c

(2)

T(1), (2) suy ra :

c/ + n nh khng?

( n nh

+ ton nh khng?

Gi s

Vy

( x khng lin tc (v l v )

( khng l ton nh1.7/ Gi s X, Y l hai kgc. CMR

a/

b/ S hi t trong X x Y tng ng vi s hi t theo ta , ngha l trong X x Y trong X v trong Y

c/ l dy Cauchy trong X x Y ln lt l dy Cauchy trong X v YGii

a/ +

+

=

+

=

=

b/

c/ Cauchy trong X x Y

1.10/ Cho l mt khng gian nh chun trn K. Tm iu kin ca cng thc , cng l chun trn EGii

*

Do tha mn yu cu

*

khng l chun khng tha mn

+ l chun trn ETht vy:

2.8/

Bit rng khvt vi hai php ton thng thng ca dy l khng gian Banach vi chun . Xt nh x xc nh bi cng thc:

a/ CMR nh x f l tuyn tnh lin tc

b/ Tm

c/ Chng minh rng f l nh x m

Gii

a/

+ Chng minh f tuyn tnh:

Ta cn chng minh

=

=

=

=

Vy

( f tuyn tnh

+ Chng minh A lin tc:

(

( f b chn

Kt hp vi f tuyn tnh ta c f lin tc

+ Tnh

Vi M =

Vy

(1)

Cho

(

(

(2)T(1), (2) suy ra:

c/ Chng minh rng f l nh x m

f l nh x tuyn tnh lin tc vy chng minh f l ax m th ta cn chng minh f l ton nh

( f l ton nh

Vy f l mt nh x m

Bi 3.8 Xt KG Banach vi chun v t

a/ CMR cng thc x.nh 1 tch v hng trn b/ CMR vi tch v hng trong cu a l 1 kg Hilbertc/ CMR vi l phm hm tuyn tnh, lin tc trn . Tnh theo cu a

a/ +Dng Hilbert:

+ Dng

+ Xc nh dng

Vy l mt tch v hng trn

b/ l khng gian tin Hilmite vi tch v hng

Ta chng minh y vi chun sinh bi tch v hng

Ta c:

Vy l kg HB

c/ Ta c vi

( f l mt phm hm tuyn tnh lin tc

V

Vy

3.4/ Gi s E1, E2 l hai khng gian tin Hilbert

a/ CMR E1x E2 l khng gian tin Hilbert vi tch v hng ,

b/ CMR nu E1, E2 l hai khng gian Hilbert th E1x E2 cng l khng gian Hilbert

Gii

a/

+Dng Hilbert:

+ Dng

+ Xc nh dng

b/ chun sinh bi tch v hng

Gi s l dy Cauchy

Ta chng minh

Hilbert

3.5/ Cho l mt gii trc chun trong kgHb E cn l dy s b chn. CMR

a/ Chui hi t vi mi

b/ nh x l tuyn tnh lin tc. Tnh chun

Gii

a/ Ta c

Vi

Vy

hi tb/

+ Chng minh f tuyn tnh:

Ta cn chng minh

=

Vy

( f tuyn tnh

+ Chng minh f lin tc:

(

( f b chn

Kt hp vi f tuyn tnh ta c f lin tc

+ Tnh

T chng minh f b chn ta c:

(1)

(2)T(1), (2) suy ra:

3.3 Xt K khng gian vect

Vi mi

t

a/ CMR hi t

b/ CMR cng thc

d/ CMR vi l mt dy ltt trong

e/ p dng cng thc tnh trc giao ha Gram-Schmidt xc nh mt c s trc chun ca khng gian con vi c xc nh trong cu trn

gii

a/ Ta c

Vy

hi t

b/

+Dng Hilbert:

+ Dng

+ Xc nh dng

Vy l mt tch v hng trn

d/ CMR ltt trong

Ta c

Vy ltt trong

e/

m

Vy l c s trc chun ca

Bi 3.6: Xt KG CMR

a/ Cng thc vi mi xc nh 1 tch v hng trn

b/ vi tch v hng trn l 1 KG Hilbert

c/ Cng thc x.nh mt phim hm t. tnh lin tc trn . Tm

gii

a/ +Dng Hilbert:

+ Dng

+ Xc nh dng

Vy l mt tch v hng trn

b/

Vy l kg HB vi tch v hng cho

c/ Ta c

l mt phm hm tuyn tnh lin tc

Tm

l mt phm hm tuyn tnh lin tc

Cu kim tra:

Trnh by s gii bi tp: Cho nh x , CMR l mt nh x tuyn tnh lin tc v tm . Minh ha chi tit vi nh x vi mi

Gii

- S gii bi tp

* Chng minh f tuyn tnh

Ta cn chng minh:

* chng minh f lin tc: Ta cn chng minh f b chn

f b chn kt hp vi f tuyn tnh suy ra f lin tc

* Tnh

T chng minh f b chn

(1)

Ta cn chng minh

(2)

T (1) v (2) suy ra:

- Minh ha chi tit

* Chng minh f tuyn tnh

tuyn tnh

* Chng minh f lin tc

b chn

Kt hp vi f tuyn tnh ta c f lin tc

* Tnh

T chng minh f b chn

(1)

Chn

(2)

T (1) v (2) suy ra:

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