On Cao Hoc Giai Tich

GII TCH (C S)

Phn 1. Khng gian metric1. Metric trn mt tp hp. S hi t.

Khng gian y

Phin bn chnh sa

PGS TS Nguyn Bch Huy(Typing by thuantd)

Ngy 10 thng 11 nm 2004

A. Tm tt l thuyt

1. Khng gian metric

nh ngha 1. Cho tp X 6= . Mt nh x d t X X vo R c gi l mt metric trnX nu cc iu kin sau c tha mn x, y, z X:

i. d(x, y) 0d(x, y) = 0 x = y

ii. d(x, y) = d(y, x)

iii. d(x, y) d(x, z) + d(z, y) (bt ng thc tam gic)Nu d l metric trn X th cp (X, d) gi l mt khng gian metric.

Nu d l metric trn X th n cng tha mn tnh cht sau

|d(x, y) d(u, v)| d(x, u) + d(y, v) (bt ng thc t gic)

V d 1. nh x d : Rm Rm R, nh bi

d(x, y) =

[mi=1

(xi yi)2]1/2

, x = (x1, . . . , xm), y = (y1, . . . , ym)

1

l mt metric trn Rm, gi l metric thng thng ca Rm.Khi m = 1, ta c d(x, y) = |x y|Trn Rm ta cng c cc metric khc nh

d1(x, y) =mi=1

|xi yi|

d2(x, y) = max1im

|xi yi|

V d 2. K hiu C[a,b] l tp hp cc hm thc x = x(t) lin tc trn [a, b]. nh x

d(x, y) = supatb

|x(t) y(t)|, x, y C[a,b]

l metric trn C[a,b], gi l metric hi t u.

2. S hi t

nh ngha 2. Cho khng gian metric (X, d). Ta ni dy phn t {xn} X hi t (hi ttheo metric d, nu cn lm r) v phn t x X nu lim

nd(xn, x) = 0.

Khi ta vit

limn

xn = x trong (X, d)

xnd x

xn xlimxn = x

Nh vy, limn

xn = x trong (X, d) c ngha

> 0,n0 : n N, n n0 d(xn, x) < Ta ch rng, cc metric khc nhau trn cng tp X s sinh ra cc s hi t khc nhau.

Tnh cht

1. Gii hn ca mt dy hi t l duy nht.

2. Nu dy {xn} hi t v x th mi dy con ca n cng hi t v x.3. Nu lim

nxn = x, lim

nyn = y th lim

nd(xn, yn) = d(x, y)

V d 3. Trong Rm ta xt metric thng thng. Xt phn t a = (a1, . . . , am) v dy {xn} vixn = (xn1 , . . . , x

nm). Ta c

d(xn, a) =

mi=1

(xni ai)2 |xni ai|, i = 1, . . . ,m

2

T y suy ra:

limn

xn = a trong (Rm, d) limn

xni = ai trong R, i = 1, . . . , n

V d 4. Trong C[a,b] ta xt "metric hi t u". Ta c

xnd x ( > 0,n0 : n n0 sup

atb|xn(t) x(t)| < )

dy hm {xn(t)} hi t u trn [a, b] v hm x(t)= lim

nxn(t) = x(t), t [a, b]

Nh vy, limn

xn(t) = x(t), t [a, b] l iu kin cn limxn = x trong C[a,b] vi metrichi t u.

Ch ny gip ta d on phn t gii hn.

3. Khng gian metric y

nh ngha 3. Cho khng gian metric (X, d). Dy {xn} X c gi l dy Cauchy (dy cbn) nu lim

n,md(xn, xm) = 0

hay > 0,n0 : n,m n0 d(xn, xm) <

Tnh cht

1. Nu {xn} hi t th n l dy Cauchy.2. Nu dy {xn} l dy Cauchy v c dy con hi t v x th {xn} cng hi t v x.

nh ngha 4. Khng gian metric (X, d) gi l y nu mi dy Cauchy trong n u ldy hi t.

V d 5. Khng gian Rm vi metric d thng thng l y .Tht vy, xt ty dy Cauchy {xn}, xn = (xn1 , . . . , xnm). V

{d(xn, xk) |xni xki | (i = 1, . . . ,m)lim

n,kd(xn, xk) = 0 limn,k |x

ni xki | = 0,

nn ta suy ra cc dy {xni }n (i = 1, . . . ,m) l dy Cauchy trong R, do chng hi t v Ry .

t ai = limn

xni (i = 1,m) v xt phn t a = (a1, . . . , am), ta c limn

xn = a trong

(Rm, d).

V d 6. Khng gian C[a,b] vi metric hi t u d l y .Gi s {xn} l dy Cauchy trong (C[a,b], d).

3

Vi mi t [a, b], ta c |xn(t) xm(t)| d(xn, xm). T gi thit limn,m

d(xn, xm) = 0 ta

cng c limn,m

|xn(t) xm(t)| = 0Vy vi mi t [a, b] th {xn(t)} l dy Cauchy trong R, do l dy hi t. Lp hm x xc nh bi x(t) = limxn(t), t [a, b].Ta cn chng minh x C[a,b] v lim d(xn, x) = 0.Cho > 0 ty . Do {xn} l dy Cauchy, ta tm c n0 tha

n,m n0 d(xn, xm) < Nh vy ta c

|xn(t) xm(t)| < , n n0,m n0, t [a, b]C nh n, t v cho m trong bt ng thc trn ta c

|xn(t) x(t)| , n n0, t [a, b]Nh vy, ta chng minh rng

> 0,n0 : n n0 supatb

|xn(t) x(t)|

T y suy ra:

Dy hm lin tc {xn(t)} hi t u trn [a, b] v hm x(t), do hm x(t) lin tc trn[a, b].

limn

d(xn, x) = 0.

y l iu ta cn chng minh.

B. Bi tp

Bi 1. Cho khng gian metric (X, d). Ta nh ngha

d1(x, y) =d(x, y)

1 + d(x, y), x, y X

1. Chng minh d1 l metric trn X.

2. Chng minh xnd1 x xn d x

3. Gi s (X, d) y , chng minh (X, d1) y .

Gii

1. Hin nhin d1 l mt nh x t X X vo R. Ta kim tra d1 tha mn cc iukin ca metric

4

(i) Ta c: d1(x, y) 0 do d(x, y) 0d1(x, y) = 0 d(x, y) = 0 x = y

(ii) d1(y, x) =d(y, x)

1 + d(y, x)=

d(x, y)

1 + d(x, y)= d(x, y)

(iii) Ta cn chng minh

d(x, y)

1 + d(x, y) d(x, z)

1 + d(x, z)+

d(z, y)

1 + d(z, y)

gn, ta t a = d(x, y), b = d(x, z), c = d(z, y).Ta c a b+ c; a, b, c 0 (do tnh cht ca d)

a1 + a

b+ c1 + b+ c

(do hm

t

1 + ttng trn [0,))

a1 + a

b1 + b+ c

+c

1 + b+ c

b1 + b

+c

1 + c(pcm)

2. Gi s xn d x. Ta clim d(xn, x) = 0

d1(xn, x) =d(xn, x)

1 + d(xn, x)

Do , lim d1(xn, x) = 0 hay xnd1 x

Gi s xn d1 x. Tlim d1(xn, x) = 0

d(xn, x) =d1(xn, x)

1 d1(xn, x)ta suy ra lim d(xn, x) = 0 hay xn

d x.3. Xt ty dy Cauchy {xn} trong (X, d1), ta cn chng minh {xn} hi t trong

(X, d1).

Ta clim

n,md1(xn, xm) = 0

d(xn, xm) =d1(xn, xm)

1 d1(xn, xm) lim

n,md(xn, xm) = 0 hay {xn} l dy Cauchy trong (X, d)

{xn} l hi t trong (X, d) (v (X, d) y ) t x = lim

nxn (trong (X, d)), ta c x = lim

nxn trong (X, d1) (do cu 2).

5

Bi 2. Cho cc khng gian metric (X1, d1), (X2, d2). Trn tp X = X1 X2 ta nh ngha

d((x1, x2), (y1, y2)) = d1(x1, y1) + d2(x2, y2)

1. Chng minh d l metric trn X

2. Gi s xn = (xn1 , xn2 ) (n N), a = (a1, a2). Chng minh

xnd a

{xn1

d1 a1xn2

d2 a2

3. Gi s (X1, d1), (X2, d2) y . Chng minh (X, d) y .

Bi 3. K hiu S l tp hp cc dy s thc x = {ak}k. Ta nh ngha

d(x, y) =k=1

1

2k.|ak bk|

1 + |ak bk| , x = {ak}, y = {bk}

1. Chng minh d l metric trn X

2. Gi s xn = {ank}k, n N, x = {ak}k. Chng minh

xnd x lim

nank = ak , k N

3. Chng minh (S, d) y .

Bi 4. Trn X = C[0,1] xt cc metric

d(x, y) = sup0x1

|x(t) y(t)|

d1(x, y) =

10

|x(t) y(t)| dt

1. Chng minh: (xnd x) (xn d1 x)

2. Bng v d dy xn(t) = n(tn tn+1), chng minh chiu "" trong cu 1) c th

khng ng.

3. Chng minh (X, d1) khng y .

6

GII TCH (C S)

Phn 1. Khng gian metricPhin bn chnh sa - c phn b sung ca bi trc

PGS TS Nguyn Bch Huy


Ni dung chnh ca mn C sChuyn ngnh: Ton Gii tch

Phng php Ging dy Ton

Phn 1: Khng gian metric

1. Metric trn mt tp hp. S hi t. Khng gian y .

2. Tp m. Tp ng. Phn trong, bao ng ca tp hp.

3. nh x lin tc gia cc khng gian metric. Cc tnh cht:

Lin h vi s hi t Lin h vi nh ngc ca tp m, tp ng. nh x m, nh x ng, nh x ng phi.

4. Tp compc. Cc tnh cht cn bn:

H c tm cc tp ng. Tnh cht compc v s hi t. nh ca tp compc qua nh x lin tc.

Phn 2: o v tch phn.

1. i s trn tp hp.

o v cc tnh cht cn bn.

2. Cc tnh cht ca o Lebesgue trn R (khng xt cch xy dng).

3. Hm s o c. Cc tnh cht cn bn.

1

Cc php ton s hc, ly max,min trn 2 hm o c. Ly gii hn hm o c (khng xt: hi t theo o, nh l Egoroff, Lusin).

4. Tch phn theo mt o. Cc tnh cht cn bn (khng xt tnh lin tc tuyt i).

5. Cc nh l Levi, Lebesgue v qua gii hn di du tch phn.

Phn 3: Gii tch hm.

1. Chun trn mt khng gian vect. Chun tng ng. Khng gian Banach.

2. nh x tuyn tnh lin tc. Khng gian cc nh x tuyn tnh lin tc (khng xt nhx lin hp, nh x compc, cc nguyn l c bn).

3. Khng gian Hilbert. Phn tch trc giao. Chui Fourier theo mt h trc chun. H trcchun y .

1 Metric trn mt tp hp. S hi t.Khng gian y

Phn ny c thm phn b sung ca bi trc

1. Tm tt l thuyt

1.1 Khng gian metric

nh ngha 1 Cho tp X 6= . Mt nh x d t X X vo R c gi l mt metric trnX nu cc iu kin sau c tha mn x, y, z X:

i. d(x, y) > 0d(x, y) = 0 x = y

ii. d(x, y) = d(y, x)

iii. d(x, y) 6 d(x, z) + d(z, y) (bt ng thc tam gic)

Nu d l metric trn X th cp (X, d) gi l mt khng gian metric.

Nu d l metric trn X th n cng tha mn tnh cht sau

|d(x, y) d(u, v)| 6 d(x, u) + d(y, v) (bt ng thc t gic)

V d. nh x d : Rm Rm R, nh bi

d(x, y) =

[mi=1

(xi yi)2]1/2

, x = (x1, x2, . . . , xm), y = (y1, y2, . . . , ym)

2

l mt metric trn Rm, gi l metric thng thng ca Rm.Khi m = 1, ta c d(x, y) = |x y|. Trn Rm ta cng c cc metric khc nh

d1(x, y) =mi=1

|xi yi|

d2(x, y) = max16i6m

|xi yi|

V d. K hiu C[a,b] l tp hp cc hm thc x = x(t) lin tc trn [a, b]. nh x

d(x, y) = supa6t6b

|x(t) y(t)|, x, y C[a,b]

l metric trn C[a,b], gi l metric hi t u.

1.2 S hi t

nh ngha 2 Cho khng gian metric (X, d). Ta ni dy phn t {xn} X hi t (hi ttheo metric d, nu cn lm r) v phn t x X nu lim

nd(xn, x) = 0.

Khi ta vit

limn

xn = x trong (X, d)

xnd x

xn xlimxn = x

Nh vy, limn

limxn = x trong (X, d) c ngha

> 0,n0 : n N, n > n0 d(xn, x) < Ta ch rng, cc metric khc nhau trn cng tp X s sinh ra cc s hi t khc nhau.

Tnh cht.

1. Gii hn ca mt dy hi t l duy nht.

2. Nu dy {xn} hi t v x th mi dy con ca n cng hi t v x.3. Nu lim

nxn = x, lim

nyn = y th lim

nd(xn, yn) = d(x, y)

V d. Trong Rm ta xt metric thng thng. Xt phn t a = (a1, . . . , am) v dy {xn} vixn = (xn1 , x

n2 , . . . , x

nm). Ta c

d(xn, a) =

mi=1

(xni ai)2 > |xni ai|, i = 1, 2, . . . ,m

T y suy ra:

limn

xn = a trong (Rm, d) limn

xni = ai trong R, i = 1, 2, . . . , n

3

V d. Trong C[a,b] ta xt metric hi t u. Ta c

xnd x ( > 0,n0 : n > n0 sup

a6t6b|xn(t) x(t)| < )

dy hm {xn(t)} hi t u trn [a, b] v hm x(t)= lim

nxn(t) = x(t), t [a, b]

Nh vy, limn

xn(t) = x(t), t [a, b] l iu kin cn limxn = x trong C[a,b] vi metric hit u. Ch ny gip ta d on phn t gii hn.

1.3 Khng gian metric y

nh ngha 3 Cho khng gian metric (X, d). Dy {xn} X c gi l dy Cauchy (dy cbn) nu

limn,m

d(xn, xm) = 0

hay > 0,n0 : n,m > n0 d(xn, xm) <

Tnh cht.

1. Nu {xn} hi t th n l dy Cauchy.2. Nu dy {xn} l dy Cauchy v c dy con hi t v x th {xn} cng hi t v x.

nh ngha 4 Khng gian metric (X, d) gi l y nu mi dy Cauchy trong n u ldy hi t.

V d. Khng gian Rm vi metric d thng thng l y .Tht vy, xt ty dy Cauchy {xn}, xn = (xn1 , . . . , xnm). V

{d(xn, xk) > |xni xki | (i = 1, . . . ,m)lim

n,kd(xn, xk) = 0 limn,k |x

ni xki | = 0,

nn ta suy ra cc dy {xni }n (i = 1, . . . ,m) l dy Cauchy trong R, do chng hi t v Ry .

t ai = limn

xni (i = 1, 2, . . . ,m) v xt phn t a = (a1, . . . , am), ta c limn

xn = a

trong (Rm, d).

V d. Khng gian C[a,b] vi metric hi t u d l y .Gi s {xn} l dy Cauchy trong (C[a,b], d).Vi mi t [a, b], ta c |xn(t) xm(t)| 6 d(xn, xm). T gi thit lim

n,md(xn, xm) = 0 ta

cng c limn,m

|xn(t) xm(t)| = 0.Vy vi mi t [a, b] th {xn(t)} l dy Cauchy trong R, do l dy hi t.

4

Lp hm x xc nh bi x(t) = limxn(t), t [a, b]. Ta cn chng minh x C[a,b] vlim d(xn, x) = 0.

Cho > 0 ty . Do {xn} l dy Cauchy, ta tm c n0 tha

n,m > n0 d(xn, xm) <

Nh vy ta c|xn(t) xm(t)| < , n > n0,m > n0, t [a, b]

C nh n, t v cho m trong bt ng thc trn ta c

|xn(t) x(t)| , n > n0, t [a, b]

Nh vy, ta chng minh rng

> 0,n0 : n > n0 supa6t6b

|xn(t) x(t)| 6

T y suy ra:

Dy hm lin tc {xn(t)} hi t u trn [a, b] v hm x(t), do hm x(t) lin tc trn[a, b].

limn

d(xn, x) = 0.

y l iu ta cn chng minh.

2. Bi tp

Bi 1 Cho khng gian metric (X, d). Ta nh ngha

d1(x, y) =d(x, y)

1 + d(x, y), x, y X

1. Chng minh d1 l metric trn X.

2. Chng minh xnd1 x xn d x

3. Gi s (X, d) y , chng minh (X, d1) y .

Gii.

1. Hin nhin d1 l mt nh x t X X vo R. Ta kim tra d1 tha mn cc iu kin cametric

(i) Ta c: d1(x, y) > 0 do d(x, y) > 0d1(x, y) = 0 d(x, y) = 0 x = y

5

(ii) d1(y, x) =d(y, x)

1 + d(y, x)=

d(x, y)

1 + d(x, y)= d(x, y)

(iii) Ta cn chng minh

d(x, y)

1 + d(x, y)6 d(x, z)

1 + d(x, z)+

d(z, y)

1 + d(z, y)

gn, ta t a = d(x, y), b = d(x, z), c = d(z, y).

Ta c a 6 b+ c; a, b, c > 0 (do tnh cht ca metric d)

a1 + a

6 b+ c1 + b+ c

(do hm

t

1 + ttng trn [0,))

a1 + a

6 b1 + b+ c

+c

1 + b+ c6 b

1 + b+

c

1 + c(pcm)

2. Gi s xn d x. Ta clim d(xn, x) = 0

d1(xn, x) =d(xn, x)

1 + d(xn, x)

Do , lim d1(xn, x) = 0 hay xnd1 x

Gi s xn d1 x. Tlim d1(xn, x) = 0

d(xn, x) =d1(xn, x)

1 d1(xn, x)ta suy ra lim d(xn, x) = 0 hay xn

d x.3. Xt ty dy Cauchy {xn} trong (X, d1), ta cn chng minh {xn} hi t trong (X, d1). Ta c

limn,m

d1(xn, xm) = 0

d(xn, xm) =d1(xn, xm)

1 d1(xn, xm) lim

n,md(xn, xm) = 0 hay {xn} l dy Cauchy trong (X, d)

{xn} l hi t trong (X, d) (v (X, d) y ) t x = lim

nxn (trong (X, d)), ta c x = lim

nxn trong (X, d1) (do cu 2).

Bi 2 Cho cc khng gian metric (X1, d1), (X2, d2). Trn tp X = X1 X2 ta nh ngha

d ((x1, x2), (y1, y2)) = d1(x1, y1) + d2(x2, y2)

6

1. Chng minh d l metric trn X.

2. Gi s xn = (xn1 , xn2 ), (n N), a = (a1, a2). Chng minh xn d a

{xn1

d1 a1xn2

d2 a23. Gi s (X1, d1), (X2, d2) y . Chng minh (X, d) y .

Gii.

1. Ta kim tra tnh cht i), iii) ca metric. Gi s x = (x1, x2), y = (y1, y2), z = (z1, z2), tac:

i) d(x, y) = d1(x1, y1) + d2(x2, y2) > 0

d(x, y) = 0{

d1(x1, y1) = 0d2(x2, y2) = 0

{

x1 = y1x2 = y2

x = y

iii) Cng tng v cc bt ng thc:

d1(x1, y1) 6 d1(x1, z1) + d1(z1, y1)d2(x2, y2) 6 d2(x2, z2) + d2(z2, y2)

ta cd(x, y) 6 d(x, z) + d(z, y)

2. Ta cd1(x

n1 , a1), d2(x

n2 , a2) 6 d(xn, a) = d1(xn1 , a1) + d2(xn2 , a2)

Do :

lim d(xn, a) = 0{

lim d1(xn1 , a1) = 0

lim d2(xn2 , a2) = 0

3. Gi s {xn} l dy Cauchy trong (X, d), xn = (xn1 , xn2 ). Ta c {xni } l dy Cauchy trong(Xi, di) (v di(x

ni , x

mi ) 6 d(xn, xm)). Suy ra

ai Xi : xni di ai (do (Xi, di) y ) xn d a := (a1, a2) (theo cu 2))

Bi 3 K hiu S l tp hp cc dy s thc x = {ak}k. Ta nh ngha

d(x, y) =k=1

1

2k.|ak bk|

1 + |ak bk| , x = {ak}, y = {bk}

1. Chng minh d l metric trn X.

2. Gi s xn = {ank}k, n N, x = {ak}k. Chng minh

xnd x lim

nank = ak , k N

7

3. Chng minh (S, d) y .

Gii.

1. u tin ta nhn xt rng chui s nh ngha s d(x, y) l hi t v s hng th k nhhn 1/2k.

Vi x = {ak}, y = {bk}, z = {ck}, cc tnh cht i), iii) kim tra nh sau:i) Hin nhin d(x, y) > 0,

d(x, y) = 0 ak = bk k N x = y

iii) T l lun bi 1 ta c

|ak bk|1 + |ak bk| 6

|ak ck|1 + |ak ck| +

|ck bk|1 + |ck bk| k N

Nhn cc bt ng thc trn vi 1/2k ri ly tng, ta c

d(x, y) 6 d(x, z) + d(z, y)

2. Ta c

d (xn, x) =k=1

1

2k.|ank ak|

1 + |ank ak|n N

Gi s xn x. Ta c: k N

1

2k.|ank ak|

1 + |ank ak|6 d(xn, x) ()

|ank ak| 62kd (xn, x)

1 2kd (xn, x)(

khi n ln d (xn, x) 0 ty . Ta chn s k0 sao cho

k=k0+1

12k

< 2. Xt dy s:

sn =

k0k=1

1

2k.|ank ak|

1 + |ank ak|, n N

Do lim sn = 0 nn c n0 sao cho sn n0.

Vi n > n0, ta c

d(xn, x) = sn +

k=k0+1

(. . . ) 6 sn +

k=k0+1

1

2k<

8

Nh vy ta chng minh

> 0 n0 : n > n0 d(xn, x) <

hay lim d(xn, x) = 0.

3. Xt ty dy Cauchy {xn} trong (S, d), xn = {ank}k. L lun tng t () ta c

|ank amk | 62kd(xn, xm)

1 2kd(xn, xm) 0 khi m,n

Suy ra {ank}n l dy Cauchy trong R, do hi t.t ak = lim

nank v lp phn t a := {ak}. p dng cu 2) ta c xn a trong (S, d).

Bi 4 Trn X = C[0,1] xt cc metric

d(x, y) = sup06x61

|x(t) y(t)|

d1(x, y) =

10

|x(t) y(t)| dt

1. Chng minh: (xnd x) (xn d1 x)

2. Bng v d dy xn(t) = n(tn tn+1), chng minh chiu trong cu 1) c th khng

ng.

3. Chng minh (X, d1) khng y .

Gii.

1. Ta c

|x(t) y(t)| 6 d(x, y) t [0, 1]

10

|x(t) y(t)| dt 6 d(x, y) 10

dt = d(x, y)

d1(x, y) 6 d(x, y) x, y C[0,1]Do , nu lim d(xn, x) = 0 th cng c lim d1(xn, x) = 0.

2. K hiu x0 l hm hng bng 0 trn [0, 1]. Ta c:

d1(xn, x0) = 10|xn(t) x0(t)| dt =

10n (tn tn+1) dt = n

(n+1)(n+2) 0 khi n.

9

d(xn, x0) = sup06t61

n(tn tn+1) = n ( nn+1

)n. 1n+1

(hy lp bng kho st hm n(tn tn+1)trn [0, 1]). Do

limn

d(xn, x0) = limn

(n

n+ 1

)n.

n

n+ 1=

1

e6= 0

Suy ra xnd

/ x0.3. Xt dy {xn} C[0,1] xc nh nh sau:

xn(t) =

0 t [0, 1

2]

n(t 12) t [1

2, 12+ 1

n]

1 t [12+ 1

n, 1]

(n > 2)

Trc tin ta chng minh {xn} l dy Cauchy trong(C[0,1], d1

).

Tht vy, vi m < n, ta c:

d1(xn, xm) = 10|xn(t) xm(t)| dt

= 1/2+1/m1/2

|xn(t) xm(t)| dt6 1/2+1/m1/2

1.dt = 1m

Do limm,n

d1(xn, xm) = 0

Ta chng minh {xn} khng hi t trong(C[0,1], d1

).

Gi s tri li:x C[0,1] : lim d1(xn, x) = 0

Khi

d1(xn, x) > 1/20

|xn(t) x(t)| dt = 1/20

|x(t)| dt , n N

1/20

|x(t)| dt = 0 x(t) 0 trn [0, 12].

Mt khc, vi mi a (12, 1)ta c 1

2+ 1

n< a khi n ln.

Do

d1(xn, x) > 1a

|xn(t) x(t)| dt = 1a

|1 x(t)| dt

x(t) = 1 t [a, 1] (l lun nh trn)Do a > 1

2ty , ta suy ra x(t) = 1 t (1

2, 1].

Ta gp mu thun vi tnh lin tc ca hm x.

10

2 Tp m, tp ng. Phn trong, baong ca mt tp hp

1. Tp m. Phn trong

Cho khng gian metric (X, d).Vi x0 X, r > 0, ta k hiu B(x0, r) = {x X : d(x, x0) < r}gi l qu cu m tm x0, bn knh r.

nh ngha 1 Cho tp hp A X.1. im x c gi l im trong ca tp hp A nu r > 0 : B(x, r) A

2. Tp hp tt c cc im trong ca A gi l phn trong ca A, k hiu IntA hayA. Hin

nhin ta c IntA A.3. Tp A gi l tp m nu mi im ca n l im trong. Ta qui c l m. Nh vy,

A m A = IntA (x A r > 0 : B(x, r) A)Tnh cht.

1. H cc tp m c ba tnh cht c trng sau:

i) , X l cc tp m.ii) Hp ca mt s ty cc tp m l tp m.

iii) Giao ca hu hn cc tp m l tp m.

2. Phn trong ca A l tp m v l tp m ln nht cha trong A.

Nh vy:(B A,B m) B IntA

V d. Qu cu m B(x0, r0) l tp m.Tht vy, x B(x0, r0) ta c r = r0 d(x, x0) > 0. Ta s ch ra B(x, r) B(x0, r0). Vi

y B(x, r), ta c d(y, x0) 6 d(y, x) + d(x, x0) < r + d(x, x0) = r0 nn y B(x0, r0).V d. Trong R vi metric thng thng, cc khong m l tp m.

Tht vy, trong R ta c B(x, r) = (x r, x+ r). Mi khong hu hn (a, b) l qu cu tm a+b

2, bn knh ba

2nn l tp m.

(a,+), (a R) l tp m v x (a,+) ta t r = xa th (x r, x+ r) (a,+).V d. Trong R2 vi metric thng thng mi hnh ch nht m A = (a, b) (c, d) l tp m.

Tht vy, xt ty x = (x1, x2) A. Ta t r = min{x1 a, b x1, x2 c, d x2} th cB(x, r) A.nh l 1

1. Mi tp m trong R l hp ca khng qu m c cc khong m i mt khng giaonhau.

2. Mi tp m trong R2 l hp ca khng qu m c cc hnh ch nht m.

11

2. Tp ng. Bao ng ca mt tp hp

nh ngha 2

1. Tp hp A X gi l tp ng nu X \ A l tp m.2. im x c gi l mt im dnh ca tp A nu A B(x, r) 6= ,r > 0.3. Tp tt c cc im dnh ca A gi l bao ng ca A, k hiu l A hay ClA.

Hin nhin ta lun c A A.

Tnh cht.

1. , X l cc tp ng.Giao ca mt s ty cc tp ng l tp ng.

Hp ca hu hn tp ng l tp ng.

2. A l tp ng v l tp ng nh nht cha A.

Nh vy (B A,B ng) B A3. A ng A = A.

nh l 2

1. x A ({xn} A : limxn = x)2. Cc tnh cht sau l tng ng:

a) A l tp ng;

b) {xn} A (limxn = x x A).

V d. Qu cu ng B(x0, r) := {x X : d(x, x0) 6 r} l tp ng.

Chng minh. Do s tng ng ca tnh cht a), b) nn ta chng minh B(x0, r) c tnhcht b). Xt ty dy {xn} m {xn} B(x0, r), xn x, ta phi chng minh x B(x0, r).Tht vy: {

d(xn, x0) 6 r n = 1, 2, . . .lim d(xn, x0) = d(x, x0) (do tnh cht 3) ca s hi t)

d(x, x0) 6 r (pcm)

12

Bi tp

Bi 1 Chng minh rng trong mt khng gian metric ta c

1. A B A B;2. A B = A B;

3. A = A

Gii.

1. Ta c: (B l tp ng, B A) B A.2. Ta c: A A B,B A B (do cu 1)) nn A B A B

Mt khc: {A B l tp ng (do A,B ng)

A B A B A B A B (do tnh cht nh nht ca bao ng)

3. Ta c A l tp ng nn n bng bao ng ca n.

Bi 2 Trong C[a,b] ta xt metric hi t u. Gi s x0 C[a,b]. Ta xt cc tp sau:M1 ={x C[a,b] : x(t) > x0(t)t [a, b]}M2 ={x C[a,b] : x(t) > x0(t)t [a, b]}M3 ={x C[a,b] : t [a, b] : x(t) > x0(t)}

Chng minh M1 m, M2 v M3 ng.

Gii.

Chng minh M1 m. Xt ty x M1, ta cx(t) x0(t) > 0 t [a, b]

r := infa6t6b

[x(t) x0(t)] > 0 (v t0 [a, b] : r = x(t0) x0(t0) > 0)

Ta s chng minh B(x, r) M1. Tht vy, vi y B(x, r) ta c:supa6t6b

|y(t) x(t)| < r

|y(t) x(t)| < r t [a, b]y(t) > x(t) r t [a, b]y(t) x0(t) > x(t) x0(t) r > r r = 0 t [a, b]y M1

13

Chng minh M2 ng.Gi s {xn} M2, xn d x, ta cn chng minh x M2. Ta c limnxn(t) = x(t) t [a, b]

(do xn

d x)

xn(t) > x0(t) t [a, b], n N (do xn M2)

Suy ra x(t) > x0(t)t [a, b] , do x M2. Chng minh M3 ng.

Cch 1. t M4 ={x C[a,b] : x(t) < x0(t) t [a, b]

}. Ta c M3 = C[a,b] \M4 v M4 l

tp m (chng minh tng t M1 m) nn M3 ng.

Cch 2. Gi s {xn} M3, xn d x ta cn chng minh x M3.Do xn M3 nn tn ti tn [a, b] tha xn(tn) > x0(tn). Dy {tn} b chn nn c dy con{tnk}k hi t v mt t0 [a, b]. Ta s chng minh x(t0) > x0(t0). u tin ta chng minh

limk

xnk(tnk) = x(t0) (1)

Tht vy:

|xnk(tnk)x(t0)| 6 |xnk(tnk)x(tnk)|+ |x(tnk)x(t0)| 6 d(xnk , x)+ |x(tnk)x(t0)| (2)

v v v phi ca (2) hi t v 0 khi k nn (1) ng.T xnk(tnk) > x0(tnk) v (1) ta c x(t0) > x0(t0). Ta chng minh t0 [a, b] : x(t0) >x0(t0) hay x M3.

Bi 3 Trong C[a,b] vi metric hi t u ta xt cc tp hp sau:

M1 ={x C[a,b] : x l n nh, 0 6 x(t) 6 1 t [a, b]

}M2 =

{x C[a,b] : x l ton nh, 0 6 x(t) 6 1 t [a, b]

}Chng minh M1 khng l tp ng, M2 l tp ng.

14

GII TCH (C S)

Chuyn ngnh: Gii Tch, PPDH Ton

Phn 1. Khng gian metric

3. nh x lin tc(Phin bn chnh sa)



Tm tt l thuyt

1 nh ngha

Cho cc khng gian metric (X, d), (Y, ) v nh x f : X Y

Ta ni nh x f lin tc ti im x0 X nu > 0, > 0 : x X, d(x, x0) < = (f(x), f(x0)) <

Ta ni f lin tc trn X nu f lin tc ti mi x X

2 Cc tnh cht

Cho cc khng gian metric (X, d), (Y, ) v nh x f : X Y .

nh l 1. Cc mnh sau tng ng

1. f lin tc ti x0 X

2. {xn} X (limxn = x0) = lim f(xn) = f(x0)

1

H qu. Nu nh x f : X Y lin tc ti x0 v nh x g : Y Z lin tc ti y0 = f(x0)th nh x hp g f : X Z lin tc ti x0.

nh l 2. Cc mnh sau tng ng

1. f lin tc trn X

2. Vi mi tp m G Y th tp nghch nh f1(G) l tp m trong X.

3. Vi mi tp ng F Y th tp f1(F ) l tp m trong X.

3 nh x m, nh x ng, nh x ng phi

Cho cc khng gian metric X, Y v nh x f : X Y .

nh x f gi l nh x m (ng) nu vi mi tp m (ng) A X th nh f(A) ltp m (ng).

nh x f gi l nh x ng phi nu f l song nh lin tc v nh x ngc f1 : Y Xlin tc.

4 Mt s cc h thc v nh v nh ngc

Cho cc tp X, Y khc trng v nh x f : X Y . Vi cc tp A,Ai X v B,Bi Y , tac

1. f(iI

Ai) =iI

f(Ai), f(iI

Ai) iI

f(Ai)

2. f1(iI

Bi) =iI

f1(Bi), f1(iI

Bi) =iI

f1(Bi)

f1(B1 \B2) = f1(B1) \ f1(B2)

3. f(f1(B)) B ("=" nu f l ton nh)f1(f(A)) A ("=" nu f l n nh)

Bi tp

Bi 1. Trong khng gian C[a,b], ta xt metric d(x, y) = supatb

|x(t) y(t)| v trong R ta xtmetric thng thng. Chng minh cc nh x sau y lin tc t C[a,b] vo R.

2

1. f1(x) = infatb

x(t)

2. f2(x) =ba

x2(t)dt

Gii. 1. Ta s chng minh |f1(x) f1(y)| d(x, y) (*)Tht vy

f1(x) x(t) = y(t) + (x(t) y(t)) y(t) + d(x, y) t [a, b]= f1(x) d(x, y) y(t), t [a, b]= f1(x) d(x, y) f1(y) hay f1(x) f1(y) d(x, y)Tng t, ta c f1(y) f1(x) d(x, y) nn (*) ng. T y, ta thy{xn}, lim

nxn = x = lim

nf1(xn) = f1(x)

2. Xt ty x C[a,b], {xn} C[a,b] m limxn = x, ta cn chng minh lim f2(xn) = f2(x)Ta c

|x2n(t) x2(t)| = |xn(t) x(t)|.|xn(t) x(t) + 2x(t)| d(xn, x).[d(xn, x) +M ] (M = sup

atb2|x(t)|)

= |f2(xn) f2(x)| b

a

|x2n(t) x2(t)|dt

d(xn, x)[d(xn, x) +M ](b a)

Do lim d(xn, x) = 0 nn t y ta c lim f2(xn) = f2(x) (pcm)

Ghi ch. Ta c th dng cc kt qu v nh x lin tc gii bi tp 3 (2). V d, chngminh tp

M = {x C[a,b] : x(t) > x0(t), t [a, b]} (x0 C[a,b] cho trc )

l tp m, ta c th lm nh sau. Xt nh x

f : C[a,b] R, f(x) = infatb

(x(t) x0(t))

Ta c:

f lin tc (l lun nh khi chng minh f1 lin tc)

3

M = {x C[a,b] : f(x) > 0} = f1((0,+)), (0,) l tp m trong R

Bi 2. Cho cc khng gian metric X, Y v nh x f : X Y . Cc mnh sau l tngng

1. f lin tc trn X

2. f1(B) f1(B) B Y

3. f(A) f(A) A X

Gii. 1) 2) Ta c{f1(B) l tp ng (do f lin tc v B Y l tp ng)f1(B) f1(B)

= f1(B) f1(B) (do tnh cht "nh nht" ca bao ng)

2) 3) t B = f(A) trong 2), ta c f1(f(A) ) f1(f(A)) ADo f(f1(f(A) )) f(A) = f(A) f(A)

3) 1) Xt ty tp ng F Y , ta cn chng minh f1(F ) l tp ng.t A = f1(F ), ta c

f(A) f(A) = f(f1(F )) F = F (do F ng)= f1(f(A)) f1(F )= A AVy A = A nn A l tp ng.

Bi 3. Trong C[a,b] ta xt metric d(x, y) = sup{|x(t) y(t)|, a t b}. Cho : [a, b]R Rl hm lin tc. Chng minh nh x sau y lin tc

F : C[a,b] C[a,b], F (x)(t) = (t, x(t))

Gii. C nh x0 C[a,b], ta s chng minh F lin tc ti x0.t M = 1 + sup

atb|x0(t)|. Cho > 0 ty .

Hm lin tc trn tp compact D := [a, b] [M,M ] nn lin tc u trn D. Do ,tn ti s 1 > 0 sao cho

(t, s), (t, s) D, |t t| < 1, |s s| < 1 = |(t, s) (t, s)| <

4

t = min(1, 1). Vi mi x C[a,b], d(x, x0) < , ta c|x(t) x0(t)| < t [a, b]x(t) [M,M ] (do |x(t) x0(t)| < 1, t [a, b])

Do , |(t, x(t)) (t, x0(t))| < , t [a, b]= |F (x)(t) F (x0)(t)| < , t [a, b]= d(F (x), F (x0)) <

Nh vy, ta chng minh

> 0, > 0 : x C[a,b], d(x, x0) < d(F (x), F (x0)) < hay F lin tc ti x0.

Bi 4. Cho cc khng gian metric X, Y v song nh f : X Y . Chng minh cc mnh sau tng ng

1. f1 : Y X lin tc

2. f l nh x ng

Gii. Ta c (f1 : Y X lin tc) (A X,A ng (f1)1(A) ng trong Y ) (A X,A ng f(A) ng) (f : X Y l nh x ng)

Bi 5. Cho khng gian metric (X, d). Vi x X, 6= A X, ta nh ngha

d(x,A) = infyA

d(x, y)

Chng minh cc khng nh sau y

1. nh x f : X R, f(x) = d(x,A) lin tc

2. x A d(x,A) = 0

3. Nu F1, F2 l cc tp ng, khc v F1F2 = th tn ti cc tp m G1, G2 sao cho

F1 G1, F2 G2, G1 G2 =

Gii. 1. Ta s chng minh |f(x) f(x)| d(x, x) (*)Tht vy, ta c d(x, y) d(x, x) + d(x, y) y A

= infyA

d(x, y) d(x, x) + infyA

d(x, y)

= d(x,A) d(x, A) d(x, x)

5

2. Ta c

d(x,A) = 0 ({xn} A : limn

d(x, xn) = 0) (do tnh cht ca inf v d(x,A) 0) ({xn} A : limxn = x) x A

3. Ta xt nh x g : X R, g(x) = d(x, F1) d(x, F2)Ta c g lin tc theo cu 1)

t G1 = {x X : g(x) < 0}, G2 = {x X : g(x) > 0}, ta c

G1 G2 = G1, G2 l cc tp m (do G1 = g1((, 0)), G2 = g1((0,+)), (0,+),(, 0)l cc tp m v g lin tc).

F1 G1 v x F1 {

d(x, F1) = 0

d(x, F2) > 0 (do x / F2 v kt qu cu 2)) g(x) < 0

Tng t, F2 G2

Bi tp t gii c hng dn

Bi 6. Cho cc khng gian metric X, (Y1, d1), (Y2, d2). Trn Y1 Y2, ta xt metricd((y1, y2), (y

1, y

2)) = d1(y1, y

1) + d2(y2, y

2)

Gi s rng f1 : X Y1, f2 : X Y2 l cc nh x lin tc. Chng minh rng nh xf : X Y1 Y2, f(x) = (f1(x), f2(x)) lin tc.

Hng dn

S dng nh l 1 v iu kin hi t trong khng gian metric tch trong bi tp 1.

Bi 7. Cho cc khng gian metric X, Y v nh x f : X Y . Chng minh cc mnh sautng ng:

1. f lin tc trn X

2. f1(IntB) Int f1(B) B Y

6

Hng dn

1) 2) p dng nh l 2 v tnh cht "ln nht" ca phn trong.

2) 1) p dng nh l 2 v tnh cht G = IntG nu G m.

Bi 8. Cho cc khng gian metric (X, d), (Y, ) v cc nh x lin tc f, g : X Y . Ta nhngha nh x

h : X R, h(x) = (f(x), g(x)), x X

1. Chng minh h lin tc

2. Suy ra rng tp A := {x X : f(x) = g(x)} l tp ng.

Hng dn

1. Chng minh rng nu dnd x th h(xn) h(x) trong R, s dng tnh cht yn y,

zn z th (yn, zn) (y, z)

2. A = h1({0}), {0} l tp ng trong R

Bi 9. Cho khng gian metric (X, d) v A, B l cc tp ng khc , khng giao nhau. Chngminh rng tn ti nh x lin tc f : X R sao cho

0 f(x) 1, x X,f(x) = 0, x A,f(x) = 1, x B

Hng dn

Chng minh hm f(x) =d(x,A)

d(x,A) + d(x,B)cn tm.

7

GII TCH (C S)


Phn 1. Khng gian metric

4. Tp compact, khng gian compact(Phin bn chnh sa)



Tm tt l thuyt

1 nh ngha

Cho cc khng gian metric (X, d)

1. Mt h {Gi : i I} cc tp con ca X c gi l mt ph ca tp A X nu A iI

Gi

Nu I l tp hu hn th ta ni ph l hu hn.

Nu mi Gi l tp m th ta ni ph l ph m.

2. Tp A X c gi l tp compact nu t mi ph m ca A ta lun c th ly ra cmt ph hu hn.

3. Tp A c gi l compact tng i nu A l tp compact.

1

2 Cc tnh cht

2.1 Lin h vi tp ng

Nu A l tp compact trong khng gian metric th A l tp ng.

Nu A l tp compact, B A v B ng th B l tp compact.

2.2 H c tm cc tp ng

H {Fi : i I} cc tp con ca X c gi l h c tm nu vi mi tp con hu hn J IthiJ

Fi 6= .

nh l 1. Cc mnh sau l tng ng:

1. X l khng gian compact.

2. Mi h c tm cc tp con ng ca X u c giao khc .

nh l 2. Gi s f : X Y l nh x lin tc v A X l tp compact. Khi , f(A) ltp compact.

H qu. Nu f : X R l mt hm lin tc v A X l tp compact th f b chn trn Av t gi tr ln nht, nh nht trn A, ngha l:

x1, x2 A : f(x1) = inf f(A), f(x2) = sup f(A)

nh l 3 (Weierstrass). Trong khng gian metric X, cc mnh sau l tng ng:

1. Tp A X l compact.

2. T mi dy {xn} A c th ly ra mt dy con hi t v phn t thuc A.

2.3 Tiu chun compact trong Rn

Trong khng gian Rn (vi metric thng thng), mt tp A l compact khi v ch khi n ngv b chn.

2.4 Tiu chun compact trong C[a,b]

nh ngha. Cho tp A C[a,b].

2

1. Tp A c gi l b chn tng im trn [a, b] nu vi mi t [a, b] tn ti s Mt > 0sao cho |x(t)| Mt, x A.Tp A c gi l b chn u trn [a, b] nu tn ti s M > 0 sao cho

|x(t)| M , t [a, b], x A.

2. Tp A gi l ng lin tc tc trn [a, b] nu vi mi > 0, tn ti s > 0 sao cho vi

mi t, s [a, b] m |t s| < v vi mi x A th ta c |x(t) x(s)| < .

V d. Gi s A C[a,b] l tp cc hm x = x(t) c o hm trn (a, b) v |x(t)| 2,t (a, b).

Tp A l lin tc ng bc. Tht vy, do nh l Lagrange ta c|x(t) x(s)| = |x(c)(t s)| 2.|t s|

Do , cho trc > 0, ta chn =

2th c:

x A, t, s [a, b], |t s| < |x(t) x(s)| <

Nu thm gi thit A b chn ti im t0 [a, b] th A b chn u trn [a, b]. Tht vy|x(t)| |x(t) x(t0)|+ |x(t0)| = |x(c).(t t0)|+ |x(t0)|

2(b a) +Mt0 t [a, b],x A

nh l 4 (Ascoli - Arzela). Tp A C[a,b] (vi metric hi t u) l compact tng i khiv ch khi A b chn tng im v ng lin tc trn [a, b].

Bi tp

Bi 1. 1. Cho X l khng gian metric compact, {Fn} l h cc tp ng, khc rng, thamn Fn Fn+1 (n = 1, 2, . . . ). Chng minh

n=1

Fn 6=

2. Gi s {Fn} l h c tm cc tp ng, b chn trn R. Chng minhn=1

Fn 6=

Gii. 1. Ta chng minh {Fn} l h c tm. Nu J N l tp hu hn, ta t n0 = max Jth s c

nJ

Fn = Fn0 6=

Ghi ch. Dng khc ca cu 1) l: Cho F1 l tp compact, Fn (n 2) l cc tp ngkhc v F1 F2 . Khi

n=1

Fn 6=

3

2. Ta xy dng dy tp hp {Kn} nh sau:

K1 = F1, Kn =n

k=1

Fk (n 2)

Th th ta c

Kn compact, Kn 6= (do h {Fn} c tm)

F1 F2 ,n=1

Kn =n=1

Fn

Do , theo ghi ch trn ta cn=1

Kn 6=

Bi 2. Cho X l khng gian compact v f : X R lin tc. Chng minh f b chn trn Xv t gi tr nh nht.

Gii. t a = inf f(x), ta c a (ta hiu cn di ng ca tp khng b chn di l). Ta lun c th tm c dy s {an} sao cho an > an+1, lim an = a. Ta t Fn = {x X : f(x) an} (n 1), ta c

Fn l tp ng (do Fn = f1((, an]))

Fn 6= (do an > a = inf f(X)

Fn Fn+1 (do an > an+1)

Do , theo bi 1) th tn ti x0 n=1

Fn. Ta c

f(x0) an n = 1, 2, . . . f(x0) a

Vy f(x0) = a, ni ring a 6= . Ta c pcm.

Bi 3. Cho khng gian metric (X, d) v A, B l cc tp con khc ca X. Ta nh nghad(A,B) = inf

xA,yBd(x, y)

1. Gi s A, B l cc tp compact, chng minh tn ti x0 A, y0 B sao chod(A,B) = d(x0, y0)

2. Gi s A ng, B compact v A B = , chng minh d(A,B) > 0.Nu v d chng t kt lun khng ng nu thay gi thit B compact bng B ng.

4

Gii. 1. Tn ti cc dy {xn} A, {yn} B sao cho lim d(xn, yn) = d(A,B). Do Acompact nn {xn} c dy con {xnk}k hi t v mt phn t x0 A. Xt dy con tngng {ynk}k ca {yn}. Do B compact nn {ynk}k c dy con {ynki}i hi t v mt phnt y0 B.Ta c:

limi

xnki = x0 (v l dy con ca {xnk}) lim

id(xnki , ynki ) = d(A,B) (v l dy con ca {d(xn, yn)})

limi

d(xnki , ynki ) = d(x0, y0) (h qu ca bt t gic)

Do , d(x0, y0) = d(A,B)

2. Gi s tri li, d(A,B) = 0. Khi , ta tm c cc dy {xn} A, {yn} B saocho lim d(xn, yn) = 0.

Do B compact nn {yn} c dy con {ynk}k hi t v y0 B. Td(xnk , y0) d(xnk , ynk) + d(ynk , y0)

ta suy ra limk

xnk = y0

Do A l tp ng, {xnk} A nn ta suy ra y0 A, mu thun vi gi thitA B = .

Trong R2 ta xt metric thng thng v tA = {(t, 0) : t R},B =

{(t,1

t

): t > 0

}Ta c A, B l cc tp ng, A B = t x = (t, 0), y =

(t,1

t

)(t > 0)

Ta c d(x, y) =1

t 0 (t +)

Do , d(A,B) = 0

Bi 4. Cho khng gian metric (X, d) v A X, l tp compact, V l tp m cha A. Ta khiu B(A, ) := {x X : d(x,A) < }

Chng minh tn ti s > 0 sao cho B(A, ) V .

Gii. Cch 1Do A V v V l tp m nn x A,rx > 0 : B(x, 2rx) V

5

H {B(x, rx) : x A} l mt ph m ca tp compact A nn tn ti x1, . . . , xn sao cho

A n

k=1

B(xk, rxk)

t = min{rx1 , . . . , rx2}, ta s chng minh B(A, ) V .Xt ty y B(A, ), ta c

d(y, A) <

x A : d(y, x) < k = 1, n : x B(xk, rxk)

Khi , d(y, xk) d(y, x) + d(x, xk) < + rxk 2rxkDo , y B(xk, 2rxk) V

Cch 2t B = X \ V , ta c B ng v A B = nn theo bi 3 ta c d(A,B) > 0. Chn = d(A,B). Ta s chng minh B(A, ) V hay ch cn chng t B(A, ) B = Tht vy, nu c y B(A, ) B, th ta c

d(y, A) < x A : d(y, x) < Mt khc x A, y B nn d(x, y) d(A,B) = . V l.

Bi 5. Cho X, Y l cc khng gian metric, vi X l khng gian compact v f : X Y lsong nh lin tc. Chng minh f l nh x ng phi.

Gii. Ta cn chng minh nh x ngc f1 lin tc. Do mt bi tp 3, ch cn chng t fl nh x ng.

Vi A X l tp ng, ta cA compact f(A) compact

f(A) ngVy f l nh x ng.

Cc bi tp t gii

Bi 6. Cho cc khng gian metric compact X, Y v nh x f : X Y . Chng minh cc mnh sau tng ng:

1. f lin tc

2. f1(K) l tp compact vi mi tp compact K Y

6

Hng dn

S dng lin h gia tnh compact v tnh ng.

Bi 7. Cho khng gian metric (X, d) v cc tp A, B khc , trong A compact. Chngminh tn ti im x0 A sao cho d(x0, B) = d(A,B).

Hng dn

S dng d(A,B) = infxA

d(x,B)

Bi 8. Cho khng gian metric (X, d) v f : X X l nh x lin tc. im x gi l im btng ca f nu f(x) = x.

1. Chng minh tp im bt ng ca f l tp ng.

2. Gi s X l compact v f khng c im bt ng no. Chng minh tn ti s c > 0 sao

cho d(f(x), x) c x X

Hng dn

t h(x) = d(f(x), x), x X th h : X R lin tc.

1. Ch rng: x bt ng h(x) = 0

2. Cn chng minh infxX

h(x) > 0

Ngoi ra, cu 1) c th chng minh trc tip da vo lin h gia tnh cht ng v s hi

t, cu 2) c th dng phn chng gii.

7

GII TCH (C S)Ti liu n thi cao hc nm 2005

Phin bn chnh sa



5. Bi n tpBi 1:

Trn X = C[0,1] ta xt metric hi t u. Cho tp hp A = {x X : x(1) = 1, 0 x(t) 1 t [0, 1]} v nh x f : X R, f(x) =

10

x2(t) dt.

1. Chng minh inf f(A) = 0 nhng khng tn ti x A f(x) = 0.2. Chng minh A khng l tp compact.

Gii1. t = inf f(A). Ta c f(x) 0 x A nn 0.Vi xn(t) = t

n, ta c xn A

f(xn) = 10

t2n dt =1

2n+ 1 0 (n)

Do = 0.

Nu f(x) = 0, ta c:( 10

x2(t) dt = 0, x2(t) 0, x2(t) lin tc trn [0, 1])

= x(t) = 0 t [0, 1]= x / A.

2. Ta c: {f lin tc trn X, nhn gi tr trong R (xem bi tp 3)f(x) 6= inf f(A) x A

= A khng compact (xem l thuyt 4).

1

Bi 2:Cho (X, d) l khng gian metric compact v nh x X X tha mn

d(f(x), f(y)) < d(x, y) x, y X, x 6= y. (1)Chng minh tn ti duy nht im x0 X tha mn x0 = f(x0) (ta ni x0 l im bt ng canh x f).

GiiTa xt hm g : X R, g(x) = d(f(x), x), x X. Ta ch cn chng minh tn ti duy nht

x0 X sao cho g(x0) = 0.p dng bt ng thc t gic v iu kin (1), ta c

|g(x) g(y)| = |d(f(x), x) d(f(y), y)| 2d(x, y)nn g lin tc. T y v tnh compact ca X ta c:

x0 X : g(x0) = inf g(X) (2)Ta s chng minh g(x0) = 0. Gi s g(x0) 6= 0; ta t x1 = f(x0) th x1 6= x0, do :

d(f(x1), f(x0)) < d(x1, x0) d(f(x1), x1) < d(f(x0), x0) g(x1) < g(x0), mu thun vi (2).

Vy g(x0) = 0 hay f(x0) = x0. chng minh s duy nht ta gi s tri li, c x 6= x0 v x = f(x). Khi :

d(x, x0) = d(f(x), f(x0)) < d(x, x0)

Ta gp mu thun.

Bi 3:Cho cc khng gian metric (X, d), (Y, ) v nh x f : X Y . Trn X Y ta xt metric

d1((x, y), (x, y)) = d(x, x) + (y, y), (x, y), (x, y) X Y.

v xt tp hp G = {(x, f(x)) : x X}.

1. Gi s f lin tc, chng minh G l tp ng.

2. Gi s G l tp ng v (Y, ) l khng gian compact, chng minh f lin tc.

Gii1. Xt ty dy {(xn, f(xn))} G m lim(xn, f(xn)) = (a, b) (1)Ta cn chng minh (a, b) G hay b = f(a).

T (1), ta c

limxn = a (2), lim f(xn) = b (3).

2

T (2) v s lin tc ca f ta c lim f(xn) = f(a); kt hp vi (3) ta c b = f(a) (pcm).

2. Xt ty tp ng F Y , ta cn chng minh f1(F ) l tp ng trong X: chng minh f1(F ) ng, ta xt ty dy {xn} f1(F ) m limxn = a v cn chng t

a f1(F ).Ta c: {

f(xn) F, n NF l tp compact (do F ng, Y compact)

= {xnk} : limk

f(xnk) = b F .Khi :

limk

(xnk , f(xnk)) = (a, b), (xnk , f(xnk)) G,G ng= (a, b) G hay b = f(a).

Vy f(a) F hay a f1(F ) (pcm).

Bi 4:Cho khng giam metric compact (X,d) v cc nh x lin tc fn : X R (n N) tha mn

cc iu kin sau:

f1(x) f2(x) . . . , limn

fn(x) = 0 x X ()

Chng minh dy {fn} hi t u trn X v khng, ngha l:

> 0 n0 : n n0 = supxX

|fn(x)| < ()

p dng phng php sau: vi > 0 cho, t

Gn = {x X : fn(x) < }, n NCh cn chng minh tn ti n0 sao cho Gn0 = X.

GiiTrc tin t gi thit (*) ta suy ra rng fn(x) 0 x X, n N. Ta c:Gn l tp m (do fn lin tc v Gn = f

1n (, ))

Gn Gn+1, (do fn(x) fn+1(x))X =

n=1

Gn (do x X nx : n nx fn(x) < )

Do X l khng gian compact ta tm c n1, n2, . . . , nk sao cho

X =k

i=1

Gni

3

t n0 = max{n1, . . . , nk} ta c X = Gn0 . Khi n n0 ta c Gn Gn0 nn Gn = X. T yta thy (**) ng.

Bi 5:Cho khng gian metric compact (X, d) v nh x lin tc f : X X. Ta nh ngha

A1 = f(X), An+1 = f(An), n = 1, 2, . . . , A =n=1

An.

Chng minh A 6= v f(A) = A.

GiiTa c

6= A1 X,A1 compact (do X compact v f lin tc).Dng quy np, ta chng minh c rng

6= An An+1, An compact n = 1, 2, . . .T y ta c {An} l h c tm cc tp ng trong khng gian compact. Do A 6= 0.

Bao hm thc f(A) A c suy t

f(A) f(An1) = An n = 1, 2, . . . ( do A An1, vi quy c A0 = X).

chng minh A f(A), ta xt ty x A. V x An+1 = f(An) nn

n = 1, 2, . . . xn An : x = f(xn).

Do X compact nn c dy con {xnk}, limk

xnk = a. Khi

x = limk

f(xnk) (do cch xy dng {xn})= f(a) (do f lin tc)

Ta cn phi chng minh a A. C nh n, ta c

xnk An khi nk n (do xnk Ank An)= a = lim

kxnk An (do An ng).

Vy a An n = 1, 2, . . . ; do a A v x = f(a) f(A). (pcm).

4

GII TCH (C S)


Phn 2. Khng gian nh chun

nh x tuyn tnh lin tc

1. Khng gian nh chun(Phin bn chnh sa)



L thuyt

1 Chun

Gi s X l mt khng gian vect (k.g.v.t) trn trng s K (K = R hoc K = C). Mt nhx p : X R c gi l mt chun trn X nu tha mn cc iu kin sau cho mi x, y X,mi K:

i) p(x) 0p(x) = 0 x = ( ch phn t khng trong X)

ii) p(x) = ||p(x)

iii) p(x+ y) p(x) + p(y)

S p(x) gi l chun ca phn t x.

Thng thng, ta dng k hiu ||x|| thay cho p(x).

Mnh 1. Nu p l mt chun trn k.g.v.t X th ta c:

1

1. |p(x) p(y)| p(x y) (hay |||x|| ||y||| ||x y||) x, y X.

2. d(x, y) := p(x y) l mt mtric trn X, gi l mtric sinh bi chun p (hay d(x, y) =||x y||)

V d 1. Trn Rn nh x

x = (x1, . . . , xn) 7 ||x|| =(

nk=1

x2k

)1/2l chun, gi chun Euclide. Mtric sinh bi chun ny chnh l mtric thng thng ca Rn.

V d 2. Trn C[a, b], nh x x 7 ||x|| := supatb |x(t)| l mt chun mtric sinh bi chunny l mtric hi t u trn C[a, b]

2 Khng gian nh chun

nh ngha 1.

Khng gian vect X cng vi chun || || trong n, c gi l mt khng gian nh chun(kgc), k hiu (X, || ||).

Cc khi nim hi t, tp m, ng, compact, dy Cauchy, trong (X, || ||) c hiul cc khi nim tng ng i vi mtric sinh bi chun.

Ni ring, trong (X, || ||) ta c

B(x0, r) = {x X : ||x x0|| < r}

( limn

xn = x(cng vit xn|||| x)) lim

n||xn x|| = 0

({xn} l dy Cauchy) limn,m

||xn xm|| = 0.

nh ngha 2. Kgc (X, || ||) c gi l khng gian Banach nu X vi mtric sinh bi || ||l khng gian y .

V kgc l trng hp c bit ca khng gian mtric nn tt c cc kt qu v khng gian

mtric cng ng cho kgc. Ngoi ra, ta c cc kt qu sau v kgc.

Mnh 2. Cho Kgc (X, .) trn trng s K v cc dy {xn}, {yn} X, {n} K,limxn = x, lim yn = y, limn = . Khi :

1. lim xn = x

2. lim(xn + yn) = x+ y, limnxn = x.

H qu. Cc nh x f, g : X X, f(x) = x0 + x, g(x) = 0x (0 K\{0}) l ng phi.

2

3 Chun tng ng

nh ngha 3. Hai chun .1, .2 trn kgvt X gi l tng ng (vit .1 .2) nu tnti cc hng s dng a, b sao cho

x1 ax2 , x2 bx1 x X

Mnh 3. Gi s .1, .2 l hai chun tng ng trn kgvt X. Khi :

1. (limxn = x theo .1) (limxn = x theo .2)

2. (X, .1) y (X, .2) y .

4 Mt s khng gian nh chun

4.1 Khng gian nh chun con

Cho kgc (X, .) v X0 l mt kgvt con ca X. K hiu .0 l thu hp ca . trn X0 th.0 l mt chun trn X0. Cp (X0, .0) gi l kgc con ca (X, .).

4.2 Tch ca hai kgc

Cho cc kgc (X1, .1), (X2, .2). Tch cc X1 X2 s tr thnh kgvt nu ta nh nghacc php ton

(x1, x2) + (y1, y2) = (x1 + y1, x2 + y2) (x1, x2) = (x1, x2)

Kgvt X1 X2 vi chun(x1, x2) := x11 + x22 ()

hoc vi chun tng ng vi (*), gi l kgc tch ca cc kgc (X1, .1), (X2, .2).Ta d dng kim tra c cc tnh cht sau:

Dy (xn1 , xn2 ) hi t v phn t (x1, x2) trong kgc tch khi v ch khi cc dy {xni } hi tv xi trong kgc (Xi, .i), i = 1, 2.

Nu (Xi, .i)(i = 1, 2) l cc khng gian Banach th kgc tch cng l khng gian Banach.

4.3 Kgc hu hn chiu

Gi s X l kgvt m chiu v e = {e1, . . . , em} l mt c s ca X. Khi nh x

x =mk=1

kek 7 xe :=(

mk=1

|k|2)1/2

l mt chun, gi l chun Euclide sinh bi c s e.

3

Mnh 4.

1. Trn mt khng gian hu hn chiu, hai chun bt k lun tng ng vi nhau.

2. Trn kgc hu hn chiu, mt tp l compact khi v ch khi n ng v b chn.

3. Mt khng gian nh chun hu hn chiu lun l khng gian y . Do , mt kgvt

con hu hn chiu ca mt kgc l tp ng trong khng gian .

nh l 1 (Riesz). Nu qu cu B(, 1) := {x X : x 1} ca cc kgc X l tp compactth X l khng gian hu hn chiu.

5 Chui trong kgc

Nh c php ton cng v ly gii hn, trong kgc ta c th a ra khi nim chui phn t

tng t khi nim chui s.

nh ngha 4. Cho kgc (X, .) v dy {xn} cc phn t ca X. Ta ni chui phn tn=1

xn ()

hi t v c tng bng x nu nh x = limn sn, trong : s1 = x1, sn = x1+ +xn (n N) Nu chui s n=1 xn hi t th ta ni chui (**) hi t tuyt i.

Mnh 5. Nu X l khng gian Banach th mi chui hi t tuyt i l chui hi t

4

Bi tpBi 1. K hiu C1[a,b] l khng gian cc hm thc x = x(t) c o hm lin tc trn [a, b]. C

1[a,b]

l kgvt trn R vi cc php ton thng thng v cng hai hm v nhn hm vi s thc. Tanh ngha p1(x) = |x(a)|+ sup

atb|x(t)| , p2(x) = sup

atb|x(t)|, p3(x) = sup

atb{|x(t)|+ |x(t)|}

1. Chng minh p1, p2, p3 l cc chun trn C1[a,b].

2. Chng minh p2 6 p33. Chng minh p1 p3

Gii.

1. lm v d, ta kim tra p1 l chun.

i) Hin nhin ta c p1(x) 0 x C1[a,b]; hn na

p1(x) = 0{

x(a) = 0

x(t) = 0 t [a, b] {

x(a) = 0

x(t) l hm hng s x(t) = 0t [a, b].

ii) p1(x) = |x(a)|+ supatb

|x(t)| = ||(|x(a)|+ sup

atb|x(t)|

)= ||p1(x)

iii) Vi x, y C1[a,b] ta c

|x(a) + y(a)|+ |(x(t) + y(t))| |x(a)|+ |y(a)|+ |x(t)|+ |y(t)| p1(x) + p1(y) t [a, b]

= p1(x+ y) p1(x) + p1(y).

2. D thy p2(x) p3(x) x C1[a,b]. Ta s chng minh khng tn ti s c > 0 sao cho

p3(x) cp2(x) x C1[a,b] ()

Xt dy xn(t) = (t a)n, n N. D dng tnh c:p2(xn) = (b a)np3(xn) = (b a)n + n(b a)n1

Do , nu tn ti c > 0 (*) ng th ta c

(b a)n + n(b a)n1 c(b a)n n = 1, 2, b a+ n c(b a) n = 1, 2,

Ta gp mu thun.

5

3. Ta d dng kim tra p1(x) p3(x) x C1[a,b] Mt khc ta c:|x(t)| |x(a)|+ |x(t) x(a)| = |x(a)|+ |x(c)(t a)|(p dng nh l Lagrange)

|x(a)|+ (b a) supatb

|x(t)|Mp1(x) t [a, b] (M = max{1, b a})

|x(t)| p1(x) t [a, b].Do p3(x) (M + 1)p1(x) x C1[a,b].Vy p1 p3.

Bi 2. K hiu l2 l khng gian cc dy s thc x = {k}k tha mn iu kink=1

2k <

vi cc php ton thng thng v cng hai dy s v nhn dy s vi s thc. Trn l2 ta xt

chun x =(

k=1

2k

)1/2nu x = {k} l2

1. Xt cc dy s en = {n,k}k (n N) trong n,k = 1 nu n = k, n,k = 0 nu n 6= k.Chng minh rng nu x = {k} l2 th x =

n=1

nen

2. Chng minh l2 y .

Gii.

1. t sn = 1e1 + + nen, ta cn chng minh limn

sn = x

Ta c:

sn = (1, , n, 0, 0, )

x sn = (0, , 0, n+1, n+2, ), x sn =( k=n+1

2k

)1/2

V chuik=1

2k hi t nn limn

k=n+1

2k = 0.

Vy limn

x sn = 0 (pcm).

2. Gi s {xn} l dy Cauchy trong l2, xn = {nk}k, n N.

Vi mi k N, ta c:

|nk mk | (

k=1

|nk mk |2)1/2

= xn xm (1)

6

v {xn} l dy Cauchy nn {nk}n l dy Cauchy trong R, do hi t.t ak = lim

nnk (k N) v lp dy s a = {ak}

Tip theo ta s chng minh a l2 v limn

xn a = 0Cho > 0 ty . Do {xn} l dy Cauchy ta c n0 tha mn

n,m N, n,m n0 xn xm < . (1)T (1) ta c

Nk=1

|nk mk |2 < 2 N N,n,m n0

Nk=1

|nk ak|2 2 N N,n n0(ta cho m trong bt trn)

k=1 |nk ak|2 2 n n0 (2)T (2) ta suy ra xn a l2 (n n0) v do a = xn (xn a) cng thuc l2. Hnna, ta chng minh:

> 0n0 : n n0 = xn a hay l lim xn a = 0

Ghi ch

trn ta khng kim tra l2 l kgvt v cc iu kin ca chun. lm v d, ta s chng

minh rng nu x = {k} l2, y = {k} l2 th x + y l2 v x + y x + y. Tht vy,ta c theo bt ng thc Bunhiakowski:N

k=1(k + k)2 =

Nk=1

2k + 2

Nk=1 kk +

Nk=1

2k

x2 + 2x.y+ y2 N N.Cho N ta c pcm.Bi 3. Gi m l khng gian cc dy s thc x = {k}k b chn vi chun x = sup{|k| : k N}.

1. Chng minh m l khng gian Banach.

2. K hiu C l tp hp cc dy s hi t. Chng minh C l khng gian con ng ca m.Gii. 1. Gi s {xn} l dy Cauchy trng m,xn = {nk}k, n N

Vi mi k N, ta c:

|nk mk | sup{|nk mk | : k N} = xn xmv do {xn} l dy Cauchy nn {nk}n l dy Cauchy trong Rv do vy, hi t.t ak = lim

nnk v lp dy s a = {ak}k.

7

Ta chng minh a m v lim xn a = 0Cho > 0, ta tm c n0 sao cho

n,m n0 xn xm <

Ta c:

|nk nk | < k N,n,m n0 |nk ak| k N,n n0(cho m trong bt trn) sup

k|nk ak| n n0.

Nh vy, ta chng minh:

* (xn a) m, do a = xn (xn a) m.* > 0 n0 : n n0 xn a hay lim xn a = 0.

2. Gi s ta c dy {xn} C, xn = {nk}k m xn hi t v a = {ak} m ta cn chng minha C. Mun vy, ta ch cn chng minh a l dy Cauchy.Cho > 0, ta tm c n sao cho

supk|nk ak| = xn a < /3(do a = lim xn trong m)

V xn = {nk }k C nn n l dy Cauchy, do c k0 sao cho:

k, l k0 |nk n

l | < /3.

Vi k0 ny, ta c:

k, l k0 |ak al| |ak nk |+ |nk nl |+ |nl al|< /3 + /3 + /3 =

Vy {ak} l dy Cauchy (pcm).

Bi 4. Cho kgc X v cc tp A,B X khc . Chng minh

1. Nu A m th A+B m

2. Nu A,B compact th A+B compact.

3. Nu A ng, B compact th A+B ng

Gii.

8

1. Trc tin ta chng minh rng b B th A+ b l tp m.Tht vy, nh x f : X X, f(x) = x+ b l ng phi nn

A m f(A) m hay A+ b m

Do A+B =bB

(A+ b) nn A+B m.

2. Xt ty dy {xn} A + B, ta chng minh {xn} c dy con hi t v phn t thucA+B.

Ta c: xn = an + bn vi an A, bn B.Do A compact nn {an} c dy con {ank}k hi t v mt a A. Do B compact nn dy{bnk}k c dy con {bnkl}l hi t v b B. Tng ng vi dy {bnkl}l ta c dy {ankl}lvn hi t v a.

Suy ra dy con xnkl = ankl + bnkl hi t v a+ b (pcm).

Ghi ch: Cu ny c th gii nh sau:

Xt kgc tch X X v nh x f : X X X, f(x, y) = x+ y. Ta c:(f lin tc, AB l tp compact trong X X) = f(AB) l tp compact trong X.Do f(AB) = A+B ta c pcm.

3. Xt dy ty {xn} A+ B, xn = an + bn, an A, bn B m limxn = x, ta cn chngminh x A+BDo B compact nn {bn} c dy con {bnk} hi t v mt b B. Khi ank = xnk bnkhi t v x b v v A ng nn x b A.Ta c x = (x b) + b nn x A+B (pcm).

Bi 5. Cho kgc (X, .) v X0 l khng gian con hu hn chiu ca X. Chng minh tn tix0 X0 sao cho

a x0 = infxX0

a x

Gii. t d = inf{a x : x X0} v chn dy {xn} X0 tha mn lim a xn = d.Ta c: xn a+ a xn nn {xn} b chn

M > 0 : {xn} B(,M)

Tp B(,M)X0 compact (do dimX0

Ghi ch: Bi ny cn c th gii bng cch tm s M > 0 sao cho

infxX0

a x = infxX0B(,M)

a x

Sau s dng tnh compact ca tp X0 B(,M) v tnh lin tc ca hm x 7 a x

Bi 6. Cho kgc X v A X l tp li. Chng minh tc tp A, IntA cng li.

Gii (Hng dn). C nh s t (0, 1)

chng minh tA+ (1 t)A A ta dng lin h gia im dnh v s hi t.

chng minh t IntA + (1 t) IntA IntA ch cn kim tra v tri l tp m, chatrong A.

Bi 7. Gi s trong kgc X, tp S = {x X : x = 1} l compact. Chng minh dimX

GII TCH (C S)




2. nh X Tuyn Tnh Lin Tc(Phin bn chnh sa)



PHN L THUYT

1. S lin tc ca ca nh x tuyn tnh :

nh x tuyn tnh lin tc gia cc khng gian nh chun c tt c cc tnh cht camt nh x lin tc gia cc khng gian metric. Ngoi ra n cn c cc tnh cht c bitnu trong nh l sau :nh l 1 :Gi s X, Y l cc khng gian nh chun trn cng mt trng s v A : X Y lmt nh x tuyn tnh. Cc mnh sau l tng ng :

(a) A lin tc ti mt im no ca X.

(b) A lin tc trn X.

(c) Tn ti s M > 0 sao cho A(x)Y 6M ||x||X x X2. Chun ca nh x tuyn tnh lin tc. Khng gian L(X,Y )

(a) Nu A : (X, ||.||X) (Y, ||.||Y ) l nh x tuyn tnh lin tc th ta nh nghachun ca A bi :

||A|| = supxXx 6=0

||A(x)||Y||x||X

T nh ngha ny, ta d thy cc tnh cht sau :

i. ||A|| = sup||x||X1

||A(x)||Y = sup||x||X=1

||A(x)||Y

1

ii. Nu A tuyn tnh lin tc th ||A(x)||Y 6 ||A||.||x||X , x Xiii. Nu A tuyn tnh v tn ti s dngM sao cho ||A(x)||Y 6M.||x||X , x X

th A lin tc v ||A|| 6M(b) Ta k hiu L(X, Y ) l tp tt c cc nh x tuyn tnh lin tc t X vo Y .

L(X, Y ) tr thnh khng gian nh chun nu ta nh ngha chun ca mi A L(X, Y ) nh trn v cc php ton nh sau :

(A+B)(x) = A(x) +B(x)

(A)(x) = A(x), x Xnh l 2 :Nu Y l khng gian Banach th L(X, Y ) l khng gian Banach.

3. Phim hm tuyn tnh lin tc

Mt nh x tuyn tnh t khng gian nh chun X vo trng s K cng cn gil mt phim hm tuyn tnh.nh l 3 :Cho f : (X, ||.||) K l phim hm tuyn tnh. Cc mnh sau l tng ng:

(a) f lin tc ti mt im no ca X.

(b) f lin tc trn X.

(c) M > 0 : |f(x)| 6M.||x|| x X(d) Kerf = {x X : f(x) = 0} l khng gian con ng.

Khng gian L(X,K) tt c cc phim hm tuyn tnh lin tc trn X thng khiu l X v gi l khng gian lin hp ca X. T nh l 2 ta c X l khnggian Banach vi chun

||f || = supx 6=

|f(x)|||x||

PHN BI TP

Bi 1Cho cc khng gian nh chun (X, ||.||X), (Y, ||.||Y ) vi dimX = n v A : X Y l nhx tuyn tnh. Chng minh :

1. A lin tc.

2. Tn ti dim xo X sao cho :

||xo||X = 1, ||A|| = ||A(xo)||Y

Gii

2

1. Gi s e = {e1, . . . , en} l mt c s ca X v ||.||e l chun Euclide sinh bi c s e(xem1). Vi x =

nk=1

kek, ta c :

||A(x)||Y n

k=1

|k|.||A(ek)|| (

nk=1

|k|2) 1

2

||x||e

.

(n

k=1

||A(ek)||2) 1

2

M

Nh vy tn ti s M 0 tha mn :||A(x)||Y M.||x||e

V X hu hn chiu nn ||.||e ||.||X v c s a > 0 sao cho :||x||e a||x||X , x X.

T y ta c :||A(x)||Y Ma||x||X , x X

Do A lin tc.

2. Ta c :||A|| = sup

||x||X=1||A(x)||Y ,

nh x x 7 ||A(x)||Y lin tc trn (X, ||.||X), tp S = {x X : ||x||X = 1} l tpcompc trong (X, ||.||X) (V X hu hn chiu).Do tn ti xo S sao cho ||A(xo)||Y = sup

xS||A(x)||Y (pcm)

Bi 2Cho cc khng gian nh chun (X1, ||.||1), (X2, ||.||2), (Y, ||.||Y ) v cc nh x tuyn tnhlin tc Ak : Xk Y, k = 1, 2. Trn khng gian nh chun tch X1 X2 ta xt chun||(x1, x2)|| = ||x1||1 + ||x2||2 v xt nh x

A :X1 X2 YA(x1, x2) = A1(x1) + A2(x2), (x1, x2) X1 X2.

Chng minh A tuyn tnh lin tc v ||A|| = max(||A1||, ||A2||)Gii

t M = max(||A1||, ||A2||) Vi x = (x1, x2), x = (x1, x2) X1 X2, , K, ta c :

x+ x = (x1 + x1, x2 + x2)

A(x+ x) = A1(x1 + x1) + A2(x2 + x2)= A1(x1) +

A1(x1) + A2(x2) + A2(x2)

= [A1(x1) + A2(x2)] + [A1(x1) + A2(x

2)]

= A(x) + A(x)

Vy A l nh x tuyn tnh.

3

||A(x1, x2)||Y = ||A1(x1) + A2(x2)||Y

||A1(x1)||Y + ||A2(x2)||Y ||A1||.||x1||1 + ||A2||.||x2||2 (do A1, A2 lin tc )M(||x1||1 + ||x2||2)

||A(x1, x2)||Y M ||(x1, x2)|| (x1, x2) X1 X2 A lin tc v ||A|| M.

Tip theo ta chng minh ||A|| M . Ta c :||A1(x1)||Y = ||A(x1, )||Y ||A||.||(x1, )||

hay ||A1(x1)||Y ||A||.||x1|| x1 X1Do ||A1|| ||A||. Tng t, ||A2|| ||A||. Vy M ||A|| (pcm)

Bi 3

Gi l1 l tp hp cc dy s thc x = {k} sao cho : ||x|| =k=1

|k|

T () ta suy ra |f(x)| M ||x|| x l1Do f lin tc v ||f || M chng minh ||f || M ta xt cc dy en = {kn}k vi kn = 0 nu k 6= n, nn = 1Ta c

||en|| = 1, f(en) = n, ||f(en)|| ||f ||.||en||nn ||f || |n| n N. Suy ra ||f || M = sup

n|n|.

Vy ||f || = M2. Vi en c nh ngha trn ta t n = f(en). Ta c

|n| ||f ||.||en|| = ||f || n = 1, 2, . . .

nn {k}k l dy b chn. Ta s chng minh vi k nh ngha nh trn th () ng.C nh x = {k} l1 ta t xn = 1e1+ . . .+nen, n N Ta d dng thy limxn = xtrong l1 (xem mt bi tp 1), do theo tnh lin tc v tuyn tnh ca f ta c :

f(x) = limn

f(xn)

= limn

nk=1

kf(ek)

= limn

nk=1

kk

T y ta c ()

Bi 4Gi X l khng gian nh chun cc hm thc x = x(t) lin tc trn [0,) vi chun

||x|| = supt[0,)

eat|x(t)| 0 cho trc)

Chng minh phim hm f sau l tuyn tnh lin tc trn X v tnh chun ca n :

f(x) =

+0

tx(t)dt x X.

Gii

Trc tin ta cng kim tra f(x) xc nh. Vi x X ta c

|tx(t)| = eat.|x(t)|.teat ||x||.teat t [0,) (1)

Hm teat kh tch trn [0,) (d tnh+0

teatdt =1

a2) nn hm tx(t) cng kh tch trn

[0,).

5

D dng kim tra c f l tuyn tnh. T bt ng thc () ta c :

|f(x)| +0

teatdt.||x||

=1

a2||x|| x X (2).

Do f lin tc v ||f || 1a2

Ta s chng minh ||f || = 1a2. Trong (1), (2) ta thy du = t c khi x = eat. t

xo = eat, ta c :

xo X, ||xo|| = 1, f(xo) = 1a2

Mt khc |f(xo)| ||f ||.||xo||. Do ta suy ra ||f || 1a2. Vy ||f || = 1

a2

Bi 5Trn C[1, 1] vi chun hi t u ta xt phim hm :

f(x) =

10

x(t)dt0

1

x(t)dt x C[1, 1]

Chng minh f tuyn tnh lin tc v tnh ||f ||.Gii :

D dng kim tra f l tuyn tnh v

|f(x)| 1

0

|x(t)|dt+0

1

|x(t)|dt 2||x|| x C[1, 1]

Do vy f lin tc v ||f || 2. Ta s chng minh ||f || = 2.Ta thy trong bt ng thc trn du = t c khi xo(t) = 1 trn (0, 1], xo(t) = 1 trn[1, 0], nhng hm xo ny khng thuc C[1, 1]. Ta xt dy hm {xn} nh sau :

xn(t) =

1, nu t [1, 1

n]

nt, nu t [ 1n,1

n] (n 2)

1, nu t [ 1n, 1]

(nu v th ca hm xn v hm xo ta s thy ngha ca vic chn xn). Ta c :

xn C[1, 1], ||xn|| = 1, f(xn) = 2 1n.

6

M ta cng c |f(xn)| ||f ||.||xn||.Do ta c ||f || 2 1

nn N

Cho n ta c ||f || 2. Vy ||f || = 2Bi 6Cho cc khng gian nh chun (X, ||.||X), (Y1, ||.||1), (Y2, ||.||2) v cc nh x tuyn tnh lintc Ak : X Yk, k = 1, 2.Ta xt nh x

A : X Y1 Y2A(x) = (A1(x), A2(x)), x X.

Chng minh A tuyn tnh, lin tc v :

1. max(||A1||, ||A2||) ||A|| ||A1||+ ||A2||nu trong Y1 Y2 ta xt chun :

||(y1, y2)|| = ||y1||1 + ||y2||2, (y1, y2) Y1 Y2.2. ||A|| = max(||A1||, ||A2||) nu trong Y1 Y2 ta xt chun :

||(y1, y2)|| = max(||y1||1, ||y2||2)Hng dn

1. Ta c :

||A(x)|| = ||A1(x)||1 + ||A2(x)||2

{ ||A(x)|| (||A1||+ ||A2||)||x||||Ak(x)|| ||A(x)|| ||A||.||x||

{ ||A|| ||A1||+ ||A2||||Ak|| ||A||

2. ||A(x)|| = max(||A1(x)||, ||A2(x)||), s dng cc nh gi tng t.Bi 7Trn C[1, 1] ta xt chun hi t u v xt phim hm :

f(x) =

11

x(t)dt x(0), x C[1, 1].

Tnh ||f ||.Hng dn

|f(x)| 1

1

|x(t|dt+ |x(0)| 3||x||, x C[1, 1].

Do ||f || 3. chng minh ||f || 3 ta ch rng trong bt ng thc trn, du = t c ti hm xo(t) = 1 nu t 6= 0, xo(0) = 1, nhng xo / C[1, 1].

7

GII TCH (C S)




3. Khng gian Hilbert(Phin bn chnh sa)



I. Phn l thuyt

1 Tch v hng, khng gian Hilbert

1.1 nh ngha

nh ngha 1 1. Cho khng gian vect X trn trng s K(K =

R hoc K = C).Mt nh x t X X vo K, (x, y) x, yc gi l mt tch v hng trn X nu n tha mn cc iu

kin sau:

(a) x, x 0 x Xx, x = 0 x =

(b) y, x = x, y (y, x = x, y nu K = R), x, y X(c) x + x, y = x, y + x, y x, x, y X(d) x, y = x, y x, y X, K

1

T cc tnh cht i) - iv) ta cng c:

x, y + y = x, y + x, y, x, y = x, y2. Nu ., . l mt tch v hng trn X th nh x xx, x

l mt chun trn X, gi l chun sinh bi tch v hng.

3. Nu ., . l tch v hng trn X th cp(X, ., .) gi l mtkhng gian tin Hilbert (hay khng gian Unita, khng gian vi

tch v hng). S hi t, khi nim tp m,...,trong (X, ., .)lun c gn vi chun sinh bi ., .. Nu khng gian nhchun tng ng y th ta ni (X, ., .) l khng gianHilbert.

1.2 Cc tnh cht

1. Bt ng thc Cauchy - Schwartz: |x, y| x.y2. x + y2+x y2 = 2(x2+y2) (ng thc bnh hnh).3. Nu lim xn = a, lim yn = b th limxn, yn = a, b

V d 1 1. Trong C[a, b] cc hm thc lin tc trn [a, b] th nh

x

(x, y) 7 x, y = ba

x(t)y(t)dt

l mt tch v hng. Khng gian (C[a, b], ., .) khng l khnggian Hilbert.(xy dng v d tng t phn khng gian met-

ric)

2. Trong l2, vi x = {k}, y = {k}, ta nh ngha

x, y =k=1

kk

th ., . l tch v hng, (l2, ., .) l khng gian Hilbert.2

2 S trc giao

2.1 nh ngha

nh ngha 2 Cho khng gian vi tch v hng (X, ., .) vx, y X, 6= M X.1. Ta ni x trc giao vi y (vit xy) nu x, y2. Nu xy y M th ta vit xM . Ta k hiu

M = {x X : x M}.

2.2 Cc tnh cht

1. Nu x M th x M(M ch khng gian con sinh bi M)2. Nu x yn n N v lim yn = y th x y. Suy ra nux M th cng c x M .

3. M l mt khng gian con ng.

4. Nu x1, . . . , xn i nt trc giao th

x1 + . . . + xn2 = x12 + . . . + xn2(ng thc Pythagore)

nh l 1 (v phn tch trc giao) Nu M l mt khng gian

con ng ca khng gian Hilbert (X, ., .) th mi x X c duynht phn tch dng

x = y + z, y M, z M (1)Phn t y trong (1) gi l hnh chiu trc giao ca x ln M v

c tnh cht

x y = infyM

x y.

3

3 H trc chun. Chui Fourier

3.1 nh ngha

Cho khng gian Hilbert (X, ., .)1. H {e1, e2, . . .} X gi l mt h trc chun nu

ei, ej ={

0 nu i 6= j1 nu i = j

Nh vy, {en} l h trc chun nu en = 1 n N vei ej(i 6= j).

2. H trc chun {en} gi l y , nu n c tnh cht sau:(x en n = 1, 2, . . .) x = .

3. Nu {en} l h trc chun th chui

n=1x, en en gi l chuiFourier ca phn t x theo h chun {en}.

nh l 2 Cho {en} l h trc chun trong khng gian Hilbert(X, ., .) v {n} l mt dy s. Ta xt chui

n=1

nen (2)

Ta c:

1. Chui (2) hi t khi v ch khi

n=1 |n|2

nh l 3 Chui Fourier ca mi phn t x X theo h trcchun {en} l hi t v ta c

n=1

|x, en|2 x2 (bt dng thc Bessel).

ngha ca h trc chun y c lm r trong nh l sau.

nh l 4 Cho {en} l h trc chun. Cc mnh sau l tngng:

1. H {en} y 2.

x =

n=1

x, enen, x X.

3.

x2 =n=1

|x, en|2 x X (ng thc Parseval)

II. Phn Bi tp

Bi tp 1 Trong khng gian l1 cc dy s thc x = {k},

k=1 |k| < ta nh ngha

x, y =k=1

k k, x = {k} l1, y = {k} l1

1. Chng minh ., . l mt tch v hng trn l1.2. (l1, ., .) khng l khng gian Hilbert.

Gii

5

1. Trc tin ta cn kim tra x, y xc nh x, y l1. Tht vy,v limk = 0 nn {k} b chn: M R, |k| Mk N.Do

k=1

|kk| Mk=1

|k| m:

xn xm = {0, . . . , 0, 1m + 1

, . . . ,1

n, 0, 0, . . .}

xn xm = (n

k=m+1

1

k2)12 0 (khi n,m)

Ta chng minh {xn} khng hi t.Gi s tri li tn ti a = {k} l1 sao cho lim xna = 0.C nh k N, khi n k, ta c

|1k k| xn a

T y ta c k =1k k N, v l v dy {1k}k / l1.

Vy l1 vi tch v hng trn khng l khng gian Hilbert.

Bi tp 2 Cho khng gian Hilbert X v X0 l khng gian con

ng ca X, A : X0 Y l nh x tuyn tnh lin tc(Y l mtkhng gian nh chun). Chng minh tn ti nh x tuyn tnh

lin tc B : X Y sao cho B(x) = A(x) x X0, B = A6

Gii

Ta nh ngha nh x A nh sau. Theo nh l v phn tch trcgiao, mi x X c duy nht phn tch

x = y + z, y X0, z X0 (3)v ta t B(x) := A(y).

V phn tch dng (3) ca x X0 l x = x + nn ta c ngayB(x) = A(x) x X0

Ta kim tra B l tuyn tnh: vi x, x X,, K ta vitphn tch (3) v

x = y + z, y X0, z X0Khi :

x + x = (y + y) X0

+ (z + z) X0

B(x + x) = A(y + y)= A(y) + A(y)= B(x) + B(x).

Tip theo ta chng minh B lin tc v B = A.T (3) v nh l Pythagore ta c x2 = y2 + z2, do :

B(x) = A(y) A y (Do A lin tc) B(x) A x, x X

Vy B lin tc v B A. Mt khc ta c:B = supxX,x=1 B(x) supxX0,x=1 B(x)= supxX0,x=1 A(x) = AVy B = A.

7

Bi tp 3 Cho h trc chun {en} trong khng gian HilbertX. Xt dy nh x

Pn :X X

Pn(x) =n

k=1

x, ekek, x X,n N.

1. Chng minh Pn(x) l hnh chiu trc giao ca x ln Xn :=

e1, . . . , en.2. Gi s h {en} y . Chng minh limn ||Pn(x)I(x)|| =

0 x X nhng ||PnI||9 0(I : X X l nh x ngnht)

Gii

1. Ta c: x = Pn(x) + (x Pn(x)), Pn(x) Xn.Do ch cn phi chng minh x Pn(x) Xn hay x Pn(x) Xn. V Xn sinh bi {e1, . . . , en} nn ch cn chngminh x Pn(x) ei i = 1, . . . , n.Tht vy:

xPn(x), ei = x, ein

k=1

x, ekek, ei = x, eix, ei = 0

2. Do ng thc Parseval, ta c x X :

x =k=1

x, ek ek = limn

nk=1

x, ek ek = limnPn(x)

t

Qn(x) = I(x)Pn(x) =

k=n+1

x, ekek, x X,n = 1, 2 . . .

8

Ta c Qn(x) tuyn tnh v

||Qn(x)||2 =

k=n+1

|x, ek|2 ||x||2 (bt Bessel)

Qn lin tc, ||Qn|| 1Mt khc, Qn(en+1) = en+1 v ||Qn|| ||Qn(en+1||||en+1|| = 1 nnta c ||Qn|| = 1 hay ||I Pn|| = 1

Bi tp 4 Cho {en} l h trc chun trong khng gian Hilbert Xv {n} l dy s.

1. Gi s {n} l dy b chn. Chng minh rng

A(x) =

n=1

nx, en en x X (4)

l nh x tuyn tnh lin tc t X vo X v ||A|| = supnN |n|2. Gi s chui trong (4) hi t x X. Chng minh {n} l

dy b chn.

Gii

1. t M = supnN |n|u tin ta kim tra A xc nh hay chng minh chui trong (4)

hi t. Ta cn=1

|nx, en|2 M 2n=1

|x, en|2 M 2 ||x||2 (5)

nn theo nh l 2, chui trong (4) hi t.

D kim tra A l nh x tuyn tnh. T nh l 2 v (5) ta c

9

||A(x)||2 =n=1

|x, en|2 M 2 ||x||2 x X

A lin tc, ||A|| MMc khc ta c:

A(ek) = kek v ||A(ek)|| ||A|| k Nnn ||A|| |k| k N. Do ||A|| M . Vy ||A|| = Mpcm.

2. T gi thit v nh l 2, ta c

n=1

|n|2 |x, en|2 k(k N). Ta c

k=1

1

|nk|2

k=1

1

k2 a :=

k=1

1

nkenk (theo nh l 2)

Ta c

a, en ={

1nk

nu n = nk

0 nu n / {n1, n2, . . .}do :

n=1

|n|2 |a, en|2 =k=1

|nk|2 1

|nk|2=

Ta gp mu thun vi (6).

10

GII TCH (C S)

Phn 3. o V Tch PhnChuyn ngnh: Gii Tch, PPDH Ton

1. o(Phin bn chnh sa)



1 PHN L THUYT

1. Khng gian o cnh ngha :

1) Cho tp X 6= ; mt h F cc tp con ca X c gi l mt i s nu n thamn cc iu kin sau :

i. X F v nu A F th Ac F , trong Ac = X \ A.ii. Hp ca m c cc tp thuc F cng l tp thuc F .

2) Nu F l i s cc tp con ca X th cp (X,F ) gi l mt khng gian o c; mi tp A F gi l tp o c (o c i vi F hay F o c)

Tnh chtGi s F l i s trn X. Khi ta c :1) X.

Suy ra hp ca hu hn tp thuc F cng l tp thuc F .2) Giao ca hu hn hoc m c cc tp thuc F cng l tp thuc F .3) Nu A F , B F th A \B F .

2. onh ngha :Cho mt khng gian o c (X,F )

1) Mt nh x : F [0,] c gi l mt o nu :i. () = 0

ii. c tnh cht cng, hiu theo ngha

{An}n F, (An Am = , n 6= m) (n=1

An) =n=1

(An)

2) Nu l mt o xc nh trn i s F th b ba (X,F, ) gi l mt khnggian o

1

Tnh cht :Cho l mt o xc nh trn i s F ; cc tp c xt di y u gi thitl thuc F .

1) Nu A B, th (A) (B), hn na nu (A)

3) Tp A R l (L)o c khi v ch khi vi mi > 0, tn ti tp ng F , tp mG sao cho

F A G, (G \ F ) < 4) Nu A l tp (L)o c th cc tp x+ A, xA cng l (L)o c v :

(x+ A) = (A) (xA) = |x|(A)

5) o Lebesgue l , hu hn

2 PHN BI TP

1. Bi 1 Cho khng gian o (X,F, ), tp Y 6= v nh x : X Y Ta nh ngha :

A = {B Y : 1(B) F}

(B) = (1(B))

Chng minh A l i s trn Y v l o xc nh trn A

Gii

Ta kim tra A tha hai iu kin ca i s :i. Ta c Y A v 1(Y ) = X F

Gi s B A, ta cn chng minh Bc = Y \B A. Tht vy, ta c{1(Y \B) = 1(Y )\1(B) = X\1(B)1(B) F ( do B A) nn X \ 1(B) F

1(Y \B) F hay Y \B Aii. Gi s Bn A(n N) v B =

n=1

Bn. Ta c

1(B) =n=1

1(Bn)

1(Bn) F (n N)

1(B) F hay B A. Tip theo ta kim tra l o.Vi B A ta c 1(B) F nn s [1(B)] xc nh, khng m. Vy s (B) 0,xc nh.

i. Ta c () = [1()] = () = 0

ii. Gi s Bn A (n N), Bn Bm = (n 6= m) v B =n=1

Bn.Ta c

1(B) =n=1

1(Bn),

1(Bn) 1(Bm) = 1(Bn Bm) = (n 6= m). [1(B)] =

n=1

[1(Bn)] (do tnh cng ca )

(B) =n=1

(Bn)

3

2. Bi 2 Cho khng gian o (X,F, ) v cc tp An F (n N). t :B =

k=1

( n=k

An

)(Tp cc im thuc mi An t mt lc no )

C =k=1

( n=k

An

)(Tp cc im thuc v s cc An).

Chng minh

1) (B) limn

(An)

2) (C) limn

(An) Nu c thm iu kin (n=1

An)

t

Ck =n=k

Bn (k = 1, 2 . . .),

ta cCk F,C1 C2 . . .

vB = B1 . . . Bn Cn+1

k=1

Ck = (Xem ngha tp C, bi 2 v gi thit v cc Bn)

(B) =

nk=1

(Bk) + (Cn+1) (2) ( do tnh cht ii.)

limm

(Cn) = 0 ( do tnh cht iii.)

Cho n trong (2) ta c (1).4. Bi 4 : K hiu l o Lebesgue trn R. Cho A [0, 1] l tp (L)o c v

(A) = a > 0. Chng minh rng trong A c t nht mt cp s m hiu ca chng l shu t.

Gii

Ta vit cc s hu t trong [0, 1] thnh dy {rn}n v t An = rn +A (n N). Ta chcn chng minh tn ti n 6= m sao cho An Am 6= . Gi s tri li, iu ny khngng. Khi ta c

(n=1

An) =n=1

(An) (1)

Mt khc, ta c

(An) = (A) = a,n=1

An [0, 2]

Do v phi ca (1) bng + cn v tri 2, v l5. Bi 5 : Cho tp (L) o c A R. Chng minh A c th vit thnh dng A = B \C

vi B l giao ca m c tp m v C l tp (L)o c, c o Lebesgue bng 0.Gii

Do tnh cht 3) ca o Lebesgue, vi mi n Nta tm c tp m Gn A sao cho(Gn \ A) < 1

n

t B =n=1

Gn v C = B \ A.Ta c B l (L) o c v do C cng l (L) o c. V

C Gn \ A n = 1, 2, . . .nn ta c :

(C) 1n

n = 1, 2, . . .Vy (C) = 0.

5

6. Bi 6 : Cho tp L o c A [0, 1] vi (A) = a > 0. Chng minh:1) Hm f(x) = (A [0, x]) lin tc trn [0, 1].2) b (0, a) B A : B (L) o c, (B) = b

Gii

1) Vi 0 x < y 1 ta c

f(y) =(A [0, y])=(A [0, x]) + (A (x, y])

f(y) f(x) = (A (x, y]) 0 f(y) f(x) y x

Do f lin tc trn [0, 1]2) Ta c f(0) = 0, f(1) = a v f lin tc nn tn ti xo (0, 1) tha f(xo) = b hay

(A [0, xo]) = b. Tp B := A [0, xo] cn tm.

6

GII TCH (C S)

Phn 3. o V Tch PhnChuyn ngnh: Gii Tch, PPDH Ton

2. HM O C(Phin bn chnh sa)



PHN L THUYT

1. nh ngha:Cho mt khng gian o c (X,F), tp A F v hm f : A R. Vi a R, ta s khiu:

A[f < a] = {x A : f(x) < a}Cc tp hp A[f a], A[f > a], A[f a] c nh ngha tng t.Ta ni hm f o c trn A (o c i vi -i s F hay F -o c) nu:

A[f < a] F , a Rnh l 1:Cc mnh sau tng ng

1) f o c trn A2) A[f a] F , a R3) A[f > a] F , a R4) A[f a] F , a R

2. Mt s lp hm o cCho khng gian o c (X,F). Cc tp hp c xt di y lun gi thit l thucF .1) Hm hng s l o c. Hm c trng 1A ca tp A l o c khi v ch khi A F .2) Nu f o c trn A v B A th f o c trn B

Nu f o c trn mi An (n N) th f o c trnn=1

An

3) Gi s cc hm f, g o c trn A v ch nhn cc gi tr hu hn. Khi cc hmsau cng o c trn A :

|f |, |f | ( > 0), f + g, f.g, fg(nu g(x) 6= 0 x A)

4) Gi s cc hm fn o c trn A (n N). Khi cc hm sau cng o c trn Aa) g(x) = sup{fn(x) : n N}, h(x) = inf{fn(x) : n N}b) f(x) = lim

nfn(x), nu gii hn tn ti ti mi x A.

1

3. Hm o c theo LebesgueHm o c i vi -i s cc tp (L) o c gi l hm o c theo Lebesgue hay(L) o cnh l 2Nu A R l tp (L)-o c v hm f : A R lin tc th f l hm (L)-o c.

4. Hm n ginnh ngha : Cho khng gian o c (X,F) v tp A S.Hm f : A R gi l hm n gin nu n c dng

f(x) =ni=1

ai1Ai(x)

trong : Ai F , (i = 1, n), Ai Aj = (i 6= j),ni=1

An = A v 1Ai l hm c trng

ca tp AiNh vy, hm n gin l hm o c, ch nhn hu hn gi tr.nh l 3Nu f l hm khng m, o c trn A th tn ti dy {sn} cc hm n gin trn Asao cho

i) 0 sn(x) sn+1(x), x Aii) lim

nsn(x) = f(x), x A

2

PHN BI TP

Bi 1 : Cho hm f : X R o c v cc s a, b R, a < b. Chng minh rng hm

g(x) =

f(x) nu a f(x) ba nu f(x) < ab nu f(x) > b

l o c trn X.

GII:Cch 1:t A1 = X[a f b], A2 = X[f < a], A3 = X[f > b], ta c:

Ak F , k = 1, 2, 3, A1 A2 A3 = X

g(x) =

f(x) x A1a x A2b x A3

g o c trn A2 v A3 v l hm hng trn cc tp nyg o c trn A1 v f o c trn A1

Do g o c trn A1 A2 A3 = X.Cch 2:Ta d dng kim tra rng g(x) = min{b,max{a, f(x)}}T cc hm o c qua php ly max, min ta nhn c hm o c. Do go c.

Bi 2 : 1) Cho cc hm f, g : X R o c. Chng minh tpA := {x X : f(x) = g(x)} l o c (ngha l thuc F).

2) Cho dy hm {fn} o c trn X. Chng minh rng tpB := {x X : lim

nfn(x) tn ti} o c.

GII:1) Cch 1:t A1 = {x X : f(x) < g(x)}, A2 = {x X : g(x) < f(x)}Ta chng minh A1, A2 F .. Ta vit tp Q thnh dy {rn}. Ta thy

f(x) < g(x) n : f(x) < rn < g(x)Do :

A1 =n=1

{x X : f(x) < rn < g(x)}

=n=1

(X[f < rn] X[g > rn])nn A1 F . Chng minh A2 F tng t.. Do A = X\(A1 A2) nn A FCch 2:t

A1 = {x X : f(x) = +}, A2 = {x X : f(x) = }A3 = {x X : g(x) = +}, A4 = {x X : g(x) = }Y = X\

4k=1

Ak

Ta c th chng minh Ak, Y F v

3

A = (A1 A3) (A2 A4) Y [f g = 0]Ch rng trn Y th f, g o c, ch nhn gi tr hu hn nn f g o c trnY v do Y [f g = 0] F .

2) t f(x) = limn

fn(x), g(x) = limn

fn(x)

. Theo nh ngha, ta cf(x) = lim

n(infkn

fk(x)), g(x) = limn

(supkn

fk(x))

Cc hm Fn(x) := infkn

fk(x) o c nn f(x) = limn

Fn(x) o c

Tng t, ta c g o c. Ta c B = {x X : f(x) = g(x)} nn p dng cu 1) c B F

Bi 3 : Cho khng gian o (X,F , ), A F v hm f : A R o c.1) t An = {x A : |f(x)| n}, n N. Chng minh

limn

(Bn) = (A).

2) Gi s (A) < . Chng minh rng vi mi > 0, tn ti tp B A, B Fsao cho

(A\B) < , f b chn trn BGII:1) Ta c: An F (v |f | o c), An An+1

A =n=1

An (do f ch nhn gi tr hu hn)

Do limn

(An) = (A)

2) Do (A)

l (L)-o c trn (a, b)

GIIXt cc hm fn : (a, b) R xc nh nh sau

fn(x) =

{n[f(x+ 1

n

) f(x)] , nu x (a, b 1n

)c , nu x [b 1

n, b) , n N

Ta c(1) lim

nfn(x) = f(x) x (a, b)

Tht vy vi x (a, b) ta c x < b 1nkhi n ln, do

limn

fn(x) = limn

n[f(x+ 1

n

) f(x)] = f (x)(2) fn l (L)-o c trn (a, b)Tht vy, trn (a, b 1

n) hm fn lin tc (v f kh vi nn f lin tc) trn

[b 1

n, b)

fn cng l hm lin tc nn fn l (L)- o c)T (1),(2) ta c f l (L)-o c.

5

GII TCH (C S)

Phn 3. o V Tch Phn3. TCH PHN THEO LEBESGUE

Chuyn ngnh: Gii Tch, PPDH Ton(Phin bn chnh sa)



1 PHN L THUYT

1. iu kin kh tch theo RiemannNu hm f kh tch trn [a, b] theo ngha tch phn xc nh th ta cng ni f kh tchtheo Riemann hay (R)kh tch.nh l 1Hm f kh tch Riemann trn [a, b] khi v ch khi n tha mn hai iiu kin sau :

i. f b chn.ii. Tp cc im gin on ca f trn [a, b] c o Lebesgue bng 0.

2. nh ngha tch phn theo LebesgueCho khng gian o (X,F, ) v A F, f : A R l hm o c

(a) Nu f l hm n gin, khng m trn A v f =n

i=n

ai.1Ai vi Ai F,Ai Aj =

(i 6= j) vni=1

Ai = A th ta nh ngha tch phn ca f trn A theo o bi :

A

fd :=n

i=n

ai(Ai)

(b) Nu f l hm o c, khng m th tn ti dy cc hm n gin, khng m fnsao cho

fn(x) fn+1(x), limn

fn(x) = f(x) x AKhi ta nh ngha

A

fd = limn

A

fnd

Ch rng, tch phn hm o c khng m lun tn ti, l s khng m v cth bng +

1

(c) Nu f l hm o c th f+(x) = max{f(x), 0}, f(x) = max{f(x), 0} l cchm o c, khng m v ta c f(x) = f+(x) f(x). Nu t nht mt trong cctch phn

A

f+d,

A

fd l s hu hn th ta nh ngha

A

fd =

A

f+dA

fd

Ta ni f kh tch trn A nu

A

fd tn ti v hu hn (hay c hai tch phnA

f+d,

A

fd l s hu hn).

3. Cc tnh chtCho khng gian o (X,F, )

3.1 Mt s cc tnh cht quen thuc :Gi s A F v f, g l cc hm o c, khng m trn A hoc kh tch trn A.Khi ta c

A

(f + g)d =

A

fd+

A

gdA

cfd = c

A

fd c R

Nu f(x) g(x) x A thA

fd A

gd

Nu A = A1 A2 vi A1, A2 F,A1 A2 = thA

fd =

A1

fd+

A2

fd

3.2 S khng ph thuc tp o O. Khi nim "hu khp ni"nh nghaGi s P (x) l mt tnh cht pht biu cho mi x A sao cho x A th hoc P (x)ng hoc P (x) sai. Ta ni tnh cht P (x) ng (hay xy ra) hu khp ni (vit tthkn) trn tp A nu tp

B = {x A : P (x) khng ng}

c cha trong mt tp C F m (C) = 0 (hoc (B) = 0 nu bit B F ).V d1) Gi s f, g o c trn A. Ta c

B := {x A : f(x) 6= g(x)} F

Do vy ta ni "f(x) = g(x) hkn trn A " th c ngha l (B) = 0.2) Nu f o c trn A th tp B = {x A : |f(x)| = +} thuc F . Ta ni "f

hu hn hkn trn A" th c ngha (B) = 0.

2

3) Cho cc hm o c fn, f (n = 1, 2, . . .). Ta ni "Dy {fn} hi t hkn trn Av F th c ngha B = {x A : fn(x) 6 f(x)} c o 0.

S khng ph thuc tp o 0

Nu (A) = 0 v f o c trn A th

A

fd = 0. Do :

Nu f c tch phn trn A B v (B) = 0 thAB

gd =

A

fd

Nu f, g o c trn A, f(x) = g(x) hkn trn A v f c tch phn trn A thA

gd =

A

fd

3.3 Nu f o c, khng m trn A v

A

fd = 0 th f(x) = 0 hkn trn A.

3.4 Nu f kh tch trn A th f(x) hu hn hkn trn A

3.5 Tnh cht cngGi s An F (n N), An Am = (n 6= m) v f l hm o c, khng mhoc kh tch trn A =

n=1

An. Khi

A

fd =n=1

An

fd

3.6 Mt s iu kin kh tch:

Nu f o c trn A th f kh tch trn A khi v ch khi |f | kh tch trn A. Nu f o c, g kh tch trn A v |f(x)| g(x) x A th f cng kh tchtrn A.

Nu f o c, b chn trn A v (A)

ii. limn

fn(x) = f(x) x AKhi

A

fd = limn

A

fd

Ghi chDo s khng ph thuc vo tp o 0 ca tch phn, ta c th gi thit cc diu kini., ii. trong nh l Levi v Lebesgue ch cn ng hkn trn A.

5. Lin h gia tch phn Riemann v tch phn LebesgueNu A R l tp (L)o c th tch phn theo o Lebesgue cng k hiu

(L)

A

f(x)dx hoc (L)

ba

f(x)dx nu A = [a, b].

nh l

1) Nu f kh tch Riemann trn [a, b] th f cng kh tch theo ngha Lebesgue trn [a, b]v ta c

(L)

ba

f(x)dx = (R)

ba

f(x)dx

2) Nu f kh tch Riemann suy rng trn [a, b] (hoc trn [a,]) v l hm khng mth f kh tch theo ngha Lebesgue trn [a, b] (trn [a,]) v ta c :

(R)

ba

f(x)dx = (L)

ba

f(x)dx

(R) a

f(x)dx = (L)

a

f(x)dx

2 PHN BI TP

Trong cc tp di y ta lun gi thit c mt khng gian o (X,F, ). Cc tp c xtlun thuc FBi 1Cho hm f o c trn A, hm g, h kh tch trn A sao cho g(x) f(x) h(x) x A.Chng minh f kh tch trn A.

Gii

Ta c

f+ h+, f g ( v g f h)A

f+d A

h+d,

A

f A

gd

Cc tch phn v phi hu hn nn

A

fd < . Suy ra f kh tch. (Bi ny cng c th

gii da vo bt ng thc |f(x)| |g(x)|+ |h(x)|)Bi 2

4

1. Cho hm s f 0, o c trn A. Xt cc hm

fn(x) =

{f(x), nu f(x) nn, nu f(x) > n

(n N)

Chng minh limn

A

fnd =

A

fd

2. ng dng kt qu trn tnh (L)

10

dxx

Gii

1. Ta d dng kim tra rng fn(x) = min{n, f(x)}. Do : fn(x) o c, khng m. fn(x) = min{n, f(x)} min{n+ 1, f(x)} = fn+1(x) lim

nfn(x) = min{ lim

nn, f(x)} = min{+, f(x)} = f(x)

p dng nh l Levi ta c pcm.

2. t f(x) =1x, x (0, 1], f(0) = +. Ta d dng tm c

fn(x) =

1x, nu x [ 1

n2, 1]

n nu x [0, 1n2]

(L)

10

fn(x)dx = (R)

10

fn(x)dx = 2 1n

Theo cu 1) ta c

(L)

10

f(x)dx = limn

10

fn(x)dx = 2.

Bi 3Cho hm f kh tch trn A. Ta xy dng cc hm fn nh sau :

fn(x) =

f(x), nu |f(x)| nn, nu f(x) > nn, nu f(x) < n

Chng minh limn

A

fnd =

A

fd

Gii

Ta d thy fn(x) = min{n,max{n, f(x)}}. T y ta suy ra :

5

fn o c, |fn| |f | n N

limn

fn(x) = min{+,max{, f(x)}} = f(x) x A

p dng nh l Lebesgue ta c pcm.Bi 4Cho l hm o c, khng m trn X. Ta nh ngha :

(A) =

A

d, A F

1. Chng minh l o.

2. Gi s f l hm o c, khng m trn X. Chng minhX

fd =

X

fd

Gii

1. V l hm o c, khng m nn

A

d tn ti, khng m.

Ch rngA

d = 0 nu (A) = 0, ta c () = 0

S dng tnh cht cng ca tch phn ta suy ra c tnh cng2. u tin ta kim tra rng ng thc ng khi f l hm n gin, khng m :

f =ni=1

ai1Ai , Ai Aj = (i 6= j),ni=1

Ai = X.

Tht vy X

fd =ni=1

ai(Ai)

X

fd =ni=1

ai

X

1Aid =ni=1

ai

Ai

d

T y ta c pcm.

Nu f o c, khng m th tn ti dy cc hm n gin fn0 fn fn+1, lim fn = f.

Ta c : X

fnd =

X

fnf n N (do bc trn)

lim

X

fnd =

X

fd, lim

X

fnd =

X

fd

(Do nh l Levi)T y ta c pcm.

6

Bi 5Cho cc hm f, g kh tch trn A. Vi n N ta t : An = {x A : n |f(x)| < n+ 1}Bn = {x A : |f(x)| n}Chng minh :

1. limn

An

gd = 0

2.n=1

n(An) < +

3. limn

n(Bn) = 0

Gii

Ta d kim tra c An Am = (n 6= m)n=0

An = A

1. Do tnh cht cng, ta c:n=0

An

gd =

A

gd R.

T y ta c pcm (do iu kin cn ca s hi t ca chui)

2. Cng do tnh cht cng, ta c:n=0

An

|f |d =A

|f |d

2. Gi s limn

fn(x) = f(x). Chng minh limn

fn = f trong (M,d).

Gii

1. Trc ht ta kim tra s d(f, g) hu hn vi mi cp f, g M . Tht vy, hm h =|f g|

1 + |f g| o c, b chn trn tp X v (X) < nn l hm kh tch. Kim traiu kin i), iii) ca metric nh sau :

i) Hin nhin d(f, g) 0

d(f, g) = 0 |f(x) g(x)|1 + |f(x) g(x)| = 0 hkn trn X

f(x) = g(x) hkn trn X f = g trong M

iii) Vi f, g, h M ta c :

|f(x) g(x)| |f(x) h(x)|+ |h(x) g(x)|

|f(x) g(x)|1 + |f(x) g(x)|

|f(x) h(x)|1 + |f(x) h(x)| +

|h(x) g(x)|1 + |h(x) g(x)|

(Phng php chng minh bit)Ly tch phn hai v ta c d(f, g) d(f, h) + d(h, g)

2. Ta cn chng minh limn

d(fn, f) = 0

t hn =|fn f |

1 + |fn f | (n N), ta c :

hn o c trn X, |hn| = hn 1, hm g(x) = 1 kh tch trn X (do (X)

V 2k1(Ak) Ak

fd 2k(Ak) ta c

1

2

+k=

2k(Ak) A

fd +

k=2k(Ak)

T y ta c iu phi chng minh.Bi 8Cho dy cc hm {fn} kh tch, hu hn trn A, hi t u trn A v hm f v (A) < .Chng minh f kh tch trn A v

limn

A

fnd =

A

fd

Gii

V cc hm fn o c nn f o c.V dy {fn} hi t u trn A v f nn c s no N tha mn

|fn(x) f(x)| 1 x A,n no (1). T (1) ta c |f(x)| 1 + |fn(x)|. V (A)

|fn(x)| = fn(x) 1 + x2 n Np dng nh l Lebesgue, ta c :

limn

20

fn(x)dx =

20

f(x)dx =10

3

2. t fn(x) l hm trong du tch phn th ta c

limn

fn(x) = f(x) vi f(x) = x, x [1, 0], f(x) = x2, x (0, 1].

|fn(x)| |x|+ x2enx

1 + enx 1 n N,x [1, 1].

3. t

fn(x) =

{ (1 + x

n

)n.e2x , x [0, n]

0 , x (n,+) fn (L)o c trn [0,). Vi mi x [0,) th x [0, n] khi n ln, do :

limn

fn(x) = limn

(1 +

x

n

)n.e2x = ex.e2x = ex.

|fn(x)| (1 +

x

n

)n.e2x ex.e2x = ex. n N,x [0,).

(ta s dng 1 + t et, t 0)Hm g(x) = ex l (L)kh tch trn [0,)p dng nh l Lebesgue ta c

limn

n0

(1 +

x

n

)n.e2x.dx = lim

n

+0

fn(x).dx =

+0

ex = 1

Bi 10

Chng minh limn

n

10

xn

1 + x.dx = 1

2

Gii

y ta khng th p dng nh l Lebesgue cho dy hm fn(x) =nxn

1 + xv khng tm c

hm g kh tch sao cho |fn(x)| g(x) n.Ta tch phn tng phn v c :

n

10

xn

1 + x.dx =

n

n+ 1

xn+11 + x

|10 +1

0

xn+1

(1 + x)2.dx

=

n

n+ 1

(1

2+ In

)p dng nh l Lebesgue ta chng minh c lim

nIn = 0.

10

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