Tìm hiểu tích phan LEBESGUE và không gian Lp

7/30/2019 Tm hiu tch phan LEBESGUE v khng gian Lp

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I HC QUC GIA H NITRNG I HC KHOA HC T NHIN

KHOA TON - C - TIN HC

Trnh Thu Trang

TM HIU V TCH PHN LEBESGUEV KHNG GIAN Lp

KHA LUN TT NGHIP I HC H CHNH QUY

Ngnh: Ton - Tin ng dng

Ngi hng dn: TS. ng Anh Tun

H Ni - 2011


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3/60

LI CM N

Trc khi trnh by ni dung chnh ca kha lun, em xin by t lng bit

n su sc ti Tin s ng Anh Tun ngi thy tn tnh hng dn em

c th hon thnh kha lun ny.

Em cng xin by t lng bit n chn thnh ti ton th cc thy c gio

trong khoa Ton - C - Tin hc, i hc Khoa Hc T Nhin, i Hc Quc

Gia H Ni dy bo em tn tnh trong sut qu trnh hc tp ti khoa.

Nhn dp ny em cng xin c gi li cm n chn thnh ti bn b nhng

ngi lun bn cnh c v, ng vin v gip em.

c bit cho em gi li cm n chn thnh nht ti gia nh nhng ngi

lun chm lo, ng vin v c v tinh thn cho em.

H Ni, ngy 16 thng 05 nm 2011

Sinh vin

Trnh Thu Trang


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Mc lc

M u 1

1 Tch phn Lebesgue 3

1.1 i s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 o trn -i s tp hp . . . . . . . . . . . . . . . . . . 7

1.2.2 o Lebesgue . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Hm o c Lebesgue . . . . . . . . . . . . . . . . . . . . . . . . . 161.3.1 Hm o c Lebesgue . . . . . . . . . . . . . . . . . . . . . 16

1.3.2 Cc php ton v hm s o c . . . . . . . . . . . . . . 17

1.3.3 Cu trc hm o c . . . . . . . . . . . . . . . . . . . . . 19

1.3.4 Hi t hu khp ni . . . . . . . . . . . . . . . . . . . . . . 20

1.3.5 S hi t theo o . . . . . . . . . . . . . . . . . . . . . . 22

1.3.6 Mi lin h gia hi t . . . . . . . . . . . . . . . . . . . . . 24

1.4 Tch phn Lebesgue . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.4.1 Tch phn ca hm n gin . . . . . . . . . . . . . . . . . 28

1.4.2 Tch phn ca hm khng m . . . . . . . . . . . . . . . . . 29

1.4.3 Tch phn ca hm c du bt k . . . . . . . . . . . . . . 29

1.4.4 Cc tnh cht s cp . . . . . . . . . . . . . . . . . . . . . . 30

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MC LC

1.4.5 Qua gii hn di du tch phn . . . . . . . . . . . . . . . 33

1.4.6 Mi lin h gia tch phn Lebesgue v Rie mann . . . . . 36

2 Khng gian Lp 382.1 Khng gian Lp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2 Tnh tch c ca Lp . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.3 Bin i Fourier . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.3.1 Bin i Fourier trong L1 . . . . . . . . . . . . . . . . . . . 51

2.3.2 Bin i Fourier trong Lp . . . . . . . . . . . . . . . . . . . 52

Kt lun 55

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M u

Tch phn Lebesgue xut hin vo th k XX nhm gii quyt mt vi

nhc im ca tch phn Riemann, chng hn hm Dirichlet l hm n gin

nhng khng kh tch Riemann. C mt iu th v v tng xy dng hai

loi tch phn ny. Hai loi tch phn ny c xy dng da trn hai cch nhn

khc nhau v hm s: Bernhard Riemann nhn hm s bt u t min xc nh

cn Henri Lebesgue nhn hm s t tp gi tr. Kha lun ca em nhm tm

hiu cch xy dng tch phn Lebesgue v cc lp hm kh tch Lebesgue cng

nh c nhng so snh vi cc kt qu hc trong tch phn Riemann. Khalun c chia thnh hai chng.

Trong Chng 1, em trnh by cch thc xy dng tch phn Lebesgue t

o Lebesgue, hm o c Lebesgue ri tch phn Lebesgue v hm kh tch

Lebesgue. Trong chng ny c khi nim hi t hu khp ni v hi t theo

o l s m rng ca khi nim hi t im v hi t u. Em a vo

cc v d cho thy s khc nhau gia cc khi nim hi t ny. Phn gn cui

chng c cp n cc kt qu quan trng v vic chuyn gii hn qua du

tch phn ca Beppo Levi, Pierre Fatou, c bit ca Henri Lebesgue v hi t

chn. Em a v d cho thy kt qu hc Gii tch v vic chuyn gii hn

qua du ly tch phn c m rng thc s. Kt thc chng ny l kt qu

v mi quan h gia tch phn Lebesgue v tch phn Riemann.

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M u

Trong Chng 2, em trnh by khng gian Lp, 1 p v cc tnh cht.

y l lp khng gian Banach (nh chun, y ) hn na cn tch c (c

tp con m c tr mt) ngoi tr trng hp p =

. Sau khi trnh by cc

tnh cht c bn ny, em trnh by php bin i Fourier trong Lp, 1 p 2.

xy dng c php bin i Fourier em da vo Bt ng thc Hausdorff-

Young. Trong trng hp p > 2 em a vo v d cho thy Bt ng thc

ny khng cn ng.

Do thi gian c hn cng nh vic nm bt kin thc cn hn ch nn

trong Kha lun khng trnh khi thiu st, chng hn em cha a vo chngminh Bt ng thc Hausdorff -Young v chng minh ny i hi kh nhiu kin

thc chun b (L thuyt ni suy khng gian). Rt mong c s ch bo ca

thy c v bn b khp ni.

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Chng 1

Tch phn Lebesgue

1.1 i s

nh ngha 1.1.1. [1]Cho tp X l mt tp ty khc rng. Mt h C cc tp

con ca X c gi l i s cc tp con ca X, nuC tha mn ba iu kin:

i) X C,

ii) A C th X\A C,

iii) A1, A2, A3, . . . An C thn

k=1

Ak C.

Mnh 1.1.1. Cho C l i s tp con ca X th:

i) C,

ii) A1, A2, . . . An C thn

k=1

Ak C,

iii) A C, B C th A\B C.

Chng minh.

i) Do C l i s tp con ca X nn theo iu kin (i) ca i s X C.

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Chng 1. Tch phn Lebesgue

M i s kn vi php ly phn b nn X\X = C.

ii) Do A1, A2, . . . An C nn X\A1, X\A2, . . . X \An C. V C kn vi php hp hu

hn nnn

k=1

(X

\Ak)

C. Mt khc

n

k=1

(X

\Ak) = X

\(

n

k=1

Ak) nn X\

(n

k=1

Ak)

C.

M C kn vi php ly phn b nn X\(X\n

k=1

Ak) =n

k=1

Ak C. Vyn

k=1

Ak C.

iii) Ta c A\B = A (X\B). M A, X\B C nn A (X\B) C (theo tnh cht

2 va chng minh). Vy A\B C.

Mnh 1.1.2. Cho X = R, C = {n

i=1

i : i l gian, i = 1, 2,...,n,n N,

i

j =

vi i

= j

}l i s cc tp con caR.

Trong , gian trnR l mt tp im c mt trong cc dng sau

(a, b), [a, b], (a, b], [a, b), (, a), (, a], (a, +), [a, +), (, +) via, b R v

= [a, b] th || = a b c gi l di ca trnR.

Chng minh.

i)Chn 1 = (

, 0), 2 = [0, +

), 3 = (a, a) th R = 1

2

C v

= 3

C.

ii)Gi sA C th khi A l hp ca hu hn ca cc gian khng giao nhau.

Trng hp A l hp hu hn ca cc gian c dng i = (ai, ai+1) vi ai, ai+1 R.

Khng mt tnh tng qut, gi sa1 < a2 < .. . < a2n. Khi A =n

i=1

i v

R\A = (, a1] [a2, a3] ... [a2n, +)

=

n1i=1

[a2i, a2i+1] (, a1] [a2n, +),

cng l hp hu hn ca cc gian.

Mt cch xy dng tng t vi cc trng hp cn li ca tp A ta cng c

R\A cng l hp hu hn ca cc gian. Vy C kn vi php ly phn b.

iii) Gi sP, Q C. Trc ht ta chng minh P Q C.

t P =n

i=1Ii , Ii l mt gian Ii

Ii = vi i = i

.

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Q =k

j=1

Jj , Jj l mt gian Jj

Jj = vi j = j

. Khi

P

Q = P

(

k

j=1

Jj) =

k

j=1

(P

Jj) =

k

j=1

[(

n

i=1

Ii)

Jj ] =

k

j=1

n

i=1

(Ii

Jj).

M Ii Jj = Lij(i = 1, . . . n;j = 1, . . . k) l cc gian khng giao nhau i mt nnk

j=1

ni=1

Lij C hay P Q C.

Theo chng minh trn th R\P, R\Q C nn (R\P) (R\Q) C,

hay R\(P Q) C.

T chng minh (ii) trn c P

QC. S dng quy np ta c nu A1, A2, . . . An

C

thn

i=1

Ai C.

nh ngha 1.1.2. [1]Cho X l mt tp hp khc rng, mt h F cc tp con

ca X c gi l -i s, nuF tha mn ba iu kin:

i) X F,

ii) A F th X\A F,

iii) A1, A2, ...An, . . . F th+k=1

Ak F.

V d 1.1.1. Cho X = R, C = {n

i=1

i : i l cc gian ri nhau, i = 1,...n, n N}

khng l -i s.

Tht vy, t Ak = [2k, 2k + 1], kN th Ak

C. Ta cn i chng minh

k=1

Ak khng c dngn

i=1

i, vi i l mt gian.

S dng phn chng, gi s rng

k=1

Ak =n

i=1

i vi i l gian v ij = (i = j ).

Gi s1 c u mt l a1, a2 ; 2 c u mt l a3, a4; . . . ; n c u mt l

a2n1, a2n.

Do cc gian ri nhau nn khng mt tnh tng qut, gi sa1 < a2 < . . . < a2n1 < a2n.

Nu a2n < +, chn k0 sao cho 2k0 > a2n. Nh vy 2k0

k=1

[2k, 2k + 1] nhng

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2k0 /n

i=1

i. iu ny v l.

Nu a2n = +, chn k0 sao cho 2k0 > a2n1.

Nh vy 2k0 +3

2 n nhng 2k0 +

3

2/

k=1

[2k, 2k + 1]. iu ny v l.

Vy iu gi s l sai, C khng l -i s.

Ta s xy dng mt -i s nh nht cha C.

nh ngha 1.1.3. [1]-i s nh nht bao hm lp cc tp m trong khng

gianR c gi l -i s Borel ca khng gianR v nhng tp thuc-i s

ny c gi l tp Borel trong khng gianR.

Tp Borel l nhng tp xut pht t tp m v thc hin mt s hu hn hay

m c php ton hp, giao trn tp .

Theo nh ngha -i s mt tp l tp Borel th phn b ca n cng l

tp Borel. Do tp m l tp Borel nn tp ng cng l tp Borel.

Do -i s ng vi php hp v giao m c nn hp ca mt s m

c cc tp ng l mt tp Borel v giao ca mt s m c tp m cng

l tp Borel.

Mnh 1.1.3.

i) -i s Borel trong khng gianR cng l-i s nh nht bao hm lp cc

tp ng.

ii) -i s Borel trnR cng l -i s nh nht bao hm lp cc khong.

iii) -i s Borel trnR cng l -i s nh nht bao hm lp cc gian.

Chng minh. i) Cho M l lp cc tp m trong R. Gi F(M) l -i s nh

nht bao hm lp M hay -i s Borel. N l lp cc tp ng, F(N) l -i

s nh nht bao hm N. Ta c N F(M) nn F(N) F(M).

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Mt khc v mi tp m l phn b ca tp ng nn M F(N). Do

F(M) F(N). Vy F(M) = F(N) hay -i s nh nht bao hm lp cc tp

ng cng l -i s Borel.

ii) Cho M l lp cc tp m trong R, N l lp cc khong. V mi khong

u l tp m nn N F(M) vi F(M) l -i s nh nht bao hm M v

F(N) F(M).

M mi tp m l hp hu hn hay m c cc khong nn M F(N) v

F(M) F(N).

Vy F(M) = F(N) hay -i s nh nht bao hm lp cc khong cng l -i

s Borel.

iii) Cho G l lp cc gian, N l lp cc khong. Gi F(G), F(N) l -i s nh

nht bao hm mi tp . Do gian cha cc khong m nn F(N) F(G).

M mi gian li biu din c thnh hp hu hn hoc m c ca cc tp

m hoc ng v -i s nh nht bao hm lp cc tp m cng l -i s

nh nht bao hm cc tp ng. Do F(G) F(N).

Vy F(G) = F(N).

1.2 o

1.2.1 o trn -i s tp hpCho X l tp bt k trong khng gian R, F l -i s cc tp con ca X.

Xt hm tp : F [0, +].

nh ngha 1.2.1. [1] c gi l cng tnh nu

A, B

F, A

B =

, A

B

F th (A

B) = (A) + (B).

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nh ngha 1.2.2. [1] c gi l cng tnh hu hn nu c mt h hu hn

cc tp hp i mt ri nhau A1, A2, . . . An F th

(

ni=1

Ai) =

ni=1

(Ai).

nh ngha 1.2.3. [1] c gi l -cng tnh nu c mt h m c cc

tp hp i mt ri nhau A1, A2, . . . An,... F th

(

+i=1

Ai) =

+i=1

(Ai).

Mt hm -cng tnh th cng tnh nhng ngc li khng ng.

nh ngha 1.2.4. [1] l o trn -i s nu tha mn hai iu kin sau:

i) () = 0,

ii) l -cng tnh.

Tnh cht ca o

Vi l o trn F ta c cc tnh cht sau:

1. A, B F, A B th (A) (B).

V A B nn B = (B\A) A, B\A A = .

Do (B) = (B\A) + (A) (A).

2. Nu A, B F, A B, (A) < + th (B\A) = (B) (A).

V (B) = ((B\A) A) = (B\A) + (A) hay (B\A) = (B) (A).

3. Hp ca mt h m c cc tp c o bng 0 l tp c o bng 0.

Ta c (Ak) = 0 vi k = 1, 2, . . . , n . . . v l -cng tnh nn

(

k=1

Ak) =

k=1

(Ak) = 0.

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nh ngha 1.2.5. o c gi l o nu mi tp con ca tp c

o bng 0 u l tp o c v c o bng 0.

nh ngha 1.2.6. [1] Mt hm

xc nh trn mt lp tt c cc tp conca khng gianR, c gi l o ngoi nu:

i) (A) 0 vi mi A X,

ii) () = 0,

iii) A

k=1

Ak th (A)

k=1

(Ak).

nh l 1.2.1. [1](Caratheodory) Cho l o ngoi trn X, k hiu L l

lp tt c cc tp con A ca X sao cho

(E) = (E A) + (E\A) vi mi E X. (1.2.1)

Khi yL l -i s v hm = /L (thu hp ca trn L) l o trn L.

Chng minh. Trc ht ta chng minh L l mt -i s.

D nhin L v vi mi E X : (E) = () + (E) = (E ) + (E\).

Lp L cng kn i vi php ly phn b, v nu A L th vi mi E X ta c

(E) = (E A) + (E\A) = (E\(X\A)) + (E (X\A)).

chng minh L l -i s ta cn chng minh L kn vi php hp m c.

Cho Ai L, i = 1, 2, . . . v tp bt k E X. p dng ng thc 1.2.1, ta c:

(E) = (E A1) + (E\A1)

= (E A1) +

(E\A1) A2

+

(E\A1)\A2

= ...

=

k

j=1

(E\

j1

i=1 Ai) Aj

+

(E\

k

j=1 Aj).

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Do

(E) k

j=1

(E\j1i=1

Ai) Aj

+(E\

j=1

Aj).

V iu ny ng vi mi k nn

(E)

j=1

(E\j1i=1

AiAj) + (E\

j=1

Aj). (1.2.2)

Mt khc d dng nhn thy

E (

j=1

Aj) =

j=1

(E\

j1i=1

Ai) Aj

,

(v nu c mt j vi x E Aj th ly j l ch s nh nht nh vy ta c

x E\Ai vi mi i = 1, . . . , j 1).

Vy theo tnh cht di cng tnh (iii) ca :

(E) (E (

j=1

Aj)) + (E\

j=1

Aj)

j=1

((E\j1i=1

Ai) Aj) + (E\

j=1

Aj)

(E) (theo 1.2.2),

suy ra

j=1

Aj L, chng t L l -i s.

Cho Ai L, i = 1, 2, . . . l cc tp ri nhau. Ly E =

j=1

Aj. Khi E\

j=1

Aj =

v (E\j1i=1

Ai) Aj = Aj .

Ta c (

j=1

Aj)

j=1

(Aj ) theo (1.2.2),

M theo iu kin (iii) ca o ngoi ta c (

j=1

Aj)

j=1

(Aj).

Vy (

j=1

Aj) =

j=1

(Aj) hay trn L l mt o.

Nh vy nu xy dng mt o ngoi trn R tha mn mn nh l

Caratheodory th ta c mt o trn R. Ta xy dng o ngoi nh sau.

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Cho hm : R [0, +]

(A) = inf{+i=1

|i| :+i=1

i A, i l gian, i = 1, 2, . . .},

khi l mt o ngoi trn R.

Tht vy, hin nhin (A) 0 vi mi A R, () = 0.

Vi > 0 bt k, vi mi i = 1, 2, . . . ta ly mt h khong m k,i , k = 1, 2, . . .

sao chok,i

k,i Ai v

k

|k,i| (Ai) + 2i

. V A k,i

k,i ta c

(A)k,i |

k,i| i (

(Ai) +

2i) =

i

(Ai) + .

Do > 0 ty nn (A)

i=1

(Ai). Vy l o ngoi trn R.

1.2.2 o Lebesgue

nh ngha 1.2.7. [1]Cho hm : R [0, +]

(A) = inf{+i=1

|i| :+i=1

i A, i l gian, i = 1, 2, 3, . . .},

c gi l o ngoi Lebesgue trnR.

Hm tp l mt o ngoi trn R nh vy ta c th p dng nh l

Caratheodory xy dng mt o trn R, chnh l o Lebesgue.

nh ngha 1.2.8. Hm : L [0, ] trong L l lp tt c cc tp con A

caR sao cho

(E) = (E A) + (E\A) vi mi E R,

l o Lebesgue trnR, k hiu l v A c gi l tp o c Lebesgue.

Theo nh l Caratheodory th lp cc tp o c Lebesgue L l mt -i s.

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nh ngha 1.2.9. Tp A R c gi l tp o c Lebesgue trongR nuA

thuc -i s Lebesgue.

Vy tp khng o c Lebesgue s nh th no? Ta ly v d sau y t tiliu [4]

V d 1.2.1. Vi mi tp Ax = {y [0, 1] : x y = r, r Q} chn mt im.

Tp tt c cc im ny gi l P th P l mt tp khng o c.

nh ngha 1.2.10. [1] Tp N bt k c gi l tp c o 0 nu(N) = 0,

tc l sao cho

inf{

k=1

|k| :

k=1

k N, k l gian} = 0. (1.2.3)

nh l 1.2.2. [1] Mt tp N c o 0 khi v ch khi vi mi > 0 c th tm

c mt h (hu hn hay m c) giank ph N v c di tng cng nh

hn

+

k

k N, +k=1

|k| < .

Chng minh. Tht vy, nu (N) = 0 th theo cng thc (1.2.3) vi > 0 cho

trc c mt h khong m k ph N sao cho

k=1

|k| < .

Ngc li, nu vi mi > 0 u c mt ph nh vy th

inf{

k=1|k| :

k=1

k N, k l gian} = 0.

Vy N l tp c o 0.

V d 1.2.2.

1. Tp N = 1, 2, . . . , n l tp c o 0.

2. Tp cc s hu t c o 0.

3. Tp Cantor P trn [0, 1] xy dng theo cch di y c o 0.

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Xt tp hp [0, 1].

Bc 1. Chia [0, 1] thnh ba khong bng nhau, b i khong gia G1 = (1

3,

2

3).

Bc 2. Chia ba mi on cn li l [0,1

3

] v [2

3

, 1] b i khong gia ca chng.

t G2 = (1

9,

2

9) (7

9,

8

9) . . . Gi Gn l hp ca 2n1 cc khong b i bc th

n, G =

k=1

Gk l hp ca tt c cc khong b i, P = [0, 1]\G.

Ta c (Gn) = 2n1.(1

3)

n

=1

2.(

2

3)n.

Khi (G) =

n=1 (Gn) =1

2

n=1(

23)

n = 1.

M [0, 1] = ([0, 1]\G) G = P G nn ([0, 1]) = (P) + (G).Vy (P) = ([0, 1]) (G) = 1 1 = 0.

Ta thy tp c o 0 c th c lc lng l hu hn, m c hay khng

m c. Tp Cantor l mt tp c bit. Lc lng ca tp Cantor trn R l

khng m c nhng o ca n vn bng 0.

nh l 1.2.3. [1] o Lebesgue l o .Chng minh. Gi s(A) = 0 ta cn chng minh mi tp con ca A u o c

v c o bng 0.

Gi N l tp con ca A th 0 (N) (A). M (A) = 0 th (N) = 0. Li

c E = (E N) (E\N) nn (E) (E N) + (E\N) vi mi E R.

Do (E

N)

N nn (E

N)

(N) = 0 v (E)

(E

\N).

Mt khc (E\N) E nn (E\N) (E). Do (E) = (E\N), tc l

(E) = (E N) + (E\N).

Vy N l tp o c Lebesgue v (N) = (N) = 0.

nh l 1.2.4. Mi tp Borel u o c Lebesgue.

Chng minh. Trc ht ta i chng minh mi khong m u o c Lebesgue.

Ly mt khong m bt k. Xt mt tp E R ty v mt h gian k

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ph E. R rng vi mi k th k = k l gian v k\ =k,i

k,i l hp

cc gian.

Cho nn k

k = (k

k)(k

k,i) v k |

k

|=

k |

k

|+

k,i |

k,i

|.

Do

(E) = inf{

k

|k|}

= inf{

k

|k| +

k,i

|k,i|}

inf

{k |

k

|}+ inf

{k,i |

k,i

|},

Suy ra (E) (E ) + (E\), E R, hay o c Lebesgue.

Do l khong m bt k nn mi khong m u o c Lebesgue. M mi

tp m trong R l mt hp m c nhng khong m, nn -i s nh nht

bao hm lp cc khong m cng l -i s nh nht bao hm lp cc tp m,

tc l -i s Borel. M -i sL

l -i s bao hm lp cc khong. Vy

-i s L cha -i s Borel, hay tp Borel o c Lebesgue.

nh l 1.2.5. Mi tp o c Lebesgue l mt tp Borel thm hay bt mt

tp c o 0.

Chng minh. B l tp Borel v N l tp c o 0 th B, N L nn vi tp

A = B\N v A = B N cng o c Lebesgue.Ngc li gi sA L. Ta i chng minh tn ti tp Borel B sao cho (B) = (A).

V A L nn c th tm c cho mi k = 1, 2, ..., nhng khong m Pik sao cho

A

i=1

Pik v

i=1

(Pik) (A) + 1/k = (A) + 1/k.

t B =

k=1

i=1

Pij ta thy B A v B thuc i s Borel.

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Mt khc vi mi k, B

i=1

Pij nn

(B)

i=1

(Pik) (A) + 1/k.

Do (B) (A) m B A nn (B) = (A).

t N = B\A ta c (N) = (B\A) = 0.

V A L nn R\A L. Tn ti hai tp B l tp Borel v N l tp c o 0

sao cho R\A = B\N. Suy ra A = (R\B) N, hay A = B N vi B = R\B

l tp Borel.

Vy mi tp o c Lebesgue chng qua l mt tp Borel thm hay bt mt

tp c o 0.

nh l 1.2.6. i vi mt tp A trnR ba iu kin sau l tng ng:

i) A o c Lebesgue.

ii) Vi mi > 0 c th tm c tp m G A sao cho (G\A) < .

iii) Vi mi > 0 c th tm c mt tp ng F A sao cho

(A\F) < .Chng minh. (i) (ii). Trc ht ta xt trng hp (A) < . T nh ngha

o ngoi, vi > 0 cho trc c th tm c mt h khong m k ph

A sao cho

k

|k| < (A) + . ng nhin G l tp m bao hm A v c

(G) k

|k| < (A) + . T (G\A) = (G) (A), suy ra (G\A) < .

Trong trng hp tng qut, A =

n=1 A [n, n] v mi tp An = A [n, n] c

(An) < , nn theo trn c nhng tp m Gn An vi (Gn\An) < 1/2n. Khi

y tp G =

n=1

Gn m, bao hm A v tha mn

(G\A)

n=1

(Gn\An) 0 c th tm

c mt tp m G (R\A) sao cho (G\(R\A)) < . D nhin vi F l phn

b ca G th F A v (A\F) = (G\(R\A)) < . T suy ra (i) (iii).

1.3 Hm o c Lebesgue

1.3.1 Hm o c Lebesgue

nh ngha 1.3.1. Hm s f : A [, +] c gi l o c trn A vi

A l mt tp o c Lebesgue nu

a R, E1 = {x A : f(x) < a} L. (1.3.4)

nh l 1.3.1. iu kin (1.3.4) trong nh ngha tng ng vi cc ng

thc sau:

a R, E2 = {x A | f(x) > a} L. (1.3.5)

a R, E3 = {x A | f(x) a} L. (1.3.6)

a

R, E4 =

{x

A

|f(x)

a

} L. (1.3.7)

Chng minh. (1.3.4) (1.3.7) v E2 v E4 b nhau nn E4 L v L kn i vi

php ly phn b.

Tng t (1.3.5) (1.3.6) v E2, E3 b nhau.

(1.3.4) (1.3.6). Tht vy f(x) a khi v ch khi vi mi n c f(x) < a + 1n

.

Nn vi mi n {x A : f(x) a} =+

n=1{x A : f(x) < a +1

n} L,

v {x A : f(x) < a + 1n} L.

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Ngc li (1.3.6) (1.3.4). Tht vy f(x) < a khi v ch khi mi n c f(x) a 1n

.

Nn {x A : f(x) < a} =+n=1

{x A : f(x) a 1n} L, (v {x A : f(x)

a

1

n} L).

1.3.2 Cc php ton v hm s o c

Mnh 1.3.1. Cho A l tp o c Lebesgue.

i) Nu f(x) o c trn A th vi mi > 0 hm s |f(x)| cng o c.

ii) Nu f(x), g(x) o c trn A v hu hn th cc hm s

f(x) g(x), f(x).g(x),max{f(x), g(x)},min{f(x), g(x)}

cng o c, v nu g(x) khng trit tiu th hm s 1/g(x) cng o c.

Chng minh. i) Nu f(x) o c th vi mi a > 0

{x A : |f(x)| < a} = {x A : |f(x)| < a 1}

= {x A : a 1 < f(x) < a 1}

= {x A : f(x) < a 1} {x A : f(x) > a 1} L,

v mi tp {x A : f(x) < a 1} v {x A : f(x) > a 1} u thuc L.

Nu a

0 th{

x

A :

|f(x)

| < a

}=

L. Vy

|f(x)

| o c.

ii) Cho a l mt s thc bt k, r1, r2, r3, . . . , rn, . . . l dy cc s hu t. Khi

f(x) + g(x) < a f(x) < a g(x).

Do tp hu t tr mt trong tp s thc nn tn ti s hu t rn sao cho

f(x) < rn < a g(x). Nh vy

{x A : f(x) g(x) < a} =

n{x A : f(x) < rn < a g(x)}

=

n=1

{x A : f(x) < rn} {x A : g(x) < a rn} L,17


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v mi tp {x A : f(x) < rn}, {x A : g(x) < a rn} u thuc L.

Vy f(x) + g(x) l o c. Tng t ta c f(x) g(x) l o c.

Ta c cc h thc sau

f(x).g(x) =1

4[(f(x) + g(x))2 (f(x) g(x))2],

max{f(x), g(x)} = 12

(f(x) + g(x) + |f(x) g(x)|),

min{f(x), g(x)} = 12

(f(x) + g(x) |f(x) g(x)|).

Vy cc hm s f(x).g(x),max{

f(x), g(x)

},min

{f(x), g(x)

}cng o c.

nh l 1.3.2. Cho A l mt tp o c Lebesgue, fn : A R, n = 1, 2, 3 . . . l

nhng hm o c v hu hn trn A th cc hm

supn

fn(x), infn

fn(x), limn

fn(x), limn

fn(x)

cng o c trn A, v nu hm s limn

fn(x) tn ti th n cng o c.

Chng minh. Chn s thc a bt k c

{x A : supn

fn(x) a} =

n=1

{x A : fn(x) a} L,

{x A : infn

fn(x) a} =

n=1

{x A : fn(x) a} L.

Suy ra cc hm s supn

fn(x), infn

fn(x) o c.

Do ta c

limn

fn(x) = infn

{supk

fn+k(x)},

limn

fn(x) = supn

{infk

fn+k(x)},

cng o c.

Nu dy fn(x) hi t th limn fn(x) = limn fn(x). Vy limn fn(x) o c.

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1.3.3 Cu trc hm o c

nh ngha 1.3.2. Cho A l mt tp bt k trong khng gianR, ta gi hm

c trng ca A l hm sXA(x) xc nh nh sau

XA(x) =

0 nu x / A ,

1 nu x A .

nh ngha 1.3.3. Cho A l tp o c Lebesgue, hm f : A R c gi l

hm n gin nu n hu hn, o c v ch ly mt s hu hn gi tr. Gi

f1, f2, . . . , f n l cc gi tr khc nhau ca f(x) v Ai = {x : f(x) = fi} th tp Aio c, ri nhau v ta c

f(x) =

ni=1

fiXAi(x).

nh l 1.3.3. Mi hm f(x) o c trn tp o c A l gii hn ca mt

dy hm n gin fn(x),f(x) = lim

nfn(x).

Nu f(x) 0 vi mi x A th c th chn cc fn cho

fn(x) 0, fn+1(x) fn(x),

vi mi n v mi x

A.

Chng minh. t f(x) = 0 vi mi x / A ta c th coi nhf(x) xc nh v o

c trn ton R.

Nu f(x) 0. t

fn(x) =

n nu f(x) n,

i 12n

nu i 12n

f(x) < i2n

(i = 1, 2, . . . , n .2n).

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R rng fn(x) l hm n gin v fn(x) 0, fn+1(x) fn(x).

Ta cn chng minh f(x) = limn

fn(x).

Nu f(x) 0 limn+

({x A : |fn(x) f(x)| }) = 0.

Ni cch khc > 0, > 0 tn ti n0 N sao cho

n N : n > n0 th ({x A : |fn(x) f(x)| }) < .

nh l 1.3.5.

i) Nu f(x), g(x) o c v f(x), g(x) bng nhau h.k.n trn A, fn f trn A

th fn g trn A.

ii) Nu fn f trn A v fn g trn A th f(x), g(x) bng nhau h.k.n trn A.

Chng minh. i) V f(x), g(x) bng nhau h.k.n trn A nn tn ti mt tp

B = {x A : f(x) = g(x)} c o (B) = 0 (v f(x), g(x) o c nn B L).

Vi mi > 0 ta c:

An = {x A : |fn(x) g(x)| }

= {x A\B : |fn(x) f(x)| } {x B : |fn(x) g(x)| }

{x A\B : |fn(x) g(x)| } B.

M {x A\B : |fn(x) g(x)| } = {x A\B : |fn(x) f(x)| },

nn An {x A : |fn(x) f(x)| } B. Suy ra

(An) ({x A : |fn(x) f(x)| }) + (B)

= ({x A : |fn(x) f(x)| }) 0 khi n ,

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v fn f trn A. Do lim

n(An) = 0. Vy fn

g trn A.

ii) t A0 = {x A : |f(x) g(x)| > 0} = {x A : f(x) = g(x)},

A =

{x

A :

|f(x)

g(x)

|

}, > 0,

Ak = {x A : |f(x) g(x)| 1k}, k N,

Bn = {x A : |fn(x) f(x)| 2}, n N,

Cn = {x A : |fn(x) g(x)| 2}, n N.

Cc tp hp ny u o c v fn(x), f(x), g(x) u o c trn A.

Ta cn chng minh (A0) = 0.Trc ht ta chng minh

A0 =

k=1

Ak (1.3.8)

Ly x0 A0, ta c x A v |f(x) g(x)| > 0.

Theo tnh tr mt ca tp s thc s tn ti s t nhin k0 sao cho

|f(x) g(x)| 1k0

> 0,

suy ra x Ak0. Do x

k=1

Ak

Ngc li ly x

k=1

Ak th tn ti s t nhin k0 sao cho x0 A.

Suy ra x A v |f(x) g(x)| 1k0

nn |f(x) g(x)| > 0, do x A0.

Vy ng thc (1.3.8) c chng minh. Khi ta c

(A0)

k=1(Ak) (1.3.9)

By gi ta chng minh

A Bn Cn, n N, > 0 (1.3.10)

hay A\A A\(Bn Cn) = (A\Bn) (A\Cn). Tht vy, ly x (A\Bn) (A\Cn)

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Ta c x A v |fn(x) f(x)| < 2

, |fn(x) g(x)| < 2

. Suy ra

|f(x) g(x)| = |f(x) fn(x) + fn(x) g(x)|

|fn(x) f(x)| + |fn(x) g(x)| < .

Do x A\A. Vy (1.3.10) c chng minh.

Khi

(A) (Bn) + (Cn). (1.3.11)

M limn

(Bn) = 0, limn

(Cn) = 0.

V fn f, fn g trn A, nn ly gii hn hai v ca (1.3.11) ta c (A) = 0,

vi mi > 0.

Suy ra (Ak) = 0, khi =1

k> 0, vi mi k N.

T (1.3.9) ta c (A0) = 0.

1.3.6 Mi lin h gia hi tnh l 1.3.6. (Egorov) Cho mt dy hm{fn} o c, hu hn h.k.n, trn

mt tp o c A c o (A) < +. Vi mi > 0 tn ti mt tp o c

B A sao cho (A\B) < v dy hm{fn} hi t u trn tp B.

Trc ht ta chng minh b sau

B 1.3.1. Cho , > 0 th c mt tp ng B l con ca A v mt s thc

K sao cho (A\B) < v |f(x) fk(x)| < vi mi x F v k > K.

Chng minh. C nh , > 0. Cho m bt k, t Am = {x A : |f(x)

fk(x)| < vi mi k > m}. Nh vy Am =

k>m{x A : |f(x) fk(x)| < }

v Am l o c.

R rng Am Am+1. Ngoi ra fk(x) hi t h.k.n n hm s f(x) trn A v f(x)

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l hu hn nn Am tng n A\Z vi (Z) = 0. Do (Am) (A\Z) = (A).

T(A) < ta thy rng (A\Am) 0.

Chn m0 sao cho (A\

Am0) 0 p dng b 1.3.1 chn tp ng

Bm A, m 1 v s thc Km, sao cho (A\Bm) < 12m

v |f(x) fk(x)| < 1m

trong Bm nu k > Km,. B =m

Bm l tp ng v B Bm vi mi m nn fk(x)

hi t u ti f(x) trn B.

Suy ra A\B = A\m

Bm =m

(A\Bm). Vy (A\B)

(A\Bm) < .

nh l 1.3.7. Nu mt dy hm{fn} o c trn mt tp A hi t h.k.n ti

mt hm s f(x) th f(x) o c v nu (A) < th fn f.

Chng minh. {fn(x)} hi t h.k.n ti f(x) trn A nn tn ti B = {x A :

fn(x) f(x)}, (B) = 0 v mi tp con ca B cng o c v c o 0 (v

l o ). Do f(x) o c trn B.

Mt khc fn(x) f(x) vi mi x A\B nn theo nh l 1.3.2 f(x) o c

trn A\B.

Vy f(x) o c trn B (A\B) = A.

Chn > 0 ty . Vi mi n tn ti i sao cho |fn+i(x) f(x)| suy ra x B,

cho nn n=1

i=1

{x A : |fn+i(x) f(x)| } B,

do

n=1

i=1

{x A : |fn+i(x) f(x)| }

= 0.

t En =

i=1{x A : |fn+i(x) f(x)| }, ta c

E1 E2 . . . En . . .

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v E1 A nn (E1) (A) < nn (

n=i

En) = limn

(En),

do (

i=1

{x A : |fn+i(x) f(x)| }) (En) 0.

Vy fn

f.

nh l 1.3.8. Nu dy hm s o c fn(x) hi t theo o ti f(x), th c

mt dy con fnk(x) hi t h.k.n ti f(x).

Chng minh. Chn dy gk 0 v dy tk > 0 sao cho

k=1 tk < . Vi mi k

tn ti mt s t nhin n(k) sao cho vi mi n n(k)

({x : |fn(x) f(x)| gk}) < tk.

t n1 = n(1), n2 = max{n1 + 1, n(2)},... ta s c n1 < n2 < . . . v dy ny hi t

ti +. Vi mi k ta c

({x : |fnk(x) f(x)| gk}) < tk.

Xt tp B =

i=1

Qi vi Qi =

k=1{x : |fnk(x)f(x)| gk}. Vi mi i ta c B Qicho nn

(B) (Qi)

k=i

({x : |fnk(x) f(x)| gk}) n0 th x [0, 1 1n

] suy ra fn(x) = x = f(x)

Do fn(x) hi t im n f(x) trn [0, 1)

sup[0,1)

|fn(x)

f(x)

|= 1 +

1

n

1 khi n

.

fn(x) khng hi t u. Vy fn(x) hi t theo o nhng khng hi t u.

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Ta xt v d v mt hm hi t h.k.n nhng c o l v cng th s khng

hi t theo o.

V d 1.3.3. Chof

n xc nh trnR

fn(x) =

1 khi x [n, n + 1],

0 ti cc im khc,

v hm f(x) = 0.

Ly x R. Chn n0 = [x]+1 th vi mi n > n0 tc l n > [x]+1 > x th fn(x) = 0.

Nn limn fn(x) = 0 hay fn(x) hi t h.k.n n f(x) = 0 trn R v (R) = Chn =

1

2th

|fn(x) 0| 12

khi x [n, n + 1].

Do Bn = {x R : |fn(x) 0| 12} = [n, n + 1].

(Bn) = 1 1 khi n , hay fn(x) khng hi t theo o n f(x) = 0.

Vyf

n(x)

hi t h.k.n trnR

nhngf

n(x)

khng hi t theo o.

1.4 Tch phn Lebesgue

1.4.1 Tch phn ca hm n gin

nh ngha 1.4.1. Cho A l tp o c, f : A [, +] l hm n gin,

o c trn A. Gi f1, f2, . . . f n l cc gi tr khc nhau i mt ca f(x).

t Ak = {x A : f(x) = fk}, k = 1, ...n.

A =

nk=1

Ak v f(x) =n

k=1

fkXAk , x A.

Khi tch phn ca hm n gin f(x) trn A vi o l s

A

f(x)d =

n

k=1

fk(Ak).

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V d 1.4.1. Cho hm s f : [0, 1] R

f(x) =

1 khi x [0, 1] Q,

0 khi x [0, 1]\Q.

Khi

[0,1]

f(x)d = 1. ([0, 1] R) + 0. ([0, 1]\Q) = 1.0 + 0.1 = 0.

1.4.2 Tch phn ca hm khng m

Cho A l tp o c Lebesgue, hm f : A [0, +] l hm o c. Khi

tn ti dy n iu tng cc hm n gin o c fn(x) 0 hi t h.k.n v

f(x) trn A.

nh ngha 1.4.2. Tch phn ca hm f(x) trn A i vi o o l

A

f(x)d = limn+

(

A

fn(x)d).

1.4.3 Tch phn ca hm c du bt k

nh ngha 1.4.3. Cho A l tp o c Lebesgue, hm f : A R l hm o

c trn A. Khi ta c

f(x) = f+(x) f(x) vi f+(x), f(x) 0.

Cc hm sf+(x), f(x) c tch phn tng ng trn A lA

f+(x)d,A

f(x)d.

Nu hiuA

f+(x)d A

f(x)d c ngha th tch phn ca hm o c f(x)

trn A vi o l

A

f(x)d =

A

f+(x)d A

f(x)d.

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1.4.4 Cc tnh cht s cp

1. Cng tnh

Nu A B = th

ABf(x)d =

A

f(x)d +B

f(x)d.

Chng minh. Nu f(x) l hm n gin trn A B.

Tn ti f1, f2, . . . f n l cc gi tr khc nhau i mt ca f(x).

t Ek = {x A B : f(x) = fk}, k = 1, . . . n th thn

k=1

Ek = A B.

Ta c f(x) =n

k=1 fkXEk(x), Ek = (A B) Ek = (A Ek) (B Ek).

V A B = nn A Ek, B Ek ri nhau. Do AB

f(x)d =

nk=1

fk(x)(Ek)

=

nk=1

fk(x)(A Ek) +n

k=1

fk(x)(B Ek)

=

A

f(x)d +

B

f(x)d.

Nu f(x) > 0 trn tp A B, fn(x) l dy hm n gin khng m v hi

t ti f(x) ti mi im x A B.

Theo chng minh trn

AB

fn(x)d =

A

fn(x)d +

B

fn(x)d.

Suy ra limn

AB

fn(x)d = limn

A

fn(x)d + limn

B

fn(x)d.

Hay

ABf(x)d =

A

f(x)d +B

f(x)d.

Nu f(x) c du bt k.

Ta t f(x) = f+(x) f(x)

AB

f(x)d = AB

f+(x)d

ABf(x)d.

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Theo chng minh trn th

AB

f+(x)d =

A

f+(x)d +

B

f+(x)d. (1.4.12)

AB

f(x)d =

A

f(x)d +

B

f(x)d. (1.4.13)

Ly (1.4.12)-(1.4.13) ta c

AB

f(x)d =

A

f+(x)d

A

f(x)d +

B

f+(x)d

B

f(x)d

=

A

f(x)d +

B

f(x)d.

2. Bo ton th t

Nu f(x), g(x) bng nhau h.k.n trn A th

Af(x)d =

A

g(x)d.

3. Tuyn tnh

i)

Ac f(x)d = c

Af(x)d (c l hng s).

ii) f(x) + g(x) xc nh h.k.n trn A th

A

f(x) + g(x)

d =

A

f(x)d +

A

g(x)d.

4. Kh tch

i) Nu

A f(x)d c ngha th

A f(x)d A |f(x)|d.

ii)f(x) kh tch khi v ch khi |f(x)| kh tch.

iii) Nu |f(x)| g h.k.n trn A v g(x) kh tch th f(x) cng kh tch.

iv) Nu f(x), g(x) kh tch th f(x) g(x) cng kh tch. Nu f(x) kh tch,

g(x) b chn th f(x).g(x) cng kh tch.

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Chng minh. i)Ta c

|

A

f(x)d| = |

A

f+(x)d

A

f(x)d|

A

f+(x)d +

A

f(x)d

=

A

(f+(x) + f(x))d =

A

|f(x)|d.

ii)Nu |f(x)| kh tch th A

|f(x)|d < + suy ra | A

f(x)d| < + hay

f(x) kh tch.

Nu f(x) kh tch, A

f+(x)d A f

d < +

nn A

f+(x)d < +

,

Af(x)d < +.

Vy

A|f(x)|d =

A

f+(x) + f(x)

d < + hay |f(x)| kh tch.

iii) V |f(x)| g(x) h.k.n trn A, g(x) kh tch nn

A

|f(x)|d

A

g(x)d < +.

Do |f(x)| kh tch nn theo chng minh trn f(x) kh tch.iv) Nu f(x), g(x) kh tch th

A

f(x)d,

Ag(x)d hu hn.

A

(f(x) + g(x)) d =

Af(x)d +

A

g(x)d hu hn hay f(x) + g(x) kh tch.

Tng t ta c f(x) g(x) kh tch.

Nu f(x) kh tch, g(x) b chn. Gi s |g(x)| m.

Ta c|f(x).g(x)

| m.

|f(x)

|nn

A

|f(x).g(x)|d A

(m.|f(x)|)d = m. A

f(x)d.

Mt khc f(x) kh tch nn |f(x)| cng kh tch (chng minh trn).

Do A

|f(x)|d < suy ra A

|f(x)g(x)|d < hay |f(x)g(x)| kh tch. Vy

f(x)g(x) cng kh tch.

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1.4.5 Qua gii hn di du tch phn

nh l 1.4.1. (hi t n iu Beppo Levi) Nu fn(x) 0 v fn(x) n iu

tng n f(x) trn A th

limn

A

fn(x)d =

A

f(x)d.

Chng minh. Nu fn(x) l hm n gin th y chnh l nh ngha tch phn

ca hm n gin.

Nu hm fn(x) bt k v o c. Vi mi n c mt dy hm n gin, khng

m g(n)m (x) fn(x). V fn+1(x) > fn(x) nn c th coi g(n+1)m (x) g(n)m (x).

Vy vi k n ta c

g(k)n (x) g(n)n (x) fn(x),

A

g(k)n (x)d

A

g(n)n (x)d

A

fn(x)d,

cho n ta c fk(x) limn

g(n)n (x) f(x) v

A

fk(x)d

A

limn

g(n)n (x)d lim

n

A

fn(x)d.

Cho k ta c f(x) limn

g(n)n (x) f(x) v

limk

A

fk(x)d

A

limn

g(n)n (x)d lim

n

A

fn(x)d.

Nh vy limn

g(n)n (x) = f(x), lim

nA

fn(x)d = A

f(x)d.

Ch rng c th thay iu kin fn(x) 0 bi iu kin f1(x) kh tch.

nh l 1.4.2. (nh l Dini) Nu fn(x) l dy hm lin tc, n iu, hi t

im n mt hm f(x) lin tc trnR th fn(x) hi t u ti f(x).

Chng minh. Gi sfn(x) l dy hm n iu gim.

Cho mi n N, t gn(x) = fn(x)f(x) th gn(x) l hm lin tc v limn

gn(x) = 0.

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Hn na gn(x) > gn+1(x) > 0. Gi Mn = sup gn(x) : x R. Ta cn chng minh

limn

Mn = 0. Ly > 0, gi On = {xn : g1n (x) < }. V gn(x) l hm lin tc nn

tp On l tp m. T gn(x) > gn+1n (x) c On

On+1.

Vi mi x R, limn

gn(x) = 0 s c n N vi gn(x) < th x On. Vy

n=1

= R.

Ta c R l compact nn h tp m On ph R s cha mt h ca On hu hn

vn ph R. TOn On+1 nn s c mt tp hu hn ln nht ph R. Vy tn

ti s N N sao cho ON = R. Do gN(x) < vi mi x R. Vy MN . T

Mn gim vi mi n N nn vi mi Mn 0 ta s c limn

Mn = 0. Nh vy nu

fn(x) hi t u n f(x) trn tp A o c th limn

A fn(x)d =

A f(x)d.

V d 1.4.2. Cho fn(x) =x + 2

xn + 1vi x [0, 1].

Ta c fn(x) 0 vi x [0, 1] v fn(x) tng n hm f(x) = x + 2 trn [0, 1] nn

limn

[0,1]

fnd =

[0,1]

(x + 2)d.

nh l 1.4.3. (B Fatou) Nu fn(x) 0 trn A th

A

limn

fn(x)d limn

A

fn(x)d.

Chng minh. t gn(x) = inf{fn(x), fn+1(x), . . .}.

Ta c gn(x) 0 v gn(x) limn

fn(x), cho nn theo nh l 1.4.1 c

limn

A

gn(x)d =

A

limn

fn(x)d.

Nhng gn(x) fn(x) nnA

gn(x)d A

fn(x)d v limn

A

gn(x)d limn

A

fn(x)d.

Do A

limn

fn(x)d limn

A

fn(x)d.

nh l 1.4.4. (hi t chn Lebesgue) Nu |fn(x)| g(x), g(x) kh tch v

fn(x) f(x) (h.k.n hay theo o) trn A th

A

fn(x)d A

f(x)d.

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Chng minh. Trng hp fn(x) f(x) h.k.n trn A.

Ta c g(x) fn(x) g(x) v g(x) kh tch. Theo b Fatou cho cc hm

g(x)

fn(x)

0 v fn(x) + g(x)

0 ta c

A

limn

fn(x)d limn

A

fn(x)d,

A

limn

fn(x)d limn

A

fn(x)d.

Nhng fn(x) f(x) h.k.n trn A nn limn

fn(x) = limn

fn(x) = f(x), nn

A

f(x)d limn

A

fn(x)d limn

A

fn(x)d

A

f(x)d,

suy ra limn

A fn(x)d =

A f(x)d.

Trng hp fn f theo o. Theo nh ngha gii hn trn ta c mt dy nksao cho

A

fnk(x)d limn

A

fn(x)d.

Mt khc do fn(x) f(x) theo o nn c th trch ra mt dy con nki hi t

h.k.n n f(x) (theo nh l 1.3.8) hay fnki (x) f(x) h.k.n.Do

limn

Afn(x)d = limk

A

fnk(x)d = limi

Afnki (x)d =

A

f(x)d.

Tng t ta c

limn

A

fn(x)d =

A

f(x)d,

cho nn limn

A

fn(x)d =

Af(x)d.

Nh hc, chuyn gii hn qua du tch phn th iu kin cn l dy hm

{fn} hi t u n f(x) trn mt on [a, b]. y l mt iu kin ngt ngho.

Trong khi iu kin hi t n iu v hi t chn th rng ri hn.

V d 1.4.3. Cho hm fn : [0, 1] R, fn(x) =

nxenx2

.

Vi mi x [0, 1] ta c f(x) = limn

nxenx

2

= 0.

sup

[0,1]

|fn(x) f(x)| = sup[0,1]

nxenx

2

= (2e)1 0 khi n .

Nh vy fn(x) khng hi t u n f(x) trn [0, 1]. Mt khc10

f(x)dx = 0.

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limn

10

fn(x)dx = limn

10

nxenx

2

= limn

1

2(

1 enn

) = 0.

Do limn

10

fn(x)dx =10

f(x)dx.

L do c du = xy ra l v 0

fn(x)

g(x) = (2e)1 x

[0, 1], g(x) l mt

hm kh tch v fn(x) f(x) trn [0, 1] nn fn(x), f(x) tha mn iu kin ca

nh l hi t chn. Nh vy ta c th chuyn gii hn qua du tch phn.

1.4.6 Mi lin h gia tch phn Lebesgue v Rie mann

nh l 1.4.5. Cho f : [a, b] R

Nu f kh tch Riemann trn [a, b] th kh tch Lebesgue trn [a, b] v

[a,b]

f(x)d =

ba

f(x)dx

Chng minh. Nu f(x) kh tch Riemann trn on [a, b] th f(x) b chn v lin

tc h.k.n trn [a, b].

Xt dy phn hoch Dn ca on [a, b] vi||

Dn||

0. Gi Sn l tng Darboux

di ng vi phn hoch Dn.

Gi i l phn hoch th i ca phn hoch Dn trn [a, b].

Sn =

ni=1

ti|i| ti = infxi

f(x).

t fn(x) =n

i=1tiXi(x). Ta c f(x) lin tc h.k.n trn [a, b], vi n ln th

||Dn|| 0 v do i s nh Vi > 0 bt k, |f(x) fn(x)| = |f(x) ti| < . Suy ra fn(x) f(x) h.k.n trn

[a, b]. Do theo nh l v hi t chn ta c

Sn =

[a,b]

fn(x)d [a,b]

f(x)d.

Mt khc Sn

b

af(x)dx.

Vy[a,b]

f(x)d =b

af(x)dx.

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Tuy nhin mt hm kh tch Lebesgue th khng kt lun c s kh tch

Riemann. Ta c v d 1.4.1 v hm Dirichlet kh tch Lebesgue trn [0, 1] nhng

khng kh tch Riemann trn .

V d 1.4.4. Cho hm s D : [0, 1] R

D(x) =

1 khi x [0, 1] Q,

0 khi x [0, 1]\Q.

Chng minh. Ta s i chng minh hm Dirichlet khng kh tch Riemann.

Tht vy, vi mi phn hoch T ca on [0, 1], gi i l phn hoch th i ca

T. Nu ly l im c ta l nhng s hu t th D(T, ) =n

i=1

D(i)|i| = 1

cn nu ly

l im c ta l nhng s v t th D(T,

) = 0 nn hm

Dirichlet khng kh tch Riemann.

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Chng 2

Khng gian Lp

2.1 Khng gian Lp

Cho khng gian R, E l tp o c Lebesgue v mt o .

nh ngha 2.1.1. [1] H cc hm s f(x) c ly tha bc p (1 p < ) ca

modun kh tch trn E, tc l sao choE

|f(x)|pd < ,

gi l khng gian Lp(E).

Hm s f(x) o c trn E gi l b chn ct yu nu tn ti mt tp hp

P c o 0, sao cho f(x) b chn trn tp hp E\

P, tc l tn ti s K sao cho

|f(x)| K vi mi x E\P.

Cn di ng ca tp hp cc s K tha mn bt ng thc trn gi l cn

trn ng ct yu ca hm f(x), c k hiu l ess supE

|f(x)|.

nh ngha 2.1.2. [2] H tt c cc hm f(x) b chn ct yu trn E c gi

l khng gian L(E).

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Chng 2. Khng gian Lp

Mnh 2.1.1. [2] Nu hm f(x) L(E) th

|f(x)| ess supE

|f(x)| h.k.n trn E.

Chng minh. Gi s{Kn} l mt dy s thc n iu gim n K = ess supE

|f(x)|.

Khi , tp hp

Pn = {x E : |f(x)| > Kn},

c o 0 vi mi n. Hin nhin tp hp P =

n=1

Pn c o 0 v

|f(x)| K vi mi x E\P.

Vy |f(x)| ess supE

|f(x)| h.k.n trn E.

nh l 2.1.1. [1] (Bt ng thc Holder) Nu f(x), g(x) o c, xc nh trn

mt tp o c E v p, q l hai s thc sao cho 1 < p < v 1p

+1

q= 1 th

E

|f(x) g(x)|d E

|f(x)|pd1p

E|g(x)|qd

1q

.

Cch chng minh da vo b sau

B 2.1.1. Nu a,b khng m v p, q l hai s thc sao cho 1 < p < v1

p+

1

q= 1 th ta c

ab

ap

p

+bq

q

.

Chng minh. Xt f(t) =t

p+

1

q t1/p (t 0).

Ta thy f(1) = 0 v f

(t) =1

p 1

p t1/(p1) dng vi t > 1, m vi t < 1, chng

t rng f(t) t cc tiu ti t = 1.

Do vi mi t 0 ta c tp

+1

q t1/p 0. Vi t = apbq 0 ta c

apbq

p

+1

q abp/q

0 hay

ap

p

+bq

q ab

0.

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Chng minh. ( Chng minh bt ng thc Holder)

NuE

|f(x)|d, E

|g(x)|d hu hn v dng. p dng bt ng thc va chng

minh vi

a =|f(x)|

E|f(x)|pd 1p , b =

|g(x)|E

|g(x)|qd 1q .Ta c

|f(x).g(x)|E

|f(x)|pd 1p E

|g(x)|qd 1q |f(x)|p

p

E|f(x)|pd +

|g(x)|qq

E|g(x)|qd .

Ly tch hai v

E

|f(x)g(x)|dE

|f(x)|pd 1p

E|g(x)|qd

1q

E|f(x)|pd

p

E|f(x)|pd

+

E|g(x)|qd

q

E|g(x)|qd

=1

p+

1

q= 1.

NuE

|f(x)|d hoc E

|g(x)|d bng v cng th bt ng thc ng.

Nu E |

f(x)|d hoc

E |g(x)

|d bng 0, chng hn

E |f(x)

|d = 0 th f(x) = 0

h.k.n trn E nn f(x)g(x) = 0 h.k.n trn E. Do bt ng thc ng.

nh l 2.1.2. [1](Bt ng thc Minkowski) Nuf(x), g(x) l hai hm o c

trn E v 1 p < th

E

|f(x) + g(x)|pd 1

p

E

|f(x)|pd 1

p

+

E

|g(x)|pd 1

p

.

Chng minh. Vi p=1, bt ng thc ng.

Xt vi 1 < p < . Chn q sao cho 1p

+1

q= 1.

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p dng bt ng thc Holder ta c

E

|f(x) + g(x)|pd

E

|f(x)| + |g(x)|

.|f(x) + g(x)|p1d

=

E

|f(x)|.|f(x) + g(x)|p1d +

E

|g(x)|.|f(x) + g(x)|p1d

E

|f(x)|pd 1

p

E

|f(x) + g(x)|(p1)qd 1

q

+

E

|g(x)|pd 1

p

E

|f(x) + g(x)|(p1)qd 1

q

=

E

|f(x)|pd 1

p

+

E

|g(x)|pd 1

p

E

|f(x) + g(x)|pd 1

q

.

T

E

|f(x) + g(x)|pd1 1

q

E

|f(x)|pd 1

p

+

E

|g(x)|pd 1

p

.

v 1 1q

=1

pnn

E

|f(x) + g(x)|pd 1p E

|f(x)|pd 1p + E

|g(x)|pd 1p .

nh l 2.1.3. Tp hp Lp

(E), trong khng phn bit cc hm bng nhauh.k.n l mt khng gian vector nh chun, vi cc php ton thng thng v

cng hm s, nhn hm s vi s, v vi chun

||f||p =

E

|f(x)|pd 1

p

khi 1 p < ,

||f|| = ess supE

|f(x)| khi p = .

Chng minh. Vi 1 p < .

Gi sf(x), g(x) Lp(E). Ta c |f(x) + g(x)| 2max{|f(x)|, |g(x)|}

suy ra |f(x) + g(x)|p 2p(max{|f(x)|, |g(x)|})p 2p|f(x)|p + |g(x)|p.Do |f(x)|p, |g(x)|p kh tch th |f(x) + g(x)|p cng kh tch, hay

f(x) + g(x)

Lp(E) (2.1.1)

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Mt khc, nu f(x) Lp(E), l s thc bt k th |f(x)|p = ||p.|f(x)|p kh

tch, nn

f(x)

Lp(E) (2.1.2)

T (2.1.1) v (2.1.2) suy ra Lp(E) l khng gian vector.

Ta li c||f||p > 0 khi f(x) = 0 v ||f||p = 0 khi f(x) = 0 h.k.n v khng phn bit

hai hm bng nhau h.k.n nn tha mn tin 1 v chun.

||.f||p =

E

|f(x)|pd 1

p

= ||

E

|f(x)|pd 1

p

= ||.||f||p,

tha mn iu kin thun nht ca chun.

p dng bt ng thc Minkowski ta c bt ng thc tam gic

||f + g||p =

E

|f(x) + g(x)|pd 1

p

E

|f(x)|pd 1

p

+

E

|g(x)|pd 1

p

= ||f||p + ||g||p.

Vi p = +.Gi sf(x) L(E).Ta c ||f|| > 0 nu f(x) = 0 v ||f|| = 0 nu f(x) = 0 h.k.n trn E nn tha

mn tin 1 v chun.

Ta i chng minh iu kin thun nht ca chun. Vi l s thc bt k ta

cn chng minh

||.f

|| =

|

|.

||f

||. (2.1.3)

Nu = 0 th iu kin (2.1.3) lun ng.

Nu = 0. Gi s phn chng ||.f|| < M < ||.||f||. Ta c

||.|f(x)| = |.f(x)| ess supE

|.f(x)| < M.

Do .f(x) L(E) v ||.||f|| < M. Nh vy th ||.||f|| < M < ||.||f||.

iu ny v l. Chng minh mt cch tng t ta cng c ||.||f|| < M < ||.f||

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l iu v l hay ||.f|| = ||.||f||.

Cui cng ta i chng minh iu kin v bt ng thc tam gic.

Tht vy, gi sf(x), g(x)

L(E).

|f(x)+g(x)| |f(x)|+|g(x)| ess supE

|f(x)|+ess supE

|g(x)| nn f(x)+g(x) L(E).

V ess supE

|f(x)+g(x)| ess supE

|f(x)|+ess supE

|g(x)| hay ||f+g|| ||f||+ ||g||.

nh l 2.1.4. Lp(E) l khng gian vector nh chun .

Chng minh. Cho fn(x) l dy c bn trong Lp

(E), tc l ||fn fm||p 0. Nhvy lun tm c n1 ln cho ||fn fm||p < 1/2 vi mi n, m n1. Tip

tc tm c n2 > n1 sao cho ||fn fm||p < 1/22 vi mi n, m n2. Do ta c

th chn c mt dy n1 < n2 < .. . < nk < . . . cho vi mi k ta c

n, m nk ||fn fm||p < 1/2k.

Nh vy ||fnk+1 fnk ||p < 1/2k. p dng b Fatou (1.4.3) cho dy hm khng

m gs(x) = |fn1(x)| +s

k=1 |fnk+1(x) fnk(x)| Lp(E).

Vi mi x c nh gs(x) khng gim theo s nn tn ti limn

gs(x).

Do E

lims

(gs(x))pd lim

s

E

(gs(x))pd = lim

s(||gs||p)p.

Mt khc

||gs||p ||fn1||p +s

k=1

||fnk+1 fnk ||p

< ||fn1||p +s

k=1

(1/2k) < ||fn1||p + 1.

nn lims

(||gs||p)p < . Do

Elim

s(gs(x))

p d < .

iu ny chng t lims

|gs(x)|p < h.k.n, hay tn ti lims

gs(x) v hu hn

h.k.n trn E.

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Suy ra fn1(x) +

k=1

fnk+1(x) fnk(x)

hi t tuyt i h.k.n.

Nh vy khi s s tn ti gii hn hu hn h.k.n ca hm

fns+1(x) = fn1(x) +

sk=1

fnk+1(x) fnk(x) .

Ta gi gii hn ny l f0(x), fns+1 f0(x) h.k.n. V |fns+1(x)| lims

gs(x) Lp(E).

Theo nh l hi t chn c

E

|f0(x)|pd = lims

E

|fns+1(x)|pd,

tc l f0(x) Lp(E). p dng b Fatou ta c

||f0 fnk ||p =

E

lims

|fns+1(x) fnk(x)|pd

limn

E

|fns+1(x) fnk(x)|pd = lims

||fns+1 fnk ||p

= lims

||s

t=k

fnt+1 fnk ||p lims

st=k

||fnt+1 fnk ||p

lims

st=k

12t

=

t=k

12t

.

Suy ra limk

||f0 fnk ||p = 0. Cui cng v {fn} l dy c bn nn vi n, nk

ln ta s c ||fnk fn||p < . Khi ta chn k ln va c nk n0 v

||f0 fnk ||p < th s c vi mi n n0

||f0

fn||p ||

f0

fnk ||

+||

fnk

fn||

+ ,

chng t dy fn(x) hi t ti f0(x).

H qu 2.1.1. Nu mt dy{fn} hi t trong Lp(E) th n cha mt dy con

{fnk} hi t h.k.n trn E.

Tht vy, nu

{fn

}hi t th n l dy c bn nn theo nh l 2.1.4 c th

trch ra mt dy con {fnk} hi t h.k.n.

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nh l 2.1.5. Nu 1 p < p < v (E) < th Lp

(E) Lp(E).

Chng minh. p dng bt ng thc Holder ta c

||f.g||1 ||f||p.||g||q vi 1 p < , 1p

+1

q= 1.

Ta ly g(x) = 1, thay f(x) bi |f(x)|p, thay p, q bi p/p,p/(p p) ta c

|| |f| ||1 || |f|p || pp

.(E)p

p

p .

Do (E) 0, t B = {x : |fn(x) f(x)| }, ta cE

|fn(x) f(x)|pd

B

|fn(x) f(x)|pd

B

pd = p.(B),

iu ny chng t rng (B) 0 khi n , nu fn(x) Lp

f(x).

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Mi quan h gia hi t

Hi t hu khp ni (A) < Hi t theo o

Hi t trung bnh

( ch ra rng c th trch ra mt dy con hi t).

Mt hm hi t trung bnh th hi t theo o nhng iu ngc li khng ng.

V d 2.1.3. Xt dy hm fn(x) trn on [0, 1] c xc nh nh sau

fn(x) =

n nu x 0,

1

n

0 nu x

1

n, 1

.

v hm f(x) = 0.

Ta c

({x [0, 1] : |fn(x) f(x)| }) = ([0, 1n

]) =1

n 0 khi n .

Do fn(x) f(x).Tuy nhin,

[0,1]

|fn(x) f(x)|d = n.

0,1

n

= 1 0 khi n nn fn(x)

khng hi t trung bnh n f(x) trn [0, 1].

V d sau cho thy mt hm hi t h.k.n nhng khng hi t trung bnh.

V d 2.1.4. Cho hm fn(x) xc nh trn [0, 1]

fn(x) =n

1 + n2x2,

v hm f(x)=0.

Ta c limn

fn(x) = limn

n

1 + n2x2= lim

n

11

n+ nx2

= 0. Do fn(x) hi t

h.k.n n f(x) = 0 trn [0, 1]. Tuy nhin

limn

10

fn(x)dx = limn

10

n1 + n2x2

dx.

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t u = nx ta c

limn

n0

1

1 + u2du = lim

n(arctan u)|n0 = lim

narctan n =

2.

Vy fn(x) khng hi t trung bnh v f(x) = 0.

2.2 Tnh tch c ca Lp

nh l 2.2.1. Mi h hm sau y l tr mt trong Lp(E), 1 p < .

1. Cc hm n gin.

2. Cc hm lin tc.

Chng minh. H cc hm gi l tr mt trong Lp(E) nu vi mi f(x) Lp(E)

v mi > 0, u tn ti mt hm g(x) thuc h sao cho ||f g||p < .

a) Xt mt hm bt k f(x)

Lp(E).

Ta c f(x) = f+(x) + f(x), vi f+(x), f(x) 0.

Tn ti mt dy hm n gin khng m f+n (x) f+(x).

V

f+(x) f+n (x)p 0 nn

E

(f+(x) f+n (x))pd

E

0d = 0,

ngha l ||f+ f+n ||p 0.Vy vi n ln ta s c mt hm n gin fn (x) vi ||f fn ||p 0 tn ti mt tp m G A v mt tp ng F A sao cho (G\A) < p

2,

(A\F) < p

2, tc l (G\F) < p. Ta ly

g(x) =(x,R

\G)

(x,R\G) + (x, F) .

Trong (x, M) = infyM

||x y|| ch khong cch t x n M. R rng x R\G

th (x,R\G) = 0 nn g(x) = 0. Vi x F th (x,R\G) = 0, (x, F) = 0 nn

g(x) = 1. Hm (x,R\G), (x, F) u lin tc v c tng khc 0 nn g(x) cng

lin tc. Hiu XA(x) g(x) c gi tr gm 0 v 1 trn tp G\F v bng 0 ngoi

tp . Cho nn

||XA g||p =

E

XA(x) g(x)pd 1

p

(G\F) 1p < .Do mi hm c dng XA(x) vi A o c u c th xp x ty bi

mt hm lin tc. Suy ra h cc hm lin tc tr mt trong h cc hm n

gin, v mi hm n gin c dng

vi=1 iXAi(x) ch vic chn hm lin tc

gi(x) sao cho ||XAi gi||p < v|i| , th s c hm lin tc g(x) =

igi(x) vi

||iXAi g||p v1 i||XAi g||p < .Mt khc mt h M tr mt trong Lp(E) v mt h N tr mt trong M th h

N cng tr mt trong Lp(E). Vy h cc hm lin tc tr mt trong Lp(E).

nh l 2.2.2. Khng gian Lp(R), 1 p < tch c (ngha l cha mt tp

con m c tr mt trong n).

Chng minh. Ly f(x) Lp(R). Gi fn0 : R R xc nh

fn0(x) =

f(x) khi |x| < n0,

0 khi |x| n0.

th fn0(x) f(x) Lp(R) khi n .

Nh vy tn ti s n0 sao cho ||f fn0 ||p 3 v tn ti hm lin tc

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gn0 : [n0, n0] R sao chon0n0

|fn0(x) gn0(x)|pdx 1

p

0 sao cho |z| < 1 th

|ez 1| < 2 2||f||1 ,

nn vi |x| < R0 v |( 0)| < 1R0

th

|x| 0 sao cho

|

0

|< th

|Ff() Ff(0)| < 12

2

2+

2

2

= .

Vy Ff L(R) C(R).

2.3.2 Bin i Fourier trong Lp

Ta xy dng bin i Fourier trong Lp(R) nh sau. t

X[R,R](x)f(x) =

0 nu |x| > R

f(x) cn li .

Ta c |X[R,R](x)f(x) f(x)| 0 khi R nn |X[R,R](x)f(x) f(x)|p 0 khi

R

.

M |X[R,R](x)f(x) f(x)|p |f(x)|p, |f(x)|p Lp(R).

Theo nh l v hi t chn ta c

R

|X[R,R](x)f(x) f(x)|pd 0 khi R .

Do X[R,R](x)f(x) Lp(R) l dy Cauchy hi t n f(x) trong Lp(R).

Ta cn chng minh {F(X[R,R]f)} hi t trong Lq(R) vi 1 p 2, 1p + 1q = 1.

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Theo bt ng thc Hausdorff-Young ta c

Vi g(x) L1(R) Lp(R), 1 p 2, 1p

+1

q= 1 th Fg Lp(R) v

||Fg||q C||g||p.

Nh vy vi g(x) = X[R,R](x)f(x) ta c {F(X[R,R]f)} l dy Cauchy v hi t

n mt hm Ff trong Lq(R) vi 1 p 2, 1p

+1

q= 1.

Nhng vi 2 < p < vic xy dng nh th ny khng cn ng. Ngi ta phi

xy dng bin i Fourier theo cch khc. V d sau cho thy vi 2 < p < bt

ng thc Hausdorff-Young nh trn khng cn ng.

V d 2.3.1. [3] Chn dy fk(x) = e(1ik)x2

, k = 1, 2 . . .. Bin i Fourier ca

f(x) l

Ff() = (2)12

R

eixe(1ik)x2

d.

Bin i ny tha mn bi ton Cauchy

2(1 ik)dFfkd

() Ffk() = 0,

vi iu kin ban u lFfk(0) =1

2(1 ik)nnFfk() =

12(1 ik)

e (1+ik)

2

4(1+k2) .

Khi chun trong Lp(R) l

||fk||p = 1/2p

p1/2p

Cn chun trong Lq(R) l

||Ffk||q = 21q 12 (1 + k2)

12q

14

12p

q12q

.

V 2 < p < nn 1 q < 2 nn ||Ffk||q khi k .

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Kt lun

Kha lun trnh by hai ni dung chnh l:

Xy dng tch phn Lebesgue t o Lebesgue, hm o c Lebesgue,tch phn Lebesgue, chuyn gii hn qua du ly tch phn.

Khng gian Lp, 1 p , tnh cht y v tch c, php bin i

Fourier trong Lp, 1 p 2.

Tuy nhin do thi gian lm kha lun cn hn ch khng th trnh c nhng

nhm ln, sai st nn em mong nhn c s gp ca thy c v bn c.

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Ti liu tham kho

Ting Vit

[1] Hong Ty (2005), Hm thc v gii tch hm, Nh xut bn i hc Quc

gia H Ni.

[2] Nguyn Xun Lim (2007), Gii tch hm, Nh xut bn Gio dc.

Ting Anh

[3] Elliott H. Lieb and Michael Loss (2001), Analysis (second edition), American

Mathematical, Society.

[4] Frank Burk (1998), Lebesgue Measure and Integral An Introduction, John

Wiley & Sons, Inc.

[5] Richard L. Wheeden and Antoni Zygmund (1977), Measure and Integral an

Introduction to Real Analysis, Marcel Dekker, Inc. New York.

Ting Php

[6] H. Brezis (1983), Analyse Fonctionnelle, Masson.

Tìm hiểu tích phan LEBESGUE và không gian Lp

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