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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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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
i
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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
ii
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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.
1
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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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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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Chng 1. Tch phn Lebesgue
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 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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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Chng 2. Khng gian Lp
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.