Li Thuyet Ham Suy Rong & Khong Gian Sobolev - Dang Anh Tuan

L thuyt Hm suy rng v Khng gian Sobolev

ng Anh Tun

H Ni, ngy 20- 11- 2005

Chng 1

Cc khng gian hm c bn v khng

gian hm suy rng

1.1 Mt s kin thc b sung

1.1.1 Mt s k hiu

N = {1, 2, . . . } l tp cc s t nhin, Z+ = {0, 1, 2, . . . } l tp cc s nguyn khngm, R l tp cc s thc, C l tp cc s phc. n v o

1 = i.Vi mi s t nhin n N, tp Zn+ = { = (1, . . . , n)

j Z+, j = 1, . . . , n}, tpRn = {x = (x1, x2, . . . , xn)

xj R, j = 1, 2, . . . } l khng gian thc n chiu vi chunEuclid

x = ( nj=1

x2j) 12 .

Nu khng c g c bit, k hiu l tp m trong Rn.Vi mi k Z+ k hiu cc tp nh sau:

Ck() = {u : Cu kh vi lin tc n cp k}, C() = C0() = {u : lin tc C},Ck0 () = {u : C

u Ck(), suppu l tp compact}, C0() = C00(),C() = k=1Ck(), C0 () = k=1Ck0 (),

trong , suppu = cl{x u(x) 6= 0}.Vi mi s thc 1 p

2trong , ess supx |u(x)| = inf{M > 0m{x |u(x)| > M} = 0}.Vi 1 p , k hiu

Lploc() = {u : Lebesgue

Cu Lp(), vi mi tp con o c }

Lpcompact() = {u : Lebesgue

Cu Lp(), : u(x) = 0 h.k.n. trong \ },trong , ngha l bao ng cl() l tp compact trong .Vi mi hm u C(), = (1, 2, . . . , n) Zn+ k hiu

Du = D11 D22 . . . D

nn u,D

jj =

j

xjj

, j = 1, 2, . . . .

Khi , vi u, v C(), = (1, 2, . . . , n) Zn+ c cng thc Leibnitz

D(uv) =

(

)DuDv,

trong ,

(

)=n

j=1

(jj

),(jj

)=

j !

j !(jj)! ,l tng ly trn tp cc a ch s Zn+m , ngha l 0 j j, j = 1, 2, . . . , n.

1.1.2 Phn hoch n v

nh ngha 1.1. Cho l mt tp trong Rn.Mt h m c cc cp {(j, j)}j=1, trong j l tp m trong Rn, j l hm thuc lp cc hm kh vi v hn trn Rn, c gi lmt phn hoch n v ca tp nu cc tnh cht sau c tho mn:

(i) {j}=1 l mt ph m ca , ( j=1j,j l tp m),(ii) 0 j(x) 1,x , j = 1, 2, . . . ,(iii) j C0 (Rn), suppj j, j = 1, 2, . . . ,(iv)

j=1 j(x) = 1,x .Ta cn gi {j}j=1 l phn hoch n v ng vi ph m {j}j=1 ca tp .nh l 1.1. Cho K l mt tp compact trong Rn, h hu hn {Uj}Nj=1 l mt ph m caK. Khi , tn ti mt h hu hn cc hm kh vi v hn {j}Nj=1 xc nh mt phn hochn v ng vi ph m {Uj}Nj=1 ca tp K. chng minh nh l ta cn mt s kt qu sau.

T y tr i, k hiu hm : Rn R l hm c xc nh nh sau

(x) :=

{Ce

1||x||21 , nu||x|| < 1,

0, nu ||x|| 1,trong , C l hng s sao cho

Rn (x)dx = 1.

rng, hm c cc tnh cht sau

3(i) C0 (Rn), supp = B1(0) = {x Rn||x|| 1}, (x) 0,x Rn,(ii)

Rn (x)dx = 1, l hm ch ph thuc vo ||x||(radial function).Vi mi > 0, t (x) =

n(x). Hm cng c cc tnh cht ca hm :

(i) C0 (Rn), supp = B(0) = {x Rn||x|| 1}, (x) 0,x Rn,(ii)

Rn (x)dx = 1, l hm ch ph thuc vo ||x||(radial function).Vi mi hm f L1loc(Rn), t

f(x) = (f )(x) =Rnf(y)(x y)dy.

Vic t ny l c ngha vRnf(y)(x y)dy =

Rnf(x y)(y)dy =

B(0)

f(x y)(y)dy.

Mnh 1.2. Cho f L1loc(Rn). Khi , ta c cc kt lun sau.(i) f C(Rn).(ii) Nu supp f = K Rn th f C0 (Rn), supp f K = {x Rn

d(x,K) }.(iii) Nu f C(Rn) th lim

0+supxK

|f(x) f(x)| 0,K Rn.

(iv) Nu f Lp(Rn)(1 p hay ||xy|| > ,y K.M supp B(0)nn (x y) = 0,y K. Do , f(x) = 0 khi x 6 K hay supp f K.(iii) D thy

f(x) f(x) =Rn

(f(x y) f(x))(y)dy =

B1(0)

(f(x y) f(x))(y)dynn |f(x) f(x)| sup

yB(0)|f(x y) f(x)|m f C(Rn) c f lin tc u trn tng tp compact K Rndo lim

0+supxK

|f(x) f(x)| 0,K Rn.

4Mnh 1.3. Cho K Rn. Khi , vi mi > 0, c mt hm C0 (Rn) tho mn

(i) 0 (x) 1,x Rn,

(ii) supp K,

(iii) (x) = 1,x K 2.

Chng minh. Ly hm c trng ca tp K 34

(x) :=

{1, nu x K 3

4,

0, nu x 6 K 34.

C L1(Rn) L1loc(Rn), supp = K 34, nn theo Mnh 1.2 c

(i) 4 C0 (Rn),

(ii) supp( 4) K,

(iii)0 ( 4)(x),x Rn.

rng,

( 4)(x) =

B 4(0)

(x y) 4(y)dy

nn

(i) ( 4)(x)

B 4(0) 4(y)dy = 1,

(ii) Nu x K 2th (x y) K 3

4,y B

4, do (

4)(x) =

B 4(0) 4(y)dy = 1.

Chng minh. Chng minh nh l 1.1. T gi thit K l tp compact, {Uj}Nj=1 l mt phm ca K c

W1 := K\(Nj=2Uj) U1nn tn ti 1 > 0 sao cho

W1 W1 +B1(0) U1.Theo Mnh 1.3 c mt hm nhn gi tr trong khong (0, 1) l 1 C0 (Rn) sao cho

V1 := W1 +B 12(0) supp1 W1 +B1 U1.

Li c, W1 := K\(Nj=2Uj) V1 m V1 l tp m nn

W2 := K\(V1 (Nj=3Uj)) U2.

5Do , tn ti 2 > 0 sao cho

W2 W2 +B2(0) U2.Theo Mnh 1.3, c mt hm nhn gi tr trong khong (0, 1) l 2 C0 (Rn) sao cho

V2 := W2 +B 22(0) supp2 W2 +B2 U2.C nh th ta xy dng c dy cc hm {j}Nj=1 tho mn(i) j C0 (Rn),(ii) Vj := Wj +B j

2(0) suppj Wj +Bj Uj,

(iii)

Nj=1 j(x) > 0,x Nj=1Vj( K),(iv)

Nj=1 j(x) < N + 1,x Rn.C K Nj=1Vj nn tn ti mt s > 0 sao cho

K K +B(0) Nj=1Vj.Theo Mnh 1.3 c mt hm khng m tho mn

(i) C0 (Rn),(ii) K K +B

2(0) supp K +B Nj=1Vj,(iii)0 (x) 1,x Rn, (x) = 1,x K +B(0).t

j(x) :=j(x)

(x)(N

k=1 k(x))+ (1 (x))

(N + 1Nk=1 k(x))c

(i) 0 j(x) 1,x K, j = 1, 2, . . . ,(ii) j C0 (Rn), suppj Uj, j = 1, 2, . . . ,(iii)

j=1 j(x) = 1,x K.

Ch . xy dng cc hm j t j ta c th dng mt trong hai cch sau:

(i) th nht

j(x) :=

{(x)j(x)Nk=1 k(x)

, nu x supp,0, nu x 6 supp,(ii) th hai

1(x) = 1(x), 2(x) = (1 1(x))2(x), . . . , N(x) = N(x)N1j=1

(1 j(x)).

61.2 Khng gian hm c bn D(), khng gian hm suy

rng D()

1.2.1 Khng gian hm c bn D()

nh ngha 1.2. Khng gian D() l khng gian gm cc hm C0 () vi khi nimhi t sau: dy {j}j=1 cc hm trong C0 () c gi l hi t n hm C0 () nu(i) c mt tp compact K m suppj K, j = 1, 2, . . . ,(ii) lim

jsupx

|Dj(x)D(x)| = 0, Zn+.

Khi , ta vit l = D limj

j.

Ch . 1.Nu = D limj

j th supp K.2. Khi nim hi t trn D() l ph hp vi cu trc tuyn tnh trn D(), ngha l, nu

, C, k, k, , D(), k = 1, 2, . . . , c

nu D limk

k = ,D limk

k = th D limk

(k + k) = + .

3. Hn th, ta cn c th chng minh nu C(), v = D limj

j th =

D limj

j. Tht vy, do nu k(x) = 0 th (x)k(x) = 0 nn supp(k) suppk,v theo cng thc Leibnitz c vi mi Zn+

D(k)(x) =

(

)D(x)Dk(x)

m = D limj

j ngha l

(i) c mt tp compact K m suppk K, k = 1, 2, . . . ,(ii) lim

ksupx

|Dk(x)D(x)| = 0, Zn+,

do ,

(i)supp(Dk) K, k = 1, 2, . . . , Zn+ nn

supp(k) K, supp(D(k)) K, k = 1, 2, . . . ,

(ii) supx

|D(k)(x) D()(x)| C(

supxK

|D(x)|)supx

|Dk(x) D(x)|

nn

limk

supx

|D(k)(x)D()(x)| = 0, Zn+.

7Vi mi Zn+, php ton o hm D l nh x tuyn tnh lin tc trong D(), ngha l(i) D D(), suppD supp,(ii) nu , C, , D() th D(+ ) = D+ D,(iii) nu D lim

kk = 0 th D lim

kDk = 0.

Nh vy, ton t vi phn tuyn tnh P =||m

a(x)D, a C() l ton t vi phntuyn tnh lin tc trn D() m suppPu suppu, u D(). Peetre, J. chng minhc rng nu ton t tuyn tnh P trn C0 () tho mn tnh cht suppPu suppu, u C0 () th P l ton t vi phn.4. Dy {j}j=1 c gi l mt dy Cauchy trong D() nu(i) c mt tp compact K Rn m suppj K, j = 1, 2, . . . ,(ii) lim

jk

supxK

|Dj(x)Dk(x)| = 0, Zn+.

5. Cho k D(), k = 1, 2, . . . , chui hnh thc

k=1 k c gi l hi t trong D()

nu dy cc tng ring {kj=1 j}k=1 hi t trong D().Mnh 1.4. Khng gian D() l .

Chng minh. Ly dy {j}j=1 l mt dy Cauchy trong D() th(i) c mt tp compact K m suppDj K, j = 1, 2, . . . ,,(ii) lim

jk

supxK

|Dj(x)Dk(x)| = 0, Zn+

nn vi mi dy {Dj}j=1 l dy Cauchy trong khng gian C(K) vi chun sup, mkhng gian C(K) vi chun sup l khng gian , do c mt hm C() sao cho(i) supp K,(ii) lim

jsupxK

|Dj(x) (x)| = 0.

Ta s chng minh = D0. Khi , 0 C0 () v(i) suppD0 K,(ii) lim

jsupx

|Dj(x)D0(x)| = limj

supxK

|Dj(x)D0(x)| = 0.

hay 0 = D limj

j.

chng minh iu ny ta ch cn chng minh khi = (1, 0, . . . , 0). Cc trng hp

= (0, . . . , 0,

j1 , 0, . . . , 0) ta chng minh tng t. Sau , bng qui np ta chng minh

cho cc trng hp cn li.

iu ny l hin nhin v D1j hi t u n (1,0,...,0)trong K, v j hi t u n

0

trong K.

81.2.2 Khng gian hm suy rng D()

nh ngha 1.3. Ta ni rng f l mt hm suy rng trong nu f l mt phim hm tuyn

tnh lin tc trn D().

Hm suy rng f D() tc ng ln mi D() c vit l f, . Hai hm suyrng f, g D() c gi l bng nhau nu

f, = g, , D().Tp tt c cc hm suy rng trong lp thnh khng gian D().

Ch . Trn D() c th xy dng mt cu trc khng gian vect trn C, ngha l ta cth nh ngha cc php ton tuyn tnh nh sau

(i) php cng: vi f, g D() tng f + g c xc nh nh sauf + g : 7 f + g, = f, + g, , D(),khi , f + g D(), ngha l, f + g l phim hm tuyn tnh lin tc trn D(),(ii) php nhn vi s phc: vi C, f D() tch f c xc nh nh sau

f : 7 f, = f, , D(),khi , f D(), ngha l, f l phim hm tuyn tnh lin tc trn D().Hn th, ta cn c th nh ngha php nhn vi mt hm trong C().Vi C(), f D() tch f D() c xc nh nh sau

f : 7 f, = f, , D(),khi , f D().Tht vy, do f D() nn d thy f : D() C l nh x tuyn tnh. chng minh f lin tc ta ly dy {k}k=1 mD lim

kk = 0 ta chng minh lim

kf, k =

limk

f, k = 0. iu ny l hin nhin v f lin tc v D limk

k = 0.

V d 1. Vi mi f L1loc() c coi l mt hm suy rng bng cch sau

f : 7 f, =

f(x)(x)dx, D().

Nh vy, c th coi L1loc() l tp con ca D(). Hm suy rng f L1loc() c gi lhm suy rng chnh quy.

Vi f, g L1loc(), th s bng nhau theo ngha hm suy rng v theo ngha thng thngl nh nhau, ngha l

f, g L1loc(),

f(x)(x)dx =

g(x)(x)dx, D() th f = g, h.k.n trong .

V d 2. Hm Dirac:

: 7 , = (0), D().

91.2.3 o hm suy rng

Trong trng hp mt bin, p dng cng thc tch phn tng phn cho f C1(R), C0 (R) c +

f (x)(x)dx = f(x)(x)|+

+

f(x)(x)dx = (1) +

f(x)(x)dx.

Nh vy, ta c th nh ngha o hm ca mt hm nh mt hm suy rng. Ngoi ra, bng

cch nh ngha nh vy ta c th nh ngha o hm cho hm f L1loc(R).

nh ngha 1.4. Cho f D(), = (1, . . . , n) Zn+. o hm suy rng cp cahm suy rng f trong , k hiu l Df, l nh x t D() vo C c xc nh bi

Df : 7 (1)||f,D, D().

Ch . Vi mi Zn+, f D(), o hm suy rng cp ca hm suy rng f trong l mt hm suy rng, ni cch khc, o hm suy rng Df l phim hm tuyn tnh lin

tc t D() vo C, v

vi mi , C, , D() c

Df, + = (1)||f,D(+ ) = (1)||(f,D+ f,D)= Df, + Df,

vi k D(), k = 1, 2, . . . ,D limk

k = 0 th D limk

Dk = 0, Zn+ nn

limk

Df, k = limk

f,Dk = 0.

Vi mi , Zn+, f D(Rn) c o hm suy rng cp , , + l Df,Df,D+fv

D+f = D(Df) = D(Df).

Do , D = D11 D22 . . . D

nn , vi D

jj = D

(0,...,0,

j1 ,0,...,0) . . . D(0,...,0,

j1 ,0,...,0)

j ln

, v th t

c th thay i.

V d 3. Nu f L1loc() c o hm cp theo ngha thng thng Df L1loc() tho hm theo ngha suy rng ca hm suy rng f cng l Df.

V d 4. Hm Heaviside

(t) :=

{1, nu t > 0,

0, nu t 0.c o hm suy rng D(t) = (t).

10

V d 5. Cho f D(), C() c

D(f) =

(

)DDf, trong

(

)=

nj=1

(jj

),

(jj

)=

j!

j!(j j)! .

V d 6. t E(x) = (2pi)1 ln ||x||, nu x R2\{0}, cn vi n 3 t

E(x) = 1(n 2)cn ||x||

2n, x Rn\{0},

vi cn l din tch mt cu n v trong trong gian Rn.Khi , E = trong D(Rn), = D21 + . . . D2n.Tht vy, trc ht ta chng minh E L1loc(Rn). D dng thy E kh vi v hn ti miim x 6= 0, v vi x 6= 0 c

DjE(x) =1

cnxj||x||n, D2jE(x) =

1

cn(||x||2 nx2j)||x||n ch c2 = 2pi

E(x) = D21E(x) + . . . D2nE(x) = 0.

Nh vy chng minh E L1loc(Rn) ta ch cn chng minh E kh tch trong hnh cu nv B1(0). Bng cch chuyn sang h to cu ta c

B1(0)

E(x)dx =

{ 2pi0

10

1c2ln(r)rdrd nu n = 2,

||x||=1 10 1(n2)cn r2nrn1drdS nu n 3,hay

B1(0)

E(x)dx =

{ 10ln(r)rdr = r

2 ln r2

10 1

0r2dr = 1

4nu n = 2,

10

1(n2)rdr =

12(n2) nu n 3.Vi C0 (Rn) c mt s R > 0 supp BR(0), khi , theo cng thc Gausscho hnh { ||x|| R} vi hai bin { = ||x||}, {||x|| = R}

DjE, = E,Dj = lim0+

||x||R

E(x)Dj(x)dx

= lim0+

||x||R

1

cnxj||x||n(x)dx+ lim

0+

=||x||

E(x)(x)xj||x||dS

lim0+

||x||=R

E(x)(x)xj||x||dS

m (x) = 0, ||x|| R, v trn bin { = ||x||} th |E(x)(x) xj||x|| | l v cng b O(ln(1 ))nu n = 2 v O(2n) nu n 3 nn

=||x||E(x)(x)xj||x||dS l v cng b O( ln(

1)) nu

n = 2 v O() nu n 3 khi 0+ nn

DjE, = lim0+

||x||R

1

cnxj||x||n(x)dx.

11

nn o hm suy rng DjE c th vit di dng mt hm kh tch a phng

DjE(x) =1

cnxj||x||n.

Li c

D2jE, = DjE,Dj = lim0+

||x||R

DjE(x)Dj(x)dx

= lim0+

||x||R

1

cn(||x||2 nx2j)||x||n(x)dx

lim0+

||x||=R

DjE(x)(x)xj||x||dS

+ lim0+

||x||=

(x)x2j

cn||x||n+1dS

m (x) = 0, ||x|| R nn

E, = lim0+

1

cnn1

||x||=

(x)dS = (0) = , ,

hay E = .

V d 7. Trong Rn+1, k hiu (x, t) Rn R v

E(x, t) = (4pit)n2 e

||x||24t , t > 0, , E(x, t) = 0, t 0.

Khi , E C(Rn+1\{0}) L1loc(Rn+1), v

(Dt x)u = .

V d 8. Trong R2, k hiu (x, t) R R v

E1(x, t) =1

2(t |x|).

Khi , (D2t D2x)E1(x, t) = .Trong R3, k hiu (x, t) R2 R v

E2(x, t) =(t ||x||)

2pit2 ||x||2 , t 6= ||x|| , E(x, t) = 0, t = ||x||.

Khi , (D2t x)E2(x, t) = .Trong R4, k hiu (x, t) R3 R v

E3(x, t) =1

2pi(t)(t2 ||x||2).

Khi , (D2t x)E3(x, t) = .

12

Trong trng hp = R, vi f, F D(R), ta ni F l nguyn hm suy rng ca hmsuy rng f nu o hm suy rng ca F l f, ngha l DF = f.

Mnh 1.5. Mi hm suy rng f D(R) u c nguyn hm suy rng.Chng minh. Vi mi C0 (R) t

(x) = (x) (x) +

(t)dt

(x) =

x

(t)dt.

C (x) C0 (R) nn vi mi hm suy rng f D(R),ta c th tF, = f,.Khi , F D(R) v

DF, = F, = f, (x) x

(y)

+

(t)dtdy = f, .

Nu hm suy rng F c o hm suy rng DF = 0 th

F, = F, +( +

(t)dt

)F,

= DF,+( +

(t)dt

)F,

=( +

(t)dt

)F, .

Do , nu hm suy rng F c nguyn hm suy rng DF = 0 th F tng ng vi hm

hng F F, trong lp hm kh tch a phng L1loc(R).Khi , vi mi hm suy rng f D(R), lun c mt h cc nguyn hm suy rng mhai nguyn hm trong h sai khc nhau mt hm suy rng c th biu din di dng hm

kh tch a phng hng.

1.2.4 Cp ca hm suy rng

nh ngha 1.5. Cho K , f D(). Ta ni hm suy rng f c cp hu hn trn Knu c mt s nguyn khng m k v mt s dng C sao cho

|f, | C||k

supxK

|D(x)|, C0 (), supp K. (1.1)

S nguyn khng m k nh nht trong cc s nguyn khng m m ta c bt ng thc (1.1)

c gi l cp ca hm suy rng f trn tp K.

Nu khng c mt s nguyn khng m k no c (1.1) vi s dng C no , th ta ni

rng, hm suy rng f c cp v hn trn tp K.

n gin, ta ni rng, hm suy rng f D() c cp k nu n c cp k trn .

13

V d 9. Mi hm suy rng f L1() u c cp 0.V d 10. Trn Rn, hm Dirac (x) D(Rn) c cp 0. Vi Zn+, o hm suy rng cp ca hm Dirac D c cp ||. Tht vy, chn C0 (Rn) sao cho (0) = 1, supp B1(0). t (x) = x

(x) c

D, = (1)||,D = (1)||D(x(x))(0) = (1)||!.

Li c do nu ||x|| th (x) = 0 nn

supxRn

|D(x)| C|| 0 khi 0, < ,

do vi s nguyn khng m k < ||, vi bt k s c > 0 ta u tm c s > 0

|D, | = ! > c||k

supxRn

|D(x)|,

cn vi k = || th

|D, | = |D(0)| C||k

supxRn

|D(x)|, C0 (Rn).

V d 11. Trn R, hm suy rng c xc nh nh sau

f, =+j=0

(j)(j)

c cp v hn.

Tht vy, chn C0 (R) m (x) = 1, x [12 , 12 ], supp (1, 1). t j(x) =(x j)j(xj

j), j chn sau. C D

kj(k) = 0, k 6= j, v Djj(j) = j! nn f, j = j!.Nhng, do nu |x j| j th j(x) = 0 nn

supxR

|Dkj(x)| cjkj , k < j,

ta chn j > 0 sao cho

|f, j| = j! > jj1k=1

supxR

|Dkj(x)|.

Do , vi mi k > 0, c > 0 chn j = max{k + 1, c+ 1} c

|f, j| = j! > jj1l=1

supxR

|Dlj(x)| > ckl=1

supxR

|Dkj(x)|

hay cp ca f l v hn.

14

nh l 1.6. Mi phim hm tuyn tnh f trn D() l mt hm suy rng khi v ch khi,

trn mi tp compact K , c mt s nguyn khng m k v mt s dng C sao cho

|f, | C||k

supx

|D(x)| = CCk(), C0 (), supp K.

Chng minh. chng minh iu kin ta ch cn chng minh tnh lin tc ca f ti gc,

ngha l nu c mt dy {j}j=1 trong C0 () m D limj

j = 0 th limj

f, j = 0.iu ny l d thy t gi thit.

chng minh iu kin cn ta dng phn chng, ngha l gi s c mt tp compact

K vi mi k Z+ ta u c

supC0 ()

suppK, 6=0

|f, |Ck()

= +

do , tn ti k C0 (), supp K, kCk() > 0 sao cho |f, k| > kkCk().Chn k(x) =

1

k12 kCk()

k(x) c

k C0 (), suppk K, D lim

kk = 0, |f, k| k 12 ,

nn f 6 D(), tri vi gi thit.Nh vy, iu gi s sai hay ta c iu phi chng minh.

1.2.5 S hi t trong khng gian hm suy rng D()

nh ngha 1.6. Cho fk, f D(), k = 1, 2, . . . . Ta ni rng, dy {fk}k=1 hi t n ftrong D() khi k tin ra v cng nu

limk

fk, = f, , D().

Khi , ta vit D limk

fk = f.

V d 12. D limk

1k= .

Tht vy, vi mi C0 (Rn) c

| 1k, (0)|

Rn 1k(y)|(y) (0)|dy =

B 1k(0)

1k(y)|(y) (0)|dy

supyB 1

k(0)

|(y) (0)|

nn limk

| 1k, (0)| = 0 hay ta c iu phi chng minh.

15

V d 13. S hi t trong D() trng vi s hi t yu v trong L1loc(), ngha l nufk, f L1loc(),D lim

kfk = f, th

limk

fk(x)(x)dx =

f(x)(x)dx, C0 ().

Ch . 1. Khi nim hi t c nh ngha trn l ph hp vi cu trc tuyn tnh trn

D(), ngha l vi , C, fk, gk, f, g D(), k = 1, 2, . . . v

D limk

fk = f,D limk

gk = g

th

D limk

(fk + gk) = f + g.

2. Cho a(.) C() php ton nhn vi a(.) bin f D() thnh af D() l nh xtuyn tnh lin tc, ngha l

(i) a(f + g) = af + ag,, C, f, g D(),

(ii) Nu fk, f D(), k = 1, 2, . . . v D limk

fk = f th D limk

afk = af.

3. Vi mi Zn+, php ton o hm suy rng D cng l nh x tuyn tnh lin tctrong D(), ngha l

(i) D(f + g) = Df + Dg,, C, f, g D(),

(ii) Nu fk, f D(), k = 1, 2, . . . v D limk

fk = f th D limk

Dfk = Df.

4. Cho fk D(), k = 1, 2, . . . , chui hnh thc

k=1 fk c gi l hi t trong D()nu dy tng ring {kj=1 fj}k=1 hi t trong D(). Khi , chuik=1Dfk cng hit trong D() v

D( k=1

fk

)=

k=1

Dfk.

5. Dy {fk}k=1 c gi l dy Cauchy trongD() nu vi mi D() dy {fk, }k=1l dy Cauchy trong C.

nh l 1.7. D() l khng gian .

chng minh nh l ta cn n B sau.

B 1.8. Cho dy {k}k=1 trong D() m D limk

k = 0, v {fk}k=1 l dy Cauchytrong D(). Khi , lim

kfk, k = 0.

16

Chng minh. Ta chng minh bng phn chng, gi s fk, k 6 0 khi k , ngha lc mt s c > 0 v mt dy con, n gin k hiu, ta c th gi s

|fk, k| > c, k = 1, 2, . . . .

Bng cch ly ra mt dy con ca dy con trn, n gin k hiu, ta c th c

|Dk(x)| 14k,x , || k, k = 1, 2, . . . .

t k = 2kk c

(i) k C0 (), suppk suppk,

(ii) D limk

k = 0, limk

fk, k = +.

Ta i xy dng dy {f k, k}k=1 bng cch quy np nh sau.Do lim

lfl, l = + nn c mt s t nhin l1 sao cho |fl1 , l1| > 1.t f 1 = fl1 ,

1 = l1 .

Do D liml

l = 0 nn c mt s t nhin k1 > l1 sao cho |f 1, l| < 1,l k1. M dy{fl, 1}l=1 l dy Cauchy nn b chn, cn lim

lfl, l = + nn c mt s t nhin

l2 > k1 sao cho |fl, l| > |fl, 1|+ 1,l l2.t f 2 = fl2 ,

2 = l2 . C

(i)|f 1, 2| < 12 ,

(ii)|f 2, 2| > |f 2, 1|+ 1.

Gi s ta c f 1, . . . , fk1,

1, . . . , k1(k > 2, f

j = flj , l1 < l2 < < lk1) m

(i)|f j, k1| < 12k1j , j = 1, . . . , k 2,

(ii)|f k1, k1| >k2

j=1 |f k1, j|+ k 1.

Do D liml

l = 0 nn c mt s t nhin k2 > lk1 sao cho

|f j, l| k2 sao cho

|fl, l| >k2j=1

|fl, j|+ k,l lk.

t f k = flk , k = lk . C

17

(i)|f j, k| < 12kj , j = 1, . . . , k 1,

(ii)|f k, k| >k1

j=1 |f k, j|+ k.

C dy {k}k=1 = {lk}k=1 l dy con ca dy {k}k=1 m k = 2kk nn

(i) c mt tp compact K sao cho suppk K, k = 1, 2, . . . ,

(ii)vi mi Zn+,m2,m1 Z+,m2 > m1 > || cm2

k=m1

supx

|Dk(x)| =m2

k=m1

supx

|Dlk(x)| 0 v mt dy con, n gink hiu ta c th gi s |f, k| = lim

l|fl, k| > c, k = 1, 2, . . . . Do , vi mi k cmt s lk sao cho |flk , k| > c.t f k = flk c

(i) {f k}k=1 l dy Cauchy trong D(),

18

(ii) D limk

k = 0,

(iii) |f k, k| > c, k = 1, 2, . . . ,m theo B 1.8 c lim

k|f k, k| = 0 nn xy ra iu mu thun. Do iu gi s saihay f lin tc.

1.2.6 a phng ho

Cho 1,2 l cc tp m trong Rn v 1 2. Vi mi hm C0 (1) c th coil hm trn 2 bng cch sau

2(x) =

{(x) , nu x 1,

0 , nu x 2\1,th C0 (2).Khi , vi mi f D(2) ta coi l mt hm suy rng trn 1 bng cch sau

f |1 , = f, 2, D(1).Nu f, g D(2), f 6= g th cha chc f |1 6= g|1 hay nu f |1 = g|1 th cha chcf = g. Nu C0 (2), supp 1 th c th coi C0 (1) v f, = f |1 , .nh ngha 1.7. Cho l tp m trong Rn, im x , cc hm suy rng f, g D().Ta ni rng f = g ti x nu c mt ln cn m ca x

f | = g|.Ch . 1. Cho f, g D(). Khi , f 6= g ti mt im x nu vi mi ln cn m ca x u c mt hm D(), supp sao cho

f, 6= g, hay c mt dy hnh cu Brk(x) m rk 0 khi k v mt dy hm k C0 ()m suppk Brk(x) sao cho

f, k 6= g, k.2. Cho f, g D(). Nu f = g trong D() th f = g ti mi im x .nh l sau cho ta thy iu ngc li cng ng.

nh l 1.9. Cho f, g D(). Nu vi mi x u c f = g ti x th f = g trongD().

Chng minh. Vi mi D() c K = supp l tp compact trong . T gi thit, vimi x K c mt ln cn m x ca x m f |x = g|x .C K xKx m K compact nn c mt s hu hn im x1, . . . , xm K m K mj=1xj . Theo nh l 1.1 (nh l phn hoch n v) c mt h hu hn cc hm {j}mj=1trong D() sao cho

19

(i) 0 j(x) 1,x K, j = 1, 2, . . . ,(ii) suppj xj , j = 1, 2, . . . ,(iii)

mj=1 j(x) = 1,x K.

Khi , c

f, = f, (mj=1

j)

=mj=1

f |j , j( v j D(xj))

=mj=1

g|j , j

= g, ,

nn ta c f = g trong D().

20

1.3 Khng gian hm c bn E(), khng gian hm suy rng

vi gi compact E()

1.3.1 Khng gian hm c bn E()

nh ngha 1.8. Khng gian E() l khng gian gm cc hm C() vi khi nimhi t sau:

dy {k}k=1 trong C() c gi l hi t n hm C() trong E() nulimk

supxK

|Dk(x)D(x)| = 0, Zn+, K .

Khi , ta vit E limk

k = .

Ch . 1. Ta c th tch thnh = k=1k vi km k

compact k+1 (chng hnk = {x | ||x|| < k & d(x, ) > 1k}). Do , mt dy {k}k=1 trong C() cgi l hi t n hm C() trong E() nu mt trong cc trng hp sau xy ra

limk

supxj

|Dj(x)D(x)| = 0, Zn+, j = 1, 2, . . . ,

limk

k Cj(j) = 0, j = 1, 2, . . . ,

trong k Cj(j) =||j

supxj

|Dk(x)D(x)|.Khi , mt dy {k}k=1 trong C() c gi l dy Cauchy trong E() nu mt trongcc trng hp sau xy ra

limkl

supxj

|Dk(x)Dl(x)| = 0, Zn+, j = 1, 2, . . . ,

limkl

k lCj(j) = 0, j = 1, 2, . . . .

2. Khi nim hi t trong E() l ph hp vi cu trc tuyn tnh ca n, ngha l vi

, C, k, , k, E(), k = 1, 2, . . . ,nu E lim

kk = ,E lim

kk = th lim

k(k + k) = + .

3. Nu a(.) C() th php nhn vi a(.) bin E() thnh a E() l nh xtuyn tnh lin tc.

Vi mi Zn+, php ton o hm D l nh x tuyn tnh lin tc trong E().4. Tp C0 () l tr mt trong E(). Tht vy, do k l tp compact trong nn theoMnh 1.3 c mt hm k C0 () m k(x) = 1, x k. Khi , vi mi E()c (k) C0 () v E lim

kk = .

5. Nu k, C0 (), k = 1, 2, . . . , v D limk

k = th E limk

k = . Do , ta c

php nhng lin tc D() E().

21

V d 14. Trn R, t hm

(x) =

{e

1|x|21nu |x| < 1,

0 nu |x| 1,

v (k)(x) = (x k), th , (k) C0 (R) v

E limk

(k) = 0, limk

supx[k,k]

|(k)(x)| = e 43 > 0,

khng c gii hn D limk

(k).

V d 15. Trn R, vi k N, do [ 13k, 23k] l tp compact nn theo Mnh 1.3 c mt hm

k C0 (R) m k(x) = 1, x [ 13k , 23k ], suppk [0, 1k ]. t k(x) = (x 12k )kk(x) c

|Djk(x)| = |jl=0

(j

l

)k!

(k l)!(x1

2k)klDklk(x)|

jl=0

(j

l

)k!

(k l)!(1

2k)kl|Dklk(x)|,

m limk

(jl

)k!

(kl)!(12k)kl = 0, j, l = 1, 2, . . . , nn E lim

kk = 0,

li c

|Dkk(x)| = |kl=0

(k

l

)k!

(k l)!(x1

2k)klDklk(x)|,

nn khi x = 12kc |Dkk( 12k )| = |k( 12k )| = 1.

Mnh 1.10. Khng gian E() l .

Chng minh. Ly {k}k=1 l dy Cauchy trong E(). C

limkl

supxj

|Dk(x)Dl(x)| = 0, Zn+, j = 1, 2, . . . .

Do , trn tng j dy {k}k=1 hi t u n mt hm (j).Khi , vi mi x c mt s t nhin j sao cho x j, ta t (x) = (j)(x).D dng chng minh C() v

limk

supxj

|Dk(x)D(x)| = 0, Zn+, j = 1, 2, . . . ,

hay E limk

k = .

22

1.3.2 Khng gian hm suy rng E()

nh ngha 1.9 (Gi ca hm suy rng). Cho f D(). Gi ca hm suy rng f cxc nh nh sau

supp f = {x f 6= 0 ti x}.Hm suy rng f c gi l c gi compact nu gi supp f l tp compact. Tp hp tt c

cc hm suy rng c gi compact c k hiu l E().

Ch . 1. S 0 trong nh ngha l hm suy rng khng c xc nh nh sau:

0, = 0, D().

2. Hm suy rng f 6= 0 ti x ngha l vi mi ln cn m ca x u c mt hm C0 () m f, 6= 0. Do , ta cn c th nh ngha gi ca hm suy rng nh sau.t W l hp tt c cc tp m trong sao cho f, = 0, C0 (). Khi , gica hm suy rng f s l supp f = \W. R rng, supp f l tp ng trong .3. Cho f E(), C0 () v supp f supp = th f, = 0.4. Cho f D(), C() th supp(f) supp supp f. Nu f, g E() thsupp(f + g) (supp f supp g) v Df E(), suppDf supp f, Zn+.5. Khng gian E() l khng gian ng i vi cc php ton tuyn tnh, php nhn vimt hm C(), php ly o hm suy rng.6. Vi mt hm f : C o c th khi nim gi theo ngha thng thng l khngc ngha. thy c iu ny ta xt v d sau, vi = (0, 1) cn hm f : (0, 1) Cc xc nh nh sau

f(x) :=

{1, nu x hu t,

0, nu x v t,

c f = 0 h.k.n trn (0, 1) v supp f = (0, 1) 6= = supp 0.Vi mt hm lin tc f C() th gi thng thng v gi ca hm suy rng tng ngvi f l bng nhau.

V d 16. supp = {0}.V d 17. supp = [0,+).V d 18. Tng

cD

, trong ch c mt s hu hn m c 6= 0 c gi {0}. Mthm suy rng f E(), 0 m c supp f = {0} th f c dng nh trn.Tht vy, do f E() nn f c gi compact nn theo nh l 1.6 f c cp hu hn, nghal c mt s t nhin m, mt s c > 0 sao cho

|f, | c||m

supx

|D(x)|, C0 ().

Vi mi C0 () c gi supp = K l tp compact trong nn theo Mnh 1.3 cmt hm h C0 () m h(x) = 1,x K. Khi , c

23

(x) = h(x)( ||m

11!...n!

D(0)x) C0 (),

= C0 (), |D(x)| = O(||x||m+1||) khi ||x|| nh, (|| m),nn f, = f, + f, =

||mcD

(0) + f, , c = 11!...n!f, h(x)x.t (x) =

B2(0)

(x y)dy c vi > 0 nh C0 (), supp {0} = , (x) = 1 khi ||x|| , nn f, = f, .Li c

|f, | c||m

supx

|D((x))|,

D()(x) =||||

(

)D(x)D(x),

|D(x)| =B2(0)

(x y)dy = ()||B2(0)

D(x y)dy c||,

nn vi mi > 0 nh |f, | c

|||||O(m+1||)|

|| do , khi 0 th |f, | =|f, | 0 hay |f, | = 0.Nh vy

f, =||m

cD(0) =

||m

cD, .

nh l 1.11 (nh ngha khc ca khng gian hm suy rng c gi compact). (i) Cho

hm suy rng vi gi compact f E(). Khi , ta c th thc trin f ln thnhphim hm tuyn tnh lin tc trn E().

(ii) Gi s f l mt phim hm tuyn tnh b chn trn E(). Khi , ta c th thu hp f

thnh hm suy rng c gi compact.

(iii) Cc tng ng trn cho ta mt song nh gia khng gian hm suy rng vi gi compact

E() v khng gian cc phim hm tuyn tnh lin tc trn E().

Nhn xt. T nh l 1.11, mi hm suy rng c gi compact c th c xem nh mt

phim hm tuyn tnh lin tc trn E(), khng gian cc hm suy rng c gi compact

E() c th coi l khng gian cc phim hm tuyn tnh lin tc trn E().T nh l 1.6 mi hm suy rng c gi compact f E() u c cp hu hn trn .Hn na, c mt tp compact K , mt s nguyn khng m m v mt s dng C saocho

|f, | C||m

supxK

|D(x)|, C(), (1.2)

Mt phim hm tuyn tnh f : E() C tho mn bt ng thc (1.2) c f : E() Cl lin tc.

24

Chng minh. (i) Gi s f E() c supp f = K l tp compact trong . Theo Mnh 1.3, c mt hm C0 () m (x) = 1 vi x nm trong ln cn no ca K.t f l phim hm t E() vo C c xc nh nh sau

f , = f, , E().

D thy

(a) f khng ph thuc vo vic chn hm , ngha l nu c 1, 2 C0 () m

1(x) = 1 khi x nm trong ln cn K1 ca K, 2(x) = 1 khi x nm trong ln cn K2 ca K,

th supp(1 2) K = nn f, 1 = f, 2, E(),

(b) f E(), ngha l

nu , C, , E() th f , + = f , + f , , nu E lim

kk = 0 th lim

kf , k = 0,

(c) fD()

= f, ngha l vi mi D() c K1 = (K supp) nn theo Mnh 1.3 c mt hm C0 () m (x) = 1 vi x nm trong ln cn no ca K1,do (x)(x) = (x), x nn f , = f, = f, .

Gi s f, g E() m f = g th f = f D()

= gD()

= g, hay nh x f 7 f l n nht E() vo khng gian cc phim hm tuyn tnh lin tc trn E().(ii) Ly g l phim hm tuyn tnh lin tc trn E(), t f = g|D(). Do php D() E() l lin tc nn f D(). Ta ch cn phi chng minh supp f l tp compact.Ta chng minh bng phn chng, ngha l gi s supp f khng compact, m c th vit

thnh = k=1k vi km k

compact k+1 nn supp f 6 k, k = 1, 2, . . . , hay vimi k c xk \k m f 6= 0 ti xk. Khi , vi mi k c mt ln cn m k (\k),mt hm k C0 (k) sao cho f, k = 1.Vi mi tp compact K u c mt s t nhin k0 sao cho K k,k k0. Msupp k (\k) nn k(x) = 0, x K, k k0. Do , E lim

kk = 0.

Nh vy, c

(a)E limk

k = 0,

(b)g, k = f, k = 1,

nn g khng lin tc trn E().

iu ny tri vi gi thit hay iu gi s sai ngha l supp f compact.

(iii) Di y, ta s chng minh nh x f 7 f l ton nh t E() vo khng gian ccphim hm tuyn tnh lin tc trn E(), c th l ta chng minh nh x g 7 g|D() l nh

25

x ngc ca n.

Ly g l phim hm tuyn tnh lin tc trn E(), t f = g|D(). Theo (i), (ii) c f E(), f l phim hm tuyn tnh lin tc trn E() v f |D() = g|D().Do C0 () l tp tr mt trong E() v f , g l cc phim hm tuyn tnh lin tc trn E()nn f = g, ngha l vi mi g l phim hm tuyn tnh lin tc trn E() u c mt hm

f = g|D() E() sao cho f = g.Nh vy, ta c mt song nh t khng gian hm suy rng vi gi compact E() n khnggian cc phim hm tuyn tnh lin tc trn E().

1.3.3 S hi t trong khng gian hm suy rng E()

nh ngha 1.10. Cho fk, f E(), k = 1, 2, . . . . Ta ni rng dy {fk}k=1 hi t n ftrong E() nu

(i) c mt tp compact K m supp fk K, k = 1, 2, . . . ,

(ii)dy {fk}k=1 hi t n f trong D() ngha l D limk

fk = f.

Khi , ta vit E limk

fk = f.

V d 19. E limk

1k= .

V d 20. Cho dy {k}k=1 trong C0 () m suppk+1 suppk,k=1 suppk = {0}v

|k(x)|dx C,

k(x)dx = 1 th E

limk

k = .

Ch . 1. Khi nim hi t trn E() l ph hp vi cu trc tuyn tnh. Cc php tonnhn vi mt hm trong C() v o hm suy rng D, vi mi Zn+, l nh x tuyntnh lin tc trong E().2. Dy {fk}k=1 trong E() c gi l dy Cauchy trong E() nu


(ii)dy {fk}k=1 l dy Cauchy trong D().

3. Nu E limk

fk = f th D limk

fk = f. Do , php nhng E() D() l lin tc.Ngoi ra, lim

kfk, = f, , E().

nh l 1.12. Khng gian E() l .

Chng minh. Ly {fk}k=1 l dy Cauchy trong E(). Khi , c


(ii)dy {fk}k=1 l dy Cauchy trong D().

26

M D() l khng gian nn tn ti f D() D limk

fk = f. Ta ch cn phi

chng minh supp f l tp compact trong .

Gi s, supp f 6 K, ngha l c mt im x \K m f 6= 0 ti x hay c mt ln cnm x (\K), mt hm C0 (x) m f, = 1.Li c, do x K = hay supp supp fk = nn fk, = 0.Nn lim

kfk, = 0 6= 1 = f, , hay D lim

kfk 6= f.iu ny dn n iu mu thun, hay supp f K.

nh l 1.13 (nh ngha khc v khi nim hi t trn E()). Cho fk E(), k =1, 2, . . . . Khi , {fk}k=1 l dy Cauchy trong E() khi v ch khi {fk, }k=1 l dyCauchy trong C, vi mi E().

chng minh nh l 1.13 ta cn n B sau.

B 1.14. Cho fk : E() C, k = 1, 2, . . . , l cc phim hm tuyn tnh lin tc saocho dy {fk, }k=1 l dy Cauchy trong C, vi mi E(); dy {k}k=1 trong E()m E lim

kk = 0. Khi , lim

kfk, k = 0.

Chng minh. Ta chng minh bng phn chng. Gi s, fk, k 6= 0 khi k ngha l cmt s c > 0 v mt dy con, n gin k hiu ta c th gi s |fk, k| > c, k = 1, 2, . . . .Do E lim

kk = 0 nn c mt dy con, n gin k hiu ta c th gi s

supxl

|Dk(x)| < 14k,max{||, l} k, k = 1, 2, . . . .

t k = 2kk c E lim

kk = 0 v lim

k|fk, k| = +. Ta i xy dng dy {f k, k}k=1bng cch quy np nh sau.

Do liml

fl, l = + nn c mt s t nhin l1 sao cho |fl1 , l1| > 1.t f 1 = fl1 ,

1 = l1 .

Do E liml




2 = l2 . C

(i)|f 1, 2 < 12 ,

(ii)|f 2, 2| > |f 2, 1|+ 1.

Gi s ta c f 1, . . . , fk1,

1, . . . , k1(k > 2, f

j = flj , l1 < l2 < < lk1) m

(i)|f j, k1| < 12k1j , j = 1, . . . , k 2,

(ii)|f k1, k1| >k2

j=1 |f k1, j|+ k 1.

27

Do E liml


|f j, l| k2 sao cho

|fl, l| >k2j=1

|fl, j|+ k,l lk.


(i)|f j, k| < 12kj , j = 1, . . . , k 1,(ii)|f k, k| >

k1j=1 |f k, j|+ k.

C dy {k}k=1 = {lk}k=1 l dy con ca dy {k}k=1 m k = 2kk nn vi mi Zn+, l,m1,m2 Z+,m2 > m1 > max{||, l} c

m2k=m1

supxl

|Dk(x)| =m2

k=m1

supxl

|Dlk(x)|

l1 v x2 6 2 sao cho fl2 6= 0 tix2 ngha l c mt ln cn m b chn 2 2 = ca x2 v mt hm 2 C0 () mfl2 , 2 = 1. t g2 = fl2 .C th, ta s xy dng c dy {gk, k}k=1 c tnh cht sau

dy {gk}k=1 l dy con ca dy {fk}k=1 nn dy {gk, k}k=1 l dy Cauchy trongC vi mi E(),

vi mi k c k(x) = 0, x k nn E limk

k = 0,

nn theo B 1.14 c limk

fk, k = 0. Nhng fk, k = 1 nn ta c iu mu thun.Do iu gi s sai hay K l tp compact.

29

1.4 Khng gian cc hm gim nhanh S(Rn) v khng giancc hm tng chm S(Rn)

1.4.1 Khng gian cc hm gim nhanh (Schwartz) S(Rn)

nh ngha 1.11. Khng gian S(Rn) l tp hp

S(Rn) = { C(Rn)|xD(x)| < c,,x Rn,, Zn+}vi khi nim hi t c nh ngha nh sau

dy {k}k=1 trong S(Rn) c gi l hi t n S(Rn) trong S(Rn) nulimk

supxRn

|xDk(x) xD(x)| = 0,, Zn+.

Khi , ta vit S limk

k = .

Ch . 1. Hm C(Rn) l hm gim nhanh, ngha l vi mi , Z+ c|xD(x)| C,,x Rn khi v ch khi(a) vi mi m Z+, Zn+ c (1 + x2)m|D(x)| Cm,,x Rn

(b) hay vi mi m Z+ c (1 + x2)m

||m |D(x)| Cm,x Rn.Vi k, S(Rn), k = 1, 2, . . . , c S lim

kk = khi v ch khi vi mi s t nhin m

c

limk

supxRn

(1 + x2)m|Dk(x)D(x)| = 0, Zn+,hay

limk

supxRn

(1 + x2)m||m

|Dk(x)D(x)| = 0.

Mt dy {k}k=1 trong S(Rn) c gi l dy Cauchy trong S(Rn) nu mt trong cctrng hp sau xy ra

limkl

supxRn

(1 + x2)m|Dk(x)Dl(x)| = 0, Zn+,

limkl

supxRn

(1 + x2)m ||m

|Dk(x)Dl(x)| = 0.

Ch rng, nu dy {k}k=1 trong S(Rn)m limm

supxRn

(1+x2)m||m |Dm(x)| = 0th S lim

kk = 0, nhng iu ngc li khng ng.

Chng hn, trn R hm ex2 S(R) v

ex2

= 1 + 0 + (x2) + 0 + 12!(x2)2 + ...

= 1 +D1(ex

2)(0)

1!x+

D2(ex2)(0)

2!x2 +

D3(ex2)(0)

3!x3 +

D4(ex2)(0)

4!x4 + ...

30

nn supxR

(1 + |x|2)2m 12m|D2m(ex2)| (2m1)!

m!v S lim

m1mex

2= 0.

2. Khi nim hi t trong S(Rn) l ph hp vi cu trc tuyn tnh trn ngha l vi mi, C, k, k, , S(Rn), k = 1, 2, . . .

nu S limk

k = , S limk

k = th S limk

(k + k) = + .

3. Nu a(.) C(Rn) sao cho vi mi Zn+ c mt s thc m = m(), v mt sdng c = c() c |Da(x)| < c(1 + x)m, th nh x bin mi thnh a l nh xtuyn tnh lin tc t S(Rn) vo S(Rn).Vi mi Zn+, php ton o hm D l nh x tuyn tnh lin tc t S(Rn) vo S(Rn).4. Tp C0 (Rn) tr mt trong khng gian S(Rn). Tht vy, vi mi s t nhin k hnh cung B1(0) l tp compact trong Rn nn theo Mnh 1.3 c mt hm C0 (Rn) m(x) = 1, x B1(0) v suppk B2(0). t k(x) = ( 1kx), k = k c

Dk(x) = D(k)(x) = k(x)D

(x) +1

k

1||

1

k||1D(

1

kx)D(x).

nn S limk

k = .

5. Cho k, D(Rn), k = 1, 2, . . . , v D limk

k = th S limk

k = . Do , ta c

php nhng lin tc D(Rn) S(Rn).Cho k, S(Rn), k = 1, 2, . . . , v S lim

kk = th E lim

kk = . Do , ta c php

nhng lin tc S(Rn) E(Rn).

V d 21. Trn R, hm : R R c xc nh nh sau

(x) =

{e

1|x|21nu |x| < 1,

0 nu |x| 1,

v ((k))(x) =(xk)

(1+|x|2)k , th , ((k)) C0 (R) v S lim

k((k)) = 0,

khng c gii hn D limk

((k)).

nh l 1.15. S(Rn) l khng gian y ,

Chng minh. Ly {k}k=1 l mt dy Cauchy trong S(Rn) c

limkl

supxRn

|Dk(x)Dl(x)| = 0,

nn dy {Dk}k=1 hi t u trn tng compact trong Rn n mt hm ().D dng chng minh = (0) C(Rn) v D(0) = ().Ta cn phi chng minh

31

(i) vi mi m Z+, Zn+ c (1 + ||x||2)m|D(x)| Cm,,x Rn,(ii) vi mi m Z+, Zn+ c lim

ksupRn(1 + ||x||2)m|Dk(x)D(x)| = 0.

Vi mi > 0,m Z+, Zn+, c mt s t nhin k0 sao cho vi mi k k0, l k0 tac

supxRn

(1 + ||x||2)m|Dk(x)Dl(x)| < 4.

Do k0 S(Rn) nn (1 + ||x||2)m+1|Dk0(x)| Cm, nn vi R0 >

2Cm,sao cho

(1 + ||x||2)m|Dk0(x)| R0.Khi

(1 + ||x||2)m|Dk(x)| < (1 + ||x||2)m|Dk0(x)|+ (1 + ||x||2)m|Dk(x)Dk0(x)| R0.

Li c do limk

sup||x||R0

|Dk(x) D(x)| = 0 nn c mt s k1 > k0 sao cho vi mik k1 c

sup||x||R0

|Dk(x)D(x)| < (1 +R20)

m

do

sup||x||R0

(1 + ||x||2)m|Dk(x)D(x)| < .

Vi ||x|| > R0, do limk

|Dk(x)D(x)| = 0 nn c mt s k(x) > k0 sao cho

(1 + ||x||2)m|Dk(x)(x)D(x)| < 2

do vi mi k k0 c(1 + ||x||2)m|Dk(x)D(x)| 0,m Z+, Zn+ u c v c mt s t nhin k0 vi mi k k0

supxRn

(1 + ||x||2)m|Dk(x)D(x)| <

hay S(Rn) v S limk

k = .

32

1.4.2 Khng gian cc hm suy rng tng chm S(Rn)

nh ngha 1.12. Cho hm suy rng f D(Rn). Hm suy rng f c gi l hm suyrng tng chm nu c mt s t nhin m v mt s dng c sao cho

|f, | c supxRn

(1 + ||x||2)m||m

|D(x)|, D(Rn).

Khng gian cc hm suy rng tng chm S(Rn) l tp tt c cc hm suy rng tng chm.

Ch . 1. Khng gian cc hm suy rng tng chm S(Rn) l ng i vi cc php tontuyn tnh, php nhn vi mt hm a(.) C(Rn) m vi mi Zn+ u c mt s thcm v mt s dng c sup

xRn|Da(x)| c(1+ ||x||2)m, v php ly o hm suy rng D.

V d 22. Cho f L1loc(Rn) sao choRn

|f(x)|(1 + ||x||)N dx < +,

vi N > 0 no th n tng ng vi hm suy rng tng chm.

Nh vy, vi 1 p , f Lp(Rn) th f cng tng ng vi hm suy rng tng chm v

vi p = 1 chn N = 0, cn vi p = chn N = n+ 1,

vi 1 < p

33

Nhn xt. T nh l 1.16 ta c th coi hm suy rng tng chm l phim hm tuyn

tnh lin tc trn S(Rn), khng gian cc hm suy rng tng chm S(Rn) l khng gian ccphim hm tuyn tnh lin tc trn S(Rn).

Chng minh. (i) Gi s f S(Rn), ngha l f D(Rn) v c mt s t nhin m v sdng C sao cho

|f, | c supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn).

Do tp C0 (Rn) tr mt trong S(Rn) nn vi mi S(Rn) u c mt dy {k}k=1 trongC0 (Rn) sao cho = S lim

kk. Khi

limkl

supxRn

||m

(1 + ||x||2)m|Dk(x)Dl(x)| = 0

do {f, k}k=1 l dy Cauchy trong C hay c mt s phc, k hiu f, sao chof, = lim

kfk, k.Trc tin, ta chng minh s f, khng ph thuc vo vic chn dy {k}k=1, ngha lnu c hai dy {k}k=1, {k}k=1 trong C0 (Rn) m = S lim

kk = S lim

kk th

limk

supxRn

||m

(1 + ||x||2)m|Dk(x)Dk(x)| = 0

do limk

f, k k = 0 hay limk

f, k = limk

f, k.Nh vy vi mi S(Rn) c mt tng ng, k hiu f , vi mt s phc f, . D thytng ng f l nh x tuyn tnh t S(Rn) vo C v vi mi S(Rn) c

|f, | C supxRn

(1 + ||x||2)m||m

|D(x)|,

hay f lin tc.

Nh vy, vi mi f S(Rn) u c thc trin lin tc ln S(Rn).(ii) Gi s f : S(Rn) R l phim hm tuyn tnh lin tc. Ta s chng minh rng hnch ca f trn D(Rn) l hm suy rng tng chm bng phn chng ngha l vi mi s tnhin m u c hm m D(Rn)\{0} m

|f, m| > m supxRn

(1 + ||x||2)m||m

|Dm(x)|,

t m(x) =1

m supyRn

(1+||y||2)m ||m

|Dm(y)|m(x) c

m C0 (Rn),S limm

m = 0,

f, m m,

nn f khng lin tc ti 0. Tri vi gi thit. Do , ta c iu phi chng minh.

34

1.4.3 S hi t trong khng gian cc hm suy rng tng chm S(Rn)

nh ngha 1.13. Cho fk, f S(Rn), k = 1, 2, . . . . Dy {fk}k=1 c gi l hi t trongS(Rn) n f, vit S lim

kfk = f, nu

(i) c mt s t nhin m v mt s dng C sao cho

|fk, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn), k = 1, 2, . . . ,

(ii) dy {fk}k=1 l hi t trong D(Rn) n f.Ch . 1. Khi nim hi t trong S(Rn) l ph hp vi cu trc tuyn tnh trn , nghal vi , C, fk, f, gk, S(Rn), k = 1, 2, . . . , c

nu S limk

fk = f, S limk

gk = g th S limk

(fk + gk) = f + g.

Mt dy {fk}k=1 c gi l dy Cauchy trong S(Rn) nu(i) c mt s t nhin m v mt s dng C sao cho

|fk, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn), k = 1, 2, . . . ,

(ii) dy {fk}k=1 l Cauchy trong D(Rn).2. Nu a(.) C(Rn) sao cho vi mi Zn+ c mt s thc m = m(), v mt sdng c = c() c |Da(x)| < c(1 + x)m, th nh x bin mi f thnh af l nh xtuyn tnh lin tc t S(Rn) vo S(Rn).Vi mi Zn+, php ton o hm D l nh x tuyn tnh lin tc t S(Rn) vo S(Rn).3. C cc php nhng lin tc E(Rn) S(Rn) D(Rn).V d 25. Cho (x) = (2pi)

n2 e

||x||22 . t (x) =

n(x), > 0, th , S(Rn)

S(Rn), vRn (x)dx =

Rn (x)dx = 1,

lim0+

||x||>R

(x)dx = lim0+

||x||> R

(x)dx = 0

nn S lim0+

= .

nh l 1.17. S(Rn) l khng gian y .

Chng minh. Ly {fk}k=1 l dy Cauchy trong S(Rn) ngha l(i) c mt s t nhin m v mt s dng C sao cho

|fk, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn), k = 1, 2, . . . ,

35

(ii) dy {fk}k=1 l Cauchy trong D(Rn).

DoD(Rn) l khng gian y nn c mt hm suy rng f D(Rn) sao choD limk

fk =

f. Ta ch cn phi chng minh f S(Rn) hay c mt s t nhin m v mt s dng Csao cho

|f, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn).

Tht vy, do vi mi C0 (Rn) c limk

fk, = f, v

|fk, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn), k = 1, 2, . . . ,

nn

|f, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn).

nh l 1.18 (nh ngha khc v s hi t trong S(Rn)). Cho fk S(Rn). Khi , dy{fk}k=1 l dy Cauchy trong S(Rn) khi v ch khi vi mi S(Rn) dy {fk, }k=1 ldy Cauchy trong C.

chng minh nh l ta cn n B sau.

B 1.19. Cho fk : S(Rn) C, k = 1, 2, . . . , l cc phim hm tuyn tnh lin tc saocho dy {fk, }k=1 l dy Cauchy trong C vi mi S(); dy {k}k=1 trong S(Rn)sao cho S lim

kk = 0. Khi , lim

kfk, k = 0.

Chng minh. Ta chng minh bng phn chng. Gi s dy {fk, k}k=1 khng hi t ti0 khi k , ngha l c mt s dng c v mt dy con, n gin k hiu ta gi s

|fk, | > c, k = 1, 2, . . . .

Do S limk

k = 0 nn c mt dy con, n gin k hiu ta c th gi s

supxRn

(1 + ||x||2)m|Dk(x)| 14k,max{m, ||} k, k = 1, 2, . . . .

t k(x) = 2kk(x) c

S limk

k = 0,

limk

|fk, k| = +.

36

Ta i xy dng dy {f k, k}k=1 bng cch quy np nh sau.Do lim

lfl, l = + nn c mt s t nhin l1 sao cho |fl1 , l1| > 1.t f 1 = fl1 ,

1 = l1 .

Do S liml




2 = l2 . C

(i)|f 1, 2 < 12 ,(ii)|f 2, 2| > |f 2, 1|+ 1.

Gi s ta c f 1, . . . , fk1,

1, . . . , k1(k > 2, f

j = flj , l1 < l2 < < lk1) m

(i)|f j, k1| < 12k1j , j = 1, . . . , k 2,(ii)|f k1, k1| >

k2j=1 |f k1, j|+ k 1.

Do S liml


|f j, l| k2 sao cho

|fl, l| >k2j=1

|fl, j|+ k,l lk.


(i)|f j, k| < 12kj , j = 1, . . . , k 1,(ii)|f k, k| >

k1j=1 |f k, j|+ k.

C dy {k}k=1 = {lk}k=1 l dy con ca dy {k}k=1 m k = 2kk nn vi mi Zn+,m,m1,m2 Z+,m2 > m1 > max{m, ||} c

m2k=m1

supxRn

(1 + ||x||2)m|Dk(x)| =m2

k=m1

supxRn

(1 + ||x||2)m|Dlk(x)|

0. Do tp C0 (Rn) tr mt trong S(Rn) nn vi mi S(Rn) c mt hm C0 (Rn) sao cho

supxRn

(1 + ||x||2)m||m

|D(x)D(x)| 2C

.

Do dy {fk}k=1 l Cauchy trongD(Rn) nn c mt s t nhin k0 sao cho vi mi k, l k0c

|fk fl, | 2.

Do vi mi > 0 u c mt s t nhin k0 k, l k0|fk fl, | |fk fl, |+ |fk fl, |

C supxRn

(1 + ||x||2)m||m

|D(x)D(x)|+ 2

38

hay vi mi S(Rn) dy {fk, }k=1 l dy Cauchy trong C.Ta chng minh iu kin . Gi s vi mi S(Rn) dy {fk, }k=1 l dy Cauchytrong C. Do C0 (Rn) S(Rn) nn dy {fk}k=1 l dy Cauchy trong D(Rn).Ta ch cn phi chng minh c mt s t nhin m v mt s dng C sao cho

|fk, | C supxRn

(1 + ||x||2)m||m

|D(x)|, C0 (Rn), k = 1, 2, . . . .

Ta chng minh bng phn chng, ngha l vi mi s t nhin m u c mt hm m C0 (Rn)

|fm, m| > m supxRn

(1 + ||x||2)m||m

|Dm(x)|, C0 (Rn).

t m(x) =1

m supyRn

(1+||y||2)m ||m

|Dm(y)|m(x) c

m C0 (Rn),S limm

m = 0,

fm, m m,

m theo B 1.19 th limk

fm, m = 0. Nh vy xy ra iu mu thun. Do ta ciu phi chng minh.

Chng 2

Tch chp v Php bin i Fourier

2.1 Tch chp

2.1.1 Tch chp gia cc hm trong L1loc(Rn)

Nu f, g C0 (Rn) ta nh ngha

f g(x) =Rnf(x y)g(y)dy =

Rnf(y)g(x y)dy (2.1)

xc nh vi mi x Rn. T nh l Fubini cRn|f g(x)|dx =

Rn|Rnf(x y)g(y)dy|dx

Rn|g(y)|

(Rn|f(x y)|dx

)dy

||f ||L1(Rn)||g||L1(Rn), (2.2)

nn f g L1(Rn) v

||f g||L1(Rn) ||f ||L1(Rn)||g||L1(Rn).

M C0 (Rn) tr mt trong L1(Rn) nn ta c th nh ngha

f g(x) = L1 limk

k k(x)

vi f, g L1(Rn), f = L1 limk

k, g = L1 limk

k, k, k C0 (Rn), k = 1, 2, . . . .Ta gi f g l tch chp ca hm f theo hm g. R rng, trong trng hp ny, tch chpca hm f theo hm g v tch chp ca hm g theo hm f l nh nhau.

Vi f L1(Rn), g Lp(Rn)(1 p ) bng cch tng t trn vi bt ng thc YoungRn

Rnf(x y)g(y)dy

pdx ||f ||pL1(Rn)||g||pLp(Rn)c f g Lp(Rn). Ta cng cha th ni f g L1(Rn), nu 1 < p. V vi 1 < p ta lunchn c s k (n

p, n). Khi , hm f = B1(0) L1(Rn), g(x) = (||x||+1)k Lp(Rn)

37

38

v f g(x) > C(||x||+ 2)k 6 L1(Rn).Vi f L1loc(Rn), g L1compact(Rn) tch chp f g ca hm f theo hm g, v tch chpg f ca hm g theo hm f , theo cng thc (2.1) l xc nh vi hu ht x Rn. Hn na,f g = g f L1loc(Rn) v supp(f g) (supp f + supp g) (v (f g)(x) 6= 0 y Rn : f(y) 6= 0, g(x y) 6= 0). Tuy nhin, ta cha th ni f g L1(Rn), chng hn lyf = B1(0), g = 1.

Vi f Lp(Rn), g Lq(Rn)(1 < p

39

m nu t

h1,x(y) =1

h1

((x1 + h1 y1, x2 y2, . . . , xn yn) (x y)

)D(1,0,...,0)(x y)=

1

h1

h10

t0

D(2,0,...,0)(x1 y1 + , x2 y2, . . . , xn yn)ddt

h1,x(y) =1

h1

((x1 + h1 y1, x2 y2, . . . , xn yn) (x y)

)D(1,0,...,0)(x y)=

1

h1

h10

t0

D(2,0,...,0)(x1 y1 + , x2 y2, . . . , xn yn)ddt

th h1,x, h1,x hi t v 0 trong E(Rn), (D(Rn), S(Rn)) khi h1 tin v 0, tu theo , C(Rn)(C0 (Rn), S(Rn)), nn

D(1,0,...,0)( )(x) = limh10

1

h1

(( )(x1 + h1, x2, . . . , xn) ( )(x)

)=((D(1,0,...,0)) )(x)

=( (D(1,0,...,0)))(x).Vi , C0 (Rn) th C0 (Rn). Vi , S(Rn) nn vi mi m Z+, c(1 + ||y||2)m L1(Rn), v s cm > 0 sup

xRn(1 + ||x||2)m|D(x)| < cm, do

supxRn

(1 + ||x||2)m|D( )(x)| Rn

supxRn

(1 + ||x||2)m|D(x y)||(y)|dy

supxRn

(1 + ||x y||2)m|D(x y)|Rn(1 + ||y||2)m|(y)|dy. (2.3)

Tnh tuyn tnh ca cc nh x l d thy.

Ta chng minh tnh lin tc ca cc nh x nh sau.

(i) Nu C(Rn), k C0 (Rn), k = 1, 2, . . . ,D limk

k = 0 th

c mt tp compact K Rn suppDk K, k = 1, 2, . . . , |D( k)(x)| = |

K(xy)Dk(y)dy| sup

yK|Dk(y)|

K|(xy)|dy,

trn mi tp compact K Rn, c

supxK

K

|(x y)|dy KK

|(z)|dz < +

nn limk

supxK

|D( k)(x)| = 0,

nu C0 (Rn), c supxRn

K|(x y)|dy

supp|(y)|dy < +, nn

limk

supxRn

|D( k)(x)| = 0,

40

(ii) Nu C0 (Rn), k C(Rn), k = 1, 2, . . . ,E limk

k = 0 th

|D( k)(x)| = |supp

(y)Dk(x y)dy|

supysupp

|Dk(x y)|supp

|(y)|dy,

m trn mi tp compact K Rn csupxK

supysupp

|Dk(x y)| supz(Ksupp)

|Dk(z)|

nn

limk

supxK

|D( k)(x)| = 0.

(iii) Nu , k S(Rn), k = 1, 2, . . . , S limk

k = 0 th p dng (2.3).

Mnh 2.2. (i)Nu D(Rn) th D lim0+

= .(ii)Nu E(Rn) th E lim

0+ = .(iii)Nu S(Rn) th S lim

0+ = .

Chng minh. Ch rng supp( ) (supp+ B(0)) v theo Mnh 2.1 c D( ) = (D

) nn vi chng minh (i), (ii) ta ch cn phi chng minh hi tu n trn tng tp compact K khi gim dn v 0, cn (iii) ta chng minh vi mi s

t nhin k c (1 + ||x||2)k( )(x) hi t u n (1 + ||x||2)k(x) trn Rn.C

Rn((x y) (x))(y)dy =

B(0)

((x y) (x))(y)dy,

m C(Rn) nn vi 0 < < 1, y B(0) theo nh l Lagrange c|(x y) (x)| sup

zB(0)||(x z)||, vi l o nh ca ,

nn supxK

|( )(x)(x)| supyK

||(y)||, hay ta c iu phi chng minh cho (i), (ii).Nu S(Rn) th

supxRnzB1(0)

(1 + ||x||2)k||(x z)||

C1 supxRnzB1(0)

(1 + ||z||2)k(1 + ||x z||2)knj=1

|Dj(x z)|

C2 supyRn

(1 + ||y||2)knj=1

|Dj(y)| < +

nn ta c iu phi chng minh cho (iii).

41

Nhn xt. Nu thay bi cc hm k trong C0 (Rn) m

Rnk(x)dx = 1,

Rn|k(x)|dx C, suppk+1 suppk,k=1 suppk = {0}

th ta cng c cc kt lun nh Mnh 2.2.

t (x) = (2pi)n2 e

||x||22 , (x) =

n(x). Khi , nu thay bng ta cng c kt

lun (iii) ca Mnh 2.2, vRn(x)dx =

Rn(x)dx = 1, lim

0+

||x||R

(x)dx = lim0+

||x|| R

(x)dx = 0.

2.1.2 Tch chp gia hm suy rng v hm c bn

nh ngha 2.1. Cho f D(Rn), C0 (Rn) hay f E(Rn), C(Rn) hocf S(Rn), S(Rn). Tch chp, k hiu f , ca hm suy rng f theo hm l hmc xc nh nh sau

f : x 7 (f )(x) = f, x, x(y) = (x y).

Ch . Vi mi x Rn hm x C0 (Rn) hay (C(Rn), S(Rn)) tu theo C0 (Rn)hay (C(Rn), S(Rn)) nn nh x trong nh ngha hon ton xc nh. nh x ny c mts tnh cht sau.

Mnh 2.3. Cho f D(Rn), C0 (Rn) hay f E(Rn), C(Rn) hoc f S(Rn), S(Rn). Khi , ta c cc kt lun sau.

(i) f C(Rn), D(f)(x) = ((Df))(x) = (f(D))(x),x Rn, Zn+.(ii) supp(f ) (supp f + supp).

(iii) Nu f D(Rn), k D(Rn), k = 1, 2, . . . ,D limk

k = 0 th E limk

f k = 0.Nu f E(Rn), k E(Rn), k = 1, 2, . . . ,E lim

kk = 0 th E lim

kf k = 0.Nu f E(Rn), k D(Rn), k = 1, 2, . . . ,D lim

kk = 0 th D lim

kf k = 0.Nu f S(Rn), k S(Rn), k = 1, 2, . . . , S lim

kk = 0 th E lim

kf k = 0.Nu f E(Rn), k S(Rn), k = 1, 2, . . . , S lim

kk = 0 th S lim

kf k = 0.

(iv) Nu fk D(Rn), D(Rn), k = 1, 2, . . . ,D limk

fk = 0 th E limk

fk = 0.Nu fk E(Rn), E(Rn), k = 1, 2, . . . ,E lim

kfk = 0 th E lim

kfk = 0.Nu fk E(Rn), D(Rn), k = 1, 2, . . . ,E lim

kfk = 0 th D lim

kfk = 0.Nu fk S(Rn), S(Rn), k = 1, 2, . . . , S lim

kfk = 0 th E lim

kfk = 0.Nu fk E(Rn), S(Rn), k = 1, 2, . . . ,E lim

kfk = 0 th S lim

kfk = 0.

42

Ch . Vi f S(Rn), C0 (Rn) th cha th ni rng f S(Rn). Chng hn,f(x) 1 l hm b chn tng ng vi mt hm suy rng tng chm, C0 (Rn),(f )(x) =

Rn (x y)dy = 1 l hm kh vi v hn, nhng khng gim nhanh.Nhn xt. 1. Nu g D(Rn), D(Rn) th

nh x 7 g l nh x tuyn tnh lin tc t D(Rn) vo E(Rn),

nh x f 7 f l nh x tuyn tnh lin tc t D(Rn) vo E(Rn).

2. Nu g E(Rn), E(Rn) th

nh x 7 g l nh x tuyn tnh lin tc t E(Rn) vo E(Rn),

nh x f 7 f l nh x tuyn tnh lin tc t E(Rn) vo E(Rn).

3. Nu g E(Rn), D(Rn) th

nh x 7 g l nh x tuyn tnh lin tc t D(Rn) vo D(Rn),

nh x f 7 f l nh x tuyn tnh lin tc t E(Rn) vo D(Rn).

4. Nu g S(Rn), S(Rn) th

nh x 7 g l nh x tuyn tnh lin tc t S(Rn) vo E(Rn),

nh x f 7 f l nh x tuyn tnh lin tc t S(Rn) vo E(Rn).

5. Nu g E(Rn), S(Rn) th

nh x 7 g l nh x tuyn tnh lin tc t S(Rn) vo S(Rn),

nh x f 7 f l nh x tuyn tnh lin tc t E(Rn) vo S(Rn).

Chng minh. (i)Do

C(Rn), f E(Rn) th D C(Rn), Df E(Rn),

C0 (Rn), f D(Rn) th D C0 (Rn), Df D(Rn),

S(Rn), f S(Rn) th D S(Rn), Df S(Rn),

nn chng minh (i) ta ch cn chng minh tch chp f c o hm ring theo binth nht v

D(1,0,...,0)(f )(x) = ((D(1,0,...,0)f) )(x) = (f (D(1,0,...,0)))(x),x Rn.

43

Vi x Rn, h1 R c

(f )(x1 + h1, x2, . . . , xn) (f )(x)= fy, (x1 + h1 y1, x2 y2, . . . , xn yn) (x y)= fy, (x1 + h1 y1, x2 y2, . . . , xn yn) (x y) h1D(1,0,...,0)x (x y)+ fy, h1D(1,0,...,0)x (x y),

m nu t

h1,x(y) =1

h1

((x1 + h1 y1, x2 y2, . . . , xn yn) (x y)

)D(1,0,...,0)x (x y),th h1,x hi t v 0 trongD(Rn) hay (E(Rn), S(Rn)), tu theo D(Rn) hay (E(Rn), S(Rn))khi h1 tin v 0, nn

(D(1,0,...,0)f )(x) = limh10

(f )(x1 + h1, x2, . . . , xn) (f )(x)h1

= limh10

fy, h1+ fy, D(1,0,...,0)x (x y)= fy, D(1,0,...,0)x (x y) = D(1,0,...,0)y fy, (x y)= (f D(1,0,...,0))(x) = (D(1,0,...,0)f )(x).

(ii) Ly x Rn. Gi s (f )(x) 6= 0 m (f )(x) = f, , x(y) = (x y)nn suppx supp f 6= hay tn ti y supp f (x y) supp, do x (supp f+supp).M tng ca hai tp ng l ng nn supp(f ) (supp f+supp).(iii) T (i) v (ii) nu C(Rn), f E(Rn) hay C0 (Rn), f D(Rn) hoc S(Rn), f S(Rn) th f C(Rn), v C0 (Rn), f E(Rn) th f C0 (Rn).Nu S(Rn), f E(Rn), do D(f ) = (Df) , Zn+, c f S(Rn)ta ch cn phi chng minh vi m Z+ c hng s cm > 0 khng ph thuc x sao cho

(1 + ||x||2)m|(f )(x)| cm.

Tht vy v f E(Rn), nn c mt s dng R0, mt hm C0 (Rn) m (x) = 1, x supp f, supp BR0(0). Khi , c mt s t nhin l0, mt s dng c (cc s ny khngph thuc x) sao cho

(1 + ||x||2)m|(f )(x)| = (1 + ||x||2)m|f, | c1 sup

yBR0 (0)(1 + ||x||2)m

||l0

|D(y)| ( D(Rn), (y) = (y)(x y))

c2 supyBR0 (0)

||l0

((1 + ||y||2)m|D(y)|(1 + ||x y||2)m|D(x y)|)

c3||l0

(1 + ||x||2)m|D(x)|. (2.4)

Phn cn li c chng minh nh sau.

44

1. Nu f D(Rn), k D(Rn), k = 1, 2, . . . ,D limk

k = 0 th vi mi x Rn

c mt tp compact K Rn suppk K, k = 1, 2, . . . , lim

ksupyRn

||l

|Dk(x y)| = 0, l = 1, 2, . . . ,

vi tp compact K Rn bt k, f c cp hu hn trn tp compact (K K),ngha l c mt s t nhin l0 v s dng c

|(f k)(x)| = |f, k,x| c||l0

supyRn

|Dk(x y)|, k = 1, 2, . . . ,

trong , k,x(y) = k(x y), suppk,x = x suppk (K K), x K .2. Nu f E(Rn), k E(Rn), k = 1, 2, . . . ,E lim

kk = 0 th vi mi x Rn

limk

supyBR(0)

||l

|Dk(x)| = 0, l = 1, 2, . . . , R > 0,

do supp f = K l tp compact nn c mt s R0 > 0, v mt hm C0 (Rn)m (y) = 1, y K, supp BR0(0). Khi f, = f, , E(Rn),

f c cp hu hn trn Rn, ngha l c mt s t nhin l0 v s dng c vimi s t nhin k c

|(f k)(x)| = |f, k| c||l0

supyBR0 (0)

|Dk(y)|, k = 1, 2, . . . ,

trong , k(y) = (y)k(x y), m trn mi tp compact K Rn, c

supxK

supyBR0 (0)

|Dk(y)| (

supyBR0 (0)

|

D(y)|) supxKR0

|Dk(x)|

3. Nu f S(Rn), k S(Rn), k = 1, 2, . . . , S limk

k = 0 th vi mi x Rn

limk

supyRn

(1 + ||y||2)l ||l

|Dk(y)| = 0, l = 1, 2, . . . ,

c mt s t nhin l0 v mt s dng c vi mi k, l = 1, 2, . . . ,

|(f k)(x)| = |f, k,x| c supyRn

||l0

(1 + ||y||2)l0|Dk(x y)|

trong , k,x(y) = k(x y), m

supyRn

||l0

(1 + ||y||2)l0 |Dk(x y)|

(1 + ||x||2)l0 supyRn

||l0

(1 + ||x y||2)l0|Dk(x y)|.

45

4. Nu f E(Rn), k C0 (Rn), k = 1, 2, . . . ,D limk

k = 0 th c mt tp compact

K Rn supp(f k) K, k = 1, 2, . . . .

5. Nu f E(Rn), k S(Rn), k = 1, 2, . . . , S limk

k = 0

limk

supyRn

(1 + ||y||2)l ||l

|Dk(y)| = 0, l = 1, 2, . . . ,

c mt s dng R0, mt hm C0 (Rn) m (x) = 1, x supp f, supp BR0(0). Khi , c mt s t nhin l0, mt s dng c (cc s ny khng ph

thuc x) c bt ng thc (2.4)

(1 + ||x||2)m|(f k)(x)| c(

supyBR0 (0)

||l0

|D(y)|) sup

zRn

||l0

(1 + ||z||2)m|Dk(z)|.

(iv)

1. Nu fk D(Rn), D(Rn), k = 1, 2, . . . ,D limk

fk = 0 th

limk

|fk (x)| = limk

|fk, | = 0, (y) = (x y),x Rn, gi s c mt tp compact K Rn m dy {fk }k=1 khng hi t un 0 trn , ngha l c mt s dng , mt dy con, n gin ta gi s

{fk }k=1 v mt dy cc im {xk}k=1 trong tp compact K sao cho

|fk (xk)| = |fk, k| > 2, k(y) = (xk y),

m dy {xk}k=1 nm trong tp compact K nn n c dy con, n gin tagi s dy {xk}k=1 hi t n x0 trong K, khi d thy

D limk

k = 0, 0(y) = (x0 y)


fk, k 0 = 0 do , vi k ln |fk, 0| > ,iu ny tri vi gi thit D lim

kfk = 0,

nh vy, iu gi s sau hay trn mi tp compact K Rn m dy {fk }k=1hi t u n 0,

D(fk )(x) = (fk D)(x).

2. Nu fk E(Rn), E(Rn), k = 1, 2, . . . ,E limk

fk = 0 th

limk

|fk (x)| = limk

|fk, | = 0, (y) = (x y),x Rn,

46

gi s c mt tp compact K Rn m dy {fk }k=1 khng hi t un 0 trn , ngha l c mt s dng , mt dy con, n gin ta gi s


|fk (xk)| = |fk, k| > 2, k(y) = (xk y),


E limk

k = 0, 0(y) = (x0 y)


fk, k0 = 0 do , vi k ln |fk, 0| > ,iu ny tri vi gi thit E lim

kfk = 0,


D(fk )(x) = (fk D)(x).3. Nu fk S(Rn), S(Rn), k = 1, 2, . . . , S lim

kfk = 0 th

limk

|fk (x)| = limk

|fk, | = 0, (y) = (x y),x Rn, gi s c mt tp compact K Rn m dy {fk }k=1 khng hi t un 0 trn , ngha l c mt s dng , mt dy con, n gin ta gi s


|fk (xk)| = |fk, k| > 2, k(y) = (xk y),


S limk

k = 0, 0(y) = (x0 y)


fk, k0 = 0 do , vi k ln |fk, 0| > ,iu ny tri vi gi thit S lim

kfk = 0,


D(fk )(x) = (fk D)(x).4. Nu fk E(Rn), C0 (Rn), k = 1, 2, . . . ,E lim

kfk = 0 th c mt tp compact

K Rn supp(fk ) K, k = 1, 2, . . . .5. Nu fk E(Rn), S(Rn), k = 1, 2, . . . ,E lim

kfk = 0, th

c mt tp compact K supp fk K, k = 1, 2, . . . ,

47

c mt s dng R0, mt hm C0 (Rn) m (x) = 1, x K, supp BR0(0). M E

limk

fk = 0 c S limk

fk = 0 hay c mt s t nhin m v mt

s dng c

|(fk )(x)| = |fk, | supyRn

(1 + ||y||2)m||m

|D(y)|, k = 1, 2, . . . ,

trong , (y) = (y)(x y), supp BR0(0) v vi l = 1, 2, . . . ,

(1 + ||x||2)l supyRn

(1 + ||y||2)m||m

|D(y)|

c( ||m

(1 + ||x y||2)2l|D(x y)|2) 12 sup

yBR0 (0)(1 + ||y||2)l+m(

||m|D(y)|2) 12

do , vi k, l = 1, 2, . . . , c

(1 + ||x||2)l|(fk )(x)| Cl,m( ||m

(1 + ||x||2)2l|D(x)|2) 12 cl,m, (2.5) vi ||x||2 2cl+1,m

do E lim

kfk = 0 nn D

limk

fk = 0 m C0 (Rn),do c mt s k0 vi k k0 c (1 + ||x||2)l|(fk )(x)| 2 ,

vi ||x||2 2cl+1,mc (1 + ||x||2)l|(fk )(x)| (1+||x||2)l+1|(fk)(x)|1+||x||2 2 ,

D(fk )(x) = (fk D)(x).

Nhn xt. 1. Nu , D(Rn), f D(Rn) th f E(Rn), D(Rn). Do ,(f ) , f ( ) l hon ton xc nh.2. Nu E(Rn), D(Rn), f E(Rn) th f E(Rn), f D(Rn), D(Rn). Do , (f ) , (f ) , f ( ) l hon ton xc nh.3. Nu , S(Rn), f S(Rn) th f E(Rn), S(Rn). Do , (f ) , f ( ) l hon ton xc nh. Ngoi ra, c mt s t nhin l0, mt s dng c sao cho

|(f )(x)| c(1 + ||x||2)l0 ,x Rn.

Mnh 2.4. Cho , D(Rn), f D(Rn) hay E(Rn), D(Rn), f E(Rn)hoc , S(Rn), f S(Rn). Khi ,

(f ) = f ( ) = f ( ) = (f ) .

48

Chng minh. Trong trng hp c gi compact, tch chp (x), ((f ) )(x) uc dng tch phn Riemann trn hnh lp phng P cha supp

( )(x) =P

(x y)(y)dy, v ((f ) )(x) =P

(f )(x y)(y)dy,

nn tng Riemann

h(x) = hnk

(x kh)(kh), v gh(x) = hnk

(f )(x kh)(kh)

trong tng

k

ly trn cc im c to nguyn trong Rn, tng ny l tng hu hn

v supp l compact, hi t n ( )(x), ((f ) )(x) trong D(Rn) hay E(Rn), tutheo , f D(Rn) hay E(Rn) khi h gim dn v 0,m (f h)(x) = gh(x) nn nu D(Rn), f D(Rn) hay E(Rn), f E(Rn) th

(f ) = E limh0+

gh = E limh0+

f h = f ( ).

Trong trng hp D(Rn) c

(f ) = f ( ) = f ( ) = (f ) .

Trong trng hp E(Rn), f E(Rn), do C0 (Rn) tr mt trong E(Rn) nn c mt dy{k}k=1 trong C0 (Rn) hi t n trong E(Rn) do

f = E limk

f k, (f ) = E limk

(f ) k, = = E lim

k k = E lim

kk ,

m (f k) = f (k ) = f ( k) = (f ) k nn

(f ) = f ( ) = f ( ) = (f ) .

Trong trng hp , S(Rn), f S(Rn), do C0 (Rn) tr mt trong S(Rn) nn c mtdy {k}k=1 trong C0 (Rn) hi t n trong S(Rn) do

f = E limk

f k, (f ) = E limk

(f ) k, = = S lim

k k = S lim

kk ,

m (f k) = f (k ) = f ( k) = (f ) k nn

(f ) = f ( ) = f ( ) = (f ) .

49

Ch . Vi f S(Rn), , S(Rn) c

f, ( ) = (f ( ))(0) = ((f ) )(0) =Rn(f )(y)(y 0)dy

nn f c th coi l mt hm suy rng tng chm.Mnh 2.5. (i) Nu f D(Rn) th f = D lim

0+f .(ii) Nu f E(Rn) th f = E lim

0+f .(iii) Nu f S(Rn) th f = S lim

0+f .Chng minh. Ly D(Rn) khi f D(Rn), hay E(Rn) khi f E(Rn), S(Rn)khi f S(Rn). t (x) = (x) c

f , = ((f ) )(0) = (f ()))(0) = f, ( ),m theo Mnh 2.2 ( ) hi t n trong D(Rn) hay E(Rn), S(Rn) tu theo D(Rn) hay E(Rn), S(Rn) nn ta c iu phi chng minh.

Nhn xt. T Mnh 2.5 c tp C0 (Rn) tr mt trong E(Rn), C(Rn) tr mt trongD(Rn). Khi , vi a(.) C(Rn), f D(Rn) c mt dy {k}k=1 trong C(Rn) hit n f trong D(Rn). Do , vi mi Zn+ c

D limk

DaDk = DaDf,D limk

D(ak) = D(af), ,

m D(ak) =

(

)DaDk

nn D(af) =

(

)DaDf.T nhn xt ca Mnh 2.2, nu ta thay bi cc hm k C0 (Rn) m

Rnk(x)dx = 1, suppk+1 suppk,k=1 suppk = {0}

th ta cng cc kt lun nh Mnh 2.5.

nh l 2.6. C0 (Rn) l tp tr mt trong S(Rn),D(Rn).

Chng minh. Ly u D(Rn), C0 (Rn) hay u S(Rn), S(Rn).Do Bk(0) l tp compact nn c mt hm k C0 (Rn) m k(x) = 1, x Bk(0).t uk = (ku) 1

kc uk C0 (Rn) v

nu S(Rn) th = S limk

k( 1k ),

nu C0 (Rn) th = D limk

k( 1k ),nn lim

kuk, = lim

k(ku) 1

k, = lim

ku, k( 1

k ) = u, hay

nu u S(Rn), S(Rn) th u = S limk

uk,

nu u D(Rn), C0 (Rn) th u = D limk

uk.

50

2.1.3 Tch chp gia cc hm suy rng

Vi f D(Rn), g E(Rn) c mt dy {gk}k=1 trong C0 (Rn) m E limk

gk = g.

Vi mi D(Rn) c D limk

gk = g . Do ,limk

f gk, = limk

(f (gk ))(0) = (f (g ))(0),nn dy {f gk}k=1 l dy Cauchy trong D(Rn), m D(Rn) l , nn hi t n mthm suy rng, k hiu f g, xc nh bi cng thc (khng ph thuc vo vic chn dy{gk}k=1)

f g, = (f (g ))(0), D(Rn).Vi f E(Rn), g D(Rn) c mt dy {gk}k=1 trong C0 (Rn) m D lim

kgk = g. Vi

mi D(Rn) c E limk

gk = g . Do ,limk

f gk, = limk

(f (gk ))(0) = (f (g ))(0),nn dy {f gk}k=1 l dy Cauchy trong D(Rn), m D(Rn) l , nn hi t n mthm suy rng, k hiu f g, xc nh bi cng thc (khng ph thuc vo vic chn dy{gk}k=1)

f g, = (f (g ))(0), D(Rn).nh ngha 2.2. Cho f, g D(Rn), m t nht mt hm suy rng c gi compact. Hmsuy rng f g c gi l tch chp ca hm suy rng f theo hm suy rng g.V d 1. Vi f D(Rn) c f = f = f.Nhn xt. 1. Vi D(Rn), f, g D(Rn),(t nht mt hm suy rng c gi compact) c((f g) )(x) = f g, = lim

k(f (gk ))(0) = (f (g ))(x), (y) = (x y)nn (f g) = f (g ) v f g, = f, (g ).2. Vi E(Rn), g D(Rn),(t nht mt trong chng c gi compact) th nu coi nhmt hm suy rng c

g = limk

gk = limk

gk = g ..Khi , vi f, g D(Rn), (t nht mt hm suy rng c gi compact), do 1

k 1

k

C0 , supp( 1k 1

k) = B 2

k(0),

Rn( 1k

1k)(x)dx = 1 nn theo nhn xt ca Mnh 2.5

c

f g = D limk

f (g ( 1k 1

k))

= D limk

f ((g 1k) 1

k)( do 1

k C0 (Rn))

= D limk

f ( 1k (g 1

k))( do g 1

k C(Rn))

= D limk

(f 1k) (g 1

k)( do g 1

k C(Rn))

= D limk

(g 1k) (f 1

k)( do f 1

k C(Rn)) = g f.

51

Vi f, g D(Rn), (t nht mt hm suy rng c gi compact) do f g = D limk

f (g 1k)

m supp(f (g 1k)) (supp f + supp g + B 1

k(0)) nn nu x 6 (supp f + supp g) m

(supp f +supp g) l tp ng, c mt s kx x 6 (supp f +supp g+ B 1k(0)), khi k > kxdo f (g 1

k) = 0 ti x khi k > kx hay f g = 0 ti x. Nh vy, nu f g 6= 0 ti xth x ((supp f + supp g) hay supp(f g) (supp f + supp g).Vi f, g, h D(Rn), (nhiu nht mt hm suy rng khng c gi compact) th

(f g) h = D limk

(f g) (h 1k) = D lim

kf (g (h 1

k))

= D limk

f ((g h) 1k) = f (g h). (2.6)

Nu trong f, g, h c nhiu nht mt hm suy rng c gi compact th ng thc (2.6) ni

chung khng cn ng, chng hn trn R, vi 1 l hm hng bng 1 c

1 D = 0, 0 = 0, nn (1 D) = 0, D = , 1 = 1, nn 1 (D ) = 1.

Vi mi Zn+, do o hm suy rng D l nh x tuyn tnh lin tc trong D(Rn) nn

D(f g) = D limk

D(f (g 1k)) = (Df) g = f (Dg).

3. Vi f S(Rn), g E(Rn), S(Rn) th (g ) S(Rn) nn f, (g ) l xc nhhay f g l phim hm tuyn tnh trn S(Rn). Hn na, nu k S(Rn) m S lim

kk = 0

c S limk

(g k ) = 0 nn limk

f, (g k ) = 0 hay f g S(Rn).Vi f, g E(Rn) do supp(f g) (supp f + supp g) nn f g c gi compact hayf g E(Rn).Mnh 2.7. (i) Nu f D(Rn), g E(Rn) th

nh x h 7 f h = h f l nh x tuyn tnh lin tc t E(Rn) vo D(Rn), nh x h 7 h g = g h l nh x tuyn tnh lin tc t D(Rn) vo D(Rn), nh x h 7 h g = g h l nh x tuyn tnh lin tc t E(Rn) vo E(Rn).(ii) Nu f E(Rn), g S(Rn) th

nh x h 7 f h = h f l nh x tuyn tnh lin tc t S(Rn) vo S(Rn), nh x h 7 h g = g h l nh x tuyn tnh lin tc t E(Rn) vo S(Rn).Chng minh. Tnh tuyn tnh ca cc nh x l do cc khng gian D(Rn),E(Rn) l tuyntnh v vi mi , R, D(Rn)

f + h, (g ) = f, (g )+ h, (g ), f, ((g + h) ) = f, (g )+ f, (h ).

52

Gi s {hk}k=1 l mt dy trong D(Rn) v D limk

hk = 0 th

nu g E(Rn), D(Rn) c (g ) D(Rn) nn limk

hk, (g ) = 0.

Gi s {hk}k=1 l mt dy trong E(Rn) v E limk

hk = 0 th

nu g D(Rn), D(Rn) c (g ) E(Rn) nn limk

hk, (g ) = 0.

nu g E(Rn) c supp(g hk) (supp g + supphk), k = 1, 2, . . . ,

nu g S(Rn), hoc vi D(Rn) c s l,m Z+, l > m v cm > 0 sao cho(i)|g, | cm sup

xRn(1 + ||x||2)m

||m|D(x)|, D(Rn)(ii) t bt ng thc (2.5), vi mi k N c

supxRn

(1 + ||x||2)m||m

|D(hk )(x)| c supyRn

(1 + ||x||2)l||l

|D(x)|,

v c mt tp compact K Rn, supp(hk ) K,nn |g, (hk )| c sup

xRn(1 + ||x||2)l

||l|D(x)|, k = 1, 2, . . . ,hay vi S(Rn) c g E(Rn) nn lim

khk, (g ) = 0.

Gi s {hk}k=1 l mt dy trong S(Rn) v S limk

hk = 0 th vi g E(Rn), D(Rn)c (g ) D(Rn) v, t bt ng thc (2.4), mt s t nhin l0, mt s dng c sao cho

(1 + ||x||2)m||m

|D(g )(x)| c(1 + ||x||2)m||l0

|D(x)|,m N

m S(Rn) v S limk

hk = 0 nn c mt s t nhin m > l0 v mt s dng c

|hk, (g )| c supxRn

(1+ ||x||2)m||m

|D(g )(x)| supxRn

(1+ ||x||2)m||m

|D(x)|,

hay vi S(Rn) c g S(Rn) nn limk

hk, (g ) = 0.

nh l 2.8. (i) Cho L l mt nh x tuyn tnh lin tc t D(Rn) vo E(Rn) tho mn

Lh = hL, h Rn, D(Rn),

vi php dch chuyn h c xc nh bi h(x) = (x h).Khi , c duy nht mt hm suy rng T D(Rn) L = T , D(Rn)

(ii)Cho T D(Rn). Khi , nh x bin mi D(Rn) thnh T l nh x tuyntnh lin tc t D(Rn) vo E(Rn).

53

Chng minh. (i) Do nh x 7 l nh x tuyn tnh lin tc trong D(Rn) nn phimhm 7 L()(0) xc nh mt hm suy rng T D(Rn).Do vi mi D(Rn) c Lx = xL, x Rn nn

L()(x) = (xL())(0) = (L(x))(0) = T (x) = (T )(x)

nn L() = T .(ii) Vic chng minh da vo Mnh 2.3.

Nhn xt. T nh l 2.8 c th coi mi hm suy rng (phim hm tuyn tnh lin tc t

D(Rn) vo R) l mt nh x tuyn tnh t D(Rn) vo E(Rn) giao hon vi php ton dchchuyn, do tch chp ca hai hm suy rng (t nht mt trong chng c gi compact) c

th coi l mt nh x tuyn tnh t D(Rn) vo E(Rn) giao hon i vi php dch chuyn,v ngc li.

54

2.2 Php bin i Fourier

2.2.1 Php bin i Fourier trong S(Rn) v S(Rn)

nh ngha 2.3. Cho S(Rn). Bin i Fourier ca hm , k hiu l F, l hm cxc nh bi

F() = (2pi)n2

Rneix,(x)dx,

v bin i Fourier ngc ca hm , k hiu l F1, l hm c xc nh bi

F1() = (2pi)n2

Rneix,(x)dx.

Ch . Do |eix,| = |eix,| = 1 v L1(Rn) nn bin i Fourier F v bin iFourier ngc F1 l xc nh trn Rn. Ngoi ra, F() = F1(),F1() = F().K hiu B0(Rn) = {f L(Rn) C(Rn) lim

||x|||f(x)| = 0}.Khi , nu f L1(Rn) th Ff,F1f B0(Rn).V d 2. Cho (x) = e

||x||22 , bin i Fourier bng bin i Fourier ngc ca v

F() =nk=1

(2pi)12

+

eixkkx2k2 dxk =

nk=1

e2k2 (2pi)

12

+

e12(xkik)2dxk,

m hm e12z2l hm gii tch nn

0 =

CR

e12z2dz =

RR

e12t2dt

RR

e12(t+ik)2dt+

k0

e12R2eiRe

122d

k0

e12R2eiRe

122d,

trong CR l ng cong kn i theo chiu ngc chiu kim ng h gm cc on

[R,R], [R,R + ik], [R + ik,R + ik], [R + ik,R],m

limR

RR

e12t2dt =

+

e12t2dt, lim

R

RR

e12(t+ik)2dt =

+

e12(t+ik)2dt,

limR

k0

e12R2eiRe

122d = 0 = lim

R

k0

e12R2eiRe

122d

nn (2pi)12

+ e

12(xkik)2dxk = (2pi)

12

+ e

12x2kdxk = 1 hay F() = ().

Mnh 2.9. (i) Cho S(Rn). Khi , F,F1 S(Rn) v

DF() = (i)||F(x(x))(), DF1() = i||F1(x(x))(), F() = (i)||F(D(x))(), F1() = i||F1(D(x))().

55

T ta c, php bin i Fourier F v php bin i Fourier ngc F1 l nh x tuyntnh lin tc trn S(Rn).(ii)FF1 = F1F = , S(Rn).T ta c, php bin i Fourier F l mt ng cu tuyn tnh trn S(Rn) vi nh xngc chnh l php bin i Fourier ngc F1.(iii) Vi , S(Rn) c

RnF(x)(x)dx =

Rn()F()d,

Rn|(x)|2dx =

Rn|F()|2d.

T ta c, php bin i Fourier F l mt ng cu tuyn tnh trn S(Rn) theo tpL2(Rn) vi nh x ngc chnh l php bin i Fourier ngc F1. Khi , ta c th thctrin php bin i Fourier F ln thnh mt ng cu tuyn tnh ng c, t lin hp trn

L2(Rn) vi nh x ngc F1.(iv) Vi , S(Rn) c

F( )() = (2pi)n2F()F(), (2pi)n2F((x)(x))() = F() F(),F1( )() = (2pi)n2F1()F1(), (2pi)n2F1((x)(x))() = F1() F1().

Chng minh. (i) Gi s S(Rn). C ng thc

D (F)() = D

((2pi)

n2

Rneix,(x)dx

)= (2pi)

n2

Rn(ix)eix,(x)dx

= (i)||F(x(x))()

do eix,x(x) c tch phn trn Rn hi t u theo . Do , F C(Rn).Do vi mi Rn, , Zn+ c Dx(eix,(x)) tin v 0 khi ||x|| tin ra v cng nnbng php tnh tch phn tng phn c

F() = (2pi)n2

Rn(iDx)

eix,(x)dx = (2pi)n2

Rneix,(iDx)(x)dx

= (i)||F(D)().

Nh vy, vi mi , Zn+ c

D (F)() = (2pi)n

2

Rneix,(iDx)

((ix)(x))dxnn

supRn

|D (F)()| (2pi)n2 supxRn

|Dx((x)(x))|(1 + ||x||)n+1

Rn

1

(1 + ||x||)n+1dx

C supxRn

(1 + ||x||2)n+1+||

|D(x)|

56

F S(Rn), v php bin i Fourier l nh x tuyn tnh lin tc trn S(Rn).i vi php bin i Fourier ngc F1 ta chng minh tng t.(ii) Vi , S(Rn) do

Rneix,

((2pi)

n2

Rneiy,(y)dy

)F()d =

Rneix,()()d =

=

Rneix,F()

((2pi)

n2

Rneiy,(y)dy

)d

nn theo nh l Fubini cRn(y)

((2pi)

n2

Rneixy,F()d

)dy =

=

Rn(y)

((2pi)

n2

Rneixy,F()d

)dy

hay Rn(y)F1(F)(x y)dy =

Rn(y)F1(F)(x y)dy.

Chn (x) = (2pi)n2 e

||x||22 , (x) =

n(x), > 0 c

F() = F1() = (),

nn Rn(y)(x y)dy =

Rn(y)F

1(F)(x y)dy

khi , theo Nhn xt 2 sau Mnh 2.2, ta cho gim dn v 0 th (x) = F1(F)(x).Nh vy, F1(F) = , S(Rn). Do , F l ng cu tuyn tnh trn S(Rn) vi nh xngc F1.(iii) Vi , S(Rn) theo Fubini c

Rn(y)

((2pi)

n2

Rneiy,()d

)dy =

Rn()

((2pi)

n2

Rneiy,(y)dy

)d

nn Rn(y)F(y)dy =

Rn()F()d,

nu thay bi F1 = F c = F vRn|(y)|2dy =

Rn|F()|2d.

Nh vy, php bin i Fourier F l mt php ng cu tuyn tnh, t lin hp, ng c trn

khng gian S(Rn) vi metric L2(Rn).Do tp S(Rn) tr mt trong L2(Rn) nn c th thc trin php bin i Fourier F thnh mt

57

php ng cu tuyn tnh, t lin hp, ng c trn L2(Rn).(iv) Vi , S(Rn) t nh l Fubini c

(2pi)n2

Rneix,

( Rn(y)(x y)dy)dx =

=

Rneiy,(y)

((2pi)

n2

Rneixy,(x y)dx)dy,

(2pi)n2

Rneix,

( Rn(y)(x y)dy)dx =

=

Rneiy,(y)

((2pi)

n2

Rneixy,(x y)dx)dy,nn

F( )() = (2pi)n2F()F(),F1( )() = (2pi)n2F1()F1(),

do

(F F)() = (2pi)n2F(F1(F)F1(F))() = (2pi)n2F()(),(F1 F1)() = (2pi)n2F1(F(F1)F(F1))() = (2pi)n2F1())().

Mnh 2.10. Mt s tnh cht ca php bin i Fourier.

(i) F( h) = F[eihx(x)](), , h Rn, S(Rn).

(ii) F[(x h)]() = eihF(), , h Rn, S(Rn).

(iii)F[(tx)]() = |t|nF( t), t 6= 0, Rn, S(Rn).

(iv) Nu A GL(Rn) th F[(Ax)]() = 1| detA|F((A1)t).

nh ngha 2.4. Cho f S(Rn). Bin i Fourier ca hm suy rng f, k hiu Ff, l hmsuy rng tng chm c xc nh bi

Ff, = f,F, S(Rn),

v bin i Fourier ngc, k hiu F1f, l hm suy rng tng chm c xc nh bi

F1f, = f,F1, S(Rn).

Ch . 1. T phn (i) ca Mnh 2.9 php bin i Fourier F, v php bin i Fourier

ngc F1 l nh x tuyn tnh lin tc trn S(Rn) nn nh x 7 f,F v 7

58

f,F1 l nh x tuyn tnh lin tc t S(Rn) vo R, vi mi f S(Rn).Ngoi ra, vi S(Rn) c(D(Ff)), = (1)||fx,F(D) = fx, (ix)(F)(x) = F((ix)fx), (Ff), = fx,F() = fx, (iD)(F)(x) = F((iD)fx), nn D(Ff) = (i)||F(xf), (Ff) = (i)||F(Df).Tng t, c D(F1f) = i||F1(xf), (F1f) = i||F1(Df).Hn na, vi php bin i Fourier F, v php bin i Fourier ngc F1 l nh x tuyntnh lin tc trn S(Rn).2. T phn (ii) ca Mnh 2.9 c

F1Ff, = f,FF1 = f, = f,F1F = FF1f, , S(Rn),nn FF1f = F1Ff, f S(Rn). Do , php bin i Fourier F l ng cu tuyn tnhlin tc trn S(Rn) vi nh x ngc l php bin i Fourier ngc F1.3. Vi f S(Rn), S(Rn) theo Ch ca Mnh 2.4 th f S(Rn). Khi ,bin i Fourier F(f ) l hm suy rng tng chm c xc nh nh sau

F(f ), = f ,F = ((f ) (F))(0)= f, ( F1) = f, (F1(F) F1)= f, (2pi)n2F((F)) = (2pi)n2FFf, trong S(Rn), nn F(f ) = (2pi)n2FFf.Tng t, c F1(f ) = (2pi)n2F1F1f.Khi , c

Ff F = F((2pi)n2F1(F)F1(Ff)) = (2pi)n2F(f),F1f F1 = F1((2pi)n2F(F1)F(F1f)) = (2pi)n2F1(f).V d 3. F l hm hng (2pi)

n2 .

V d 4. Hm hng 1 (l hm suy rng tng chm) c bin i Fourier F1 = (2pi)n2 .

V d 5. Trong khng gian Rn+1, bin i Fourier ca hm Dirac theo bin x c xcnh nh sau

Fx, = (2pi)n2Rnxeix,0(x, 0)dx = (2pi)

n2

Rnx(x, 0)dx.

V d 6. Tm nghim c bn E ca phng trnh Parabolic:

(

tx)E = . (2.7)

Hm suy rng E tho mn phng trnh (2.7) khi v ch khi bin i Fourier (theo bin x)

FxE tho mn phng trnh vi phn thng

tFxE + ||||2FxE = Fx. (2.8)

59

C FxE(, t) = (2pi)n

2 e||||2t, t > 0 v FxE(, t) = 0, t 0, tho mn phng trnh(2.8)v vi C0 (Rn+1) c

tFxE + ||||2FxE, =

+0

Rn(2pi)

n2 e||||

2t(

t(, t) ||||2(, t))ddt

=Rn

+0

(2pi)n2 e||||

2t

t(, t)dtd

+

Rn

+0

(2pi)n2 e||||

2t||||2(, t)dtd

=(2pi)n2

Rn(, 0)d = Fx,

do E(x, t) = (2pit)n2 e

||x||22t , t > 0 v E(x, t) = 0, t 0 nghim c bn ca phngtrnh Parabolic (2.7).

V d 7. Trn R, vi mi R > 0, C0 (R) c

F[(R |x|)](), = (R |x|),F = RR

(2pi)12

+

eix(x)dxd

nn theo nh l Fubini, F[(R |x|)]() = RR eixd = ( 2pi ) 12 sin(R) .V d 8. Vi (x, t) R R tm nghim c bn ca phng trnh hyperbolic trn ngthng

(2

t2

2

x2)E = . (2.9)

Hm suy rng E tho mn phng trnh (2.9) khi v ch khi bin i Fourier theo bin x

ca n FxE tho mn phng trnh vi phn thng cp 2

2

t2FxE + ||2FxE = Fx. (2.10)

C FxE(, t) = (2pi) 12sin(t)tho mn phng trnh (2.10)v vi C0 (R2) c

2

t2FxE + ||2FxE, =

+0

R(2pi)

12sin(t)

(2

t2(, t) + ||2(, t))ddt

=

R

+0

(2pi)12sin(t)

2

t2(, t)dtd

+

R

+0

(2pi)12sin(t)

||2(, t)dtd

=(2pi)12

R(, 0)d = Fx,

do , E(x, t) = 12(t |x|).

60

2.2.2 Cc nh l Paley- Wiener

Cho C0 (Rn), supp BR(0), bin i Fourier F ca hm l mt hm gimnhanh, do C0 (Rn) S(Rn). Hn na, ta cn c th thc trin F ln trn Cn

F : 7 F() = (2pi)n2BR(0)

eix,(x)dx,

vi, x, =nk=1 xkk =nk=1 xkk + ink=1 xkk, k = k + ik.D thy, F() l hm kh vi v hn trn Cn. Ngoi ra, ta c

F() = (2pi)n2

||x||R

eix,(iD)(x)dx

= (2pi)n2

||x||R

ex,ix,(iD)(x)dx ( = + i)

nn |F()| CeR|| do , vi mi N > 0 u c mt s CN > 0

|F()| < CN(1 + ||||)NeR||=||, Cn. (2.11)

Tng t, c

|D(F)()| = |F(x)()| CR||eR||=||, C

nn F() l hm gii tch trn Cn.nh l Paley- Wiener s chng minh bt ng thc (2.11) l iu kin cn v mt

hm gii tch trn Cn l bin i Fourier ca mt hm kh vi v hn c gi compact trnRn.

nh l 2.11. Cho : Cn C l hm gii tch. Khi , iu kin cn v c mt sR > 0 mt hm C0 (Rn), supp BR(0) sao cho () = F() ltn ti mt s R > 0, vi mi N > 0 u c mt s CN > 0 sao cho

|()| < CN(1 + ||||)NeR||=||, Cn. (2.12)

Chng minh. iu kin cn c chng minh trn. Ta ch cn phi chng minh iu

kin . T bt ng thc (2.12) tch phn (x) = (2pi)n2

Rn e

ix,()d l xc nhvi mi x, ng thi n l hm kh vi theo x, do eix,() l hm c tch phn trn Rn

hi t u theo x. Vi mi Rn, do hm eix,() l hm gii tch trn Cn nn n

61

gii tch theo tng bin, tng t nh V d 2, c

(x) = (2pi)n2

Rn

eix,()d

= (2pi)n2

Rn

ei(x11+n

j=2 xjj)(1, 2, . . . , n)d

= (2pi)n2

Rn

ei(x11+x22+n

j=3 xjj)(1, 2, 3, . . . , n)d

= (2pi)n2

Rn

ei(n

j=1 xjj)(1, . . . , n)d

= (2pi)n2

Rn

eix,+i( + i)d.

Khi , t bt ng thc (2.12) c (.+ i) L2(Rn) nn t Mnh 2.9, phn (iii) c

F( + i) = (2pi)nxRn

eix,+iRn

eix,+i( + i)ddx

= (2pi)nxRn

eix,Rn

eix,( + i)dxdx = F(F1[(.+ i)])()

= ( + i)

v |(x)| CneR||||x,Rn(1 + ||||)n1dm

Rn(1 + ||||)n1d hi t,v nu ||x|| > R, = 1

tx th lim

t0+eR||||x, = lim

t0+e(R||x||)||x||

t = 0

nn (x) = 0, ||x|| > R, do C0 (Rn), supp BR(0).

T nh l Paley- Wiener, ta c th nh ngha bin i Fourier cho hm suy rng bng

cch sau.

nh ngha 2.5. Khng gian S(Cn) l khng gian bao gm tt c cc hm gii tch A(Cn) vi tnh cht

R > 0,N > 0,C = CN > 0 : |()| C(1 + ||||)NeR||=||, Cn,

vi khi nim hi t nh sau

mt dy {k}k=1 trong S(Cn) c gi l hi t n hm S(Cn) trong S(Cn) nu

(i) R > 0,N > 0,C = CR > 0 :

|k()| C(1 + ||||)NeR||=||, Cn, k = 1, 2, . . . ,

(ii) limk

supCn

|k() ()| = 0, Cn.

62

T nh l Paley- Wiener ta c php bin i Fourier F l mt ng cu tuyn tnh t

D(Rn) vo S(Cn), m c php nhng lin tc D(Rn) S(Rn)nn, theo Mnh 2.9, cnhng lin tc t S(Cn) vo S(Rn). T , mi hm suy rng tng chm c th coi l mtphim hm tuyn tnh lin tc trn S(Cn).

nh ngha 2.6. Cho f D(Rn). Bin i Fourier ca hm suy rng f, k hiu Ff, lmt nh x t S(Cn) vo C c xc nh nh sau,

7 Ff, = f, , S(Cn),

trong , C0 (Rn) c xc nh, theo nh l Paley- Wiener, t sao cho () =F().

Ch . 1. Php bin i Fourier F l mt ng cu tuyn tnh t D(Rn) vo S(Cn) nn Ffl mt phim hm tuyn tnh lin tc trn S(Cn).2. Vi f S(Rn), th bin i Fourier Ff c th thc trin ln thnh mt phim hm tuyntnh lin tc trn S(Rn) bng cch

7 f,F, S(Rn).

Do , nh ngha 2.6 khng mu thun vi nh ngha 2.4.

Phn u ca mc ny l nh l Paley -Wiener cho hm c bn. Tip theo, ta s trnh

by nh l Paley-Wiener- Schwartz cho hm suy rng.

Cho f E(Rn). Do E(Rn) S(Rn) nn bin i Fourier Ff c xc nh nh sau

7 f,F, S(Rn).

Do S(Rn) c mt dy {k}k=1 trong C0 (Rn) hi t n trong S(Rn).Vi mi k, gi suppk l tp compact trong Rn,Fk E(Rn), nn cc tng Riemann

h() = (2pi)n

2

j

eijh,k(jh),

trong ,

j

l tng ly trn cc im c to nguyn trong Rn,tng ny l hu hn v

gi suppk tp compact, hi t n Fk() trong E(Rn) khi h gim dn v 0 nn

f,Fk = limh0+

f, h = (2pi)n2 limh0+

j

f, eijh,k(jh)

= (2pi)n2 limh0+

j

k(jh)f, eijh,

m tng Riemann

h() = (2pi)n

2

j

k(jh)f, eijh,

63

hi t n (2pi)n2

xRn k(x)f, eix,dx khi h gim dn v 0, do

f,Fk = (2pi)n2xRn

k(x)f, eix,dx.

Li c do S limk

k = nn S limk

Fk = F, m f E(Rn) c cp 0, vi mix Rn hm eix, E(Rn) nn c s dng C v s t nhin m khng ph thuc x saocho

|f, eix,| C supRn

||m

|D (eix,)| C(1 + ||x||)m

do cho k ra v cng c

f,F = (2pi)n2xRn

(x)f, eix,dx

do , hm suy rng Ff c th vit di dng hm thng thng (2pi)n2 f, eix,.V d 9. Trn R3, R > 0 hm suy rng (R ||x||) c xc nh nh sau

7 (R ||x||), =||x||=R

(x)dS.

D thy gi supp (R ||x||) = SR l tp compact hay (R ||x||) E(R3). Khi , bini Fourier ca (R ||x||) l hm

F[(R ||x||)]() = (2pi) 32 (R ||x||), eix, = (2pi)n2||x||=R

eix,dS,

chuyn sang h to cc (vi trc (0, 0, t) = )c

F[(R ||x||)]() = (2pi) 32R2 pi0

2pi0

sin()eiR cos()||||dd

= (2pi)32R4pi

sin(R||||)|||| .

V d 10. Trong (x, t) R3R, tm nghim c bn E ca phng trnh sng trong khnggian

(2

t2x)E = . (2.13)

Hm suy rng E l nghim ca phng trnh (2.13) khi v ch khi bin i Fourier ca E

(theo bin x)FxE tho mn phng trnh vi phn thng

2

t2FxE + ||||2FxE = Fx. (2.14)

64

C FxE(x, t) = (2pi)32

sin(t||||)|||| , t > 0 v FxE(x, t) = 0, t 0 tho mn phng trnh(2.14), v vi C0 (R2) c

2

t2FxE + ||||2FxE, =

+0

R3(2pi)

32sin(t||||)|||| (

2

t2(, t) + ||||2(, t))ddt

=

R3

+0

(2pi)32sin(t||||)||||

2

t2(, t)dtd

+

R3

+0

(2pi)32sin(t||||)|||| ||||

2(, t)dtd

=(2pi)32

R(, 0)d = Fx,

do , E(x, t) = 14pit(t ||x||), t > 0, E(x, t) = 0, t 0.Nh vy, vi f E(Rn) th bin i Fourier Ff l mt hm t Rn vo C c xc nhbi

7 fx, eix,.Hm Ff() l hm kh vi v hn v c th thc trin ln thnh mt hm gii tch trn Cn

nh sau

7 fx, eix,.Do f E(Rn), nn c mt s R > 0 supp f BR(0). Khi , c mt hm C0 (R)m (t) = 0, t 1, (t) = 1, t 1

2.

t (x) = eix,(||||(||x|| R)) c

vi mi th C0 (Rn),

vi = 0 th 0(x) = eix,0, D(x) = 0,x Rn, 6= 0,

vi 6= 0 th (x) = eix,,||x|| R + 12|||| , supp BR+ 1|||| (0), v

D(x) =

(i)eix,Dx (||||(||x|| R)),

m f E(Rn), supp f BR(0)

nn fx, eix, = f, , v c mt s N Z+, v mt s c > 0 sao cho

|f, | c supxRn

||N

|D(x)|

c(1 + ||||)NeR||=||. (2.15)

nh l Paley -Wiener- Schwartz khng nh iu kin cn v mt hm gii tch l

bin i Fourier ca mt hm suy rng c gi compact l bt ng thc (2.15).

65

nh l 2.12. Cho : Cn C l hm gii tch. Khi , iu kin cn v c mt sR > 0, mt hm suy rng f E(Rn), supp f BR(0) sao cho () = Ff() ltn ti s R,N,C > 0 sao cho

|()| C(1 + ||||)NeR||=||, Cn. (2.16)Chng minh. iu kin cn c chng minh trn. Ta ch cn phi chng minh iu

kin . T bt ng thc (2.16) Vi mi Rn, hm (. + i) S(Rn) nn bin iFourier ca n f = F1[()] S(Rn) v Ff() = (), Cn.Do D(Rn)( S(Rn)), f S(Rn) nn nu t f = f th hm

F(f) = (2pi)n2F()F(f) = (2pi)

n2F()

l hm gii tch trn Cn.Li c, do C0 (Rn), supp() B(0) nn theo nh l Paley- Wiener vi miN1 > 0,u c mt s CN1 > 0

|F()()| CN1(1 + ||||)N1e||=||, Cn

do , t bt ng thc (2.11), c

|F(f)()| CN1C(1 + ||||)N1+Ne(R+)||=||, Cn.T nh l Paley- Wiener c supp(f) BR+(0),m S lim

0+f = f nn supp f BR(0).

Cho f S(Rn), g E(Rn). T nh l Paley- Wiener- Schwartz, bin i Fourier cahm suy rng c gi compact g l Fg() = gx, eix, l hm kh vi v hn trn Rn vc s N,C > 0 sao cho

|Fg()| C(1 + ||||)N , Rn,nn vi mi Zn+ u c s N, C > 0 sao cho

|D(Fg)()| = |||Fg()| C(1 + ||||)N , Rn,do , tch FgFf S(Rn). Ngoi ra f g S(Rn) nn F(f g) S(Rn).Khi , vi mi S(Rn) c

FgFf, = Ff, (Fg) = f,F((Fg))= (2pi)

n2 f,F(Fg) F() (Fg S(Rn))

= (2pi)n2 f, g F() (Fg = F1g)

= (2pi)n2 f (g F()) = (2pi)n2 f (g (F))

= (2pi)n2 f g,Fnn F(f g) = (2pi)n2FgFf.Tng t, F1(f g) = (2pi)n2F1gF1f.

Chng 3

Khng gian Sobolev

3.1 nh ngha v mt s tnh cht

3.1.1 Khng gian Sobolev cp nguyn khng m

Cho l mt tp m trong Rn. Cho f L2()( L2loc() L1loc()), khi f c thcoi nh mt hm suy rng c xc nh nh sau

7 f, =

f(x)(x)dx, D().

T bt ng thc Cauchy- Schwartz c nh gi

|f, | (

|f(x)|2dx) 1

2(

|(x)|2dx) 1

2, D(). (3.1)

Ta s chng minh bt ng thc (3.1) l iu kin cn v mt hm suy rng l mt

hm trong L2(). Vi ch , khng gian i ngu (gm cc phim hm tuyn tnh lin tc)

ca L2() cng l L2() nn ta s pht biu Mnh di dng sau.

Mnh 3.1. (i) Cho f D(). Nu c mt s dng C sao cho

|f, | C(

|(x)|2dx) 12 , D() (3.2)

th ta c th thc trin f ln thnh mt phim hm tuyn tnh lin tc trn L2().

(ii) Cho f L2(). Khi , thu hp ca f trn D() l mt hm suy rng tho mn btng thc (3.2) vi hng s C = ||f ||L2 .

Chng minh. (i) Do C0 () tr mt trong L2() nn nu hm suy rng f tho mn bt ng

thc (3.2) th ta c th thc trin f ln thnh phim hm tuyn tnh lin tc trn L2(). Thc

trin ny l thc trin duy nht.

(ii) Phn ny c chng minh t bt ng thc Cauchy- Schwartz.

65

66

Ch . Nh vy, ta c th coi L2() l khng gian tt c cc hm suy rng tho mn bt

ng thc (3.1), vi ch f, g L2() bng nhau theo ngha suy rng khi v ch khi bngnhau trong L2().

Nu coi mt hm bnh phng kh tch l mt hm suy rng th s hi t theo ngha suy

rng chnh l s hi t yu. S hi t yu ny khng dn n s hi t trong L2()

ngay c khi thm c tnh b chn u. Chng hn, ta xt v d sau trn ng thng R,ly fk(x) = [k,k+1](x) hi t yu v 0 khi k tin ra v cng v vi mi D(Rn)c mt s k0 N supp [k0, k0], nn

R [k,k+1](x)(x) = 0, vi k > k0. v

||[k,k+1]||L2 = 1, k = 1, 2, . . . , nn dy {[k,k+1]}k=1 khng hi t trong L2(R).Khc vi cc khng gian hm suy rng khc, nu f D(S,E,E, S,D) th Df D(S,E,E, S,D), Zn+, mt cch tng ng; nu f L2() th cha chc Df L2(), ngay c khi f c o hm suy rng cp cao hn nm trong L2().

Chng hn, trn = [1, 1], hm Heaviside H(t) L2(1, 1) nhng DH = 6L2(1, 1). Hoc hm du sgn(t) L2(1, 1) nhng D sgn(t) 6 L2(1, 1) v gi s khngphi th 1

1D sgn(t)(t)dt =

11

sgn

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Li Thuyet Ham Suy Rong & Khong Gian Sobolev - Dang Anh Tuan