EXTREME POINTS AND ROTUNDITY OF ORLICZ-SOBOLEV SPACES

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EXTREME POINTS AND ROTUNDITY OF ORLICZ-SOBOLEV SPACES

SHUTAO CHEN, CHANGYING HU, and CHARLES XUEJIN ZHAO Received 28 February 2002

It is well known that Sobolev spaces have played essential roles in solving nonlinear par- tial differential equations. Orlicz-Sobolev spaces are generalized from Sobolev spaces. In this paper, we present sufficient and necessary conditions of extreme points of Orlicz- Sobolev spaces. A sufficient and necessary condition of rotundity of Orlicz-Sobolev spaces is obtained.

2000 Mathematics Subject Classification: 47L10.

Definition1. LetA(u)=_|u|

0 p(t)dt, wherep(t)satisfies the following proper- ties:

(1) p(t)is right-continuous and nondecreasing;

(2) p(t) >0(t >0);

(3) p(0)=0, lim_t→∞p(t)= ∞.

ThenA(u)is called anN-function andp(t)is called the right derivative ofA(u).

Definition2. LetA(u)be anN-function,p(t)the right derivative ofA(u). Let

q(v)=sup

u≥0 :p(u)≤v

=inf

u≥0 :p(u)≥v

. (1)

Then ¯A(v)=_|v|

0 q(t)dtis called the complementary function ofA(u).

Definition3. LetA(u)be anN-function,u∈R, ifv, w∈R,v+w=2u,u≠v, impliesA((v+w)/2) < (1/2)(A(v)+A(w)). Thenuis called a strictly convex point ofA. The set of strictly convex points ofAis denoted bySA.

Definition4. LetA(u)be anN-function,Ω⊂Rⁿ, Orlicz space is defined as fol- lows:

LA(Ω)=

u(t):∃λ >0,such that

ΩA λu(t)

dt <∞

. (2)

Definition5. LetA(u)be anN-function, andΩbe a bounded and connected field ofRⁿ. Orlicz-Sobolev space is defined as follows:

W_m,A⁰ =

u∈LA(Ω):∂^αu∈LA(Ω),|α| ≤m

, (3)

whereα=(α1, α2, . . . , αn),|α| =α1+α2+···+αn,∂^αuis a distribution ofu.

Foru∈W_m,A⁰ , its norm is defined as

u⁰m,A=

0≤|α|≤m

∂^α(u)⁰p

1/p

, 1≤p <∞. (4)

Orlicz-Sobolev spaces with the norm defined above are Banach spaces, see [1].

Definition6. For anyx≠0,x∈LA(Ω), let

K^∗_x=inf

K >0 :

Ω

A¯ p

kx(t) dt≥1

, K_x^∗∗=sup

K >0 :

Ω

A¯ p

kx(t) dt≤1

.

(5)

Thenk^∗_x≤k^∗∗_x . We setK(x)=[k^∗_x, k^∗∗_x ].

Definition7. LetXbe a Banach space,B(X)the closed unit ball ofX, andS(X) its unit sphere. Letx∈S(X). Ify, z∈B(X), y+z=2x impliesx=y=z, thenx is called an extreme point ofB(X). The set of extreme points ofB(X)is denoted by extB(X). IfS(X)=extB(X), thenXis called a rotund space.

Lemma8. For anyx∈L⁰_A,x⁰A=(1/k){1+

ΩA(kx(t))dt}if and only ifk∈K(x). Theorem 9. Letx∈S(W_m,A⁰ ). Ifµ{t∈Ω:kx(t)∉SA} =0, k∈K(x), thenx∈ extB(W_m,A⁰ ).

Proof. Lety, z∈B(W_m,A⁰ ), andy+z=2x. By the convexity off (u)=u^p,(1≤ p <∞)

1=

y⁰m,A

p

+ z⁰m,A

p

2 =

0≤|α|≤m

∂^αy⁰p

+∂^αz⁰p

2

≥

0≤|α|≤m

∂^αy⁰+∂^αz⁰ 2

p

≥

0≤|α|≤m

∂^αy+∂^αz 2

⁰p

=

0≤|α|≤m

∂^αx⁰p

=1^p=1.

(6)

So the equality holds in the above inequalities. Since for any 0≤ |α| ≤m, we have ∂^αy⁰p

+∂^αz⁰p

2 ≥ ∂^αy⁰+∂^αz⁰ 2

p

≥

∂^αy+∂^αz 2

⁰p

. (7)

From (6) and (7), we know that the equality holds in (7). In particular, whenp >1, ∂^αy⁰+∂^αz⁰=2∂^αx⁰. (8)

Takeh∈K(y),l∈K(z), and letk=hl/(h+l). Then

2x⁰= y⁰+z⁰

= 1 h

1+

ΩA hy(t)

dt

+1 l

1+

ΩA lz(t)

dt

=h+l hl +1

h

ΩA hy(t)

dt+1 l

ΩA lz(t)

dt

=h+l hl

1+

Ω

l h+lA

hy(t) + h

h+lA lz(t)

dt

≥h+l hl

1+

ΩA hl

h+l

y(t)+z(t) dt

≥2· 1 2k

1+

ΩA 2kx(t)

dt

≥2x⁰.

(9)

So the equality holds in the above inequalities. Hence 2k∈K(x)and for a.e.t∈Ω, (l/(h+l))A(hy(t))+(h/(h+l))A(lz(t))=A(2kx(t)). By the known conditions, for almost allt∈Ω,hy(t)=lz(t)=2kx(t). Therefore,

l=lz⁰m,A= lz⁰m,A= hy⁰m,A=hy⁰m,A=h. (10)

This impliesx=y=z. Sox∈extB(W_m,A⁰ ).

Theorem10. Letx∈S(W_m,A⁰ ). If for anyi=1,2, . . . , n,µ{t∈Ω:ki∂ix(t)∉SA} = 0, whereKi∈K(∂ix(t)). Thenx∈extB(W_m,A⁰ ).

Proof. Lety, z∈B(W_m,A⁰ ), and y+z=2x. By the proof ofTheorem 9, for any 0≤ |α| ≤mwe have

2∂^αx⁰=∂^αy⁰+∂^αz⁰. (11) In particular, if|α| =1, then 2∂ix⁰= ∂iy⁰+∂iz⁰. Takehi∈K(∂iy),li∈K(∂iz), and letki=hili/(hi+li). By the proof ofTheorem 9, we have

hi∂iy(t)=li∂iz(t)=2ki∂ix(t), i=1,2, . . . , n (12)

andli=hi=2ki. Hence∂iy(t)=∂ix(t)=∂iz(t). Thus there exists a constantcsuch thaty(t)=x(t)+c,z(t)=x(t)−c. Now, we show thatc=0. If not,c≠0. Without loss of generality, we may assume thatc >0. If|x|< c, theny(t) >0,z(t) <0. Since 0∈SA, whena >0,b <0, for anyλ∈(0,1), we haveA(λa+(1−λ)b) < λA(a)+(1−λ)A(b).

By (9),|x(t)|< cdoes not hold. Then for a.e.t∈Ω,|x(t)| ≥c.

LetE1= {t∈Ω:x(t)≥c},E2= {t∈Ω:x(t)≤ −c}. Thenµ(E1∪E2)=µΩ. SinceΩ is connected, for anyp∈E1,q∈E2,pcan continuously move toqinΩby a transform of finite single-variable. IfµE1>0 andµE2>0, there exists at least ap∈E1,q∈E2

such that the connecting line betweenp and q over E1∪E2 is condense. So there exists a linel= {(t1, t2, . . . , t_i−1, λt_i+1, . . . , tn) λ∈[a, b]}on that connecting line, such thatl∩E1≠∅,l∩E2≠∅. Butx(t)≥coverE1andx(t)≤ −coverE2whereasE1∪E2

is condense ofl. This is a contradiction to the fact that∂ix(t)∈LA⊂L1implies that x(t)is absolutely continuous with respect toti. So, eitherµE1=0 orµE2=0. Without loss of generality, letµE2=0. Then for almost allt∈Ω,x(t)≥c. So,y(t) > x(t).

Thusy⁰m,A>x⁰m,A=1. This contradictsy∈B(W_m,A⁰ ). From above, we know that c=0. Sox(t)=y(t)=z(t). This meansx∈extB(Wm,A).

Theorem11. Letx∈S(W_m,A⁰ ). For anyi=1,2, . . . , n, µ

t∈Ω:kx(t)∉SA

∩

t∈Ω:ki∂ix(t)∉SA

=0, ki∈K

∂ix

, k∈K(x), (13)

thenx∈extB(W_m,A⁰ ).

Proof. Lety, z∈B(W_m,A⁰ )andy+z=2x. LetB= {t∈Ω:kx(t)∉SA},Bi= {t∈ Ω:ki∂ix(t)∉SA}, andy(t)=x(t)+δ(t).

Case1. For almost allt∈Ω\B,δ(t)=0 byTheorem 10. Thereforex(t)=y(t)= z(t).

Case2. For anyi=1,2, . . . , n,µ(B∩Bi)=0, so for almost allt∈B,t∉Bi. Hence

∂ix(t)∈SA. By the proof ofTheorem 10, we know that∂iδ(t)=0, whenδ(t)=c. Sim- ilarly,x(t)=y(t)=z(t)byTheorem 10. By Cases1and2we knowx∈extB(W_m,A⁰ ).

Theorem12. Letx∈S(W_m,A⁰ ). If there exists an affine interval(aα, bα)and >0 such that

int

0≤|α|≤m

t∈Ω:∂^αkαx(t)∈

aα+, bα−

≠∅, (14)

thenx∉extB(W_m,A⁰ ). Proof. LetG=

0≤|α|≤m{t∈Ω:kα∂^αx(t)∈(aα+, bα−)}and intG≠∅. Take t, t∈intG,r >0 such thatB(t, r )=B1⊂G,B(t, r )=B2⊂G, andB1∩B2= ∅. For anyt^∗∈ΩsatisfyingB(t^∗, r )⊂Ω. Define

Jt^∗(t)=





e^{−1/(r2−ni=1(ti−t∗i)2)}, t∈B t^∗, r

,

0, t∈Ω\B

t^∗, r

. (15)

ThenJt^∗(t)is an infinitely differentiable function on Ω and for any 0≤ |α| ≤m,

∂^αJt^∗(t)=0 onΩ\B(t^∗, r ). Let

c= min

0≤|α|≤m

1

maxt∈Ω∂^αJt∗(t)

. (16)

Thenc >0 and for allt∈Ω,c∂^αJt^∗(t)≤. Define

y(t)=x(t)+cJt(t)−cJt, z(t)=x(t)−cJt(t)+cJt. (17)

Theny, z∈W_m,A⁰ , andy+z=2x,y≠z. LetA(u)=hαu+bαon(aα+, bα−). For anykα∈K(∂^αx),

∂^αy⁰= 1 kα

1+

ΩA

kα∂^αy(t) dt

= 1 kα

1+

Ω\(B₁∪B₂)A

kα∂^αx(t) dt+

B₁A

kα∂^αx(t)+kα∂^α

cJt(t) dt +

B₂A

kα∂^αx(t)−kα∂^α

cJt(t) dt

= 1 kα

1+

Ω\(B1∪B2)

A

kα∂^αx(t) dt +

B1

hαkα∂^αx(t)+bα

dt+

B1

hαkα∂^α cJt(t)

dt +

B₂

hαkα∂^αx(t)+bα

dt−

B₂

hαkα∂^α

cJt(t) dt

= 1 kα

1+

ΩA

kα∂^αx(t) dt

=∂^αx⁰.

(18)

Hence for any 0≤ |α| ≤m, we have∂^αy⁰= ∂^αx⁰.

Likewise, for any 0≤ |α| ≤m, we have∂^αz⁰= ∂^αx⁰. Then

y⁰m,A= z⁰m,A= x⁰m,A=1. (19)

Thereforey, z∈S(W_m,A⁰ ).We know thatx∉extB(W_m,A⁰ )sincey≠z.

Theorem13. We show thatW_m,A⁰ is rotund if and only ifAis strictly convex.

Proof

Sufficiency. It is immediately obtained fromTheorem 9.

Necessity. SupposeAis not strictly convex. Then there exists 0< a < bsuch that A(u)is an affine function on(a, b). SinceΩis bounded, we can taket∈Ω¯,t∈Ω¯ such that

n i=1

t_i= inf

(t₁,t₂,...,t_n)∈Ω

n i=1

ti, n i=1

t_i= sup

(t₁,t₂,...,tn)∈Ω

n i=1

ti. (20)

(1) When

ΩA(p((a¯ +b)/2))dt <1, we setg(c)=

ΩA(p(((a¯ +b)/2)e^{cni=1(ti−ti)}))dt.

Then by the continuity of ¯A and the right continuity ofp,g(c)is right continuous

with respect toc and g(0)=

ΩA(p((a¯ +b)/2))dt <1, lim_c→∞g(c)= ∞. Takec0= inf{c >0 :g(c)≥1}, then the following two statements hold:

(a) g(c0)≥1, soc0>0;

(b) for anyl∈(0,1),

ΩA(p(((a¯ +b)/2)le^{c0ni=1(ti−ti)}))dt <1.

Indeed, takecnc0such thatg(cn)≥1. Theng(c0)=lim_n→∞g(cn)≥1 sinceg(c)is right continuous. So (a) holds.

Letλ=sup_{(t1,t2,...,tn)}n

i=1(ti−t_i). Then for anyt∈Ω,λ≥n

i=1(ti−t_i) >0. For any 0< l <1, since lnl <0,

0< la+b

2 e^{c0ni=1(ti−ti)}=a+b

2 e^{lnl+c0ni=1(ti−ti)}≤a+b

2 e^{(c0+lnl/λ)ni=1(ti−ti)}. (21) By the definition ofc0,

Ω

A¯

p

la+b

2 e^{c0ni=1(ti−ti)}

dt≤g

c0+lnl λ

<1. (22)

Letx(t)=((a+b)/2)e^{c0ni=1(ti−ti)}. By the above discussion, 1∈K(x). Thenx⁰= 1+

ΩA(x(t))dt. Letx0(t)=x(t)/x⁰m,A. Thenx0(t)∈S(W_m,A⁰ )and x0⁰= x⁰

x⁰m,A

= 1 x⁰m,A

1+

ΩA x(t)

dt

= 1 x⁰m,A

1+

ΩA

x⁰m,Ax0(t) dt

.

(23)

Therefore x⁰m,A ∈K(x0(t)). Set 1/b0= x⁰m,A. Since (t1, t2, . . . , tn)∈Ω, x(t)→ (a+b)/2 asti→t_i, we can choose a ballB⊂Ωsuch thatx(B)⊂(a, b). It means that

t∈Ω:x(t)∉SA

⊃B. (24)

Therefore,

t∈Ω: 1 b0

x0(t)∉SA

⊃B. (25)

On the other hand, as 1≤ |α| ≤m,

∂^αx0(t)= ∂^αx(t) x⁰m,A

= c^|α|₀ x⁰m,A

x(t)=bαx(t), (26)

wherebα=c₀^|α|/x⁰m,A. ByLemma 8, 1/bα∈K(∂^αx(t)). So

t∈Ω: 1 bα

∂^αx0(t)∉SA

⊃B. (27)

Then,

int

0≤|α|≤m

t∈Ω: 1 bα

∂^αx0(t)∉SA

≠∅. (28)

ByTheorem 12, we knowx0∉extB(W_m,A⁰ ). This is a contradiction.

(2) When

ΩA(p((a+¯ b)/2))dt≥1.

Set g(c)=

ΩA(p(((a+¯ b)/2)e^{cni=1(ti−ti)}))dt. Then g(c)is left-continuous with respect to c. For any (t1, t2, . . . , tn)∈Ω, n

i=1(ti−t_i) <0, and g(0)=

ΩA(p((a¯ + b)/2))dt≥1, lim_c→∞g(c)=0. Takec0=sup{c >0 :g(c)≤1}. As in (1), we can prove g(c0)≤1 and for anyl >1,

Ω

A¯

p

la+b

2 e^{c0ni=1(ti−ti)}

dt≥1. (29)

Letx(t)=((a+b)/2)e^{c0ni=1(ti−ti)},x0(t)=x(t)/x⁰m,A. Then x0∈S(W_m,A⁰ ). Like- wise, we can showx0∉extB(W_m,A⁰ ). This is also a contradiction.

By (1) and (2) we know thatAis strictly convex.

Acknowledgment. This work was supported by the Chinese Science Foundation and Heilongjiang Province Science Foundation.

References

[1] R. A. Adams,Sobolev Spaces, Pure and Applied Mathematics, vol. 65, Academic Press, 1975.

Shutao Chen: Department of Mathematics, Harbin Normal University, Harbin 150080, China

Changying Hu: Department of Mathematics, Harbin Normal University, Harbin 150080, China

Charles Xuejin Zhao: Department of Mathematics, P.O. Box20047, Savannah State University, Savannah, GA31404, USA

E-mail address:[email protected]

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