J Smoothness of Orlicz spaces

ANNALES SOCIETAT1S MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE XXVII (1987) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXVII (1987)

Chen Shutao

(Harbin)

Smoothness of Orlicz spaces

It is well known that the smoothness of Banach spaces is important. It is applied in optimal control ([

]), best approximation ([

]), prediction operator ([3]) and other theories.

M. M. Rao ([9]—[10]) discusses the smoothness of Orlicz spaces with respect to Luxemburg norm. But those papers have not solved the problem completely and an incorrect assertion that (EN, ||j|(JV))* = (L*M, II

(should be (L*M, || •||M)) appears in one of them. This error leads to some wrong results in the paper.

This paper deals with the smoothness of Orlicz spaces equipped with Orlicz norm which does not seem to have been investigated up to now.

Criteria for the smoothness, very smoothness, strong smoothness and uniform smoothness of the spaces are given.

A Banach space X is said to be smooth at x in X if there exists a unique f x in X* such that f x (x) = ||/x||||x|| = ||x||. The mapping xt-> /x is called a support mapping. X is said to be smooth if it is smooth at every point in its unit sphere S(X). X is said to be very (strongly or uniformly) smooth if it is smooth and its support mapping is norm to weak (norm or norm uniformly) continuous from S(X) to S(X*) (see [4]).

Throughout this paper, we denote by M(u), N{u) a pair of conjugate N- functions; p(u), q(u) their right derivatives; L*M, Ц М) the Orlicz spaces generated by M(u) and equipped with Orlicz norm and Luxemburg norm, respectively; EM, E{M) the closure of bounded measurable functions in them, respectively. Moreover, we assume the support set G of elements of the spaces is an atomless set with finite measure.

For convenience, we denote R M{u) = j M (u(t))dt; by M

A2 that M(u)

G

satisfies d 2-condition for large u; by M

V2 that N

A2.

The Orlicz norm and the Luxemburg norm is indicated respectively by

(

) ||м||м = sup J u(t)v(t)dt,

R N (v) ^ 1 g

(2) - ||м||(М) = inf {k> 0: R M(u/k) ^ l j .

4 — Prace Matematyczne 27.1

P

1 (see [1]). For any f in (L*M)*, the dual of L*M, there exist a unique v in I?N and a singular functional Ф (i.e.,for every x in Е м, Ф(х) = 0), such that

(3) f( u ) = г(и) + Ф{и) = $u{t)v(t)dt + 0(u) (ueL*M)

G

and when f e(L*M))*,

ii / ii = imu = iM u+№

when /e(L*M)*,

(4) Ц/ll = \\f\\iN) = т Ц Щ > 0: R N(çv) + ï 1|ФЩ < 1 /

Proposition

2 (see [1]). ||/||^ = ||/ ||(N) iff f is a singular functional.

By Proposition 2, \\u\\M >\\u\\{m for any non-nu]l element и in 1?л/.

Proposition 3.

Any non-null singular functional Ф in cannot attain its norm on S(F?M).

P ro o f. Otherwise, there exists и in S(L*M) such that Ф(и) = ||Ф

Ф 0.

By Proposition 2, ||w||M > ||

||(W) and therefore

\\Пм = ЦФЩ = Ф(и) ^ ||Ф |Ы |и |и < ||Ф|Ы|

<|и ^ ||Ф|и, a contradiction.

For any finite function w(f), we always write

Gn(u) = {teG: |u(f)| ^ n], u„(t) = M(f)xG\G » (0 .

where %E is the characteristic function of the set E. Clearly, ме!*м implies u„(t)eEM and lim m esG„(u) = 0. Now, we introduce notations

di{u) = d(u, EM) = in f { ||« - w ||M: w e E M), d2{u) = d(u, E(M)) = in f{||

—w||(A#): w e £ Mj,

£o = £o («) = inf {<* > 0: R M (u/Ç) < oo j . Obviously, u e E M iff d t (и) = d2 (w) = £o = 0.

Proposition 4.

For any и in F?M

lim \ \ u - u n\\(M) = lim ||u - u „ ||M = £0.

П GO П -+ CO

P ro o f. It is trivial by [7], Lemma 10.1 when u e E M. Now we assume

u e ! f M\ E M; then

> 0. Since \\un- u \\(m and ||ия —и||м are non-increasing

Smoothness o f Orlicz spaces 51

when n increases, the limits in the proposition exist. For any given £e(0, ç0), by the definition of £0,

R M и

4£o + £ (it < 00 .

( и — un \

Notice that lim mes Gn (u) = 0, we have RM --- < 1 for all n large

n

\S o T £ /

enough. By (2), ||м -м и||(М)

+ £ for all large n. Thus, lim ||u -u „||(M) ^ f 0;

hence e is arbitrary.

/ и \ (u — un \

On the other hand, since R M - --- ) = со, we have R M - ---=

Ц о - s J \Qo- £ /

for all 1. It implies by (2) that \\u — u„||(M) ^

— £ for all n ^ 1. Therefore, lim \\ u - u n\\(M) = £0-

« “♦00

Since ||

-

„||

^ \ \ u - u n\\(M) for all n ^ 1, we have lim \\u — u„\\M ^ ç0-

л -> 00

( U \ л r n ( U~ Un

Conversely, for any

0, we have RM

--- I <

therefore R M I -- ---

\ so + £/ Vso + £

-> 0 as n-+oo. Recall [7], Theorem 10.5

N | M = inf \k(\ + R M{u/k))}.

k > 0

We have

1

|м -и „

м < (Co + £)

-F R

''u — un

vCo + £ £o + £ as n —>oo. Thus, lim \\u ^ u n\\M = £0.

n -* CO

Proposition 5. F o r an y и in L*M, ^ ( u )

=

2{

) =

0.

P ro o f. Without loss of generality we may assume that

?

\

M.

Since for any

^ 1,

M, by Proposition 4,

£ 0

= lim \ \ u - u n\\{M) ^

2(

).

П

Conversely, for any

in

and £ e (

, i £ 0), there exists

0 >

such that ||w ' wn

IU) < £, where

_ fw(r), when |w(f)| <

0,

"

^ ~~ |o, when |w(r)| ^

0.

Choose a >

such that (1-a )/(£

-2£) >

~ £)- Since \

(

)\ ^

for all t

in G„(u), we understand \u(t) — w„ (f)| > (1 —a)|w(f)| for all n >'n0/oc and all t in Gn(u). Thus, for all n > n0/oc,

R M u — wn0

£

~ 2 £

> fu(t)-W „ 0(t)\ r / u (Л M l —

----;---)dt ^ I M | - ---)dt

V <э0 ~ • .^o-e

G „ ( u ) Gn(u)

M ( u{ T - f ) d , = a ° -

This shows that \\u — vv„ ||(M) ^

— 2e. Therefore

I|

-

||(M) > l|w-H’no||(M)-||w„

- w ||(M) ^ ^ 0 — 2e — e.

e and w being arbitrary, we get d2(u) ^ £0, thus d2(u) — £0.

Finally, combining [7], Lemma 10.1 with Proposition 4, we have

£o = lim \ \ u - u n\\M = d 1(u).

n~* 00

Proposition 6.

For any functional f = v + Ф as in

||/ ||(/V)

if\\f\\(N) is attainable on S(L*M), then

(5) JJV(o(r))* + ||<P||w = l.

G

P ro o f. By the hypothesis, there exists a point и in S(I?M) such that 1

= l l / i u = /(«)• Observe that

9(О =

^ ( » (

) * + П|Ф||(»>

G

is a left-continuous function of £ on (0, + oo), by (4), one immediately gets that

\ N ( v ( t ) ) d t + \ m w ^ i . G

If я л ( » ) + ||ф |и < 1 , by

1

= ll/IU) = v(u) + 0{u) = f u(t)v(t)dt + 4>(u)

G

and Proposition 3, we have §u(t) v{t)dt = d > 0. Thus, there exists n0 > 1

G

such that

J u(t)v(t)dt > jd .

G \G „ 0 (v)

Notice that

f N(£v{t))dt+ f N(v(t))dt + \ m m

G \ G nQ(v) G„q(v)

h(Z) =

Smoothness o f Orlicz spaces 53

is a continuous function of £ and h( 1) = R M{v) + ||Ф||(ЛГ) < 1, there exists a constant

> 1 such that h(Ç2) < 1- Define

M O ^

*

, when te G \G „ 0(v),

when feG „

(

,

— V

+ Ф,

then by the choice of £2, ^ ( M + ll^llw = M£2) ^ ^ imPhes by (4) that ll/oll(jv) ^

- But this leads to a contradiction

1 ^ НУоЩ1М1м ^ /o(w) = v0(u)+(p{u)

—

j u(t)v(t)dt+ f u(t)v(t)dt + <P(u)

G \G nQ(v) G „ 0 ( 0

= /(w) + (<^

- l) J u ( t) v { t ) d t> l+ ( Ç 2- l ) i d > l :

Thus, (5) holds completing the proof.

Pr o p o s it io n

7. Let {Gfcj be a sequence of non-overlapping measurable

subsets of G, Iuk) a sequence, u(t) = X uk XGk(t)e Цм) and there exists

k = 1

v0eS (IfN) such that ||u||(M) = §u(t)v0(t)dt; then there exists a sequence \vk)

g •

00

such that v(t) = X vkXck(t)£

*

, INU =

and

k = 1 •

00

(

) \\u\\m = $u(t)v(t)dt = X uk vk mesGk.

G k = 1

P ro o f. By the assumption,

00

(7) ||

||(M) = $u{t)v0(t)dt = X uk I v0(t)dt.

G к = 1 Gjç

Without loss of generality we may assume that mes Gk > 0 for all к

= 1 , 2 , . . . Define

vk

mes Gk MO

Gk

GO

H t ) = X Vk %Gk ( t ) -

1

Then by (7), v(t) satisfies (

). Only \\v\\N = 1 remains to be shown.

For any w*eL*M, R M{u*) < 1, write

Uk

mes Gt

k=

then

G

v(t)u*(t)dt = Y vk \u*{t)dt = k=

J

Gk

GO

00

= Y \ u'kvo (t)d t= \u'(t)v0{t)dt.

k=

G

By Jensen’s inequality (see [7], Chapter 2, §

),

Combine with (

),

§v(t)u*(t)dt = J‘u'{t)v0{t)dt ^ ||®olU =

-

G G

u*(t) being arbitrary, we obtain ||i;||N ^ 1. On the other hand, from (

), IMU

<

is impossible.

Th e o r e m

1. Let p(u) be continuous; then

is smooth at ueS(L?M) iff и attains its norm on S(L*N)).

P ro o f. S u ffic ie n c y . By the condition, there exists v0eS{I?iN)) such that

By Proposition

,

= 1. Suppose / = v +Ф (where v e B N, Ф is singular) satisfies \\f\\{N) =

and

Therefore ||т(/+Ео)11(ло = 1- Observe (5) and the convexity of N(-), M l M = v0(u) = $u(t)v0{t)dt.

G

||u||

— f ( u) = v(u) + <P(u) = §u(t)v{t)dt + <P(u);

G

then the functional i ( / + y 0) satisfies

This shows that

(9) 2

Smoothness o f Orlicz spaces 55

Notice the continuity of p(u) is equivalent to the strict convexity of N(-), i.e., x

y implies N((x + y)/2) < ^ N ( x ) + j N ( y ) , from (9), for almost all t in G we have i;

(r) = u(f). In other words, v0 = v in I?{N). Combine R N(v0) = 1 with (5), Ф = 9. Thus, / = v0, i.e., L*M is smooth at u.

N e c e s s ity . Choose f = v + Ф satisfying (3) such that ||/ ||(N) — 1 and 1

= ||

||

= f(u ) = v(u) + Ф{u). If ||u||M is not attainable on S(Ü{N)), then Ф

0 and ифЕм or

>

.

00

Denote Gn = \teG : n — 1 ^ |u(r)| < n); then G = [j G„ and for any n, n=

|

(^

)I < И < ---г И г

)1, t l , t 2eG„.

n

—

Select G'n, G" a Gn such that G'n u G" = G„, G'n n G" = 0 and mes G'„ = mes G” =. \ mes G„.

Define

MO = £

= £

n= 1

tl —

For any ee(0, ^ £ 0), choose n

large enough so that

« o - l

then

n

—

s

— e ’

u(r)

d t> Y >

n= 1

G ' G'„

G „ G „

u (t)-u„ 0(t)

£o — e

d t = с о

£ being arbitrary and by Proposition 5, d 1(hi) = £0. Similarly, d l (h2) = £0- Denote = span (fy, E M} {i = 1,2); then any x in £ M(/ii) can be uniquely expressed as x = ahl +w, where a is a real number, w e E M.

Observe that

= sup { j

«J

U G'n

n=

\\h2 — x\\M

+ j [ - a h 1 ( t ) - w ( t ) ' ] v , ( t ) d t } 0 0

U G'n n= 1

^ s u p f [h 2 ( t ) - w { t ) 2 v ' ( t ) d t R N ( v ' ) ^ l »

U G 'n n= 1

=

oo IIM ^

U Gn

n=

d i ( h 2 , E M { h l ) ) = i n S \ \ \ h 2 - x \ \ M : x e E M { h i ) } = £ 0 .

S im ila r ly ,

d i ( / 7 i , = i n f { | | ^ i - x | | M: ^{x e E} M { h 2 ) } = £ 0 -

B y H a h n - B a n a c h ’s T h e o r e m , th e r e e x is t Ф ,е(Т *м )* s u c h t h a t ||Ф (||(ЛГ) = 1 a n d s u c h t h a t fo r a n y х, е £ м (/г,) ^{( i =} 1, 2) w e h a v e ^{Ф 1} ( x 2) = 0, Ф2 ( х i) — 0 a n d

Ф1 (^ i) = d i (/?!, E M ( h 2)) = d>2 ( h 2) = { h 2 , £m(/?i)) = £ 0 - C le a r ly , Ф 15 Ф 2 a r e s in g u la r f u n c t io n a ls . N o w , d e fin e

/ = » + ЦФ|иФ|

=

,

);

t h e n / l5 / 2 е(Т*м )* , Л # / 2 a n d , b y P r o p o s i t i o n 6,

ll/lll(N) = ll/2ll(W) = 1 = ||/||(N)-

S in c e Ф is s in g u la r a n d fo r a ll n ^ 1, u „ { t ) e EM, w e s e e Ф(м„) = 0 fo r a ll n

^ 1. It f o l lo w s b y P r o p o s i t i o n 4 th a t

£ о 1 |Ф Щ = lim \ \ u - u n \ \ M \ \ < P \ \ { N) ^ lim ^Ф ( и - и п) = Ф (м ).

n - *• 0 0 И “ ► 0 0

T h u s

Ф ( и )

Ф 1 ( и ) = Ф 1 { к 1) = £ (/ = 1 , 2 ) . 11

ФП(Х)

T h e r e fo r e

i / ( “ ) = *>(«) + I I ^ I U Ф /(и) ^ и(м) + Ф (и ) = / ( u ) = ||м ||м

(г = 1, 2). T h is c o n t r a d i c t s t h e a s s u m p t i o n o f L*M b e in g s m o o t h a t u .

Th e o r e m 2. E M i s s m o o t h i f f p ( u ) i s c o n t i n u o u s .

P r o o f . S u f f i c i e n c y . S in c e (^E M)* = I f ( N ) , fo r e v e r y ^и in ^{S ( E M \} it s n o r m is a t t a in e d o n S ( I ? { N ) ) . B y T h e o r e m 1, E M is s m o o t h a t и t h e r e f o r e is s m o o t h . N e c e s s i t y . I f ^{p ( u )} is n o t c o n t i n u o u s , t h e n ^{N ( v )} is lin e a r o n s o m e

Smoothness o f Orlicz spaces 57

interval [a, b]: N

E n F = 0 , 0 < mes E = mes F ^ ^ mes G and N (a)mes E +N(b) mes F ^ l.

Moreover, select a constant c > 0 such that

N (a) mes E + N(b) mes F + N (c) mes ( G \ £ u F ) = 1.

Define

a + b

= ^N (a) + ^N(b). Choose E, F cz G such that

Vl

(0 =

(0 +

(0 +

(О,

V2

(

) =

+

( г )

;

then vx Ф v ,a + b N

R, 2 Vi + V>2

2

and

) = 1 implies \\Vi\\(N) = 1

= 1, 2). Notice that

^iV(a) + iiV(b) and that mes E = mes F, it is easy to compute

=

therefore

I

[|(ло = 1. By Proposition 7, there exists м

= ^ХЕиг

+ ^Хо\£ч,г(

е £ м

such that \\u\\M =

and

V i + V 2

2 (AO U

vl (t) + v2 (t) u(t) dt.

It follows that

{ V i ( t ) u ( t ) d t

= 1 =

(i = 1,2).

Thus, E M is not smooth at и contradicting the smoothness of EM.

Theorem 3. L*M

is smooth iff p(u) is continuous and M

A2.

P ro o f. The sufficiency follows immediately from Theorem 2 since M

A 2 implies l?M = E M.

Conversely, since the smoothness of I?M implies the smoothness of EM, by Theorem 2, p(u) is continuous. If М ф А 2, then (L*M)* p I?{N). By Bishop- Phelps’ Theorem (see [4]), the collection of functionals that attain their norm on S(L*M) is dense in (L*M)*. Therefore, there exists / = v + Ф in (L*M)*, \\f\\(N)

= 1 (where

L*n, Ф Ф 0 is singular) attaining its norm on S(L*M). In others words, there exists и in S(L*M) such that f (u) = ||/ ||w = 1. On the other hand, by the smoothness of L*M and Theorem 1, there exists v0 in S(L*N)) such that v0(u) = ||u||M = 1. Clearly, v0 Ф f contradicting the smoothness of Rw •

Theorem

4. The following are equivalent : (1) I?M is strongly smooth,

(

) ITM is very smooth,

(3) p(u) is continuous, M e l l

and M e V 2.

P ro o f. (1)=>(2) is trivial.

(2) =>(3). Since the very smoothness implies the smoothness, by Theorem 3, p(u) is continuous and M e A 2. Hence X is reflexive when X* is very smooth (see [4]), it follows from (£(JV))* = £*м that E{N) is reflexive. Thus,

Ц,ю

£(v) ^

*

) ~

£<ло • This means that E(N) = L?(N) or M e V 2.

(3) =>(1). Since p{u) is continuous and M e V2 means that N(v) is strictly convex and N e A 2, by [2], I?(N) is locally uniformly convex. From [4], the fact that X* is locally uniformly convex implies that X is Fréchet differentiable, therefore strongly smooth and combine M e A 2 implying

= (EM)* = I?(N), one yields that L*M is strongly smooth.

Theorem 5. L*M

is uniformly smooth iff N e A 2, N(v) is strictly convex and is uniformly convex for large u.

P ro o f. It follows immediately from [4], [5].

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[10] - , J. Math. Anal. Appl. 37 (1972), 228-234.