On weak sequential compactness of generalized Orlicz spaces

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXIV (1984)

J. Szelmeczka (Lublin)

On weak sequential compactness of generalized Orlicz spaces

In [3] there are given necessary and sufficient conditions for relative а(Ьф, Lr )-compactness of a subset of an Orlicz space Ьф. Applying the method of the proof of this theorem and making use of a criterion of weak compactness of a set in (see [2]), we shall prove a theorem of this type for generalized Orlicz spaces L0 generated by convex functions depending on a parameter.

Let ju be a finite atomless measure in an abstract set Q. A function Ф: Q x R + -+ R + — [0, oo] is called a convex (p-function, depending on a parameter f eO if Ф(г, и) is measurable with respect t in Q for every и ^ 0, and is a convex ^-function of и ^ 0 for every t e Q and Ф(г, u)/u ->0 as и -► 0 +, Ф(£, u)/u -> oo as и -> oo for a.e. te Q .

Let Ф be for every t e Q the function complementary in the sense of Young to the function Ф, with respect to u. Then, considering in the Young inequality u-v < Ф(г, u)+ u) the case of equality and writing tpx = \tp, we have

u 'tp (t,u ) = Ф(£, u)+ 4*(t, (p{t, u)) = 2u(px (t , m), where tp and ф are such that

U V

Ф(Г, m) = j (p(t, s)ds, W(t,v) = ^ ( t , s)ds

о о

and

№ (t, и) = Ф ^ м ) = $q>1(t,s)ds.

0

The function *Р15 complementary to Ф1 in the sense of Young satisfies the equality (f, v) = ^ ( t , 2v).

We have

2utp1 (t, u) = 0 (t,u)+ W (t, (p{t,u)), and so

utp(t,u) = Ф (t , u) + W(t, (p (t, u)) = 2u(p1 (t , и)

^ 2Ф(г,и) + 2Ф(Г, (p1{t,u)).

Hence

W(t, ( p ( t,u ))-2 ¥ (t, ( P i ( t , u ) ) ^ Ф(г, и), and we get

(*)

for и ^ 0, t e Q . Now, we show that the condition ] w ( t , 4 v0{t))dp < 00

Si

implies f Ф(г, ф (t, 2v0(t)))dp < со.

h

This follows from inequality (2) from [1], which may be applied to functions depending on a parameter t for every t, separately:

Ф(1,ф(и)) < ¥(t,2u).

Taking in this inequality и = 2v0(t) and integrating over Q, we obtain the required condition.

The purpose of this paper is to prove the following

Th e o r e m. Let Ф and ¥ be convex cp-functions depending on the parameter t e Q mutually complementary in the sense of Young. Let p be a finite, atomless measure on Q. Moreover, let us assume the following condition: there exists a non­

negative, integrable over Q function h(t) such that

(**) V V Э V ¥ (t, 2v) + h(t) > R ¥ { t , v), R> 0 teS2 uqU) >0p >Vÿ(t)

where v0(t) is a measurable function such that J ¥ ( t, 4v0{t))dp < 00.

S2

Then we have the following equivalence :

A set A 6 L0 is bounded in the sense of the norm of L0 and (1) sup \( f - g ) d p -> 0 as p ( E ) -+ 0 for every g e L p

feA E if and only if

(2) there exists a number A0 > 0 such that the family offunctions {#(t, A0 1/|) : f e A ] is a weakly sequentially compact subset of L 1.

P ro o f. First, we shall prove the sufficiency, i.e. that (2)=>(1).

Let us write В — (Ф(г, A0 1/|): f e A ) . The set В <=: L x is a weakly compact subset of Lj if lim \gdp = 0 uniformly with respect to g e B (see [2],

H ( E ) ^ 0e

Theorem 8.11 (corollary), p. 294) hence condition (2) says that lim |Ф (г, Я0 \f\)dg = 0 uniformly with respect to f e A . ц(Е) -*0 e

(i) We first prove that A is bounded in Ьф. By the assumption, we have (3) V 3 V V ( ^ ( £ ) < <5(e) =*> |Ф ( ( , Я0 |/|) 4 /1 < e).

г > О «5(e) > O f e A E £

Let us take e= 1 and the respective <5j = <5(1). Since the measure g. is atomless, there exist measurable and pairwise disjoint sets E u E 2, ..., EnQ such that Q = E x и E2u . . .u E„0 and д(£;) < <5j for i = 1, 2 ,..., n0. Then

f &(t, Ao\f\)dfi < 1 for every f e A , i = 1, 2 ,..., n0.

Ei Hence

Ф(1, X0\ f i ) d a < ^ U ( t , l „ | / | ) ^ < — •«„ = l

J « 0 J « 0

a q

for every f e A . Thus ||/ ||ф < и0/Я0 for every f e A , and this means the boundedness of A in Ьф.

(ii) Now, we are going to prove that

su p \\f-g \ dg, -* 0, as fi(E)-+0 for every g e L F.

feA E

Applying the Young inequality, we obtain

A J |/ *0№ < J W , X0\f\)dg + jV ^ t, y l g l 'j d g

E E E

for every Я > 0 and f e A . Let us choose Я > 0 so small that )dg < oo;

Aq

Q

such а Я exists, because g e L y . By absolute continuity of the integral as a set function, for every e > 0 there exists а <5'(г) > 0 such that if g(E) < <5'(e), then

Thus

4> L 7 -I0I

Aq dg < jXe.

\ f - g \ &(t, A0 | / l № + lfi

E E

for every f e A .

10 — Prace Matematyczne 24.2

But applying (3) and taking g(E) < 0 (^Х-е), we have J<P(r,A0| / № < i àe for every f e A .

E

Hence, writing 6 = min (b'(г), <5 (4Яе)) (<S does not depend on / ) , we obtain 1 Xe

( \ f - g \ d g < ~ - ---for g(E) < Ô and every f e A .

к A 2

Thus

s u p j j f- g jd g < £ for ju(E) < 3.

f e A E

Hence

sup Jl/'gfl dg -*■ 0 as g(E)-> 0 for every g e L y .

f e A E

This concludes the proof of the implication (2)=>(1).

Now, we proceed to the proof of necessity, i.e. (1) =>(2). We may suppose without loss of generality that дф( / ) < 1 for f e A . We indirectly prove that В is a weakly sequentially compact subset of Lb i.e. that lim J Ф(Г, X0\f\)dg = О

ц(Е ) - * Ое

uniformly with respect to f e A : this means that (3) holds for some 0 < A() ^ 1.

So, let us suppose that the family [4>(t, X0\f\): f e A } is for all 0 < A0 < 1 not uniformly absolutely continuous. Thus, writing g(t) = <P(t, X0 \ f (t)\) for f e A , the relation sup J g (t) dg -> 0 as g ( E ) -> 0 does not hold. In other words,

д е в E

(4) 3 V 3 3 {/i(E) < 0 (e) and \gdg ^ e}.

e > O^(fi) > 0 geB E £

We shall show that

(5) lim sup I (g(t) = u„(t))+ dg > e0 > 0

n — 'M geB Q

for every sequence of integrable functions 0 un(t) t oo.

Let us suppose that if (5) is not true, then there is a sequence of measurable functions such that lim sup f \g(t) — un(t)\ + dg — 0, and so there

n--> oo geB n

exists a sequence 0 ^ u „ ( f ) t o o for which sup §(g(t) — un(t))+ dg -* 0, i.e.

веВ Q

V \(g(t) — u,,(t))+dg < rj for n > N^, where Nn does not depend on geB . Let v>o‘n

us write h„(t) = (g(t) — u„(t))+ : then p i ^ d g < g, n > N n uniformly with b

respect to g e B . But

h i \ = if ^>

nU (0 if g (t) < un(t), whence

r} > j h n(t)dg ^ j h n(t)d g = j g ( t ) d g - J u„(t)dg.

Q E E n { t : g(t)>u„(t)} £ n { f: g ( t ) > u n(t)}

However, we have for t e E n {t : g(t) > u„ (r) j = G„, g (t) > un (t), and the right- hand side of the last inequality is non-negative. Hence

I j' g(t)dg — I' un(t)dg\ < ц for n >

Gn Gn

and for an arbitrary measurable set E a Q. Let us denote by Fn(E) the expression under the sign of absolute value on the left-hand side of the last inequality. Then

.1 'gdg = f gdg+ j gdg ^ |' u„(t)dg +F„(E) + |' u„{t)dg

E E —G n G n E — Gn G n

= iu„(t)dg + Ffl(E).

E

Let us take any rj > 0 and a respective N n such that \Fn(E)\ < ц for n > Nq. Let us denote by nn the least index n such that n > N„. Then \Fn(E)\

< ц for every E. Fix nn\ then

\gdg < j u„(t)d/i + r].

E E

From the integrability of un(t) it follows that is an absolutely

E

continuous set function. Hence there exists a ô(rj) > 0 such that if g(E) < ô (rj), then \u„ (t)d(.i < //. Thus, if //(£) < <5(/у), then igdg < 2g for every g

e n E

(uniformly with respect to g). Consequently, we obtain a contradiction to (4). So we have proved condition (5).

Let a + — a for a ^ 0, a + = 0 for a * 0. Since u„(t) Î we get (g(t) — -u„{t))+ > (g{t)-u„+1{t))+, and hence

Thus

implies

sup $(g(t )-u„(t))+ dg > sup $ ( g ( t ) - u n{ t ) y dg.

geB q geB n

lim sup §(g(t) — un(t))+ dg > e0 > 0

n - » » geB Q

s u p $ ( g ( t) - u n(t))+ dg > e0 > 0 for every n.

geB Q

Hence for each n there exists an f ne A such that

(7) ]’(ф(1,/1()|/,( г ) |) - и >1(г))+ ^ > £() > 0 for n = 1, 2 ,...

О

Taking R = 2n+1 in assumption (**), we may find measurable functions vn(t) > 0 such that F(t, 2v) + h(t) > 2n+1 F (t,v) for v ^ v„(t). Let us write

E„ = {t: Ф(г,Яо |/«(01) > un(t)},

g n ( t)

=

[

0|/„(

)|x£n(r)] =

)•

Moreover, let us choose, by definition, integrable functions u„(t) ^ 0 such that u„(t)îoo, un(t) ^ 2", un(t) ^ <p(t, ij/(2vn(t)))- We have

Ё?ф(Л<0 l/J) < M / J ) ^ 1*

Hence

1 ^ $<P(t,À0\fn\)dn ^ j u„(t)dfi > 2"(£„),

n E n

and so /i( £ J < ?n; thus fi(En)-+ 0 as n-+oo.

Taking и = À0\fn{t)\ and t e E n, we have, by inequality (*), that

2j « 4>M„|/„WI) Ф(М/,(01).

On the other hand, we apply the inequalities appearing in the definition of u„(t). Then, taking t e E n, we obtain

[M ol/„(O I] > w„(0 ^ Ф ^ М ^ О ) ] ; hence

'U/.WI г ^ М м о ) .

Since (p is a non-decreasing function, we get

<P {t , Яо |/„(01} ^ <P {Ф(*> 2u„(0)} = 2u„(0>

0

ЛО =

(f, 2о|/я(01) = (f, Ао1Л(01) ^

Hence

lf/^ ' ^ 2”+1 > 2”+ 2,

for t e E n. Consequently,

2 " Ч > и ,д М < P (t,\m \) + h«) for t e E n.

We integrate both sides of this inequality over Q:

2"

= 2” j Ф(г,0„(г))й^ ^ j Ф(г, |/„(0I)^+j/z(t)^

E„ E n Q

< ^ ф Ш + f h(t)du ^ 1 + J h{t)dn = C < 00,

h «

since the function #„(f) is equal to zero outside the set En. Hence Qy(gn) ^ 2~ " C for n = 1 ,2 ,...

Let us write g(t) = sup g n(t); then

Hence

g(t) = lim sup g^t).

n~+ao i < n

Y (t, g (0) < Urn sup W (t, gt (r)), t e Q,

n -* ao i < n

and so, by the Fatou lemma, we obtain

Qvi9) = I 4 ' ( t , g ( t ) ) d f i < lim js u p W(t, g ^ d g

П n->ao Q i ^ n

= lim Qy(sup(gu g2, . . . , g n))

n~*-ao

^ Qr (91) + Qv (9 2) ^ ■ •• ^ X 2 n ‘ C — С < 00 . n= 1

Thus 9 e bp. Applying the case of equality in the Young inequality, we get M-0\fH{t)xEH(t)\gH(t) = Ф(г,Яо|/„(01Х£„(0)+^(^520[„(0)-

Hence

2A0 f \fn(t)-9n(t)\dg

= j < P ( t a 0 \ f n ( t ) \ ) d g +

J

En E n E n

= J [<P(t, À0 \ fn( t ) \ ) - u „ ( t ) ] d g + J u n( t )d g

En En

> j [& (t,*o\fn(t)\)-u„(t)]dg

E n

= Ao|/„(t)|)-M„(f)]+ dg > e0 > 0 (see (7)).

x

But g (t ) = sup g n(t), and so

r, (8)

which yields

2Л-0 j If n'9\dg > 2X0 J \fn-gn\dg > e0 > 0,

E n E„

S\fn '9 \d g > -Hjy- > ° for и = 1 ,2 ,...

E n

However, we have g e b y . Moreover, g(En)-> 0. Since we assume (1), i.e.

sup J | / -g\dg -*0, as g(E)-> 0 for every g e L y ,

f e A E

sup j \f-g[dp->0,

f e A E „

we obtain

and since / пе Л , we get

J \fn'9\dn sup j \f-g\dfi^>0, as и ^ о о ,

E n f e A E„

a contradiction to (8). This proves the theorem completely.

Let us remark that assumption (**) in the theorem is needed only in the proof of necessity, i.e. (1)=>(2).

References

[1] R. B o ja n ic , J. M u s ie la k , An inequality fo r functions with derivatives in an Orlicz space, Proc.

Amer. Math. Soc. 15 (1964), 902-906.

[2] M. D u n fo r d , J. T. S c h w a r tz , Linear operators, Part I, New York, London 1958.

[3] W. A. L u x e m b u r g , A remark on a theorem o f W. Orlicz on weak convergence, Comment.

Math., Tomus specialis in honorem Ladislai Orlicz I (1978), 211-220.