On the existence of a weak unit in the quotient space

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE XXVIII (1988) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXVIII (1988)

Wi t o l d Wn u k (Poznan)

On the existence of a weak unit in the quotient space

Abstract. T h is p a p e r is a co m p letio n o f [5 ] an d p resen ts new classes o f O rlicz spaces 1?

such th a t th e q u o tie n t I f / l f i satisfies o n e o f m utu ally exclusive co n d itio n s:

(A) If/L^ has n o w eak unit, (B) I? is o rd e r is o m o rp h ic to 1®

{l?A d en o tes th e su bspace consisting o f elem ents w ith o rd e r c o n tin u o u s F-norm ).

We shall assume that the reader is familiar with the definition and elementary properties of an Orlicz space I f = L f(S , I , p) (basic facts about Orlicz spaces can be found in [4]).

Distinguish the subspace LVA = LfA (S, Г, pi) consisting of elements with order continuous F-norm, i.e., lfA = \ x e l f : |x| ^ x„|0 implies ||x„|| —► 0], where || • || is the standard F-norm on I f . We shall always assume that <p is finite-valued and strictly positive on (0, oo). Then the following equality holds

LfA = ' { x e l f : mv (rx) = §(p(r\x(s)\)dpi < oo for all r > 0}.

s

It is clear that every Orlicz space over a ^-finite measure space has a weak unit (for definitions of terms from the theory of Riesz spaces [ = vector lattices] see [3]). But there exist many Orlicz spaces I f whose quotient l f / I f A does not have any weak unit:

Theorem 1 (see [5]). (a) Let q> be an Orlicz function not satisfying the A 2-condition at infinity and assume that pi\Z n В is non-atomic for some set В o f positive and finite measure. Then the quotient Lf/LfA has no weak unit.

(b) Let <p be an Orlicz function not satisfying the A 2-condition at zero and assume that I contains a sequence (Bn) o f atoms with 0 < inf pi (B„) ^ sup pi(B„)

n n

< oo. Then the quotient I f I IfA has no weak unit.

This theorem does not cover the case of <r-finite purely atomic measures where the measures of atoms tend to zero or infinity. In this paper, we shall partially answer the question about the existence of a weak unit in the quotient l f l l f A, where I f is over a measure space mentioned above. Since we want to investigate Orlicz spaces over cr-finite purely atomic measures, so we can assume that S — IS, I = 2N, ju({n}) = ane(0, oo). Here N denotes, of

168 W. Wnuk

course, the set of natural numbers. We shall write F(an) (respectively FA (aj) instead of Lv(iV, 2N, p) (respectively L ^(N , 2N, p)). If an = 1 for all n, then we shall write shortly F. The notation F (a„k) means the subspace of F(an) consisting of these sequences (x n) e F ( a n) whose support is included in the set

('Чс i к — 1 •t°°

Now we can formulate and prove the following theorem.

Th e o r e m 2. Let (p be an Orlicz function and let (an) be a sequence such that F(a„k) ф lA(a„k) for every subsequence (nk). Then F(an)/%(an) has no weak unit or F(a„) is order isomorphic to /°° (so the quotient has a strong unit).

Proof. It is sufficient to show that if the quotient has a weak unit then F(an) is order isomorphic to /°°. Let Q denote the canonical homomorphism from F(an) onto F(a„)/l^(a„), and let l c be the characteristic function of the set C cz N.

Suppose that Q (x )e F (a n)/l^(an) is a weak unit. We can assume x > 0. It is clear that the following equivalence holds:

(*) x l c e/^(aj iff C is finite.

Indeed, if C = [nkj was infinite, then by the assumption there would exist an element z eFl (a„k)\l^ ( a nk) and therefore Q(z) > 0, but Ô (z) л Q (x) = 0, a contradiction. Using the same argument, we obtain N4 = N \s upp x is a finite or empty set. Put (e„) = e = x v l Ni . The operator T : /°° -*F (an) defined by the equality

T((cJ) = (c„e„)

is an order isomorphism onto. T is of course one-to-one homomorphism. We will prove that T is onto. Take у — (yk) e F (a n). If the sequence ( y je k) was unbounded, then the inequality \yk.\ > 2‘ ek. would hold for some subsequence (kt). Denoting N2 = {k,}, we would obtain e lN2 eF4(an). Indeed, fix t > 0 and r > 0 such that m^iry) < oo. We have t < 2'r for all fs bigger than some i0.

Thus

CO 0 0 00

Z 9 (tek.) ak. < Z 9 (r2* %) % < Y , 9 (ryki) ak. < (ry) < oo.

•0 ‘0 *0

Finally, we get a contradiction to (*). Therefore, ( y je ^ must be bounded and T ((y je k)) = (y*).

Now we give some examples of Orlicz spaces satisfying the assumption of Theorem 2. Before doing this we shall mention a certain lemma:

Le m m a. F (a„) = (a„) holds if and only if the following conditions are satisfied: there exist constants К > 0, a > 0 and sequences 0 < b„ < oo, 0 < cn

< oo such that

On the existence o f a weak unit in the quotient space I?/I% 169

(i) (p(2r)an ^ K(p(r)a„ + cn if r e [0 , b„], (ii) (p(r)a„> a if r > bn (when b„ < oo),

00

(iü) X e" < °°- l

The lemma is due to Drewnowski [1] and it has been proved in more general case (for generalized Orlicz sequence spaces).

The condition

P (a„k) Ф t]4 (a„k) for every subsequence (nk) is satisfied in the following situations:

(A) tp does not satisfy the d 2"condition at zero and 0 < infa„ < supa„

n n

< 00,

(B) lim (p(2r)/(p(r) = oo and a„ -*0,

r-*oo

(C) lim (p(2r)/tp(r) = oo and an -+ oo.

r-* 0

Indeed, the proof of part (A) is known, because if {a,) = Now we shall check part (B). Without loss of generality we can assume that (p is strictly increasing and continuous (if not we replace tp by ф, where ф (r)

r

= (1/r) f (p(s)ds, the inequalities 2~ 1 (p(2~ 1 r) < ф(г) ^ (p(r) give the equiv-

b

alence: F(a„) = iff /^(aj = Denote by cp~l the inverse function of (p. Suppose /^ (u j = /^(a„fc) for some subsequence (nk). Denote dk = anfc.

Let K, a (bk), (ck) be as in lemma. We have to consider two cases:

(a) bk is finite for almost all k, (b) bk is infinite for infinitely many k.

Let (a) hold. Then by (ii) bk ^ tp~1(a/dk) and, using (i), we obtain а Я>(2 (p 1 (a/dk))

(a/dk)) = (p(2q> 1 {a/dk)) dk ^ K a + ck for almost all k. Therefore, lim (p(2r)/(p(r) < oo, a contradiction.

r ->oo

Let (b) hold, i.e., bk. = oo for some subsequence ( k f Choose sequence /*,•-> oo such that tp (2rf) > (K/dk.) (p (r,). This inequality together with con­

dition (i) imply

K(p{rt) < (p(2ri)dk. < К(р(г№к. + ск..

Thus X ( 1 — dk.) tp (rf) < ck. and we have a contradiction again because the left-hand side of the inequality tends to infinity and the right-side tends to zero.

The proof of part (C) is analogous.

170 W. Wnuk

Theorems 1 and 2 do not account for all Orlicz spaces. They do not include the following case: q> satisfies the ^ “Condition at infinity but does not satisfy the ^"Condition at zero and (a „ ) e c 0 \ l l. Then Р ( а п) Ф /^(a„) because Р(а„) contains an isomorphic copy of P* (see [2], p. 582), but for every subsequence (a„k) there exists a subsequence (a'„k) of (a„k) such that Pia'nJ = l \ (a'„k) (it is sufficient to choose a subsequence (a’^ e l 1).

Pr o b l e m. Does there exist an Orlicz space I f such that the quotient l f / l f A has a weak unit and has no strong unit?

References

[1] L. D r e w n o w s k i, On the generalized Orlicz sequence spaces l(f„)(private communication).

[2] F. L. H e r n a n d e z , Some classes o f non-locally convex Orlicz sequence spaces with weights,

Bull. Acad. Polon. Sci. Sér. Sci. Math. 29 (1981), 579-584.

[3] W. A. J. L u x e m b u r g , A. C. Z a a n e n , Riesz spaces I, Amsterdam-London; North Holland, 1971.

[4] J. M u s ie la k , Orlicz spaces and modular spaces, Lecture Notes in Math. 1034, Springer- Verlag, 1983.

[5] W. W nuk, On the order-topological properties o f the quotient space L jL A,Studia Math. 79 (1984), 139-149.

[6 ] —, Orlicz spaces which are Riesz isomorphic to 1®, Rocky Mountain J. Math, (in print).