ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE XXIX (1990) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO
Séria I: PRACE MATEMATYCZNE XXIX (1990)
M a r ia n N o w a k (Poznan)
Some equalities among Orlicz spaces, I
Abstract. For any Orlicz function q> let U and Ev be the Orlicz space and the space of finite elements respectively, -on an arbitrary measure space. Let Фх be the set of all Orlicze functions taking only finite values and such that (p (u) -+ oo as и -> oo and let Ф2 be the set of all Orlicz functions vanishing only at zero and such that cp(u) -» oo as и -* oo. In this paper, for each cp\ e Фх and ф2еФ 2 we find the*sets of ^-functions and f ' f 2 respectively, such that the following equalities hold: EVI = E* = Ù and L4>2 = 1* — [ ) феЧ,* 2 Е* and moreover, we obtain that if the measure space is atomless or purely atomic, then the strict inclusions hold:
Ü £ EV1 for each 'EX' and I f 2 £ E* for each ф e '¥ \2. In [9] by means of these equalities appropriate inductive and projective topologies on EVl and I f 2 respectively, are studied.
1. Introduction. Some partial results concerning equalities among Orlicz spaces have been obtained by Krasnosel’skii and Rutickii ([2], p. 60), R. Welland ([11], Theorem 1), Ph. Turpin ([10], Proposition 1.2.2), R. Lesniewicz ([3]) and the author ([7]).
Now we give some definitions. A function q>: [0, oo) -*■ [0, oo] is called an Orlicz function if it is vanishing and continuous at zero, nondecreasing and not identically equal to zero (see [10], p. 24). We shall say that an Orlicz function (p jumps, whenever there is a number u0 (0 < u0 < oo) such that (p(u) = oo for и > u0 (see [4], p. 51). An Orlicz function q> is called p-convex (0 < p ^ 1) if
(p (ecu + fiv) ^ olp (p (u) + f p(p (y), where a, f ^ 0 , ap + (}p = 1
(see [9], p. 161). A 1-convex function is called convex. An Orlicz function (p continuous for all и ^ 0 , taking only finite values, vanishing only at zero and tending to oo as и -> oo is called a (p-function ([5]). By Ф we will denote the set of all ^-functions. A convex (^-function is called an N -function if (p (u)/u -* 0 as и — » 0 and q> (u)/u -» oo as и -*■ oo (see [2], p. 9).
We say that an Orlicz function <p increases essentially more rapidly than another ф for all и (resp. for small u; resp. for large u), in symbols ф a <p (resp.
Ij/ < s(p‘, resp. ф j <p), if for an arbitrary c > 0 , ф (cu)/(p (u )-> 0 as и -+ 0 and u oo (resp. as и -*■ 0; resp. as м oo) (see [2], p. 114).
For any Orlicz functions ф and (p we define new Orlicz functions ф v (p
and ф a (p by (ф v (p)(u) = max(^(w), cp(u)) and (ф л (p)(ü) = тт(ф(и), (р(и))
for u ^ O .
Throughout this paper we assume that (Q, I , //) is a measure space, where // is a e-additive and positive measure on сг-algebra X. For a real valued, //-measurable function x defined on Q we write (x) = </> (\x(t)\)dp.
For any Orlicz function <p the Orlicz space I f consists of all func
tions x satisfying р^(Ях) < oo for some Я > 0, the Orlicz class L% consists of all functions x satisfying q v ( x ) < oo and the space of finite elements Ev consists of all x for which р^(Ях) < oo for all Я > 0 ([5], [ 6 ]). If Q = {1, 2, ...}
and //({w}) = 1 , then we will denote the spaces L<p and E* by /ф and respectively. By L°° we will denote the space of all //-essentially bounded functions on Q.
T h e o r e m 1.1. The following equalities hold: l f sjq> = l f r \ U ' and L+** = L+ + IS (see [1], Theorem 1).
We assume that 0 ■ oo = 0 .
2. On some equalities for Ev. Let Фх be the set of all Orlicz functions taking only finite values and tending to oo as и -* oo. In this section we shall prove that for each (реФх and an arbitrary measure // there exists a set of ^-functions Wf such that
E* = (J Еф = (J I f.
феЧ'Х фбЧ'Т
Moreover, we shall show that if // is an atomless measure or a purely atomic measure with measure of atoms bn satisfying 0 < inf„ bn ^ supn bn < oo, then the strict inclusion I f $ E* holds for each феФ^.
Denote
Фп = {(ре Ф{. <p(u) > 0 for и > 0}, Фi 2 — {феФi : (p{u) = 0 near zero}.
Then
Ф, = Ф ц и Ф ц .
Assuming that // is an arbitrary measure we shall now prove the following two theorems.
T h e o r e m 2.1. Suppose (реФ11. Let Wh be the set of all (p-functions ф such that (p < аф, i.e. = {феФ: (p ^ аф}.
Then
E* = (J E*= (J I f.
T h e o r e m 2.2. Suppose (реФ12, i.e. q>(u) = 0 for 0 ^ и ^ u0, (p(u)> 0 for
и > u0 > 0. Let be the set of all (p-functions ф such that q> ^ \ф and
ф (cu)/(p (и -f- U q ) — ► 0 as и — > 0 for all c > 0 .
Equalities among Orlicz spaces, 1 257
Then
E*= (J E* = (J I f.
фе'РХг № 1 2
R em ark 1. In [3] (resp. ([7], Theorem 1)) it is proved that if sp is a q>-function and ц is the Lebesgue measure on a compact subset of a finite-dimensional Euclidean space (resp. on a finite dimensional Euclidean space), then E9 — where the union is taken over all ^-functions ф such that (p ^цф (resp. sp < аф). Moreover, in [11], Theorem 1, it is proved that if sp is an iV-function, which satisfies the d 2-condition for large и (see [5]) and the Orlicz spaces are defined on the interval [0, 1] with the Lebesgue measure, then I f is the union of the Orlicz spaces which it contains strictly.
P r o o f of T h e o re m 2.1. In order to get the inclusion L? <= E9, it suffices to apply the following:
L em ma 2.3. Let <реФп and let ф be a (p-function such that for an arbitrary
c > 0, lim sup (p (си)/ф (и) <oo as и -> 0 and и -> oo. Then I f cz E <p.
The above lemma is proved in [ 8 ], Theorem 1.1, for sp being a (^-function, but the proof applies to (реФ1X.
Now, to prove the inclusion E9 c E*, it suffices to modify the proof of Theorem 1 in [7], in which for(p being a ^-function the inclusion E9
c zis proved. Indeed, let us assume that x e E 9. Writing
Qm = {teQ: 2m~l ^ |x(ï)| < 2 m}, m = 0 , ± 1 , ± 2 , . . . , we obtain for any natural number k
t 2"+2t+i)»(QJ< t s < f> (22t+ 2\x(t)\)df,
m — — 00
m= — 00 Sim
^ jsp(22k + 2 \x(t)\)dn < 00 . Q
Hence it follows that there exists a strictly increasing sequence of natural numbers (mk) such that
00
( 1 ) I <p (2~m*2k+k) /i((2 _m) ^ 2~l‘-2~k~2,
m = mic
( 2 ) £ <p(2m+2k+t)\i(Qm) < 2~k‘2~k~3.
m = m>«
It is seen that a sequence (mk) can be constructed in such a way that
<p( 2 ~mi) < q>(2mf , (p{2~mk+1) ^ <p(2~mk)/2,
(p( 2 Mk) < sp( 2 mk+% 2mk < mfc+1, mk + 2k < mk+l.
Now we construct a ^-function ф such that (p < аф and x e E As in the proof of Theorem 1 in [7], let us set
fc= 1 , 2 , ro for f = 0 ,
2~k for 2 -mfc+1 ^ t < 2~
1 2 _1 for 2 ~mi ^ r < 2 mi, . 2 k for 2mk^ t < 2 mk+\
q(s)= <
0 2~k 1 2k
for s = 0 ,
for q>{ 2 “mk+1) ^ s < (p{2~mk), к = 1 , 2 , for q>(2~mi) ^ s < (p(2mi),
for (p(2mk) ^ s < (p(2mk + 0 ,. к = 1 , 2 , . . . •
Define
*( m ) = x'(«) for м е[ 0 , м0],
*"(«) for u g [ u 0, oo ),
where u0 = J;p p (f)4 t, [ 0 , u0] [ 0 , 2 mi], *'(u) = a ^w), a(u) = $u 0p(t)dt and [u0, oo) -► [ 2 mi, oo), x"(u) = J" 0 p(t)dt + 2 mi.
Next, let us define
£(») = ? ( v ) for i>e[ 0 , <p( 2 mi)], for ue[<p( 2 mi)> oo),
where £'(u) = rj 1 (г), ц\ [ 0 , u0] -> [ 0 , <p( 2 mi)], r\{v) = Joq{s)ds and t >0 is the number such that §v0°q(s)ds = (p(2mi), £"(v) = (s)ds + y0. We have
(p ( 2 M1) ^ v0 < 2(p(2mi) < (p ( 2 mz) (see [7], p. 72).
At last, we define
= for « e [ 0 ,
|^ " (u ) for u e [u 0, oo), where ф' (и) = (<p (*' (u))) and ф” (u) = £" (<p (*" (u))).
It is seen that ф is a <p-function and, as in the proof of Theorem 1 in [7], we get (p < аф. Now we shall prove that for an arbitrary natural number k0 the following inequalities are satisfied:
( 3 ) ф(2~т+ко) < 2k + 2 (p{2~m+2k+l), where к ^ k0, mk < m < m k+1 — 1 ; ( 4 ) ф(2т+ко) ^ 2k+3 (p( 2 m + 2fc+1), where /с ^ /с 0 + т 15 m* ^ m m k+1 — 1 .
First we shall prove that (3) holds. For к ^ k0 and mk ^ m ^ mk + 1 — 1 we have
ос(2~m+2*,+1) > p(2~m+2k) ‘ 2~m+2k ^ p(2_mk+1)-2_m+2k = 2~m+k 2~m+ko
Equalities among Orlicz spaces, I 259
Hence, since x' (u) — ot 1 (и) we get
^ 2 ~m+k°) < 2~m + 2k + 1 which implies
'(5) ф(2~т+ко) = Ç’((p(x'{2-m+ko))) ^ (<p(2~m + 2k + 1)).
Since (p(2~mk+2) < 2 _1 <p(2_mk+1) we have
rj((p{2~m+2k + 1)) ^ 2~l q(2~k (p(2~m+2k + 1))(p(2~m + 2k + 1)
^ 2 ' 1 q(2~1 (p{2-mk+i))(p(2-m + 2k+l)
^ 2 - 1 q((p(2~mk+2))(p(2~m + 2k + 1) = 2~k~2- (p(2~m + 2k + 1) and hence, since Ç (t>) = r\ ~ 1 (v), by virtue of concavity of the function £' we obtain
( 6 ) (p(2~m + 2k + 1) ^ Ç'(2~k~2 (p(2~m + 2k + 1)) ^ 2~k~2 (<p{2~m+2k + 1)).
From (5) and ( 6 ) it follows that
ф(2~т+ко) ^ 2k+2(p{2~m+M+i).
Next, we shall prove that (4) holds. For к ^ к о + т^ and mk ^ m ^ mk + 1 — 1 we have m + k0 ^ mfc + 1 — l + fc 0 ^ mk+1+/c < mk + 2; hence
(2m+k°).^ p(2m+ko)2m+k°-j-2mi < 2k+12m+ko + 2mi < 2m + 2fe + 1 Further, since m + 2fc + l ^ mk + 1+2k < mk + 2 and v0 < 2<p(2mi), we get
q>(2m + 2k+1)
ф(2т+ко) = (2m*k°))) {"(<P.(2"+2I‘ + I)) ^ J «(s)ds + 2 <? ( 2 " ‘) 0
< 4 (<p (2"1+2k + *)) (P (2m + 2k +1) + 2q> (2mi)
^ g (cp (2mk+2)) (p (2m+2k+1) + 2(p(2mi)
= 2 k + 2 <p( 2 ", + 2k + 1) + 2 <p( 2 mi) < 2 k + 3 <p( 2 M+2k + 1).
We now show that x £ E*. Given- Я > 0, we choose a natural number k0 such that Я ^ 2ko. Then by (1), (2) and (3), (4) we get
$i/j(À\x{t)\)dp= £ J ф(2 \x(t)\)dp+ X J ij/(X\x{t)\)dp
O ’ ■
m = 1SI - m
m = 0Q m
00 XI
« 2: * ( 2 - +*°),*(û _J + X Ф0-"+hM O J
m =1 m = 0
mfco ~ 1 oo mic + i - 1
« £ l/i( 2 - ” +‘»)^(£ 2 _ J + £ ( £
m — m i fe = feo m = mk
Wfc0 + - 1" 1
+ Z ф(2~т+к‘>)р(й_т) т = О
00 Т П к + 1 - 1
+ Z ( I iA( 2 " +fc> )M fiJ )< 00 .
к = к о + m i m = m k
and this means that х е Е ф. Thus the proof is completed. ■
P r o o f o f T h e o re m 2.2. In order to get the inclusion (J^€Ï/I 2 L^ 0 E9, if suffices to apply the following:
L e m m a 2.4. Let (реФ12 and let \J/ be a (p-functions such that for all c > 0 limsup (p ( cm )/ if/ (u) <oo as и -+ о о. 77ien с E’’.
P r o o f of L em m a 2.4. It suffices to show that с Д , because E9 is the maximal subspace of L9 contained in L$. Let х е Ь ф, i.e.
Jo Ф (A 0 к (01) dp < oo for some A 0 > 0 . Then there exists a 0 > 0 and d > 0 such that
( 1 ) (p ( m ) ^ dij/ (Л 0 и)
for u ^ v 0. Without loss of generality we can assume that <p(u) — 0 for 0 ^ u ^ 1 and (p{u) > 0 for и > 1. Let us assume that v0 > 1. Then for w e[l, V q ] we get
(p («К < P W ^ (<P Ы/Ф (^o)) Ф 0*o «)■
Thus for и ^ 0 we get
(p (u) ^ d0 ф (A 0 u) where d0 = max(d, cp (v0)/\j/ (A0)). Hence
J ( p (|x (t)|) dp < d0 J ф (A0 \ x (01) dp < oo,
о о
i.e. xeL%. Thus the proof of our lemma is finished. ■
Now, it suffices to show that E9 <= Еф. Let us assume that x e E9.
Writing
Q m = { t e Q : 2 m ~ 1
< |x(f)| < 2m},
m= 0, ±1, ±2,...,
we obtain for any natural number k
00 00
Z (p(2m+2k)p (Q J ^ X J (p{22k + 1 \x(t)\)dp ^ J (p(22k + 1 \x(t)\)dp < oo.
m = — oo m — — со Qm Q
As in the proof of the above lemma we can assume that q> (u) = 0 for 0 ^ и ^ 1 and (p{u)> 0 for и > 1 (i.e. u 0 = 1). Therefore we get
2 k — 1 oo
£ <р( 2 - “ + 2 ‘)д( 0 -« )< < » , £ v ( 2 " + 2 l‘)/i (a m) < œ .
Equalities among Orlicz spaces, 1 261
From this it follows that p (Q -m) < oo,m = 1 ,2 , . . . , and there exists a strictly increasing sequence of natural numbers (mk) such that
00
( 1)
m = mjc
Next, let us put (p0 (u) = q>(u + l). It is seen that the sequence (mk) can be constructed in such a way that
<PoV~ml) < U (Po(2~mi) < <Po(2Wl), <Po(2 ~mk+i) < <Po(2~mk)/2, 2 <p 0 ( 2 mi) < (p0(2m2), mk + 2k < mk+1.
Now, we shall construct a ^-function ф such that q> <£хф, ф (cu)/(p (u +1) -► 0 as u-> 0 for all c > 0 and xeE*. For this purpose, we define the function p in the same way as in the proof of Theorem 2.1. Then, putting
Xk = m ax(l, p iQ -i) , ..., (0 _ fc)), к = 1, 2 ,..., we define
"0 for s = 0 ,
(Xmk + 22k + i mk + 2)~1 for <p0(2~mk+') ^ s < (p0(2~mk), k = 1 , 2
1 for (p0(2~mt) ^ s < (p0(2mi),
J k for (p0(2mk) ^ s < (p0{ 2 mk+1), fc = 1 , 2 , ..
Next let us set
U
x (u) — fp (t)d t for и ^ 0 ,
о
V
Ç{v) = J q(s)ds for v ^ 0 .
о
Finally, we define
<Po(x(u ))
•Ж = £ (<Po (x («))) = S 4 ( s) d s for u > °- 0
It is seen that ф is a ^-function. Similarly as in the proofs of Theorem 1 and Theorem 2 in [7] we prove that (p0 <^ф and ф < s(p0, respectively. Thus (p <€хф and ф(cu)/(p(u+ 1 ) -* 6 as u -> 0 for all c > 0 .
Now, we shall show that for any natural number k0 the following inequalities are satisfied
( 2 ) ф(2~т+ко)р (й _ т) ^ ( 2 km k+1)- 1 for к ^ k0 + 2, mk ^ m < mk + 1 - l , (3) ф(2т+ко) < 2k + 2 (p(2m+2k) for к ^ k0 + 2, mk < m ^ mk+l — 1.
10 — Roczniki PTM — Prace Matematyczne XXIX
First we prove that (2) holds. For к ^ fc 0 + 2 and mk ^ m ^ mk + i — 1, by the properties of (mk) we have
— m -rk0 ^ —mk + k0 ^ Hence
^( 2 -m+fco) ^ p^ 2 ~m+feo^ 2 -m+k° ^ p( 2 -mk_*)2~m+ko = 2~k+22~m+ko ^ 2 _m Further
, A ( 2 - ” + ‘ ») = « ( v o O c P ' ” * * 0))) e ( < P o ( 2 " " ) ) « > o ( 2 ^ )
« g(<p„(2-“-)) = l / ( 2 * ^ „ m t+1) «: l/(2‘ 2ramt+1).
Hence
<A( 2 - " +fco)p( 0 _ J ^ l / ( 2 *mfc+1).
Now we prove that (3) holds. For k ^ k0 + 2 and mk ^ m ^ m k+1 — 1, by the properties of (mfc) we have
m + fc0 ^ wk + i + ^o —1 ^ Щ + 1 + к < mk+2- Hence
X ( 2 m + k o ) ^ p ( 2 m + k o ) 2 m + k o ^ 2k + 1 2m+ko ^ 2m + 2k_1.
Since m + 2 k < m k + l + 2 k < m k + l , we get
! H 2 " + ‘ °) = £ ( < * > o ( * ( 2 " + ‘ 0) )) « Я ( ‘P o ( 2 m + 2 k ” x))< p 0 ( 2 m + 2ft~ *)
< q ( q > 0 ( 2 m k + 2 ) ) ( p 0 ( 2 m + 2 k - i ) ^ 2 k + 1 ( p 0 ( 2 m + 2 k - 1 )
= 2k+2 <p(2w+2k).
Finally, we shall show that x e JE*. Given Я > 0, we choose a natural number fc 0 such that X ^ 2ko. According to (1), (2) and (3) we get
00 oo
f il/(X\x(t)\)dp = X J ^(A|x(t)|)dp+ £ I Ф(A\x{t)\)dp
Q m= \ Q - m m = 0 Qo
Wko + 2 _ 1 oo mk + 1 - 1
« £ ф ( 2 £ ( £ ^ ( 2 - " +‘»)M£ 2 - J )
m=l fe = ко + 2 m = m k
mko + 2~ ^ oo oo
+ £ * (2 " +‘»),i(G J + £ ( £ ^ (2" +‘»)/i( O J ) < o o
m = О
k = ко + 2
m = m kand this means that xe£*. Thus the proof is completed. ■ By virtue of the proof of Theorem 2.2 we get the following:
T h e o r e m 2 . 5 . L e t < р е Ф 1 2 , i .e . ( p { u ) = 0 f o r 0 ^ и < w0 , ( p ( u ) > 0 f o r
и > u 0 > 0 a n d l e t <p b e c o n v e x . L e t 2 ( c ) b e t b e s e t о / a l l c o n v e x ( p - f u n c t i o n s
ф s u c h t h a t ( p 4 ^ ф a n d ф ( с и ) / ( р ( и + и 0 ) -»> 0 a s и - + 0 f o r a l l c > 0.
Equalities among Orlicz spaces, I 263
Then
E* = (J E* = (J Lf.
./«=4P<P ir\ ,lic'Vq‘ (r\
феЧ'^с) феЧ'^с)
Now, we shall show that if pi is an atomless measure or a purely atomic measure, then in the equalities in Theorem 2.1 (resp. Theorem 2.2) the strict inclusions hold: L f £ E9 for each ф е Т ^ t , whenever (реФХ1 (resp. for each ф еТ^х, whenever <реФ12). This follows from the following two theorems:
T h e o r e m 2.6. Let ф еФ ,,. Then for a cp-function ф the strict inclusion Ü % E* holds, if
(a) (p <^аф and pi is an atomless infinite measure;
(b) (p <^гф and pi is a finite atomless measure;
(c) (p <^&ф and pi is a purely atomic measure with measure of atoms bn satisfying 0 < inf„b„ ^ sup„b„ < oo.
P ro o f. This is proved in [ 8 ], Theorem 3.1, for ip being a (^-function, but the proof can be applied to each (реФ11. ■
T h e o r e m 2.7. Let (реФ12. Then for a (p-function ф the strict inclusion L* 5 = Ev holds, if (p <^хф and pi is an atomless measure or a purely atomic measure as in Theorem 2.6.
P ro o f. First, suppose pi is a finite atomless measure. Let us take a ^-function (p0 such that <p 0 ~ xip (see [5]). Since q>0 < хф, by Theorem 2.6. we get Lf £ £7° = E*.
Next, suppose pi is an infinite atomless measure or a purely atomic measure.
Since cp <^1ф,Ъу Lemma 2.4 the inclusion L f a Ev holds. Moreover, there exist /г-measurable pairwise disjoint sets Qt ci Q, i = 1 , 2 , . . . , such that
It is easy to verify that x e E* and хфЕ^. ■
3. On some equalities for L(p. Let Ф2 be the set of all Orlicz functions (p vanishing only at zero and tending to oo as и ->• oo.
In this section, we shall prove that for each <реФ2 and an arbitrary measure pi there exists a set of (^-functions such that
Define
1/i for t E Ü t,
i = 1
u = p) ü = n e *.
Moreover, we shall show that if p is an atomless measure or a purely atomic measure with measure of atoms bn satisfying 0 < inf„ bn < sup„ bn < оо, then the strict inclusion L9 £ holds for each i
Write
Ф 21 = {<рбФ2; ф ( м ) < oo for м > 0 }, Ф22 = {феФ2: ф jumps}.
Then
Ф 2 = Ф 2 1 и Ф 22.
Assuming that р is an arbitrary measure, we shall now prove the following two theorems:
T heorem 3.1. Suppose (реФ21. Let be the set of all (p-functions ф such that ф < z (p, le. Щ х = {феФ: ф < а(р}.
Then
L9 = f ) L*= П E
Il ie V t i феЧ'гг
T heorem 3.2. Suppose (реФ22. Let Т%2 be the set of all (p-functions ф such that ф < s(p, le . Щ 2 = {феФ: ф < s(p}.
Then
L9 = f l L*= П E*•
Ф<еЧ * 2 2 ф е Ч '2 2
R e m ark 1. In [10], Proposition 1.2.2, it is proved that if феФ 21, q> satisfies the ^-condition for large и and p is a finite atomless measure, then the equality I ? = L f holds, where the intersection is taken over all Orlicz functions ф satisfying the d 2-condition for large и and such that ф(и)/(р(и) -> 0 as и — » со.
Moreover, in [3] (resp. ([7], Theorem 2)) it is proved that if (p is a ф-function and p is the Lebesgue measure on a compact subset of a finite dimensional Euclidean space (resp. on a finite dimensional Euclidean space), then L9 = Еф, where the intersection is taken over all ф-functions ф such that ф <x(p (resp. ф < я(р).
P r o o f o f T h e o re m 3.1. By Lemma 2.3 we get that.the inclusion L? <= 0 ^ 2 1 Еф holds. To prove that the inclusion Lf c L9 holds, it suffices to modify the proof of Theorem 2 in [7] in which the inclusion
Еф a L9 is proved. Let us assume that x $ L 9. We shall construct a ф-function ф such that ф < &q> and x $ l f . Write
Qm = {teQ: 2m_1 ^ |x(t)| < 2 m}, m = 0 , + 1 , ± 2 , . . .
If p(Qmo) — oo for some integer m0, then for an arbitrary ф-function ф we
Equalities among Orlicz spaces, I 265
have (p (Я |x (t)\) d/л = oo for all À > 0, i.e. x ф L*. Therefore we can assume that fi ( Q m) < oo for each integer m. Since x ф L for any natural number n we have
Z q>(2m 2" Z f q>(2 2n l \x{t)§dn
m = — oo m = — oo Qm
= j (p ( 2~2n~ 1 \x{t)\)dfi = oo.
Q
Hence it follows that there exists a strictly increasing sequence of natural numbers (kn) such that
( n + l ) * „ + i - l
(1) £ <г)( 2 - " - 2 ”- ‘)А( й _ „ ) » 2”+3
m = nk„
or
( n + l ) k „ + i - l
(2) £ <p(2m~ 2n~ l ) p ( Q „ ) > 2” +3.
m = nk„
It is seen that the sequence (kn) can be chosen in such a way that
p p - c + i K b . + i - n ) < 2 _ 1 < p ( 2 ~ " ( k n - 1 ) ) , 2 < p ( l ) < 2 ~ " ~ 3 <p(2nkn~ 2") ,
<p( 2 n(k"_1)) < (p(2ln + m "+1- ly), (n + l)fc„ ^ nkn + 1, As in the proof of Theorem 2 in [7], let us set
0 for t = 0 ,
2 ~" for 2 ' ("+1)(k"+1" 1> < t ^ 2~n{kn~1\ 1 , 2 , . . . , 1 for 2 _kl + 1 < * < 2 kl_i,
2 " for 2 "(kn_1) < t ^ 2 ("+1)(kn+1_1), n = 1 , 2 , for s = 0 ,
2 -и for <p( 2 ~("+1)(kn+1_1)) < s ^ <p( 2 - "(kn_1)), n = l , 2 , ..., 1 for <p( 2 -kl + 1) < s < <p( 2 kl_1),
. 2 " for </> ( 2 "<k”~l)) < s ^ (p ( 2 (n+ 1)(k"+ 1 " г)), n = 1 , 2 , . . . P(t) = -s
Г0 4(0 = i
Define
X(«) = Z'( m ) . X »
for м е[ 0 , 1 ], for u e [ 1 , oo),
where [ 0 , 1 ] -» [ 0 , u j , *'(и) = SoP(t)dt, u 1 = x '( l) < U and f \ [ 1 , со] — ► [Mj,
o o) , *” («) = a - 1 (w), *(u) = $ t p{t)dt+ l.
Next, let us put
’?(v) for ve[0, (p(uj],
£"(v) for ve[(p(ut), oo),
m =
where [ 0 , <p(uj] -> [ 0 , r j , Ç'(v) = $v0-q{s)ds, Vi = I < 4 (<?(«!)) <?K) ^
О
and <*;": [ç>(Wi), oo) -► O l5 g o ) , £ ' » = r j'1 (и), rj(v) = f t q(s)ds + (p(u1).
Finally, we define
, n = f»A/ (w) for mg [ 0 , 1 ], l*"{u) for mg [ 1 , oo),
where if/' (u) = ç' (<p ( / (м))) and ф"{и) = Ç” ((p(x" (и)))- It is seen that ф is a <p-function and as in the proof of Theorem 2 in [7] we get ф <^a <p. Moreover, using the properties of the sequence (kn) we can prove that for any natural number n0 the following inequalities are satisfied
(3) ф{2~т~П0+1) ^ 2 ~ п~3 (p{2-m' 2n~1),
(4) ф(2т~по + }) ^ 2~n~3 q)(2~m~2n~i),
where n ^ t n 0 and nkn ^ m ^ (n + 1 ) kn+1 — 1 .
For example, we shall show that (4) holds. For n~^n0 and nkn ^ m ^ (n+ l)/c„ + 1 — 1 we have
2 m~ 2"
a ( 2 m-2«)= j p (i)^ + l ^ p(2m~2n)2m~2n+ l и i
< 2" 2m — 2n _|_ j < 2m — ” + 1 < 2m — no + 1 and hence, since x” (M) = a_1 (MX
^ 2 m-no + > 2m~2n Thus
ф(2т~по+1) = Г (<p(x"( 2 m~"°H ' 1))) ^ Г (q>( 2 m“n)).
Since ' ( m ) = */“ 1 (u), it suffices to show that q(2~n~3 <p{2m~2n)) ^ (p(2m~2n).
We have
2 ~ n ~ 3<p(2m ~ 2n)
rj(2~n~3 (p(2m~2n)) = f ^( s )^/ s + ç )( m 1).
» i
Since m — 2 n ^ nk„~2n, by virtue of the properties of the sequence (kn), we obtain
2 " 3 <p (2 m 2") ^ 2 ” 3(p(2nkn 2n) > 2 <p( 1 ) > (p(ut) > vt . Hence
2 ~ n ~ 3<p(2m ~ 2n) 2 - n _ 3 </>(2n(kn ~ 2))
f q{s)ds ^ f 4 («Ms ^ <?(<p(l))<p(l) = <p(l) > ^ ( m J .
о 0
Equalities among Orlicz spaces, I 267
Therefore
2 ~ n ~ 3<p(2m ~ 2n) 2 ~ n ~ 3</>(2m ~ 2n)
tj(2~n~3(2m~2n)) = J q(s)ds + (p(ul) ^ 2 J <?(s)ds
i>i о
^ 2q(2~n~3 (p(2m~2n))2~n~3 (p(2m~2n)
^ 2q (<p ( 2 (” + D*" + i ~ *)) 2 ~ 3 <p ( 2 M~ 2")
= 2 -2 <р( 2 т_2") < (p{2m~2n).
Similarly we can prove that (3) holds.
We show now that x ф Ьф. Given X > 0, we choose a natural number n0 such that X ^ 2“no+2. According to (1), (3) or (2), (4) we get
J ijf(X\x(t)\)dfi ^ JiA(2_no+ 2 |x (0 l)^
Q Q
00 00
» £ + £ ф(2’"-"°+1) ^ ( 1 т)
. m = 1 m = 0
00 (n + l)fc„ + 1 — 1
> I ( I ф ( 2 - " - - ^ 1) ц ( й . т))
n = no m = nk„
oo ( n + l ) k „ + i — 1
+ I ( I — 00
п = п о m = nk„
and this means that хфй*. Thus the proof is finished. ■
P r o o f of T h e o re m 3.2. In order to get the inclusion L? c: it suffices to apply the following:
L emma 3.3. Let (реФ22 and let ф be a (p-function such that for all c > 0 limsup ф(си)/(р(и) < oo as и -> 0. Then Lv cz Ef.
P r o o f o f L em m a 3.3. It suffices to show that W <z L%. Let x e L ?, i.e.
(1) $(p(X0\x(t)\)dn< oo for some A 0 > 0, Q
( 2 ) |x(t)| ^ c0 a.e. on Q for some c0 > 0 .
By our assumption there exist u0 > 0 and d > 0 such that ф (и) ^ dtp (A 0 и) for и < w0. Without loss of generality we can assume that u0 < c0. Then for
u e [ u 0,C q ] we have
dtp (X0 и) ^ dtp (Л0 u0) ^ ф (u0) ^ (ф (и0)/ф ( cq )J ф (и).
Hence
By (1), (2) and (3) we have
f ф (|x ( 01 ) dn ^ (dij/ (с0)/ф (u0)) f (p (A 0 |x ( 01 ) dfi < 00 ,
Q Si
i.e. xeL%. Thus the proof of our lemma is finished. ■
Now, we show that ^ c ^ P• Let us assume that x ф I f. We shall construct a (^-function ф such that ф <€s(p and хф!У. Define
Q m = [ t e Q : 2m_1 < |x(01 ^ 2W} , m = 0, ± 1 , ± 2 , . . .
We can assume that n(Qm) < oo for each integer m. Since x ф I f , for any natural number n we have
00 00
I <p[ 2~m~ l n ~ ' ) t i ( Q - J + X < p ( 2 " - 2" - ‘ ) ^ ( f l J > f < ( . ( 2 - 2" - M x ( t ) l ) d i U = o o .
m = 1 m = 0 Si
Without loss of generality we- can assume that sp (u) < oo for .0 < и ^ 1 and (p(u) = oo for и > 1. Therefore we have ь '
(1) X =
m — 1
or
( 2 ) £ oo-/z(Om) = oo.
m = 2n + 2
First suppose that the condition (2) is satisfied. Then there exists a sequence of natural numbers (/„) such that /i (Qln) > 0 .
Let us assume that ju(Æin) ^ b > 0 for some h > 0, n — 1 ,2 ,... Then for any феФ and any X > 0 we have ф(X|x (01)dfi ~ oo, i.e. хф1У.
Therefore we can assume that there exists a strictly increasing sequence of natural numbers (kn) such that
(n+ l)(k„+ 1 - 1 )
(3) X
m = nk„
or
(4) fi( f lJ jO as n -> oo, fi{QkJ > 0 , Ц(Qkl) < 1.
Let (p0 be an arbitrary ^-function such that ф 0 (1) == <P(1)- It is seen that a sequence (kn) can be chosen is such a way that nkn+i ^ (n + l)k n, kn ^ 2n,
<р( 2 ~(и+1)(к"+1~1)) с 2 _1 <p( 2 ~"(kn-1)), q>{2-k' + 1)<q>0(2k%
1 < <Po( 2 k") < 2 _1 <p 0 ( 2 kn+1)- Set an = n(QkX n = 1, 2 ,...
Let us put
"0 for t = 0 ,
2~n for 2~{n+l){kn+1~1) <$ 2 ~n(k"_1), n = 1 , 2 , . . . , 1 for 2 " kl + 1 < r ^ 2 ,
2kn+2 for 2 " < t ^ 2 rt+1, n = 1 , 2 ,...;
P (0 = *S
Equalities among Orlicz spaces, I 269
"0 f o r s = 0 ,
J 2~n f o r (p(2~(n+1){kn^ ~ 1)) < s ^ <f>(2~n(kn~X)),
л
1 f o r (p{2~kl + 1) < s ^ (p0(2kl),
s. 2/^n + 1 f o r (p0(2kn) < s < (p0(2kn+1), n = 1 , 2 , . . .
Next, let us set
U
X(u) = J p{t)dt for и ^ 0, о
V
Ç (v) = J q (s) ds for v ^ 0 . о
Finally, we define
<Po(*(u))
ф(и) = £ (ф 0 (/ (u))) = J q (s) ds for и ^ 0 . 0
It is seen that ф is a «^-function. Moreover, proceeding similarly as in the proofs of Theorem 2 and Theorem 1 in [7] we get ф < s(p and (p0 < гф. As in the proof of Theorem 3.1, we can show that for an arbitrary natural number n0 the following inequalities hold
(5) ф(2~т~по+1) ^ 2 ~”~3 <p(2~m-2"~1),
where n ^ n 0 and nk„ ^.m ^ (n+ l)k n+1 — 1. Moreover, it is easy to check that
( 6 ) Ф{2п) > \ / а п
for n > 2 .
Now, we shall show that x фЬф. Given X > 0, we choose a natural number n0 such that X ^ 2~no+2. According to (3), (5) or (4), ( 6 ) we get
№(X\x{t)\)dp ^ £ J ^ ( 2 -no+ 2 |x (r )|)^ + f j ф(2~По+2\х(1)^р
П m = l Q - m m = 1 Qm
» £ * (2— "°+‘)„(O ..J + £ ф (2t"~no)ii(Q kJ
m — 1 m = no
oo (n + 1 )fc„ + i - 1
> E ( E * ( 2 - » — +V ( 0 - J )
n = n о m = nfcn
+ X ^ ( 2 ")/*(fi»J = oo,
m = /io
i.e. хфЬф. Thus the proof is completed. ■
By virtue of the proof of Theorem 3.2 we obtain the following two
theorems:
T heorem 3.4. Suppose (реФ22, he- (p(u) < oo for и 4 u0, (p(u) = oo for и > u0 > 0. Let (p0 be an arbitrary (p-function such that q>0 ( h 0) = cp (u0) and let
be the set o f all (p-functions such that ф 4 s<p and (p0 4\ф- Then
L* = p) L+ = П E *•
феЧ’Ч^0 фе'Р$2Ч>0
T heorem 3.5. Suppose (реФ22, lc . ф (м) < °o f or u ^ Mo> <P(U) ~ 00 f or и > u0 > 0 and moreover, suppose tp is p-convex, p > 0. Let q>0 be a p-convex (p-function such that <p0 (u) — q> (u) for 0 4 u 4 u0 and let T 2f° (p) be the set of all p-convex (p-functions ф such that ф 4 s(p and (p0 4 хф.
Then
LE = p| L* = f | E*.
феЧ,22Ч>0(Р) феЧ'^Чр)
E xam ple . Let us put
II for 0 < и 4 1 , и > 1 ,
w here p > 0 and let
fo
< poo (m) = < 1 00
for 0 ^ U 5 $ 1 ,
for и > 1 .
Taking (p0(u) = up for и ^ 0, we have (p{u) = max(<p 0 (u), (p^iu)) for и ^ 0;
hence LE = I / n L 00 (see Theorem 1.1). Therefore by Theorem 3.5 we obtain C orollary 3.6. Let p > 0. Then
Lp сл L® = P |L ^ = P ^ , lp = p / ^ = p | ^ ,
Ф Ф ф ф
where the intersections are taken over all p-convex (convex if p ^ 1 ) (p-functions such that ф(и)/ир -+0 as u -* 0 and ф(и)/ир -*■ oo as и -> oo.
In particular
L l n L x = f ] L f = p |£ * Z 1 = p /<" = p A*
Ф Ф ф ф
where the intersections are taken over all N-functions.
Now, we shall show that if p is an atomless measure or a purely atomic measure, then in the equalities in Theorem 3.1 (resp. Theorem 3.2) the following strict inclusions hold: LE ç Еф for each ф е Т < ^1, whenever (реФ21 (resp. for each ф e ФЧ2, whenever q> e Ф22). This follows from the following two theorems:
T heorem 3.7. Let (реФ21. Then for a (p-function ф the strict inclusion
LE ç E* holds, if
Equalities among Orlicz spaces, I 271
(a) ф a <p and p is an infinite atomless measure;
(b) ф <^x(p and p is a finite atomless measure;
(c) ф <^s(p and p is a purely atomic measure with measure of atoms bn satisfying 0 < infnbn ^ supnbn < oo.
P ro o f. Since Ф21 = Фи , this follows from Theorem 2.6. ■
T heorem 3.8. Let (реФ22• Then for a (p-function ф the strict inclusion
jj > ^ holds, if ф <^s(p and p is an atomless measure or a purely atomic measure as in Theorem 3.7.
P ro o f. First, suppose p is an atomless measure. Let us take a (^-function (p0 such that (p0 ~ s(p (see [5]). Since ф < s(p0, by Theorem 3.7 we get
JJP = l<PO £ Е Ф
Next, suppose p is an atomless measure. Since ф s q>, by Lemma 3.3 we get Lv cz E^. Let (um) be a sequence of positive numbers such that мш|оо as m-+oо, ф(и1)> 1. Then there exists a sequence (Om) of pairwise disjoint
^-measurable subsets of Ü such that
M(&m) = l/( 2 mф(muj) if p{Q) = oo, (®m) = H W ( 2 m ф (muj) if p(Q) < oo.
Let us set
x{t) = 0
for te Q m,
CO
for t e Q \ \ J Qm, m= 1
m = 1 , 2 , . . .
It is seen that хфЬ<р, because (p jumps and um joo as m -*■ oo. We shall show that x e E*. Indeed, given X > 0, we choose a natural number m 0 such that m 0 ^ A. First, let us assume that p(Q) = сю. Then
m= 1
mo — 1 oo
^ I Ф(т0ит)р {й т)+ X (^(т 0 Мт)/( 2 ">(™«т))) <
m — 1 m = mo
i.e. x e E f. Similarly we get xeE * if p(Q) < oo. ■
4. On some equalities for E^ + L00. We shall need the following:
• L em m a 4.1. Let p be an arbitrary measure and let q> and ф be Orlicz
functions such that for an arbitrary c > 0 lim su p ,,^ (р{си)/ф(и) < oo.
Then L* c E^ + L*.
P ro o f. Let x t l f , i.e. Ф(A 0 \x(t)|)dp < oo for some A 0 > 0. Let u0 be a number such that ф (A 0 u0) > 0. Denote
= {te ü : |x(t)| > u0}, Q2 = {teQ: |x(t)| < u0}.
Then x = X!+X 2 , where x t = х / Й1 and x 2 = Х 7 Й 2 б 1 ®. (Here X q { denotes the characteristic function of the set Qt.) It suffices to show that x l e E (p. Indeed, for an arbitrary c > 0 there exist d > 0 and uc > 0 such that
(1) (p(cu) < d\f/(A0u) for и ^ uc.
Without loss of generality we can assume that uc > u0 and ф (A 0 u0) < oo. Then, by ( 1 ) we get for u e [n 0, nc]
(p (cu) ^ (p (cuc) < dif/ (A 0 uc) < d (ф (A 0 ис)/ф (A 0 u0)) ф (A 0 и) and thus
(2) (р (си) ^ Я 0ф (А 0 и) for и ^ м0, where d0 = dф(X0uc)/ф(X0u0). By (2) we get
J (p (c \x (01) dp, = J (p(c\x (01) dp ^ d0 J ф (A0 |x (01) dp < oo,
o n 1 о
i.e. x ^ E * . Ш
Assuming that p is an arbitrary measure, we shall prove:
T heorem 4.2. Let (р€Ф1 ( resp. let <реФх and let q> be p-convex, p > 0). Let (resp. (p)) be the set of all continuous Orlicz functions (resp. p-convex Orlicz functions) vanishing near zero, taking only finite values and such that (p < хф.
Then
E^ + L™ = 1J L* (resp. Ev + L™ = 1J t f ) .
i/'e'f'o феЧ'о(Р')
P ro o f. By Lemma 4.1. we have the inclusion (J i f cz E^ + U 0.
Now, let x e f ^ + L®. We shall show that there exists ф е ¥ $ (resp.
феФ%(р)) such that x e l f . We have x = Xj + x2, where х^еЕ* and x 2 eL°°.
Then
( 1 ) Jç)( 2 |x 1 ( 0 l)d/i < oo for all A > 0 , о
(2) A 0 |x 2 (01 < 1 a.e. on (2 for some 0 < A 0 < 1.
Writing
Q0 = {teQ: 0 < \x1 (01 < 1 },
Qm= {teQ : 2 m~ 1 < (x 2 ( 0 ( < 2 "}, m = 1 , 2 , . . . , by ( 1 ) we get for any natural number к
( p ( 2 m + k ) p ( Q J ^ £ J <p(2*+1|xi(0l)d/* < f ( p ( 2 k + 1 l x 1 ( t ) l ) d p < oo.
m = 1 m = 1 O m О
Equalities among Orlicz spaces, I 273
Hence there exists a strictly increasing sequence of natural numbers (mk) such that
(3)
Let us put
m = nth
q> (2mk) < q> {2mk+*), mk + k < m k + 1.
f ° for 0 ^ t < 1 , p(t) = l 1 for 1 ^ t < 2 Wl,
for 2mk^ t < 2 mk+‘, fc = 1 , 2 , . . . r ° for s = 0 ,
t) = \ 1 for 0 < s < <p( 2 mi),
for q>(2mk) ^ s < q>( 2 mfe+1), к = 1 , 2 Next, let us set
X(u) = j p(t )dt for и ^ 0, о
V
