0.1.3. 4 0.1.2. 0 0 0 0 0.1. Orlicz spaces 6 0. Preliminaries

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXVI (1986)

M. No w a k (Poznan)

On some linear topology on Orlicz spaces L ^ i n ) . I

Abstract. Let ф and (p be «^-functions. We shall say that cp increases essentially more rapidly than ф, in symbols ф <| <p, if for an arbitrary c > 0

In [5 ] and [6 ] we have considered some linear topology, denoted by on the Orlicz spaces L f f p ) , where (E , E , p) is a finite dimensional Euclidean space with the Lebesgue measure ([5 ]) or (E , E, p) is an arbitrary measure space with a positive measure ([6 ]). The topology has a base o f neighbourhoods o f 0 consisting o f all sets o f the form: К Д г ) n L * f{ p ), where r > 0 and ф is such that ф <! <p.

The present paper is a continuation o f papers [5 ] and [6 ].

In Section 1 we prove the main lemma:

I f a set Z c z L * v (p) is such that sup ЩхЦ,,: x e Z j = oo, then there exists a cp-function ф such that ф <^(p and sup {||x||^: x e Z ) = oo.

In Section 2, using the main lemma, we show that a set Z c L f^ ip ) is bounded in .У"** if and only if Z c K V(R ) for some R > 0. Next, we give some criterion o f compactness o f sets in (L|v (/i), M oreover, we prove that if ( L f v (p), is a locally bounded space, then is identical with the mixed topology y (,T v, 0) on L f i p ) .

In Section 3 we consider L f f i p ) as two-norm space. Finally, in Section 4 we prove that if (£ , E, p) is a measure space with a positive, atomless, ст-finite, separable measure, then the weight o f the topology 3 ~<ч> in 0 is equal continuum.

0. Preliminaries

Throughout this paper we assume, as in [

6

], that (E , Г, ц) is a measure space with a positive measure. Henceforth we shall denote the Orlicz spaces and spaces of finite elements over (E , Г, ц) respectively by L*4* and L04f>.

0.1. Orlicz spaces

0.1.1. It is said that a function <p: [0, oo) -> [0, oo) is a (p-function if it is continuous, non-decreasing and such that <p(

0

^{) =}

0

, <p(u) >

0

for и >

0

^and

(p(u) —► oc for m —►oo.

0.1.2. ^Th e o r e m. Let ф and cp be cp-functions such that ф <^(p. Then

L*<pf— ^

4

-] the proof of Theorem 1).

0.1.3. ^Th e o r e m. In L** an F -norm can be defined as follows:

ф(си)

lim - 0 ([4 ], p. 69, [5 ], p. 72).

u -*0 <p(u) «->* (p(u)

IM U = in f [e > 0: qv(x/e) ^ e}.

The space L*v is complete with respect to the F-norm ||-||ф.

We shall denote by the topology on L** generated by the F-norm Il II,.

0.2. Topology of convergence in measure

0.2.1. ^{Let S}

0

be the linear space consisting of all real valued functions, defined and /л-measurable on E, which are almost everywhere finite valued and bounded outside a set of finite measure. Then in S0 an F-norm ||

||0

can be defined as follows:

IMIo = inf {e > 0: p {te E : | x (t)| > e }< e } ([7], p. 30).

We shall denote by ,T 0 the topology on S0 generated by the F-norm || -||0. It is seen that a sequence x„ in S0 is convergent to x e S 0 if and only if a sequence xn is convergent to x in measure, in symbols х „^ х . Moreover, for every ф-function (p we have L*v c= S0 and that is strictly finer than restricted to L*v ([7], p. 30).

0.3. Linear topology /7<<tp on ^{L*^}

The definition of the topology ^ <tp is given in [5] and [

6

]. Now, we recall some theorems from [5] and [

6

] which we need in this paper.

0.3.1. ^Theorem. The topology 7 ' <<p has a base of neighbourhoods of 0 consisting of all sets of the form: K^{r) n L*^, where r > 0 and ф (p ([5], P- 74).

0.3.2. Theorem. The topology 7 <<p is coarser than the usual topology 7 ^ on L** ([5], p. 78).

The two theorems above are proved in [5] in the case where (F, I , p) is a finite-dimensional Euclidean space with the Lebesgue measure, but they remain true when (F, I , p) is a measure space with a positive measure.

0.3.3. Theorem. Let Z a L*<p be a set such that sup {||х||ф: x e Z ) < oo.

Then on a set Z the topology 7 <<f> is identical with the topology 0 ~0, i.e., Р <щ\z = ^olz ( [

6

], Theorem 1.6).

0.4. Linear spaces with mixed topology

0.4.1. ^Definition. Suppose that in a linear space X there are defined two linear Hausdorff topologies: 7 and 7 * . Let 3 ^ and 3 $-* be bases of neighbourhoods of 0 in topologies 7 and 7 * , respectively. Neighbourhoods in 3 ^ will be denoted by U, V, ..., and neighbourhoods in 3$-* will be denoted by U*, V*, ... The mixed topology on X determined by the topologies 7 and 7 * we call the linear topology, which has a base of neighbourhoods of

0

consisting of all sets of the form

00

^N

U XO/Jnnl/)

IV- 1 n- 1

for any sequence ( U *) in 0$ $-* and any U e We shall denote the mixed topology by y (.r t T * ) ([9], p. 49).

0.4.2. ^Theorem. I f the neighbourhoods of 0 belonging to j- are bounded in the topology 0Г, then for every linear topology т defined on X, the condition

t\z <= 3~*\z for each set Z bounded in 0Г, implies the inclusion

TCz y { r , ЗГ*) ([9], p. 51).

In particular, the mixed topology y(&~, 0Г*) is the finest of all linear topologies t on X which satisfy the condition:

i|z = 0Г*\z for each set Z bounded in .

0.4.3. ^Theorem. I f the neighbourhoods of 0 belonging to 0 $ are bounded in the topology 0Г, then for every linear topological space (Y, t) a linear operation А: X -> Y is (y(0~, 0~*), z)-continuous if and only if the function A\z: Z - + Y is (0~*\z, z)-continuous for each set Z bounded in ([9], p. 52).

0.4.4. ^Definition. We say that a linear topology on X satisfies condition (*) with respect to the pair {0~, 0Г*) if the following condition is satisfied:

(*) for every linear topological space (Y, t) and a linear operation A: X -> Y if for each set Z c X bounded in the function A\z: Z -> У is (^~*|z , i)-continuous, then the operation A: X - + Y is

continuous.

0.4.5. ^Theorem. I f the neighbourhoods of 0 belonging to 0&у are bounded in the topology 0Г and the linear topology ЗГ' on X satisfies condition (*), then

Г у(0Г, .Г*) ([9], p. 52).

0.5. Two-norm spaces

0.5.1. ^Definition. Let in a linear space X two F-norms || || and || ||* be defined. We say that a sequence (xn) in X is y(|| ||, || ||*)-convergent to x0gX,

in symbols if

ll*n—*о

1

^{Г - + о}

and

the set (x

„}®=1

is bounded in the topology 0 ц ц

generated by the F-norm || ||.

The space X with y-convergence is denoted by <X , || ||, || ||*> and called a two-norm space ([1], p. 49).

0.5.2. Definition. Let <X, || ||, || ||*> be a two-norm space. Let (Y, 3F) be a linear topological space. A linear operation A: X -+ (Y ,;T ) is called

(

7

,

.T)-linear if

xn^ x

0

^implies A (x „ )^ A (x 0).

In particular, a linear functional F : X -> R is called у-linear if

x^ X q implies F (x n)- + F (x 0) ([9], p. 59).

1. The main lemma

1.1. Lemma. Let q> be a (p-function and let Z c: be such that SUP {IWU: x e Z } —

00

. Then there exists a (p-function ф such that

1

j/ < (p and sup{||x||^: x e Z } =

00

^.

Proof. Let Z c be such that

(*) sup [||x||„: xeZJ =

00

^.

For a measurable function

x:

E - * R and a ,b e R such that —

go

^ a < b

^

+00

let us denote

Eba{x) = {teE : 2e < |x(f)| < 2b}

and let Fb(x): R -+ R be the function defined as Fba{x){t) = x(t)

0

for te E b(x), for teE \ E b(x).

First, we shall show that from (*) it follows that there exist a sequence (x„) of functions from Z and a strictly increasing sequence of natural numbers (kn) such that

( 1) (**)

or or

+ > 4" for

n =

1, 2, (2) l|F-Ü;;.‘- llW IU > 4 " for n = 1,2,..., (3) 1"-”! 1 + 2 >4” for

n

= 1, 2, ...

We shall prove this by induction. Let /cj > 0. Assume that for every x e Z there holds

11*4” Mil, < 4 and || F : J ; W | | ^ 4 and ||fI « Mil, « 4.

Then, from these

inequalities it would

follow that

11*11, < \\F ~ * (*)IU + I I * \ WIU + ll ^ ( * ) l l , < 12 for x e Z and this contradicts (*).

Hence, there exists x xe Z such that

ll^-

1

ao(*i)ll4> > 4 or (*i)IL

> 4

^or \\F% (*i)ll, > 4-

~ кл .

Since \\F_a and ||Fk (х^Ц^ are non-decreasing functions of variable a > kx and b > kx and

lim ||F:;i (x 1)||<p = ||Ffc_

100

(x1)||(/,> 4 a -* i

Hm||Fj

1

(x 1)||, = ||Fk® (x

1

^)||,>4,

b-*ao

there exists a natural number k2 > kl such that

Н ^м

2

+ <(*,)11„>4 or ||Fl“

2

^-

1

^(x

1

^)||„>4.

Now, we assume that there exist already natural numbers k{ < k2 < ... < kn and functions x

l5

xn_ xe Z such that there holds (1) or (2) or (3).

Let us assume that for every x gZ there holds

"w ii,, « 4» and i i F ^ ; ; ; ,

1

. „ м п , « r

and l|F„G£n+„_

1

(x)||(p ^ 4".

Then from these inequalities it would follow that

IMU « ||f: ^ ”+"‘

1

^\x)||» + ||F*"

1

^{I ; ;i,,-}

1

^,(x)||, + ||FS|i + „ - 1(x)||, « 3-4", and this is contradicts (*).

Therefore there exists x„gZ such that

||F:';‘ “+"_ U(xJ , > 4" or ||Frt(”* ; +" » - ( * J „ > 4”

or l|FSB+n-i W IU >4".

Since уF _ (”kn+n

1

,(x„)||v and ||Fjk||+„_

1

(x„)||v are non-decreasing functions of variable a > nk„ + n —

1

and b > nk„ + n —

1

^and

lim = \ \ F

I

* n + ”~1 (xJII^ > 4”

a -»-ao

lim ||F»kn+n_

1

(xn)||<, = \\Fnkn+n_ 1 (х„)||ф > 4", .

b-* oo

so there exists a natural number kn+l > k„ such that

\\FZt^m!+i ^ - 2 ) ( xn)\\v > ^ n or l l ^ +

1

^ïk-"î

1

^+"‘

2

W I| ,> 4 -.

Thus, we proved (**).

Now, we shall show that there exists a ^-function ф such that ift < (p and

SUP {N1*: x e Z} = oo.

It is clear that for this purpose, it suffices to show that there exists a

</>-function ф such that ф q>and for every n ^ 2 k 1+2 there holds (***) e A 2 ~ ’ * 4 x „ ) > 2 ”" 5/n.

From inequality (**) we have

(i) 4" < e, (

4

' " F : (!îï+iMT,+

1

^{+ -}

2

i <*"»

( n+ l ) k „ + 1 + П - 3

« £ v (4 - " - 2 - " )/ i(E :;- ,(x j)

m = nkn + n — 1 ( n + l ) k „ + 1 - 2

m = nkn

or

*1

(II)

4

"< e ,(

4

- F '*

1

"J;;+ "»-

11

W ) = I

4

«(

4

-"-

2

”)/

1

(E:_

1

(

x

J) +

m = — kj

nkn + n — 1 nkn + n— 2

+ X V(

4

-*-

2

")

ai

(

e

; -

1

(

xj

)+ I <p(

4

-"-

2

-")/j(E:;.

1

(xJ)

m = k j + 1 m — к j

or

(HI)

4

" <e,(

4

" F £ +

1

?-*î

1

""~

2

(x11))

( и+ l)fc„ + i + и - 3

« £ v (4 -” -2” ) /I(E :_,(x„))

m = nk„ + n

(n+l№„+l

-2

£ <г>(

2

" - " )д (£ ” : ; _

1

^(х,))

m = nk„

It is seen that a sequence (k„) can be constructed as in the proof of Theorem 2 (see [4 ]), i.e.,

(k„) is strictly increasing,

<p(

2

”(fc"-1)) < q)(2ln+1)kn+l~ \ <p(l/

2

("+1)fcn+1_1) < cp(

2

"(fe"_1)), (р(2"1кя~1))/2н+2 > <p{

1

^{), <p(l/}

2

("+1)k"+1_1) ^ <p{l/2nkn+n))/2.

We define the ^-function ф as in the proof of Theorem 2 ([4]). Then, from the proof of Theorem 2 ([4]) for any natural number n we have

(a) ф{2~т+2) ^ 2~n~3 (p(2~m~n), where nk„ ^ m ^ (n+ l)k„+k —

1

^,

(b) ф(2т + 2) ^ 2~n~l (p(2m~n), where пк„ ^ m ^ (и + l)/c

n+1

^—

1

^.

Now, we shall show that

^ ( 2 и+4хи) > 2" 5/и for n ^ 2/cj.

Let us assume that there holds condition (I). Then from (a) we get

0

^{* (}

2

- " +4xJ 5* M

2

" " +

4

f : i ^ +m J

1

^+B-

4

^{>W )}

(n + l)k „ + i + n - 4

v ^ (

2

- - ' +> ( £ : Г ‘ м )

m = nkn + n — 2 (" + l)k„ + i - 2

x <И

2

^{- " +}

2

^{)m( £ : :î; : Hx„D}

m = nkn

(n+ l)kn+ i - 2

m = nkn

^

2~"~3

^-

4

n > 2n~5/n.

Now, let us assume that there holds condition (II). Then from (b) we get

^ (

2

^{- ”+}

4

x„) > e *(

2

^{- ”+}

3

x„) > e

4

⁽

2

^{- " +}

3

^F(;

4

⁺

1

^’‘_"t

1

+”' :!(x„))

( « + l)fc„ + 1 + n- 3

» X ^ (

2

^{” - ”+}

3

)д (£ " +

1

^(х„))

m = nkn + n— l

(n + l)/cn + j -

2

X i/'(2”+3)m( £ : : ; -iW )

m = nkn (n+ l)k „ + i ~ 2

m = nkn

>

2

^-П-

1

^.

4

» > 2n~5/n.

At last, let us assume that there holds condition (III). Then

(Ci)

or

( c 2)

*1

X

<p(4-” -2m)n(E” -i(Xn))>4n- 1

m = — к j + 1

nkn + n — 2

X <?>(4-” -2")M(£ " _

1

(x J )> 4

" - 1

m = k j + 1

(c3) X 4>(4-"-2^ ^ ( £ : ; . , (xJ )> 4 '

m = к j

Let there hold condition (ct) and let n ^ 2kx. Then

*i

4

"-1

^{< X} <p(4-n-2”') n ( E : - 1(x„))^<p(2k' - 2n) fl(EÜ\l (x„))

m = — к J + 1

й*(2-"+2х .)> е*(2 ""+2f ï ; w ) > Z ^(2-" + 2-2")^(£:+1(x„))

m = — fe j

> ^ (

2

^{- ,” +}

,‘-1

⁺

2

^{) /}

1

^{(£‘Л |(х„|).}

But lkt ^ n + /ci ^ (/ + l)/c,+ i ~ 1, where l ^ n — 2/q. Hence

ф(2~(п+к1) + 2) ^ 2~l~3 (p(2~{n+kl) l) ^ 2 - " - 3 (p(2~2n+kl), and therefore finally we obtain

дф{2~п + 2хп) ^ 2 - а- 3-4п- 1 = 2 n~5 > 2 n~5/n.

Now, let there hold condition (c2) and let n ^ 2kl +2. Then there exists n0 such that

1

< nn ^ n—

1

^and

(n0+

1

^)*и0+

1

⁺ⁿ⁰

4 "-‘/(и-1|< £ V (4-"-2” ) / i ( £ r ‘ (x,)) m="

0

^k/i

0

⁺

"0

(n0+ l)fc„0+ 1

= I o<2-2"+”+" V ( £ : ; ; r ‘ (x„,).

m=n

0

^kn

0

Then

в»

(2

■- '■ ■♦ ' x„) 5* Q, (

2

^{■- '■ ■-}

1

f i r '£ ? + ' + "° + ‘ M"

0

^kn

0

⁺

n0

<no + 1>fe«0+ 1 +»o

> £ (2 - " +1 - 2")

JU + 1 (X . ))

m==«

0

^{kno +}

"0

(n 0 + l ) k „ 0 + !

= £ ^

2

- " +" +"°+‘ м

£ ::;» +1

^{m .}

m=n0kn0

Let n

0

^/c

„0

^ m ^ (n

0

+ l)^n0+i • We have — n + m + n0+ 1 = (m + {n0 — n)— 1) + + 2.

(ot) First, let us consider the case where m + (n0 — 1) — 1 ^ —k1. We have m + {n0- n ) - 1 = — ((n -n 0) — m+ 1). Then there exists a natural number lm such that

1

^ lm ^ n —

1

^and

lmkim^ { n - n 0) - m + l

<(/m+l)fcIm+1- l.

Hence from (a) we obtain

(_l_) ф(2т+{п°~п)+1) = ф(2 ^п~По) т+1^ + 2) ^ (р(2 ^п~п°) т+

1

^~i",)/

2

^{<m +}

3

ÿ ^ ,2- (," - ”o|- ” + 1, + "+1)/2"+2 ^{= <p}( 2 ~ 2" * " 0 + m) / 2 ” + 2 .

(P) Next, let us consider the case where m + (n0 — n)— 1 ^ kl . Then there

exists a natural number lm such that

1

< lm ^ n—

1

and lmkim < m + (n0 — n)~

— 1 ^(/m + l)/c,

+1

— 1. Hence from (b) we get (+ + ) i/,(

2

”, + ("0_"’ + ') =

0

(

2

<“ + ‘-o- " ’ - 1)+2)

> v(

2

” + " ° " ”“

1

"'")/

2

^{'", +}

1

> <j)(

2

^{” + "}

0

“ " ' ' “ "+')/

2

^"

= <p(

2

-2" + m+"°)/

2

^{” ÿ ç>(}

2

“ 2" + "' + "

0

^)/

2

^{" + 2-}

(y) At last, let us consider the case when — kx < m + (n0 — n)— 1 < kx Then —3kl < m + (n0 — n) — 1 — 2Ay < — kx and hence kx < — (m + (n

0

^{—n) —}

— 1—2/cj) < 3/cj ^ 2A

:2

— 1 (lm = 1). Therefore from (a) we obtain ф(

2

m+("o~n)+ ^ _ ^ (

2

_(_(m + (”0_”)_

1

^_

2

fci)) + 2fel + 2)

( - (m + (n0 - и) - 1 - 2k i )) + 2^

m + («Q — n) — 1 — 2fci — l j

> 2~"~3 ^{~ O m} + ("

0

^~")~n')

> 2~n~3 ™/'">~2n+m + no\

(+ + + )

^ ф{2

^ 2~n~3c p(2

<P( 2

(p(2 ...‘u).

Thus from ( + ), ( + + ) , (+ + +), it follows that

дф(2~n+1x„) ^ 2~n~3 -4n~ 1/(n— 1) = 2n~5/(n— 1).

Finally, let there hold (c3) and let n ^ 2кл . Then similarly as in case (c2) there exists a natural number n0 such that

1

^ n0 ^ n —

1

^and

(n0+ l)k„0+ ! +n0-

1

4 " - 7 ( n - l K I <p(4-"-2-")/i(£::^I (xll))

m= n0kn0 + n0 - 1

("

0

^{+ Vkn0+}

1

2n — m +itQ— 1. /—. — m + riQ— 1

X <p(2 zn0kn0

)m(E - m + hq —

2

(*«))•

Then

e*(

2

^{” "+}

3

^{x„) +}

(n0+

1

^)fe„0+

1

^{+n0- 2}

> X ф(2~n+3 ■ 2~m) n(El™ +1 (x„)) m=n0kn0 + n0~2

(n0+ i)fc„0+ i m=n0kn0

We have — m + n0 — n + \ = — (ra-f и —n

0

+ l)-f 2. Then there exists a natural number lm such that lm ^ n and lmkt ^ m -Ь n — n0

+1

^ (L + l ) ^ m+i ~

1

^-

Hence from (a) it follows that

ф(

2

_m + "°_ "+1) _ ^ (

2

~(m+"~"0+1)+2) > 2~1щ~ъ (p(2~mn+n°~ 1~lm)

Hence we get

0ф(2~п+3 xn) ^ 2~"~2 3 -4n~ i/(n~

1

) = 2n~5/(n—

1

^).

Thus we proved (***) and this proves the lemma.

Using Lemma 1.1 we obtain the following

1.2. ^Th e o r e m. A sequence (x„) in L*v is convergent to xeL*''’ in the topology .TT<(? if and only if simultaneously (x„) is convergent to x in measure and sup {||х„У < oo.

П

Proof. Let x„ -r4<p >x. Since 2Г0 <= 3T<tp on L*q>, we have that x„^>x, 1. e., xn is convergent to x in measure. On the other hand хп^^->х if and only if ||x„ — x||^ —> 0 for every ф е Т 4*. Hence sup{||x„||^} < go for every

n

ф е Т ^ , and therefore by Lemma 1.1 we have sup

ÏIW U

< oo. Now, let

n

xn^ x 0 and sup{||x„m < oo.

П

Denote by Z = { х , , } ^ u {x }. Then x„->x in the topology 3 ~0\2. But since ,9~<<p\z = ^~o\z ( [

6

], Theorem 2.3), we obtain x„ -> x in the topology

^~4<p\z and hence xn^ <<P >x.

2. Bounded sets in (L*v, J~4<p)

2.1. ^Th e o r e m. Let q> be a (p-function. Then the balls K^iR)

= {x eL*v: ЦхЦ^, ^ R ], R > 0 are bounded sets in (L*(p, . T <4>).

Proof. It suffices to show that for every ф е Т 4* and for every e > 0 there exists X > 0 such that

I M * < e for ||x||„ ^ R.

Indeed, let ф be a ^-function such that ф <^q> and let e be an arbitrary positive number. Then there exist ut > 0 , u2 > 0, u2 < ut such that

(1) ф^и/е) ^ E(p(u)/3R for

(2) ф^и/г) ^ e(p(u)/3R for u ^ u 2.

For xeK y iR ) write

E lx = (t e E : \x{t)\/R ^ щ}, E l = {teE : \x{t)\/R ^ u2}, El = {te E : u2 < |x(t)|/Æ < Ui}.

Let 1 > X > 0 be such that Xux = u2. Then from (1) and (2), for we obtain

вф (Àx/e) = \ф (À \x (t)\/e) dp + ф(Л \x (f)|/e) dp + ф{1 |x(t)|/e) du

< U M L u ï » « l w

«à - ( ? K И6?

^{(p, \~Y~ m^ I /}

< ^ £>

î.e.,

||Лх||*<е for x e K ^ R ).

2.2. ^Th e o r e m. Let <p be a (p-function. Then a s et Z cz L*9 is bounded in the topology ^~<9 if and only if Z cz K^iR) for some R > 0.

Proof. From Theorem 2.1 the balls K^iR) are bounded sets in the topology ' T <<p.

On the other hand, if sup {||x||v: x e Z ) = oo, then from the proof of Lemma 1.1 it follows that there exists a sequence (k„) of natural numbers, a sequence (x„) in Z and a (^-function ф е Т <9 such that

Q^(xJ2n~A) > 2n~5/n for n ^ 2 k x + 2.

Hence it follows that Z is not bounded in the topology , T <9.

Now, using Theorem 2.2, we shall prove some criterion of compactness of subsets of L*9 in the topology , T <9.

2.3. ^Th e o r e m. A set Z c L*9 is compact in the topology , T <9 if and only if simultaneously Z is compact in the topology of convergence in measure and sup{||x||v : x e Z } < oo.

Proof. Let Z be compact in 3T<9. Since $~0 cz ^~<9 on L*9, so Z is obviously compact in &~0. But, if Z is compact in then Z is bounded in 3~<9, and this means by Theorem 2.2 that sup{||x||v: xeZJ c o o . Conversely, let Z be compact in 3T0 and supdlxl^: x e Z } < oo. Then by Theorem 2.3 ( [

6

]) it follows that &~0\z = 3~<9\z and hence Z is compact in

cy-<<p

On the other hand, in [7] and [

8

] is proved:

2.4. Th e o r e m, (a) Let ф be an Orlicz function and let Z cz Ь*ф be compact in (Ь*ф, -Tf). Then there exists an Orlicz function qy such that

hrn —— =

0

u -*0 <P\U)

and _{и сю}hm “П = _{(p (w)}

0

and Z cz Ky (R) for some R > 0 (see [

8

], Definition 3 and Proposition 3c).

(b) Let ф be an Orlicz function which satisfies the A 2-condition and let set Z cz L*ф be compact in (L*^, Then there exists an Orlicz function <p such that ф <4 (p and Z cz KV(R) for some R > 0 (see [7 ], Definition 0.3.8.2, Proposition 0.3.8.4b).

Henceforth, in this section we will suppose that (E, Z, p) is an atomless measure space.

2.5. Le m m a. Let (L*(f>, Т Г be a locally bounded space. Then a set Z cz L*4* such that sup {||x||v: x e Z ] < oo is bounded in the topology

Proof. Let a set Z cz L*v be such that sup {ЦхЦ^: x e Z } <oo. Then from Theorem 0.3.10.2 ([7]) it follows that Z is additive bounded in the topology i.e., for every neighbourhood F of

0

for the topology there exists a natural number N such that

Z c:,F+ ... + V . .

JV-times

Let V0 be a bounded neighbourhood of 0 for the topology ^~v>. Then (1) Z c V0+ ... + V0, where N 0 is a natural number.

«---.--- '

Ng-times

Now, we shall show that the set Z is bounded in i.e., for every e > 0 there exists X0 > 0 such that XoZczK^is}. Indeed, let e > 0. Since V0 is bounded in ^~(p, so there exists A

0

^>

0

such that

(

2

⁾ lo V o cK ^ e / N o ).

Now, let x e Z. Then from (1) we have

(3) x = * ! + . . . +xNq, where x,e V0 and 1 ^ i ^ N 0.

Hence from (2) we obtain

(4) UoxX ^ e/Nq.

From (3) and (4) it follows that

Uo

*11, = ||/0 *!+ ...+ Ac

xNo\\, ^ No

•

e/N0 = e,

i.e, А0хе^(е).

2.6. Co r o l l a r y. I f (L*q>, .JT f ) is a locally bounded space, then the balls (r), where r >

0

constitute a base of neighbourhoods of

0

, which consist of bounded sets in

Proof. The balls К ф(г), where r > 0 constitute a base of neighbourhoods of 0 in Т ф. In virtue of Lemma 2.5 the balls К ф(г) are bounded in Т ф.

2.7. ^Th e o r e m. Let (L*v, Т ф) be a locally bounded space. Then a set Z cz L*<p is bounded in the topology T <<p if and only if Z is bounded in the topology J V

Proof. From Theorem 2.2 and Lemma 2.5 it follows that a set Z о L**

bounded in T <(p is bounded in Т ф. Since <= Т ф, the converse is obvious.

Now, let у {-Ту, T о) be a mixed topology in L*cp. Assume that (L*tp, T f ) is locally bounded. Then, in virtue of Lemma 2.5, Corollary 2.6 and Theorem 0.1.2, we have у {Т ф, T 0) = T i, where T l is the topology of the strict inductive limit of the sequence of balanced topological spaces

\Кф(2п), T 0L l7n n ^ O ) . Since from Theorem 3.4 ([6]) we have T x

= T <<p, we obtain the following

2.8. ^Th e o r e m. I f (L*ф, Т ф) is a locally bounded space, then

■ Г ** = у ( . Г 9 , T 0).

3. Orlicz spaces L^*v as two-norm spaces

Throughout this section we will assume that (E, Z, p) is an atomless measure space.

Let (L **, || H^, || ||o) be a two-norm space, where || ||0 is the F-norm of convergence in measure in the space S0 defined as follows:

IMIo = inf fe > 0: p(\teE: |x(t)| > e}) ^ e }.

It is said that a sequence (x„) in L*<p is y-convergent to x0 eL**, in symbols хп- ^ х 0, if and only if:

(1) ||x„ —Xq||o -> 0,

(2) the set (x„]®=1 is bounded in the topology Т ф (see Definition 0.2.1).

3.1. ^Th e o r e m. Let (L*4*, Т ф) be a locally bounded space. Then хп- ^ х 0 if and only if x„-y( /<p, > 0> >x0.

Proof. It follows from Lemma 2.5, Theorem 1.2 and Theorem 2.8.

3.2. Th e o r e m. Let (L**, Т ф) be a locally bounded space. Then for any linear operation A: —► (У, i), where (У, i) is any linear topological space we have: an operation A is (у, z)-linear if and only if is (у {Т ф, T 0), z)-continuous.

Proof. Let A: L** -> Y be (y, i)-linear. Then by definition, for any sequence (x„) in L*v, x0eLw the condition x^ -^ Xq implies A (xx) — >A (x0) . Therefore the function A\7: Z —> Y is (.:T0|Z, H-continuous for every Z bounded in Т ф. Hence, from Theorem 0.1.3 and Corollary 2.6 it follows that the operation A: L*^—>■ Y is у {Т ф, ,^0)-continuous.

Conversely, let A: L*^ -» Y be y(.y (p, y 0)-continuous. Then from Theorem 3.1 it follows that the operation A is (y, r)-linear.

33. ^Co r o l l a r y. L e t (L*tp, be a locally bounded space. Then (L*\ y(.Tv, = (L «\ y)*,

where (L*tp, y)* denotes the set of linear functionals F: L*(p -► R which are y- linear.

3.4. ^Th e o r e m. Let (L*<p, У be locally bounded space. Then, if a linear topology T in L*<p satisfies the conditions:

(i)

( ü )

•*iV "> = * olv /or n > 0’

for an arbitrary linear-topological space (Y, z) and a linear operation A : I * <p -> Y an operation A is (y, z)-linear if and only if A is {.T, z)- continuous,

then

Г = y(i% , 3T0).

Proof. Let us assume that a linear topology .T in L** satisfies condition (i). Then from the equality уфТ^, ,T 0) = 3T<(p = 3Tj and from definition of the topology , it follows that

(1) c

f o) -

On the other hand, if a topology tT satisfies condition (ii), then it satisfies condition (*) from Definition 0.1.4. Indeed, let A\z: (Z, jT 0|z) ->• У be a continuous function for an arbitrary set Z bounded in Then an operation A : L -* Y is (y, i)-linear; and hence by virtue of (ii) it is {,T, i)- continuous. Therefore from Theorem 0.1.5 we obtain

(2) / з | ( ^ Д 0).

Remark. In this section we assumed that (L*tp, & v) is a locally bounded space. The following theorem is known.

Th e o r e m. Let (E , Z, p) be an atomless measure space, where p(E)

= oo (F(E) < oc). Then (L*<p, y) is a locally bounded space if and only if

<P X (<P ~ x), where x(u) = ij/{us), ф is a convex (p-function, 0 < s ^ 1 ([3]).

In the case when cp is a convex ^-function the results of this section can be obtained applying the tools of the paper [9].

4. The weight of the topology <T<<P in 0

4.1. ^{Le m m a.} Let cp be a (p-function and let (E , Z, p) be a separable measure space. Then the space (L*<p, $~<(р) is separable.

Proof. It is known that if the measure p is separable, then the space of

simple functions 2P with the topology is separable, i.e., there exists a countable set .#0 c= & such that

(1) * = # , where the closure is taken in

On the other hand the set is dense in (L**, ([6], Theorem 2.3), i.e., (2) — i*<p^ where the closure is taken in , T <ip.

Since .T <q> cz from (1) and (2) we obtain

& C * = {& % **)*** => = L**.

4.2. ^Th e o r e m. Let tp be a (p-function and let (E , I , p) be a measure space with an atomless, а-finite, separable measure. Then the weight of the topology c?~<tp in 0 is equal to continuum.

Proof. We know that the space (L**, 3T<<P) is separable. Let <= L*v denote a countable dense set in (L*9, ^ <<p). Let I be a base of neighbourhoods of 0 for 3T<q>. Let us take into consideration the family of the sets:

{U n J > 0: Ue.tf) c 2 y°.

Then the cardinality of this family is not greater than continuum. There holds the equality

Ü = U n < ? о, where the closure is taken in ^~<<p ([2], p. 44).

On the other hand, the family M = {17: U e& } constitutes a base of neighbourhoods of 0 for ^ <<p. Therefore the space (L*(p, 3T<(p) has a base of neighbourhoods of 0 which the cardinality is not greater than continuum.

Now, let us assume that the space (L*v, , T <<P) has a countable base of neighbourhoods of 0. Then, it is not difficult to verify that the system 08o of all sets:

(1) и к = Щфк, г к) = Ь **п У (ф к, г к), where

CO _ /

К = У(Фк> rk) = {xe Ç) L ||x||* < r k}, n— \

фкеЧ, ^,р, rk > 0, к = 1 ,2 ,... constitutes a base of neighbourhoods of 0 for Since фп (p, from the proof of Theorem 2 ([4]) we have the inclusion

oo

(2) с П I ", where ФпеТ <<р.

5 — Prace matematyczne 26.1

The family i r = { J£}£L x forms a base of neighbourhoods of 0 for some linear oo 0,

topology on the set П L • In virtue of (1) we have that M0 — У "!^ , n = 1

hence : Т <Ф = ,T*\u^. But since the space (L*v, . Т <ф) is complete ([5], p. 82), so from the above equality of topologies it follows that L*v is a closed subset of f] L Vn with the topology

n = 1

Now, we shall show that there holds the inclusion:

(3)

0° n.

Ь*ф 0 L ", where

In fact, let xe П Then (Ях) < oo for all A > 0. Since the measure /i П= 1

is cr-finite, then there exists a sequence of measurable sets Ek in E such that 00

E = (J Ek and £j c £ 2 c ... and f*{Ek) < oo k= 1

for к = 1, 2, ...

Let xk: E -> R be the functions defined as:

xk{t) =

^{x(t) for} |x(r)| ^ к and te E k, 0 for |x(f)| > к or t<£Ek.

Then, as in the proof of Lemma 4.1 ([5]) we prove that j IJ/„ (A |x (t) - xk (r)j) dfi 0 for all A > 0,

E

i.e.,

ll*-*jk ll*„F ^ 0 for « = 1 .2 ,...

Hence it follows that xk^->x, where xke L * <p. Since, as we above noted L*v is

oo

a closed subset of f] L 4'” with the topology .T*, so х е Ь * ф. Thus we proved

n — 1

(3). From (2) and (3) we obtain the equality 00

L*v = П Г " , where

n= i

Now, we shall show that inclusion (3) implies that there exist: a natural number m and constants a, b, u0 > 0 such that:

(4) (p(u) ^ афт(Ьи) for u ^ u 0.

This means that the condition фт 4 q>does not hold, and it is contradicts фтеЧ,<(р. Thus we shall prove that the space (Ь*ф, , T <tp) has no countable base of neighbourhoods of 0, i.e., that the weight of the topology , T <4> is equal continuum.

Indeed, let us assume that condition (4) does not hold. Then there would exist a sequence м„|оо such that

<p(u„) > 2"ip„(n2u„), фп(п2ип) > 1 for n = ¹, 2, ...

Since the measure p is atomless, then there exists a sequence of pairwise disjoint, /г-measurable sets in E such that

P

(£„) =

1/2" фп(n2 ^u„

)

(p(En) = p (E)/2" ф„ (n^{2 u„)} if /*(£)< oo).

Define

x(t) = nu„

0

for te E n, n = for

00

U

_En-

n= 1

Let us note that we can assume that the sequence (ipn) is non-decreasing. In fact, otherwise, it suffices to take the sequence ф'п{и) = т а х (^ (м ), ..., ф„(и)), where n — 1 ,2 ,... Then the base Ж0 — {U'k = U (ф'к, rk)} is equivalent to the base m0. Now, let к be an arbitrary natural number and let Я > 0. Taking a natural number n0 such that max(/c, Я) ^ n0 we get

в*к(Л х)= Z Фк(Лпи„)/2пфп(п2ип)

n= 1

oc n0

< Z Ф п ( п 2 и п) / 2 п ф „ ( п 2 и „ ) + Z Ф к ( п 1 и п) < GO,

n = HQ n= 1

SO

oo

(5) X6 П f " = LW .

n= 1

On the other hand, for an arbitrary Я > 0 we take a natural number nt such that Я ^ 1 fnx. Then by virtue of negation of condition (4) there holds

GO 0 0 0 0

Z 4>(^пип)/2пфп{п2ип) ^ Z Я>{ип)12пфп{п2ип) ^ Z 1 = 00’

n— 1 n = n\ n= Лj

i.e., x $ L *v and this contradicts (5).

This contradiction proves that inclusion (3) implies condition (4), but this condition contradicts фт^ Т <<р.

4.3. Corollary. Let q> be a cp-function and let (E , I , p) be a measure space with an atomless, о-finite, separable measure. Then the space (L*(p, ,T <tp) is not metrizable.

References

[1 ] A. A l e x i e w i c z , On the two-norm convergence, Studia Math. 14 (1954), 49-56.

[2 ] R. E n g e lk in g , General topology, Warszawa 1977.

[3 ] W . M a t u s z e w s k a , W. O r lic z , A note on the theory o f s-normed spaces o f (p-integrable functions, Studia Math. 21 (1961), 107-115.

[4 ] M . N o w a k , On two equalities f o r Orlicz spaces L * <p, Funct. et Approx. 10 (1980), U A M , 61-81.

[5 ] —, On two linear topologies on Orlicz spaces L * <p. I, Comment. Math. 23 (1983), 71-84.

[6 ] —, Inductive limit o f a sequence o f balanced topological spaces in Orlicz spaces L * v(jj), ibidem 25 (1986), 295-313.

[7 ] Ph. T u r p in , Convexités dans les espaces vectoriels topologiques généraux, Dissert. Math.

131 (1976).

[8 ] —, Opérateurs linéaires entre espaces d'Orlicz non localement convexes, Studia Math. 46 (1973), 153-165.

[9 ] A. W iw e g e r , Linear spaces with mixed topology, ibidem 20 (1961), 47-68.

INSTITUTE OF MATHEMATICS A. MICKIEWICZ UNIVERSITY, POZNAN INSTYTI T MATE MATY КI

UNIWERSYTET im. A. M1CKIEWICZA, POZNAN