On a decomposition of some functions

Series I: COMMENTATIONES MATHEMATICAE XXX (1991) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXX (1991)

С. E.

and L.

(Caracas)

On a decomposition of some functions

Abstract. A representation of submultiplicative and supermultiplicative functions on (0, 1) is given with some applications to Orlicz spaces.

1. Decomposition theorem. Let I be a subset of R+ = [0,

such that x y e l whenever x , y e l , i.e.,

(

) .

and let / : /-> R + be a measurable positive function which is zero at zero if OeJ satisfying the inequality

(2) / (xy) < / (x )f (y) for all x , y e I .

Then / will be called submultiplicative on I. If the reverse inequality holds, then we say that / is supermultiplicative on I.

Examples of submultiplicative functions on (0, 1), (0,

and [1,

may be found in [7], [12] and [6]. They appear in many places and are related to diverse subjects.

Theorem 1. (a)

I f f is a submultiplicative function on I =

then

exists and

a = lim

ln /(x )

„

+ lnx /(x ) = xag(x)

with g ( x ) ^ 1 for

and limx_0+ x£g(x) = 0 for every e > 0.

(b) I f f is a supermultiplicative function on I = (0, 1), then

exists and

fi = lim x->0 +

ln /(x ) lnx

(4)

/(x ) = xph(x)

with h(x) ^ 1 for x e l and lim*_>0+ x eh(x) =

for every e > 0.

5 — Comment. Math. 30.2

P ro o f. Note that if / is submultiplicative on /, then l / / i s supermulti­

plicative on I and vice versa. Therefore, it is enough to prove (a). Let /(xy)

< /( x ) /( y ) for x , y e l and let F(x) = \n f(e~ x). Then F(x + y) ^ F(x) + F(y) for x, ye(0, oo).

Hence F is a measurable subadditive function on (0, oo). A result from [7], p.

244, asserts that

lim F(x)/x = inf F(x)/x = — a.

x->oo x > 0

Replacing — x by In y yields

,• In f ( e ~ x) In f(y) a = lim --- = lim

x-»oo y - > 0 + I n У

If 0 < g(x0) < 1 for some 0 < x0 < 1, then for any n

In f ( x o)/lnxo = a + lng(xo)/lnxô ^ а + 1п0(хо)Дпхо > а, and so

lim In f(x )/\n x ^ а + 1пд(х0)Дпх0 > а.

x -+ 0 +

This contradiction means that g{x) ^ 1 for x e l .

If lim*_>0 + xE g (x) > 0 for some e > 0, then there exist constants c > 0, x0 > 0, with g(x) ^ cx~e for 0 < x < x 0, and

In f{x)/\n x = а + ln g (x)/lnx < a + (lnc —е1пх)Дпх so that

lim ln/(x)/lnx ^ a —e,

x - 0 +

a contradiction. This completes the proof.

Note that if limJC_>0+ /(x ) = 0 then а > 0. Indeed, if 0 < / (x0) < 1 for some 0 < x0 < 1, then

In f (xô)/ln X

^ 1п/(х0)"ДпХо = ln /( x 0)/lnx0 and hence

а = lim ln/(xo)/lnxo ^ In / ( x 0)/lnx0 > 0.

X-*00

R e m ark 1. The above representation for a bounded supermultiplicative function was stated in [3], p. 147, and used to obtain some estimate of the modulus of convexity of Lorentz sequence spaces. In [1] a representation theorem for supermultiplicative functions on (0, 1) was proved with a nonde­

creasing factor h. But, as we will see in some examples (submultiplicative case)*»

the factor h is not always a monotonie function.

R em ark 2. The above theorem is also true for functions on / = (1,

because / is submultiplicative or supermultiplicative on (1,

if and only if /*(x) = l / / ( 1/x) is supermultiplicative or submultiplicative on (0, 1), respec­

tively.

Let us give some examples (always p ^ 1).

Example

1. Let / be an interval such that (1) holds and let / (x) be xp for x rational from I and 2xp for x irrational from /. Then / is submultiplicative on I, a = p and <?(x) = x~p/(x ) is not monotonie on any subinterval of I.

Example

2 (see [12], Ex. 5). Let /(x ) = xp(l + |sinlnx|) or /(x ) = xpe|sinlnx|.

Then / is a submultiplicative continuous increasing function on (0,

with / ( 1) = 1 and with g(x) = x~p/(x ) not monotonie on any interval (0, e), 0 < e < 1.

Example 3.

Define,

^

and n =

x/n

n(n + l)(n + 2) — x

if {n — l)n2 < x ^ n2{n + 2), if n2(n + 2) < x ^ n(n+1)2.

Then u(x)/x is nonincreasing and so и is subadditive on (0,

This example was introduced in [3], p. 141. Let, for xe(0, 1) and p ^ 2,

/ ( x ) = XP0(X)

with #(x) = exp(w( ln -

Then / is a submultiplicative increasing convex function on (0, 1), a = p, Ит*_0+ gf(x) =

and g is not monotonie on any interval (0, e), 0 < г < 1.

Example 4.

Define, for x ^ 0 and n =

v(x) =

n2 — (n— l)x/n !

if 0 < x < 2, if n! ^ x ^ nn!, if nn! < x < ( n + 1)!.

Then v(x)/x is nonincreasing and so v is subadditive on

Let, for

and p ^ 2,

/(x ) = xpg(x) with #(x) = exp( id In

Then / is a submultiplicative increasing convex function on (0, 1), a = p, liminfx-*0+ g(x) = expflim ,,^ v(n\)) = e, 1нп8ирх_>0+ g(x) = expflim ^^ v(nn\))

=

and g is not monotonie on any interval (0, e), 0 < e < 1.

2. Vector-valued Orlicz spaces and Orlicz spaces on product spaces. Let us

recall some notations from the theory of Orlicz spaces. An Orlicz function (p is

a continuous convex increasing function on [0,

so that (p(0) = 0.

The Orlicz space Lv = L ^S ) on a сг-finite measure space (S, I , p) is the Banach space of Z-measurable functions (with the usual identification) defined by

L v = {x: S->R measurable | т ф(гх) = J (p(r\x(s)\)dp(s) < oo for some r >0}

S

with the Luxemburg-Nakano norm

I N , = inf{a > 0: m^x/a) ^ 1}.

In recent years a number of papers have appeared in which spaces of vector-valued functions are considered. In the proof of theorems about vector-valued Lp spaces it is often used that Lp(S1? Lp(S2)) = Lp(S2 x 5Д that is, it is possible to consider that space as an Lp space on a product space. An essential limitation to the extension for Orlicz spaces will be the content of the following theorem.

Theorem 2.

Let I

and let (рх,(р2,(р be Orlicz functions. Let L91(I, Lyfl)) and LV(I x I) be the vector-valued Orlicz space on I and the Orlicz space on I x 7. Then

(5)

L Vi( I , L n (I)) = L „ ( I x I )

if and only if L VI(I) = L n (I) = L J I ) = L f (I) for some p > 1.

P ro o f. It is sufficient to prove the necessity. Assume that for some positive constants a and b,

Я IMI l * i (J,!•„(/)) ^ ^ b\\x\\L<f>i(hL(f)2(I)).

Then, in particular, for any measurable subsets A and В of I,

i.e.,

(6) a

<Pi à

1

’ Ï '

mAmB -У тВ)

First, putting mA = 1/и and тВ = 1 in (6), and then mA = 1 and mB = 1/u we get

— ^ 1(u) ^ ^ (pï 1(м) f o r w > l , and

<Pi 41) - u ^ ^ - i , * ^ <Pi 41) - i

---Я>2 ( и ) ^ ( Р (n ) ^ ---0 2

(u)

b a for и > 1,

respectively. The above inequalities mean that 7^(7) = 7^(7) = L92{I) and

1 d

9 mA 9 - 1 1

mB

П

_\mAmB __ 1 ^{sc 9 - l} _mA ^1_

with c = aq>ï1 (1)92 1 (l)/b2, d = cb3/a3. Therefore, /(u) = c/<p_1(l/u) is a su­

permultiplicative function on 7 = (0, 1) and f(u) = d/q>~1(l/м) is a submulti­

plicative function on 7. From the representation theorem:

9 (1/ .-1/

) and so

= u1/ph(u) with h(u) ^ 1,

9 (1/

) = м1/рбг(ы) with #(

) ^ 1,

= ullph(u) ^ u1/p ^ ui,pg(u) = for all м е7.

9 Ч Щ 9 4 W

Hence, d~pvp ^ q>(v) < c~pvp for ve[d, oo) and so 7^,(7) = Lp(7).

R em ark 3. If (S Г{, /х£), i = 1,2 , are nonatomic c-finite measure spaces, then L(pi(Sl , L (P2(S2)) = L9(S2 x S 1) if and only if L9l(S} = L ^ i ) = L9{Sd

= LpiSi), i = 1, 2, for some p ^ 1. The proof is the same as that of the above theorem.

R em ark 4. The first part of our Theorem 2, i.e., only the equalities Api = L92 = L^, was also proved in [2], [5], [9], [13].

3. Strictly singular inclusions between some Orlicz spaces. Now we will

consider the case of Orlicz sequence spaces l9. Kalton [8] proved that if an Orlicz function q> satisfies the A2-condition at zero (i.e. limsupM _ 0+ (p(2u)/(p{u)

< oo) and q>(u) ^ Cup for some C > 0 , l ^ p c o o and for every 0 ^ и ^ 1, then the imbedding i: l9 <^ lp is a strictly singular operator (i.e. there is no infinite-dimensional subspace E of 19 such that i\E is an isomorphism) if and only if

(7)

s((p, p) = liminf

E-*0 + inf

0 < s $ l

1 f At

ln(l/e)J sptp+1 00.

For example, if (pPtq(u) = up( l+ |lnu|)9 with q ^ 1 and p ^ ' i q then (рРг9 is an Orlicz function and

s(9P,q>

P) >

P) - Hminf inf ( ln - + ^ l n i ) = oo,

e - > 0 + 0 < s ^ 1 \ S £ /

and so the imbedding lVp q c+ lp is strictly singular.

Note that (ppq is a submultiplicative function on (0, oo). In general it is not

a simple matter to verify when (7) holds, but using the above representation

theorem we prove the strict singularity of the imbedding 19 <+ I provided that

9 is a submultiplicative or supermultiplicative function on (0, 1).

Th e o r e m 3.

Let (p be an Orlicz function satisfying the A2-condition at zero and such that (p(u) ^ Cup for some C > 0, 1 < p < со and every 0 ^ и ^ 1. Let limsupH->0 + (p(u)/up

i.e., (p is not equivalent at zero to up. I f (p is either a supermultiplicative function on (0, 1) or a submultiplicative function on (0, 1) with limu_>0+ (p(u)/up

then the imbedding i: l^

lp is strictly singular.

P ro o f. First, if (p(u) is equivalent at zero to uq with 1 ^ q < p then i: ly c; lp is strictly singular.

If q> is a submultiplicative function on (0, 1) not equivalent at zero to uq for any q ^ 1, then by assumption and the representation theorem

Cup < <p{u) = uph(u) ^ u p for 0 ^ и < 1,

and so f < p (if (I = p, then (p(u) is equivalent at zero to up). The exponent j8 is precisely the Matuszewska-Orlicz index (see [4], [12]). Therefore, being the intervals associated to the functions q> and up disjoint, it follows from a result of Lindberg [10] that lv and lp are totally incomparable, i.e., they have no isomorphic infinite-dimensional subspaces and so any bounded linear operator from lv into lp is strictly singular. In particular, the imbedding c; lp is strictly singular.

If (p is a submultiplicative function on (0, 1) not equivalent at zero to uq for any q ^ 1, then by Theorem 1(a) we have

ua ^ uag(u) = (p(u) for 0 ^ и ^ 1, and by the assumption,

a = lim и-*0 +

ln (p(u) In (Cup)

< lim = p.

Ill U w_*q+ 111 Ы

If a < p then the imbedding /a c; lp is strictly singular and since the imbedding lv

la is continuous, it follows that the imbedding lv

lp is strictly singular.

We have used here the well-known fact that the composition of a strictly singular operator with a bounded operator is also strictly singular.

Let a = p and limu_>0 + 9(u) = 00 • The function g(u) = inf gf(su) = inf g(t)

0<s<1 0<t^u

is nonincreasing, g(u) < g(u) and limu_>0+ §(u) =

. Therefore,

J (g{t)/t)dt

1 1 1 1

s{(p, p) = lim inf inf — — f (g(st)/t)dt ^ lim inf——

e- > 0

+

<a« iln (l/e)i v E^ 0+ ln(l/e)

= lim --- - 0+ jd t/t

£

lim

8 -> 0 +

д {Ф

1/8

= OO,

Hence (7) holds and the imbedding 19

lp is strictly singular.

References

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