REPRESENTATION OF KB SPACES AS K ¨OTHE SPACES

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AS K ¨ OTHE SPACES

ION CHIT¸ ESCU and R ˘AZVAN SAVA

We prove that any KB space is algebraically, isometrically and lattice isomorphic to a normed K¨othe space of measurable functions.

AMS 2010 Subject Classification: 46B40, 46B42, 46E30, 46B04.

Key words: Measure space, direct sum property, normal measure, strictly positive measure, linear lattice, normed lattice, KB space, function norm, K¨othe space.

1. PRELIMINARY PART

Throughout the paper, we shall adopt mostly the terminology in [4].

References to the terminology in [2], [3] and [1] will also appear.

Let X be a linear lattice (vector lattice in [2]). We shall write X+ to denote the set of all positive elements in X. For a sequence (xn)n in X we shall writex_n↑(respectively x_n↓) to denote the fact that (x_n)_n is increasing (respectively decreasing). When we write xn ↑ x (respectively xn ↓ x), this means that x_n ↑ (respectively x_n ↓) and there exists x ∈ X such that x = sup

n

xn (respectively x= inf

n xn).

A linear lattice X is called Dedekind complete (respectively Dedekind σ-complete) if it has the following property: for every non empty set E ⊂X which is bounded from above (respectively for every non empty set which is at most countable and bounded from above) there exists sup(E) ∈ X.

From this definition, it follows easily that (at most countable) sets which are bounded from below possess infimum. In [2] the Dedekind complete (respectively Dedekind σ-complete) lattices are called complete (respectively σ- complete) vector lattices.

Anormed lattice (in [2]: normed vector lattice) is a linear latticeXwhich is simultaneously a normed space in which the norm satisfies the following con- dition

|x| ≤ |y| ⇒ kxk ≤ kyk for any x, y inX.

MATH. REPORTS13(63),3 (2011), 265–270

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In case the normed latticeX is complete with respect to the norm, we call it a Banach lattice (in [2]: Banach vector lattice).

A KB space is a Dedekindσ-complete normed lattice in which the norm satisfies the following additional conditions for any sequence (xn)n inX:

(A) If x_n ↓ 0 in X, then x_n −→

n 0 (the last convergence is in norm and means that kx_nk −→

n 0).

(B) Ifxn↑ and sup

n

kx_nk<∞, there existsx∈X such thatxn↑x.

It can be shown that under the conditions from above, X is Dedekind complete and X is a Banach lattice.

Throughout the paper, ameasure space will be a triple (T,T, µ), where T is a non empty set, T is a σ-algebra of subsets of T and µ : T → R+ is a complete measure. We say that the measure µ has the direct sum property if there exists a family (T_i)i∈I of mutually disjoint sets T_i ∈ T such that 0 < µ(Ti) < ∞ for all i ∈ I, having the following property: for any A ∈ T with µ(A)<∞, there existsJ ⊂I,J at most countable, such that

A=

"

[

i∈J

(A∩T_i)

#

∪M

where M ∈ T,µ(M) = 0 (of course,M changes with A). It is clear that any σ-finite measure has the direct sum property.

Let us consider a fixed measure space (T,T, µ).

The spaceLof allµ-measurable functionsf :T →Rgenerates the space L of all equivalence classes of functions in L, where the equivalence of f and g meansf =g µ-almost everywhere (µ-a.e.).

ThenL becomes a linear lattice with order ˜f ≤g˜⇔f(t)≤g(t) µ-a.e.

In case µ has the direct sum property, L is Dedekind complete (see Theorem 17 in [3, p. 55]).

Assume T is a totally disconnected compact and separated space (the open and closed setsO form a basis for the topology of T). Denote byN the first Baire category sets in T.

We shall consider that T is theσ-algebra of all sets A having the form A=G4N, whereG∈ O andN ∈ N. Then T contains all the Borel sets of T (see Lemma 6 in [3, p. 315]).

The measure µ is called normal if µ(N) = 0 for any N ∈ N and has the additional property that for any G ∈ O with µ(G) = ∞ there exists O 3 G1 ⊂ G with 0 < µ(G1) < ∞. A normal measure is called strictly positive if for any∅6=G∈ O one hasµ(G)>0.

Let us write M₊(µ) to denote all positive µ-measurable functions u:T →R+.

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A function norm is a function ρ : M+(µ) → R+ having the following properties:

(i)ρ(u) = 0⇔u(t) = 0µ-a.e.;

(ii)u≤v⇒ρ(u)≤ρ(v);

(iii) ρ(u+v)≤ρ(u) +ρ(v);

(iv)ρ(αu) =αρ(u)

for u,v inM₊(µ) and R3α≥0 (with the convention 0· ∞= 0).

We say thatρis ofabsolutely continuous typeifρ(un)↓0 for any sequence u_n∈M₊(µ) such thatρ(u₁)<∞and u_u↓0µ-a.e.

We say thatρhas the weak Fatou property (and writeρwF) ifρ(u)<∞ for any sequence un inM+(µ) such that un↑u µ-a.e. and sup

n

ρ(un)<∞.

We say thatρhas the Fatou property (and writeρ F) ifρ(u) = sup

n

ρ(u_n) for any sequence un inM+(µ) such that un↑u µ-a.e..

Assumingρ is a function norm on the measure space (T,T, µ), we define the K¨othe spaceLρas follows. Firstly, we write

L_ρ=

f :T →R|f isµ-measurable and ρ(|f|)<∞ .

It is seen that L_ρ becomes a seminormed space, with seminormf 7→ρ(|f|).

Secondly, using the canonical procedure we obtain the associated normed space Lρ using the equivalence on L_ρ given via f ∼ g ⇔ ρ(|f −g|) = 0 ⇔ f(t) =g(t)µ-a.e..

Then the quotient spaceLρ=L_ρ/∼ becomes a normed space with norm kfk˜ =ρ(|f|) for anyf ∈f˜(we callL_ρaK¨othe space, see [1]). In caseρwF, L_ρ is a Banach space.

Any K¨othe spaceLρ becomes a normed lattice with order given by ˜f ≤

˜

g ⇔ f(t) ≤g(t) µ-a.e. for any representatives f ∈ f˜, g ∈ g. It is seen that˜ Lρ is a sublattice of L. Assume X and Y are linear lattices. We say that X and Y are algebraically and lattice isomorphic if there exists an isomorphism Ω :X→Y, i.e., Ω is bijective, linear and isotone (x≤yinX⇔Ω(x)≤Ω(y) in Y). Assuming, additionally, that (X,k k), (Y,k| k|) are normed lattices, we say that X and Y arealgebraically, isometrically and lattice isomorphic if k|Ω(x)k|=kxk for any x∈X (where Ω is the aforementioned isomorphism).

2. RESULTS

I. Let us consider a measure space (T,T, µ) and a function norm ρ on (T,T, µ), which gives the K¨othe space L_ρ.

Definition1. We shall say thatLρ isdistinguished if it has the following properties:

1. The measureµhas the direct sum property.

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2. The function normρ is of absolutely continuous type.

3. We have ρwF.

Remark. Under the conditions from above, we have actuallyρF. Indeed, this is precisely Theorem 59 in [1, p. 264].

Theorem 2. Assume L_ρ is distinguished. Then L_ρ is a KB space.

Proof. a) Because µ has the direct sum property, it follows that L is Dedekind complete. Now, let us consider a subset E ⊂(L_ρ)₊ which is upper bounded. We can find sup(E) = ˜f inL.

If ˜h∈L_ρ is an upper bound forE, we have ˜f ≤˜h and kf˜k ≤ khk˜ <∞.

Consequently, ˜f = sup(E) in L_ρ.

b) Let (˜u_n)_n a sequence of positive elements inL_ρ such that ˜u_n↓0. Be- causeρis of absolutely continuous type, it follows that (taking representatives) ρ(u_n)↓0, i.e.,k˜u_nk ↓0.

c) Let (˜un)n be an increasing sequence of positive elements in Lρ such that sup

n

k˜u_nk< ∞. Taking representatives, letu_n ↑u pointwise, µ-a.e.. We have sup

n

ρ(un) = sup

n

k˜unk<∞.

Because ρwF, we get ρ(u) < ∞. Hence 0 ≤u ≤ ∞µ-a.e. and we got

˜

u∈L_ρ. It is seen that ˜u_n↑u˜ inL_ρ.

Corollary 3. Assume X is a normed lattice which is algebraically, isometrically and lattice isomorphic to a distinguished K¨othe space L_ρ. Then X is a KB space.

II.Conversely, we shall prove that any KB space is isomorphic to some (distinguished)L_ρ.

Theorem 4. Let X be a KB space. Then, there exists a measure space (T,T, µ) and a function norm ρ on(T,T, µ) such that:

1.T is a totally disconnected compact and separated space, µ is normal and strictly positive,µhas the direct sum property(see the previous definitions concerning T and µ).

2.The function normρis of absolutely continuous type andρwF (equiva- lently ρ F).

3.The spaceXis algebraically, isometrically and lattice isomorphic toLρ. Proof. A. Our proof relies heavily on Theorem 5 in [3, p. 318].

We shall take the foundation of X to be precisely equal to X. Indeed, according to [4, p. 259] we have X⁰ = ˜X_n. Here X⁰ = the space of linear and continuous functionals x⁰ : X → R and ˜X_n = the space of linear and (o)- continuous functionals x⁰ :X→R(in [4, p. 247] one writes H_o(X) instead of X˜_n). It follows that ˜X_n is total in X.

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According to Theorem 5 in [3, p. 318], one can find a measure space (T,T, µ), whereµis normal and strictly positive,µhas the direct sum property and such that X is algebraically and lattice isomorphic to an ideal space Y over (T,T, µ).

B. We shall construct a function norm ρ on (T,T, µ) such that Y = Lρ. To this end, let Ω : X → Y be the linear and lattice above mentioned isomorphism.

We defineρ:M+(µ)→R+ as follows

ρ(u) =

( kxk ifu= Ω(x), for some (unique)x∈X+,

∞ ifu /∈Ω(X).

Let us show thatρ is a function norm.

a)ρ(u) = 0⇔u(t) = 0µ-a.e.. Ifρ(u) = 0, it follows that u= Ω(x) with kxk= 0, hence x= 0 andu= Ω(0) = 0.

Conversely, letu(t) = 0µ-a.e.. Using Theorem 4 in [3, p. 317], it follows that u(t) = 0 everywhere, henceu= 0 = Ω(0) and ρ(u) =k0k= 0.

b) 0≤u ≤v⇒ ρ(u) ≤ρ(v). Indeed, the result is obvious if ρ(v) =∞.

In case ρ(v) < ∞, we find y ∈ X+ such that Ω(y) = v. The space Y being ideal and 0 ≤u ≤ v, it follows that u ∈ Y. Let x ∈X such that Ω(x) =u.

Because u ≤ v, we get Ω(x) ≤ Ω(y), hence 0 ≤ x ≤ y and it follows that kxk ≤ kyk, i.e., ρ(u)≤ρ(v).

c) For R 3 α ≥ 0 and u ∈ M₊(µ), one has ρ(αu) = αρ(u) (with the convention 0· ∞= 0).

So, letα >0. In caseρ(u) =∞, we have ρ(αu) =∞. Indeed, otherwise ρ(αu) < ∞, hence αu = Ω(x) for some x ∈X+. It follows that u = Ω(_α¹x), hence ρ(u) =k_α¹xk= _α¹kxk<∞, contradiction.

In case ρ(u) <∞, let x ∈X+ such that u = Ω(x), which implies αu= Ω(αx) andρ(αu) = kαxk=αkxk =αρ(u) a.s.o.. We made systematical use of the linearity of Ω (see Lemma V.3.2, Theorem V.3.1 and Theorem V.4.2 in [4]).

d) For anyu,v inM₊(µ), one hasρ(u+v)≤ρ(u) +ρ(v).

Indeed, if ρ(u) =∞, the result is obvious. If ρ(u) <∞ and ρ(v) <∞, letx,y inX₊such thatu= Ω(x),v = Ω(y)⇒u+v= Ω(x+y)⇒ρ(u+v) = kx+yk ≤ kxk+kyk=ρ(u) +ρ(v).Returning to point a), we actually proved that Y =Lρ (identificationL_ρ≡Lρ).

Indeed, a class ˜f ∈L_ρconsists exactly in one element: ˜f ={f}. This is seen as follows. If f(t) =g(t)µ-a.e. andf,g are inL_ρ, then|f−g| ∈M+(µ) and |f−g|= 0µ-a.e.. Then, we have seen that|f−g|= 0, which means that f =g.

C. The end of our proof consists in showing thatρ is of absolutely continuous type and ρwF.

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a)ρ is of absolutely continous type. Indeed, letu∈M+(µ) withρ(u)<

∞ and letu_n∈M₊(µ),u_n↓0 such thatu_n↓0µ-a.e.,u_n≤u. It follows that

˜

u_n ∈L_ρ, ˜u_n ↓0 in L_ρ (It is clear that ˜0 = inf ˜u_n in L_ρ. Otherwise, we could find A ∈ T with µ(A) >0 and α >0 such that u(t) ≥α for allt∈A, where

˜

u = inf ˜un.) Hence ˜un = un (see the above identification) and un = Ω(xn), with x_n↓0 in X, which implies kx_nk ↓0, i.e.,ρ(u_n)↓0.

b)ρwF. Indeed, letun∈M+(µ),un↑, such that sup

n

ρ(un)<∞(we can consider u_n to be finite everywhere, hence u_n ∈L_ρ). For any n, let x_n inX such that Ω(xn) =un. It follows that 0≤xn↑and sup

n

kx_nk= sup

n

ρ(un)<∞.

We can therefore find x∈X such thatx_n↑x.

Let u = Ω(x). Hence u ≥ Ω(xn) = un for any n, which implies ρ(u) = kxk<∞. Lett= sup

n

un(pointwise). Hencet≤u µ-a.e. andρ(t)≤ρ(u)<∞.

We have proved that Ω : X → Y = L_ρ is an algebraic and lattice isomorphism. At the same time, this isomorphism is also an isometry, because, for any x∈X, one haskΩ(x)k=k˜uk=kxk, identifying u= ˜u.

III.Theorem 5(Synthesis Theorem).

Let X be a normed lattice.The following assertions are equivalent:

1.X is a KB space.

2. There exists a measure space (T,T, µ), µ having the direct sum property and a function norm ρ on (T,T, µ), ρ of absolutely continuous type, ρ wF such that X is algebraically, isometrically and lattice isomorphic to the distinguished K¨othe space L_ρ.

REFERENCES

[1] I. Chit¸escu,Spat¸ii de funct¸ii. Ed. S¸t. Encicl., Bucure¸sti, 1983.

[2] R. Cristescu, Ordered Vector Spaces and Linear Operators. Ed. Acad. Bucure¸sti, Roma- nia. Abacus Press, Tunbridge Wells, Kent, England, 1976.

[3] L.V. Kantorovici and G.P. Akilov,Analiz˘a funct¸ional˘a. Ed. S¸t. Encicl., Bucure¸sti, 1986.

[4] B.Z. Vulikh, Introduction to the Theory of Partially Ordered Spaces. Wolters-Noordhof Scientific Publications LTD, Gr¨oningen, 1967.

Received 1 January 2011 University of Bucharest

Faculty of Mathematics and Computer Science ionchitescu@yahoo.com

and

School Bogd˘ane¸sti, Suceava sava raz@yahoo.com