LINEAR AND CONTINUOUS OPERATORS ON KB SPACES

ON KB SPACES

ION CHIT¸ ESCU and R ˘AZVAN SAVA

Using the representation of KB spaces as K¨othe spaces of measurable functions, we represent all the linear and continuous operators on a given KB space as vector integrals. Supplementary considerations and two examples are presented, too.

AMS 2010 Subject Classification: 46B40, 46B42, 46E30, 46G10, 47B38, 47B60.

Key words: Banach lattice, KB space, direct sum property, product measure and integration, function norm, K¨othe space, associated space, vector measure, integral representation of a linear operator.

1. PRELIMINARY PART

Throughout the paper, we shall adopt the terminology in [1] and [2]. All the vector spaces will be real.

Concerning linear lattices (see [7]) or vector lattices (see [3]) the reader can use [4] too. Let X be a linear lattice. We shall write X+ to denote the set of all positive elements in X. For a sequence (x_n)_n in X, we shall write xn ↑ (respectively xn ↓) to denote the fact (x_n)n is increasing (respectively decreasing). When we writex_n↑x(respectivelyx_n↓x), this means thatx_n↑ (respectively xn↓) and there existsx∈X such that x= sup

n

xn (respectively x= inf

n x_n).

A linear latticeX is called Dedekind complete(respectively Dedekindσ- complete) if it has the following property: for any non empty set (respectively non empty at most countable set)E ⊂X which is bounded from above, there exists sup(E) ∈ X. From this definition, it follows that sets (respectively at most countable sets) which are bounded from below possess infimum.

A normed lattice (normed vector lattice in [3]) is a linear lattice X in which the norm satisfies the following condition (for any x,y inX):

|x| ≤ |y| ⇒ kxk ≤ kyk.

In case the normed lattice is complete (with respect to the norm) we call it Banach lattice(Banach vector lattice in [3]).

MATH. REPORTS14(64),3 (2012), 279–290

Assume X and Y are linear lattices. We say that X and Y are alge- braically and lattice isomorphic if there exists an isomorphism Ω : X → Y, i.e., Ω is bijective, linear and isotone (x≤y inX ⇔Ω(x)≤Ω(y) inY). As- suming, additionally, that X andY are normed lattices (with norms k kinX and||| |||inY) we say thatX andY are algebraically, isometrically and lattice isomorphic, if |||Ω(x)|||=kxk for any x ∈X, where Ω is the aforementioned isomorphism.

A KB space is a Dedekind σ-complete normed lattice X in which the norm satisfies the following additional conditions, for any sequence (x_n)_n in X₊:

(A) Ifxn↓ 0, thenxn−→

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

n 0).

(B) Ifx_n↑ and sup

n

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

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

Throughout the paper, ameasure space will be a triple (T,Σ, µ), where T is a non empty set, Σ is a σ-algebra of subsets of T and µ : Σ → R+

is a complete and non null measure. We shall say that µ has the direct sum property if there exists a family (Ti)i∈I of mutually disjoint sets Ti ∈ Σ with 0 < µ(T_i) < ∞ for any i ∈ I, and having the following property: for any A ∈Σ with µ(A)<∞, there exists J ⊂I,J at most countable andM ∈Σ, µ(M) = 0, such that

A=

[

i∈J

(A∩Ti)

∪M

(where, of course,JandMchange with A). It is clear that anyσ-finite measure has the direct sum property.

Let us consider a fixed measure space (T,Σ, µ). The space L of all µ- measurable functions f : T → R generates the space L of all equivalence classes of functions in L, where the equivalence of f and g means that f =g µ-almost everywhere (µ-a.e.). Then Lbecomes a linear lattice with order ˜f ≤

˜

g ⇐⇒^D f(t) ≤g(t) µ-a.e. and linear operations on representatives. In case µ has the direct sum property, L is Dedekind complete (see Theorem 17 in [4, p. 55]).

WriteM+(µ) to denote all positive µ-measurable functions u:T →R+. A function norm(on the measure space (T,Σ,µ)) is a functionρ: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) andR3α≥0 (with the convention 0· ∞= 0).

We say thatρis of absolutely continuous typeifρ(un)↓0 for any sequence (un) inM+(µ) such thatρ(u1)<∞ and uu ↓ 0µ-a.e. We say that ρ has the Fatou property ifρ(u) = sup

n

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

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

Firstly, we define

L_ρ={f ∈ L |ρ|f|<∞}

(we wroteρ|f|=^D ρ(|f|)). It is seen thatL_ρbecomes a seminormed space, with seminorm f 7→ρ|f|.

Secondly, using the canonical procedure, we obtain the associated normed spaceL_ρusing the equivalence onL_ρgiven viaf ∼g⇔ρ|f−g|= 0⇔f(t) = g(t)µ-a.e. Then the quotient space L_ρ/∼

=D Lρ becomes a normed space with norm kf˜k = ρ|f| for any f ∈ f. We call˜ L_ρ a K¨othe space (of measurable functions), see [2]. In case ρhas the Fatou property, Lρis a Banach space.

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

˜

g ⇐⇒^D f(t)≤g(t)µ-a.e. for any representatives f ∈f,˜ g∈g. Clearly,˜ L_ρ is a sublattice of L.

For any A ∈ Σ, we write ρ(A) =^D ρ(ϕ_A) where ϕ_A ∈ M₊(µ) is the characteristic (indicator) function of A.

Considering again a function normρ on the measure space (T,Σ, µ), let us define

ρ⁰ :M₊(µ)→R via ρ⁰(u) = sup Z

uvdµ|v∈M₊(µ), ρ(v)≤1

. One can prove that, in caseρ is a saturated function norm(i.e., for any A ∈Σ with ρ(A)>0, one can find Σ3B ⊂A such that 0< ρ(B) <∞), it follows thatρ⁰is a function norm (see Theorem 40 in [2, p. 249]). For instance, in case µ isσ- finite, allk k_p, 1≤p≤ ∞, are saturated function norms.

In case ρ⁰ is a function norm, if is called the associated function norm (toρ). In this case, the K¨othe spaceLρ⁰ is usually denoted byL⁰_ρand is called the associated K¨othe space (toLρ), see [2]. For instance, take any 1≤p≤ ∞ and consider the (saturated) function normρ=k k_p (e.g. in caseµisσ-finite).

Then, the associated function norm is ρ⁰ = k k_p^∗, where p^∗ is the conjugate exponent ofp, namely 1≤p^∗ ≤ ∞is such that ¹_p+_p¹∗ = 1 with the conventions

1

∞ = 0, ¹₀ =∞(see [2]).

Finally, let us recall that, in caseXandY are normed spaces,L(X, Y) = {F : X → Y | F is linear and continuous} becomes a normed space with the usual operator norm (it is a Banach space when Y is Banach). Write L(X,R) =X⁰.

2. RESULTS

Our aim is to describe the space L(L, Y), whereL is a KB space andY is a Banach space. To this end, we shall “represent” L as a K¨othe space Lρ

using a previous result (see [1]) as follows.

Definition 1. Let (T,Σ, µ) be a measure space, ρ a function norm on (T,Σ, µ) and Lρ the corresponding K¨othe space. We say that Lρ is distin- guished if the following conditions are fulfilled:

a)µhas the direct sum property;

b) ρ is of absolutely continuous type;

c)ρ has the Fatou property.

For instance, in caseµ has the direct sum property and 1≤p <∞, the function norm ρ given via ρ(u) = kuk_p = R

u^pdµ1/p

furnishes the distin- guished K¨othe spaceL_ρ=L^p(µ)=^D L^p.

In [1] we proved

Theorem 2. Let L be a real normed space. The following assertions are equivalent:

1)L is a KB space.

2) There exists a distinguished K¨othe space L_ρ and a linear, isometric and lattice isomorphism Ω :Lρ→L.

Using this fact, let us consider a KB spaceLand a (real) Banach spaceY. We have a linear and isometric isomorphism H : L(L, Y) → L(L_ρ, Y) given via H(U) = U ◦Ω which identifies the spaces L(L, Y) and L(L_ρ, Y). So, instead of describing the elementsU ∈ L(L, Y), one can describe the elements F =H(U)∈ L(L_ρ, Y) and this will be done in the sequel.

So, letL_ρbe a distinguished K¨othe space on the measure space (T,Σ, µ) and let Y be a Banach space.

Before presenting the main result (i.e., Theorem 4) we shall introduce some preliminary facts.

a)C(ρ)=^D {A∈Σ|ρ(A)<∞}.

It is seen thatC(ρ) is aδ-ring (in [2], p. 179, one writesC1(ρ) insted ofC(ρ)).

b) E_ρ={f :T →R|f isC(ρ)-simple}.

So, the elementsf inE_ρ are of the formf = Pn

i=1

a_iϕ_Ai witha_i ∈R,A_i ∈C(ρ) mutually disjoint. Because ρ|f| ≤ Pⁿ

i=1

|a_i|ρ(A_i) < ∞, one has E_ρ ⊂ L_ρ and, consequently, E_ρ⊂L_ρ, whereE_ρ={f˜|f ∈ E_ρ}.

BecauseLρis distinguished, it follows thatEρ is dense inLρand we say that property (P₃) is fulfilled (see Theorem 57 in [2], p. 262 and [2], p. 188).

Using (P₃), for any ˜f ∈ L_ρ, we find a sequence ( ˜f_n)_n in E_ρ such that f˜n −→

n

f˜inLρ (we say that (fn)n is a P3-determinant sequence for f˜, see [2, p. 189]).

c) Letm:C(ρ)→Y be an additive measure. For anyf =

n

P

i=1

aiϕAi ∈ E_ρ, write

Z

fdm=

n

X

i=1

a_im(A_i), kmk= sup

Z

fdm

|f ∈ E_ρ, ρ|f| ≤1

≤ ∞ and MY(ρ) ={m:C(ρ)→Y |m is additive andkmk<∞}(in caseY =R, we shall write M(ρ) instead of M_R(ρ)).

It follows that for any ˜f ∈ Lρ and any P3-determinant sequence (fn)n

for ˜f, the sequence (R

f_ndm)_nis convergent. More precisely, one can see that in case E_ρ 3f˜_n −→

n

f˜and f_n ∈f˜_n, g_n ∈f˜_n, one has R

f_ndm = R

g_ndm and limn

R f_ndm exists and does not depend upon the sequence ( ˜f_n)_n.

Definition 3. An element m ∈ MY(ρ) is called a representing measure.

For any ˜f, ( ˜f_n)_n andm as above, we shall write Z

f˜dm= lim^D

m

Z

fndm.

Now, we are in position to present

Theorem 4 (Integral Representation Theorem). The Banach spaces L(L_ρ, Y) and M_Y(ρ) are linearly and isometrically isomorphic, via the mu- tually inverse isomorphisms:

(a:L(L_ρ, Y)→M_Y(ρ),

a(F) =m, where m(A) =F( ˜ϕA), ∀A∈C(ρ);

(

b:MY(ρ)→ L(L_ρ, Y),

b(m) =F, where F( ˜f) =R f˜dm, ∀f˜∈Lρ. OnL(L_ρ, Y) we have the usual operator norm

kFk₀ = sup{kF( ˜f)k |f˜∈Lρ, kf˜k ≤1}

and on MY(ρ) we have the previously defined norm kmk.

Proof. We shall divide the proof into several steps.

Step1. Definition of the mapa.

Let F : Lρ → Y linear and continuous. Define m : C(ρ) → Y via m(A) =F( ˜ϕ_A). We show thatm∈M_Y(ρ).

1)mis additive (becauseF is additive: forA, B∈C(ρ) withA∩B =∅, one has m(A∪B) =F( ˜ϕA∪B) =F( ˜ϕ_A+ ˜ϕ_B) a.s.o.).

2) Takef =

n

P

i=1

a_iϕ_Ai ∈ E_ρ withρ|f| ≤1. Then

Z

fdm

=

n

X

i=1

a_im(A_i)

=

n

X

i=1

a_iF( ˜ϕ_Ai)

= F

n

X

i=1

a_iϕ˜_Ai

=kF( ˜f)k ≤ kFk₀. It follows that kmk ≤ kFk₀ <∞ and m∈MY(ρ).

Consequently, we can define a : L(L_ρ, Y) → M_Y(ρ) via a(F) = m.

Obviously, ais linear andka(F)k=kmk ≤ kFk₀, henceais continuous.

Asamatter of fact,ais an isometry. In order to see this fact, we write B ={f˜∈L_ρ| kf˜k ≤1}.

Then, according to the definition

(1) kmk= sup{kF( ˜f)k |f˜∈B∩E_ρ}.

Taking the closure inLρ, we have

(2) B∩Eρ=B∩Eρ=B∩Lρ=B

(the equality B∩C=B∩C holds for any positive cone C⊂L_ρ).

From (1) and (2) it follows (using the continuity of F) that kmk = sup{kF( ˜f)k |f˜∈B}=kFk₀ and ais an isometry.

Step2. Definition of the mapb.

Letm∈MY(ρ). DefineF1:Eρ→Y viaF1

f˜=

n

P

i=1

aiϕ˜Ai

=R

fdm=

n

P

i=1

aim(Ai).

The definition of F₁ is coherent: if f and g are inE_ρ and f = g µ-a.e., then

Z

fdm− Z

gdm

= Z

(f −g) dm

≤ρ|f−g| kmk= 0 because kR

udmk ≤ρ|u| kmk for any u∈ E_ρ.

The mapF1 is linear (obvious) and (uniformly) continuous:

(∗) kF₁( ˜f)k= Z

fdm

≤ρ|f| kmk=kmk kf˜k.

Using the uniform continuity ofF₁, we extendF₁onE_ρ=L_ρ, getting the (unique) uniformly continuous extension F :Lρ→Y which is linear (because F₁ is linear). One can see thatkFk₀≤ kmk.

Indeed, for any ˜f ∈ Lρ, there exists a sequence ( ˜fn)n in Eρ such that f˜_n −→

n

f˜and F₁( ˜f_n)−→

n F( ˜f), hencekF( ˜f)k = lim

n kF₁( ˜f_n)k ≤lim

n kmk kf˜_nk= kmk kf˜k (using (∗)).

Now, we are in position to defineb :M_Y(ρ)→ L(L_ρ, Y) viab(m) =F. Of course,bis linear and continuous (kb(m)k=kFk₀≤ kmk, as we have seen).

As a matter of fact, b is an isometry. To see this, we shall use previous notations and facts. Let m∈M_Y(ρ) and putF =b(m),

kFk₀= sup{kF( ˜f)k |f˜∈B}= sup{kF( ˜f)k |f˜∈E_ρ∩B}.

Forf =

n

P

i=1

a_iϕ_Ai ∈ E_ρ, one hasF( ˜f) =R

fdm=

n

P

i=1

a_im(A_i) hence kFk₀= sup{kF( ˜f)k |f˜∈E_ρ∩B}

= sup

Z

fdm

|f ∈ E_ρ, ρ|f| ≤1

=kmk.

Step3. aand bare mutually inverse.

A.b◦a= 1_L(Lρ,Y).

TakeF ∈ L(L_ρ, Y) and letm=a(F)∈M_Y(ρ). Then writeF₁ =b(m) = (b◦a)(F).

For any ˜f ∈L_ρ, one hasF₁( ˜f) = lim

n F₁( ˜f_n), where ˜f_n∈E_ρ, ˜f_n−→

n

f˜. In casefn=

pn

P

i=1

aⁿ_iϕAⁿ_i, we have F₁( ˜f_n) =

Z

f_ndm=

pn

X

i=1

aⁿ_im(Aⁿ_i) =

pn

X

i=1

aⁿ_iF( ˜ϕ_Aⁿ

i) =F( ˜f_n) hence F₁( ˜f) = lim

n F( ˜f_n) =F( ˜f) and this impliesF₁=F.

B.a◦b= 1_MY(ρ).

Takem∈MY(ρ) and letF =b(m)∈ L(L_ρ, Y). Then writem1 =a(F) = (a◦b)(m).

For anyA∈C(ρ),m1(A) =F( ˜ϕA) =m(A) hencem1=m.

Final remark concerning the proof. The spaceM_Y(ρ) is indeed a Banach space, being isometric to the Banach space L(L_ρ, Y), as we have seen.

On the basis of Theorem 4, we give

Definition5. AssumeF ∈ L(L_ρ, Y) andm∈M_Y(ρ) are as in Theorem 4 (i.e., a(F) = m, b(m) = F, b = a⁻¹ and a = b⁻¹). We shall say that the (representing)measurem representsF (ormis a representing measure forF) and the representation is integral.

Comment. Theorem 4 can be obtained directly from Theorem 9 in [2], p. 189. We preferred this self-contained presentation.

Now, let us apply Theorem 4 in the particular case whenY =R, in order to characterize more precisely the dual of L_ρ.

Theorem 6. Accept the following supplementary assumptions:

1.The measure µis σ-finite.

2.The function normρ is saturated.

Then, the dual (L_ρ)⁰, the space M(ρ) and the associated K¨othe space L⁰_ρ are linearly and isometrically isomorphic Banach spaces. More precisely:

A. The map H : L⁰_ρ → (L_ρ)⁰, given via H(˜g) = F_g˜, is a linear and isometric isomorphism.

Here F_˜g :L_ρ → R acts via F_g˜( ˜f) =R

f gdµ for any g ∈g, any˜ f˜∈ L_ρ and any f ∈f˜.

B. The map G : L⁰_ρ → M(ρ), given via G(˜g) = m˜g, is a linear and isometric isomorphism.

Here, m_g˜ : C(ρ) → R acts via m_˜g(A) =R

Agdµ for any A ∈ C(ρ) and any g∈g.˜

In other words, the representing measure for F_g ism_˜g =gµ= the mea- sure with basis µ and density g (for any g∈g).˜

Proof. A. follows from Theorem 65 in [2], p. 278 and Theorem 70 in [2], p. 288.

B. is obtained as follows. LetF ∈(Lρ)⁰. Hence, using A., we find ˜g∈L⁰_ρ such that F = F_˜g. Considering the isomorphism a from Theorem 4, we find m =a(F˜g) = the representing measure for Fg˜. It is seen that m:C(ρ) →R acts via m(A) =F_˜g( ˜ϕ_A) =R

Af gdµ=R

Agdµ= (gµ)(A) =m_˜g(A).

Remark.One can see that mg˜ is countably additive and absolutely con- tinuous with respect to µ. As a matter of fact, this is true in the general case too.

Theorem 7. The representing measure inMY(ρ) are countably additive and absolutely continuous with respect to µ.

Proof. Let m ∈ MY(ρ). It is sufficient to prove that m µ (m is ab- solutely continuous with respect to m), i.e., ∀A ∈ C(ρ), ∀ > 0, ∃δ > 0, (A⊃B ∈C(ρ),µ(B)< δ)⇒ km(B)k< .

So, let us takeA∈C(ρ) and >0. We must find δ2 >0 such that (3) (A⊃B ∈C(ρ), µ(B)< δ₂)⇒ km(B)k< .

The first step in proving (3) consists in considering (the unique) F ∈ L(L_ρ, Y) which is represented by m. For the aforementioned > 0, one can find δ1 >0 such that for anyB ∈C(ρ) withρ(B) =kϕ˜Bk< δ1, one has (4) km(B)k=kF( ˜ϕ_B)k< .

The second step in proving (3) consists in proving that ρ µ, which means that

(5) ∀₀ >0,∀A₁∈C(ρ),∃δ₀>0, (A₁⊃B∈C(ρ), µ(B)< δ₀)⇒ρ(B)< ₀. Indeed, because Lρ is distinguished, it follows that ρ is of absolutely continuous type and this implies that ρ ^f µ, which means that ∀₀ > 0,

∀u∈M₊(µ) withρ(u)<∞,∃δ₀>0, (B ∈C(ρ),µ(B)< δ₀)⇒ρ(uϕ_B)< δ₀. The fact thatρ^f µfollows from Theorem 49 in [2], p. 256. Fromρ^f µ, it follows thatρµ, using Lemma 53 in [2], p. 49. The second step is proved.

We takeδ1 instead of 0 and A instead of A1 in (5) and we find δ2 >0 (instead of δ0) such that (A⊃B ∈C(ρ),µ(B)< δ2)⇒ρ(B)< δ1.

Using (4), it follows thatkm(B)k< and this is precisely (3).

We would like to close with two examples of integral operators onL^p(µ) spaces (which are distinguished K¨othe spaces in case µ has the direct sum property and 1≤p <∞).

Practically, the first example is a particular case of the second example.

Example 8. Let (X,τ) be a compact topological space with Borel sets B and let µ : B → R+ be a non null and a finite measure. We consider a continuous function P :X×X →R. For any 1≤p <∞, we shall define the linear and continuous operator F :L^p → L^p (we wrote L^p instead of L^p(µ)) given via F(˜x) = ˜y, where

(6) y(s) =

Z

P(s, t)x(t) dµ(t) for any x∈x,˜ y∈y.˜

The definition is correct. More precisely, we shall show that, for any

˜

x∈L^p, one has

(7) kF(˜x)k_p ≤µ(X)kPk k˜xk_p, where kPk= sup{|P(s, t)| |s, t∈X}.

Before anything, it is worth to say that, in case µ is not complete, we can consider its completion and the space L^p obtained for the completion is isometrically and linearly isomorphic to the initial L^p (in order to remain into the accepted framework).

Concerning the correctness of the definition: let ˜x∈L^p and x∈x.˜ a) For any s∈ X, the function t 7→ P(s, t)x(t) is µ-integrable, because t7→P(s, t) is µ-measurable and bounded andt7→x(t) is integrable (we have L^p⊂L¹,µbeing finite). So, the integral (6) exists.

b) Using (6), we defined the function y : X → R, given via y(s) = R P(s, t)x(t) dµ(t) andy depends only upon ˜x(using differentx∈x,˜ y(s) will be the same ).

It is seen that the functionV :X×X→R, given viaV(s, t) =P(s, t)x(t) is µ⊗µ-integrable, whereµ⊗µis the product measure defined on B ⊗ B.

Indeed, clearly, V is µ⊗ µ-measurable. Then t 7→ P(s, t)x(t) is µ- integrable (for any s ∈ X) and R

|P(s, t)x(t)|dµ(t) ≤ kPkR

|x(t)|dµ(t) ≤ kPk kxk₁ which impliesR

(R

|V(s, t)|dµ(t)) dµ(s)≤ kPk kxk₁µ(X)<∞.

Tonelli’s theorem says thatV is µ⊗µ-integrable. Fubini’s theorem says that the function (defined everywhere, as we have seen)

s7−→

Z

P(s, t)x(t) dµ(t)

is µ-integrable. Consequently,y is µ-integrable, hencey isµ-measurable.

c) The correspondence ˜x→y˜is clearly linear. We shall prove that (8) kyk_p ≤µ(X)kPk kxk˜ _p

and this will prove (7) (and the correctness of the definition of F).

Writing, for anys∈X,C(s) = sup{|P(s, t)| |t∈X}we get (9) |y(s)| ≤C(s)

Z

|x(t)|dµ(t) =C(s)kxk₁. Using H¨older’s inequality, one has

kxk₁ = Z

1· |x(t)|dµ(t)≤ k1k_qkxk_p=µ(X)^1q · kxk_p,

where q = p^∗ = the conjugate exponent of p: ¹_p + ¹_q = ¹_p + _p¹∗ = 1 (with q = ∞, ¹_q = 0 and µ(X)^1q = 1 in case p = 1). From (9) we get, for all s∈X : |y(s)| ≤C(s)µ(X)^1q kxk_p.

It follows that |y(s)|^p ≤ C(s)^pµ(X)^p/qkxk^p_p ≤ kPk^pµ(X)^p/qkxk^p_p ⇒ R |y(s)|^pdµ(s) ≤ kPk^pµ(X)^p/qkxk^p_pµ(X) ⇒ kyk_p ≤ kPkµ(X)^1q+1pkxk_p = kPkµ(X)kxk_p and everything is proved.

The representing measure forF ism:B →L^p given viam(A) =F( ˜ϕ_A), which means m(A) = ˜y wherey:X →R is given viay(s) =R

AP(s, t) dµ(t).

The operator F : L^p → L^p is regular (see [7]), because one can write F = F₊−F− where F₊ (respectively F−) is given by P₊ (respectively P−), which means

F₊(˜x) = ˜y, y(s) = Z

P₊(s, t)x(t) dµ(t), F−(˜x) = ˜z, z(s) =

Z

P−(s, t)x(t) dµ(t).

The regularity ofF can be proved alternatively, observing that the ope- rator given by the kernel |P| acts continuosly from L^p to L^p, because |P| is continuous, too. See Theorem 4.2 in [5], p. 68

Example 9. Let 1 ≤ p ≤ ∞, 1 ≤ q ≤ ∞ and let p^∗ be the conjugate exponent of p(with the conventions _∞¹ = 0, ¹₀ =∞).

Consider the σ-finite measure spaces (S,Σ, µ), (T,T, ν) and a µ⊗ν- measurable function P :S×T → R. Assume that P is of absolute operator type (p, q) (see [6], p. 313 and [6], p. 315) which means

kPk_p,q= sup^D Z Z

|P(s, t)|u(s)υ(t) d(µ⊗ν)(s, t)<∞,

where the supremum is taken for all u ∈ M₊(µ) with kuk_p^∗ ≤ 1 and all v ∈M+(ν) withkvk_q≤1.

Then, it can be shown (see Corollary 22 in [6], p. 315) that for any ν-measurable functionx:T →Rwith kxk_q <∞, the integral

(10) y(s) =

Z

P(s, t)x(t) dν(t)

exists for µ-almost alls∈S. Defining arbitrarily y(s) for those s∈S, where the integral (10) does not exist, we obtain a function y:S →R.

One can prove thaty isµ-measurable and

(11) kyk_p ≤ kPk_p,qkxk_q

(see Corollary 22 in [6], p. 315)

So, using (11), we see that we succeded in defining the linear and con- tinuous operator F :L^q(ν) → L^p(µ), given via F(˜x) = ˜y where ˜y is given by (10) and we have kFk₀≤ kPk_p,q.

We can see two more supplementary facts.

A. The operator F is regular. Indeed, the operator given by |P| is of the same type (because |P| is of absolute type (p, q), too), therefore acting continuously from L^q(ν) to L^p(µ). We use again Theorem 4.2 in [5], p. 68.

B. In the particular case whenS=T =X= a compact space, Σ =B= the Borel sets of X,µ=ν = a finite measure onB andp=q, 1≤p <∞ one can see that Example 8 can be obtained as a particular case of Example 9.

Indeed, letP :X×X→Rbe continuous (henceP isµ⊗µ-measurable).

We shall see that P is of absolute operator type (p, p). To this aim, take u, v in M+(µ) such thatkuk_p^∗≤1 andkvk_p ≤1. Tonelli’s theorem says that

V P =^D Z Z

|P(s, t)|u(s)v(t) d(µ⊗µ)(s, t) (12)

= Z Z

|P(s, t)|u(s)v(t) dµ(s)

dµ(t).

But Z

|P(s, t)|u(s)v(t) dµ(s)≤v(t)kPk Z

u(s) dµ(s)≤v(t)kPk kuk_p^∗k1k_p

≤v(t)kPk k1k_p=v(t)kPkµ(X)^1/p. Using (12) we get

V P ≤ kPkµ(X)^1/p Z

v(t) dµ(t)≤ kPkµ(X)^1/pkvk_pk1k_p^∗

≤ kPkµ(X)^1/pk1k_p^∗ =kPkµ(X)^1/pµ(X)^1/p∗ =kPkµ(X).

Hence

(13) kPk_p,q ≤ kPkµ(X)<∞

and P is of absolute operator type (p, p). At the same time, (13) confirms (7).

REFERENCES

[1] I. Chit¸escu and R. Sava, Representation of KB spaces as K¨othe spaces.Math. Reports 13(63)(2011),3, 265–270.

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

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

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

[5] M.A. Kranosel’skii, P.P. Zabreiko, E.I. Pustil’nik and P.E. Sbolevskii,Integral Operators in Spaces of Summable Functions.Noordhof, Leyden, 1976.

[6] H.L. Royden,Real Analysis(third edition). Macmillan Publishing Company, New York.

Collier Macmillan Publishing, London, 1988.

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

Received 17 June 2011 University of Bucharest

Faculty of Mathematics and Computer Science [email protected]

School Bogd˘ane¸sti, Suceava sava [email protected]