Compact-Like Operators in Lattice-Normed Spaces

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by  A. Aydın

COMPACT-LIKE OPERATORS IN LATTICE-NORMED
SPACES
7
1 A.AYDIN1,4, E. YU.EMELYANOV1,2, N. ERKURS¸UN O¨ZCAN3,M. A.A.
0 MARABEH1
2

n
Abstract. A linear operator T between two lattice-normed spaces is
a
J said to be p-compact if, for any p-bounded net xα, the net Txα has
a p-convergent subnet. p-Compact operators generalize several known
3
2 classesofoperatorssuchascompact,weaklycompact,orderweaklycom-
pact, AM-compact operators, etc. Similar to M-weakly and L-weakly

] compact operators, we definep-M-weakly and p-L-weakly compact op-
A
eratorsandstudysomeoftheirproperties. Wealsostudyup-continuous
F and up-compact operators between lattice-normed vector lattices.
.
h
t
a
m
1. Introduction
[

2 It is known that order convergence in vector lattices is not topological
v in general. Nevertheless, via order convergence, continuous-like operators
3
(namely, order continuous operators) can be defined in vector lattices with-
7
out using any topological structure. On the other hand, compact operators
0
3 play an important role in functional analysis. Our aim in this paper is to
0 introduce and study compact-like operators in lattice-normed spaces and in
.
1 lattice-normed vector lattices by developing topology-free techniques.
0 Recall that a net (x ) in a vector lattice X is order convergent (or o-
7 α α∈A
convergent, forshort)tox ∈X,ifthereexistsanothernet(y ) satisfying
1 β β∈B
v: yβ ↓ 0, and for any β ∈ B, there exists αβ ∈ A such that |xα −x| ≤ yβ
o
i for all α ≥ αβ. In this case we write xα−→x. In a vector lattice X, a net
X
x is unbounded order convergent (or uo-convergent, for short) to x ∈ X if
α
r o uo
a |xα −x|∧u−→0 for every u ∈ X+; see [10]. In this case we write xα−→x.
In a normed lattice (X,k·k), a net x is unbounded norm convergent to
α
un
x ∈ X, written as x −→x, if k|x −x|∧uk → 0 for every u ∈ X ; see
α α +
[7]. Clearly, if the norm is order continuous then uo-convergence implies
un-convergence. Throughout the paper, all vector lattices are assumed to
be real and Archimedean.
Let X be a vector space, E be a vector lattice, and p : X → E be a
+
vector norm (i.e. p(x) = 0 ⇔ x = 0, p(λx) = |λ|p(x) for all λ ∈ R, x ∈ X,
Date: 20.01.2017.
2010 Mathematics Subject Classification. 47B07, 46B42, 46A40.
Keywordsandphrases. compactoperator,vectorlattice,lattice-normedspace,lattice-
normed vector lattice, up-convergence,mixed-normed space.
1
2 A.AYDIN1,4,E.YU.EMELYANOV1,2,N.ERKURS¸UNO¨ZCAN3,M.A.A.MARABEH1
and p(x + y) ≤ p(x) + p(y) for all x,y ∈ X) then the triple (X,p,E) is
called a lattice-normed space, abbreviated as LNS. The lattice norm p in an
LNS (X,p,E) is said to be decomposable if for all x ∈ X and e ,e ∈ E , it
1 2 +
follows fromp(x) = e +e ,thatthereexistx ,x ∈X suchthatx = x +x
1 2 1 2 1 2
and p(x ) = e for k = 1,2. If X is a vector lattice, and the vector norm
k k
p is monotone (i.e. |x| ≤ |y| ⇒ p(x) ≤ p(y)) then the triple (X,p,E) is
called a lattice-normed vector lattice, abbreviated as LNVL. In this article
weusually usethepair (X,E) or justX to refer to an LNS(X,p,E) if there
is no confusion.
o p
We abbreviate the convergence p(x −x)−→0 as x −→x and say in this
α α
case that x p-converges to x. A net (x ) in an LNS (X,p,E) is said
α α α∈A
to be p-Cauchy if the net (xα −xα′)(α,α′)∈A×A p-converges to 0. An LNS
(X,p,E) iscalled (sequentially) p-complete ifevery p-Cauchy (sequence)net
inX isp-convergent. InanLNS(X,p,E)asubsetAofX iscalledp-bounded
if there exists e ∈ E such that p(a)≤ e for all a ∈A. An LNVL (X,p,E) is
o o
called op-continuous if x −→0 implies that p(x )−→0.
α α
A net x in an LNVL (X,p,E) is said to be unbounded p-convergent to
α
up o
x ∈ X (shortly, x up-converges to x or x −→x), if p(|x −x|∧u)−→0 for
α α α
all u∈ X ; see [4, Def.6].
+
Let (X,p,E) be an LNS and (E,k·k ) be a normed lattice. The mixed
E
norm on X is defined by p-kxk = kp(x)k for all x ∈ X. In this case the
E E
normed space (X,p-k·k ) is called a mixed-normed space (see, for example
E
[13, 7.1.1, p.292]).
A net x in an LNS (X,p,E) is said to relatively uniformly p-converge
α
rp
to x ∈ X (written as, x −→x) if there is e ∈ E such that for any ε > 0,
α +
there is α satisfying p(x −x) ≤ εe for all α ≥ α . In this case we say that
ε α ε
x rp-converges to x. A net x in an LNS (X,p,E) is called rp-Cauchy
α α
if the net (xα −xα′)(α,α′)∈A×A rp-converges to 0. It is easy to see that for
rp
a sequence x in an LNS (X,p,E), x −→x iff there exist e ∈ E and a
n n +
numerical sequence ε ↓ 0 such that for all k ∈ N and there is n ∈ N
k k
satisfying p(x −x) ≤ ε e for all n ≥ n . An LNS (X,p,E) is said to be
n k k
rp-complete if every rp-Cauchy sequence in X is rp-convergent. It should
benoticed that in a rp-complete LNS every rp-Cauchy net is rp-convergent.
Indeed,assumex is a rp-Cauchy netin a rp-complete LNS(X,p,E). Then
α
an element e ∈ E exists such that, for all n ∈ N, there is an α such that
+ n
p(xα′−xα) ≤ n1e for all α,α′ ≥ αn. We select a strictly increasing sequence
α . Then it is clear that x is rp-Cauchy sequence, and so there is x ∈ X
n αn
such that x −r→p x. Let n ∈ N. Hence, there is α such that for all
αn 0 n0
α≥ α we have p(x −x ) ≤ 1 e and, for all n ≥ n p(x−x ) ≤ 1 e,
n0 α αn0 n0 0 αn0 n0
rp
from which it follows that x −→x.
α
Werecallthefollowingresult(seeforexample[13,7.1.2,p.293]). If(X,p,E)
is an LNS such that (E,k·k ) is a Banach space then (X,p-k·k ) is norm
E E
complete iff the LNS (X,p,E) is rp-complete. On the other hand, it is
COMPACT-LIKE OPERATORS IN LATTICE-NORMED SPACES 3
not difficult to see that if an LNS is sequentially p-complete then it is rp-
complete. Thus, the following result follows readily.
Lemma 1. Let (X,p,E) be an LNS such that (E,k·k ) is a Banach space.
E
If (X,p,E) is sequentially p-complete then (X,p-k·k ) is a Banach space.
E
Consider LNSs (X,p,E) and (Y,m,F). A linear operator T : X → Y is
said to be dominated if there is a positive operator S : E → F satisfying
m(Tx) ≤ S(p(x)) for all x ∈ X. In this case, S is called a dominant for T.
The set of all dominated operators from X to Y is denoted by M(X,Y).
In the ordered vector space L∼(E,F) of all order bounded operators from
E into F, if there is a least element of all dominants of an operator T then
such element is called the exact dominant of T and denoted by |||T|||; see [13,
4.1.1,p.142].
By considering [13, 4.1.3(2),p.143] and Kaplan’s example [2, Ex.1.17],
we see that not every dominated operator possesses an exact dominant.
On the other hand if X is decomposable and F is order complete then
every dominated operator T : X → Y has an exact dominant |||T|||; see [13,
4.1.2,p.142].
We refer the reader for more information on LNSs to [5, 8, 12, 13] and
[4]. It should be noticed that the theory of lattice-normed spaces is well-
developed in the case of decomposable lattice norms (cf. [12, 13]). In [6]
and [17] the authors studied some classes of operators in LNSs under the
assumption that the lattice norms are decomposable. In this article, we
usually do not assume lattice norms to be decomposable.
Throughout this article, L(X,Y) denotes the space of all linear operators
betweenvectorspacesX andY. FornormedspacesX andY weuseB(X,Y)
for the space of all norm boundedlinear operators from X into Y. We write
L(X) for L(X,X) and for B(X) for B(X,X). If X is a normed space then
X∗ denotes the topological dual of X and B denotes the closed unit ball
X
of X. For any set A of a vector lattice X, we denote by sol(A) the solid hull
of A, i.e. sol(A) = {x ∈ X : |x| ≤ |a| for some a ∈A}.
The following standard fact will be used throughout this article.
k·k
Lemma 2. Let (X,k·k) be a normed space. Then x −−→x iff for any sub-
n
k·k
sequence x there is a further subsequence x such that x −−→x.
n n n
k kj kj
Thestructureof this paperis as follows. Insection 2, werecall definitions
of p-continuous and p-boundedoperators between LNSs. We study the rela-
tion between p-continuous operators and norm continuous operators acting
in mixed-normed spaces; see Proposition 3 and Theorem 1. We show that
every p-continuous operator is p-bounded. We end this section by giving a
generalization of the fact that any positive operator from a Banach lattice
into a normed lattice is norm bounded in Theorem 2.
In section 3, we introduce the notions of p-compact and sequentially p-
compact operator between LNSs. These operators generalize several known
4 A.AYDIN1,4,E.YU.EMELYANOV1,2,N.ERKURS¸UNO¨ZCAN3,M.A.A.MARABEH1
classes of operators such as compact, weakly compact, order weakly com-
pact, and AM-compact operators; see Example 5. Also the relation be-
tween sequentially p-compact operators and compact operators acting in
mixed-normed spaces are investigated; see Propositions 7 and 8. Finally
we introduce the notion of a p-semicompact operator and study some of its
properties.
In section 4, we define p-M-weakly and p-L-weakly compact operators
which correspond respectively to M-weakly and L-weakly compact opera-
tors. Several properties of these operators are investigated.
Insection5,thenotionsof(sequentially)up-continuousand(sequentially)
up-compact operators acting between LNVLs, are introduced. Composition
of a sequentially up-compact operator with a dominated lattice homomor-
phism is considered in Theorem 8, Corollary 4, and Corollary 5.
2. p-Continuous and p-Bounded Operators
In this section we recall the notion of a p-continuous operator in an LNS
which generalizes thenotion of order continuous operator in a vector lattice.
Definition 1. Let X, Y be two LNSs and T ∈ L(X,Y). Then
p p
(1) T is called p-continuous if x −→0 in X implies Tx −→0 in Y. If
α α
the condition holds only for sequences then T is called sequentially
p-continuous.
(2) T is called p-bounded if it maps p-bounded sets in X to p-bounded
sets in Y.
Remark 1.
(i) The collection of all p-continuous operators between LNSs is a vector
space.
(ii) Using rp-convergence one can introduce the following notion:
A linear operator T from an LNS (X,E) into another LNS (Y,F) is
rp rp
called rp-continuousif x −→0 inX implies Tx −→0 inY. But this
α α
notion is not that interesting because it coincides with p-boundedness
of an operator (see [5, Thm. 5.3.3 (a) ]).
(iii) Ap-continuous(respectively, sequentiallyp-continuous)operator be-
tween twoLNSsisalso knownas bo-continuous(respectively, sequen-
tially bo-continuous) see e.g. [13, 4.3.1,p.156].
(iv) Let (X,E) be a decomposable LNS and let F be an order complete
vector lattice. Then T ∈ M (X,Y)iff itsexactdominant|||T|||isorder
n
continuous [13, Thm.4.3.2], where M (X,Y) denotes the set of all
n
dominated bo-continuous operators from X to Y.
(v) Every dominated operator is p-bounded. The converse not need be
true, for example consider the identity operator I : (ℓ ,|·|,ℓ ) →
∞ ∞
(ℓ ,k·k,R). It is p-bounded but not dominated (see [5,Rem.,p.388]).
∞
COMPACT-LIKE OPERATORS IN LATTICE-NORMED SPACES 5
Next we illustrate p-continuity and p-boundedness of operators in partic-
ular LNSs.
Example 1.
(i) Let X and Y be vector lattices then T ∈ L(X,Y) is (σ-) order con-
tinuous iff T : (X,|·|,X) → (Y,|·|,Y) is (sequentially) p-continuous.
(ii) LetX andY bevectorlatticesthenT ∈L∼(X,Y)iffT :(X,|·|,X) →
(Y,|·|,Y) is p-bounded.
(iii) Let (X,k·k ) and (Y,k·k ) be normed spaces then T ∈ B(X,Y) iff
X Y
T : (X,k·k ,R)→ (Y,k·k ,R)isp-continuousiffT : (X,k·k ,R)→
X Y X
(Y,k·k ,R) is p-bounded.
Y
(iv) Let X be a vector lattice and (Y,k·k ) be a normed space. Then T ∈
Y
o
L(X,Y) is called order-to-norm continuous if x −→0 in X implies
α
Tx −k−·k−→Y 0, see [15, Sect.4,p.468]. Therefore, T : X → Y is order-
α
to-norm continuous iff T : (X,|·|,X) → (Y,k·k ,R) is p-continuous.
Y
Lemma 3. Given an op-continuous LNVL (Y,m,F) and a vector lattice X.
If T : X → Y is (σ-) order continuous then T : (X,|·|,X) → (Y,m,F) is
(sequentially) p-continuous.
p o
Proof. Assume that X ∋ x −→0 in (X,|·|,X) then x −→0 in X. Thus,
α α
o
Tx −→0 in Y as T is order continuous. Since (Y,m,F) is op-continuous
α
o p
then m(Tx )−→0 in F. Therefore, Tx −→0 in Y and so T is p-continuous.
α α
The sequential case is similar. (cid:3)
Proposition 1. Let (X,p,E) be an op-continuous LNVL, (Y,m,F) be an
LNVL and T : (X,p,E) → (Y,m,F) be a (sequentially) p-continuous posi-
tive operator. Then T :X → Y is (σ-) order continuous.
Proof. We show only the order continuity of T, the sequential case is analo-
gous. Assume x ↓ 0 in X. Since X is op-continuous then p(x ) ↓0. Hence,
α α
p o
x −→0 in X. By the p-continuity of T, we have m(Tx )−→0 in F. Since
α α
o
0≤ T then Tx ↓. Also we have m(Tx )−→0, so it follows from [4, Prop.1]
α α
that Tx ↓ 0. Thus, T is order continuous. (cid:3)
α
Corollary 1. Let (X,p,E) be an op-continuous LNVL, (Y,m,F) be an
LNVL such that Y is order complete. If T : (X,p,E) → (Y,m,F) is p-
continuous and T ∈ L∼(X,Y) then T :X → Y is order continuous.
Proof. SinceY is order complete andT is order boundedthen T = T+−T−
by Riesz-Kantorovich formula. Now, Proposition 1 implies that T+ and T−
are both order continuous. Hence, T is also order continuous. (cid:3)
Proposition 2. Let(X,k·k )beaσ-order continuous Banach lattice. Then
X
T ∈B(X) iff T :(X,|·|,X) → (X,k·k ,R) is sequentially p-continuous.
X
p
Proof. (⇒) Assume that T ∈ B(X), and let x −→0 in (X,|·|,X). Then
n
o
x −→0 in X. Since (X,k·k ) is σ-order continuous Banach lattice then
n X
6 A.AYDIN1,4,E.YU.EMELYANOV1,2,N.ERKURS¸UNO¨ZCAN3,M.A.A.MARABEH1
x −k−·k−→X 0 and hence Tx −k−·k−→X 0. Therefore, T : (X,|·|,X) → (X,k·k ,R)
n n X
is sequentially p-continuous.
(⇐) Assume T : (X,|·|,X) → (X,k·k ,R) to be sequentially p-continuous.
X
k·kX k·kX
Suppose x −−−→0 and let x be a subsequence. Then clearly x −−−→0.
n n n
k k
Since (X,k·k ) is a Banach lattice, there is a subsequence x such that
X n
kj
o p
x −→0inX (cf. [18,Thm.VII.2.1]), andsox −→0in(X,|·|,X). SinceT
n n
kj kj
k·kX
issequentiallyp-continuousthenTx −−−→0. Thus,itfollowsfromLemma
n
kj
2 that Tx −k−·k−→X 0. (cid:3)
n
Proposition 3. Let (X,p,E) be an LNVL with a Banach lattice (E,k·k )
E
and (Y,m,F) be an LNS with a σ-order continuous normed lattice (F,k·k ).
F
If T :(X,p,E) → (Y,m,F) is sequentially p-continuous then T : (X,p-k·k )
E
→ (Y,m-k·k ) is norm continuous.
F
Proof. Let x be a sequence in X such that x −p−-k−·−k→E 0 (i.e. kp(x )k →
n n n E
0). Given a subsequence x then kp(x )k → 0. Since (E,k·k ) is a
n n E E
k k
o
Banach lattice, there is a further subsequence x such that p(x )−→0
n n
kj kj
p
in E (cf. [18, Thm.VII.2.1]). Hence, x −→0 in (X,p,E). Now, the p-
n
kj
o
continuity of T implies m(Tx )−→0 in F. But F is σ-order continuous
n
kj
and so km(Tx )k → 0 or m-kTx k → 0. Hence, Lemma 2 implies
n F n F
kj kj
m-kTx k → 0. So T is norm continuous. (cid:3)
n F
The next theorem is a partial converse of Proposition 3.
Theorem 1. Suppose (X,p,E) to be an LNS with an order continuous
(respectively, σ-order continuous) normed lattice (E,k·k ) and (Y,m,F) to
E
be an LNS with an atomic Banach lattice (F,k·k ). Assume further that:
F
(i) T : (X,p-k·k ) → (Y,m-k·k ) is norm continuous, and
E F
(ii) T : (X,p,E) → (Y,m,F) is p-bounded.
Then T : (X,p,E) → (Y,m,F) is p-continuous (respectively, sequentially
p-continuous).
Proof. We assume that (E,k·k ) is an order continuous normed lattice and
E
p
show the p-continuity of T, the other case is similar. Suppose x −→0 in
α
o
(X,p,E) then p(x )−→0 in E and so there is α such that p(x ) ≤ e for
α 0 α
all α ≥ α . Thus, (x ) is p-bounded and, since T is p-bounded then
0 α α≥α0
(Tx ) is p-bounded in (Y,m,F).
α α≥α0
o
Since(E,k·k )isordercontinuousandp(x )−→0inE thenkp(x )k → 0
E α α E
or p-kx k → 0. The norm continuity of T : (X,p-k·k ) → (Y,m-k·k ) en-
α E E F
suresthatkm(Tx )k → 0orm-kTx k → 0. Inparticular,km(Tx )k →
α F α F α F
0 for α≥ α .
0
Let a ∈ F be an atom, and f be the biorthogonal functional corre-
a
spondingto a then f m(Tx ) → 0. Since m(Tx ) is order boundedfor all
a α α
(cid:0) (cid:1)
COMPACT-LIKE OPERATORS IN LATTICE-NORMED SPACES 7
α≥ α andf m(Tx ) → 0foranyatoma ∈ F, theatomicity ofF implies
0 a α
that m(Tx )−→o(cid:0) 0 in F (cid:1)as α ≤ α → ∞. Thus, T : (X,p,E) → (Y,m,F) is
α 0
p-continuous. (cid:3)
The next result extends the well-known fact that every order continuous
operator between vector lattices is order bounded, and its proof is similar
to [1, Thm.2.1].
Proposition 4. Let T be a p-continuous operator between LNSs (X,p,E)
and (Y,m,F) then T is p-bounded.
Proof. Assume that T : X → Y is p-continuous. Let A ⊂ X be p-bounded
(i.e. there is e ∈ E such that p(a) ≤ e for all a ∈ A). Let I = N × A
be an index set with the lexicographic order. That is: (m,a′) ≤ (n,a) iff
m < n or else m = n and p(a′) ≤ p(a). Clearly, I is directed upward.
Define the following net as x = 1a. Then p(x ) = 1p(a) ≤ 1e. So
(n,a) n (n,a) n n
o p o
p(x )−→0inE orx −→0. Byp-continuity ofT,wegetm(Tx )−→0.
(n,a) (n,a) (n,a)
So there is a net (z ) such that z ↓ 0 in F and for any β ∈ B, there
β β∈B β
exists (n′,a′) ∈ I satisfying m(Tx ) ≤ z for all (n,a) ≥ (n′,a′). Fix
(n,a) β
β ∈ B. Then there is (n ,a ) ∈ I satisfying m(Tx ) ≤ z for all
0 0 0 (n,a) β0
(n,a) ≥ (n ,a ). In particular, (n +1,a) ≥ (n ,a ) for all a ∈ A. Thus,
0 0 0 0 0
m(Tx ) = m( 1 Ta) ≤ z or m(Ta) ≤ (n +1)z for all a ∈ A.
(n0+1,a) n0+1 β0 0 β0
Therefore, T is p-bounded. (cid:3)
Remark 2.
(i) It is known that the converse of Proposition 4 is not true. For
example, let X = C[0,1] then X∗ = X∼ and X∼ = X∼ = {0}.
c n
Here X∼ denotes the σ-order continuous dual of X and X∼ denotes
c n
the order continuous dual of X. So, for any 0 6= f ∈ X∗ we have
f :(X,|·|,X) → (R,|·|,R) is p-bounded, which is not p-continuous.
(ii) If T : (X,E) → (Y,F) between two LNVLs is p-continuous then
T : X → Y as an operator between two vector lattices need not
be order bounded. Let’s consider Lozanovsky’s example cf. [2,
Exer.10,p.289] . If T :L [0,1] → c is defined by (cid:0)
1 0
(cid:1) 1 1
T(f)= f(x)sinx dx, f(x)sin2x dx,... .
(cid:18)Z Z (cid:19)
0 0
Then it can be shown that T is norm bounded which is not order
bounded. So T : (L [0,1],k·k ,R) → (c ,k·k ,R) is p-continuous
1 L1 0 ∞
and T :L [0,1] → c is not order bounded.
1 0
Recall that T ∈ L(X,Y); where X and Y are normed spaces, is called
w k·k
Dunford-Pettis if x −→0 in X implies Tx −−→0 in Y.
n n
Proposition 5. Let (X,k·k ) be a normed lattice and (Y,k·k ) be a normed
X Y
space. Put E := RX∗ and define p : X → E by p(x)[f] = |f|(|x|) for
+
f ∈ X∗. It is easy to see that (X,p,E) is an LNVL (cf. [4, Ex.4]).
8 A.AYDIN1,4,E.YU.EMELYANOV1,2,N.ERKURS¸UNO¨ZCAN3,M.A.A.MARABEH1
(i) If T ∈ L(X,Y) is a Dunford-Pettis operator then T : (X,p,E) →
(Y,k·k ,R) is sequentially p-continuous.
Y
(ii) The converse holds true if the lattice operations of X are weakly
sequentially continuous.
p o
Proof. (i) Assume that x −→0 in X. Then p(x )−→0 in E, and hence
n n
p(x )[f] → 0 or |f|(|x |) → 0 for all f ∈ X∗. From which, it follows
n n
w w
that |x |−→0 and so x −→0 in X. Since T is a Dunford-Pettis operator
n n
then Tx −k−·k−→Y 0.
n
w
(ii) Assume that x −→0. Since the lattice operations of X are weakly
n
sequentially continuous then we get |x |−→w 0. So, for all f ∈ X∗, we have
n
p
|f|(|x |) → 0 or p(x )[f] → 0. Thus, x −→0 and, since T is sequentially
n n n
p-continuous, we get Tx −k−·k−→Y 0. Therefore, T is Dunford-Pettis. (cid:3)
n
Remark 3. It should be noticed that there are many classes of Banach
lattices that satisfy condition (ii) of Proposition 5. For example the lat-
tice operations of atomic order continuous Banach lattices, AM-spaces and
Banach lattices with atomic topological dual are all weakly sequentially con-
tinuous (see respectively, [16, Prop. 2.5.23], [2, Thm. 4.31] and [3, Cor.
2.2])
ItisknownthatanypositiveoperatorfromaBanachlatticeintoanormed
lattice is norm continuous or, equivalently, is norm bounded (see e.g., [2,
Thm.4.3]). Similarly we have the following result.
Theorem 2. Let (X,p,E) be a sequentially p-complete LNVL such that
(E,k·k ) is a Banach lattice, and let (Y,k·k ) be a normed lattice. If T :
E Y
X → Y is a positive operator then T is p-bounded as an operator from
(X,p,E) into (Y,k·k ,R).
Y
Proof. Assume that T : (X,p,E) → (Y,k·k ,R) is not p-bounded. Then
Y
there is a p-bounded subset A of X such that T(A) is not norm bounded
in Y. Thus, there is e ∈ E such that p(a) ≤ e for all a ∈ A, but T(A) is
+
not norm bounded in Y. Hence, for any n ∈ N, there is an x ∈ A such
n
that kTx k ≥ n3. Since |Tx | ≤ T|x |, we may assume without loss of
n Y n n
∞
generality that x ≥ 0. Consider the series 1 x in the mixed-norm
n n2 n
nP=1
space (X,p-k·k ), which is a Banach lattice due to Lemma 1. Then
E
∞ ∞ ∞
1 1 1
p-k x k = kp(x )k ≤ kek < ∞.
n E n E E
n2 n2 n2
nX=1 nX=1 nX=1
∞
Since the series 1 x is absolutely convergent, it converges to some ele-
n2 n
nP=1
∞
ment, say x, i.e. x = 1 x ∈ X. Clearly, x ≥ 1 x for every n ∈ N and,
n2 n n2 n
nP=1
COMPACT-LIKE OPERATORS IN LATTICE-NORMED SPACES 9
since T ≥ 0 then T(x) ≥ 1 Tx , which implies kTxk ≥ 1 kTx k ≥ n for
n2 n Y n2 n Y
all n ∈ N; a contradiction. (cid:3)
Example 2. (Sequential p-completeness in Theorem 2 can not be removed)
∞
Let T :(c ,|·|,ℓ )→ (R,|·|,R) be defined by T(x ) = nx . Then T ≥ 0
00 ∞ n n
n=1
P
and clearly the LNVL (c ,|·|,ℓ ) is not sequentially p-complete.
00 ∞
Consider the p-bounded sequence e in (c ,|·|,ℓ ). Since Te = n for
n 00 ∞ n
all n ∈ N, the sequence Te is not norm bounded in R. Hence, T is not
n
p-bounded.
Example 3. (Norm completeness of (E,k·k ) can not be removed in The-
E
∞
orem 2) Consider the LNVL (c ,p,c ), where p(x ) = ( |x |)e . It
00 00 n n 1
nP=1
can be seen easily that (c ,p,c ) is sequentially p-complete. Note that
00 00
(c ,k·k ) is not norm complete. Define S : (c ,p,c ) → (R,|·|,R) by
00 ∞ 00 00
∞
S(x ) = nx . Then S ≥ 0, p(e ) ≤ e for each n ∈ N. But Se = n is
n n n 1 n
not boundnPe=d1in R.
It is well-known that the adjoint of an order bounded operator between
two vector lattices is always order bounded and order continuous (see, for
example [2, Thm.1.73]). The following two results deal with a similar situ-
ation.
Theorem 3. Let (X,k·k ) be a normed lattice and Y be a vector lattice.
X
Let Y∼ denote the σ-order continuous dual of Y. If 0 ≤ T :(X,k·k ,R)→
c X
(Y,|·|,Y) is sequentially p-continuous and p-bounded then the operator T∼ :
(Yc∼,|·|,Yc∼)→ (X∗,k·kX∗,R) defined by T∼(f) := f ◦T is p-continuous.
k·k
Proof. First, we prove that T∼(f) ∈ X∗ for each f ∈ Y∼. Assume x −−→0.
c n
o
Since T is sequentially p-continuous then Tx −→0 in Y. Since f is σ-order
n
continuousthenf(Tx )→ 0or(f◦T)(x ) → 0. Hence, wehave f◦T ∈ X∗.
n n
Next, we show that T∼ is p-continuous. Assume 0 ≤ f −→o 0 in Y∼, we
α c
showkT∼fαkX∗ → 0orkfα◦TkX∗ → 0. Now, kfα◦TkX∗ = sup |(fα◦T)x|.
x∈BX
Since B is p-bounded in (X,k·k ,R) and T is p-bounded operator then
X X
T(B )isorderboundedinY. Sothereexistsy ∈ Y suchthat−y ≤ Tx ≤ y
X +
for all x ∈ B . Hence −f y ≤ (f ◦T)x ≤ f y for all x ∈ B and for all α.
X α α α X
Sokfα◦TkX∗ ⊆ [−fαy,fαy]forallα. Itfollowsfrom[18,Thm.VIII.2.3]that
limfαy = 0. Thus, limkfα◦TkX∗ = 0. Therefore, T∼ is p-continuous. (cid:3)
α α
Theorem 4. Let X be a vector lattice and Y be an AL-space. Assume
0 ≤ T : (X,|·|,X) → (Y,k·k ,R) is sequentially p-continuous. Define T∼ :
Y
(Y∗,k·kY∗,R) → (X∼,|·|,X∼) by T∼(f) = f ◦T. Then T∼ is sequentially
p-continuous and p-bounded.
Proof. Clearly, if f ∈ Y∗ then f ◦T is order bounded, and so T∼(f)∈ X∼.
10A.AYDIN1,4,E.YU.EMELYANOV1,2,N.ERKURS¸UNO¨ZCAN3,M.A.A.MARABEH1
We prove that T∼ is p-bounded. Let A ⊆ Y∗ be a p-bounded set in
(Y∗,k·kY∗,R) then there is 0 < c < ∞ such that kfkY∗ ≤ c for all f ∈ A.
Since Y∗ is an AM-space with a strong unit then A is order bounded in Y∗;
i.e., there is a g ∈ Y∗ such that −g ≤ f ≤ g for all f ∈ A. That is, −g(y) ≤
+
f(y)≤ g(y) for any y ∈ Y , which implies −g(Tx) ≤ f(Tx)≤ g(Tx) for all
+
x ∈ X . Thus, −g◦T ≤ f ◦T ≤ g◦T or −g◦T ≤ T∼f ≤ g◦T for every
+
f ∈ A. Therefore, T∼(A) is p-bounded in (X∼,|·|,X∼).
Next,weshowthatT∼ issequentiallyp-continuous. Assume0≤ f −k−·k−Y→∗
n
0 in (Y∗,k·kY∗). Since Y∗ is an AM-space with a strong unit, say e, then
f −k−·k→e 0. It follows from [14, Thm.62.4] that f e-converges to zero in Y∗.
n n
Thus, there is a sequence ε ↓ 0 in R such that for all k ∈N there is n ∈ N
k k
satisfying f ≤ ε e for all n ≥ n . In particular, f (Tx) ≤ ε e(Tx) for all
n k k n k
x ∈ X and for all n ≥ n . From which it follows that f ◦T e-converges
+ k n
to zero in X∼ and so f ◦T −→o 0 in X∼. Hence, T∼(f )−→o 0 in X∼ and T∼
n n
is sequentially p-continuous. (cid:3)
3. p-Compact Operators
Given normed spaces X and Y. Recall that T ∈ L(X,Y) is said to be
compact if T(B ) is relatively compact in Y. Equivalently, T is compact
X
iff for any norm bounded sequence x in X there is a subsequence x such
n n
k
that the sequence Tx is convergent in Y. Motivated by this, we introduce
n
k
the following notions.
Definition 2. Let X, Y be two LNSs and T ∈ L(X,Y). Then
(1) T is called p-compact if, for any p-bounded net x in X, there is a
α
p
subnet x such that Tx −→y in Y for some y ∈ Y.
α α
β β
(2) T is called sequentially p-compact if, for any p-bounded sequence x
n
p
in X, there is a subsequence x such that Tx −→y in Y for some
n n
k k
y ∈ Y.
Example 4. (A sequentially p-compact operator need not be p-compact)
Let’s take the vector lattice
c (R):= f : R → R :∃a ∈ R,∀ε> 0, card {x ∈ R :|f(x)−a| ≥ ε} < ℵ .
ℵ1 1
Consider t(cid:8)he identity operator I on c (R(cid:0)),|·|,c (R) . Let f b(cid:1)e a p-(cid:9)
ℵ1 ℵ1 n
bounded sequence in c (R),|·|,c (R(cid:0)) . So there is g (cid:1)∈ c (R) such that
ℵ1 ℵ1 ℵ1
0≤ fn ≤ g for all n ∈(cid:0)N. (cid:1)
For any n ∈ N, there is a ∈ R such that for all ε > 0, card {x ∈ R :
n +
|f(x)−a | ≥ ε} < ℵ . Clearly the sequence a is bounded in R, so(cid:0)there is a
n 1 n
subsequence an (cid:1)and a ∈R such that an → a as k → ∞. For each m,k ∈ N,
k k
∞ ∞
let A := {x ∈ R : |f (x)−a | ≥ 1}. Put A = A and let
m,nk nk nk m m,nk
mS=1kS=1
h = aχR\A then fnk−→o h, since order convergence in cℵ1(R) is pointwise
convergence. Thus, I is sequentially p-compact.