# On Sums of Strictly Narrow Operators Acting from a Riesz Space to a Banach Space

Authors: Oleksandr Maslyuchenko; Mikhail Popov
Journal: Journal of Function Spaces
 (2019)

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## Abstract

We prove that ifE is a Dedekind complete atomless Riesz space and X is a Banach space, then the sum of two laterally continuous orthogonally additive operators from E to X, one of which is strictly narrow and the other one is hereditarily strictly narrow with finite variation (in particular, has finite rank), is strictly narrow. Similar results were previously obtained for narrow operators by different authors; however, no theorem of the kind was known for strictly narrow operators.


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## Body

## 1. Introduction

Narrow operators were introduced in 1990 [1]; however, some deep results on these operators were obtained earlier; see [2]. Generalizing compact operators on function spaces, narrow operators gave new geometric facts. The most unusual thing about narrow operators is that, on the space L1, the sum of two continuous linear narrow operators is narrow [2, Theorem 7.46]; however, if a rearrangement invariant space E on [0,1] has an unconditional basis, then every operator on E is a sum of two narrow operators [2, Theorem 5.2].A result of Mykhaylyuk and the second named author asserts that, for every Köthe Banach spaceE on [0,1], there exist a Banach space X and narrow operators from E to X with nonnarrow sum [3].If the norm of the domain Köthe Banach spaceE is not absolutely continuous (for instance, if E=L∞), then the usual technique does not work. So, there are nonnarrow continuous linear functionals on L∞. However, questions about narrowness of the sum of two narrow operators are still interesting. A sum of two narrow operators on L∞ need not be narrow [4]. Moreover, if 1<p≤∞, then there are regular narrow operators S,T:Lp→L∞ with nonnarrow sum S+T [3].Now let a pair of spacesE,X be such that there are narrow operators S,T:E→X with nonnarrow sum S+T. Is the sum of a narrow operator and a compact (or even finite rank) operator narrow? It is known [2, Corollary 11.4] that if E is a Köthe Banach space with an absolutely continuous norm, then for any Banach space X the sum of a narrow operator and a “small” operator (like compact, AM-compact, Dunford-Pettis operators, etc.) is narrow.If the norm ofE is not absolutely continuous and a compact operator need not be narrow, a weaker question naturally arises: is the sum of two narrow operators, at least one of which is compact, narrow? The strongest result in this direction was obtained by Mykhaylyuk [5]: if E is a Köthe F-space, X is a locally convex F-space, and S,T∈L(E,X) are narrow operators such that T maps the set of all signs to a relatively compact subset of X (in particular, if T is compact), then the sum S+T is narrow.In 2014, narrow operators were generalized to nonlinear maps, more precisely to orthogonally additive operators [6], which were studied by Mazón, S. Segura de León in [7, 8]. In different contexts, when dealing with narrow linear operators, the linearity has been used for orthogonal pairs of elements only. This allowed generalizing results on narrow operators obtained in [9] from linear to orthogonally additive operators. For example, a result of [6] asserts that every laterally continuous C-compact orthogonally additive operator acting from an atomless Dedekind complete Riesz space is narrow. Recently, the latter theorem was essentially generalized in [10] by proving that if E is a Dedekind complete atomless Riesz space and X is a Banach space, then the sum of narrow and C-compact laterally continuous orthogonally additive operators from E to X is narrow.However, no result is known concerning a sum of two strictly narrow operators. Notice that in every known example of two narrow operators with nonnarrow sum, the summands are not strictly narrow. To be more precise, we recall necessary definitions.By aKöthe Banach space on a finite atomless measure space (Ω,Σ,μ), we mean a Banach space E which is a linear subspace of L1(μ) possessing the following properties: 1Ω∈E, and for every x∈E and y∈L1(μ) the condition |y|≤|x| implies that y∈E and y≤x (by 1A we denote the characteristic function of a set A∈Σ, and the inequality u≤v in E means that u~(t)≤v~(t) holds for μ-almost all t∈Ω, where u~∈u and v~∈v are some/any representatives of the classes u,v). A Köthe Banach space E on a finite atomless measure space (Ω,Σ,μ) is said to have anabsolutely continuous norm if limμ(A)→0x·1A=0 for all x∈E.IfX,Y are Banach spaces, by L(X,Y) we denote the Banach space of all continuous linear operators T:X→Y, and L(X) stands for L(X,X). By x⊔y we denote the disjoint sum x+y in a Köthe Banach space, that is, under the assumption suppx∩suppy=∅ or, more generally, in a Riesz space under the assumption x⊥y. In a Boolean algebra, x⊔y means the disjoint supremum x∨y, that is, under the assumption x∧y=0.For familiarly used information on Riesz spaces, the reader can refer to [11]. Let E be a Riesz space and X be a linear space. A function T:E→X is called anorthogonally additive operator (OAO in short) if T(x⊔y)=T(x)+T(y) for all disjoint elements x,y∈E. Simple examples of OAOs are the positive, negative parts and the modules of an element: T1(x)=x+, T2(x)=x-, T3(x)=|x|, and x∈E. For more examples of OAOs including integral Uryson operators, see [6–8].An elementx of a Riesz space E is called afragment of y∈E (write x⊑y) provided x⊥y-x. The set of all fragments of an element e∈E is denoted by Fe. Observe that if z=x⊔y, then x and y are disjoint fragments of z. We say that an element a≠0 of a Riesz space E is anatom if the only fragments of a are 0 and a itself. A Riesz space having no atom is said to beatomless.LetE be an atomless Riesz space and let X be a Banach space. An OAO T:E→X is called(i)
narrow at a pointe∈E if for every ε>0 there is a decomposition e=e′⊔e′′ such that T(e′)-T(e′′)<ε;(ii)
narrow if it is narrow at each point e∈E;(iii)
strictly narrow at a pointe∈E if there is a decomposition e=e′⊔e′′ such that T(e′)=T(e′′);(iv)
strictly narrow if it is strictly narrow at each point e∈E.The atomlessness assumption in the above definition serves to avoid triviality, because otherwise every narrow or strictly narrow operator must send an atom to zero.Observe thatT(0)=0 for every OAO T; hence, every OAO is strictly narrow at zero. Every strictly narrow (at a point e) is narrow (at a point e); however, the converse is not true [2, Proposition 2.2]. Under mild assumptions on the domain Riesz space, every operator with finite-dimensional range is strictly narrow and every operator from an atomless Banach lattice to a purely atomic Banach lattice is strictly narrow [12].IfE is a Köthe Banach space with an absolutely continuous norm on a finite atomless measure space (Ω,Σ,μ) and X is a Banach space, then an OAO T:E→X is narrow if and only if for every ε>0 every A∈Σ admits a decomposition A=B⊔C, B,C∈Σ such that μ(B)=μ(C) and T(1B)-T(1C)<ε [2, Proposition 10.2], and a similar statement holds for strictly narrow OAOs. Remark that the latter property of narrow (strictly narrow) operators was initially considered as a definition.One more definition for Köthe Banach spaces is essential for our investigation. LetE be a Köthe Banach space on a finite atomless measure space (Ω,Σ,μ), and let X be a Banach space. An operator T∈L(E,X) is calledhereditarily narrow if for every A∈Σ, μ(A)>0 and every atomless sub-σ-algebra F of Σ(A) the restriction of T to E(F) is narrow (here E(F)={x∈E(A):xisF-measurable}). The following proposition gives lots of examples of pairs of narrow operators with narrow sum.Proposition 1 ([13], [2], Proposition 11.2).
LetE be a Köthe Banach space on [0,1] with an absolutely continuous norm, and let X be a Banach space. Then the sum T=T1+T2 of a narrow operator T1∈L(E,X) and a hereditarily narrow operator T2∈L(E,X) is narrow. In particular, the sum of two hereditarily narrow operators is hereditarily narrow.Questions on the strict narrowness of sums of strictly narrow operators seem to be much more involved than similar questions on narrow operators. So, no example is known of strictly narrow operators with nonstrictly narrow sum.Problem 2.
LetE be an atomless Riesz space, and let X be a Banach space. Is the sum S+T of strictly narrow operators S,T:E→X strictly narrow or, at least, narrow?Our main result, which is an analogue of Proposition1 for strictly narrow operators, is the first result in this direction. The idea of the proof, inspired by paper [12], is to consider the set Fe of all fragments of a fixed element of the domain Riesz space E as the main object for investigation. This becomes possible because the definitions of all notions from the main theorem could be equivalently restricted to Fe. Since the set Fe is a Boolean algebra with respect to the natural operations, we come to analogous questions for functions defined on a Boolean algebra.


## 2. Dividing Measures on Boolean Algebras

Let(uα) be a net in a Boolean algebra B. The notation uα↓0 means that the net (uα) decreases and infαuα=0. We say that a net (xα) in Border converges to an element x∈B if there exists a net (uα) in B with the same index set such that xα△x≤uα for all indices α and uα↓0. In this case, we write xα→x and say that x is theorder limit of (xα).A Boolean algebraB is said to beorder complete if any nonempty subset of B has the supremum. A Boolean algebra B is said to be σ-complete if any countable subset of B has the supremum. By apartition (of unity) in a Boolean algebra B we mean a maximal disjoint subset A⊆B, that is, (∀x∈B)((∀a∈Aa∩x=0)⇒(x=0)). Adisjoint union⋃A (i.e., the union of a disjoint system A⊆B), if exists, is denoted by ∐A. Although in some cases an infinite union in a Boolean algebra does not exist, it is immediate that if A is a partition then ∐A=1 exists. Conversely, if ∐A=1 then A is a partition.LetB be a Boolean algebra, and let X be a linear space. A function f:B→X is said to be ameasure provided f(x⊔y)=f(x)+f(y) for every pair of disjoint elements x,y∈B. Obviously, f(0)=0 for a measure. An element a∈B is called anatom of a measure f:B→X provided f(a)≠0 and for any x∈B with x≤a one has either f(x)=0 or f(x)=f(a). A measure f:B→X is calledatomless provided there is no atom of f.A measuref:B→X is said to havefinite rank if the closed linear span [f(B)] is a finite-dimensional subspace of X. Let B be a σ-complete Boolean algebra, and let X be a Banach space. A measure f:B→X is said to be σ-additive provided for every disjoint sequence (xn)n=1∞ in B one has f(⋁n=1∞xn)=∑n=1∞f(xn), where the series converges unconditionally in X.LetB be a Boolean algebra, and let X be a set. A function f:B→X is said to bedividing provided every element b∈B has a two-point partition b=b′⊔b′′ with f(b′)=f(b′′). We say that a pair of functions f,g:B→X isuniformly dividing if every element b∈B has a two-point partition b=b′⊔b′′ with f(b′)=f(b′′) and g(b′)=g(b′′).Next we define a hereditarily dividing measure, which takes an important place in our investigation. Given a Boolean algebraB and any b∈B, we set Bb={x∈B:x≤b}, which is a Boolean algebra with the induced operations and unity b.Definition 3.
LetB be a σ-complete Boolean algebra and let X be a Banach space. An atomless σ-additive measure T:B→X is calledhereditarily dividing if, for every b∈B and every σ-complete subalgebra U of Bb, the atomlessness of the restriction T|U of T to U implies that T|U is dividing on U.Obviously, a hereditarily dividing measure is dividing. By [12, Theorem 2.11] and Lemma 9, every atomless σ-additive measure with finite-dimensional range is dividing. Hence, as a consequence, we obtain the following example of hereditarily dividing measures.Theorem 4.
LetB be a σ-complete Boolean algebra, and let X be a Banach space. Then every atomless σ-additive measure T:B→X with finite-dimensional range is hereditarily dividing.The following theorem brings an important tool for the main result.Theorem 5.
LetB be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:B→X be σ-additive measures. If S is dividing and T is hereditarily dividing and has finite variation, then S+T is dividing.Actually, we prove more.Theorem 6.
LetB be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:B→X be σ-additive measures. If S is dividing and T is hereditarily dividing and has finite variation, then the pair S,T is uniformly dividing.It is an obvious observation that Theorem6 yields Theorem 5. For the proof, we need several lemmas. Let B be a σ-complete Boolean algebra. A sequence τ=(an)n=1∞ in B is called atree if a1=1 and an=a2n⊔a2n+1 for all n∈N. The minimal σ-complete subalgebra including a tree τ is called atree subalgebra of B generated by τ.Lemma 7.
Letτ=(an)n=1∞ be a tree in a σ-complete Boolean algebra B and let Uτ be the tree subalgebra generated by τ. If μ:Uτ→[0,+∞) is a σ-additive measure and μ(a2n+i)≤(3/4)μ(an) for all n∈N and i∈{0,1} then μ is atomless.Proof of Lemma7.
Fix anya∈Uτ with μ(a)>0. By [14, Lemma 1.2.14], the smallest subalgebra B(τ) of Ba including τ equals the set of all finite disjoint unions p=∐i=1mani of elements of τ. By [15, 313F(c)], the σ-order closure of a subalgebra is a subalgebra. Hence, the σ-order closure of the subalgebra B(τ) equals U. Since μ is a σ-additive measure on U, it is σ-continuous. Hence, we may and do choose a finite disjoint union p=∐j∈Jaj such that(1)μa△p≤16μa.
Set(2)p′=∐j∈Ja2jandp′′=∐j∈Ja2j+1.
Then(3)μp′=∑j∈Jμa2j≤34∑j∈Jμaj=34μp.
Similarly,μ(p′′)≤(3/4)μ(p). Hence, μ(p′)=μ(p)-μ(p′′)≥(1/4)μ(p). Thus, (4)14μp≤μp′≤34μp.
Observe that(5)μa≤μp+μa△p≤by1μp+16μa.
Hence,μ(a)≤(6/5)μ(p) and therefore, by (1), μ(a△p)≤(1/5)μ(p). Then we obtain (6)μa∩p′=μp′-μp′-a≥μp′-μp△a≥14μp-15μp=120μp>by10.
Similarly,μ(a∩p′′)>0. Hence (7)0<μa∩p′<μa∩p′+μa∩p′′≤μa, which yields that a is not an atom for μ.Lemma 8.
LetS:B→X be a dividing σ-additive measure and let μ:B→[0,+∞) be a σ-additive atomless measure. Then there is a tree subalgebra U of B such that S|U is a rank-one atomless measure and μ|U is an atomless measure.Proof of Lemma8.
First we prove the followingclaim: for every b∈B there exist b′,b′′∈B such that b=b′⊔b′′, S(b′)=S(b′′)=(1/2)S(b), and μ(b′),μ(b′′)≤(3/4)μ(b).
Using the atomlessness ofμ, we choose a partition b=b1⊔b2 with μ(b1)=μ(b2) (formally we can apply [12, Theorem 2.11] to get this). Using the fact that S is dividing, we choose partitions bi=bi′⊔bi′′ so that S(bi′)=S(bi′′)=(1/2)S(b) for i=1,2. With no loss of generality, we may and do assume that μ(bi′)≤μ(bi′′) for i=1,2. Then μ(bi′)≤(1/2)μ(bi) for i=1,2. Set b′=b1′⊔b2′′ and b′′=b1′′⊔b2′. Then (8)Sb′=Sb1′+Sb2′′=12Sb1+Sb2=12Sband similarly S(b′′)=(1/2)S(b)=S(b′). Moreover, (9)μb′=μb1′+μb2′′≤12μb1+μb1=32μb1=34μb.
Similarly,μ(b′′)≤(3/4)μ(b). Hence (10)μb′=μb-μb′′≥μb-34μb=14μb,which completes the proof of the claim.
To prove the lemma, we setb1=1. Assume for a given n∈N that bn has been already defined. Using the claim with b=bn, we choose b2n=b′ and b2n+1=b′′ such that bn=b2n⊔b2n+1; (11)Sb2n=Sb2n+1=12Sbnandμb2n+i≤34μbnforalln∈Nandi∈0,1.
LetUτ be the tree subalgebra generated by τ=(bn)n=1∞. Observe that the image S(B(τ)) (the subalgebra B(τ) was defined in the proof of Lemma 7) is the set of all vectors of the form (12)l12k1+…+lj2kjSb,wherej∈N,l12k1+…+lj2kj∈0,1,and so one has that S(Uτ)={tS(t):t∈[0,1]}. Hence, there exists a scalar probability measure (i.e., σ-additive with nonnegative values and maximal value 1) μ0:Uτ→[0,1] such that(13)∀x∈Uτ,Sx=μ0x·S1.In particular, for every n∈N, one has μ0(bn)=2-k, where n=2k+l with l<2k.
Finally, by Lemma7, both scalar nonnegative measures μ|U and μ0 are atomless, and hence S|U is atomless.The following two lemmas seem to be well known.Lemma 9.
LetB be a σ-complete Boolean algebra, and let X be a Banach space. Then every σ-additive finite rank measure ν:B→X has finite variation |ν|:B→[0,+∞) which is σ-additive as well.To prove Lemma9, one can use Hahn’s decomposition theorem [15, 326 I] to every coordinate of an Rn-valued measure and decompose unity of B into 2n disjoint parts where every coordinate has a certain constant sign. Obviously, on every such a part ν has finite variation, which in their disjoint union gives |ν|.Lemma 10.
LetB be a σ-complete Boolean algebra, let X be a Banach space, and let ν:B→X be a σ-additive measure having finite variation |ν|:B→[0,+∞). Then ν is atomless if and only if |ν| is.Proof of Lemma10.
Let|ν| be atomless. Assume, on the contrary, that there is an atom a∈B of ν. Then ν(a)≠0 and |ν|(a)>0. Choose b≤a so that 0<|ν|(b)<|ν|(a). Then for every finite partition a=∏k=1mak one has ν(ak0)=ν(a) for some k0∈{1,…,m} and ν(ak)=0 for k≠k0 and hence ∑k=1mνak=ν(a). By the arbitrariness of the partition, |ν|(a)=ν(a). Since |ν|(b)>0, there is c≤b with ν(c)≠0, and, hence, ν(c)=ν(a) as a is an atom and c≤a. Thus, ν(a)≤|ν|(c)≤|ν|(b)<|ν|(a)=ν(a), a contradiction.Letν be atomless. Let a∈B be such that |ν|(a)>0. Choose any b≤a with ν(b)≠0. Using the atomlessness of ν, we choose a partition b=c⊔d so that ν(c)≠0≠ν(d). Hence, |ν|(c)>0 and |ν|(d)>0, which implies that 0<|ν|(c)<|ν|(a), and so a is not an atom for |ν|. Thus, |ν| is atomless as well.Proof of Theorem6.
Fix any0<b∈B. We let μ=T|Bb. By Lemma 10, μ is atomless. Using Lemma 8 for S|Bb (which is dividing and σ-additive as well) and μ, we choose a tree subalgebra U of Bb such that S|U is a rank-one measure and μ|U is an atomless measure.
Show that the measuresT|U and |S|U| are well defined and satisfy the assumptions of Lemma 8. Since μ|U is atomless, the measure T|U is atomless as well by Lemma 10. And since T is hereditarily dividing, T|U is dividing. By Lemma 8, the measure S|U is atomless. By Lemma 9, the measure |S|U| is well defined, and, by Lemma 10, |S|U| is atomless.
Applying Lemma8 to the measures T|U and |S|U|, we choose a tree subalgebra U1 of U such that T|U1 is a rank-one measure and μ|U1 is an atomless measure. Now consider the measure V=(S|U1,T|U1) which takes values in a 2-dimensional linear space Y×Z, where Y and Z are the 1-dimensional subspaces of X in which the measures S|U1 and T|U1 take values, respectively. Since both coordinates are atomless measures, the measure V is atomless as well. Indeed, let a∈U1 be such that V(a)≠0, say, S(a)≠0. Then we choose a′≤a and a′∈U1 so that 0≠S(a′)≠S(a) and obtain that 0≠V(a′)≠V(a).
By Theorem4, we can decompose b=b′⊔b′′ so that V(b′)=V(b′′); that is, S(b′)=S(b′′) and T(b′)=T(b′′).Using Theorem4, Lemma 9 and Theorem 6, we obtain the following partial result.Theorem 11.
LetB be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:B→X be σ-additive measures. If S is dividing and T is atomless and finite rank, then the pair S,T is uniformly dividing. In particular, S+T is dividing.


## 3. Implications to Orthogonally Additive Operators on Riesz Spaces

As mentioned in Introduction, there are many results on the narrowness of the sum of two narrow operators. Remark that all of them have common scheme of the proof: to prove thatS+T is narrow, it is sufficient to prove that every e∈E admits a decomposition e=e′⊔e′′ such that both vectors S(e′)-S(e′′) and T(e′)-T(e′′) are small in certain sense depending on the kind of narrowness.LetE be an atomless Riesz space, and let X be a Banach space. We say that a pair of OAOs S,T:E→X isuniformly strictly narrow if every e∈E admits a decomposition e=e′⊔e′′ such that S(e′)=Se′′ and T(e′)=T(e′′). For the first time, the uniform narrowness of operators was considered in [16].Recall that a net(xα)α∈Λ in a Riesz space Eorder converges to an element x∈E (notation xα→ox) if there exists a net (uα)α∈Λ in E such that uα↓0 and |xβ-x|≤uβ for all β∈Λ. A net (xα) in Elaterally converges to x∈E if xα⊑xβ⊑x for all indices α<β and xα⟶ox. In this case we write xα⟶lx. For positive elements xα,x the condition xα⟶lx means that xα⊑x and xα↑x.LetE be a Riesz space and let X be a Banach space. An OAO T:E→X is said to belaterally-to-norm continuous provided for every net (xα) in E and every x∈E the condition xα⟶lx implies T(xα)-T(x)→0. We say that an OAO T:E→X hasfinite variation if for every e∈E(14)sup⁡∑k=1mTek:m∈N,e=∐k=1mek<∞.It is not hard to see that ifE is Dedekind complete and T:E→X is a laterally-to-norm continuous OAO then for every e∈E the restriction T|Fe of T to the Boolean algebra Fe of all fragments of e is a σ-additive measure. If, moreover, T has finite variation then the measure T|Fe is of finite variation as well for all e∈E.For the proof of our main results, we need one more known lemma (see [12, Lemma 2.13]).Lemma 12.
LetE be an atomless Dedekind complete Riesz space, let X be a Banach space, and let T:E→X be a laterally-to-norm continuous OAO. Then for every 0≠e∈E the measure T|Fe is atomless.Remark that the original Lemma 2.13 from [12] is proven for positive elements e>0 only. However, the general case then easily follows from the decomposition e=e+-e- and the observation that the same arguments work for negative elements.Now an application of Theorem11 gives the following result.Theorem 13.
LetE be a Dedekind complete atomless Riesz space, let X be a Banach space, and let S,T be laterally continuous OAOs. If S is strictly narrow and T has finite rank, then S+T is strictly narrow. Moreover, the pair S,T is uniformly strictly narrow.In order to apply Theorem6, we first give a definition of a hereditarily strictly narrow operator. We say that an OAO T:E→X ishereditarily strictly narrow if for any element e∈E the restriction T|Fe is a hereditarily divisible measure.As a consequence of Theorem6 we obtain the following result.Theorem 14.
LetE be a Dedekind complete atomless Riesz space, let X be a Banach space, and let S,T be laterally continuous OAOs. If S is strictly narrow and T is hereditarily strictly narrow and has finite variation, then S+T is strictly narrow. Moreover, the pair S,T is uniformly strictly narrow.We conjecture that the assumption onT to have finite variation is superfluous. However, not is the sense that every hereditarily strictly narrow has finite variation (as it happened with finite rank operators), because this is not true as the following example shows.Example 15.
There exists a Dedekind complete atomless Riesz spaceE, a Banach space X, and a hereditarily strictly narrow linear bounded operator T:E→X having infinite variation.Construction. Let 1<p<+∞. We set E=X=Lp[0,1]. Consider a disjoint sequence (An) of measurable subsets of [0,1] with [0,1]=∐n=1∞An and μ(An)=n-p(∑j=1∞j-p)-1 for all n∈N. Then the conditional expectation operator with respect to the σ-algebra generated by Ans (15)Tx=∑n=1∞1μAn∫Anxdμ1An,where 1A is the characteristic function of A, possesses the desired properties.Remark that we are still far from a solution of Problem2. Another related problem is in [16] and is still unsolved.Problem 16.
LetE be a Riesz space and let X be a Banach space (or, more generally, F-space). Are the following assertions equivalent for every pair of narrow linear operators (or, more generally, OAOs) S,T:E→X? (i)
S+T is narrow;(ii)
S,T are uniformly narrow.



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