JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi 10.1155/2019/8569409 8569409 Research Article On Sums of Strictly Narrow Operators Acting from a Riesz Space to a Banach Space Maslyuchenko Oleksandr 1 2 http://orcid.org/0000-0002-3165-5822 Popov Mikhail 3 4 Martin Miguel 1 Institute of Mathematics University of Silesia in Katowice Bankowa 12 40-007 Katowice Poland us.edu.pl 2 Yuriy Fedkovych Chernivtsi National University Department of Mathematical Analysis Kotsiubynskoho 2 58012 Chernivtsi Ukraine chnu.edu.ua 3 Institute of Mathematics Pomeranian University in Słupsk ul. Arciszewskiego 22d 76-200 Słupsk Poland apsl.edu.pl 4 Vasyl Stefanyk Precarpathian National University Ukraine pu.if.ua 2019 262019 2019 06 01 2019 30 04 2019 19 05 2019 262019 2019 Copyright © 2019 Oleksandr Maslyuchenko and Mikhail Popov. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove that if E 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.

University of Silesia
1. Introduction

Narrow operators were introduced in 1990 ; however, some deep results on these operators were obtained earlier; see . 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 space E on [0,1], there exist a Banach space X and narrow operators from E to X with nonnarrow sum .

If the norm of the domain Köthe Banach space E 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 . Moreover, if 1<p, then there are regular narrow operators S,T:LpL with nonnarrow sum S+T .

Now let a pair of spaces E,X be such that there are narrow operators S,T:EX 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 of E 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 : if E is a Köthe F-space, X is a locally convex F-space, and S,TL(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 , 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  from linear to orthogonally additive operators. For example, a result of  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  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 a Köthe 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: 1ΩE, and for every xE and yL1(μ) the condition |y||x| implies that yE and yx (by 1A we denote the characteristic function of a set AΣ, and the inequality uv in E means that u~(t)v~(t) holds for μ-almost all tΩ, 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 an absolutely continuous norm if limμ(A)0x·1A=0 for all xE.

If X,Y are Banach spaces, by L(X,Y) we denote the Banach space of all continuous linear operators T:XY, and L(X) stands for L(X,X). By xy we denote the disjoint sum x+y in a Köthe Banach space, that is, under the assumption suppxsuppy= or, more generally, in a Riesz space under the assumption xy. In a Boolean algebra, xy means the disjoint supremum xy, that is, under the assumption xy=0.

For familiarly used information on Riesz spaces, the reader can refer to . Let E be a Riesz space and X be a linear space. A function T:EX is called an orthogonally additive operator (OAO in short) if T(xy)=T(x)+T(y) for all disjoint elements x,yE. 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 xE. For more examples of OAOs including integral Uryson operators, see .

An element x of a Riesz space E is called a fragment of yE (write xy) provided xy-x. The set of all fragments of an element eE is denoted by Fe. Observe that if z=xy, then x and y are disjoint fragments of z. We say that an element a0 of a Riesz space E is an atom if the only fragments of a are 0 and a itself. A Riesz space having no atom is said to be atomless.

Let E be an atomless Riesz space and let X be a Banach space. An OAO T:EX is called

narrow at a point eE if for every ε>0 there is a decomposition e=ee such that T(e)-T(e)<ε;

narrow if it is narrow at each point eE;

strictly narrow at a point eE if there is a decomposition e=ee such that T(e)=T(e);

strictly narrow if it is strictly narrow at each point eE.

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 that T(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 .

If E 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:EX is narrow if and only if for every ε>0 every AΣ admits a decomposition A=BC, B,CΣ 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. Let E be a Köthe Banach space on a finite atomless measure space (Ω,Σ,μ), and let X be a Banach space. An operator TL(E,X) is called hereditarily narrow if for every AΣ, 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)={xE(A):xisF-measurable}). The following proposition gives lots of examples of pairs of narrow operators with narrow sum.

Proposition 1 ([<xref ref-type="bibr" rid="B6">13</xref>], [<xref ref-type="bibr" rid="B16">2</xref>], Proposition 11.2).

Let E 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 T1L(E,X) and a hereditarily narrow operator T2L(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.

Let E be an atomless Riesz space, and let X be a Banach space. Is the sum S+T of strictly narrow operators S,T:EX strictly narrow or, at least, narrow?

Our main result, which is an analogue of Proposition 1 for strictly narrow operators, is the first result in this direction. The idea of the proof, inspired by paper , 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 uα0 means that the net (uα) decreases and infαuα=0. We say that a net (xα) in B order converges to an element xB if there exists a net (uα) in B with the same index set such that xαxuα for all indices α and uα0. In this case, we write xαx and say that x is the order limit of (xα).

A Boolean algebra B is said to be order 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 a partition (of unity) in a Boolean algebra B we mean a maximal disjoint subset AB, that is, (xB)((aAax=0)(x=0)). A disjoint union A (i.e., the union of a disjoint system AB), 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.

Let B be a Boolean algebra, and let X be a linear space. A function f:BX is said to be a measure provided f(xy)=f(x)+f(y) for every pair of disjoint elements x,yB. Obviously, f(0)=0 for a measure. An element aB is called an atom of a measure f:BX provided f(a)0 and for any xB with xa one has either f(x)=0 or f(x)=f(a). A measure f:BX is called atomless provided there is no atom of f.

A measure f:BX is said to have finite 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:BX is said to be σ-additive provided for every disjoint sequence (xn)n=1 in B one has f(n=1xn)=n=1f(xn), where the series converges unconditionally in X.

Let B be a Boolean algebra, and let X be a set. A function f:BX is said to be dividing provided every element bB has a two-point partition b=bb with f(b)=f(b). We say that a pair of functions f,g:BX is uniformly dividing if every element bB has a two-point partition b=bb 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 algebra B and any bB, we set Bb={xB:xb}, which is a Boolean algebra with the induced operations and unity b.

Definition 3.

Let B be a σ-complete Boolean algebra and let X be a Banach space. An atomless σ-additive measure T:BX is called hereditarily dividing if, for every bB 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.

Let B be a σ-complete Boolean algebra, and let X be a Banach space. Then every atomless σ-additive measure T:BX with finite-dimensional range is hereditarily dividing.

The following theorem brings an important tool for the main result.

Theorem 5.

Let B be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:BX 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.

Let B be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:BX 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 Theorem 6 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 a tree if a1=1 and an=a2na2n+1 for all nN. The minimal σ-complete subalgebra including a tree τ is called a tree subalgebra of B generated by τ.

Lemma 7.

Let τ=(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 nN and i{0,1} then μ is atomless.

Proof of Lemma <xref ref-type="statement" rid="lem2.5">7</xref>.

Fix any aUτ 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=jJaj such that(1)μap16μa.

Set (2)p=jJa2jandp=jJa2j+1.

Then (3)μp=jJμa2j34jJμaj=34μp.

Similarly, μ(p)(3/4)μ(p). Hence, μ(p)=μ(p)-μ(p)(1/4)μ(p). Thus, (4)14μpμp34μp.

Observe that (5)μaμp+μapby1μp+16μa.

Hence, μ(a)(6/5)μ(p) and therefore, by (1), μ(ap)(1/5)μ(p). Then we obtain (6)μap=μp-μp-aμp-μpa14μp-15μp=120μp>by10.

Similarly, μ(ap)>0. Hence (7)0<μap<μap+μapμa, which yields that a is not an atom for μ.

Lemma 8.

Let S:BX 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 Lemma <xref ref-type="statement" rid="lem2.6">8</xref>.

First we prove the following claim: for every bB there exist b,bB such that b=bb, S(b)=S(b)=(1/2)S(b), and μ(b),μ(b)(3/4)μ(b).

Using the atomlessness of μ, we choose a partition b=b1b2 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=bibi 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=b1b2 and b=b1b2. Then (8)Sb=Sb1+Sb2=12Sb1+Sb2=12Sband similarly S(b)=(1/2)S(b)=S(b). Moreover, (9)μb=μb1+μb212μb1+μb1=32μb1=34μb.

Similarly, μ(b)(3/4)μ(b). Hence (10)μb=μb-μbμb-34μb=14μb,which completes the proof of the claim.

To prove the lemma, we set b1=1. Assume for a given nN that bn has been already defined. Using the claim with b=bn, we choose b2n=b and b2n+1=b such that bn=b2nb2n+1; (11)Sb2n=Sb2n+1=12Sbnandμb2n+i34μbnforallnNandi0,1.

Let Uτ 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,wherejN,l12k1++lj2kj0,1,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)xUτ,Sx=μ0x·S1.In particular, for every nN, one has μ0(bn)=2-k, where n=2k+l with l<2k.

Finally, by Lemma 7, 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.

Let B be a σ-complete Boolean algebra, and let X be a Banach space. Then every σ-additive finite rank measure ν:BX has finite variation |ν|:B[0,+) which is σ-additive as well.

To prove Lemma 9, 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.

Let B be a σ-complete Boolean algebra, let X be a Banach space, and let ν:BX be a σ-additive measure having finite variation |ν|:B[0,+). Then ν is atomless if and only if |ν| is.

Proof of Lemma <xref ref-type="statement" rid="lem2.8">10</xref>.

Let |ν| be atomless. Assume, on the contrary, that there is an atom aB of ν. Then ν(a)0 and |ν|(a)>0. Choose ba 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 kk0 and hence k=1mνak=ν(a). By the arbitrariness of the partition, |ν|(a)=ν(a). Since |ν|(b)>0, there is cb with ν(c)0, and, hence, ν(c)=ν(a) as a is an atom and ca. Thus, ν(a)|ν|(c)|ν|(b)<|ν|(a)=ν(a), a contradiction.

Let ν be atomless. Let aB be such that |ν|(a)>0. Choose any ba with ν(b)0. Using the atomlessness of ν, we choose a partition b=cd so that ν(c)0ν(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 Theorem <xref ref-type="statement" rid="thm2.4">6</xref>.

Fix any 0<bB. 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 measures T|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 Lemma 8 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 aU1 be such that V(a)0, say, S(a)0. Then we choose aa and aU1 so that 0S(a)S(a) and obtain that 0V(a)V(a).

By Theorem 4, we can decompose b=bb so that V(b)=V(b); that is, S(b)=S(b) and T(b)=T(b).

Using Theorem 4, Lemma 9 and Theorem 6, we obtain the following partial result.

Theorem 11.

Let B be a σ-complete Boolean algebra, and let X be a Banach space. Let S,T:BX 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 that S+T is narrow, it is sufficient to prove that every eE admits a decomposition e=ee such that both vectors S(e)-S(e) and T(e)-T(e) are small in certain sense depending on the kind of narrowness.

Let E be an atomless Riesz space, and let X be a Banach space. We say that a pair of OAOs S,T:EX is uniformly strictly narrow if every eE admits a decomposition e=ee such that S(e)=Se and T(e)=T(e). For the first time, the uniform narrowness of operators was considered in .

Recall that a net (xα)αΛ in a Riesz space E order converges to an element xE (notation xαox) if there exists a net (uα)αΛ in E such that uα0 and |xβ-x|uβ for all βΛ. A net (xα) in E laterally converges to xE if xαxβx for all indices α<β and xαox. In this case we write xαlx. For positive elements xα,x the condition xαlx means that xαx and xαx.

Let E be a Riesz space and let X be a Banach space. An OAO T:EX is said to be laterally-to-norm continuous provided for every net (xα) in E and every xE the condition xαlx implies T(xα)-T(x)0. We say that an OAO T:EX has finite variation if for every eE(14)supk=1mTek:mN,e=k=1mek<.

It is not hard to see that if E is Dedekind complete and T:EX is a laterally-to-norm continuous OAO then for every eE 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 eE.

For the proof of our main results, we need one more known lemma (see [12, Lemma 2.13]).

Lemma 12.

Let E be an atomless Dedekind complete Riesz space, let X be a Banach space, and let T:EX be a laterally-to-norm continuous OAO. Then for every 0eE the measure T|Fe is atomless.

Remark that the original Lemma 2.13 from  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 Theorem 11 gives the following result.

Theorem 13.

Let E 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 Theorem 6, we first give a definition of a hereditarily strictly narrow operator. We say that an OAO T:EX is hereditarily strictly narrow if for any element eE the restriction T|Fe is a hereditarily divisible measure.

As a consequence of Theorem 6 we obtain the following result.

Theorem 14.

Let E 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 on T 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 space E, a Banach space X, and a hereditarily strictly narrow linear bounded operator T:EX 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=1An and μ(An)=n-p(j=1j-p)-1 for all nN. Then the conditional expectation operator with respect to the σ-algebra generated by Ans (15)Tx=n=11μAnAnxdμ1An,where 1A is the characteristic function of A, possesses the desired properties.

Remark that we are still far from a solution of Problem 2. Another related problem is in  and is still unsolved.

Problem 16.

Let E 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:EX?

S+T is narrow;

S,T are uniformly narrow.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The authors are grateful to V. Kadets for valuable remarks. This research was supported by the University of Silesia Mathematics Department (Iterative Functional Equations and Real Analysis program).

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