#### Abstract

We prove that if is a Dedekind complete atomless Riesz space and is a Banach space, then the sum of two laterally continuous orthogonally additive operators from to , 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.

#### 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 , the sum of two continuous linear narrow operators is narrow [2, Theorem 7.46]; however, if a rearrangement invariant space on has an unconditional basis, then every operator on 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 on , there exist a Banach space and narrow operators from to with nonnarrow sum [3].

If the norm of the domain Köthe Banach space is not absolutely continuous (for instance, if ), then the usual technique does not work. So, there are nonnarrow continuous linear functionals on . However, questions about narrowness of the sum of two narrow operators are still interesting. A sum of two narrow operators on need not be narrow [4]. Moreover, if , then there are regular narrow operators with nonnarrow sum [3].

Now let a pair of spaces be such that there are narrow operators with nonnarrow sum . 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 is a Köthe Banach space with an absolutely continuous norm, then for any Banach space the sum of a narrow operator and a “small” operator (like compact, AM-compact, Dunford-Pettis operators, etc.) is narrow.

If the norm of 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 is a Köthe F-space, is a locally convex F-space, and are narrow operators such that maps the set of all signs to a relatively compact subset of (in particular, if is compact), then the sum 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 is a Dedekind complete atomless Riesz space and is a Banach space, then the sum of narrow and C-compact laterally continuous orthogonally additive operators from to 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 , we mean a Banach space which is a linear subspace of possessing the following properties: , and for every and the condition implies that and (by we denote the characteristic function of a set , and the inequality in means that holds for -almost all , where and are some/any representatives of the classes ). A Köthe Banach space on a finite atomless measure space is said to have an* absolutely continuous norm* if for all .

If are Banach spaces, by we denote the Banach space of all continuous linear operators , and stands for . By we denote the disjoint sum in a Köthe Banach space, that is, under the assumption or, more generally, in a Riesz space under the assumption . In a Boolean algebra, means the disjoint supremum , that is, under the assumption .

For familiarly used information on Riesz spaces, the reader can refer to [11]. Let be a Riesz space and be a linear space. A function is called an* orthogonally additive operator* (*OAO* in short) if for all disjoint elements . Simple examples of OAOs are the positive, negative parts and the modules of an element: , , , and . For more examples of OAOs including integral Uryson operators, see [6–8].

An element of a Riesz space is called a* fragment* of (write ) provided . The set of all fragments of an element is denoted by . Observe that if , then and are disjoint fragments of . We say that an element of a Riesz space is an* atom* if the only fragments of are and itself. A Riesz space having no atom is said to be* atomless*.

Let be an atomless Riesz space and let be a Banach space. An OAO is called(i)*narrow at a point * if for every there is a decomposition such that ;(ii)*narrow* if it is narrow at each point ;(iii)*strictly narrow at a point * if there is a decomposition such that ;(iv)*strictly narrow* if it is strictly narrow at each point .

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 for every OAO ; hence, every OAO is strictly narrow at zero. Every strictly narrow (at a point ) is narrow (at a point ); 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].

If is a Köthe Banach space with an absolutely continuous norm on a finite atomless measure space and is a Banach space, then an OAO is narrow if and only if for every every admits a decomposition , such that and [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 be a Köthe Banach space on a finite atomless measure space , and let be a Banach space. An operator is called* hereditarily narrow* if for every , and every atomless sub--algebra of the restriction of to is narrow (here ). The following proposition gives lots of examples of pairs of narrow operators with narrow sum.

Proposition 1 ([13], [2], Proposition 11.2). *Let be a Köthe Banach space on with an absolutely continuous norm, and let be a Banach space. Then the sum of a narrow operator and a hereditarily narrow operator 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 be an atomless Riesz space, and let be a Banach space. Is the sum of strictly narrow operators 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 [12], is to consider the set of all fragments of a fixed element of the domain Riesz space as the main object for investigation. This becomes possible because the definitions of all notions from the main theorem could be equivalently restricted to . Since the set 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 be a net in a Boolean algebra . The notation means that the net decreases and . We say that a net in * order converges* to an element if there exists a net in with the same index set such that for all indices and . In this case, we write and say that is the* order limit* of .

A Boolean algebra is said to be* order complete* if any nonempty subset of has the supremum. A Boolean algebra is said to be -*complete* if any countable subset of has the supremum. By a* partition* (of unity) in a Boolean algebra we mean a maximal disjoint subset , that is, A* disjoint union * (i.e., the union of a disjoint system ), if exists, is denoted by . Although in some cases an infinite union in a Boolean algebra does not exist, it is immediate that if is a partition then exists. Conversely, if then is a partition.

Let be a Boolean algebra, and let be a linear space. A function is said to be a* measure* provided for every pair of disjoint elements . Obviously, for a measure. An element is called an* atom* of a measure provided and for any with one has either or . A measure is called* atomless* provided there is no atom of .

A measure is said to have* finite rank* if the closed linear span is a finite-dimensional subspace of . Let be a -complete Boolean algebra, and let be a Banach space. A measure is said to be -*additive* provided for every disjoint sequence in one has , where the series converges unconditionally in .

Let be a Boolean algebra, and let be a set. A function is said to be* dividing* provided every element has a two-point partition with . We say that a pair of functions is* uniformly dividing* if every element has a two-point partition with and .

Next we define a hereditarily dividing measure, which takes an important place in our investigation. Given a Boolean algebra and any , we set , which is a Boolean algebra with the induced operations and unity .

*Definition 3. *Let be a -complete Boolean algebra and let be a Banach space. An atomless -additive measure is called* hereditarily dividing* if, for every and every -complete subalgebra of , the atomlessness of the restriction of to implies that is dividing on .

Obviously, a hereditarily dividing measure is dividing. By [12, Theorem 2.11] and Lemma 9, every atomless -additive measure with finite-dimensional range is dividing. Hence, as a consequence, we obtain the following example of hereditarily dividing measures.

Theorem 4. *Let be a -complete Boolean algebra, and let be a Banach space. Then every atomless -additive measure with finite-dimensional range is hereditarily dividing.*

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

Theorem 5. *Let be a -complete Boolean algebra, and let be a Banach space. Let be -additive measures. If is dividing and is hereditarily dividing and has finite variation, then is dividing.*

Actually, we prove more.

Theorem 6. *Let be a -complete Boolean algebra, and let be a Banach space. Let be -additive measures. If is dividing and is hereditarily dividing and has finite variation, then the pair is uniformly dividing.*

It is an obvious observation that Theorem 6 yields Theorem 5. For the proof, we need several lemmas. Let be a -complete Boolean algebra. A sequence in is called a* tree* if and for all . The minimal -complete subalgebra including a tree is called a* tree subalgebra* of generated by .

Lemma 7. *Let be a tree in a -complete Boolean algebra and let be the tree subalgebra generated by . If is a -additive measure and for all and then is atomless.*

*Proof of Lemma 7. *Fix any with . By [14, Lemma 1.2.14], the smallest subalgebra of including equals the set of all finite disjoint unions of elements of . By [15, 313F(c)], the -order closure of a subalgebra is a subalgebra. Hence, the -order closure of the subalgebra equals . Since is a -additive measure on , it is -continuous. Hence, we may and do choose a finite disjoint union such thatSet Then Similarly, . Hence, . Thus, Observe that Hence, and therefore, by (1), . Then we obtain Similarly, . Hence which yields that is not an atom for .

Lemma 8. *Let be a dividing -additive measure and let be a -additive atomless measure. Then there is a tree subalgebra of such that is a rank-one atomless measure and is an atomless measure.*

*Proof of Lemma 8. *First we prove the following* claim*: for every there exist such that , , and .

Using the atomlessness of , we choose a partition with (formally we can apply [12, Theorem 2.11] to get this). Using the fact that is dividing, we choose partitions so that for . With no loss of generality, we may and do assume that for . Then for . Set and . Then and similarly . Moreover, Similarly, . Hence which completes the proof of the claim.

To prove the lemma, we set . Assume for a given that has been already defined. Using the claim with , we choose and such that ; Let be the tree subalgebra generated by . Observe that the image (the subalgebra was defined in the proof of Lemma 7) is the set of all vectors of the form and so one has that . Hence, there exists a scalar probability measure (i.e., -additive with nonnegative values and maximal value ) such thatIn particular, for every , one has , where with .

Finally, by Lemma 7, both scalar nonnegative measures and are atomless, and hence is atomless.

The following two lemmas seem to be well known.

Lemma 9. *Let be a -complete Boolean algebra, and let be a Banach space. Then every -additive finite rank measure has finite variation which is -additive as well.*

To prove Lemma 9, one can use Hahn’s decomposition theorem [15, 326 I] to every coordinate of an -valued measure and decompose unity of into disjoint parts where every coordinate has a certain constant sign. Obviously, on every such a part has finite variation, which in their disjoint union gives .

Lemma 10. *Let be a -complete Boolean algebra, let be a Banach space, and let be a -additive measure having finite variation . Then is atomless if and only if is.*

*Proof of Lemma 10. *Let be atomless. Assume, on the contrary, that there is an atom of . Then and . Choose so that . Then for every finite partition one has for some and for and hence . By the arbitrariness of the partition, . Since , there is with , and, hence, as is an atom and . Thus, , a contradiction.

Let be atomless. Let be such that . Choose any with . Using the atomlessness of , we choose a partition so that . Hence, and , which implies that , and so is not an atom for . Thus, is atomless as well.

*Proof of Theorem 6. *Fix any . We let . By Lemma 10, is atomless. Using Lemma 8 for (which is dividing and -additive as well) and , we choose a tree subalgebra of such that is a rank-one measure and is an atomless measure.

Show that the measures and are well defined and satisfy the assumptions of Lemma 8. Since is atomless, the measure is atomless as well by Lemma 10. And since is hereditarily dividing, is dividing. By Lemma 8, the measure is atomless. By Lemma 9, the measure is well defined, and, by Lemma 10, is atomless.

Applying Lemma 8 to the measures and , we choose a tree subalgebra of such that is a rank-one measure and is an atomless measure. Now consider the measure which takes values in a -dimensional linear space , where and are the -dimensional subspaces of in which the measures and take values, respectively. Since both coordinates are atomless measures, the measure is atomless as well. Indeed, let be such that , say, . Then we choose and so that and obtain that .

By Theorem 4, we can decompose so that ; that is, and .

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

Theorem 11. *Let be a -complete Boolean algebra, and let be a Banach space. Let be -additive measures. If is dividing and is atomless and finite rank, then the pair is uniformly dividing. In particular, 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 is narrow, it is sufficient to prove that every admits a decomposition such that both vectors and are small in certain sense depending on the kind of narrowness.

Let be an atomless Riesz space, and let be a Banach space. We say that a pair of OAOs is* uniformly strictly narrow* if every admits a decomposition such that and . For the first time, the uniform narrowness of operators was considered in [16].

Recall that a net in a Riesz space * order converges* to an element (notation ) if there exists a net in such that and for all . A net in * laterally converges* to if for all indices and . In this case we write . For positive elements the condition means that and .

Let be a Riesz space and let be a Banach space. An OAO is said to be* laterally-to-norm* continuous provided for every net in and every the condition implies . We say that an OAO has* finite variation* if for every

It is not hard to see that if is Dedekind complete and is a laterally-to-norm continuous OAO then for every the restriction of to the Boolean algebra of all fragments of is a -additive measure. If, moreover, has finite variation then the measure is of finite variation as well for all .

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

Lemma 12. *Let be an atomless Dedekind complete Riesz space, let be a Banach space, and let be a laterally-to-norm continuous OAO. Then for every the measure is atomless.*

Remark that the original Lemma 2.13 from [12] is proven for positive elements only. However, the general case then easily follows from the decomposition and the observation that the same arguments work for negative elements.

Now an application of Theorem 11 gives the following result.

Theorem 13. *Let be a Dedekind complete atomless Riesz space, let be a Banach space, and let be laterally continuous OAOs. If is strictly narrow and has finite rank, then is strictly narrow. Moreover, the pair 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 is* hereditarily strictly narrow* if for any element the restriction is a hereditarily divisible measure.

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

Theorem 14. *Let be a Dedekind complete atomless Riesz space, let be a Banach space, and let be laterally continuous OAOs. If is strictly narrow and is hereditarily strictly narrow and has finite variation, then is strictly narrow. Moreover, the pair is uniformly strictly narrow.*

We conjecture that the assumption on 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 , a Banach space , and a hereditarily strictly narrow linear bounded operator having infinite variation.

*Construction.* Let . We set . Consider a disjoint sequence of measurable subsets of with and for all . Then the conditional expectation operator with respect to the -algebra generated by s where is the characteristic function of , possesses the desired properties.

Remark that we are still far from a solution of Problem 2. Another related problem is in [16] and is still unsolved.

*Problem 16. *Let be a Riesz space and let be a Banach space (or, more generally, -space). Are the following assertions equivalent for every pair of narrow linear operators (or, more generally, OAOs) ? (i) is narrow;(ii) 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).