Multiplication operators in BV spaces

The aim of this paper is to provide necessary and sufficient conditions on the generator of a multiplication operator acting in the spaces of functions of bounded Young and Riesz variation so that it is, among other things, invertible, continuous, finite rank, compact, Fredholm or has closed range. Furthermore, we characterize various spectra of such operators and give some estimates on their measure of non-compactness.


Introduction
Multiplication operators are among the most basic linear operators acting in function spaces, or, in a more general setting, in (normed) algebras. Because of that and because of the fact that they are also building blocks for other important linear and nonlinear operators (such as, for example, weighted composition and superposition operators), multiplication operators have attracted great interest of analysts.
It seems impossible to summarize all the recent research concerning multiplication operators and related topics. Therefore, let us only draw the readers' attention to a few articles. Multiplication operators between two classical spaces of Lebesgue integrable functions were studied by, for example, Takagi et al.in [37]. Those results were extended to 1 3 Orlicz spaces by Chawziuk et al. (see [17][18][19] and the references therein). On the other hand, properties of multiplication operators in Köthe spaces were investigated, for example, by Drewnowski et al. in [22] as well as by Castillo et al. in [15] (see also [34]). Let us also mention the papers by Bonet et al. [11] and de Jager et al. [25], where such operators were studied on, respectively, weighted Banach spaces of analytic functions and non-commutative spaces.
It may, therefore, come as a surprise that there are only two papers dealing with both the function-theoretic and topological properties of multiplication operators in the spaces of functions of bounded variation. Multiplication operators in the space BV of functions of bounded Jordan variation were exhaustively studied in [8], while those acting in the space WBV p of functions of bounded Wiener variation were investigated in [10] (for the definitions of the Jordan, Wiener and other variations see Section 2).
One of the reasons for this situation may be the fact that the multiplier classes Y/X, consisting of those elements g for which the multiplication x ↦ gx is a well-defined operator from X to Y (for a formal definition of Y/X see Section 3), have been fully characterized for various BV-type spaces only very recently (see [12,13]; cf. also [16]). This said, let us make a small digression. If X is a linear space of real-valued functions defined on the interval [0, 1] which is closed under multiplication and contains constant functions (for example, if X coincides with C, that is, the space of continuous functions on [0, 1], or with B, that is, the space of bounded functions on the same interval), then it is easy to check that X∕X = X . However, in other classes of functions which are not necessarily closed under multiplication, a characterization of X/X can be much harder. For instance, the proof that D/D consists only of constant functions requires a lot more work; here D stands for the class of Darboux functions 1 . (An elementary proof of this fact, together with some brief historical background concerning this result, can be found in [12]). Adding another space Y to the mix complicates things even more. In some cases, a full determination of Y/X is extremely difficult. For instance, the class D/C satisfies the chain of inclusions C ⊊ D ∩ B 1 ⊊ D∕C ⊊ D , where B 1 denotes the class of Baire-1 functions, but its exact characterization is-at least to our knowledge-unknown. A detailed discussion of these and many more multiplier classes can be found in [12,13].
The aim of this paper is twofold. First is to extend the results of [8,10] to multiplication operators acting between not necessarily equal spaces YBV of functions of bounded Young variation and to provide necessary and sufficient conditions guaranteeing that such operators are, among other things, bijective, continuous, finite rank, compact, Fredholm or have closed range. Second is to check whether multiplication operators in different BV spaces also enjoy similar properties. Therefore, beside the spaces YBV we chose to investigate also the spaces RBV p of functions of bounded Riesz variation, which, roughly speaking, are situated on the other end of the "spectrum" of BV-type spaces ( RBV p spaces are contained in BV, while YBV spaces contain BV; all functions in RBV p are continuous, when 1 < p < +∞ , while YBV contain also some functions which are discontinuous; finally, RBV p spaces are decreasing with respect to p, while WBV p spaces, which are a special case of YBV for (u) = u p , increase with respect to the parameter). Finally, whenever it was possible we decided to prove abstract results concerning multiplication operators acting in general linear/normed spaces of real-valued functions defined on the interval [0, 1].
As it would only obscure the paper and make the readers experience the déjà vu phenomenon, we decided not to include the information concerning the relation between our results and the results of [8,10] after each theorem. However, following the saying "give credit where credit is due," we would like to underline once again that invertibility, continuity, compactness, Fredholmness and several other properties of multiplication operators M g ∶ BV → BV and M g ∶ WBV p → WBV p were first characterized in, respectively, [8] and [10]. Furthermore, let us add that although we clearly used and adapted some ideas and methods introduced in [8,10], we also employed several new techniques (especially when multiplication operators in the spaces RBV p were considered).
The paper is organized as follows. In Sect. 2, we gather basic definitions and facts concerning functions of bounded Young and Riesz variation which will be needed throughout the article. In Sect. 3, we briefly discuss multiplier classes of some BV-type spaces. Sect. 4 is devoted to studying function-theoretic properties (such as injectivity, surjectivity and bijectivity) of multiplication operators in spaces of functions of bounded Young and Riesz variation as well as in some abstract function spaces. Finally, in Sect. 5 we discuss in detail topological properties of such operators. We begin with continuity and compactness; in particular, we provide some estimates on the measure of non-compactness of multiplication operators and explain how similar estimates for the essential norm can be obtained. Furthermore, we describe multiplication operators in the spaces of bounded Young and Riesz variation which have closed range and are Fredholm operators. Finally, we provide also a full characterization of various spectra of those operators.

Added in the proof
After we had submitted our paper to the journal, we found that Astudillo-Villalba et al. had just published a paper concerning multiplication operators between different spaces of functions of Wiener bounded variation (see [9]). It is worth noting that the results established in [9] are contained in our results on multiplication operators acting in the spaces of functions of Young bounded variation.

Preliminaries
The aim of this section is to introduce the notation used in the paper and recall some basic definitions and facts concerning functions of bounded variation. Since we will try to be as brief as possible, we refer readers who are not familiar with various generalizations of the classical Jordan variation to a very nice monograph on that subject [6].

Notation
Let us begin with some notation and conventions. If A ⊆ ℝ , then by A we will denote the characteristic function of the set A, that is, A (t) = 0 for t ∈ ℝ ⧵ A and A (t) = 1 for t ∈ A . Moreover, if A is finite, then by #A we will denote the number of elements of the set A; clearly, #� = 0 . We will also set #A = +∞ if A is infinite. Let us also recall that a set is called countable if it is equinumerous with some subset of positive integers ℕ . In particular, the empty set is both countable and finite. We will call a closed interval I ⊆ ℝ degenerate if I = [a, a] = {a} for some a ∈ ℝ . Often in the paper, we will write that X is a linear space of real-valued functions defined on the interval [0, 1]. We will always assume that the linear structure of X is inherited from the field of the real numbers ℝ , that is, for x, y ∈ X and ∈ ℝ the functions x + y and x will be given by t ↦ x(t) + y(t) and t ↦ x(t) , respectively. In particular, the zero function, i.e., the constant function taking the value 0, will be always the additive identity. Similarly, (xy)(t) = x(t)y(t) , although the product xy may not always be an element of X even though x, y will be. As regards division, by x/y we will mean the function t ↦ x(t)∕y(t) , provided that it is well-defined. We will also understand that two functions in X are equal if they attain the same values at each point in the interval [0, 1]. As usual by C and B, we will denote the Banach spaces of all, respectively, continuous and bounded real-valued functions defined on the interval [0, 1], endowed with the supremum norm ‖x‖ ∞ ∶= sup t∈[0,1] �x(t)� . Furthermore, the symbol L p will stand for the Banach space of all (equivalence classes of) functions which are Lebesgue integrable with p-th power on the interval [0, 1], where 1 ≤ p < +∞ , endowed with the norm If (X, ‖⋅‖ X ) and (Y, ‖⋅‖ Y ) are two normed spaces, we will say that X is embedded into Y (and write X ↪ Y ) when X ⊆ Y (as sets) and the identity mapping I ∶ X → Y is continuous, that is, when there is a constant c > 0 such that ‖x‖ Y ≤ c‖x‖ X for all x ∈ X . We will call the constant c an embedding constant. Throughout the paper by Ker L and Im L we will denote, respectively, the kernel and range of the linear operator L.

Support of a function
We follow the notation introduced in [8] and for a function x ∶ [0, 1] → ℝ we write supp (x) ∶= {t ∈ [0, 1]|x(t) ≠ 0} for its support. Note that in contrast to the standard definition of the support of a function, we do not take the closure here. For ≥ 0 we also write supp (x) ∶= {t ∈ [0, 1]||x(t)| > } . Observe that supp (x) decreases with respect to for any fixed function . In particular, this implies that if supp (x) is countable for each > 0 , then so is supp (x) (as a countable union of countable sets). In other words, if supp (x) is uncountable, then supp (x) is uncountable (and hence infinite) for some > 0.
In the sequel, we will also need a notion complementary to the concept of the support of a function. For a given function x ∶ [0, 1] → ℝ by Z x , we will denote the set of zeros of x, that is, Z x ∶= [0, 1] ⧵ supp (x).

Functions of bounded Young variation
Before we are able to define a function of bounded variation in the sense of Young, we need to recall the definition of a Young function.
Definition 1 A function ∶ [0, +∞) → [0, +∞) is said to be a Young function (or -function) if it is convex and such that (t) = 0 if and only if t = 0.
With this definition at hand, one can define the variation in the sense of Young, which, as the name suggests, was introduced by Laurence Chisholm Young in 1937 (see [39]). Definition 3 Let x be a real-valued function defined on [a, b] with a < b and let be a given Young function. The (possibly infinite) number where the supremum is taken over all finite partitions a = t By YBV , we will denote the class of all functions x ∶ [0, 1] → ℝ such that x is of bounded Young variation for some > 0 , that is, It is well-known that YBV is a linear space and becomes a Banach space when endowed with the norm ‖x‖ YBV ∶= ‖x‖ ∞ + �x� YBV , where (cf. [32]). Note that the definition of the norm ‖⋅‖ YBV is meaningful as each function of bounded Young variation is bounded.

Remark 4
The special case when p (t) = t p for p ≥ 1 will be of particular interest for us. If p = 1 , the Young variation coincides with the classic variation, which goes back to Camille Jordan (see [26,27]). In this case, we will simply write BV instead of YBV 1 , var (x) instead of var 1 (x) and ‖⋅‖ BV instead of ‖⋅‖ YBV 1 . For p ≥ 1 , the Young variation becomes the Wiener variation, which was introduced by Wiener in [38]; we will write WBV p instead of YBV p , var p (x) instead of var p (x) and ‖⋅‖ WBV p instead of ‖⋅‖ YBV p . It is easy to check that in those special cases, |x| YBV p = ( var p (x)) 1∕p for p ≥ 1 ; in particular, To study multiplication operators acting between different spaces of functions of bounded Young variation, we need to introduce a relation between Young functions. For two Young functions and , we will write ≺ if and only if (see [13,Section 6]). Equivalently, ≺ if and only if Although the following embedding result is known in the literature (see, for example, [21, Theorem 4.1.1]), the estimates on the embedding constant are often omitted. Therefore, for readers' convenience, we will provide its short proof.
Proof Fix x ∈ YBV . Notice that we may assume that |x| YBV > 0 , since otherwise x is a constant function and x ∈ YBV with ‖x‖ YBV = ‖x‖ YBV . Set ∶= max{2T −1 ‖x‖ ∞ , �x� YBV , d�x� YBV } , where is an arbitrary number greater than 1. If 0 = t 0 < … < t n = 1 is a finite partition of the interval [0, 1], then to obtain the second inequality one has to consider two cases: d ∈ (0, 1] and d > 1 . Therefore, var (x∕ ) ≤ 1 , and so x ∈ YBV . Moreover, in view of the arbitrariness of and the continuity of the max function, this also implies that The proof is complete. ◻

Remark 6
The estimate for the embedding constant appearing in Proposition 5, in general, may be not optimal. For example, it is easy to check that the condition (1) is satisfied with = d = T = 1 for (t) = t 2 and (t) = t , and so we get the estimate 1 ≤ c ≤ 3 . However, it can be proved that ( var q (x)) 1∕q ≤ ( var p (x)) 1∕p for x ∈ WBV p , where 1 ≤ p ≤ q (see, for example, [6,Proposition 1.38], [20,Remark 2.5] or [29, p. 55]), which means that WBV p ↪ WBV q with the embedding constant c = 1.
Among all Young functions especially important are those which satisfy a certain growth condition; namely, the so-called 2 -condition. We say that the Young function It can be shown that to every Young function satisfying the 2 -condition one can assign the non-decreasing function Λ ∶ (0, +∞) → [1, +∞) given by (see [6, p. 115]); for simplicity, we also extend the function Λ over the whole nonnegative half-axis putting Λ(0) ∶= 0 . The significance of Young functions satisfying the 2

Functions of bounded Riesz variation
The second type of variation we are going to deal with in this paper is the Riesz variation. It was introduced in 1910 by Frigyes Riesz (see [35]), and its definition reads as follows. By RBV p , we will denote the space of all functions x ∶ [0, 1] → ℝ with bounded Riesz variation, that is, It can be proved that RBV p is a Banach space when endowed with the norm (cf. [6, Proposition 2.51]). As in the case of the Young variation, the definition of the norm ‖⋅‖ RBV p is meaningful, since it is easy to show that each function x ∈ RBV p is bounded.
In the sequel, we will frequently use the following result characterizing functions of bounded Riesz variation which was first proved in [35] (see also [6,Theorem 3.34]).
proper subset of [0, 1] is such an example). However, it still holds for the class of all absolutely continuous functions which is a subclass of BV; of course, then var R 1 (x) reduces to var (x)-cf. Remark 4.
Using the Riesz theorem we can also prove some embedding results for spaces of functions of bounded Riesz variation.
Proof If p = q , then the claim is trivial. So, we may assume that 1 ≤ q < p . The proof follows from Theorem 9, Remark 10 and the well-known estimates between L p -norms holding for any y ∈ L p (see [24,Theorem 13.17]). ◻

Banach algebras
A natural habitat for multiplication operators is algebras. Let us recall that an algebra X is called a normed algebra if it is a normed space with a norm ‖⋅‖ satisfying the estimate of the form ‖xy‖ ≤ ‖x‖‖y‖ for all x, y ∈ X . If, additionally, the norm ‖⋅‖ is complete, then X is called a Banach algebra. Some authors assume also that a Banach algebra X must contain a unit e, that is, an element of norm 1 such that xe = ex = x for all x ∈ X (cf., for example, [36,Part III]). It turns out that the spaces 2 YBV and RBV p are Banach algebras when endowed with the norms ‖⋅‖ YBV and ‖⋅‖ RBV p , respectively (as a unit we take the constant function e ∶ [0, 1] → ℝ given by e(t) = 1 for t ∈ [0, 1] ). This result was first proved by Maligranda and Orlicz (see [29,Theorems 2 and 3]). The main ingredient in the proof are the estimates of the form and which hold for any functions x, y belonging to an appropriate space. It is worth noting here that the formula (2.91) in [6], which reads as follows is in general incorrect. To see this let us take a look at the following example.

Multiplier spaces
The main objects of our study are multiplication operators and their properties. If X and Y are linear spaces of real-valued functions defined on the interval [0, 1], and g ∶ [0, 1] → ℝ is a given function, then the multiplication operator M g ∶ X → Y , generated by the function g, is given by In order to guarantee that M g is welldefined, we have to make sure (by imposing appropriate conditions on g) that M g (X) ⊆ Y , that is, the product gx must belong to Y, whenever x belongs to X. If we write The set Y/X is often called the multiplier class of Y over X (or simply, the multiplier class, when the starting and target spaces are known). Notice that Y/X is a non-empty set, as it always contains the zero function.
In the recent paper [13] multiplier classes of various BV-type spaces were characterized. Let us briefly recall a few of those results: where S c denotes the set of real-valued functions which are zero everywhere on [0, 1] except at countably many points. For completeness, let us also add that the proofs of the above-mentioned characterizations of multiplier classes presented in [13] contain some minor flaws, which can be fixed with almost no effort.

Function-theoretic properties
We are going to start our investigations by giving general criteria for injectivity, surjectivity and bijectivity of the multiplication operator M g ∶ X → Y.
Since M g is a linear operator, we immediately obtain a criterion for injectivity. The above criterion is too broad to be useful, but it shows that the injectivity of M g in general does not only depend on g but also on X. In some cases, namely if the space X is sufficiently "large," it turns out that the dependency on X is redundant. To make our considerations as general as possible, let us state the following somewhat technical definition.

Definition 14
We say that a linear space X of real-valued functions defined on [0, 1] • separates points if for each t ∈ [0, 1] there is some x ∈ X such that x(t) ≠ 0, • strongly separates points if X contains all characteristic functions of singletons, • uniformly separates points if X ⊆ C and if for each t ∈ [0, 1] and each > 0 there is

Remark 15
Note that each space which separates points uniformly or strongly also separates points. Other relations, however, do not hold. For instance, the spaces C, B, BV, WBV p , YBV and RBV p separate points. On the other hand, the spaces B, BV, WBV p and YBV separate points strongly, but not uniformly, whereas the spaces C and RBV p (with 1 < p < +∞ ) separate points uniformly, but not strongly. Finally, the space of constant functions defined on the interval [0, 1] only separates points, but neither strongly nor uniformly. One of the reasons that this space cannot separate points either strongly or uniformly is that it is one-dimensional, and spaces which strongly/uniformly separate points are necessarily infinite-dimensional. Now, we are ready to prove the injectivity criterion for multiplication operators, which associates the injectivity of a given operator M g with the number of zeros of its generator g.

Theorem 16 Let X, Y be two linear spaces of real-valued functions defined on the interval
[0, 1] and let M g ∶ X → Y be the multiplication operator generated by g ∈ Y∕X .
(a) If X strongly separates points (especially, if X is one of the spaces BV, WBV p or YBV ),

Remark 17
Before we proceed to the proof of Theorem 16 let us notice that, since supp (g) is a closed subset of [0, 1], there are in fact only two possibilities in (b): either dim Ker M g = 0 , or dim Ker M g = +∞ , depending on whether supp (g) = [0, 1] or not. 16 We begin with the proof of (a). First, we show that dim Ker M g ≥ #Z g . Clearly, we may assume that Z g ≠ ∅ . Fix any n ∈ ℕ such that n ≤ #Z g . Then, there exist n (distinct) points t 1 , … , t n ∈ [0, 1] such that g(t i ) = 0 for i = 1, … , n . So,

Proof of Theorem
As the set {x 1 , … , x n } is linearly independent, this means that dim Ker M g ≥ n . Since this is true for any number n ≤ #Z g , we get dim Ker M g ≥ #Z g . Now, we will prove the opposite inequality. This time we may assume that #Z g < +∞ . If #Z g = 0 , that is supp (g) = [0, 1] , then clearly dim Ker M g = 0 . Suppose now that #Z g = n for some n ∈ ℕ . Then, we can write Z g = {t 1 , … , t n } for some distinct points here and throughout the paper by lin A we denote the linear span (or hull) of the set A. So, dim Ker M g ≤ n = #Z g . This ends the first part of the proof.
The proof of (b) is slightly different from the above one. Suppose that there is a sequence (t n ) n∈ℕ of elements of the support of g which converges to t. Then, clearly, x(t n ) = 0 for each n ∈ ℕ . But the function x is continuous by definition, and so x(t) = 0 . Consequently, x = 0 , which means that dim Ker M g = 0.
Now, let us assume that supp (g) is a proper subset of [0, 1], and let us fix n ∈ ℕ . Since supp (g) is closed in [0, 1], we can find n points t 1 , … , t n together with open and pairwise disjoint subsets U 1 , … , U n of [0, 1] such that t i ∈ U i and U i ∩ supp (g) = � for i = 1, … , n . As X uniformly separates points, this means that there exist n continuous functions As this is true for any positive integer n, we get dim Ker M g = +∞ . In view of Remark 17, this ends the proof. ◻ Note that there is no simple analogue of the above result in the case when X is assumed to separate points, but neither strongly nor uniformly.

Example 18
To see this it suffices to take X to be the space of all constant real-valued func- The following corollary for the spaces of our interest follows immediately from Theorem 16.

Corollary 19
(a) Let X be any of the spaces BV, WBV p or YBV and let g ∈ X . Then,

3
Our next task is to characterize surjectivity of multiplication operators. We begin with some abstract results. It should not come as a surprise that this time we will need to assume some additional conditions on the target space.

Theorem 20 Let X, Y be two linear spaces of real-valued functions defined on the interval
As M g is surjective, we can find some x ∈ X such that M g (x) = y . In particular, g(t) ≠ 0 , and consequently g(t) ≠ 0 for all t ∈ [0, 1] as t was arbitrary. This shows that supp (g) = [0, 1] . In particular, the function 1/g is well-defined. To prove the second condition, note that for any y ∈ Y we can again find x ∈ X such that gx = y . Hence, y∕g = x ∈ X . This proves that 1∕g ∈ X∕Y.
For the converse assume that supp (g) = [0, 1] and 1∕g ∈ X∕Y . For y ∈ Y the function x ∶= y∕g belongs to X and satisfies gx = y , i.e., M g is a surjection.
The fact that surjectivity of M g implies its injectivity is a consequence of the first part of the proof and Proposition 13. ◻ It turns out that if Y = X , then surjectivity of the multiplication operator always implies its injectivity; in other words, we do not need the additional assumption that Y (which in this case coincides with X) separates points. Indeed, let X be any space of real-valued functions defined on [0, 1] and let M g ∶ X → X be a surjective operator generated by a function g ∈ X∕X . Further, suppose on the contrary that there is some x ∈ X with M g (x) = 0 , but x ≢ 0 . That means there is some t ∈ [0, 1] such that x(t) ≠ 0 . Due to the fact that gx ≡ 0 , we get g(t) = 0 . But since M g is assumed to be surjective, we must find some z ∈ X such that M g (z) = x ; in particular, 0 = g(t)z(t) = x(t) ≠ 0 , a contradiction.
In the general case, however, that is, when X and Y do not necessarily coincide, if we do not assume that Y separates points, it is easy to give an example of a multiplication operator M g ∶ X → Y which is surjective but not injective.
Then, Y cannot separate points as for any t ∈ (0, 1] and any y ∈ Y we have y(t) = 0 . Moreover, M g cannot be injective, because for the two constant functions From Theorem 20 we obtain two corollaries for the BV-type spaces we are interested in.

Corollary 22 Let X be a linear space of real-valued functions defined on the interval
Proof Let us assume that M g is surjective. Then, from Theorem 20 we obtain that supp (g) = [0, 1] and 1∕g ∈ X∕Y . Note that y ≡ 1 belongs to all of the considered spaces BV, WBV p , YBV , RBV p , and so we obtain X∕Y ⊆ X ⊆ B . Thus, 1/g is bounded, and this is possible Conversely, assume that inf t∈[0,1] |g(t)| > 0 . Then, supp (g) = [0, 1] . It is also easy to see that 1∕g ∈ Y . Since Y is closed under multiplication, Y ⊆ Y∕Y ⊆ X∕Y , and hence 1∕g ∈ X∕Y . Again from Theorem 20, we obtain that M g is surjective. ◻

Corollary 23
Let X be one of the spaces BV, WBV p , YBV or RBV p and let M g ∶ X → X be the multiplication operator generated by g ∈ X . Then, the following conditions are equivalent: As functions in X are bounded, passing with n → +∞ in the above inequality yields 1 Let us also take a look at the following qualitative version of Theorem 20.

Theorem 24 Let X, Y be two linear spaces of real-valued functions defined on the interval
[0, 1]. Assume that X separates points strongly or uniformly (in particular, X is one of the spaces BV, WBV p , YBV and RBV p ). Moreover, let M g ∶ X → Y be the multiplication operator generated by g ∈ Y∕X . Then, dim Im M g = # supp (g).
Proof First, we show the inequality dim Im M g ≥ # supp (g) which is obviously true for g ≡ 0 . Thus, we assume that g ≢ 0 , which implies that # supp (g) ≥ 1 . Fix n ∈ ℕ with # supp (g) ≥ n . Then, we can find n distinct numbers t 1 , On the other hand, if X separates points uniformly, then we can find n continuous functions , n} ; and we set y j = gx j . Now, for j = 1, … , n , let j ∈ ℝ be so that ∑ n j=1 j y j ≡ 0 . By evaluating this equation at each t = t k , we get that This implies that k = 0 for 1 ≤ k ≤ n . Thus, {y 1 , … , y n } is a linearly independent subset of Im M g ; in particular, dim Im M g ≥ n . Since this is true for each n such that # supp (g) ≥ n , we obtain dim Im M g ≥ # supp (g).
In order to show the remaining inequality dim Im M g ≤ # supp (g) , we may assume that # supp (g) < +∞ , because otherwise this inequality is clearly true. Moreover, if supp (g) = � , then dim Im M g = 0 . Hence, we may assume that n = # supp (g) for some positive integer n and write supp (g) = {t 1 , … , t n } . Since X separates points strongly/ uniformly, we can find n functions x 1 , … , x n in X such that x j (t j ) = 1 for j = 1, … , n and supp (x i ) ∩ supp (x j ) = � for i ≠ j . Define y j ∶= gx j . Let y ∈ Im M g . Then, there is some x ∈ X such that y = gx . Moreover, y = gx = ∑ n j=1 x(t j )y j , which shows that the linear span of {y 1 , … , y n } contains Im M g ; in particular, dim Im M g ≤ n = # supp (g) . This completes the proof. ◻

Remark 25
Note that we cannot drop the phrase "strongly or uniformly" in Theorem 24. For instance, let X be the space of constant functions and let Y = C . Consider

Topological properties
We now turn to analytic properties of the multiplication operators M g ∶ X → Y . Here, we are particularly interested in continuity and compactness for X and Y being one of the spaces BV, WBV p , YBV or RBV p .

Continuity
Recall that for a linear operator L ∶ X → Y between two normed spaces (X, ‖⋅‖ X ) and Although multiplication operators are one of the simplest operators one can imagine, they are not always bounded. In particular, in [8] the authors remarked (see page 106) that multiplication operators in Köthe spaces are (well-defined) and continuous if and only if they are generated by an essentially bounded function (we refer to the paper [22] for more information on the boundedness of multiplication operators in such spaces; see also [15]). For readers' convenience we provide yet another example of a discontinuous multiplication operator acting from a linear subspace of C 1 (which is not a Köthe space); here by C 1 we denote the space of all real-valued continuously differentiable functions defined on the interval [0, 1].

Example 26
Consider the space C 1 0 ∶= {x ∈ C 1 |x(0) = 0} equipped with the norm ‖⋅‖ ∞ , and the function g ∶ [0, 1] → ℝ defined by Then, g ∈ L 1 ∕C 1 0 , because for x ∈ C 1 0 the function gx is almost everywhere equal to the continuous (and thus Lebesgue integrable) function y ∶ [0, 1] → ℝ given by Consequently, the operator M g ∶ C 1 0 → L 1 is well-defined. However, M g is not bounded. To see this let us consider the functions x n ∶ [0, 1] → ℝ , where n ∈ ℕ , defined by It is easy to check that x n ∈ C 1 0 and ‖x n ‖ ∞ = 1 for every n ∈ ℕ . But which means that the sequence M g (x n ) n∈ℕ is unbounded in L 1 . Thus, M g cannot be continuous.
On a more positive note, we have the following simple, but quite general, result, which we will use to prove some norm estimates for the multiplication operator acting in the BV spaces.

Proposition 27
Let (X, ‖⋅‖ X ) and (Y, ‖⋅‖ Y ) be two normed spaces of real-valued functions defined on [0, 1] and assume that the constant function e ≡ 1 is contained in X with ‖e‖ X = 1 . Moreover, let M g ∶ X → Y be the multiplication operator generated by a function g ∈ Y∕X . Then, the following statements hold.
Proof Note that since e ∈ X , we have g = M g (e) ∈ Y , which ensures that the quantity ‖g‖ Y appearing in (a) and (b) makes sense. The proof of (a) is obvious, because Now, let us move to (b). If X ↪ Y with the embedding constant c > 0 (i.e., ‖x‖ Y ≤ c‖x‖ X for all x ∈ X ) and if Y is a normed algebra, then shows (b) and completes the proof.

Remark 28
Observe that Example 26 does not contradict Proposition 27, because e ≡ 1 is not contained in C 1 0 .
All the BV spaces considered in this paper are Banach algebras and contain the constant function e ≡ 1 , which has norm 1. Therefore, in the special case when X and Y coincide and are one of our BV spaces, we have the following corollary.

Corollary 29
Let X be one of the spaces BV, WBV p , YBV or RBV p and let M g ∶ X → X be the multiplication operator generated by a function g ∈ X . Then, the operator M g is bounded and x n (t) = 2nt − n 2 t 2 for 0 ≤ t ≤ 1∕n, 1 for 1∕n < t ≤ 1.
Now, we would like to study the continuity of the multiplication operator acting between different spaces of functions of bounded variation. We start with the Young variation and a technical lemma.

Lemma 30
Assume that ∶ [0, +∞) → [0, +∞) is a given Young function. Moreover, let x ∶ [0, 1] → ℝ be a real-valued function with countable support and let > 0 . Then, Before we provide a proof of Lemma 30 one remark is in order: throughout the paper, we adopt a useful convention that summing over an empty set always gives zero.

Proof of Lemma 30
Of course, we may assume that supp (x) ≠ � , because otherwise there is nothing to prove. First, we will show the left inequality. Let 0 ≤ t 1 < … < t n ≤ 1 be arbitrary n points in the support of x. Moreover, if # supp (g) = 1 , take any s 1 ∉ supp (g) . Similarly, if # supp (g) ≥ 2 , take any n points s 1 , … , s n not belonging to supp (x) such that s i ∈ (t i , t i+1 ) for i = 1, … , n − 1 and s n ∈ (s n−1 , t n ) ; this is clearly possible as the support of x is countable. Then, So now, let ⊀ . Then, YBV ∕YBV = YBV ∩ S c . Clearly, we may assume that supp (g) ≠ � ; then, in particular, |g| YBV > 0 . Fix a nonzero function x in YBV and let ∶= 2 ‖x‖ ∞ �g� YBV , where > 1 . Applying Lemma 30, we get Thus, |M g (x)| YBV ≤ . In view of the arbitrariness of , we obtain ‖M g (x)‖ YBV ≤ ‖x‖ ∞ ‖g‖ ∞ + 2‖x‖ ∞ �g� YBV ≤ 2‖x‖ YBV ‖g‖ YBV , and so ‖M g ‖ YBV →YBV ≤ 2‖g‖ YBV . The other inequality follows from Proposition 27 (a). ◻ Observe that from the proof it follows that for the constant c in Theorem 31 we have: c ≤ 2 if ⊀ and c ≤ max{1 + 2T −1 , , d } when ≺ , where the constants , d, T appear in (1). In some cases, however, it is possible to give optimal estimates on the constant c. Let us look at one such situation.
Then, the multiplication operator M g ∶ WBV p → WBV q generated by a function g ∈ WBV q is continuous and Proof We use Proposition 27, but this time instead of the general embedding result, that is Proposition 5, we use the fact that WBV p ↪ WBV q with the embedding constant c = 1 (cf. Remark 6). ◻ Finally, we will deal with the continuity of multiplication operators acting in the spaces of functions of bounded Riesz variation. This time, however, the situation (and the proofs) will be much simpler.
Theorem 33 Let 1 ≤ p, q < +∞ and let M g ∶ RBV p → RBV q be the multiplication operator generated by a function g ∈ RBV q ∕RBV p . Then, M g is continuous and Proof First, let us assume that 1 ≤ p < q . Then, RBV q ∕RBV p = {0} . Consequently, M g is the zero operator. In particular, it is continuous and the formula for its norm holds. If, on the other hand, 1 ≤ q ≤ p , then the proof follows from Propositions 11 and 27 and the fact that {\text{RBV}} spaces are Banach algebras. ◻

Remark 34
It is worth noting that other cases that are not covered by Proposition 27 are sometimes also known. For instance, one can show with the help of Hölder's inequality that for g ∈ L pq∕(p−q) , where 1 ≤ q < p < +∞ , the multiplication operator M g ∶ L p → L q is well-defined and continuous with ‖M g ‖ L p →L q = ‖g‖ L pq∕(p−q) .

Spectra
In this short section, we will show how to apply the results established in the previous parts of the paper to characterize various spectra of multiplication operators acting in BV-type spaces. Let L ∶ X → X be a continuous linear operator acting in a real Banach space X. Set The above sets are, respectively, called the spectrum, point spectrum, residual spectrum and continuous spectrum of the operator L. It is well-known that (L) is a disjoint union of p (L) , r (L) and c (L) . (For more information on various spectra of linear operators we refer the reader to, for example, [7,Chapter 1] or [23,Chapter VI].) In the case when X is one of our BV spaces and L is a multiplication operator, then L is continuous by Corollary 29, and we get the following two results.

Theorem 35
Let X be one of the spaces BV, WBV p or YBV , and let M g ∶ X → X be the multiplication operator generated by a function g ∈ X . Then, Proof Notice that I − M g = M −g for any ∈ ℝ ; by a slight abuse of notation, we identify the constant function with its value. To prove the above equalities, we will use Corol- To show that c (M g ) is empty, it suffices to observe that if the range of I − M g is dense in X, then this operator must be bijective by Corollary 23. This, in turn, together with the inverse mapping theorem and Corollary 29, implies that its inverse ( I − M g ) −1 must be continuous on X. And so the conditions " Im ( I − M g ) is dense in X" and " ( I − M g ) −1 is not bounded" cannot be satisfied simultaneously.
Finally, the equality for r (M g ) follows from the fact that (M g ) is a disjoint union of Theorem 36 Let 1 < p < +∞ and let M g ∶ RBV p → RBV p be the multiplication operator generated by a function g ∈ RBV p . Then,

Remark 37
It is worth noting here that the spectral behavior of the multiplication operator acting in RBV p spaces for 1 < p < +∞ is identical to the behavior of the multiplication operator acting in the space of continuous functions C (for more details see [7, Example 1.6]).

Compactness
Now, we turn our attention to studying compactness of multiplication operators. Let us recall that a linear operator L ∶ X → Y between Banach spaces is compact if the image L(B X (0, 1)) of the closed unit ball (or, in fact, any bounded set) in X is a relatively compact subset of Y. Clearly, not every multiplication operator is compact. The simplest example is probably the identity operator on an infinite-dimensional normed space of real-valued functions defined on [0, 1] (cf. also Example 26). One important family of compact operators is the class of operators of finite rank. A continuous operator L ∶ X → Y between Banach spaces is said to be of finite rank if the range Im L is a finite-dimensional subspace of Y. As the properties of compact and finite-rank operators are classical and well-known, we will not dwell on this issue any longer. We refer the reader to, for example, the monograph [30] for more information. Without further ado, let us move to the main topic of this section. A simple rewording of Theorem 24, together with the continuity results from Section 5.1, leads to the following characterization of finite-rank operators for functions of bounded Young variation.
So far, the properties of multiplication operators acting in YBV and RBV p spaces have been similar. This is the first time when the two theories depart slightly from one another.
Theorem 39 Let 1 < p, q < +∞ and let M g ∶ RBV p → RBV q be the multiplication operator generated by a function g ∈ RBV q ∕RBV p . Proof In view of the fact that RBV q ∕RBV p = {0} if q > p , we need to address the case q ≤ p only. According to Theorem 24 the operator M g has finite-dimensional range if and only if # supp (g) < +∞ . But since RBV q ∕RBV p = RBV q ⊆ C for 1 < q ≤ p < +∞ , the support of g consists of at most finitely many elements if and only if it is empty. Thus, if M g ∶ RBV p → RBV q for 1 < q ≤ p < +∞ is of finite rank, then g ≡ 0 . The other implication is obvious. ◻ Note that in the preceding theorem we excluded the situation when either the starting or the target space coincides with RBV 1 . We did this because the nature of the spaces RBV p is different for p > 1 and p = 1 (cf. Remark 10), and it turns out that those two cases need to be treated separately.
Theorem 40 Let 1 ≤ p < +∞ . The multiplication operator M g ∶ RBV p → BV generated by a function g ∈ BV has finite rank if and only if # supp (g) < +∞.
Proof The proof follows from Theorem 24 and the fact that the multiplication operator M g ∶ RBV p → BV , where 1 ≤ p < +∞ , is continuous (see Theorem 33). ◻

Remark 41
Let us explain why we did not study the multiplication operator M g ∶ BV → RBV p for 1 < p < +∞ in the above theorem. The reason is simple. Such an operator must be generated by the zero function (cf. Section 3), so it has trivially finite rank.
Now, let us move to the study of compactness. We begin with abstract results providing a necessary condition for a multiplication operator to be compact.

Proposition 42 Let X be a normed space of real-valued functions defined on [0, 1] which strongly separates points and in which the set of all characteristic functions of singletons is bounded. Moreover, let Y be another normed space of real-valued functions defined on
[0, 1] such that Y ↪ B . If the multiplication operator M g ∶ X → Y , generated by a function g ∈ Y∕X , is compact, then supp (g) is countable.
Proof Suppose on the contrary that M g ∶ X → Y is compact but supp (g) is not countable. This implies that for some > 0 there is a sequence (t n ) n∈ℕ of distinct points in the interval [0, 1] such that |g(t n )| ≥ for all n ∈ ℕ (cf. Section 2.2). The functions x n ∶= {t n } form a bounded sequence in X, but for m, n ∈ ℕ with m ≠ n we have where the positive constant c is such that ‖y‖ ∞ ≤ c‖y‖ Y for all y ∈ Y . Hence, (M g (x n )) n∈ℕ cannot have a subsequence converging in Y, and thus M g cannot be compact-contradiction. ◻ It turns out that the necessary condition described in Proposition 42 is also a sufficient one in many situations. However, before we will be able to prove this in the case of the spaces of functions of bounded Young variation we need the following technical lemma.

Lemma 43
Proof We may clearly assume that supp (x) ≠ � as otherwise the claim is trivial. (Let us recall that in this case the quantity Λ(‖x‖ ∞ ) is also meaningful since we put Λ(0) ∶= 0 .) If 0 = 0 < … < n = 1 is an arbitrary finite partition of the interval [0, 1], then reasoning as in the proof of Lemma 30 we can show that for any j ∈ ℕ and use the fact that Λ is non-decreasing. ◻ Now, we are in position to prove a characterization of those multiplication operators acting in the spaces of functions of bounded Young variation which are compact. Note that we will require the -function of the target space to satisfy the 2 -condition. Proof Note that YBV ∕YBV = YBV ∩ S c if ⊀ . This, together with Proposition 42, implies that we need to show only that the countability of supp (g) guarantees the compactness of M g . We will prove both cases ≺ and ⊀ simultaneously. If supp (g) is finite, then M g has finite rank by Theorem 38, and hence is compact. On the other hand, if supp (g) is infinite, we can write E ∶= {t 1 , t 2 , t 3 , …} = supp (g) ⊆ [0, 1] . Setting E n ∶= {t 1 , t 2 , … , t n } , we see that the functions g n ∶= E n g have finite support and thus belong to YBV ∩ S c . By Theorem 38 the operators M g n ∶ YBV → YBV have finite rank, and hence are compact. To end the proof it suffices now to show that M g n → M g with n → +∞ in the operator norm. But, in view of Theorem 31, there exists a constant c ≥ 1 such that ‖g n − g‖ YBV ≤ ‖M g n − M g ‖ YBV →YBV ≤ c‖g n − g‖ YBV for all n ∈ ℕ . So, equivalently, we need to show that ‖g n − g‖ YBV → 0 as n → +∞.
This shows that the operator M g is compact and ends the proof. ◻ Remark 45 It is worth noting here that from the proof of Theorem 44 it follows that each compact multiplication operator between YBV and YBV with ∈ 2 is the limit of a sequence of finite-rank multiplication operators.
The following example shows that in general the requirement ∈ 2 cannot be dropped. The idea to use the Young function given by (5) comes from Example 2.3 in [6]. There, Appell et al. used the same function to illustrate that without the 2 -condition the set {x ∶ [0, 1] → ℝ| var (x) < +∞} is not linear. Also, the function g below and f in [6] are similar. However, for readers' convenience we decided to provide all the details, not only those connected with the lack of compactness of the multiplication operator generated by g.

Example 46
Let us consider the Young function ∶ [0, +∞) → [0, +∞) given by It can be easily checked that does not satisfy the 2 -condition. Furthermore, let g ∶ [0, 1] → ℝ be defined by Using Lemma 30 we obtain Hence, g ∈ YBV ∩ S c . In particular, the multiplication operator M g ∶ YBV → YBV is well-defined and continuous. However, as we are going to show, it is not compact. Consider the sequence (x n ) n∈ℕ , where x n ∶= 1 4 (0, 1 n ) for n ∈ ℕ . It is not hard to check that x n ∈ YBV and |x n | YBV ≤ 1 2 (and thus ‖x n ‖ YBV ≤ 3 4 ). Now, let us suppose that (M g (x n )) n∈ℕ has a subsequence (M g (x n k )) k∈ℕ convergent to some y ∈ YBV . Since the norm convergence in YBV is stronger than the uniform convergence and the sequence (x n ) n∈ℕ converges pointwise to zero on the interval [0, 1], the sole candidate for y is the zero function. Let K ∈ ℕ be such that ‖M g (x n k )‖ YBV < 1 4 for all k ≥ K . Then, it is not difficult to check that var (4gx n k ) ≤ 1 for all k ≥ K . As var (4gx n k ;[0, 1]) ≥ var (4gx n k ;[0, 1∕n k ]) , this means that var (4gx n k ;[0, 1∕n k ]) ≤ 1 for all k ≥ K ; here by var (f ;[a, b] ln n , if t = 1 n for some n ∈ ℕ with n ≥ e 4 , 0, otherwise. The obtained contradiction shows that the sequence (M g (x n )) n∈ℕ does not contain a convergent subsequence. In other words, the multiplication operator M g ∶ YBV → YBV , generated by g, is not compact.
We will end this part with a result providing a sufficient condition for a multiplication operator between general spaces of functions of bounded Young variation to be compact. Naturally, this time, we will not require the -function of the target space to satisfy the 2 -condition. Instead, we will require the generator g of the multiplication operator not to oscillate too much; namely, we will assume that var ( g) < +∞ for each > 0 . Note that there are plenty of such functions; for example, each nonzero function x in BV satisfies this condition, as for any finite partition 0 = t 0 < … < t n = 1 of the interval [0, 1]. Moreover, the condition in question is also satisfied by any g ∈ YBV if ∈ 2 (cf. Proposition 7). Unfortunately, we do not know whether the assumption " var ( g) < +∞ for each > 0 ," besides being sufficient, is also necessary in the general setting.
Proof Note that the assumption g ∈ YBV ∩ S c guarantees that the multiplication operator M g ∶ YBV → YBV is well-defined regardless of whether ≺ or not. Of course, we may additionally assume that g is nonzero, since otherwise there is nothing to prove.
Let (x n ) n∈ℕ be an arbitrary sequence in YBV with elements in the closed unit ball. By Helly's selection theorem (cf. [32,Theorem 1.3]) the sequence (x n ) n∈ℕ has a subsequence (x n k ) k∈ℕ pointwise convergent on [0, 1] to a function x ∈ YBV . We are going to show that ‖M g (x n k ) − M g (x)‖ YBV → 0 as k → +∞ , which would clearly mean that the operator M g is compact. Fix any > 0 and > 0 . By Lemma 30, we have ∑ t∈ supp (g) (4 �g(t)�) ≤ var (4 g) , which in view of the assumption implies that the series ∑ t∈ supp (g) (4 �g(t)�) is (absolutely) convergent. In particular, there is a finite set T ∶= {t 1 , … , t m } ⊆ supp (g) of distinct points such that ∑ t∈ supp (g)⧵T (4 �g(t)�) ≤ 1 2 . Let N ∈ ℕ be such that for all k ≥ N and i = 1, … , m . Note also that supp (gx n k − gx) ⊆ supp (g) for all k ∈ ℕ . Thus, by Lemma 30,for To end the proof it suffices now to show that ‖M g (x n k ) − M g (x)‖ ∞ → 0 . Take any t ∈ supp (g) . As supp (g) is countable, and in particular Z g ≠ ∅ , we get If t ∉ supp (g) , the above inequality is trivially satisfied. Thus, which, in view of the first part of the proof, shows that ‖M g (x n k ) − M g (x)‖ ∞ → 0 as k → +∞ . This completes the proof. ◻ For the Riesz spaces RBV p we have a result of a similar (yet distinct) flavor. Since each function in RBV p for 1 < p < +∞ is continuous, compactness of M g leads to a stronger degeneracy. Notice the resemblance of Theorem 39 and the following theorem. However, now the proof will require a bit more work and a compactness result proved recently by Bugajewski and Gulgowski in [14], which says that if a non-empty subset A of BV is relatively compact, then it is equivariated, meaning that for each > 0 there is a finite partition Theorem 48 Let 1 < p, q < +∞ and let M g ∶ RBV p → RBV q be the multiplication operator generated by a function g ∈ RBV q ∕RBV p .
(a) If q > p , then M g is always compact (as the zero operator).
Proof Since RBV q ∕RBV p = {0} for 1 < p < q , we need only to prove the necessity part of (b). So, let M g ∶ RBV p → RBV q be a compact operator and let us assume that 1 < q ≤ p . Then, in particular, RBV q ∕RBV p = RBV q . Suppose now that g is not identically zero. Since it is continuous, there must exist some interval [a, b] ⊆ [0, 1] of positive length and a positive constant such that |g(t)| > for all t ∈ [a, b] . For any fixed n ∈ ℕ let s k ∶= a + k 2n ⋅ (b − a) for k = 0, 1, … , 2n and define x n ∶ [0, 1] → ℝ to be the real-valued function whose graph is a simple polygonal line with nodes at the points (0, 0), (1, 0), (s k , 0) for k ∈ {0, … , 2n} even, and (s k , 1 2n ) for k ∈ {0, … , 2n} odd. It can be easily checked that x n ∈ RBV p with ‖x n ‖ RBV p = 1 2n + (b − a) 1∕p−1 for n ∈ ℕ , meaning that the set A ∶= {x n |n ∈ ℕ} ⊆ RBV p is bounded. In view of our assumption, this implies that M g (A) var ).
is a relatively compact subset of RBV q . Since RBV q is continuously embedded into RBV 1 = BV (see Proposition 11), M g (A) is a relatively compact subset of BV. Therefore, M g (A) is equivariated, that is, for each > 0 there exists a finite partition for every n ∈ ℕ . In particular, for = 1 8 , we can find a finite partition 0 = t 0 < … < t l = 1 such that for every positive integer n ≥ 8l −1 ‖g‖ ∞ we have But, for any n ∈ ℕ , we have which leads to a contradiction. Thus, g is identically equal to zero. ◻ In the special cases when p = q , it is also possible to give another proof of Theorem 48, which does not require any knowledge about (relatively) compact subsets of BV; for more details consult [33].
We end this subsection with a compactness criterion for multiplication operators acting from the space RBV p for 1 < p < +∞ into BV.
Theorem 49 Let 1 < p < +∞ and let M g ∶ RBV p → BV be the multiplication operator generated by a function g ∈ BV . Then, M g is compact if and only if supp (g) is countable; moreover, M g is then the limit of finite-rank multiplication operators acting between RBV p and BV.
Proof A reasoning similar to the one used in the proof of Theorem 44 shows that if supp (g) is countable then either M g has finite rank (if supp (g) is finite), or is the limit of finite-rank operators (if supp (g) is infinite). In either case, M g is compact.
Now, let us assume that g ∈ BV and that the multiplication operator M g ∶ RBV p → BV is compact. If supp (g) were uncountable, then it would contain a point of continuity of g (because the set of points of discontinuity of a function of bounded Jordan variation is countable). In particular, there would exist an interval [a, b] ⊆ [0, 1] of a positive length and a constant > 0 such that |g(t)| ≥ for all t ∈ [a, b] . Now, it remains to repeat the reasoning presented in the first proof of Theorem 48 to obtain a contradiction. Thus, supp (g) must be countable. ◻

Measure of non-compactness and essential norm
In the previous section, we studied conditions guaranteeing compactness of multiplication operators in certain BV-type spaces. Now, we would like to look at the same problem from a more qualitative point of view. To estimate how far an operator is from being compact, one

3
can use, for example, measures of non-compactness. Let us recall that the measure of noncompactness (or, the -norm) [L] of a bounded linear operator L ∶ X → Y between Banach spaces is given by the formula here X and Y denote the Kuratowski measure of non-compactness in X and Y, respectively. (In the sequel we will be omitting the subscripts X and Y and simply write , as the spaces involved will always be clear from the context.) Although various measures of noncompactness are widely known and used throughout nonlinear analysis, for readers' convenience let us recall the definition of the measure . If A is a bounded subset of a Banach space (or, in general, a metric space) X, then its Kuratowski measure of non-compactness is given by the formula (see [28]). For basic properties of the index we refer the reader to [2]. It is well-known that [L] ≤ ‖L‖ X→Y and that the operator L is compact if and only if [L] = 0 . Furthermore, it can be shown that Other properties of the -norm of a bounded linear operator with some illustrative examples can be found in, for example, [5] (cf. also [7, Section 1.2]). Now, let us state and prove the main result of this section concerning the lower bound for [M g ] in the case of the YBV spaces.

Theorem 51 Let
, ∶ [0, +∞) → [0, +∞) be two Young functions and let M g ∶ YBV → YBV be the multiplication operator generated by a function g ∈ YBV ∕YBV . Then, Proof Notice that we may assume that the set { > 0|# supp (g) = +∞} is nonempty, since otherwise there is nothing to prove. So, let > 0 be an arbitrary number such that the set supp (g) is infinite. Then, there exists a sequence (t n ) n∈ℕ in (0, 1) of distinct points with the property that |g(t n )| > for n ∈ ℕ . Let x n ∶= {t n } and A ∶= {x n |n ∈ ℕ} . Clearly, A ⊆ YBV . Moreover, by Lemma 30 applied with . Let us also note that ‖x n − x m ‖ YBV ≥ ‖x n − x m ‖ ∞ = 1 for distinct n, m ∈ ℕ . Consequently, the sequence (x n ) n∈ℕ does not contain a Cauchy subsequence, and so the set A cannot be relatively compact in YBV . In other words, (A) > 0.
Now, let us suppose that there exists a finite covering of A with sets of diameter less than or equal to and for simplicity let us denote the quantity on the right-hand side of the above inequality by . This means that there is a finite collection of subsets B 1 , … , B k of YBV such that M g (A) ⊆ B 1 ∪ … ∪ B k and diam B l < for l = 1, … , k . Since there are only finitely many sets B l and all the elements of M g (A) are distinct, one of those sets (say, B 1 ) contains infinitely many distinct elements of M g (A) , and particularly two of them, say M g (x i ) and M g (x j ) . If is a positive number such that then we have , and a contradiction, since we know that diam B 1 < . Therefore, We have thus shown that if > 0 does not belong to the set { > 0|# supp (g) < +∞} , then does not belong to the set of all positive numbers such that (M g (A)) < (A) for every bounded subset A of YBV with (A) > 0 . Consequently, This ends the proof. ◻

Remark 52
It is worth noting that in the special case of the Jordan variation, that is, when (u) = (u) = u , the constant appearing on the left-hand side of (6) equals 1.
To obtain an upper estimate for [M g ] (better than ‖M‖ YBV →YBV ) we will additionally assume that satisfies the 2 -condition. Proof Before we proceed to the main part of the proof, let us recall that functions of bounded Young variation have one-sided limits at each point and countably many points of discontinuity (cf. [31,Theorem 10.9]). Thus, the right and left regularizations of g are well-defined. Moreover, it is fairly easy to check that g r , g l ∈ YBV and max � ‖g r ‖ YBV , ‖g l ‖ YBV � ≤ ‖g‖ YBV . As the functions g, g r and g l differ at at most countably many points, we also have g r , g l ∈ YBV ∕YBV . Now, let h r , h l ∶ [0, 1] → ℝ be given by h r ∶= g − g r and h l ∶= g − g l . Clearly, h r , h l ∈ YBV ∩ S c ⊆ YBV ∕YBV . Therefore, by Theorem 44, the multiplication operators M h r , M h l ∶ YBV → YBV are compact. Using the properties of the Kuratowski measure of non-compactness, for any bounded subset A of YBV we get and similarly (M g (A)) ≤ ‖M g r ‖ YBV →YBV ⋅ (A) . To end the proof it suffices now to apply Theorem 31. ◻ −1 1 2

Remark 54
From the proof of Proposition 53 follows an even stronger version of the estimate (7). Namely, the minimum on the right-hand side of (7) may be replaced by the infimum of all the YBV -norms of functions g * ∈ YBV which differ from g at at most countably many points in [0, 1]. We chose to state Proposition 53 for the left and right regularization of g, because in many cases g r and g l make the function g nicer; for example, they smooth out isolated jumps of g. Moreover, in the special case when ⊀ , that is, when g ∈ YBV ∩ S c , if g(0) = g(1) = 0 , we get g r ≡ g l ≡ 0 , which implies that the multiplication operator generated by g is compact (cf. Theorem 44). Finally, let us note that similar to what we did in Section 5.1 (see Corollary 29 as well as Proposition 32 and the paragraph before it), it is possible to give some estimates on the constant c appearing in (7).
Although in some cases the upper estimate of [M g ] provided in the above proposition gives satisfactory results (cf. Remark 54), in general it seems far from being optimal even when c = 1 . It would be interesting to find the exact formula for [M g ] . A closer look at the results of this and the previous section may suggest that the -norm of M g ∶ YBV → YBV and the quantity inf{ > 0|# supp (g) < +∞} are equivalent, at least in the situation when ∈ 2 . In other words, we conjecture that in such a case there exist positive constants a, b such that It turns out that there are some clues indicating that the above estimate may be true. Let us take a look at two such instances.
Example 55 Let ∶ [0, +∞) → [0, +∞) be a Young function and let (r n ) n∈ℕ be an arbitrary sequence of distinct numbers in (0, 1). Fix a nonnegative number and consider the multiplication operator M g ∶ YBV → YBV generated by the function g ∶ [0, 1] → ℝ given by Note that the operator M g is well-defined as g ∈ YBV . Using a similar approach to the one we used in the proof of Proposition 53 it can be shown that (M g (A)) = (M g r (A)) = (M g l (A)) , and hence (M g (A)) = ( A) = (A) for each bounded subset A of YBV (the functions g − g r and g − g l belong to BV ∩ S c , and so we can use Theorem 47). This implies that [M g ] = . Furthermore, observe that supp (g) = [0, 1] if < . On the other hand, if ≥ , then supp (g) consists of those points r n whose indices n ∈ ℕ satisfy the inequality 1∕n 2 > − . Thus, # supp (g) < +∞ if and only if > , and so inf{ > 0|# supp (g) < +∞} = . In other words, Now, let us move to the second result connected with the conjecture (8); this time, we will assume that we work with functions of bounded Jordan variation only. However, before we will be able to state it let us recall that we call g ∶ [0, 1] → ℝ a step function if and only if there exist a finite collection of (not necessarily distinct) real numbers 1 , … , n , 1 , … , m together with a finite collection of pairwise disjoint subintervals I 1 , … , I n of [0, 1] with non-empty interiors and a finite collection of distinct points g(t) = + 1∕n 2 for t = r n , otherwise.
In the proof of the following result, we will aim at expressing g in the simplest form possible, reducing the number of points t i to the minimum zero.

Proposition 56 If M g ∶ BV → BV is the multiplication operator generated by a step function g, then
Proof The step function g is clearly of bounded Jordan variation, and so the multiplication Moreover, if we take x, y ∈ A k ∩ B , using the properties of the Jordan variation, we obtain let us recall that var (z;I) denotes the (Jordan) variation of the function z over the closed interval I. And so, Therefore, (M g (A)) ≤ | | (A) + . Since the number > 0 was arbitrary, we finally get Openproblem 2 Provide any non-trivial estimates for the measure of non-compactness (and/or the essential norm) of the multiplication operator M g ∶ RBV p → RBV q .

Multiplication operators with closed range
In this section, we are going to discuss conditions guaranteeing that multiplication operators in BV-type spaces have closed range. We begin with a necessary condition and spaces of functions of bounded Young variation.
Note that the condition "there exists a constant > 0 such that |g(t)| ≥ for each t ∈ supp (g) " is satisfied vacuously when supp (g) = � . So, we do not need to exclude the situation when g ≡ 0 from our considerations.

Proof of Theorem 58
Clearly, we may assume that supp (g) ≠ � . Since M g is bounded and has closed range there is a constant c > 0 such that for each y ∈ Im M g there is some ∈ YBV satisfying y = M g ( ) and ‖ ‖ YBV ≤ c‖y‖ YBV (see [1,Corollary 2.15]). Put ∶= (c + c∕ −1 ( 1 2 )) −1 . Let us fix t ∈ supp (g) and consider x ∶= {t} . Clearly, x ∈ YBV . Now, take a function ∈ YBV such that M g (x) = M g ( ) and ‖ ‖ YBV ≤ c‖M g (x)‖ YBV . As t ∈ supp (g) , the equality M g (x) = M g ( ) implies that (t) = 1 , and therefore ) , whence our claim follows. ◻ To obtain necessary and sufficient conditions for the operator M g ∶ YBV → YBV to have closed range we need to distinguish two cases: ⊀ and ≺ . As we will see, the former one is relatively easy to handle, whereas the latter one is more technical.
Proof If # supp (g) < +∞ , then by Theorem 38 the range of the multiplication operator M g is finite-dimensional, and hence closed in YBV (see [3,Lemma 4.9]). Now, let us assume that M g has closed range. By Proposition 58 this means that there is a constant > 0 such that |g(t)| ≥ for all t ∈ supp (g) . Note also that since YBV ∕YBV = YBV ∩ S c when ⊀ , the support of g is countable. If it were infinite, then by Lemma 30 for each > 0 we would have This would, however, contradict the fact that g ∈ YBV . Hence, supp (g) is finite. ◻ If ∈ 2 we can also provide an alternative proof of Theorem 59, which does not use Proposition 58.
Alternative proof of Theorem 59 when ∈ 2 . In view of Theorem 44 the multiplication operator M g is compact. To end the proof it suffices now to apply Theorem 38 and a well-known result in functional analysis saying that a compact linear operator between Banach spaces has closed range if and only if it is of finite rank (see [30, for some > 0 . Moreover, let M g ∶ YBV → YBV be the multiplication operator generated by a function g ∈ YBV ∕YBV . Then, M g has closed range if and only if there exists a constant > 0 such that |g(t)| ≥ for all t ∈ supp (g).

Proof
In view of Proposition 58, we need to prove only the sufficiency part. So let us assume that supp (g) ≠ � (otherwise there is nothing to show) and that there exists > 0 such that |g(t)| ≥ for all t ∈ supp (g) and let (y m ) m∈ℕ be a sequence in Im M g which converges in YBV to a function y. Then, in particular, y(t) = 0 for each t ∈ Z g . Now, let us define x ∶ [0, 1] → ℝ by the formula Our aim is to show that x ∈ YBV . Let 0 = t 0 < … < t n = 1 be an arbitrary finite partition of the interval [0, 1] and let ∶= min , where the number > 0 is such that var ( g) < +∞ and var ( y) < +∞ . (Note that such a number exists since YBV ∕YBV = YBV .) To estimate the sum ∑ n k=1 ( �x(t k ) − x(t k−1 )�) let us consider four cases.
In particular, x ∈ YBV . Since M g (x) = y , this shows that Im M g is closed and ends the proof. ◻ The situation is significantly different when there is a constant > 0 such that This time, however, we will need to assume the 2 -condition.
The proof of Theorem 61 is a somewhat mixture of the proofs of the previous results in this section.
Putting all the pieces together, for any k ∈ ℕ we finally have And, passing to the limit with k → +∞ , yields ∕‖g‖ ∞ ≤ 0 , which is impossible. Thus, supp (g) must be finite. The proof is complete. ◻ Finally, let us discuss closed range multiplication operators in spaces of functions of bounded Riesz variation.
Theorem 62 Let 1 < p, q < +∞ and let M g ∶ RBV p → RBV q be the multiplication operator generated by a function g ∈ RBV q ∕RBV p . Proof Note that (a) is trivial as in this case RBV q ∕RBV p = {0} , and so Im M g = {0}.
Now, let q = p . In this case RBV q ∕RBV p = RBV p . If g = 0 , there is nothing to prove. On the other hand, if supp (g) = [0, 1] , then, since g is continuous, there is a constant > 0 such that |g(t)| ≥ for all t ∈ [0, 1] . This, by Corollary 23 means that M g is surjective, and hence has closed range.
Finally, let us show (c). Clearly, we need only to prove the necessity part. So let us assume that M g has closed range, which implies that there exits a constant c > 0 such that for each y ∈ Im M g there is some ∈ RBV p satisfying y = M g ( ) and ‖ ‖ RBV p ≤ c‖y‖ RBV q (see [1,Corollary 2.15]). However, on the contrary, let us suppose that supp (g) ≠ � . Since RBV q ∕RBV p = RBV q ⊆ C for q < p , there must exist a non-degenerate interval [a, b] ⊆ [0, 1] and a constant > 0 such that |g(t)| ≥ for t ∈ [a, b] . Let us fix a point s ∈ (a, b) and for each n ∈ ℕ let us define x n ∶ [0, 1] → ℝ by (1 + k ) ∕‖g‖ ∞ ≤ ‖ k ‖ YBV ≤ c‖M g (x k )‖ YBV .

Remark 64
The case M g ∶ BV → RBV p for 1 < p < +∞ is trivial since then RBV p ∕BV = {0} , and so M g , as the zero operator, has closed range.

Fredholm multiplication operators
We devote this last section to studying Fredholm multiplication operators in certain BV spaces. Let us recall that a bounded linear operator L ∶ X → Y acting between Banach spaces is called a Fredholm operator if dim Ker L < +∞ , Im L is closed in Y and dim(Y∕ Im L) < +∞ . The index of a Fredholm operator is defined by ind L ∶= dim Ker L − dim(Y∕ Im L) . (For more information on Fredholm operators see [7] or [3, Section 11.6].)

Remark 65
As it is not possible that both the sets Z g and supp (g) are finite, in view of Theorems 16, 59 and 61, there are no Fredholm multiplication operators M g ∶ YBV → YBV , when either ⊀ , or ∈ 2 and for some > 0.
On a more positive note, we have the following result.
Theorem 66 Let , ∶ [0, +∞) → [0, +∞) be two Young functions such that for some > 0 . Moreover, let M g ∶ YBV → YBV be the multiplication operator generated by a function g ∈ YBV ∕YBV . Then, M g is a Fredholm operator if and only if #Z g < +∞ and there exists a constant > 0 such that |g(t)| ≥ for all t ∈ supp (g) ; in addition, in such a case, ind M g = 0.
Proof Note that we need to only prove the sufficiency, because the necessity part follows from Theorems 16 and 60. So, let us assume that #Z g = n for some n ∈ ℕ ∪ {0} and that there exists a constant > 0 such that |g(t)| ≥ for all t ∈ supp (g) . If n = 0 , then supp (g) = [0, 1] , and so the operator M g is an isomorphism of YBV and YBV (cf. the proof of Theorem 60), so it is clearly a Fredholm operator. Now, let us assume that n ≥ 1 , and let Z g = {t 1 , … , t n } . We may assume that those points are arranged in an ascending order, that is, 0 ≤ t 1 < t 2 < … < t n ≤ 1 . From Theorems 16 and 60 we know that dim Ker M g = n < +∞ and that Im M g is closed. It remains to show that dim(YBV ∕ Im M g ) < +∞ . We will show that dim(YBV ∕ Im M g ) = n . To this end we will prove that Im M g coincides with the set {y ∈ YBV |y(t i ) = 0 for i = 1, … , n} . As g(t i ) = 0 for i = 1, … , n , it is clear that Im M g ⊆ {y ∈ YBV |y(t i ) = 0 for i = 1, … , n}. Now, let y ∈ YBV be an arbitrary function which vanishes at the points t i , i = 1, … , n . Define x ∶ [0, 1] → ℝ by Using the same approach as in the proof of Theorem 60 it can be shown that x ∈ YBV . Since y = M g (x) , we see that y ∈ Im M g . This, in turn, implies that {y ∈ YBV |y(t i ) = 0 for i = 1, … , n} ⊆ Im M g .
To end the proof it suffices to note that YBV ∕ Im M g is linearly isomorphic with ℝ n by the map y + Im M g ↦ (y(t 1 ), … , y(t n )) . ◻ We end this section and the whole paper with a remark concerning Fredholm multiplication operators in RBV p spaces.
Remark 67 Let 1 < p, q < +∞ . Note that due to Theorem 62 and the characterization of the multiplier class RBV q ∕RBV p , there are no Fredholm multiplication operators M g ∶ RBV p → RBV q when q ≠ p.
Similarly, there are no Fredholm multiplication operators acting to BV of from BV. The latter case follows from the fact that any multiplication operator M g ∶ BV → RBV p , where 1 < p < +∞ , is the zero operator. To prove the former case, let us notice that if M g ∶ RBV p → BV , where 1 < p < +∞ , is a Fredholm operator generated by g ∈ BV , then it has closed range, and so, by Theorem 63, # supp (g) < +∞ . But this, in view of Theorem 16, implies that dim Ker M g = +∞ and leads to a contradiction (recall that a Fredholm operator has a finite-dimensional kernel).
The problem whether there are Fredholm multiplication operators M g ∶ RBV p → RBV p , where 1 < p < +∞ , other than automorphisms is still open.