Multiplication operators on the Banach algebra of bounded Φ-variation functions on compact subsets of ℂ

Mireya Bracamonte; Jurancy Ereú; Luz Marchan

doi:10.1515/dema-2022-0162

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Multiplication operators on the Banach algebra of bounded Φ-variation functions on compact subsets of ℂ

Mireya Bracamonte , Jurancy Ereú and Luz Marchan

Published/Copyright: November 7, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

Let A Φ ( K ) be the Banach algebra of bounded Φ -variation functions defined on a compact set K in the complex plane, h a function defined on K , and M h a multiplication operator induced by h . In this article, we determine the conditions that h must satisfy for M h to be an operator that has closed range, finite rank or is compact. We also characterize the conditions that h must satisfy for M h to be a Fredholm operator.

Keywords: Φ-variation; multiplication operator

MSC 2010: 26A45; 47B07

1 Introduction and preliminaries

Bounded variation functions represent an area of functional analysis on which research has been carried out for some decades and continues to generate scientific concerns and applications in modern disciplines such as probability, statistics, and economics or in more traditional disciplines such as differential equations or mathematical physics. In [1], we have proven that A Φ ( K ) , the space of bounded Φ -variation functions defined on a compact K of the complex plane C is a Banach Algebra. This last result combined with a particular type of linear operators such as multiplication operators constitutes the central topic of this article, characterizing these operators through the conditions under which they act on A Φ ( K ) .

As in [1], throughout this article, K will denote a compact subset of the complex plane C , if z , z ′ ∈ C , then [ z , z ′ ] denotes the line segment joining z and z ′ .

Definition 1.1

(See [1,2,3]). Let ξ = [ z 0 , z 1 , … , z n ] be an ordered finite list of elements of C (not necessarily different), where n ≥ 1 and suppose that ℓ is a line on the plane. The line segment [ z i , z i + 1 ] is a crossing segment of ξ on ℓ if any one of the following holds:

z i and z i + 1 lie on (strictly) opposite sides of ℓ .
i = 0 and z i ∈ ℓ .
i > 0 , z i ∈ ℓ and z i − 1 ∉ ℓ .
i = n − 1 , z i ∉ ℓ and z i + 1 ∈ ℓ .

As in [1,4], we shall denote by N the set of all functions Φ : [ 0 , + ∞ ) → [ 0 , + ∞ ) such that Φ is unbounded, continuous, convex, and nondecreasing with Φ ( 0 ) = 0 .

Definition 1.2

(See [1]). Let f : K → C and ξ = [ z 0 , z 1 , … , z n ] be an ordered finite list of element of K , and the curve variation of f on the set S is

cvar Φ ( f , ξ ) = ∑ i = 1 n Φ ( ∣ f ( z i ) − f ( z i − 1 ) ∣ ) ,

and the variation of f is defined to be

(1.1) var Φ ( f , K ) = sup ξ cvar Φ ( f , ξ ) vf ( ξ ) ,

where the supremum is taken over all finite ordered lists of elements of K , vf ( ξ ) = max ℓ vf ( S , ℓ ) and vf ( S , ℓ ) is the number of crossing segments of S on ℓ .

In [1], it is shown that the class of functions defined by

A Φ ( K ) = f : K → C : var Φ f λ , K < ∞ for some λ > 0

is a Banach algebra under pointwise operations and having norm

‖ f ‖ A Φ ≔ ‖ f ‖ ∞ + ρ Φ ( f ) , where ρ Φ ( f ) ≔ inf λ > 0 : var Φ f λ , K ≤ 1 .

As simple examples of functions found in this algebra are the constant functions defined on any compact subset of C , as well as the function χ w : K → C , the characteristic function defined by

χ w ( z ) = 1 if z = w 0 otherwise ,

with var Φ ( χ w , K ) ≤ 4 Φ ( 1 ) .

Consequently, if S is a finite set, the function

χ S ( z ) = 1 if z ∈ S 0 otherwise

can be seen as a finite sum of functions in the space; thus, χ S ∈ A Φ .

Throughout this article, ideals in this algebra are studied, and to carry out this study, the following result will be useful.

Theorem 1.3

[1]. Let f ∈ A Φ ( K ) , then,

If inf z ∈ K ∣ f ( z ) ∣ > 0 , then 1 f ∈ A Φ ( K ) .
If inf z ∈ K ∣ f ( z ) ∣ = 0 , then 1 f ∉ A Φ ( K ) .

2 Multiplication operator

When studying operators, it is always interesting to know if these operators are of closed rank, especially in the case of Fredholm operators. Therefore, it is always natural to try to characterize multiplication operators with closed ranges.

Let G , H be complex Banach spaces, and ℬ ( G , H ) the algebra of all bounded linear operators from G into H , we shall denote by ker ( T ) the kernel of T . In the particular case, where G = H , we will write ℬ ( H ) instead of ℬ ( H , H ) .

Also, for vector spaces G , H of K -valued functions on a set S , let T be an operator from G into H , and let h be a K -valued function on S . T is said to be the operator M h induced by h if

M h ( f ) ≔ h f = T ( f ) ∈ H

for all f ∈ G . The function h is called symbol (see [5]).

It is always important to determine under what conditions M h acts on A Φ ( K ) , so in what follows the necessary and sufficient condition is specified.

Theorem 2.1

The multiplication operator M h maps A Φ ( K ) into itself if and only if h ∈ A Φ ( K ) .

Proof

First, let us assume that the multiplication operator maps A Φ ( K ) into itself, then, since the constant function 1 ∈ A Φ ( K ) (see [1]), h = M h ( 1 ) belongs to A Φ ( K ) .

Reciprocally, if h ∈ A Φ ( K ) , which is a commutative Banach algebra, the multiplication operator M h maps the space A Φ ( K ) into itself.□

As an immediate consequence, we have that

‖ M h ‖ ≔ sup ‖ f ‖ A Φ ( K ) ≤ 1 ‖ M h ( f ) ‖ A Φ ( K ) = sup ‖ f ‖ A Φ ( K ) ≤ 1 ‖ h f ‖ A Φ ( K ) = ‖ h ‖ A Φ ( K ) .

So, since A Φ ( K ) is a Banach algebra, then M h is defined for any h ∈ A Φ ( K ) . Also, M h is an isometry if and only h is a function of modulus one.

Given the importance of this type of operator, we now determine under what conditions M h is injective and closed.

In the remainder of this article, we consider M h : A Φ ( K ) → A Φ ( K ) ; therefore, we will only use M h .

Lemma 2.2

The operator M h is injective if and only if supp ( h ) = K , where supp ( h ) = { z ∈ K : h ( z ) ≠ 0 } .

Proof

Suppose that M h is injective and supp ( h ) ≠ K , then there exists w ∈ K such that h ( w ) = 0 . Note that h ⋅ χ w is the null function; thus, χ w ∈ ker M h , which contradicts the initial assumption; therefore, the first part of the theorem is concluded.

Reciprocally, if supp ( h ) = K and f ∈ ker ( M h ) , then h ( z ) ⋅ f ( z ) = 0 for all z ∈ K , and hence, f ( z ) = 0 for all z ∈ K . So M h is injective.□

Lemma 2.3

If M h is onto then is injective.

Proof

Suppose that M h is not injective, then exists w ∈ K such that h ( w ) = 0 . Since χ w ∈ A Φ ( K ) and M h is onto, there exists g ∈ A Φ ( K ) such that M h ( g ) = χ w ; thus, 1 = χ w ( w ) = g ( w ) h ( w ) = 0 , which is a contradiction. Therefore M h must be injective.□

Therefore, if M h is onto, it will be invertible. Under what conditions will that inverse be continuous?

Theorem 2.4

Suppose that h ∈ A Φ ( K ) . M h is bijective with continuous inverse if and only if there exists a δ > 0 such that ∣ h ( z ) ∣ > δ for all z ∈ K .

Proof

Suppose that M h is bijective, then exists a linear operator T : A Φ ( K ) → A Φ ( K ) such that M h ∘ T = T ∘ M h = I , where I is a identity operator on A Φ ( K ) . Thus, for each f ∈ A Φ ( K ) ,

f = ( M h ∘ T ) ( f ) = M h ( T ( f ) ) = h ⋅ ( T ( f ) ) .

By virtue of Lemma 2.2, we have that h ( z ) ≠ 0 for all z ∈ K . Then T ( f ) = f h = M 1 h ( f ) .

Since 1 h is a bounded function, there exists M > 0 such that 1 h ( z ) < M , for all z ∈ K . So, the statement for δ = 1 M is fulfilled.

Reciprocally, if there exists a δ > 0 such that ∣ h ( z ) ∣ > δ for all z ∈ K then, by Theorem 1.3, 1 h ∈ A Φ ( K ) . So the operator M h is bijective with continuous inverse M 1 h .□

Corollary 2.5

Suppose that h ∈ A Φ ( K ) . M h is onto with continuous inverse if and only if there exists a δ > 0 such that ∣ h ( z ) ∣ > δ for all z ∈ K .

T ∈ ℬ ( X ) is bounded below if T is injective and has closed range. This is equivalent to saying that there exist γ > 0 such that ‖ T ( x ) ‖ X ≥ γ ‖ x ‖ X for all x ∈ X .

Theorem 2.6

The following statements are equivalent:

M h is bijective with continuous inverse,
R ( M h ) = A Φ ( K ) ,
M h is bounded below,
inf z ∈ K ∣ h ( z ) ∣ > 0 .

Proof

Clearly, (A) implies (B).

If (B), then M h has a closed range, and Lemma 2.3 guarantees that M h is injective, so we have that M h is bounded below. That is (B) implies (C).

On the other hand, if inf z ∈ K ∣ h ( z ) ∣ > 0 , then by Theorem 2.4, we have that M h is bijective with continuous inverse ((D) implies (A)).

This means that to complete the proof is sufficient to verify that (C) implies (D). In fact, suppose that inf z ∈ K ∣ h ( z ) ∣ = 0 , then for each n ∈ N , there exists a z n ∈ K such that 0 ≤ ∣ h ( z n ) ∣ ≤ 1 n .

So we have that

‖ M h ( χ z n ) ‖ A Φ ( K ) = ‖ h χ z n ‖ A Φ ( K ) = ∣ h ( z n ) ∣ ‖ χ z n ‖ A Φ ( K ) ≤ 1 n ‖ χ z n ‖ A Φ ( K ) ,

in whose case, M h is not bounded below.□

Let us now consider the set, which will open the way to characterize the properties that must be fulfilled by the symbols h so that M h is a closed-range operator on A Φ ( K ) . With this objective, we introduce the following set:

X h = { f ∈ A Φ ( K ) : f ( z ) = 0 , ∀ z ∉ supp ( h ) } .

In the case where [ supp ( h ) ] c is a non-empty set, we have that

The null function is at X h , this means that X h ≠ ∅ .
If α , β ∈ C and f , g ∈ X h , then, for all z ∉ supp ( h ) , we have
( α f + β g ) ( z ) = α f ( z ) + β g ( z ) = 0 .
This means that X h is a subspace of A Φ ( K ) .
Also any non-null constant function belong to A Φ ( K ) but does not belong to the set X h . That is, X h is a proper subspace.
If f ∈ X h and g ∈ A Φ ( K ) , then
( f g ) ( z ) = f ( z ) g ( z ) = 0 ,
which means that f g ∈ X h . Thus, X h is an ideal, because the aforementioned is also a proper ideal.
If g ∈ R ( M h ) , then there is a function f ∈ A Φ ( K ) such that M h ( f ) = g . Thus, for all z ∉ supp ( h ) , we have that
g ( z ) = ( M h ( f ) ) ( z ) = h ( z ) f ( z ) = 0 .
i.e., g ∈ X h and therefore R ( M h ) ⊆ X h . The question is whether
X h ⊆ R ( M h )
or what conditions must be met for R ( M h ) = X h ?
As an immediate consequence of that mentioned earlier, X h is an invariant subspace of M h .
If f ∈ X h ¯ , then there exists a sequence { f n } in X h such that ‖ f n − f ‖ A Φ ( K ) → 0 as n → + ∞ . Hence, ‖ f n − f ‖ ∞ → 0 as n → + ∞ , which implies that ∣ f n ( z ) − f ( z ) ∣ → 0 as n → + ∞ for all z ∉ supp ( h ) . Thus, f ∈ X h ; therefore, X h is a closed subspace of A Φ ( K ) .

Closed range on A Φ ( K ) . Let us now give an answer to the question of when R ( M h ) = X h is fulfilled.

Lemma 2.7

If inf z ∈ supp ( h ) ∣ h ( z ) ∣ > 0 , then R ( M h ) = X h .

Proof

Let f ∈ X h and define the function

g ( z ) = f ( z ) h ( z ) if z ∈ supp ( h ) 0 otherwise .

Let us now see that g is also a function in A Φ ( K ) . Let ξ = [ z 0 , z 1 , … , z n ] be an ordered finite list of element of K .

First, note that there exist constants α > 0 and β > 0 such that

var Φ f α , K < + ∞ and var Φ h β , K < + ∞ .

Now, we consider the following cases:

Case I: If z i , z i − 1 ∉ supp ( h ) , then

∣ g ( z i ) − g ( z i − 1 ) ∣ = 0

and

Φ ( ∣ g ( z i ) − g ( z i − 1 ) ∣ ) = 0 .

Case II: If z i , z i − 1 ∈ supp ( h ) , then for

λ ≔ ‖ h ‖ ∞ β ( inf z ∈ supp ( h ) ∣ h ( z ) ∣ ) 2 + ‖ f ‖ ∞ α ( inf z ∈ supp ( h ) ∣ h ( z ) ∣ ) 2 = α ‖ h ‖ ∞ + β ‖ f ‖ ∞ α β ( inf z ∈ supp ( h ) ∣ h ( z ) ∣ ) 2 ,

we have

g λ α β ( z i ) − g λ α β ( z i − 1 ) = 1 λ α β f h ( z i ) − 1 λ α β f h ( z i − 1 ) = h ( z i − 1 ) f ( z i ) − h ( z i ) f ( z i − 1 ) λ α β h ( z i ) h ( z i − 1 ) .

However, since ∣ h ( z i ) ∣ ≥ inf z ∈ supp ( h ) ∣ h ( z ) ∣ > 0 for all i = 0 , … , n , we have

g λ α β ( z i ) − g λ α β ( z i − 1 ) ≤ ∣ h ( z i − 1 ) f ( z i ) − h ( z i ) f ( z i − 1 ) ∣ λ α β ( inf z ∈ supp ( h ) ∣ h ( z ) ∣ ) 2 = ∣ h ( z i − 1 ) f ( z i ) − h ( z i − 1 ) f ( z i − 1 ) + h ( z i − 1 ) f ( z i − 1 ) − h ( z i ) f ( z i − 1 ) ∣ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ ≤ ‖ h ‖ ∞ ∣ f ( z i ) − f ( z i − 1 ) ∣ + ‖ f ‖ ∞ ∣ h ( z i − 1 ) − h ( z i ) ∣ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ .

This implies that

Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) ≤ Φ ‖ h ‖ ∞ ∣ f ( z i ) − f ( z i − 1 ) ∣ + ‖ f ‖ ∞ ∣ h ( z i − 1 ) − h ( z i ) ∣ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ = Φ ‖ h ‖ ∞ α f α ( z i ) − f α ( z i − 1 ) + ‖ f ‖ ∞ β h β ( z i − 1 ) − h β ( z i ) α ‖ h ‖ ∞ + β ‖ f ‖ ∞ ≤ ‖ h ‖ ∞ α α ‖ h ‖ ∞ + β ‖ f ‖ ∞ Φ f α ( z i ) − f α ( z i − 1 ) + ‖ f ‖ ∞ β α ‖ h ‖ ∞ + β ‖ f ‖ ∞ Φ h β ( z i − 1 ) − h β ( z i ) .

Case III: If z i ∈ supp ( h ) and z i − 1 ∉ supp ( h ) , then

g λ α β ( z i ) − g λ α β ( z i − 1 ) = g λ α β ( z i ) = 1 λ α β f ( z i ) h ( z i ) ≤ 1 λ α β ∣ f ( z i ) ∣ inf z ∈ supp ( h ) ∣ h ( z ) ∣ = 1 λ β inf z ∈ supp ( h ) ∣ h ( z ) ∣ f α ( z i ) − f α ( z i − 1 ) .

Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) = Φ 1 λ β inf z ∈ supp ( h ) ∣ h ( z ) ∣ f α ( z i ) − f α ( z i − 1 ) = Φ α inf z ∈ supp ( h ) ∣ f ( z ) ∣ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ f α ( z i ) − f α ( z i − 1 ) ≤ Φ α ‖ h ‖ ∞ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ f α ( z i ) − f α ( z i − 1 ) ≤ α ‖ h ‖ ∞ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ Φ f α ( z i ) − f α ( z i − 1 ) .

Case IV: If z i ∉ supp ( h ) and z i − 1 ∈ supp ( h ) , then similar to the previous case, and it is obtained that

Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) ≤ α ‖ h ‖ ∞ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ Φ f α ( z i ) − f α ( z i − 1 ) .

Therefore,

cvar Φ g λ α β , ξ = ∑ i = 1 n Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) = ∑ Case I is satisfied Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) + ∑ Case II is satisfied Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) + ∑ Case III is satisfied Φ g λ α β ( z i ) − g λ α β ( z i − 1 ) + ∑ Case IV is satisfied Φ g λ α β ( z i ) − g λ α β ( z i − 1 )

≤ ‖ h ‖ ∞ α α ‖ h ‖ ∞ + β ‖ f ‖ ∞ ∑ i = 1 n Φ f α ( z i ) − f α ( z i − 1 ) + ‖ f ‖ ∞ β α ‖ h ‖ ∞ + β ‖ f ‖ ∞ ∑ i = 1 n Φ h β ( z i − 1 ) − h β ( z i ) + 2 α ‖ h ‖ ∞ α ‖ h ‖ ∞ + β ‖ f ‖ ∞ ∑ i = 1 n f α ( z i ) − f α ( z i − 1 ) .

cvar Φ g λ α β , ξ ≤ 3 var Φ f α , K + var Φ h β , K ,

and therefore, g ∈ A Φ ( K ) . Now

M h ( g ) = f .

That is, f ∈ R ( M h ) . □

3 Main results

The following theorem provides a well-known characterization for closed-range operators on a Banach algebra.

Theorem 3.1

Let T : X → Y be a bounded linear operator from a Banach space X into a Banach space Y. Then T has closed range in Y if and only if there is a constant c > 0 such that, for a given x ∈ X , there is an element y ∈ X such that T ( x ) = T ( y ) and ‖ y ‖ ≤ c ‖ T ( x ) ‖ .

By using this theorem, we will demonstrate the following characterization of the multiplication operator on A Φ ( K ) .

Theorem 3.2

The operator M h has closed range if and only if there exists a δ > 0 such that ∣ h ( z ) ∣ ≥ δ for all z ∈ supp ( h ) .

Proof

Suppose that M h has closed range and for all n > 0 , then there exist z n ∈ supp ( h ) such that ∣ h ( z ) ∣ ≤ 1 n . For each n , define

g n = n χ z n .

However, if for each n , there is a f n ∈ A Φ ( K ) such that

f n = M h ( g n ) ,

and this means that

n ≤ ‖ f n ‖ ∞ ≤ ‖ f n ‖ A Φ ( K ) and ‖ M h ( g n ) ‖ A Φ ( K ) ≤ 4 Φ ( 1 ) .

Under these conditions, there exists a constant c > 0 such that

c ‖ f n ‖ A Φ ( K ) ≤ ‖ M h ( f n ) ‖ A Φ ( K ) ,

for every n . Therefore, M h does not have a closed range in A Φ ( K ) . From this contradiction, it can be concluded that there exists a δ > 0 such that ∣ h ( z ) ∣ ≥ δ for all z ∈ supp ( h ) .

Reciprocally, let’s suppose now that there exists δ > 0 such that ∣ h ( z ) ∣ ≥ δ for all z ∈ supp ( h ) , then from Lemma 2.7, R ( M h ) = X h , which is a closed subspace of A Φ ( K ) .□

Theorem 3.3

The operator M h has finite range if and only if supp ( h ) is a finite set.

Proof

Suppose that supp ( h ) is not a finite set, then there exists a sequence { z n } ⊆ supp ( h ) such that z n ≠ z m for all n ≠ m . Note that

h ⋅ χ z n ( z ) = h ( z ) if z = z n 0 if z ≠ z n .

Let us show that the set { M h ( χ z n ) } is linearly independent in A Φ ( K ) , in effect, and let α n 1 , α n 2 , … , α n k be scalars and

∑ i = 1 k α n i M h ( χ z n i ) = 0 .

However,

∑ i = 1 k α n i M h ( χ z n i ) ( z ) = ∑ i = 1 k α n i h ( z n i )

since h ( z n i ) ≠ 0 , and we obtain that α n i = 0 . Then the operator M h has an infinite range.

Conversely, suppose that supp ( h ) = { z 1 , z 2 , … , z k } ⊆ K . It will be shown that { M h ( χ z n ) : 1 ≤ n ≤ k } is a basis for M h ( A Φ ( K ) ) .

We know that M h ( χ z n ) ∈ A Φ ( K ) for every 1 ≤ n ≤ k and { M h ( χ z n ) : 1 ≤ n ≤ k } is a linearly independent set. Then, it is sufficient to show that this set generates the set M h ( A Φ ( K ) ) . Indeed, if f ∈ M h ( A Φ ( K ) ) , so there exists g ∈ A Φ ( K ) such that M h ( g ) = h g = f .

Thus, for each z ∈ supp ( h ) , let us say that z = z n , we have

f ( z n ) = h ( z n ) g ( z n ) = g ( z n ) ( h ⋅ χ z n ) ( z ) = ∑ i = 1 k g ( z i ) ( h ⋅ χ z i ) ( z ) .

On the other hand, if z ∉ supp ( h ) , we can write

f ( z ) = 0 = ∑ i = 1 k g ( z i ) ( h ⋅ χ z i ) ( z ) .

In other words,

f ( z ) = ∑ i = 1 k g ( z i ) ( h ⋅ χ z i ) ( z )

for all z ∈ K , and it is therefore verified that { M h ( χ z n ) : 1 ≤ n ≤ k } is a basis, and consequently, M h has finite range.□

Let’s study further the compactness of this operator. Given ε > 0 , we define the set

E ε = { z ∈ K : ∣ h ( z ) ∣ ≥ ε }

and the associated set

X E ε = { f ∈ A Φ ( K ) : f ( z ) = 0 ∀ z ∈ K \ E ε } .

Is easy to verify that X E ε is a subspace of A Φ ( K ) . Also, if { f n } n is a sequence of functions in X E ε such that f n → f , then ‖ f n − f ‖ ∞ → 0 when n → ∞ , and consequently, f ( z ) = 0 for all z ∈ K − E ε .

Furthermore, if f ∈ X E ε , we have to

M h ( f ) ( z ) = h ( z ) f ( z ) = 0

for all z ∈ K − E ε . This means that X E ε is M h -invariant.

After all this, we will be able to present the following result.

Theorem 3.4

Suppose that f ∈ A Φ ( K ) . The following statements are equivalent:

The operator M h is compact,
dim ( X E ε ) < + ∞ for all ε > 0 ,
E ε is a finite set for all ε > 0 .

Proof

First, suppose that (A) holds, and consider ε > 0 . As indicated earlier, X E ε is a closed subspace of A Φ ( K ) ; consequently, if U ⊆ A Φ ( K ) is an open set, then i X E ε − 1 ( U ) = X E ε ∩ U is open in the subspace topology X E ε , where i X E ε denote the inclusion operator X E ε → A Φ ( K ) ; that is, the inclusion operator i X E ε ( f ) = f is continuous. Given that M h ∘ i X E ε results from the composition of a compact operator and a continuous operator, we have that M h ∘ i X E ε is a compact operator.

Thus R ( M h ∘ i X E ε ) = X E ε .

If f ∈ R ( M h ∘ i X E ε ) , there is g ∈ X E ε such that M h ∘ i X E ε ( g ) = f . However, since g ( z ) = 0 for all z ∈ K − E ε , it is clear that f ( z ) = 0 for all z ∈ K − E ε , that is, f ∈ X E ε , as R ( M h ∘ i X E ε ) ⊆ X E ε .

On the other hand, if f ∈ X E ε , we define the function

g ( z ) = f ( z ) h ( z ) if z ∈ E ε 0 otherwise .

Proceeding similarly as in Lemma 2.7, using ε instead of inf z ∈ supp ( h ) ∣ h ( z ) ∣ , we obtain that g ∈ A Φ ( K ) and, by definition, g ∈ X E ε . Also,

M h ∘ i X E ε ( g ) = M h g = h g = f .

With this, we have verified that R ( M h ∘ i X E ε ) = X E ε .

Let us now verify that M h ∘ i X E ε : X E ε → X E ε is an invertible operator, and if this were true, since it is compact, its domain X E ε must be of finite dimension. Therefore, it only remains to verify that the operator is injective.

If f ∈ ker ( M h ∘ i X E ε ) , then f ( z ) = 0 for all z ∈ K − E ε since f ∈ X E ε . While, if z ∈ E ε , then f ( z ) = 0 since f ( z ) h ( z ) = 0 and h ( z ) ≠ 0 . Therefore, (A) implies (B).

Suppose that E ε is infinite for some ε > 0 , then we can choose a sequence { z n } n ⊆ E ε with z i ≠ z j for i ≠ j . For each n , is clear that χ z n ∈ X E ε . Furthermore, if α n 1 , α n 2 , … , α n k are scalars and suppose that

∑ i = 1 k α n i χ n i = 0 ,

and then by evaluating at z = z n j , we conclude that α n j = 0 , which means { χ n : n ∈ N } is linearly independent, and therefore, dim ( X E ε ) = ∞ . It has been verified that (B) implies (C).

To continue, suppose that (C) is satisfied and knowing that

supp ( h ) = { z ∈ K : h ( z ) ≠ 0 } = ⋃ n = 1 ∞ z ∈ K : ∣ h ( z ) ∣ ≥ 1 n

is a countable union of finite sets, so it is also countable. If supp ( h ) is a finite set, Theorem 3.3 guarantees that the operator M h : A Φ ( K ) → A Φ ( K ) has finite range and therefore is compact. In the case, where supp ( h ) is countable but not finite, let’s say that supp ( h ) = { z 1 , z 2 , … , z n , … } .

Note that,

h n ( z ) ≔ h ( z ) if z ∈ E n 0 if z ∈ K − E n = h χ E n ( z )

is the sum of a finite number of functions in A Φ ( K ) and therefore a function in A Φ ( K ) . Also, by virtue of Theorem 3.3, { M h n } n is a sequence of compact operators. Let us now prove that M h n → M h when n → ∞ .

Indeed,

‖ M h n ( f ) − M h ( f ) ‖ A Φ ( K ) = ‖ h n f − h f ‖ A Φ ( K ) = ‖ ( h n − h ) f ‖ A Φ ( K ) = ‖ h n − h ‖ A Φ ( K ) ‖ f ‖ A Φ ( K ) .

This means that M h n − M h is a bounded operator and

‖ M h n − M h ‖ ≤ ‖ h n − h ‖ A Φ ( K ) ,

then when n → ∞ , we obtain that M h is the limit of compact operators, so M h is compact. With this, it has been proven that (C) implies (A), thus finishing the proof of the theorem.□

Using operator theory, the following corollary is obtained.

Corollary 3.5

Suppose that f ∈ A Φ ( K ) . The following statements are equivalent:

The operator M h ∗ : A Φ ( K ) ∗ → A Φ ( K ) ∗ is compact,
dim ( X E ε ) < + ∞ for all ε > 0 ,
E ε is a finite set for all ε > 0 .

where A Φ ( K ) ∗ denotes the dual of A Φ ( K ) and M h ∗ the dual operator of M h .

4 Applications

4.1 Fredholm multiplication operators on A Φ ( K )

When studying operators, it is always interesting to know if they have closed rank, especially in the case of Fredholm operators.

Given two Banach spaces X and Y , in [6], the set of all upper semi-Fredholm operator is defined by

F + ( X , Y ) ≔ { T ∈ ℬ ( X , Y ) : dim ( k e r ( T ) ) < ∞ and R ( T ) is closed } ,

while the set of all lower semi-Fredholm operator is defined by

F − ( X , Y ) ≔ { T ∈ ℬ ( X , Y ) : dim ( Y / R ( T ) ) < ∞ } .

The set of all semi-Fredholm operator is defined by

F ± ( X , Y ) ≔ F + ( X , Y ) ∪ F − ( X , Y ) ,

and the class F ( X , Y ) of all Fredholm operator is defined by

F ( X , Y ) = F + ( X , Y ) ∩ F − ( X , Y ) .

If T is bounded below, then T is upper semi-Fredholm with dim ( Y / R ( T ) ) ≤ 0 , while any surjective operator is lower semi-Fredholm with dim ( Y / R ( T ) ) ≥ 0 .

We shall F + ( A Φ ( K ) , A Φ ( K ) ) ≔ F + ( A Φ ( K ) ) and F − ( A Φ ( K ) , A Φ ( K ) ) ≔ F − ( A Φ ( K ) ) , while F ( A Φ ( K ) , A Φ ( K ) ) ≔ F ( A Φ ( K ) ) and F ± ( A Φ ( K ) , A Φ ( K ) ) ≔ F ± ( A Φ ( K ) ) .

Now, let’s see what happens with the multiplication operator.

Theorem 4.1

Suppose that h ∈ A Φ ( K ) . The following statements are equivalent:

M h ∈ F ( A Φ ( K ) )
M h ∈ F + ( A Φ ( K ) )
M h ∈ F − ( A Φ ( K ) )
[ supp ( h ) ] c is a finite set and there exists δ > 0 such that ∣ h ( z ) ∣ ≥ δ for all z ∈ supp ( h ) .

Proof

Clearly, (A) implies (B).

Now, suppose (B), that is, the operator M h ∈ F + ( A Φ ( K ) ) , then by definition, dim ( ker ( M h ) ) < ∞ and R ( M h ) is a closed subspace of A Φ ( K ) . By Theorem 3.2, there exists δ > 0 such that ∣ h ( z ) ∣ > δ for all z ∈ supp ( h ) ; hence, it is enough to show that [ supp ( h ) ] c is a finite set. If this is not the case, then there exists a sequence { z n } n of elements in [ supp ( h ) ] c , with z i ≠ z j for i ≠ j . Then { χ z n : n ∈ N } is an infinite linearly independent set contained into ker ( M h ) , which is a contradiction. Therefore, [ supp ( h ) ] c is a finite set, and the verification that (B) implies (D) is completed.

Suppose ( D ) , in which case [ supp ( h ) ] c = { z 1 , z 2 , … , z k } , and recall that

A Φ ( K ) / R ( M h ) = { [ f ] = f + R ( M h ) : f ∈ A Φ ( K ) } .

We are about to show that B = { [ χ z i ] : i = 1 , 2 , … , k } is a basis for A Φ ( K ) / R ( M h ) . Indeed, if α 1 , α 2 , … , α k are scalars and

∑ i = 1 k α i χ z i = ∑ i = 1 k α i [ χ z i ] = [ 0 ] .

This means that ∑ i = 1 k α i χ z i − 0 ∈ R ( M h ) , or in other words, there exists a function g ∈ A Φ ( K ) such that ∑ i = 1 k α i χ z i = M h ( g ) = g h .

In particular, evaluating at z = z j ∈ [ supp ( h ) ] c , we conclude that α j = 0 . As this is valid for all j , it is concluded that B is a linearly independent set of A Φ ( K ) / R ( M h ) .

On the other hand, if [ f ] ∈ A Φ ( K ) / R ( M h ) . For i = 1 , 2 , … , k , we consider the scalars α i = f ( z i ) with z i ∈ [ supp ( h ) ] c , and the function

g ( z ) = f ( z ) h ( z ) if z ∈ supp ( h ) 0 otherwise .

Since there exists δ > 0 such that ∣ h ( z ) ∣ > δ for all z ∈ supp ( h ) , and proceeding similarly as in the proof of Lemma 2.7, it is obtained that g ∈ A Φ ( K ) . Furthermore, for z ∈ supp ( h ) , we have

f ( z ) − ∑ i = 1 k α i χ z i ( z ) = f ( z ) = h ( z ) g ( z ) ,

while if z ∈ [ supp ( h ) ] c , then z = z j for some j = 1 , 2 , … , k ; so

f ( z ) − ∑ i = 1 k α i χ z i ( z ) = f ( z j ) − α j χ z j ( z j ) = 0 = h ( z ) g ( z ) .

Hence, f − ∑ i = 1 k α i χ z i = M h ( g ) , which means f − ∑ i = 1 k α i χ z i ∈ R ( M h ) and [ f ] = ∑ i = 1 k α i [ χ z i ] . Therefore, B is a basis for A Φ ( K ) / R ( M h ) and dim ( A Φ ( K ) / R ( M h ) ) < ∞ . With this, we conclude that (D) implies (C).

Note that (C) implies (B), indeed, suppose dim ( A Φ ( K ) / R ( M h ) ) < ∞ . Then R ( M h ) is a closed subspace of A Φ ( K ) . On the other hand, proceeding as in (D) implies (C), and is verified that the set { [ χ z ] : z ∉ supp ( h ) } is a linearly independent subset of A Φ ( K ) / R ( M h ) . So [ supp ( h ) ] c must be finite, say that { z 1 , z 2 , … , z n } . If f ∈ ker ( M h ) , then f ( z ) = 0 for all z ∈ supp ( h ) and f = ∑ i = 1 n f ( z i ) χ z i . Therefore, ker ( M h ) is generated by a finite amount of functions, so dim ( ker ( M h ) ) < ∞ , which means that M h ∈ F + ( A Φ ( K ) ) .

Finally, (B) implies (A), since (B) implies (C). Also (B) and (C) imply (A).□

As an immediate consequence of the aforementioned result, we have the following:

Corollary 4.2

Suppose that h ∈ A Φ ( K ) . Then,

The operator M h is Fredholm if and only if [ supp ( h ) ] c is a finite set and R ( M h ) is a closed subspace of A Φ ( K ) .
The operator M h is invertible (with continuous inverse) if and only if dim ( A Φ ( K ) / R ( M h ) ) < ∞ and supp ( h ) = K .

Proof

The proof of (A) is obtained from theorems 4.1 and 3.2 applied consecutively.

It is enough to show (B). If the operator M h have continuous inverse then, by Lemma 2.2 supp ( h ) = K , and Theorem 2.6 ensures that R ( M h ) = A Φ ( K ) , so dim ( A Φ ( K ) / R ( M h ) ) < ∞ .

Reciprocally, if dim ( A Φ ( K ) / R ( M h ) ) < ∞ and supp ( h ) = K , then M h is lower semi-Fredholm, and by Theorem 4.1, there exists δ > 0 such that ∣ h ( z ) ∣ ≥ δ for all z ∈ supp ( h ) . Therefore, by Theorem 2.4, we have that M h is an invertible operator, with continuous inverse.□

Is well known that every Fredholm operator is relatively regular; therefore, as a consequence of Corollary 4.2, we have the following corollary.

Corollary 4.3

Suppose that h ∈ A Φ ( K ) . If [ supp ( h ) ] c is a finite set and R ( M h ) is a closed subspace of A Φ ( K ) , then M h is relatively regular.

The study performed generalizes the results obtained in [7] for the case of functions of bounded φ -variation.

4.2 What are we looking for?

The spectral theory of operators is an important part of functional analysis that has application in several areas, such as in differential and integral equations. In particular, the multiplication operator is well studied, in function spaces such as Berman, among others. And this attention is due to the fact that every normal operator is similar to a multiplication operator. Therefore, it is always natural to try to characterize multiplication operators with closed ranges. In our case, we are mainly interested, in the future, in studying some properties of the linear integral operators of Fredholm type defined on the space A Φ ( K ) , seeking to generalize the work done for the real case in [8].

mrbracam@espol.edu.ec

Funding information: They also declare that they have not received specific funding from any source, whether commercial, public, or non-profit organizations.
Conflict of interest: The authors declare that they have no conflict of interest related to this research, either by authorship or publication.

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Received: 2021-12-09

Revised: 2022-08-09

Accepted: 2022-08-17

Published Online: 2022-11-07

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2022-0162

Keywords for this article

Φ-variation; multiplication operator

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