𝑓-Orthomorphisms and 𝑓-Linear Operators on the Order Dual of an 𝑓-Algebra Revisited

Abstract

We give a necessary and sufficient condition on an 𝑓-algebra 𝐴 for which orthomorphisms, 𝑓-linear operators, and 𝑓-orthomorphisms on the order dual 𝐴^∼ are the same class of operators.

1. Introduction and Preliminaries

First of all, we point out that the standard book [1] is adopted in this paper as a unique source of unexplained terminology and notation.

Let 𝐿 and 𝑀 be Riesz spaces. The operator 𝑇∶𝐿→𝑀 is called order bounded if the image under 𝑇 of an order bounded set in 𝐿 is again an order bounded set in 𝑀. An order bounded operator 𝜋∶𝐿→𝐿 is said to be an orthomorphism if |𝜋𝑥|∧|𝑦|=0 whenever |𝑥|∧|𝑦|=0 for all 𝑥,𝑦∈𝐿. The set of all orthomorphisms on 𝐿 is denoted by Orth(𝐿).

If 𝐿 is a Riesz space 𝐿^∼, the first order dual of 𝐿 will be the Riesz space of all order bounded linear functionals on 𝐿. The second order dual (or order bidual) is denoted by 𝐿^∼∼. It is well known that 𝐿^∼ is a Dedekind complete Riesz space. Define for each 𝑥∈𝐿 the element 𝑥^∼∼∈𝐿^∼∼ by 𝑥^∼∼(𝑓)=𝑓(𝑥) for all 𝑓∈𝐿^∼ and then define the mapping 𝜅∶𝐿→𝐿^∼∼ by 𝜅(𝑥)=𝑥^∼∼ for all 𝑥∈𝐿. If 𝐿^∼ separates the points of 𝐿 then 𝜅 is injective and we can identify 𝐿 with the Riesz subspace 𝜅(𝐿). Throughout the paper, we only consider Archimedean Riesz spaces 𝐿 with point separating order dual 𝐿^∼.

From now on, 𝐴 denotes an 𝑓-algebra, that is, a lattice ordered algebra in which 𝑎∧𝑏=0 implies that 𝑎⋅𝑐∧𝑏=0 for all 0≤𝑐∈𝐴. Following constructions in [2], a multiplication can be introduced in the order bidual 𝐴^∼∼ of 𝐴. This is accomplished in three steps as explained next. For every 𝑓∈𝐴^∼ and 𝑥∈𝐴, we define 𝑓⋅𝑥∈𝐴^∼ by the following: 𝑓⋅𝑥𝑦=𝑓𝑥𝑦𝑦∈𝐴. (1) Then, for 𝐺∈𝐴^∼∼ and 𝑓∈𝐴^∼, we introduce 𝐺⋅𝑓∈𝐴^∼ by putting the following: 𝐺⋅𝑓(𝑥)=𝐺𝑓⋅𝑥(𝑥∈𝐴). (2) Finally, let 𝐹,𝐺∈𝐴^∼∼ and define 𝐹𝐺∈𝐴^∼∼ by the following (𝐹𝐺)𝑓=𝐹𝐺⋅𝑓𝑓∈𝐴^∼. (3)

The latter equality defines the Arens multiplication in 𝐴^∼∼. Bernau and Huijsmans in [3] that 𝐴^∼∼ is an 𝑓-algebra with respect to the Arens multiplication. The band of all order continuous linear functionals on 𝐴^∼ is denoted by (𝐴^∼)^∼_𝑛 and its disjoint complement in 𝐴^∼∼ by (𝐴^∼)^∼_𝑠. Observe that 𝐴^∼∼=(𝐴^∼)^∼_𝑛⊕(𝐴^∼)^∼_𝑠, (4) as 𝐴^∼∼ is Dedekind complete.

For each 𝐹∈(𝐴^∼)^∼_𝑛, we define the mapping 𝑉_𝐹∶𝐴^∼→𝐴^∼ by the following: 𝑉_𝐹𝑓=𝐹⋅𝑓∀𝑓∈𝐴^∼. (5) It is shown in [4, Theorem 5.2] that 𝑉_𝐹∈Orth(𝐴^∼) and the mapping 𝑉∶𝐹→𝑉(𝐹)=𝑉_𝐹 are an algebra and lattice homomorphism between (𝐴^∼)^∼_𝑛 and Orth(𝐴^∼).

The final paragraph of this introduction is devoted to the definition of the so-called 𝑓-orthomorphism and 𝑓-linear operator on the order dual.

Let 𝐴 be an 𝑓-algebra. For all 𝑓∈𝐴^∼ define the set 𝑅(𝑓) by the following: 𝑅𝑓=𝐹⋅𝑓∶𝐹∈(𝐴^∼)^∼_𝑛. (6)

An order bounded operator 𝑇∶𝐴^∼→𝐴^∼ is said to be as follows:(1)an 𝑓-orthomorphism if 𝑅(𝑇𝑓)⊆𝑅(𝑓) for each 𝑓∈𝐴^∼, the collection of all 𝑓-orthomorphisms on 𝐴^∼ will be denoted by Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛);(2)an 𝑓-linear with respect to (𝐴^∼)^∼_𝑛 if 𝑇(𝐹⋅𝑓)=𝐹⋅𝑇𝑓 for all 𝑓∈𝐴^∼ and 𝐹∈(𝐴^∼)^∼_𝑛, the set of all 𝑓-linear operators on 𝐴^∼ will be denoted by ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).

Turan showed [5, Proposition 2.6] that Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛) whenever 𝐴 is unital 𝑓-algebra (since 𝐴 is topologically full with respect to itself). Recently, Feng et al. [6, Theorem 3.4] proved that for any 𝑓-algebra 𝐴 we have the following: Orth(𝐴^∼)⊆ℒ_𝑏𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛⊆Orth𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛. (7)

This leads to a very natural question, namely, when does any 𝑓-orthomorphisms is an orthomorphism. A partial answer was already obtained by Feng et al. in [6, Theorem 3.4]. In this regards they proved that if 𝐴 has the factorization property, then Orth(𝐴^∼)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)=Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛). In this paper, we give a complete answer to this question. In fact, we give a necessary and sufficient condition on an 𝑓-algebra 𝐴 for which orthomorphisms, 𝑓-orthomorphisms, and 𝑓-linear operators on the order dual are precisely the same class of operators.

2. When 𝑓-Orthomorphisms Are Orthomorphisms

Let 𝐴 be an 𝑓-algebra with separating order dual. Let 𝐽 be the order ideal generated by all products of 𝐴; that is, 𝐽=𝑥∈𝐴∶|𝑥|≤𝑦⋅𝑧forsome𝑦,𝑧∈𝐴. (8)

It is easy to see that 𝐽 is an ℓ-ideal (ring and order ideal) in 𝐴. The annihilator 𝐽⁰=𝑓∈𝐴^∼∶𝑓(𝑥)=0∀𝑥∈𝐽 (9) is always a band in 𝐴^∼. Since 𝐴^∼ is Dedekind complete Riesz space, then 𝐴^∼=𝐽⁰⊕(𝐽⁰)^𝑑. We denote by 𝑃₀∶𝐴^∼→𝐽⁰ the band projection.

As we will see later, the set 𝐽⁰ plays a key role in the proof of the main result of this paper. We start our discussion by a study of the connection between 𝑓-orthomorphism and orthomorphism in a special case.

Example 1. Suppose that the set 𝐽={0}; hence 𝐽⁰=𝐴^∼. This implies that the product in 𝐴 is given by 𝑥⋅𝑦=0 for all 𝑥,𝑦∈𝐴. It is easy to see that ℒ_𝑏𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛=Orth𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛=ℒ_𝑏(𝐴^∼). (10)
Thus, Orth(𝐴^∼)=Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛) if and only if every order bounded operator is an orthomorphism. Hence, dim𝐴^∼=dim𝐴=1.

From now on, we may assume that the ideal 𝐽≠{0}.

It is obvious that if 𝐴 has the factorization property, then 𝐽⁰={0}. In the next example we show that the condition “𝐽⁰={0}” is strictly weaker than the condition “has the factorization property.” Recall that the condition “𝐽⁰={0}” is equivalent to the condition “(𝐴^∼)^∼_𝑛 is a semi-prime 𝑓-algebra” (see [4, Corollary 6.3]).

Example 2. Let 𝐶([0,1)) be the Archimedean unital 𝑓-algebra of all real-valued continuous functions on [0,1). A function 𝑓∈𝐶([0,1)) is an eventual-polynomial if there is 𝑟_𝑓∈[0,1) and a (unique) polynomial 𝑃_𝑓 such that 𝑓(𝑟)=𝑃_𝑓(𝑟) for all 𝑟∈[𝑟_𝑓,1). Let 𝑌 be the set of all eventual-polynomials in 𝐶([0,1)). Now, put the following: 𝐴=𝑓∈𝑌∶𝑃_𝑓(0)=0, (11) and observe that 𝐴 is a semiprime 𝑓-subalgebra of  𝑌. Clearly, 𝐴 does not satisfy the factorization property. It is shown in [7, Example 7] that (𝐴^∼)^∼_𝑛 is a unital 𝑓-algebra, hence semiprime, 𝑓-algebra. So 𝐽⁰={0}.

We list some simple properties of the band 𝐽⁰. The proof of the next lemma is straightforward and therefore omitted.

Lemma 3. Let 𝐴 be an f-algebra and 𝐹∈𝐴^∼∼. Then(1)𝑓∈𝐽⁰ if and only if 𝑓⋅𝑥 for all 𝑓∈𝐴^∼and 𝑥∈𝐴.(2)𝐹⋅𝑓=0 for all 𝑓∈𝐽⁰.(3)𝐹⋅𝑓∈(𝐽⁰)^𝑑 for all 𝑓∈𝐴^∼.

The next result is important in the context of this work; it is already proved in [6] but for the sake of completeness we partially reproduce the proof.

Proposition 4. Let 𝐴 be an 𝑓-algebra. ThenOrth(𝐴^∼)⊆ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)⊆Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).

Proof. Let 𝜋∈Orth(𝐴^∼); we have to show that 𝜋(𝐹⋅𝑓)=𝐹⋅𝜋(𝑓) for all 𝐹∈(𝐴^∼)^∼_𝑛 and 𝑓∈𝐴^∼. Since 𝑉_𝐹∈Orth(𝐴^∼) which is a commutative algebra we get the following: 𝜋𝐹⋅𝑓=𝜋∘𝑉_𝐹𝑓=𝑉_𝐹∘𝜋𝑓=𝐹⋅𝜋𝑓. (12) Thus, 𝜋∈ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).
Let 𝑇∈ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛) and 𝑓∈𝐴^∼. Pick 𝑥∈𝐴 and observe that 𝑇𝑓⋅𝑥=𝑇𝑥^∼∼⋅𝑓=𝑥^∼∼⋅𝑇𝑓=𝑇𝑓⋅𝑥. (13) Then for all 𝐹∈(𝐴^∼)^∼_𝑛 we get the following: (𝑥)𝑇(𝑥)𝐹⋅𝑇𝑓=𝐹𝑇𝑓⋅𝑥=𝐹𝑓⋅𝑥=𝐹∘𝑇𝑓⋅𝑥=𝐹∘𝑇⋅𝑓∀𝑥∈𝐴. (14)
Hence, 𝐹⋅𝑇𝑓=𝑇^∼(𝐹)⋅𝑓 where 𝑇^∼ is the order adjoint mapping of 𝑇 defined by 𝑇^∼(𝐹)=𝐹∘𝑇. Consequently, 𝑇∈Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).

Another lemma turns out to be useful for later purposes.

Lemma 5. Let 𝐴 be an 𝑓-algebra and 𝑇∈Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛). Then |𝑇𝑓|∧𝑔∈𝐽⁰ whenever 𝑓∧𝑔=0 in 𝐴^∼.

Proof. It suffices to show that (|𝑇𝑓|∧𝑔)⋅𝑥=0 for all 0≤𝑥∈𝐴. Let 0≤𝑥∈𝐴. Since 𝑉_𝑥∼∼∈Orth(𝐴^∼) we get the following: 𝑇𝑓∧𝑔⋅𝑥=𝑥^∼∼⋅𝑇𝑓∧𝑔=𝑥^∼∼⋅𝑇𝑓∧𝑥^∼∼=𝑥⋅𝑔_∼∼⋅𝑇𝑓∧𝑥^∼∼⋅𝑔. (15) There exists 𝐺∈(𝐴^∼)^∼_𝑛 such that 𝑥^∼∼⋅𝑇𝑓=𝐺⋅𝑓. Consequently, 𝑇𝑓∧𝑔⋅𝑥=𝐺⋅𝑓∧𝑥^∼∼𝑓⋅𝑔≤𝐺⋅∧𝑥^∼∼≤(⋅𝑔𝐺+𝑥^∼∼)⋅𝑓∧𝑔=0. (16)

We have now gathered all of the ingredients for the proof of the principal theorem of this paper.

Theorem 6. Let 𝐴 be an 𝑓-algebra. Then the following are equivalents:(i)Orth(𝐴^∼)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)=Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛);(ii)𝐽⁰={0}.

Proof. (i)⇒(ii) Arguing by contradiction, suppose that 𝐽⁰≠{0}. There exists a nonzero positive element ℎ∈𝐴^∼ such that ℎ(𝑎)=0 for all 𝑎∈𝐽. Let 0≤𝑔∈(𝐽⁰)^𝑑 and 0≤𝑥∈𝐴 such that 𝑔(𝑥)≠0. Define the mapping 𝑇∶𝐴^∼→𝐴^∼ by 𝑇(𝑓)=𝑓(𝑥)ℎ. It is not hard to see that 𝑇 is positive and for all 𝐹∈(𝐴^∼)^∼_𝑛, 𝑓∈𝐴^∼ we have the following: 𝐹⋅𝑇𝑓=𝑓(𝑥)𝐹⋅ℎ=0∈𝑅𝑓. (17)
Thus, 𝑇∈Orth(𝐴^∼,𝐴^∼,𝐴^∼). But 𝑇 is not an orthomorphism since 𝑔∧ℎ=0 and 𝑇𝑔∧ℎ=𝑔(𝑥)ℎ∧ℎ≠0.
(ii)⇒(i) In view of Proposition 4, it suffices to show that Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)⊆Orth(𝐴^∼). To do this, pick 𝑇∈Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛) and 𝑓,𝑔∈𝐴^∼ such that 𝑓∧𝑔=0. According to Lemma 5,we obtain |𝑇𝑓|∧𝑔∈𝐽⁰={0}; that is, |𝑇𝑓|∧𝑔=0, as required.

It follows from Theorem 6 that if Orth(𝐴^∼)=Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛) then (𝐴^∼)^∼_𝑛; hence 𝐴, is a semiprime 𝑓-algebra.

We end this section with a consequence of Theorem 6. The notion of weak approximate unit plays a key role in the context of this study. Recall that an upward directed net {𝑎_𝜏∶𝜏∈𝑇} of positive elements in an 𝑓-algebra 𝐴 is said to be an approximate unit if sup_𝜏𝑎_𝜏⋅𝑥=𝑥 for all 0≤𝑥∈𝐴. An approximate unit {𝑎_𝜏} is called weak approximate unit if sup_𝜏𝑓(𝑎_𝜏𝑥)=𝑓(𝑥) for all 0≤𝑓∈𝐴^∼ and 0≤𝑥∈𝐴 (see [4, Definitions 2.2 and 7.1]). It is well known that if 𝐴 is a semi-prime 𝑓-algebra then 𝐴 can be embedded as a ring and 𝑓-subalgebra in the unital 𝑓-algebra Orth(𝐴).

In Theorem 6, we gave two necessary and sufficient conditions for which 𝑓-orthomorphisms are orthomorphisms. In the following proposition, we shall present a second one in terms of approximate unit. We have to impose, however, on 𝐴 an extra condition, namely, the Stone condition (i.e., 𝑥∧𝐼∈𝐴 whenever 𝑥 is positive in 𝐴, where 𝐼 is the identity of Orth(𝐴)). It should be pointed out here that every uniformly complete semi-prime 𝑓-algebra 𝐴 satisfies the Stone condition.

Proposition 7. Consider in a semi-prime 𝑓-algebra 𝐴 the following conditions.

Theorem 8. (i)  𝐴 has a weak approximate unit.
(ii)  Orth(𝐴^∼)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛)=Orth(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).
(i)⇒(ii) Moreover, if 𝐴 satisfies the Stone condition then (i)⇔(ii).

Proof. (i)⇒(ii) Assume that (𝑎_𝜏)_𝜏 is a weak approximate unit. Let 0≤𝑓∈𝐽⁰, it follows from 𝑓(𝑎_𝜏𝑥)=0 for all 0≤𝑥∈𝐴 that 𝑓(𝑥)=0.
(ii)⇒(i) It follows from Theorem 6, that (𝐴^∼)^∼_𝑛 is semi-prime. The result follows from [4, Theorem 7.2].

3. When 𝑓-Linear Operators Are Orthomorphisms

In this section, we study the connection between 𝑓-linear operators and orthomorphisms. From Proposition 4 and Theorem 6 if 𝑓-linear operators on the dual 𝐴^∼ of an 𝑓-algebra 𝐴 are orthomorphisms and if 𝐽⁰={0} then the two classes coincide. Next, we give an example of an 𝑓-algebra such that 𝐽⁰≠{0} in which 𝑓-linear operators are orthomorphisms.

Example 9. Let 𝐴=ℝ² equipped with the coordinatewise operations and ordering. Consider the multiplication defined by (𝑎,𝑏)⋅(𝑐,𝑑)=(𝑎𝑐,0). It is easy to see that for all 𝑓=(𝑓₁,𝑓₂)∈𝐴^∼ and 𝐹=(𝐹₁,𝐹₂) we have 𝑓⋅(𝑎,𝑏)=(𝑓₁𝑎₁,0) and 𝐹⋅𝑓=(𝐹₁⋅𝑓₁,0). Note that 𝐽⁰={𝑓=(0,𝑓₂)∶𝑓₂=0}. Let 𝑇∶=^𝛼𝛽_𝛾𝜎 an 𝑓-linear operator. An easy computation shows that 𝛽=𝛾=0. Thus, 𝑇 is an orthomorphism.

It seems natural therefore to ask under what condition we have Orth(𝐴^∼)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛). The answer is given in the next theorem. First, let us discuss the ideal of nilpotents elements in the bidual of an 𝑓-algebra 𝐴. Let 𝐹∈𝐴^∼∼ the absolute kernel or null ideal 𝑁_𝐹 of 𝐹 is defined by the following: 𝑁_𝐹=𝑓∈𝐴^∼∶|𝐹|𝑓=0. (18) It is evident that 𝑁_𝐹 is an order ideal. The disjoint complement of 𝐶_𝐹=𝑁^𝑑_𝐹 is called the carrier of 𝐹 and is always a band in 𝐴^∼. Given 0≤𝐹∈𝐴^∼∼ such that 𝐹⋅𝑓=0 for all 𝑓∈𝐴^∼ (𝐹 is nilpotent in 𝐴^∼∼), then (𝐽⁰)^𝑑⊆𝑁_𝐹. Indeed, let 𝑓∈𝐶_𝐹, since 𝐶_𝐹 is a band and 𝑉_𝑥∼∼ is an orthomorphism in 𝐴^∼, then 𝑓⋅𝑥∈𝐶_𝐹 for all 𝑥∈𝐴. But, by hypothesis 𝑓⋅𝑥∈𝑁_𝐹, 𝑓⋅𝑥∈𝑁_𝐹, so 𝑓⋅𝑥=0 for all 𝑥∈𝐴. This implies that 𝑓∈𝐽⁰. Consequently, 𝐶_𝐹⊆𝐽⁰ and (𝐽⁰)^𝑑⊆𝑁_𝐹. For more information about the nilpotent elements in the bidual of 𝑓-algebra, the reader is referred to [4].

We are now in position to prove the main theorem in this section.

Theorem 10. Let 𝐴 be an 𝑓-algebra. The followings are equivalents.(i)Orth(𝐴^∼)=ℒ_𝑏(𝐴^∼,𝐴^∼,(𝐴^∼)^∼_𝑛).(ii)dim𝐽⁰=1.

Proof. (i)⇒(ii) Suppose that dim𝐽⁰≥2. We can find therefore positive elements ℎ,𝑘∈𝐽⁰ such that ℎ∧𝑘=0. Let 0≤𝑎∈𝐴 such that 𝑘(𝑎)≠0. Define the mapping 𝑇∶𝐴^∼→𝐴^∼ by the following: 𝑇𝑓=𝑃₀∘𝑓(𝑎)⋅ℎ. (19) Since 𝐹⋅𝑓∈(𝐽⁰)^𝑑 for all 𝐹∈(𝐴^∼)^∼_𝑛 and 𝑓∈𝐴^∼, we get 𝑇(𝐹⋅𝑓)=𝐹⋅𝑇𝑓=0. Thus 𝑇 is an 𝑓-linear operator. Clearly, 𝑇 is not an orthomorphism since ℎ∧𝑘=0 and 𝑇(𝑘)∧ℎ=𝑘(𝑎)ℎ∧ℎ≠0.
(ii)⇒(i) Let 𝑇 be an 𝑓-linear mapping on 𝐴^∼ and 0≤ℎ∈𝐽⁰. First, we show that 𝑇((𝐽⁰)^𝑑)⊂(𝐽⁰)^𝑑 and 𝑇(𝐽⁰)⊂𝐽⁰. There exists 𝜑∈𝐴^∼∼ such that 𝑃₀∘𝑇(𝑓)=𝜑(𝑓)⋅ℎ. Since 𝑃₀∘𝑇 is an 𝑓-linear operator, we derive that 𝜑(𝐹⋅𝑓)=0 for all 𝐹∈(𝐴^∼)^∼_𝑛 and 𝑓∈𝐴^∼. That is 𝜑⋅𝑓=0 for all 𝑓∈𝐴^∼. By the remark above, we get 𝜑(𝑓)=0 for all 𝑓∈(𝐽⁰)^𝑑. This implies that 𝑇(𝑓)∈(𝐽⁰)^𝑑 whenever 𝑓∈(𝐽⁰)^𝑑. Now, If 𝑓∈𝐽⁰ then 𝑇𝑓⋅𝑥=𝑥^∼∼⋅𝑇𝑓=𝑇𝑥^∼∼⋅𝑓=𝑇𝑓⋅𝑥=0∀𝑥∈𝐴. (20) This shows that 𝑇(𝑓)∈𝐽⁰, in particular 𝑇(ℎ)=𝑡⋅ℎ where 𝑡∈ℝ. Now, let 𝑓∧𝑔=0 in 𝐴^∼. We have to show that |𝑇𝑓|∧𝑔=0. Decompose 𝑓 and 𝑔 as 𝑓=𝛼ℎ+𝑓₁ and 𝑔=𝛽ℎ+𝑔₁ with 𝛼,𝛽∈ℝ and 𝑓₁,𝑔₁∈(𝐽⁰)^𝑑. The decomposition of 𝑇𝑓 is given by the following: 𝑇𝑓=𝛼𝑡⋅ℎ+𝑇𝑓₂. (21) Since 𝑓∧𝑔=0, then 𝛼⋅𝛽=0 and 𝑓₁∧𝑔₁=0. On the other hand, according to Lemma 5, we have |𝑇𝑓|∧𝑔∈𝐽⁰. So 𝑇𝑓∧𝑔=𝛼𝑡∧𝛽⋅ℎ=0 (22) as 𝛼⋅𝛽=0. This completes the proof of the theorem.