summary:Extensions of order bounded linear operators on an Archimedean vector lattice to its relatively uniform completion are considered and are applied to show that the multiplication in an Archimedean lattice ordered algebra can be extended, in a unique way, to its relatively uniform completion. This is applied to show, among other things, that any order bounded algebra homomorphism on a complex Archimedean almost $f$-algebra is a lattice homomorphism

Triki, Abdelmajid

English

Institute of Mathematics AS CR, v. v. i.

Commentationes Mathematicae Universitatis CarolinaeAbdelmajid TrikiOn algebra homomorphisms in complex almost f -algebrasCommentationes Mathematicae Universitatis Carolinae, Vol. 43 (2002), No. 1, 23--31Persistent URL: http://dml.cz/dmlcz/119297Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 2002Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.czComment.Math.Univ.Carolin. 43,1 (2002)23–31 23On algebra homomorphisms in complex almost f-algebrasAbdelmajid TrikiAbstract. Extensions of order bounded linear operators on an Archimedean vector latticeto its relatively uniform completion are considered and are applied to show that themultiplication in an Archimedean lattice ordered algebra can be extended, in a uniqueway, to its relatively uniform completion. This is applied to show, among other things,that any order bounded algebra homomorphism on a complex Archimedean almost f -algebra is a lattice homomorphism.Keywords: vector lattice, order bounded operator, lattice ordered algebra, f -algebra,almost f -algebraClassification: 06F20, 06F251. IntroductionIn this paper, we show that any order bounded linear operator from an Archi-medean vector lattice A into a uniformly complete vector lattice B has a uniqueorder bounded extension to the relatively uniform completion of A. As an appli-cation, we show that the multiplication in an Archimedean lattice ordered algebraA can be extended, in a unique way, into a lattice ordered algebra multiplicationon Ā, the relatively uniform completion of A, in such a manner that A becomesa sub algebra of Ā. Moreover, Ā is an f -algebra (respectively almost f -algebra,d-algebra) whenever A is an f -algebra (respectively almost f -algebra, d-algebra).In Section 4 we are mainly concerned with generalizing a theorem, due toE. Scheffold (cf. [8, Theorem 2.2]), which states that if A is a Banach almostf -algebra (FF-algebra in his terminology), then any order bounded multiplica-tive functional on AC , the complexification of A, is a lattice homomorphism.We show in a different manner that in fact the latter result holds for an arbi-trary order bounded algebra homomorphism between AC and BC respectivelythe complexifications of the Archimedean almost f -algebras A and B, providedB is semiprime. Also, a Nagasawa-like theorem is proved for complex f -algebrasin order to characterize algebra homomorphisms.2. PreliminariesFor unexplained terminology and the basic results on vector lattices and f -algebras we refer to [4], [7] and [9]. All vector lattices and lattice ordered algebras24 A.Trikiunder consideration are supposed to be Archimedean and the only topology weconsider on these spaces is the relatively uniform topology (cf. [4, Sections 16and 63]). If Â is the Dedekind completion of the Archimedean vector lattice A,then Ā, the closure of A in Â with respect to the relatively uniform topology,is a relatively uniformly complete vector lattice which is, with respect to ([6,Definition 2.12]), the relatively uniform completion of A.The real algebra A is called a lattice ordered algebra if A is a vector latticewith the following property(i) ab ∈ A+ for all a, b ∈ A+.The lattice ordered algebra A is called an almost f -algebra if A verifies thefollowing property:(ii) ab = 0 for all a, b ∈ A such that a ∧ b = 0.The lattice ordered algebra A is called an f -algebra if A verifies the followingproperty:(iii) a ∧ b = 0 implies ac ∧ b = ca ∧ b = 0 for all c ∈ A+.The lattice ordered algebra A is called a d-algebra if A verifies the followingproperty:(iv) a∧ b = 0 implies that c(a∧ b) = ca∧ cb and (a∧ b)c = ac∧ bc for all c ∈ A+.The lattice ordered algebra A is said to be semiprime if 0 is the only nilpotentelement of A.The linear mapping T defined on the vector lattice A with values in the vectorlattice B is called order bounded (notation T ∈ Lb(A, B) or T ∈ Lb(A) if A = B)if T maps order intervals into order intervals.The mapping T ∈ Lb(A) is said to be an orthomorphism if it follows from|a| ∧ |b| = 0 that |T (a)| ∧ |b| = 0. The collection Orth(A) of all orthomorphismson A is, with respect to the usual vector spaces operations and composition asmultiplication, an Archimedean f -algebra with the identity mapping IA on A asa unit element. If A is a semiprime Archimedean f -algebra, then the mappingρ : A −→ Orth(A), which assigns to a ∈ A the operator Ta defined on A byTa(b) = ab for all b ∈ A, is an injective f -algebra homomorphism.Throughout this paper, a semiprime Archimedean f -algebra will be identifiedwith ρ(A) in Orth(A). If A is in addition relatively uniformly complete then Averifies the so-called “Stone condition”; i.e., a ∧ I ∈ A holds for all a ∈ A+. Weshall denote by Ab the subalgebra of all bounded elements in A, i.e.,Ab = {a ∈ A : |a| ≤ λIA, λ ∈ R+}.Let A be an Archimedean semiprime f -algebra, as agreed upon, we consider A asa sub algebra of Orth(A) and we denote by AC = A+ iA the complexification ofA. If, in addition, A is relatively uniformly complete then AC and Orth(A)C arecomplex f -algebras (cf. [1, Section 5]).On algebra homomorphisms in complex almost f -algebras 25Note, in this connection, that even if A is not relatively uniformly complete,AC can be regarded as a sub algebra of the complex f -algebra Orth(Ā)C andhence |a| exists in Orth(A)C for all a ∈ AC .The linear mapping T = T1 + iT2 : AC −→ BC is called order bounded ifT1, T2 ∈ Lb(A, B). T is said to be contractive if |T (a)|2 ≤ |T (a)| whenever |a|2 ≤|a| or equivalently if |T (a)| ≤ IB whenever |a| ≤ IA (the absolute value is taken,if necessarily in Orth(A)C or in Orth(B̄)C). A bijective operator T is said to bebicontractive if T and T−1 are contractive.3. An extension theoremLet A and B be Archimedean vector lattices and assume, in addition, that B isrelatively uniformly complete and let T ∈ Lb(A, B). In this section we show thatT has a unique extension into an element T ′ ∈ Lb(Ā, B), where Ā is the relativelyuniform completion of A. Moreover, we use this result to prove that the relativelyuniform completion of an Archimedean lattice ordered algebra is again a latticeordered algebra.To a given 0 ≤ T ∈ Lb(A, B) we associate the mappings U and V definedon Ā (the relatively uniform completion of A) with values in B̂, the Dedekindcompletion of B, as follows:U(x) = sup{T (a) : a ∈ A, a ≤ x for all x ∈ Ā};V (x) = inf{T (a) : a ∈ A, x ≤ a for all x ∈ Ā}.In the following lemma we collect some properties of U and V , the easy proofof which we leave to the reader.Lemma 3.1. (1) U and V are increasing mappings and U(x) ≤ V (x)for all x ∈ Ā.(2) U(a) = V (a) = T (a) for all a ∈ A.(3) U(x+ y) ≥ U(x) + U(y) and V (x+ y) ≤ V (x) + V (y) for all x, y ∈ Ā.(4) U(λx) = λU(x) for all 0 ≤ λ ∈ R and U(λx) = λV (x) for all λ < 0.Next we show that U = V and that U is a positive extension of T to Ā.Lemma 3.2. U ∈ Lb(Ā, B̂) and it is the (unique) positive extension of T to Ā.Proof: Since A is relatively uniformly dense in Ā and since U(a) = V (a) for alla ∈ A, in order to show that U(x) = V (x) for all x ∈ Ā, it is therefore sufficientto prove that U and V are relatively uniformly continuous on Ā. To this end,let {xn : n = 1, 2, . . .} be a relatively uniformly convergent sequence in Ā withlimit x. Then, there is v ∈ A+ such that, for every ε > 0 there exists a naturalnumber Nε such that |xn − x| ≤ εv for all n ≥ Nε, i.e., x − εv ≤ xn ≤ x + εv.We have to show that U(xn) and V (xn) converge respectively to U(x) and V (x).U(x − εv) = sup{T (a) : a ∈ A, a ≤ x − εv}. If we put b = a + εv then we have26 A.Trikib ∈ A and b ≤ x. It follows that{T (a) : a ∈ A, a ≤ x − εv} = {T (b − εv) : b ∈ A, b ≤ x}.Hence,U(x − εv) = sup{T (b)− εT (v) : b ∈ A, b ≤ x}= sup{T (b) : b ∈ A, b ≤ x} − εT (v)= U(x)− eT (v).Analogously we have that U(x+ εv) = U(x) + εT (v).It follows from Lemma 3.1(1) thatU(x)− εT (u) ≤ U(xn) ≤ U(x) + εT (v).In other words, U(xn) converges relatively uniformly to U(x) and hence U isrelatively uniformly continuous. Similarly, V is relatively uniformly continuouson Ā and it follows that U(x) = V (x) for all x ∈ Ā. From Lemma 3.1(2), (3)and (4) we deduce easily that U is a positive linear mapping in Lb(Ā, B̂) whichextends T to Ā. Finally it is not difficult to see that U is the unique positiveextension of T to Ā. This completes the proof. Now, let T be an element in Lb(A, B). T can be regarded as an element ofLb(A, B̂). Hence, there exist two positive elements T1 and T2 in Lb(A, B̂) suchthat T = T1 − T2. By the preceding lemma, T1 and T2 have unique positiveextensions T ′1 and T′2 to Ā the relatively uniform completion of A. Hence, T′ =T ′1 − T′2 ∈ Lb(Ā, B̂) is the unique order bounded extension of T to Ā. Moreover,the relatively uniform continuity of T implies that the range of T′is contained inB since B is relatively uniformly closed in B̂.Summarizing, we have the following theorem:Theorem 3.3. Let A and B be two Archimedean vector lattices. Assume inaddition that B is relatively uniformly complete. Then any order bounded linearoperator T : A −→ B has a unique order bounded linear extension T ′ : Ā −→ B,where Ā is the relatively uniform completion of A.Let A be an Archimedean lattice ordered algebra and let a ∈ A+. If we defineTa : A −→ Ā by Ta(y) = ay, then Ta ∈ Lb(A, Ā) and hence, by the precedingtheorem, Ta extends to T′a ∈ Lb(Ā). For a fixed y ∈ (Ā)+ we put ay = Ta(y).Now, for a fixed y ∈ (Ā)+ we define Ry : A −→ Ā by Ry(a) = T′a(y). ThenRy ∈ Lb(A, Ā) and hence, Ry extends into R′y ∈ Lb(Ā).If for arbitrary x, y ∈ (Ā)+ we define on Ā a multiplication by putting xy =R′y(x), then it is an easy task to verify that this multiplication is the uniqueOn algebra homomorphisms in complex almost f -algebras 27lattice ordered algebra multiplication in Ā which extends the multiplication inA in such a manner that A becomes a subalgebra of Ā with respect to thismultiplication. Moreover, it is easy to verify that such a multiplication is an f -algebra (respectively, almost f -algebra, d-algebra) multiplication whenever A isan f -algebra (respectively, almost f -algebra, d-algebra).Thus we obtain the following theorem:Theorem 3.4. Let A be an Archimedean lattice ordered algebra. Then the mul-tiplication in A can be extended in a unique way into a lattice ordered algebramultiplication on Ā in such a manner that A becomes a sub algebra of Ā. More-over, Ā is an f -algebra (respectively almost f -algebra, d-algebra) whenever A isan f -algebra (respectively almost f -algebra, d-algebra).4. The main resultsLet A and B be Archimedean semiprime f -algebras with unit elements eA andeB . The linear operator T : AC −→ BC is contractive if |T (a)| ≤ eB whenever|a| ≤ eA. In order to make things surveyable, we first show that if T ∈ Lb(AC , BC)is contractive and verifies T (eA) = eB, then T is positive (i.e., T (a) ∈ B+ for alla ∈ A+). To this end, we need the following inequalities:(∗) If T ∈ Lb(AC , BC) and u ∈ A+ then|T (u)| ≤ |T |(u) ≤ sup{|T (a)| : a ∈ AC ; |a| ≤ u}(cf. [9, Lemma 92.5 and Theorem 92.6]).Proposition 4.1. If T ∈ Lb(AC , BC) is contractive and verifies T (eA) = eB,then T is positive.Proof: Note first that there exist T1, T2 ∈ Lb(A, B) such that T = T1+ iT2. So,T1(eA) = eB since T (eA) = eB. Consider, now, T as an element of Lb(AC , B̂C).Then, the contractivity of T together with the inequalities (∗) imply that|T |(eA) = eB = T1(eA).It follows from the fact that T1 ≤ |T |, that0 ≤ (|T | − T1)(a) ≤ (|T | − T1)(eA) = 0 for all a ∈ A+ verifying a ≤ eA.Consequently, the restriction of |T | − T1 to Ab is the null operator on Ab. Therelatively uniform continuity of |T | − T1 together with the relatively uniformdensity of Ab in A imply now that |T |−T1 is the null operator on A, i.e., |T | = T1.Finally, the identity |T (a)|2 = T1(a)2 + T2(a)2 for all a ∈ A+ ([1, Theorem 5.2])already implies that T2 = 0 and hence |T | = T , which is the desired result. Before proving our main theorem we need the following lemma28 A.TrikiLemma 4.2. If (a+ αIA) ∈ A#C verifies |a+ αIA| ≤ IA then |α| ≤ 1.Proof: It follows from the inequalities |a| − |α|IA ≤ |a+αIA| ≤ IA and |α|IA −|a| ≤ |a + αIA| ≤ IA that (|α| − 1)IA ≤ |a| ≤ (1 + |α|)IA. As a consequenceof the Stone condition we have that |α| ≤ 1. Indeed, |α| > 1 would lead toIA < (|α| − 1)−1|a| and hence IA = (|α| − 1)−1|a| ∧ IA ∈ A, contradictory to ourassumption that IA /∈ A. Hence, we necessarily have |α| ≤ 1. We are now in a position to prove the main result of this paper which extendsTheorem 5.3 of [3], proved in the real case under the additional assumption thatA has a unit element.Theorem 4.3. Let A and B be Archimedean semiprime f -algebras. If T :AC −→ BC is an order bounded algebra homomorphism then T is a latticehomomorphism (i.e., T is real and the restriction of T to A is a lattice homomor-phism).Proof: First, we show that T is contractive. To this end, let a ∈ AC be suchthat |a| ≤ IA. This implies that 0 ≤ |an| ≤ |a| for n = 1, 2, . . . , so we have that|an| ∈ [0, |a|] and it follows from the order boundedness of T that there existsb ∈ B+ such that|T (an)| = |Ta|n ≤ b for all n.Now, if we put|T (a)| ∨ IB = IB + hthen h is positive and we have0 ≤ |T (an)| ∨ IB = (|T (a)| ∨ IB)n = (IB + h)n ≤ b ∨ IB for all n,from this we deduce that0 ≤ IB + nh ≤ b ∨ IB for all n.Multiplying these inequalities by 1n we find0 ≤1nIB + h ≤1n(b ∨ IB) for all nand so0 ≤ h ≤1n(b ∨ IB − IB).Therefore, it follows from the Archimedean assumption that h = 0. Consequently|T (a)| ≤ IB and hence T is contractive.Now we consider two cases.First case: A has a unit element eA.On algebra homomorphisms in complex almost f -algebras 29In this case T (eA) is an idempotent in B. Moreover, it follows from the factthatT (a) = T (aeA) = T (a)T (eA) for all a ∈ Athat T (a) is an element of the band F generated by T (eA) in B. F is an f -algebrawith unit element T (eA), so if we consider T as an algebra homomorphism fromAC into BC it follows from Proposition 4.1 that T is positive and hence T is alattice homomorphism (see e.g. [3, Section 5]).Second case: A does not possess a unit element.Observe first that if Ā is the relatively uniform completion of A then it followsfrom Theorem 3.3 that T has a unique order bounded linear extension T ′ to ĀCwith values in B̄C . Moreover, it takes a little effort to verify that T′ is an algebrahomomorphism. So, we shall assume without loss of generality that A and Bare relatively uniformly complete. Assume now that A does not possess a unitelement and let Ab be the f -algebra of all bounded elements in A. Since T iscontractive, the restriction of T to (Ab)C (which we shall denote again by T ) is acontractive algebra homomorphism from (Ab)C into BC . Consider now T# theextension of T to (A#b )C , then, an easy computation shows that T# is an algebrahomomorphism. Moreover, if |a+αIA| ≤ IA then it follows from Lemma 4.2 that|a| ≤ 2IA and hence that|T#(a+ αIA)| ≤ |T (a)|+ |α|IB ≤ 3IB .Consequently, T# is an order bounded algebra homomorphism. Again, T# iscontractive and since T#(IA) = IB , it follows from Proposition 4.1 that T# ispositive and, hence, the restriction of T to Ab is positive. Now the relativelyuniform continuity of T together with the relatively uniform density of Ab in Aimply that T is positive on A and, hence, that T is a lattice homomorphism,which is the desired result. Remark 4.1. Example 5.2 of [3] shows that the condition that T is orderbounded cannot be dropped in Theorem 4.3.Next, we give a characterization of algebra isomorphisms in terms of contractiveoperators. This result is, in some respect, an f -algebra version of the well knownNagasawa’s result for Banach algebras ([5, Theorem 1]).Theorem 4.4. Let A and B be Archimedean f -algebras with unit elementseA and eB and let T : AC → BC be an order bounded bijection such thatT (eA) = eB . The following properties are equivalent:(i) T is an algebra isomorphism;(ii) T is a lattice isomorphism;(iii) T is bicontractive.30 A.TrikiProof: (i)⇔(ii). Follows from Theorem 4.3 and ([3]; Theorem 5.4).(ii)⇒(iii). Obvious.(iii)⇒(ii). Since T is contractive and T (eA) = eB , it follows from Proposi-tion 4.1 that T is positive. T ′, the restriction of T to Ab, is a bijection ontoBb, so, T′−1 is an order bounded and contractive operator on Bb such thatT ′−1(eB) = eA. It follows again from Proposition 4.1 that T′−1 is positive and,hence, T ′ is a lattice homomorphism. Now, let a and b be elements in A suchthat a ∧ b = 0 and define the sequences an and bn by putting:an = a ∧ neA and bn = b ∧ neA for n = 1, 2, . . . .an and bn are elements of Ab and converge relatively uniformly respectively to aand b. Moreover, we have that an ∧ bn = 0, so,T (an ∧ bn) = T′(an ∧ bn) = 0.The relative uniform continuity of T now implies that T (a ∧ b) = 0. This showsthat T is a lattice isomorphism and we are done. Assume now that A and B are Archimedean almost f -algebras and that B issemiprime. Let T : AC −→ BC be an order bounded algebra homomorphism.N(A), the set of all nilpotent elements in A, is a relatively uniformly closed algebraideal and order ideal in A. It follows that A/N(A) is a semiprime Archimedeanf -algebra.Let s : AC −→ (A/N(A))C be the canonical surjection. Then s is a lattice andalgebra homomorphism.The linear operator T ′ : (A/N(A))C −→ BC defined by T′(s(a)) = T (a) for alla ∈ AC is an order bounded algebra homomorphism from (A/N(A))C into BC .So, by Theorem 4.3, T ′ is a lattice homomorphism and hence T = T ′ ◦ s is alattice homomorphism since s is a lattice homomorphism.This leads to the above mentioned generalization of Scheffold’s result (cf. [8,Theorem 2.2]) which we state as a theorem.Theorem 4.5. Let A and B be Archimedean almost f -algebras. Assume, inaddition, that B is semiprime. If T : AC −→ BC is an order bounded algebrahomomorphism then T is a lattice homomorphism.References[1] Beukers F., Huijsmans C.B., Pagter B. de, Unital embedding and complexification off-algebras, Math. Z. 183 (1983), 131–144.[2] Huijsmans C.B., Pagter, B. de, Averaging operators and positive contractive projections,J. Math. Anal. Appl. 113 (1986), 163–184.[3] Huijsmans C.B., Pagter B. de, Subalgebras and Riesz subspaces of an f-algebra, Proc.London Math. Soc. (3) 48 (1984), 161–174.On algebra homomorphisms in complex almost f -algebras 31[4] Luxemburg W.A.J., Zaanen A.C., Riesz Spaces I, North Holland, Amsterdam, 1971.[5] Nagasawa M., Isomorphisms between commutative Banach algebras with an application torings of analytic functions, Kodai Math. Semin. Rep. 11 (1959), 182–188.[6] Quinn J., Intermediate Riesz spaces, Pacific J. Math. 56 (1975), 225–263.[7] Schaefer H.H., Banach Lattices and Positive Operators, Springer, Berlin, 1974.[8] Scheffold E., FF-Banachverband algebren, Math. Z. 177 (1981), 193–205.[9] Zaanen A.C, Riesz Spaces II, North-Holland, Amsterdam, 1983.Département de Mathématiques, Faculté des sciences de Tunis CampusUniversitaire, 1060 Tunis, TunisieE-mail : Abdelmajid.Triki@fst.rnu.tn(Received February 8, 2001)

On algebra homomorphisms in complex almost $f$-algebras

http://dml.cz/bitstream/handle/10338.dmlcz/119297/CommentatMathUnivCarolRetro_43-2002-1_3.pdf

On algebra homomorphisms in complex almost $f$ -algebras

Abstract

Similar works

Full text

Available Versions

Institute of Mathematics AS CR, v. v. i.

On algebra homomorphisms in complex almost fff-algebras

Abstract

Similar works

Full text

Available Versions

Institute of Mathematics AS CR, v. v. i.

On algebra homomorphisms in complex almost $f$ -algebras