# Biseparating Maps on Fréchet Function Algebras

Authors: M. S. Hashemi; T. G. Honary; M. Najafi Tavani
Journal: Journal of Function Spaces and Applications
 (2012)

---

## Abstract

LetA and B be strongly regular normal Fréchet function algebras on compact Hausdorff spaces X and Y, respectively, such that the evaluation homomorphisms are continuous on A and B. Then, every biseparating map T:A→B is a weighted composition operator of the form Tf=h·(f∘φ), where φ is a homeomorphism from Y onto X and h is a nonvanishing element of B. In particular, T is automatically continuous.


---

## Body

## 1. Introduction and Preliminaries

Assume thatA and B are spaces of complex functions on topological spaces X and Y, respectively. A linear map T:A→B is called separating or disjointness preserving whenever coz(f)∩coz(g)=∅ implies coz(Tf)∩coz(Tg)=∅, for all f,g∈A, where the cozero set of an element f∈A is defined by coz(f)={x∈X:f(x)≠0}. Equivalently, a linear map T:A→B is separating if for every f,g∈A, the equality f·g=0 implies the equality Tf·Tg=0. Moreover, T is called biseparating if it is bijective and both T and T-1 are separating.The concept of disjointness preserving operators was introduced for the first time in 1940s (see [1, 2]). Since then, many authors have extended this concept to various kinds of Banach algebras. For example in [3], Jarosz has studied separating maps between spaces of continuous scalar-valued functions. He showed that if X and Y are compact Hausdorff spaces, A=C(X), the space of all continuous scalar-valued functions on X and B=C(Y), then every bijective separating map T:A→B is a weighted composition operator of the form Tf(y)=h(y)f(φ(y)), y∈Y and f∈A, where φ is a homeomorphism from Y onto X and h is a nonvanishing continuous complex-valued function on Y. Later, Font extended this result to the case where A and B are regular commutative semisimple Banach algebras satisfying Ditkin’s condition [4]. On the other hand, Gau et al. used an algebraic method to study separating maps between spaces of continuous scalar as well as vector-valued functions in [5, 6]. For more information about separating maps one, can refer to [7–15].In this paper, we generalize the results of Jarosz in [3] to Fréchet function algebras using a similar method as in [5, 6]. Then, we define the concept of a cozero preserving map and show that if A and B are Banach function algebras on compact Hausdorff spaces X and Y, respectively, and T:A→B is a unital cozero preserving map, then T is automatically continuous. Finally, we will find the relation between cozero preserving, separating and biseparating maps between certain Fréchet function algebras. Recently, Li and Wong have obtained several Banach-Stone type theorems for the vector-valued functions, specially in the case that the bijective linear map T:C(X,E)→C(Y,F) preserves zero set containments, that is,
(1.1)Z(f)⊆Z(g)⟺Z(T(f))⊆Z(T(g))(f,g∈C(X,E)),
where X,Y are realcompact or metric spaces and E,F are locally convex spaces [16]. In fact, T preserves zero set containments if and only if T and T-1 are cozero preserving. In Corollary 2.6, we obtain similar results for Ff-algebras.We now present some definitions and known results which we need in the sequel.A Fréchet algebra (F-algebra) is a locally multiplicatively convex algebra (LMC-algebra) A which is also a complete metrizable space. The topology of a Fréchet algebra can be defined by an increasing sequence (pn) of submultiplicative seminorms and without loss of generality; we may assume that pn(1)=1, for all n∈ℕ, if A has unit. An F-algebra A with a defining sequence of seminorms (pn) is denoted by (A,(pn)). The set of all characters (nonzero complex homomorphisms) of an F-algebra (A,(pn)) is denoted by SA, and the continuous character space, or the spectrum of (A,(pn)), denoted by MA, is the set of all continuous characters on A. We always endow SA and MA with the Gelfand topology, and A^ is the set of all Gelfand transforms f^ of elements f in A. The algebra A is called functionally continuous whenever SA=MA.Note that a sequence(fk)k in an F-algebra (A,(pn)) converges to an element f∈A if and only if for each n∈ℕ,pn(fk-f)→0 as k→∞.Definition 1.1.
LetX be a nonempty topological space. A subalgebra A of C(X) is a function algebra on X if A contains the constants and separates the points of X. The algebra A is a Fréchet function algebra (Ff-algebra) or a Banach function algebra (Bf-algebra) on X if A is a function algebra which is also an F-algebra or a Banach algebra, respectively, with respect to some topology.Clearly every Bf-algebra is a Ff-algebra. LetA be an Ff-algebra (Bf-algebra) on X such that the evaluation homomorphisms δx:A→ℂ are all continuous, where δx(f)=f(x) for f∈A and x∈X. It is clear that the map J:X→MA,x↦δx is continuous and injective. If this map is also surjective and its inverse is continuous, then it is a homeomorphism, and in this case, we say that A is a natural Ff-algebra (Bf-algebra) on X, and we identify X with MA, through this map.Note that the evaluation homomorphisms are always continuous in Bf-algebras, but they may not be continuous in Ff-algebras. By [17, Lemma 3.2.5], the class of natural Ff-algebras and the class of unital commutative semisimple Fréchet algebras are the same. Moreover, all Ff-algebras as well as Bf-algebras are semisimple.Example 1.2.
Let(X,d) be a compact metric space and α>0. The algebra of all complex-valued functions f on X for which
(1.2)pα(f)=sup{|f(x)-f(y)|dα(x,y):x,y∈X,x≠y}<∞
is denoted by Lip(X,α), and its subalgebra of those functions with the property limd(x,y)→0(|f(x)-f(y)|/dα(x,y))=0 is denoted by lip(X,α). It is known that Lip(X,α) for 0<α≤1 and lip(X,α) for 0<α<1 are Bf-algebras on X under the norm ∥f∥α=∥f∥X+pα(f).
In the case thatX is a perfect compact plane set which is a finite union of regular sets (see [18] for the definition), the algebra of all functions f with derivatives of all orders (resp., f(k)∈Lip(X,α)f(k)∈lip(X,α) for all k∈ℕ) is denoted by D∞(X) (resp., Lip∞(X,α)lip∞(X,α)) (see, e.g., [19]). It is interesting to note that D∞(X),Lip∞(X,α) and lip∞(X,α) are natural Ff-algebras on X which are not Bf-algebras.
For a function algebraA on a nonempty topological space X and for each nonempty closed subset S of X, we consider the following subsets of A:
(1.3)A00={f∈A:suppfiscompactinX},I(S)={f∈A00:(suppf)∩S=∅},M(S)={f∈A:f(S)=0}.
For x∈X, we usually write Ix for I({x}) and Mx for M({x}). Note that A00 is an ideal in A and A=A00 whenever X is compact.Definition 1.3.
LetA be an Ff-algebra on a topological space X.
(
i
) Ais said to be regular on X if for any nonempty closed subset S of X and each x∈X∖S, there exists f∈A such that f(x)=1 and f(S)={0}, and it is normal if for each nonempty closed subset E and nonempty compact subset F of X with E∩F=∅, there exists f∈A such that f(E)={1} and f(F)={0}.
A commutativeF-algebra is regular (normal) if SA≠∅ and A^ is a regular (normal) Ff-algebra on SA.
(
ii
)
A is said to be a strongly regular algebra if for every f∈A and x∈X with f(x)=0, there exists a sequence {fn} in A00 and open neighborhoods Vn of x such that fn∣Vn=0 for all n∈ℕ, and fn→f as n→∞, or equivalently, Mx=Ix¯ for each x∈X.
(
iii
) An Ff-algebra A on X is said to satisfy Ditkin’s condition if for every f∈A and x∈X with f(x)=0, there exists a sequence {fn} in A00 and open neighborhoods Vn of x such that fn∣Vn=0 for all n∈ℕ, and fnf→f as n→∞, or equivalently, f∈fIx¯ for all x∈X and f∈Mx.It is clear that every strongly regular algebra is regular. Moreover, if an Ff-algebra satisfies Ditkin’s condition, thenA is strongly regular. In general, the converse is not true as the following example shows.Recall that a Banach sequence algebra on a nonempty setS is a Banach algebra A such that c00(S)⊂A⊂ℂS, where c00(S) is the linear span of the set {χs:s∈S} consisting of all characteristic functions χs of the singleton subsets {s} of S.Example 1.4 (see [20, Example 4.5.33]).
Consider the Banach space(L2(𝕋),∥·∥2) where 𝕋=[-π,π]. Let S=[-π/2,π/2] and set
(1.4)W={f∈L2(𝕋):f∣S∈C(S)}.
For each f,g∈W and θ∈[-π,π], set
(1.5)(f⋆g)(θ)=12π∫-ππ‍f(θ-ϕ)g(ϕ)dϕ,∥f∥=∥f∥2+∥f∥S=12π(∫-ππ‍|f(θ)|2dθ)1/2+∥f∥S.
Then, (W,⋆,∥·∥) is a commutative Banach algebra (for an equivalent norm).
IdentifyingW with its algebra of Fourier transforms on ℤ,W is a strongly regular Banach sequence algebra on ℤ. Moreover, W#, the unitization of W, is a strongly regular Bf-algebra on ℤ∞, the one point compactification of ℤ. Now, we show that W# does not satisfy Ditkin’s condition. In the following, we write rℤ+s for the subset {rn+s:n∈ℤ} of ℤ. Set F1=4ℤ∪{∞}, F2=(4ℤ+2)∪{∞}, and F=F1∪F2=2ℤ∪{∞}. Define g0 on 𝕋 by
(1.6)g0(θ)=1(|θ|≤π2),g0(θ)=-1(π2<|θ|≤π).
Then, g0∈W and
(1.7)g0^(k)=1π∫0π‍g0(θ)coskθdθ=2kπsin(kπ2)(k∈ℤ,k≠0),
with g0^(0)=0, and so g0^∈M(F). By [20, Example 4.5.33(v)], g0^∈M(F)∖I(F)¯. Since g0^∈M(F), necessarily g0^I∞¯⊂I(F)¯, and so g0^∉g0^I∞¯, where I∞ is the set of all functions in A00 which are zero on a neighborhood of ∞.


## 2. Main Results

We first state the following useful result, which is, in fact, the generalization of [6, Lemma 2.1 and Theorem 2.2].Lemma 2.1.
LetX and Y be compact Hausdorff spaces, A and B normal Ff-algebras on X and Y, respectively, and T:A→B a biseparating map. Then, for each x∈X, there exists a unique y∈Y such that TIx=Iy. If we define φ:Y→X by φ(y)=x, then φ is a homeomorphism.Proof.
We omit the proof, since it is similar to the proofs of [6, Lemma 2.1 and Theorem 2.2].We now bring the following theorem, which is an extension of the results of Jarosz and Font.Theorem 2.2.
Let(A,(pn)) and (B,(qn)) be strongly regular normal Ff-algebras on compact Hausdorff spaces X and Y, respectively, such that the evaluation homomorphisms on A and B are continuous. Then, every biseparating map T:A→B is a weighted composition operator of the form
(2.1)Tf=h·(f∘φ),(f∈A),
where φ is a homeomorphism from Y onto X and h is a nonvanishing element of B. In particular, T is automatically continuous.Proof.
By Lemma2.1, there exists a homeomorphism φ from Y onto X defined by φ(y)=x, where TIx=Iy. We first show that TMx⊆My. Suppose on the contrary that there exists f∈Mx such that Tf(y)≠0. If x belongs to the interior of f-1(0), then f∈Ix, and thus Tf(y)=0, since TIx=Iy. Therefore, we may assume that there exists a net {xλ}λ of distinct elements of X converging to x such that f(xλ) is never zero. Consider the net {yλ}λ in Y such that φ(yλ)=xλ. Clearly, yλ converges to y, and by passing through a subnet if necessary, we may assume that there exists a constant ε such that
(2.2)|Tf(yλ)|≥ε>0,
for all λ. Since f(xλ)→0, we can find a subsequence {f(xn)} such that f(xn)→0. Since xλ→x, it follows that xn→x. By the normality of X, there exists a neighborhood Wn of xn such that Wn∩Wm=∅ if n≠m. Let Vn be a neighborhood of xn such that Vn⊆V¯n⊆Wn. Consider the sequence {yn} in Y such that φ(yn)=xn, for each n∈ℕ. By (2.2), without loss of generality, we may assume that |T(f-f(xn)1)(yn)|≥δ for some positive δ and for all n∈ℕ. Since A is normal, for each n∈ℕ, there exists sn∈A such that sn=1 on Vn and sn=0 on X∖Wn. If we take fn=n(f-f(xn)1), then fn(xn)=0, and since A is strongly regular, we can find hn in A and a neighborhood Un of xn in X such that Un⊆Vn, hn=0 on Un and pn(fn-hn)≤1/n2(pn(sn)+1). If we set ψn=(fn-hn)sn, then pn(ψn)≤1/n2, and hence for each k∈ℕ, k≥2, we have
(2.3)∑n=1∞‍pk(ψn)≤∑n=1k-1‍pk(ψn)+∑n=k∞‍pn(ψn)≤∑n=1k-1‍pk(ψn)+∑n=k∞‍1n2<∞.
Therefore, ∑n=1∞‍ψn converges to an element ψ∈A. On the other hand, for each n∈ℕ, ψn=fn on Un, which implies that ψn-fn∈Ixn and T(ψn-fn)∈Iyn. Consequently, Tψn(yn)=Tfn(yn). Since the evaluation homomorphisms are continuous on A, the series ∑n=1∞‍ψn(x) converges to ψ(x) for each x∈X. Hence, ψ=ψn on Wn, since the elements of the sequence {Wn} are pairwise disjoint and cozψn⊂Wn. Therefore,
(2.4)|Tψ(yn)|=|Tψn(yn)|=|Tfn(yn)|=n|T(f-f(xn)1)(yn)|≥nδ,
for all n∈ℕ, which is a contradiction, since yn=φ-1(xn)→φ-1(x) and Tψ(yn)→Tψ(φ-1(x)). Therefore, TMx⊆My.
By a similar argument, we can show thatT-1My⊆Mx and hence TMx=My. Thus kerδx=ker(δy∘T), and so there exists a scalar h(y) such that δy∘T=h(y)δx. Equivalently, Tf(y)=h(y)f(φ(y)) for all f in A and y in Y. In particular, when f=1, we have h=T1, which is a nonvanishing element of B, since T is surjective.Definition 2.3.
LetA and B be Ff-algebras on compact Hausdorff spaces X and Y respectively. A linear map T:A→B is called cozero preserving, whenever coz(f)⊆coz(g) implies coz(Tf)⊆coz(Tg).In [21], Font has studied the automatic continuity of cozero preserving maps between Fourier algebras. In the following theorems, we generalize the results of Font to Bf-algebras as well as Ff-algebras.Theorem 2.4.
LetA and B be Bf-algebras on compact Hausdorff spaces X and Y, respectively, such that B is inverse closed. If T:A→B is a unital cozero preserving surjective map, then T is automatically continuous.Proof.
Letλ∉spA(f). Then, (λ1-f)∈Inv(A), and so coz(λ1-f)=X. Therefore, we have X=coz(1)⊆coz(λ1-f). Since T1=1 and T is cozero preserving, we conclude that Y=coz(1)⊆coz(λ1-Tf). Since B is inverse closed and (λ1-Tf)(y)≠0 for all y∈Y, it follows that λ∉spB(Tf), which implies that spB(Tf)⊆spA(f), for every f∈A. Thus by [20, Theorem 5.1.9(iii)], 𝔖(T)⊆rad(B)={0}, and hence T is automatically continuous.We now adopt a similar method as in the proof of [6, Lemma 3.3] to obtain the following results.Theorem 2.5.
LetA and B be function algebras on compact Hausdorff spaces X and Y, respectively, and T:A→B a cozero preserving injection. If A is regular, then T-1 is separating.Proof.
Suppose on the contrary that there existf and g in A such that TfTg=0 but f(x)g(x)≠0 for some x∈X. So, we can find an open neighborhood V of x such that V⊆coz(f)∩coz(g). Since A is regular, there exists h∈A such that h(x)=1 and h|X∖V=0. It is clear that coz(h)⊆coz(f)∩coz(g). So by hypothesis, coz(Th)⊆coz(Tf)∩coz(Tg). On the other hand, TfTg=0 implies that coz(Tf)∩coz(Tg)=∅. It follows that coz(Th)=∅, that is, Th=0. Now injectivity of T shows that h=0, which is a contradiction. Therefore, T-1 is separating.Corollary 2.6.
LetA and B be strongly regular normal Ff-algebras on compact Hausdorff spaces X and Y, respectively, such that evaluation homomorphisms are continuous on A and B. If T:A→B is a linear bijection, then the following statements are equivalent: (i)
Tis separating and cozero preserving;(ii)
T is biseparating;(iii)
T and T-1 are both cozero preserving;(iv)
T and T-1 are weighted composition operators.Proof.
It suffices to prove(ii)⇒(i). The other implications are direct consequences of Theorems 2.2 and 2.5. If (ii) is satisfied, then all hypotheses of Theorem 2.2 are satisfied. Let for f,g∈A and y∈Y, we have coz(f)⊆coz(g) and y∉coz(Tg). By Theorem 2.2, TMx=My. Since Tg(y)=0, it follows that g(x)=0, and hence f(x)=0, that is, f∈Mx. Thus, Tf(y)=0 and consequently y∉coz(Tf).



---

*Source: 947080-2012-12-27.xml*