Journal of Mathematical Analysis and Applications

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1 J. Math. Anal. Appl Contents lists available at ScienceDirect Journal of Mathematical Analysis Applications Generalizations of weakly peripherally multiplicative maps between uniform algebras Kristopher Lee, Aaron Luttman Department of Mathematics, Bo 5815, Clarkson University, Potsdam, NY, USA article info abstract Article history: Received 1 May 2010 Available online 26 August 2010 Submitted by K. Jarosz Keywords: Spectral preserver problems Uniform algebra Weak peripheral multiplicativity Algebra isomorphism Let A B be uniform algebras on first-countable, compact Hausdorff spaces X Y, respectively. For f A, theperipheral spectrum of f,denotedby f ={λ σ f : λ = f }, is the set of spectral values of maimum modulus. A map T : A B is weakly peripherally multiplicative if T f T g fg for all f, g A. WeshowthatifT is a surjective, weakly peripherally multiplicative map, then T is a weighted composition operator, etending earlier results. Furthermore, if T 1, T 2 : A B are surjective mappings that satisfy T 1 f T 2 g fg for all f, g A, thent 1 f T 2 1 = T 1 1T 2 f for all f A, themap f T 1 f T 2 1 is an isometric algebra isomorphism Elsevier Inc. All rights reserved. 1. Introduction background The study of a map, not assumed to be linear, between Banach algebras that preserves some property or subset of the spectrum of elements has become known as a spectral preserver problem. If X is a first-countable, compact Hausdorff space, CX is the space of comple-valued, continuous functions on X, T : CX CX is a surjective map that satisfies σ T f T g = σ fg for all f, g CX, Molnár [1] showed that there eists a homeomorphism ϕ : X X such that T f = T 1 f ϕ for all f CX all X, i.e.t is a weighted composition operator. In particular, if T 1 = 1, then T is an isometric algebra isomorphism. In [2], Rao Roy proved that the underlying domain need not be first-countable that similar results hold when the mapping T is a mapping from a uniform algebra A CX where X is the maimal ideal space of A onto itself. This was generalized further by Hatori et al. in [3], to the case where the underlying domains of A B need not be the maimal ideal spaces. Throughout, A CX B CY refer to uniform algebras on an arbitrary compact Hausdorff spaces. In fact, the full spectrum need not be preserved to achieve results of this type; it can be replaced by the peripheral spectrum, f = { λ σ f : λ = f }, 1 the set of spectral values of maimum modulus, where f denotes the uniform norm of f A. Mappings that satisfy T f T g = fg are called peripherally multiplicative, it was shown in [4] that if A B are uniform algebras on compact Hausdorff spaces X Y, respectively, T : A B is a surjective, peripherally multiplicative mapping, then T is a weighted composition operator. If, in addition, T 1 = 1, then T is an isometric algebra isomorphism. Related work on peripherally multiplicative mappings in settings outside of uniform algebras can be found in [5]. * Corresponding author. addresses: leekm@clarkson.edu K. Lee, aluttman@clarkson.edu A. Luttman X/$ see front matter 2010 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 K. Lee, A. Luttman / J. Math. Anal. Appl The proofs of each of the results above proceed by constructing a homeomorphism between the Choquet boundaries of the algebras A B, this is done by analyzing the effect of T on the peaking functions. The set of peaking functions in a uniform algebra A is the collection PA = { h A: h ={1} }, i.e. the set of functions h such that h 1forall X h =1 if only if h = 1. Again it is possible to generalize these results ask whether or not the entire peripheral spectrum must be multiplicatively preserved, it turns out that the answer is no. In [6], it was shown that mappings T : A B between uniform algebras that satisfy T PA = PB, T 1 = 1, T f T g σπ fg 2 must be isometric algebra isomorphisms. A mapping that satisfies 2 is called weakly peripherally multiplicative, the first goal of this work is to show that preserving the peaking functions is an unnecessary assumption to guarantee that such a map is a weighted composition operator, in the case that the underlying domains are first-countable. Whereas A B refer to uniform algebras on arbitrary compact Hausdorff spaces, we denote by A B uniform algebras on first-countable compact Hausdorff spaces. Theorem 1. Let A B be uniform algebras on first-countable, compact Hausdorff spaces X Y, respectively, let T : A B be surjective weakly peripherally multiplicative. Then the map Φ : A B defined by Φ f = T 1T f is an isometric algebra isomorphism. In fact, in this case it is again true that T is a weighted composition operator. This etends the results in [6], in the case that the underlying spaces X Y are first-countable; related results have been shown in algebras of Lipschitz functions [7,8] in function algebras without units [9]. A natural net step is to analyze pairs of mappings that jointly satisfy criteria such as 2. Hatori et al. have shown [10] that if T 1, T 2 : A B are surjections between uniform algebras such that T 1 f T 2 g = fg, thent 1 T 2 must be weighted composition operators. We etend their results under the assumption of first-countability the results above by the following: Theorem 2. Let A B be uniform algebras on first-countable, compact Hausdorff spaces X Y, respectively, let T 1, T 2 : A B be surjective maps satisfying T1 f T 2 g fg 3 for all f, g A. ThenT 1 f T 2 1 = T 1 1T 2 f holds for all f A, themappingφ : A B defined by Φ f = T 1 f T 2 1 is an isometric algebra isomorphism. The maimizing set of f A, denotedm f ={ X: f = f }, is the set of points where f attains its maimum modulus; the maimizing set of a peaking function h is often called its peak set. The collection of peaking functions that contain a point 0 X in their peak set is denoted by P 0 A ={h PA: 0 Mh}, a point X is called a weak peak point or strong boundary point or p-point if h P A Mh ={}. Equivalently, X is a weak peak point if only if for every open neighborhood U of, there eists a peaking function h P A such that Mh U. It is well known that in uniform algebras the Choquet boundary of A, denotedδ A, is the set of all weak peak points. An important relationship between δ A the peripheral spectrum of functions in a uniform algebra is that, given λ f, there eists an δ A such that f = λ. A fundamental use of the peaking functions is that they can multiplicatively isolate the values of functions at points of the Choquet boundary. This fact, originally due to Bishop [11, Theorem 2.4.1], has been generalized in many directions e.g. [6, Lemma 3], [3, Lemma 2.3], among several others. This result is essential to the work here, so we include the version given by Hatori et al. in [12]. Lemma 1. See [12, Proposition 2.2]. Let A be a uniform algebra on a compact Hausdorff space X, 0 δ A, f A. If f 0 0, then there eists h P 0 A such that fh ={f 0 }. If f 0 = 0, then for any ε > 0 there eists a peaking function h P 0 A such that fh < ε. A strong peak point is a point X such that M f ={} for some f A. Clearly,if X is a strong peak point, then δ A, but the converse is not true in general, as uniform algebras need not have any strong peak points [13, E. 10, p. 54]. If X is first-countable, however, then all weak peak points are strong peak points [14, Lemma 12.1, p. 56], leading to the following corollary:

3 110 K. Lee, A. Luttman / J. Math. Anal. Appl Corollary 1. Let A be a uniform algebra on a first-countable, compact Hausdorff space X; let 0 δa; let f A be such that f 0 0. Then there eists h PA such that Mh = M fh ={ 0 }.Inparticular, fh ={f 0 }. Proof. Let 0 δa, then, since X is first-countable, { 0 } is a G δ set. Thus there eists a countable collection of open sets {U n } n=1 such that n=1 U n ={ 0 }. Since 0 δa, for each n there eists an h n P 0 A such that Mh n U n. Thus n=1 Mh n ={ 0 }. Set h = h n n=1 2 n, then h PA Mh ={ 0 }. By Lemma 1, there eists a k PA such that fk ={f 0 }. Therefore kh PA is the peaking function we seek, M fkh ={ 0 }. Note that the difference between this Lemma 1 is that this result ensures the peak set of h consists solely of the point 0. Following the arguments in [8], for each X we define the set F A = { f A: f = f = 1 }. Notice that P A F A that f, g F A imply fg F A. A useful property of these sets is that they can identify elements of the Choquet boundary, as shown by the following lemma. Lemma 2. Let A be a uniform algebra on a compact Hausdorff space X, δ A, X. Then = if only if F A F A. Proof. The forward direction is clear, so we suppose F A F A.SinceX is Hausdorff, there eist open sets U, V such that U, V, U V =. Thus there eists a peaking function h P A such that Mh U,hence h < 1, which contradicts F A F A. The assumption that δ A is essential to the result of Lemma 2. Consider the disk algebra, AD, the set of continuous functions on the closed unit disk that are analytic on the interior of the disk. It is well known that δ AD = T ={z C: z =1}. Ifz D, then, by the maimum modulus principle, F z AD consists precisely of the constant functions of modulus one. Hence for any pair z 1, z 2 D, we have that F z1 AD = F z2 AD. 2. Weak peripheral multiplicativity As described above, a mapping T : A B that satisfies T f T g σπ fg 4 for all f, g A is called weakly peripherally multiplicative. In general, a weakly peripherally multiplicative map need not be an algebra isomorphism, as is shown in [6, Eample 2] where also the following proposition is proven: Proposition 1. See [6, Theorem 3]. Let A B be uniform algebras on compact Hausdorff spaces X Y, respectively. If T : A B is a weakly peripherally multiplicative map such that PB ={T 1T h: h PA}, then the map Φ : A BdefinedbyΦ f = T 1T f is an isometric algebra isomorphism. In the case that X Y are first-countable, Theorem 1 is more general than Proposition 1, as it shows that the requirement that PB ={T 1T h: h PA} is superfluous. The proof of Theorem 1 will follow from Proposition 1 by showing that if T is a weakly peripherally multiplicative map, then the map f T 1T f automatically preserves the peaking functions General results on weakly peripherally multiplicative maps In this section we assume that A CX B CY are uniform algebras on first-countable, compact Hausdorff spaces that T : A B is a surjective, weakly peripherally multiplicative map. Note that 4 implies T f T g = fg 5 for all f, g A. A mapping that satisfies 5 is called norm multiplicative, the following proposition shows that such a mapping, when restricted to the Choquet boundary, is a composition operator in modulus. Proposition 2. See [15, Theorem 4.1.2]. If Ψ : A B is surjective norm multiplicative, then there eists a homeomorphism τ : δa δb such that Ψf τ = f 6 holds for all δa.

4 K. Lee, A. Luttman / J. Math. Anal. Appl Note that any surjective, weakly peripherally multiplicative map satisfies the hypothesis of Proposition 2,, as is shown net, must be injective. Lemma 3. A surjective, weakly peripherally multiplicative map T : A B is injective. Proof. Let 0 δa, f, g A, suppose that T f = T g. If f 0 = 0, then, by 6, 0 = f 0 = T f τ 0 = T gτ 0 = g 0, sog 0 = f 0. If f 0 0, then 6 implies that T f τ 0 0. Hence, by Corollary 1, there eists a peaking function k P τ 0 B such that Mk = MT f k ={τ 0 }, thus T f k ={T f τ 0 }. Let h A be such that T h = k, choose M fh δa. By 5 6, T f τ k τ = T f τ T h τ = f h = fh = T f T h = T f k. Since MT f k ={τ 0 }, τ = τ 0, the injectivity of τ yields = 0.ThusM fh δa ={ 0 }, which gives that fh ={f 0 h 0 }. A similar argument shows gh ={g 0 h 0 }. As T f = T g, 4 implies that T f T h σπ fh, T f T h = σπ T gt h σπ gh. Since T f T h ={T f τ 0 }, we have that f 0 h 0 = T f τ 0 g 0 h 0 = T f τ 0. Thus f 0 h 0 = g 0 h 0, which implies that f 0 = g 0,since h 0 = Thτ 0 = kτ 0 =1. Therefore f = g for all δa, i.e.t is injective. If T preserves the peaking functions, then it is straightforward to show that T 1 2 = 1 [6], but the following lemma demonstrates that this is true even when the peaking functions are not assumed to be preserved. Lemma 4. A surjective, weakly peripherally multiplicative map T satisfies T 1 2 = 1. Proof. Firstly note that T 1τ = 1 =1forall δa. Choose y 0 δb, let 0 δa be such that τ 0 = y 0. Corollary 1 implies that there eists a peaking function k PB such that Mk = MT 1 2 k ={y 0 },so T 1 2 k = {T 1 2 y 0 }. Notice also that T 1 2 k 2 ={T 1 2 y 0 }. Since T is surjective, there eists h A such that T h = T 1k, so choose Mh δa. By 6, k τ = T 1 τ k τ = T h τ = h = h = T 1T h = T 1 2 k = T 1 2 y 0 = 1. Now ky =1 if only if y = y 0,soτ = y 0 = τ 0, which yields that = 0. Therefore Mh δa ={ 0 }, h ={h 0 }. Note further that h 2 ={h 2 0 }. By 4 T 1 2 k h, T 1 2 k 2 h 2, so h 0 = T 1 2 y 0 h 2 0 = T 1 2 y 0.Nowh 0 0, since h0 = T h τ 0 = T 1 τ 0 k τ 0 = T 1 τ 0 ky0 = T 1 τ 0 = 1. Thus T 1 2 y 0 = T 1 2 y 0 2, which implies that T 1 2 y 0 = 1. Since T 1 2 is identically 1 on δb, it is identically 1 on all of Y.

5 112 K. Lee, A. Luttman / J. Math. Anal. Appl Proof of Theorem 1 Given a surjective, weakly peripherally multiplicative map T : A B, defineφ : A B by Φf = T 1T f. 7 Lemma 5. The mapping Φ is surjective, unital, weakly peripherally multiplicative, satisfies Φ f f. Proof. Lemma4impliesthatT 1 2 = 1, so Φ1 = T 1 2 = 1, i.e. Φ is unital. In addition, Φ f Φg = T 1T f T 1T g = T f T g holds for all f, g A. Hence, the weak peripheral multiplicativity of T implies that Φ is weakly peripherally multiplicative. The fact that Φ is weakly peripherally multiplicative unital immediately give Φ f σπ f 8 for all f A. Let g B, then, by the surjectivity of T, there eists an f A such that T f = T 1g. ThusΦ f = T 1T f = T 1 2 g = g, which implies that Φ is surjective. Since Φ is surjective weakly peripherally multiplicative, all of the results of Section 2.1 hold for Φ. Thus, by Lemma 3, Φ is injective has a formal inverse Φ 1 : B A, which satisfies Φ 1 f Φ 1 g fg 9 for all f, g B. ThusΦ 1 is a bijective, unital, weakly peripherally multiplicative map. As noted before, Theorem 1 follows if the map Φ preserves the peaking functions. We first show that Φ preserves functions whose maimizing sets are singletons. Lemma 6. Let h A 0 δa.thenmh δa ={ 0 } if only if MΦh δb ={τ 0 },inwhichcaseh 0 = Φhτ 0. Proof. Note that Mh δa ={ 0 } if only if h < h for all δa \{ 0 }, which holds if only if Φhτ < Φh for all τ δb \{τ 0 }, i.e. if only if MΦh δb ={τ 0 }. In this case, h ={h 0 } Φh ={Φhτ 0 }, so 8 implies h 0 = Φhτ 0. Note that this implies a certain class of peaking functions are preserved the peaking functions that peak at a single point but not necessarily that all peaking functions are preserved. We now proceed with the proof of Theorem 1. Theorem 1. Let A B be uniform algebras on first-countable compact Hausdorff spaces X Y, respectively, let T : A B be surjective weakly peripherally multiplicative. Then the map Φ : A B defined by Φ f = T 1T f is an isometric algebra isomorphism. Proof. By Lemma 5, Φ is a surjective, unital, weakly peripherally multiplicative operator, so it is only to show that Φ satisfies Φ[PA]=PB. Choose h PA y 0 MΦh δb. SinceΦ preserves norm, 1 = h = Φh, so Φhy 0 =1. By Corollary 1, there eists k PB such that Mk = MΦhk ={y 0 }. Set 0 = τ 1 y 0 g = Φ 1 k, note that Mg ={ 0 } g 0 = 1, by Lemma 6. If Mhg δa, then Φh τ Φg τ = h g = hg = ΦhΦg = Φhk, which implies that τ = y 0 = τ 0,i.e. = 0. Therefore Mhg δa ={ 0 }, h0 = Φh τ 0 = Φhy 0 = 1. Since h is a peaking function, h 0 = 1, so hg ={1}. Thus1 Φhk ={Φhy 0 }, i.e.φhy 0 = 1. Since this holds for any y 0 MΦh δb, Φh ={1}, which is to say Φh PB. This shows that Φ[PA] PB, a similar argument with Φ 1 proves the reverse inclusion. Since Φ is a surjective, unital, weakly peripherally multiplicative operator that preserves the peaking functions, Proposition 1 gives that Φ is an isometric algebra isomorphism.

6 K. Lee, A. Luttman / J. Math. Anal. Appl Jointly weakly peripherally multiplicative maps Given the above results on single weakly peripherally multiplicative operators, it is natural to analyze pairs of maps that jointly satisfy related conditions. Throughout this section we assume that A CX B CY are uniform algebras on first-countable, compact Hausdorff spaces X Y, T 1, T 2 : A B are surjective mappings that satisfy T1 f T 2 g fg for all f, g A. Notice that 10 implies that T1 f T 2 g = fg holds for all f, g A General results on T 1 T 2 WebeginwithsomepropertiesofT 1 T 2. Lemma 7. Let f, g A, then the following are equivalent: a f g for all δa, b T 1 f y T 1 gy for all y δb, c T 2 f y T 2 gy for all y δb. Proof. a b. Let f, g A be such that f g on δa, then fh gh for h A. Ifk PB h A is such that T 2 h = k, then, by 11, T1 f k = T1 f T 2 h = fh gh = T1 gt 2 h = T1 gk. Since k PB was chosen arbitrarily, T 1 f y T 1 gy for all y δb see e.g. [4, Lemma 2]. b c. Let f, g A be such that T 1 f y T 1 gy for all y δb, then T 1 f k T 1 gk for all k B. If k PB h A is such that T 1 h = k, then T2 f k = fh = T1 f T 2 h T1 gt 2 h = gh = T2 gk. As k PB was arbitrarily chosen, T 2 f y T 2 gy for all y δb. c a. Suppose that f, g A are such that T 2 f y T 2 gy for all y δb, then T 2 f k T 2 gk for any k B. Ifh PA, then 11 implies that fh = T1 f T 2 h T2 gt 2 h = gh. By the liberty of the choice of h, f g for all δa. Given h, k F A, 11 implies that T 1 ht 2 k = hk =1, hence MT 1 ht 2 k ={y Y : T 1 ht 2 ky =1}. Following the argument pioneered by Molnár [1], for each δa, wedefinetheset A = M T 1 ht 2 k. 12 h,k F X Lemma 8. For each δa,theseta is non-empty. Proof. We will show that the family {MT 1 ht 2 k: h, k F A} has the finite intersection property. Let h 1,...,h n, k 1,...,k n F A set h = h 1 h n F A k = k 1 k n F A. Since h i ζ 1 k i ζ 1forall1 i n all ζ δa, hζ h i ζ kζ k i ζ for any 1 i n all ζ δa. Lemma 7 implies that T 1 hη T 1 h i η T 2 kη T 2 k i η for any 1 i n all η δb. Since maimizing sets meet the Choquet boundary [15, Lemma 3.2.3], there eists a y MT 1 ht 2 k δb. Hence1= T 1 hyt 2 ky T 1 h i yt 2 k i y 1foreach 1 i n, thus T 1 h i yt 2 k i =1foreach1 i n. Thusy n i=1 MT 1h i T 2 k i, {MT 1 ht 2 k: h, k F A} has the finite intersection property as claimed. Since maimizing sets are compact subsets of the compact set Y, A is non-empty. Since A is a non-empty intersection of maimizing sets, it meets the Choquet boundary [15, Lemma 3.2.3]. Lemma 9. Let f, g A. Then for each δa each y A δb, fg F A if only if T 1 f T 2 g F y B.

7 114 K. Lee, A. Luttman / J. Math. Anal. Appl Proof. Fi δa, choose y A δb. IfT 1 f T 2 g F y B, then, by 11, 1 = T 1 f T 2 g = fg. Thusweneed only show that f g =1. If f g = 0, then, without loss of generality, assume f = 0. Hence Lemma 1 implies that there eists a peaking function h P A such that fh < 1 g.ash F A, Lemma8yieldsT 1 ht 2 h F y B, so 1 = T1 f T 2 gt 1 ht 2 h T1 f T 2 h T1 ht 2 g = fh gh < 1 g g =1, which is a contradiction. Hence f g 0, i.e. f, g 0, thus, by Corollary 1, there eist peaking functions h 1, h 2 P A such that Mh 2 = M fh 2 ={} Mh 1 = Mgh 1 ={}. As h 1, h 2 F A, again Lemma 8 implies that T 1 h 1 T 2 h 2 F y B, thus f g = fh 2 gh 1 = T1 f T 2 h 2 T1 h 1 T 2 g T1 f T 2 gt 1 h 1 T 2 h 2 = 1. Therefore f g =1, yielding fg F A. Conversely, suppose that fg F A, then1= fg = T 1 f T 2 g, it is only to show that T 1 f yt 2 gy =1. If T 1 f yt 2 gy = 0, then, without loss of generality, assume T 1 f y = 0. Lemma 1 implies that there eists an k P y B such that T 1 f k < 1 T 2 g.leth 1, h 2 A be such that T 1 h 1 = T 2 h 2 = k, then, as T 1 h 1 T 2 h 2 F y B, we have that h 1 h 2 F A, thus 1 = fgh 1 h 2 fh 1 gh 2 = T1 f k T2 gk < 1 T 2 g T2 g = 1, which is a contradiction. Hence T 1 f yt 2 gy 0, thus T 1 f y, T 2 gy 0, so by Corollary 1, there eist peaking functions k 1, k 2 P y B such that Mk 1 = MT 1 f k 1 ={y} Mk 2 = MT 2 gk 2 ={y}. Leth 1, h 2 A be such that T 1 h 1 = k 2 T 2 h 2 = k 1,thenh 1 h 2 F A, hence T1 f yt 2 gy = T1 f k 1 T2 gk 2 = fh2 gh 1 fgh 1 h 2 =1, which implies that T 1 f yt 2 gy =1. Therefore T 1 f T 2 g F y B. Not only is A δb non-empty, but the following lemma shows that it is, in fact, a singleton. Lemma 10. For each δa,theseta δb is a singleton. Proof. Fi δa, let y, y A δb. Ify y, then there eist open sets U V such that y U, y V, U V =. Since y δb, there eists a peaking function k P y B such that Mk U. If h 1, h 2 A are such that T 1 h 1 = T 2 h 2 = k, then Lemma 9 implies that h 1 h 2 F A. Hence, by Lemma 9, k 2 = T 1 h 1 T 2 h 2 F y B, which is a contradiction. Thus no two distinct points y y are elements of A δb, i.e.a δb is a singleton. Given that A δb is a singleton for each δa, we define the map τ : δa δb by { τ } = A δb. 13 Lemma 11. The map τ : δa δb defined by 13 is injective. Proof. Let, δa, choose h F A. By Lemma 9, T 1 ht 2 h F τ B. Ifτ = τ,thent 1 ht 2 h F τ B, which, again by Lemma 9, shows h 2 F A thus h F A. Lemma2thengives =. Lemma 12. Let f, g A,then T 1 f τ T 2 gτ = f g holds for all δa. Proof. If any of f, g, T 1 f, T 2 g is identically 0, then the result follows by 11, thus we can assume that f, g, T 1 f, T 2 g 0. Let δa. If f g = 0, then, without loss of generality, we can assume that f = 0. Given ε > 0, Lemma 1 implies that there eists a peaking function h P A such that fh < ε g.ash2 F A, Lemma 9 gives T 1 ht 2 h F τ B, so T1 f τ T 2 g τ T1 f T 2 gt 1 ht 2 h T1 f T 2 h T1 ht 2 g = fh gh < ε g g =ε. Therefore T 1 f τ T 2 gτ = 0, by the liberty of the choice of ε. A symmetric argument shows that T 1 f τ T 2 gτ = 0implies f g = 0.

8 K. Lee, A. Luttman / J. Math. Anal. Appl If f g 0, then f, g 0. Hence, by Corollary 1, there eist peaking functions h 1, h 2 P A such that Mh 2 = M fh 2 ={} Mh 1 = Mgh 1 ={}. Since h 1 h 2 F A, Lemma 9 implies that T 1 h 1 T 2 h 2 F τ B, hence T1 f τ T 2 g τ T1 f T 2 gt 1 h 1 T 2 h 2 T1 f T 2 h 2 T1 h 1 T 2 g = fh 2 gh 1 = f g. Since T 1 f τ T 2 gτ = 0 if only if f g = 0, the assumption that f g 0 implies that T 1 f τ T 2 gτ 0, then T 1 f τ, T 2 gτ 0. Corollary 1 implies that there eist k 1, k 2 P τ B such that Mk 2 = MT 1 f k 2 ={τ } Mk 1 = MT 2 gk 1 ={τ }. Let h 1, h 2 A be such that T 1 h 1 = k 1 T 2 h 2 = k 2.Ask 1 k 2 F τ B, Lemma9yieldsthath 1 h 2 F A, thus f g fgh1 h 2 fh 2 gh 1 = T1 f k 2 T2 gk 1 = T1 f τ T 2 g τ. Therefore T 1 f τ T 2 gτ = f g holds for all δa. Lemma 13. The mappings T 1 T 2 are injective. Proof. Let f, g A be such that T 1 f = T 1 g let 0 δa. If f 0 = 0, then, by Lemma 12, f 0 = T 1 f τ 0 T 2 1τ 0 = T 1 gτ 0 T 2 1τ 0 = g 0, thus f 0 = g 0. If f 0 0, then Lemma 12 implies that T 1 f τ 0 0. Hence, by Corollary 1, there eists a peaking function k P τ 0 B such that MT 1 f k ={τ 0 } T 1 f k ={T 1 f τ 0 }. Let h A be such that T 2 h = k. We claim that fh ={f 0 h 0 }. Indeed,let M fh δa, then, by 11 Lemma 12, T1 f τ k τ = T1 f τ T 2 h τ = f h = fh = T1 f T 2 h = T1 f k. Since MT 1 f k ={τ 0 }, τ = τ 0,so = 0.ThusM fh δa ={ 0 },i.e. fh ={f 0 h 0 }. A similar argument shows that gh ={g 0 h 0 }. As T 1 f = T 1 g, 10 implies that T1 f T 2 h fh, T1 f T 2 h gh. Thus f 0 h 0 =T 1 f τ 0 = g 0 h 0.Letl A be such that T 1 l = k, then l 0 h 0 = T 1 lτ 0 T 2 hτ 0 = kτ 0 2 = 1, thus h 0 0, which yields that f 0 = g 0. Therefore f = g for all δa, i.e.t 1 is injective. A similar argument applies to T 2. Since T 1, T 2 are injective, there eist formal inverses T 1 1, T 1 2 : B A. In addition, T 1 1 T 1 2 are bijective mappings that satisfy T f T2 g fg for all f, g B, thus all of the previous results for T 1 T 2 hold for T 1 1 T 1 2. Hence there eists an injective mapping ψ : δb δa such that T f yt2 gy = f ψy g ψy 15 holds for all y δb f, g B. Lemma 14. The map τ : δa δb defined by 13 is surjective. Proof. Let y δb let k F τ ψy B. Ifh 1, h 2 A are such that T 1 h 1 = T 2 h 2 = k, then Lemma 9 yields that h 1 h 2 F ψy A. Hence, by 15, 1 = h 1 ψyh 2 ψy = ky 2,thusk F y B. Therefore, by Lemma 2, y = τ ψy. 14 Of course a similar argument applies to ψ, showing that τ ψ are mutual inverses of each other. As τ is a bijective mapping between δa δb, Lemma 12 implies that T 1 1T 2 1 =1 onδb. In fact, as the following lemma shows, T 1 1T 2 1 = 1.

9 116 K. Lee, A. Luttman / J. Math. Anal. Appl Lemma 15. The mappings T 1 T 2 satisfy T 1 1T 2 1 = 1. Proof. Let y 0 δb 0 = ψy 0. Lemma 12 implies that T 1 1y 0 T 2 1y 0 =1, thus, by Corollary 1, there eists a peaking function k P y0 B such that MT 1 1T 2 1k ={y 0 }. Notice that T 1 1T 2 1k ={T 1 1y 0 T 2 1y 0 } T 1 1T 2 1k 2 ={T 1 1y 0 T 2 1y 0 }. Let h 1, h 2 A be such that T 1 h 1 = T 1 1k T 2 h 2 = T 2 1k. If Mh 1 δa, then, by Lemma 12, T1 1T 2 1k = T1 h 1 T 2 1 = h 1 = h1 = T1 h 1 τ T 2 1 τ = T1 1 τ T 2 1 τ k τ. Since MT 1 1T 2 1k ={y 0 },wehavethatτ = y 0 = τ 0, so the injectivity of τ implies that Mh 1 δa ={ 0 }. A similar argument yields that Mh 2 δa ={ 0 },hence h 1 ={h 1 0 }, h 2 ={h 2 0 }, h 1 h 2 ={h 1 0 h 2 0 }. By 10, h 1 T1 1T 2 1k, h 2 T1 1T 2 1k, h 1 h 2 T1 1T 2 1k 2, so h 1 0 = h 2 0 = h 1 0 h 2 0 = T 1 1y 0 T 2 1y 0, which yields that T 1 1y 0 T 2 1y 0 = T 1 1y 0 T 2 1y 0 2. Therefore, T 1 1y 0 T 2 1y 0 = 1, i.e. T 1 1yT 2 1y = 1forally δb Proof of Theorem 2 Define the mappings Φ 1,Φ 2 : A B by Φ 1 f = T 1 f T 2 1 Φ 2 f = T 1 1T 2 f. Lemma 16. The mappings Φ 1,Φ 2 : A B are surjective, unital, satisfy Φ 1 f Φ 2 g fg for all f, g A. Proof. By Lemma 15, Φ 1 1 = Φ 2 1 = T 1 1T 2 1 = 1, hence Φ 1 Φ 2 are unital. Additionally, Φ 1 f Φ 2 g = T 1 f T 2 1T 1 1T 2 g = T 1 f T 2 g holds for all f, g A. Thus, by 10, Φ 1 f Φ 2 g fg. Letg B, then the surjectivity of T 1 implies that there eists an f A such that T 1 f = T 1 1g. ThusΦ 1 f = T 1 f T 2 1 = T 1 1T 2 1g = g, which implies that Φ 1 f is surjective. The surjectivity of Φ 2 is proved similarly. As Φ 1 Φ 2 are unital satisfy Φ 1 f Φ 2 g fg, we have also that Φ1 f f, Φ2 f f, Φ1 f = Φ2 f = f holds for all f A. In addition, Φ 1 Φ 2 are surjective, thus the results of Section 3.1 hold for Φ 1 Φ 2.Inparticular, Lemma 12 yields that Φ 1 f τ = f = Φ 2 f τ 19 holds for all f A δa. Lemma 17. Let h A 0 δa, then the following are equivalent: a Mh δa ={ 0 }, b MΦ 1 h δb ={τ 0 }, c MΦ 2 h δb ={τ 0 }. In any case Φ 1 hτ 0 = h 0 = Φ 2 hτ 0. The proof of Lemma 17 is similar to the proof of Lemma 6. Notice that if Φ 1 = Φ 2,thenΦ 1 is a surjective, unital, weakly peripherally multiplicative map, hence Theorem 1 implies that Φ 1 is an isometric algebra isomorphism. Therefore the proof of Theorem 2 follows from showing that Φ 1 f = Φ 2 f holds for all f A.

10 K. Lee, A. Luttman / J. Math. Anal. Appl Theorem 2. Let A B be uniform algebras on first-countable compact Hausdorff spaces X Y, respectively, let T 1, T 2 : A B be surjective mappings that satisfy T 1 f T 2 g fg for all f, g A. ThenT 1 f T 2 1 = T 1 1T 2 f holds for all f A, the mapping Φ : A B defined by Φ f = T 1 f T 2 1 is an isometric algebra isomorphism. Proof. Let Φ 1 Φ 2 be defined as above, let f A y 0 δb. If 0 δa satisfies τ 0 = y 0 Φ 1 f y 0 = 0, then by 19, Φ 1 f y 0 = f 0 = Φ 2 f y 0, soφ 1 f y 0 = Φ 2 f y 0. Suppose that Φ 1 f y 0 0, then 19 implies that f 0 0, hence Corollary 1 implies that there eists a peaking function h P 0 A such that Mh = M fh ={ 0 } fh ={f 0 }. Ify MΦ 1 f Φ 2 h δb, then, by 19, f τ 1 y h τ 1 y = Φ 1 f y Φ 2 h y = Φ 1 f Φ 2 h = fh. Since M fh ={ 0 }, τ 1 y = 0,thusy = τ 0 = y 0. Therefore MΦ 1 f Φ 2 h δb ={y 0 },hence Φ 1 f Φ 2 h = {Φ 1 f y 0 Φ 2 hy 0 }. As Mh δa ={ 0 }, Lemma 17 gives MΦ 2 h δb ={y 0 } Φ 2 hy 0 = h 0 = 1, which implies that Φ 1 f Φ 2 h ={Φ 1 f y 0 }. A similar argument shows that Φ 1 hφ 2 f ={Φ 2 f y 0 }. By Lemma 16, Φ1 f Φ 2 h fh Φ1 hφ 2 f fh. Thus Φ 1 f y 0 = f 0 = Φ 2 f y 0, therefore Φ 1 f y = Φ 2 f y for all y δb, which yields that Φ 1 f = Φ 2 f. Since Φ 1 = Φ 2,wehavethatΦ = Φ 1 is a unital, surjective, weakly peripheral multiplicative map. By Theorem 1, Φ is an isometric algebra isomorphism. Acknowledgments The authors would like to thank Scott Lambert Dillon Ethier for helpful discussions on the results, as well as the anonymous reviewers for their comments, which helped to improve the manuscript. References [1] L. Molnár, Some characterizations of the automorphisms of BH CX, Proc. Amer. Math. Soc [2] N.V. Rao, A.K. Roy, Multiplicatively spectrum preserving maps of function algebras, Proc. Amer. Math. Soc [3] O. Hatori, T. Miura, H. Takagi, Characterization of isometric isomorphisms between uniform algebras via non-linear range preserving properties, Proc. Amer. Math. Soc [4] A. Luttman, T. Tonev, Uniform algebra isomorphisms peripheral multiplicativity, Proc. Amer. Math. Soc [5] T. Tonev, A. Luttman, Algebra isomorphisms between stard operator algebras, Studia Math [6] S. Lambert, A. Luttman, T. Tonev, Weakly peripherally multiplicative operators between uniform algebras, in: Contemp. Math., vol. 435, 2007, pp [7] M. Burgos, A. Jiménez-Vargas, M. Villegas-Vallecillos, Non-linear conditions for weighted composition operators between Lipschitz algebras, J. Math. Anal. Appl [8] A. Jiménez-Vargas, A. Luttman, M. Villegas-Vallecillos, Weakly peripherally multiplicative surjections of pointed Lipschitz algebras, Rocky Mountain J. Math , in press. [9] T. Tonev, Weak multiplicative operators on function algebras without units, Banach Center Publ. 2010, in press. [10] O. Hatori, T. Miura, R. Shindo, H. Takagi, Generalizations of spectrally multiplicative surjections between uniform algebras, Rend. Circ. Mat. Palermo [11] A. Browder, Introduction to Function Algebras, W.A. Benjamin Inc., [12] O. Hatori, K. Hino, T. Miura, H. Oka, Peripherally monomial-preserving maps between uniform algebras, Mediterr. J. Math [13] G. Leibowitz, Lectures on Comple Function Algebras, Scott, Foresman Company, [14] T.W. Gamelin, Uniform Algebras, AMS Chelsea Publishing, [15] S. Lambert, Spectral preserver problems in uniform algebras, PhD thesis, 2008.