Surjective isometries on a Banach space of analytic functions on the open unit disc
aa r X i v : . [ m a t h . F A ] J a n SURJECTIVE ISOMETRIES ON A BANACH SPACEOF ANALYTIC FUNCTIONS ON THE OPEN UNITDISC
TAKESHI MIURA
Abstract.
Let H ( D ) be the linear space of all analytic functionson the open unit disc D . We define S ∞ by the linear subspaceof all f ∈ H ( D ) with bounded derivative f ′ on D . We give thecharacterization of surjective, not necessarily linear, isometries on S ∞ with respect to the following two norms: k f k ∞ + k f ′ k ∞ and | f ( a ) | + k f ′ k ∞ for a ∈ D , where k·k ∞ is the supremum norm on D . Introduction and Main results
Let ( M, k·k M ) and ( N, k·k N ) be real or complex normed linear spaces.A mapping T : M → N is said to be an isometry if and only if k T ( f ) − T ( g ) k N = k f − g k M ( f, g ∈ M ) . Here we notice that isometries need not be linear. A linear map S : M → N is an isometry if and only if it preserves the norm inthe following sense: k S ( f ) k N = k f k M for f ∈ M . Banach [1] gavethe characterization of surjective isometries on the Banach space of allcontinuous real valued functions C R ( K ), K a compact metric space,with the supremum norm. Many mathematicians have been investi-gating linear isometries on various subspaces of all analytic functions Mathematics Subject Classification.
Key words and phrases. bounded analytic function, disc algebra, extreme point,isometry, Lipschitz space, uniform algebra.The author is supported by JSPS KAKENHI Grant Numbers 15K04921 and16K05172. H ( D ) on the open unit disc D . Nagasawa [18] characterized surjectivecomplex linear isometries on uniform algebras. By the characterizationof Nagasawa, we see that every surjective complex linear isometry onthe Hardy space H ∞ ( D ) is a weighted composition operator. deLeeuw,Rudin and Wermer [5] obtained the form of surjective complex linearisometries on H ∞ ( D ) and H ( D ). Forelli [8] extended their results tothe characterization of complex linear isometries, which need not besurjective, on the Hardy space H p ( D ) for 1 ≤ p < ∞ with p = 2. Re-sults for isometries on other spaces were obtained by Botelho [2], Cimaand Wogen [4], Hornor and Jamison [13] and Kolaski [14].Let f ′ be the derivative of f ∈ H ( D ). Novinger and Oberlin [19]obtained the characterization of complex linear isometries on the space S p = { f ∈ H ( D ) : f ′ ∈ H p ( D ) } for 1 ≤ p < ∞ , p = 2 with respectto the norms | f (0) | + k f ′ k H p and k f k ∞ + k f ′ k H p , where k·k H p is thenorm of H p ( D ). They assume no surjectivity of isometries, while theydo assume that p is finite for S p . The purpose of this paper is to givethe form of all surjective, not necessarily linear, isometries on S ∞ withthe complete norms k f k ∞ + k f ′ k ∞ and | f ( a ) | + k f ′ k ∞ for a ∈ D .The following are the main results of this paper. Theorem 1. If T : S ∞ → S ∞ is a surjective, not necessarily linear,isometry with the norm k f k Σ = sup z ∈ D | f ( z ) | + sup w ∈ D | f ′ ( w ) | for f ∈ URJECTIVE ISOMETRIES ON A BANACH SPACE 3 S ∞ , then there exist constants c, λ ∈ C with | c | = | λ | = 1 such that T ( f )( z ) = T (0)( z ) + cf ( λz ) ( f ∈ S ∞ , z ∈ D ) , or T ( f )( z ) = T (0)( z ) + cf ( λz ) ( f ∈ S ∞ , z ∈ D ) . Conversely, each of the above forms is a surjective isometry on S ∞ with the norm k·k Σ , where T (0) is an arbitrary element of S ∞ . Theorem 2.
Let a ∈ D . If T : S ∞ → S ∞ is a surjective, not necessar-ily linear, isometry with the norm k f k σ = | f ( a ) | + sup w ∈ D | f ′ ( w ) | for f ∈ S ∞ , then there exist constants c , c , λ ∈ C with | c | = | c | = | λ | =1 and b ∈ D such that, for the function ρ ( z ) = λ z − bbz − defined on D , T ( f )( z ) = T (0)( z ) + c f ( a ) + Z [ a,z ] c f ′ ( ρ ( ζ )) dζ , ( f ∈ S ∞ , z ∈ D ) , or T ( f )( z ) = T (0)( z ) + c f ( a ) + Z [ a,z ] c f ′ ( ρ ( ζ )) dζ , ( f ∈ S ∞ , z ∈ D ) , or T ( f )( z ) = T (0)( z ) + c f ( a ) + Z [ a,z ] c f ′ ( ρ ( ζ )) dζ , ( f ∈ S ∞ , z ∈ D ) , or T ( f )( z ) = T (0)( z ) + c f ( a ) + Z [ a,z ] c f ′ ( ρ ( ζ )) dζ ( f ∈ S ∞ , z ∈ D ) . Conversely, each of the above forms is a surjective isometry on S ∞ with the norm k·k σ , where T (0) is an arbitrary element of S ∞ . T. MIURA
We may consider the space S A of all f ∈ H ( D ), whose derivative f ′ can be extended to a continuous function on D . Surjective isometrieson S A with the norm | f (0) | + k f ′ k ∞ are characterized in [17].2. Preliminaries and an embedding of S ∞ into C ( X )Let H ( D ) be the algebra of all analytic functions on the open unitdisc D = { z ∈ C : | z | < } in the complex number field C . The constantfunction in H ( D ), which takes the value only 1, is denoted by . Wedenote by H ∞ ( D ) the commutative Banach algebra of all boundedanalytic functions on D with the supremum norm k v k D = sup z ∈ D | v ( z ) | for v ∈ H ∞ ( D ). The maximal ideal space of H ∞ ( D ) is denoted by M .The Gelfand transform b v of v ∈ H ∞ ( D ) is a continuous function on M with k b v k M = sup η ∈M | b v ( η ) | = k v k D for every v ∈ H ∞ ( D ). In the restof this paper, we denote by A the uniform algebra { b v ∈ C ( M ) : v ∈ H ∞ ( D ) } on M , i.e., A is a uniformly closed subalgebra of C ( M ), thealgebra of all continuous complex valued functions on M , that containsthe constant functions and separates points of M . It is well-known thatthe Choquet boundary Ch( A ) for A is the Shilov boundary ∂ A of A .The Shilov boundary ∂ A for A is the smallest closed subset of M withthe property that k b v k ∂ A = sup η ∈ ∂ A | b v ( η ) | = k b v k M for b v ∈ A . We definea Banach space S ∞ by S ∞ = { f ∈ H ( D ) : f ′ ∈ H ∞ ( D ) } with norms k f k Σ and k f k σ , defined by k f k Σ = k f k D + k f ′ k D and k f k σ = | f ( a ) | + k f ′ k D URJECTIVE ISOMETRIES ON A BANACH SPACE 5 for f ∈ S ∞ , where a ∈ D . We see that S ∞ coincides with the set ofall Lipschitz functions in H ( D ) (see Remark 4.1). Let A ( D ) be theBanach algebra of all analytic functions on D which can be extendedto continuous functions on the closed unit disc D . We see that S ∞ is asubset of A ( D ) by [6, Theorem 3.11]. It is well-known that the maximalideal space of A ( D ) is D . We denote by b f the Gelfand transform of f ∈ A ( D ); the Gelfand transforms of H ∞ ( D ) and A ( D ) are representedby the same symbol b · . Let T be the unit circle in C . Then T is theShilov boundary for A ( D ), that is, T is the smallest closed subset of D with the property that k b f k T = sup z ∈ T | b f ( z ) | = k f k D for all f ∈ A ( D ).The identity function on D is denoted by id. We refer the reader to [3]for information about uniform algebras.Let a ∈ D and T ∈ { T , { a }} . We define X T = T × ∂ A × T . Then X T is a compact Hausdorff space with the product topology. Foreach f ∈ S ∞ , we define e f by(2.1) e f ( z, η, w ) = b f ( z ) + c f ′ ( η ) w ( z, η, w ) ∈ X T . As we mentioned, b f ∈ “ A ( D ) and “ f ′ ∈ A for f ∈ S ∞ . We define anormed linear subspace B T of C ( X T ) by B T = { e f ∈ C ( X T ) : f ∈ S ∞ } T. MIURA with the supremum norm k e f k X T = sup x ∈ X T | e f ( x ) | for e f ∈ B T . Notethat k v k D = k b v k M = k b v k ∂ A for v ∈ H ∞ ( D ). Thus, we observe that(2.2) k e f k X T = k b f k T + k “ f ′ k ∂ A = k f k Σ and k e f k X { a } = | f ( a ) | + k “ f ′ k ∂ A = k f k σ for every f ∈ S ∞ .For each x ∈ X T , we define the point evaluational functional δ x : B T → C at x by δ x ( e f ) = e f ( x ) ( e f ∈ B T ) . We define the complex linear map I : H ∞ ( D ) → H ∞ ( D ) by(2.3) I ( v )( z ) = Z [0 ,z ] v ( ζ ) dζ ( z ∈ D ) , where [0 , z ] is the straight line interval from 0 to z .To analyze isometries on S ∞ , we need some properties of the maximalideal space M of H ∞ ( D ). We refer the reader to the following booksfor information about the maximal ideal space M of H ∞ ( D ), [10] and[12]. Proposition 2.1.
Let
T ∈ { T , { a }} for a ∈ D . For each pair ofdistinct points x , x ∈ X T = T × ∂ A × T there exists e f ∈ B T suchthat e f ( x ) = e f ( x ) .Proof. Let x j = ( z j , η j , w j ) ∈ X T for j = 1 , x = x . Supposethat z = z , and we set ξ j = “ id( η j ) ∈ T for j = 1 ,
2. Then we havethree cases to consider.
URJECTIVE ISOMETRIES ON A BANACH SPACE 7
Case 1. ξ , ξ
6∈ { z , z } . Let f ( z ) = ( z − z )( z − ξ ) ( z − ξ ) ∈ S ∞ .Then we obtain b f ( z ) = 0 = b f ( z ) and “ f ′ ( ξ ) = 0 = “ f ′ ( ξ ). Equality(2.1) shows that e f ( x ) = f ( z ) = 0 = e f ( x ).Case 2. ξ j ∈ { z , z } and ξ k
6∈ { z , z } for j, k ∈ { , } with j = k .Without loss of generality, we may and do assume that ξ ∈ { z , z } and ξ
6∈ { z , z } . For functions g ( z ) = (2 z − z + z )( z − z ) ∈ S ∞ and g ( z ) = (2 z − z + ξ )( z − ξ ) ∈ S ∞ , we see that c g ( z ) = 0 = c g ( z ) = ” g ′ ( z ) = ” g ′ ( z )and c g ( z ) = 0 = c g ( ξ ) = ” g ′ ( z ) = ” g ′ ( ξ ) . We set g = g g ∈ S ∞ , and then b g ( z ) = 0 = b g ( z ). The multiplicationlaw, g ′ = g ′ g + g g ′ , shows that “ g ′ ( z ) = 0 = “ g ′ ( z ) = “ g ′ ( ξ ). Hence “ g ′ ( ξ ) = “ g ′ ( ξ ) = 0. Since g ′ is a polynomial, it is constant on each fiber { η ∈ M : “ id( η ) = ξ } of M over ξ ∈ T . Therefore, “ g ′ ( η j ) = “ g ′ ( ξ j ) = 0for j = 1 ,
2. It follows from (2.1) that e g ( x ) = b g ( z ) = 0 = e g ( x ).Case 3. ξ , ξ ∈ { z , z } . Let h ( z ) = (2 z − z + z )( z − z ) ∈ S ∞ ,and then we see that b h ( z ) = 0 = b h ( z ) and “ h ′ ( z ) = 0 = “ h ′ ( z ). Bythe choice of ξ j , we have “ h ′ ( ξ j ) = 0 for j = 1 ,
2. By (2.1), we get e h ( x ) = b h ( z ) = 0 = e h ( x ).Now we assume that z = z and w = w . We have that e id( x ) = z + w = z + w = e id( x ) by (2.1).If z = z and w = w , then we have η = η . Since the Gelfandtransform A of H ∞ ( D ) separates points of M , there exists v ∈ H ∞ ( D )such that b v ( η ) = b v ( η ). As z = z and w = w , we see that ] I ( v )( x ) = ] I ( v )( x ) by (2.1) and (2.3). T. MIURA
Consequently, e f ( x ) = e f ( x ) for some e f ∈ B T , as is claimed. (cid:3) Essential ideas of Proof of Lemmas 2.2 and 2.3 are due to OsamuHatori. The author is grateful for his comments and suggestions onthem.
Lemma 2.2.
Let W = { z ∈ D : 1 − ε < | z |} and W = { z ∈ D : | z | < − ε } for each ε with < ε < . Then M \ W is the closureof W in M .Proof. Let cl( W ) be the closure of W in M . Since M \ W is a closedsubset of M containng W , we see that cl( W ) ⊂ M \ W .Now we shall show that M \ D ⊂ cl( W ). Let η ∈ M \ D . By theCorona Theorem, the open unit disc D is dense in M , and then thereexists a net { z ν } in D such that { z ν } converges to η with respect to theweak*-topology. We assert that η ∈ cl( W ); in fact, if η were not incl( W ), then we would have that η is in the open set M \ cl( W ). Since { z ν } converges to η , we may and do assume that z ν ∈ M \ cl( W ) forall ν . Then we get z ν ∈ D \ W for all ν . Let “ id − ( ξ ) be the fiber over ξ ∈ T . Since η ∈ M \ D = ∪ ξ ∈ T “ id − ( ξ ), there exists a unique ξ ∈ T such that η ∈ “ id − ( ξ ). Since z ν W , we have | z ν | ≤ − ε , andthen | z ν − ξ | ≥ ε for all ν . By the choice of η and { z ν } , we obtain z ν = “ id( z ν ) → “ id( η ) = ξ , which contradicts | z ν − ξ | ≥ ε for all ν . Thisimplies that η ∈ cl( W ). Consequently, we have that M \ D ⊂ cl( W ). URJECTIVE ISOMETRIES ON A BANACH SPACE 9
It is obvious that { z ∈ D : 1 − ε ≤ | z |} ⊂ cl( W ). As M \ D ⊂ cl( W ), we get M \ W = { z ∈ D : 1 − ε ≤ | z |} ∪ ( M \ D ) ⊂ cl( W ).Thus we have that cl( W ) = M \ W . (cid:3) Lemma 2.3.
Let ε > , η ∈ ∂ A and V an open neighborhood of η in M . There exists f ∈ S ∞ such that k f k D ≤ ε , “ f ′ ( η ) = 1 = k “ f ′ k M and | “ f ′ | < ε on M \ V .Proof. Let ε > η ∈ ∂ A and let V be an open neighborhood of η in M . Let 2 ε = min { ε, } , and we set W = { z ∈ D : 1 − ε < | z |} and W = { z ∈ D : | z | < − ε } . By Lemma 2.2, cl( W ) = M \ W andhence η ∈ cl( W ). We shall prove that η is an interior point of cl( W ).If O were not contained in cl( W ) for all open neighborhood O of η ,there would exist a net { η ν } in M \ cl( W ) such that { η ν } convergesto η with respect to the weak*-topology. Since cl( W ) = M \ W ,we get η ν ∈ W ⊂ D , and hence | η ν | < − ε for all ν . Therefore, η ν = “ id( η ν ) → “ id( η ) = 1, which contradicts | η ν | < − ε for all ν . Wehave proven that η is an interior point of cl( W ).Let O be an open neighborhood of η in M such that O ⊂ cl( W ).We set V = V ∩ O , and then V is an open neighborhood of η with V ∩ D ⊂ cl( W ) ∩ D = { z ∈ D : 1 − ε ≤ | z |} . Since η ∈ ∂ A = Ch( A ),the Choquet boundary for A , there exists v ∈ H ∞ ( D ) such that b v ( η ) =1 = k b v k M and | b v | < ε on M \ V . Define f ∈ S ∞ by f = I ( v ), andthen f ′ = v on D . By the choice of V , “ f ′ ( η ) = 1 = k “ f ′ k M . Since V ⊂ V and ε < ε , we obtain | “ f ′ | < ε on M \ V . We show that k f k D ≤ ε . Let z be an arbitrary point of D . If z ∈ W = { z ∈ D : | z | < − ε } , then the line interval [0 , z ] from 0to z lies in D \ V since V ∩ D ⊂ { z ∈ D : 1 − ε ≤ | z |} . It follows that | v | < ε on [0 , z ], and thus | f ( z ) | = | I ( v )( z ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z [0 ,z ] v ( ζ ) dζ (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | sup ζ ∈ [0 ,z ] | v ( ζ ) | < ε . Hence, | f ( z ) | < ε for z ∈ W .If z ∈ D \ W , then we set z r = (1 − rε ) z / | z | for each r with1 < r <
2. Note that ε < rε < ε ≤
1. Since | z r | = 1 − rε < − ε ,we obtain z r ∈ W . Then we have that | f ( z r ) | ≤ ε by the fact provedin the last paragraph. Recall that | z r | = 1 − rε and | z | ≥ − ε bythe choice of z r and z . Then we get | z r − z | < rε . It follows that | f ( z ) | = | I ( v )( z ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z [0 ,z r ] v ( ζ ) dζ (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z [ z r ,z ] v ( ζ ) dζ (cid:12)(cid:12)(cid:12)(cid:12) ≤ | f ( z r ) | + | z r − z | sup ζ ∈ [ z r ,z ] | v ( ζ ) | ≤ ε + rε = (1 + r ) ε . Since r > | f ( z ) | ≤ ε ≤ ε for z ∈ D \ W .We thus conclude that | f | ≤ ε on D , and consequently k f k D ≤ ε . (cid:3) Lemma 2.4.
Let x = ( z , η , w ) ∈ X T for T ∈ { T , { a }} with a ∈ D .Then µ ( T × { η } × T ) = 1 for each representing measure µ for δ x ,Proof. Let x = ( z , η , w ) be an arbitrary point of X T = T × ∂ A × T .Let µ be a representing measure for δ x , that is, µ is a regular Borelmeasure on X T such that δ x ( e f ) = Z X T e f dµ for every e f ∈ B T : Such ameasure do exist by the Riesz representation theorem. Since δ x ( e ) =1 = k δ x k , we see that µ is a probability measure (see, for example, [3,p. 81]), where k·k is the operator norm. URJECTIVE ISOMETRIES ON A BANACH SPACE 11
Let ε > V an openneighborhood of η ∈ ∂ A in M . We set V c = M \ V , P V = T × V × T and P V c = T × V c × T . Then we see that(2.4) ( z , η , w ) ∈ P V , P V ∪ P V c = X T and P V ∩ P V c = ∅ . By Lemma 2.3, there exists f ∈ S ∞ such that k f k D ≤ ε, ” f ′ ( η ) = 1 = k ” f ′ k M and | ” f ′ | < ε on V c . Recall that e f ( z, η, w ) = c f ( z ) + ” f ′ ( η ) w by (2.1). By the choice of f ,we have | e f | ≤ ε + 1 on P V and | e f | < ε on P V c . Since µ is a representing measure for δ x , we get Z X T e f dµ = δ x ( e f ) = c f ( z ) + ” f ′ ( η ) w = c f ( z ) + w . It follows from the above equalities with (2.4) that1 − ε ≤ | c f ( z ) + w | = (cid:12)(cid:12)(cid:12)(cid:12)Z X T e f dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z P V e f dµ (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z P V c e f dµ (cid:12)(cid:12)(cid:12)(cid:12) < ( ε + 1) µ ( P V ) + 2 εµ ( P V c ) . By the liberty of the choice of ε >
0, we get 1 ≤ µ ( P V ) ≤ µ ( X T ) = 1.Hence 1 = µ ( P V ) = µ ( T × V × T ) for all open neighborhood V of η .Since µ is a regular measure, we have that µ ( T × { η } × T ) = 1. (cid:3) Lemma 2.5. If x = ( z , η , w ) ∈ T × ∂ A × T with η “ id − ( z ) , then µ ( { z } × { η } × T ) = 1 for each representing measure µ for δ x . Proof.
Let ε > W an open neighborhood of z ∈ T in T . We set W c = T \ W , Q W = W × { η } × T and Q W c = W c × { η } × T . Thenwe see that Q W ∪ Q W c = T × { η } × T and Q W ∩ Q W c = ∅ . By Lemma 2.4, we obtain µ ( Q W ∪ Q W c ) = 1. For each m ∈ N , we set g m ( z ) = { ( z z + 1) / } m for z ∈ D . Then we see that g m ∈ S ∞ , ” g m ( z ) = 1 = k ” g m k D and | ” g m ( z ) | < z ∈ T \ { z } . Let ξ = “ id( η ) ∈ T . Since ξ = z , we have | ( z ξ + 1) / | < m { ( z ξ + 1) / } m − converges to 0 as m →∞ . In addition, since | ” g m ( z ) | < z ∈ T \ { z } , we can find m ∈ N such that | ‘ g m | < ε on W c and | ‘ g m ′ ( η ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m z (cid:18) z ξ + 12 (cid:19) m − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ε. We set f = g m ∈ S ∞ , and hence c f ( z ) = 1 = k c f k D , | c f | < ε on W c and | ” f ′ ( η ) | < ε. Equality (2.1) with the above shows that | δ x ( e f ) | = | c f ( z ) + ” f ′ ( η ) w | ≥ − ε. Since µ ( Q W ∪ Q W c ) = 1, it follows that1 − ε ≤ | δ x ( e f ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z Q W ∪ Q Wc e f dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z Q W n c f ( z ) + ” f ′ ( η ) w o dµ (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z Q Wc n c f ( z ) + ” f ′ ( η ) w o dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ (1 + ε ) µ ( Q W ) + 2 εµ ( Q W c ) . URJECTIVE ISOMETRIES ON A BANACH SPACE 13
By the liberty of the choice of ε >
0, we have that 1 ≤ µ ( Q W ) ≤ µ ( Q W ) = µ ( W × { η } × T ) for all open neighborhood W of z . Since µ is regular, we obtain µ ( { z } × { η } × T ) = 1. (cid:3) Lemma 2.6.
Let x = ( z , η , w ) ∈ X T for T ∈ { T , { a }} with a ∈ D .If η “ id − ( z ) , then the Dirac measure concentrated at x is the uniquerepresenting measure for δ x .Proof. Let x = ( z , η , w ) ∈ X T be such that η “ id − ( z ). Let µ bea representing measure for δ x . We shall prove that µ ( { x } ) = 1. Set R = { z } × { η } × T . We have two possible cases to consider: If T = T ,then µ ( R ) = 1 by Lemma 2.5, and if T = { a } , then R = T × { η } × T and hence µ ( R ) = 1 by Lemma 2.4. In any case, we obtain µ ( R ) = 1.Here, we remark that we need the assumption that η “ id − ( z ) forthe case when T = T ; in fact, if T = { a } , then z = a . Therefore, | “ id( η ) | = 1 > | a | = | z | , which shows that “ id( η ) = z .Let f = id − z ∈ S ∞ . Then c f ( z ) = 0 and ” f ′ ( η ) = 1. By (2.1), δ x ( e f ) = c f ( z ) + ” f ′ ( η ) w = w . Since µ is a representing measure for δ x , the fact that µ ( R ) = 1 impliesthat w = Z R e f dµ = Z R n c f ( z ) + ” f ′ ( η ) w o dµ = Z R w dµ. Since µ ( R ) = 1, we have that Z R ( w − w ) dµ = 0. Thus, there exists Z ⊂ R = { z } × { η } × T such that µ ( Z ) = 0 and w − w = 0 on ( { z } × { η } × T ) \ Z. This shows that µ ( { z } × { η } × { w } ) = 1, and hence µ ( { x } ) = 1.We have that µ is a Dirac measure concentrated at x = ( z , η , w ),as is claimed. (cid:3) Recall that B T = { e f ∈ C ( X T ) : f ∈ S ∞ } for X T = T × ∂ A × T with T ∈ { T , { a }} and a ∈ D . Let Ch( B T ) be Choquet boundary for B T ,that is, the set of all x ∈ X T such that δ x is an extreme point of ( B T ) ∗ ,the closed unit ball of the dual space ( B T ) ∗ of ( B T , k · k X T ). We denoteby ext(( B T ) ∗ ) the set of all extreme points of ( B T ) ∗ . By the Arens-Kelly theorem, we see that ext(( B T ) ∗ ) = { λδ x : λ ∈ T , x ∈ Ch( B T ) } (cf. [7, Corollary 2.3.6 and Theorem 2.3.8]). Lemma 2.7.
Let K = { ( z , η , w ) ∈ X T : η “ id − ( z ) } . Then K ⊂ Ch( B T ) .Proof. Let x = ( z , η , w ) ∈ K , and then η “ id − ( z ). We showthat δ x ∈ ext(( B T ) ∗ ). Let δ x = ( χ + χ ) / χ , χ ∈ ( B T ) ∗ . Then χ ( e ) + χ ( e ) = 2 δ x ( e ) = 2 by (2.1). Since χ j ∈ ( B T ) ∗ , | χ j ( e ) | ≤ χ j ( e ) = 1 = k χ j k for j = 1 ,
2, where k·k is the operatornorm on ( B T ) ∗ . Let σ j be a representing measure for χ j , that is, χ j ( e f ) = Z X T e f dσ j for e f ∈ B T . We see that σ j is a probability measureon X T (cf. [3, p. 81]). Then ( σ + σ ) / δ x . It follows from Lemma 2.6 that ( σ + σ ) / τ x concentrated at x . Since σ j is a positive measure, σ j ( E ) = 0 foreach Borel set E with x E . Hence σ j = τ x for j = 1 ,
2, andconsequently χ = χ . Therefore, δ x is an extreme point of ( B T ) ∗ ,and thus x ∈ Ch( B T ). This shows that K ⊂ Ch( B T ). (cid:3) URJECTIVE ISOMETRIES ON A BANACH SPACE 15
Let k·k ς = k·k Σ or k·k ς = k·k σ and T a surjective, not necessar-ily linear, isometry from ( S ∞ , k·k ς ) onto itself. Define a mapping T : S ∞ → S ∞ by(2.5) T = T − T (0) . By the Mazur-Ulam theorem [15, 21], T is a surjective, real linear isometry from ( S ∞ , k·k ς ) onto itself. Let U : ( S ∞ , k·k ς ) → ( B T , k · k X T )be a mapping defined by U ( f ) = e f ( f ∈ S ∞ ) . Then we see that U is a surjective complex linear isometry (see (2.1)and (2.2)). Denote U T U − by S ; the mapping S : B T → B T is well-defined since U : ( S ∞ , k·k ς ) → ( B T , k·k X T ) is a surjective complex lin-ear isometry. S ∞ T −−−→ S ∞ U y y U B T −−−→ S B T The equality S = U T U − is equivalent to(2.6) S ( e f ) = ] T ( f ) ( e f ∈ B T ) . We define a mapping S ∗ : ( B T ) ∗ → ( B T ) ∗ by(2.7) S ∗ ( χ )( e f ) = Re [ χ ( S ( e f ))] − i Re [ χ ( S ( i e f ))]for χ ∈ ( B T ) ∗ and e f ∈ B T , where Re z denotes the real part of acomplex number z . The mapping S ∗ is a surjective real linear isometrywith respect to the operator norm on ( B T ) ∗ (cf. [20, Proposition 5.17]). This shows that S ∗ preserves extreme points, that is, S ∗ (ext(( B T ) ∗ )) =ext(( B T ) ∗ ). Lemma 2.8.
The set K = { ( z , η , w ) ∈ X T : η “ id − ( z ) } is densein X T = T × ∂ A × T for T ∈ { T , { a }} with a ∈ D .Proof. If T = { a } , then the condition that η “ id − ( z ) is automati-cally satisfied (see Proof of Lemma 2.6). Then we see that K = X T if T = { a } . We now assume that T = T . Let x = ( z , η , w ) ∈ X T and O an open neighborhood of x in X T . We show that O ∩ K = ∅ .To this end, we may assume that x K , and hence η ∈ “ id − ( z ).By the definition of the product topology, we can choose open sets W , W ⊂ T and V ⊂ ∂ A such that z ∈ W , w ∈ W , η ∈ V and W × V × W ⊂ O . Choose z ∈ W with z = z , and then we obtain( z , η , w ) ∈ O and id( η ) = z = z . Hence η “ id − ( z ). Therefore,( z , η , w ) ∈ O ∩ K . This yields that K is dense in X T . (cid:3) The idea of the following lemma is due to Professor Kazuhiro Kawa-mura. The author is grateful for his suggestion.
Lemma 2.9.
Let B = { λδ x : λ ∈ T , x ∈ X T } be a topological subspaceof ( B T ) ∗ with the relative weak*-topology. Let T × X T be the compactHausdorff space with the product topology for T ∈ { T , { a }} , a ∈ D .Then the map h : T × X T → B defined by h ( λ, x ) = λδ x for ( λ, x ) ∈ T × X T is a homeomorphism. In particular, h ( T × Ch( B T )) = ext(( B T ) ∗ ) .Proof. By the definition of the map h we see that h is surjective. Since B T contains the constant function e and separates points of X T , we see URJECTIVE ISOMETRIES ON A BANACH SPACE 17 that h is injective. By the definition of the weak*-topology, we observethat h is a continuous map from the compact space T × X T onto theHausdorff space B . Hence, h : T × X T → B is a homeomorphism. Bythe Arens-Kelly theorem, ext(( B T ) ∗ ) = { λδ x : λ ∈ T , x ∈ Ch( B T ) } ,and thus ext(( B T ) ∗ ) = h ( T × Ch( B T )). (cid:3) Lemma 2.10. S ∗ ( B ) = B holds for B = { λδ x : λ ∈ T , x ∈ X T } .Proof. Let h be the homeomorphism as in Lemma 2.9. Then we havethat h ( T × Ch( B T )) = ext(( B T ) ∗ ) = S ∗ (ext(( B T ) ∗ )). By Lemma 2.7, K ⊂ Ch( B T ) ⊂ X T , and hence we observe that S ∗ ( h ( T × K )) ⊂ S ∗ ( h ( T × Ch( B T ))) = S ∗ (ext(( B T ) ∗ ))= ext(( B T ) ∗ ) = h ( T × Ch( B T )) ⊂ h ( T × X T ) . By the definition of the map h , we have h ( T × X T ) = B , and hence S ∗ ( h ( T × K )) ⊂ B . We denote by cl E ( M ) the closure a subset M ina topological space E . By Lemma 2.8, K is dense in X T , and thuscl X T ( K ) = X T . Since the map h is a homeomorphism, we get B = h ( T × X T ) = h ( T × cl X T ( K )) = cl B ( h ( T × K )) , and thus B = cl B ( h ( T × K )). Since S ∗ : ( B T ) ∗ → ( B T ) ∗ is a surjectiveisometry with the operator norm, it is a homeomorphism with respectto the weak*-topology on ( B T ) ∗ . Since S ∗ ( h ( T × K )) ⊂ B , we obtain S ∗ ( B ) = S ∗ (cl B ( h ( T × K ))) = cl B ( S ∗ ( h ( T × K ))) ⊂ B , which implies S ∗ ( B ) ⊂ B . By the same arguments, applied to S − ∗ , wehave that S − ∗ ( B ) ⊂ B , and thus S ∗ ( B ) = B is proven. (cid:3) Characterization of the induced map S In this section, we use the same notation as in the previous sec-tion. We assume that S : B T → B T is a surjective real linear isometrythat satisfies (2.6), and S ∗ : ( B T ) ∗ → ( B T ) ∗ is a surjective real linearisometry defined by (2.7). Definition 3.1.
Let h : T × X T → B be a homeomorphism definedin Lemma 2.9. Let p : T × X T → T and p : T × X T → X T be thenatural projections from T × X T onto the first and second coordinate,respectively. We define maps α : T × X T → T and Φ : T × X T → X T by α = p ◦ h − ◦ S ∗ ◦ h and Φ = p ◦ h − ◦ S ∗ ◦ h . T × X T −−−→ T × X T h y y h B −−−→ S ∗ B Recall that S ∗ is a real linear isometry with S ∗ ( B ) = B , and thus S ∗ isa homeomorphism with respect to the relative weak*-topology. Hence α and Φ are both well-defined, surjective continuous maps.By the definitions of α and Φ, ( h − ◦ S ∗ ◦ h )( λ, x ) = ( α ( λ, x ) , Φ( λ, x ))for ( λ, x ) ∈ T × X T . Hence ( S ∗ ◦ h )( λ, x ) = h ( α ( λ, x ) , Φ( λ, x )), whichimplies S ∗ ( λδ x ) = α ( λ, x ) δ Φ( λ,x ) for every ( λ, x ) ∈ T × X T . Set, foreach λ ∈ T , α λ ( x ) = α ( λ, x ) for x ∈ X T . Then α λ satisfies S ∗ ( λδ x ) = α λ ( x ) δ Φ( λ,x ) for every ( λ, x ) ∈ T × X T . Lemma 3.1.
For each x ∈ X T , α i ( x ) = iα ( x ) or α i ( x ) = − iα ( x ) . URJECTIVE ISOMETRIES ON A BANACH SPACE 19
Proof.
Let x ∈ X T , and we set λ = (1 + i ) / √ ∈ T . By the reallinearity of S ∗ , we obtain √ α λ ( x ) δ Φ( λ ,x ) = S ∗ ( √ λ δ x ) = S ∗ ( δ x ) + S ∗ ( iδ x )= α ( x ) δ Φ(1 ,x ) + α i ( x ) δ Φ( i,x ) . Hence √ α λ ( x ) δ Φ( λ ,x ) = α ( x ) δ Φ(1 ,x ) + α i ( x ) δ Φ( i,x ) . Evaluating thisequality at e ∈ B T , we get √ α λ ( x ) = α ( x )+ α i ( x ). Since | α λ ( x ) | = 1for λ ∈ T , we have √ | α ( x ) + α i ( x ) | = | α i ( x ) α ( x ) | . Thenwe see that α i ( x ) α ( x ) = i or α i ( x ) α ( x ) = − i , which implies that α i ( x ) = iα ( x ) or α i ( x ) = − iα ( x ). (cid:3) Lemma 3.2.
There exists a continuous function s : X T → {± } suchthat S ∗ ( iδ x ) = is ( x ) α ( x ) δ Φ( i,x ) for every x ∈ X T .Proof. For each x ∈ X T , α i ( x ) = iα ( x ) or α i ( x ) = − iα ( x ) byLemma 3.1. We define K + and K − by K + = { x ∈ X T : α i ( x ) = iα ( x ) } and K − = { x ∈ X T : α i ( x ) = − iα ( x ) } . Then we have X T = K + ∪ K − and K + ∩ K − = ∅ . By the continuityof the functions α = α (1 , · ) and α i = α ( i, · ), we see that K + and K − are both closed subsets of X T . Hence, the function s : X T → {± } ,defined by s ( x ) = ( x ∈ K + − x ∈ K − , is continuous on X T . Then we have that α i ( x ) = is ( x ) α ( x ) for every x ∈ X T . This shows S ∗ ( iδ x ) = is ( x ) α ( x ) δ Φ( i,x ) for all x ∈ X T . (cid:3) Lemma 3.3.
Let s be the continuous function as in Lemma 3.2. Foreach λ = r + it ∈ T with r, t ∈ R and x ∈ X T , (3.1) λ s ( x ) e f (Φ( λ, x )) = r e f (Φ(1 , x )) + its ( x ) e f (Φ( i, x )) for all e f ∈ B T .Proof. Let λ = r + it ∈ T with r, t ∈ R and x ∈ X T . Recall that S ∗ ( δ x ) = α ( x ) δ Φ(1 ,x ) , and S ∗ ( iδ x ) = is ( x ) α ( x ) δ Φ( i,x ) by Lemma 3.2.Since S ∗ is real linear, we have α λ ( x ) δ Φ( λ,x ) = S ∗ ( λδ x ) = rS ∗ ( δ x ) + tS ∗ ( iδ x )= rα ( x ) δ Φ(1 ,x ) + its ( x ) α ( x ) δ Φ( i,x ) , and thus α λ ( x ) δ Φ( λ,x ) = α ( x ) { rδ Φ(1 ,x ) + its ( x ) δ Φ( i,x ) } . The evaluationof this equality at e ∈ B T shows that α λ ( x ) = α ( x )( r + its ( x )).Since λ = r + it ∈ T and s ( x ) ∈ {± } , r + its ( x ) = λ s ( x ) and hence α λ ( x ) = λ s ( x ) α ( x ). We obtain λ s ( x ) δ Φ( λ,x ) = rδ Φ(1 ,x ) + its ( x ) δ Φ( i,x ) ,which implies that λ s ( x ) e f (Φ( λ, x )) = r e f (Φ(1 , x )) + its ( x ) e f (Φ( i, x ))for all e f ∈ B T . (cid:3) Recall that Φ( λ, x ) ∈ X T = T × ∂ A × T for ( λ, x ) ∈ T × X T byDefinition 3.1. Definition 3.2.
Let q j be the projection from X T = T × ∂ A × T ontothe j -th coordinate of X T for j = 1 , ,
3. Let Φ : T × X T → X T be themap as in Definition 3.1. We define φ : T × X T → T , ψ : T × X T → ∂ A and ϕ : T × X T → T by φ = q ◦ Φ, ψ = q ◦ Φ and ϕ = q ◦ Φ,respectively.
URJECTIVE ISOMETRIES ON A BANACH SPACE 21
For each λ ∈ T , we also denote φ λ ( x ) = φ ( λ, x ), ψ λ ( x ) = ψ ( λ, x ) and ϕ λ ( x ) = ϕ ( λ, x ) for all x ∈ X T .By the definition of φ , ψ and ϕ , Φ( λ, x ) = ( φ ( λ, x ) , ψ ( λ, x ) , ϕ ( λ, x ))for every ( λ, x ) ∈ T × X T . By (2.1),(3.2) e f (Φ( λ, x )) = b f ( φ ( λ, x )) + “ f ′ ( ψ ( λ, x )) ϕ ( λ, x )for all f ∈ S ∞ and ( λ, x ) ∈ T × X T . Note that φ , ψ and ϕ are surjectiveand continuous since so is Φ (see Definition 3.1). Lemma 3.4.
The function φ : X T → T is a surjective function with φ λ ( x ) = φ ( x ) for all x ∈ X T and λ ∈ T .Proof. Let x ∈ X T . First, we show that φ λ ( x ) ∈ { φ ( x ) , φ i ( x ) } for all λ ∈ T . Suppose, on the contrary, that φ λ ( x )
6∈ { φ ( x ) , φ i ( x ) } for some λ ∈ T \ { , i } . Then there exists a polynomial f ∈ S ∞ such that c f ( φ λ ( x )) = 1 , c f ( φ ( x )) = 0 = c f ( φ i ( x ))and ” f ′ ( ψ µ ( x )) = 0 ( µ = λ , , i ) . By (3.2), we have e f (Φ( λ , x )) = 1 and e f (Φ(1 , x )) = 0 = e f (Φ( i, x )).Substituting these equalities into (3.1) to get λ s ( x )0 = 0, which con-tradicts λ ∈ T . Consequently, we have φ λ ( x ) ∈ { φ ( x ) , φ i ( x ) } for all λ ∈ T and x ∈ X T .We next prove that φ ( x ) = φ i ( x ) for all x ∈ X T . To this end,suppose not. Then there exists x ∈ X T such that φ ( x ) = φ i ( x ).Let λ = (1 + i ) / √ ∈ T . Then φ λ ( x ) ∈ { φ ( x ) , φ i ( x ) } by theprevious paragraph. We consider the case when φ λ ( x ) = φ ( x ). We choose a polynomial f ∈ S ∞ so that c f ( φ i ( x )) = 1 , c f ( φ ( x )) = 0 and ” f ′ ( ψ µ ( x )) = 0 ( µ = λ , , i ) . Substituting these equalities into (3.2), we get e f (Φ( i, x )) = 1 and e f (Φ( λ , x )) = 0 = e f (Φ(1 , x )). By (3.1), we obtain 0 = is ( x ),which contradicts s ( x ) ∈ {± } . By a similar arguments, we can reacha contradiction even when φ λ ( x ) = φ i ( x ). Thus, we get φ ( x ) = φ i ( x ) for all x ∈ X T , and consequently φ λ ( x ) = φ ( x ) for all λ ∈ T and x ∈ X T .We show that φ is surjective. Let ξ ∈ T . Since φ is surjective, thereexists ( λ, x ) ∈ T × X T such that φ ( λ, x ) = ξ . By the fact proved in thelast paragraph, we obtain φ ( x ) = φ λ ( x ) = φ ( λ, x ) = ξ . This yieldsthe surjectivity of φ . (cid:3) By a similar argument to Lemma 3.4, we can prove that ψ λ ( x ) = ψ ( x ) for all λ ∈ T and x ∈ X T . Just for the sake of the completeness,here we give its proof. Lemma 3.5.
The function ψ : X T → ∂ A is a surjective continuousfunction with ψ λ ( x ) = ψ ( x ) for all x ∈ X T and λ ∈ T .Proof. Let x ∈ X T . By Lemma 3.4, Φ( λ, x ) = ( φ ( x ) , ψ λ ( x ) , ϕ λ ( x )) for λ ∈ T . Equality (3.2) is reduced to(3.3) e f (Φ( λ, x )) = b f ( φ ( x )) + “ f ′ ( ψ λ ( x )) ϕ λ ( x )for all f ∈ S ∞ and λ ∈ T .First, we show that ψ λ ( x ) ∈ { ψ ( x ) , ψ i ( x ) } for all λ ∈ T . Suppose,on the contrary, that ψ λ ( x )
6∈ { ψ ( x ) , ψ i ( x ) } for some λ ∈ T \ { , i } . URJECTIVE ISOMETRIES ON A BANACH SPACE 23
Then there exists a polynomial f ∈ S ∞ such that c f ( φ ( x )) = 0 , ” f ′ ( ψ λ ( x )) = 1 and ” f ′ ( ψ ( x )) = 0 = ” f ′ ( ψ i ( x )) . By (3.3), e f (Φ( λ , x )) = ϕ λ ( x ) and e f (Φ(1 , x )) = 0 = e f (Φ( i, x )). If wesubstitute these equalities into (3.1), we obtain λ s ( x )0 ϕ λ ( x ) = 0, whichcontradicts λ , ϕ λ ( x ) ∈ T . Consequently, ψ λ ( x ) ∈ { ψ ( x ) , ψ i ( x ) } forall λ ∈ T and x ∈ X T .We next prove that ψ ( x ) = ψ i ( x ) for all x ∈ X T . Suppose that thereexists x ∈ X T such that ψ ( x ) = ψ i ( x ). Let λ = (1 + i ) / √ ∈ T .Then ψ λ ( x ) ∈ { ψ ( x ) , ψ i ( x ) } by the previous paragraph. If weassume ψ λ ( x ) = ψ ( x ), then we can choose a polynomial f ∈ S ∞ so that c f ( φ ( x )) = 0 = c f ′ ( ψ ( x )) and c f ′ ( ψ i ( x )) = 1 . Applying these equalities to (3.3), we obtain e f (Φ( i, x )) = ϕ i ( x ) and e f (Φ(1 , x )) = 0 = e f (Φ( λ , x )) since ψ λ ( x ) = ψ ( x ). By (3.1), wehave 0 = is ( x ) ϕ i ( x ), which is impossible. We can reach a similarcontradiction even if ψ λ ( x ) = ψ i ( x ). Therefore, we conclude that ψ ( x ) = ψ i ( x ) for all x ∈ X T . Consequently ψ λ ( x ) = ψ ( x ) for all λ ∈ T and x ∈ X T .Finally, since ψ is surjective, for each η ∈ ∂ A there exists ( λ, x ) ∈ X T such that ψ ( λ, x ) = η . By the last paragraph, we see that η = ψ λ ( x ) = ψ ( x ), which shows the surjectivity of ψ . (cid:3) Lemma 3.6.
Let ϕ and ϕ i be functions from Definition 3.2. There ex-ists a continuous function s : X T → {± } such that ϕ i ( x ) = s ( x ) ϕ ( x ) for all x ∈ X T .Proof. Let x ∈ X T . According to Lemmas 3.4 and 3.5, we can writeΦ( λ, x ) = ( φ ( x ) , ψ ( x ) , ϕ λ ( x )) for all λ ∈ T . Let λ = (1 + i ) / √ ∈ T and f = id − φ ( x ) ∈ S ∞ . Then c f ( φ ( x )) = 0 and ” f ′ = 1 on ∂ A . By(3.3), e f (Φ( µ, x )) = ϕ µ ( x ) for µ = λ , , i . If we apply these equalitiesto (3.1), then we obtain √ λ s ( x )0 ϕ λ ( x ) = ϕ ( x ) + is ( x ) ϕ i ( x ). As ϕ λ ( x ) ∈ T for all λ ∈ T , we have that √ | ϕ ( x ) + is ( x ) ϕ i ( x ) | = | is ( x ) ϕ i ( x ) ϕ ( x ) | . Then we get is ( x ) ϕ i ( x ) ϕ ( x ) = i or is ( x ) ϕ i ( x ) ϕ ( x ) = − i . Thus, foreach x ∈ X T , we obtain ϕ i ( x ) = s ( x ) ϕ ( x ) or ϕ i ( x ) = − s ( x ) ϕ ( x ).By the continuity of ϕ and ϕ i , there exists a continuous function s : X T → {± } such that ϕ i ( x ) = s ( x ) ϕ ( x ) for all x ∈ X T . (cid:3) In the rest of this paper, we denote a + ibs by [ a + ib ] s for a, b ∈ R and s ∈ {± } . Thus, for each λ ∈ C , [ λ ] s = λ if s = 1 and [ λ ] s = λ if s = −
1. Therefore, [ λµ ] s = [ λ ] s [ µ ] s for all λ, µ ∈ C . If, in addition, λ ∈ T , then [ λ ] s = λ s . Lemma 3.7.
For each f ∈ S ∞ and x ∈ X T , (3.4) S ( e f )( x ) = [ α ( x ) b f ( φ ( x ))] s ( x ) + [ α ( x ) “ f ′ ( ψ ( x )) ϕ ( x )] s ( x ) s ( x ) . Proof.
Let f ∈ S ∞ and x ∈ X T . By the definition (2.7) of S ∗ , we haveRe [ S ∗ ( χ )( e f )] = Re [ χ ( S ( e f ))] for every χ ∈ ( B T ) ∗ . Taking χ = δ x and URJECTIVE ISOMETRIES ON A BANACH SPACE 25 χ = iδ x into the last equality, we getRe [ S ∗ ( δ x )( e f )] = Re [ S ( e f )( x )] and Re [ S ∗ ( iδ x )( e f )] = − Im [ S ( e f )( x )] , respectively, where Im z is the imaginary part of a complex number z .Therefore,(3.5) S ( e f )( x ) = Re [ S ∗ ( δ x )( e f )] − i Re [ S ∗ ( iδ x )( e f )] . Recall that S ∗ ( δ x ) = α ( x ) δ Φ(1 ,x ) by definition, and S ∗ ( iδ x ) = is ( x ) α ( x ) δ Φ( i,x ) by Lemma 3.2. Substitute these two equalities into (3.5) to obtain S ( e f )( x ) = Re [ α ( x ) e f (Φ(1 , x ))] + i Im [ s ( x ) α ( x ) e f (Φ( i, x ))] . Lemmas 3.4, 3.5 and 3.6 imply that Φ(1 , x ) = ( φ ( x ) , ψ ( x ) , ϕ ( x )) andΦ( i, x ) = ( φ ( x ) , ψ ( x ) , s ( x ) ϕ ( x )). Applying these two equalities tothe above formula of S ( e f )( x ), it follows from (2.1) that S ( e f )( x ) = Re [ α ( x ) { f ( φ ( x )) + “ f ′ ( ψ ( x )) ϕ ( x ) } ]+ i Im [ s ( x ) α ( x ) { f ( φ ( x )) + “ f ′ ( ψ ( x )) s ( x ) ϕ ( x ) } ]= [ α ( x ) f ( φ ( x ))] s ( x ) + [ α ( x ) “ f ′ ( ψ ( x )) ϕ ( x )] s ( x ) s ( x ) . This completes the proof. (cid:3) Proof of the main results
In this section, we shall simplify equality (3.4). By (2.6), S ( e f ) = ] T ( f ) for f ∈ S ∞ . Applying (2.1), we can rewrite equality (3.4) as(4.1) ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) w = [ α ( x ) b f ( φ ( x ))] s ( x ) + [ α ( x ) “ f ′ ( ψ ( x )) ϕ ( x )] s ( x ) s ( x ) for all f ∈ S ∞ and x = ( z, η, w ) ∈ X T = T × ∂ A × T for T ∈ { T , { a }} with a ∈ D . Lemma 4.1.
Let x = ( z, η, w ) ∈ X T . The value φ ( x ) is independentof w ∈ T .Proof. Let z ∈ T , η ∈ ∂ A . For each triple of ξ , ξ , ξ ∈ T , we showthat the set G = { φ ( z, η, ξ j ) : j = 1 , , } contains at most two points.Assume that the set G contains three distinct points. Then w , w and w are mutually distinct. We set x j = ( z, η, ξ j ). By our hypothesis, φ ( x ), φ ( x ) and φ ( x ) are distinct points. There exists a polynomial f ∈ S ∞ such that b f ( φ ( x )) = 1 , b f ( φ ( x )) = 0 = b f ( φ ( x ))and “ f ′ ( ψ ( x j )) = 0 ( j = 1 , , . Substituting the above equalities into (4.1), we obtain ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ = [ α ( x )] s ( x ) ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ = 0 = ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ . Since ξ = ξ , the last equalities yield ◊ T ( f ) ′ ( η ) = 0 = ÷ T ( f )( z ). There-fore, [ α ( x )] s ( x ) = 0, which is impossible since α ( x ) ∈ T . Hence,the set G contains at most two points, as is claimed.We now prove that the value φ ( z, η, w ) is independent of w ∈ T .Since φ is continuous, so is the function w φ ( z, η, w ) for w ∈ T .Thus, the set φ ( z, η, T ) = { φ ( z, η, w ) : w ∈ T } is connected. Supposethat φ ( z, η, w ) = φ ( z, η, w ) for some w , w ∈ T . Then we have URJECTIVE ISOMETRIES ON A BANACH SPACE 27 φ ( z, η, w ) ∈ { φ ( z, η, w ) , φ ( z, η, w ) } for all w ∈ T by the fact provedin the last paragraph. This contradicts the connectedness of the set φ ( z, η, T ). Hence φ ( z, η, w ) = φ ( z, η, w ) for all w , w ∈ T . Thisimplies that the value φ ( z, η, w ) is independent of w ∈ T . (cid:3) Lemma 4.2.
Let x = ( z, η, w ) ∈ X T . The value ψ ( z, η, w ) is inde-pendent of w ∈ T .Proof. Let z ∈ T and η ∈ ∂ A . For each triple of ξ , ξ , ξ ∈ T , weprove that the set H = { ψ ( z, η, ξ j ) : j = 1 , , } contains at mosttwo points. Suppose, on the contrary, that H has three points. Then ξ , ξ and ξ are mutually distinct. We set x j = ( z, η, ξ j ) for j = 1 , , φ ( x j ) is independent of j . Since H has three distinctpoints, there exists a polynomial f ∈ S ∞ such that b f ( φ ( x )) = 0 , “ f ′ ( ψ ( x )) = 1 and “ f ′ ( ψ ( x )) = 0 = “ f ′ ( ψ ( x )) . Equality (4.1) with the above equalities shows that ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ = [ α ( x ) ϕ ( x )] s ( x ) s ( x ) ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ = 0 = ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) ξ . As ξ = ξ , the last equalities show that ◊ T ( f ) ′ ( η ) = 0 = ÷ T ( f )( z ).Hence, [ α ( x ) ϕ ( x )] s ( x ) s ( x ) = 0, which contradicts α ( x ) ψ ( x ) ∈ T . We thus conclude that the set H has at most two points.We show that the value ψ ( z, η, w ) is independent of w ∈ T . The set ψ ( z, η, T ) = { ψ ( z, η, w ) : w ∈ T } is connected on T since ψ is con-tinuous. By the last paragraph, we see that ψ ( z, η, w ) = ψ ( z, η, w ) for each w , w ∈ T . Hence, the value ψ ( x, η, w ) does not depend on w ∈ T , as is claimed. (cid:3) Lemma 4.3.
Let s and s are functions as in Lemmas 3.2 and 3.6,respectively. For x = ( z, η, w ) ∈ X T , the values s ( x ) and s ( x ) areindependent of z ∈ T and w ∈ T .Proof. Let k = 0 , η ∈ ∂ A . The function s k ( · , η, · ), which sends( z, w ) to s k ( z, η, w ), is continuous on T × T . Since T ∈ { T , { a }} , theimage s k ( T , η, T ) of T × T under the mapping s k ( · , η, · ) is a connectedsubset of {± } . Then we deduce that the value s k ( z, η, w ) does notdepend on z ∈ T and w ∈ T . (cid:3) By Lemmas 4.1, 4.2 and 4.3 we may write φ ( z, η, w ) = φ ( z, η ), ψ ( z, η, w ) = ψ ( z, η ) and, for k = 0 , s k ( z, η, w ) = s k ( η ) for ( z, η, w ) ∈ X T . Then we can rewrite equality (4.1) as follows:(4.2) ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) w = [ α ( x ) b f ( φ ( z, η ))] s ( η ) + [ α ( x ) “ f ′ ( ψ ( z, η )) ϕ ( x )] s ( η ) s ( η ) for all f ∈ S ∞ and x = ( z, η, w ) ∈ X T . Proposition 4.4.
Let λ, µ ∈ C . If | λ + µw | = 1 for all w ∈ T , then λµ = 0 and | λ | + | µ | = 1 .Proof. Suppose, on the contrary, that λµ = 0. Choose w ∈ T so that µ w = λ | µ || λ | − , and set w = − w . By hypothesis, | λ + µ w | = 1 = | λ + µ w | , that is, (cid:12)(cid:12)(cid:12)(cid:12) λ + λ | µ || λ | (cid:12)(cid:12)(cid:12)(cid:12) = 1 = (cid:12)(cid:12)(cid:12)(cid:12) λ − λ | µ || λ | (cid:12)(cid:12)(cid:12)(cid:12) . URJECTIVE ISOMETRIES ON A BANACH SPACE 29
These equalities yield that | λ | + | µ | = (cid:12)(cid:12) | λ | − | µ | (cid:12)(cid:12) . This implies that λ = 0 or µ = 0, which contradicts the hypothesis that λµ = 0. Thuswe have λµ = 0, and then | λ | + | µ | = 1. (cid:3) Proof of Theorem 1.
In this subsection, we assume that T = T ,and give proof of Theorem 1. In this case, equality (4.2) holds for all f ∈ S ∞ and x = ( z, η, w ) ∈ T × ∂ A × T = X T . Lemma 4.5. (1)
The function s as in Lemma 3.2 is constant on ∂ A ; we will write s instead of s ( η ) . (2) There exists a constant c ∈ T such that (a) ÷ T ( )( z ) = c s for all z ∈ D , (b) ◊ T ( i )( z ) = is ÷ T ( )( z ) for all z ∈ D , (c) α ( x ) = c for all x ∈ T × ∂ A × T .Proof. Let λ ∈ { , i } . If we substitute f = λ ∈ S ∞ into (4.2), thenwe have that(4.3) ◊ T ( λ )( z ) + ÿ T ( λ ) ′ ( η ) w = [ λα ( x )] s ( η ) for all ( z, η, w ) ∈ X T . We show that ÿ T ( λ ) ′ ( η ) = 0 for all η ∈ ∂ A .To this end, suppose not, and then there exists η ∈ ∂ A such that ÿ T ( λ ) ′ ( η ) = 0. Let z ∈ T . By (4.3), | ◊ T ( λ )( z ) + ÿ T ( λ ) ′ ( η ) w | = 1for all w ∈ T . Since ÿ T ( λ ) ′ ( η ) = 0, Proposition 4.4 shows that ◊ T ( λ )( z ) = 0. By the liberty of the choice of z ∈ T , we deduce that ◊ T ( λ ) = 0 on T . Since T ( λ ) is an analytic function on D , we havethat T ( λ ) is identically zero on D . Hence ÿ T ( λ ) ′ = 0 on M , which contradicts ÿ T ( λ ) ′ ( η ) = 0. We thus obtain ÿ T ( λ ) ′ ( η ) = 0 for all η ∈ ∂ A .Since ∂ A is a boundary for A , ÿ T ( λ ) ′ is identically zero on M .We see that T ( λ ) is constant on D since D is connected. We set T ( λ ) = c λ ∈ C , and then c λ = [ λα ( x )] s ( η ) for all x = ( z, η, w ) ∈ X T by (4.3). Then we obtain | c λ | = 1. It follows that c i = [ iα ( x )] s ( η ) =[ i ] s ( η ) [ α ( x )] s ( η ) = is ( η ) c , and hence c i = is ( η ) c for all η ∈ ∂ A .This shows that s is a constant function on ∂ A . We set c = c s , andthen ÷ T ( ) = c = c s and ◊ T ( i ) = c i = is c s on D , and α = c s = c on T × ∂ A × T . (cid:3) By Lemma 4.5, equality (4.2) is reduced to(4.4) ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) w = [ c b f ( φ ( z, η ))] s + [ c “ f ′ ( ψ ( z, η )) ϕ ( x )] s s ( η ) for every f ∈ S ∞ and x = ( z, η, w ) ∈ T × ∂ A × T . Lemma 4.6.
Let c ∈ T be the constant as in Lemma 4.5. Then ◊ T (id)( z ) = [ cφ ( z, η )] s for all z ∈ T and η ∈ ∂ A ; in particular,the function φ is independent of the variable η ∈ ∂ A .Proof. Let z ∈ T and η ∈ ∂ A . We set ξ = φ ( z , η ) and g = id − ξ ∈S ∞ . By the definition of φ , we see that ξ ∈ T . Then b g ( φ ( z , η )) = 0and g ′ = , and thus, by (4.4),(4.5) ÷ T ( g )( z ) + ÷ T ( g ) ′ ( η ) w = [ cϕ ( z , η , w )] s s ( η ) URJECTIVE ISOMETRIES ON A BANACH SPACE 31 for all w ∈ T . Hence | ÷ T ( g )( z ) + ÷ T ( g ) ′ ( η ) w | = 1 for all w ∈ T .Proposition 4.4 shows that ÷ T ( g )( z ) ÷ T ( g ) ′ ( η ) = 0. We shall provethat ÷ T ( g )( z ) = 0. Suppose, on the contrary, that ÷ T ( g )( z ) = 0, andthen ÷ T ( g ) ′ ( η ) = 0. By (4.5), we obtain ÷ T ( g )( z ) = [ cϕ ( z , η , w )] s s ( η ) for all w ∈ T . By the surjectivity of T , there exists h ∈ S ∞ such that ÷ T ( h )( z ) = 0 and ◊ T ( h ) ′ ( η ) = 1. Applying these two equalities to(4.4), we get w = ÷ T ( h )( z ) + ◊ T ( h ) ′ ( η ) w = [ c b h ( φ ( z , η ))] s + [ cϕ ( z , η , w )] s s ( η ) [ “ h ′ ( ψ ( z , η ))] s s ( η ) = [ c b h ( φ ( z , η ))] s + ÷ T ( g )( z )[ “ h ′ ( ψ ( z , η ))] s s ( η ) for all w ∈ T . The rightmost hand side of the above equalities isindependent of w , and thus we arrive at a contradiction. Therefore, wehave ÷ T ( g )( z ) = 0 as is claimed.Since T is real linear, 0 = ÷ T ( g )( z ) = ◊ T (id)( z ) − ◊ T ( ξ )( z ), andthus ◊ T (id)( z ) = ◊ T ( ξ )( z ). Here, we note that ÷ T ( ) = c s and ◊ T ( i ) = is ÷ T ( ) by Lemma 4.5. These imply that ◊ T (id)( z ) = ◊ T ( ξ )( z ) = (Re ξ ) ÷ T ( )( z ) + (Im ξ ) ◊ T ( i )( z )= (Re ξ + is Im ξ ) ÷ T ( )( z )= [ ξc ] s = [ cφ ( z , η )] s , where we have used the real linearity of T . By the liberty of the choiceof z ∈ T and η ∈ ∂ A , we have that ◊ T (id)( z ) = [ cφ ( z, η )] s for all z ∈ T and η ∈ ∂ A . This shows that the function φ is independent ofthe variable η ∈ ∂ A . (cid:3) Lemma 4.7.
For each ( z, η, w ) ∈ T × ∂ A × T , ◊ T (id) ′ ( η ) = [ cϕ ( z, η, s s ( η ) and ϕ ( z, η, w ) = w s s ( η ) ϕ ( z, η, . In particular, the function ϕ does not depend on the variable z ∈ T .Proof. Let ( z, η, w ) ∈ T × ∂ A × T . Then ◊ T (id)( z ) = [ cφ ( z, η )] s byLemma 4.6. Substituting this equality and f = id into (4.4), we obtain ◊ T (id) ′ ( η ) w = [ cϕ ( z, η, w )] s s ( η ) . If we take w = 1 in the last equality,we have that ◊ T (id) ′ ( η ) = [ cϕ ( z, η, s s ( η ) . This implies that thefunction ϕ is independent of z ∈ T . Note that ◊ T (id) ′ ( η ) ∈ T since c, ϕ ( z, η, ∈ T (see Definition 3.2 and Lemma 4.5). We have that w = w ◊ T (id) ′ ( η ) ◊ T (id) ′ ( η ) = [ cϕ ( z, η, w )] s s ( η ) [ cϕ ( z, η, s s ( η ) = [ ϕ ( z, η, w )] s s ( η ) [ ϕ ( z, η, s s ( η ) , and hence [ ϕ ( z, η, w )] s s ( η ) = w [ ϕ ( z, η, s s ( η ) . We thus concludethat ϕ ( z, η, w ) = w s s ( η ) ϕ ( z, η,
1) for all ( z, η, w ) ∈ X T . (cid:3) Proof of Theorem 1.
Let f ∈ S ∞ , z ∈ T and η ∈ ∂ A . By Lemma 4.7, ϕ ( z , η , w ) = w s s ( η ) ϕ ( z , η ,
1) for all w ∈ T . Lemmas 4.6 and 4.7show that φ and ϕ are independent of η ∈ ∂ A and z ∈ T , respectively.Thus we may write φ ( z, η ) = φ ( z ) and ϕ ( z, η, w ) = w s s ( η ) ϕ ( η ) forall ( z, η, w ) ∈ X T . Now equality (4.4) is reduced to ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) w = [ c b f ( φ ( z ))] s + w [ cϕ ( η )] s s ( η ) [ “ f ′ ( ψ ( z , η ))] s s ( η ) URJECTIVE ISOMETRIES ON A BANACH SPACE 33 for all w ∈ T . The above equality holds for every w ∈ T , and then ÷ T ( f )( z ) = [ c b f ( φ ( z ))] s , (4.6) ◊ T ( f ) ′ ( η ) = [ cϕ ( η )] s s ( η ) [ “ f ′ ( ψ ( z , η ))] s s ( η ) . (4.7)Let ρ = c − s T (id) ∈ A ( D ), and then c s b ρ ( z ) = ◊ T (id)( z ) = [ cφ ( z )] s for all z ∈ T by (4.6). This shows that(4.8) φ ( z ) = [ b ρ ( z )] s for all z ∈ T . Consequently,(4.9) ÷ T ( f )( z ) = [ c b f ([ b ρ ( z )] s )] s for all z ∈ T . Since T is a boundary for A ( D ), equality (4.9) holds forall z ∈ D . Note that ρ ∈ A ( D ) satisfies | b ρ ( z ) | = | φ ( z ) | = 1 for all z ∈ T , and thus b ρ ( D ) ⊂ D by the maximal modulus principle.Since T − is a surjective real linear isometry on S ∞ with k·k Σ , theabove arguments can be applied to T − . Then there exist d ∈ T , θ ∈ A ( D ) with b θ ( D ) ⊂ D and s ∈ {± } such that(4.10) ÿ T − ( f )( z ) = [ d b f ([ b θ ( z )] s )] s for all f ∈ S ∞ and z ∈ D . Let z ∈ D . If we substitute f = T ( )into (4.10), then we have 1 = [ d ÷ T ( )([ b θ ( z )] s )] s . As ÷ T ( ) = c s by(4.9), we obtain 1 = [ dc s ] s . Substituting f = T (id) into (4.10) to get z = [ d ◊ T (id)([ b θ ( z )] s )] s . By (4.9), ◊ T (id)( z ) = c s b ρ ( z ), and hence z = [ d ◊ T (id)([ b θ ( z )] s )] s = [ dc s b ρ ([ b θ ( z )] s )] s = [ b ρ ([ b θ ( z )] s )] s , where we have used that 1 = [ dc s ] s . By the liberty of the choice of z ∈ D , we have that z = [ b ρ ([ b θ ( z )] s )] s for all z ∈ D . This shows that D ⊂ b ρ ( D ), and therefore, b ρ ( D ) = D .Recall that φ ( z ) = φ ( z, η, w ) is surjective by Lemma 3.4. Equality(4.8) shows that b ρ ( T ) = [ φ ( T )] s = T . We shall prove that b ρ : D → D is injective. Suppose that b ρ ( z ) = b ρ ( z ) for z , z ∈ D . Since T issurjective, there exists f ∈ S ∞ such that T ( f ) = id ∈ S ∞ . Then, by(4.9), z = ◊ T ( f )( z ) = [ c c f ([ b ρ ( z )] s )] s = [ c c f ([ b ρ ( z )] s )] s = ◊ T ( f )( z ) = z , and thus b ρ is injective, as is claimed. The function ρ = c − s T (id) ∈ A ( D ) is an analytic function on D , which is a homeomorphism on D aswell. It is well-known (cf. [20, Theorem 12.6]) that for such a function ρ there exist λ ∈ T and b ∈ D such that(4.11) b ρ ( z ) = λ z − b − bz ( z ∈ D ) . Since ◊ T (id)( z ) = c s b ρ ( z ) for z ∈ D , we have T (id) ′ ( z ) = c s ρ ′ ( z ) for z ∈ D . Differentiating both sides of (4.11), we see that ρ ′ ( z ) = λ (1 −| b | ) / (1 − bz ) for z ∈ D . Hence c − s T (id) ′ = ρ ′ ∈ A ( D ). Substituting f = id into (4.7), we get | ◊ T (id) ′ | = | cϕ | = 1 on ∂ A . It follows that | “ ρ ′ ( z ) | = 1 for all z ∈ T , that is, 1 − | b | = | − bz | for all z ∈ T . Thisimplies that b = 0: In fact, if b = 0, then we substitute z = ± b/ | b | intothe equality 1 − | b | = | − bz | to obtain(1 − | b | ) = (cid:12)(cid:12)(cid:12)(cid:12) − b b | b | (cid:12)(cid:12)(cid:12)(cid:12) = 1 − | b | = (cid:12)(cid:12)(cid:12)(cid:12) − b − b | b | (cid:12)(cid:12)(cid:12)(cid:12) = (1 + | b | ) , URJECTIVE ISOMETRIES ON A BANACH SPACE 35 which is impossible. Therefore, we have b = 0, and consequently b ρ ( z ) = λz for all z ∈ D . By (4.9) we conclude that T ( f )( z ) = cf ( λz ) for all f ∈ S ∞ and z ∈ D , or T ( f )( z ) = cf ( λz ) for all f ∈ S ∞ and z ∈ D . By(2.5), T = T − T (0), and thus T ( f ) = T (0) + T ( f ) is of the desiredform.Conversely, if T is of the above form, then it is easy to check that T is a surjective isometry on S ∞ with k·k Σ . (cid:3) Proof of Theorem 2.
In this subsection, we consider the casewhen T = { a } for a ∈ D . Recall that ÷ T ( f )( z ) + ◊ T ( f ) ′ ( η ) w = [ α ( x ) b f ( φ ( z, η ))] s ( η ) + [ α ( x ) “ f ′ ( ψ ( z, η )) ϕ ( x )] s ( η ) s ( η ) for all x = ( z, η, w ) ∈ T × ∂ A × T by (4.2). By definition, φ is amap from T × ∂ A × T to T . Since T = { a } , we have that φ is aconstant function which takes the value a . In addition, we may write ψ ( η ) = ψ ( z, η ) and ϕ ( z, η, w ) = ϕ ( η, w ) for ( z, η ) ∈ { a } × ∂ A and w ∈ T . Now we obtain the following equality(4.12) T ( f )( a ) + ◊ T ( f ) ′ ( η ) w = [ α ( x ) f ( a )] s ( η ) + [ α ( x ) “ f ′ ( ψ ( η )) ϕ ( η, w )] s ( η ) s ( η ) for all f ∈ S ∞ and x = ( a, η, w ) ∈ X { a } = { a } × ∂ A × T . Lemma 4.8.
There exists c ∈ T such that ÷ T ( ) = c on D and α ( a, η, w ) = [ c ] s ( η ) = α ( a, η, for all ( η, w ) ∈ ∂ A × T . In particu-lar, the value α ( a, η, w ) is independent of the variable w ∈ T . Proof.
Let id a = id − a . If we apply f = , id a ∈ S ∞ to (4.12), thenwe have T ( )( a ) + ◊ T ( ) ′ ( η ) w = [ α ( a, η, w )] s ( η ) , (4.13) T (id a )( a ) + ÿ T (id a ) ′ ( η ) w = [ α ( a, η, w ) ϕ ( η, w )] s ( η ) s ( η ) (4.14)for all ( η, w ) ∈ ∂ A × T . We shall prove that T ( )( a ) = 0. To this end,we assume T ( )( a ) = 0, and then ◊ T ( ) ′ ( η ) w = [ α ( a, η, w )] s ( η ) for all( η, w ) ∈ ∂ A × T . Substituting this equality and (4.14) into (4.12) toobtain T ( f )( a ) + ◊ T ( f ) ′ ( η ) w = ◊ T ( ) ′ ( η ) w [ f ( a )] s ( η ) + { T (id a )( a )+ ÿ T (id a ) ′ ( η ) w } [ “ f ′ ( ψ ( η ))] s ( η ) s ( η ) for all f ∈ S ∞ and ( η, w ) ∈ ∂ A × T . By the liberty of the choice of w ∈ T , we get(4.15) T ( f )( a ) = T (id a )( a )[ “ f ′ ( ψ ( η ))] s ( η ) s ( η ) for all f ∈ S ∞ and η ∈ ∂ A . Taking f = id ∈ S ∞ in (4.15), wehave T (id )( a ) = T (id a )( a )[2 “ id( ψ ( η ))] s ( η ) s ( η ) for all η ∈ ∂ A . Since ψ : ∂ A → ∂ A is surjective, for each j ∈ {± } there exists η j ∈ ∂ A such that “ id( ψ ( η j )) = j . Then we obtain T (id a )( a ) = 0. By (4.15),we get T ( f )( a ) = 0 for all f ∈ S ∞ . This is impossible since T issurjective, which shows that T ( )( a ) = 0, as is claimed.By equality (4.13) with Proposition 4.4, we see that ◊ T ( ) ′ ( η ) = 0for all η ∈ ∂ A . Since ∂ A is a boundary for A , we have ◊ T ( ) ′ = 0 on M . Then T ( ) is constant on D , say c ∈ C . Since D is dense in M ,we obtain ÷ T ( ) = c on M . Substituting ◊ T ( ) ′ ( η ) = 0 into (4.13) to URJECTIVE ISOMETRIES ON A BANACH SPACE 37 get c = T ( )( a ) = [ α ( a, η, w )] s ( η ) for all ( η, w ) ∈ ∂ A × T , and thus c ∈ T . Hence α ( a, η, w ) = [ c ] s ( η ) = α ( a, η, (cid:3) By Lemma 4.8, α ( a, η, w ) = [ c ] s ( η ) for all ( η, w ) ∈ ∂ A × T . Thenequality (4.12) is reduced to(4.16) T ( f )( a ) + ◊ T ( f ) ′ ( η ) w = c [ f ( a )] s ( η ) + [ c ] s ( η ) [ “ f ′ ( ψ ( η )) ϕ ( η, w )] s ( η ) s ( η ) for every f ∈ S ∞ and ( η, w ) ∈ ∂ A × T . Lemma 4.9. (1)
Let id a = id − a ∈ S ∞ . Then ÿ T (id a ) ′ ( η ) =[ c ] s ( η ) [ ϕ ( η, s ( η ) s ( η ) for all η ∈ ∂ A . (2) ϕ ( η, w ) = ϕ ( η, w s ( η ) s ( η ) for all ( η, w ) ∈ ∂ A × T .Proof. Let η ∈ ∂ A . Taking f = id a in (4.16), we obtain(4.17) T (id a )( a ) + ÿ T (id a ) ′ ( η ) w = [ c ] s ( η ) [ ϕ ( η, w )] s ( η ) s ( η ) for every w ∈ T . Then we have | T (id a )( a ) + ÿ T (id a ) ′ ( η ) w | = 1 forall w ∈ T . We shall prove that T (id a )( a ) = 0. To this end, sup-pose on the contrary that T (id a )( a ) = 0. Proposition 4.4 showsthat ÿ T (id a ) ′ ( η ) = 0. Substituting this equality into (4.17), we obtain T (id a )( a ) = [ c ] s ( η ) [ ϕ ( η, w )] s ( η ) s ( η ) for all w ∈ T . Since T is sur-jective, there exists g ∈ S ∞ such that T ( g )( a ) = 0 and ÷ T ( g ) ′ ( η ) = 1.Applying these equalities and T (id a )( a ) = [ c ] s ( η ) [ ϕ ( η, w )] s ( η ) s ( η ) to(4.16), we get w = T ( g )( a ) + ÷ T ( g ) ′ ( η ) w = c [ g ( a )] s ( η ) + T (id a )( a ) [ “ g ′ ( ψ ( η ))] s ( η ) s ( η ) for all w ∈ T . This is impossible since the rightmost hand side ofthe above equalities is independent of w ∈ T . Consequently, we have T (id a )( a ) = 0. By the liberty of the choice of η ∈ ∂ A , we get ÿ T (id a ) ′ ( η ) w = [ c ] s ( η ) [ ϕ ( η, w )] s ( η ) s ( η ) for all ( η, w ) ∈ ∂ A × T by(4.17). Taking w = 1 in the last equality, we obtain ÿ T (id a ) ′ ( η ) =[ c ] s ( η ) [ ϕ ( η, s ( η ) s ( η ) for all η ∈ ∂ A .Note that ÿ T (id a ) ′ ( η ) ∈ T for η ∈ T . It follows that w = ÿ T (id a ) ′ ( η ) w ÿ T (id a ) ′ ( η ) = [ ϕ ( η, w )] s ( η ) s ( η ) [ ϕ ( η, s ( η ) s ( η ) , and thus, ϕ ( η, w ) = ϕ ( η, w s ( η ) s ( η ) for all ( η, w ) ∈ ∂ A × T . (cid:3) Proof of Theorem 2.
For the sake of simplicity, we will write ϕ ( η ) = ϕ ( η, T ( f )( a ) + ◊ T ( f ) ′ ( η ) w = c [ f ( a )] s ( η ) + w [ c ] s ( η ) [ ϕ ( η ) “ f ′ ( ψ ( η ))] s ( η ) s ( η ) for all f ∈ S ∞ and ( η, w ) ∈ ∂ A × T . By the liberty of the choice of w ∈ T , we obtain(4.18) ◊ T ( f ) ′ ( η ) = [ c ] s ( η ) [ ϕ ( η ) “ f ′ ( ψ ( η ))] s ( η ) s ( η ) and T ( f )( a ) = c [ f ( a )] s ( η ) for all f ∈ S ∞ and η ∈ ∂ A . Substituting f = i ∈ S ∞ into the last equality, we get T ( i )( a ) = c [ i ] s ( η ) for all η ∈ ∂ A . This shows that s is a constant function. For the sake ofsimplicity of notation, we will write s ( η ) = s . Then we have(4.19) T ( f )( a ) = c [ f ( a )] s for all f ∈ S ∞ . For each v ∈ H ∞ ( D ), we define I a ( v ) by I a ( v )( z ) = Z [ a,z ] v ( ζ ) dζ ( z ∈ D ) , URJECTIVE ISOMETRIES ON A BANACH SPACE 39 where [ a, z ] denotes the straight line interval from a to z in D . Then I a ( v ) ∈ H ∞ ( D ) satisfies I a ( v )( a ) = 0 and(4.20) I a ( v ) ′ = v on D , and hence I a ( v ) ∈ S ∞ . Then ÷ I a ( v ) ′ = b v on M . We may regard I a as acomplex linear operator from H ∞ ( D ) onto S ∞ .Recall that A = { b v ∈ C ( M ) : v ∈ H ∞ ( D ) } . Define W : A → A by(4.21) W ( b v )( η ) = ¤ T ( I a ( v )) ′ ( η ) ( v ∈ H ∞ ( D ) , η ∈ M ) . Since the Gelfand transform from H ∞ ( D ) onto A is an isometric iso-morphism, we see that the mapping W is well-defined. A W −−−→ A b · x x b · H ∞ ( D ) H ∞ ( D ) I a y y I a S ∞ −−−→ T S ∞ Since T : S ∞ → S ∞ is real linear, so is W : A → A . Recall that ÷ I a ( v ) ′ = b v on M for v ∈ H ∞ ( D ). Substituting f = I ( v ) into (4.18) toobtain ⁄ T ( I ( v )) ′ ( η ) = [ c ] s ( η ) [ ϕ ( η ) b v ( ψ ( η ))] s s ( η ) for all v ∈ H ∞ ( D )and η ∈ ∂ A , where we have used that s is a constant function. Thelast equality, combined with (4.21), shows that W ( b v )( η ) = [ c ] s ( η ) [ ϕ ( η ) b v ( ψ ( η ))] s s ( η ) for b v ∈ A and η ∈ ∂ A . Note that c , ϕ ( η ) ∈ T by definition (seeLemma 4.8 and Definition 3.2), and thus |W ( b v )( η ) | = | b v ( ψ ( η )) | for all b v ∈ A and η ∈ ∂ A . Since ψ : ∂ A → ∂ A is surjective by Lemma 3.5, we see that kW ( b v ) k ∞ = k b v k ∞ for all b v ∈ A , where k·k ∞ denotes thesupremum norm on M . Thus W is a real linear isometry on ( A , k · k ∞ ).We show that W is surjective. By the surjectivity of T : S ∞ → S ∞ ,for each v ∈ H ∞ ( D ) there exists g ∈ S ∞ such that T ( g ) = I a ( v ) on D , and hence(4.22) T ( g ) ′ ( z ) = v ( z )for every z ∈ D by (4.20). Since I a ( g ′ ) ′ = g ′ on D , we see that I a ( g ′ ) − g is constant on D , say c g ∈ C . Equality (4.18) shows that ◊ T ( c g ) ′ = 0on ∂ A . Since ∂ A is a boundary for A , we see that ◊ T ( c g ) ′ = 0 on M ,and hence T ( c g ) ′ = 0 on D . Therefore, T ( I a ( g ′ ) − g ) ′ = T ( c g ) ′ = 0on D , and consequently T ( I a ( g ′ )) ′ ( z ) = T ( g ) ′ ( z ) for every z ∈ D .Combining this equality with (4.22), we deduce T ( I a ( g ) ′ ) ′ ( z ) = v ( z )for all z ∈ D . Since D is dense in M , we have that ¤ T ( I a ( g ′ )) ′ ( η ) = c v ( η )for all η ∈ M . Equality (4.21) shows that W ( “ g ′ )( η ) = ¤ T ( I a ( g ′ )) ′ ( η ) = c v ( η ) for all η ∈ M , which yields the surjectivity of W : A → A .Hence W is a surjective real linear isometry on the uniform algebra( A , k·k ∞ ). By [16, Theorem 1.1] (see also [11, Theorem 3.3]), there exista continuous function u : ∂ A → T , a closed and open set G ⊂ ∂ A and a homeomorphism ̺ : ∂ A → ∂ A such that(4.23) W ( b v )( η ) = ( u ( η ) b v ( ̺ ( η )) η ∈ G u ( η ) b v ( ̺ ( η )) η ∈ ∂ A \ G for all b v ∈ A . Then W ( b ) = u on ∂ A .Since W is surjective, there exists v ∈ H ∞ ( D ) such that W ( c v ) = b on M . Then k c v k ∞ = kW ( c v ) k ∞ = 1 since W is a real linear isometry. URJECTIVE ISOMETRIES ON A BANACH SPACE 41
Equality (4.23) implies that W ( b ) W ( c v ) = u W ( c v ) = W ( c v ) on ∂ A , and thus W ( b ) W ( c v ) = b on ∂ A . Since ∂ A is a boundary for A , we have W ( b ) W ( c v ) = b on M . Note that |W ( c v )( η ) | ≤ η ∈ M since kW ( c v ) k ∞ = k c v k ∞ = 1. By the same reasoning, |W ( b )( η ) | ≤ η ∈ M . Combining these inequalities with theequality W ( b )( η ) W ( c v )( η ) = 1 for η ∈ M , we deduce |W ( b ) | = 1on M . Since the function W ( b ) is analytic on D ⊂ M , it must be aconstant function of modulus one. Hence, W ( b ) = c on M for some c ∈ T .Since u = W ( b ) = c on ∂ A , the mapping c W is a bijective reallinear map on A . By (4.23), we see that c W ( c v c v ) = c W ( c v ) c W ( c v )on ∂ A for all c v , c v ∈ A . Since ∂ A is a boundary for A , c W ismultiplicative, and hence it is a bijective real-algebra automorphism on A . By [11, Themrem 2.1], there exist a homeomorphism ρ : M → M and a closed and open set G of M such that c W ( b v )( η ) = ( b v ( ρ ( η )) η ∈ G b v ( ρ ( η )) η ∈ M \ G for all b v ∈ A . By the connectedness of M , we see that G = M or G = ∅ , and hence there exists s ∈ {± } such that(4.24) W ( b v )( η ) = c [ b v ( ρ ( η ))] s for all b v ∈ A and η ∈ M .Since W ( “ id) ∈ A , there exists h ∈ H ∞ ( D ) such that W ( “ id) = c h .Then h is a non-constant function with | “ id ◦ ρ | = | c h | on M by (4.24).Thus k c h k ∞ = k “ id ◦ ρ k ∞ = 1, which implies h ( D ) ⊂ D . Since h is a non-constant analytic function on D , the open mapping theorem showsthat h ( D ) is an open set, and hence h ( D ) ⊂ D . Since | “ id | = 1 on M \ D (see [12, p. 161]), the inequalities | “ id ◦ ρ | = | c h | < D implythat ρ ( D ) ⊂ D .The above arguments can be applied to the surjective real linearisometry W − . Then W − ( b ) is constant on M , say c − , and thus themapping c − W − is a bijective real-algebra automorphism on A . Thereexist a homeomorphism ρ − : M → M and s ∈ {± } such that(4.25) W − ( b v )( η ) = c − [ b v ( ρ − ( η ))] s for all b v ∈ A and η ∈ M . Choose h − ∈ H ∞ ( D ) so that W − ( “ id) = ‘ h − .Thus | “ id ◦ ρ − | = | ‘ h − | , and hence h − ( D ) ⊂ D by the open mappingtheorem. In addition, we observe ρ − ( D ) ⊂ D .By (4.24) and (4.25), the equality b v = W − ( W ( b v )) shows that b v ( η ) = W − ( c · [ b v ◦ ρ ] s )( η ) = c − h c [ b v ◦ ρ ] s ( ρ − ( η )) i s for all v ∈ H ∞ ( D ) and η ∈ M , where we define [ b v ◦ ρ ] s ( η ) =[ b v ( ρ ( η ))] s . For v = , the above equalities show that 1 = c − [ c ] s ,and hence b v = [ b v ◦ ρ ◦ ρ − ] s s for all v ∈ H ∞ ( D ). Substituting v = idinto this equality to obtain “ id = [ “ id ◦ ρ ◦ ρ − ] s s on M . Since ρ j ( D ) ⊂ D for j = ±
1, we see that z = [ ρ ( ρ − ( z ))] s s for all z ∈ D . This showsthat D ⊂ ρ ( D ), and therefore ρ | D : D → D is a homeomorphism. Foreach z ∈ D , c [ ρ ( z )] s = c [ “ id( ρ ( z ))] s = c h ( z ) = h ( z ) by (4.24)with W ( “ id) = c h , and thus [ ρ | D ] s = c h is analytic on D . Since ρ | D : D → D is a homeomorphism, there exist λ ∈ T and b ∈ D such URJECTIVE ISOMETRIES ON A BANACH SPACE 43 that [ ρ ( z )] s = λ z − b − ¯ bz ( z ∈ D )(see. [20, Theorem 12.6]).Let f ∈ S ∞ , and then I a ( f ′ ) ′ = f ′ on D by (4.20). Thus I a ( f ′ ) − f is constant on D . By the definition of I a , we have I a ( f ′ )( a ) = 0, andhence(4.26) I a ( f ′ ) = f − f ( a ) on D Since T is real linear, we have T ( I a ( f ′ )) = T ( f ) − T ( f ( a )) on D .Therefore, ¤ T ( I a ( f ′ )) ′ = ◊ T ( f ) ′ − ⁄ T ( f ( a )) ′ on M . Equality (4.18)shows ⁄ T ( f ( a )) ′ = 0 on ∂ A , and thus ⁄ T ( f ( a )) ′ = 0 on M since ∂ A is aboundary for A . We deduce ¤ T ( I a ( f ′ )) ′ = ◊ T ( f ) ′ on M . This equality,combined with (4.21) and (4.24), shows that ◊ T ( f ) ′ ( η ) = ¤ T ( I a ( f ′ )) ′ ( η ) = W ( “ f ′ )( η ) = c [ “ f ′ ( ρ ( η ))] s for every η ∈ M . In particular, since ρ ( D ) = D , T ( f ) ′ ( z ) = c [ f ′ ( ρ ( z ))] s ( z ∈ D ) . Equality (4.26), applied to T ( f ) instead of f , shows that I a ( T ( f ) ′ ) = T ( f ) − T ( f )( a ) on D , that is, T ( f )( z ) = T ( f )( a ) + I a ( T ( f ) ′ )( z )for all z ∈ D . Recall that T ( f )( a ) = c [ f ( a )] s by (4.19), and that[ ρ ( z )] s = λ ( z − b ) / (1 − ¯ bz ) for z ∈ D . By the definition of I a with T ( f ) ′ = c [ f ′ ◦ ρ ] s , we get T ( f )( z ) = T ( f )( a ) + I a ( T ( f ) ′ )( z )= c [ f ( a )] s + Z [ a,z ] c [ f ′ ( ρ ( ζ ))] s dζ = c [ f ( a )] s + Z [ a,z ] c (cid:20) f ′ (cid:18)(cid:20) λ ζ − b − ¯ bζ (cid:21) s (cid:19)(cid:21) s dζ for all f ∈ S ∞ and z ∈ D . Since T = T − T (0) by (2.5), we obtain theforms displayed in Theorem 2.Conversely, if T is of the above form, then we can check that it is asurjective isometry on S ∞ . This completes the proof. (cid:3) Remark 4.1.
We define L ( f ), with 0 ≤ L ( f ) ≤ ∞ , by L ( f ) = sup z ,z ∈ D z = z | f ( z ) − f ( z ) || z − z | for each f ∈ H ( D ). A function f ∈ H ( D ) is said to be of Lipschitz ifand only if L ( f ) is finite. We set Lip A ( D ) = { f ∈ H ( D ) : L ( f ) < ∞} .We shall prove that Lip A ( D ) = S ∞ and L ( f ) = sup z ∈ D | f ′ ( z ) | for all f ∈ Lip A ( D ). In fact, if f ∈ Lip A ( D ), then | f ′ ( z ) | ≤ L ( f ) < ∞ foreach z ∈ D by the definition of the derivative of f . Hence f ′ is boundedon D , that is, Lip A ( D ) ⊂ S ∞ with sup z ∈ D | f ′ ( z ) | ≤ L ( f ). Conversely,if f ∈ S ∞ , then sup z ∈ D | f ′ ( z ) | < ∞ . Choose z , z ∈ D with z = z .We have that R [ z ,z ] f ′ ( z ) dz = f ( z ) − f ( z ), where [ z , z ] is the lineinterval from z to z . Then we get | f ( z ) − f ( z ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z [ z ,z ] f ′ ( z ) dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z − z | sup z ∈ D | f ′ ( z ) | , and hence | f ( z ) − f ( z ) | / | z − z | ≤ sup z ∈ D | f ′ ( z ) | for all z , z ∈ D with z = z . By the definition of L ( f ), we have that L ( f ) ≤ sup z ∈ D | f ′ ( z ) | , URJECTIVE ISOMETRIES ON A BANACH SPACE 45 and therefore S ∞ ⊂ Lip A ( D ). Consequently, we obtain Lip A ( D ) = S ∞ ,as is claimed. Moreover, L ( f ) = sup z ∈ D | f ′ ( z ) | for all f ∈ Lip A ( D ). References [1] S. Banach, Theory of linear operations, Translated by F. Jellett, Dover Publi-cations, Inc. Mineola, New York, 2009.[2] F. Botelho,
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Real-linear isometries between function algebras , Cent. Eur. J. Math. (2011), 778–788.[17] T. Miura and N. Niwa, Surjective isometries on a subspace of analytic functionson the open unit disc , Nihonkai Math. J. (2018), No. 1, pp. 53–67.[18] M. Nagasawa, Isomorphisms between commutative Banach algebras with anapplication to rings of analytic functions , K¯odai Math. Sem. Rep. (1959),182–188.[19] W.P. Novinger and D.M. Oberlin, Linear isometries of some normed spaces ofanalytic functions , Can. J. Math. (1985), 62–74.[20] W. Rudin, Real and complex analysis, Third Edition. McGraw-Hill Book Co.,New York, 1987. [21] J. V¨ais¨al¨a, A proof of the Mazur-Ulam theorem , Amer. Math. Monthly, (2003), 633–635.(Takeshi Miura)
Department of Mathematics, Faculty of Science, Ni-igata University, Niigata 950-2181 Japan
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