MULTIPLICATIVE BIJECTIONS OF C(X,I)

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 134, Number 4, Pages S (05)08069-X Article electronically published on July 20, 2005 MULTIPLICATIVE BIJECTIONS OF C(X,I) JANKO MAROVT (Communicated by Joseph A. Ball) Abstract. Let X be a compact Hausdorff space which satisfies the first axiom of countability, let I =[0, 1] and let C(X,I) be the set of all continuous functions from X to I. If ϕ : C(X,I) C(X,I) is a bijective multiplicative map, then there exist a homeomorphism µ : X X and a continuous map k : X (0, ), such that ϕ(f)(x) =f(µ(x)) k(x) for all x X and for all f C(X,I). 1. Introduction and statement of the result The problem we consider in this paper has been motivated by a recent result of L. Molnár in [7], where he proved a structural result for the so-called sequential isomorphisms between the sets of von Neumann algebra effects. To explain the content of that result we recall the following notions. Let A be a unital C -algebra. The effects in A are the positive elements of A which are less than or equal to the unit of A. The set of all effects in A is denoted by E (A). If A equals the algebra B (H) of all bounded linear operators on the complex Hilbert space H, then the corresponding effects are called Hilbert space effects. These objects play a very important role in certain parts of quantum mechanics, e.g., in the quantum theory of measurement; see, for example, [2]. For more general concepts of effects we refer to the monograph [3]. Motivated by physical considerations, in their paper [5] Gudder and Nagy introduced the concept of the sequential product between effects; see also [6]. This operation is defined by A B = A 1/2 BA 1/2, A,B E (A). In the paper [4], Gudder and Greechie described the general form of the sequential automorphisms of the set of all Hilbert space effects assuming that the underlying Hilbert space is at least three dimensional. If A, B are unital C -algebras, then the bijective map ϕ : E (A) E (B) is called a sequential isomorphism if it satisfies ϕ (A B) =ϕ(a) ϕ(b), A,B E (A). The result of Gudder and Greechie says that for Hilbert space effects every such transformation ϕ is implemented by either a unitary or an antiunitary operator U of the underlying Hilbert space, i.e., ϕ is of the form ϕ(a) =UAU. Received by the editors September 10, 2004 and, in revised form, October 27, Mathematics Subject Classification. Primary 46J10; Secondary 46E05. Key words and phrases. Preserver, multiplicative map c 2005 American Mathematical Society Reverts to public domain 28 years from publication

2 1066 JANKO MAROVT In the paper [7] this result was significantly generalized from the case of effects in B (H) to the case of effects in general von Neumann algebras. Namely, Molnár proved the following result. Let A, B be von Neumann algebras and let ϕ : E (A) E (B) beasequential isomorphism. Then there are direct decompositions A = A 1 A 2 A 3 and B = B 1 B 2 B 3 within the category of von Neumann algebras and there are bijective maps ϕ 1 : E (A 1 ) E (B 1 ), Φ 2 : A 2 B 2, Φ 3 : A 3 B 3, such that (i) A 1, B 1 are commutative von Neumann algebras and the algebras A 2 A 3, B 2 B 3 have no commutative direct summands; (ii) ϕ 1 is a multiplicative bijection, Φ 2 is an algebra *-isomorphism, Φ 3 is an algebra *-antiisomorphism and ϕ = ϕ 1 Φ 2 Φ 3 holds on E (A). Much information is available on algebra *-isomorphisms and algebra *-antiisomorphisms. For example, we know their general forms in many particular cases. In contrast to this, nothing seems to be known about the third factor in the above decomposition, i.e., about the bijective multiplicative maps between the sets of effects in commutative von Neumann algebras or, more generally, in commutative unital C -algebras. The aim of this paper is to somehow try to fill this gap by presenting a structural result concerning these latter transformations. It is well known that every commutative C -algebra is isomorphic to the algebra of all continuous complex-valued functions on a compact Hausdorff space X. Therefore, it is enough to consider the structure of all continuous functions from X to the unit interval I which we denote by C(X,I). The main result of our paper describes the general form of all bijective multiplicative maps of C(X,I) under the technical condition that X satisfies the first axiom of countability. Theorem. Let X be a compact Hausdorff space which satisfies the first axiom of countability and let I =[0, 1]. If ϕ : C(X,I) C(X,I) is a bijective multiplicative map, then there exist a homeomorphism µ : X Xand a continuous map k : X (0, ), such that for x X, ϕ(f)(x) =f(µ(x)) k(x) for all f C(X,I). We believe that the same result also holds without the countability assumption, and it would be interesting to find a proof of this conjecture. 2. Proof of the Theorem Our first auxiliary result gives us the form of multiplicative functions on the interval I. Lemma 2.1. Let m : I I be a multiplicative function. Then m(x) =0for all x I, or m(x) =1for all x I, or there exists k>0 such that m(x) =x k for all x I, or { 0, x =0, m(x) = 1, x (0, 1],

3 MULTIPLICATIVE BIJECTIONS OF C(X,I) 1067 or { 0, x [0, 1), m(x) = 1, x =1. Proof. The function m is multiplicative, therefore m(0)(m(0) 1) = 0 which yields m(0) = 0 or m(0) = 1. If m(0) = 1, then m(x) =m(x)m(0) = m(0) = 1 for every x X.Inthiscasemis of the form m(x) = 1 for all x I. Let us assume that m(0) = 0 and let us define h(x) =m(e x ),x [0, ). Then h(x + y) =m(e x y )=m(e x )m(e y )=h(x)h(y), x,y [0, ). The function h is also bounded from above by 1. Therefore by [1, Chapter 3] h has one of the following forms: h(x) = 0 for all x [0, ), or h(x) =e cx for all x [0, ), where c is a nonpositive constant, or { 0, x > 0, h(x) = 1, x =0. For x (0, 1] we obtain h( ln x) = m(x) and therefore m(x) = 0 for all x (0, 1], or m(x) =e c ln x = x c for all x (0, 1],c 0, or { 0, x (0, 1), m(x) = 1, x =1. Considering that m sends 0 to 0, we conclude the proof of Lemma 2.1. From now on let ϕ : C(X,I) C(X,I) be a multiplicative bijective map. Also, let 1 X (x) = 1 for all x X and 0 X (x) = 0 for all x X. We will next show that ϕ(0 X ) = 0 X and ϕ(1 X ) = 1 X. Since ϕ is multiplicative we obtain ϕ(0 X )(ϕ(h) 1 X )=0 X for every h C(X,I). Suppose ϕ(0 X )( ) 0forsome X. Because ϕ is surjective there exists h C(X,I) such that ϕ(h)( ) 1. Since this is contrary to the upper equation we get ϕ(0 X )=0 X. Similarly, we obtain ϕ(1 X )(ϕ(1 X ) 1 X )=0 X. If ϕ(1 X )( ) 1 and therefore ϕ(1 X )( )=0forsome X,thenϕ(h)( )=ϕ(1 X )( )ϕ(h)( ) = 0 for every ϕ(h). This is a contradiction since ϕ is surjective. Hence, ϕ(1 X )=1 X. Lemma 2.2. Let f,g C(X,I) with f(x) g(x) for all x X and f(x) 0for all x X.Thenϕ(f)(x) ϕ(g)(x) for all x X. Proof. Let h = g f. Then h C(X,I)andhf = g. The map ϕ is multiplicative, therefore ϕ(h)ϕ(f) =ϕ(g) which yields ϕ(f)(x) ϕ(g)(x) for all x X. In the following lemma we will need the notion of the so-called maximal open set of the function from C(X,I). Let W be a closed subset of X with W X and IntW = U. If there exists f C(X,I), such that f 1 (1) = W,thenU will be called the maximal open set of the function f. Lemma 2.3. Let U be an open nonempty subset of X where U X. Then there exists f C(X,I), f 1 X, such that f ( U ) = {1}. Furthermore, for every f C(X,I), f 1 X,withf ( U ) = {1} there exists the maximal open set V of the function ϕ(f).

4 1068 JANKO MAROVT Proof. Let U be a nonempty open subset of X with U X. Every compact Hausdorff space is normal, so by Urysohn s lemma there exists a continuous function f from X to I such that f(x) = 1 for all x U and f 1 X. Let f C(X,I), f 1 X and f ( U ) = {1}. Since X is normal and U there exists a closed nonempty set U 1 U. The sets U c and U 1 are closed disjoint subsets of X, therefore by Urysohn s lemma there exists a continuous function g : X Isuch that g(x) = 0 for all x U c and g(x) = 1 for all x U 1. Then gf = g which, by the multiplicativity of the map ϕ, yields ϕ(g)(ϕ(f) 1 X )=0 X. This implies that if ϕ(g)(x) 0, thenϕ(f)(x) = 1andifϕ(f)(x) 1, then ϕ(g)(x) =0forx X. Let us define V =Int ( ϕ(f) 1 (1) ). The set ( ϕ(g) 1 (0) ) c is a nonempty open set since g 0 X and ϕ is injective. Also, ( ϕ(g) 1 (0) ) c ϕ(f) 1 (1) and therefore by the definition of V, V. Observe that ϕ(f) 1 (1) X since f 1 X and ϕ is injective. This yields that V is the maximal open set of the function ϕ(f). The next lemma is a consequence of Lemma 2.3. Lemma 2.4. Let U be an open nonempty subset of X where U X. Then there exists f C(X,I), f 0 X, such that f ( U ) = {0}. Furthermore, for every f C(X,I), f 0 X,withf ( U ) = {0} there exists an open set V such that ϕ(f)(v )={0}. Proof. By Urysohn s lemma there exists a continuous function f from X to I such that f(x) = 0 for all x U and f 0 X. Since X is normal there exists an open nonempty set U 1,whereU 1 U. Again by Urysohn s lemma there exists a function g : X I such that g(x) = 0 for all x U c and g(x) = 1 for all x U 1. Then gf =0 X and therefore ϕ(g)ϕ(f) =0 X. This implies that if ϕ(g)(x) 0, thenϕ(f)(x) = 0andifϕ(f)(x) 0, then ϕ(g)(x) =0forx X. By Lemma 2.3 there exists the maximal open set V of the function ϕ(g) which yields ϕ(f) ( V ) =0. Lemma 2.5. Let U be the maximal open set of the functions f 1 and f 2. If V 1 is the maximal open set of the function ϕ(f 1 ) and V 2 is the maximal open set of the function ϕ(f 2 ),thenv 1 = V 2. Proof. Let us assume that V 1 V 2.ThenV 1 V 2 because V 1 and V 2 are maximal. Without loss of generality we may assume that V 1 is not a subset of V 2. Then c there exists a nonempty open set V 3 V 1 V 2. So, V3 V 1 and V 2 V 3 =. By Urysohn s lemma and the surjectivity of ϕ there exists a function ϕ(h) 0 X such that ϕ(h)(v3 c )={0}. Then on one hand we obtain ϕ(h)ϕ(f 1 )=ϕ(h) and therefore by the multiplicativity and injectivity of ϕ h(f 1 1 X )=0 X. It follows that h(x) = 0 for all x U c which yields hf 2 = h. On the other hand we get ϕ(h)ϕ(f 2 ) ϕ(h) and therefore hf 2 h, which means that our assumption was wrong. Therefore V 1 = V 2, which completes the proof.

5 MULTIPLICATIVE BIJECTIONS OF C(X,I) 1069 Lemma 2.6. Let U 1,U 2,..., U n be the maximal open sets of the functions f 1,f 2,..., f n, respectively. Then U 1 U 2... U n if and only if V 1 V 2... V n, where V 1,V 2,..., V n are the maximal open sets of the functions ϕ(f 1 ),ϕ(f 2 ),..., ϕ(f n ), respectively. Proof. Suppose U 1 U 2... U n. Because of Lemma 2.5 we may assume without loss of generality that f i (x) 0forallx X and for all i {1, 2,..., n}. The finite intersection of open sets is an open set, so by Urysohn s lemma there exist a function h C(X,I) and an open set U U 1 U 2... U n, where h(x) =0 for every x (U 1 U 2... U n ) c and U is the maximal open set of h. This yields h(x) f i (x) for all x X, i =1, 2,..., n, and therefore by Lemma 2.2 (2.1) ϕ(h)(x) ϕ(f i )(x) for all x X, i =1, 2,..., n. By Lemma 2.3 there exist the maximal open set V of the function ϕ(h) andthe maximal open sets V 1,V 2,..., V n of the functions ϕ(f 1 ),ϕ(f 2 ),..., ϕ(f n ), respectively. From (2.1) we may conclude that V V 1 V 2... V n. This implication is also true in the converse direction since ϕ 1 has the same properties as ϕ. In the next step we will construct the homeomorphism µ : X X. Almost to the end of the proof we will assume that X > 1. For the point X let A x0, A x0 X be an arbitrary open neighbourhood of. Then by Urysohn s lemma there exists an open neighbourhood U of the point, U A x0, and a function f C(X,I), where U is the maximal open set of f. Let U Ax0 be the family of all pairs (U, f), where U is the maximal open set of f, U and U A x0. Then (U,f ) U A U.Letx 1 X, x 1. Then there exist open sets A 1,A 2 such that A 1 A 2 = and A 1,x 1 A 2. Again by Urysohn s lemma there exists (U, f) U Ax0 with U A 1 A x0. So, U A 2 = and hence x 1 / (U,f ) U A U. This gives us U = { }. (U,f ) U A By Lemma 2.3 there exists for every (U, f) U Ax0 the maximal open set V of the function ϕ(f). Let V Ax0 be the family of all (V,ϕ(f)), where (U, f) U Ax0 and V is the maximal open set of ϕ(f). We will next show that there exists a point x 1 X such that V = {x 1 }. (V,ϕ(f)) V A Letusfirstassumethat (V,ϕ(f)) V A V =. Then x (V,ϕ(f)) V A V c = X. 0 Each open covering in a compact space has a finite subcovering, hence there exist (V 1,ϕ(f 1 )), (V 2,ϕ(f 2 )),..., (V n,ϕ(f n )) V Ax0

6 1070 JANKO MAROVT such that V 1 c V2 c... Vn c = X. This yields V 1 V 2... V n =. Let (U 1,f 1 ), (U 2,f 2 ),..., (U n,f n ) U Ax0. Since U 1 U 2... U n we obtain by Lemma 2.6 that V 1 V 2... V n, a contradiction. We have proved that V. (V,ϕ(f)) V A Letusnextassumethat (V,ϕ(f)) V A V =. Then there exist x λ (V,ϕ(f)) V A V and (V λ,ϕ(f λ )) V Ax0,wherex λ V λ and x λ / V λ. Let (U λ,f λ ) U Ax0. By Urysohn s lemma there exist a function f µ and an open set U µ, U µ U λ,where(u µ,f µ ) U Ax0 and f µ (Uλ c )={0}. This yields f µ f λ = f µ and therefore ϕ(f µ )(ϕ(f λ ) 1 X )=0 X. If ϕ(f λ )(x) 1,thenϕ(f µ )(x) =0,x X, which implies by the continuity of ϕ(f µ ) that ϕ(f µ )(Vλ c )={0}. Therefore, since x λ Vλ c, we obtain that x λ / V µ where V µ is the maximal open set of ϕ(f µ ). This is a contradiction, since (V µ,ϕ(f µ )) V Ax0 and x λ (V,ϕ(f)) V A V. We have proved that V. (V,ϕ(f)) V A Let us now assume that there exist x 1,x 2 X, x 1 x 2, such that {x 1,x 2 } (V,ϕ(f)) V A V. Let V and V be disjoint open neighbourhoods of the points x 1 and x 2. There exists by Urysohn s lemma and the surjectivity of ϕ a function ϕ(h 1 ) with maximal open set V 1, where V 1 V,x 1 V 1, and ϕ(h 1 )(V c ) = {0}. Similarly, there exists a function ϕ(h 2 ) with maximal open set V 2,whereV 2 V,x 2 V 2, and ϕ(h 2 )(V c )={0}. Clearly, V 1 V 2 =, ϕ(h 1 )ϕ(h 2 )=0 X and hence by the multiplicativity and injectivity of ϕ, (2.2) h 1 h 2 =0 X. By Lemma 2.3 and (2.2) there exist U 1 and U 2 which are the maximal open sets of h 1 and h 2 such that U 1 U 2 =. Without loss of generality we may assume that / U 2. Since X is normal we may find an open neighbourhood U 3, U 3 A x0 of the point such that U 2 U 3 =. By Urysohn s lemma there exists a function h 3 andanopensetu 4 such that U 4 U 3,whereU 4 is the maximal open set of h 3, U 4, and h 3 (U3)={0}. c So, (U 4,h 3 ) U Ax0. This yields that there exists V 3 where (V 3,ϕ(h 3 )) V Ax0. Then on one hand we establish that x 2 V 3 V 2,

7 MULTIPLICATIVE BIJECTIONS OF C(X,I) 1071 and on the other hand, since U 4 U 2 =, we obtain by Lemma 2.6 V 3 V 2 =, which is a contradiction. Therefore V = {x 1 }. (V,ϕ(f)) V A Next, we will prove that this intersection is independent of the selection of an open neighbourhood of the point.nowletb x0, B x0 X, B x0 A x0, be an open neighbourhood of and let (V,ϕ(f)) V B V = {x 2 }. Suppose C x0 = A x0 B x0 and let (V,ϕ(f)) V C V = {x 3 }. Clearly, if (U, f) U Cx0,then(U, f) U Ax0 and (U, f) U Bx0. Hence (V,ϕ(f)) V Ax0 and (V,ϕ(f)) V Bx0 for all (V,ϕ(f)) V Cx0. This yields x 1 = x 3 = x 2. Now let ψ : X X be the function for which x 1. We will prove that ψ is a homeomorphism. Let x a x b,x a,x b X. Then there exist functions f a and f b and disjoint maximal open sets U a and U b of f a,f b, respectively, that are neighbourhoods of the points x a,x b.letv a,v b be the corresponding maximal open sets of the functions ϕ(f a ),ϕ(f b ). Therefore (V a,ϕ(f a )) V and (V Ax a b,ϕ(f b )) V Bxb. Since U a U b = we get by Lemma 2.6 V a V b =. This yields {ψ(x a )} = V V = {ψ(x b )}. (V,ϕ(f)) V B x b (V,ϕ(f)) V A xa So, ψ(x a ) ψ(x b ) which means that ψ is injective. Since ϕ 1 has the same properties as ϕ it follows that ψ is also surjective. Now let V X be the maximal open set of some function ϕ(f) andletx v be an arbitrary point in V.LetUbethe corresponding maximal open set of the function f. The set U is then a neighbourhood of the point ψ 1 (x v ). So, ψ 1 (V ) U. Similarly, for each x u U we can conclude that ψ(x u ) V which yields ψ(u) V and therefore ψ 1 (V )=U. We have proved that the inverse image of any maximal open set of some function ϕ(f) is the maximal open set of the function f. Now let A X be any nonempty open set. We may find, by Urysohn s lemma, that for every a A a function f a and the maximal open set V a of f a, V a A, such that V a is a neighbourhood of the point a. Therefore A = a A V a which gives us ψ 1 (A) = ψ 1 (V a ). a A The inverse image of any maximal open set of some function ϕ(f) is the maximal open set of the function f, hence ψ 1 (A) isanopenset. This yields that ψ is a continuous function. Since X is a compact Hausdorff space and ψ is a continuous bijection, we can conclude that ψ is a homeomorphism. Let us denote the homeomorphism µ = ψ 1. In the conclusion of the proof of the theorem we will need another two auxiliary results. We will show that the same result as in Lemma 2.2 is also valid locally.

8 1072 JANKO MAROVT Lemma 2.7. Let f,g C(X,I). Suppose there exists an open neighbourhood U of µ(x 1 ),x 1 X, where f(x) =g(x) for all x U and there is no x X such that f(x) =g(x) =0. Then ϕ(f)(x 1 )=ϕ(g)(x 1 ). Proof. Let j =max{f,g}. Then j C(X,I). Let h 1 = g j and h 2 = f j. Suppose h 1 1 X and h 2 1 X. Then there exist maximal open sets U 1 U and U 2 U of the functions h 1 and h 2, respectively, which are neighbourhoods of µ(x 1 ). Hence, ϕ(h 1 )(x 1 )=ϕ(h 2 )(x 1 )=1. Also, since ϕ is multiplicative, ϕ(h 1 )ϕ(j) =ϕ(g) and ϕ(h 2 )ϕ(j) =ϕ(f). Therefore ϕ(j)(x 1 )=ϕ(g)(x 1 )andϕ(j)(x 1 )=ϕ(f)(x 1 ), hence ϕ(g)(x 1 )=ϕ(f)(x 1 ). Clearly, if h 1 = h 2 =1 X,thenf = g and therefore ϕ(f) =ϕ(g). If we assume that h 1 =1 X and h 2 1 X,theng = j and ϕ(h 2 )(x 1 )=1. So,ϕ(j)(x 1 )=ϕ(g)(x 1 ), ϕ(j)(x 1 )=ϕ(f)(x 1 ) and therefore ϕ(f)(x 1 )=ϕ(g)(x 1 ). Similarly, we obtain the same result if h 1 1 X and h 2 =1 X. Lemma 2.8. Let f,g C(X,I) such that f(µ(x 1 )) >g(µ(x 1 )), x 1 X, and there is no x X such that f(x) =g(x) =0. Then ϕ(f)(x 1 ) ϕ(g)(x 1 ). Proof. By the continuity of the functions f and g there exists an open neighbourhood U of µ(x 1 ), where f(x) >g(x) for all x U. Let j =max{f,g}. Then by Lemma 2.2 ϕ(j)(x) ϕ(f)(x) andϕ(j)(x) ϕ(g)(x) for all x X. Also, f(x) =j(x) for all x U and therefore by Lemma 2.7 ϕ(f)(x 1 )=ϕ(j)(x 1 ). Hence, ϕ(f)(x 1 ) ϕ(g)(x 1 ). Now we are in the position to conclude the proof of the theorem. First, we will discuss the form of the image of any constant function. Let X and let m 0 : I I be the function defined in the following way: m x0 (c) =ϕ(c)( ), where on one hand c I and on the other hand c C(X,I) is a constant function. Then for c i,c j I m x0 (c i c j )=ϕ(c i c j )( )=ϕ(c i )( ) ϕ(c j )( )=m x0 (c i ) m x0 (c j ), hence m x0 is a multiplicative function from I to I. Letx X. Then we obtain by Lemma 2.1 and since ϕ(0 X )=0 X and ϕ(1 X )=1 X that { 0, c [0, 1), (2.3) ϕ(c)(x) = 1, c =1, or { 0, c =0, (2.4) ϕ(c)(x) = 1, c (0, 1], or there exists k(x) > 0 such that (2.5) ϕ(c)(x) =c k(x) for all c [0, 1]. Let X 1 be the set of all x X that satisfy (2.5), let X 2 be the set of all x X that satisfy (2.3), and let X 3 be the set of all x X that satisfy (2.4). Note that

9 MULTIPLICATIVE BIJECTIONS OF C(X,I) 1073 X 1,X 2,X 3 are disjoint and X =X 1 X 2 X 3. Let us prove that if x X 1,then there exists k(x) > 0 such that for every f C(X,I) ϕ(f)(x) =f(µ(x)) k(x). We will discuss several different options. Let x 1 X 1 and first let f C(X,I) be any function for which f(µ(x 1 )) (0, 1). Then there exists ɛ>0 such that (f(µ(x 1 )) ɛ, f(µ(x 1 )) + ɛ) (0, 1). Since we obtain by Lemma 2.8 that f(µ(x 1 )) ɛ<f(µ(x 1 )) <f(µ(x 1 )) + ɛ, ϕ(f(µ(x 1 )) ɛ)(x 1 ) ϕ(f)(x 1 ) ϕ(f(µ(x 1 )) + ɛ)(x 1 ). Since x 1 X 1 we have ϕ(f(µ(x 1 )) + ɛ)(x 1 )=(f(µ(x 1 )) + ɛ) k(x 1),k(x 1 ) > 0, and ϕ(f(µ(x 1 )) ɛ)(x 1 )=(f(µ(x 1 )) ɛ) k(x 1). So, (f(µ(x 1 )) ɛ) k(x 1) ϕ(f)(x 1 ) (f(µ(x 1 )) + ɛ) k(x 1) for every ɛ>0forwhich(f(µ(x 1 )) ɛ, f(µ(x 1 )) + ɛ) (0, 1). This yields ϕ(f)(x 1 )=f(µ(x 1 )) k(x 1), k(x 1 ) > 0, for every f C(X,I), where f(µ(x 1 )) (0, 1). Now let f(µ(x 1 )) = 0. Then ϕ(f(µ(x 1 ))+ɛ)(x 1 )=(f(µ(x 1 ))+ɛ) k(x 1),k(x 1 ) > 0, ɛ (0, 1). By Lemma 2.8 we obtain ϕ(f)(x 1 ) (f(µ(x 1 )) + ɛ) k(x 1) = ɛ k(x 1) and therefore ϕ(f)(x 1 )=0. If f(µ(x 1 )) = 1, then ϕ(f(µ(x 1 )) ɛ)(x 1 )=(f(µ(x 1 )) ɛ) k(x1),k(x 1 ) > 0, ɛ (0, 1), and similarly as before we obtain ϕ(f)(x 1 )=1. Next, we will prove that X 1 = X. Let us assume that X 2 is a nonempty set and let X 2. First, we will show that there exists for each open neighbourhood U of apointx U X 1. Let U be a nonempty open set such that U X 2 and c (0, 1). Then by (2.3) ϕ(c)(x) = 0 for all x U. Since ϕ 1 has the same properties as ϕ there exists by Lemma 2.4 an open set U such that c(x) =0for all x U, a contradiction. So, there does not exist a nonempty open set U where U X 2 and similarly by Lemma 2.3 there does not exist a nonempty open set U X 3. Let us now assume that there is an open neighbourhood U 0 of such that U 0 X 2 X 3. Since X satisfies the first axiom of countability there exists an at most countable family {U n ; n N,U n U 0 } of neighbourhoods of where for each open set G, G, there is some U n G. Let k V k = U i. i=1 Then V k, V k is an open set for every k N and V 1 V 2 V We may find x 1 V 1,x 2 V 2,..., where x i X 3 for all i N. If V is an arbitrary open neighbourhood of, then there exists U i such that U i V. This yields {x i,x i+1,x i+2,...} V i U i V hence = lim i x i.

10 1074 JANKO MAROVT Note that the sequence {x i } converges only to since X is a Hausdorff space. Let c (0, 1). Then by (2.4) ϕ(c)(x i ) = 1 for all i N. By the continuity of the function ϕ(c) weobtain ϕ(c)( ) = lim ϕ(c)(x i )=1, i but ϕ(c)( )=0since X 2, a contradiction. We may conclude that for each open neighbourhood U of there exists x U X 1. If we define V i,i N, in a similar way as before, then there exist x 1 V 1, x 2 V 2,..., where x i X 1 for all i N and = lim x i. Let c (0, 1). By the i continuity of the function ϕ(c) weobtain ϕ(c)( ) = lim ϕ(c)(x i ) = lim c k(xi), k(x i ) > 0. i i Since ϕ(c)( ) = 0 we may conclude that k(x i ) when i. So, there exists i 0 N such that k(x j ) > 1 for all j i 0. Now let f r : {µ(x j ),j N, j i 0 } {µ( )} [0, 1] be the function defined in the following way: { f r (µ(x j )) = 1, j =0orj i 0 and j odd, j i 0 and j even. 1 1 k(x j ), Since µ is continuous we obtain µ ( ) = lim µ(x j). The space X is first countable, j therefore for an arbitrary A X, A = {x X; x is a limit of a sequence from A}. This yields that {µ(x j ),j i 0 } {µ( )} is a closed subset of X. By Tietze theorem, since f r is continuous, there exists a continuous extension f : X I. Furthermore, since ϕ(f) is continuous, we get ϕ(f)( ) = lim ϕ(f)(x j) = j lim f(µ(x j)) k(xj),k(x j ) > 0. But j lim j f(µ(x 2j 1)) k(x 2j 1) =1 and ( lim f(µ(x 2j)) k(x2j) = lim 1 1 ) k(x2j ) = e 1, j j k(x 2j ) a contradiction. This yields that X 2 is the empty set. Similarly, we obtain that X 3 is also empty. It follows that X 1 = X. We have proved that for X > 1andx X there exists k(x) > 0 such that ϕ(f)(x) =f(µ(x)) k(x). Since for every constant function c (0, 1) we have ϕ(c)(x) =c k(x) and ϕ(c) is a continuous function, it follows that the function k : X (0, ) is continuous. Finally, let X =1. Then ϕ(f)(x) =m(f), where m : I I. The function m is multiplicative and bijective, so by Lemma 2.1, ϕ(f)(x) =f(x) k,k>0.

11 MULTIPLICATIVE BIJECTIONS OF C(X,I) 1075 References [1] J. Aczél, J. Dhombres, Functional equations in several variables, Cambridge University Press, Cambridge, MR (90h:39001) [2] P. Busch, P.J. Lahti, P. Mittelstaedt, The Quantum Theory of Measurement, Springer-Verlag, MR (93m:81014) [3] A. Dvurečenskij, S. Pulmannová, Recent Trends in Quantum Structures, Kluwer Academic Publisher, [4] S. Gudder, R Greechie, Sequential products on effect algebras, Rep. Math. Phys. 49 (2002), MR (2003c:81015) [5] S. Gudder, G. Nagy, Sequential quantum measurements, J. Math. Phys. 42 (2001), MR (2002h:81032) [6] S. Gudder, G. Nagy, Sequentially independent effects, Proc. Amer. Math. Soc. 130 (2002), MR (2002i:81014) [7] L. Molnár, Sequential isomorphisms between the sets of von Neumann algebra effects, Acta Sci. Math. (Szeged) 69 (2003), MR EPF-University of Maribor, Razlagova 14, 2000 Maribor, Slovenia address: janko.marovt@uni-mb.si