Preserving affine Baire classes by perfect affine maps
aa r X i v : . [ m a t h . F A ] J a n PRESERVING AFFINE BAIRE CLASSES BY PERFECT AFFINEMAPS
ONDˇREJ F.K. KALENDA AND JIˇR´I SPURN ´Y
Abstract.
Let ϕ : X → Y be an affine continuous surjection between com-pact convex sets. Suppose that the canonical copy of the space of real-valuedaffine continuous functions on Y in the space of real-valued affine continuousfunctions on X is complemented. We show that if F is a topological vectorspace, then f : Y → F is of affine Baire class α whenever the composition f ◦ ϕ is of affine Baire class α . This abstract result is applied to extend knownresults on affine Baire classes of strongly affine Baire mappings. Introduction and the main results
Let ϕ : X → Y be a continuous surjection between compact Hausdorff spaces.If f : Y → T is a mapping with values in a topological space T , some properties of f can be deduced from the properties of f ◦ ϕ . In particular, since ϕ is a closedmapping, it is easy to check that f is continuous if and only if f ◦ ϕ is continuous.Analogous statements hold for Borel measurable, Baire measurable and resolvablymeasurable mappings due to [2]. We formulate them in the following theorem wherewe use the notation from [7]. Theorem A.
Let ϕ : X → Y be a continuous surjection between compact Hausdorffspaces, f : Y → T be a mapping with values in a topological space T and α < ω . (1) f is Borel measurable if (and only if ) f ◦ ϕ is Borel measurable. Moreover, f is Σ α (Bos( Y )) -measurable if (and only if ) f ◦ ϕ is Σ α (Bos( X )) -measurable. (2) f is resolvably measurable if (and only if ) f ◦ ϕ is resolvably measurable.Moreover, f is Σ α (Hs( Y )) -measurable if (and only if ) f ◦ ϕ is Σ α (Hs( X )) -measurable. (3) f is Baire measurable if (and only if ) f ◦ ϕ is Baire measurable. Moreover, f is Σ α (Bas( Y )) -measurable if (and only if ) f ◦ ϕ is Σ α (Bas( X )) -measurable. Let us explain it in more detail. The notation and terminology follow [7]. Amapping f : Y → T is Borel measurable if the inverse image of any open set is aBorel set. Similarly, f is Baire measurable if the inverse image of any open setbelongs to the Baire σ -algebra (i.e., to the σ -algebra generated by cozero sets);and f is resolvably measurable if the inverse image of any open set belongs to the σ -algebra generated by resolvable sets. Further, f is Σ α (Bos( Y ))-measurable if f − ( U ) ∈ Σ α (Bos( Y )) for any open set U ⊂ T . Similarly we define Σ α (Bas( Y ))-measurable and Σ α (Hs( Y ))-measurable mappings. Mathematics Subject Classification.
Key words and phrases. vector-valued Baire function; strongly affine function.Our investigation was supported by the Research grant GAˇCR P201/12/0290. The secondauthor was also supported by The Foundation of Karel Janeˇcek for Science and Research.
The ‘only if’ parts of all the three assertions of Theorem A are obvious. The‘if part’ of the assertion (1) follows from [2, Theorem 10] and that of the assertion(2) follows from [2, Theorem 4]. To show the ‘if part’ of the assertion (3) we firstobserve that by [1, Theorem 2] the mapping f is Baire measurable whenever f ◦ ϕ isBaire measurable and we conclude by using [7, Theorem 3.6] saying that a mappingis Σ α (Bas( Y ))-measurable if and only if it is Baire measurable and Σ α (Bos( Y ))-measurable.If T is a convex subset of a Fr´echet space, the hierarchy of Baire measurablemappings corresponds to the hierarchy of Baire functions. Since we will need moresuch hierarchies, we will introduce it in an abstract setting:Given a set S , a topological space T and a family of mappings F from S to T ,we define the Baire classes of mappings as follows. Let ( F ) = F . Assuming that α ∈ [1 , ω ) is given and that ( F ) β have been already defined for each β < α , we set( F ) α = { f : S → T ; there exists a sequence ( f n ) in [ β<α ( F ) β such that f n → f pointwise } . In particular, if S and T are topological spaces, by C α ( S, T ) we denote the set( C ( S, T )) α , where C ( S, T ) is the set of all continuous functions from S to T . Further,by C α ( S ) we mean C α ( S, R ).It is known (see, e.g. [4, Lemma 3.2]) that, whenever Y is a compact spaceand T is a convex subset of a Fr´echet space, then a mapping f : Y → T belongsto C α ( Y, T ) if and only if it is Σ α +1 (Bas( Y ))-measurable. Therefore we have thefollowing equivalence: Theorem B.
Let ϕ : X → Y be a continuous surjection between compact spacesand T be a convex subset of a Fr´echet space. Let f : Y → T be a mapping and α < ω . Then f ∈ C α ( Y, T ) if (and only if ) f ◦ ϕ ∈ C α ( X, T ) . Now, suppose that X and Y are, moreover, compact convex sets (i.e., compactconvex subsets of locally convex spaces) and the continuous surjection ϕ is affine.We address the following question: In this setting, does an analogue of Theorem B hold for affine Baire classes?
Let us first recall the definition of affine Baire classes: If Y is a compact convexset and T is a convex subset of a topological vector space, by A ( Y, T ) we denotethe set of all affine continuous functions defined on Y with values in T and, for α < ω we set A α ( Y, T ) = ( A ( Y, T )) α . Further, A ( Y ) stands for the space A ( Y, R )and A α ( Y ) means A α ( Y, R ).Hence, the precise question we address is the following: Assuming that f ◦ ϕ ∈ A α ( X, T ) , is necessarily f ∈ A α ( Y, T ) ? It is clear that f is affine whenever f ◦ ϕ is affine. Therefore the answer is positivein case α = 0. It is further positive if T = R and α = 1, since in this case theclass A ( Y, R ) coincides with the class of all affine functions belonging to C ( Y, R )by a result of Mokobodzki [6, Theorem 4.24]. However, the answer is negative ingeneral. More precisely, for α ≥ α = 1 it is negative for vector-valued functions.Let us explain it more detail. Let Y be an arbitrary compact convex set. If µ isa Radon probability measure on Y , then there is a unique point x ∈ Y such that u ( x ) = R u d µ for any u ∈ A ( Y ). This unique point is called the barycenter of µ RESERVING AFFINE BAIRE CLASSES BY PERFECT AFFINE MAPS 3 and is denoted by r ( µ ) (see [6, Definition 2.26]). Further, a function f : Y → R iscalled strongly affine if for each Radon probability µ on X f is µ -integrable and R f d µ = f ( r ( µ )). The same definition can be used for f : Y → F , where F is aFr´echet space (see [4]). Since the vector integral used in the definition is the Pettisone, such f is strongly affine if and only if τ ◦ f is strongly affine for each τ ∈ F ∗ (see [4, Fact 1.2]). We will use this characterization as a definition.If we set X = M ( Y ), the set of all Radon probability measures on Y equippedwith the weak ∗ topology, then X is again a compact convex set and, by [6, Propo-sition 2.38] the mapping r : µ r ( µ ) is a continuous affine surjection. Moreover,by [6, Proposition 6.38] the set X is a Bauer simplex (i.e., a Choquet simplex withclosed set of extreme points), therefore by [4, Theorem 2.5] any strongly affine map g : X → F with values in a Fr´echet space F which belongs to C α ( X, F ) in factbelongs to A α ( X, F ).Now, by [9] there is a compact convex set Y and a strongly affine function f ∈ C ( Y ) such that f / ∈ S α<ω A α ( Y ). If we choose X and r as in the previousparagraph, then f ◦ r ∈ A ( X ). Similarly, by [4, Theorem 2.2] there is a compactconvex set Y , a Banach space F and a strongly affine f ∈ C ( Y, F ) such that f / ∈ S α<ω A α ( Y, F ). If we choose X and r as above, we get f ◦ r ∈ A ( X, F ).The main result of the present paper is a sufficient condition for a positiveanswer. To formulate it we need the following notation. Let ϕ : X → Y be acontinuous affine surjection between compact convex sets. We define a mapping ϕ ∗ : A ( Y ) → A ( X ) by ϕ ∗ ( f ) = f ◦ ϕ . Then ϕ ∗ is an isometric embedding of A ( Y )into A ( X ). Theorem 1.1.
Let X and Y be compact convex sets, ϕ : X → Y a continuousaffine surjection such that ϕ ∗ ( A ( Y )) is a complemented subspace of A ( X ) . Let F be a topological vector space. Let f : Y → F be such that f ◦ ϕ ∈ A α ( X, F ) forsome α < ω . Then f ∈ A α ( X, F ) . We point out that this result is quite abstract and that the range space F isjust a topological vector space – no local convexity, metrizability or completenessis required. The proof is given in the next section and is rather elementary. Ofcourse, the most interesting case is that of Fr´echet range. In this setting we get thefollowing corollary. Theorem 1.2.
Let either (a) X = ( B E ∗ , w ∗ ) , where E is a (real or complex) Banach space which isisomorphic to a complemented subspace of a (real or complex) L -predual E , or (b) X is a compact convex set such that A ( X, F ) is isomorphic to a comple-mented subspace of an L -predual E over F .If F is a Fr´echet space, then any strongly affine f ∈ C α ( X, F ) belongs to A α ( X, F ) .If ext B E ∗ is moreover F σ in the weak ∗ topology, α can be replaced by α . Recall that a (real or complex) L -predual is a (real or complex) Banach spacewhose dual is isometric to a space of the form L ( µ ) for some non-negative measure µ . The letter F stands for R or C .As an immediate consequence we get the following result. Corollary 1.3.
ONDˇREJ F.K. KALENDA AND JIˇR´I SPURN´Y • The Banach space E constructed in [9] is not isomorphic to a complementedsubspace of an L -predual. • Let X be any of the simplices constructed in [8, Theorem 1.1] , then A ( X ) is not isomorphic to a complemented subspace of a C ( K ) space. (In fact, A ( X ) is even not isomorphic to a complemented subspace of A ( Y ) , where Y is a simplex with ext Y being F σ .)Proof. By [9] there is a strongly affine function on ( B E ∗ , w ∗ ) which is in the class C but not in A α for any α < ω . If X is any of the simplices from [8, Theorem1.1], then there is a strongly affine function f ∈ C ( X ) \ A ( X ). Hence both casesindeed follow from Theorem 1.2. (cid:3) Remark 1.4.
The complementability condition in Theorem 1.1 is sufficient but notnecessary. For example, if K is any compact space and F is a Fr´echet space, thenthe space A α ( M ( K ) , F ) coincide with the subspace of C α ( M ( K ) , F ) consistingof strongly affine mappings (by [4, Theorem 2.5]). Therefore, if both X and Y are of the form M ( K ) for some compact space K , the conclusion follows fromTheorem B and [6, Proposition 5.29]. Further, one can easily find X and Y ofthis form and choose ϕ such that ϕ ∗ ( A ( Y )) is not complemented in A ( X ). Indeed,choose compact spaces K and L and a continuous surjection ψ : K → L such that ψ ∗ ( C ( L )) is not complemented in C ( K ). (One can take K to be the Cantor set { , } N , L to be the unit interval and ψ : K → L be the standard surjection. Then ψ ∗ ( C ( L )) is not complemented in C ( K ), which follows e.g. from [5, Lemma 2.7].)Further, let ϕ : M ( K ) → M ( L ) assign to each µ ∈ M ( K ) its image under ψ .Then ϕ is an affine continuous surjection and ϕ ∗ ( A ( M ( L ))) is not complementedin A ( M ( K )). 2. Proof of the main abstract result
The aim of this section is to prove Theorem 1.1. To do that we need severallemmata. We point out that all vector spaces in this section are supposed to bereal. The results of this section can be also used for complex vector spaces if weforget the complex multiplication and look at them as at real spaces.If X is a compact convex set and x ∈ X , we will denote by ε x the respectiveevaluation functional, i.e., ε x ( u ) = u ( x ) , u ∈ A ( X ) . Then clearly ε x ∈ A ( X ) ∗ and k ε x k = 1 for each x ∈ X . Moreover, we have thefollowing lemma. Lemma 2.1.
Let X be a compact convex set. Then the following assertions hold. (i) The mapping ε : x ε x is an affine homeomorphism of X into ( A ( X ) ∗ , w ∗ ) .Moreover, ε ( X ) = { η ∈ A ( X ) ∗ ; η ≥ η (1) = 1 } . (ii) For any η ∈ A ( X ) ∗ there are x , x ∈ X and a , a ≥ such that η = a ε x − a ε x and k η k = a + a . (iii) The weak topology on A ( X ) coincides with the topology of pointwise con-vergence on X .Proof. The assertions (i) and (ii) are proved in [6, Proposition 4.31(a,b)]. Theassertion (iii) follows immediately from (ii).Since we will use it later, we indicate how to find x , x ∈ X and a , a ≥ η ∈ A ( X ) ∗ be given. By the Hahn-Banach theorem the RESERVING AFFINE BAIRE CLASSES BY PERFECT AFFINE MAPS 5 functional η can be extended to some ˜ η ∈ C ( X ) ∗ with the same norm. By the Rieszrepresentation theorem the functional ˜ η is represented by a signed Radon measure µ on X . We set a = µ + ( X ) and a = µ − ( X ). If a = 0, let x ∈ X be arbitrary;if a >
0, let x be the barycenter of µ + a . The point x is defined analogously. (cid:3) In the sequel we will use the following notation. If X and Y are two convexsubsets of some vector spaces, by Aff( X, Y ) we denote the set of all affine mappingsdefined on X with values in Y . Further, if E and F are vector spaces, Lin( E, F )will denote the vector space of all linear operators from E to F . Lemma 2.2.
Let X be a compact convex set and F a vector space. For anyaffine mapping f : X → F there is a unique linear map L f : A ( X ) ∗ → F such that L f ( ε x ) = f ( x ) for each x ∈ X . Moreover, the operator L : f L f is a linearbijection of the space Aff(
X, F ) onto the space Lin( A ( X ) ∗ , F ) .Proof. This essentially follows from [6, Exercise 4.48], where a similar assertion isshown for scalar functions. Let us indicate the proof. If η ∈ A ( X ) ∗ , let x , x ∈ X and a , a ≥ L f ( η ) = a f ( x ) − a f ( x ) . Hence the uniqueness is clear. It remains to observe that this formula correctlydefines a linear mapping. To see that the definition of L f is correct it is enough tocheck that x , x , y , y ∈ X, a , a , b , b ≥ , a ε x − a ε x = b ε y − b ε y ⇒ a f ( x ) − a f ( x ) = b f ( y ) − b f ( y ) , which easily follows from the fact that both ε and f are affine maps. Now, L f isclearly affine and L f (0) = 0, hence L f is linear.It is clear that the operator L is linear. Moreover, if T : A ( X ) ∗ → F is a linearmapping, then T = L T ◦ ε , hence L is a linear bijection. (cid:3) Lemma 2.3.
Let X be a compact convex set and F a topological vector space. Let L : f L f be the operator from Lemma 2.2. Then the following assertions hold. (i) L is a homeomorphism of Aff(
X, F ) onto Lin( A ( X ) ∗ , F ) , when both spacesare equipped with the pointwise convergence topology. (ii) L f is bounded if and only if f is bounded. (iii) L f is weak ∗ -continuous on B A ( X ) ∗ if and only if f is continuous. (iv) L f ∈ L α ( A ( X ) ∗ , F ) if and only if f ∈ A α ( X, F ) . Here L α ( A ( X ) ∗ , F ) =( L ( A ( X ) ∗ , F )) α , where L ( A ( X ) ∗ , F ) is the subspace of Lin( A ( X ) ∗ , F ) con-sisting of maps which are weak ∗ -continuous on the unit ball.Proof. (i) Given η ∈ A ( X ∗ ) fix x , x ∈ X and a , a ≥ f ∈ Aff(
X, F ) we have L f ( η ) = a f ( x ) − a f ( x ), thus the mapping f L f ( η ) is continuous in the pointwise convergence topology. It follows that L is continuous.Conversely, given x ∈ X we have L − ( T )( x ) = T ( ε x ) for T ∈ Lin( A ( X ) ∗ , F ),hence T L − ( T )( x ) is continuous in the pointwise convergence topology. Itfollows that L − is continuous.(ii) If L f is bounded, i.e., if L f ( B A ( X ) ∗ ) is bounded in F , then f = L f ◦ ε is alsobounded as ε ( X ) ⊂ B A ( X ) ∗ . ONDˇREJ F.K. KALENDA AND JIˇR´I SPURN´Y
Conversely, let f be bounded, i.e., let f ( X ) be bounded. We will show that L f ( B A ( X ) ∗ ) is bounded as well. To this end let U be any neighborhood of zero in F . Let V be a balanced neighborhood of zero in F such that V + V ⊂ U . Since f ( X ) is bounded, there is λ > f ( X ) ⊂ λV . Let η ∈ B A ( X ) ∗ be arbitrary. Fix x , x ∈ X and a , a ≥ a + a ≤
1, we have a f ( x ) ∈ λV and − a f ( x ) ∈ λV . Then L f ( η ) = a f ( x ) − a f ( x ) ∈ λV + λV ⊂ λU . Hence L f ( B A ( X ) ∗ ) ⊂ λU and the proof is completed.(iii) The ‘only if’ part is trivial. Let us show the ‘if’ part. Suppose that f iscontinuous. Set K = X × X × { ( a , a ) ∈ R ; a ≥ , a ≥ , a + a ≤ } . Then K is a compact space. Moreover, define mappings ψ : K → B A ( X ) ∗ and g : K → F by the formulas ψ ( x , x , a , a ) = a ε x − a ε x ,g ( x , x , a , a ) = a f ( x ) − a f ( x ) . Then ψ is a continuous mapping of K onto ( B A ( X ) ∗ , w ∗ ), g is continuous and g = L f ◦ ψ . It follows that L f is continuous on ( B A ( X ) ∗ , w ∗ ) which was to beproved.(iv) This assertion follows by transfinite induction from (iii) since, due to (i), f n → f pointwise on K if and only if L f n → L f pointwise. (cid:3) Lemma 2.4.
Let X be a compact convex set and let L : f L f be the operatorprovided by Lemma 2.2 in case F = R . Then L is a linear isometry of A b ( X ) , thespace of all bounded affine functions on X equipped with the supremum norm, onto A ( X ) ∗∗ . Moreover, it is a homeomorphism from the pointwise convergence topologyto the weak ∗ topology and its restriction to A ( X ) is the canonical embedding of A ( X ) into its second dual.Proof. This is an easy consequence of Lemma 2.3. (It also follows from [6, Propo-sition 4.32].) (cid:3)
In the sequel we will identify A ( X ) ∗∗ and A b ( X ). Proof of Theorem 1.1.
Fix a bounded linear projection P : A ( X ) → ϕ ∗ ( A ( Y )) andset P = ( ϕ ∗ ) − ◦ P . Then P : A ( X ) → A ( Y ) is a bounded linear operator suchthat P ◦ ϕ ∗ is the identity map on A ( Y ). Recall that ϕ ∗ is defined by ϕ ∗ ( f ) = f ◦ ϕ , f ∈ A ( Y ).The dual operator ϕ ∗∗ : A ( X ) ∗ → A ( Y ) ∗ satisfies(2.1) ϕ ∗∗ ( ε x ) = ε ϕ ( x ) , x ∈ X. Indeed, for any x ∈ X and g ∈ A ( Y ) we have ϕ ∗∗ ( ε x )( g ) = ε x ( ϕ ∗ ( g )) = ε x ( g ◦ ϕ ) = g ( ϕ ( x )) = ε ϕ ( x ) ( g ) . Let F be a topological vector space and let f : X → F be affine. We define anaffine map Qf : Y → F by the formula(2.2) Qf ( y ) = L f ( P ∗ ε y ) , y ∈ Y. It is clear that Qf is affine since it is a composition of three affine maps. Moreover, Qf ∈ A α ( Y, F ) whenever f ∈ A α ( X, F ). Indeed, the mapping ε is continuous from Y to ( A ( Y ) ∗ , w ∗ ), P ∗ is weak ∗ -to-weak ∗ continuous and P ∗ ( ε ( Y )) ⊂ k P k B A ( X ) ∗ . RESERVING AFFINE BAIRE CLASSES BY PERFECT AFFINE MAPS 7
Further, if f ∈ A α ( X, F ), then by Lemma 2.3(iv), L f ∈ L α ( A ( X ) ∗ , F ), hence, inparticular, L f ∈ A α (( k P k B A ( X ) ∗ , w ∗ ) , F ). Thus we get in this case Qf ∈ A α ( Y, F ).The proof will be completed if we show that Q ( g ◦ ϕ ) = g for each affine g : Y → F . So, fix an affine map g : Y → F and y ∈ Y . Then P ∗ ( ε y ) ∈ A ( X ) ∗ , hence, byLemma 2.1, we can find x , x ∈ X and a , a ≥ P ∗ ( ε y ) = a ε x − a ε x and a + a ≤ k P k . We have Q ( g ◦ ϕ )( y ) = L g ◦ ϕ ( P ∗ ε y ) = a ( g ◦ ϕ )( x ) − a ( g ◦ ϕ )( x )= a g ( ϕ ( x )) − a g ( ϕ ( x )) = a L g ( ε ϕ ( x ) ) − a L g ( ε ϕ ( x ) )= L g ( a ε ϕ ( x ) − a ε ϕ ( x ) ) = L g ( a ϕ ∗∗ ( ε x ) − a ϕ ∗∗ ( ε x ))= L g ( ϕ ∗∗ ( a ε x − a ε x )) = L g ( ϕ ∗∗ ( P ∗ ( ε y )))= L g (( P ◦ ϕ ∗ ) ∗ ( ε y )) = L g ( ε y ) = g ( y ) . Indeed, the first four equalities follow from the definitions. The fifth one followsfrom the linearity of L g . The sixth equality is a consequence of (2.1), the seventhone follows from the linearity of ϕ ∗∗ . The next one follows from the choice of a , a , x , x . In the following two equalities we use that ϕ ∗∗ ◦ P ∗ = ( P ◦ ϕ ∗ ) ∗ =(id A ( Y ) ) ∗ = id A ( Y ) ∗ . Finally, the last one follows from the definition of L g . (cid:3) Applications to affine classes of strongly affine Baire mappings
The aim of this section is to prove Theorem 1.2. We will need several lemmata.The first one shows an easy correspondence between complementability of Ba-nach spaces and complementability of spaces of affine continuous functions.
Lemma 3.1.
Let E be a complemented subspace of a Banach space E . Let π : E ∗ → E ∗ be the restriction mapping. Then π ∗ ( A ( B E ∗ , w ∗ )) is complemented in A ( B E ∗ , w ∗ ) .Proof. Let us first give the proof for real spaces. Let P : E → E be a boundedlinear projection. For any f ∈ A ( B E ∗ , w ∗ ) there is a unique x f ∈ E such that f ( x ∗ ) = f (0) + x ∗ ( x f ) for any x ∗ ∈ B E ∗ . Hence we can define a bounded linearoperator Q : A ( B E ∗ , w ∗ ) → A ( B E ∗ , w ∗ ) by Qf ( x ∗ ) = f (0) + x ∗ ( P x f ) , x ∗ ∈ B E ∗ , f ∈ A ( B E ∗ , w ∗ ) . Then π ∗ ◦ Q is the required projection.Now suppose that E and E are complex spaces. Again, let P : E → E be a bounded linear projection. If f ∈ A (( B E ∗ , w ∗ ) , C ), then there are uniquelydetermined x f , y f ∈ E such that f ( x ∗ ) = f (0) + x ∗ ( x f ) + x ∗ ( y f ) , x ∗ ∈ B E ∗ , and, moreover, the assignment f x f is a bounded linear map and f y f is abounded conjugate-linear map (this follows, for example, from [4, Lemma 3.11(a)]).Hence, we can define a bounded linear operator Q : A (( B E ∗ , w ∗ ) , C ) → A (( B E ∗ , w ∗ ) , C )by Q f ( x ∗ ) = f (0) + x ∗ ( P x f ) + x ∗ ( P y f ) , x ∗ ∈ B E ∗ , f ∈ A (( B E ∗ , w ∗ ) , C ) . Then a required bounded linear projection Q : A ( B E ∗ , w ∗ ) → π ∗ ( A ( B E ∗ , w ∗ )) canbe defined by Qf = π ∗ (Re Q f ). (cid:3) ONDˇREJ F.K. KALENDA AND JIˇR´I SPURN´Y
The second lemma is a trivial observation on isomorphic spaces.
Lemma 3.2.
Let T : E → E be a surjective isomorphism of Banach spaces and F be a vector space. Then there is a unique linear map ˜ T : Aff( B E ∗ , F ) → Aff( B E ∗ , F ) such that for any real-linear map L : E ∗ → F we have ˜ T ( L | B E ∗ ) = ( L ◦ T ∗ ) | B E ∗ .Moreover, ˜ T is a linear bijection and, if F is a topological vector space and f ∈ Aff( B E ∗ , F ) , the following assertions hold. • ˜ T is a homeomorphism when both spaces are equipped with the pointwiseconvergence topology. • ˜ T f is continuous if and only if f is continuous. • ˜ T f ∈ C α (( B E ∗ , w ∗ ) , F ) if and only if f ∈ C α (( B E ∗ , w ∗ ) , F ) . • ˜ T f ∈ A α (( B E ∗ , w ∗ ) , F ) if and only if f ∈ A α (( B E ∗ , w ∗ ) , F ) .Further, if F is Fr´echet space, • ˜ T f is strongly affine if and only if f is strongly affine.Proof. Let f ∈ Aff( B E ∗ , F ). Then there is a real-linear mapping U f : E ∗ → F suchthat f = f (0) + U f | B E ∗ . It is enough to set ˜ T f = f (0) + ( U f ◦ T ∗ ) | B E ∗ . (cid:3) We will need also the following lemma which summarizes complex versions ofsome lemmata from the previous section.
Lemma 3.3.
Let X be a compact convex set. (i) For any η ∈ A ( X, C ) ∗ there exist x , x , x , x ∈ X and a , a , a , a ≥ such that a + a + a + a ≤ k η k and η = a ε x − a ε x + i ( a ε x − a ε x ) . (ii) If F is a complex vector space and f ∈ Aff(
X, F ) , then there is a unique L f ∈ Lin( A ( X, C ) ∗ , F ) such that L f ( ε x ) = f ( x ) for each x ∈ X . Moreover,the operator L : f L f is a linear bijection. (iii) If F is a complex topological vector space, then the operator L has theproperties from Lemma 2.3.Proof. (i) This assertion can be derived from Lemma 2.1(ii). However, we presenta direct proof, since we will later use the construction. Let η ∈ A ( X, C ) ∗ . By theHahn-Banach theorem we can choose ˜ η ∈ C ( X, C ) ∗ extending η and having thesame norm. Fix a complex Radon measure µ on X which represents ˜ η (by theRiesz representation theorem). We set a = (Re µ ) + ( X ) , a = (Re µ ) − ( X ) , a = (Im µ ) + ( X ) , a = (Im µ ) − ( X ) . If a = 0, let x ∈ X be arbitrary. If a >
0, let x be the barycenter of (Re µ + ) /a .Similarly we define x , x and x .(ii) Let f ∈ Aff(
X, F ). Given η ∈ A ( X, C ) ∗ , fix a representation of η by (i). Wehave to set L f ( η ) = a f ( x ) − a f ( x ) + i ( a f ( x ) − a f ( x )) . Therefore, the uniqueness is clear. To see that L f is well defined, it is enough tocheck that x , x , x , x , y , y , y , y ∈ X, a , a , a , a , b , b , b , b ≥ ,a ε x − a ε x + i ( a ε x − a ε x ) = b ε y − b ε y + i ( b ε y − b ε y ) ⇓ a f ( x ) − a f ( x ) + i ( a f ( x ) − a f ( x ))= b f ( y ) − b f ( y ) + i ( b f ( y ) − b f ( y )) , RESERVING AFFINE BAIRE CLASSES BY PERFECT AFFINE MAPS 9 which easily follows from the fact that ε and f are affine. It is clear that L f is affine.Since L f (0) = 0 and L f ( iη ) = iL f ( η ) for η ∈ A ( X, C ) ∗ , we get that L f is linear.Further, the operator L is obviously linear. Since for any T ∈ Lin( A ( X, C ) ∗ , F ) wehave T = L T ◦ ε , L is a linear bijection.(iii) The proof is analogous to the real case. (cid:3) Lemma 3.4.
Let X be a compact convex set, F a Fr´echet space over F and f : X → F a strongly affine mapping which belongs to C α ( X, F ) for some α < ω . Let L f : A ( X, F ) ∗ → F be the linear map provided by Lemma 2.1 or Lemma 3.3. Then L f | ( B A ( X, F ) ∗ ,w ∗ ) is a strongly affine mapping which belongs to C α (( B A ( X, F ) ∗ , w ∗ ) , F ) .Proof. Let us denote by M ( X, F ) the space of all F -valued Radon measures on X considered as the dual space to C ( X, F ) equipped with the weak ∗ topology. Forany bounded Baire function g : X → F let us define a mapping ˜ g : B M ( X, F ) → F by setting ˜ g ( µ ) = R X g d µ . The integral is considered in the Pettis sense, see [4,Section 1.3]. The mapping ˜ g is well defined by [4, Lemma 3.6].Moreover, ˜ g ∈ A α ( B M ( X, F ) , F ) whenever g ∈ C α ( X, F ). We will prove it bytransfinite induction on α . Let α = 0, i.e., let g be continuous. Then for each τ ∈ F ∗ the function τ ◦ g is continuous, thus the mapping µ τ (˜ g ( µ )) = Z X τ ◦ g d µ, µ ∈ B M ( X, F ) , is continuous by the very definition of the weak ∗ topology. Thus ˜ g is continuousfrom B M ( X, F ) to the weak topology of F . Further, the range of ˜ g is contained in theclosed absolutely convex hull of g ( X ) (by [4, Lemma 3.5(c)]). Since g ( X ) is compact,we deduce that ˜ g ( X ) is relatively compact by [3, Proposition 6.7.2]. Hence the weaktopology coincides with the original one on ˜ g ( X ), so ˜ g is continuous. Finally, it isclear that ˜ g is affine.Let α ∈ (0 , ω ) be such that the statement hold for each β < ω . Let g ∈C α ( X, F ) be a bounded function. Then there is a uniformly bounded sequence ( g n )in S β<α C β ( X, F ) pointwise converging to g . Indeed, by [4, Lemma 3.1(c)] we getthat g ∈ C α ( X, co g ( X )) and the set co g ( X ) is bounded. Now it follows from [4,Theorem 3.7] that ˜ g n → ˜ g pointwise. Since by the induction hypothesis we have˜ g n ∈ S β<α A β ( B M ( X, F ) , F ), we conclude ˜ g ∈ A α ( B M ( X, F ) , F ).We return to the proof of the lemma. Let X, F, f, α be as in the statement.Since f is bounded by [4, Lemma 4.1], we get ˜ f ∈ A α ( B M ( X, F ) , F ). Further, let π : M ( X, F ) → A ( X, F ) ∗ be the restriction map (recall that M ( X, F ) is identifiedwith C ( X, F ) ∗ ). We claim that(3.1) ˜ f = ( L f ◦ π ) | B M ( X, F ) . Since the mappings on both sides are restrictions of linear operators, it is enough tocheck that they agree on probability measures. Let µ ∈ M ( X ). Then π ( µ ) = ε r ( µ ) ,hence L f ( π ( µ )) = L f ( ε r ( µ ) ) = f ( r ( µ )) = Z f d µ = ˜ f ( µ ) , where we used the definitions and the assumption that f is strongly affine.Finally, π is a continuous affine surjection. Since ˜ f ∈ A α ( B M ( X, F ) , F ), by Theo-rem B we conclude that L f | ( B A ( X, F ) ∗ ,w ∗ ) ∈ C α (( B A ( X, F ) ∗ , w ∗ ) , F ) and L f | ( B A ( X, F ) ∗ ,w ∗ ) is strongly affine by [6, Proposition 5.29] and [4, Fact 1.2]. (cid:3) Proof of Theorem 1.2. (a) Let E be an L -predual, E a Banach space isomorphicto a complemented subspace of E and X = ( B E ∗ , w ∗ ). Further, let F be a Fr´echetspace, α < ω and f ∈ C α ( X, F ) a strongly affine function. Fix T : E → E anisomorphism of E onto a complemented subspace of E . Denote E = T ( E ) andlet π : E ∗ → E ∗ denote the restriction mapping.Let g = ˜ T f (see Lemma 3.2). Then g is strongly affine and g ∈ C α (( B E ∗ , w ∗ ) , F )(by Lemma 3.2). Hence g ◦ π is a strongly affine function in C α (( B E ∗ , w ∗ ) , F ). By[4, Theorem 2.5] we get g ◦ π ∈ A α (( B E ∗ , w ∗ ) , F ). Therefore, by Lemma 3.1and Theorem 1.1 we get g ∈ A α (( B E ∗ , w ∗ ) , F ). By Lemma 3.2 we conclude that f ∈ A α ( X, F ).If ext B E ∗ is moreover a weak ∗ F σ -set, from [4, Theorem 2.5] we obtain g ◦ π ∈ A α (( B E ∗ , w ∗ ) , F ), therefore we conclude f ∈ A α ( X, F ).(b) Let us start by the real case. Hence, let E be a real L -predual, X acompact convex space, T : A ( X ) → E an isomorphism onto a complementedsubspace. Let F be a Fr´echet space and f : X → F a strongly affine functionwhich belongs to C α ( X, F ). Without loss of generality we can suppose that F is real (if F is complex, we can forget complex multiplication and consider F asa real space). By Lemma 3.4 the mapping L f | B A ( X, F ) ∗ is strongly affine and be-longs to C α (( B A ( X, F ) ∗ , w ∗ ) , F ). Therefore by the already proved case (a) we get L f | B A ( X, F ) ∗ ∈ A α (( B A ( X, F ) ∗ , w ∗ ) , F ), hence f ∈ A α ( X, F ).If F = C and F is complex, the proof can be done in the same way as in the realcase.Finally, suppose that F = C and F is real. Let F C be the complexification of F .By the previous case we get f ∈ A α ( X, F C ). Since the canonical projection of F C onto F is real-linear and continuous, we get f ∈ A α ( X, F ).If ext B E ∗ is weak ∗ F σ , 1 + α can be replaced by α as in the case (a). (cid:3) References [1]
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Charles University in Prague, Faculty of Mathematics and Physics, Department ofMathematical Analysis, Sokolovsk´a 83, 186 75, Praha 8, Czech Republic
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