aa r X i v : . [ m a t h . GN ] M a r ON THE MACKEY PROBLEM FOR FREE LOCALLY CONVEX SPACES
SAAK GABRIYELYAN
Abstract.
We show that the free locally convex space L ( X ) over a Tychonoff space X is a Mackeygroup iff L ( X ) is a Mackey space iff X is discrete. Introduction
Let (
E, τ ) be a locally convex space (lcs for short). A locally convex vector topology ν on E iscalled compatible with τ if the spaces ( E, τ ) and (
E, ν ) have the same topological dual space. Theclassical Mackey–Arens theorem states that for every lcs (
E, τ ) there exists the finest locally convexvector space topology µ on E compatible with τ . The topology µ is called the Mackey toplogy on E associated with τ , and if µ = τ , the space E is called a Mackey space .An analogous notion in the class of locally quasi-convex (lqc for short) abelian groups wasintroduced in [3]. For an abelian topological group (
G, τ ) we denote by b G the group of all continuouscharacters of ( G, τ ) (for all relevant definitions see the next section). Two topologies µ and ν on anabelian group G are said to be compatible if \ ( G, µ ) = \ ( G, ν ). Following [3], an lqc abelian group(
G, µ ) is called a
Mackey group if for every lqc group topology ν on G compatible with τ it followsthat ν ≤ µ .Not every Mackey lcs is a Mackey group. In [6] we show that the space C p ( X ), which is aMackey space for every Tychonoff space X , is a Mackey group if and only if it is barrelled. Inparticular, this result shows that there are even metrizable lcs which are not Mackey groups thatgives a negative answer to a question posed in [4]. Only very recently, answering Question 4.4 of[5], Außenhofer [1] and the author [7] independently have shown that there are lqc groups whichdo not admit a Mackey group topology. For historical remarks, references and open questions wereferee the reader to [5, 12]. In Question 4.3 of [5], we ask: For which Tychonoff spaces X the freelcs L ( X ) is a Mackey space or a Mackey group? Below we give a complete answer to this question. Theorem 1.1.
For a Tychonoff space X , the following assertions are equivalent: (i) L ( X ) is a Mackey group; (ii) L ( X ) is a Mackey space; (iii) X is discrete. In particular, Theorem 1.1 essentially strengthen Theorem 6.4 of [9] which states that L ( X ) isquasibarrelled if and only of X is discrete.2. Proof of Theorem 1.1
We start from some necessary definitions and notations. Let X be a Tychonoff space. The space X is called a k R -space if every real-valued function on X which is continuous on every compactsubset of X is continuous on X . A subset A of X is called functionally bounded in X if everycontinuous real-valued function on X is bounded on A , and X is a µ -space if every functionallybounded subset of X has compact closure. The Dieudonn´e completion µX of X is always a µ -space. Mathematics Subject Classification.
Primary 46A03; Secondary 54H11.
Key words and phrases. free locally convex space, Mackey topology.
Denote by S the unit circle group and set S + := { z ∈ S : Re( z ) ≥ } . Let G be an abeliantopological group. A character χ ∈ b G is a continuous homomorphism from G into S . A subset A of G is called quasi-convex if for every g ∈ G \ A there exists χ ∈ b G such that χ ( g ) / ∈ S + and χ ( A ) ⊆ S + . The group G is called locally quasi-convex if it admits a neighborhood base atthe neutral element 0 consisting of quasi-convex sets. Every real locally convex space is a locallyquasi-convex group by Proposition 2.4 of [2].Following [11], the free locally convex space L ( X ) on a Tychonoff space X is a pair consisting ofa locally convex space L ( X ) and a continuous map i : X → L ( X ) such that every continuous map f from X to a locally convex space E gives rise to a unique continuous linear operator ¯ f : L ( X ) → E with f = ¯ f ◦ i . The free locally convex space L ( X ) always exists and is essentially unique. The set X forms a Hamel basis for L ( X ) and the map i is a topological embedding, see [13, 14].Let X be a Tychonoff space. For χ = a x + · · · + a n x n ∈ L ( X ) with distinct x , . . . , x n ∈ X and nonzero a , . . . , a n ∈ R , we set k χ k := | a | + · · · + | a n | , and supp( χ ) := { x , . . . , x n } . For an lcs E , we denote by E ′ the topological dual space of E . For a cardinal number κ ,the classical Banach space c ( κ ) consists of all bounded functions g : κ → R such that the set { i ∈ κ : | g ( i ) | ≥ ε } is finite for every ε > k · k ∞ .We denote by C k ( X ) the space C ( X ) of all real-valued continuous functions on X endowed withthe compact-open topology τ k . The support of a function f ∈ C ( X ) is denoted by supp( f ). Denoteby M c ( X ) the space of all real regular Borel measures on X with compact support. It is well-knownthat the dual space of C k ( X ) is M c ( X ), see [10, Proposition 7.6.4]. For every x ∈ X , we denote by δ x ∈ M c ( X ) the evaluation map (Dirac measure), i.e. δ x ( f ) := f ( x ) for every f ∈ C ( X ). Denoteby τ e the polar topology on M c ( X ) defined by the family of all equicontinuous pointwise boundedsubsets of C ( X ). We shall use the following deep result of Uspenski˘ı [14]. Theorem 2.1 ([14]) . Let X be a Tychonoff space and let µX be the Dieudonn´e completion of X .Then the completion L ( X ) of L ( X ) is topologically isomorphic to (cid:0) M c ( µX ) , τ e (cid:1) . We need also the following corollary of Theorem 2.1 noticed in [8].
Corollary 2.2 ([8]) . Let X be a µ -space. Then the topology τ e on M c ( X ) is compatible with theduality ( C k ( X ) , M c ( X )) .Proof. It is well-known that L ( X ) ′ = C ( X ), see [13]. Now Theorem 2.1 implies ( M c ( X ) , τ e ) ′ = L ( X ) ′ = C ( X ). (cid:3) The next lemma follows from Proposition 2.5 of [5], we give its proof for the sake of completenessof the paper.
Lemma 2.3.
If a real lcs ( E, τ ) is a Mackey group, then it is a Mackey space.Proof. Let ν be a locally convex vector topology on E compatible with τ . Applying Proposition 2.3of [2] we obtain \ ( E, ν ) = \ ( E, τ ). Hence ν is a locally quasi-convex group topology (see Proposition2.4 of [2]) compatible with τ . Therefore ν ≤ τ since ( E, τ ) is a Mackey group. Thus (
E, τ ) is aMackey space. (cid:3)
We need the following characterization of non-discrete Tychonoff spaces.
Proposition 2.4.
A Tychonoff space X is not discrete if and only if there exist an infinite cardinal κ , a point z ∈ X , a family { g i } i ∈ κ of continuous functions from X to [0 , and a family { U i } i ∈ κ ofopen subsets of X such that (i) supp( g i ) ⊆ U i for every i ∈ κ ; (ii) U i ∩ U j = ∅ for all distinct i, j ∈ κ ; N THE MACKEY PROBLEM FOR FREE LOCALLY CONVEX SPACES 3 (iii) z U i for every i ∈ κ and z ∈ cl (cid:0) S i ∈ κ { x ∈ X : g i ( x ) ≥ } (cid:1) .Proof. The sufficiency follows from (i)-(iii) which cannot hold simultaneously for discrete spaces.To prove the necessity we consider two cases.
Case 1. There is a continuous function h : X → [0 , such that the set L := { x ∈ X : h ( x ) > } is not closed. So there is a z ∈ cl( L ) such that h ( z ) = 0. We distinguish between two subcases. Subcase 1.1. For every neighborhood W of z , the closure h ( W ) of h ( W ) contains an interval ofthe form [0 , ε ) for some ε > . For every n ∈ N , set t n ( x ) := max (cid:26) h ( x ) − n + 1 , (cid:27) · max (cid:26) , n − − h ( x ) (cid:27) and A n := sup { t n ( x ) : x ∈ X } . Let m be the least natural number such that A n > n > m . For every n > m , set U n := h − (cid:18) n + 1 , n − (cid:19) and g n ( x ) := 2 A n · t n ( x ) , Then, for every n > m , we have g n ( X ) ⊆ [0 , g n ) ⊆ U n and z U n . Clearly, (i) and(ii) are fulfilled and z U n for every n > m . So to check that the sequences { U n : n > m } and { g n : n > m } and the point z satisfy (i)-(iii) we have to show that z ∈ cl (cid:0) S n>m g − n (cid:0) [1 , (cid:1)(cid:1) .Fix arbitrarily a neighborhood W of z in X . Then, by assumption, h ( W ) contains [0 , ε ) forsome ε ∈ (0 , n > (1 + 3 ε ) / (3 ε ) there is a y ∈ W such that t n ( y ) ≥ (1 / A n , andhence g n ( y ) ≥
1. Thus g − n (cid:0) [1 , (cid:1) ∩ W is not empty and hence z ∈ cl (cid:0) S n>m g − n (cid:0) [1 , (cid:1)(cid:1) . Subcase 1.2. There is a neighborhood W of z such that the closure h ( W ) of h ( W ) does notcontain an interval of the form [0 , ε ). Then there exist sequences { a n } n ∈ N and { b n } n ∈ N in (0 , b n +1 < a n < b n , [ b n +1 , a n ] ∩ h ( W ) = ∅ and ( a n , b n ) ∩ h ( W ) = ∅ , ∀ n ∈ N . Set a := 1 and c n := 12 ( b n +1 + a n ) and d n := 12 ( b n + a n − ) , ∀ n ∈ N . Then c n < a n < b n < d n <
1. For every n ∈ N , let r n ( x ) be the piecewise linear continuous functionfrom [0 ,
1] to [0 ,
1] such that r n (cid:0) [0 , c n ] ∪ [ d n , (cid:1) = { } and r n (cid:0) [ a n , b n ] (cid:1) = { } , and set U n := h − ( c n , d n ) and g n ( x ) := r n (cid:0) h ( x ) (cid:1) , x ∈ X. By construction, the sequences { U n : n ∈ N } and { g n : n ∈ N } and the point z satisfy (i) and(ii) and z U n for every n ∈ N . Let us show that every neighborhood U of z contains elementsof S n ∈ N g − n (cid:0) { } (cid:1) . We can assume that U ⊆ W . Since [ b n +1 , a n ] ∩ h ( W ) = ∅ we obtain that h ( W ) ⊆ { } ∪ S n ∈ N ( a n , b n ). Therefore, if y ∈ U and n ∈ N is such that h ( y ) ∈ ( a n , b n ) (such a y exists because z ∈ cl( L )), then g n ( y ) = 1. Case 2. For every continuous function h : X → [0 , the set { x ∈ X : h ( x ) > } is closed. We claim that X has a neighborhood base containing closed-and-open sets. Indeed, since X isTychonoff, for every point x ∈ X and each open neighborhood U of x there is a continuous function h : X → [0 ,
1] such that h ( x ) = 1 and h ( X \ U ) = { } . It remains to note that, by assumption, theopen neighborhood h − (cid:0) (0 , (cid:1) ⊆ U of x also is closed.Now, by the assumption of the proposition, there is a non-isolated point z ∈ X . By the Zornlemma, there exists a maximal (under inclusion) family U = { U i : i ∈ κ } of pairwise disjointclosed-and-open sets such that z U i for every i ∈ κ . The maximality of U and the claim imply S. GABRIYELYAN that z ∈ cl (cid:0) S U (cid:1) . For every i ∈ κ , let g i be the characteristic function of U i . Clearly, the families U and { g i : i ∈ κ } and the point z satisfy conditions (i)-(iii). (cid:3) The following proposition is crucial for the proof of Theorem 1.1.
Proposition 2.5.
Let X be a Dieudonn´e complete space. If (cid:0) M c ( X ) , τ e (cid:1) is a Mackey space, then X is discrete.Proof. Suppose for a contradiction that X is not discrete. Then, by Proposition 2.4, there existan infinite cardinal κ , a point z ∈ X , a family { g i } i ∈ κ of continuous functions from X to [0 , { U i } i ∈ κ of open subsets of X satisfying (i)-(iii) of that proposition. Define a map R : M c ( X ) → c ( κ ) by R ( µ ) := (cid:0) µ ( g i ) (cid:1) , µ ∈ M c ( X ) . Claim 1. The map R is well-defined .Indeed, let µ ∈ M c ( X ) be a positive measure. Since µ is finite and σ -additive, the condition (ii)of Proposition 2.4 implies that for every ε > i ∈ κ for which µ ( U i ) ≥ ε is finite. Now the claim follows from the inclusion supp( g i ) ⊆ U i (see (i)) and the inequalities0 ≤ µ ( g i ) ≤ µ ( U i ).Consider a map T : M c ( X ) → (cid:0) M c ( X ) , τ e (cid:1) × c ( κ ) defined by T ( µ ) := (cid:0) µ, R ( µ ) (cid:1) , ∀ µ ∈ M c ( X ) . The map T is well-defined by Claim 1. Denote by T the locally convex vector topology on M c ( X )induced from the product (cid:0) M c ( X ) , τ e (cid:1) × c ( κ ). Claim 2. The topology T is compatible with τ e . First we note that for every ( λ i ) ∈ (cid:0) c ( κ ) (cid:1) ′ = ℓ ( κ ), the function P i λ i g i belongs to C ( X ). TheHahn–Banach extension theorem implies that every χ ∈ ( M c ( X ) , T ) ′ has the form χ = (cid:0) F, ( λ i ) (cid:1) , where F ∈ ( M c ( X ) , τ e ) ′ and ( λ i ) ∈ ℓ ( κ ) . By Corollary 2.2, we have F ∈ C ( X ) and hence G := F + P i ∈ κ λ i f i ∈ C ( X ). Therefore χ ( µ ) = µ ( F ) + X i ∈ κ λ i · µ ( g i ) = µ F + X i ∈ κ λ i g i ! = µ ( G ) , ∀ µ ∈ M c ( X ) . Applying Corollary 2.2 once again we obtain χ = G ∈ ( M c ( X ) , τ e ) ′ as desired. Claim 3. We claim that τ e < T . Indeed, it is clear that τ e ≤ T . Set S := { δ x : there is an i ∈ κ such that g i ( x ) ≥ } ⊆ M c ( X ) . To show that τ e = T , we shall prove that (1) δ z ∈ cl τ e ( S ), and (2) δ z cl T ( S ) . To prove that δ z ∈ cl τ e ( S ), fix arbitrarily a standard neighborhood[ K ; ε ] := { µ ∈ M c ( X ) : | µ ( f ) | < ε ∀ f ∈ K } of zero in ( M c ( X ) , τ e ), where K is a pointwise bounded equicontinuous subset of C ( X ) and ε > U of z such that | f ( x ) − f ( z ) | < ε, ∀ f ∈ K, ∀ x ∈ U. By (iii) of Proposition 2.4, take an i ∈ κ and x i ∈ U such that g i ( x i ) ≥
1. Then δ x i ∈ S and (cid:12)(cid:12) ( δ x i − δ z )( f ) (cid:12)(cid:12) = | f ( x i ) − f ( z ) | < ε, ∀ f ∈ K. Thus δ x i ∈ δ z + [ K ; ε ] and hence δ z ∈ cl τ e ( S ).To show that δ z cl T ( S ), consider the neighborhood W := M c ( X ) × U of zero in T , where U = { g ∈ c ( κ ) : k g k ∞ ≤ / } . Fix arbitrarily δ x ∈ S and choose j ∈ κ such that g j ( x ) ≥
1. Then
N THE MACKEY PROBLEM FOR FREE LOCALLY CONVEX SPACES 5 the j th coordinate δ x ( g j ) of R ( δ x ) satisfies the following (in the last equality we use (i) and (ii) ofProposition 2.4) (cid:12)(cid:12) δ x ( g j ) − δ z ( g j ) (cid:12)(cid:12) = | g j ( x ) − g j ( z ) | = g j ( x ) ≥ > / . Therefore R ( δ x ) − R ( δ z ) U and hence δ x − δ z W . As x was arbitrary we obtain δ z cl T ( S ).Finally, Claims 2 and 3 imply that (cid:0) M c ( X ) , τ e (cid:1) is not a Mackey space. This contradiction showsthat X must be discrete. (cid:3) Theorem 1.1 follows from the next more general result.
Theorem 2.6.
For a Tychonoff space X the following assertions are equivalent: (i) L ( X ) is a Mackey group; (ii) L ( X ) is a Mackey space; (iii) (cid:0) M c ( µX ) , τ e (cid:1) is a Mackey group; (iv) (cid:0) M c ( µX ) , τ e (cid:1) is a Mackey space; (v) X is discrete.Proof. (i) ⇒ (ii) and (iii) ⇒ (iv) follow from Lemma 2.3.(ii) ⇒ (iv) It is well known that the completion of a Mackey space is a Mackey space, see Propo-sition 8.5.8 of [10]. Therefore, by Theorem 2.1, the space (cid:0) M c ( µX ) , τ e (cid:1) is a Mackey space.(iv) ⇒ (v) By Proposition 2.5, µX is discrete. Thus X is discrete as well.(v) ⇒ (i),(iii) Since X is discrete Theorem 2.1 implies L ( X ) = (cid:0) M c ( X ) , τ e (cid:1) . Therefore L ( X ) is abarrelled space by [9, Theorem 6.4], and hence L ( X ) is a Mackey group by [3, Theorem 4.2]. (cid:3) References
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