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This topology is called the uniform topology and is denoted by u. N Following [2], denote by c0(G) the following subgroup of G N c0(G) := (xn)n∈N ∈ G : lim xn =0 . n n N o The uniform group topology on c0(G) induced from (G , u) we denote by u0 and set F0(G) := (c0(G), u0). Clearly, the group F0(R) coincides with the classical c0. For a subset A of a topological E, we denote the of A by conv(A), so n conv(A) := λ1a1 + ··· + λnan : λ1,...,λn ≥ 0, λi =1, a1,...,an ∈ A, n ∈ N . ( i=1 ) X If n ∈ N, we set (1)A := A and (n + 1)A := (n)A + A. Let X be a Tychonoff space and e ∈ X a distinguished point. Recall that the free locally convex space L(X) over X is a pair consisting of a 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 (lcs) E gives rise to a unique continuous linear operator f¯ : L(X) → E with f = f¯◦ i. For every Tychonoff space X, the free locally convex space L(X) exists, is unique up to isomorphism of locally convex spaces, and L(X) is algebraically the free vector space on X. It is more convenient for our purpose to consider Graev free locally convex spaces instead of free locally convex spaces. Recall that the Graev free locally convex space LG(X) over X is a pair consisting of a locally convex space LG(X) and a continuous map i : X → L(X) such that i(e) = 0 and every continuous map f from X to a locally convex space E gives rise to a unique continuous linear operator f¯ : LG(X) → E with f = f¯ ◦ i. Also the Graev free locally convex space LG(X) exists, is unique up to isomorphism of locally convex spaces, and is independent of the choice of e in X. Further, LG(X) is algebraically the free vector space on X \{e}. We shall use the following fact which can be obtained exactly as in [7] or Proposition 2.10 in [6]: For every Tychonoff space X, one holds

L(X)= R ⊕ LG(X).

Therefore L(X) is a if and only if LG(X) is a Mackey space, see Corollary 8.7.5 and Theorem 8.8.5 of [10]. 1 Let s = { n }n∈N ∪{0} be a convergent sequence with the usual topology induced from R. Set s∗ := s \{1}, so s is the disjoint union of {1} and s∗. Then ∗ ∗ (2.1) LG(s)= LG({1}) ⊕ LG(s )= R ⊕ LG(s )= R ⊕ LG(s)= L(s). Dealing with the concrete case X = s it is convenient and useful to describe G the space LG(s) and its topology ν s explicitly that we do below. We shall use this description also in forthcoming papers. The free vector space Vs over s is (N) just the direct sum R of ℵ0-many copies of the reals R. Denote by πn the projection of R(N) onto the nth coordinate, and we denote by 0 the zero vector THE FREE LOCALLY CONVEX SPACE L(s)ISNOTAMACKEYSPACE 3 of Vs. For every n ∈ N, set en = (0,..., 0, 1, 0,... ) ∈ Vs, where 1 is placed in G position n. Below we also find the maximal vector topology µs and the maximal G locally convex vector topology ν s on Vs in which the sequence s converges to 0. G Then (see [6]) the VG(s) := (Vs,µµs ) is called the Graev G free topological vector space over s, and the space LG(s) := (Vs,ννs ) is the Graev free locally convex space on s. G G Below we describe the µs and ν s . Let A be the family of all sequences a = (an)n∈N such that an > 0 for all n ∈ N and limn an = ∞. Denote by Ss the family of all subsets of Vs of the form

S(a) := (−an,an) × {0} , where a ∈ A.   n∈N j∈N, j6 n [ Y=   For every sequence S = {S(ak)}k∈ω ⊂ Ss, we put

U(S) = S(ak) := S(a0)+ S(a1)+ ··· + S(ak) . k∈ω k∈ω X [  If ak = (an,k)n∈N for every k ∈ ω, then v ∈ conv U(S) if and only if there are m ∈ N and r ∈ ω such that v has the form  m r v = λi tj,ie ,  l(j,i) i=1 j=0 X X where  

(a) λ1,...,λm ∈ (0, 1] and λ1 + ··· + λm = 1; (b) l(j, i) ∈ N for every 0 ≤ j ≤ r and 1 ≤ i ≤ m; (c) |tj,i|

Ns := {conv(V ): V ∈ Ns} G G is a base at 0 for ν s . In particular, for every neighborhood U of zero in ν s and for every t ∈ N and each a> 0, thereb exists q ∈ N such that every vector of the form

v = λ1aem1 + ··· + λtaemt , where q

Proof. First we check that the family Ns satisfies the following conditions: ′ ′ ′ (a) for each U(S) ∈ Ns there is a U(S ) ∈ Ns with U(S )+ U(S ) ⊆ U(S); ′ ′ (b) for each U(S) ∈ Ns there is a U(S ) ∈ Ns such that λU(S ) ⊆ U(S) for all |λ|≤ 1; (c) for each U(S) ∈ Ns and every x ∈ Vs, there is n ∈ N such that x ∈ nU(S); (d) Ns = {0}. Indeed, let S = {S(a )} , where a = (a ) N ⊆ (0, ∞). Set T k k∈ω k n,k n∈ 1 S′ = {S(a′ )} , where a′ := min{a ,...,a } . k k∈ω k 2 n,0 n,2k+1  n∈N ′ ′ Then S(ai)+ S(ai) ⊆ S(a2i)+ S(a2i+1) for every i ∈ ω. Therefore ′ ′ ′ ′ (2) S(a0)+ ··· + S(ak) ⊆ S(a0)+ ··· + S(a2k+2) and U(S )+ U(S ) ⊆ U(S),   4 S. GABRIYELYAN that proves (a). Items (b)-(d) are trivial. So, by [11, §15, 2(1)]), Ns forms a base for some vector topology τ on Vs. Moreover, en → 0 in τ, since en ∈ S(a0) for all sufficiently large n. G ′ To show that τ = µs , fix an arbitrary vector topology τ on Vs in which en → 0 and let V ∈ τ ′. Choose inductively a sequence of symmetric open circled neighbor- ′ hoods {Vk}k∈ω of 0 in τ such that V0 + V0 ⊆ V and Vk+1 + Vk+1 ⊆ Vk for each N k ∈ ω, so k∈ω Vk ⊆ V . For every k ∈ ω and each n ∈ , set P ank := sup λ ∈ (0,n] : if |x| < λ, then x ∈ Vk ∩ Rn × {0} .    j∈N, j6 n  Y=    We show that lim n ank = ∞ for every k ∈ ω. Indeed, if anik < A for some sequence 1 N n1

Below we describe the dual space of LG(s).

Proposition 2.2. The dual space E of LG(s) is linearly isomorphic to the space c0 of all real-valued sequences converging to zero.

G Proof. From the description of the base Ns of the topology ν s of LG(s) given in G Proposition 2.1, it immediately follows that ν s is strictly weaker than the box (N) (N) topology Tb on R . It is well known that the dual space of R , Tb is linearly isomorphic to the direct product RN. Now the continuity of the identity map (N) N  R , Tb → LG(s) implies that E ⊆ R . Thus, to prove the proposition we have to show that χ = (yn) ∈ E if and only if yn → 0.  Claim 1. If χ = (yn) ∈ E, then yn → 0. Indeed, suppose for a contradiction that there is a sequence 0 < r1 < r2 < . . . of indices such that |yri | > a for some a > 0. We assume that all yri are positive. Let U be a neighborhood of zero in LG(s) such that χ(U) ⊆ [−1, 1] and take t ∈ N such that ta > 1. By Proposition 2.1, there is q ∈ N such that every vector of the form

v = em1 + ··· + emt , where q q and set w := eri+1 + ··· + eri+t . Then w ∈ U and

χ(w)= yri+1 + ··· + yri+t >t · a> 1, a contradiction. Thus yn → 0. THE FREE LOCALLY CONVEX SPACE L(s)ISNOTAMACKEYSPACE 5

Claim 2. If yn → 0, then χ ∈ E. Indeed, since χ is linear, by Proposition 2.1, it is sufficient to show that there is a sequence S = {S(ak)}k∈ω ⊂ Ss such that χ U(S) ⊆ [−1, 1]. To this end, for every i ∈ ω, we shall find ai ∈ A such that 1 1 (2.2)  χ S(a ) ⊆ − , . i 2i+1 2i+1    Set I1 := {n ∈ N : yn =0} and I2 := N \ I1. If n ∈ I1, set ai(n) := n, and set 1 ai(n) := i+1 , for every n ∈ I2. 2yn · 2

It is clear that ai(n) → ∞ at n → ∞, so ai ∈ A. The inclusion (2.2) holds trivially by the construction of ai and S(ai). Now, if w ∈ U(S) has a decomposition w = v0 + ··· + vk with vi ∈ S(ai), we obtain k k 1 |χ(w)|≤ |χ(v )|≤ < 1. i 2i+1 i=0 i=0 X X Thus χ ∈ E.  The next important result is proved in [9, 12], see also [8, 23.32]. Fact 2.3. Let E be a locally convex space. Then the mapping p : E′ → E, defined by the equality p(f) = exp{2πif}, for all f ∈ E′, b is a group isomorphism between E′ and E. Now Theorem 1.1 immediately follows from (2.1) and the following result. b Theorem 2.4. The space LG(s) is not a Mackey space. Consequently LG(s) is not a Mackey group. Proof. First we describe a method for constructing of compatible group topologies on LG(s). Let h be a continuous homomorphism from R to a locally quasi-convex (N) abelian group H. Define an algebraic monomorphism TH : R → LG(s) × F0(H) by

(N) (2.3) TH (xk) := (xk), h(xk) , ∀ (xk) ∈ R ,     (N) and let TH be the locally quasi-convex group topology on R induced from LG(s)× F0(H). (N) Step 1. The topology TH is weaker than the box topology Tb on R . As we G explained in the beginning of the proof of Proposition 2.2, the topology ν s is weaker than Tb. Therefore we have to show only that the map (N) p : R , Tb → F0(H), p (xk) := h(xk) , is continuous at zero. Fix a neighborhood U of the identity in H. Choose a neighborhood V of zero in R such that h(V ) ⊆ U. Set W := R(N) ∩ V N. Then W is (N) a neighborhood of zero in R , Tb such that for every (xk) ∈ W (recall that (xk) has finite support) we have  N p (xk) = h(xk) ∈ c0(H) ∩ U .

Hence p is continuous. Thus TH is weaker than Tb. 6 S. GABRIYELYAN

G G Step 2. We claim that the topology TH is compatible with ν s such that ν s ≤ TH . (N) Indeed, set G := R , TH . Step 1 implies that

∧ ∧ (N) ∧ LG(s) ⊆ G ⊆ R , Tb . Taking into account Proposition  2.2 and Fact 2.3, we can identify G∧ with a sub- (N) ′ N group Y of R , Tb = R containing c0. Thus, to prove the claim we have to show that Y = c . 0  Fix arbitrarily χ = (yn) ∈ Y . We have to show that yn converges to zero. Suppose for a contradiction that yn 6→ 0. So there is a sequence 0 1 for every i ∈ . We assume that all ymi are positive. Since χ is TH -continuous, there exists a standard neighborhood −1 N W = TH U × V of zero in G, where U is a neighborhood of zero in LG(s) and V is a neighborhood of the identity in H, such that χ(W ) ⊆ S . Observe that, by  + (2.3), (xk) ∈ W if and only if

(2.4) (xk) ∈ U and h(xk) ∈ V for every k ∈ N. Choose l ∈ N such that h [−1/2l, 1/2l] ⊆ V . Set 1  1 t := lA, a := and λj := for j =1, . . . , t. 2l A · ymq+j Now Proposition 2.1 implies that there exists q ∈ N such that the vector 1 1 v := λ · e + ··· + λ · e 1 2l mq+1 t 2l mq+t belongs to U. Since 0 < λj < 1, the choice of l implies that h(λj /2l) ∈ V for every j =1,...,t. Hence, by (2.4), we obtain v ∈ W . Therefore

t 1 t πi S χ(v) = exp 2πi ymq+j · = exp 2πi = e = −1 ∈ +,  2lA · ym  2lA j=1 q+j  X    G a contradiction. Thus yn → 0 and Y = c0. The inequality ν s ≤ TH holds trivially. Step 3. LG(s) is not a Mackey space. Indeed, let H = R and h : R → H be the identity map. Then F0(R) is the classical Banach space c0. Therefore the G compatible topology TH is a locally convex vector topology such that ν s ≤ TH . G G G We show that ν s 6= TH . Indeed, by the definition of ν s , we have en → 0 in ν s . However, en 6→ 0 in the Banach space c0, and hence en 6→ 0 in TH . The space LG(s) is not a Mackey group by (i) of Proposition 2.5 of [4]. 

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Department of Mathematics, Ben-Gurion University of the Negev, Beer-Sheva, P.O. 653, Israel E-mail address: [email protected]