Extended seminorms and extended topological vector spaces

Topology and its Applications 210 (2016) 317–354

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Topology and its Applications www.elsevier.com/locate/topol

Extended seminorms and extended topological vector spaces David Salas a,∗ , Sebastián Tapia-García b a

Université de Montpellier, Institut Montpelliérain Alexander Grothendieck, Case Courrier 051 Place Eugène Bataillon, 34095 Montpellier cedex 05, France b Universidad de Chile, Departamento de Ingeniería Matemática, Beauchef 851, Torre Norte – Piso 4, Santiago, Chile

a r t i c l e

i n f o

Article history: Received 1 August 2015 Received in revised form 21 July 2016 Accepted 1 August 2016 Available online 4 August 2016 MSC: 46A03 46A17 46A22 54H11

a b s t r a c t We introduce the notions of extended topological vector spaces and extended seminormed spaces, following the main ideas of extended normed spaces, which were introduced by G. Beer and J. Vanderwerff. We provide a topological study of such structures, giving a unifying theory with main applications in the study of spaces of continuous functions. We also generalize classical results of functional analysis, as open mapping theorem and closed graph theorem. © 2016 Elsevier B.V. All rights reserved.

Keywords: Topological vector spaces Seminorms Bornologies Extended norms Extended seminorms Topological groups Projective limits

1. Introduction and problem formulation In the papers [2,4,5], G. Beer and J. Vanderwerff introduced and studied the notion of extended norm on a vector space X over a field K (which is R or C), as a functional  ·  : X → [0, +∞] which satisfies 1. x = 0 ⇔ x = 0; 2. For all x ∈ X and α ∈ K, αx = |α|x (with the convention 0 · (+∞) = 0); 3. For all x, y ∈ X, x + y ≤ x + y. * Corresponding author. E-mail addresses: [email protected] (D. Salas), [email protected] (S. Tapia-García). http://dx.doi.org/10.1016/j.topol.2016.08.001 0166-8641/© 2016 Elsevier B.V. All rights reserved.

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The space X endowed with the extended norm  ·  was called the extended normed space (X,  · ). Of course,  ·  induces a topology over X, which happens to fail to be compatible with the vectorial structure of X: it is compatible with the sum but not with the scalar multiplication. In the same papers, many classical results concerning normed spaces are extended to this new framework, showing that this new object is in some sense “well-behaved”. Motivated by these works, a natural generalization arises: Extended seminorms and Extended seminormed spaces. Definition 1.1. Let X be a vector space over a field K (which is R or C). A function ρ : X → [0, +∞] is called an extended seminorm if, adopting the convention 0 · (+∞) = 0, it satisfies (a) ρ(λx) = |λ|ρ(x), for all λ ∈ K and x ∈ X. (b) ρ(x + y) ≤ ρ(x) + ρ(y), for all x, y ∈ X. For a vector space X over R or C and a family of extended seminorms P := {ρi : i ∈ I} over X, the induced topology of P on X, denoted by T(P), is the coarsest topology on X for which all functions ρi,x : X → [0, +∞] x → ρi (x − x) with i ∈ I and x ∈ X are continuous. It is easily verified that T(P) is a group topology over X (see, e.g. [8, Ch. III, §1, Definition 1], or section 2 below), and so we are led to the following definition: Definition 1.2. Let X be a vector space over K = R or K = C and τ be a group topology over X. We say that (X, τ ) is an extended seminormed space (esns, for short) if there exists a family P = {ρi : i ∈ I} of extended seminorms over X such that its induced topology T(P) coincides with τ . Extended seminormed spaces appear naturally in functional analysis when we study the convergences in functional spaces: Let (M, d) be a metric space and B be a bornology on M , namely, a family of subsets of M such that: 1. For each B ∈ B , each subset of B also belongs to B . 2. For each B1 , B2 ∈ B , B1 ∪ B2 ∈ B .  3. B forms a cover of M , namely, M = B∈B B. On the space of continuous functions over M with values in a normed space (Y,  ·), C(M, Y ), the uniform convergence over the elements of B is posed as follows: a net (fα ) ⊆ C(M, Y ) converges B -uniformly to a function f ∈ C(M, Y ) if sup fα (x) − f (x) → 0,

x∈B

∀B ∈ B .

This type of convergence induces a topology τ (B ) which, unless B is a sub-bornology of the Hadamard bornology H (which is given by all relatively compact sets of (M, d)), fails to be a locally convex topology in the usual sense. Nevertheless, if we consider the family P = {ρB : B ∈ B }, where ρB (f ) = supx∈B f (x), then the functions ρB are extended seminorms, and so (C(M, Y ), τ (B )) is an extended seminormed space. For further information on bornologies, we refer the reader to [13].

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Usual functional analysis doesn’t allow us to fully understand this kind of structures, and therefore it is necessary to expand the theory. This is our main goal: We will provide a full picture of extended structures giving a unifying framework to work with, from both a functional analysis perspective as well as a topological one. It is worth mentioning that topologies induced by functionals that can take the value +∞ have been previously studied by G. Beer and M. J. Hoffman in [1,3]. Toghether with [2,4,5], these papers contain the seminal ideas that motivated this work. The paper is structured as follows: Section 2 gives all necessary preliminaries, specifically about linear projections, and fixes some notation. In Section 3, suitable notions of extended topological vector spaces and extended locally convex spaces are introduced and fully studied. This section provides the fundamental topological structure of extended seminormed spaces, since it will be proven that they coincide with the extended locally convex spaces. Section 4 studies the special properties of esns. Also, in Subsection 4.2, the particular case of esns for which the topology is given by countable many extended seminorms is studied. The work ends with Subsection 4.3, where the main question of the paper is posed: When can an extended seminormed space be split into its finite space and a topological complement? In what follows, we will assume that the reader is familiar with the basics in topological groups, including topological fields. We refer to [8, Ch. III] for further information on these topics. The notation and the definitions are based mainly on [8] and [15]. 2. Preliminaries and notation First, throughout this paper, for any topological space (T, τ ) and any subset S of T , we will simply write (S, τ ) to denote the topological subspace S endowed with the induced topology from τ . If the notation is  ambiguous, we will write τ S to specify the induced topology. For a point t0 ∈ T , we will denote the family of neighborhoods of t0 as NT (t0 , τ ). If there is no ambiguity, we will omit the space, the topology or both writing simply NT (t0 ), N (t0 , τ ) or N (t0 ), respectively.  For a family {τα : α ∈ A } of topologies on the same set T , we will denote by α∈A τα the topology  generated by α∈A τα , namely the coarsest topology on T containing each τα . We say that the family {τα : α ∈ A } is directed by inclusion if for every two elements α, β ∈ A there exists a third one γ ∈ A such that τα ∪ τβ ⊆ τγ . The following proposition is well-known in the literature (see, e.g., [8, Ch. II] and [10, Ch. III]) and we will use it many times during this work. Proposition 2.1. Let {τα : α ∈ A } be a family of topologies on X directed by inclusion. Denote τ :=  α∈A τα . Then (a)



τ is a basis of the topology τ ; that is, every element of τ can be written as an union of elements α∈ A α in α∈A τα .  (b) For each t0 ∈ T , Bt0 = α∈A N (t0 , τα ) is a fundamental system (of neighborhoods) of N (t0 , τ ); that is, for every V ∈ NX (t0 , τ ), there exist α ∈ A and Vα ∈ N (t0 , τα ) such that Vα ⊆ V . Also, G and K will always denote an abelian group and a field, respectively. For G, we will always denote by + its group law and by 0G its identity element. Respectively, for K we will always denote by + its first law, · its second law, 0K its zero and 1K its unit. We denote by End(G) the set of endomorphisms of G, and by Aut(G) the set of automorphisms of G. If two groups G and H are algebraically isomorphic (namely, there exists a bijective homomorphism f : G → H), we will write G ≈ H. Recall that a topological space (G, τ ) is said to be a topological group if G is a group and the mappings (g, h) ∈ G × G → g + h ∈ G and g ∈ G → −g ∈ G are continuous. In such a case τ is said to be a group

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topology on G. Two topological groups (G, τG ) and (H, τH ) are topologically isomorphic if there exists a bijective homomorphism f : G → H which is bicontinuous. In such a case, we will write (G, τG ) ∼ = (H, τH ) there is no confusion). (or simply G ∼ H if = Recall that a topological space (T, τ ) is connected if T cannot be written as a disjoint union of two nonempty open sets. Also, a nonempty subset S of T is connected if it is connected as a topological subspace of (T, τ ). The space (T, τ ) is said to be locally connected, if each point t ∈ T has a fundamental system of connected neighborhoods. For t0 ∈ T we will denote the connected component of t0 by C[t0 , τ ] (or simply C[t0 ] if there is no ambiguity). An easy observation is that a topological group (G, τ ) is locally connected if and only if 0G has a fundamental system of connected neighborhoods. In the sequel, X will stand for a vector space over K and θ will be a field topology on K. For k ∈ K, we denote by ϕk the algebraic endomorphism induced by k, namely ϕk : X → X x → kx.

(1)

The following definition contains the primal structures that we will work with: Topological groups with operators and topological vector spaces. The second notion is well known and the first one can be found in [8, Ch. III, §6]. We present them as a definition since we will use them many times in the following sections. Definition 2.2 (Topological group with operators and topological vector space). Given a group topology τ on X and a field topology θ on K, we say that 1. (X, τ ) is a topological group with operators in K if for each k ∈ K, the function ϕk is τ -τ -continuous; 2. (X, τ ) is a topological vector space (tvs, for short) over K if the scalar multiplication ·:K×X →X (λ, x) → λx is (θ × τ )-τ -continuous. For two subspaces M, N of X, and a group topology τ over X, we say that M and N are algebraic complements if X = M ⊕ N . In such a case, for x ∈ X we will denote by PM,N (x) and PN,M (x) the unique elements of M and N respectively such that x = PM,N (x) + PN,M (x). The mappings PM,N , PN,M : X → X, which are linear and idempotent, are called parallel projections. We will say that M and N are τ -supplements if they are algebraic complements and PM,N and PN,M are τ -τ -continuous. In such a case, we will write X = M ⊕τ N . In some contexts, it will be useful to denote the mapping from X to M which maps x → PM,N (x) also as PM,N . If there is any confusion we will specify which of those functions we refer to. Definition 2.3 (Complement spaces w.r.t. a Hamel basis). Let X be a vector space over a field K and let H be a Hamel basis of X. For a subspace Z of X we will say that H generates Z if Z ∩ H is a Hamel basis of Z. If Z is a subspace of X generated by H we will denote by CH (Z) := span{H \ Z}, which is a complement subspace of Z in X, namely a subspace of X such that X = Z ⊕ CH (Z). In such a case, PZ,H will stand for the parallel projection of X to Z (parallel to CH (Z)).

(2)

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Fixing M as a subspace of X, we will denote by X/M the quotient group, by πM (or simply π, if there is no ambiguity) the canonical quotient map, and by πM (τ ) the quotient topology. The equivalence class of x ∈ X is denoted by [x]M , or simply [x] is there is no confusion. We will say that  : X/M → X is a lifting, if πM ◦  = idX/M . Recall that a mapping P : X → X is called a linear projection if it is linear and idempotent. We end this section with some simple results on projections, many of which are contained in [8, Ch. III, §6] (written in different words) and are well known in the literature. Nevertheless we will sketch the arguments of some of them again in order to fix a common context and make the reading of this article easier. First, since PM,N + PN,M = idX , we have the following proposition. Proposition 2.4. Let τ be a group topology over X and M, N be two complement subspaces in X. The following assertions are equivalent: (i) (ii) (iii) (iv)

PM,N is τ -τ -continuous. PN,M is τ -τ -continuous. M and N are τ -supplement subspaces in X. The mapping     + : M × N, τ M × τ N → (X, τ ) (m, n) → m + n is a topological isomorphism, where its inverse is given by the mapping D : x → (PM,N (x), PN,M (x)).

In the above proposition, taking M = P (X) and N = Ker(P ) for a linear projection P : X → X, we obtain the following corollary: Corollary 2.5. Let P : X → X be a linear projection. We have that P is τ -τ -continuous if and only if P (X) and Ker(P ) are τ -supplements. Proposition 2.6. Let (M1 , N1 ) and (M2 , N2 ) be two pairs of τ -supplement subspaces in X. Assume that there exists a Hamel basis H of X such that M1 , M2 , N1 and N2 are generated by subsets of H. Then, M = M1 ∩ M2 and N = span{N1 ∪ N2 } are also τ -supplement subspaces in X. Proof. Let H = {bj : j ∈ Δ} and define the sets Δ0,i , Δ1,i ⊂ Δ (with i = 1, 2) such that span{bj : j ∈ Δ0,i } = Mi and Δ1,i = Δ \ Δ0,i (thus, span{bj : j ∈ Δ1,i } = Ni ). Defining Δ0 = Δ0,1 ∩ Δ0,2 and Δ1 = Δ \ Δ0 , it is easy to see that M = span{bj : j ∈ Δ0 }

and

N = span{bj : j ∈ Δ1,1 ∪ Δ1,2 } = span{bj : j ∈ Δ1 },

with the convention span{∅} = {0}. On the other hand, noting that PMi ,Ni = PMi ,H (with i = 1, 2) we have that PM,N = PM,H = PM1 ,H ◦ PM2 ,H , and so, PM,N is τ -τ -continuous. The conclusion follows from Proposition 2.4.

2

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D

M ×N

D

M ×N

X +

X +

ι ˆN πM

PN

βN



PN,M

N

h=πM ◦ˆ ιN

X/M

N

h−1

h=πM ◦ˆ ιN

X/M

h−1

Fig. 1. Parallel diagram of (M, N ).

When M and N are two complement spaces in X, it is well known that if we define ˆιN : N → X as the canonical injection mapping, PN : M × N → N as the mapping given by PN (m, n) = n, D : X → M × N as the mapping given by D(x) = (PM,N (x), PN,M (x)) (which is the inverse of +) and β : N → M × N as the mapping given by βN (n) = (0, n), the two diagrams in Fig. 1 commute, where  : X/M → X is the linear lifting such that for any [x] ∈ X/M , ([x]) is the unique element in N ∩ [x]. In particular, πM N is an isomorphism, and its inverse is given by 

 −1 πM  N = PN ◦ D ◦ .

In the following, these two diagrams will be called the parallel diagram of (M, N ) and the lifting  just defined in it will be called the parallel lifting and will be denoted by M,N or simply by N if there is no ambiguity. In these diagrams, if we endow X with a group topology τ , X/M with the quotient topology πM (τ ),  M × N with the product topology τ M × τ N and N with the induced topology τ N , then we get that the functions +, PN , βN , πM , ˆιN and h are continuous. Proposition 2.7. Let M and N be two complement subspaces in X. The following assertions are equivalent: (i) M and N are τ -supplements. (ii) In the parallel diagram of (M, N ), h = ˆιN ◦ πM is a topological isomorphism. (iii) The parallel lifting N : (X/M, π(τ )) → (X, τ ) is continuous. Proof. (i)⇒(ii): If M and N are τ -supplements, then + is a topological isomorphism in the parallel diagram of (M, N ), and therefore D is continuous. We only need to prove that h is open. Let then A ⊂ N be an open set. It is easy to see that h(A) = {[n] : n ∈ A} = πM (A + M ). On the other hand, denoting by s : M × N → X the sum mapping + in the parallel diagram, we see that A + M = s ◦ PN−1 (A), and so, it is open. Since πM is an open mapping, we conclude that h(A) is open, which proves the implication. (ii)⇒(iii): From the parallel diagram of (M, N ), we know that N = s ◦ βN ◦ h−1 , and since h−1 is continuous, the conclusion follows.

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(iii)⇒(i): It is easy to see that PN,M = N ◦ πM . Since N is continuous, the conclusion follows from Proposition 2.4. 2 Corollary 2.8. Let M be a subspace of X. Then, the following assertions are equivalent: (i) There exists a subspace N of X such that M and N are τ -supplements. (ii) There exists a continuous linear lifting  : X/M → X. (iii) There exists a continuous projection P : X → X such that P (X) = M . Proof. (i) ⇒ (ii) follows from Proposition 2.7, and (iii) ⇒ (i) is Corollary 2.5. Finally, to prove (ii) ⇒ (iii) we take P = idX − ◦ πM . Noting that ( ◦ πM ) ◦ ( ◦ πM ) =  ◦ (πM ◦ ) ◦ πM =  ◦ πM , we conclude that P is idempotent, and therefore it is a continuous projection with P (X) = Ker( ◦ πM ) = Ker(πM ), where the last equality follows from the fact that every lifting is injective. Since Ker(πM ) = M , the proof is finished. 2 3. Extended topological vector spaces A simple but central observation of Beer in [2] is that every extended normed space (X,  · ) can be decomposed algebraically as X = Xfin ⊕ N, where Xfin = {x ∈ X : x < +∞}. Clearly, (Xfin ,  · ) is a normed vector space and (N,  · ) is a discrete extended normed space, where the discrete extended norm  · d is given by md =

 0

if m = 0

+∞ otherwise.

(3)

In fact, provided by the simple fact that Xfin is an open subspace, it is easy to prove that such decomposition is also topological (see [2, Theorem 3.13]). In order to carry the notion of “extended structure” to the more general context, we will use this kind of decomposition. 3.1. Fundamental etvs Definition 3.1. Let X be a vector space over a field K, τ be a group topology over X and θ be a field topology over K. We say that (X, τ ) is a fundamental extended topological vector space (fundamental etvs, for short) if there exists an algebraic decomposition X = M ⊕ N satisfying (A1) (A2) (A3)

   M, τ M is a tvs over (K, θ);    is a discrete space; N, τ 

N       (X, τ ) is topologically isomorphic to M, τ M × N, τ N .

In such a case, we will say that (M, N ) is a (K, θ)-compatible decomposition (or simply a compatible decomposition, if there is no confusion) of X.

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   In the above definition, if (K, θ) = (R, | ·|) or (C, | ·|) and M, τ M is a locally convex space (lcs, for short) over (K, θ) we will rather say that (X, τ ) is a fundamental extended locally convex space (fundamental elcs, for short) over (K, θ). This particular object will be further studied in Section 4. The first natural question about fundamental etvs is whether there exists a unique compatible decomposition, in the sense that the space M in Definition 3.1 is unique (which clearly is the case of extended normed spaces since M must be Xfin ). To answer this question, we will need some lemmas. Lemma 3.2. Let (X, τ ) be a fundamental etvs over (K, θ). Then (X, τ ) is a topological group with operators in K. Proof. Let (M, N ) be a compatible decomposition and let α ∈ K. From Definition 2.2, we need to prove that ϕα is continuous, where ϕα is the endomorphism given in equation (1). Since (M, τ ) is a tvs over (K, θ),  we get that ϕα M is continuous. Also, since (N, τ ) is a discrete space, ϕα N is continuous. Finally, we can write       ϕα = ϕα M ◦ PM,N + ϕα N ◦ PN,M , and since PM,N and PN,M are both continuous, ϕα is continuous, which finishes the proof. 2 Lemma 3.3. Let τ be a group topology over X and θ be a field topology over K. Assume that (X, τ ) is a topological group with operators in K. Then, C[0X ] (the connected component of 0X ) is a vector space. Moreover, if (X, τ ) is a fundamental etvs over (K, θ), then for every compatible decomposition (M, N ), C[0X ] ⊆ M . Proof. For the first part, let z ∈ C[0X ] and α ∈ K with α = 0K . Since C[0X ] is already a subgroup of X (see e.g. [8, Ch. III, §2, Proposition 8]), we only need to prove that αz ∈ C[0X ]. Let us consider then the set C := α · C[0X ]. Since the induced endomorphism ϕα (see equation (1)) is τ -continuous, the set C has to be connected. Also, since 0X ∈ C, we get that C ⊆ C[0X ]. Then, αz ∈ C ⊆ C[0X ], which is what we wanted to prove. Now, let us prove that C[0X ] ⊆ M . Since M and N are τ -supplements, we have that N ∼ = X/M , and so X/M is discrete. Therefore, the connected component of 0X/M in X/M for the quotient topology is the singleton {0X/M }. Finally, since π : X → X/M is continuous, it maps connected sets into connected sets and so π(C[0X ]) ⊆ {0X/M }. Thus, C[0X ] ⊆ π −1 ({0X/M }) = M, which finishes the proof. 2 Lemma 3.4. Let (X, τ ) be a topological group with operators in K, Z be any open subspace of X and H be a Hamel basis which generates Z. Then,  1. (CH (Z), τ ) ∼ = (X/Z, πZ (τ )) where the topological isomorphism is πZ CH (Z) . 2. PCH (Z),Z = CH (Z) ◦ πZ , where CH (Z) is the parallel lifting of the decomposition X = Z ⊕ CH (Z) (see Fig. 1).

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3. Z and CH (Z) are τ -supplements. Moreover, if (X, τ ) is a fundamental etvs over (K, θ) and (M, N ) is a compatible decomposition, then, each subspace Z of X such that M ⊆ Z is an open set of (X, τ ). In particular, for each subspace N  of X such that X = M ⊕N  , (M, N  ) is also a compatible decomposition (and therefore, N  ∼ = N ). Proof. For the first part, let Z be an open subspace of X. Applying again [8, Ch. III, §2, Proposition 18], X/Z is a discrete space. Let then H be a Hamel basis which generates Z. Since X/Z is a discrete space, the parallel lifting CH (Z) : X/Z → X is continuous and therefore, by Proposition 2.7, the proof is finished. For the second part, since M and N are τ -supplements, we have that N ∼ = X/M . Therefore, X/M endowed with the quotient topology is a discrete space. Applying [8, Ch. III, §2, Proposition 18], we get that M is an open set. Then, for each subspace Z ⊇ M we can write  Z= z + M, z∈Z

and therefore, Z is an open set.

2

Proposition 3.5. Let (X, τ ) be a fundamental etvs over (K, θ) and (M, N ) be a compatible decomposition. Then: 1. If (K, θ) is non-discrete, then (M, N ) is the unique compatible decomposition, in the sense that if (M  , N  ) is another compatible decomposition, then M = M  and N ∼ = N . 2. If θ is the discrete topology, then for each subspace Z of X containing M and each Hamel basis H which generates Z, (Z, CH (Z)) is a compatible decomposition. In particular, (X, τ ) is a tvs over (K, θ). Proof. 1. Let be (M  , N  ) be another compatible decomposition and H be a Hamel basis which generates the subspaces M , M  and M  = M ∩ M  (it is not hard to see that such a basis exists). Provided by Lemma 3.4, we may replace N and N  by CH (M ) and CH (M  ) respectively, and therefore we can assume without losing generality that H also generates N and N  . Thus, if we define N  = span{N ∪ N  }, we can apply Proposition 2.6 to conclude that M  and N  are τ -supplements. We will prove first that (N  , τ ) is discrete. Observe that, since H generates span{M ∪ M  }, then it also generates the subspace N ∩ N  . Thus, we can write PN ∩N  ,H = PN,H ◦ PN  ,H , and also, we have the equality PN  ,H = PN,H + PN  ,H − PN ∩N  ,H . Now, let (nλ )λ∈Λ be a net in N  converging to n ∈ N  . Since (N, τ ), (N  , τ ) and (N ∩ N  , τ ) are discrete spaces and all three parallel projections with respect to H are continuous, we conclude that there exists λ0 ∈ Λ such that for each λ ≥ λ0 , PN,H (nλ ) = PN,H (n), PN  ,H (nλ ) = PN  ,H (n) and PN ∩N  ,H (nλ ) = PN ∩N  ,H (n). Thus, for each λ ≥ λ0 nλ = PN  ,H (nλ ) = PN,H (nλ ) + PN  ,H (nλ ) − PN ∩N  ,H (nλ )

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= PN,H (n) + PN  ,H (n) − PN ∩N  ,H (n) = PN  ,H (n) = n. Then, all converging nets in (N  , τ ) are stationary, and so the space is discrete. We conclude that (M  , N  ) is also a compatible decomposition by just observing that (M  , τ ) is a tvs since M  is a subspace of M . We will prove now that M = M  which, by symmetry, will imply that M = M  , finishing the proof. Let H0 be the subset of H which generates M . Then, if we consider the topological group (M, τ ), we have that (M  , N0 ) is a compatible decomposition of M which makes it a fundamental etvs over (K, θ) (considering N0 = CH0 (M  ) = N  ∩ M ). Let us suppose that there exists n ∈ N0 with n = 0. Since (K, θ) is non-discrete, there exists α ∈ K \ {0} such that {α} is not an open set. On the other hand, since N0 is a discrete space, {αn} is open in N0 . Let (αV ) be a net in K indexed by the neighborhoods V ∈ N (α, θ) such that ∀V ∈ N (α, θ), αV ∈ V \ {α}. Since {α} is not open, such a net exists, and even more αV → α. Now, since αV − α = 0K , we have that the map m → (αV − α)m is an automorphism of M , and therefore αV n = αn. This implies that αV n → αn in N0 . But noting that (αV , n) → (α, n) in K × N0 , we conclude that (N0 , τ ) is not a tvs over (K, θ). This last statement contradicts the fact that M is tvs, since N0 is a subspace of M and each subspace of a tvs over (K, θ) endowed with the induced topology has to be also a tvs over (K, θ). Therefore, N0 = {0M } and so, M  = M , which is what we wanted to prove. 2. Assume that (M, N ) is a compatible decomposition and that Z is a subspace of X with M ⊆ Z. Without losing generality we may assume that N is such that there exists a Hamel basis H which generates M , N and Z. Then, since CH (Z) is a subspace of N , we get that (CH (Z), τ ) is a discrete space. On the other hand, let (αλ , zλ )λ∈Λ be a net in K × Z converging to (α, z) ∈ K × Z. Since (K, θ) is discrete, there exists λ0 ∈ Λ such that for each λ ≥ λ0 , αλ = α. Then, applying Lemma 3.2, we get that for λ ≥ λ0 αλ zλ = αzλ → αz, and therefore, (Z, τ ) is a tvs over (K, θ). Finally, applying Lemma 3.4, we conclude that Z and CH (Z) are τ -supplements. Then, (Z, CH (Z)) is a compatible decomposition, which finishes the proof. 2 Remark 1. When a vector space X over a field K is endowed with a group topology τ such that (X, τ ) is a topological group with operators in K, it is easy to realize that (X, τ ) is in fact a tvs over (K, θ) when θ is the discrete topology. Such kinds of spaces have been already introduced in the literature by the name of topological vector groups (see [14]) and further developed during the 70’s. Therefore, we won’t work on those objects and so we will always assume that (K, θ) is a non-discrete field. Corollary 3.6. Let (X, τ ) be an etvs over a non-discrete field (K, θ) and (M, N ) be a compatible decomposition. Then M=



{Z : Z is a τ -open subspace of X}.

Proof. Let us denote M  = {Z : Z is a τ -open subspace of X}. By Lemma 3.4, we get that M ⊇ M  . Suppose that the latter inclusion is strict. Then, there exists an open subspace Z such that Z  M . Replacing Z by Z ∩ M , we may assume that Z  M , and therefore, (Z, τ ) is a tvs over (K, θ). Applying again Lemma 3.4, we get that for any Hamel basis which generates Z, the pair (Z, CH(Z)) is a compatible

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decomposition. Then, by Proposition 3.5, M = Z, which is a contradiction. We conclude that M = M  , finishing the proof. 2 Observe that in the case of an extended normed space (X,  · ), the identification Xfin = {Z : Z is a  · -open subspace of X}, is somehow direct, since Xfin is itself open and it doesn’t contain any open proper subspaces provided that it is a connected subspace. Therefore, to adopt the notation of [2], for every vector space X endowed with a group topology τ , we will define its finite space as Xfin =



{Z : Z is a τ -open subspace of X}.

(4)

Remark 2. Note that, provided by Corollary 3.6 and Lemma 3.4, we get that, if (K, θ) is a non-discrete topological field, then (X, τ ) is a fundamental etvs over (K, θ) if and only if (Xfin , τ ) is a tvs over (K, θ) and Xfin is τ -open in X. Note also that, since every open subgroup of a topological group is also a closed subgroup (see [8, Ch. III, §2, Corollary of Proposition 4]), we have that Xfin is always τ -closed. In the general case, the finite space of a fundamental etvs may not coincide with the connected component of zero. A counterexample arrives easily by just considering the vector space X = Q2 over the topological field (Q, | · |) (where Q denotes the group of rational numbers) and the group topology τ over Q2 induced by the function ρ : Q2 → [0, +∞] (q1 , q2 ) → ρ(q1 , q2 ) =

 |q1 | +∞

if q2 = 0 otherwise.

Here, the finite space is Xfin = Q × {0}, but C[0X ] = {(0, 0)}. This motivates the following definition. Definition 3.7. Let (X, τ ) be a fundamental etvs over a non-discrete topological field (K, θ). We will say that (X, τ ) is a proper fundamental etvs over (K, θ) if Xfin = C[0X , τ ]. Since every tvs over (R, | · |) or (C, | · |) has to be connected, we have the following corollary: Corollary 3.8. If (X, τ ) is a fundamental etvs over (R, | · |) or (C, | · |), then it is proper. 3.2. Structure of etvs Knowing the definition of fundamental etvs, we are ready to introduce what will be the most suitable notion of extended topological vector space. The main utility of such a definition will be fully appreciated in Theorem 4.3, which gives a topological characterization of esns in terms of what we will call “extended locally convex spaces”. For the sake of deductive order of the paper, we will come back to this subject in Section 4 and in this section we will restrict our attention to the study of extended topological vector spaces. Definition 3.9. Let X be a vector space over K, τ be a group topology on X and θ be a non-discrete field topology on K. We say that (X, τ ) is an extended topological vector space (etvs, for short) if there exists a family {τα : α ∈ A } of group topologies on X such that 1. τ =

 α∈A

τα .

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2. For each α ∈ A , (X, τα ) is a fundamental etvs over (K, θ). In such a case, the family {τα : α ∈ A } is called a generating family of τ . As before, the finite space Xfin of (X, τ ) is defined as Xfin =



{M : M τ -open subspace of X}

and (X, τ ) is said to be a proper etvs if Xfin = C[0X , τ ]. Observe first that, as in the case of fundamental etvs, if (X, τ ) is an etvs then C[0X ] ⊆ M , whenever M is a τ -open subspace. Therefore, the notion of proper etvs is well-defined. Also, if in Definition 3.9 (K, θ) = (R, | · |) or (C, | · |), and {τα : α ∈ A } is a locally convex generating family (i.e., (X, τα ) is a fundamental elcs for each α ∈ A ), then we will say that (X, τ ) is an extended locally convex space (elcs, for short). Note also that the generating family used in Definition 3.9 can be assumed to be directed by inclusion. Indeed, it is enough to consider instead of {τα : α ∈ A } the family 



τα : A ∈ P(A ) with Card(A) < ∞ ,

α∈A

which generates the same topology τ . In what follows, we will simply say that {τα : α ∈ A } is a directed generating family, whenever it is a generating family of τ and it is directed by inclusion. Finally, recall that we will always assume that (K, θ) is a non-discrete topological field. Proposition 3.10. Let (X, τ ) be an etvs over (K, θ). Then: 1. For every generating family {τα : α ∈ A } of τ , we have that Xfin =



α Xfin ,

α∈A α where, for each α ∈ A , Xfin denotes the finite space of (X, τα ). 2. (X, τ ) is a topological group with operators in K. 3. Xfin is a tvs over (K, θ).

Proof. 1. Let F be the directed generating family induced by {τα : α ∈ A }, namely  F =



τα : A ∈ P(A ) with Card(A) < ∞ .

α∈A σ Noting that for each σ ∈ F there exist α1 , . . . , αn ∈ A such that the finite space of (X, σ) is Xfin =

n αi X , we can write i=1 fin

σ∈F

σ Xfin =



α Xfin .

α∈A

Then, we may assume without loss of generality that {τα : α ∈ A } is directed by inclusion. Let us

α denote Z = {Xfin : α ∈ A }. It is direct that Z ⊇ Xfin . On the other hand let M be a τ -open

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subspace of X. By Proposition 2.1, there exist α ∈ A and V ∈ N (0X , τα ) such that V ⊆ M . Then, M is τα -open by [8, Ch. III, §2, Corollary of Prop. 4]. Since (K, θ) is non-discrete there is no τα -open α proper subspace of Xfin , and therefore α Z ⊆ Xfin ⊆ M.

By arbitrariness of M , we conclude that Z ⊆ Xfin . 2. Let k ∈ K and let {τα : α ∈ A } be a directed generating family. Fix any V ∈ N (0X , τ ). Since, by  Proposition 2.1, B = α∈A N (0X , τα ) is a fundamental system of N (0X , τ ), we get that there exist α ∈ A and Vα ∈ N (0X , τα ) such that Vα ⊆ V . Then, since ϕk is τα -τα -continuous, we get that there exists Uα ∈ N (0X , τα ) such that ϕk (Uα ) ⊆ Vα ⊆ V. Finally, recalling that Uα ∈ N (0X , τ ), we conclude that ϕk is τ -τ -continuous, which finishes the proof.  3. Let {τα : α ∈ A } be a directed generating family of τ . Since B = α∈A N (0X , τα ) is a fundamental system of NX (0X , τ ), we may consider Bfin = {V ∩ Xfin : V ∈ B} as a fundamental system of NXfin (0X , τ ). It is clear that (Xfin , τ ) is a topological group and so, we only need to verify that the scalar multiplication · : K × Xfin → Xfin is (θ × τ )-τ -continuous, according to Definition 2.2. Choose then V ∩ Xfin ∈ Bfin . There exists α ∈ A such that V ∈ N (0X , τα ). Without α α losing generality, we may assume that V ⊆ Xfin . Therefore, since (Xfin , τα ) is a tvs over (K, θ), there exist O ∈ NK (0, θ) and U ∈ N (0X , τα ) such that O · U ⊆ V . Then, since Xfin is trivially a vector space over K, we have that O · (U ∩ Xfin ) ⊆ V ∩ Xfin and so, provided by U ∩ Xfin ∈ NXfin (0X , τ ), the proof is complete. 2 As a direct consequence of Lemma 3.4 and Proposition 3.10, we get the following corollary: Corollary 3.11. Let (X, τ ) be an etvs over (K, θ). Then, (X, τ ) is a fundamental etvs if and only if Xfin is τ -open. Remark 3. Note that, as in Remark 2, Xfin is a τ -closed subspace of X, since every τ -open subspace of X is also τ -closed. Nevertheless, the above corollary shows that it is not necessarily τ -open. Lemma 3.12. Let Z be a subspace of X, (X, τ ) be a topological group with operators in K and θ be a non-discrete field topology over K. Then: 1. If (X, τ ) is a fundamental etvs over (K, θ), then (Z, τ ) is also a fundamental etvs over (K, θ) with Zfin = Z ∩ Xfin . 2. If (X, τ ) is an etvs over (K, θ), then (Z, τ ) is also a etvs over (K, θ) with Zfin = Z ∩ Xfin . Proof. Observe first that (Z, τ ) is already a topological group with operators in K (see [8, Ch. III, §6]). 1. If (X, τ ) is a fundamental etvs, then the space Z  = Xfin ∩ Z is a τ -open subspace of Z. Since Z  is also a subspace of Xfin , we get that (Z  , τ ) is a tvs over (K, θ). Applying Lemma 3.4 we get that (Z, τ ) is a fundamental etvs and (Z  , N ) is a compatible decomposition, where N is any algebraic complement of Z  in Z. By Proposition 3.5, we conclude that Zfin = Z  , finishing the first part of the proof.

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2. Assume now that (X, τ ) is an etvs and let {τα : α ∈ A } be a directed generating family of τ . By the α α first part of the proof, we have that (Z, τα ) is a fundamental etvs with Zfin = Z ∩ Xfin . On the other  hand, since by Proposition 2.1 the set α∈A τα is a basis of the topology τ , it is direct that    τ Z = τα Z , α∈A

and therefore (Z, τ ) is an etvs. Finally, by Proposition 3.10, we have that Zfin =





α Zfin =

α∈A

α (Z ∩ Xfin ) = Z ∩ Xfin ,

α∈A

which finishes the proof. 2 Lemma 3.13. Let (X, τ ) be an etvs over (K, θ). Then Xfin is the maximal subspace of X which is a tvs over (K, θ), namely, for each subspace Z of X such that Xfin ∩ Z  Z, we have that (Z, τ ) fails to be a tvs over (K, θ). Proof. Let Z be subspace of X such that Xfin ∩Z  Z. Suppose that (Z, τ ) is a tvs over (K, θ). By definition of Xfin , there exists a τ -open subspace M such that M  = Z ∩ M  Z. Note that M  is an open subspace of (Z, τ ). Then, by [8, Ch. III, §2, Proposition 18], we have (Z/M  , π(τ )) is a discrete space and therefore, since (K, θ) is not discrete, (Z/M  , π(τ )) is not a tvs. This last statement contradicts the assumption that (Z, τ ) is a tvs, since the quotient space of a tvs endowed with the quotient topology is also a tvs. 2 The following theorem will give us a more intuitive characterization of etvs: An extended topological vector space is a “limit” of fundamental etvs. To establish such a theorem we need to recall the suitable notion of limit that we will use. Motivated by the known notion of projective limits for topological groups and topological vector spaces (see, e.g., [8] and [15]), we will apply the same notion for topological groups with operators in a field. Recall that for a directed set of indexes (A , ), a projective system of topological groups with operators in K, (Xα , τα , fαβ )α∈A , is a family {(Xα , τα )}α∈A of topological groups with operators in the same field K which has an associated family of continuous linear mappings {fαβ : Xβ → Xα : α, β ∈ A and α  β} such that: (i) Whenever α  β  γ in A , fαβ ◦ fβγ = fαγ . (ii) For each α ∈ A , fαα = idXα . For a projective system (Xα , τα , fαβ )α∈A , its projective limit X = lim Xα is the vector space ←−

 lim Xα := ←−

(xα )α∈A ∈

Xα : ∀α  β, fαβ (xβ ) = xα

,

(5)

α∈A

 endowed with the induced topology of the product topology τ = α∈A τα . This space is also a topological group with operators in K, since any subspace of the product space is a topological group with operators in K (see [8, Ch. III, §6]). Theorem 3.14. Let (K, θ) be a non-discrete topological field. A topological group (X, τ ) is an etvs over (K, θ) if and only if there exists a projective system (Xα , τα , fαβ )α∈A such that

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(a) for each α ∈ A , (Xα , τα ) is a fundamental etvs over (K, θ); and (b) (X, τ ) is topologically isomorphic to lim Xα . ←−

α α In such a case, we have that Xfin is topologically isomorphic to lim Xfin , where Xfin is the finite space of Xα . ←−

Proof. ⇒) Assume that (X, τ ) is an etvs over (K, θ) and let {τα : α ∈ A } be a directed generating family of τ . We will consider the set of indexes as A , endowed with the order  given by α  β ⇐⇒ τα ⊆ τβ . Clearly, (A , ) is a directed set. Now, for each α ∈ A we will consider the space Xα = X, endowed with the topology τα , and for each α  β ∈ A we will consider the map fαβ = idX : (Xβ , τβ ) → (Xα , τα ), ˜ = lim Xα and τ˜ =  which is clearly continuous. Let us denote X α∈A τα . It is easy to see that X is ←− ˜ algebraically isomorphic to X, where the isomorphism is the function h : X → lim Xα where h(x) is ←−

the constant net (xα ) with xα = x. We will denote this latter net as x ˜. We only need to verify that h  is bicontinuous. Indeed, consider the fundamental system B = α∈A N (0X , τα ) of NX (0X , τ ) given by Proposition 2.1, and choose any V ∈ B. Let α ∈ A be the index such that V ∈ N (0X , τα ). It is not hard to realize that ˜ ∩ p−1 (V ), h(V ) = {˜ x : x∈V}=X α where pα is the αth canonical projection of the product space, given by pα (x) = xα . Since h is bijective and 

 ˜ ∩ p−1 (V ) : α ∈ A , V ∈ N (0X , τα ) ⊆ N (0 ˜ , τ˜), X α X

we conclude that h is bicontinuous, finishing this part of the proof. ⇐) Let (Xα , τα , fαβ )α∈A be a projective system of fundamental etvs, such that X is topologically isomor˜ τ˜) is an etvs, where τ˜ is the product topology ˜ = lim Xα . We only need to show that (X, phic to X ←−

restricted to the projective limit. Define then τ˜α = p−1 α (τα ), where pα is the αth canonical projec tion. Clearly, τ˜ = α∈A τ˜α , so, by Lemma 3.12, it is sufficient to show that (X, τ˜α ) is a fundamental  α etvs, where X = α Xα . Let us consider the space Zα = p−1 α (Xfin ) as subspace of X. Since pα is τ˜α -τα -continuous, we have that Zα is τ˜α -open and therefore, by Lemma 3.4, we only need to show that (Zα , τ˜α ) is a tvs over (K, θ). Fix then V˜ ∈ N (0Zα , τ˜α ) which, by construction of τ˜α , can be written as V˜ = V ×

Xβ ,

β=α α α , τα ). Since (X where V ∈ N (0Xfin fin , τα ) is a tvs, there exist a neighborhood O ∈ N (0K , θ) and a α , τα ) such that neighborhood U ∈ N (0Xfin

O · U ⊆ V. −1 ˜ Then, is not hard to see that O · p−1 ˜α ). α (U ) ⊆ V , which finishes the proof since pα (U ) ∈ N (0Zα , τ

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α It only remains to prove that, whenever conditions (a) and (b) hold, Xfin ∼ . Since it is not hard to = lim Xfin ←− α ∼ ˜ ˜ ˜ realize that Xfin = Xfin (where X = lim Xα , as before), it is enough to show that Xfin = lim Xfin . Observe ←− ←− ˜ first that, if x ∈ Xfin , then for each α ∈ A , α ˜ x ∈ p−1 α (Xfin ) ∩ X

˜ Therefore pα (x) ∈ X α which proves that since the latter is an open subspace of X. fin ˜ fin ⊆ lim X α . X fin ←−

α ˜ which, endowed with τ˜, is a tvs over (K, θ) since it is also the Note now that lim Xfin is a subspace of X ←− α ˜ fin , which projective limit of a projective system of tvs over (K, θ). Thus, by Lemma 3.13, lim Xfin ⊆ X ←−

finishes the proof. 2

Theorem 3.15. Let (K, | · |) be a non-discrete Archimedean valuated field and τ be a group topology over X. Then, (X, τ ) is an etvs over (K, | · |) if and only if there exists a fundamental system B of neighborhoods of 0X such that: (i) For each V ∈ B, K · V is a subspace of X. (ii) For each V ∈ B, there exists U ∈ B such that K · V = K · U and that U + U ⊆ V . (iii) Each element V ∈ B is balanced. Proof. First, we will prove the necessity. Assume then that (X, τ ) is an etvs over (K, | · |) and let {τα : α ∈ α A } be a directed generating family of τ . Fix α ∈ A . Since (Xfin , τα ) is a tvs over (K, | · |), by [15, Ch. I, α (0X , τα ) satisfying (i), (ii) and (iii). Also, since (X, τα ) 1.2], there exists a fundamental system Bα of NXfin α (0X , τα ) is a fundamental system of N (0X , τα ), and therefore so is Bα . Finally, is a fundamental etvs, NXfin   considering the set B = α∈A Bα , the conclusion follows, since α∈A N (0X , τα ) is a fundamental system of N (0X , τ ). Now, to prove the sufficiency, assume the existence of a fundamental system B of neighborhoods of 0X satisfying (i), (ii) and (iii). We need to construct a generating family {τα : α ∈ A } of τ . To do so, for each V ∈ B, we will define the subspace MV = K · V . Then, we define BV = {U ∈ B : K · U = MV }. By (ii) and (iii), BV induces a group topology τV on X. It is not hard to see that in (MV , τV ), conditions (i) and (iii) imply that each element of BV is absorbing and balanced, and that condition (ii) implies that for each element W ∈ BV , there exists U ∈ BV such that U + U ⊆ W . Therefore, applying [15, Ch. I, 1.2], we conclude that (MV , τV ) is a tvs over (K, | · |), and therefore (X, τV ) is a fundamental etvs over (K, | · |).  The proof is concluded noting that τV ⊆ τ and the fact that τ ⊆ V ∈B τV , provided by the equality  B = V ∈B BV . 2 In [2], many of the results obtained for extended normed spaces came from the fact that the finite space of these is always open or, in our terminology, that every normed space is a fundamental etvs. We would like then to have a similar splitting theorem for etvs. But we will show that even in the most elemental cases, we can find examples of spaces where the finite space is not topologically complementable. Lemma 3.16. Let (X, τ ) be an etvs over (R, | ·|) or (C, | ·|) such that (Xfin , τ ) is Hausdorff and dim[Xfin ] < ∞. Suppose that Xfin has a τ -supplement space in X. Then, if {τα : α ∈ A } is any directed generating family α of τ , there exists α ∈ A such that Xfin has a τα -supplement space in Xfin .

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Proof. Let dim[Xfin ] = n with {e1 , . . . , en } a Hamel basis of Xfin , and let {τα : α ∈ A } be a directed generating family of τ . Since X = Xfin ⊕τ N (for some subspace N of X) we have  that the unique linear functionals δi : X → R defined by the relations δi (ei ) = 1, δi (ej ) = 0 for j = i and δi N = 0, are τ -continuous  for each i = 1, . . . , n. Recalling that B = α∈A N (0X , τα ) is a fundamental system of N (0X , τ ) we get that for each i ∈ {1, . . . , n} there exist αi ∈ A and Ui ∈ N (0X , ταi ) such that δi (Ui ) ⊆ {k ∈ K : |k| < 1}, (where K is R or C) and so δi is ταi -continuous (according to the fact that for each λ > 0, λUi ∈ N (0X , ταi )). Since the generating family is directed for the inclusion, there exists α ∈ A such that δi is τα -continuous for each i ∈ {1, . . . , n}. n α α Define then the function F : Xfin → Xfin given by F (x) = i=1 δi (x)ei . It is clear that F is a linear α projection such that F (Xfin ) = Xfin . Therefore, it is sufficient to prove that F is τα -continuous. But this is α α direct since each δi restricted to Xfin is τα -continuous and (Xfin , τα ) is a tvs. 2 Example 3.17 (An etvs with Xfin lacking of τ -supplements). Consider p ∈ (0, 1) and fix X = Lp [0, 1]. Let us denote by Θ the usual topology in Lp [0, 1] and let H be a Hamel Basis of X such that 1 ∈ H, where 1 stands for the constant function 1(t) = 1. Let A = {A ⊂ H \ {1} : Card(A) < +∞}, and for each A ∈ A consider the subspace X A = span{H \ A}, and define the topology ΘA on X as the group topology induced by {O ∩ X A : O ∈ Θ}, namely the unique topology on X such that the family BA = {O ∩ X A : O ∈ Θ and 0X ∈ O} is a fundamental system of neighborhoods of 0X . It is easy to see that, with the latter construction, (X, ΘA ) A is a proper fundamental etvs where X A is its finite space; so we will write Xfin instead of X A . Also, the family {ΘA : A ∈ A } is directed by inclusion: Indeed, let be I, J ∈ A and consider A = I ∪ J which, since both I and J are finite, is also finite and therefore A ∈ A . We will show that ΘI , ΘJ ⊂ ΘA . Let O ∈ ΘI . I I Without losing generality we assume that 0X ∈ O and, since Xfin is open, O ⊆ Xfin . Then, by construction,   I there exists O ∈ Θ such that O = O ∩ Xfin and so, since I ⊆ A we have that A I O ∩ Xfin ⊆ O ∩ Xfin = O. A Finally, since O ∩ Xfin ∈ ΘA , we have that O ∈ N (0X , ΘA ). Therefore, N (0X , ΘI ) ⊆ N (0X , ΘA ) which implies that ΘI ⊂ ΘA . Repeating the same reasoning for ΘJ instead of ΘI we conclude the desired inclusions, proving that {ΘA : A ∈ A } is directed by inclusion as we claimed.  Define then τ = A∈A ΘA . We have then, that (X, τ ) is an etvs over (R, | · |) with

Xfin = R · 1. We claim that Xfin doesn’t have any τ -supplement space in X. Let us suppose the contrary. Applying A Lemma 3.16, there exists A ∈ A such that Xfin has a ΘA -supplement space in Xfin . Let us denote by N such a subspace. Now, we can write X = Z ⊕ N,

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where Z = span{A ∪ {1}}. Since Z is a finite dimensional subspace of Lp [0, 1] we get that N is Θ-dense in X (see [15, Ch. I, Exercises 6 and 7]) and therefore there exists a net (nλ ) ⊆ N such that nλ →Θ 1. But A since N ⊆ Xfin we get that nλ →ΘA 1, which contradicts the fact that Xfin and N are ΘA -supplements A in Xfin . Therefore, Xfin cannot have any τ -supplement space in X, as we claimed. 3.3. Subspaces, products and quotients of etvs Proposition 3.18. Let (X, τ ) be an etvs over (K, θ), {τα : α ∈ A } be a generating family of τ and Z be a subspace of X. The following assertions are equivalent: (i) Z is τ -closed. α (ii) For each α ∈ A , Z ∩ Xfin is τ -closed. α (iii) There exists α ∈ A such that Z ∩ Xfin is τ -closed. α is τ -closed (see Remark 2), (i) ⇒ (ii) follows immediately, and Proof. Noting that for each α ∈ A , Xfin α (ii) ⇒ (iii) is direct. To prove (iii) ⇒ (i), let α ∈ A such that Z ∩ Xfin is τ -closed, and consider a Hamel α α basis H of X generating Xfin and Z. Then, denoting N = CH (Xfin ) and applying Lemma 3.4, we can write α X = Xfin ⊕τ N and also α α ,N (Z) ⊕τ PN,X α (Z) = (Z ∩ X Z = PXfin fin ) ⊕τ (Z ∩ N ). fin α α α ,N (zλ ) → PX α ,N (z) ∈ Z ∩ X Let (zλ ) be a net in Z converging to z ∈ X. Then, PXfin fin since Z ∩ Xfin fin α (zλ ) → PN,X α (z) ∈ Z ∩ N thanks to the discreteness of N (recall that (N, τα ) is is τ -closed, and PN,Xfin fin discrete by definition of fundamental etvs and by Lemma 3.4, and that τα is coarser than τ ). Therefore, α ,N (z) + PN,X α (z) ∈ Z, finishing the proof. z = PXfin 2 fin

Proposition 3.19. Let (X, τ ) be an etvs over (K, θ), {τα : α ∈ A } be a directed generating family of τ and Z be a subspace of X. The following assertions are equivalent: (i) (ii) (iii) (iv)

Z is τ -open. α For each α ∈ A , Z ∩ Xfin is τ -open. α There exists α ∈ A such that Z ∩ Xfin is τ -open. α There exists α ∈ A , such that Xfin ⊆ Z.

Proof. The implications (i) ⇒ (ii) ⇒ (iii) and (iv) ⇒ (i) are direct, and therefore we only need to prove α (iii) ⇒ (iv). Assume then that there exists α ∈ A such that Z ∩ Xfin is τ -open. Then, there exist β ∈ A α and V ∈ N (0X , τβ ) such that V ⊆ Z ∩ Xfin . Without losing generality, we may assume that τα ⊆ τβ and β β β that V ⊆ Xfin . Since (Xfin , τβ ) is a tvs over (K, θ) and (K, θ) is non-discrete, V must generate Xfin and so, since V ⊆ Z we conclude β Xfin = span(V ) ⊆ Z,

which finishes the proof. 2 Proposition 3.20. Let (Xi , τi )i∈I be a family of etvs over (K, θ). The product space X =  the product topology τ = τi is also an etvs, with

i Xfin = Xfin , i i where Xfin stands for the finite space of (Xi , τi ).

 i

Xi endowed with

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Proof. It is known (see [8, Ch. III, §6]) that (X, τ ) is a topological group with operators in (K, θ). For each i ∈ I, let us denote by pi : X → Xi the canonical projection given by pi (x) = xi , and let {τi,α : α ∈ Ai } be a generating family of τi . For i ∈ I and α ∈ Ai , it is easy to see that τ (i, α) := p−1 i (τi,α ) is a group topology over X such that (X, τ (i, α)) is a fundamental etvs over (K, θ) where its finite space is τ (i,α)

Xfin

i,α i,α = p−1 i (Xfin ) = Xfin ×

τ

τ

Xj .

j=i

Also, we have that p−1 i (τi ) =

 α∈Ai

τ=



τ (i, α) and therefore p−1 i (τi ) =



{τ (i, α) : i ∈ I, α ∈ Ai },

i∈I

which proves that (X, τ ) is an etvs. Now, by Proposition 3.10, we have that Xfin =



τ (i,α)

Xfin

.

Let x = (xi ) ∈ Xfin . Then, for each i ∈ I we have that pi (x) ∈



τ

i Xfini,α = Xfin ,

α∈A

 i  i τ and so Xfin ⊆ Xfin . Now, if x ∈ i Xfin , we have that for each i ∈ I and each α ∈ Ai , pi (x) ∈ Xfini,α and τ (i,α) therefore x ∈ Xfin . Then, x ∈ Xfin , which finishes the proof. 2    Since open subspaces of the product space X = i∈I Xi are of the form j∈J Mj × i∈I\J Xi where Mj is an open subspace of Xj and Card(J) < ∞, we get the following direct corollary: Corollary 3.21. Let (Xi , τi )i∈I be a family of fundamental etvs over (K, θ). The product space (X, τ ) is a fundamental etvs if and only if the set of indexes 

i  Xi i ∈ I : Xfin



is finite. Corollary 3.22. Let (Xi , τi , fij ) be a projective system of topological groups with operators in K, and θ be a non-discrete field topology over K. If (Xi , τi ) is an etvs over (K, θ) for each i ∈ I, then X = lim Xi endowed ←−  with the product topology τ = i τi is also an etvs over (K, θ). Moreover, i Xfin = lim Xfin , ←−

i where Xfin denotes the finite space of (Xi , τi ).

Proof. Since lim Xi is a subspace of



Xi , the first conclusion follows immediately from Proposition 3.20   i i and Lemma 3.12. Also, since (Xfin , τi ) is a tvs over (K, θ) for each i ∈ I, we have that lim Xfin , τ is also ←−

i

i a tvs over (K, θ). Therefore, lim Xfin ⊆ Xfin , by Lemma 3.13. ←−

←−

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Now, let x ∈ Xfin and fix i ∈ I. Let {τα : α ∈ Ai } be a generating family of τi . For α ∈ Ai , we have α α that Xfin is a τi -open subspace of (Xi , τi ) and therefore, Mα = X ∩ p−1 i (Xfin ) is an open subspace of (X, τ ). α Then, x ∈ Mα and so, pi (x) ∈ Xfin . Then, pi (x) ∈



α i Xfin = Xfin ,

α∈Ai

where the last equality is provided by Proposition 3.10. Since the latter relation holds for every i ∈ I,  i x ∈ i Xfin and, since  i lim Xfin ←−

=

 i Xfin

∩ X,

i

i we conclude that x ∈ lim Xfin , which finishes the proof. 2 ←−

Lemma 3.23. Let (X, τ ) be a fundamental etvs over (K, θ) and Z be a subspace of X such that Z ⊆ Xfin . Then, (X/Z, π(τ )) is a fundamental etvs over (K, θ) and (X/Z)fin = Xfin /Z. Proof. Let us note first that (X/Z, π(τ )) is a topological group with operators in K (see [8, Ch. III, §6]). Also, since π is an open mapping, we have that Xfin /Z = π(Xfin ) is open. By Lemma 3.4, it only remains to prove that Xfin /Z is a tvs over (K, θ). This last statement is known to hold (see for example [7, Ch. I, §1, Part 3]) and therefore, the proof is complete. 2 Lemma 3.24. Let X, Y be two vector spaces over a field K, τX , τY be two group topologies over X and Y respectively and θ be a field topology over K. Assume that there exists a linear mapping h : X → Y which is an isomorphism of topological groups. Then, (X, τX ) is a topological group with operators in K if and only if (Y, τY ) is a topological group with operators in K. Moreover, (X, τX ) is a tvs over (K, θ) ⇐⇒ (Y, τY ) is a tvs over (K, θ). Proof. Assume first that (X, τX ) is a topological group with operators in K and fix k ∈ K. We will denote by ϕk both endomorphisms induced by k, the one defined over X and the other over Y (see equation (1)). It is not hard to see, according to the linearity of h, that we can write ϕk = ϕk ◦ h ◦ h−1 = h ◦ ϕk ◦ h−1 , and so, ϕk is τY -τY -continuous. Thus (Y, τy ) is a topological group with operators in K and, by symmetry, the first equivalence is proven. Assume now that (X, τX ) is a tvs over (K, θ) and let (kλ , yλ ) be a net in K×Y converging to (k, y) ∈ K×Y . We have that kλ · yλ = kλ · h(h−1 (yλ )) = h(kλ · h−1 (yλ )) → h(k · h−1 (y)) = k · y, where the convergence is given by the continuity of h−1 and the fact that (X, τX ) is a tvs over (K, θ). We conclude that (Y, τY ) is a tvs over (K, θ) and, by symmetry, the second equivalence is proven. 2 Proposition 3.25. Let (X, τ ) be a fundamental etvs over (K, θ) and Z be a subspace of X. Then, (X/Z, π(τ )) is a fundamental etvs over (K, θ) and (X/Z)fin = π(Xfin ).

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Proof. As in Lemma 3.23, (X/Z, π(τ )) is a topological group with operators in K and π(Xfin ) is an open subspace of X/Z. Therefore, we only need to prove that π(Xfin ) is a tvs over (K, θ). Let us consider the space Y = Xfin + Z. By [8, Ch. III, §2, Proposition 20], we have that π(Xfin ) = π(Y ) and Y /Z are isomorphic as topological groups, where the isomorphism is given by the mapping ˜ 1 : Y / Ker(h1 ) → π(Y ) induced by the mapping h1 : Y → π(Y ) defined as h1 = π ◦ ˆι, where ˆι is the h canonical injection from Y into X (see the comments before [8, Ch. III, §2, Proposition 20] and [6, Ch. I, ˜ 1 meets the hypothesis of Lemma 3.24. §4, Theorem 3]). Observing that h1 is a linear mapping, we have that h On the other hand, by Lemma 3.12, we know that (Y, τ ) is a fundamental etvs over (K, θ) with Yfin = Y ∩ Xfin = Xfin . We can consider then a Hamel basis H of Y which generates Xfin and Z and denote N = CH (Xfin ). By construction, we get that N is a subspace of Z and then, applying the corollary of [8, Ch. III, §7, Proposition 22], we get that π(Xfin ) ∼ = Y /Z ∼ = (Y /N )/(Z/N ), ˜ 1 described before and the isomorphism where the isomorphism between π(Xfin ) and Y /Z is the mapping h ˜ 2 : Y /Z → (Y /N )/(Z/N ) induced by h2 : Y → between Y /Z and (Y /N )/(Z/N ) is given by the mapping h ˜2 (Y /N )/(Z/N ) defined as h2 = πZ/N ◦ πN . Observing that h2 is also a linear mapping, the isomorphism h meets the hypothesis of Lemma 3.24. Noting that Y /N ∼ = Xfin where the isomorphism is the parallel lifting N,Xfin (which is linear) we conclude, thanks to Lemma 3.24 applied to N,Xfin , that Y /N is a tvs over (K, θ) and therefore (Y /N )/(Z/N ) is also a tvs over (K, θ). Finally, π(Xfin ) also is a tvs over (K, θ), by Lemma 3.24 applied to the isomorphism ˜ −1 . 2 ˜ : (Y /N )/(Z/N ) → π(Xfin ) given by h ˜=h ˜1 ◦ h h 2 Proposition 3.26. Let (X, τ ) be an etvs over (K, θ) and Z be a subspace of X. Then, (X/Z, π(τ )) is an etvs over (K, θ) and we have   π(τ ) τ π(Xfin ) ⊆ π Xfin + Z ⊆ π(Xfin ) ⊆ (X/Z)fin . Moreover, if Z ⊆ Xfin , then the latter inclusions hold with equality. Proof. Let {τα : α ∈ A } be a generating family of τ . By Proposition 3.25, we have that for each α ∈ A , (X/Z, π(τα )) is a fundamental etvs over (K, θ). On one hand, applying [8, Ch. III, §2, Proposition 17] we have that  π(τ ) = π(τα ). α∈A

  Then, (X/Z, π(τ )) is an etvs over (K, θ). On the other hand, it is easy to see that π(Xfin ) ⊆ π Xfin + Z   α ) and, by continuity of π, that π Xfin + Z ⊆ π(Xfin ). Note also that for each α ∈ A we have that π(Xfin is an open subspace of X/Z, therefore it is also closed (see Corollary of [8, Ch. III, §2, Proposition 4]), and it contains π(Xfin ). Thus we can write π(Xfin ) ⊆

α∈A



α π(Xfin )=



(X/Z)α fin = (X/Z)fin ,

α∈A



α where the equality α∈A π(Xfin ) = α∈A (X/Z)α fin is given by Proposition 3.25. Thus, the inclusions in the statement hold. Assume now that Z ⊆ Xfin and let [x] ∈ (X/Z)fin . Then,   −1 α α α [x] ⊆ π π(Xfin ) = (Xfin + Z) = Xfin = Xfin . α∈A

α∈A

α∈A

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Then, π −1 ((X/Z)fin ) ⊆ Xfin and so, (X/Z)fin ⊆ π(Xfin ). Thus, the inclusions in the statement hold with equality, which was what we wanted to prove. 2 3.4. Linear operators Definition 3.27 (Bounded sets). Let (X, τ ) be an etvs over a topological field (K, θ) and A be a subset of X. We say that A is τ -bounded if for every neighborhood V ∈ NX (0X , τ ) there exist a finite set of centers {x1 , . . . , xn } ⊆ A and a finite set {α1 , . . . , αn } ⊆ K such that A⊆

n 

(xj + αj V ).

j=1

Observe that this definition of bounded sets is exactly the same of [2, Definition 4.1] when (X, τ ) is an extended normed space. Proposition 3.28. Let (X, τ ) and (Y, σ) be two etvs over the same topological field (K, θ), and T : X → Y be a linear operator. Then, T is continuous if and only if it is continuous at 0X . In such a case, we have that: 1. If (X, τ ) is proper, then T (Xfin ) ⊆ Yfin . 2. T is bounded, namely, T maps τ -bounded sets into σ-bounded sets. Proof. The first part is clear, since for any net (xλ ) ⊆ X converging to x ∈ X, we can write T (xλ ) → T (x) ⇐⇒ T (xλ − x) → 0Y . Also, if T is continuous and (X, τ ) is proper, we have that T (Xfin ) = T (C[0X ]) ⊆ C[0Y ] ⊆ Yfin . For the second part, let A ⊆ X be a nonempty τ -bounded set and let V ∈ NY (0Y , σ). Since T is continuous at 0X and T (0X ) = 0Y , there exists a neighborhood U ∈ NX (0X , τ ) such that T (U ) ⊆ V . Since A is bounded, there exist sets {x1 , . . . , xn } ⊆ A and {α1 , . . . , αn } ⊆ K such that A⊆

n 

(xi + αi U ).

i=1

Then, by linearity of T we have that T (A) ⊆

n n   (T (xi ) + αi T (U )) ⊆ (T (xi ) + αi V ), i=1

i=1

and therefore, noting that {T (x1 ), . . . , T (xn )} ⊆ T (A), we conclude that T (A) is σ-bounded. 2 Proposition 3.29. Let (K, θ) = (R, | · |) or (C, | · |), (X, τ ) be an etvs over (K, θ) and f : X → K be a linear functional. We have that f is continuous if and only if Ker(f ) is τ -closed. Proof. The necessity is trivial since {0K } is closed. To prove the sufficiency, assume that Ker(f ) is τ -closed. Then, applying Proposition 3.26, we have that the quotient space X/ Ker(f ) endowed with the quotient

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topology is an etvs over (K, θ) of dimension 1. Also, since Ker(f ) is τ -closed, we have that (X/ Ker(f ), π(τ )) is a Hausdorff space. Thus, there are only two possibilities: either π(τ ) is discrete or (X/ Ker(f ))fin = X/ Ker(f ). In both cases any linear function g : X/ Ker(f ) → K is continuous. In particular, the function g : X/ Ker(f ) → K given by g([x]) = f (x) is continuous. The conclusion follows, since f = g ◦ π. 2 Definition 3.30 (Dual space). Let (X, τ ) be an etvs over (R, | · |) (or (C, | · |)). We define the dual space (X, τ )∗ (or simply X ∗ if there exists no confusion) as the vector space over R (or C) of all linear functionals from X to R (or C) which are continuous. Proposition 3.31. Let (X, τ ) be a real (or complex) fundamental etvs. If N is a τ -supplement space of Xfin in X, then we have that ∗ X ∗ ≈ Xfin × N alg ,

where N alg denotes the algebraic dual of N . Proof. Let H be a Hamel basis of X generating Xfin and N . We will consider the identification φ : X ∗ → ∗ × N alg given by Xfin     φ(x∗ ) = x∗ X , x∗ N . fin

Clearly, φ is well-defined and it is an algebraic homomorphism. Also, it is injective. Indeed, if x∗ ∈ X ∗ is such that φ(x∗ ) = (0, 0), then ∀h ∈ H, x∗ , h = 0, ∗ and therefore x∗ = 0. Finally, if we consider (x∗f , n∗ ) ∈ Xfin × N alg , we can define

x∗ = x∗f ◦ PXfin ,N + n∗ ◦ PN,Xfin . Since the parallel projections are continuous, we get that x∗ ∈ X ∗ and that φ(x∗ ) = (x∗f , n∗ ). Thus, φ is bijective, which finishes the proof. 2 Remark 4. Note that, following the notation of the Parallel Diagram described in Fig. 1, we get that φ ∗ and N alg to N . Furthermore, from the preceding proof, we get corresponds to the mapping D, M to Xfin  ∗ ∗ ∗ ,N alg is the restriction to Xfin , that is PX ∗ ,N alg (x ) = x  that PXfin . Also, we would like to mention Xfin fin that the notation N alg is not very common. Usually, the algebraic dual of a space N is noted by N  (or by N ∗ , when the topological dual is noted by N  ). Nevertheless, to avoid confusions, we prefer to use this notation. Recall that, for a directed set of indexes (I, ), an inductive system of vector spaces over a field K, is a family (Xi , fji )i∈I where Xi is a vector space over K and for each i, j, k ∈ I: • Whenever j  i, fji : Xj → Xi is a linear operator. • fii = idXi . • Whenever k  j  i, fji ◦ fkj = fki .

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For an inductive system (Xi , fji )i∈I , the inductive limit X = lim(Xi , fji ) is given by −→

 lim Xi = −→



 Xi

/R,

i∈I

where R is the equivalence relation given by xi Ryj ⇔ ∃k ∈ I, i, j  k and fik (x) = fjk (y). The algebraic structure in lim Xi (which is known to be a vectorial structure over the field K) is the following: −→ For xi ∈ Xi , yj ∈ Xj and λ ∈ K • [xi ]R + [yj ]R = [fik (xi ) + fjk (yj )]R , for any k ∈ I such that i, j  k. • λ[xi ]R = [λxi ]R . We want now to compute the dual space of an etvs (X, τ ). The following proposition gives us a characterization of it in terms of a generating family {τα : α ∈ A }, as an inductive limit. Observe that, whenever τβ , τα are two topologies on X with τβ ⊆ τα , then necessarily (X, τβ )∗ ⊆ (X, τα )∗ and so, the canonical embedding ˆιβα : (X, τβ )∗ → (X, τα )∗ is well defined. Proposition 3.32. Let (X, τ ) be a real (or complex) etvs, and {τα : α ∈ A } be any directed generating family of τ . Then, X∗ =



α (X, τα )∗ ≈ lim[(Xfin , τα )∗ × Nαalg , fβα ], −→

α∈A

β α with, whenever τβ ⊆ τα , fβα : (Xfin , τβ )∗ × Nβalg → (Xfin , τα )∗ × Nαalg given by

fβα = φα ◦ ˆιβα ◦ φ−1 β , α where φα is the isomorphism between (Xfin , τα )∗ × Nαalg and Xα∗ given in Proposition 3.31, and ˆιβα is the ∗ canonical embedding of (X, τβ ) into (X, τα )∗ .

Proof. To simplify the proof, let us denote X1∗ =



α (X, τα )∗ and X2∗ = lim[(Xfin , τα )∗ × Nαalg , fβα ]. Let

α∈A

−→

us also denote by K the field over which X is a vector space (which is R or C). Observe that, since the generating family is directed, we have that X1∗ is a subspace of the algebraic dual X alg . Now, we can define the mapping φ : X1∗ → X2∗ given by φ(x∗ ) = [φα (x∗ )],

whenever x∗ ∈ (X, τα )∗ .

Note that whenever x∗ belongs to (X, τα )∗ and to (X, τβ )∗ , there exists γ ∈ A such that τα ∨ τβ ⊆ τγ , and therefore x∗ ∈ (X, τγ )∗ . Thus, [φα (x∗ )] = [φγ (x∗ )] = [φβ (x∗ )] and so, φ is well defined. Also, by definition, φ is an onto homomorphism. Finally, if φ(x∗ ) = 0 for some x∗ ∈ X1∗ , then there exist α, β, γ ∈ A with α  γ and β  γ such that x∗ ∈ (X, τα )∗ and fαγ (φα (x∗ )) = fβγ (φβ (0)). Since φγ is an isomorphism, we have that ˆιαγ (x∗ ) = ˆιβγ (0) = 0. Since ˆιαγ is injective, we conclude that x∗ = 0, proving that φ is injective. Thus, X1∗ ≈ X2∗ .

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To prove that X ∗ = X1∗ , we only need to show that X ∗ ⊆ X1∗ (the other inclusion is direct, since every  τα -continuous functional is also τ -continuous, for each α ∈ A ). Let x∗ ∈ X ∗ . Since α∈A N (0X , τα ) is a fundamental system of neighborhoods of N (0X , τ ), we have that there exist α ∈ A and V ∈ N (0X , τα ) such that x∗ (V ) ⊆ {λ ∈ K : |λ| < 1}. Since for each r > 0, rV ∈ N (0X , τα ), we have that the latter inclusion implies that x∗ is τα -continuous, and therefore x∗ ∈ X1∗ . So, X ∗ ⊆ X1∗ , finishing the proof. 2 There is not much more that we can say about the dual space of a general etvs or about the linear operators defined over it. We will return to these subjects in the context of elcs, in Subsection 4.1. Remark 5. The general strategy when we study extended topological vector spaces is to reduce them to their fundamental etvs. This approach may have applications in Lie group theory (see [11]). For example, if we have an abelian matrix Lie group (G, τ ), induced by a commutative matrix Lie algebra g, it is known that exp(g) = C[0G , τ ], and even more, the exponential map is a local homeomorphism (see Proposition 2.3, Proposition 2.16, Corollary 2.29 and Corollary 2.31 of [11]). Therefore, C[0G , τ ] is an open subgroup of G and so, G can be decomposed as a topological sum of a connected abelian Lie group and a discrete Lie group. Thus, many of the techniques exposed in this section could be applied to the study of projective limits of Lie groups. This subject is outside of the scope of the paper. 4. Extended seminormed spaces The notions of fundamental etvs and etvs gave us a global context to work with, but our attention, as in the classical theory, arrives quickly at the convex case. In this section we will study the properties of the extended locally convex spaces and the extended seminormed spaces, providing one of our principal results: (X, τ ) is an esns if and only if it is an elcs. In the following, K will be always R or C, endowed with the usual absolute value. Recall that for a family P = {ρi : i ∈ I} of extended seminorms over X, T(P) denotes the induced topology of P on X, which was described in the comments above Definition 1.2. Proposition 4.1. Every fundamental elcs is an esns. Proof. Let (X, τ ) be a fundamental elcs over K. Then, (Xfin , τ ) is a lcs over K and therefore there exists a family P = {ρi : i ∈ I} of seminorms on Xfin such that  τ X = T(P). fin

Let us write X = Xfin ⊕τ N and, for each i ∈ I, set ρ˜i : X → [0, +∞] given by ρ˜i (x) = ρi (PXfin (x)) + PN (x)d , where  · d is the discrete norm (see Equation (3) at the beginning of Section 3). We have that P˜ = {˜ ρi : i ∈ I} is a family of extended seminorms on X and ˜ τ = T(P), which proves that (X, τ ) is an esns (see Definition 1.2).

2

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Lemma 4.2. Let (X, τ ) be a topological group and ρ : X → [0, +∞] be a τ -continuous extended seminorm. Then, the sets ρ−1 ([0, +∞)) and ρ−1 ({+∞}) are both open and closed. Proof. Clearly, ρ−1 ([0, +∞)) is open by continuity. On the other hand, choose x0 ∈ X with ρ(x0 ) = +∞. Since ρ is continuous and τ is a group topology, the set U = {x ∈ X : ρ(x0 − x) < 1} is open. Applying the triangle inequality of ρ we have that for each x ∈ U ρ(x0 ) ≤ ρ(x0 − x) + ρ(x), and since ρ(x0 − x) < 1, we conclude that ρ(x) = +∞. Thus, U ⊆ ρ−1 ({+∞}). Since x0 is an arbitrary element of ρ−1 ({+∞}), we conclude that ρ−1 ({+∞}) is open, finishing the proof. 2 Theorem 4.3. (X, τ ) is an esns if and only if it is an elcs. Proof. Assume first that (X, τ ) is an esns and let P = {ρi : i ∈ I} be a family of extended seminorms such that τ = T(P). Then, for each i ∈ I let us consider the topology τi := T({ρi }). i It is easy to see that (X, τi ) is a fundamental elcs with Xfin = ρ−1 i (R). Noting that

τ=



τi ,

i∈I

we conclude that (X, τ ) is an elcs. For the other direction, assume that (X, τ ) is an elcs and let {τα : α ∈ A } be a locally convex generating family (see comments after Definition 3.9) of τ . Since (X, τα ) is a fundamental elcs, we can apply Proposition 4.1 to find a family Pα = {ρi : i ∈ Iα } of extended seminorms such that τα = T(Pα ). It is not hard to see that τ=

 α∈A

τα =

 α∈A

 T(Pα ) = T



 Pα

,

α∈A

which finishes the proof. 2 Remark 6. Observe that the study of projective limits of Section 3.2 and all stability results of Section 3.3 are still valid if we fix K = R or K = C with their usual topology and replace fundamental etvs and etvs by fundamental elcs and elcs, respectively. The extended seminormed spaces inherit this structure. 4.1. Structure of esns In the following, motivated by Theorem 4.3, we will denote by S(X, τ ) the set of all τ -continuous extended seminorms on X, and for each ρ ∈ S(X, τ ) we will denote τρ := T({ρ}) and ρ Xfin := {x ∈ X : ρ(x) < +∞}. ρ Note that (X, τρ ) is a fundamental elcs and its finite space coincides with Xfin . So this notation is not ambiguous.

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Proposition 4.4. Let (X, τ ) be an esns, P be a family of extended seminorms with τ = T(P), and ρ : X → [0, +∞] be an extended seminorm. The following assertions are equivalent: (i) ρ ∈ S(X, τ ). (ii) There exist C > 0 and {pi }ni=1 ⊆ P such that ρ ≤ C max{pi : i = 1, . . . , n}. (iii) ρ is continuous at 0X . Proof. Note first that (ii) ⇒ (iii) is direct thanks to the positivity of ρ. Let us prove (iii) ⇒ (i). Assume then that ρ is τ -continuous at 0X and let (xλ )λ∈Λ be a net in X converging to x ∈ X. Then, xλ − x → 0X and so, there exists λ0 ∈ Λ such that for each λ ≥ λ0 , ρ(xλ − x) < 1. If ρ(x) < +∞, we can apply the triangle inequality getting that for each λ ≥ λ0 , |ρ(xλ ) − ρ(x)| ≤ ρ(xλ − x) → 0. Then, in this case, ρ(xλ ) → ρ(x). On the other hand, if ρ(x) = +∞, then since ρ(x) ≤ ρ(xλ ) + ρ(xλ − x), we can assure that, for each λ ≥ λ0 , ρ(xλ ) = +∞, and therefore ρ(xλ ) = ρ(x). Then, ρ(xλ ) → ρ(x) and so the conclusion follows. To prove (i) ⇒ (ii), note that if for each F ⊆ P with |F | < ∞ we write V (F ) = {x ∈ X : p(x) ≤ 1, ∀p ∈ F }, then B = {C · V (F ) : C > 0, F ⊆ P with |F | < ∞} is a fundamental system of neighborhoods of N (0 X , τ ): the reasoning is the same as  the one after Defini tion 3.9, namely, we replace {τp : p ∈ P} by as generating family p∈F τp : F ⊂ P, Card(F ) < ∞ of τ , which is directed by inclusion. Therefore, since ρ is a continuous extended seminorm, we have that there exists an element V ∈ B such that V ⊆ ρ−1 ((−1, 1)). Since V = C · V (F ) for some C > 0 and some F ⊆ P finite, we conclude that ρ ≤ C −1 · max{p : p ∈ F }, finishing the proof. 2 Theorem 4.5. Let (X, τ ) be an esns. Then, for each family of extended seminorms P = {ρi : i ∈ I} such that τ = T(P) we have that Xfin = C[0X ] = {x ∈ X : ρi (x) < +∞, ∀i ∈ I}. Moreover, the following assertions are equivalent: (i) (X, τ ) is a fundamental elcs. (ii) (X, τ ) is a locally connected esns. ρ (iii) (X, τ ) is an esns and there exists ρ ∈ S(X, τ ) such that Xfin = Xfin .

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Proof. For the first part, let P = {ρi : i ∈ I} be a family of extended seminorms such that τ = T(P). Since {τρ : ρ ∈ P} is a generating family of τ , we can apply Proposition 3.10 to get that Xfin =



ρ Xfin = {x ∈ X : ρi (x) < +∞, ∀i ∈ I},

ρ∈P

and that (Xfin , τ ) is a locally convex space. Therefore, (Xfin , τ ) is connected and so, by the inclusion C[0X ] ⊆ Xfin , we get that C[0X ] = Xfin . For the second part, (i) ⇒ (ii) follows directly from the fact that (Xfin , τ ) is connected. For (ii) ⇒ (iii), assume that there exists a connected neighborhood V ∈ N (0X , τ ). Since B = {ρ−1 ((−ε, ε)) : ρ ∈ S(X, τ ), ε > 0} is a fundamental system of N (0X , τ ), we get that there exist ρ ∈ S(X, τ ) and ε > 0 such that ρ−1 ((−ε, ε)) ⊆ V . Then, we have that ρ Xfin =





ρ−1 ((−nε, nε)) ⊆

n∈N

ρ nV ⊆ C[0X ] = Xfin ⊆ Xfin .

n∈N

ρ It only remains to prove (iii) ⇒ (i). Let ρ ∈ S(X, τ ) be such that Xfin = Xfin . In particular, by Lemma 4.2, we have that Xfin is τ -open and so the conclusion follows from Lemma 3.4 and the fact that (Xfin , τ ) is a locally convex space (for example, by Proposition 3.10.3, adapted to elcs). 2

Corollary 4.6. Let X be a vector space over K. The following assertions are equivalent: (i) For any family of extended seminorms P = {ρi : i ∈ I} on X, (X, T(P)) is a fundamental elcs. (ii) For any family of extended seminorms P = {ρi : i ∈ I}, the finite space Xfin of (X, T(P)) is open. (iii) X is finite dimensional. Proof. (i) ⇒ (ii) Direct by Corollary 3.11. (ii) ⇒ (iii) Let us suppose that X has infinite dimension and let H = {bi : i ∈ Δ} be a Hamel basis of X. Consider the canonical embedding of X in c0 (Δ) given by ˆι : x =



αi bi → (αi )i∈Δ ∈ c0 (Δ),

i∈Δ

and for each i ∈ Δ define the extended seminorm ρi : X → [0, +∞] given by  ρi (x) =

ˆι(x)∞

if ˆι(x)(i) = 0.

+∞

otherwise.

It is easy to realize that, for τ = T({ρi : i ∈ Δ}), Xfin = {0} and therefore, for Xfin to be τ -open (X, τ ) must be a discrete space. But this last statement doesn’t hold, since for each   infinite sequence (in ) in Δ with no repeated elements, the non-stationary sequence n1 bin τ -converges to 0X : Indeed, for each j ∈ Δ there exists n0 ∈ N such that in = j for each n ≥ n0 and therefore, starting from n0 , we have that ρj

1

n bin



=

1 ι(bin )∞ n ˆ

=

1 → 0. n

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(iii) ⇒ (i) Let P = {ρi : i ∈ I} be a family of extended seminorms on X and let H = {bj }nj=1 be a Hamel basis of X generating Xfin . By reordering H if necessary, we may and do assume that Xfin = span({b1 , . . . , bk }) for some k ∈ {1, . . . , n}. If k = n then X = Xfin and the result follows directly. If k < n, then for each j ∈ {k + 1, . . . , n} there exists ρij ∈ P such that ρij (bj ) = +∞. Considering ρ = max{ρik+1 , . . . , ρin } we have that ρ ∈ S(X, T(P)) and ρ Xfin ⊆ Xfin . ρ Then, since the reverse inclusion always holds, Xfin = Xfin and the conclusion follows from Theorem 4.5. 2

Proposition 4.7. Let τ be a group topology over X. Then, (X, τ ) is an esns over K if and only if there exists a neighborhood basis B of 0X such that each element V ∈ B is absolutely convex (i.e. convex and balanced). Proof. For the necessity, assume that (X, τ ) is an esns over K and consider the family   B = Vρ,ε = ρ−1 ((−ε, ε)) : ρ ∈ S(X, τ ), ε > 0 . It is easy to see that for each ρ ∈ S(X, τ ) and each ε > 0 the set Vρ,ε is absolutely convex. Also, since τ = T(S(X, τ )) and {τρ : ρ ∈ S(X, τ )} is directed by inclusion, we have that for each V ∈ N (0X , τ ) there exist ρ ∈ S(X, τ ) and ε > 0 such that ρ−1 ((−ε, ε)) ⊆ V, and therefore B is a fundamental system of N (0X , τ ), which proves the necessity. Assume now that there exists a fundamental system B of N (0X , τ ) such that each V ∈ B is absolutely convex. Then, for each V ∈ B consider ρV as the Minkowski functional of V , namely the function given by ρV (x) = inf{λ > 0 : x ∈ λV }, with the convention inf ∅ = +∞. Since V is absolutely convex, it is not hard to realize that ρV is an extended seminorm on X. Also, it is direct that T({ρV : V ∈ B}) ⊆ τ , since for each V ∈ B, ρV is τ -continuous. The other inclusion follows from the fact that N (0X , τ ) = N (0X , T({ρV : V ∈ B})) and that τ is already a group topology. 2 Proposition 4.8. Let (X, τ ) and (Y, σ) be two esns, P ⊆ S(X, τ ) such that τ = T(P) and T : X → Y be a linear operator. Then, T is continuous if and only if for all q ∈ S(Y, σ) there exist C > 0 and a finite set {pi }ni=1 ⊆ P such that q(T (x)) ≤ C max{pi (x) : i = 1, . . . , n}, ∀x ∈ X. Proof. The necessity is direct, since, if T is continuous, then for each q ∈ S(Y, σ), we have that q◦T ∈ S(X, τ ) and we can apply Proposition 4.4. For the sufficiency, fix q ∈ S(Y, σ) and let C > 0 and {pi }ni=1 ⊆ P be such that q ◦ T ≤ C max{pi : i = 1, . . . , n}. Then, T −1 ({y ∈ Y : q(y) ≤ 1}) ⊇

n 

x ∈ X : pi (x) ≤

1 C



,

i=1

where the latter set is known to be a neighborhood of 0X . Since the family {q −1 ([−1, 1]) : q ∈ S(Y, σ)} is a fundamental system of N (0Y , σ), the conclusion follows. 2

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We will end this section characterizing the dual space of Xfin when (X, τ ) is an extended seminormed space. To do so, we will use a suitable adaptation of Hahn–Banach extension theorem (see Theorem 4.10 below) in order to understand what are the duals of the subspaces of X. Recall first that, in a real vector space X, a function f : X → R ∪ {+∞} is said to be an extended sublinear function if f (0) = 0 and it satisfies (a) for all x ∈ X and λ > 0, f (λx) = λf (x); and (b) for all x, y ∈ X, f (x + y) ≤ f (x) + f (y). Lemma 4.9. Let (X, τ ) be a real esns and f : X → R ∪ {+∞} be an extended sublinear function. Then, the following assertions are equivalent: (i) f is τ -continuous. (ii) f is τ -continuous at 0X . (iii) There exists ρ ∈ S(X, τ ) such that f ≤ ρ. Proof. (i) ⇒ (ii): Direct. (ii) ⇒ (iii): Since f is continuous at 0X , f −1 ((−1, 1)) ∈ N (0X , τ ). Then, by Proposition 4.7, there exists an absolutely convex neighborhood U of 0X with U ⊆ f −1 ((−1, 1)). Therefore, the Minkowski functional ρU is τ -continuous and, since f is positively homogeneous, f ≤  ρU . (iii) ⇒ (i): Let ρ ∈ S(X, τ ) such that f ≤ ρ. By classical analysis we have that f X ρ is continuous in fin ρ ρ (Xfin , τρ ), and therefore it is also continuous in (Xfin , τ ). Now, let (xλ )λ∈Λ be a net in X ρ τ -converging to x ∈ X. Without loss of generality, we may assume that (xλ )λ∈Λ ⊂ x + Xfin . Noting that for each λ ∈ Λ we have f (xλ ) ≤ f (x) + f (xλ − x) and f (x) ≤ f (xλ ) + f (x − xλ ) and recalling that f (x − xλ ) → 0 and f (xλ − x) → 0, we have that lim sup f (xλ ) ≤ f (x) ≤ lim inf f (xλ ), and therefore, f (xλ ) → f (x), which finishes the proof. 2 Theorem 4.10. Let (X, τ ) be a real esns and let M be a subspace of X. Then, for any linear functional φ : M → R and any continuous extended sublinear function f : X → R ∪ {+∞} such that Z := f −1 (R) is  a subspace of X and φ ≤ f M , there exists a continuous linear functional φˆ : X → R such that  φˆM = φ

and

φˆ ≤ f.

Proof. Since f is τ -continuous, we have that Z is a τ -open subspace. Choose a Hamel basis H generating Z and M and denote N := CH (Z). Clearly, by Lemma 3.4 we get that (N, τ ) is a discrete space. Denote φ1 = φM ∩Z and φ2 = φM ∩N . We will extend both functionals separately. On one hand, note that φ1 ≤ f M ∩Z and so, by the classic Hahn–Banach extension theorem, there exists   a linear functional φˆ1 : Z → R such that φˆ1  = φ1 and φˆ1 ≤ f  . M ∩Z

Z

On the other hand, since (N, τ ) is discrete, we can extend continuously φ2 to N directly: Just choose any subspace N0 such that N = (M ∩ N ) ⊕ N0 and define the extension as φˆ2 = φ2 ◦ PM ∩N,N0 , which clearly is a τ -continuous linear functional. The proof is finished considering the extension of φ as φˆ := φˆ1 ◦ PZ,N + φˆ2 ◦ PN,Z . 2

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Remark 7. Observe that the above result cannot be extended just erasing the hypothesis that f −1(R) is a subspace of X. Indeed, in [16, Counterexample to (3)], Simons provided a counterexample in R2 of an extended linear functional f dominating a linear functional φ in a subspace, for which there is no algebraic linear extension φˆ preserving the inequality φˆ ≤ f . Therefore, endowing R2 with the discrete topology, Simons’ counterexample applies to our context as well. Corollary 4.11. Let X be a vector space over K and M be a subspace of X. Then, for any linear functional  φ : M → K and any extended seminorm ρ on X (not necessarily continuous) such that |φ| ≤ ρM , there exists a linear functional φˆ : X → K such that  φˆM = φ

ˆ ≤ ρ. |φ|

and

In particular, if (X, τ ) is an esns, then each element φ ∈ M ∗ admits a linear extension φˆ ∈ X ∗ . Proof. If K = R, the proof follows directly from Theorem 4.10 endowing X with the topology τρ and ˆ ≤ ρ. Assume then that K = C observing that, since ρ is an extended seminorm, φˆ ≤ ρ is equivalent to |φ|  and let us denote by (φ) the real part of φ. It is easy to see that (φ) ≤ |φ| ≤ ρM and therefore, since ˆ any vector  space over C is also a vector space over R, there exists a real linear functional φr : X → R such ˆ ˆ  that φr M = (φ) and |φr | ≤ ρ. Then, considering the endomorphism ϕi induced by i given in equation (1) ˆ ˆ ˆ and  applying [9, Lemma 6.3], we get that φ := φr − i(φr ◦ ϕi ) is a complex linear functional on X satisfying ˆ ˆ  φ M = φ and |φ| ≤ ρ. The second part of the corollary follows directly from Lemma 4.9. 2 Corollary 4.12. Let (X, τ ) be an esns and M be a subspace of X. Then, M ∗ ≈ X ∗ /M ⊥ , where M ⊥ stands for the annihilator of M .  Proof. Let us consider first the map φ : X ∗ → M ∗ given by φ(x∗ ) = x∗ M . By Corollary 4.11, we have that φ is an onto homomorphism. Also, it is direct that Ker(φ) = M ⊥ . Therefore, taking any linear lifting  : X ∗ /M ⊥ → X ∗ , we have that φ ◦  is an isomorphism between X ∗ /M ⊥ and M ∗ , which finishes the proof. 2 Proposition 4.13. Let (X, τ ) be an esns and {τα : α ∈ A } be a directed locally convex generating family of τ . Then,  α  ∗ (Xfin , τ )∗ ≈ lim (Xfin , τα ) , fβα , −→

 β α where, whenever τβ ⊆ τα , fβα : (Xfin , τβ )∗ → (Xfin , τα )∗ is given by fβα (x∗β ) = x∗β X α . fin

Proof. By Corollary 4.12, we know that (Xfin )∗ ≈ X ∗ /(Xfin )⊥ . Now, let  : X ∗ /(Xfin )⊥ → X ∗ be a linear  lifting. Since {τα : α ∈ A } is directed we have, by Proposition 3.32, that X ∗ = α∈A (X, τα )∗ . We define   ∗ α the assignation φ : X ∗ → lim (Xfin , τα ) , fβα as follows: Whenever x∗ ∈ (X, τα )∗ , −→

   φ(x∗ ) = x∗ X α . fin

 To simplify notation, for x∗ ∈ X ∗ we will denote x∗α = x∗ X α , for each α ∈ A . Let us prove now that φ fin is well defined. Assume that x∗ ∈ X ∗ is τα -continuous and τβ -continuous. Since the generating family is

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directed, there exists γ ∈ A such that τα ⊆ τγ and τβ ⊆ τγ and so, x∗ is also τγ -continuous. Further, we γ γ β α have that Xfin ⊆ Xfin and Xfin ⊆ Xfin and therefore fαγ (x∗α ) = x∗γ = fβγ (x∗β ). So, [x∗α ] = [x∗β ], showing that φ is well defined. Also, it is not hard to see that φ is an onto homomorphism. To finish the proof, we will show that φ ◦  is bijective. To do so, it is sufficient to prove that Ker(φ) = (Xfin )⊥ . Fix first x∗ ∈ (Xfin )⊥ \ {0}. Then, since x∗ is τ -continuous, we have that Ker(x∗ ) is a closed subspace of codimension 1. Let x0 ∈ X be any vector such that x∗ , x0  = 1. Therefore, since Xfin ⊆ Ker(x∗ ), then there exists ρ ∈ S(X, τ ) such that ρ(x0 ) = +∞. Noting that the projection P : X → K · x0 given by ρ˜ P (x) = x∗ , xx0 is continuous, we have that ρ˜ = ρ ◦ P ∈ S(X, τ ) and then Xfin ⊆ Ker(x∗ ). Therefore,  ∗ Ker(x ) is τ -open. Now, since by Proposition 2.1 the set α∈A N (0X , τα ) is a fundamental system of N (0X , τ ), we get that there exists α ∈ A such that Ker(x∗ ) is τα -open (by [8, Ch. III, §2, Corollary α of Proposition 4]). Therefore, Xfin ⊆ Ker(x∗ ) and so, x∗α = 0. Finally, noting that x∗ ∈ (X, τα )∗ (see Proposition 3.31) we can write φ(x∗ ) = [x∗α ] = [0]. Thus, (Xfin )⊥ ⊆ Ker(φ). For the other inclusion, it is enough to note that if x∗ ∈ Ker(φ), then there exists α α ∈ A such that x∗α = 0. Then, since Xfin ⊆ Xfin , we have that x∗ ∈ (Xfin )⊥ , finishing the proof. 2 4.2. Countable esns In this section we will focus on the special structure of extended seminormed spaces for which their topology can be induced by a countable family of extended seminorms. Definition 4.14. Let (X, τ ) be an esns over K. We say that (X, τ ) is a countable extended seminormed space if there exists a countable family P = {ρn : X → [0, +∞] : n ∈ N} of extended seminorms such that τ = T(P). If P = {ρn : n ∈ N} is a family of extended seminorms, we can define the equivalent family P˜ = { ρ˜n : n ∈ N} where ρ˜n = max{ρ1 , . . . , ρn }. ˜ and therefore, when (X, τ ) is a countable esns, we may assume that the countable Clearly T(P) = T(P), family of extended seminorms is pointwise nondecreasing. In the following, when the countable family P has been fixed, we will use the notation ρ˜n n Xfin := Xfin = {x ∈ X : ρ1 (x) < +∞, . . . , ρn (x) < +∞}.

(6)

For any metric d on X, we will write Bd (x, r) to denote the closed d-ball centered in x ∈ X and of radius r > 0 and by τd the topology on X induced by d. Also, for each extended seminorm ρ on X, each point x ∈ X and each r > 0 we will write B(X,ρ) (x, r) to denote the closed ball centered in x of radius r induced by ρ, that is B(X,ρ) (x, r) := {x ∈ X : ρ(x − x) ≤ r}.

(7)

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This notation will be natural after Lemma 4.15 below, on which we show that any countable esns is metrizable. Finally, recall that a metric d over X is said to be translation-invariant if ∀x, y, z ∈ X, d(x, y) = d(x + z, y + z). Lemma 4.15. Let (X, τ ) be a Hausdorff countable esns. Then, there exists a translation-invariant metric d : X × X → R+ such that τ = τd and the balls Bd (0, r) := {x ∈ X : d(0, x) ≤ r} are absolutely convex. Proof. We will proceed as in [15, Ch. I, Theorem 6.1]. Since (X, τ ) is a countable esns, there exists a countable basis {Vn : n ∈ N} of N (0X , τ ) such that each Vn is absolutely convex and, for all n ∈ N, Vn+1 + Vn+1 ⊆ Vn .   Now, for each nonempty finite subset F ⊆ N we define VF = n∈F Vn and pF = n∈F 2−n . In [15, Ch. I, Theorem 6.1] it is proven that the translation-invariant function d(x, y) = min(1, inf{pF : F ⊆ N finite, x − y ∈ VF }) is a metric on X with τ = τd and with balanced balls. We will prove now that the balls induced by d are convex. Let r > 0, x, y ∈ Bd (0X , r) and λ ∈ (0, 1). Observing that for r ≥ 1 the ball Bd (0X , r) = X, we only need to prove the convexity for r ∈ (0, 1). Let us denote by D the dyadic numbers in [0, 1]. Since the dyadic numbers are dense in [0, 1], there exists a decreasing sequence (rn) ⊆ D such that rn  r. Now, for each n ∈ N we have that d(x, 0X ) < rn and d(0X , y) < rn . Therefore there exist two finite subsets F (n, x), F (n, y) ⊆ N such that x ∈ VF (n,x) ,

y ∈ VF (y,n) ,

and

pF (n,x) < rn , pF (n,y) < rn .

Also, since rn is dyadic, there exists a finite subset F (n) ⊆ N such that pF (n) = rn . The latter inequality shows that VF (n,x) ⊆ VF (n) and VF (n,y) ⊆ VF (n) and therefore x, y ∈ VF (n) . Since VF (n) is absolutely convex, we obtain that λx + (1 − λ)y ∈ VF (n) and therefore d(0X , λx + (1 − λ)y) < rn . We conclude that d(0X , λx + (1 − λ)y) ≤ r and so Bd (0X , r) is convex.

2

Theorem 4.16. Let (X, τ ) be a Hausdorff countable esns, (Y, σ) be an esns and T : X → Y be a linear operator. The following assertions are equivalent: (i) T is τ -σ-continuous. (ii) T maps τ -bounded sets into σ-bounded sets. (iii) For each sequence (xn ) ⊂ X τ -converging to 0X , the set {T (xn ) : n ∈ N} is σ-bounded. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are direct, and therefore we only need to prove (iii) ⇒ (i). Reasoning by contradiction, suppose that (iii) holds but T is not continuous. Since (X, τ ) is metrizable, the continuity of T is characterized by sequences and therefore, by Proposition 3.28, there exists a sequence (xn ) ⊆ X converging to 0X such that T (xn ) → 0Y . Without loss of generality, we may assume that there exists q ∈ S(Y, σ) such that for all n ∈ N, q(T (xn )) > 1.

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Since A := {T (xn ) : n ∈ N} is bounded, by Definition 3.27 there exist a finite set {yi }ki=1 ⊆ Y and a finite set {αi }ki=1 ⊆ R with αi > 0 for each i ∈ {1, . . . , k} such that A⊆

n 

B(Y,q) (yi , αi ),

i=1

where, for y ∈ Y and α > 0, the set B(Y,q) (y, α) is given as in equation (7). We have that at least for one i ∈ {1, . . . , k}, the set of indexes {n ∈ N : T (xn ) ∈ B(Y,q) (yi , αi )} is infinite. Thus, up to subsequences, we may assume that A ⊆ B(Y,q) (y, α) for some y ∈ Y and some α > 0. q Suppose first that q(y) = +∞. Then, for each n ∈ N, q(T (xn )) = +∞, but A ⊆ y + Xfin . Therefore, if we q 1 consider the new sequence (˜ xn ) ⊆ X given by x ˜n = n xn we will get that for each n ∈ N, T (˜ xn ) ∈ n1 y + Xfin and so, the set {T (˜ xn ) : n ∈ N} is unbounded (since y = 0Y ). This latter statement cannot hold, since x ˜n → 0. We conclude then that q(y) < +∞, and therefore there exists β > 0 such that A ⊆ B(Y,q) (0Y , β). Now, let P = {ρk : k ∈ N} be a countable family of extended seminorms over X such that τ = T(P). Since xn → 0X , we can build a subsequence (xnj )j∈N such that for each k ∈ N ρk (xnj ) <

1 , j2

∀j ≥ k.

Therefore, the new sequence (˜ xj ) ⊆ N given by x ˜j = jxnj is still convergent to 0X . But, q(T (˜ xj )) = jq(T (xnj )) > j, and so {T (˜ xj ) : j ∈ N} is unbounded, leading to a contradiction and finishing the proof. 2 Lemma 4.17. Let (X, τ ) and (Y, σ) be two Hausdorff countable esns with (X, τ ) complete, and let T : X → Y be a continuous map satisfying ∀r > 0, ∃p(r) > 0, T (BX (0X , r)) ⊇ BY (0Y , p(r)). Then, for all ε > 0, T (BX (0X , r + ε)) ⊇ BY (0Y , p(r)). Proof. The proof follows exactly as the Lemma previous to [15, Ch. III, Theorem 2.1]. 2 Theorem 4.18 (Open mapping theorem). Let (X, τ ) and (Y, σ) be two complete Hausdorff countable esns, and let T : X → Y be a continuous linear operator such that p q ∀p ∈ S(X, τ ), ∃q ∈ S(Y, σ), T (Xfin ) = Yfin .

Then T maps τ -open sets of X to σ-open sets of Y . Proof. Fix first r > 0 and let V = BX (0X , 2r ). We have that V is absolutely convex and that the Minkowski ρV q ρV functional ρV ∈ S(X, τ ). Therefore, there exists q ∈ S(Y, σ) such that T (Xfin ) = Yfin . Also, since Xfin =  q nV , we have that, by the linearity of T and the σ-closedness of Y , fin n∈N q Yfin =



nT (V ) =

n∈N



nT (V ).

n∈N

q Since Yfin is a closed metric space of a complete metric space, we have that it is a Baire space, and so there q exists n ∈ N such that nT (V ) has nonempty σ Y q -interior. Thus, since Yfin is also σ-open, we have that fin

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nT (V ) has nonempty σ-interior. Since ϕn (as given in equation (1)) is a topological isomorphism, we have that T (V ) has nonempty σ-interior and therefore, there exists p(r/2) > 0 such that T (V ) ⊇ BY (0Y , p(r/2)). We can apply Lemma 4.17 for ε = 2r , concluding that T (BX (0X , r)) ⊇ BY (0Y , p(r/2)), which finishes the proof. 2 Remark 8. In Theorem 4.18, the condition p q ∀p ∈ S(X, τ ), ∃q ∈ S(Y, σ), T (Xfin ) = Yfin , p is also necessary. Indeed, if T : X → Y is open, then for any p ∈ S(X, τ ), the set T (Xfin ) is a σ-open subspace of Y . Therefore, the map q : Y → [0, +∞] given by

 q(y) =

0

p ) if y ∈ T (Xfin

+∞

otherwise,

q p is a σ-continuous extended seminorm with Yfin = T (Xfin ).

Corollary 4.19 (Closed graph theorem). Let (X, τ ) and (Y, σ) be two Hausdorff complete countable esns and u : X → Y be a linear operator. Then, u is τ -σ-continuous if and only if q p (i) for each q ∈ S(Y, σ), there exists p ∈ S(X, τ ) such that u−1 (Yfin ) = Xfin ; and (ii) the graph of u

gph(u) = {(x, u(x)) : x ∈ X} is (τ × σ)-closed in X × Y . Proof. The necessity is direct. For the sufficiency, let us suppose conditions (i) and (ii) hold. It is not hard to realize that (X × Y, τ × σ) is a Hausdorff complete countable esns and that G = gph(u) is a closed subspace. Therefore, (G, τ × σ) is also a Hausdorff complete countable esns. Let us consider now the map T : G → X given by T (x, u(x)) = x. Clearly T is bijective and, since it is the restriction of the parallel projection PX,Y to G, T is also continuous. Now, let us consider an extended seminorm ρ ∈ S(G, τ × σ). The finite space Gρfin is open, and therefore there exists a (τ × σ)-open set V = VX × VY ⊆ X × Y such that Gρfin = V ∩ G. Moreover, since Gρfin is a vector space, we have that span(V ) ∩ G = Gρfin . Let us write span(V ) = MX × MY , where MX = span(VX ) and MY = span(VY ). We can compute T (Gρfin ) = MX ∩ u−1 (MY ). p1 Since MX and MY are open subspaces, there exist p1 ∈ S(X, τ ) and q ∈ S(Y, σ) such that MX = Xfin and q MY = Yfin : For example, we can define p1 (resp. q) as p1 (x) = 0 (resp. q(y) = 0) if x ∈ MX (resp. y ∈ MY )

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and p1 (x) = +∞ (resp. q(y) = +∞) otherwise. Therefore, by condition (ii), there exists p2 ∈ S(X, τ ) such p2 that u−1 (MY ) = Xfin . Finally, we get that max(p1 ,p2 )

T (Gρfin ) = Xfin

.

Now, by Theorem 4.18, we get that T is also (τ ×σ)-τ -open, which implies that the map T −1 : x → (x, u(x)) is τ -(τ × σ)-continuous. Noting that u = PY,X ◦ T −1 , the proof is complete. 2 4.3. Splittable esns The main structural principles of fundamental etvs, etvs and esns have been already established. We will end this work with what is perhaps the main structural remaining question: Can an etvs or an esns be split into two τ -supplement spaces where one of them is the finite space? The problem of splitting has been largely studied in the context of normed spaces and also has been treated for general topological groups (see, e.g., [12]). Example 3.17 shows that in general etvs, such a decomposition cannot be always performed. Nevertheless, it is still an open question what happens in the case of esns. Theorem 4.10 gives us the first partial answer: Whenever we are able to write X = Xfin ⊕τ N , we also can perform extensions of linear functionals from (Xfin )∗ to X ∗ , and so the extended seminormed spaces are the framework to search for an answer. Definition 4.20. An esns (X, τ ) is said to be a splittable esns, if there exists a subspace N of X such that X = Xfin ⊕τ N. Of course, every fundamental elcs is a splittable esns, since each algebraic complement of the finite space is also a topological supplement. Unfortunately, this situation doesn’t hold in esns. Moreover, Proposition 4.21 shows that, in the context of countable esns, we always can find an algebraic complement of the finite space which fails to be a topological supplement. Proposition 4.21. Let (X, τ ) be a countable esns of infinite dimension such that Xfin  {0X }. Then, one of the following holds: (i) (X, τ ) is a fundamental elcs. (ii) There exists a subspace N of X such that Xfin and N are algebraic complements in X but not τ -supplements. Proof. Let P = {ρn : n ∈ N} be a nondecreasing sequence of extended seminorms on X generating the topology τ . If we assume that (i) doesn’t hold, then we can find a subsequence P1 = {ρnk : k ∈ N} such that n

nk Xfin  Xfink+1 , ∀k ∈ N.

Note that, since the sequence P is nondecreasing, we have that T(P) = T(P1 ). Therefore, without loss of generality, we may assume that P = P1 . Now, we can construct inductively a sequence of subspaces (Nn )n∈N 0 such that, denoting Xfin = X, we have that n−1 n ∀n ∈ N, Xfin = Xfin ⊕τn Nn ,

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where τn = T(ρn ). For each n ∈ N we can select xn ∈ Nn+1 \ {0X } such that ρn (xn ) < n1 . Therefore, since P is nondecreasing, we have that xn → 0X . Also, we have that the set {xn : n ∈ N} is linearly independent by construction. Now, since Xfin = {0X }, there exists xf ∈ Xfin \{0X } and we can consider the set H = {xn +xf : n ∈ N} which is still linearly independent. Since H ∩ Xfin = ∅, there exists a Hamel basis H of X generating Xfin and with H ⊆ H. Finally, if we fix N = CH (Xfin ), we have that N cannot be a τ -supplement with Xfin , since it is not closed: by construction, the point xf belongs to N \ N . 2 Proposition 4.22. Let (X, τ ) be an esns such that (Xfin , τ ) is Hausdorff and dim[Xfin ] < ∞. Then, there exists a subspace N of X such that X = Xfin ⊕τ N . Proof. Let dim[Xfin ] = n and {bj }nj=1 be a Hamel basis of Xfin . Since (Xfin , τ ) is Hausdorff, for each j ∈ {1, . . . , n} there exists ρj ∈ S(X, τ ) such that ρj (bj ) = 1. Define ρ = max{ρ1 , . . . , ρn }. We then have that ρ is a norm on Xfin and therefore Xfin is τρ -closed. ρ , τρ ) is a locally convex space and Xfin is a τρ -closed subspace of finite dimension, we have Since (Xfin ρ ρ that there exists a subspace N0 of Xfin such that Xfin = Xfin ⊕τρ N0 . Then, since (X, τρ ) is an elcs, there ρ exists a subspace N1 of X such that X = Xfin ⊕τρ N1 . If we denote N = N0 ⊕ N1 , we get that X = Xfin ⊕τp N.   Now, since Xfin is of finite dimension and ρ is a norm on Xfin , we can conclude that τρ X = τ X , fin fin according to the fact that there exists a unique Hausdorff topology compatible with the vector structure over a finite dimensional vector space. Then, since the parallel projection PXfin ,N is τρ -τρ -continuous, we get that it is also τ -τ -continuous, and the conclusion follows from Corollary 2.8. 2 5. Concluding remarks Clearly, this work opens a new field for researchers in functional analysis and topology. There are still many questions to work through and we hope this to become a rich field in the future. There are some topics we are able to propose for further development: 1. There is still a necessity to study a suitable theory of duality. An interesting way to do this is, for an esns (X, τ ), to study the “finest locally convex topology that is still coarser than τ ” and apply classical duality theory to topologize the dual space. 2. In the context of extended normed and seminormed spaces, the subdifferential calculus and the optimization theory related to it remains undeveloped. Along this line, it is important to study Hahn–Banach-like separation theorems in this framework. 3. Further relations between the Bornology theory and the extended topologies presented in this work would be a very interesting research. Acknowledgements We would like to thank firstly Gerald Beer, for proposing this topic of research to us and for explaining the link between the rich research concerning to Bornologies. Also, we thank Lionel Thibault for the conversations and many of the corrections that made this work what it is now. Finally, we thank the CMM of the University of Chili for the financial support of the visit of the second author to Montpellier.

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