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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}.

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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

,

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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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