Lineability criteria, with applications

arXiv:1309.3656v1 [math.FA] 14 Sep 2013 Lineability criteria, with applications Luis Bernal-Gonz´alez and Manuel Ord´o˜nez Cabrera Departamento de An´alisis Matem´atico. Facultad de Matem´aticas. Apdo. 1160. Avda. Reina Mercedes, 41080 Sevilla, Spain. E-mails: [email protected], [email protected] Abstract Lineability is a property enjoyed by some subsets within a vector space X. A subset Aof Xis called lineable whenever Acontains, except for zero, an infinite dimensional vector subspace. If, additionally, Xis endowed with richer structures, then the more stringent notions of dense-lineability, maximal dense-lineability and spaceability arise naturally. In this paper, several lineability criteria are provided and applied to specific topological vector spaces, mainly function spaces. Sometimes, such criteria furnish unified proofs of a number of scattered results in the related literature. Families of strict-order integrable functions, hypercyclic vectors, non-extendable holomorphic mappings, Riemann non-Lebesgue integrable functions, sequences not satisfying the Lebesgue dominated convergence theorem, nowhere analytic functions, bounded variation functions, entire functions with fast growth and Peano curves, among others, are analyzed from the point of view of lineability. 2010 Mathematics Subject Classification: 15A03, 26A46, 28A25, 30B40, 46E10, 46E30, 47A16. Key words and phrases: Lineability, maximal dense-lineability, spaceability, strict-order integrability, hypercyclicity, non-continuable holomorphic functions, fast growth entire functions, Peano curves. 1 Introduction In the last two decades there has been a crescent interest in the search of nice algebraic-topological structures within sets (mainly sets of functions or sequences) that do not enjoy themselves such structures. This paper wants to contribute to shed light on this recent trend, by providing a number of general criteria that guarantee the existence of the mentioned structures, 1 with emphasis in maximal dense-lineability and spaceability. Definitions are given below. For a recent survey on lineability, see [31]. To this respect, let us recall some recent terminology introduced in [7], [12], [24] and [61]. Assume that Xis a vector space (over K:= the real line Ror the complex plane C) and that αis a cardinal number. Then a subset Aof Xis called •lineable if A∪ {0}contains an infinite dimensional vector subspace, •α-lineable if A∪{0}contains an α-dimensional vector subspace (hence lineable means ℵ0-lineable, where ℵ0= card (N) and Nstands for the set of positive integers), •maximal lineable if Ais dim (X)-lineable. If, in addition, Xis a topological vector space, then we say that Ais •dense-lineable or algebraically generic whenever A∪ {0}contains a dense vector subspace of X, •maximal dense-lineable whenever A∪ {0}contains a dense vector subspace Mof Xwith dim (M) = dim (X), •spaceable if A∪ {0}contains some infinite dimensional closed vector subspace. Other interesting properties –such as algebrability, introduced in [8], additivity, introduced in [78, 79] (see also [52]), and moduleability [55]– will not be considered here. Note that if Xis an infinite dimensional separable Baire topological vector space then c, the cardinality of the continuum, is the maximal dimension allowed to any vector subspace of X. In particular, spaceability implies maximal lineability in this case. In the subsequent sections of this paper, a number of sufficient conditions for maximal dense-lineability and spaceability will be stated, see Sections 2- 3. The results that are obtained turn to be improvements of known criteria. Finally, in Section 4, our results will be applied to obtain lineability statements, mainly in the setting of function spaces. It is also shown how a number of known assertions about lineability can be proved by using our theorems. 2 2 Maximal dense-lineability Many examples of nonlinear sets containing large vector spaces have been given in the literature. Perhaps one of the most outstanding is the Herrero- Bourdon theorem (see [43, 62]) asserting that the set HC(T) of hypercyclic vectors of a (continuous, linear) operator T:X→Xon a complex Banach space Xis dense-lineable (moreover, the dense subspace obtained is T-invariant; the result was extended by B`es [32] and Wengenroth [87] to any real or complex topological vector space). Recall that an operator T:X→Xis said to be hypercyclic whenever it admits a dense orbit, that is, whenever there is a vector x0∈X(called hypercyclic for T) such that the set {Tnx0:n∈N}is dense in X(see [15] and [60] for excellent surveys on this subject). Another nice example was established by Aron, Garc´ıa and Maestre [5] in 2001. Namely, if G⊂Cis a domain (i.e. Gis nonempty, open and connected) and H(G) is the space of holomorphic functions in G(endowed with the compact-open topology) then Mittag-Leffler discovered in 1884 that the subfamily He(G) of functions which are holomorphic exactly at G–that is, which are non-extendable holomorphically across the boundary ∂G of G– is nonempty. The authors of [5] showed that He(G) is both dense-lineable and spaceable in H(G) (in fact, the result is given in [5] for domains of holomorphy in CN). We will go back on these subjects later. By adopting a wider point of view, one might believe that large topological size always entails algebraic genericity (for instance, HC(T) is residual if Tis hypercyclic on an F-space Xand, as proved by Kierst and Szpilrajn [69] in 1933, He(G) is residual in H(G)). This is far from being true. As an example, let N0:= N∪ {0}and α= (ak)∈CN0be a sequence with lim supk→∞ |ak|1/k <+∞, and define the associated diagonal operator ∆α as ∆α:P∞ k=0 fkzk∈H(C)7→ P∞ k=0 akfkzk∈H(C). If {αn= (ak,n)k≥0: n∈N}is dense in CN0then the set A:= {f∈H(C) : (∆αnf)n≥1is dense in H(C)}is residual in H(C), but Ais not even 2-lineable [29]. In 2005, Bayart [12] gave several useful dense-lineability criteria, but focused on divergence and universality of operators. With the aim to include more general situations, Aron et al. [6] and the first author [22, 24] proved respectively the following theorems. According to [6], if Aand Bare subsets of a vector space X, then Ais said to be stronger than Bprovided that A+B⊂A. Theorem 2.1. Assume that Xis a metrizable separable topological vector space. If Aand Bare subsets of Xsuch that Ais lineable, Bis dense lineable and Ais stronger than B, then Ais dense-lineable. 3 Theorem 2.2. Assume that Xis a metrizable separable topological vector space. Suppose that Γis a family of vector subspaces of Xsuch that TS∈ΓS is dense in X. We have: (a) If αis an infinite cardinal number such that TS∈Γ(E\S)is α-lineable then it contains, except for zero, a dense vector subspace of dimension α. (b) In particular, if TS∈Γ(E\S)is lineable then it is dense-lineable. And if TS∈Γ(E\S)is maximal lineable then it is maximal dense-lineable. The idea which is in the core of both results above is to obtain the desired dense subspace by adding small vectors coming from a known lineable set to the vectors of a dense subset. Theorems 2.1 and 2.2 have been used in [6, 22, 24, 26] to show the following assertions (each space Cp[0,1], C∞[0,1], Lp, H(D) is endowed with its natural topology): •The set ND[0,1] of continuous nowhere differentiable functions on [0,1] as well as the set DNM[0,1] of differentiable nowhere monotone functions on [0,1] are dense in C[0,1] [6]. •Let p∈N0. Then the class of functions f∈Cp[0,1] such that f(p)is nowhere differentiable on [0,1] is dense-lineable in Cp[0,1] [6, 22]. •The set of C∞-functions on [0,1] which are nowhere analytic is denselineable in C∞[0,1] [6, 22]. •Let (Ω,M, µ) be a measure space and p∈[1,∞) such that the Lebesgue space Lp:= Lp(µ, Ω) is separable. Denote Lp r-strict := Lp\Sq∈(p,∞]Lq (p≥1), Lp l-strict := Lp\Sq∈[1,p)Lq(p > 1), and Lp strict := Lp\ Sq∈[1,∞]\{p}Lq(p > 1). We have: ◮Lp r-strict is maximal dense-lineable if inf{µ(S) : S∈ M, µ(S)>0}= 0. [α] ◮Lp l-strict is maximal dense-lineable if sup{µ(S) : S∈ M, µ(S)<∞} =∞. [β] ◮Lp strict is maximal dense-lineable if both [α] and [β] hold. (In fact, conditions [α], [β], [α] + [β] are respectively necessary in the just mentioned assertions, because they are respectively equivalent to the non-vacuousness of Lp l-strict, Lp l-strict, Lp strict, thanks to a result by Romero [81] and Subramanian [84]; see also [80, Section 14.8] and [24, proof of Theorem 3.4].) In particular, for the Lebesgue measure on [0,1] we obtain for all p > 1 that Lp[0,1] \Sq∈[1,p)is maximal denselineable [24] (see also [6] and [77]). 4 •Let D:= {z∈C:|z|<1}, the open unit disc. The set of functions f∈H(D) that are strongly annular (i.e. lim supr→1min{|f(z)|:|z|= r}= +∞) is maximal dense-lineable [26]. The approach of Theorems 2.1–2.2 can be used to discover (maximal) dense-lineability in more (already known or new) cases. In order to undertake the task in a more systematic way, we are going to strengthen the above theorems. Note that in the following Theorem 2.3 and Corollary 2.4 neither metrizability nor separability are needed as a general assumption. On the contrary, we need disjointness of the subsets A, B in order to estimate the dimension of the subspaces obtained. Theorem 2.3. Assume that Xis a topological vector space. Let A⊂X. Suppose that there exists a subset B⊂Xsuch that Ais stronger than B and Bis dense-lineable. We have: (a) If Ais α-lineable and Xhas an open basis Bfor its topology such that card(B)≤α, then Ais dense-lineable. If, in addition, A∩B=∅, then A∪ {0}contains a dense vector space Dwith dim(D) = α. (b) If Xis metrizable and separable and αis an infinite cardinal number such that Ais α-lineable, and A∩B=∅, then A∪ {0}contains a dense vector space Dwith dim(D) = α. (c) If the origin possesses a fundamental system Uof neighborhoods with card(U)≤dim(X),Ais maximal lineable and A∩B=∅, then Ais maximal dense-lineable. In particular, the same conclusion follows if Xis metrizable, Ais maximal lineable and A∩B=∅. Proof. Observe that (b) is derived from (a) because if Xis metrizable and separable then it is second countable, hence it has a countable open basis Bfor its topology. Therefore card(B) = card(N) = ℵ0≤αbecause αis infinite, and (a) applies. Let us show that (c) is also a consequence of (a). For this, assume that A, B and Uare as in the hypothesis of (c). Let Cdenote a dense countable subset of K, and let {ui}i∈Ian algebraic basis of X, so that card(I) = dim(X). Denote by Pf(I) the family of nonempty finite subsets of I. Since card(U)≤dim(X), we must have that dim(X) is not finite, hence card(Pf(I)) = card(I) = dim(X)≥card(C). Moreover, card(IF) = card(I) for any nonempty finite set F, and card(C×I) = card(I). Now, it is easy to see that the family B:= (U+X i∈F αiui:U∈ U, αi∈Cfor all i∈F, F ∈ Pf(I)) 5 is an open basis for the topology of X. We have that card(B)≤cardU × [ F∈Pf(I) (C×I)F= cardU × [ F∈Pf(I) IF ≤card(U×Pf(I)×I) = card(U×I×I) = max{card(U),card(I)}= dim(X). Since Ais dim(X)-lineable, by applying (a) again we obtain the first part of (c). As for the second part, simply observe that if Xis metrizable then Ucan be chosen countable, so card(U)≤dim(X) if dim(X) is infinite. If dim(X) is finite then the conclusion is evident because A∪{0}=X; indeed, every vector subspace Mof a finite dimensional vector space Xsuch that dim(M) = dim(X) must equal X. Thus, our only task is to prove (a). Suppose that Ais α-lineable and that card(B)≤αfor some open basis Bof X. We are also assuming that A+B⊂Aand Bis dense-lineable. It follows that there exist vector spaces A1, B1such that A1⊂A∪ {0},B1⊂B∪ {0},B1is dense in Xand dim(A1) = α≥card(B). Hence there are sets I, J, vectors ai(i∈I) and open sets Uj(j∈J), such that card(I) = α,{ai}i∈Iis a linearly independent system contained in A1,B={Uj}j∈Jand there exists a surjective mapping ϕ:I→J. By density, we can assign to each j∈Ja vector bj∈Uj∩B1. Fix j∈J. As Uj−bjis a neighborhood of 0 and multiplication by scalars is continuous on X, for each i∈ϕ−1({j}) there is εi>0 satisfying εiai∈ Uj−bj, or εiai+bj∈Uj. Define D:= span {εiai+bϕ(i):i∈I}. Then Dis a vector subspace of X. Since ϕis surjective, we can pick for each j∈Jand index i(j)∈Iwith ϕ(i(j)) = j. As {Uj}j∈Jis an open basis and vi(j)ai(j)+bj∈Uj(j∈J), these vectors form a dense subset of X. But Dcontains these vectors, so Dis also dense. Furthermore, if x∈Dthen there are p∈N, (λ1,...,λp)∈Kp\ {(0,...,0)}and i1,...,ip∈Iwith x=λ1εi1ai1+···+λpεipaip+λ1bϕ(i1)+···+λpbϕ(ip). Let define u:= λ1εi1ai1+···+λpεipaipand y:= λ1bϕ(i1)+···+λpbϕ(ip). Then y∈B1⊂B∪{0}, and u∈A1\{0}because of the linear independence of the ai’s. Hence u∈Aand x=u+y∈A+ (B∪ {0})⊂A∪A=A. Consequently, D\ {0} ⊂ Aand Ais dense-lineable. Finally, we suppose further that A∩B=∅. We want to prove that dim(D) = αor, that is the same, the vectors xi:= εiai+bϕ(i)(i∈I) are 6 linearly independent. With this aim, consider a p∈Nand two p-tuples (λ1,...,λp)∈Kpand (i1,...,ip)∈Ipsuch that Pp j=1 λjxij= 0. Assume, by way of contradiction, that (λ1,...,λp)6= (0,...,0). Then u+y= 0, where uand yare as in the preceding paragraph. Hence y∈A(because y=−u∈A1\ {0} ⊂ A) and y∈B(because y=−u6= 0, so y∈B1\ {0} ⊂ B), which implies A∩B6=∅. This contradicts the assumption A∩B=∅, and we are done. Corollary 2.4. Let Xbe a topological vector space. Suppose that Γis a family of vector subspaces of Xsuch that TS∈ΓSis dense in X. We have: (a) If TS∈Γ(X\S)is α-lineable and Xhas an open basis Bfor its topology such that card(B)≤αthen TS∈Γ(X\S)is dense-lineable and, moreover, it contains a dense vector space Dwith dim(D) = α. (b) If Xis metrizable and separable and αis an infinite cardinal number such that TS∈Γ(X\S)is α-lineable, then TS∈Γ(X\S)contains, except for zero, a dense vector space Dwith dim(D) = α. (c) If the origin possesses a fundamental system Uof neighborhoods with card(U)≤dim(X)then TS∈Γ(X\S)is maximal dense-lineable. The same conclusion holds if Xis metrizable and TS∈Γ(X\S)is maximal lineable. Proof. In order to apply Theorem 2.3, it is enough to check that A:= TS∈Γ(X\S) is stronger than B:= TS∈ΓS, that Bis dense-lineable and that A∩B=∅. The last property is obvious, whereas the dense-lineability of Bis trivial in view of its denseness and the fact that Bis itself a vector space. As for the property A+B⊂A, consider x∈A,y∈Band z:= x+y. If z /∈Athen there exists S∈Γ with z∈S. Then x=z+ (−y)⊂S−B⊂S−S=S as Sis a vector subspace. Thus x /∈A, a contradiction, which concludes the proof. Note that the same technique shows that the first part of Theorem 2.2(b) is a special instance of Theorem 2.1. We want to conclude this section by establishing a simple characterization of dense-lineability for the complement of a subspace. Theorem 2.5. Let Xbe a metrizable separable topological vector space and Ybe a vector subspace of X. If X\Yis lineable then X\Yis denselineable. Consequently, both properties of lineability and dense-lineability for X\Yare equivalent provided that Xhas infinite dimension. 7 Proof. It is evident that X\Yis lineable if and only Yhas infinite algebraic codimension. The assumptions imply that Xhas a countable open basis {Gn:n≥1}. Assume that X\Yis lineable. In particular, Y(X. Then Y0=∅, hence X\Yis dense. Therefore there is x1∈G1\Y. Since codim(Y) = ∞, we have span(Y∪ {x1})(X. Then (span(Y∪{x1}))0=∅. It follows that there exists x2∈G2\span(Y∪ {x1}). With this procedure, we get recursively a sequence of vectors {xn}n≥1satisfying xn∈Gn\span(Y∪ {x1, ..., xn−1}) (n≥1). In particular, the set {xn:n≥1}is dense. Now, if we define M:= span{xn:n≥1}then Mis a dense vector space and M\ {0} ⊂ X\Y. Plainly, the scope of this result is shorter than that of the remaining criteria of this paper, so its use yields weaker assertions on lineability in the diverse examples given in Section 4. Yet Theorem 2.5 is easy to apply. For instance, the set A:= {f∈C(R) : fis unbounded}is dense-lineable because A=X\Ywith X=C(R), Y={bounded continuous functions R→R},Yis a vector subspace and Ais lineable; indeed, Acontains the vector space of all non-zero polynomials Pwith P(0) = 0. Remark 2.6. A partial complement of Theorem 2.5 is possible in the nonseparable case. Namely, by assuming the Continuum Hypothesis, we have: Let Xbe a non-separable F-space and Ybe a closed separable vector subspace of X. Then X\Yis maximal lineable. Indeed, let Zbe a vector space that is an algebraic complement of Y, so that Z\ {0} ⊂ X\Y. Note that dim (Y)≤c≤dim (X) = dim (Y) + dim (Z). If dim (Z)≤ ℵ0then Z, and so X(= Y+Z), would be separable (a contradiction). Hence dim (Z)≥c, which implies dim (Z) = dim (X), and we are done. 3 Spaceability Up to date, there not exist many explicit general criteria of existence of large closed subspaces within a subset of a topological vector space. In fact, most spaceability proofs on specific settings have been done directly and constructively. One has to go back to Wilansky ([88], 1975) to find what maybe was the first general criterium. He proved that if Yis a closed vector subspace of a Banach space X, then X\Yis spaceable if and only if Yhas infinite codimension (compare to Theorem 2.5). An improved version of this result, where Xis allowed to be a Fr´echet space, is ascribed by Kitson and Timoney 8 [70, Theorem 2.2] to Kalton. The authors of [70] exploit it to obtain the following assertion (see [70, Theorem 3.3]). Theorem 3.1. Let Zn(n∈N)be Banach spaces and Xa Fr´echet space. Let Tn:Zn→Xbe continuous linear mappings and Ythe linear span of SnTn(Zn). If Yis not closed in Xthen the complement X\Yis spaceable. Among other applications, the last result is used in [70] to show spaceability of the set of non-absolutely convergent power series in the disk algebra A(D) and of the family of non-absolutely p-summing operators between certain pairs of Banach spaces. Recently, the authors of [30] in their Theorem 2.2 have stated a sufficient condition for spaceability on function Banach spaces. Then this theorem is applied to prove that conditions [α],[β],[α] + [β] given in Section 2 are respectively equivalent to the spaceability of Lp r-strict (if p≥1), Lp l-strict (if p > 1) and Lp strict (if p > 1) (hence equivalent to the respective nonvacuousness of these sets). It is also used to show that, if CBV [0,1] denotes the Banach space (under the norm kfk=|f(0)|+ Var (f)) of continuous functions [0,1] →Rwith bounded variation and AC[0,1] represents the subset of absolutely continuous functions, then set CBV [0,1] \AC[0,1] is spaceable in CBV [0,1]. Incidentally, Wilansky’s theorem provides us with a shorter proof of this fact: AC[0,1] is a closed subspace of CBV [0,1] (see e.g. [1]) and has infinite codimension, because CBV [0,1] = AC[0,1]⊕S[0,1], where S[0,1] stands for the subspace of continuous bounded variation singular (that is, with derivative 0 almost everywhere) functions. The main ingredient in the proof of [30, Theorem 2.2] is Nikolskii’s theorem of characterization of basic sequences. But Nikolskii’s theorem turns to be true in the setting of F-spaces (recall that an F-space is a complete metrizable topological vector space). Namely, if Xis an F-space and k · k is an F-norm defining the topology of X, then a sequence (xn)⊂X\ {0}is basic if and only if there is a constant α∈(0,+∞) such that, for every pair r, s ∈Nwith s≥rand every finite sequence of scalars a1,...,as, one has   r X n=1 anxn ≤α  s X n=1 anxn  (see [68, Theorem 5.1.8, p. 67]). Recall that an F-norm on a vector space X is a functional k·k :X→[0,+∞) satisfying, for all x, y ∈Xand λ∈K, the following properties: kx+yk ≤ kxk+kyk;kλxk ≤ kxkif |λ| ≤ 1; kλxk → 0 if λ→0; kxk= 0 only if x= 0. Then we can establish the following theorem, that is an improvement of Theorem 2.2 in [30]. By P(Ω) we represent, as usual, the family of subsets 9 (b) The set L∞ l-strict is spaceable if and only if [β]holds. The Banach version of Theorem 3.2 given in [30] has been recently used by Akbarbaglu and Maghsoudi [2] to discover spaceability in certain related subsets of Orlicz spaces. In their paper [58] (see also [57]) Glab, Kaufmann and Pellegrini proved, among other results, the following, which improves [24, Theorem 4.1]. Theorem 4.5. Assume that µis an atomless, outer regular, positive Borel measure on Ωwith full support, where Ωis a topological space admitting a countable family (Un)of nonvoid open subsets such that every nonvoid open subset Aof Ωcontains some Uj. Let p∈(0,∞)and consider the set Sp:= {f∈Lp:fis nowhere Lqfor each q∈(p, ∞]}. We have: (a) The set Spcontains, except for zero, an ℓp-isometric subspace of Lp. In particular, Spis spaceable. (b) The set Spis maximal dense-lineable. We recall that fis nowhere Lqmeans that, given a nonvoid open subset Uof Ω, the restriction f|Udoes not belong to Lq(U). We notice that, once (a) is achieved, the proof of (b) given in [58] can be considerably shortened by using Theorem 2.3(b): take X=Lp,α=c,A=Spand B={the step functions}. Remark 4.6. In view of the last argument (and others along this paper) one might believe that maximal dense-lineabiity can only happen when there is spaceability. This is far from being true. For instance, for X=c0or ℓp (1 ≤p≤+∞) Cariello and Seoane [45] have recently proved that the subset Z(X) := {x= (xn)∈X:xn= 0 only for finitely many n∈N} is c-lineable (so maximal lineable) but not spaceable. Now, if we take A= Z(X) and B=c00 = the space of sequences with only finitely many nonzero entries, then A+B⊂A, and Theorem 2.3 yields that Z(X)is maximal dense-lineable. 4.2 Spaces of continuous and differentiable functions. Theorem 2.3 can also be applied to reinforce other statements given after Theorem 2.2. Recall that a C∞-function f: [0,1] →Kis said to have a Pringsheim singularity at a point x0∈[0,1] whenever the radius of convergence of the Taylor series of fat x0is zero. Obviously, in such a case, fis not analytic at x0; but the converse is false. 16 Theorem 4.7. Consider the spaces Cp[0,1] (p∈N0∪ {∞})endowed with their natural topologies. We have: (a) Let p∈N0. The set A1:= {f∈Cp[0,1] : f(p)is differentiable at no point of [0,1]}is maximal dense-lineable in Cp[0,1]. (b) The set A2:= {f∈C∞[0,1] : fis analytic at no point of [0,1]}is maximal dense-lineable in C∞[0,1]. (c) If K=C, the set A3:= {f∈C∞[0,1] : fhas a Pringsheim singularity at every point of [0,1]}is maximal dense-lineable in C∞[0,1]. Proof. Since the set B:= {polynomials}is dense in Cp[0,1] for all p∈ N0∪ {∞}, it is enough, according to Theorem 2.3(b), to prove that A1, A2 and A3are c-lineable (because Cp[0,1] is metrizable, separable and complete, and c= dim(Cp[0,1]) for all p∈N0∪ {∞}). It is plain that Ai∩B=∅for i= 1,2,3. (a) From the spaceability of the set of nowhere differentiable functions in C[0,1] (obtained by Fonf et al. in [49]) it follows the c-lineability of A1 in the case p= 0 (alternatively, see a constructive proof in [66]). Let C be a c-dimensional vector subspace of C[0,1] consisting, except for zero, of nowhere differentiable functions on [0,1]. If p∈Nand we let ϕpdenote the unique antiderivative of order pof a continuous function ϕ: [0,1] →Rsuch that ϕ(k) p(0) = 0 (k∈ {0, ..., p −1}), then it is easy to see that {ϕp:ϕ∈C} is a c-dimensional vector space contained in A1∪ {0}. (b) The maximal dense-lineability of A2is in fact established explicitly in [23, Theorem 3.1], where the c-lineability of A2is part of the proof: if ϕis nowhere analytic and eα(x) := eαx (α > 0) then span{eαϕ:α > 0} ⊂ A2 and dim(span{eαϕ:α > 0}) = c(also Cater [46] had obtained in 1984 such ac-dimensional space). (c) The c-lineability of A3is stated in [23, Theorem 3.2] for K=C. Let us briefly turn our attention to divergent Fourier series. The existence of continuous functions f:T→C(T:= {z=eit :t∈[0,2π]}, the unit circle) whose Fourier series P∞ k=−∞ ˆ f(k)eikt diverges at some points is well known. Denote Sn(f, t) := Pn k=−nˆ f(k)eikt (n∈N), the partial Fourier sums. If E⊂T, let FE:= {f∈C(T) : {Sn(f, t)}n≥1is unbounded for each eit ∈E}. In 2005, Bayart [12, 13] proved that, given E⊂Twith Lebesgue measure zero, the set FEis dense-lineable and spaceable (Aron et al. showed in [8] 17 that FE∪ {0}contains, in fact, an infinitely generated dense algebra). If E⊂T, consider the smaller set FpE := {f∈C(T) : {(Sn(f, ·)|E}n≥1is dense in CE}, where CEis endowed with the topology of pointwise convergence. In 2010, M¨uller [75] proved that if Eis countable then FpE is residual in C(T), while the first author [25] demonstrated that FpE is spaceable and maximal denselineable. We remark that, once the spaceability is established, the maximal dense-lineability can be obtained from Theorem 2.3: just choose X=C(T), A=FpE and B={the trigonometric polynomials}. 4.3 Holomorphic functions regular up to the boundary. A similar result holds for the family of non-continuable boundary-regular holomorphic functions. Assume that Gis a domain in Cand consider the Fr´echet space H(G) (a Fr´echet space is a locally convex F-space). Recall that for f∈H(G) we have that f∈He(G) if and only if, for all z0∈G, the radius of convergence ρ(f, z0) of the Taylor series of fwith center z0 equals the euclidean distance d(z0, ∂G) between z0and the boundary ∂G of G. It was mentioned at the beginning of Section 2 that Kierst and Szpilrajn showed the residuality of He(G) (this result was extended by Kahane [67] in 2000 to certain subspaces of H(G)) and that in [5] the dense-lineability and the spaceability of He(G) were established (in [85] additional topological properties are found for the dense subspace within He(G), and in [20] spaces of holomorphic functions in Dare investigated). With more sophisticated methods –including the use of Arakelian’s approximation theorem– the first author (see [21]) was able to state the maximal dense-lineability of He(G) in H(G). Consider now the space A∞(G) of boundary-regular holomorphic functions in G, that is, f∈A∞(G) if and only if f∈H(G) and each derivative f(N)(N≥0) extends continuously on the closure Gof G. Then A∞(G) can be endowed with a natural topology, namely, the topology of uniform convergence of functions and all their derivatives on each compact subset of G. In 1980 Chmielowski [47] proved that if Gis regular (i.e. G0=G) then A∞(G)∩He(G) is nonempty, and finally Valdivia [86] showed in 2009 that A∞(G)∩He(G) is in fact dense-lineable in A∞(G). By assuming additional conditions on G(under which the authors of [27] had obtained dense-lineability in 2008), we are going to see that the last conclusion can be reinforced. We say that a domain G⊂Cis finite-length provided that 18 there is M∈(0,+∞) such that for any pair a, b ∈Gthere exists a curve γ⊂Gjoining ato bfor which length(γ)≤M. Theorem 4.8. If G⊂Cis a regular finite-length domain such that C\G is connected then A∞(G)∩He(G)is maximal dense-lineable in A∞(G). Proof. Firstly, let us prove that A∞(G)∩He(G) is maximal lineable. Since Gis regular, we can choose ϕ∈A∞(G)∩He(G). Consider the functions eα(z) := eαz (α > 0). The functions eαϕare linearly independent. Indeed, let (c1, ..., cN)∈CN\ {(0, ..., 0)}and different α1,...,αN∈(0,+∞) such that PN i=1 cieαiϕ= 0 on G. Without loss of generality, we can assume that N≥2, c16= 0 and α1> αiif i≥2. Since ϕ6= 0 and H(G) is an integrity domain, we have PN i=1 cieαi= 0 on G. By the Identity Principle, we have PN i=1 cieαiz= 0 for all z∈C. In particular, c1+c2e(α2−α1)x+···+cNe(αN−α1)x= 0 for all x > 0. Letting x→+∞, we get c1+0 = 0, a contradiction, which shows the desired linear independence. Now, set M:= span{eαϕ:α > 0} ⊂ A∞(G). Then dim(M) = c= dim(A∞(G)). If f∈M\{0}then there are (c1, ..., cN)∈ CN\{(0, ..., 0)}and different α1,...,αN∈(0,+∞) such that f=PN i=1 cieαiϕ. Suppose, by way of contradiction, that f /∈He(G). Let us denote by Sz0the sum of the Taylor series of fwith center at z0. Then there are a point a∈G and a number r > d(a, ∂G) such that Sa∈H(B(a, r)). Of course, Sa=fin B(a, |a−b|), where bis a point on ∂G such that |a−b|=d(a, ∂G). Therefore there are a point c∈∂G and a number ε > 0 with B(c, ε)⊂B(a, r) and PN i=1 cieαiz6= 0 for all z∈B(c, ε); indeed, B(a, r) is a neighborhood of b, the point bis not isolated in ∂G (by the regularity of G), and the set of zeros of PN i=1 cieαiin Cis discrete. Now take a point ζ∈B(c, ε/2) ∩G. Then B(ζ, ε/2) ⊂B(c, ε)⊂B(a, r) and PN i=1 cieαiz6= 0 for all ζ∈B(ζ, ε/2). The function Sζequals fin a neighborhood of ζ, whence Sζ/PN i=1 cieαiequals ϕin a neighborhood of ζ. We get from the non-extendability of ϕthat ε 2> d(ζ, c)≥d(ζ, ∂G) = ρ(ϕ, ζ) = ρSζ PN i=1 cieαi , ζ≥ε 2. This contradiction shows that f∈He(G), so M\ {0} ⊂ A∞(G)∩He(G) and the maximal lineability of the last set is guaranteed. According to [71, Proof of Theorem 4], under the assumptions on G (specifically, Gis finite-length and C\Gis connected) the set of of polynomials is dense in A∞(G). Now, it is sufficient to apply Theorem 2.3 with X=A∞(G), A=A∞(G)∩He(G) and B={polynomials}. 19 4.4 Sets of hypercyclic vectors. Our next application concerns hypercyclicity. The notion can be easily extended to sequences of operators, see [60]: given two (Hausdorff) topological vector spaces X, Y , a sequence (Tn)⊂L(X, Y ) := {continuous linear mappings X→Y}is said to be hypercyclic provided that there is a vector x0∈X(called hypercyclic for (Tn)) such that the orbit {Tnx0:n∈N}of x0under (Tn) is dense in Y. We denote HC((Tn)) = {x∈X:xis hypercyclic for (Tn)}. Note that if X=Yand T:X→Xis an operator (that is, T∈L(X) := L(X, X)), then Tis hypercyclic if and only if the sequence (Tn) of powers of Tis hypercyclic; moreover, HC(T) = HC((Tn)). Only separable infinite dimensional topological vector spaces can support hypercyclic operators, see [60]. At the beginning of Section 2 we mentioned the Herrero-Bourdon-B`es- Wengenroth theorem asserting the dense-lineability of HC(T). The first author [19] proved that HC(T) is maximal dense-lineable provided that Tis hypercyclic on a Banach space (again, the dense subspace obtained in [19] is T-invariant). As for sequences (Tn)⊂L(X, Y ), it was demonstrated in [18] that if Yis metrizable and each subsequence (Tnk) (with n1< n2<···) is hypercyclic then HC((Tn)) is lineable, and that if Xand Yare metrizable and separable and HC((Tnk)) is dense for each subsequence (Tnk) of (Tn) then HC((Tn)) is dense-lineable. In Theorem 4.9 below it can be seen how spaceability, when it happens, comes in our help to obtain maximality. But, prior to this, let us recall a recent, quantified version of hypercyclicity. According to Bayart and Grivaux [14], an operator Ton a topological vector space Xis said to be frequently hypercyclic provided there exists a vector x0∈Xsuch that lim inf n→∞ card{k∈ {1,2, ..., n}:Tnx0∈U} n>0 for every nonempty open subset Uof X. In this case, x0is called a frequently hypercyclic vector for T, and the set of these vectors will be denoted by F HC(T). The extension of the notion of frequent hypercyclicity to sequences (Tn)⊂L(X, Y ) is obvious: replace Tnby Tnin the display above, and fix Uamong the nonempty open subsets of Y. The corresponding set of frequent hypercyclic vectors in Xis denoted by FHC((Tn)). In [14] it is shown that if Tis a frequent hypercyclic operator on a separable F-space X then FHC(T) is dense-lineable (once more, the dense subspace obtained is T-invariant). 20 Theorem 4.9. (a) Let Xbe an infinite-dimensional separable F-space and Ybe a metrizable separable topological vector space. Assume that (Tn)⊂L(X, Y )and that there is a dense subset D⊂Xsuch that the (Tn)converges pointwise on D. If HC((Tn)) (resp. FHC((Tn))) is spaceable then it is maximal dense-lineable. (b) Let Xbe an F-space. Assume that T∈L(X). If HC(T)is spaceable and there is a sequence {n1< n2<· · ·} ⊂ Nsuch that (Tnk)converges pointwise on some dense subset of Xthen HC(T)is maximal denselineable. If FHC(T)is spaceable and (Tn)converges pointwise on some dense subset of Xthen FHC(T)is maximal dense-lineable. Proof. Part (b) follows from (a) by considering Tn:= Tnfor frequent hypercyclicity, and taking Tk:= Tnkfor mere hypercyclicity, together with the trivial fact HC(T)⊃HC((Tnk)). Observe also that if Tis hypercyclic then Xmust be separable and infinite-dimensional. Let us prove (a). Suppose first that HC((Tn)) is spaceable. Then there is a closed infinite dimensional vector space M⊂HC((Tn)) ∪ {0}. We have dim(X) = c= dim(M) due to Baire’s theorem. Therefore A:= HC((Tn)) is maximal lineable. Note that B:= {x∈X: (Tnx) converges}is a vector space, and it is dense because B⊃D. Hence Bis dense-lineable. Now, trivially, if a vector x0has dense orbit and y0∈Bthen {Tn(x0+y0) = Tnx0+Tny0}n≥1is also dense, so x0+y0∈A. In other words, A+B⊂A. Then HC((Tn)) is maximal dense-lineable by Theorem 2.3. Now, assume that FHC((Tn)) is spaceable. Take A:= FHC((Tn)) and Bas before. Again by Theorem 2.3, the only property to show is A+B⊂A. Fix x0∈Aand y0∈B. Given a nonempty open set U⊂Y, choose any u0∈Uand a neighborhood Vof 0 in Ysuch that V+V⊂U−u0. Then lim infn→∞ card{k∈ {1,2, ..., n}:Tnx0∈V+u0−z0}/n > 0, where z0:= limn→∞ Tny0. Since Tny0∈V+z0for nlarge enough, we obtain that Tn(x0+y0) = Tnx0+Tny0∈(V+u0−z0) + (V+z0)⊂Uwhenever Tnx0∈V+u0−z0and nis large enough. This yields lim infn→∞ card{k∈ {1,2, ..., n}:Tn(x0+y0)∈U}/n > 0, that is, x0+y0∈A. Now, we can establish a general existence result for Fr´echet spaces. Corollary 4.10. Let Xbe a separable infinite dimensional Fr´echet space. Then Xsupports an operator Tsuch that HC(T)is maximal dense-lineable. Proof. In 1998, Bonet and Peris [35] proved that if Xis as in the hypothesis then there exists T∈L(X) such that Tis hypercyclic. Recently, Menet [73, Theorem 2.4] has shown that Tcan be chosen such that HC(T) is spaceable. If Xis not isomorphic to ω:= KN, it is observed in [73, Proof 21 of Theorem 2.4] that the operator Tobtained there (which is based on the construction in [35]) satisfies that (Tn) converges pointwise on a dense set, so Theorem 4.9 applies. If Xis isomorphic to ω, let S:X→ωbe such an isomorphism. B`es and Conejero [34] demonstrated that for the backward shift B: (x1, x2, ...)∈ω7→ (x2, x3, ...)∈ωone has that HC(B) is spaceable. Trivially, Bn→0 poinwise on the dense subset D0:= {(xn)∈ω:xn6= 0 only for finitely many n}. It follows that the operator T:= S−1BS :X→X satisfies that HC(T) is spaceable and Tn→0 pointwise on the dense set S−1(D). A new application of Theorem 4.9 yields the conclusion. Of course, in order to apply Theorem 4.9, it is important to have to our disposal a number of results on spaceability of the set of hypercyclic/frequently hypercyclic vectors: the interest reader is referred to [15], [38] and [60]. As a first example, note that the maximal dense-lineability of the family FpE given in Subsection 4.2 may be obtained by using the mentioned theorem. Let us give examples of operators on non-Banach spaces whose sets of hypercyclic vectors are maximal dense-lineable. To start with, we consider the space H(CN) of entire functions CN→C, endowed with the compactopen topology. Recall that each a∈CNgenerates a translation operator τa: f∈H(CN)7→ f(·+a)∈H(CN). Also, if Ddenotes the derivative operator on H(C) (i.e. Df =f′), then every polynomial P(z) = a0+a1z+···+anzn generates a finite order differential operator P(D) := a0I+a1D+···+anDn, where Iis the identity operator. Proposition 4.11. Assume that T:H(CN)→H(CN)is an operator that commutes with translations, that is, Tτa=τaTfor all a∈CN. Assume also that Tis not a scalar multiple of the identity. We have: (a) The set HC(T)is maximal dense-lineable. (b) If N > 1then FHC(T)is maximal dense-lineable. (c) If N= 1 and Tis not a finite order differential operator then FHC(T) is maximal dense-lineable. Proof. In Corollary 2 of [38] it is shown that if N > 1 (if N= 1, resp.) then any non-scalar convolution operator (any non-scalar convolution operator that is not P(D) for any polynomial P, resp.) T∈L(H(CN)) satisfies that FHC(T) is spaceable. Since an operator is of convolution if and only if it commutes with translations (see e.g. [59]) and since HC(T)⊃FHC(T), we get spaceability for all sets in (a), (b), (c). Indeed, the only case to consider in order to complete this claim is the spaceability of HC(P(D)) whenever N= 1 and Pis a nonconstant polynomial. But this has been 22 recently proved by Menet [72]. According to Theorem 4.9, to conclude the proof it is enough to show that, for any operator Tas in the statement of the theorem, (Tn) converges pointwise on some dense subset of X:= H(CN). Bonilla and Grosse-Erdmann (see [37] and [38]) have proved that such a T satisfies the so-called Frequent Hypercyclicity Criterion, one of whose items is the existence of a dense subset D⊂Xsuch that Pn≥1Tnxconverges unconditionally for every x∈D. Clearly, this implies Tnx→0 for all x∈D, and we are done. A second example is provided by composition operators. Suppose that G⊂Cis a domain and that ϕ:G→Gis a holomorphic self-mapping. Then ϕgenerates the composition operator Cϕ:f∈H(G)7→ f◦ϕ∈H(G). B`es [33, Theorem 1] has proved that if ϕis one-to-one and has no fixed point in Gthen, for every nonconstant polynomial P, the set FHC(P(Cϕ)) is spaceable. Since in [33] it is shown that every such P(Cϕ) satisfies the Frequent Hypercyclic Criterion, we get P(Cϕ)n→0 on a dense set and, by Theorem 4.9, the set FHC(P(Cϕ)) is maximal dense-lineable. Finally, similar arguments allow us to assert (under appropriate conditions) maximal dense-lineability for HC(Bw), where Bw: (xn)∈X7→ (wnxn+1)∈Xis the backward shift with weight sequence (wn) acting on a K¨othe sequence space X=λp(A) or c0(A) (1 ≤p < ∞), where A= (aj,k)j,k≥1is a matrix such that aj,k >0 and aj,k ≤aj+1,k for any j, k ≥1 (see [72] for conditions guaranteeing spaceability of HC(Bw) in these spaces). 4.5 Functions of bounded variation. In Section 3 we considered the space CBV [0,1] of functions f: [0,1] →R which are continuous and of bounded variation, endowed with the norm kfk:= |f(0)|+ Var (f). Recall the decomposition CBV [0,1] = AC[0,1] ⊕ S[0,1]. Observe that the latter norm is strictly finer that the maximum norm. In fact, the Banach space (CBV [0,1],k · k) is nonseparable (see e.g. [1]; see also a nice proof in [10, Section 1]). Recall that a function f∈CBV [0,1] is said to be strongly singular whenever f∈S[0,1] (that is, f′= 0 a.e. on [0,1]) and fis nonconstant on any subinterval of [0,1]. In particular, every strongly singular function is not absolutely continuous in any subinterval of [0,1]. The set of these functions will be denoted by SS[0,1]. Recently, Jim´enez-Rodr´ıguez [64] has shown that c0is isometrically isomorphic to a subspace of continuous functions [0,1] →Rall of whose nonzero members are non-Lipschitz and have a.e. null derivative, so improving a result due to Jim´enez et al. [65] asserting the c-lineability of this family of 23 functions. In particular, this family is spaceable in C[0,1]. Notice that every member of SS[0,1] belongs to the described family. In [10], Balcerzak el al. have demonstrated the spaceability of SS[0,1] in CBV [0,1] (in fact, a nonseparable closed subspace is found in SS[0,1]∪{0}). This improves the result of spaceability of CBV [0,1] \AC[0,1], see Section 3. In [10, Theorem 10] it is proved an assertion containing the maximal dense-lineability of SS[0,1] in C[0,1]. In particular, the family Aof functions f∈CBV [0,1] being not absolutely continuous in any subinterval of [0,1] is maximal dense-lineable in C[0,1]. This result can also be deduced from the mentioned spaceability of SS[0,1] (which implies the spaceability of A, and so the c-lineability of A) together with the density of the set Bof polynomials in C[0,1]: just observe that A+B⊂ A and apply Theorem 2.3. Remark 4.12. In [22, Proposition 3.3] it is asserted the dense-lineability in CBV [0,1] of the family of functions f∈CBV [0,1] which are differentiable on no interval in [0,1], while in [30, Theorem 4.2] it is established the (stronger) assertion of maximal dense-lineability of Ain CBV [0,1]. Unfortunately, both proofs were based on the density of the set of polynomials in CBV [0,1], which is false. Consequently, the mentioned assertions are not proved (nor disproved, as far as we know) up to date. We apologize for this. 4.6 Riemann-integrable functions on unbounded intervals. Let I⊂Rbe an unbounded interval. Consider the Lebesgue space L1(I), the Banach space B(I) of all bounded functions I→R(endowed with the supremum norm), and the vector space R(I) of all Riemann-integrable functions on I. On the one hand, Garc´ıa-Pacheco, Mart´ın and Seoane [54] established in 2009 the spaceability in B(I) of the set of all continuous bounded functions on Iwhich are not Riemann-integrable, as well as the spaceability in L1(I) of L1(I)\R(I) (see also [53], [57] and [31, Section 2.4]; the last result has been recently improved in [58], as mentioned in Theorem 4.5 above: take Ω = I,p= 1). On the other hand, the existence of Riemann-integrable functions on a given unbounded interval being not Lebesgue-integrable on it is well known: consider the classical example f(x) = sin x xon (0,+∞). In fact, in [54] it is proved the lineability of R(I)\L1(I). In this context, an arising natural question is whether this lineability can be enriched within some appropriate topological structure. Theorem 4.14 below provides a positive answer, but we need a preliminary lemma. If I⊂Ris an unbounded interval, we denote by C0(I) the space of all continuous functions f:I→R such that limx→∞ f(x) = 0. For the sake of simplicity, we will only consider 24 the case I= [0,+∞), the remaining ones being analogue. Lemma 4.13. Let I= [0,+∞). Then the expression kfk:= sup x≥0 |f(x)|+ sup x≥0Zx 0 f(t)dt defines a norm on the space C0(I)∩R(I)which makes it a separable Banach space. The set Bof continuous functions [0,+∞)→Rwith bounded support is a dense vector subspace of this space. Proof. The linearity of the integral together with the fact that supx≥0|f(x)| is a norm on C0(I) yields that k · k is a norm on X:= C0(I)∩R(I). Let us prove that (X, k· k) is complete. If (fn) is a k·k-Cauchy sequence in Xthen it is, trivially, a Cauchy sequence in the Banach space C0(I) endowed with the supremum norm. Hence there is f∈C0(I) such that fn→funiformly on I. We need to show that f∈R(I) and fn→ffor k · k. To this end, fix ε > 0. There is N∈Nsuch that kfm−fnk< ε/3 for all m≥n≥N. Then |Rx 0(fm(t)−fn(t)) dt|< ε/3 for all x > 0 and all m≥n≥N. In particular, setting n=Nand letting m→ ∞ one gets by invoking uniform convergence on [0, x] that |Rx 0(f(t)−fN(t)) dt| ≤ ε/3 for all x > 0. Since fN∈R(I) there is a > 0 such that |Rc bfN(t)dt|< ε/3 for all c > b > a. It follows from the triangle inequality that Zc b f(t)dt≤Zc b fN(t)dt+Zb 0 (f(t)−fN(t)) dt+ Zc 0 (f(t)−fN(t)) dt<ε 3+ε 3+ε 3=ε, and Cauchy’s criterion for improper Riemann integrals guarantees that f∈ R(I). Now, we have |Rx 0(f(t)−fn(t)) dt| ≤ ε/3 for all n≥Nand all x > 0, and Ncan be chosen so that supx>0|f(x)−fn(x)|< ε/2 for all n≥N. Therefore kfn−fk< ε if n≥N, which shows that fn k·k −→ nf. Hence (X, k · k) is complete. Next, consider the set Bdefined in the statement of this lemma. It is clear that Bis a vector subspace of X. Fix a function f∈Xas well as an ε > 0. Then there is a > 0 with |f(x)|< ε/6 (x > a) and |Rc bf(t)dt|< ε/6 for all c > b > a. Define fa: [0,+∞)→Ras fa=fon [0, a], fa= 0 on (a+ 1,+∞), and faaffine-linear on [a, a + 1] with fa(a+ 1) = 0. It follows that fa∈Band kf−fak ≤ supa<x<a+1 |f(x)−fa(x)|+ supx≥a+1 |f(x)|+ |Ra+1 af(t)dt|+|Ra+1 afa(t)dt|+ supx>a+1 |Rx a+1 f(t)dt| ≤ 5·ε 6< ε. Thus B is dense in X. Finally, by using the Weierstrass polynomial approximation 25 bounded support σ(f) (see (1)). On the one hand, Bis a dense vector subspace of C(RM,RN). On the other hand, SM,N,∞+B⊂ SM,N,∞: indeed, if f−1({y}) is unbounded and g∈Bthen (f+g)−1({y})⊃f−1({y})\σ(g), and the last set is still unbounded because σ(g) is bounded. An application of Theorem 2.3 yields the maximal dense-lineability of SM,N,∞. Consequently, our only task is to show the spaceability of SM,N,∞. For this, we will use Theorem 3.2 with Ω = RM,S(A) = A(i.e. S(A) is the closure of Ain RM, so that S(σ(h)) = σ(h), the topological support of a function h:RM→RN), X=C(RM,RN), K=R,Z=RN,S= C(RM,RN)\ SM,N,∞={f∈C(RM,RN) : f−1({y}) is bounded for some y∈RN}(we agree that ∅is bounded), and kfk= ∞ X k=1 1 2k supx∈Bkkf(x)k2 1 + supx∈Bkkf(x)k2 , where k · k2denotes Euclidean norm in RN. Let us check conditions (i) to (v) in Theorem 3.2: •(i) holds because uniform convergence on compacta implies pointwise convergence. •(ii) is true (with C= 1) since the map t∈[0,+∞)7→ x 1+x∈[0,+∞) is increasing and kf(x) + g(x)k2≥ kf(x)k2for all x∈RNwhenever σ(f)∩σ(g) = ∅. Here σ(h) denotes the support of has defined in (1). •(iii) is satisfied because αf = 0 if α= 0 and, provided that α6= 0, then (αf)−1({y}) = f−1({α−1y}) for all y∈RN. •Assume that f, g ∈Xand σ(f)∩σ(g) = ∅. Then, in particular, σ(f)∩σ(g) = ∅, from which it follows that (f+g)−1({y}) = f−1({y})∪ g−1({y}) for all y∈RN\ {0}. Suppose that f+g∈S. Then either there is y∈RN\ {0}such that (f+g)−1({y}) is bounded, or (f+ g)−1({y}) is unbounded for all y6= 0 but (f+g)−1({0}) is bounded. In the first case, the last set identity forces f−1({y}) to be bounded, so f∈S. Assume now that (f+g)−1({y}) is unbounded for all y6= 0 but (f+g)−1({0}) is bounded. We can suppose that f−1({y}) is unbounded for all y6= 0 (otherwise, f∈Sand we would be done). Therefore σ(f) is unbounded. Let us prove that f−1({0}) is bounded (in which case f∈S). By way of contradiction, assume that f−1({0}) is unbounded. Then ∂f−1({0}) is also unbounded [indeed, if ∂f−1({0}) is bounded then there is α > 0 such that f(x)6= 0 for all xwith kxk2> α due to the unboundedness of σ(f) and the closedness of f−1({0}); 32 hence f−1({0}) would be bounded, which is absurd]. Now, we have: ∂f−1({0}) = ∂(RM\f−1({0})) = ∂σ(f)⊂σ(f)⊂RM\σ(g) = g−1({0}). We derive that if x∈∂f−1({0}) then (since f−1({0}) is closed) f(x) = 0 = g(x), so (f+g)(x) = 0. Therefore ∂f−1({0})⊂ (f+g)−1({0}), so (f+g)−1({0}) is also unbounded, which contradicts our assumption. This yields (iv). •The idea underlying the proof of (v) is to construct continuous functions by shifting and scaling appropriately the function Φ given in (3). Firstly, it is plain that there is sequence of points (aj)⊂RNsatisfying RN=Sj≥1(aj+IN). For each k∈N0and each a∈RNwe consider the mapping Φk,a :{k, k + 1} ∪ k+1 3, k +2 3×RM−1→RN given by Φk,a = 0 on {k, k+1}×RM−1=∂(k+S0) and Φk,a(x1,...,xM) = a+ϕ(3(x1−k)−1) if (x1,...,xM)∈[k+1 3, k +2 3]×RM−1. Tietze’s extension theorem comes in our help to provide a continuous extension Φk,a :k+S0→RN(observe that Tietze’s theorem can be applied to each component of Φk,a). Note that Φk,a(k+S0)⊃a+INfor all k≥0. Since card (N3) = card (N), we can select N2-many pairwise disjoint sequences {p(n, m, 1) < p(n, m, 2) <··· < p(n, m, j)<· · ·} (n, m ∈N) of natural numbers. For each n∈N, define fn:RM→RN by fn(x) = Φp(n,m,j),aj(x) if x∈p(n, m, j) + S0(m, j ∈N) 0 otherwise, Since fn= 0 on each boundary ∂(p(n, m, j) + S0), we have that each fnis well defined and continuous. Furthermore, for every n∈Nand every y∈RN=Sj≥1(aj+IN), the set f−1 n({y}) possesses at least one point in every set Sj≥1(p(n, m, j)+S0) (m= 1,2,...), so f−1 n({y}) is unbounded and fn∈ SM,N,∞. Finally, the supports of the functions fn(n= 1,2,...) satisfy σ(fk)∩σ(fn) = ∅for all k6=n, because σ(fn)⊂Sm,j≥1(p(n, m, j) + S0 0) and the numbers p(n, m, j) are pairwise different. This had to be shown. Theorem 4.18 is best possible in terms of dimension because, as noticed in [3], there is no surjective continuous function R→RN(see [76]), RNbeing the space of real sequences endowed with the product topology. 33 Acknowledgements. 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