Linear and algebraic structures in function sequence spaces

UNIVERSIDAD DE SEVILLA Facultad de Matem´aticas Departamento de An´alisis Matem´atico LINEAR AND ALGEBRAIC STRUCTURES IN FUNCTION SEQUENCE SPACES Memoria presentada por Pablo Jos´e Gerlach Mena para optar al grado de Doctor por la Universidad de Sevilla. Madel Carmen Calder´on Moreno Prof. Titular Universidad Dpto. An´alisis Matem´atico Universidad de Sevilla Jos´e Antonio Prado Bassas Prof. Titular Universidad Dpto. An´alisis Matem´atico Universidad de Sevilla V◦B◦de la Directora V◦B◦del Director Sevilla, Febrero de 2020 Contents Resumen vii Abstract ix Introduction xi 1 Preliminaries 1 1.1 Basic concepts and definitions . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Someclassicalspaces............................ 3 1.2.1 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2.2 Sequencespaces........................... 4 1.2.3 The space L0(X, µ)......................... 6 1.2.4 Lp(X, µ) spaces of functions for 1 ≤p < +∞........... 7 1.2.5 L∞(X, µ) space of functions . . . . . . . . . . . . . . . . . . . . 7 1.3 Topology of sequence spaces . . . . . . . . . . . . . . . . . . . . . . . . 9 i Contents Lin. Alg. Struc. Func. Seq. Spaces 2 Lineability 15 2.1 Introduction................................. 15 2.2 Maindefinitions............................... 16 2.3 On the relation between lineability, spaceability and denseness . . . . . 18 3 Modes of convergence 27 3.1 Definitions and relations between convergences . . . . . . . . . . . . . . 27 3.2 Counterexamples on the reciprocals . . . . . . . . . . . . . . . . . . . . 32 4 Lineability and modes of convergence 41 4.1 Measure vs. Pointwise a.e. Convergence . . . . . . . . . . . . . . . . . 42 4.1.1 Algebrability and spaceability . . . . . . . . . . . . . . . . . . . 48 4.2 Pointwise vs. Uniform Convergence . . . . . . . . . . . . . . . . . . . . 54 4.2.1 Definitions and first examples . . . . . . . . . . . . . . . . . . . 54 4.2.2 Lineability results . . . . . . . . . . . . . . . . . . . . . . . . . . 57 5 The space L1([0,+∞)) 61 5.1 Convergence in L1(X, µ) .......................... 62 5.1.1 Measure vs. L1convergence .................... 62 5.1.2 Uniform vs L1convergence..................... 63 5.2 Unbounded functions in CL1([0,+∞)) .................. 71 5.3 Sequences of continuous, unbounded and integrable functions . . . . . . 82 ii Lin. Alg. Struc. Func. Seq. Spaces Contents 5.4 Finalremarks................................ 87 5.4.1 Maximal possible convergence . . . . . . . . . . . . . . . . . . . 87 5.4.2 Onthegrowth ........................... 89 5.4.3 Smoothness ............................. 93 6 Anti M-Weierstrass sequences of functions 97 6.1 Concepts and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.2 Lineability within AMW .........................104 6.3 Algebrability within AMW ........................107 Bibliography 113 iii Contents Lin. Alg. Struc. Func. Seq. Spaces iv Agradecimientos A mis directores, Mar´ıa del Carmen Calder´on Moreno y Jos´e Antonio Prado Bassas, sin los cuales esta Tesis no habr´ıa sido posible. Gracias por confiar en m´ı durante todos estos a˜nos de carrera, m´aster y doctorado. Gracias por toda la ayuda acad´emica durante la realizaci´on y escritura de la Tesis, as´ı como por todos los consejos que me hab´eis dado a nivel personal durante todo el proceso. Os aseguro que no caen en saco roto, y espero ser un poco mejor cada d´ıa gracias a vosotros. A los profesores del Proyecto ESTALMAT, por introducirme en el mundo de las matem´aticas a tan temprana edad, con el consiguiente reto que ello conlleva. Gracias a magnificos docentes y apasionados de las Matem´aticas como Antonio Aranda, Antonio P´erez Jim´enez, Manolo Delgado, Ram´on Piedra o Francisco Castro, entre muchos m´as, he seguido en el mundo de las Matem´aticas. En especial a Alfonso Carriazo, quien, adem´as de haber sido mi profesor en ESTALMAT, la carrera y el m´aster, ha sido un amigo inestimable a nivel personal, apoy´andome en todo y mostrando su m´as sincero inter´es por m´ı. Al profesor Luis Bernal, por atraerme al campo de An´alisis Matem´atico en segundo de carrera, y darme la oportunidad de conocer a mis actuales directores, Carmen y Jos´e. Gracias por toda la ayuda y apoyo dado durante la realizaci´on de mi Erasmus en Berl´ın. v Agradecimientos Lin. Alg. Struc. Func. Seq. Spaces Al Departamento de An´alisis Matem´atico, por todo el apoyo brindado durante estos 4 a˜nos, tanto en el ´ambito acad´emico como personal, as´ı como a todo el PAS de la Facultad y del IMUS por hacerme m´as agradables todos los a˜nos que ha durado este viaje. A mis amigos de los Altos Colegios, Fran, Bel´en, Isaac, Elena, Jes´us, Mercedes, Mery, Alejandro, Jon (y muchos m´as), y a la Yonkozona, V´ıctor, Fran, Josemi, Javi, Julia y Adolfo, por aguantarme, animarme y ense˜narme a ser mejor persona. Gracias a todos vosotros he seguido firme en este viaje por las Matem´aticas, aprendiendo de mis (muchos) errores, y tratando de dar lo mejor de m´ı. A mis padres, Lola y Frank, y a Javier, quien ha sido como un segundo padre para m´ı, a mis hermanos, Mauro, Violeta y Mateo, as´ı como a mi familia “iliturgitana”, Isi, Pepa, Juanma y el peque˜no Nacho, y mi familia “bejarana”, Lola, Juli´an y Blanca, por apoyarme en todas mis decisiones, y darme el apoyo necesario para seguir adelante pese a todos los contratiempos sufridos, los agobios, las noches de estudio, las indecisiones y dem´as. Gracias por estar siempre a mi lado. A Teresa, mi pareja, y uno de los pilares fundamentales en todo este proceso junto a mi familia. Desde primero de carrera has estado a mi lado, vi´endome crecer y madurar como persona. T´u me has hecho mucho mejor de lo que era, y has sufrido estos a˜nos de doctorado conmigo de la mano. Todos las tardes de estudio en casa, los agobios por los ex´amenes y las notas, la alegr´ıa de los premios y, por fin, el doctorado, tienen su recompensa, y buena parte de ella es gracias a t´ı. Por ´ultimo, a mi Lela, quien, a pesar de haberse ido antes de poder ver este proceso terminado, tuvo el tiempo necesario para ense˜narme que, en la vida, lo m´as importante que existe al final es el amor hacia tus seres queridos. vi Resumen Hist´oricamente han sido muchos los matem´aticos de todas las ´epocas que se han sentido atra´ıdos y fascinados por la existencia de grandes estructuras algebraicas que satisfacen ciertas propiedades que, a priori, pueden contradecir a la intuici´on matem´atica. El objetivo de la presente Memoria es el estudio de la lineabilidad de diversas familias de sucesiones de funciones con propiedades muy espec´ıficas. La Memoria se divide en 6 cap´ıtulos, donde los Cap´ıtulos 1, 2 y 3 se centran en introducir la notaci´on b´asica y la terminolog´ıa principal de la teor´ıa de la Lineabilidad y de los modos de convergencia que usaremos a lo largo de esta Memoria. En el Cap´ıtulo 4 comenzamos el estudio del tama˜no algebraico de dos familias de sucesiones de funciones con distintos modos de convergencia en el intervalo unidad cerrado [0,1]: convergencia en medida pero no puntual en casi todo y convergencia puntual pero no uniforme. En el Cap´ıtulo 5 centramos nuestra atenci´on en el marco de las funciones integrables (Lebesgue). Comenzamos con sucesiones de funciones integrables y distintos modos de convergencia en comparaci´on con la convergencia en norma L1, y finalizamos el cap´ıtulo con el tama˜no algebraico de las familias de funciones no acotadas, continuas e integrables en [0,+∞), y las sucesiones de ellas. vii Introduction Lin. Alg. Struc. Func. Seq. Spaces considered the family of sequences of continuous bounded and integrable functions fn:R→R(n∈N) such that kfnk∞→+∞(n→ ∞), sup kfnk1:n∈N<+∞ but kfnk16→ 0 (n→ ∞), and proved its maximal lineability. In the setting of series of functions, we can find a wide plethora of results concerning the study of the divergence of the series and its algebraic size. Its origin goes back to E. du Bois-Reymond (1873, see [40]), who was the first one to exhibit an example of a continuous function fon the unit circle Twhose Fourier series diverges at a point. This was later improved in 1966 by J.P. Kahane and V. Katznelson (see [36],[38]), extending the divergence to a set E⊂Tof (Lebesgue) measure zero. If we drop away the continuity of the functions, in 1926 A. Kolmogorov [41]) was able to find a function f∈L1(T) such that its Fourier series diverges everywhere on T. In 2005 Bayart [13] showed that the set of continuous functions on Twhose Fourier series diverges on a set E⊂Tof measure zero is dense-lineable. A year later, Aron et al. stated its dense-algebrability. When Eis countable, results on lineability of divergent Fourier series with additional properties are obtained in [16] or [43]. Again Bayart [12], [13] showed that the set of Dirichlet series f(s) = ∞ X n=1 ann−s that are bounded in the right half-plane and diverge everywhere in the imaginary axis is lineable and spaceable. Together with Quarta [14], they were also able to establish the algebrability. In the setting of Banach spaces, A. Aizpuru, C. P´erez-Eslava and Seoane-Sep´ulveda in 2006 [1], asserted that the set of unconditionally convergent but not absolutely convergent series of any infinite dimensional Banach space is c-lineable. Moreover, the set of sequences such that its partial sums are bounded but the series diverges is also c-lineable. If we focus on the complex plane, in 2011 A. Bartoszewicz, S. Glab and T. Poreda proved in [11] that the set of non-absolutely convergent complex series and the set of divergent complex series with bounded partial sums are c-algebrable. xiv Lin. Alg. Struc. Func. Seq. Spaces Introduction Following with sequences of real numbers, in 2013 Bartoszewicz and Glab [10] showed that c0\[ p≥1 `pis densely strongly c-algebrable, where c0denotes the set of all real sequences converging to zero, and `pis the set of all p-sumable real sequences. Later, in 2017, Ara´ujo el al. [4] studied the linear structure of sequences of real numbers such that its series fails the root and ratio test. Concretely, they show that the set of sequences in `1that generate series for which the ratio or the root tests fail and the set of sequences in ω(the vector space of all real sequences endowed with the product topology) that generate divergent series for which the ratio and the root tests fail are c-dense-lineable in `1or ω. In the present Dissertation we generalize and extend some of the previous results. In Chapter 1 we establish the main notation and basic definitions that are needed to the correct development of this Dissertation. In Chapter 2, definitions of the theory of Lineability are introduced, as well as different relations among them, with some general results about the existence of these linear structures. In Chapter 3 we recall different modes of convergence in the setting of function sequence spaces. We show the different relations among them, including counterexamples when the reciprocals are not true. Chapter 4 brings modes of convergence into the setting of Lineability. We focus our attention on sequences of functions with different modes of convergence, and study the size of these families under the point of view of Lineability, providing positive results about the existence of both linear and algebraic structures. In particular, we show the spaceability and strong c-algebrability of the family of sequences of (Lebesgue) measurable functions fn: [0,1] →R(n∈N) such that they converge to zero in measure but not pointwise almost everywhere on [0,1], extending the work in [4], and the maximal dense-lineability, spacebility and strong c-algebrability of the family of sequences pointwise but not uniformly convergent to zero on [0,1], satisfying that for xv Introduction Lin. Alg. Struc. Func. Seq. Spaces any ε > 0 there is a measurable set E⊂[0,1] with m(E)< ε such that ess sup [0,1]\E |fn| → 0 (n→ ∞). In Chapter 5 we focus our attention on the setting of (Lebesgue) integrable functions. First, we recall the convergence in L1-norm, and complete the results of Chapter 3 with the existence of sequences of functions fn: [0,+∞)→Rconverging to zero in measure but not in L1-norm, converging to zero in L1-norm but not uniformly on [0,+∞), and converging to zero uniformly on [0,+∞) but not in L1-norm. In fact, we study the algebraic size of this last family, obtaining its maximal dense-lineability, spaceability and strong c-algebrability. Later, we consider unbounded, continuous and integrable functions and sequences of them, and show that they are ℵ0-algebrable and maximal dense-lineable. We finish this chapter with some final remarks about the maximal possible convergence of the sequences of functions, its growth and smoothness. Finally, in Chapter 6 we deal with series of functions, inspired in the previous existing results about convergence of the series. In particular, we study the existence of vector spaces and linear algebras inside the family of uniformly and absolutely convergent series on a closed interval [a, b]⊂Rthat do not fulfill all the hypothesis of the Weierstrass M-test. Most of all the results exposed in Chapters 3, 4, 5 and 6 are original. The results from Chapters 3, 4 and 5 are published in [24] and [25]. The ones from Chapter 6 are collected in [26]. xvi Chapter 1 Preliminaries 1.1 Basic concepts and definitions In order for this Dissertation to be as self-contained as possible, in this first Chapter we include basic concepts and definitions with which we deal in this work, as well as many results that will become useful while developing the upcomings chapters. By N,Q,Rand Cwe denote the set of all natural numbers, the set of rational numbers, the real line and the complex plane, respectively. By Kwe will denote indistinctly the set of real or complex numbers, that is, K=Ror K=C. The cardinality of Nwill be ℵ0, and the continuum cwill be the cardinality of the real line R. The different concepts that we introduce next can be briefly consulted in [3], [49], [53]. Let X6=∅be a set. An application d:X×X→[0,+∞) is a metric over Xif for any x, y, z ∈Xwe have d(x, y)=0⇔x= 0; d(x, y) = d(y, x); and d(x, y)≤d(x, z) + d(z, y). The pair (X, d) is called metric space. A topologic space (X, τ) is metrizable if there exists a metric dover Xwhich induces the topology τ, 1 Preliminaries Lin. Alg. Struc. Func. Seq. Spaces that is, the family of open balls B(x, ε) := {y∈X:d(x, y)< ε},(x∈X, ε > 0) is a basis of τ. When Xis a vector space and τis a topology on Xwhere the singletons are closed and compatible with the linear structure, that is, the applications of addition + : (x, y)∈X×X7→ x+y∈Xand product by scalars ·: (λ, x)∈K×X7→ λ·x∈X are continuous, we say that (X, τ) is a topological vector space. A metric ddefined on a vector space Xis translation-invariant if d(x+z, y +z) = d(x, y) for every x, y, z ∈X. A metric space (X, d) is complete if every Cauchy sequence (xn)n⊂Xis convergent on X, that is, there exists x∈Xsuch that for every ε > 0 there is N∈Nsuch that d(xn, x)< ε for every n≥N. We denote by lim n→∞ xn=xor xn→x(n→ ∞) the convergence of a sequence (xn)nto an element x. A topological vector space Xwhich is metrizable by a translation-invariant metric and complete is called an F-space. If, in addition, the F-space Xis locally convex, we say that Xis a Fr´echet space. Recall that a topological vector space is locally convex if it is Hausdorff and the zero element has a neighbourhood basis formed by convex sets. For any Fr´echet space (X, d), the application k · k :X→[0,+∞) defined as kxk:= d(x, 0) is a seminorm, that is, kx+yk≤kxk+kykand kλxk=|λ| · kxk. If, in addition, kxk= 0 if and only if x= 0, we have that k · k is a norm and (X, k · k) is a complete normed space, that is (X, k·k) is a Banach space. Let (X, τ) be a topological space. Given A⊂Xwe denote by int(A) and Athe interior and, respectively, the closure of A. We will use the following category concepts to study the topological size of some sets. A set A⊂Xis said to be: (1) dense in Xif A=X; (2) nowhere dense in Xif intA=∅; 2 Lin. Alg. Struc. Func. Seq. Spaces Preliminaries (3) of first category if A= ∞ [ n=1 Fn, where each Fnis nowhere dense in X; (4) of second category if Ais not of first category; and (5) residual if its complement Ac:= X\Ais of first category. A topological space (X, τ) is a Baire space if any countable intersection of dense open sets is also dense in X. Theorem 1.1 (Baire’s Theorem).Every complete metric space (X, d)is a Baire space. In particular, we have the following result. Corollary 1.2. Let (X, d)be a complete metric space and A⊂Xbe a subset. If A is of first category, then X\Ais dense in X. 1.2 Some classical spaces In this Section we recall some well-known spaces which we will use in this work. 1.2.1 Continuous functions Let [a, b]⊂Rbe a closed interval of the real line, where in Rwe consider always the euclidean norm. We define C([a, b]) as the space of all functions f: [a, b]→Kthat are continuous on [a, b]. This space becomes a separable Banach space when endowed with the supremum norm k·k∞given by kfk∞:= sup x∈[a,b] |f(x)|,for every f∈C([a, b]). 3 Preliminaries Lin. Alg. Struc. Func. Seq. Spaces The space of all sequences of continuous functions on [a, b] will be denoted by C([a, b])N, and it becomes a separable Banach vector space when endowed with the norm k(fn)nk:= sup n∈N kfnk∞. For the interval [0,+∞), we can consider the space C([0,+∞)) of all functions f: [0,+∞)→Kthat are continuous on [0,+∞). Although it is not a Banach space, it does become a Fr´echet space when endowed with the compact-open topology. Recall that for this topology we can consider the metric d(f, g) := ∞ X n=1 1 2n·kf−gk∞,n 1 + kf−gk∞,n , where khk∞,n := sup |h(x)|:x∈[0, n]for any h∈C([0,+∞)) and n∈N. We have that fn→f(n→ ∞) in the metric dif and only if fn→f(n→ ∞) uniformly on compacta if and only if kfn−fk∞,n →0 (n→ ∞). 1.2.2 Sequence spaces The `psequence spaces for 1≤p < +∞ Let p∈[1,+∞). Given a sequence (xn)n⊂Kwe define its `p-norm or simply p-norm as kxnkp= ∞ X n=1 |xn|p!1/p . The sequence space `pis defined as those sequences (xn)n⊂Kfor which its `p-norm is finite, that is, `p:= (xn)n⊂K:kxnkp<+∞, and its becomes a (separable) Banach space when endowed with the k · kpnorm for every 1 ≤p < +∞. 4 Lin. Alg. Struc. Func. Seq. Spaces Preliminaries The `∞sequence space For the case p= +∞, the `∞-norm is taken slightly different. We define the `∞-norm, ∞-norm or supremum norm as kxnk∞:= sup n∈N |xn|. The sequence space `∞is defined as those sequences (xn)n⊂Kfor which its ∞-norm is finite, that is, `∞:= (xn)n⊂K:kxnk∞<+∞. The pair (`∞,k · k∞) becomes a Banach space, whose elements are precisely those sequences (xn)n⊂Kthat are bounded. Moreover, the space (`∞,k·k∞) is not separable. The c0and c00 sequence spaces The family of all sequences (xn)n⊂Kconverging to zero is denoted by c0, that is, c0:= (xn)n⊂K:xn→0 (n→ ∞). Observe that c0is a subspace of `∞, since convergence of a sequence implies its boundedness. Furthermore, if we endow c0with the supremum norm of `∞, we obtain that (c0,k·k∞) is a closed subspace of (`∞,k·k∞), and hence it is also a Banach space. If we consider the family c00 of all vanishing sequences, that is, c00 := {(xn)n⊂K: there exists N∈Nsuch that xn= 0 for every n≥N}, then c00 is a dense subspace of (c0,k·k∞). Indeed, for any x= (xn)n∈c0, the sequences ωk= (ωk n)n= (x1, x2, . . . , xk,0,0, . . .)∈c00 verify kωk−xk∞= sup n∈N |wk n−xn|= sup n≥k |xn|. 5 Preliminaries Lin. Alg. Struc. Func. Seq. Spaces But x∈c0, so given ε > 0, there exists N∈Nsuch that |xn|< ε for every n≥N. Hence, kωk−xk∞< ε for any k≥N. In fact, the set {(rn)n⊂Q: there exits N∈Nsuch that rn= 0 for n≥N} is also dense in c0, and hence, (c0,k·k∞) is separable. 1.2.3 The space L0(X, µ) Let X6=∅be a non-empty set. A family Mof subsets of Xis a σ-algebra on X if ∅, X ∈ M;X\A∈ M whenever A∈ M; and every countable union of sets of M is again a member of M. An application µ:M → [0,+∞) is a measure on the σ-algebra Mif µ(∅) = 0; and µ ∞ [ n=1 An!= ∞ X n=1 µ(An) for every collection (An)n⊂ M of pairwise disjoints elements. The triplet (X, M, µ) is called measure space over X. A function f:X→Kis measurable (with respect to the measure µ) on Xif for every open subset B⊂Kwe have that f−1(B)∈ M. We denote by L0(X, µ) the set of all measurable functions from Xinto K. Given two functions f, g ∈ L0(X, µ) we say that f=galmost everywhere, and we denote it by f∼g, if there exists a measurable set A∈ M such that µ(A) = 0 and f=gon X\A. This is a relation of equivalence, and we denote by [f] := {g∈ L0(X, µ) : f∼g}the equivalence class of a measurable function f, and by L0(X, µ) the quotient space under this relation, that is, L0(X, µ) := L0(X, µ)/∼. It is known that for any pair fand gof measurable functions such that f∼g, the Lebesgue integrals for fand gin any A∈ M with respect to µare equals. So, the 6 Lin. Alg. Struc. Func. Seq. Spaces Preliminaries application k·k1:L0(X, µ)→[0,+∞] given by kfk1:= ZX |f|dµ, is well-defined. We define the space L1(X, µ) as all the classes of measurable functions that are integrable over X, that is, L1(X, µ) := f∈L0(X, µ) : kfk1<+∞. In L1(X, µ), the application k·k1defines a norm. In fact, there is no problem if we consider powers of the functions. 1.2.4 Lp(X, µ)spaces of functions for 1≤p < +∞ Let f:X→Kbe a measurable function on X. For 1 ≤p < +∞we define the Lp-norm or simply p-norm as kfkp:= ZX |f|pdµ1/p . The space Lp(X, µ) will denote the collection of all measurable (classes of) functions fsuch that kfkp<+∞. The pair (Lp(X, µ),k·kp) becomes a separable Banach space for every p∈[1,+∞). 1.2.5 L∞(X, µ)space of functions Notwithstanding, in order to study the case p= +∞we have to be more precise. If we try to reason as in the case of sequences of scalars, we see that the supremum over the whole space is not well-defined among classes of measurable functions. For instance, the functions f(x) = 1, and g(x) = 1 if x∈[0,1] \ {1/n}(n∈N) and g(1/n) = n(n∈N) represent the same class in L0([0,1], m) (where mdenotes the 7 Preliminaries Lin. Alg. Struc. Func. Seq. Spaces 14 Chapter 2 Lineability 2.1 Introduction Many mathematicians over History have been attracted and fascinated over the existence of different mathematical objects fulfilling properties that are, a priori, antiintuitive. This thought was plausible with the idea that was held until the end of the 19th century, where the geometric behaviour of the graph of a given function was thought to characterize the analytic behaviour of the same function. It was then commonly believed that it was impossible to have a continuous function not being differentiable at some set of points “big enough”. Nonetheless, this idea was disproved by the German mathematician K. Weierstrass [52], providing the existence of the so-called Weierstrass’ Monsters. A new trend that has been growing attention of mathematicians all around the globe for the past few decades focus on the study of large algebraic structures inside a (commonly) non-linear setting, where those “pathological” objects usually lie. In fact, in [30], Gurariy provided a first (and very famous) example of this new branch. 15 Lineability Lin. Alg. Struc. Func. Seq. Spaces Theorem 2.1. There exists an infinite dimensional linear space such that its non-null elements are continuous nowhere differentiable functions on [0,1]. Observe that the above theorem states that what was initially supposed to be an isolated phenomenon, as Weierstrass pointed out with the existence of his Monsters, has actually a big linear structure supporting it. This new approach receives the name of Lineability, and the main terminology was firstly introduced by Aron, Gurariy and Seoane-Sep´ulveda in [7], [32], [51], although the essence of them can be found in [30] and [31]. In this new terminology Gurariy showed that the set of Monsters is lineable. 2.2 Main definitions Vector spaces and linear algebras are an example of some mathematical structures which are, at first, unexpected to appear in spaces with a strong non-linear setting, that is, their existence within families with strange or pathological properties is rather unlikely. As commented before, Gurariy was the first one who provided such an example, with the existence of an infinite dimensional vector subspace in the family of continuous nowhere differentiable functions on [0,1]. In this Section, the basic concepts and definitions about existence of such linear structures are presented. The concepts of lineability and spaceability are due to Aron, Gurariy and Seoane-Sep´ulveda, although these notions were treated before, under another terminology (see [7], [32], [51]). The notion of maximal lineability (see Definition 2.2) was introduced by Bernal in [15]. Later, Gurariy partnered with Aron et al. coined the concepts of algebrability (see Definition 2.4), which has its roots in the papers [8], [9], [51]. Finally, this definitions were completed with the notion of strong algebrability, introduced by Bartoszewicz and Glab (see [10]). 16 Lin. Alg. Struc. Func. Seq. Spaces Lineability Definition 2.2. Let Xbe a vector space and let A⊂Xbe a subset. Let κbe a (finite or infinite)cardinal number. We say that Ais: (1) κ-lineable if there is a vector space Msuch that M⊂A∪ {0}and dim(M) = κ; (2) maximal lineable if Ais dim(X)-lineable. Although we speak about lineability of some subset A⊂X, this set does not need to have a linear structure itself. In fact, the neutral or 0 element of X, or even the sum of any of these objects would not have the “strange” behaviour. The definitions of lineability can be completed if a topological structure is taken into account, as the next definition shows. Definition 2.3. Let Xbe a topological vector space and let A⊂Xbe a subset. Let κ be a cardinal number. We say that Ais: (1) spaceable if there is a closed infinite dimensional vector space Msuch that M⊂ A∪ {0}; (2) κ-dense-lineable if there is a dense vector space Min Xwith M⊂A∪ {0}, and dim(M) = κ; (3) maximal dense-lineable if Ais dim(X)-dense-lineable. If we replace linear spaces by algebras we have the notion of algebrability. Definition 2.4. Let Xbe a vector space contained in some (linear)algebra. Let κbe a cardinal number. We say that A⊂Xis: (1) algebrable if there is an algebra Bso that B ⊂ A∪{0}and Bis infinitely generated, that is, the cardinality of any system of generators of Bis infinite; (2) κ-algebrable if there is a κ-generated algebra Bwith B ⊂ A∪ {0}; 17 Lineability Lin. Alg. Struc. Func. Seq. Spaces (3) strongly κ-algebrable if, in addition, the algebra Bcan be taken free. If the structure of the algebra is commutative, the notion of strong κ-algebrability is equivalent to the existence of a generating system Cof the subalgebra B ⊂ A∪ {0} with card(C) = κand such that for any positive integer N∈N, any non-constant polynomial Pin Nvariables, and any distinct f1, f2, . . . , fN∈ C, we have P(f1, f2, . . . , fN)∈ B \ {0}. 2.3 On the relation between lineability, spaceability and denseness Given a subset A⊂X, whenever Ais spaceable or algebrable, it is trivial that Ais also lineable. For the other definitions of lineability, in 2009 Aron et al. [6] showed that there is actually no relation between the concepts of spaceability and dense-lineability. We include the proof as appears in [5, Theorem 7.2.1]. Theorem 2.5. (a) Let Xbe an infinite dimensional locally convex space. There exists a subset A⊂X such that Ais spaceable and dense in X, although it is not dense-lineable in X. (b) Let Xbe an infinite dimensional F-space. There exists a subset A⊂Xwhich is lineable and dense in X, but which is not spaceable. If Xis separable, then Acan also be chosen to be dense-lineable in X. Proof. We will prove only the real case. For K=C, just change Qby Q+ iQin the proofs. (a) Take any vector u∈X\ {0}and let Z:= span({u}) = {cu :c∈R}. Since dim(Z) = 1 and Xis infinite dimensional, there is an algebraic complement Yof Z 18 Lin. Alg. Struc. Func. Seq. Spaces Lineability (that is, a linear subspace Yof Xsuch that Z∩Y={0}and X=Z+Y), satisfying that Yis a closed subspace. Take A:= Y+{λu :λ∈Q}. We will now show that Ais our candidate of dense and spaceable but not denselineable subset of X. Since Y⊂Aand Yis a closed infinite dimensional subspace of X(because Xis infinite dimensional and dim(Z) = 1), the set Ais spaceable. The set Ais also dense. Indeed, {λu :λ∈Q}is dense in Z, so A=Y+{λu :λ∈Q}is dense in Y+Z=X. Finally, let us prove that Ais not dense-lineable. By way of contradiction, suppose that Mis a dense vector subspace of Xcontained in A∪ {0}=A. Take w∈Mand write w=y+qu with y∈Yand q∈Q. Suppose that q6= 0. Then π qw∈M⊂A; that is, π qw=y0+q0u with y0∈Yand q0∈Q. Since X=Y+Zand Zis the algebraic complement of Y, we infer that y0=π qyand q0u=π qqu, but this implies that π=q0∈Q, which is clearly false. Thus, q= 0, and we get that w=y+qu =y∈Y, so, M⊂Y, which (again) is a contradiction because Yis closed and proper, and Mis dense. (b) Let us consider a Hamel basis Bfor X. Because dim(X) = ∞, we always can take a countably infinite subset {bn:n∈N}of B. Let Y= span{bn:n∈N}and Z= spanB \ {bn:n∈N}. It is clear that Zis the algebraic complement of Yand that X=Y+Z. Let A:= Y+ spanQB \ {bn:n∈N}, 19 Lineability Lin. Alg. Struc. Func. Seq. Spaces where spanQ(C) denotes the set of all finite linear combinations of elements of Cwith coefficients in Q. We claim that Asatisfies the required properties. First, Ais lineable since Y⊂A. Note that spanQB \ {bn:n∈N}is dense in Z, so A=Y+ spanQB \ {bn:n∈N}is dense in Y+Z=X. Finally, Ais not spaceable. Indeed, suppose that Mis an infinite dimensional closed vector subspace of Xcontained in A∪ {0}=A. Now, we proceed similarly as we did in part (a). Take w∈Mand write: w=y+q1b0 1+· · · +qrb0 r with y∈Y,q1, . . . , qr∈Qand b0 1,· · · , b0 r∈ B \ {bn:n∈N}. Our aim is to show that qj= 0 for every j∈ {1,2, . . . , r}. Suppose not, and assume, without loss of generality, that q16= 0. Then π q1 w∈M⊂A, that is, π q1 w=y0+q0 1b0 1+· · · +q0 sb0 s, with s≥r,y0∈Y,q0 1, . . . , q0 s∈Q, and b0 1, . . . , b0 s∈ B \ {bn:n∈N}. Therefore, y0=π q1 y, and πb0 1+q2π q1 b0 2+· · · +qrπ q1 b0 r=q0 1b0 1+· · · +q0 sb0 s, obtaining again the contradiction that π=q0 1∈Q. Thus, all qj’s are 0, and, consequently, M⊂Y. But this is impossible because dim(Y) = ℵ0and the cardinality of any Hamel basis of Mis uncountable (any infinite dimensional closed subspace of a complete metrizable space has the cardinality of continuum). To finish the proof, notice that if Xis separable, then we can choose Yto be dense in X, and therefore Ais dense-lineable. Nonetheless, if we ask for some conditions, it is possible to find a positive relation between lineability and dense-lineability. In order to do so, we introduce the following definition (see [6]). 20 Lin. Alg. Struc. Func. Seq. Spaces Lineability Definition 2.6. Let Xbe a vector space and let A, B ⊂Xbe two subsets. We say that Ais stronger than Bif A+B⊂A. Under this situation, if we have lineability of some subset A⊂Xand denselineability of a subset B⊂X, it is possible to transfer the density from Bto A. This result was proved by Bernal [15]. Previously, Aron et al. [6] gave a version for separable Banach spaces and, in fact, is often presented in the current literature in many and vary forms (see [5], [20]). The proof included is an adaptation of the proof on [15]. Theorem 2.7. Let Xbe a metrizable separable topological vector space and κbe an infinite cardinal. Let A, B ⊂Xbe two subsets such that Ais κ-lineable and Bis denselineable in X. If Ais stronger than Band A∩B=∅, then Ais κ-dense-lineable in X. Proof. Since Xis separable, there exists a sequence (xn)n⊂Xsuch that the set {xn:n∈N}is dense in X. Now, Ais κ-lineable in X, so there exists a vector subspace A1of A∪ {0}with dim(A1) = κ. Because dim(A1) = κ, there exists {vi:i∈I} ⊂ A1\ {0}linearly independent such that A1= span{vi:i∈I}and card(I) = κ. Moreover, κis an infinite cardinal, which means that we can split Iinto infinitely many pairwise disjoint non-empty sets Jn(n∈N), that is, I= ∞ [ n=1 Jn. Fix n∈Nand i∈Jn. Because multiplication by scalars is a continuous operation in a topological vector space, there exists εi>0 such that d(εivi,0) <1/n, where dis the translation-invariant distance that generates the topology of X. On the other hand, Bis dense-lineable in X, so there exists a vector subspace B1 such that B1⊂B∪ {0}and B1is dense in X. For every n∈N, the denseness of B1 guarantees the existence of yn∈B1such that d(yn, xn)<1/n. 21 Lineability Lin. Alg. Struc. Func. Seq. Spaces Now, we define the elements xn,i := yn+εivi for every n∈Nand i∈Jn, and consider the vector space M:= span{xn,i :n∈N, i ∈Jn}. We are going to see that Mis dense in X,M⊂A∪ {0}and dim(M) = κ. (1) Fix n∈Nand take some in∈Jn. Let un:= xn,in. We have that d(un, xn)≤d(un, yn) + d(yn, xn) = d(yn+εinvin, yn) + d(yn, xn) =d(εinvn,0) + d(yn, xn)<1 n+1 n=2 n→0 (n→ ∞). But (xn)nis dense in X, so (un)nis also dense, and the same is true for M. (2) Fix x∈M\ {0}. There are scalars c1, c2, . . . , cswith cs6= 0, and indices jr∈I (r= 1,2, . . . , s) such that x=c1x1,j1+c2x2,j2+. . . +csxs,js. But by the definition of (xn,i)n,i we have that x=c1y1+c2y2+. . . +csys+c1εj1vj1+c2εj2vj2+. . . +csεjsvjs=: y0+z0, where y0=c1y1+c2y2+. . . +csysand z0=c1εj1vj1+c2εj2vj2+. . . +csεjsvjs. Recall that y1, y2, . . . , ys∈B1, which is a vector space, so y0∈B1⊂B∪ {0}. Analogously, vj1, vj2, . . . , vjs∈A1, they are linearly independent, and csεjs6= 0, so z0∈A1\{0} ⊂ A. Finally, x=y0+z0∈(B∪ {0}) + A⊂Abecause Ais stronger than B, and we have M⊂A. (3) It only remains to prove that dim(M) = κ. For this, it is clear that card{(n, i) : n∈N, i ∈Jn}= card ∞ [ n=1 Jn!= card(I) = κ. 22 Lin. Alg. Struc. Func. Seq. Spaces Lineability So, if we prove that the vectors of {xn,i :n∈N, i ∈Jn}are linearly independent we are done. Indeed, assume by way of contradiction that c1x1,j1+c2x2,j2+. . .+csxs,js= 0 with cs6= 0. As done before (and following the same notation), we have that y0+z0= 0, where y0∈B1∪ {0}and z0∈A1\ {0}. But then, y0=−z0∈A1\ {0}, since A1is a vector space. Hence, we have that y0∈A1\ {0}∩B1∪ {0}⊂A∩B=∅, which is a contradiction. Recall that, given two vector spaces Xand Ywith Y⊂X, the codimension of Y, denoted by codim(Y), is defined as the dimension of the algebraic complement of Yin X, that is, the dimension of the vector space Zsuch that Z+Y=Xand Z∩Y={0}. Having this definition in mind, the next result by Bernal and Ord´o˜nez-Cabrera (see [19]) assures that, in the separable case, there is nothing to add in order to obtain the dense-lineability from mere lineability. We include the proof as it appears in [5, Theorem 7.3.3]. Theorem 2.8. Let Xbe a metrizable separable topological vector space and Ybe a vector subspace of X. If X\Yis lineable, then X\Yis dense-lineable. Consequently, both properties of lineability and dense-lineability for X\Yare equivalent provided that Xhas infinite dimension. Proof. It is evident that X\Yis lineable if and only Yhas infinite codimension. Since Xis metrizable and separable, it has a countable open basis {Gn:n∈N}. Assume that X\Yis lineable. In particular, Yis a proper vector space of X, so int(Y) = ∅. Hence X\Yis dense and there is x1∈G1\Y. Since codim(Y) = ∞, we have span(Y∪ {x1}) X. Then, int(span(Y∪ {x1})) = ∅, and it follows that there exists x2∈G2\span(Y∪ {x1}). With this procedure, we get recursively a sequence of vectors (xn)n∈Nsatisfying xn∈Gn\span(Y\ {x1, . . . , xn−1}) (n∈N). 23 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces Yet, the reciprocal of this implications are not true (see Section 3.2 for counterexamples). In fact, when we speak about convergence a.e. we only have uniqueness of the limit if we understand it as for the exception of a measure zero subset. Now, we would like to see if there is possible to find a relation among these new modes of convergence, as we have done with the uniformly and pointwise convergence. Recall that given a sequence (En)n⊂ M, we define the limit superior and the limit inferior, respectively, as lim sup n→∞ En:= ∞ \ m=1 ∞ [ n=m En,lim inf n→∞ En:= ∞ [ m=1 ∞ \ n=m En, which are both measurable sets. Theorem 3.4. Let (X, M, µ)be a measure space. Let fn, f :X→K(n∈N)be measurable functions and A∈ M. We have the following implications: (1) If fn→funiformly a.e. on A, then fn→falmost uniformly on A. (2) If fn→falmost uniformly on A, then (2.1) fn→fpointwise a.e. on A; (2.2) fn→fin measure on A. Proof. (1) Since fn→funiformly a.e. on Awe have that there exists E∈ M with µ(E)=0 such that fn→funiformly on A\E. So, for any ε > 0 we have that µ(E)=0< ε, and fn→funiformly on A\E. Hence, fn→falmost uniformly on A. (2.1) Now, let’s suppose that fn→falmost uniformly on A. Then, for every N∈N there exists a measurable set EN⊂Awith µ(EN)<1 2Nsuch that fn→funiformly on A\EN. 30 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence Consider the set F= lim sup N→∞ ENand denote by Fm:= ∞ [ N=m EN,so F= ∞ \ m=1 Fm. Thanks to the countable subadditivity of the measure µwe obtain that µ(Fm)≤ ∞ X N=m µ(EN)< ∞ X N=m 1 2N=1 2m−1. Hence, µ(F)≤µ(Fm)<1 2m−1for every m∈N, which leads to µ(F) = 0. Furthermore, we also know that A\F= lim inf N→∞ (A\EN). Since fn→funiformly on A\EN, and then, in particular, also converges pointwise to fon each A\EN, we conclude that fn→fpointwise on A\F. (2.2) Now let’s again assume that fn→falmost uniformly on A. For the convergence in measure, given any ε > 0 we have to prove that µ{x∈A:|fn(x)−f(x)| ≥ ε}→0 (n→ ∞). For this, fix ε > 0 and take δ > 0. Since fn→falmost uniformly on A, there exists some set E∈ M with µ(E)< δ such that fn→funiformly on A\E. Thus, for the prefixed ε > 0, there exists N∈Nsuch that |fn(x)−f(x)|< ε for any n≥Nand x∈A\E. But this would imply, in particular, that: {x∈A:|fn(x)−f(x)| ≥ ε} ⊂ Efor every n≥N, and hence µ{x∈A:|fn(x)−f(x)| ≥ ε}≤µ(E)< δ. Thus, we have shown that given any δ > 0, there exists N∈Nsuch that for every n≥Nwe have that µ{x∈A:|fn(x)−f(x)| ≥ ε}< δ, 31 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces that is, µ{x∈A:|fn(x)−f(x)| ≥ ε}→0 (n→ ∞), and the convergence in measure is obtained. 3.2 Counterexamples on the reciprocals The reciprocals of the different implications among the modes of convergence that have been introduced are not true in general. This will be clarified during this Section with appropiated counterexamples on the modes of convergence. We start with the easiest one, namely that pointwise convergence does not imply uniform convergence. Example 3.5 (Pointwise ;Uniform).Let fn:R→R(n∈N) be the sequence of functions given by fn(x) := x n,(x∈R, n ∈N). 0 1 1 1 2 1 2 2 2 2 1 2 f1(x) = x 1 0 1 1 1 2 1 2 2 2 2 1 2 f2(x) = x 2 1 Figure 3.2: First terms of the sequence fn(x) = x n 32 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence Firstly, we have that fn(x)→0 =: f(x) (n→ ∞) pointwise on R. Yet, if we apply Theorem 3.2, we obtain that sup x∈R |fn(x)−f(x)|= sup x∈R |x| n≥1, which means that we cannot have uniformly convergence to the null function on R. Although this is of the most typical examples of a sequence of functions converging pointwise (to zero) but not uniformly, we want to point out a different example, since it will be the germ of the proofs of some results in Chapter 4. Recall that if (X, M, µ) is a measure space and A⊂X, the characteristic function of Ais the function χA:X→R defined as: χA(x) :=    1 if x∈A, 0 if x6=A. Clearly, χAis a measurable function if and only if A∈ M. Example 3.6 (Pointwise ;Uniform).For fixed n∈N, denote by En=1 n+1,1 n. Let fn: [0,1] →R(n∈N) be the sequence of functions given by fn(x) := χEn(x) (x∈[0,1], n ∈N). 0 1 1 3 1 1 2 f1(x) = χ[1 2,1](x) 0 1 1 3 1 1 2 f2(x) = χ[1 3, 1 2](x) Figure 3.3: First terms of the sequence fn(x) = χ[1 n+1 ,1 n](x) 33 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces Clearly, we have that fn(x)→0 =: f(x) (n→ ∞) pointwise on [0,1], because fn(0) = 0 for all n∈N, and for x∈(0,1] there is Nx∈Nsuch that 1 Nx < x, thus fn(x) = 0 for all n≥Nx. However, sup x∈[0,1] |fn(x)−f(x)|= sup x∈[0,1] |χEn(x)|= 1, and an application of Theorem 3.2 gives us the non-uniform convergence to the null function on [0,1]. Now we will focus on the modes of convergence that arises from the framework of measures spaces. In order to present the different examples we will consider the measure space given by (R,L, m), where Ldenotes the σ-algebra of Lebesgue measurable sets and mthe Lebesgue measure on R. From Theorem 3.4 we obtained that almost uniform convergence is a mode of convergence between pointwise a.e. and uniformly a.e. convergence, but the reciprocal of none of these implications are true. This will be shown in the next two examples. Example 3.7 (Pointwise a.e ;Almost uniform).Let fn:R→R(n∈N) be the sequence of functions given by fn(x) := x nχR\Q(x) + χQ(x),(x∈R, n ∈N). Clearly, we have that fn(x)→0 pointwise on R\Q, and since m(Q) = 0, we can state the pointwise a.e. convergence of the sequence of functions (fn)nto 0 on R. Now we will prove that the almost uniform convergence to the null function cannot be achieved. For this, let E∈ L be any measurable set with m(E)<+∞. Then, we have that m(R\(E∪Q)) = +∞. Hence, for any n∈N,m((R\(E∪Q∪[−n, n])) = +∞, and in particular, there exists xn∈R\(E∪Q) with |xn|> n, so sup x∈R\E |fn(x)|≥|fn(xn)|=|xn| n>1. 34 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence Thus, (fn)ncannot converge uniformly to 0 on R\E, which means that the almost uniformly convergence of (fn)nto 0 on Rcannot be obtained. Example 3.8 (Almost uniformly ;Uniformly a.e).For fixed n∈N, denote by En=−1 n,1 n. Let fn:R→R(n∈N) be the sequences of functions given by fn(x) := χEn(x),(x∈R, n ∈N). 0 1 1 1 2 1 2 1 f1(x) = χ[−1,1](x) 0 1 1 1 2 1 2 1 f2(x) = χ[− 1 2,1 2](x) Figure 3.4: First terms of the sequence fn(x) = χ[−1 n,1 n](x) Trivially fn→χ{0}pointwise on R. Let’s see that fn→χ{0}almost uniformly but not uniformly a.e. on R. Fix ε > 0 and take N∈Nsuch that 2 N≤ε, so that m(EN) = m−1 N,1 N=2 N< ε. Furthermore, for every n≥Nit holds that En⊂EN, and then fn(x) = 0 for all x∈R\EN. Thus, fn→0 uniformly on R\EN, and fnconverges almost uniformly to χ{0}on R(observe that χ{0}= 0 on R\EN). On the other hand, if fn→χ{0}uniformly a.e. on R, there is E∈ L with m(E)=0 such that fn→0 uniformly on R\E. In this case, for ε∈(0,1), there is N∈Nsuch that for any n≥N,|fn(x)|<ε<1 if x∈R\E. But fn(x) = χEn(x), so this implies that |fn(x)|= 0, and, hence, x∈R\En. Thus R\E⊂\ n≥N (R\En) = R\EN, 35 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces and EN⊂E, which is a contradiction with m(E) = 0. F. Riesz showed in 1909, and at the same time he introduced the convergence in measure, the existence of a relation between the convergence in measure and the pointwise a.e. convergence, as next the result states (see [48], [33] or [45, Theorem 21.9]). Theorem 3.9. Let (X, M, µ)be a measure space. Let fn, f :X→K(n∈N)be measurable functions. If fn→f(n→ ∞)in measure, then there exists a subsequence (fnk)k⊂(fn)nsuch that fnk→f(k→ ∞)pointwise a.e. on X. Since we can already obtain from the convergence in measure of a sequence (fn)nto fthe existence of some subsequence (fnk)ksuch that it converges pointwise a.e. to the function f, our first logical question would be if this convergence could be extended to the whole sequence instead of just only for some subsequence. But the so-called “Typewriter sequence” already states that this cannot be done, that is, in general, the whole sequence (Tn)n⊂ L0([0,1]) cannot inherit the pointwise a.e. convergence to 0 of some subsequence (Tnk)k. Example 3.10 (Measure ;Pointwise a.e. (Typewriter sequence)).Let divide the unit interval [0,1] into infinitely many subintervals Inof the form In:= jn 2kn,jn+ 1 2kn, where for each n∈N, the non-negative integers jnand knare uniquely determined by n= 2kn+jnand 0 ≤jn<2kn. We define the “Typewriter sequence” Tn: [0,1] →R (n∈N) by Tn(x) := χIn(x) = χ[jn 2kn,jn+1 2kn](x),(x∈[0,1], n ∈N).(3.1) 36 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence 0 1 1 1 2 T1(x) = χ[0,1](x) 3 4 1 4 0 1 1 1 2 T2(x) = χ[0,1 2](x) 3 4 1 4 0 1 1 1 2 T3(x) = χ[1 2,1](x) 3 4 1 4 0 1 1 1 2 T4(x) = χ[0,1 4](x) 3 4 1 4 0 1 1 1 2 T5(x) = χ[1 4,1 2](x) 3 4 1 4 0 1 1 1 2 T6(x) = χ[1 2,3 4](x) 3 4 1 4 0 1 1 1 2 T7(x) = χ[3 4,1](x) 3 4 1 4 Figure 3.5: Firsts iterations of the “Typewriter sequence” 37 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces Consider ([0,1],L, m). It is clear that each Tnis a measurable function for each n∈N. Observe that m{x∈[0,1] : Tn(x)6= 0}=m(In)→0 (n→ ∞). In particular, we obtain the convergence in measure of the “Typewriter sequence” to the null function. Yet, for any fixed x∈[0,1], the sequence (Tn(x))ntakes infinitely many times the value 0 and infinitely many times the value 1. Hence, the sequence (Tn(x))n can not be convergent to zero, and thus, the “Typewriter sequence” cannot converge pointwise a.e. on [0,1] to the null function. In fact, the sequence (Tn)ndoes not converge pointwise at any point on [0,1]. Observe that as Riesz’s Theorem assures, there is a subsequence of (Tn)nconvergent pointwise a.e. to zero, just taking (Tnk)kfor nk= 2k(k∈N). Indeed, Tnk(x)→0 (k→ ∞) for all x∈(0,1] and Tnk(0) = 1 for every k∈N. Example 3.11 (Pointwise a.e. ;Measure).Let fn:R→R(n∈N) be the sequence of functions given by fn(x) := χ[n,n+1](x) (x∈R, n ∈N). 0 4 1 2 f1(x) = χ[1,2](x) 3 1 5 0 4 1 2 f2(x) = χ[2,3](x) 3 1 5 0 4 1 2 f3(x) = χ[3,4](x) 3 1 5 Figure 3.6: First terms of the sequence fn(x) = χ[n,n+1](x) We clearly have that fn(x)→0 (n→ ∞) pointwise on R, and hence pointwise a.e. on R. On the other hand, for any ε∈(0,1) and for any n∈N, m{x∈R:|fn(x)| ≥ ε}=m[n, n + 1]= 1, 38 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence and m{x∈R:|fn(x)| ≥ ε}6→ 0 (n→ ∞). Thus, (fn)ndoes not converge to zero in measure. In order to finish this Chapter, we will provide an useful diagram where all the different relations and implications among the studied concepts of convergence are established. Measure Pointwise a.e. Almost uniformUniform a.e. Pointwise Uniform X X X X X X X X X Figure 3.7: Relation among modes of convergences 39 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces From now on, in this Section we will work with sequences of (classes of) functions on the unit interval, that is, our framework will be the space L0([0,1])N. Recall from Section 1.3 that the metric Ddefined on L0([0,1])Nby D((fn)n,(gn)n) = ∞ X n=1 1 2n·ρ(fn, gn) 1 + ρ(fn, gn) is compatible with the product topology in L0([0,1])N. Recall that c0(L0([0,1])) = {(fn)n∈L0([0,1])N:ρ(fn,0) →0 (n→ ∞)} ={(fn)n∈L0([0,1])N:fn→0 in measure} Define the family nPaeM([0,1]) of all sequences of functions in L0([0,1])Nconverging to zero in measure but not pointwise a.e. on [0,1], that is, nPaeM([0,1]) := (fn)n∈c0(L0([0,1])) : fn6→ 0 pointwise a.e. on [0,1]. In 2017, Ara´ujo et al. [4] provided a first result about the topological and linear size of nPaeM([0,1]). We include here the original proof for the sake of completeness. Theorem 4.3. The family nPaeM([0,1]) is maximal dense-lineable in L0([0,1])Nwith the product topology. Proof. Let (Tn)n∈L0([0,1])Nbe the “Typewriter sequence” defined in Example 3.10. Recall that we already know that (Tn)n∈nPaeM([0,1]). In order to find our desired vector space we will start by extending each Tnfrom the unit interval [0,1] into the real line R, just by defining Tn(x) = 0 for every x /∈[0,1]. Now, for every n∈N, consider the translated-dilated sequence (Tn,t)ngiven by Tn,t(x) := Tn(2(x−t)),0,1 2, x ∈R). It is clear that (Tn,t)n∈L0([0,1])N, and that also converges to zero in measure for every t∈(0,1/2). Let us consider the vector space Mgiven by M:= span (Tn,t)n:t∈0,1 2. 46 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence In a first step we are going to show that M⊂nPaeM([0,1]) ∪ {0}, and dim(M) = dim(L0([0,1])N) = c(and the last equality holds because L0([0,1])Nis a complete metrizable topological vector space, hence a Baire space). In order to see that the elements of Mare linearly independent, assume that there exist 0 < t1< t2< · · · < ts<1 2and real numbers c1, c2, . . . , csnot simultaneously zero (take cs6= 0 for example), such that c1Tn,t1+c2Tn,t2+· · · +csTn,ts= 0,(n∈N). In particular, by choosing n= 1 we have that c1T1,t1(x) + c2T1,t2(x) + · · · +csT1,ts(x) = 0, for almost every x∈[0,1]. Since T1,t(x) = T1(2(x−t)) = χ[0,1](2(x−t)) = χ[t,t+1/2](x), we have that the last equation can be written as c1χ[t1,t1+1 2]+c2χ[t2,t2+1 2]+· · · +csχ[ts,ts+1 2]= 0,for almost every x∈[0,1]. But, for (almost) every x∈ts−1+1 2, ts+1 2, we would obtain that c1·0 + c2·0 + · · · +cs−1·0 + cs·1 = cs= 0, which would be a contradiction with cs6= 0. This shows the required linear independence. As dim 0,1 2=c, we obtain that dim(M) = c. Moreover, since L0([0,1])Nis a topological vector space endowed with the topology of convergence in measure, and (Tn,t)nconverges to zero in measure for each t∈0,1 2, we get that every element (Fn)n:= (c1Tn,t1+c2Tn,t2+· · · +csTn,ts)n∈M, is a sequence converging to zero in measure. Next, fix any sequence (Fn)n∈Mas above, with 0 < t1< t2<· · · < ts<1 2and cs6= 0. For all x∈ts−1+1 2, ts+1 2, we have that Fn(x) = s X j=1 cjTn,tj(x) = s X j=1 cjTn(2(x−tj)) = csTn(2(x−ts)). 47 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces For each x∈(ts−1+1/2, ts+1/2], we have w:= 2(x−ts)∈(2(ts−1−ts)+1,1] ⊂[0,1]. Hence, since cs6= 0 and the sequence (Tn(w))ndoes not converge (as a sequence of scalars) for any w∈[0,1],we derive that, for each x∈ts−1+1 2, ts+1 2, the sequence (Fn)ndoes not converge pointwise a.e. to zero. This shows that M⊂nPaeM([0,1]) ∪ {0}. Thus nPaeM([0,1]) is maximal lineable. Moreover, since the set c00(L0([0,1])) is dense in (L0([0,1])N, D) (see Section 1.3), nPaeM([0,1]) ∩c00(L0([0,1])) = ∅(because the elements of c00(L0([0,1])) are pointwise convergent to 0 on [0,1]), and nPaeM([0,1]) is stronger than c00(L0([0,1])), an application of Theorem 2.7 with A=nPaeM([0,1]), B=c00(L0([0,1])) and κ=c puts an end on the proof. Recall that c00(L0([0,1]) is also dense in c0(L0([0,1]) when endowed with the supremum metric d∞. In particular, analogously we can state: Theorem 4.4. The family nPaeM([0,1]) is maximal dense-lineable in (c0(L0([0,1])), d∞). 4.1.1 Algebrability and spaceability With the maximal dense-lineability of this family we already have that these sequences of functions maintain a good structure under linear combinations. In the next result we find that it also has a nice algebraic structure. But firstly, we will introduce some useful notation and two technical lemmas. Let us represent each N-tuple (r1, r2,· · · , rN)∈RNby r, and set |r|:= r1+r2+ · · · +rNand r·s:= r1s1+r2s2+· · · +rNsN. Lemma 4.5. Let I⊂Rbe an interval, and let k1, k2, . . . , kNbe mutually different real numbers (N∈N). Then, the set of exponential functions {ek1x, ek2x, . . . , ekNx}(x∈I) is a linearly independent set of functions. 48 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence Proof. Observe that we can assume without loss of generality that 0 ∈I(if not, just use a translation). Assume that a linear combination of these functions is zero, that is, there exist λ1, λ2, . . . , λN∈Rsuch that λ1ek1x+λ2ek2x+· · · +λNekNx= 0. If we derive this equation N−1 times, we get λ1kp 1ek1x+λ2kp 2ek2x+· · · +λNkp NekNx= 0,(0 ≤p≤N−1). Now, by taking x= 0, we get the following Vandermonde system              λ1+λ2+· · · +λN= 0 λ1k1+λ2k2+· · · +λNkN= 0 . . .. . . λ1kN−1 1+λ2kN−1 2+· · · +λNkN−1 N= 0 Thus, since {k1, k2, . . . , kN}are mutually distinct, the only solution of the above system is the trivial one and we are done. Lemma 4.6. Let I⊂Rbe an interval. Let H⊂(0,+∞)be a Q-linearly independent set. Then, the algebra Agenerated by the set of exponential functions {e−cx :c∈H}(x∈I) is free. Proof. Let Fbe an element of A \ {0}. Then, there exists a non-zero polynomial P in Nvariables without constant term and mutually different c1, c2, . . . , cN∈Hsuch that F(x) = P(e−c1x, e−c2x, . . . , e−cNx). Specifically, there exist a natural number N∈N, a non-empty finite set J⊂NN 0\ {(0,(N) . . ., 0)}, scalars αj∈R\{0}, for j= (j1, j2, . . . , jN)∈J, and c= (c1, c2, . . . , cN)∈ HNsuch that F(x) = X j∈J αj(e−c1x)j1·(e−c2x)j2· · · (e−cNx)jN=X j∈J αje−(c·j)x 49 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces Since His Q-linearly independent and J⊂NN 0\ {(0,(N) . . ., 0)}, the numbers {c·j: j∈J, c∈HN}are non-null and mutually different. But Lemma 4.5 assures that the family {e−(c·j)x:j∈J, c∈HN}is linearly independent, so F6≡ 0 (because αj6= 0 for all j∈J). It is well-known that the field R, as seen as a vector space over Q, has dimension c. Hence, we always can find a Q-linearly independent set in R(or even inside (0,+∞)) with cardinal c. Theorem 4.7. The family nPaeM([0,1]) is strongly c-algebrable. Proof. Let H⊂(0,+∞) be a linearly Q-independent set with card(H) = c. For each c∈H, we define the sequence fc= (fc,n)nas fc,n(x) := e−cx ·Tn(x),(x∈[0,1], n ∈N), where (Tn)n∈L0([0,1])Nis the “Typewriter sequence” defined in Example 3.10. Recall that m{x∈[0,1] : Tn(x)6= 0}→0 (n→ ∞),(4.1) and that (Tn)ndoes not converge to 0 pointwise a.e. on [0,1]. Let Bbe the algebra generated by the family of sequences {fc:c∈H}, that is, Bis the family of all sequences (Fn)nfor which there exists N∈N, mutually different c1, c2, . . . , cN∈Hand a non-zero polynomial Pin Nreal variables without constant term such that Fn=P(fc1,n, fc2,n, . . . , fcN,n),(n∈N). Specifically, there exist a natural number N∈N, a non-empty finite set J⊂NN 0\ {(0,(N) . . ., 0)}, scalars αj∈R\{0}, for j= (j1, j2, . . . , jN)∈J, and c= (c1, c2, . . . , cN)∈HN 50 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence such that Fn(x) = X j∈J αjfc1,n(x)j1· · · fcN,n(x)jN =X j∈J αje−(c·j)xTn(x)|j|= X j∈J αje−(c·j)x!Tn(x),(4.2) where in the last equality is crucial the fact that Tn(x) is a characteristic function, so Tn(x)β=Tn(x) for any β > 0. For each subset J⊂NN 0\ {(0,(N) . . ., 0)}and each c∈HN, consider the function ϕc,J : [0,1] →Rgiven by ϕc,J (x) := X j∈J αje−(c·j)x. Since His a Q-linearly independent set and αj6= 0 (j∈J), by Lemma 4.6, ϕc,J (x) is a non-null function. So, by (4.2), Fnis non-null for each n∈N, and the algebra B is free. It only rests to prove that every non-zero element of our algebra Bconverges in measure to zero but not pointwise a.e. on [0,1]. Observe that the sequence (Fn(x))n= (ϕc,J (x)Tn(x))nconverges to zero in measure because, for any ε > 0, we have the following inclusion {x∈[0,1] : |ϕc,J (x)Tn(x)|> ε}⊂{x∈[0,1] : Tn(x)6= 0}. By (4.1), the measures of the above sets go to zero when n→ ∞, and then Fn→0 in measure. Finally, as ϕc,J (x) is a finite linear combination of exponentials, we have that ϕc,J (x) = 0 for finitely many points (specifically, at most card(J)−1 points, see for instance, [47, p. 46]). Hence, (4.2) and the non-pointwise a.e. convergence of (Tn)n to zero give us that (Fn)ndoes not converge to zero pointwise a.e. on [0,1]. Observe that while searching for a free algebra inside our family of sequences of functions, we did not need to pay much attention to the topological structure of 51 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces L0([0,1])N. This is not the case anymore if we speak about dense-lineability, as Ara´ujo et al. did previously in [4]. Now, by taking into account the topological structure of L0([0,1])N, we are able to get spaceability of our family of sequences of functions. Theorem 4.8. The family nPaeM([0,1]) is spaceable in (c0(L0([0,1])), d∞). Proof. The proof presented here is a constructive proof, which also heavily relies on the “Typewriter sequence” defined previously on Example 3.10. First, we start by dividing the set Ninto infinitely many strictly increasing and disjoints subsequences (i(k, n))n∈N(k∈N) such that N=[ k,n∈N {i(k, n)}. We can take for instance the subsequences (i(k, n))ngiven by i(k, n) := k(k+ 1) 2+ (n−1)k, (k, n ∈N). For each k∈N, define the sequence T(k) = (T(k, n))nas follows T(k, n)(x) :=    χhj 2i(k,m),j+1 2i(k,m)i(x) if n=j+ 2i(k,m),0≤j < 2i(k,m), 0 elsewhere. Roughly speaking, for fixed k∈N, we preserve every term of the “Typewriter sequence” where the support has length 1 2i(k,m)for m∈N, and change the rest to be 0. Similarly as the “Typewriter sequence”, it is straightforward that every sequence T(k) is convergent to zero in measure, since for every ε > 0 it holds {x∈[0,1] : |T(k, n)(x)|> ε} ⊂ {x∈[0,1] : T(k, n)(x)6= 0}, and the measure of the lasts sets goes to zero when n→ ∞. Moreover, by construction of the sequences T(k), given any x∈[0,1], there are infinitely many terms of T(k)(x) where the sequence takes the value 0, and infinitely many terms where it takes the 52 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence value 1, thus making it impossible for the sequence (T(k, n)(x))nto be convergent to zero. By construction, if k6=k0, then the sequences T(k) and T(k0) cannot be both simultaneously non-zero. This implies that, if we choose scalars λ1, λ2, . . . , λN∈R, and consider the linear combination λ1T(k1) + λ2T(k2) + · · · +λNT(kN)=0, then for any 1 ≤j0≤Nand x∈[0,1] we can always find some nj0∈Nsuch that T(kj0, nj0)(x) is non-zero but T(kj, nj0) = 0 for j6=j0. So, writing the linear combination at the nj0-coordinate, we get that λj0T(kj0, nj0)(x) = 0, whence λj0= 0, and we get that the set {T(k) : k∈N}is linearly independent. Now, in order to prove the spaceability of our family of sequences of functions we need to find some closed infinite dimensional vector space inside our family. For this, let us define M:= span{T(k) : k∈N}. It is clear that Mis a closed infinite dimensional subspace of (c0(L0([0,1])), d∞). It only rests to prove the non-pointwise a.e. convergence to zero of the members of M. For that purpose, observe that every non-zero member of Mis a finite or infinite linear combination of sequences T(k). More precisely, if F∈M\ {0}, there exists a subset J⊂Nand scalars λj∈R\ {0}for every j∈J, such that we can write F=X j∈J λjT(kj). Fix j0∈Jand let J0:= 2i(kj0,m)+j:m∈N,0≤j < 2i(kj0,m). By construction, Fn=λj0T(kj0, n) for every n∈J0, and so, for fixed x∈[0,1], there are infinitely many natural numbers (at least every number in J0) such that 53 Modes of convergence Lin. Alg. Struc. Func. Seq. Spaces Fn(x) = λj06= 0, and infinitely many natural numbers such that Fn(x) = 0. Hence (Fn)nis not pointwise a.e. convergent to zero. 4.2 Pointwise vs. Uniform Convergence 4.2.1 Definitions and first examples In this Section we focus on two well-known different types of convergence, the pointwise and the uniform convergence on [0,1]. The fact that these two types of convergence are the most commonly studied when working with sequences of functions allows us to provide easily many examples of sequences of functions converging pointwise but not uniformly on [0,1]. For instance, we can just recall the sequence (fn)n⊂L0([0,1]) defined on Example 3.6 given by fn(x) = χEn(x),for every n∈N, where En=1 n+1,1 n. Observe that, in particular, lim sup n→∞ En=∅. This is not only a coincidence for the sequence of functions (fn)n, but rather a shared property of many sequences of scalar multiples of characteristics functions that are pointwise convergent to zero but not uniformly convergent on [0,1]. Proposition 4.9. Let (X, M, µ)be a measure space, and (αn)nbe a sequence of nonzero real numbers such that either (αn)n∈c0or there exists M > 0such that |αn|> M for nlarge enough. Let En∈ M \ {∅}and fn=αnχEn(n∈N). Then: (1) fn→0pointwise on Xif and only if (αn)n∈c0or lim sup n→∞ En=∅. (2) fn→0pointwise a.e. on Xif and only if (αn)n∈c0or µlim sup n→∞ En= 0. (3) fn→0uniformly on Xif and only if (αn)n∈c0. 54 Lin. Alg. Struc. Func. Seq. Spaces Modes of convergence Proof. (1) Suppose that fn→0 pointwise on Xand lim sup n→∞ En6=∅. Let x0∈lim sup n→∞ En. Then, for every n∈Nthere exists mn∈Nwith mn≥nand x0∈Emn. So, |αmn|=|fmn(x0)| → 0 (n→ ∞).(4.3) Hence (αn)ncannot be far from zero, and by hypothesis, must converge to 0. Conversely, for any x∈X,|fn(x)|=|αn|χEn(x)≤ |αn|. So, if (αn)n∈c0, then fn(x)→0. On the other hand, if lim sup n→∞ En=∅, then there is n0∈Nsuch that x /∈Enfor all n≥n0; hence, fn(x) = 0 for all n≥n0and we are done. (2) Assume that fn→0 pointwise a.e. on Xand µlim sup n→∞ En>0. Then, we always can find x0∈lim sup n→∞ Enand fn(x0) converges to zero and we can finish as in (1). In fact, in (1) we have shown that if x /∈lim sup n→∞ En, then fn(x)→0 (n→ ∞), and if x∈lim sup n→∞ En, then fn(x)→0 (n→ ∞) if and only if (αn)n∈c0. So, it is clear that if (αn)n∈c0or µlim sup n→∞ En= 0, then fn→0 pointwise a.e. on X. (3) Suppose that fn→0 uniformly on X. Assume, by way of contradiction, that αn/∈c0. Then, by hypothesis, there exist M > 0 and n0∈Nsuch that |αn|> M > 0 for all n≥n0. But the uniform convergence of fnallows us to get m≥n0such that |fn(x)|<M 2for all x∈Xand n≥m. Therefore, fn= 0, and so En=∅for all n≥m, which is impossible by hypothesis. The reciprocal is immediate because of the fact that |fn(x)|≤|αn|for all n∈N and every x∈X. Remark 4.10. Observe that, similarly as in part (1) of the above Proposition, if we have any sequence (ϕn)nof measurable functions, it can be proved that (ϕn·χEn)n converges pointwise to zero on Xif lim sup n→∞ En=∅. Indeed, it would exists n0∈N such that x /∈Enfor every n≥n0. Hence, ϕn(x)·χEn(x) = 0 for all n≥n0. 55 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces 5.1 Convergence in L1(X, µ) If we consider a general measure space, we define the L1-convergence as follows. Definition 5.1. Let (X, M, µ)be a measure space. Let fn, f :X→K(n∈N)be measurable functions. We say that fn→f(n→ ∞)in L1-norm on X, if kfn−fk1=ZX |fn−f|dµ →0 (n→ ∞). In this section we will study the relationship among the convergence in L1-norm and other modes of convergence. 5.1.1 Measure vs. L1convergence As next result shows, the convergence in L1-norm is stronger than the convergence in measure. Theorem 5.2. Let (X, M, µ)be a measure space. Let fn, f :X→K(n∈N)be measurable functions. If fn→f(n→ ∞)in L1-norm, then fn→f(n→ ∞)in measure on X. Proof. Fix ε > 0 and n∈N. Then 0≤µ{x∈X:|fn(x)−f(x)| ≥ ε}=Z{|fn−f|≥ε} 1dµ ≤Z{|fn−f|≥ε} 1 ε· |fn−f|dµ ≤1 εZX |fn−f|dµ =1 εkfn−fkL1. Thus, if fn→f(n→ ∞) in L1-norm we have that µ{x∈X:|fn(x)−f(x)| ≥ ε}→0 (n→ ∞), that is, fn→f(n→ ∞) in measure on X. 62 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) The reciprocal is not true, as the next example shows. Example 5.3 (Measure ;L1-norm).Let fn: [0,+∞)→R(n∈N) be the sequence of functions given by fn(x) := nχ[1 n,2 n](x),(x∈[0,+∞), n ∈N). 0 1 1 2 2 1 2 f1(x) = χ[1,2](x) 0 1 1 2 2 1 2 f2(x)=2χ[1 2,1](x) Figure 5.1: First terms of the sequence fn(x) = nχ[1 n,2 n](x) Clearly, fn→0 =: fpointwise on [0,+∞). Now, given any ε∈(0,1), we have that m{x∈[0,+∞) : |fn(x)−f(x)| ≥ ε}≤m1 n,2 n=1 n. So, we also have that fn→0 in measure. Nonetheless, let us compute its L1-norm: kfn−fk1=Z+∞ 0 |fn(x)|dx =Z2 n 1 n n dx =n·1 n= 1. Thus, we do not have convergence of the sequence (fn)nto 0 in the L1-norm. 5.1.2 Uniform vs L1convergence Observe that there is no relation between uniform convergence and convergence in L1-norm, as the next two examples show. 63 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces Example 5.4 (L1-norm ;Uniform).Let fn: [0,1] →R(n∈N) be the sequence of functions of Example 3.6 given by fn(x) := χEn(x) (n∈N, x ∈[0,1]), where En=1 n+1,1 nfor all n∈N. We already know that this sequence converges pointwise but not uniformly to f:= 0 on [0,1]. But, kfn−fk1=Z1 n 1 n+1 1dx =m 1 n+1,1 n=1 n(n+ 1) →0,(n→ ∞). So, we obtain that fn→0 in L1-norm. For the other relation, observe that this cannot happen in a finite measure setting. Indeed, if we have that fn→funiformly on Xwith µ(X)<+∞, then we also have the convergence of the integrals, hence the L1-norm convergence. So, we will work with measurable functions defined on [0,+∞). Example 5.5 (Uniform ;L1-norm on infinite measure spaces).Consider the sequence of functions fn: [0,+∞)→R(n∈N) given by fn(x) := 1 nχ[0,n](x),(x≥0, n ∈N). 0 1 1 2 2 1 f1(x) = χ[0,1](x) 0 1 1 2 2 1 f2(x) = 1 2χ[0,2](x) Figure 5.2: First terms of the sequence fn(x) = 1 nχ[0,n](x) 64 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Each fnis a measurable function, and fn→0 uniformly on [0,+∞) (just take into account that |fn(x)| ≤ 1/n for all x≥0). Furthermore, if we compute the L1-norm of each fn, we obtain, kfnkL1=Z+∞ 0 |fn(x)|dx =Zn 0 1 ndx = 1, and fncannot converge to zero in L1-norm. The last sequence is a classical example of a sequence of functions in L0([0,+∞))N converging uniformly to zero but not in L1-norm, and it will be the germ of the proofs of all results in this Subsection. But before, let us collect all the relations between the different convergences. MeasurePointwise a.e. L1-norm Almost uniformUniform a.e. Pointwise Uniform X X X X X X X X X X X X (a) µ(X) = ∞ MeasurePointwise a.e. L1-norm Almost uniformUniform a.e. Pointwise Uniform X X X X X X X X X (b) µ(X)<∞ Define the family nL1U([0,+∞)) of all sequences of (classes of) Lebesgue measurable functions that are uniformly convergent to zero on [0,+∞) but not in L1-norm, that is, nL1U([0,+∞)) := (fn)n∈L0([0,+∞))N:fn→0 uniformly on [0,+∞) and kfnk16→ 0 (n→ ∞). 65 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces From the previous Example 5.5, we already have that this family is non-empty. In the next results we will state which is its algebraic size in terms of Lineability. But before we need an auxiliary result. Lemma 5.6. For any non-void set H⊂(0,+∞)the family of sequences 1 ncn :c∈H is linearly independent. Proof. Take a finite linear combination of elements of the family, that is, fix N∈N, c1, c2, . . . , cN∈H(c1< c2<· · · < cN) and λ1, λ2, . . . , λN∈R, and assume that, for all n∈N, λ1 1 nc1+λ2 1 nc2+· · · +λN 1 ncN= 0.(5.1) But it is a well-known result of P´olya (see for instance [34, Corollary 3.2] or [47, pp. 46–47]) that the “generalized polynomial” p(x) := λ1xc1+λ2xc2+· · · +λNxcN has a finite number of zeros (at most N−1). So, the only possibility for (5.1) to be true for all n∈Nis that λ1=λ2=· · · =λN= 0. Theorem 5.7. The family nL1U([0,+∞)) is strongly c-algebrable. Proof. Let H⊂(0,+∞) be a Q-linearly independent set with card(H) = c. For every c∈H, consider the sequence of functions fc:= (fc,n)n∈L0([0,+∞))Ngiven by fc,n(x) := 1 ncχ[0,en](x) (x≥0, n ∈N).(5.2) For any c∈H, sup x∈[0,+∞) |fc,n(x)|= sup x∈[0,+∞) 1 ncχ[0,en](x) =1 nc→0 (n→ ∞), 66 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) and applying Theorem 3.2, the sequence fcis uniformly convergent to zero. Furthermore, it does not converge to zero in L1-norm, since for every c∈H, we have kfc,nkL1=Z+∞ 0 1 ncχ[0,en](x) dx =en nc→+∞(n→ ∞). Let Bbe the algebra generated by the family {fc:c∈H}, which is a linearly independent family by Lemma 5.6. We claim that Bis a free algebra such that any non-zero member is a sequence uniformly convergent to zero in [0,+∞) but not in L1-norm. Let F= (Fn)n∈ B\{0}. Similarly to Theorems 4.7 and 4.12, there exist a natural number N∈N, a non-empty finite set J⊂NN 0\ {(0,(N) . . ., 0)}and scalars αj∈R\ {0}, for j= (j1, j2, . . . , jN)∈Jsuch that, for every n∈Nand x≥0, Fn(x) = X j∈J αjfc1,n(x)j1· · · fcN,n(x)jN= X j∈J αj 1 nc·j!χ[0,en](x),(5.3) where the fact of χ[0,en]being a characteristic function is crucial again in the last equality. By the Q-linearly independence of H, the numbers c·jare mutually distinct and non-null. Moreover, αj6= 0 (j∈J), hence by Lemma 5.6 the numbers X j∈J αj 1 nc·j can be zero at most for a finite number of n. In particular, (Fn)nis a non-null sequence and the algebra is free. Furthermore, sup x≥0 |Fn(x)| ≤ X j∈J |αj|1 nc·j→0 (n→ ∞). So, F= (Fn)nis uniformly convergent to zero on [0,+∞). Finally, kFnk1=Z+∞ 0 X j∈J αj 1 nc·j!χ[0,en](x) dx =X j∈J αj 1 nc·j en→+∞(n→ ∞), which concludes the proof. 67 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces Since L0([0,+∞))Nwith the product topology is also a complete metrizable (hence Baire) space and the family nL1U([0,+∞)) is c-algebrable, we obtain the next corollary about its linear size. Corollary 5.8. The family nL1U([0,+∞)) is maximal lineable. Observe that, although every non-zero sequence of the algebra Bconstructed in the proof of the previous theorem does not converge to zero in L1-norm, all the functions are, in fact, integrable on [0,+∞). Hence, B \ {0} ⊂ L1([0,+∞))N. Observe also that for any (Fn)n∈ B \ {0}we have sup n∈N kFnk1= +∞. In 2014, Bernal and Ord´o˜nez [19] considered the space CBLsof all sequences of continuous, bounded and integrable functions fn:R→R(n∈N) such that kfnk∞→0 (n→ ∞) and sup n∈N kfnk1<+∞, that is, their supremum norm converges to zero and the sequence of L1-norms is uniformly bounded. Moreover, they proved that k(fn)nk:= sup n∈N kfnk∞+ sup n∈N kfnk1 defines a norm on CBLs, under which it becomes a Banach space. Now, consider the family Fof all sequences of functions (fn)nof CBLssuch that kfnk16→ 0 (n→ ∞). In [19, Theorem 4.16 ] Bernal and Ord´o˜nez establish the spaceability of the family F in CBLs As a consequence, this family is also maximal lineable. However, the point with the family Fthat Bernal and Ord´o˜nez considered is that this family turns out to be smaller than the family nL1U([0,+∞)), that is, F ⊂ nL1U([0,+∞)), while the topology of CBLsis finer than the product topology of L1([0,+∞))N. As a consequence, the spaceability of the family nL1U([0,+∞)) cannot be directly derived from [19, Theorem 4.16], because not every open set in L1([0,+∞))Nturns out to be also open in CBLs. 68 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Again in [19], Bernal and Ord´o˜nez, just by considering the map T:`∞→CBLs given by T(an) =            2an n(x−k+ 1) if k−1≤x<k+1 2(1 ≤k≤n), 2an n(x−k) if k+1 2≤xk (1 ≤k≤n), 0 otherwise, for any sequence (an)n∈`∞, constructed an isomorphism Tbetween the non-separable space `∞and its image. Thus, T(`∞), and so CBLs, are not separable, which means that the maximal dense-lineability of the family Fcannot be obtained as an application of Theorem 2.7. But, in this setting, taking into account that the product topology of L1([0,+∞))Nactually makes L1([0,+∞))Nseparable, the next result can be proven. Theorem 5.9. The family nL1U([0,+∞)) is maximal dense-lineable in L1([0,+∞))N with the product topology. Proof. Both from Bernal and Ord´o˜nez result or directly from Theorem 5.7 (at this moment we do not care about the topology), we know that the family nL1U([0,+∞)) of the hypothesis is c-lineable. The space L1([0,+∞))N, endowed with the product topology, is a separable complete metrizable topological vector space, so its dimension is c. Moreover, as seen in Section 1.3, the vector space c00(L1([0,+∞))) of vanishing sequences of functions of L1([0,+∞)) is dense in L1([0,+∞))Nwith the product topology. In addition, nL1U([0,+∞)) ∩c00(L1([0,+∞))) = ∅because every sequence in c00(L1([0,+∞))) converges to zero in L1-norm, and nL1U([0,+∞))+c00(L1([0,+∞))) ⊂ nL1U([0,+∞)). Now, applying Theorem 2.7 with A=nL1U([0,+∞)), B=c00(L1([0,+∞))), and κ=c, we get the maximal dense-lineability. 69 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces In order to state the spaceability, it turns out that we must take into account not only the product topology but also the property of the sequences of being uniformly convergent to zero on [0,+∞). For this purpose, let Z1:= (fn)n∈L1([0,+∞))N:fn→0 uniformly on [0,+∞) endowed with the next F-norm k(fn)nkZ1:= sup n∈N kfnk∞+ ∞ X n=1 1 2n·kfnk1 1 + kfnk1 . It is straightforward that (Z1,k · kZ1) becomes a complete separable metrizable locally convex topological vector space, hence a Fr´echet space, whose topology recover both the product topology inherited by k·k1and the uniform convergence to zero. This is the right framework to state the spaceability. Theorem 5.10. The family nL1U([0,+∞)) is spaceable in (Z1,k·kZ1). Proof. Let us divide the interval [0,+∞) into infinitely many sequences of pairwise disjoint intervals (except, possibly, for the extremes). For every N∈Nand every M= 1, . . . , N, let IN,M := "N−1 X j=1 j(N−j) + M(M−1) 2, N−1 X j=1 j(N−j) + M(M+ 1) 2#. Observe that for each M∈N, the interval IN,M has always length M. For every k∈N, define the sequence G(k)=(G(k, n))n:= 1 nχIk+n−1,n n. It is straightforward that every sequence G(k) converges to zero uniformly in [0,+∞) but not in L1-norm (observe that kG(k, n)k1= 1 for all n∈N). Moreover, the family {G(k) : k∈N}is linearly independent because of the disjointness of the (interiors of the) supports of all functions included in it. Now, an application of Kalton criterion of spacebility (see Theorem 2.10) with X=Z1and Y={(fn)n∈X:kfnk1→0 (n→ ∞)}(which is closed in Xby a standard topological argument) finishes the proof. 70 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Remark 5.11. Sequences of nL1U([0,+∞)) are, by definition, not convergent to zero in L1-norm. But, in fact, by the construction of the algebra Bof the proof of Theorem 5.7, the sequences of L1-norms of every non-zero member of Bdiverges to infinity. However, every sequence of the closed vector space given by the spaceability of CBLsis uniformly bounded in L1-norm. Thus, it is natural to ask about the algebraic genericity of the family of sequences (fn)n∈L1([0,+∞))Nsuch that sup n∈N kfnk1<+∞, fn→0 uniformly on [0,+∞) and kfnk16→ 0 (n→ ∞). 5.2 Unbounded functions in CL1([0,+∞)) In this Section we will focus our attention on the space CL1([0,+∞)) of all continuous and integrable (classes of) functions on [0,+∞), that is, CL1([0,+∞)) = L1([0,+∞)) ∩C([0,+∞)). This space becomes a topological metrizable vector space when endowed with the natural translation-invariant metric given by dCL1(f, g) := kf−gk1+ ∞ X n=1 1 2n kf−gk∞,n 1 + kf−gk∞,n ,(5.4) Note that dCL1-convergence means L1-convergence plus uniform convergence on compacta. The necessary condition of convergence of series of scalars states that if a series is convergent, then its general term must tend to zero. Now, turning into the setting of integrable (and even continuous) functions, one may think that something similar should happen, namely, that if fis integrable on an unbounded interval (say [0,+∞)), then it must tend to zero somehow. However, this is far from being true. Our intuition would be actually correct if the function falso verifies that the limit of f(x) (x→+∞) exists, or even if the function is decreasing over the interval. Nonetheless, this affirmation is false in general. 71 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces under linear combinations. Let us see that this can also be extended to algebraic combinations, which would lead into the algebrability of the family nBCL1([0,+∞)). But, before we start with this, we need some notation, since the argumentation here differs a little bit from the one carried out in the proofs of Theorems 4.7 and 4.12. Definition 5.17. Let N∈N. Given a (monic)monomial in Nvariables m(x1, x2, . . . , xN) = N Y i=1 xαi i, where αi∈N∪{0}for all i= 1,2, . . . , N, and an increasing sequence of prime numbers P= (pi)i, we define the P-index of mby indP(m) := m(p1, p2, . . . , pN) = N Y i=1 pαi i. Remark 5.18. Observe that we can always state a bijection between any increasing sequence of prime numbers and the whole sequence of primes, so, by the uniqueness of the factorization theorem, given a natural number n∈N, there is only one way of expressing it as product of powers of prime numbers, that is, there is an unique factorization n= N Y i=1 pαi i, and this number is the P-index of the monomial m(x1, x2, . . . , xN) = N Y i=1 xαi i. This fact allows us to state a bijection between Nand the set of all monic monomials. So, given a monomial m, it is uniquely described by any of its P-index. Thanks to this, we will be able to find an algebra of dimension ℵ0inside the family nBCL1([0,+∞)). 78 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Theorem 5.19. The family nBCL1([0,+∞)) is strongly ℵ0-algebrable. Proof. For any n∈Nand p≥1, we consider the “triangles” on [0,+∞) given by Tn,p(x) =            np(2n+1x+ (1 −n2n+1))pif x∈n−1 2n+1 , n, np(−2n+1x+ (1 + n2n+1))pif x∈n, n +1 2n+1 , 0 otherwise, and we define the functions gp: [0,+∞)→Ras: gp(x) = ∞ X n=1 Tn,p(x). In fact, each Tn,p draws a “curved triangle” whose basis lies in the x-axis, has height np, and the other sides are convex (p≥1) functions. Because of the disjointness of the supports of the triangles Tn,p(x) (n∈N), the functions gpare well defined and continuous on [0,+∞). Furthermore, we can easily bound their L1-norm. Observe that, for each p≥1 and n∈N, the “triangle” Tn,p is continuous on n−1 2n+1 , n +1 2n+1 and infinite differentiable on n−1 2n+1 , n +1 2n+1 \ {n}. In fact, if x∈n−1 2n+1 , n, T0 n,p(x) = pnp2n+1(2n+1x+ (1 −n2n+1))p−1>0, T00 n,p(x) = p(p−1)np22n+2(2n+1x+ (1 −n2n+1))p−2>0, and if x∈n, n +1 2n+1 , T0 n,p(x) = −pnp2n+1(−2n+1x+ (1 + n2n+1))p−1<0, T00 n,p(x) = p(p−1)np22n+2(−2n+1x+ (1 + n2n+1))p−2>0, so the triangles Tn,p(x) are always “inside” the real triangles of basis n−1 2n+1 , n +1 2n+1  and height Tn,p(n) = np. Hence kgpkL1= ∞ X n=1 Z∞ 0 Tn,p(x)dx ≤ ∞ X n=1 np 2n+1 <+∞. 79 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces So, gp∈CL1([0,∞)). Now, by evaluating gpon the sequence of positive integers xn=nfor every n∈N, we get that gp(xn) = gp(n) = np→+∞(n→ ∞) which yields gp∈nBCL1([0,+∞)) for all p≥1. Let P= (pj)jbe the increasing sequence of all prime numbers which are greater than 3, and let us define for each j∈Nthe function Fjas Fj(x) := ∞ X n=1 Tn,log pj(x) = glog pj(x). Note that (Fj)j⊂nBCL1([0,+∞)). Let Bbe the algebra generated by (Fj)j, that is, B=P(F1, . . . , FN) : Pis a polynomial in Nvariables without constant term, N ∈N. We are going to prove that Bis the desired infinitely generated algebra. Let m(x1, . . . , xN) = N Y i=1 xαi i(αi∈N∪ {0}) be a non-constant monomial. Hence m(F1, . . . , FN)(x) = N Y i=1 Fi(x)αi= N Y i=1 ∞ X n=1 Tn,log(pi)(x)!αi . For each n∈Nand each x∈n−1 2n+1 , nwe have m(F1, . . . , FN)(x) = N Y i=1 Tn,log(pi)(x)αi = N Y i=1 n2n+1x+ (1 −n2n+1)log(pi)αi =n2n+1x+ (1 −n2n+1)PN i=1 log(pαi i) =n2n+1x+ (1 −n2n+1)log(indP(m)) 80 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Analogously, for each x∈n, n +1 2n+1 , we get m(F1, . . . , FN)(x) = n−2n+1x+ (1 + n2n+1)log(indP(m)) . So m(F1, . . . , FN) = glog(indP(m)).(5.11) In particular, m(F1, . . . , FN)∈nBCL1([0,+∞)), and trivially B ⊂ CL1([0,+∞)) (because continuity and integrability are stable under finite linear combinations). It only remains to prove that (Fj)jforms an algebraic independent set and that any element of Bis not bounded. In order to prove these properties, let us take a nontrivial algebraic combination F(x) := P(F1, . . . , FN)(x) = l X i=1 λimi(F1, . . . , FN)(x). For each n∈N, we have by (5.11) that F(n) = P(F1, . . . , FN)(n) = l X i=1 λinlog(indp(mi)).(5.12) By Remark 5.18, all exponents of the right hand part of (5.12) are positive and pairwise different, so |F(n)| → +∞(n→ ∞). In particular, F≡0 if and only if λi= 0 for every i= 1, . . . , l, and the algebra Bis free. Hence, F∈nBCL1([0,+∞)), and B ⊂ nBCL1([0,+∞))∪{0}. Thus, nBCL1([0,+∞)) is strongly ℵ0-algebrable. We finish this section stating that the family nBCL1([0,+∞)) is not only algebraically large, but also large in a pure topological sense. Theorem 5.20. The family nBCL1([0,+∞)) is residual in CL1([0,+∞)). Proof. Note that CL1([0,∞)) is Baire because, as it is easy to see, the distance dCL1 given in (5.4) is complete. Indeed, CL1([0,∞)) \nBCL1([0,+∞)) = ∞ [ n=1 An, 81 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces where for each n∈N, An:= f∈CL1([0,∞)) : |f(x)| ≤ nfor all x≥0. Since uniform convergence on compacta implies pointwise convergence, it is clear that each Anis closed in CL1([0,∞)). Moreover, given f∈ Anand ε > 0, it is also easy but cumbersome to construct a function g∈CL1([0,+∞)) such that dCL1(f, g)< ε but |g(x0)|> n for some x0>0 (just “cut” a small enough part of garound x0and substitute it by a linear affine function joining continuously gto the point (x0, n+1)). This implies that every Anhas empty interior, so that CL1([0,+∞))\nBCL1([0,+∞)) is of first category, which proves that nBCL1([0,+∞)) is residual. 5.3 Sequences of continuous, unbounded and integrable functions Until this point we have focused our attention on the algebraic structure of the family nBCL1([0,+∞)) of continuous, unbounded, and integrable functions. In this Section we will step up and consider sequences of functions of the family nBCL1([0,+∞)) with additional properties on its convergence. For this, let us introduce some useful notation. The family nBCL1 0([0,+∞)) will denote all the sequences of continuous, unbounded and integrable functions on [0,+∞) converging to zero in the metric dCL1, that is: nBCL1 0([0,+∞)) := (fn)n∈CL1([0,+∞))N:fn∈nBCL1([0,+∞)) for every n∈N and dCL1(fn,0) →0 (n→ ∞) =(fn)n∈c0(CL1([0,+∞))) : lim x→+∞|fn(x)|= +∞for all n∈N. Recall that convergence to zero in the metric dCL1means that kfnk1+ +∞ X k=1 1 2k·kfnk∞,k 1 + kfnk∞,k →0 (n→ ∞) 82 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) that is, we have convergence to zero in the L1-norm and uniformly convergence on compacta. The next example shows that the family nBCL1 0([0,+∞)) is not empty. Example 5.21. Let Tn(n∈N) be the triangular function already defined in Example 5.12. In order to construct our sequence of functions (fn)n∈nBCL1 0([0,+∞)) we will truncate the series given by the triangles, that is, for every n∈Nlet fn: [0,+∞)→R be defined as: fn(x) := ∞ X k=n Tk(x) (x≥0). Clearly, (fn)n⊂nBCL1([0,+∞)). On the other hand, since each fn(x) is the tail of a convergent series in CL1([0,+∞)) (see Example 5.12), we have that (fn)nconverges to zero in the metric dCL1as n→ ∞. Thus, (fn)n∈nBCL1 0([0,+∞)). As in the case of single functions, we can prove the existence of a free-generated infinite dimensional algebra inside our family nBCL1 0([0,+∞)). Theorem 5.22. The family of sequences nBCL1 0([0,+∞)) is strongly ℵ0-algebrable. Proof. Using the notation of the proof of Theorem 5.19, for each j, n ∈N, let Fj,n(x) := ∞ X k=n Tk,log(pj)(x). Let B0be the algebra generated by the sequences {(Fj,n)n:j∈N}. Following the same argument as in Theorem 5.19 and taking the monomial m(x1, . . . , xN) = N Y i=1 xαi i, we have for each n∈Nand x≥0, that m(F1,n, . . . , FN,n)(x) = ∞ X k=n Tk,log(indP(m))(x).(5.13) 83 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces Therefore, the sequence m((F1,n)n,...,(FN,n)n) is the sequence of tails of a convergent series (in the topology generated by dCL1) and we have that dCL1(m(F1,n, . . . , FN,n),0) →0 (n→ ∞). In particular, m((F1,n)n,...,(FN,n)n)∈nBCL1 0([0,+∞)). Finally, let (Fn)nbe a non-trivial algebraic combination of the sequences (Fj,n)n (j∈N), that is, for each non-zero polynomial Pin Nvariables without constant term and each n∈N, we consider Fn(x) := P(F1,n, . . . , FN,n)(x) = l X i=1 λimi(F1,n, . . . , FN,n)(x), where the λi’s are not simultaneously zero. For each n∈Nand each k≥n, by taking (5.13) into account, we obtain Fn(k) = P(F1,n, . . . , FN,n)(k) = l X i=1 λiklog(indP(mi)). Now, continuing in the same way as on the proof of Theorem 5.19, we get that |Fn(k)| → +∞(k→ ∞), hence B0⊂nBCL1 0([0,+∞)) ∪ {0}and B0is a free algebra. Thus nBCL1 0([0,+∞)) is strongly ℵ0-algebrable. An immediate consequence from the algebrability of nBCL1 0([0,+∞)) is that mere lineability can be easily obtained. Corollary 5.23. The family of sequences nBCL1 0([0,+∞)) is ℵ0-lineable. In this case we could only infer the ℵ0-lineability, since we obtained ℵ0-algebrability. Nonetheless, by a direct approach, it is easy to attain the maximal dimension of the space CL1([0,+∞))N. 84 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) Theorem 5.24. The family of sequences nBCL1 0([0,+∞)) is maximal lineable. Proof. As in the proof of Theorem 5.19, for each t∈0,1 8and each n∈Nwe define the functions fn,t(x) := ∞ X k=n Tk(x−t) (x≥0), and consider the set M0given by M0:= span (fn,t)n:t∈0,1 8. Following the same argument as in Theorem 5.14, we have that each fn,t is a continuous, unbounded and integrable function on [0,+∞), that is, fn,t ∈nBCL1([0,+∞)) for all t∈0,1 8and n∈N, and that the sequences (fn,t)n:t∈0,1 8are linearly independent. In addition, it is clear that the whole series ft(x) := ∞ X k=1 Tk(x−t) is convergent, both in L1-norm and uniformly on compact sets of [0,+∞), so in dCL1. Hence dCL1(fn,t,0) →0 (n→ ∞) for all t∈0,1 8,M0⊂nBCL1 0([0,+∞)) ∪ {0}and nBCL1 0([0,+∞)) is maximal lineable. Yet, in order to obtain the maximal dense-lineability, we need to establish a proper framework and the topology to be considered there. We consider the sequence space c0(CL1([0,+∞))) := {(fn)n⊂CL1([0,+∞)) : dCL1(fn,0) →0 (n→ ∞)}, endowed with the distance d∞given by d∞((fn)n,(gn)n) := sup n∈N dCL1(fn, gn),(5.14) for all (fn)n,(gn)n∈c0(CL1([0,+∞))). We know that (c0(CL1([0,+∞))), d∞) is a complete metric topological vector space, and thus, using again Baire’s Theorem, we have that its dimension is c. 85 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces We define the set B00 as B00 := (bn)n: exists n0∈Nsuch that bn∈Bfor all n≤n0and bn= 0 for all n > n0, where Bis the dense subset of CL1([0,+∞)) defined in Lemma 5.15. Lemma 5.25. The space B00 is dense in (c0(CL1([0,+∞))), d∞). Proof. First, it is clear that B00 ⊂c0(CL1([0,+∞))), since B⊂CL1([0,+∞)). Now, let (fn)n∈c0(CL1([0,+∞))). Given ε > 0, there is n0∈Nwith dCL1(fn,0) < ε for all n≥n0. Furthermore, the denseness of Bin CL1([0,+∞)) guarantees the existence of b1, . . . , bn0−1∈Bsuch that dCL1(fn, bn)< ε (1 ≤n≤n0−1). Finally, define bn:= 0 for n≥n0. It is clear that (bn)n∈B00 and, by construction d∞((fn)n,(bn)n) = sup n∈N dCL1(fn, bn) = max max 1≤n≤n0−1dCL1(fn, bn),sup n≥n0 dCL1(fn,0)< ε, and we are done. Theorem 5.26. The family of sequences nBCL1 0([0,+∞)) is maximal dense-lineable in c0(CL1([0,+∞))). Proof. We already have the maximal lineability of the family nBCL1 0([0,+∞)) from Theorem 5.24 and, by Lemma 5.25, B00 is a dense vector subspace of c0(CL1([0,+∞))) (hence, dense-lineable). Moreover, each function bnof a sequence (bn)n∈B00 is bounded and bn= 0 for nbig enough, so B00 +nBCL1 0([0,+∞)) ⊂nBCL1 0([0,+∞)), and B00 ∩nBCL1 0([0,+∞)) = ∅. Finally, Theorem 2.7 tells us that nBCL1 0([0,+∞)) is maximal dense-lineable in c0(CL1([0,+∞))). 86 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) 5.4 Final remarks 5.4.1 Maximal possible convergence Observe that, in the above proofs, when we deal with the family nBCL1 0([0,+∞)), and we consider the sequences (fn,t)ngiven by fn,t(x) = fn(x−t) = ∞ X m=n Tm(x−t), for t∈0,1 8,x∈[0,+∞) and n∈N, we only checked them to be in our space of functions c0(CL1 0([0,+∞))), that is, they had to converge to zero in L1-norm plus uniformly on compact sets of [0,+∞). We want to comment that the convergence to zero can be strengthen. Let ([0,+∞),L, m) be the Lebesgue measure space. Recall (see Definition 3.3) that a sequence of functions (fn)nconverges almost uniformly (on [0,+∞)) to a function fif, for every ε > 0, there exists a set E⊂[0,+∞) with m(E)< ε such that fn→f uniformly on [0,+∞)\E. In this line, all the sequences (fn,t)nconstructed in the proof of Theorem 5.19 and Theorem 5.22 not only converge to zero uniformly on compacta on [0,+∞), but also converge almost uniformly to zero on [0,+∞). Indeed, following the notation used in this proof, and denoting by Enthe support of each function fn,t, we obtain, because of the pairwise disjointness of the supports of the Tk’s, that, En:= supp(fn,t) = supp ∞ X k=n Tk(x−t)!= ∞ [ k=n supp(Tk(x−t)) = ∞ [ k=nk+t−1 2k+1 , k +t+1 2k+1  Clearly, for each natural number n∈Nwe have that En+1 ⊂En. Furthermore, 0≤m(En)≤ ∞ X k=n 1 2k→0 (n→ ∞). 87 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces Theorem 5.33. The family nBC∞L1([0,+∞)) of unbounded and integrable functions on [0,+∞)of class C∞is maximal lineable. Proof. Following the steps of the proof of Theorem 5.14, consider the vector space M given by M:= span ft:t∈0,1 8, where ftis the infinite sum of the Φn’s traslated to the right by the factor t∈01 8. It is obvious that every finite linear combination of elements of Mis of class C∞and integrable, hence it only remains to prove the unboundedness of each member of M. Let us assume that there are 0 ≤t1< t2<· · · < ts<1 8and scalars c1, c2, . . . , cs∈R such that cs6= 0, and there exists N∈Nwith min{ts−ti:i= 1,2, . . . , s −1}>1 2N. Hence, by taking for any n > N the point xn=n+ts, we have |c1ft1(xn) + c2ft2(xn) + · · · +csfts(xn)|=|cs|ne1−1 1−(2n+1ts)2→+∞(n→ ∞). Thus, the family nBC∞L1([0,+∞)) is maximal lineable. This result can be also extended to sequences of functions, using a similar construction as in the one provided in Theorem 5.24. Theorem 5.34. The family nBC∞L1 0([0,+∞)) of sequences (fn)nof unbounded and integrable functions on [0,+∞)of class C∞such that fn→0 (n→ ∞)in L1-norm and almost uniformly on [0,+∞)is maximal lineable. Proof. Following the same steps as in Theorem 5.24, and using the exponential-like functions defined at the beginning, we define the sequence fn,t for every n∈Nand t∈0,1 8as fn,t(x) = ∞ X k=n Φk(x−t). So, if we consider the vector subspace M0:= span (fn,t)n:t∈0,1 8, 94 Lin. Alg. Struc. Func. Seq. Spaces The space L1([0,+∞)) we have that each fn,t is of class C∞, unbounded and integrable on [0,+∞). In addition, it is clear that the whole series ∞ X k=1 Φk(x) converges to zero both in L1-norm and almost uniformly on [0,+∞). Thus, nBC∞L1 0([0,+∞)) is maximal lineable. 95 The space L1([0,+∞)) Lin. Alg. Struc. Func. Seq. Spaces 96 Chapter 6 Anti M-Weierstrass sequences of functions 6.1 Concepts and examples In the previous chapters we have considered sequences of functions with different properties, and we studied the linear and algebraic size of these families. Now, we will turn our attention into series of functions. Let X6=∅be a non-empty set, and fn:X→K(n∈N) be a sequence of functions. Usually, when the uniform convergence of the series of functions ∞ X n=1 fn(x) has to be studied, the first tool one thinks about is the well-known Weierstrass M-test (see for instance [3, §9.6]). Theorem 6.1 (Weierstrass M-test).In the above conditions, if there exists a sequence (Mn)n⊂(0,+∞)such that ∞ X n=1 Mn<+∞and |fn(x)| ≤ Mnfor every x∈X and every n∈N, the series ∞ X n=1 fn(x)is absolutely and uniformly convergent on X. 97 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces However, the reciprocal is false in general, that is, it is possible to find series of functions that are uniformly convergent but do not satisfy all the hypothesis of the Weierstrass M-test. More precisely, the condition about the majorant sequence can be dropped without losing the uniform convergence of the series of functions. The upcoming example (see [23, Chapter 1, Example 10]) will show this situation. Example 6.2. Let fn: [0,1] →R(n∈N) be the sequence of (continuous) functions given by fn(x) :=            1 nsin2(2n+1πx) if x∈1 2n+1 ,1 2n, 0 if x∈0,1 2n+1 ∪1 2n,1. Since fn(x)≥0 for all x∈[0,1], the normal convergence of the series and the absolute convergence are equal for this series. If x∈0,1 2\ {2−n:n∈N}, there is a unique n0∈Nsuch that x∈1 2n0+1 ,1 2n0. So, ∞ X n=1 fn(x) = fn0(x) = 1 n0 sin2(2n0+1πx). As fn(x) = 0 elsewhere, we get the absolute convergence of our series for any x∈[0,1]. In order to obtain the uniform convergence we will apply the Cauchy’s Criterion for series (see for instance [3, §8.8]). Fix n, p ∈Nand x∈[0,1]. As for the whole series, there is at most one nx∈Nwith n+ 1 ≤nx≤n+psuch that fnx(x)6= 0, and, in this case,  n+p X k=n+1 fk(x) <|fnx(x)| ≤ 1 nx ≤1 n+ 1 <1 n. for every n, p ∈Nand every x∈[0,1]. Now, given any ε > 0, there is N∈Nsuch that 1 n< ε for every n≥N. So, for 98 Lin. Alg. Struc. Func. Seq. Spaces Anti-M Weierstrass every n, p ∈Nwith n≥N,  n+p X k=n+1 fk(x) =1 n< ε, and we have the uniform convergence of the series on [0,1]. Finally, for any n∈N, fn3 2n+2 =1 nsin23 2π=1 n, so kfnk∞=1 n, and ∞ X n=1 kfnk∞= +∞. A sequence of functions as given in Example 6.2 will be called an Anti M-Weierstrass sequence. Concretely: Definition 6.3. Let fn: [a, b]→R(n∈N)be a sequence of continuous functions on [a, b]. We will denote by AMW([a, b]) (or just AMW, if there is no possibility of confusion on the interval)the family of Anti M-Weierstrass sequences on [a, b], that is, the family of sequences (fn)nsuch that ∞ X n=1 fn(x)is absolutely and uniformly convergent on [a, b], but ∞ X n=1 kfnk∞diverges. Thanks to Example 6.2 we already know that AMW([0,1]) 6=∅. In order to look for more general examples we introduce the following helpful family. Definition 6.4. Let Fbe the family of all sequences of continuous functions un: [a, b]→R such that: (a) The supports are pairwise disjoint, that is, supp(un)∩supp(um) = ∅, n 6=m, 99 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces (b) The sequence (||un||∞)nis bounded and far from zero, that is, 0<inf n∈N||un||∞≤sup n∈N ||un||∞<+∞. Observe that any element of this family Fallows us construct a series of functions with a very concrete convergence. Lemma 6.5. Let (un)n∈ F and (an)n⊂R. Then: (a) The series ∞ X n=1 anun(x)converges absolutely on [a, b]. (b) The series ∞ X n=1 anun(x)converges uniformly on [a, b]if and only if (an)n∈c0. (c) The series ∞ X n=1 kanunk∞<+∞if and only if (an)n∈`1. Proof. (a) The absolute convergence of the series is immediate, since the disjointness of the supports of the un’s implies that, for a fixed x0∈[a, b], either un(x0) = 0 for all n∈N, or there exists only one n0∈Nsuch that x0∈supp(un0), and ∞ X n=1 |anun(x0)|=|an0un0(x0)|(<+∞). (b) Firstly, consider the case when (an)n∈c0. Because (un)n∈ F, then M:= sup n∈N ||un||∞∈(0,+∞). Given any ε > 0, there exists N∈Nsuch that |an|<ε Mfor any n≥N. Thus, as for each x∈[a, b] there is at most one n0≥Nsuch that x∈supp(un0),  ∞ X n=N anun(x) =|an0un0(x)| ≤ ε M·M=ε. Hence, the uniform convergence of the series on [a, b] is obtained. 100 Lin. Alg. Struc. Func. Seq. Spaces Anti-M Weierstrass Reciprocally, as the series is uniformly convergent, we have anun(x)→0 (n→ ∞) uniformly on [a, b]. Because (un)n∈ F, then L:= inf n∈N||un||∞>0. Given any ε > 0, there is n0∈Nsuch that |anun(x)|< ε ·Lfor every x∈[a, b] and every n≥n0. But for any n≥n0, there exists xn∈[a, b] such that un(xn)≥L, hence |an|=|anun(xn)| |un(xn)|≤ε·L L=ε, and an→0 (n→ ∞). (c) As (un)n∈ F, we have 0< L := inf n∈N||un||∞≤sup n∈N ||un||∞=: M < +∞, and then, 0≤L· |an| ≤ kanunk∞≤M· |an|<+∞,(n∈N). Thus, by comparison test, (an)n∈`1if and only if (||anun||∞)n∈`1. As a consequence of this lemma, whenever we have a sequence of functions (un)n∈ Fand a sequence of coefficients (an)n∈c0\`1, the sequence of functions (anun)n belongs to the family AMW([a, b]). In particular, it allows us to provide a wide plethora of sequences of functions with this behaviour. Examples 6.6. 1. The first example of an Anti M-Weierstrass sequence given in Example 6.2 can be rewritten in terms of Lemma 6.5. For this, for every n∈Nand x∈[0,1] we just have to consider an=1 nand un(x) = sin2(2n+1πx)χ(1 2n+1 ,1 2n)(x). 2. For any interval [a, b] we can also adapt this example. For this, consider the sequence un: [a, b]→R(n∈N) of continuous functions given by un(x) =          sin 2nπx−a b−a−πif x∈In, 0 elsewhere, 101 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces where In=(2n−1)a+b 2n,(2n−1−1)a+b 2n−1for every n∈N. The In’s are pairwise disjoint, and so the un’s have disjoint support. Furthermore, since kunk∞= 1 for all n∈N, we immediately obtain that (un)n∈ F. Finally, thanks to Lemma 6.5, whenever we choose a sequence a= (an)n∈c0\`1 we will have absolute and uniform convergence on [a, b] of the series ∞ X n=1 anun(x), and there will not be a mayorant sequence, since for xn=(2n−2−1)a+b 2n−1∈In, we have that anun(xn) = ansin π 2=an, and the series ∞ X n=1 andiverges. Thus, the sequence (anun)n∈ AMW([a, b]). In the above examples, the element of Fcomes from the sinus function, but it is possible to consider any other continuous functions. Example 6.7. Let f∈C([a, b])\{0}. Let Λ := (αn)nbe any sequence of scalars such that a=α1< α2<· · · < αn−1< αn<· · · → b(n→ ∞). For each n∈Nwe define the function uΛ,f n: [a, b]→Rby uΛ,f n(x) :=                                    f(a)·x−α3n−2 α3n−1−α3n−2 if x∈[α3n−2, α3n−1] fa+ (b−a)·x−α3n−1 α3n−α3n−1if x∈[α3n−1, α3n] f(b)·α3n+1 −x α3n+1 −α3n if x∈[α3n, α3n+1] 0 otherwise. 102 Lin. Alg. Struc. Func. Seq. Spaces Anti-M Weierstrass Observe that with this construction we are adjusting the graph of finto the interval [α3n−1, α3n], and extending it continuously and affine-linearly to zero in the left and right intervals. It is clear that uΛ,f n∈C([a, b]); supp(uΛ,f n)⊂(α3n−2, α3n+1) for each n∈N, so the supports of the uΛ,f n’s are pairwise disjoint; and ||uΛ,f n||∞= ||f||∞∈(0,+∞) for any n∈N. So, trivially, (uΛ,f n)n∈ F. Thus, whenever we choose a sequence of coefficients (an)n∈c0\`1, the sequence of functions (anuΛ,f n)n∈ AMW([a, b]) for every continuous function f. In particular, card(AMW([a, b])) = c. Observe that if f≡0, with the same definition, uΛ,f n≡0 for any n∈N. But in this case, (uΛ,f n)n6∈ F. The above example let us define the following application. Proposition 6.8. Let Λ := (αn)nbe any sequence of scalars such that a=α1< α2<· · · < αn−1< αn<· · · → b(n→ ∞). Then, the application JΛ:C([a, b]) → F ∪ {0} f7→ JΛ(f) := (uΛ,f n)n, is well-defined, linear, injective, and satisfies the following properties: (1) uΛ,f n(α3n−1) = f(a)for any n∈Nand any f∈C([a, b]); (2) supp(uΛ,f n)⊂(α3n−2, α3n+1)for any f∈C([a, b]) and any n∈N. (3) For any n∈Nthere exists a linear affine transformation τnsuch that τn([α3n−1, α3n]) = [a, b]and uΛ,f n=f◦τnfor each x∈[α3n−1, α3n]and each f∈C([a, b]). (4) ||uΛ,f n||∞=||f||∞for any n∈Nand any f∈C([a, b]). Proof. By Example 6.7 we have that JΛis well-defined (even for f≡0), and we have the properties (1) to (4). The injectivity follows from the definition of uΛ,f nin [α3n−1, α3n]. Finally, the linearity is clear, because of the linearity of the piecewise 103 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces But, by the definition of JΛ(see Proposition 6.8(3)), for each n∈Nthere is a linear affine transformation τnsuch that τn[α3n−1, α3n] = [a, b] and JΛ(γigi+ 1)(x) = (γigi+ 1)(τn(x)) for any x∈[α3n−1, α3n] and any i= 1, . . . , N. So, for any n∈N, Fn(x) = X j∈J λja|j| n((γ1g1+ 1)(τn(x)))j1· · · (γNgN+ 1)(τn(x)))jNx∈[α3n−1, α3n]. Therefore, if Fn= 0 in [a, b], and in particular Fn= 0 in [α3n−1, α3n], X j∈J λja|j| n((γ1g1+ 1)(w))j1· · · ((γNgN+ 1)(w))jN= 0 w∈[a, b].(6.3) Now, as {gi}i∈Igenerates a free algebra in C([a, b]), by Lemma 6.13, {γigi+1 : i∈I} is also a generator system of a free algebra, so from (6.3) and the fact that there are infinitely many an6= 0 (this is true because an6∈ `1for example), we get λj= 0 for j∈Jand the algebra Ais free. Observe that trivially we also obtain that the dimension of Ais the dimension of the algebra generated by {gi}i∈Iin C([a, b]). It only rest to show that A ⊂ AMW ∪ {0}, that is, that any non-null sequence (Fn)nas above is Anti M-Weierstrass. Observe that, again by the definition of the application JΛ(see Proposition 6.8(2) and (4)), we have: supp(ui n)⊂[α3n−2, α3n+1],(1 ≤i≤N, n ∈N),(6.4) and, ||ui n||∞=||γigi+ 1||∞∈(0,+∞) for each i= 1, . . . , N and n∈N.(6.5) Therefore, by (6.4), supp(u1 n· · · uN n)∩supp(u1 m· · · uN m) = ∅for n 6=m, (6.6) and, by (6.2), we have the absolute convergence of ∞ X n=1 Fn. Now, by (6.5), sup n∈N ||u1 n· · · uN n||∞=||γ1g1+ 1||∞· · · ||γNgN+ 1||∞=: C∈(0,∞),(6.7) 110 Lin. Alg. Struc. Func. Seq. Spaces Anti-M Weierstrass and, since (an)n∈c0, and similarly to Lemma 6.5(b), by (6.2), (6.6), and (6.7), the series ∞ X n=1 Fn(x) converges uniformly on [a, b]. Finally, again by the definition of JΛ (see Proposition 6.8(1)), ui n(α3n−1) = γigi(a) + 1 =: δ= 1 or 2 for any i= 1, . . . , N and any n∈N.(6.8) Now, by (6.2) and (6.8), for each n∈N, |Fn(α3n−1)|=X j∈J λj(anu1 n)j1(α3n−1)· · · (anuN n)jN(α3n−1) =X j∈J λjδ|j|a|j| n . But (an)n∈c0\[ p≥1 `p, so by applying Lemma 6.14 to the polynomial p(x) = X j∈J λjδ|j|x|j|, we have ∞ X n=1 ||Fn||∞≥ ∞ X n=1 |Fn(α3n−1)|= ∞ X n=1 |p(an)|= +∞. Thus (Fn)n∈ AMW, and we are done. Theorem 6.16. Let Lbe a free algebra in c0\`1generated by {(ai n)n}i∈I. Let (un)n∈ F. Then the algebra Agenerated by {(ai nun)n}i∈Iis a free algebra in AMW ∪ {0} with the same dimension than L. Proof. Let (Fn)n∈ A. Then, for any x∈[a, b] we have, Fn(x) := X j∈J λj(a1 nun(x))j1· · · (aN nun(x))jN =X j∈J λj(a1 n)j1· · · (aN n)jN·(un(x))|j|, where N∈N,J⊂NN 0\ {(0,(N) . . ., 0)}non-empty and finite and λj∈R\ {0}for any j= (j1, . . . , jN)∈J. As (un)n∈ F,L:= inf n∈N||un||∞>0 ; moreover, any un(n∈N) takes the value 0 in [a, b]. Then, by continuity of un’s and the intermediate value property, for any n∈N 111 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces there is xn∈supp(un) such that un(xn) = L. Hence |Fn(xn)|=X j∈J λj(a1 n)j1· · · (aN n)jN·L|j| (n∈N) (6.9) If (Fn)n≡0, by (6.9), for each n∈N X j∈J λjL|j|·(a1 n)j1· · · (aN n)jN = 0. But Lis free, so λjL|j|= 0 for every j∈J, and because L > 0, we get that λj= 0 for all j∈J, and Ais also free. As L ⊂ c0, then ((a1 n)j1· · · (aN n)jN)n∈c0for any j= (j1, . . . , jN)∈J. Trivially, (up n)n∈ F for any p∈N. Hence, (Fn)nis a finite linear combination of products of sequences in c0\`1and sequences u|j| nn∈ F. Now, by Lemma 6.5, we have the absolute and uniform convergence of the series ∞ X n=1 Fn(x). It is only rest to show that (||Fn||∞)n6∈ `1to obtain (Fn)∈ AMW and finish the proof. By (6.9), ∞ X n=1 ||Fn||∞≥ ∞ X n=1 |Fn(xn)|= ∞ X n=1 X j∈J λjL|j|·(a1 n)j1· · · (aN n)jN = +∞, because Lis a free algebra in c0\`1, and so X j∈J λjL|j|·(a1 n)j1· · · (aN n)jN!n 6∈ `1. Finally, observe that (an)n∈`pif and only if (ap n)n∈`1. In particular, Lis an algebra in c0\`1if and only if Lis an algebra in c0\[ p≥1 `p. In [10], Bartoszewicz and Glab showed the existence of a free algebra Lin c0\[ p≥1 `psuch that the cardinality of any system of generators is the continuum. In fact they consider the algebra generated by 1 logcnn≥2:c∈H, where H⊂(0,+∞) is a Q-linearly independent set, and card(H) = c. So, we can obtain the next result about the algebrability of the family AMW. 112 Lin. Alg. Struc. Func. Seq. Spaces Anti-M Weierstrass Corollary 6.17. The family AMW is strongly c-algebrable. Proof. As in Corollary 6.11, we will give two different approaches. (1) By Bartoszewicz and Glab result [10, Theorem 2], take the algebra Lgenerated by the family 1 logcnn≥2:c∈H, where H⊂(0,+∞) is a Q-linearly independent set, and card(H) = c, and one of the sequences of functions (un)n∈ F (for instance any sequence of Example 6.2 or Example 6.6), an application of Theorem 6.16 gives us the strong c-algebrability of the family AMW. (2) Consider now the free algebras generated by {xc:c∈H}or {ecx :c∈H}in C([a, b]) (where H⊂(0,+∞) is a Q-linearly independent set and dim(H) = c), and the sequence of coefficients (an)n∈c0\[ p≥1 `pgiven by an=1 log(n)for n≥2. Then, an application of Theorem 6.15 gives us (again) the strong c-algebrability of AMW. 113 Anti-M Weierstrass Lin. Alg. Struc. Func. Seq. Spaces 114 Bibliography [1] A. Aizpuru, C. P´erez-Eslava and J.B. Seoane-Sep´ulveda, Linear structure of sets of divergent sequences and series, Linear Algebra Appl. 418 (2006), 595–598. [2] N. Albuquerque, L. Bernal-Gonz´alez, D. Pellegrino and J.B. 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