[S]-linear and convex structures in function families

[S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES L. BERNAL-GONZ´ ALEZ, J.A. CONEJERO, M. MURILLO-ARCILA, AND J.B. SEOANE-SEP´ ULVEDA Abstract. In this paper, the notion of [S]-lineability (originally coined by Vladimir I. Gurariy) is introduced and developed in a general abstract setting. This new notion is, then, applied to specific situations, as for instance, classes of differentiable nowhere monotone functions as well as families of vectors having dense orbit with respect to an operator. Large convex structures are also shown to exist inside the family of topologically mixing continuous selfmaps of a real compact interval. 1. Preliminaries and notation Throughout this manuscript, we will be concerned with the search for large algebraic structures inside subsets of certain topological vector spaces satisfying special properties. This enters the modern theory of lineability, whose basic concepts and results can be found in the book [2] (see, also, [3,6,7,13,14,16,17,23,36,39]). The one result that, most likely, inspired the introduction of this theory was perhaps that of Levine and Milman [30] from 1940 (and that we shall mention later on in this manuscript) that states that the subset of C[0,1] of all functions of bounded variation does not contain an infinite dimensional closed subspace. Later, in 1966, Gurariy showed an analogue of this previous result ([27]), namely he showed that the set of everywhere differentiable functions on [0,1] does not contain, either, an infinite dimensional closed subspace. This battery of negative results, among others, motivated the introduction of the notion of lineability in 2005 [5]. In this paper, we shall develop –mainly in the context of function spaces– two new notions, namely, [S]-lineability and convex lineability, see below. The following terminology of lineable and spaceable was first introduced by V.I. Gurariy in [5] and, later, in [22,28]. To avoid trivial or undesirable situations, we shall assume along this paper that a vector space Xnever collapses to {0}and that a topological vector space is always Hausdorff. A subset Mof a vector space Xis said to be lineable if M∪ {0}contains an infinite dimensional vector space. If Xis, in addition, a topological vector space, then Mis called spaceable if M∪{0}contains a closed infinite 2010 Mathematics Subject Classification. 15A03, 26A48, 46A45, 46E10, 47A16. Key words and phrases. lineability, spaceability, convexity, hypercyclic operator, K¨othe space, mixing map. 1 2 BERNAL, CONEJERO, MURILLO, AND SEOANE dimensional vector subspace. As introduced in [4], Mis called dense-lineable if M∪ {0}contains a dense vector subspace. Trivially, spaceability implies lineability and, if Xis infinite-dimensional, then dense-lineability implies lineability too. Gurariy also coined the concept contained in the next definition. As usual, we will denote by N,N0,R,C, C(R), ω, respectively, the set of natural numbers, the set N∪ {0}, the real line, the complex plane, the space of continuous functions R→R, and the space of scalar sequences KN, where K=Ror C. Definition 1.1. (V.I. Gurariy, 2004) Assume that Xis a topological vector space, Mis a subset of X,Sis a vector subspace of ω, and (un)nis a linearly independent sequence in X. Then we say that Mis: •[(un)n,S]-lineable if, for every sequence (cn)n∈ S, the series P∞ n=1 cnunconverges in Xto a vector of M∪ {0}. •[S]-lineable in Xif it is [(un)n,S]-lineable for some linearly independent sequence {un}n∈N⊂X. As far as we know, the notion of [S]-lineability was never developed in the related literature. This work would be the first time in which it appears. As usual, we denote by conv(A) the convex hull of a subset Aof a vector space X, that is, conv(A) = (n X i=1 λixi:xi’s ∈A, λi’s ∈[0,1], n X i=1 λi= 1, n ∈N). We introduce here the following concept. Definition 1.2. Let Mbe a subset of a vector space X. Then Mis said to be convex lineable if there exists an infinite linearly independent subset A⊂Xsuch that conv(A)⊂M. Note that demanding conv(A)⊂M∪ {0}instead of conv(A)⊂Min the last definition would be superfluous, because 0 is never a convex line- ar combination of linearly independent vectors. Plainly, lineability implies convex lineability, but the converse is false. For instance, consider the basic sequence {en}n∈Nin ω, where en= (0,0,...,0,1,0,0, . . . ) (with the 1 at the nth place). Trivially, the set M:= conv({en}n∈N) is convex lineable but not lineable. The aim of this paper is threefold. Specifically, in Section 2 the concept of [S]-lineability will be developed into a general context, mainly in the line of its relationship with other kinds of lineability. As concrete examples, in Sections 3 and 4 we intend to study the existence of [S]-lineability in two concrete families of vectors of certain topological vector spaces: the set of hypercyclic vectors for an operator and the set of differentiable nowhere [S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES 3 monotone real functions, respectively. Finally, in Section 5, convex lineability will be analyzed inside the set of topologically mixing continuous maps on a compact interval. 2. Properties of [S]-lineability Unless otherwise specified, throughout this section Xwill represent a given topological vector space. We start with the following trivial assertions, in which the notation of Definition 1.1 is used: •If M⊂Xis [S1]-lineable in Xand S1⊃ S2, then Mis [S2]-lineable. In particular, if Mis ω-lineable in Xthen it is also [S]-lineable for any vector space Sof scalar sequences. •A subset Mof Xis lineable if and only if it is [c00]-lineable, where c00 stands for the space of all eventually zero sequences. An initial natural question is whether in Definition 1.1 the linearly independent sequence {un}n∈Ncan be found in the set Mitself. An elementary answer is given by the following proposition. Proposition 2.1. Let Sbe a vector subspace of ωsuch that S ⊃ c00. Assume that Mis an [S]-lineable subset of X. Then there is a linearly independent sequence {un}n∈N⊂Msuch that, for every sequence (cn)n∈ S, the series P∞ n=1 cnunconverges in Xto a vector of M∪ {0}. Proof. From the assumption, there is a linearly independent sequence {un}n∈N⊂Xsatisfying the property in the conclusion. Fix N∈Nand consider the Nth basic vector (cn)n=eN, that belongs to c00, so to S. It follows that P∞ n=1 cnunconverges in Xto a vector of M∪ {0}. But P∞ n=1 cnun=uNand uN6= 0, so uN∈M. Since this holds for all N∈N, the conclusion follows.  Remark 2.2. With a similar argument it can be proved that the conclusion of the last proposition holds just by demanding S ⊃ c00kfor some k∈N, where c00k={x= (xn)n∈c00 :x1=· · · =xk= 0}. Moreover, we have in fact that a subset M⊂Xis lineable if and only if it is [c00k]-lineable for some k∈N. However, it is not at all apparent that a subset is lineable as soon as it is [S]-lineable for some infinite dimensional space S ⊂ ω. The next elementary proposition furnishes a sufficient condition in that line. Recall that a sequence (un)nof a metrizable topological vector space X is said to be basic whenever every u∈span{un:n∈N}, the closed linear span of the un’s, can be uniquely represented as a convergent series u= P∞ n=1 cnun. It is easy to see that a basic sequence is a linearly independent set. Proposition 2.3. Assume that (un)nis a basic sequence in X, where X is metrizable, and that Mis a [(un)n,S]-lineable subset of X, where Sis an infinite dimensional vector subspace of ω. Then Mis lineable. 4 BERNAL, CONEJERO, MURILLO, AND SEOANE Proof. Since (un)nis a basic sequence, Sis infinite dimensional and the mapping Φ:(cn)n∈ S 7−→ ∞ X n=1 cnun∈X is linear and one-to-one. Therefore dim(Φ(S)) = dim(S) = ∞. But Φ(S)⊂ M∪ {0}, so Mis lineable.  Trivially, the space X:= ω(endowed with the product topology) is ωlineable in itself, so [S]-lineable for all S. It is well known that ωdoes not support a continuous norm. The following proposition shows that the last maximality property “[S]-lineable for all S” does not hold if Xsupports a continuous norm. Proposition 2.4. Suppose that Xsupports a continuous norm and that M⊂X. Then Mis not ω-lineable in X. In particular, no normable space is ω-lineable in itself. Proof. It is enough to prove that Xcannot be ω-lineable in itself. For this, it suffices to exhibit, for a prescribed linear independent sequence {un}n∈N⊂ X, a vector (cn)n∈ωsuch that the series P∞ n=1 cnundoes not converge in X. Let k·k be a continuous norm on X. Since un6= 0, we get kunk>0 for all n∈N. Define cn:= kunk−1. Then P∞ n=1 cnundoes not converge in Xbecause, otherwise, it would converge in k · k, and so kcnunk → 0 as n→ ∞, which is absurd because kcnunk= 1 for all n∈N. We remark that the last proposition provides us with examples of sets that are dense-lineable or spaceable but not ω-lineable. If Gis a domain of the complex plane, that is, a nonempty connected open subset of C, then H(G) will denote, as usual, the space of all holomorphic functions G→C, endowed with the topology of uniform convergence on compacta (in the special case G=C, we obtain the space H(C) of entire functions). Under this topology, H(G) becomes a Fr´echet space, that is, a complete metrizable locally convex topological vector space. Also the spaces of differentiable functions Ck(R) up to order k(k∈N0∪ {∞}) are Fr´echet spaces under the topology of uniform convergence on compacta in Rfor functions and their derivatives up to order k. The space H(G) supports a continuous norm (indeed, take kfk:= supK|f|, where K⊂Gis any compact set with nonempty interior). Hence H(G) is not ω-lineable in itself by Proposition 2.4. However, no space Ck(R) (as ω) supports a continuous norm. Nevertheless, every Ck(R) contains a topologically isomorphic copy of ω; indeed, fix a sequence of nonzero functions {ϕn}n∈N⊂C∞(R) such that the support of each ϕnis contained in [n, n + 1]; then, for all k, the mapping (cn)n∈ω7−→ ∞ X n=1 cnϕn∈Ck(R) [S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES 5 is well-defined, linear, one-to-one and continuous. Thus, every Ck(R) is ω-lineable. An important class of vector subspaces of ωis that of K¨othe echelon spaces (see, e.g., [33, Chapter 27]). Let A= (aj,k)j,k≥1be an infinite matrix such that, for any j, k ≥1, we have 0 < aj,k ≤aj+1,k. The K¨othe echelon spaces λp(A) with 1 ≤p < ∞and c0(A) are defined by λp(A) = n(ck)k∈ω: ∞ X k=1 |ckaj,k|p<∞for all j∈No, c0(A) = n(ck)k∈ω: lim k→∞ ckaj,k = 0 for all j∈No. For instance, the sequence spaces `p:= {(ck)k∈ω:P∞ k=1 |ck|p<∞}, c0={(ck)k∈ω: limk→∞ ck= 0}are, respectively, λp(A) and c0(A) with aj,k = 1 for all j, k. Moreover, the space H(C) of entire functions can be seen as λ1(A) with aj,k =jk, while the space sof rapidly decreasing sequences may be identified with λ1(A), where aj,k =kj. The following theorem links spaceability to [S]-lineability when Sis a K¨othe space. Theorem 2.5. Assume that M⊂X. We have: (a) If Mis [S0]-lineable in Xfor some K¨othe echelon space S0, then it is [S]-lineable in Xfor every K¨othe space S. (b) If Xis a Fr´echet space supporting a continuous norm and Mis spaceable in X, then it is [S]-lineable in Xfor every K¨othe echelon space S. Proof. (a) We are assuming that Mis [S0]-lineable in Xfor some K¨othe space S0=λp(A) or c0(A), where p∈[1,∞). Our aim is to prove that M is also [S]-lineable in Xfor any other K¨othe space S=λp(B) or c0(B). Since λp(C)⊂c0(C) for every K¨othe matrix Cand every p∈[1,∞), we can assume without loss of generality that S0=λp(A) and S=c0(B). Let A= (aj,k)j,k≥1and B= (bj,k)j,k≥1. From the hypothesis, there is a linearly independent sequence {un}n∈N⊂Xsuch that, for every sequence (cn)n∈ωsatisfying ∞ X k=1 |ckaj,k|p<∞for all j= 1,2,..., (1) the series P∞ n=1 cnunconverges in Xto some point of M. Our goal is to exhibit a linearly independent sequence {vn}n∈N⊂Xsuch that, for every sequence (dn)n∈ωsatisfying lim k→∞ dkbj,k = 0 for all j= 1,2,..., 6 BERNAL, CONEJERO, MURILLO, AND SEOANE the series P∞ n=1 dnvnconverges in Xto some point of M. For this, we define for all n∈Nthe following scalars and vectors: γn:= 1 n2(a1,n +a2,n +· · · +an,n)(b−1 1,n +b−1 2,n +· · · +b−1 n,n), vn:= γnun. Note that the vn’s are still linearly independent. Fix a sequence (dn)nas above and let cn:= γndn(n∈N). It follows that dnvn=cnun, and so we would have proved that the series P∞ n=1 dnvnconverges to a point of Mas soon as we verify that (cn)n satisfies (1). With this aim, fix j∈N. Since dkbj,k →0 as k→ ∞, this sequence is bounded. Therefore we can find K∈(0,+∞) such that |dk| ≤ K b−1 j,k for all k∈N. Now, by the definition of cn, we have for all k≥jthat |ckaj,k|p≤Kp k2p. Since 2p≥2>1, the comparison test yields the desired convergence. (b) On the one hand, since Xis metrizable and locally convex, its topology is defined by an increasing sequence of seminorms p1≤p2≤ · · · ≤ pn≤ · · · (see, e.g., [38, Chapter 4]). By assumption, there exists a continuous norm k·k on X. It follows that the topology of Xis also generated by the increasing sequence of norms k·k1≤ k · k2≤ · · · ≤ k · kn≤ · · · , where k·kn:= k·k+pn. On the other hand, due to the completeness, every Cauchy sequence for each translation-invariant distance defining the topology of X is convergent. Moreover, a sequence in Xis Cauchy if and only if it is k·kn- Cauchy for every n∈N. When one applies this to a series Pnwnwith {wn}n∈N⊂X, one obtains that Pnwnconverges if all series PnkwnkN (N∈N) do. Now, assume that Mis a spaceable subset in X. Since finite dimensional subspaces are always closed, we can find a linearly independent sequence {un}n∈N⊂Xsuch that span{un}n∈N⊂M∪ {0}. Since each k·k is a norm, we can define vn:= un kunkn (n∈N). Trivially, span{vn}n= span{un}n, and so span{vn}n∈N⊂M∪ {0}. According to part (a), it suffices to prove that Mis [`1]-lineable. To this end, fix a sequence (cn)n∈`1as well as an N∈N. We have on the one hand that Sn:= Pn j=1 cjvj∈span{vn}n∈Nfor all n∈N. On the other hand, kcnvnkN=|cn|kunkN kunkn≤ |cn|for any n≥N. Since Pn|cn|<∞, the comparison test yields the convergence of PnkcnvnkN. It follows that P∞ n=1 cnvnconverges to some vector of X, say w0. In other words, Sn→w0 as n→ ∞. Consequently, w0∈span{vn}n∈Nand, thus, w0∈M∪{0}. This had to be shown.  [S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES 7 Remark 2.6. The assertion contained in Theorem 2.5(b) cannot be reversed, as we are going to prove by means of the following example (more examples will be provided in Sections 3–4). Levine and Milman [30] published in 1940 what was probably the first result on non-spaceability, namely, the set BC of continuous functions [0,1] →Rhaving bounded variation is not spaceable in the Banach space C[0,1] of continuous functions endowed with the maximum norm. Nevertheless, it is [`∞]-lineable (hence [c0]-lineable, and so [S]-lineable for every K¨othe space S, due to Theorem 2.5(a)). Indeed, let us fix a sequence (cn)n∈`∞, that is, there is K∈(0,∞) such that |cn| ≤ Kfor all n∈N. The sequence fn(x) := xn n3 (n≥1) is linearly independent, consists of C1-functions and satisfies that the series P∞ n=1(cnfn)0converges uniformly on [0,1], which in turn follows from the Weierstrass M-test, the convergence of P∞ n=1 1 n2and the inequality |cnf0 n(x)| ≤ K n2. Moreover, trivially, the series P∞ n=1 fn(0) converges. Then a well-known convergence criterion yields that P∞ n=1 cnfnconverges in C([0,1]) to a C1-function h. But any C1-function on a compact interval is of bounded variation, and so h∈BC. The next and final assertion of this section provides nontrivial spaces S (that is, Sis essentially different from ω) showing that neither lineability nor dense-lineability imply [S]-lineability. Theorem 2.7. Let Xbe an infinite dimensional Banach space. Then the following holds: (a) There is a lineable subset M⊂Xsuch that, for every K¨othe echelon space S, the set Mis not [S]-lineable. (b) If Xpossesses a Schauder basis, then there exists a dense-lineable subset M⊂Xsuch that, for every K¨othe echelon space S, the set Mis not [S]-lineable. Proof. According to a theorem due to Mazur (see, e.g., [20]), there exists a basic sequence {en}n≥1in X. Observe that we can suppose that {en}n≥1 is normalized, that is, kenk= 1 for all n∈N. Define the set M:= span({en}n≥1). Since a basic sequence is linearly independent, we have that Mis lineable. Suppose that we have already proved that Mis not [`1]-lineable. Then, thanks to Theorem 2.5(a), it cannot be [S]-lineable for any K¨othe space S. Moreover, if {en}n≥1were a Schauder basis for X, then Mwould be dense in X, so dense-lineable because Mis itself a vector space. Consequently, it is enough to show that Mis not [`1]-lineable. With this aim, assume, by way of contradiction, that Mis [`1]-lineable. Then we can find a linearly independent sequence {un}n≥1⊂Xsatisfying that, for every (cn)n∈`1, the series P∞ n=1 cnunconverges in Xto an element of M∪{0}=M. Notice that, according to Proposition 2.1 and since `1⊃c00, we may assume {un}n≥1⊂M. Therefore, we can find, for every 8 BERNAL, CONEJERO, MURILLO, AND SEOANE n∈N, a finite nonempty subset Fn⊂Nas well as scalars λn,j ∈K\ {0} (j∈Fn) satisfying un=X j∈Fn λn,jej.(2) For the sake of completeness, we set λn,j := 0 whenever j∈N\Fn(n∈N). Now, we define the following sequence of positive scalars: γn= 1 + X j∈Fn |λn,j|(n∈N). Consider the vector space `1((γn)) := (cn)n∈ω:P∞ n=1 |cn|γn<∞, which becomes a Banach space under the norm k(cn)nkγ:= P∞ n=1 |cn|γn. On the one hand, observe that, since γn>1 (n= 1,2, . . . ), we have `1((γn)) ⊂`1. On the other hand, for every (cn)n∈`1((γn)), the series P∞ n=1 cnunconverges to some vector U=U((cn)) ∈Xbecause Xis complete and the series of norms P∞ n=1 kcnunkconverges. Indeed, we have for all n∈Nthat kcnunk=     cnX j∈Fn λn,jej     ≤ |cn|X j∈Fn |λn,j| kejk≤|cn|γn, and the comparison test does the job. Due to our assumption, we have that U((cn)) ∈M. Next, we consider the linear mappings Λj: (cn)∈`1((γn)) 7−→ ∞ X n=1 λn,jcn∈K(j= 1,2, . . . ). It follows from the inequalities |λn,jcn|≤|cn|γn(j, n ≥1) that every Λj is well defined and, thanks to the Closed Graph Theorem, it is continuous. Note that, due to the convergence of each series P∞ n=1 λn,jcnand the uniqueness of coefficients of the expansion associated to a Schauder basis, we get U((cn)) = ∞ X j=1 Λj((cn)) ej.(3) Assume that there exists N∈Nsuch that Λj= 0 for all j > N. Then, by taking successively (cn)n= (1,0,0,0, . . . ),(0,1,0,0, . . . ),(0,0,1,0, . . . ), . . . and so on, and after computing Λj((cn)), we get λn,j = 0 for all j > N and all n∈N. But, according to (2), this would entail that span ({un}n≥1)⊂ span{e1, . . . , eN}, which contradicts the linear independence of {un}n≥1. Therefore there is an infinite subset S⊂Nsuch that Λj6= 0 for all j∈S and Λj= 0 for all j6∈ S. Hence, for each j∈S, the kernel Λ−1 j({0}) of Λj is a vector subspace of `1((γn)) that is closed (because Λjis continuous) and of empty interior (because the inclusion Λ−1 j({0})⊂`1((γn)) is strict). [S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES 9 Due to the Baire category theorem, the countable union Sj∈SΛ−1 j({0}) is of first category in X. Hence the set A:= `1((γn)) \[ j∈S Λ−1 j({0}) is not empty. Thus, we can select an element (cn)n∈A⊂`1((γn)) ⊂`1. It follows from (3) that for this element we have U((cn)) = X j∈S Λj((cn)) ej. Finally, due to the uniqueness of the expansion in the basis {ej}j∈Nand to the fact that Λj((cn)) 6= 0 for infinitely many j, we arrive to U((cn)) 6∈ span({en}n∈N) = M. This is the sought-after contradiction.  3. [S]-lineability and hypercyclic vectors As said in the Introduction, in this section and the next one we will focus our attention on two specific families, namely, the set of hypercyclic vectors respect to a linear operator, and the class of differentiable functions that are nowhere monotone on the real line. Just as a brief summary, let us recall that the existence of these “differentiable monsters” (differentiable nowhere monotone functions) dates back to the work by Katznelson and Stromberg (1974, [29]), although several new constructions have been appearing since then (see, e.g., [18] for a recent expository work covering this class of functions). For background on hypercyclicity, the reader is referred to the monographs [9] and [26] (see, also, [8]). This topic has been widely investigated during the last three decades. If Xis a topological vector space, an operator on Xis a continuous linear selfmap T:X→X. We say that Tis hypercyclic provided that there is a vector x0∈X(called hypercyclic for T) whose T-orbit {Tnx0:n∈N}is dense in X, where we have denoted Tn=T◦T◦· · ·◦T(ntimes). If Xsupports some hypercyclic operator, then Xis infinite dimensional. The set of hypercyclic vectors for Tis denoted by HC(T). It is well-known that, if Tis hypercyclic, then HC(T) is dense (and even residual if, in addition, Xis completely metrizable). Other important property is that, under hypercyclicity of T, the set HC(T) contains, except for zero, a dense T-invariant vector subspace of X(see [42]). In particular, HC(T) is always dense-lineable. We wonder whether HC(T) is [S]-lineable for “reasonable” subspaces S ⊂ ω(that is, subspaces containing c00 strictly). According to Theorem 2.5, if Xwere a Fr´echet space supporting a continuous norm and HC(T) were spaceable, this set would be [S]- lineable for every K¨othe space S(see [31]). However, not every set HC(T) is spaceable (for sufficient conditions for spaceability or non-spaceability of 16 BERNAL, CONEJERO, MURILLO, AND SEOANE If Ais a subset of a vector space, we denote by conv∗(A) the strict convex hull of A, that is, conv∗(A) = (n X i=1 λixi:xi’s ∈A, λi’s ∈(0,1), n X i=1 λi= 1, n ∈N). Theorem 5.4. (a) Let p∈Nbe an odd number with p≥3. Then each member of the set conv({Tn p:n∈N})is topologically mixing. (b) The set f: [0,1] −→ [0,1] : fis topologically mixing  is convex lineable. Proof. (a) Assume that pis an odd positive integer with p≥3. Let us define fn: [0,1] →[0,1] by fn:= Tn p=Tpn(n∈N). We obviously have fn◦fm=fn+mfor any n, m ∈N. As f1=Tpis totally transitive, so is every fnand therefore any composition of finitely many elements in {fn}n∈Nis totally transitive or, equivalently, topologically mixing (Theorem 5.3). We aim to show that any finite convex combination of functions in {fn}n∈N is topologically mixing. In fact, since each fnis itself topologically mixing, it suffices to consider strict convex combination of such functions. First, note that since we picked p∈Nodd, both 0 and 1 are fixed points for every function in {fn}n∈N. This ensures the necessary condition that any convex combination of these functions maps [0,1] onto [0,1] and not on some strictly smaller interval. Let us pick g∈conv∗({fn}n∈N), so that there are finitely many reals λ1, . . . , λm∈(0,1) as well as natural numbers n1< n2<· · · < nmsuch that m X i=1 λi= 1 and g= m X i=1 λifni. Set ν:= n1. It follows from the construction that, for any interval Ik pνof the form Ik pν=k pν,k+ 1 pν(k∈ {0, . . . , pν−1}), we have fm(Ik pν) = [0,1] for all m≥ν. Specifically, either fmk pν= 0 and fmk+1 pν= 1 for all m≥ν(as both 0 and 1 are fixed points of f1) or vice versa. Then gk pν= 0 and gk+1 pν= 1 [S]-LINEAR AND CONVEX STRUCTURES IN FUNCTION FAMILIES 17 or vice versa. In any case, g(Ik pν) = [0,1] for every k∈ {0, . . . , pν−1}. Applying the same argument to gN=g◦g◦ · · · ◦ g(Ntimes), we get that gN(Ik pNν ) = [0,1] for every k∈ {0, . . . , pN ν −1}. Now, given a non-degenerate interval J⊂ [0,1], we can find N∈Nand k∈ {0, . . . , pNν −1}such that Ik pNν ⊂Jand therefore we have gn(J)⊃gn(Ik pNν )⊃gN(Ik pNν ) = [0,1] for every n≥N. This implies, trivially, that gn(J)⊃[ε, 1−ε] for all ε∈ (0,1/2) and all n≥N. Consequently, gis topologically mixing according to Theorem 5.3. (b) According to part (a), we have conv({Tn 3:n∈N})⊂f: [0,1] →[0,1] : fis topologically mixing. Thus, it is enough to show that the functions Tn 3(n= 1,2, . . . ) are linearly independent. Assume, by way of contradiction, that this system {Tn 3= T3n}n≥1is linearly dependent. Then there are n∈Nand c1, . . . , cn∈R satisfying T3n+1 =Pn i=1 ciT3i. For each i∈ {1,2, . . . , n + 1}, consider the set Ni:= {x∈(0,1) : T3iis not differentiable}, that is, Ni={k/3i:k= 1,2,...,3i−1}, whose cardinality is 3i−1. But N1⊂N2⊂ · · · ⊂ Nn. By the linearity of differentiability, there are at most 3n−1 points of (0,1) in which Pn i=1 ciT3i(= T3n+1 ) is not differentiable. This is absurd, because 3n−1<3n+1 −1 = card(Nn+1). This concludes the proof.  However, the property of being topologically mixing does not hold when considering convex combinations of different maps Tpiwith piprime. This will be the content of our final result (Theorem 5.6). We first need the following auxiliary result, whose proof can be seen in [37, Proposition 2.34]. Lemma 5.5. Let f: [0,1] →[0,1] be a topologically mixing continuous selfmap. If fis piecewise monotone or C1then fis locally eventually onto, that is, given ε > 0, there is N=N(ε)∈Nsuch that, for any open subinterval J⊂[0,1] with diam(J)≥εand any n≥N, we have fn(J) = [0,1]. Theorem 5.6. There are no topologically mixing maps in the set conv∗({Tp:pis prime}). Proof. Let g∈conv∗({Tp:pis prime}). Then there are finitely many prime numbers p1< p2<· · · < pNas well as scalars α1, . . . , αN∈(0,1) such that PN i=1 αi= 1 and g=Pn i=1 αiTpi. Note that N≥2. It is then clear that gis a piecewise monotone map. According to Lemma 5.5, in order to prove that gcannot be topologically mixing it is enough to show that gis not locally eventually onto. 18 BERNAL, CONEJERO, MURILLO, AND SEOANE Let us prove first that neither 0 nor 1 are in the image of (0,1) under g. Proceeding by way of contradiction, assume the existence of a point x0∈(0,1) such that g(x0) = 0, that is, α1Tp1(x0) + · · · +αNTpN(x0) = 0. Since αi>0 and Tpi≥0 for all i, the last equality is possible only if Tpi(x0) = 0 for all i. This is absurd because, the pi’s being prime, the subsets of (0,1) where the Tpi’s vanish are mutually disjoint. A similar reasoning (using αi<1 and Tpi≤1 this time) shows that a point x0∈ (0,1) satisfying g(x0) = 1 cannot exist. Thus, g((0,1)) ⊂(0,1). Since gis continuous, we obtain that if Jis a closed interval contained in (0,1) then g(J) is also a closed interval contained in (0,1). But g2(J) = g(g(J)), so that g2(J) is also a closed interval contained in (0,1). This procedure can be recursively continued, so as to show that, for every closed interval J⊂(0,1) and every n∈N, we have gn(J)⊂(0,1) and, in particular, gn(J)6= [0,1]. Thus, gis not locally eventually onto, as required.  Remark 5.7. With the same proof, we obtain in fact the following more general result: If A⊂Nis a set with card(A)≥2consisting of mutually coprime numbers, then there are no topologically mixing maps in the set conv∗({Tp:p∈A}). Acknowledgements. The first author has been supported by the Plan Andaluz de Investigaci´on de la Junta de Andaluc´ıa FQM-127 Grant P08- FQM-03543 and by MEC grant MTM2015-65242-C2-1-P. 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Departamento de An´ alisis Matem´ atico Facultad de Matem´ aticas Instituto de Matem´ aticas Antonio de Castro Brzezicki (IMUS) Universidad de Sevilla Avenida Reina Mercedes, Sevilla, 41012 (Spain). E-mail address:[email protected] Instituto Universitario de Matem´ atica Pura y Aplicada Universitat Polit` ecnica de Val` encia Val` encia, 46022 (Spain). E-mail address:[email protected] Institut Universitari de Matem` atiques i Aplicacions de Castell´ o (IMAC) Escuela Superior de Tecnolog´ ıa y Ciencias Experimentales Universitat Jaume I Campus de R´ ıu Sec, 12071 Castell´ o de la Plana (Spain). E-mail address:[email protected] Instituto de Matem´ atica Interdisciplinar (IMI) Departamento de An´ alisis Matem´ atico y Matem´ atica Aplicada Facultad de Ciencias Matem´ aticas Plaza de Ciencias 3 Universidad Complutense de Madrid Madrid, 28040 (Spain). E-mail address:[email protected]