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# The algebraic size of the family of injective operators

Luis Bernal-González
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• Departamento de Análisis Matemático. Facultad de Matemáticas Apdo. 1160. Avda. Reina Mercedes, 41080 Sevilla, Spain
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Published Online: 2017-01-03 | DOI: https://doi.org/10.1515/math-2017-0005

## Abstract

In this paper, a criterion for the existence of large linear algebras consisting, except for zero, of one-to-one operators on an infinite dimensional Banach space is provided. As a consequence, it is shown that every separable infinite dimensional Banach space supports a commutative infinitely generated free linear algebra of operators all of whose nonzero members are one-to-one. In certain cases, the assertion holds for nonseparable Banach spaces.

MSC 2010: 47L05; 15A03; 47A05; 47A16

## 1 Introduction

In the current millenium there has been a generalized interest in the search for large algebraic structures inside nonlinear sets. This area of research is called lineability. Concepts, results and references concerning this topic can be found in the survey [1] and the monograph [2]. In this note, we focus on the family of all injective (or one-to-one) operators defined on a Banach space. Obviously, this family is not a vector space. Only recently (see below) a number of assertions have been established within this context, in the real case. Our aim is to contribute to complete the existing knowledge on lineability of the mentioned family, mainly in the complex infinite dimensional setting.

Our notation will be rather usual. The symbols ℕ, ℕ0, ℚ, ℝ, ℂ will stand for the set of positive integers, the set ℕ ⋃ {0}, the set of rational numbers, the real line, and the field of complex numbers, respectively. The cardinality of the continuum, card(ℝ), is denoted by 𝔠. If X is a Banach space on 𝕂 = ℝ or ℂ then X* and L(X) will represent, respectively, the topological dual space of X and the vector space of all operators on X, that is, the family of all continuous linear self-mappings Τ : XX. Recall that L(X) becomes a linear algebra (in fact, a Banach algebra) if we endow this vector space with the internal law of composition of operators. If TL(X) then the spectrum and the point spectrum of T will be denoted by σ(T) and σP(T), respectively. Recall that σ(T) = {λ ∈ 𝕂 : TλI is not invertible} (I = the identity) and σP(T) is the set of eigenvalues of T, that is, σp(T = {λ ∈ 𝕂 : TλI is not injective} (⊂ σ (T)). The adjoint of T is the operator T* ∈ L(X*) given by (T*φ)(x) = φ(Tx)(φX*, xX).

A few lineability concepts (see [2]) will be convenient in order to establish our findings appropriately. If X is a vector space, α is a cardinal number and AX, then A is said to be: lineable if there is an infinite dimensional vector space M such that M \ {0} ⊂ A, and α-lineable if there exists a vector space M with dim(M) = α and M \ {0} ⊂ A (hence lineability means 0-lineability, where 0 = card(ℕ)). If, in addition, X is contained in some (linear) algebra then A is called: algebrable if there is an algebra M so that M \ {0} ⊂ A and M is infinitely generated, that is, the cardinality of any system of generators of M is infinite; α-algebrable if there is an α-generated algebra M with M \ {0} ⊂ A; strongly algebrable if A ⋃ {0} contains an infinitely generated algebra that is isomorphic with a free algebra, and strongly α-algebrable if A ⋃ {0} contains an α-generated algebra that is isomorphic with a free algebra. Note that if BX is a set whose elements commute respect to the algebra operation, then B is a generating set of some free algebra contained in A ⋃ {0} if, and only if, for any Ν ∈ ℕ and any nonzero polynomial P in N variables without a constant term and any distinct y1,..., yNB we have P(y1,..., yN)∈ A \ {0}. Of course, strong (α-)algebrability implies (α-)algebrability, and this in turn implies (α-, resp.) lineability.

Even if no additional condition of linearity is imposed, there are not many injective self-mappings on ℝ in the algebraic sense, as the following result (see [3]) shows.

#### Theorem 1.1

The set of injective functions ℝ → ℝ is not 2-lineable.

As for higher dimensions, we gather in the next theorem a number of assertions, which have been recently proved by Jiménez, Maghsoudi, Muñoz and Seoane [4] in the setting of real vector spaces.

#### Theorem 1.2

Let n,m ∈ ℕ. Then the following holds:

1. The set {f : ℝn → ℝ m : f is injective} is m-lineable, but not (m + 1)-lineable.

2. The set {$f:{\mathbb{R}}^{{2}^{n}m}\to {\mathbb{R}}^{{2}^{n}m}$: f is linear and bijective} is 2n -lineable.

3. If m ≥ 3 is odd then the set {f : ℝm → ℝm : f is linear and bijective} is not m-lineable.

Recall that if X is a finite dimensional Banach space then any linear mapping XX is continuous, and any injective linear mapping XX is bijective. Consistently, for a general Banach space, we consider the family $L1−1(X):={linearcontinuousinjectivemappingsX→X}={T∈L(X):Ker(T)={0}}.$

It is well known that the group of bijective operators on a Banach space (equivalently, by the Open Mapping Theorem, the group of invertible operators) is a nonempty open set (see e.g. [5, Chap. 7]). Hence L1–1 (X) contains a nonempty open set, so it is not a too small set, in the topological sense. Thus, it is natural to raise the question of whether L1–1(X) is also large in the algebraic sense. Theorem 1.2 above gave us a partial answer in the realm of real finite dimensional spaces. In [4], the following theorem concerning an important class of infinite dimensional Banach spaces in the real case is also proved.

#### Theorem 1.3

Let X be a real Banach space with a Schauder basis. Then L1–1(X) is lineable.

In Section 2, we will briefly deal with the finite dimensional case in order to generalize Theorem 1.2. In Section 3, which is the main one, we will extend the aforementioned results to the complex setting when the Banach spaces have infinite dimension. Our main contribution consists of a criterion providing algebrability in this case. Through a modification of this criterion we will also improve Theorem 1.3 by showing that its conclusion still holds -in both real and complex cases- if X is just separable and infinite dimensional. Examples of nonseparable Banach spaces where the result remains valid are also provided.

## 2 Injective operators on finite dimensional Banach spaces

Theorem 1.1 and item (a) of Theorem 1.2 hold in the case 𝕂 = ℂ with virtually the same proofs. (In the light of the original proof in [4, Theorem 2.4], we can even afford a slight generalization in the initial set ℝn) Observe that, trivially, we have also 1-lineability for the set of bijective linear self-mappings of 𝕂 = ℝ or ℂ (any nonzero multiple of the identity is bijective). Item (c) of Theorem 1.2 also holds for 𝕂 = ℂ, not only for odd m but for all m ∈ ℕ (hence Theorem 1.2(b) is false for m = 1 if we replace ℝ by ℂ). Indeed, the proof of part (c) in [4] (see Remark 2.7, proof of Corollary 2.8 and Remark 2.9 of this reference) was based on properties of determinants and on the fact that any polynomial of odd degree with coefficients in ℝ possesses at least one (real) zero. The restriction “m odd” is not needed in ℂ by the Fundamental Theorem of Algebra.

As a consequence, we can state the following result.

#### Theorem 2.1

Let m ∈ ℕ and S be a set with card (S) ≤ 𝔠. We have:

1. For 𝕂 = ℝ or ℂ, the set {f : S → 𝕂m : f is injective} is m-lineable, but not (m + 1)-lineable.

2. The set {f : ℂm → ℂm : f is linear and bijective} is m-lineable if and only if m = 1.

A natural problem remains open: for 𝕂 = ℝ or ℂ and m ≥ 2, is the family {f : 𝕂m → 𝕂m : f is linear and bijective} (m – 1)-lineable?

## 3 Injective operators on infinite dimensional Banach spaces

We turn to the infinite dimensional case when the field is 𝕂 = ℂ. Let us establish our criterion (Theorem 3.1), with which a high level of lineability is obtained. Observe that this criterion is not applicable to the finite dimensional setting, because any operator has always (complex) eigenvalues in this case. Note also that, despite the fact that the algebra L(X) is not commutative, our approach furnishes a large commutative algebra.

#### Theorem 3.1

Let X be a complex Banach space supporting an operator without eigenvalues. Then the family L1–1(X) is strongly 𝔠-algebrable.

The proof will make use of some background about holomorphic functions of operators (see for instance [6, Chap. 1] or [7, Chap. 10]). Let X be a complex Banach space and TL(X). If f is a complex function that is analytic on a neighborhood of σ(T) then it is possible to define an operator f(T) ∈ L(X) satisfying f(T) = I if f(z) ≡ 1, f(T) = T if f(z) ≡ z, (f + g)(T) = f(T) + g(T) and f(T)g(T) = (fg)(T) (where f(T)g(T) denotes composition of f(T) and g(T), while fg denotes pointwise multiplication). Observe that, for fixed T, the operators f(T) form a commutative linear algebra. In the special case of an entire function f : ℂ → ℂ, we have that if f has Taylor expansion $f\left(z\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ then $f(T)=∑n=0∞anTn,$ the series being convergent in the norm topology of L(X). Here, T0 = I and Tn+1 = TnT (n ≥ 0). A special version of the spectral mapping theorem (see [7, Theorem 10.33]) reads as follows.

#### Theorem 3.2

Assume that X is a complex Banach space. Let TL(X) and f : Ω → ℂ be holomorphic on an open set Ω ⊃ σ(T). Then f(σP(T)) ⊂ σP (f(T)). If in addition, f is nonconstant on every connected component of Ω, then f(σP(T)) = σP(f(T)).

The following auxiliary assertion provides a free algebra consisting of entire functions. Its proof is easy and is essentially contained in the proof of Lemma 2.4 in [8], which in turn is in the same spirit as [9, Proposition 7] (see also [10, Theorem 1.5] and [11]); so it will be omitted. Let H0(ℂ) denote the set of all entire functions f with f(0) = 0.

#### Lemma 3.3

Let H ⊂ (0, +∞) be a set with card (H) = 𝔠 and which is linearly independent over the field ℚ. For each r > 0, consider the function Er(z) ≔ erz – 1. Then {Er : rΗ} is a free system of generators of an algebra contained in H0(ℂ).

In the proof of our Theorem 3.1, Lemma 3.3 plays the role of generating the appropriate algebra by superposing a fixed operator belonging to the considered class with the representatives of a well chosen algebra of functions. This method was already used in [12].

Proof of Theorem 3.1. Consider the algebra 𝓐 generated by the functions Er of Lemma 3.3, that is, the collection all finite linear combinations of products ${E}_{{r}_{1}}^{{m}_{1}}\cdots {E}_{{r}_{N}}^{{m}_{N}}\left(N\in \mathbb{N};{r}_{1},\dots ,{r}_{N}\in H;\left({m}_{1},\dots ,{m}_{N}\right)\in {\mathbb{N}}_{0}^{N}\mathrm{\setminus }\left\{\left(0,\dots ,0\right)\right\}\right).$ Note that the cardinality of the free system {Er}rH is c. By hypothesis, there exists TL(X) such that σP(T) = ∅. Let f ∈ H0(ℂ) such that f is not identically zero. Then f is not constant on Ω = ℂ which is connected. It follows from Theorem 3.2 that σp(f(T)) = fP(T)) = f(∅) = ∅. In particular, 0 ∉ σP(f (T)). Thus, the operator f(T) is injective, that is, f(T) ∈ L1–1(X). Now, consider the family $B:={f(T):f∈A},$ which is clearly an algebra. Since 𝓐 ⊂ H0(ℂ), we get 𝓑 \ {0} ⊂ L1–1(X). Clearly, the operators Er(T)(rΗ) generate the algebra 𝓑. It remains to show that they generate 𝓑 in a free way. In other words, we should prove that, if N ∈ ℕ, Q is a complex polynomial in N variables without constant term, r1,..., rN are different numbers in Η and $H\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}Q\left({E}_{{r}_{1}}\left(T\right),\dots ,{E}_{{r}_{N}}\left(T\right)\right)=0,$ then Q = 0.

To this end, observe that, under the latter assumptions, we get from the properties of holomorphic functions of operators that $F\left(T\right)=0,\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}F\left(z\right):=Q\left({E}_{{r}_{1}}\left(z\right),\dots ,{E}_{{r}_{N}}\left(z\right)\right).$. Assume, by way of contradiction, that Q ≠ 0. Since the Er ’s generate a free algebra, we have FH0(ℂ)\{0}. From Theorem 3.2, it follows that ${0}=σP(0)=σP(F(T))=F(σP(T))=F(∅)=∅,$ which is absurd. The proof is finished.

Concerning applications of Theorem 3.1 (see Theorem 3.7 below), a tool that will be used is the following strong result due to Ovsepian and Pelczynski [13] about the structure of separable (real or complex) Banach spaces.

#### Theorem 3.4

If X is an infinite-dimensional separable Banach space, then there are sequences {en}n≥1X andn}n≥1X* with the following properties:

1. φm(en) = δmn for all m, n ∈ ℕ.

2. If φn(x) = 0 for all n ∈ ℕ then x = 0.

3. ||en|| = 1 for all n ∈ ℕ and supn∈ ℕ ||φn|| < ∞.

#### Remark 3.5

It is proved in [13] that the sequence {en}n≥1 may satisfy, in addition, that $\overline{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}}$ {en : n ∈ ℕ} = Χ. Nevertheless, we do not need this property at all.

Since Theorem 3.1 is given in the complex setting, in order to apply it in the real one the technique of complexification will be needed. Recall that if X is a Banach space on ℝ then its complexification $\stackrel{~}{X}$ is the vector space on ℂ given by X2 endowed with the operations $(x,y)+(u,v)=(x+u,y+v)andc(x,y)=(ax−by,bx+ay),$

where c = a + i b, a, b ∈ ℝ, x, y, u, v ∈ X. Then $\stackrel{~}{X}$ becomes a Banach space on ℂ under, for instance, the norm ||(x, y)|| = ||x|| + ||y||. If TL(X) then the complexification of T is the operator $\stackrel{~}{T}$ on $\stackrel{~}{X}$ defined as $T~(x,y)=(Tx,Ty).$

Note that $\stackrel{~}{{T}^{n}}=\left(\stackrel{~}{T}{\right)}^{n}$for all n ≥ 0. Observe also that if $f\left(z\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ is an entire function with real Taylor coefficients an then the expression $f\left(T\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{T}^{n}$also makes sense and defines an operator on X satisfying $\stackrel{~}{f\left(T\right)}=f\left(\stackrel{~}{T}\right).$ The following is a variant of Theorem 3.1 when 𝕂 = ℝ.

#### Lemma 3.6

Let X be a real Banach space and TL(X) be an operator such that ${\sigma }_{P}\left(\stackrel{~}{T}\right)=\mathrm{\varnothing }.$ Then the set L1–1(X) is strongly 𝔠-algebrable.

#### Proof

Observe that Lemma 3.3 also works when the linear algebra 𝓐 generated by the functions Er (r ∈ Η) is considered over 𝕂 = ℝ. Then this algebra is also freely 𝔠-generated and, since H ⊂ ℝ, all its members are entire functions with real Taylor coefficients. As in the proof of Theorem 3.1 we get that the family $\mathcal{B}:=\left\{f\left(\stackrel{~}{T}\right):f\in \mathcal{A}\right\}$ is a 𝔠-generated free algebra satisfying $\mathcal{B}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}\left\{0\right\}\subset {L}_{1-1}\left(\stackrel{~}{X}\right)$Let us prove that the algebra $B1={f(T):f∈A}$

is freely generated by the operators Er (T) (rΗ). To this end, let us fix, as in the last part of the proof of Theorem 3.1, functions ${E}_{{r}_{1}},...,{E}_{{r}_{N}}$ as well as a polynomial Q with degree N, but this time with real coefficients. Let $F:=Q\left({E}_{{r}_{1}},\phantom{\rule{thickmathspace}{0ex}}\dots ,\phantom{\rule{thickmathspace}{0ex}}{E}_{{r}_{N}}\right)$ and assume that F(T) = 0. Then $F\left(\stackrel{~}{T}\right)=\stackrel{~}{F\left(T\right)}=\stackrel{~}{0}=0,$ so F = 0 as in the mentioned proof. Hence Q = 0 because 𝓐 was freely generated by the Er’s. Finally, each operator f(T) ∈ 𝓑1 \ {0} is one-to-one because, otherwise, there would exist xΧ \ {0} with f(T)x = 0. This would imply that the nonzero vector $\left(x,\phantom{\rule{thickmathspace}{0ex}}0\right)\in \stackrel{~}{X}$ satisfies $f(T~)(x,0)=f(T)~(x,0)=(f(T)x,f(T)0)=(0,0),$

which contradicts the injectivity of $f\left(\stackrel{~}{T}\right)$. The proof is finished.

We are now ready to show that separability is enough to guarantee algebrability for our family of one-to-one operators.

#### Theorem 3.7

Assume that X is a separable infinite dimensional Banach space. Then L1–1(X) is strongly 𝔠-algebrable.

Proof. Choose a pair of sequences {en}n≥1X and {φn}n≥1X* with the properties given in Theorem 3.4. Define the mapping $T:x∈X↦∑n=1∞12nφn(x)en+1∈X.$(1)

From property (c) in Theorem 3.4, it follows that $∑n=1∞∥12nφn(x)en+1∥≤supn∈N∥φn∥∥x∥<∞.$(2)

Since X is a complete space, (2) shows that the series in (1) converges to a vector of X, so T is well defined. Trivially, T is linear and, by (2), ||Tx|| ≤ C||x|| (xX) with C = supn∈ ℕ ||φn|| < . In other words, TL(X).

Let us show that T lacks eigenvalues. Assume, by way of contradiction, that there is a ∈ σP(T). Then there exists xX \ {0} such that $∑n=1∞12nφn(x)en+1=ax.$(3)

If we let φ1 act on both members of (3) then we get 0 = a φ1(x), thanks to property (a) in Theorem 3.4. Suppose first that a ≠ 0. This implies φ1(x) = 0. By making φ2 act on (3), we obtain $\frac{1}{2}{\phi }_{1}\left(x\right)=a{\phi }_{2}\left(x\right),$ hence φ2(x) = 0. With this procedure, we successively derive φn(x) = 0 for all n ∈ ℕ. It follows from property (b) in Theorem 3.4 that x = 0, which is absurd. Then a = 0 and Tx = 0. Letting φm+1(m ≥ 1) act on (3), we get $\frac{1}{{2}^{m}}{\phi }_{m}\left(x\right)=0,$ so φn(x) = 0 for all n ∈ ℕ, which again implies x = 0, a contradiction. Hence σP(Τ) = ∅. This is valid in both cases 𝕂 = ℝ and 𝕂 = ℂ.

According to Theorem 3.1, L1–1(X) is strongly 𝔠-algebrable if 𝕂 = ℂ. Finally, we will prove by using Lemma 3.6 that the conclusion also holds if 𝕂 = ℝ. All that we have to show is ${\sigma }_{P}\left(\stackrel{~}{T}\right)=\mathrm{\varnothing }.$ Assume, contrariwise, that there is c = a + ib ∈ ℂ as well as a vector $z=\left(x,\phantom{\rule{thickmathspace}{0ex}}y\right)\in \stackrel{~}{X}\mathrm{\setminus }\left\{\left(0,0\right)\right\}$ such that $\stackrel{~}{T}z=cz.$ Then Tx = axby and Ty = bx + ay. Note that if a = 0 = b then Tx = 0 = Τ y, in which case 0 ∈ σP (Τ) = ∅, which is absurd. Consequently, c ≠ 0 or, that is the same, a2 + b2 ≠ 0. Therefore we have that for some (a, b) ≠ (0, 0) and some (x, y) ≠ (0, 0) the following holds: $∑n=1∞12nφn(x)en+1=ax−byand∑n=1∞12nφn(y)en+1=bx+ay.$(4)

Letting φ1 and then φm+1(m ∈ ℕ) act on both equalities of (4) we obtain $aφ1(x)−bφ1(y)=0,bφ1(x)+aφ1(y)=0,andaφm+1(x)−bφm+1(y)=2−mφm(x),$(5) $bφm+1(x)+aφm+1(y)=2−mφm(y).$(6)

Since (5) is a homogeneous linear system in the unknowns φ1(x), φ1(y) whose determinant is a2 + b2 ≠ 0, one derives φ1(x)= 0 = φ1(y). By proceeding recursively and assuming φm(x)= 0 = φm(y) for an m ∈ ℕ, one finds that (6) is, again, a homogeneous linear system with determinant a2 + b2 ≠ 0. Hence its unique solution is φm+1(x) = 0 = φm+1(y). To summarize, φm(x)= 0 = φm(y) for all m ∈ ℕ, from which it follows that (x, y) = (0, 0) because of Theorem 3.4(b). This contradiction concludes the proof.

#### Remarks 3.8

1. In particular L1–1(X) is lineable under the assumptions of Theorem 3.7. Notice that our approach is radically different from that of Theorem 1.3 given in [4, Theorem 2.12]. Observe also that this theorem is strengthened in a double direction: the conclusion is reinforced and extended, and our assumptions are weaker because every Banach space with a Schauder basis is separable but, as Enflo proved in [14], the reverse is not true.

2. The conclusion of Theorem 3.7 is optimal, in terms of the cardinality of the generating family of operators. Indeed, since X is separable, it follows that card(X) = 𝔠. Again by separability, card(C (X)) = 𝔠. Hence card(L(X)) = 𝔠 because this cardinality lies between card(X) cand card(C(X)). Then the cardinality of the generating set of the algebra founded in heorem 3.7 is optimal.

Let us give an example of an application of Theorem 3.1 that cannot be derived from Theorem 3.7. Assume that Ω is a topological space and that X is a Banach space of continuous functions Ω → 𝕂. Assume that m : Ω → ℝ is a function such that mfX for all fX. Then the mapping $f∈X↦m⋅f∈X$

is well defined and linear, so it defines an operator MmL(X) (the multiplication operator by m) due to the closed graph theorem. Instances of such spaces are the space Χ 1of all continuous functions [0, 1] → 𝕂 (endowed with the supremum norm || · ||) and the space X2of functions [0, 1] → 𝕂 that are continuous and of bounded variation (endowed with the total variation norm ||f|| = |f(0)| + Var[0,1](f)). Notice that Χ1 is separable, while X2 is not. Since each of these spaces Xi is in fact a Banach algebra, one can choose as m any member of Xi. We impose, in addition, that every a-point set m–1({a}) (a ∈ 𝕂) has empty interior in Ω (for instance, take m(x) = x in the above examples X = X1, X2). Then $σP(Mm)=∅and,ifK=R,σP(Mm~)=∅.$(9)

Theorem 3.1 yields that L1–1(X) is strongly 𝔠-algebrable. It is enough to prove (9). With this aim, assume that there exists λσP (Mm), so that there is fX \ {0} with m f = λ f. By continuity, there is a nonempty open set G ⊂ Ω such that f(x) ≠ 0 for all xG. Then m = λ on G, hence m–1 ({λ}) has nonempty interior, a contradiction. Finally, suppose that 𝕂 = ℝ and that there exists $\lambda =a+bi\in {\sigma }_{P}\left(\stackrel{~}{{M}_{m}}\right),$ so that there are functions f, gX with (f, g) ≠ (0, 0) satisfying m f = a fbg and m g = bf + ag. Fix x ∈ Ω \ m–1 ({a}). It follows that $(m(x)−a)f(x)+bg(x)=0and−bf(x)+(m(x)−a)g(x)=0.$

The determinant of this homogeneous linear system with unknowns f(x), g(x) is (m(x) – a)2 + b2, which is nonzero because m(x) ≠ a. Then its unique solution is (f(x), g(x)) = (0, 0). Therefore f = 0 and g = 0 on the dense set Ω \ m–1({a}). By continuity, f = 0 = g, that again is a contradiction.

In order to furnish another class of Banach spaces to which Theorem 3.1 applies, we need to recall a concept and some properties coming from hypercyclicity, for whose general theory and results (updated up to 2011) we refer the reader to the excellent books [15] and [16]. If X is a (Hausdorff) topological vector space (over 𝕂 = ℝ or ℂ) then an operator TL(X) is said to be hypercyclic if it possesses a dense orbit, that is, if there is a vector x0X, called hypercyclic for T, such that the set ${Tnx0:n≥1}isdenseinX.$

It is evident that if X supports some hypercyclic operator then X must be separable. In addition, X cannot be finite dimensional. Conversely, if X is an infinite dimensional separable Fréchet (in particular, Banach) space then there is a hypercyclic operator ΤL(X); see, e.g., [16, Chap. 8].

#### Theorem 3.9

Assume that X is a complex Banach space that is the dual space of some separable infinite dimensional Banach space. Then L1–1(X) is strongly 𝔠-algebrable.

Proof. By hypothesis, there is a separable infinite dimensional Banach space Y such that X = Y*. Choose a hypercyclic operator SL(Y) and define TS*. But the adjoint of any hypercyclic operator has no eigenvalues: see, e.g., [16, Lemma 2.53(a)]. Thus, it suffices to apply Theorem 3.1.

#### Remarks 3.10

1. Theorem 3.9 covers important examples of nonseparable Banach spaces, such as 𝓁(ℂ) := {x = (xn)n≥1 ∈ ℂ: (xn)n≥1 is bounded} (under the norm ||x|| = supn≥1|xn|) and BV([0, 1],ℂ) := {f : [0, 1] → ℂ : f is of bounded variation} (under the total variation norm) because the former is the dual space of the space 𝓁1(ℂ) of all absolutely summable complex sequences, while the latter is the dual space of the space C([0, 1], ℂ) of continuous complex functions on [0, 1] (Riesz’s theorem).

2. We conjecture that Theorem 3.9 also holds for 𝕂 = ℝ. This is supported by the fact that any separable infinite dimensional Banach space Y supports a mixing operator S, that is, an operator satisfying the following property: for any pair U, V of nonempty open subsets of Y, there exists some Ν ∈ ℕ such that Sn(U)⋂ V ≠ Ø for all nΝ [16, Chap. 8]. This easily implies that $\stackrel{~}{S}$ is hypercyclic on $\stackrel{~}{Y}$. Then ${\sigma }_{P}\left(\left(\stackrel{~}{S}{\right)}^{\ast }\right)=\mathrm{\varnothing }$. The handicap in order to apply Lemma 3.6 lies in the fact that we need ${\sigma }_{P}\left({\stackrel{~}{S}}^{\ast }\right)=\mathrm{\varnothing },$ but $\left(\stackrel{~}{S}{\right)}^{\ast }$ acts on (Υ × Υ)*, while $\stackrel{~}{{S}^{\ast }}$ does so on Υ* × Υ*.

Despite the last remark, the big algebraic size of the family of injective operators on 𝓁(ℝ) happens to be true. In fact, an approach similar to that used in the proof of Theorem 3.7 shows the following.

#### Theorem 3.11

Let X be a Banach space that is a subset of the sequence space 𝕂. Assume that every member of the canonical unit sequence (en)n≥1 belongs to X and that the projections $φm:x=(xn)n≥1∈X↦xm∈K(m∈N)$

are continuous. Then L1–1(X) is strongly 𝔠-algebrable.

Proof. Define Τ by $T:x=(xn)n≥1∈X↦∑n=1∞2−n∥φn∥−1en+1∥−1xnen+1∈X$

and mimic the proof of Theorem 3.7. The details are left as an exercise.

If our Banach space is reflexive, a result due to H. Salas provides us with a dual pair of large algebras of one-to-one operators.

#### Theorem 3.12

Let X be a complex infinite dimensional separable reflexive Banach space. Then there are families 𝓕 ⊂ L(X), 𝓖 ⊂ L(X*) satisfying the following properties:

1. 𝓕 and 𝓖 are commutative linear algebras.

2. 𝓕 and 𝓖 are freely 𝔠-generated.

3. Every member of 𝓕 or 𝓖 is injective.

4. 𝓖 = {S* : S ∈ 𝓕}.

Proof. In 2007, Salas [17] proved that if X is an infinite dimensional Banach space whose dual X* is separable, then there exists a hypercyclic operator Τ on X such that its adjoint T* is also hypercyclic. Under our assumptions, X is, in addition, reflexive, so X = X** = (X*)* is separable. Hence X* is separable (because if the dual Y* of a Banach space Y is separable then Y is itself separable). Therefore we can find a hypercyclic operator ΤL(X) such that T* ∈ L(X*) is hypercyclic. From [16, Lemma 2.53(a)] we have $σP(T∗)=∅andσP(T)=σP(T∗∗)=σP((T∗)∗)=∅.$

Then Theorem 3.1 furnishes families 𝓕 ⊂ L(X), 𝓖 ⊂ L(X*) satisfying properties (a), (b) and (c). But it is known (see, e.g., [6, Chap. 1] or [7, Chap. 10]) that f(T*) = (f(T))* for every entire function f, hence by the construction given in the proof of Theorem 3.1 one obtains that (d) is also fulfilled.

We want to finish this paper by posing the following problem, which is in the same spirit as [4, Question 2.14].

#### Problem

Is L1–1(X) large –in any algebraic sense– for all infinite dimensional Banach spaces?

## Acknowledgement

The author is grateful to the referee for helpful suggestions. The author has been supported by the Plan Andaluz de Investigación de la Junta de Andalucía FQM-127 Grant P08-FQM-03543 and by MEC Grant MTM2015-65242-C2-1-P.

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Accepted: 2016-09-08

Published Online: 2017-01-03

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 13–20, ISSN (Online) 2391-5455,

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