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 L*_{1–1}(*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 *T* ∈ *L*(*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 $f(z)={\displaystyle {\sum}_{n=0}^{\mathrm{\infty}}{a}_{n}{z}^{n}}$ then
$$f(T)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{T}^{n},$$ the series being convergent in the norm topology of *L*(*X*). Here, *T*^{0} = *I* and *T*^{n+1} = *T*^{n} ∘ *T* (*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 T* ∈ *L*(*X*) *and f* : Ω → ℂ *be holomorphic on an open set* Ω ⊃ *σ*(*T*). *Then f*(*σ*_{P}(*T*)) ⊂ *σ*_{P} (*f*(*T*)). *If in addition*, *f is nonconstant on every connected component of* Ω, *then f*(*σ*_{P}(*T*)) *= σ*_{P}(*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 *H*_{0}(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 E*_{r}(*z*) ≔ *e*^{rz} – 1. *Then* {*E*_{r} : *r* ∈ *Η*} *is a free system of generators of an algebra contained in H*_{0}(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 *E*_{r} of Lemma 3.3, that is, the collection all finite linear combinations of products ${E}_{{r}_{1}}^{{m}_{1}}\cdots {E}_{{r}_{N}}^{{m}_{N}}(N\in \mathbb{N};{r}_{1},\dots ,{r}_{N}\in H;({m}_{1},\dots ,{m}_{N})\in {\mathbb{N}}_{0}^{N}\mathrm{\setminus}\{(0,\dots ,0)\}).$ Note that the cardinality of the free system {*E*_{r}}_{r}_{∈H} is c. By hypothesis, there exists *T* ∈ *L*(*X*) such that *σ*_{P}(*T*) = ∅. Let *f ∈ H*_{0}(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 *σ*_{p}(*f*(*T*)) *= f*(σ_{P}(*T*)) *= f*(∅) = ∅. In particular, 0 ∉ *σ*_{P}(*f* (*T*)). Thus, the operator *f*(*T*) is injective, that is, *f*(*T*) ∈ *L*_{1–1}(*X*). Now, consider the family
$$\mathcal{B}:=\{f(T):f\in \mathcal{A}\},$$ which is clearly an algebra. Since 𝓐 ⊂ *H*_{0}(ℂ), we get 𝓑 \ {0} ⊂ *L*_{1–1}(*X*). Clearly, the operators *E*_{r}(*T*)(*r* ∈ *Η*) 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, *r*_{1}*,..., r*_{N} are different numbers in *Η* and $H\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}Q({E}_{{r}_{1}}(T),\dots ,{E}_{{r}_{N}}(T))=0,$ then *Q =* 0.

To this end, observe that, under the latter assumptions, we get from the properties of holomorphic functions of operators that $F(T)=0,\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}F(z):=Q({E}_{{r}_{1}}(z),\dots ,{E}_{{r}_{N}}(z)).$. Assume, by way of contradiction, that *Q* ≠ 0. Since the *E*_{r} ’s generate a free algebra, we have *F* ∈ *H*_{0}(ℂ)\{0}. From Theorem 3.2, it follows that
$$\{0\}={\sigma}_{P}(0)={\sigma}_{P}(F(T))=F({\sigma}_{P}(T))=F(\mathrm{\varnothing})=\mathrm{\varnothing},$$ 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* {*e*_{n}}_{n}_{≥1} ⊂ *X and* {φ_{n}}_{n}_{≥1} ⊂ *X** *with the following properties*:

*φ*_{m}(*e*_{n}) = *δ*_{mn} for all m, n ∈ ℕ.

*If φ*_{n}(*x*) = 0 *for all n* ∈ ℕ *then x =* 0.

||*e*_{n}|| = 1 *for all n* ∈ ℕ *and* sup_{n}_{∈ ℕ} ||*φn*|| < ∞.

#### Lemma 3.6

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

#### Proof

Observe that Lemma 3.3 also works when the linear algebra 𝓐 generated by the functions *E*_{r} (*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 $\mathcal{B}:=\{f(\stackrel{~}{T}):f\in \mathcal{A}\}$ is a 𝔠-generated free algebra satisfying $\mathcal{B}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus}\phantom{\rule{thinmathspace}{0ex}}\{0\}\subset {L}_{1-1}(\stackrel{~}{X})$Let us prove that the algebra
$${\mathcal{B}}_{1}=\{f(T):f\in \mathcal{A}\}$$

is freely generated by the operators *E*_{r} (*T*) (*r* ∈ *Η*). 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 $F:=Q({E}_{{r}_{1}},\phantom{\rule{thickmathspace}{0ex}}\dots ,\phantom{\rule{thickmathspace}{0ex}}{E}_{{r}_{N}})$ and assume that *F*(*T*) = 0. Then $F(\stackrel{~}{T})=\stackrel{~}{F(T)}=\stackrel{~}{0}=0,$ so *F* = 0 as in the mentioned proof. Hence *Q* = 0 because 𝓐 was freely generated by the *E*_{r}’s. Finally, each operator *f*(*T*) ∈ 𝓑_{1} \ {0} is one-to-one because, otherwise, there would exist *x* ∈ *Χ* \ {0} with *f*(*T*)*x* = 0. This would imply that the nonzero vector $(x,\phantom{\rule{thickmathspace}{0ex}}0)\in \stackrel{~}{X}$ satisfies
$$\begin{array}{l}f(\stackrel{~}{T})(x,\phantom{\rule{thickmathspace}{0ex}}0)=\phantom{\rule{thinmathspace}{0ex}}\stackrel{~}{f(T)}(x,\phantom{\rule{thickmathspace}{0ex}}0)=(f(T)x,\phantom{\rule{thickmathspace}{0ex}}f(T)0)=(0,\phantom{\rule{thinmathspace}{0ex}}0)\phantom{\rule{thickmathspace}{0ex}},\end{array}$$

which contradicts the injectivity of $f(\stackrel{~}{T})$. 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 L*_{1–1}(*X*) *is strongly* 𝔠-*algebrable*.

*Proof*. Choose a pair of sequences {*e*_{n}}_{n}_{≥1} ⊂ *X* and {*φ*_{n}}_{n}_{≥1} ⊂ *X** with the properties given in Theorem 3.4. Define the mapping
$$\begin{array}{l}T:x\in X\mapsto \sum _{n=1}^{\mathrm{\infty}}\frac{1}{{2}^{n}}{\phi}_{n}(x){e}_{n+1}\in X.\end{array}$$(1)

From property (c) in Theorem 3.4, it follows that
$$\begin{array}{l}\sum _{n=1}^{\mathrm{\infty}}\parallel \frac{1}{{2}^{n}}{\phi}_{n}(x)\phantom{\rule{thinmathspace}{0ex}}{e}_{n+1}\parallel \phantom{\rule{thinmathspace}{0ex}}\le \underset{n\in \mathbb{N}}{\phantom{\rule{thinmathspace}{0ex}}sup}\phantom{\rule{thinmathspace}{0ex}}\parallel {\phi}_{n}\parallel \phantom{\rule{thinmathspace}{0ex}}\parallel x\parallel \phantom{\rule{thinmathspace}{0ex}}<\mathrm{\infty}.\end{array}$$(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), ||*T*_{x}|| ≤ *C*||*x*|| (*x* ∈ *X*) with *C* = sup_{n}_{∈ ℕ} ||*φ*_{n}|| < *∞*. In other words, *T* ∈ *L*(*X*).

Let us show that *T* lacks eigenvalues. Assume, by way of contradiction, that there is *a* ∈ σ_{P}(*T*). Then there exists *x* ∈ *X* \ {0} such that
$$\begin{array}{l}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{2}^{n}}{\phi}_{n}(x){e}_{n+1}=ax.\end{array}$$(3)

If we let *φ*_{1} act on both members of (3) then we get 0 = *a φ*_{1}(*x*), thanks to property (a) in Theorem 3.4. Suppose first that *a* ≠ 0. This implies *φ*_{1}(*x*) = 0. By making *φ*_{2} act on (3), we obtain $\frac{1}{2}{\phi}_{1}(x)=a{\phi}_{2}(x),$ hence *φ*_{2}(*x*) = 0. With this procedure, we successively derive *φ*_{n}(*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 *φ*_{m}_{+1}(*m* ≥ 1) act on (3), we get $\frac{1}{{2}^{m}}{\phi}_{m}(x)=0,$ so *φ*_{n}(*x*) = 0 for all *n* ∈ ℕ, which again implies *x* = 0, a contradiction. Hence *σ*_{P}(*Τ*) = ∅. This is valid in both cases 𝕂 = ℝ and 𝕂 = ℂ.

According to Theorem 3.1, *L*_{1–1}(*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 ${\sigma}_{P}(\stackrel{~}{T})=\mathrm{\varnothing}.$ Assume, contrariwise, that there is *c* = *a* + *ib* ∈ ℂ as well as a vector $z=(x,\phantom{\rule{thickmathspace}{0ex}}y)\in \stackrel{~}{X}\mathrm{\setminus}\{(0,0)\}$ such that $\stackrel{~}{T}z=cz.$ Then *Tx* = *ax* – *by* and *Ty* = *bx* + *ay*. Note that if *a* = 0 = *b* then *Tx* = 0 = *Τ y*, in which case 0 ∈ *σ*_{P} (*Τ*) = ∅, which is absurd. Consequently, *c* ≠ 0 or, that is the same, *a*^{2} + *b*^{2} ≠ 0. Therefore we have that for some (*a, b*) ≠ (0, 0) and some (*x, y*) ≠ (0, 0) the following holds:
$$\begin{array}{l}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{2}^{n}}{\phi}_{n}(x){e}_{n+1}=ax-by\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{2}^{n}}{\phi}_{n}(y){e}_{n+1}=bx+ay.\end{array}$$(4)

Letting *φ*_{1} and then *φ*_{m}_{+1}(*m* ∈ ℕ) act on both equalities of (4) we obtain
$$\begin{array}{c}a{\phi}_{1}(x)-b{\phi}_{1}(y)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}b{\phi}_{1}(x)+a{\phi}_{1}(y)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\\ a\phantom{\rule{thickmathspace}{0ex}}{\phi}_{m+1}(x)-b{\phi}_{m+1}(y)={2}^{-m}{\phi}_{m}(x)\phantom{\rule{thickmathspace}{0ex}},\end{array}$$(5)
$$\begin{array}{l}b{\phi}_{m+1}(x)+a{\phi}_{m+1}(y)={2}^{-m}{\phi}_{m}(y).\end{array}$$(6)

Since (5) is a homogeneous linear system in the unknowns *φ*_{1}(*x*), *φ*_{1}(*y*) whose determinant is *a*^{2} + *b*^{2} ≠ 0, one derives *φ*_{1}(*x*)= 0 = *φ*_{1}(*y*). By proceeding recursively and assuming *φ*_{m}(*x*)= 0 = *φ*_{m}(*y*) for an *m* ∈ ℕ, one finds that (6) is, again, a homogeneous linear system with determinant *a*^{2} + *b*^{2} ≠ 0. Hence its unique solution is *φ*_{m+1}(*x*) = 0 = *φ*_{m+}_{1}(*y*). To summarize, *φ*_{m}(*x*)= 0 = *φ*_{m}(*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.

#### Theorem 3.9

*Assume that X is a complex Banach space that is the dual space of some separable infinite dimensional Banach space. Then L*_{1–1}(*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 *S* ∈ *L*(*Y*) and define *T* ≔ *S**. 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.

#### Theorem 3.11

*Let X be a Banach space that is a subset of the sequence space* 𝕂^{ℕ}. *Assume that every member of the canonical unit sequence (e*_{n})_{n}_{≥1} *belongs to X and that the projections*
$${\phi}_{m}:x=({x}_{n}{)}_{n\ge 1}\in X\mapsto {x}_{m}\in \mathbb{K}(m\in \mathbb{N})$$

*are continuous. Then L*_{1–1}*(X) is strongly* 𝔠-*algebrable*.

*Proof*. Define *Τ* by
$$T:x=({x}_{n}{)}_{n\ge 1}\in X\mapsto \sum _{n=1}^{\mathrm{\infty}}{2}^{-n}\parallel {\phi}_{n}{\parallel}^{-1}{e}_{n+1}{\parallel}^{-1}{x}_{n}{e}_{n+1}\in X$$

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*:

𝓕 *and* 𝓖 *are commutative linear algebras*.

𝓕 *and* 𝓖 *are freely* 𝔠-*generated*.

*Every member of* 𝓕 *or* 𝓖 *is injective*.

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 *Τ* ∈ *L*(*X*) such that *T** ∈ *L*(*X**) is hypercyclic. From [16, Lemma 2.53(a)] we have
$${\sigma}_{P}({T}^{\ast})=\mathrm{\varnothing}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\sigma}_{P}(T)={\sigma}_{P}({T}^{\ast \ast})={\sigma}_{P}(({T}^{\ast}{)}^{\ast})=\mathrm{\varnothing}.$$

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 L*_{1–1}(*X*) *large –in any algebraic sense– for all infinite dimensional Banach spaces?*

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