The algebraic size of the family of injective operators

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 : R 2 n m → R 2 n m : f is linear and bijective} is 2ⁿ -lineable.

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

Recall that if X is a finite dimensional Banach space then any linear mapping X → X is continuous, and any injective linear mapping X → X is bijective. Consistently, for a general Banach space, we consider the family

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 L_1–1 (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 L_1–1(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 L_1–1(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.

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 → 𝕂^m : f is injective} is m-lineable, but not (m + 1)-lineable.

2. The set {f : ℂ^m → ℂ^m : 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 : 𝕂^m → 𝕂^m : f is linear and bijective} (m – 1)-lineable?

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 ) = ∑ n = 0 ∞ a n z n then

the series being convergent in the norm topology of L(X). Here, T⁰ = I and Tⁿ⁺¹ = Tⁿ ∘ 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 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 E_r(z) ≔ e^rz – 1. Then {E_r : r ∈ Η} 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 E_r of Lemma 3.3, that is, the collection all finite linear combinations of products E r 1 m 1 ⋯ E r N m N ( N ∈ N ; r 1 , … , r N ∈ H ; ( m 1 , … , m N ) ∈ N 0 N ∖ { ( 0 , … , 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 ∈ 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 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

which is clearly an algebra. Since 𝓐 ⊂ H₀(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, r₁,..., r_N are different numbers in Η and H and Q ( E r 1 ( T ) , … , 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 , where F ( z ) := Q ( E r 1 ( z ) , … , E r N ( z ) ) . . Assume, by way of contradiction, that Q ≠ 0. Since the E_r ’s generate a free algebra, we have F ∈ H₀(ℂ)\{0}. From Theorem 3.2, it follows that

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:

1. φ_m(e_n) = δ_mn for all m, n ∈ ℕ.

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

3. ||e_n|| = 1 for all n ∈ ℕ and sup_{n∈ ℕ} ||φn|| < ∞.

Remark 3.5

It is proved in [13] that the sequence {e_n}_n≥1 may satisfy, in addition, that s p a n ¯ {e_n : 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 X ~ is the vector space on ℂ given by X² endowed with the operations

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

Note that T n ~ = ( T ~ ) n for all n ≥ 0. Observe also that if f ( z ) = ∑ n = 0 ∞ a n z n is an entire function with real Taylor coefficients a_n then the expression f ( T ) = ∑ n = 0 ∞ a n T n also makes sense and defines an operator on X satisfying f ( T ) ~ = f ( T ~ ) . The following is a variant of Theorem 3.1 when 𝕂 = ℝ.

Lemma 3.6

Let X be a real Banach space and T ∈ L(X) be an operator such that σ P ( T ~ ) = ∅ . 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 B := { f ( T ~ ) : f ∈ A } is a 𝔠-generated free algebra satisfying B ∖ { 0 } ⊂ L 1 − 1 ( X ~ ) Let us prove that the algebra

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 , … , E r N ) and assume that F(T) = 0. Then F ( T ~ ) = F ( T ) ~ = 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) ∈ 𝓑₁ \ {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 , 0 ) ∈ X ~ satisfies

which contradicts the injectivity of f ( 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

From property (c) in Theorem 3.4, it follows that

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

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 1 2 φ 1 ( x ) = a φ 2 ( x ) , hence φ₂(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 1 2 m φ 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 σ P ( T ~ ) = ∅ . Assume, contrariwise, that there is c = a + ib ∈ ℂ as well as a vector z = ( x , y ) ∈ X ~ ∖ { ( 0 , 0 ) } such that T ~ z = c z . 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, a² + b² ≠ 0. Therefore we have that for some (a, b) ≠ (0, 0) and some (x, y) ≠ (0, 0) the following holds:

Letting φ₁ and then φ_m+1(m ∈ ℕ) act on both equalities of (4) we obtain

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 φ_m(x)= 0 = φ_m(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 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.

Remarks 3.8

1. In particular L_1–1(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 m ⋅ f ∈ X for all f ∈ X. Then the mapping

is well defined and linear, so it defines an operator M_m ∈ L(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 X_i is in fact a Banach algebra, one can choose as m any member of X_i. 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

Theorem 3.1 yields that L_1–1(X) is strongly 𝔠-algebrable. It is enough to prove (9). With this aim, assume that there exists λ ∈ σ_P (M_m), so that there is f ∈ X \ {0} with m f = λ f. By continuity, there is a nonempty open set G ⊂ Ω such that f(x) ≠ 0 for all x ∈ G. Then m = λ on G, hence m^–1 ({λ}) has nonempty interior, a contradiction. Finally, suppose that 𝕂 = ℝ and that there exists λ = a + b i ∈ σ P ( M m ~ ) , so that there are functions f, g ∈ X with (f, g) ≠ (0, 0) satisfying m f = a f – bg and m g = bf + ag. Fix x ∈ Ω \ m^–1 ({a}). It follows that

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 T ∈ L(X) is said to be hypercyclic if it possesses a dense orbit, that is, if there is a vector x₀ ∈ X, called hypercyclic for T, such that the set

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 Τ ∈ L(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 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.

Remarks 3.10

1. Theorem 3.9 covers important examples of nonseparable Banach spaces, such as 𝓁_∞(ℂ) := {x = (x_n)_n≥1 ∈ ℂ^ℕ: (x_n)_n≥1 is bounded} (under the norm ||x|| = sup_n≥1|x_n|) 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 n ≥ Ν [16, Chap. 8]. This easily implies that S ~ is hypercyclic on Y ~ . Then σ P ( ( S ~ ) ∗ ) = ∅ . The handicap in order to apply Lemma 3.6 lies in the fact that we need σ P ( S ~ ∗ ) = ∅ , but ( S ~ ) ∗ acts on (Υ × Υ)^*, while S ∗ ~ 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^ℕ. Assume that every member of the canonical unit sequence (e_n)_n≥1 belongs to X and that the projections

are continuous. Then L_1–1(X) is strongly 𝔠-algebrable.

Proof. Define Τ by

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 Τ ∈ L(X) such that T* ∈ L(X*) is hypercyclic. From [16, Lemma 2.53(a)] we have

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?