Universality of Affine Semigroups on Supercyclicity of the Sequence Operators

Mortada Saeed^1*, Simon Joseph² and Ahmed Sufyan^3
*Correspondence: Mortada Saeed, Department of Mathematics, College of Education, University of Rumbek, Sudan, Email:

Keywords

Hypercyclic semigroups • Hypercyclic operators • Supercyclic operators • Supercyclic semigroups

Introduction

Unless stated otherwise, all vector spaces in this article are over the field k, being either the field C of complex numbers or the field R of real numbers and all topological spaces are assumed to be Hausdorff. As usual, Z₊ is the set of non-negative integers, N is the set of positive integers and R₊ is the set of non-negative real numbers. The symbol L(X) stands for the space of a continuous linear sequence of operators on a topological vector space X, while X in the space of continuous linear functionals on X. As usual, for T^j ∈L(X) the dual sequence of operators T^j : X'→ X' is defined by the formula Recall that an affine map on a Tj vector space X is a map of the shapes T^j x=u^j + S^j x, where uj is fixed vector in X and Sj : X→ X is linear. Clearly, Tj are continuous if and only if S^j are continuous. The symbol A^j(X) stands for the space of continuous affine maps on a topological vector space. F^j -space is a complete metrizable topological vector space. Recall that a family of continuous maps from a topological space X to a topological space Y is called universal if there is x∈X for which is dense in Y and such an x is called a universal element for F^j. Use the symbol u(F^j) for the set of universal elements for F^j. If X is a topological space and T^j : X→ X is a continuous map, then say that x∈X is universal for T^j if x is universal for the family Denote the sets of universal elements for T^j by Series families of continuous maps from a topological space X to itself are called semigroups if and for every (1+ε)∈R₊. Say that a semigroup (1+ε)∈R₊ is strongly continuous if (1+ε)→ T_{j (1+ε)}x are continuous as a map from R₊ to X for every x∈X and say that is jointly continuous if is continuous as a map from R₊ × X to X. If X is a topological vector space, semigroup a linear semigroup if for every is called an affine semigroup if for every (1+ε)∈R₊. Recall that Tj ∈L(X) is called hypercyclic if are called hypercyclic vectors. Universal linear semigroups are called hypercyclic and its universal elements are called hypercyclic vectors for then universal elements of the family are called supercyclic vectors for Tj and Tj are called supercyclic if it has a supercyclic vector. Similarly, if are linear semigroup, then a universal element of the families are called supercyclic vector for and the semigroup is called supercyclic if it has a supercyclic vector.

Hypercyclicity and supercyclicity have been studied during the last decades [1]. The concern is the relation between the supercyclicity of a linear semigroup and the supercyclicity of the individual members of the semigroup. The hypercyclicity version of the question was treated by Conejero, Müller, and Peris [2], who proved that for every strongly continuous hypercyclic linear semigroups on an F^j-space, all is hypercyclic and Virtually the same proof works in the following much more general setting.

Theorem A

Let be a hypercyclic jointly continuous linear semigroup on all topological vector space X. Then all hypercyclic and

The stronger condition of joint continuity coincides with the strong continuity in the case when X is an Fj -space due to a straightforward application of the Banach–Steinhaus theorem. It is based on a homotopytype argument and goes through without any changes for semigroups of non-linear maps. Recall that a topological space X is called connected if it has no subsets different from and X, which are closed and open and it is called simply connected if for any continuous map f_j : T→X, there is a continuous map F^_j : T×[0,1]→X and x0∈Xsuch that F^_j(z,0)=f_j(z) and F^_j(z,1)=x0for any z∈T. Next, X is called locally path connected at χ∈X if for any neighborhood U of χ, there is a neighborhood V of χ such that for any y∈V, there is a continuous map f_j :[0,1]→X satisfying f_j (0)=x, f_j (1)=y, and f_j ([0,1]) ⊆ U. Space X is called locally path-connected if it is locally path connected at every point. Just listing the conditions to run smoothly, get the following result.

Proposition 1.1

Let X be a topological space and a jointly continuous semigroup on X such that

are nowhere dense in X for every ε>-1 and u^j ∈X ;

(2) for every ε>-1 and there is such that is connected, locally path-connected, simply connected and

The natural question of whether the supercyclicity version of Theorem A holds in [3]. They have produced the following example.

Example B: Let X be a Banach space over R, be a hypercyclic linear semigroup on X and be the linear sequence of operators with the matrices Then is supercyclic linear semigroup on while are non-supercyclics whenever is rational.

Example B shows that the natural supercyclicity version of Theorem A fails in the case k=R.

Proposition 1.2

Let X be a complex topological vector space and be a supercyclic jointly continuous linear semigroup on X such that has a dense range for every ε>-1 and every λ_j∈C. Then each has a dense range for every ε>-1 and every λ_j∈C. Then each with ε>-1 is supercyclic. Furthermore, the set of supercyclic vectors for does not depend on the choice of ε>-1 and simultaneity with the set of supercyclic vectors of the plenary semigroup.

Whatever one can obtain the same result directly by considering the induced action on subsets of the projective space and applying Proposition 1.1. show that in the case k=C, the supercyclicity version of Theorem A holds without any additional assumptions.

Theorem 1.2

Let X be a complex topological vector space and be a supercyclic jointly continuous linear semigroup on X. Then all with ε>-1 is supercyclic and the set of supercyclic vectors of coincides with the set of supercyclic vectors of

It turns out that any supercyclic jointly continuous linear semigroup on a complex topological vector X either satisfies conditions of Proposition Cor has a closed invariant hyperplane Y. Reduces the following generalization of Theorem A to affine semigroups [4].

Theorem 1.3

Let X be a topological vector space and a universal jointly continuous affine semigroup on X. Then all with ε>-1 is universal and

A Dichotomy for Supercyclic Linear Semigroups

An analogue of the following result for supercyclic sequence of operators.

Proposition 2.1

Let X be a complex topological vector space and be a supercyclic strongly continuous linear semigroup on X. Then either is dense in X for every ε>-1 and λ_j∈C or there is a closed hyperplane Hin X such that

This section is devoted to the proof of Proposition 2.1. Recall that subsets B_j of vector space X are called balanced if λ_j x∈B_j for every x∈B_j and λ_j∈k such that

Proof: Assume that there is ε>-1 and λ_j∈k such that are not dense in X. By Lemma 2.5, are closed hyperplanes in X.

Lemma 2.2

Let K be a compact subset of an infinite dimensional topological vector space and X such that . Then is a closed nowhere dense subset of X.

Proof: Closeness of Λ in X is a straightforward exercise. Assume that Λ is not anywhere dense. Since Λ is closed, it's interior Lis non-empty. Since Kis closed and , find a non-empty balanced open set U such that Obviously whensoever x ∈ L and Since U is open and balanced property of Limplies that the open set is non-empty. Taking into account the definition of Λ , inclusion and the fact that Uis balanced, every x ∈ W^j be written as x=λ_jy, where y∈k and Since both Kand is a compact subset of X. Since X ⊆ Q , Wj is a non-empty open set with compact closure. Such a set exists [5] only if X is finite dimensional. This contradiction completes the proof.

The following lemma is a particular case in [6].

Lemma 2.3

Let X be a complex topological vector space such that 2 ≤ dim X < ∞ . Then X supports no supercyclic strongly continuous linear semigroups.

Lemma 2.4

Let X be an infinite dimensional topological vector space and be a strongly continuous linear semigroup such that are not supercyclics.

Proof: Let x∈ X \ {0} . It success to show that x is not a supercyclic vector for

First, consider the case λ_j=0, it is ε>-1 such that andK is a compact subset of X. By Lemma 2.2, is nowhere dense in X. Take n∈N such that Since have Notably, Y is nowhere dense in X. Obviously, whenever ε>-1. Hence is contained in and therefore is nowhere dense in X. Thus x is not a supercyclic vector for

Suppose that λ_j≠0. Then Hence each of the compact sets with n∈Z₊ does not contain 0. The sets nowhere dense in X. On the other hand, for every and therefore for each n ∈ Z ₊ . Hence which is clearly the union of coincides with therefore is nowhere dense. Thus x is not a supercyclic vector for

Lemma 2.5

Let X be a complex topological vector space and bea supercyclic strongly continuous linear semigroup on X. Let also (ε+1)₀ >0 and λ_j∈C Then space coincides with X or is a closed hyperplane in X.

Proof: Using the semigroup property. Factoring Y out, arrive in a supercyclic strongly continuous linear semigroup acting on X/Y, where Clearly, If X/Y is infinite dimensional, arrive at a contradiction with Lemma 2.4. If X/Y is finite dimensional and dim X / Y ≥ 2 , we obtain a contradiction with Lemma 2.3. Thus dim X/Y≤ 1, as required.

Lemma 2.6

Let X be a complex topological vector space and be a strongly continuous linear semigroup on X. Assume also that there is a closed hyperplane Hin X such that and let f_j∈X’ be such that H=ker f_j. Then there exists w ∈Csuch that

Proof: Since H=ker fj is invariant for every , there is a unique function Pick u_j ∈X such that for every strongly continuous, Q_j is continuous. The semigroup property for implies the semigroup property for the dual sequence of operators: for every (1+ε)∈R₊. Together with equality it implies that Q_j(0)=1 and for every (1+ε)∈R₊. The latter and the continuity of Q_j means that there is w∈C such that for each (1+ε)∈R₊. Thus as required.

Supercyclicity Subtended Universality of Affine Maps

Start with the following general lemma.

Lemma 3.1

Let X=1 be a topological vector space, u^j∈X, \ {0} f_j∈ X′ ,f(u^j)=1 and H=ker f_j. Assume also that isa family of a continuous linear sequence of operators on X such that for each a∈A_j. Then the families are universals if and only if the families of affine maps are universals on H. Moreover, x∈X is universal for F_j if and only if x = λ_j (u_j+w), where λ_j ∈k\{0} and w is universal for G_j. Next, if A_j = Z₊ and for every a∈Z₊, then for every a∈Z₊. Finally, if A_j is strongly continuous linear semigroup, then is strongly continuous affine semigroup.

Proof: Since for every a, vectors from H cannot be universal for F^j. Clearly, they also do not have the form with

Let Since It is straightforward to see that x0 is universal for Fj if and only if 0 is dense in u^j +H. That is, x₀ is universal for F_^j if and only if x is universal for the families where each is the restriction of to the invariant subset u^j +H. Clearly, the translation map is a homeomorphism and for every a∈A_j. It follows that x₀ is universal for F^j if and only if is universal for G^j. Denoting w=x-u^j, if and only if with

Since Q_a are the restrictions of T^j_a to the invariant subset u^j +H and R_a are similar to Q_a in the same manner independent on a, {R_a} inherits all the semigroup or continuity properties from The proof is complete.

Lemma 3.2

Let X be a topological vector space, and H=ker f_j. Then is supercyclic if and only if the map is universal. Moreover, x∈X is a supercyclic vector for T_j if and only if x=λ_j (uj+w), where λ_j∈k \{0} and w∈U(R).

Lemma 3.3

Let X be a topological vector space, H=ker f_j. Then a strongly continuous linear semigroup on X satisfying is supercyclic if and only if the strongly continuous affine semigroup on H defined by universals. Furthermore, x∈X is a supercyclic vector for if and only if x=λj (uj+w), where

Universality of Affine Semigroups

The proof of the following lemma is a routine verification.

Lemma 4.1

Let X be a topological vector space, a collection of continuous affine maps on be a collection of the continuous linear sequence of operators on X and (1+ε)→w_(1+ε) be a map from R₊ to X such that for every (1+ε)∈R₊ and x∈X. Then are affines semigroup if and only if are linear semigroups,

W₀ =0 and for every (1+ε)∈R₊ . (1)

Furthermore, the semigroup is strongly continuous if and only if is strongly continuous and the map is continuous. Finally, the semigroup is jointly continuous if and only if is jointly continuous and the map (1+ε)→w_(1+ε) is continuous.

Lemma 4.2

Let X be a topological vector space and be a universal strongly continuous affine semigroup on X. Then is dense in X for every ε>-1.

Proof: Suppose the contrary. Then there is ε>-1 such that Y₀ ≠ X, where Let Y be a translation of Y₀, containing Factoring out the closed linear subspace Y, arrive in the universal strongly continuous affine semigroup on X/Y, where for every (1+ε)∈R₊ and x∈X. By definition of Y, the linear part of is I. Let β+ε∈X/Y be a universal vector for By Lemma 4.1, there is a strongly continuous linear semigroup on X/Y and a continuous map from R₊ to X/Y such that and for every β∈X/Y and r, (1+ε)∈R₊, obtain that for every n∈Z₊ and (ε+1)∈R₊. It follows that

where

Since (β+ε) is universal for hence, is dense in X/Y. Since 0 is closed as a sum of a compact set and a closed set,0=X/Y. On the other hand, 0 it does not contain any sufficiently large >-1. This contradiction completes the proof.

Lemma 4.3

Let X be a topological vector space, x∈X, ε>-1 and be a universal affine semigroup on X. Assume also that continuous linear semigroup on X and is a continuous map from R+ to X. Then is hypercyclic. Furthermore, for every ε > -1.

Proof: Let and fixε ε > -1. By Lemma 4.2, are dense in X. Hence are dense in X. Using the semigroup property of together with (1), get

for every then, 0 is exactly the St-orbit of . Since o is dense in is hypercyclic vector for and therefore

Lemma 4.4

Let X be a topological vector space and be an affine semigroup on X. Then for every satisfying commutes with every

Proof: To verify that for every affine map A_j : X→X and every x₁,….,x_n∈X,

provided z_j∈k and z₁+… z_n=1.

Let (1+ε)∈R₊. Therefore,

Since commute with each other, get

Lemma 4.5

Let X be a topological vector space, be universals strongly continuous affine semigroup on X and Then where

(2)

Proof: Let and z₁+…… z_n=1. Have to show that where A commute with all Since it suffices to verify that A_j(X) are dense in X. By Lemma 4.1, write for every y∈X, where is strongly continuous linear semigroup on X and (1+ε)→w_(1+ε) is a continuous map from R₊ to X. By Lemma 4.3, are hypercyclics. Thus has dense range. Since A_j(X) is translation, B_j(X),A_j(X) is also dense in X, which completes the proof.

Proof of Theorem 1.3

Let X be a topological vector space and be a universal jointly continuous affine semigroup on X. By Theorem A, there is a hypercyclic continuous linear operator on X. Since no such thing exists on a finite-dimensional topological vector space [7], X is infinite-dimensional. Since any compact subspace of an infinite-dimensional topological vector space is nowhere dense [4], condition (1) of Proposition 1.1 is satisfied. Let By Lemma 4.5, the set Λ(x) defined in (4.2) consists entirely of universal vectors for . Obviously, By its definition, Λ(x) is an affine subspace of X. Λ(x) satisfies all requirements for the set (for every ε > -1) from condition (2) in Proposition 1.1. By Proposition 1.1, for every ε > -1, as required.

Proof of Theorem 1.2

Let X be a complex topological vector space and be a supercyclic jointly continuous linear semigroup on X in [8]. To prove that all supercyclic and the sets of supercyclic vectors of simultaneity with the set of supercyclic vectors of . If has a dense range for every ε > -1 and every λ_j∈C, then Proposition C provides the required result. Else, by Proposition 2.1, there is a closed hyperplane H in X invariant for all By Lemma 2.6, there is f_j∈X and (β+ε)∈C such that H=kerf_j and for every (1+ε)∈R₊. Obviously is a jointly continuous supercyclic linear semigroup on X with the same sets Sj of supercyclic vectors as the original semigroup . Fix u_j∈X satisfying f_j(u^j)=1. Now fix ε>-1 and v^j ∈S^j. Have to show that v^j is supercyclic for By Lemma 3.3, applied to the semigroup write and y is a universal vector for the jointly continuous affine semigroup on H defined by the formula By Theorem 1.3, y is universal for R_(1+ε). By Lemma 3.2, is a supercyclic vector for and hence v^j is a supercyclic vector for The proof is complete.

Remarks

By Lemma 4.3, the universality of a strongly continuous affine semigroup implies hypercyclicity of the underlying linear semigroup. The following example shows that the converse is not true [4].

Example 6.1

Consider the backward weighted shift T^j∈L(l2) with the weight sequence That is, is the standard basis of l₂. Then the jointly continuous linear semigroups are hypercyclics. Furthermore, there exists a continuous map from R₊ to l₂ such that is jointly continuous non-universal affine semigroup, where for x∈l₂.

Proof: Since Tj being compacts weighted backward shift, is quasinilpotent, the sequence of operators ln is well defined and bounded and is a jointly continuous linear semigroup. Moreover, are hypercyclics [9] as a sum of the identical sequence of operators and a backward weighted shift. Hence are hypercyclics.

Let For each (1+ε)∈R₊, let Since T^j are quasinil potents, are well defined bounded linear sequence of operators and the map are a sequence of operators-norm continuous. Hence (1+ε)→w_(1+ε) is continuous as a map from R₊ to l₂, to verify that w₀=0, w₁=u_j and for every ε>-1. By Lemma 4.1, is a jointly continuous affine semigroup, where It remains to show that is non-universal. Assume the contrary. Since w₁=u^j and Lemma 4.3 implies that the coset must contain a hypercyclic vector for I+T_j. This, however, is not the case as shown in [10].

Remark 6.2

Let X be a topological vector space and be hypercyclic. If then the affine map is universal. Actually, let w∈X be such that It is easy to show that for every x∈X and n∈N. Thus x is universal for Tj if and only if x-w is universal for S^j.

If additionally X is separable metrizable and Baire, then a standard Baire category type argument shows that the set of uj∈Xfor which the affine map is universal is a dense G_δ -subset of X.

Example 6.1 shows that this set can differ from X.

Recall that a locally convex topological vector space X is called barrelled if every closed convex balanced subset Bj of X satisfying contain a neighborhood of 0. The joint continuity of a linear semigroup follows from the strong continuity if the underlying space X is an Fj-space. The same is true for wider classes of topological vector spaces. For the case, it is sufficient X to be a Baire topological vector space or a barreled locally convex topological vector space. Thus the following observation holds true.

Remark 6.3

The joint continuity condition in Theorems A, 1.2 and 1.3 can be replaced by the strong continuity, provided X is Baire or X is locally convex and barrelled.

For general topological vector spaces however strong continuity of a linear semigroup does not imply joint continuity. Furthermore, the following example shows that Theorem A fails in general if the joint continuity condition is replaced by strong continuity. Recall that the Fréchet space consists of the scalar-valued functions R₊, square-integrable on [0,(1+ε)] for each ε>-1. Its dual space can be naturally interpreted as the space duality between is provided by the natural dual pairing Clearly, the linear semigroup of backward shifts is strongly continuous and therefore jointly continuous on the Fréchet space It follows that the same semigroup is strongly continuous on endowed with the weak topology.

Example 6.4

Let be the above strongly continuous semigroup on X. Then there are f_j∈X hypercyclics for such that f

are not-hypercyclics for S^j₁.

Proof: Let H be the hyperplane in L² [0,1] consisting of the functions with zero Lebesgue integral. Fix norm-dense countable subsets A_j of H. One can easily construct such that for every n∈N, the function belongs to A

; for every n∈N and

For be the indicator function of the interval and otherwise. By (a), every n∈N and therefore f_j are not hypercyclics vector for S_j1.

It remains to show that f_j are hypercyclics vector for acting on X. Using (a) and (b), we see that the Fréchet space topology closure of the orbits is exactly the sets

In order to show that fj are hypercyclics for acting on X, it suffices to verify that 0 is dense in Assume the contrary. Then there is a weakly open set which does not intersect 0. That is, there are linearly independent and such that

Let k∈N be such that all Qj vanishes on [k,∞) . Pick any Note that for every 1≤ j ≤ m, the restrictions of the functionals Q_j to are not linearly independent, see [11]. Actually, otherwise can find such that It is easy to see that it is and and arrived at a contradiction.

The fact that Q_j is not linearly independents on implies that there is a non-zero Since they are all linearly independent, they are m +1 linearly independent vectors in the m dimensional space span completes the proof.

Conflict of Interests

The authors declare that there is no conflict of interest.

Acknowledgment

The author thanks Colleagues for numerous helpful suggestions, and the referee for carefully checking the manuscript.

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