A complete locally convex space of countable dimension admitting an operator with no invariant subspaces
aa r X i v : . [ m a t h . F A ] S e p A complete locally convex space of countable dimension admitting anoperator with no invariant subspaces
Stanislav Shkarin
Abstract
We construct a complete locally convex topological vector space X of countable algebraic dimensionand a continuous linear operator T : X → X such that T has no non-trivial closed invariant subspaces. MSC:
Keywords:
Cyclic operators; invariant subspaces; topological vector spaces
All vector spaces in this article are over the field C of complex numbers. As usual, R is the field of realnumbers, T = { z ∈ C : | z | = 1 } , N is the set of positive integers and Z + = N ∪ { } . Throughout thearticle, all topological spaces are assumed to be Hausdorff . For a topological vector space X , L ( X ) is thealgebra of continuous linear operators on X and X ′ is the space of continuous linear functionals on X . For T ∈ L ( X ), the dual operator T ′ : X ′ → X ′ is defined as usual: T ′ f = f ◦ T .We say that a topological vector space X has the invariant subspace property if every T ∈ L ( X )has a non-trivial (=different from { } and X ) closed invariant subspace. The problem whether ℓ hasthe invariant subspace property is known as the invariant subspace problem and remains perhaps thegreatest open problem in operator theory. It is worth noting that Read [10] and Enflo [8] (see also [3])showed independently that there are separable infinite dimensional Banach spaces, which do not have theinvariant subspace property. In fact, Read [11] demonstrated that ℓ does not have the invariant subspaceproperty. All existing constructions of operators on Banach spaces with no invariant subspaces are rathersophisticated and artificial. On the other hand, examples of separable non-complete normed spaces withor without the invariant subspace property are easy to construct. Proposition 1.1.
Every normed space of countable algebraic dimension does not have the invariant sub-space property. On the other hand, in every separable infinite dimensional Banach space B , there is adense linear subspace X such that X has the invariant subspace property. Thus completeness is an essential difficulty in constructing operators with no non-trivial invariantsubspaces. Countable algebraic dimension is often perceived as almost incompatible with completeness.Basically, there is only one complete topological vector space of countable dimension, most analysts areaware of. Namely, the locally convex direct sum ϕ (see [12]) of countably many copies of the one-dimensionalspace C has countable dimension and is complete. In other words, ϕ is a vector space of countable dimensionendowed with the topology defined by the family of all seminorms. It is easy to see that every linear subspaceof ϕ is closed, which easily leads to the following observation. Proposition 1.2.
The space ϕ has the invariant subspace property. Contrary to the common perception, there is an abundance of complete topological vector spaces ofcountable dimension. The main result of this paper is the following theorem.
Theorem 1.3.
There is a complete locally convex topological vector space X of countable algebraic dimen-sion such that X does not have the invariant subspace property. In other words, Theorem 1.3 provides a complete locally convex topological vector space X with dim X = ℵ and T ∈ L ( X ) such that T does not have non-trivial closed invariant subspaces.1 Proof of Propositions 1.1 and 1.2
Although Propositions 1.1 and 1.2 are certainly known facts, we do not know whether they can be foundin the literature or whether they are folklore. Granted that their proofs are fairly elementary and short,we present them for the sake of convenience.
Proof of Proposition 1.1.
First, assume that X is a normed space of countable dimension. Let B be thecompletion of X . Then B is a separable infinite dimensional Banach space. According to Ansari [1] andBernal–Gonz´ales [4], there is a hypercyclic T ∈ L ( B ). That is, there is x ∈ B such that { T n x : n ∈ Z + } isdense in B . Let Z be the linear span of T n x for n ∈ Z + . Since Z and X are both dense linear subspacesof B , according to Grivaux [9], there is an invertible S ∈ L ( B ) such that S ( Z ) = X . Since Z is invariantfor T , X is invariant for ST S − . That is, the restriction A = ST S − (cid:12)(cid:12) X belongs to L ( X ). Let u ∈ X \ { } .Then S − u is a non-zero vector in Z and therefore S − u = p ( T ) x , where p is a non-zero polynomial. Dueto Bourdon [6], S − u = p ( T ) x is also a hypercyclic and therefore cyclic vector for T . By similarity, u is acyclic vector for ST S − and therefore for A . Thus every non-zero vector in X is cyclic for A . That is, A has no non-trivial closed invariant subspaces.Next, let B be a separable infinite dimensional invariant subspace. Then there is a dense linear subspace X of B ( X can even be chosen to be a hyperplane [5]) such that every T ∈ L ( X ) has the shape λI + S with λ ∈ C and dim S ( X ) < ∞ . Trivially, such a T has a one-dimensional invariant subspace. Proof of Proposition 1.2.
Let T ∈ L ( ϕ ) and x ∈ ϕ \ { } . Then the linear span L of { T n x : n ∈ Z + } is an(automatically closed) invariant subspace of T different from { } . If L = ϕ , we are done. If L = ϕ , thenthe vectors T n x are linearly independent (otherwise L is finite dimensional). Hence the linear span L of { T n x : n ∈ N } is a hyperplane in L . Clearly L is T -invariant and non-trivial. We shall construct an operator T with no non-trivial invariant subspaces, needed in order to prove Theo-rem 1.3, by lifting a non-linear map on a topological space to a linear map on an appropriate topologicalvector space. Recall that a topological space X is called completely regular (or Tychonoff ) if for every x ∈ X anda closed subset F ⊂ X satisfying x / ∈ F , there is a continuous f : X → R such that f ( x ) = 1 and f (cid:12)(cid:12) F = 0. Equivalently, a topological space is completely regular, if its topology can be defined by a familyof pseudometrics. Note that any subspace of a completely regular space is completely regular and thatevery topological group is completely regular.Our construction is based upon the concept of the free locally convex space [14]. Let X be a completelyregular topological space. We say that a topological vector space L X is a free locally convex space of X if L X is locally convex, contains X as a subset with the topology induced from L X to X being the originaltopology of X and for every continuous map f from X to a locally convex space Y there is a uniquecontinuous linear operator T : L X → Y such that T (cid:12)(cid:12) X = f . It turns out that for every completely regulartopological space X , there is a free locally convex space L X unique up to an isomorphism leaving points of X invariant. Thus we can speak of the free locally convex space L X of X . Note that X is always a Hamelbasis in L X . Thus, as a vector space, L X consists of formal finite linear combinations of elements of X .Identifying x ∈ X with the point mass measure δ x on X ( δ x ( A ) = 1 if x ∈ A and δ x ( A ) = 0 if x / ∈ A ),we can also think of elements of L X as measures with finite support on the σ -algebra of all subsets of X .Under this interpretation L X = { µ ∈ L X : µ ( X ) = 0 } is a closed hyperplane in the locally convex space L X . If f : X → X is a continuous map, from thedefinition of the free locally convex space it follows that f extends uniquely to a continuous linear operator2 f ∈ L ( L X ). It is also clear that L X is invariant for T f . Thus the restriction S f of T f to L X belongs to L ( L X ).According to Uspenskii [14], L X is complete if and only if X is Dieudonne complete and every compactsubset of X is finite. Since Dieudonne completeness follows from paracompactness, every regular countabletopological space is Dieudonne complete. Since every countable compact topological space is metrizable, fora countable X , finiteness of compact subsets is equivalent to the absence of non-trivial convergent sequences(a convergent sequence is trivial if it is eventually stabilizing). Note also that a regular countable topologicalspace is automatically completely regular and therefore we can safely replace the term ’completely regular’by ’regular’ in the context of countable spaces. Thus we can formulate the following corollary of theUspenskii theorem. Proposition 3.1.
Let X be a regular countable topological space. Then the countably dimensional locallyconvex topological vector spaces L X and L X are complete if and only if there are no non-trivial convergentsequences in X . The above proposition provides plenty of complete locally convex spaces of countable algebraic dimen-sion. We also need the shape of the dual space of L X . As shown in [14], L ′ X can be identified with thespace C ( X ) of continuous scalar valued functions on X in the following way. Every f ∈ C ( X ) produces acontinuous linear functional on L X in the usual way: h f, µ i = Z f dµ = X c j f ( x j ) , where µ = X c j δ x j and there are no other continuous linear functionals on L X . S f with no invariant subspaces The following lemma is the main tool in the proof of Theorem 1.3.
Lemma 3.2.
Let τ be a regular topology on Z such that f : Z → Z , f ( n ) = n + 1 is a homeomorphismof Z τ = ( Z , τ ) onto itself, Z + is dense in Z τ and for every z ∈ C \ { , } , n z n is non-continuous as amap from Z τ to C . Then the operators T f and S f are invertible continuous linear operators on L Z τ and L Z τ respectively and S f has no non-trivial closed invariant subspaces.Proof. We already know that T f and S f are continuous linear operators. It is easy to see that T − f = T f − and S − f = S f − . Since f − is also continuous, T f and S f have continuous inverses.Now let µ ∈ L Z τ \ { } . It remains to show that µ is a cyclic vector for S f . Assume the contrary. Thenthere is a non-constant g ∈ C ( Z τ ) such that h S nf µ, g i = h T nf µ, g i = 0 for every n ∈ Z + . Decomposing µ as alinear combination of point mass measures, we have µ = l P k = − l c k δ k with c k ∈ C . Then µ = l P j =0 c j − k T jf δ − l = p ( T f ) δ − l , where p is a non-zero polynomial. Then 0 = h T nf µ, g i = h T nf δ − l , p ( T f ) ′ g i for n ∈ Z + . Thus thefunctional p ( T f ) ′ g vanishes on the linear span of T nf δ − l with n ∈ Z + , which contains the linear span of Z + in L Z τ . Since Z + is dense in Z τ , p ( T f ) ′ g vanishes on a dense linear subspace and therefore p ( T f ) ′ g = 0.It immediately follows that T ′ f has an eigenvector, which is given by a non-constant function h ∈ C ( X ): T ′ f h = zh for some z ∈ C . Since T f is invertible, so is T ′ f and therefore z = 0. It is easy to see that T ′ f h ( n ) = h ( n + 1) for each n ∈ Z . Thus the equality T ′ f h = zh implies that (up to a multiplication by anon-zero constant) h ( n ) = z n for each n ∈ Z . Since h is non-constant, z = 1. Thus the map n z n iscontinuous on Z τ for some z ∈ C \ { , } . We have arrive to a contradiction. Lemma 3.3.
There exists a regular topology τ on Z such that the topological space Z τ = ( Z , τ ) has thefollowing properties (a) f : Z → Z , f ( n ) = n + 1 is a homeomorphism of Z τ onto itself ;3b) Z + is dense in Z τ ;(c) for every z ∈ C \ { , } , n z n is non-continuous as a map from Z τ to C ;(d) Z τ has no non-trivial convergent sequences.Proof. Consider the Hilbert space ℓ ( Z ) and the bilateral weighted shift T ∈ L ( ℓ ( Z )) given by T e n = e n − if n T e n = 2 e n − if n >
0, where { e n } n ∈ Z is the canonical orthonormal basis in ℓ ( Z ). Symbol H σ stands for ℓ ( Z ) equipped with its weak topology σ . Clearly T is invertible and therefore T is ahomeomorphism on H σ . According to Chan and Sanders, there is x ∈ ℓ ( Z ) such that the set O = { T n x : n ∈ Z + } is dense in H σ . Then Y = { T n x : n ∈ Z } is also dense in H σ . We equip Y with the topologyinherited from H σ and transfer it to Z by declaring the bijection n T n x from Z to Y a homeomorphism.Since σ is a completely regular topology, so is the just defined topology τ on Z . Since T is a homeo-morphism on ( ℓ ( Z ) , σ ) and Y is a subspace of the topological space ( ℓ ( Z ) , σ ) invariant for both T and T − , T is a homeomorphism on Y . Since T ( T n x ) = T n +1 x , it follows that f is a homeomorphism on Z τ .Density of O in Y implies the density of Z + in Z τ .Observe that the sequence {k T n x k} n ∈ Z is strictly increasing and k T n x k → ∞ as n → + ∞ . Indeed, theinequality k T u k > k u k for u ∈ ℓ ( Z ) follows from the definition of T . Hence {k T n x k} n ∈ Z is increasing.Assume that k T n +1 x k = k T n x k for some n ∈ Z . Then, by definition of T , T n x belongs to the closedlinear span L of e n with n <
0. The latter is invariant for T and therefore T m x ∈ L for m > n , whichis incompatible with the σ -density of O . Next, if k T n x k does not tend to ∞ as n → + ∞ , the sequence {k T n x k} n ∈ Z + is bounded. Since every bounded subset of ℓ ( Z ) is σ -nowhere dense, we have again obtaineda contradiction with the σ -density of O .In order to show that X has no non-trivial convergent sequences, it suffices to show that Y has no non-trivial convergent sequences. Assume that { T n k x } k ∈ Z + is a non-trivial convergent sequence in Y . Withoutloss of generality, we can assume that the sequence { n k } of integers is either strictly increasing or strictlydecreasing. If { n k } is strictly increasing the above observation ensures that k T n k x k → ∞ as k → ∞ . Sinceevery σ -convergent sequence is bounded, we have arrived to a contradiction. If { n k } is strictly decreasing,then by the above observation, the sequence {k T n k x k} of positive numbers is also strictly decreasing andtherefore converges to c >
0. Then k T l x k > c for every l ∈ Z . Since { T n k x } σ -converges to T m x ∈ Y , theupper semicontinuity of the norm function with respect to σ implies that k T m x k c and we have arrivedto a contradiction.Finally, let z ∈ C \ { , } and f : Z → C , f ( n ) = z n . It remains to show that f is not continuousas a function on Z τ . Equivalently, it is enough to show that the function g : Y → C , g ( T n x ) = z n isnon-continuous. Assume the contrary. There are two possibilities. First, consider the case | z | 6 = 1. In thiscase the the topology on the set M = { z n : n ∈ Z } inherited from C is the discrete topology. Continuity ofthe bijection g : Y → M implies then that Y is also discrete, which is apparently not the case: the densityof Y in H σ ensures that Y has no isolated points. It remains to consider the case | z | = 1, z = 1. In thiscase the closure G of { z n : n ∈ Z } is a closed subgroup of the compact abelian topological group T . Since x is a hypercyclic vector for T acting on H σ and z generates the compact abelian topological group G ,[13, Corollary 4.1] implies that { ( T n x, z n ) : n ∈ Z + } is dense in H σ × G . It follows that the graph of g isdense in Y × G . Since G is not a single-point space, the latter is incompatible with the continuity of g .The proof is now complete. Proof of Theorem 1.3.
Let τ be the topology on Z provided by Lemma 3.3. By Proposition 3.1, E = L Z τ is a complete locally convex space of countable algebraic dimension. Let f : Z → Z , f ( n ) = n + 1 and S = S f ∈ L ( E ). By Lemmas 3.2 and 3.3, S is invertible and has no non-trivial invariant subspaces. Theproof is complete (with an added bonus of invertibility of S ).It is easy to see that the topology τ on Z constructed in the proof of Lemma 3.3 does not agree with thegroup structure. That is Z τ is not a topological group. Indeed, it is easy to see that the group operation+ is only separately continuous on Z τ , but not jointly continuous. As a matter of curiosity, it would beinteresting to find out whether there exists a topology τ on Z satisfying all conditions of Lemma 3.3 andturning Z into a topological group. 4 eferences [1] S. Ansari, Existence of hypercyclic operators on topological vector spaces,
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Soviet Math. Dokl. (1983), 771–785 Stanislav ShkarinQueens’s University BelfastDepartment of Pure MathematicsUniversity road, Belfast, BT7 1NN, UKE-mail address: [email protected]@qub.ac.uk