Mixing operators with non-perfectly spanning unimodular eigenvectors
aa r X i v : . [ m a t h . D S ] J a n MIXING OPERATORS WITH NON-PERFECTLY SPANNINGUNIMODULAR EIGENVECTORS
H.-P. BEISE, L. FRERICK, AND J. M ¨ULLER
Abstract.
For arbitrary closed countable subsets Z of the unit circle exam-ples of topologically mixing operators on Hilbert spaces are given which havea densely spanning set of unimodular eigenvectors with eigenvalues restrictedto Z . In particular, these operators cannot be ergodic in the Gaussian sense. Introduction and main result
The dynamical behaviour of linear operators acting on Fr´echet spaces X hasbeen investigated intensively in the last years. Recommended introductions are thetextbooks [3] and [9], and also the recent article [12]. It turns out that the richnessof eigenvectors corresponding to unimodular eigenvalues strongly influences themetric dynamical properties of linear operators. In particular, a linear operator ona Hilbert space admits a Gaussian invariant measure of full support if and only ifit has spanning unimodular eigenvectors and is ergodic in the Gaussian sense (thatis, ergodic with respect to a Gaussian measure of full support) if and only if it hasperfectly spanning unimodular eigenvectors. For these and corresponding resultswe refer in particular to [1], [2], [4], [10] and again to [12].Due to recent deep results of Menet ([14]) and Grivaux, Matheron and Menet([12]), in the situation of Hilbert spaces X the picture has become quite complete forthe case of chaotic operators, that is, for hypercyclic operators having eigenvectorscorresponding to roots of unity (i.e. periodic vectors) which span a dense subspaceof X . Less is known, however, in the case of absence of periodic or almost periodicvectors (cf. [12, Section 1.3]). In [11, Question 3] (see also [12, Question 7.7])it is asked if a hypercyclic operator with a densely spanning set of eigenvectorscorresponding to rationally independent eigenvalues is already ergodic. In thispaper, we give examples of topologically mixing operators on Hilbert spaces whichhave a densely spanning set of eigenvectors with unimodular eigenvalues restrictedto an arbitrary prescribed closed countable subset of the unit circle T . In particular,such operators cannot be ergodic in the Gaussian sense.For an open set Ω in the extended plane C ∞ with 0 ∈ Ω we denote by H (Ω)the Fr´echet space of functions holomorphic in Ω and vanishing at ∞ endowed withthe topology of locally uniform convergence, where, as usual, via stereographicprojection we identify C ∞ and the sphere S endowed with the spherical metric.We consider Bergman spaces on general open sets Ω ⊂ C ∞ with 0 ∈ Ω: For
Mathematics Subject Classification.
Key words and phrases. unimodular eigenvalues, Taylor shift. ≤ p < ∞ let A p (Ω) be the space of all functions f holomorphic in Ω that satisfy || f || p := || f || Ω ,p := (cid:16) Z Ω | f | p dm (cid:17) /p < ∞ , where m denotes the spherical measure on C ∞ . Then A (Ω) = H (Ω) and for p ≥ A p (Ω) , || · || p ) are Banach spaces. In the case p = 2, the normis induced by the inner product ( f, g ) R Ω f g dm .For Ω being bounded in the plane or containing ∞ , we define T = T A p (Ω) : A p (Ω) → A p (Ω) by T f ( z ) := ( f ( z ) − f (0)) /z ( z = 0) , T f (0) := f ′ (0) . If f ( z ) = P ∞ ν =0 a ν z ν , then T f ( z ) = ∞ X ν =0 a ν +1 z ν for | z | sufficiently small. We call T the Taylor (backward) shift on A p (Ω). Ifwe write S n f ( z ) := P nν =0 a ν z ν for the n -th partial sum of the Taylor expansion P ∞ ν =0 a ν z ν of f about 0, then by induction it is easily seen that(1) T n +1 f ( z ) = ( f − S n f ( z )) /z n +1 ( z = 0) , T n +1 f (0) = a n +1 , for n ∈ N . In [5] it is shown that for open sets Ω with 0 ∈ Ω the Taylor shift T istopologically mixing on H (Ω) if and only if each connected component of C ∞ \ Ωmeets T . Results concerning topological and metric dynamics of the Taylor shifton Bergman spaces are proved in [6] and [16].We write M ∗ := ( C ∞ \ M ) − for M ⊂ C ∞ . Then Ω ∗ is a compact plane set(note that 0 ∈ Ω) and the spectrum of T is contained in Ω ∗ . In the case 1 ≤ p < f ∈ A p (Ω) is an eigenfunction for T if and only if, for some α ∈ Ω ∗ , the function f is a scalar multiple of γ α , where γ α is defined by γ α ( z ) = 1 / (1 − αz )for z ∈ Ω (with 0 · ∞ := 0). In this case, α is the corresponding eigenvalue andthe spectrum as well as the point spectrum both equal Ω ∗ . If p ≥
2, then thefunctions γ α still belong to A p (Ω) for all α in the interior of Ω ∗ , but in general notfor α belonging to the boundary of Ω ∗ . If Ω has small spherical measure near aboundary point 1 /α of Ω, it may, however, happen that γ α is again an eigenfunctionof T (that is, γ α belongs to A p (Ω)). For example, γ ∈ A (Ω) in the case of thecrescent-shaped region Ω = D \ { z : | z − / | ≤ / } where D denotes the openunit disc in C . This opens up the possibility to place eigenvalues at certain pointsof the boundary of Ω ∗ . A corresponding construction leads to our main result. Wewrite E ( T ) for the set of unimodular eigenvalues of T , which for T = T A (Ω) equalsthe set of λ ∈ T such that γ λ ∈ A (Ω). Theorem 1.1.
Let Z ⊂ T be an infinite closed set. Then there is a open set Ω ⊂ D so that the Taylor shift T = T A (Ω) is topologically mixing, E ( T ) ⊂ Z and { γ λ : λ ∈ E ( T ) } spans a dense subspace of A (Ω) . Remark 1.2.
Since Hilbert spaces are of cotype 2, the main theorem from [4]implies that, in the situation of Theorem 1.1, for countable Z the Taylor shift T isnot ergodic in the Gaussian sense. So Question 3 from [11] can be answered in thenegative, at least in the weak form that ergodicity in the Gaussian sense does notalways follow from the existence of a spanning set of eigenvectors corresponding PERATORS WITH NON-PERFECTLY SPANNING UNIMODULAR EIGENVECTORS 3 to a rationally independent set of unimodular eigenvalues. We are left with theopen question whether T is ergodic with respect to some measure of full supportor (upper) frequently hypercyclic.If in the situation of Theorem 1.1 the set Z consists of roots of unity, then theTaylor shift is chaotic with unimodular eigenvalues only in Z and not ergodic inthe Gaussian sense. So we have an alternative construction for a chaotic operatoron a Hilbert space that is not ergodic in the Gaussian sense. The first constructionof such an operator on ℓ ( N ) given by Menet ([14]) even leads to an operator whichis not (upper) frequently hypercyclic. The approach via Taylor shift is, however,quite different and gives more flexibility in prescribing unimodular eigenvectors.2. Proof of Theorem 1.1
As main tool for the proof of Theorem 1.1 we seek results on rational approxi-mation in the mean.
Remark 2.1.
Let Ω be an open and bounded set in C and suppose that each pointon the boundary of Ω ∗ belongs to the boundary of some component of the interior(Ω ∗ ) ◦ of Ω ∗ . Then Theorem 4 from [13] implies that the span of { γ α : α ∈ (Ω ∗ ) ◦ } is dense in A (Ω). If Ω is open in the extended plane with ∞ ∈ Ω, this is also thecase (see [16, Remark 2.6]).As indicated in the introduction, the function γ α belongs to A (Ω) if α belongsto the boundary of Ω ∗ and Ω has small spherical measure near 1 /α . For k ∈ N , z ∈ Ω and α ∈ C we write γ α,k ( z ) := z k / (1 − αz ) k +1 = z k γ k +1 α ( z ) . We show that in the case of sufficiently small measure near 1 /α all functions γ α,k belong to A (Ω) and that under appropriate conditions they span a dense subspaceof A (Ω). The approach is strongly influenced by the proof of a result on thecompleteness of polynomials in A (Ω) (see [15, Theorem 12.1], cf. also [8, ChapterI, Section 3]).Let δ > C δ be defined as the interior of the convex hull of { t + is ( t ) : δ ≤ t ≤ δ } , where s ( t ) := exp( − exp(1 / | t | ))(with s (0) = 0). With that, we say that two components A, B of an open set U ⊂ S := R + i ( − π/ , π/
2) are directly bridged at w ∈ S , if ω ∈ T and δ > w + ωC δ ⊂ A and w − ωC δ ⊂ B . If ϕ : S → S is the standard parametrisationof S \ {± (0 , , } , that is, ϕ ( t + is ) = (cos( s ) cos( t ) , cos( s ) sin( t ) , sin( s ))for t ∈ R and − π/ < s < π/
2, we say that two components
C, D of an open set V ⊂ C ∞ with 0 , ∞ 6∈ ∂V are directly bridged, if the corresponding inverse images under ϕ in S are bridged at some w . In this case ζ = ϕ ( w ) is said to be a bridge point for( C, D ). We say that
C, D are bridged if finitely many components C , C , . . . , C m exist with C = C , C m = D and so that C j , C j − are directly bridged. If C denotesthe set of components of V , then bridging induces an equivalence relation ∼ on C .If a system D ⊂ C is a complete system of representatives for ∼ , we briefly say thatthe system is complete for V . H.-P. BEISE, L. FRERICK, AND J. M¨ULLER
Theorem 2.2.
Let Ω be an open set in C ∞ which is bounded in C or contains ∞ and suppose that each point on the boundary of Ω ∗ belongs to the boundary of somecomponent of the interior (Ω ∗ ) ◦ of Ω ∗ . Moreover, suppose D to be complete for (Ω ∗ ) ◦ and for D ∈ D let α D be either in D or a bridge point of ( C, D ) for some C ∈ C . Then both { γ α : α ∈ S D ∈D D } and { γ α D ,k : D ∈ D , k ∈ N } have densespan in A (Ω) .Proof. Let g ∈ A (Ω) = A (Ω) ′ . Then the Cauchy transform V g : (Ω ∗ ) ◦ → C of g ,defined by ( V g )( α ) = Z Ω γ α ( z ) g ( z ) dm ( z )is holomorphic with ( V g ) ( k ) ( α ) = k ! Z Ω γ α,k ( z ) g ( z ) dm ( z )for all k ∈ N . According to the Hahn-Banach theorem and Remark 2.1, it sufficesto show that V g = 0 under each of the conditions stated in the theorem.1. Suppose that g ⊥ γ α for all α ∈ S D ∈D D , that is ( V g ) | D = 0 for all D ∈ D .If D = C then V g = 0. If C ∈ C \ D then C, D are bridged for some D ∈ D . We canassume that C, D are directly bridged. Let ζ be a bridge point for ( C, D ). Usingthe assumption on the flatness of Ω near ζ it can be shown by differentiation ofthe parameter integral that ( V g ) ( k ) has a two-sided non-tangential limit at ζ forall k ∈ N and that ( V g ) ( k ) ( ζ ) = k ! Z Ω γ ζ,k ( z ) g ( z ) dm ( z )(cf. proof of Theorem 12.1 in [15]). We show that ( V g ) ( k ) ( ζ ) = 0 for all k implies( V g ) | C = 0 and ( V g ) | D = 0 (cf. again the proof of Theorem 12.1 in [15]). Since( V g ) | D = 0 we have ( V g ) ( k ) ( ζ ) = 0 for all k , and then also ( V g ) | C = 0.Without loss of generality we can assume that ζ = 1 and that C lies in D and D outside D . Considering the fat that ζ = 1 is a bridge point, we may fix 0 < r < ± C r satisfy ϕ − (Ω) ∩ ± C r = ∅ . Bythe identity theorem for holomorphic functions, it suffices to prove that V g belongsto a quasi-analytic subclass of C ∞ ( I ), where I = ϕ ( i [ − r/ , r/ I is a compact interval in R with 1 in its interior).To this end, we estimate the derivatives of V g on I . By the Cauchy-Schwarzinequality we have(2) | ( V g ) ( k ) ( x ) | ≤ k ! k g k (cid:18)Z Ω dm ( z ) | − xz | k +2 (cid:19) / for k ∈ N . So it suffices to estimate the latter integrals. We define W A ( k, x ) := Z A dm ( z ) | − xz | k +2 for k ∈ N , x ∈ I and measurable A ⊂ Ω. With Q := [ − r, r ] + i [ − r, r ], we havesup x ∈ I W Ω \ ϕ ( Q ) ( k, x ) = O ( q k ) PERATORS WITH NON-PERFECTLY SPANNING UNIMODULAR EIGENVECTORS 5 for some positive q . To estimate W ϕ ( Q ) ∩ Ω , we observe that the shape of Ω in ϕ ( Q )allows that with some constant c > | − xz | ≥ c | − z | for all x ∈ I andall z ∈ Ω ∩ ϕ ( Q ). Thus, for x ∈ I we obtain (by substituting u = e /t in the laststep) W Ω ∩ ϕ ( Q ) ( k, x ) ≤ cW Ω ∩ ϕ ( Q ) ( k, ≤ c Z r − r Z s ( t ) − s ( t ) cos s | t + is | k +2 ds dt ≤ c Z r s ( t ) t k +2 dt = 4 c Z ∞ e /r e − u u − log k ( u ) du. For k sufficiently large and u ≥ k we have log k ( u ) ≤ e u/ . Hence, by splitting upthe integral at k , one can see that Z ∞ e /r e − u u − log k ( u ) du = O ( k log k ( k )) = O (5 k log k ( k )) . Putting together we find that sup x ∈ I W Ω ( k, x ) = O (5 k log k ( k )). We now combinethe latter with (2) to obtainsup x ∈ I | ( V g ) ( k ) ( x ) | = O ( k ! 5 k/ log k ( k )) . The Denjoy-Carleman theorem now shows that
V g belongs to a quasi-analyticsubclass of C ∞ ( I ).2. Suppose that g ⊥ γ α D ,k , that is, ( V g )( α D ) ( k ) = 0, for all D ∈ D and k ∈ N ,and let C ∈ C . If C ∈ D then α C ∈ C or α C is a bridge point of C, E for some E ∈ C .In both cases, ( V g ) | C = 0. If C
6∈ D , then there is D ∈ D so that C, D are bridged.Again, we can assume that
C, D are directly bridged. Let ζ be a bridge pointfor ( C, D ). If α D ∈ D , then ( V g ) | D = 0. Hence ( V g ) ( k ) ( ζ ) = 0 for all k and then( V g ) | C = 0. If α D is a bridge point of ( D, E ) for some E ∈ C , then ( V g ) ( k ) ( α D ) = 0for all k and again ( V g ) | D = 0. As in the first case, ( V g ) | C = 0. (cid:3) Remark 2.3.
If, for some component D ∈ C , the single set system { D } is completeand if α is a point in D or a bridge point of ( C, D ) for some component C , thenthe span of { γ α,k : k ∈ N } is dense in A (Ω). In particular, in case α = 0 weconclude that the polynomials form a dense set in A (Ω) (cf. Theorem 12.1 in [15],where actually a weaker condition on the sharpness of Ω near 1 /α is proved to besufficient).From the Godefroy-Shapiro criterion (see e.g. [9, Theorem 3.1]) and Theorem2.2 we obtain Corollary 2.4.
Let Ω be an open set in C ∞ which is bounded in C or contains ∞ and suppose that each point in the boundary of Ω ∗ belongs to the boundary of somecomponent of (Ω ∗ ) ◦ . If complete systems D and D ′ exist with D ⊂ D for all D ∈ D and D ′ ⊂ C ∞ \ D for all D ′ ∈ D ′ , then T A (Ω) is topologically mixing. Example 2.5.
Let Ω = C ∞ \ ( U ∪ G ) = ((1 /U ) ∪ (1 /G )) ∗ , where G, /U ⊂ D aredomains with 0 G and so that U, G (or, equivalently, 1 /G, /U ) are bridged. If ζ is a corresponding bridge point, then for each α ∈ (1 /U ) ∪ (1 /G ) ∪ { /ζ } the linearspan of the rational functions { γ α,k : k ∈ N } forms a dense set in A (Ω) (Remark2.3 with D = { /U } or D = { (1 /G } ). By Corollary 2.4, the Taylor shift T A (Ω) ismixing. If 0 ∈ /U (e.g. for 1 /U = D , in which case Ω = D \ G ), the polynomials H.-P. BEISE, L. FRERICK, AND J. M¨ULLER are dense in A (Ω) and the Taylor shift on A ( D ) is a quasi-factor of the Taylorshift on A (Ω). Theorem 2.6.
For each infinite closed set Z ⊂ T a domain G ⊂ D bridged to C ∞ \ D exists with G ∩ T ⊂ Z and so that the span of { γ ζ : ζ ∈ G ∩ T } is dense in A ( D \ G ) .Proof. According to Example 2.5 with 1 /U = D , for the proof of the denseness ofthe span of { γ ζ : ζ ∈ G ∩ T } it suffices to show that γ ζ ∈ A ( D \ G ) for ζ ∈ G ∩ T and that, for some α ∈ D ∪ { } , all γ α,k belong to the closure of the span of { γ ζ : ζ ∈ G ∩ T } .We suppose 1 to be an accumulation point of Z and write U r := { z ∈ C : | z − | < r } ∩ D for r >
0. Moreover, we consider ”flat cups” C δ,ρ defined as above with s ( t ) replacedby ρs ( t ), where 0 < ρ ≤ < r < J ⊂ N finite. According to Runge’s theorem, a finite set F ⊂ Z ∩ U r and a family ( R j ) j ∈ J in span { γ ζ : ζ ∈ F } exist withmax z ∈ D \ U r | γ α,j ( z ) − R j ( z ) | < /n for j ∈ J .Let now 0 < r <
1. We choose a finite set Z ⊂ Z ∩ U r and a function R (0)0 ∈ span { γ ζ : ζ ∈ Z } withmax z ∈ D \ U r | γ α, ( z ) − R (0)0 ( z ) | < . With conv denoting the convex hull with respect to the plane C , we define G := U r ∩ [ ζ ∈ Z conv (cid:0) ζϕ ( C δ,ρ ) (cid:1) where we choose ρ > δ > G is connected with Z U r \ G | R (0)0 | dm < Z U r \ G | γ α, | dm < . For a second step we fix 0 < r < /
2. Then we find a finite set Z ⊂ Z ∩ U r and R (1)0 , R (1)1 ∈ span { γ ζ : ζ ∈ Z } such thatmax z ∈ D \ U r | γ α,j ( z ) − R (1)1 ( z ) | < / . for j = 0 ,
1. We define G := G ∪ (cid:16) U r ∩ [ ζ ∈ Z conv (cid:0) ζϕ ( C δ,ρ ) (cid:1)(cid:17) where we choose δ, ρ in such a way that G is connected with Z U r \ G | R (1) j | dm < / Z U r \ G | γ ,j | dm < / j = 0 ,
1. Successively, we obtain 0 < r n < /n , finite sets Z n ⊂ Z ∩ U r n ,functions R ( n ) j ∈ span { γ ζ : ζ ∈ Z n } such thatmax z ∈ D \ U rn − | γ α,j ( z ) − R ( n − j ( z ) | < /n PERATORS WITH NON-PERFECTLY SPANNING UNIMODULAR EIGENVECTORS 7 for j = 0 , . . . , n − G n ⊂ D ∩ U r with G n ∩ T = Z ∪· · ·∪ Z n ,bridged to C ∞ \ D at all ζ ∈ Z ∪ · · · ∪ Z n and such that Z U rn − \ G n | R ( n ) j | dm < / ( n + 1) and Z U rn − \ G n | γ ,j | dm < / ( n + 1)for j = 0 , . . . , n − G := S n ∈ N G n satisfies all requirements. (cid:3) Proof of Theorem 1.1.
Let Ω = D \ G where G is as in Theorem 2.6. Then G ∩ T ⊂E ( T ). According to Example 2.5 and Theorem 2.6, T is topologically mixing. Sinceeach point in T \ G is an interior point of D , no point in T \ G belongs to the pointspectrum, that is E ( T ) ⊂ Z . (cid:3) Remark 2.7.
1. By deleting sufficiently small parts from G it is possible to modify G to an open set W in such a way that Ω = D \ W is connected and the Taylorshift T A (Ω) satisfies the same conditions as in Theorem 1.1.2. The statement (and proof) of Theorem 2.6 can be modified in such a waythat Ω ∗ ∩ T ⊂ Z , i.e. the spectrum intersects T only in Z (cf. [11, Question 3]). References [1] F. Bayart, S. Grivaux, Frequently hypercyclic operators,
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