aa r X i v : . [ m a t h . L O ] A p r Unilateral weighted shifts on ℓ Miami University
Paul B. Larson
Miami University
Definition 1.
Given w ∈ ℓ ∞ , define a bounded linear operator B w : ℓ → ℓ by B w ( x )( i ) = w ( i ) · x ( i + 1) . Such a B w is called a unilateral weighted shift . A vector x ∈ ℓ is hypercyclic for B w iff the set { B kw ( x ) : k ∈ ω } of forward iterates is dense in ℓ . Let HC ( w ) denote the set of all hypercyclicvectors for B w .It is routine to check that HC ( w ) is a G δ set for any w ∈ ℓ ∞ . The ques-tion addressed in the present paper is how much the complexity of HC ( w ) canbe increased by looking at those sequences which are hypercyclic for many w simultaneously. Concretely, for W ⊆ ℓ ∞ , let X W = \ w ∈ W HC ( w ) . It turns out that X W can be made arbitrarily complicated by making W suffi-ciently complex (Theorem 4). Even for a G δ set W , however, the set X W canstill be non-Borel (Theorem 5).It is necessary to introduce a few preliminaries and some terminology beforeproceeding. Let k·k denote the usual ℓ norm. In what follows, this notationwill be used for finite sequences as well, i.e., for s ∈ R <ω , k s k = p s (0) + . . . + s ( n ) assuming s is of length n + 1.The notation | s | will be used to denote both the length of a string (if s ∈ <ω )and the length of an interval (if s ⊆ ω is an interval). The notation k x k ∞ willdenote the ℓ ∞ or sup-norm of x . Again, this definition makes sense for any string x – either finite or infinite. There is a relationship between the 2-norm and thesup-norm of a finite string which will be useful in what follows. Indeed, if s is afinite string of real numbers, having length n , a computation shows that k s k ≤ n / k s k ∞ . Definition 2. A pointclass Γ is a collection of subsets of Polish (separable com-pletely metrizable) spaces such that • Γ is closed under continuous preimages, • Γ is closed under finite unions and • Γ is closed under finite intersections.Given a pointclass Γ, the dual pointclass ¯Γ consists of those Y contained in somePolish space X such that X \ Y ∈ Γ. A pointclass is non-self-dual iff there isa Polish space X and a set Y ⊆ X such that Y ∈ Γ but
Y / ∈ ¯Γ (equivalently, X \ Y / ∈ Γ).To take a few examples, “closed” and “open” are dual pointclasses as are “ F σ ”and “ G δ ”. All four of these classes are non-self-dual. Proposition 3.
For a Borel set W ⊆ ℓ ∞ , the intersection T w ∈ W HC ( w ) is co-analytic.Proof. To see this, observe that, for y ∈ ℓ , y ∈ \ w ∈ W HC ( w ) ⇐⇒ ( ∀ w ∈ ℓ ∞ )( w ∈ W = ⇒ y ∈ HC ( w )) . The key observation is that, although ℓ ∞ is not Polish, its Borel structure isthe same as that inherited from R ω (which is Polish). Therefore, the claim that T w ∈ W HC ( w ) is co-analytic follows by regarding W and ℓ ∞ as subsets of R ω andusing the fact that the relation P ( y, w ) ⇐⇒ y ∈ HC ( w )is itself G δ .The next two theorems show that the upper bound from the last propositioncannot be improved. Theorem 4.
Given a non-self-dual pointclass Γ which contains the closed sets,there is a set W ⊆ ℓ ∞ such that T w ∈ W HC ( w ) is not in Γ . Theorem 5.
There is a Borel set W such that T w ∈ W HC ( w ) is properly co-analytic, i.e., not analytic. The key to proving Theorems 4 and 5 lies with the next three lemmas.2 emma 6. If s ∈ R n and k s k ∞ < n − / ε , then k s k < ε .Proof. Suppose that s ∈ R n and k s k ∞ < n − / ε , i.e., | s ( i ) | < n − / ε for all i < n .It follows that k s k = p s (0) + . . . + s ( n − < q n · ( n − / ε ) = ε This proves the lemma.
Lemma 7. If A is a countable set and f : 2 A → ℓ is such that1. f is continuous with respect to the product topologies on A and ℓ (inheritedfrom R ω ) and2. there exists y ∈ ℓ such that | f ( x )( i ) | ≤ y ( i ) for all x ∈ A and i ∈ ω ,then f is continuous with respect to the norm-topology on ℓ .Proof. Let y ∈ ℓ be as in the statement of the lemma. Towards the goal ofshowing that f is ℓ -continuous, fix ε > n be such that k y ↾ [ n, ∞ ) k < ε/ . Since f is continuous into the product topology on ℓ , let F ⊆ A be finite andsuch that, for x , x ∈ A , if x ↾ F = x ↾ F , then | f ( x )( i ) − f ( x )( i ) | < n − / ε/ i < n . In particular, x ↾ F = x ↾ F guarantees k f ( x ) − f ( x ) ↾ n k < ε/ x , x ∈ A and x ↾ F = x ↾ F , k f ( x ) − f ( x ) k ≤ k f ( x ) − f ( x ) ↾ n k + k f ( x ) ↾ [ n, ∞ ) k + k f ( x ) ↾ [ n, ∞ ) k < ε/ k y ↾ [ n, ∞ ) k < ε/ ε/ ε. Since ε was arbitrary this completes the proof. Note that a stronger result was infact proved: f is uniformly continuous with respect to the standard ultrametricon 2 A . 3 emma 8. Given a countable set A . It is possible to assign to each a ⊆ A ,sequences y a ∈ ℓ and w a ∈ { , } ω such that y a ∈ HC ( w b ) ⇐⇒ a + b Moreover the maps a y a and a w a are homeomorphism between A and theirranges. Before proving this lemma, it will be helpful to introduce an alternative topo-logical basis for ℓ . Given a finite string q ∈ Q <ω of rationals and a (rational)number ε >
0, let U q,ε = { x ∈ ℓ : k ( x ↾ | q | ) − q k ∞ < ε | q | − / and k x ↾ [ | q | , ∞ ) k < ε } First note that each U q,ε is open. In order to check that the U q,ε form a basis for ℓ , fix a basic open ball V = { x ∈ ℓ : k x − x k < ε } where x ∈ ℓ and ε > n ∈ ω be such that k x ↾ [ n, ∞ ) k < ε/ q ∈ Q n such that k x ↾ n − q k < ε/ . First of all, it follows from the definition of U q,ε that x ∈ U q,ε/ . To see that U q,ε/ ⊆ V , observe that if x ∈ U q,ε/ , k x − x k ≤ k ( x − x ) ↾ n k + k ( x − x ) ↾ [ n, ∞ ) k ≤ n / k ( x − x ) ↾ n k ∞ + k x ↾ [ n, ∞ ) k + k x ↾ [ n, ∞ ) k < n / ( k ( x ↾ n ) − q k ∞ + k ( x ↾ n ) − q k ∞ ) + ε/ ε/ < n / (( ε/ n − / + ( ε/ n − / ) + ε/ ε As x ∈ U q,ε/ was arbitrary, it follows that U q,ε/ ⊆ V . Since V was an arbitraryopen ball, this shows that the U q,ε form a topological basis for ℓ . Proof of Lemma 8.
Let π : ω → Q <ω be a surjection. Let A be the fixed countableset from the statement of the lemma. for “coding” purposes, fix a bijection h· , · , ·i : ω × ( Q ∩ (0 , × A → ω. Given n ∈ ω , let p n ∈ ω , ε n > i n ∈ A be such that n = h p n , ε n , i n i . ρ n = min { ε r : r < n } . The first step of the proof is to choose a suitable partition I , J , I , J , . . . of ω into consecutive intervals, i.e., such that min( J n ) = max( I n ) + 1 andmin( I n +1 ) = max( J n ) + 1. Each J n will be chosen with | J n | = | π ( p n ) | . Thelengths of the I n will be chosen recursively and, for concreteness, of minimallength satisfying1. | I n | ≥ | I n − | ,2. | I n | > max( J n − ) and3. 2 −| I n | · k π ( p n ) k ≤ − n − · ρ n · − max( J n − ) · −| I n − | .for n >
1. The length of I is arbitrary – I can even be the empty interval.The next step is to define the desired y a and w a for each a ⊆ A . For n = h p, ε, i i , define y a on each I n and J n by1. ( ∀ n )( y a ↾ I n = ¯0),2. ( ∀ n )( i ∈ a = ⇒ y a ↾ J n = ¯0) and3. ( ∀ n )( i / ∈ a = ⇒ y a ↾ J n = 2 −| I n | · π ( p ).The first important observation about the map a y a is that it is continuous.To see this, first observe that every initial segment of y a is determined by aninitial segment of a . This implies that a y a is continuous into the producttopology on ℓ (which it inherits from R ω ). Now invoke Lemma 7 and use thefact that y a is always termwise bounded by y ∅ ∈ ℓ . It now follows that a y a is in fact continuous with respect to the norm-topology on ℓ .It also follows from the definition of y a that the function a y a is injec-tive. As the domain of this map (2 A ) is compact, a y a must therefore be ahomeomorphism with its range.Now define w a ∈ { , } ω (for a ⊆ A ) by making sure that the restrictions w a ↾ I n ∪ J n satisfy1. ( ∀ n )( i n / ∈ a = ⇒ w a ↾ I n ∪ J n = ¯1),2. ( ∀ n )( i n ∈ a = ⇒ ( ∀ j ∈ J n )( |{ i < j : w a ( i ) = 2 }| = | I n | ) and3. if i, j ∈ I n with i < j and w a ( j ) = 2, then w a ( i ) = 2.5he continuity of a w a follows from the fact that initial segments of w a arecompletely determined by initial segments of a .The next three claims will complete the proof. The proofs of these threeclaims all follows similar arguments using the definitions of the y a and w a . Claim.
Each y a is in ℓ .It suffices to show that the ℓ norm of y a is finite. Indeed, by the triangleinequality and the third part of the definition of y a , k y a k ≤ X n ∈ ω k y a ↾ J n k ≤ X n ∈ ω −| I n | · k π ( p n ) k ≤ X n ∈ ω − n − · ρ n · − max( J n − ) · −| I n − | ≤ X n ∈ ω − n − ≤ Claim. If a, b ⊆ A with a ⊇ b , then y a / ∈ HC ( w b ).For this claim, it suffices to show that k B kw b ( y a ) k ≤ B kw b ( y a )(0) = 0 foreach k ∈ ω . This will establish that there is no k ∈ ω such that B kw b ( y a ) is in theopen set U = { y ∈ ℓ : k y k > y (0) = 0 } . To this end, fix k ∈ ω and let n ∈ ω be such that k ∈ I n ∪ J n . First of all, if i n ∈ a , then y a ↾ I n ∪ J n = ¯0 and hence B kw b ( y a )(0) = w b (0) · . . . · w b ( k − · y a ( k ) = 0 . On the other hand, if i n / ∈ a ⊇ b , then w b ↾ I n ∪ J n = ¯1 and hence |{ j < k : w b ( j ) = 2 }| ≤ max( J n − ) . To obtain an estimate on k B kw b ( y a ) k , a couple preliminary observations will beuseful. Suppose t ∈ ω is such that k + t ∈ I r for some r ∈ ω . In this case, B kw b ( y a )( t ) = 0since y a ( k + t ) = 0. If k + t ∈ J n (where k ∈ I n ∪ J n ), then | B kw b ( y a )( t ) | ≤ max( J n − ) · | y a ( k + t ) | w b ↾ I n ∪ J n = ¯1. Finally, if k + t ∈ J r for some r > n , then | B kw b ( y a )( t ) | ≤ k · | y a ( k + t ) |≤ max( J r − ) since k ≤ max( J n ) ≤ max( J r − ). It now follows by the triangle inequality that k B kw b ( y a ) k ≤ X r ≥ n max( J r − ) · k y a ↾ J r k ≤ X r ≥ n max( J r − ) · − r − · ρ r · − max( J r − ) · | I r − | ≤ X r ≥ n − r − ≤ Claim. If a, b ⊆ A with a + b , then y a ∈ HC ( w b ).For this final claim, it suffices to show that, for each q ∈ Q <ω and ε >
0, thereis a k ∈ ω such that B kw b ( y a ) is in the open set U q,ε = { x ∈ ℓ : k ( x ↾ | q | ) − q k ∞ < ε | q | − / and k x ↾ [ | q | , ∞ ) k < ε } as these open sets form a topological basis for ℓ by remarks preceding the proof.Indeed, fix q ∈ Q <ω and let p ∈ ω be such that π ( p ) = q . Fix i ∈ b \ a and let n = h p, ε, i i . Since i ∈ b and i / ∈ a , the second case in the definition of w b ↾ I n ∪ J n and the second case in the definition of y a ↾ J n are active. In particular, for each j ∈ J n , |{ t < j : w b ( t ) = 2 }| = | I n | . It follows that B min( J n ) w b ( y a ) = π ( p ) a y for some y ∈ ℓ . To show that B min( J n ) w b ( y a ) ∈ U q,ε (for any given ε > k y k < ε , since q ≺ B min( J n ) w b ( y a ) by choice of n . Indeed,7bserve that, again by the triangle inequality, k y k ≤ | I n | · X r>n k y a ↾ J r k ≤ | I n | · X r>n −| I r | · k π ( p r ) k ≤ | I n | · X r>n − r − · ρ r · − max( J r − ) · −| I r − | ≤ | I n | · X r>n − r − · ρ n · −| I n | ≤ ε · X r>n − r − < ε since ρ n ≤ ε = ε n . This complete the proof of the claim and proves Lemma 8. Proof of Theorem 4.
Let P ⊆ ω be a perfect set such that a + b for any twodistinct a, b ∈ P . The construction of such a set is a standard inductive argument(similar to the construction of a perfect independent set). Let y a and w a be as inthe lemma for all a ⊆ ω . It follows from the independence of P that y a ∈ HC ( w b )iff a = b for all a, b ∈ P .Given a non-self-dual pointclass Γ which contains the closed sets, fix Y ⊆ P with Y ∈ Γ \ ¯Γ. Since P is closed, it follows that P \ Y ∈ ¯Γ \ Γ. Let W = { w a : a ∈ Y } . Now consider the set X W = \ w ∈ W HC ( w ) . For a ∈ P , notice that y a ∈ X W iff a / ∈ Y . Hence, X W ∩ { y a : a ∈ P } = { y a : a ∈ P and a / ∈ Y } = { y a : a ∈ P \ Y } It follows that X W / ∈ Γ since { y a : a ∈ P } is closed and { y a : a ∈ P \ Y } ∈ ¯Γ \ Γ (because a y a is a homeomorphism). This completes the proof of thetheorem. Proof of Theorem 5.
The key to this proof is an application of Lemma 8 with thecountable set A taken to be ω <ω . With this in mind, let Wf = { T ⊆ ω <ω : T is a well-founded subtree } and C = { p ⊆ ω <ω : p is a maximal ≺ -chain } .
8n other words, C may be identified with the set of infinite branches through ω <ω .The set Wf proper co-analytic while C is G δ . Let W = { w p : p ∈ C } and noticethat W is also G δ since p w p is a homeomorphism by Lemma 8. To see that X W = \ w ∈ W HC ( w )is not analytic, observe that, for any subtree T ⊆ ω <ω ,[ T ] = ∅ ⇐⇒ ( ∀ p ∈ C )( T + p ) ⇐⇒ ( ∀ p ∈ C )( y T ∈ HC ( w p )) (by Lemma 8) ⇐⇒ ( y T ∈ X W ) . It follows that Wf is a continuous preimage of X W under the map T y T . Inturn, this implies that X W cannot be analytic. References [1] F. Bayart, E. Matheron,