Introduction to the dynamical properties of Toeplitz operators on the Hardy space of the unit disc

Definition 2.1.

More formally, let X be a Fréchet space, that is a complete topological vector space whose topology is generated by a countable family of seminorms, and let T: X → X be a continuous and linear operator on X (in short T ∈ L ( X ) ) . The operator T is called hypercyclic on X if there exists a x ∈ X such that its orbit under T,

is dense in X. In such a case, x is called a hypercyclic vector for T and the set of all hypercyclic vectors for T is denoted by HC(T).

Example 2.2.

The first known example of a hypercyclic operator was constructed by Birkhoff in 1929 [2], illustrating that such “chaotic” behavior can indeed occur even for linear operators. Let H o l ( C ) be the Fréchet space of entire functions, let a be any non zero complex number and let T a : H o l ( C ) → H o l ( C ) be the associated translation operator defined by

Then T _a is hypercyclic on H o l ( C ) .

Example 2.3.

Another early example of a hypercyclic operator, still on the Fréchet space H o l ( C ) , was given by MacLane in 1952 [3]. Let D : H o l ( C ) → H o l ( C ) be the differentiation operator defined by Df = f′. Then D is hypercyclic on H o l ( C ) as well.

Example 2.4.

On Banach spaces, the first known example is provided by Rolewicz in 1969 [4]. Let X = ℓ p ( N ) , 1 ≤ p < + ∞, or X = c 0 ( N ) . Let T be the backward shift operator on X defined by

Then λT is hypercyclic on X if and only if |λ| > 1.

Definition 2.5.

Let X be a Fréchet space and T ∈ L ( X ) . The operator T is called topologically transitive if for each pair of non-empty open sets (U, V) of X, there exists n ∈ N such that T ⁿ (U) ∩ V ≠ ∅.

Theorem 2.6

(Birkhoff’s transitivity theorem, 1922 [7]). Let X be a separable Fréchet space and T ∈ L ( X ) . The following assertions are equivalent.

1. T is hypercyclic;

2. T is topologically transitive.

Proof.

(i) ⇒ (ii): Assume that T is hypercyclic and let x be a hypercyclic vector for T. Consider U and V be two non-empty open sets of X. Since orb(T, x) is dense in X, there is some k ≥ 0 such that T ^k x ∈ U. But the space X has no isolated points, and then the set {T ^m x; m ≥ k} is also dense in X. Hence there exists m ≥ k such that T ^m x ∈ V. Observe now that

meaning that T ⁿ (U) ∩ V ≠ ∅ with n = m − k. Thus T is topologically transitive.

(ii) ⇒ (i): Assume now on the contrary that T is topologically transitive and let us prove that T is hypercyclic. Since X has a countable dense set {y _j ; j ≥ 1} (remember that X is separable), then the open balls of radius 1/m around the y _j , j, m ≥ 1, form a countable base ( V k ) k ≥ 1 of the topology of X. Hence, x ∈ HC(T) if and only if for every k ≥ 1, there is some n ≥ 0 such that T ⁿ x ∈ V _k . In other words

By continuity of T, the sets

are open sets for every k ≥ 1. Let’s prove that W _k is dense for every k ≥ 1.

Let U be an open set of X. Since T is topologically transitive, there exists n ≥ 0 such that

Therefore W _k is dense in X. By Baire category theorem, the set H C ( T ) = ⋂ k ≥ 1 W k is dense in X and in particular, HC(T) ≠ ∅.□

Remark 2.7.

Let T ∈ L ( X ) and assume that T is invertible. Then T is hypercyclic if and only if T ⁻¹ is hypercyclic. Indeed it is sufficient to observe that

and to apply Birkhoff theorem.

Theorem 2.8

(Godefroy – Shapiro citerion, 1991 [8]). Let X be a separable Fréchet space and T ∈ L ( X ) . Suppose that

are dense in X. Then T is hypercyclic on X.

Proof.

We shall prove that T is topologically transitive. Let U and V be two non-empty open subsets of X. By assumption, U ∩ H ₋(T) ≠ ∅ and V ∩ H ₊(T) ≠ ∅. In other words, we can find

with Tx _k = λ _k x _k with |λ _k | < 1, Ty _k = μ _k y _k with |μ _k | > 1 and a k , b k ∈ C , 1 ≤ k ≤ m . Observe that

and

Taking into account the fact that U and V are open subsets, it follows that there is some N ∈ N so that, for all n ≥ N,

In other words, V ∩ T ⁿ (U) ≠ ∅ meaning that T is topologically transitive. According to Birkhoff’s theorem, we then deduce that T is hypercyclic.□

Proposition 2.9.

Let T ∈ L ( X ) . If T is hypercyclic on X, then σ _p (T*) = ∅.

Proof.

Let x ∈ HC(T) and argue by contradiction assuming that T*(x*) = μx* for some μ ∈ C and x* ∈ X*, x* ≠ 0. Then, we have, for every n ∈ N

Observe that the set { x * , T n x ; n ≥ 0 } is dense in C whereas clearly the set { μ n x * , x ; n ≥ 0 } is not dense. This yields the desired contradiction.□

Proposition 2.10.

Let T ∈ L ( X ) .

1. If ‖T‖ ≤ 1 then T is not hypercyclic on X.

2. If for every x ∈ X, ‖Tx‖ ≥ ‖x‖, then T is not hypercyclic on X.

Proof.

(a) If ‖T‖ ≤ 1, then for every x ∈ X and every n ≥ 1, we have ‖T ⁿ x‖ ≤ ‖x‖. Hence the orbit orb(T, x) is bounded, and then cannot be dense in X. Since this is true for every x ∈ X, then T is not hypercyclic.

(b) If for every x ∈ X, ‖Tx‖ ≥ ‖x‖, then for every x ∈ X and every n ≥ 0, we have ‖T ⁿ x‖ ≥ ‖x‖. Hence the orbit orb(T, x) stays away from 0 (if x ≠ 0) and then cannot be dense in X. One more time, T cannot be hypercyclic on X.□

Definition 2.11.

Let H be a Hilbert space and T ∈ L ( H ) . We say that T is hyponormal if T*T − TT* ≥ 0, that is

Fact 2.12.

If T is hyponormal, then we have

Proof.

Write:

□

Theorem 2.13.

Let T ∈ L ( H ) be hyponormal. Then T is not hypercyclic on H.

Proof.

Let x ∈ H, x ≠ 0.

Assume first that the sequence ( ‖ T n x ‖ ) n ≥ 0 is decreasing. Thus ‖T ⁿ x‖ ≤ ‖x‖ for all n ≥ 0 and x cannot be a hypercyclic vector.

Now, it remains to prove that when the sequence ( ‖ T n x ‖ ) n ≥ 0 is not decreasing, then the vector x cannot be hypercyclic as well.

By assumption, there exists N ∈ N such that ‖T ^N+1 x‖ > ‖T ^N x‖. In particular, it implies that T ^N+1 x ≠ 0 and then T ^N x ≠ 0. By Fact 2.12 applied to T ^N x, we obtain ‖T ^N+1 x‖² ≤ ‖T ^N+2 x‖‖T ^N x‖, whence

An induction shows that ( ‖ T n x ‖ ) n ≥ N is strictly increasing and in particular ‖T ⁿ x‖ > ‖T ^N x‖, ∀n ≥ N + 1. Hence x cannot be a hypercyclic vector.□

Theorem 2.14.

Let X be a complex separable Banach space and T ∈ L ( X ) . Assume that T is hypercyclic. Then σ ( T ) ∩ T ≠ ∅ .

Proof.

We argue by contradiction and suppose that σ ( T ) ∩ T = ∅ .

If σ ( T ) ⊂ D then by the spectral radius formula r(T) < 1 and there exists ɛ > 0 and M > 0 such that

which contradicts the fact that T is hypercyclic.

If σ ( T ) ⊂ C \ D ̄ , then T is invertible and T ⁻¹ is also hypercyclic. But since σ ( T − 1 ) = σ ( T ) − 1 ⊂ D , we also get a contradiction.

Therefore σ 1 = σ ( T ) ∩ D and σ 2 = σ ( T ) ∩ C \ D ̄ form a partition of σ(T) into two non empty closed sets. By the Riesz decomposition theorem, there exists two non-trivial closed invariant subspaces M ₁ and M ₂ such that

But it is not difficult to see that T | M 1 should be hypercyclic, which is again a contradiction because σ ( T | M 1 ) = σ 1 ⊂ D .□

Theorem 2.15

(Kitai, 1982 [9]). Let X be a complex separable Banach space and T ∈ L ( X ) . Assume that T is hypercyclic. Then every connected component of σ(T) intersects T .

Theorem 3.1

(Hartman – Wintner, 1954 [10]). The following assertions are equivalent:

1. The operator T(a) is bounded on ℓ 2 ( N ) ;

2. There exists a function ϕ ∈ L ∞ ( T ) such that a n = ϕ ̂ ( n ) , ∀ n ∈ Z .

Moreover, UT(a)U ⁻¹ = T _ϕ , where T _ϕ f = P ₊(ϕf) f ∈ H ².

Proof of (ii) ⟹ (i).

First note that if ϕ ∈ L ∞ ( T ) , then, for every f ∈ H ², we have

Hence T _ϕ is bounded on H ² and ‖T _ϕ ‖ ≤ ‖ϕ‖_∞.

Moreover, let ( e n ) n ≥ 0 be the canonical basis of ℓ 2 ( N ) . Then, for every n, k ≥ 0, on one hand, we have T ( a ) e n , e k = a k − n , and on the other hand, we have

Hence, for every n, k ≥ 0, we have T ( a ) e n , e k = U − 1 T ϕ U e n , e k , which proves that T(a) = U ⁻¹ T _ϕ U is bounded on ℓ 2 ( N ) .

For the proof of (i) ⟹ (ii), we refer to the following book: “An introduction to operators on the Hardy-Hilbert space” by Rubén A. Martinez-Avendaño and Peter Rosenthal, 2007 [11].□

Theorem 3.2

(Brown – Halmos, 1963 [12]). Let ϕ ∈ L ∞ ( T ) . Then T ϕ ∈ L ( H 2 ) and ‖T _ϕ ‖ = ‖ϕ‖_∞.

Proof.

We have already seen that T _ϕ is bounded and ‖T _ϕ ‖ ≤ ‖ϕ‖_∞.

For the converse inequality, denote by k λ ̃ = k λ / ‖ k λ ‖ 2 the normalized reproducing kernel. We have

Moreover, using that ‖ k λ ‖ 2 2 = k λ , k λ 2 = k λ ( λ ) = 1 1 − | λ | 2 , we have

where P ( ϕ ) is the Poisson transform of ϕ. Hence we get that for every λ ∈ D , we have

We will now use a well-known property of the Poisson integral (see [11], Corollary 1.1.27]) saying that

Hence, according to [7], it follows that for almost all e i θ ∈ T , we have

Hence ‖ϕ‖_∞ ≤ ‖T _ϕ ‖, which concludes the proof.□

Some observations.

Let ϕ ∈ L ^∞. It is easy to see that

1. T ϕ * = T ϕ ̄ . In particular, if S = T _z is the shift operator on H ², then S * = T z ̄ is the backward shift operator on H ². For every f ∈ H ², we have

1. If ϕ ∈ H ^∞, the algebra of bounded and analytic functions on D , then, for every λ ∈ D , we have

Indeed, let f ∈ H ², then

Now observe that since ϕ ∈ H ^∞, then ϕf ∈ H ² and we deduce that

Therefore, T ϕ ̄ k λ = ϕ ( λ ) ̄ k λ .

1. In general, T _ϕ T _ψ ≠ T _ϕψ but Brown and Halmos proved the following.□

Theorem 3.3

(Brown – Halmos, 1963 [12]). Let ϕ , ψ ∈ L ∞ ( T ) Then

Theorem 3.4

(Coburn, 1966 [13]). Let ϕ ∈ L ∞ ( T ) , ϕ ≢ 0 . Then either kerT _ϕ = {0} or ker T ϕ * = { 0 } .

Proof.

Argue by contradiction and assume that there are two functions h ₁ and h ₂ in H ^∞, h ₁, h ₂ ≠ 0 such that T _ϕ h ₁ = 0 and T ϕ ̄ h 2 = 0 . Taking into account of [15], it means that ϕ h 1 ∈ H 0 2 ̄ and ϕ ̄ h 2 ∈ H 0 2 ̄ . It is an easy exercise to prove that if f ∈ H 0 2 and g ∈ H ², then f g ∈ L 1 ( T ) and f g ̂ ( n ) = 0 , ∀n ≤ 0.

Let now define ψ = h 1 ϕ h 2 ̄ . Since ϕ h 2 ̄ ∈ H 0 2 and h ₁ ∈ H ², then ψ ∈ L 1 ( T ) and ψ ̂ ( n ) = 0 , ∀n ≤ 0. Moreover ψ ̄ = h 1 ϕ ̄ h 2 , with h 1 ϕ ̄ ∈ H 0 2 and h ₂ ∈ H ². Hence ψ ̄ ̂ ( n ) = 0 , ∀n ≤ 0, which give that ψ ̂ ( n ) = 0 , ∀n ≥ 0. Therefore ψ ̂ ( n ) = 0 , ∀ n ∈ Z and thus ψ ≡ 0. But a standard property of Hardy spaces (see [11], Corollary 2.7.2]) shows that, since h ₁, h ₂ ≠ 0, then h ₁ and h ₂ are non-zero almost everywhere. Hence ϕ ≡ 0 a.e. which is a contradiction.□

Theorem 3.5.

Let ϕ ∈ C ( T ) .

1. T _ϕ is a Fredholm operator on H ² if and only if ϕ does not vanish on T . In this case,

1. We have

Example 3.6.

1. If S denotes the shift operator on H ², it can be easily proved that σ ( S * ) = D ̄ and σ p ( S * ) = D .

2. Let ϕ be a continuous symbol such that ϕ ( T ) is represented as in Figure 3. In this example, the spectrum is the union of the curve and the two components of index $-1$ and $-2$.

Theorem 4.1

(Godefroy – Shapiro, 1991 [8]). Let ϕ ∈ H ^∞. The following assumptions are equivalent.

1. T ϕ ̄ is hypercyclic on H ²;

2. ϕ is non-constant and ϕ ( D ) ∩ T ≠ ∅ .

Proof.

(ii) ⟹ (i). Since ϕ is non constant, by the open mapping theorem, ϕ ( D ) is an open connected set which intersects T (see Figure 4).

Let U = { z ∈ D ; | ϕ ( z ) | < 1 } and V = { z ∈ D ; | ϕ ( z ) | > 1 } . Then U and V are two open sets which are both non empty. On the other hand, it follows from [2] that

where we recall that

According to Godefroy – Shapiro Criterion, it is sufficient to prove that if Ω is a non empty open subset of D , then span{k _z ; z ∈ Ω} is dense in H ². Thus let f ∈ H ² such that f ⊥ k _z , ∀ z ∈ Ω. Hence

But since Ω is a non-empty open subset of D , the uniqueness principle for analytic functions implies that f ≡ 0. Hence H − ( T ϕ ̄ ) and H + ( T ϕ ̄ ) are both dense in H ², which implies that T ϕ ̄ is hypercyclic on H ².

(i) ⟹ (ii). Assume now that T ϕ ̄ is hypercyclic on H ². Then, of course, ϕ is non-constant and T ϕ ̄ is not a contraction. In particular, ‖ ϕ ‖ ∞ = ‖ T ϕ ‖ = ‖ T ϕ ̄ ‖ > 1 . Moreover, we also have inf z ∈ D | ϕ ( z ) | < 1 . Indeed, if inf z ∈ D | ϕ ( z ) | ≥ 1 , then 1/ϕ ∈ H ^∞ and T 1 / ϕ * is not hypercyclic because ‖ T 1 / ϕ * ‖ = ‖ T 1 / ϕ ‖ = ‖ 1 / ϕ ‖ ∞ ≤ 1 . But since T _1/ϕ T _ϕ = T _ϕ T _1/ϕ = I, the operator T _ϕ is invertible and T ϕ − 1 = T 1 / ϕ . Thus T ϕ * − 1 = T 1 / ϕ * . Now, since T 1 / ϕ * is not hypercyclic, it follows from Remark 2.7 that T ϕ * = T ϕ ̄ is not hypercyclic as well, which is absurd. Hence we have

Now, a simple connectedness argument implies that ϕ ( D ) ∩ T ≠ ∅ .□

Corollary 4.2

(Rolewicz, 1969 [4]). Let λ ∈ C .Then

Lemma 4.3.

Let F ( z ) = a z + b + c z , | a | > | c | > 0 . Then

1. F ( T ) is an ellipse;

2. The interior E of the ellipse satisfies

Proof.

Let a = |a|e^iα , c = |c|e^iγ and consider F ̃ ( z ) = | a | z + | c | z . Then

This formula shows that we can assume, without loss of generality, that b = 0 and a and c are real with a > c > 0.

(a) F(e^iθ ) = ae^−iθ + ce^iθ = (a + c)cosθ + i(c − a)sinθ. Hence

and thus F ( T ) is an ellipse centered at 0 whose foci are ( ± 2 a c , 0 ) and the length of the axis are 2(a + c) and 2(a − c).