SPECTRUM OF A WEAKLY HYPERCYCLIC OPERATOR MEETS THE UNIT CIRCLE

S. J. DILWORTH AND VLADIMIR G. TROITSKY

Abstract. It is shown that every component of the spectrum of a weakly hypercyclic operator meets the unit circle. The proof is based on the lemma that a sequence of vectors in a Banach space whose norms grow at geometrical rate doesn’t have zero in its weak closure.

Suppose that T is a bounded operator on a nonzero Banach space X. Given a vector x ∈ X, we say that x is hypercyclic for T if the orbit Orb_{T} x = {T^{n}x}_{n} is dense in X. Similarly, x is said to be weakly hypercyclic if Orb_{T} x is weakly dense in X. A bounded operator is called hypercyclic or weakly hypercyclic if it has a hypercyclic or, respectively, a weakly hypercyclic vector. It is shown in [CS] that a weakly hypercyclic vector need not be hypercyclic, and there exist weakly hypercyclic operators which are not hypercyclic. C. Kitai showed in [K] that every component of the spectrum of a hypercyclic operator intersects the unit circle. K. Chan and R. Sanders asked in [CS] if the spectrum of a weakly hypercyclic operator meets the unit circle. In this note we show that every component of the spectrum of a weakly hypercyclic operator meets the unit circle.

Lemma 1. Let X be a Banach space and let c > 1. Suppose that x_{n} ∈ X satisfies ∥x_{n}∥ ⩾ c^{n} for all n ⩾ 1. Then 0 ∉ \overline{\{x_{n}\}_{n}}^{w}.

Proof. Let N be the smallest positive integer such that c^{N} > 2. We shall prove that there exist F_{1}, …, F_{N} ∈ X^{∗} such that

(1)  max_{1⩽k⩽N}|F_{k}(x_{n})| ⩾ 1  (n ⩾ 1).

Since ∥x_{n}∥ ⩾ c^{n}, by replacing x_{n} by (c^{n}/∥x_{n}∥)x_{n}, it suffices to prove (1) for the case in which ∥x_{n}∥ = c^{n} for all n ⩾ 1. First suppose that c > 2, so that N = 1. We have to construct F_{1} ∈ X^{∗} such that |F_{1}(x_{n})| ⩾ 1 for all n ⩾ 1. First choose f_{1} ∈ X^{∗} with f_{1}(x_{1}) = 1. Then either |f_{1}(x_{2})| < 1 or |f_{1}(x_{2})| ⩾ 1. In the former case the Hahn-Banach theorem guarantees the existence of g_{2} ∈ X^{∗} such that ∥g_{2}∥ ⩽ 1/∥x_{2}∥ = c^{−2}, |g_{2}(x_{2})| = 1 − |f_{1}(x_{2})|, and |(f_{1} + g_{2})(x_{2})| = 1. In the latter case, set g_{2} = 0. Note that

|(f_{1} + g_{2})(x_{1})| ⩾ 1 − ∥g_{2}∥∥x_{1}∥ ⩾ 1 − c^{−1}.

1

2000 Mathematics Subject Classification. Primary: 47A16,47A10,47A25.

Key words and phrases. Weakly hypercyclic operator, spectrum, spectral set, weak closure.

The research of the first author was supported by a summer research grant from the University of South Carolina while on sabbatical as a Visiting Scholar at The University of Texas at Austin.

Trends in Banach Spaces and Operator Theory in Contemp. Math., AMS, 2003, vol 321, pp. 67–69.

Set f_{2} = f_{1} + g_{2}. Repeating this argument, we can find g_{3} ∈ X^{∗} such that ∥g_{3}∥ ⩽ 1/∥x_{3}∥ = c^{−3} and |(f_{2} + g_{3})(x_{3})| ⩾ 1. Note that

|(f_{2} + g_{3})(x_{1})| ⩾ |f_{2}(x_{1})| − ∥g_{3}∥∥x_{1}∥ ⩾ 1 − c^{−1} − c^{−2}

and also that

|(f_{2} + g_{3})(x_{2})| ⩾ 1 − ∥g_{3}∥∥x_{2}∥ ⩾ 1 − c^{−1}.

Set f_{3} = f_{2} + g_{3}. Continuing in this way we obtain f_{n} ∈ X^{∗} such that (setting g_{n} = f_{n} − f_{n−1}) ∥g_{n}∥ ⩽ c^{−n} and

(2)  |f_{n}(x_{k})| ⩾ 1 − ∑_{i=1}^{n−k} c^{−i}  (1 ⩽ k ⩽ n).

Thus, {f_{n}}_{n} is norm-convergent in X^{∗} to some f ∈ X^{∗}. From (2), we obtain (since c > 2)

|f(x_{k})| = lim_{n}|f_{n}(x_{k})| ⩾ 1 − ∑_{i=1}^{∞} c^{−i} = \frac{c − 2}{c − 1} > 0.

Set F_{1} = (c − 1)(c − 2)^{−1}f, to complete the proof in the case c > 2.

Now suppose that 1 < c < 2. Set α = c^{N} > 2. For each 1 ⩽ k ⩽ N, consider the sequence y_{n} = x_{k+(n−1)N} (n ⩾ 1). Then ∥y_{n}∥ = (c^{k}/α)α^{n}. Since α > 2 there exists F_{k} ∈ X^{∗} such that |F_{k}(y_{n})| ⩾ 1 for all n ⩾ 1, which proves (1). □

Remark 2. Recall that a closed subspace Y of X^{∗} is said to be norming if there exists C > 0 such that

∥x∥ ⩽ C sup{|f(x)| : f ∈ Y, ∥f∥ ⩽ 1} (x ∈ X).

The argument of Lemma 1 easily generalizes to give the following result. Suppose that Y is norming for X and that {x_{n}}_{n} is a sequence in X satisfying ∥x_{n}∥ ⩾ c^{n}, where c > 1. Then 0 does not belong to the σ(X, Y)-closure of {x_{n}}_{n}. In particular, Lemma 1 is valid for the weak-star topology when X is a dual space.

We also make use of the following simple numerical fact. If (t_{n}) is a sequence in ℝ^{+} ∪ {∞}, then

lim sup_{n→∞} \sqrt[n]{t_{n}} = inf\{ ν > 0 | lim_{n→∞} \frac{t_{n}}{ν^{n}} = 0 \} = inf\{ ν > 0 | lim sup_{n→∞} \frac{t_{n}}{ν^{n}} < ∞ \}.

In particular, if T is a bounded operator with spectral radius r, then the Gelfand formula lim_{n} \sqrt[n]{∥T^{n}∥} = r yields that \frac{∥T^{n}∥}{λ^{n}} → 0 for every scalar λ with |λ| > r. Theorem 3. If T is weakly hypercyclic, then every component of σ(T) meets {z : |z| = 1}.

Proof. Let x be a weakly hypercyclic vector for T. Let σ be a non-empty component of σ(T), denote σ′ = σ(T) \ σ. Denote by X_{σ} and X_{σ′} the corresponding spectral subspaces, then X_{σ} and X_{σ′} are closed, T -invariant, and X = X_{σ} ⊕ X_{σ′}. Also, σ(T_{|X_{σ}}) = σ and σ(T_{|X_{σ′}}) = σ′. Note that σ′ might be empty, in which case we have X_{σ} = X and X_{σ′} = {0}.

Denote by P_{σ} the spectral projection corresponding to σ, then X_{σ} = RangeP_{σ}. Denote y = P_{σ}x. Without loss of generality, ∥y∥ = 1. Since P_{σ} is bounded and, therefore, weakly continuous, and Orb_{T} y = P_{σ}(Orb_{T} x), we conclude that Orb_{T} y is weakly dense in X_{σ}. Thus, y is weakly hypercyclic for T_{|X_{σ}}.

Observe that the inclusion σ ⊆ {z : |z| < 1} is impossible. Indeed, in this case the spectral radius of T_{|X_{σ}} would be less than 1, so that T^{n}y → 0, which contradicts y being weakly hypercyclic for T_{|X_{σ}}.

Finally, we show that the inclusion σ ⊆ {z : |z| > 1} is equally impossible. In this case 0 ∉ σ = σ(T_{|X_{σ}}), so that T_{|X_{σ}} is invertible. Denote the inverse by S. Then S is a bounded operator on X_{σ} and by the Spectral Mapping Theorem

σ(S) = {λ | λ^{−1} ∈ σ(T_{|X_{σ}})} ⊂ {z : |z| < 1}.

Therefore, r(S) < a for some 0 < a < 1. This yields lim_{n} \frac{∥S^{n}∥}{a^{n}} = 0, so that ∥S^{n}∥ ⩽ a^{n} for all sufficiently large n. In particular,

1 = ∥y∥ = ∥S^{n}T^{n}y∥ ⩽ a^{n}∥T^{n}y∥,

so that ∥T^{n}y∥ ⩾ \frac{1}{a^{n}}. Lemma 1 asserts that 0 ∉ \overline{\{T^{n}y\}_{n}}^{w}, which contradicts y being weakly hypercyclic for T_{|X_{σ}}. □

Proposition 4. Suppose that Y is norming for X. If T has a hypercyclic vector for the σ(X, Y) topology, then the spectrum of T intersects {z : |z| = 1}.

Proof. Suppose, to derive a contradiction, that σ(T) does not intersect the unit circle. We use the notation introduced above with σ = σ(T) ∩ {z : |z| < 1} and σ′ = σ(T) \ σ. Let x be a hypercyclic vector for the σ(X, Y)-topology. Then x = y + z, where y ∈ X_{σ} and z ∈ X_{σ′}. Since ∥T^{n}y∥ → 0, it follows easily that z is hypercyclic, and hence that z ≠ 0. But then there exists c > 1 such that ∥T^{n}z∥ ⩾ c^{n} for all sufficiently large n, which contradicts Remark 2. □

References

[CS] Kit C. Chan and Rebecca Sanders. A weakly hypercyclic operator that is not norm hypercyclic. To appear in J. Operator Theory.

[K] Carol Kitai. Invariant Closed Sets for Linear Operators. PhD thesis, Univ. of Toronto, 1982.

Department of Mathematics, University of South Carolina, Columbia, SC 29208. USA.

Current address: Department of Mathematics, The University of Texas at Austin, Austin, Texas 78712. USA

Department of Mathematical and Statistical Sciences, 632 CAB, University of Alberta, Edmonton, AB T6G 2G1. Canada.

   E-mail address: dilworth@math.sc.edu, vtroitsky@math.ualberta.ca