ORBITS, WEAK ORBITS AND LOCAL CAPACITY OF OPERATORS

Vladimír Müller

Let T be an operator on a Banach space X. We give a survey of results concerning orbits {T^{n}x : n = 0, 1, …} and weak orbits {⟨T^{n}x, x^{∗}⟩ : n = 0, 1, …} of T where x ∈ X and x^{∗} ∈ X^{∗}. Further we study the local capacity of operators and prove that there is a residual set of points x ∈ X with the property that the local capacity cap(T, x) is equal to the global capacity capT. This is an analogy to the corresponding result for the local spectral radius.

INTRODUCTION

Let T be a bounded linear operator acting on a (real or complex) Banach space X and let x ∈ X. The orbit of x under the operator T is the sequence {T^{n}x : n = 0, 1, …}.

The properties of orbits of different points may differ very much — the orbits of some points may be ”regular” while other points may have very ”irregular” orbits.

Many deep results and problems of operator theory may be formulated using the notion of orbits, For example, T has no nontrivial invariant subspace if and only if the orbit of each non-zero vector x ∈ X spans the whole space. Similarly, T has no nontrivial closed invariant subset if and only if the orbit of each x ≠ 0 is dense; in this case all orbits are extremely irregular.

Analogously, weak orbits under T are sequences {⟨T^{n}x, x^{∗}⟩ : n = 0, 1, …} where x ∈ X and x^{∗} ∈ X^{∗} are fixed. This notion is also closely related to the invariant subspace problem — the main idea of the celebrated Scott Brown technique is the construction of a weak orbit with very definite properties.

Many results for both orbits and weak orbits have their parallel for continuous one parameter semigroups of operators. In this context, orbits are closely related to stability results for semigroups of operators.

In the last section of this paper we study also polynomial orbits. By the polynomial orbit of x ∈ X we mean the set {p(T)x : p polynomial}. Apart from the invariant subspace problem this notion is closely connected with the notion of capacity of operators introduced by P. Halmos.

1991 Mathematics Subject Classification: 47A11.

Keywords and phrases: orbits, weak orbits, local spectral radius, local capacity. The research was supported by the grant No. A1019801 of AV ČR.

The aim of this paper is to give a survey of results concerning orbits, weak orbits and polynomial orbits of operators. We always try to construct all types of orbits as regular as possible.

Many results concerning orbits of operators on complex Hilbert spaces may be found in [2]. For results concerning semigroups of operators see [14].

The author wishes to thank to P. Niemiec for careful reading of the manuscript and for discovering several gaps in the proofs.

Denote by L(X) the set of all bounded linear operators acting on a Banach space X.

We say that a subset M ⊂ X is residual if its complement X \ M is of the first category. Clearly a subset M ⊂ X is residual if and only if it contains a dense G_{δ}-set.

I. ORBITS IN COMPLEX BANACH SPACES

In this section X will be a complex Banach space and T ∈ L(X).

It is known that there is a residual set of points x ∈ X with the property that the local spectral radius r_{x}(T) = lim sup_{n→∞} ∥T^{n}x∥^{1/n} is equal to the spectral radius

r(T) = lim_{n→∞} ∥T^{n}∥^{1/n} = inf_{n} ∥T^{n}∥^{1/n},

see [15], [5]. In particular, for x in this residual set, there are infinitely many powers such that ∥T^{n}x∥ is ”large”.

Moreover, by [9], there are always points x ∈ X such that ∥T^{n}x∥ is ”large” for all powers n ≥ 0.

More precisely, we have the following results:

THEOREM 1.1. Let T ∈ L(X), let (a_{n})_{n≥0} be a sequence of positive numbers such that a_{n} → 0. Then:

(i) the set of all x ∈ X with the property that

∥T^{n}x∥ ≥ a_{n}∥T^{n}∥ for infinitely many n’s

is residual. (ii) Let k ≥ 0. Then in each ball in X of radius greater than max{a_{j} : j ≥ k} there

is a vector u such that

∥T^{n}u∥ ≥ a_{n}r(T^{n}) (n ≥ k).

In particular, there is a dense set of points x ∈ X with the property that ∥T^{n}x∥ ≥ a_{n}r(T^{n}) for all but a finite number of n’s. Further, there exist points x ∈ X such that ∥T^{n}x∥ ≥ a_{n}r(T^{n}) for all n ≥ 0.

PROOF. (i) For k ∈ N set

M_{k} = {x ∈ X : there exists n ≥ k such that ∥T^{n}x∥ > a_{n}∥T^{n}∥}.

Clearly M_{k} is an open set. We prove that M_{k} is dense. Let x ∈ X and ε > 0. Choose n ≥ k such that a_{n}ε^{−1} < 1. There exists z ∈ X of norm one such that ∥T^{n}z∥ > a_{n}ε^{−1}∥T^{n}∥. Then

2a_{n}∥T^{n}∥ < ∥T^{n}(2εz)∥ ≤ ∥T^{n}(x + εz)∥ + ∥T^{n}(x − εz)∥

so that either ∥T^{n}(x + εz)∥ > a_{n}∥T^{n}∥ or ∥T^{n}(x − εz)∥ > a_{n}∥T^{n}∥. Thus either x + εz ∈ M_{k} or x − εz ∈ M_{k} so that dist{x, M_{k}} ≤ ε. Since x and ε were arbitrary, the set M_{k} is dense.

By the Baire category theorem the intersection ⋂_{k=1}^{∞} M_{k} is a dense G_{δ}-set, hence it is residual. Clearly each x ∈ ⋂_{k=1}^{∞} M_{k} satisfies ∥T^{n}x∥ > a_{n}∥T^{n}∥ for infinitely many n’s.

In particular, for a_{n} = n^{−1} we obtain

r_{x}(T) = lim sup_{n→∞} ∥T^{n}x∥^{1/n} ≥ lim sup_{n→∞}(\frac{∥T^{n}∥}{n})^{1/n} = r(T)

for all x in a residual subset of X.

(ii) was proved in [9]; for Hilbert space operators see also [2].

COROLLARY 1.2. The set {x ∈ X : lim sup_{n→∞} ∥T^{n}x∥^{1/n} = r(T)} is residual. The set {x ∈ X : lim inf_{n→∞} ∥T^{n}x∥^{1/n} = r(T)} is dense.

In particular, there is a dense subset of points x ∈ X with the property that the limit lim_{n→∞} ∥T^{n}x∥^{1/n} exists (and is equal to r(T)).

The existence of the limit lim_{n→∞} ∥T^{n}x∥^{1/n} was also studied in [1].

In general it is not possible to replace the word ”dense” in Corollary 1.2 by ”residual”.

EXAMPLE 1.3. Let H be a separable Hilbert space with an orthonormal basis {e_{j} : j = 0, 1, …} and let S be the backward shift, Se_{0} = 0, Se_{j} = e_{j−1} (j ≥ 1). Then r(S) = 1 and the set {x ∈ H : lim inf_{n→∞} ∥S^{n}x∥^{1/n} = 0} is residual.

In particular, the set {x ∈ H : the limitlim_{n→∞} ∥S^{n}x∥^{1/n} exists} is of the first category (but it is always dense by Corollary 1.2).

PROOF. For k ∈ N let

M_{k} = {x ∈ X : there exists n ≥ k such that ∥S^{n}x∥ < k^{−n}}.

Clearly M_{k} is an open subset of X. Further, M_{k} is dense in X. To see this, let x ∈ X and ε > 0. Let x = ∑_{j=0}^{∞} α_{j}e_{j} and choose n ≥ k such that ∑_{j=n}^{∞} |α_{j}|^{2} < ε^{2}. Set y = ∑_{j=0}^{n−1} α_{j}e_{j}. Then ∥y − x∥ < ε and S^{n}y = 0. Thus y ∈ M_{k} and M_{k} is a dense open subset of X.

By the Baire category theorem the set M = ⋂_{k=0}^{∞} M_{k} is a dense G_{δ}-subset of X, hence it is residual.

Let x ∈ M. For each k ∈ N there is n_{k} ≥ k such that ∥S^{n_{k}}x∥ < k^{−n_{k}} so that lim inf_{n→∞} ∥S^{n}x∥^{1/n} = 0.

Since the set {x ∈ H : lim sup_{n→∞} ∥S^{n}x∥^{1/n} = r(S) = 1} is also residual, we see that the set {x ∈ H : the limit lim_{n→∞} ∥S^{n}x∥^{1/n} exists} is of the first category.

It is also possible to combine conditions (i) and (ii) of Theorem 1.1 and to obtain points x ∈ X with ∥T^{n}x∥ ≥ a_{n} · ∥T^{n}∥ for all n; in this case, however, there is a restriction on the sequence (a_{n}). The next lemma and its corollary essentially improve the estimates of [2], Proposition 2.B.2.

LEMMA 1.4. Let X, Y be complex Banach spaces, let T_{n} ∈ L(X, Y) (n = 1, 2, …) be a sequence of operators, let a_{n} be positive numbers such that ∑_{n=1}^{∞} (α_{n})^{2/3} < 1/4. Let x ∈ X. Then there exists u ∈ X such that ∥u−x∥ < 1/4 and ∥T_{n}u∥ ≥ a_{n}∥T_{n}∥ for all n ≥ 1.

PROOF. Without loss of generality we can assume that all operators T_{n} are non-zero.

Choose δ > 0 such that (1 + δ) ∑_{n=1}^{∞} α_{n}^{2/3} < 1/4. Set ε_{n} = (1 + δ)α_{n}^{2/3} (n = 1, 2, …) so that ∑_{n=1}^{∞} ε_{n} < 1/4. For each n find z_{n} ∈ X of norm one such that ∥T_{n}z_{n}∥ ≥ (1 + δ)^{−1}∥T_{n}∥.

The proof will be done in several steps.

A. For each k ∈ N there are complex numbers λ_{1}, …, λ_{n}, |λ_{n}| ≤ ε_{n} (n = 1, …, k) such that

∥T_{n}(x + ∑_{n=1}^{k}λ_{n}z_{n})∥ ≥ a_{n}∥T_{n}∥ (n = 1, …, k).

Proof. Fix k ∈ N. Write

Λ = {λ = (λ_{1}, …, λ_{k}) ∈ C^{k} : |λ_{n}| ≤ ε_{n} (n = 1, …, k)}.

For λ ∈ Λ set u_{λ} = x + ∑_{n=1}^{k} λ_{n}z_{n}.

For j = 1, …, k let Λ_{j} = {λ ∈ Λ : ∥T_{j}u_{λ}∥ < a_{j}∥T_{j}∥}. Let 1 ≤ j ≤ k and suppose that λ, λ′ ∈ Λ_{j} where λ = (λ_{1}, …, λ_{k}) and λ′ = (λ_{1}, …, λ_{j−1}, λ′_{j}, λ_{j+1} , …, λ_{k}). Then

|λ_{j} −λ′_{j}|(1+δ)^{−1}∥T_{j}∥ ≤ |λ_{j} −λ′_{j}|·∥T_{j}z_{j}∥ = ∥T_{j}(u_{λ}−u_{λ′})∥ ≤ ∥T_{j}u_{λ}∥+∥T_{j}u′_{λ}∥ < 2a_{j}∥T_{j}∥

so that |λ_{j} − λ′_{j}| < 2a_{j}(1 + δ). Thus, for fixed λ_{1}, …, λ_{j−1}, λ_{j+1} , …, λ_{k}, the set {ν ∈ C : (λ_{1}, …, λ_{j−1}, ν, λ_{j+1} , …, λ_{k}) ∈ Λ_{j}} is contained in a ball of radius 2a_{j}(1+δ).

Let μ be the Lebesgue measure on Λ. Then μ(Λ) = ∏_{n=1}^{k} (πε_{n}^{2}) and, by the Fubini theorem,

μ(Λ_{j}) ≤ 4π(1 + δ)^{2}a_{j}^{2} ∏_{{1≤n≤k \atop n≠j}}(πε_{n}^{2}) = \frac{4a_{j}^{2}(1 + δ)^{2}}{ε_{j}^{2}} μ(Λ) ≤ 4a_{j}^{2/3} μ(Λ).

Thus

μ(Λ \ ⋃_{j=1}^{k} Λ_{j}) ≥ μ(Λ)(1 − 4 ∑_{j=1}^{k}a_{j}^{2/3} ) > 0.

Hence there exists λ ∈ Λ \ ⋃_{j=1}^{k} Λ_{j}. In other words, u = u_{λ} satisfies ∥T_{j}u∥ ≥ a_{j}∥T_{j}∥ (j = 1, …, k) and ∥u − x∥ ≤ ∑_{n=1}^{k} |λ_{n}| ≤ ∑_{n=1}^{k} ε_{n} < 1/4.

B. The set M = {x + ∑_{n=1}^{∞} λ_{n}z_{n} : |λ_{n}| ≤ ε_{n} (n = 1, 2, …)} is totally bounded.

Proof. We must show that for each η > 0 there is a finite η-net in M.

Find k ∈ N such that ∑_{n=k+1}^{∞} ε_{n} < η/2. Set

M_{k} = {x + ∑_{n=1}^{k}λ_{n}z_{n} : |λ_{n}| ≤ ε_{n} (n = 1, 2, …, k)}.

Clearly M_{k} is compact so that there exists a finite set F ⊂ M_{k} such that dist{u, F } ≤ η/2 for all u ∈ M_{k}. Clearly F is the required η-net for M.

Proof of Lemma 1.4. By A, for each k ∈ N there is u_{k} ∈ M with

∥T_{n}u_{k}∥ ≥ a_{n}∥T_{n}∥ (n = 1, …, k).

By B, there is a convergent subsequence (u_{k_{j}}) of (u_{k}). Denote by u ∈ X its limit. Clearly ∥u − x∥ ≤ lim sup_{j→∞} ∥u_{k_{j}} − x∥ ≤ ∑_{n=1}^{∞} ε_{n} < 1/4 and

∥T_{n}u∥ ≥ a_{n}∥T_{n}∥ (n = 1, 2, …).

COROLLARY 1.5. Let T ∈ L(X). Let (a_{n})_{n≥0} be a sequence of positive numbers satisfying ∑_{n=0}^{∞} a_{n}^{2/3} < ∞. Then there is a dense subset L ⊂ X such that, for each x ∈ L, there is k ∈ N with

∥T^{n}x∥ ≥ a_{n}∥T^{n}∥ (n ≥ k).

Further, there are points x ∈ X such that ∥T^{n}x∥ ≥ a_{n}∥T^{n}∥ for all n ≥ 0.

PROOF. Let x ∈ X and ε > 0. Find k ∈ N and s such that (4 ∑_{n=k}^{∞} a_{n}^{2/3} )^{3/2} <

s < ε. Set a′_{n} = \frac{a_{n}}{s}. Then ∑_{n=k}^{∞} a′_{n}^{2/3} < 1/4 so that, by Lemma 1.4, there exists u ∈ X with ∥u − \frac{x}{s}∥ < 1/4 and ∥T^{n}u∥ ≥ a′_{n}∥T^{n}∥ (n ≥ k). Thus ∥su − x∥ < ε and ∥T^{n}(su)∥ ≥ a_{n}∥T^{n}∥ (n ≥ k).

A better estimate can be obtained using the essential norm.

For T ∈ L(X) and a closed subspace M ⊂ X denote by T |M the restriction T |M : M → X. For T ∈ L(X) let ∥T ∥_{μ} = inf{∥T |M∥ : M ⊂ X, codimM < ∞}. This quantity belongs to ”measures of non-compactness” since ∥T ∥_{μ} = 0 if and only if T is compact (for more details see [8]).

For Hilbert space operators ∥T ∥_{μ} is equal to the essential norm ∥T ∥_{e} = inf{∥T + K∥ : K ∈ K(X)} where K(X) denotes the ideal of all compact operators acting on X.

The following lemma (see [10], Lemma 1) is a useful technical tool in many constructions. It plays the role of the ”orthogonal complement” in general Banach spaces.

LEMMA 1.6. Let F be a finite dimensional subspace of a Banach space X, let ε > 0. Then there exists a closed subspace M ⊂ X of finite codimension such that

∥f + m∥ ≥ (1 − ε) max{∥f∥, ∥m∥/2}

for all f ∈ F and m ∈ M.

THEOREM 1.7. Let T ∈ L(X), let (a_{n})_{n≥0} be a sequence of positive numbers satisfying ∑_{n=0}^{∞} a_{n} < ∞. Then each ball in X of radius greater than 2 ∑_{n=0}^{∞} a_{n} contains a point u such that

∥T^{n}u∥ ≥ a_{n}∥T^{n}∥_{μ}  (n = 0, 1, …).  (1)

PROOF. The statement is trivial if dimX < ∞. Suppose that X is infinite dimensional.

Let x ∈ X and ε > 2 ∑_{n=0}^{∞} a_{n}. We show that there is u ∈ X such that ∥u − x∥ < ε and (1).

Let δ > 0 satisfy (1 + δ)a_{0} + 2(1 + δ)^{3} ∑_{n=1}^{∞} a_{n} < ε. We construct inductively a convergent sequence (u_{k})_{k≥0} whose limit will satisfy the required conditions.

Let u_{0} ∈ X be any vector satisfying ∥u_{0} − x∥ = a_{0}(1 + δ) and ∥u_{0}∥ ≥ a_{0}(1 + δ) (for example, u_{0} = x + a_{0}(1 + δ)x/∥x∥ will do).

If u_{k} ∈ X has already been constructed then set E_{k} = ⋁{T^{n}u_{k} : 0 ≤ n ≤ k + 1}. By Lemma 1.6, there is a closed subspace Y_{k} ⊂ X of finite codimension such that ∥e + y∥ ≥ (1 + δ)^{−1} max{∥e∥, ∥y∥/2} for all e ∈ E_{k}, y ∈ Y_{k}. Let Z_{k} = ⋂_{s=0}^{k+1} ⋂_{j=0}^{k} T^{−s}Y_{j}. Clearly codimZ_{k} < ∞ so that there is z_{k+1} ∈ Z_{k} of norm one such that ∥T^{k+1} z_{k+1} ∥ ≥ (1 + δ)^{−1}∥T^{k+1} ∥_{μ}. Clearly T^{s}z_{k+1} ∈ Y_{j} for all s ≤ k + 1 and j ≤ k. Set u_{k+1} = u_{k} + 2(1 + δ)^{3}a_{k+1} z_{k+1} . Then ∥u_{k+1} − u_{k}∥ = 2(1 + δ)^{3}a_{k+1} so that the sequence (u_{k}) constructed in this way is Cauchy. Denote by u its limit, u = u_{0} + ∑_{k=1}^{∞} 2(1 + δ)^{3}a_{k}z_{k}. Clearly ∥u − x∥ ≤ ∥u_{0} − x∥ + 2(1 + δ)^{3} ∑_{k=1}^{∞} a_{k} < ε and ∥u∥ ≥ (1 + δ)^{−1}∥u_{0}∥ ≥ a_{0}. For each n ≥ 1 we have

∥T^{n}u∥ = ∥T^{n}(u_{n} + ∑_{k=n+1}^{∞} 2(1 + δ)^{3}a_{k}z_{k})∥ ≥ (1 + δ)^{−1}∥T^{n}u_{n}∥

= (1 + δ)^{−1} ∥T^{n}(u_{n−1} + 2(1 + δ)^{3}a_{n}z_{n})∥ ≥ \frac{(1 + δ)^{−2}}{2} ∥T^{n}(2(1 + δ)^{3}a_{n}z_{n})∥

= (1 + δ)a_{n}∥T^{n}z_{n}∥ ≥ a_{n}∥T^{n}∥_{μ}.

If X is a Hilbert space then it is possible to take in the previous proof Y_{k} = E_{k}^{⊥}. The sequence (z_{k}) is then orthonormal and it is possible to obtain a better result, which improves [2], Theorem 2.A.7 (compare also Remark 2.A.8 of [2] with the just proved Theorem 1.7).

COROLLARY 1.8. Let T be an operator on a Hilbert space H. Let (a_{n})_{n≥0} be a sequence of positive numbers satisfying ∑_{n=0}^{∞} a_{n}^{2} < ∞. Then in each ball of radius

greater than (∑_{n=0}^{∞} a_{n}^{2})^{1/2} there exists a point x such that ∥T^{n}x∥ ≥ a_{n}∥T^{n}∥_{e} (n ≥ 0).

Further, there exists a dense subset L ⊂ H such that for each x ∈ L there is k ∈ N with

∥T^{n}x∥ ≥ a_{n}∥T^{n}∥_{e} (n ≥ k).

Another result which is true for Hilbert space operators is the following theorem, see [2], Corollary 3.6. We give an alternative proof which can be adapted to Banach space operators.

LEMMA 1.9. Let H be a Hilbert space, let T ∈ L(H) be a non-nilpotent operator. Then there exists x ∈ H such that

∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} = ∞.

PROOF. We distinguish two cases.

A. There exists a subspace M ⊂ H of finite codimension such that ∥T^{n}|M∥ ≤ \frac{1}{2}∥T^{n}∥ for infinitely many n’s.

Let {f_{1}, …, f_{r}} be an orthonormal basis in M^{⊥}. Let A = {n ∈ N : ∥T^{n}|M∥ ≤ \frac{1}{2}∥T^{n}∥\}, so that A is an infinite set. For j = 1, …, r set

A_{j} = \{n ∈ A : ∥T^{n}f_{j}∥ ≥ \frac{1}{3r}∥T^{n}∥\}.

We show that ⋃_{j=1}^{r} A_{j} = A. Suppose on the contrary that there is n ∈ A \ ⋃_{j=1}^{r} A_{j}. Let x ∈ H, ∥x∥ = 1 and ∥T^{n}x∥ > \frac{5}{6}∥T^{n}∥. Write x as x = ∑_{j=1}^{r} α_{j}f_{j} + u where α_{j} ∈ C and u ∈ M. Then |α_{j}| ≤ 1, ∥u∥ ≤ 1 and

∥T^{n}x∥ ≤ ∑_{j=1}^{r}|α_{j}| · ∥T^{n}f_{j}∥ + ∥T^{n}u∥ ≤ r \frac{1}{3r}∥T^{n}∥ + \frac{1}{2}∥T^{n}∥ = \frac{5}{6}∥T^{n}∥,

a contradiction. Thus A = ⋃_{j=1}^{r} A_{j} and there exists j ∈ {1, …, r} such that A_{j} is infinite. Hence

∑_{n=0}^{∞} \frac{∥T^{n}f_{j}∥}{∥T^{n}∥} ≥ ∑_{n∈A_{j}} \frac{∥T^{n}f_{j}∥}{∥T^{n}∥} ≥ ∑_{n∈A_{j}} \frac{1}{3r} = ∞.

B. For each subspace M ⊂ H of finite codimension, ∥T^{n}|M∥ > \frac{1}{2}∥T^{n}∥ for all but a finite number of n’s.

We construct inductively a convergent sequence (x_{k}) ⊂ H and an increasing sequence (n_{k}) of positive integers such that ∥T^{n_{j}}x_{k}∥ ≥ \frac{1}{2j} · ∥T^{n_{j}}∥ (j ≤ k). Then the limit x = lim_{k→∞} x_{k} will satisfy ∥T^{n_{j}}x∥ ≥ \frac{1}{2j} · ∥T^{n_{j}}∥ for all j, so that

∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} ≥ ∑_{j=1}^{∞} \frac{∥T^{n_{j}}x∥}{∥T^{n_{j}}∥} ≥ ∑_{j=1}^{∞} \frac{1}{2j} = ∞.

Let n_{1} = 1, x_{1} ∈ H, ∥x_{1}∥ = 1 and ∥T x_{1}∥ > ∥T ∥/2. Let k ∈ N and suppose that we have found x_{k} ∈ H and n_{1} < n_{2} < ⋯ < n_{k} such that ∥T^{n_{j}}x_{k}∥ ≥ \frac{1}{2j} ∥T^{n_{j}}∥ (j ≤ k). Let

M = ⋂_{s≤n_{k}} T^{−s}(∨{T^{j}x_{i} : 0 ≤ j ≤ n_{k}, 1 ≤ i ≤ k}^{⊥}).

Clearly codimM < ∞ so that there are n_{k+1} > n_{k} and a vector u_{k+1} ∈ M of norm one such that ∥T^{n_{k+1}} u_{k+1} ∥ > \frac{1}{2}∥T^{n_{k+1}} ∥. Then

∥T^{n_{k+1}} (x_{k} + \frac{u_{k+1}}{k + 1} )∥ + ∥T^{n_{k+1}} (x_{k} − \frac{u_{k+1}}{k + 1} )∥ ≥ \frac{2∥T^{n_{k+1}} u_{k+1} ∥}{k + 1} ≥ \frac{∥T^{n_{k+1}} ∥}{k + 1}

so that either x_{k+1} = x_{k}+ \frac{u_{k+1}}{k+1} or x_{k+1} = x_{k}− \frac{u_{k+1}}{k+1} will satisfy ∥T^{n_{k+1}} x_{k+1} ∥ ≥ \frac{∥T^{n_{k+1}} ∥}{2(k+1)}. Further T^{n_{j}}x_{k} ⊥ T^{n_{j}}u_{k+1} (j ≤ k) so that

∥T^{n_{j}}x_{k+1} ∥ ≥ ∥T^{n_{j}}x_{k}∥ ≥ \frac{1}{2j}∥T^{n_{j}}∥ (j ≤ k).

Let (x_{k}) be the sequence constructed in the above described way. For m < k we have ∥x_{k} −x_{m}∥^{2} = ∑_{i=m+1}^{k} \frac{1}{i^{2}} . Thus the sequence (x_{k}) is convergent and its limit x satisfies the required condition.

THEOREM 1.10. Let T be a non-nilpotent operator on a Hilbert space H. Then the set \{x ∈ H : ∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} = ∞\} is residual.

PROOF. For k ∈ N let M_{k} = \{x ∈ H : ∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} > k\}. Clearly M_{k} is open. To show that M_{k} is dense, let x ∈ H and ε > 0. By the previous lemma there is u ∈ H such that ∑_{n=0}^{∞} \frac{∥T^{n}u∥}{∥T^{n}∥} = ∞. Clearly we can assume that ∥u∥ = ε. Then

∑_{n=0}^{∞} \frac{∥T^{n}(x + u)∥}{∥T^{n}∥} + ∑_{n=0}^{∞} \frac{∥T^{n}(x − u)∥}{∥T^{n}∥} ≥ 2 ∑_{n=0}^{∞} \frac{∥T^{n}u∥}{∥T^{n}∥} = ∞

so that either x + u or x − u belongs to M_{k}. Thus M_{k} is an open dense subset of X and M = ⋂_{k=0}^{∞} M_{k} is residual. The points of M satisfy the required property.

For Banach space operators the previous statements are not true:

EXAMPLE 1.11. There are a Banach space X and a non-nilpotent operator T ∈ L(X) such that ∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} < ∞ for all x ∈ X.

PROOF. Let X be the ℓ_{1} space with the standard basis {e_{0}, e_{1}, …}. Let T ∈ L(X) be the weighted backward shift defined by T e_{0} = 0 and T e_{k} = (\frac{k+1}{k} )^{2}e_{k−1} (k ≥ 1). For n ∈ N we have

T^{n}e_{k} = \{ \begin{matrix}0 & (n > k), \\ \frac{(k+1)^{2}}{(k−n+1)^{2}} e_{k−n} & (n ≤ k) \end{matrix}

and ∥T^{n}∥ = (n + 1)^{2}. Thus r(T) = 1.

Let x ∈ X, x = ∑_{k=0}^{∞} α_{k}e_{k} where ∑_{k=0}^{∞} |α_{k}| < ∞. Then

∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} = ∑_{n=0}^{∞} ∑_{k=n}^{∞} \frac{|α_{k}|(k + 1)^{2}}{(n + 1)^{2}(k − n + 1)^{2}} = ∑_{k=0}^{∞}|α_{k}| ∑_{n=0}^{k} \frac{(k + 1)^{2}}{(n + 1)^{2}(k − n + 1)^{2}} .

We have

\begin{matrix}∑_{n=0}^{k} \frac{(k + 1)^{2}}{(n + 1)^{2}(k − n + 1)^{2}} = ∑_{n=0}^{[k/2]} \frac{(k + 1)^{2}}{(n + 1)^{2}(k − n + 1)^{2}} + ∑_{n=[k/2]+1}^{k} \frac{(k + 1)^{2}}{(n + 1)^{2}(k − n + 1)^{2}} \\ ≤ ∑_{n=0}^{[k/2]} \frac{4}{(n + 1)^{2}} + ∑_{n=[k/2]+1}^{k} \frac{4}{(k − n + 1)^{2}} ≤ 8 ∑_{j=1}^{∞} \frac{1}{j^{2}} = \frac{4π^{2}}{3} . \end{matrix}

Thus

∑_{n=0}^{∞} \frac{∥T^{n}x∥}{∥T^{n}∥} ≤ ∑_{k=0}^{∞}|α_{k}| · \frac{4π^{2}}{3} = \frac{4∥x∥π^{2}}{3} < ∞.

REMARK 1.12. Let H be a Hilbert space, let T ∈ L(H) be an non-nilpotent operator and let c < 2. Using the method of proof of Theorem 1.9 it is easy to check that the set

\{x ∈ H : ∑_{n=0}^{∞}(\frac{∥T^{n}x∥}{∥T^{n}∥})^{c} = ∞\}

is residual. For c = 2 the statement is not true; an example will be given later. For Banach space X and T ∈ L(X) it is possible to show that

\{x ∈ X : ∑_{n=0}^{∞}(\frac{∥T^{n}x∥}{∥T^{n}∥})^{c} = ∞\}

is residual for all c < 1. By the previous example, this is not true for c = 1.

II. ORBITS IN REAL BANACH SPACES

The main technical difficulty in generalizing the results of the previous section to the real case is the lack of approximate eigenvalues. Most of the results that do not use approximate eigenvalues remain unchanged in the real case.

This is true for Theorem 1.1 (i), Lemma 1.6, Theorems 1.7, 1.8, 1.10 and Remark 1.12. Because of different geometry of the real line Theorem 1.5 is modified in the following way:

THEOREM 2.1. Let T be an operator in a real Banach space X. Let (a_{n})_{n≥0} be a sequence of positive numbers satisfying ∑_{n=0}^{∞} a_{n}^{1/2} < ∞. Then there is a dense subset L ⊂ X such that, for each x ∈ L, there is k ∈ N with

∥T^{n}x∥ ≥ a_{n}∥T^{n}∥ (n ≥ k).

Other results can be proved in the real case by using the complexification of a real Banach space.

Let X be a real Banach space. Set X_{c} = {x + iy : x, y ∈ X}. Define a norm in X_{c} by

∥x + iy∥ = inf ∑_{j=1}^{n}|α_{j} + iβ_{j}| · ∥x_{j}∥ (x, y ∈ X)

where the infimum is taken over all n ∈ N, α_{j}, β_{j} ∈ R and x_{j} ∈ X such that ∑_{j=1}^{n} (α_{j} + iβ_{j})x_{j} = x + iy. With naturally defined algebraic operations, X_{c} is a complex Banach space called the complexification of X.

It is easy to see that

max{∥x∥, ∥y∥} ≤ ∥x + iy∥ ≤ ∥x∥ + ∥y∥ (x, y ∈ X).

Let T be an operator on X. The complexification of T is the operator T_{c} ∈ L(X_{c}) defined by T_{c}(x + iy) = T x + iT y (x, y ∈ X). Clearly ∥T ∥ ≤ ∥T_{c}∥ ≤ 2∥T ∥. By the spectrum of T we understand the spectrum of its complexification T_{c}. Similarly we define the spectral radius r(T) = max{|λ| : λ ∈ σ(T_{c})} = lim_{n→∞} ∥T^{n}∥^{1/n}. In the same way we use for operators in real Banach spaces the essential spectrum σ_{e}(T) = σ_{e}(T_{c}) = {λ ∈ C : T_{c} − λ is not Fredholm}, the essential spectral radius r_{e}(T) = r_{e}(T_{c}) = max{|λ| : λ ∈ σ_{e}(T)} and the upper semi-Fredholm spectrum σ_{πe}(T) = σ_{πe}(T_{c}) = {λ ∈ C : T_{c} − λ is not upper semi-Fredholm}. Equivalently, λ ∈ σ_{πe}(T) if and only if, for each subspace M ⊂ X_{c} of finite codimension, (T_{c} −λ)|M is not bounded below. Recall that ∂σ_{e}(T) ⊂ σ_{πe}(T), see [7].

The proof of Theorem 1 (ii) (the existence of vectors x with ∥T^{n}x∥ large for all n) is based on the existence of approximate eigenvalues and thus it can not be used in the real case. In [12] it was proved for real Banach space operators under an additional assumption that r(T) = 1 and T is power bounded (sup_{n} ∥T^{n}∥ < ∞). We prove a variant of this result in general.

LEMMA 2.2. Let X be a real Banach space, T ∈ L(X), let r(T) = 1, α ∈ σ_{e}(T), |α| = 1. Then there is a positive constant C (depending only on α) with the following property: for each n ∈ N and each subspace Y ⊂ X of finite codimension there exists a vector y ∈ Y of norm one with ∥T^{j}y∥ ≥ C (j = 0, 1, …, n).

PROOF. There is k ∈ N such that min{|α^{j}−1|, |α^{j+1} −1|, …, |α^{j+k} −1|} ≤ 1/6 for all j ∈ N. This is clear if α = e^{2πit} with t rational; if t is irrational then the set {α^{j} : j = 0, 1, …} is dense in the unit circle so that there is k ∈ N such that {1, α, α^{2}, …, α^{k}} is a 1/6-net in the unit circle so that the same is true also for the set {α^{j}, α^{j+1} , …, α^{j+k} }.

Set C = (6 max{1, ∥T ∥, ∥T^{2}∥, …, ∥T^{k}∥})^{−1}.

We have α ∈ ∂σ_{e}(T) ⊂ σ_{πe}(T). Let n ∈ N and let Y be a subspace of X of finite codimension. Let X_{c} = X + iX be the complexification of X and Y_{c} = Y + iY. Clearly Y_{c} is a subspace of finite codimension in X_{c}. Then there exists a vector z ∈ Y_{c} of norm one such that ∥T^{j}z − α^{j}z∥ ≤ 1/6 (j = 0, 1, …, n + k). Express z = u + iv for some u, v ∈ Y. Then either ∥u∥ ≥ 1/2 or ∥v∥ ≥ 1/2. Without loss of generality we can assume that ∥u∥ ≥ 1/2.

Let j ≤ n. Find j′ ∈ {j, j + 1, …, j + k} such that |α^{j′} − 1| ≤ 1/6. Then
∥T^{j′}u − u∥ ≤ ∥T^{j′}z − z∥ ≤ ∥T^{j′}z − α^{j′}z∥ + ∥α^{j′}z − z∥ ≤ 1/6 + 1/6 = 1/3

so that

∥T^{j′}u∥ ≥ ∥u∥ − ∥T^{j′}u − u∥ ≥ 1/2 − 1/3 = 1/6.

Further ∥T^{j′}u∥ ≤ ∥T^{j′−j}∥ · ∥T^{j}u∥ so that ∥T^{j}u∥ ≥ \frac{1}{6∥T^{j′−j}∥} ≥ C for all j ≤ n.

The next result is a weaker form of Theorem 1.1 (ii) for real Banach spaces.

THEOREM 2.3. Let X be a real Banach space, T ∈ L(X), let (a_{n})_{n≥0} be a sequence of positive numbers, a_{n} → 0. Then there exists a dense subset L ⊂ X with the property that for each x ∈ L there is a constant c > 0 with

∥T^{n}x∥ ≥ c · a_{n}r(T^{n}) (n = 0, 1, …).

PROOF. By replacing a_{n} by sup{a_{j} : j ≥ n} we can assume that a_{n} ↘ 0. We can also assume that r(T) = 1.

We distinguish two cases:

A. r_{e}(T) < 1.

Find α ∈ σ(T) with |α| = 1. Then α is an isolated eigenvalue of T_{c}. Let M ⊂ X_{c} be the corresponding spectral subspace and let P be the Riesz projection onto M. Then dimM < ∞ and (T_{c} − α)|M is a nilpotent operator.

Suppose that w be a non-zero vector in M. Let k ∈ N satisfy (T_{c} − α)^{k}w = 0 and (T_{c} − α)^{k−1}w ≠ 0. Let Q ∈ L(M) be a projection satisfying Qw = w and Q(ker(T_{c} − α)^{k−1}|M) = 0. Then Q(T_{c} − α)T_{c}^{j−1}w = 0 (j = 1, 2, …) so that QT_{c}^{j}w = αQT_{c}^{j−1}w. By induction we get QT_{c}^{j}w = α^{j}Qw = α^{j}w for all j ≥ 0. Thus

∥T_{c}^{j}w∥ ≥ ∥Q∥^{−1}|α^{j}| · ∥w∥ = ∥Q∥^{−1}∥w∥ (j ≥ 0).

Set Z = {z ∈ X_{c} : P z ≠ 0}. For z ∈ Z we have ∥T_{c}^{j}P z∥ = ∥P T_{c}^{j}z∥ ≤ ∥P ∥ · ∥T_{c}^{j}z∥ so that

∥T_{c}^{j}z∥ ≥ ∥P ∥^{−1}∥T_{c}^{j}P z∥ ≥ \frac{∥P z∥}{∥P ∥ · ∥Q∥} (j ≥ 0).

Clearly Z is an open dense subset of X_{c}. It is sufficient to show that Z ∩ X is dense in X since all vectors y ∈ Z ∩ X satisfy

inf\{∥T^{j}y∥ : j = 0, 1, …\} ≥ \frac{∥P y∥}{∥P ∥ · ∥Q∥} > 0.

Let x ∈ X and ε > 0. Let u, v ∈ X, u + iv ≠ 0 and T_{c}(u + iv) = α(u + iv). Then u + iv ∈ M so that P (u + iv) = u + iv ≠ 0. Thus either P u ≠ 0 or P v ≠ 0. Consequently at least one of the vectors x, x + εu, x + εv is in X ∩ Z and X ∩ Z is dense in X (in fact X ∩ Z is also open so that it is even a residual subset of X).

B. r_{e}(T) = 1.

Find α ∈ σ_{e}(T) with |α| = 1. Let C be the constant from the previous lemma. Find an increasing sequence m_{1} < m_{2} < ⋯ such that a_{m_{j}} < 2^{−(j+1)} .

Let x ∈ X and ε > 0. We construct inductively vectors x_{j} (j = 0, 1, …) in the following way:

Set x_{0} = x. If j ≥ 1 and x_{j−1} ∈ X has already been constructed then let E_{j} be the finite dimensional subspaces defined by E_{j} = ∨{T^{n}x_{s} : n = 0, …, m_{j}, s = 0, …, j − 1}. By Lemma 1.6 find a subspace Y_{j} ⊂ X of finite codimension satisfying ∥e + y∥ ≥ max{∥e∥/2, ∥y∥/4} (e ∈ E_{j}, y ∈ Y_{j}) and a vector u_{j} ∈ Y_{j} of norm one such that ∥T^{n}u_{j}∥ ≥ C (n = 0, …, m_{j}). Let x_{j} = x_{j−1} + \frac{εu_{j}}{2^{j}}. Clearly the sequence (x_{j}) constructed in this way is convergent; denote its limit by u = x + ∑_{j=1}^{∞} \frac{εu_{j}}{2^{j}}. We have ∥u − x∥ ≤ ∑_{j=1}^{∞} ∥\frac{εu_{j}}{2^{j}} ∥ = ε.

Let j ≥ 1, n ∈ N and m_{j} < n ≤ m_{j+1} . Then

∥T^{n}u∥ = lim_{k→∞} ∥T^{n}x_{k}∥ = lim_{k→∞}∥T^{n}(x_{j+1} + ε ∑_{s=j+2}^{k} \frac{u_{s}}{2^{s}})∥ ≥ lim_{k→∞} \frac{1}{2}∥T^{n}x_{j+1} ∥

= lim_{k→∞} \frac{1}{2}∥T^{n}(x + ε ∑_{s=1}^{j+1} \frac{u_{s}}{2^{s}})∥ ≥ \frac{1}{8} \frac{ε∥T^{n}u_{j+1} ∥}{2^{j+1}} ≥ \frac{Cε}{8 · 2^{j+1}} ≥ \frac{Cε}{8}a_{n}.

Thus ∥T^{n}u∥ ≥ \frac{Cε}{8}a_{n} for all n ≥ m_{1} so that there there is a positive constant c with ∥T^{n}u∥ ≥ c · a_{n} for all n ≥ 0.

Consequently, Corollary 1.2 remains true for operators in real Banach spaces.

COROLLARY 2.4. Let T be an operator acting on a real Banach space X. Then the set {x ∈ X : lim sup_{n} ∥T^{n}x∥^{1/n} = r(T)} is residual and the set {x ∈ X : lim inf_{n} ∥T^{n}x∥^{1/n} = r(T)} is dense in X.

COROLLARY 2.5. Let T be an operator acting on a real Banach space X, let (a_{n})_{n≥0} be a sequence of positive numbers such that sup_{n} a_{n} < 1 and lim_{n} a_{n} = 0. Then there exists a norm-one vector x ∈ X such that ∥T^{n}x∥ ≥ a_{n}r(T^{n}) for all n ≥ 0.

PROOF. Without loss of generality we can assume that r(T) = 1 and 1 > a_{0} ≥ a_{1} ≥ · · ·. Let ε be a positive number satisfying 1 > (1 + ε)a_{0}. For n ≥ 0 set

b_{n} = max{a_{i_{1}} ⋯ a_{i_{m}}(1 + ε)^{m} : m ∈ N, i_{1} + ⋯ i_{m} = n}.

Clearly b_{n+k} ≥ b_{n}a_{k}(1 + ε) (n, k ≥ 0). Further lim_{n} b_{n} = 0. Indeed, let δ > 0 and choose j such that a_{j}(1 + ε) < δ and (a_{0}(1 + ε))^{j} < δ. Let n ≥ j^{2} and b_{n} = a_{i_{1}} ⋯ a_{i_{m}}(1 + ε)^{m}. Then either i_{s} < j for all s or there is s with i_{s} ≥ j; in both cases it is easy to verify that b_{n} < δ. Thus b_{n} → 0 and by Theorem 2.3 there exists a positive constant C and a vector u ∈ X such that ∥T^{j}u∥ ≥ C · b_{j} (j = 0, 1, …). Set C′ = inf_{j} \frac{∥T^{j}∥}{b_{j}} > 0. Fix k such that \frac{∥T^{k}u∥}{b_{k}} < C′(1 + ε). Set x = \frac{T^{k}u}{∥T^{k}u∥}. Then ∥x∥ = 1 and

∥T^{n}x∥ = \frac{∥T^{k+n} u∥}{∥T^{k}u∥} ≥ \frac{C′ · b_{k+n}}{C′(1 + ε)b_{k}} ≥ a_{n}

for all n ≥ 0.

III. WEAK ORBITS

Some results concerning orbits remain true also for weak orbits. An example is the statement of Theorem 1.1 (i):

THEOREM 3.1. Let T be an operator in a (real or complex) Banach space X, let (a_{n})_{n≥0} be a sequence of positive numbers such that a_{n} → 0. Then the set of all pairs (x, x^{∗}) ∈ X × X^{∗} such that

|⟨T^{n}x, x^{∗}⟩| ≥ a_{n}∥T^{n}∥ for infinitely many n’s

is residual in X × X^{∗}.

In particular, the set {(x, x^{∗}) ∈ X × X^{∗} : lim sup_{n→∞} |⟨T^{n}x, x^{∗}⟩|^{1/n} = r(T)} is residual in X × X^{∗}.

PROOF. (i) For k ∈ N set

M_{k} = {(x, x^{∗}) ∈ X × X^{∗} : there exists n ≥ k such that |⟨T^{n}x, x^{∗}⟩| > a_{n}∥T^{n}∥}.

Clearly M_{k} is an open subset of X×X^{∗}. We prove that M_{k} is dense. Let x ∈ X, x^{∗} ∈ X^{∗} and ε > 0. Choose n ≥ k such that a_{n} < ε^{2}. There is a vector u ∈ X of norm one such that ∥T^{n}u∥ > \frac{a_{n}}{ε^{2}}∥T^{n}∥. Let u^{∗} ∈ X^{∗} satisfy ∥u^{∗}∥ = 1 and ⟨T^{n}u, u^{∗}⟩ = ∥T^{n}u∥. We have

|⟨T^{n}(x + εu), x^{∗} + εu^{∗}⟩| + |⟨T^{n}(x + εu), x^{∗} − εu^{∗}⟩|
+ |⟨T^{n}(x − εu), x^{∗} + εu^{∗}⟩| + |⟨T^{n}(x − εu), x^{∗} − εu^{∗}⟩|
≥ |⟨T^{n}(εu + x), εu^{∗} + x^{∗}⟩ + ⟨T^{n}(εu + x), εu^{∗} − x^{∗}⟩
+ ⟨T^{n}(εu − x), εu^{∗} + x^{∗}⟩ + ⟨T^{n}(εu − x), εu^{∗} − x^{∗}⟩|
= |4⟨T^{n}εu, εu^{∗}⟩| = 4ε^{2}∥T^{n}u∥ > 4a_{n}∥T^{n}∥.

Thus there is a pair

(y, y^{∗}) ∈ {(x + εu, x^{∗} + εu^{∗}), (x + εu, x^{∗} − εu^{∗}), (x − εu, x^{∗} + εu^{∗}), (x − εu, x^{∗} − εu^{∗})}

such that |⟨T^{n}y, y^{∗}⟩| > a_{n}∥T^{n}∥. Hence (y, y^{∗}) ∈ M_{k} and M_{k} is dense in X × X^{∗}.

By the Baire category theorem the intersection M = ⋂_{k=1}^{∞} M_{k} is a residual subset of X × X^{∗} and all pairs (y, y^{∗}) ∈ M satisfy |⟨T^{n}y, y^{∗}⟩| > a_{n}∥T^{n}∥ for infinitely many n’s.

In particular, for a_{n} = n^{−1} we obtain that

lim sup_{n→∞} |⟨T^{n}y, y^{∗}⟩|^{1/n} ≥ lim sup_{n→∞}(\frac{∥T^{n}∥}{n})^{1/n} = r(T)

for all pairs (y, y^{∗}) in a residual subset of X × X^{∗}.

The weak version of Theorem 1.1 (ii) is an open problem. It may be stated as follows:

PROBLEM 3.2. Let T be an operator on a complex Banach space X, let (a_{n}) be a sequence of positive numbers such that a_{n} → 0. Do there exist vectors x ∈ X and x^{∗} ∈ X^{∗} such that

|⟨T^{n}x, x^{∗}⟩| ≥ a_{n} · r(T^{n})

for all n = 0, 1, …?

For real Banach spaces this is not true, see [12].

A positive answer was shown in [12] for positive operators on Banach lattices and for non-unitary isometries on Hilbert spaces. We show a positive answer for C_{00} operators (i.e., T^{n} → 0 and T^{∗n} → 0 strongly) on Hilbert spaces. Operators of this class play an important role in the results concerning the existence of invariant subspaces for contractions with rich spectrum, see e.g. [3], [4]. It is an interesting question whether it is possible to obtain these results using the weak orbits instead of the Scott Brown technique.

LEMMA 3.3. Let T be an operator on a complex Hilbert space H such that 1 ∈ σ(T), ∥T^{n}x∥ → 0 and ∥T^{∗n}x∥ → 0 for all x ∈ H. Then, for all ε > 0, δ > 0, n ∈ N and each subspace M ⊂ H of finite codimension, there exists a vector z ∈ M of norm one with

\begin{matrix}Re⟨T^{j}z, z⟩ ≥ 1 − δ  (j ≤ n), \\ Re⟨T^{j}z, z⟩ ≥ −ε  (j > n). \end{matrix}  (2)

PROOF. By the uniform boundedness theorem we have sup_{n} ∥T^{n}∥ < ∞ so that r(T) = lim_{n→∞} ∥T^{n}∥^{1/n} ≤ 1. Since 1 ∈ σ(T) we have r(T) = 1. Further 1 is not an eigenvalue of T since T^{n} → 0 strongly. This implies in particular that H is infinite dimensional.

Since σ(T) \ σ_{e}(T) contains only isolated eigenvalues in the unbounded component of C\σ_{e}(T), we have 1 ∈ σ_{e}(T). Clearly 1 ∈ ∂σ_{e}(T) ⊂ σ_{πe}(T). This means that for all δ > 0, k ∈ N and M ⊂ H, codimM < ∞ there is a vector z ∈ M of norm one such that ∥T^{j}z−z∥ ≤ δ (0 ≤ j ≤ k). Hence Re⟨T^{j}z, z⟩ = Re⟨z, z⟩+Re⟨T^{j}z−z, z⟩ ≥ 1−δ.

Denote by A the set of all ε > 0 for which (2) is true for all δ > 0, n ∈ N and M ⊂ H, codimM < ∞.

Clearly ε ∈ A implies (ε, ∞) ⊂ A. Further A is non-empty since T is power bounded (clearly sup_{n} ∥T^{n}∥ ∈ A).

We show that ε ∈ A implies \frac{3ε}{4} ∈ A. Hence inf A = 0 and A = (0, ∞).

Suppose that ε ∈ A. Let n ∈ N, δ > 0 and M ⊂ H, codimM < ∞. We may assume that δ < 1. By the assumption there is z ∈ M of norm one such that

\begin{matrix}Re⟨T^{j}z, z⟩ ≥ 1 − δ  (j ≤ n), \\ Re⟨T^{j}z, z⟩ ≥ −ε  (j > n). \end{matrix}

Since lim_{j→∞} ∥T^{j}z∥ = 0 and lim_{j→∞} ∥T^{∗j}z∥ = 0 there exists m ≥ n such that ∥T^{j}z∥ ≤ ε/6 and ∥T^{∗j}z∥ ≤ ε/6 for all j ≥ m. Consider the subspace Y = (∨{z, T z, …, T^{m}z})^{⊥}. Clearly codimY < ∞. Let Y′ = M ∩ ⋂_{j=0}^{m} T^{−j}Y. Since codimT^{−j}Y < ∞ for all j, we have codimY′ < ∞. By the assumption there is u ∈ Y′ of norm one such that

\begin{matrix}Re⟨T^{j}u, u⟩ ≥ 1 − δ  (j ≤ m), \\ Re⟨T^{j}u, u⟩ ≥ −ε  (j > m). \end{matrix}

Since T^{j}u ∈ Y for j = 0, 1, …, m, we have T^{j}u ⊥ T^{i}z for all i, j ≤ m. In particular, u ⊥ z and ∥z + u∥ = \sqrt{2}. Set v = \frac{z+u}{\sqrt{2}} . Then v ∈ M and ∥v∥ = 1.

For 0 ≤ j ≤ n we have

Re⟨T^{j}v, v⟩ = \frac{1}{2}(Re⟨T^{j}z, z⟩ + Re⟨T^{j}u, u⟩) ≥ \frac{1}{2}(1 − δ + 1 − δ) = 1 − δ.

For n < j ≤ m we have

Re⟨T^{j}v, v⟩ = \frac{1}{2}(Re⟨T^{j}z, z⟩ + Re⟨T^{j}u, u⟩) ≥ \frac{1}{2}(−ε + 1 − δ) ≥ \frac{−ε}{2} > \frac{−3ε}{4}.

Finally, for j > m we have

Re⟨T^{j}v, v⟩ = \frac{1}{2}(Re⟨T^{j}z, z⟩ + Re⟨T^{j}u, u⟩ + Re⟨T^{j}z, u⟩ + Re⟨T^{j}u, z⟩)

≥ \frac{1}{2}(−∥T^{j}z∥ − ε − ∥T^{j}z∥ − ∥T^{∗j}z∥) ≥ \frac{1}{2}(−ε − \frac{3}{6}ε) = −\frac{3ε}{4}.

Since δ, n and M were arbitrary, we have \frac{3ε}{4} ∈ A. Hence (2) is true for all ε > 0.

COROLLARY 3.4. Let T be an operator acting on a complex Hilbert space H such that 1 ∈ σ(T), T^{n}x → 0 and T^{∗n}x → 0 for all x ∈ H. Let (a_{n})_{n≥0} be a sequence of positive numbers such that 1 > a_{0} ≥ a_{1} ≥ ⋯ and a_{n} → 0. Then there exists x ∈ X of norm one such that

Re⟨T^{n}x, x⟩ ≥ a_{n} (n = 0, 1, …).

PROOF. Let \sqrt{a_{0}} < d < 1 and ε = \sqrt{1 − d^{2}}.

Find m_{0} ∈ N such that a_{m_{0}} < \frac{ε^{2}}{4} . Find a vector u_{0} ∈ H of norm one such that Re⟨T^{n}u_{0}, u_{0}⟩ ≥ \frac{a_{0}}{d^{2}} for all n ≤ m_{0} and Re⟨T^{n}u_{0}, u_{0}⟩ ≥ \frac{−ε^{2}}{4} for all n > m_{0}. Set x_{0} = du_{0}.

We construct inductively sequences (u_{k}), (x_{k}) ⊂ H and an increasing sequence (m_{k}) such that the limit x = lim_{k→∞} x_{k} will satisfy the required conditions.

Suppose that k ≥ 1, m_{k−1} ∈ N and vectors u_{0}, …, u_{k−1}, x_{0}, …, x_{k−1} ∈ H have already been constructed. Choose m_{k} > m_{k−1} such that

a_{m_{k}} ≤ \frac{ε^{2}}{2^{k+3}} ,

∥T^{n}x_{k−1}∥ ≤ \frac{ε^{2}}{2^{k+3}} and

∥T^{∗n}x_{k−1}∥ ≤ \frac{ε^{2}}{2^{k+3}}

for all n > m_{k}. Let M_{k} = ⋂_{j=0}^{m_{k}} T^{−j}(∨{T^{n}u_{i} : 0 ≤ i ≤ k − 1, 0 ≤ n ≤ m_{k}}^{⊥}). Clearly codimM_{k} < ∞. Find u_{k} ∈ M_{k} of norm one such that Re⟨T^{n}u_{k}, u_{k}⟩ ≥ \frac{1}{2} (n ≤ m_{k}) and Re⟨T^{n}u_{k}, u_{k}⟩ ≥ −\frac{1}{8} (n > m_{k}). Then T^{n}u_{k} ⊥ u_{i} and u_{k} ⊥ T^{n}u_{i} (n ≤ m_{k}, i < k). In particular, u_{k} ⊥ u_{i}.

Set x_{k} = x_{k−1} + \frac{εu_{k}}{2^{k/2}} .

Clearly the sequence (x_{k}) constructed in this way is convergent. Denote by x its limit, x = du_{0} + ∑_{i=1}^{∞} \frac{εu_{i}}{2^{i/2}} . Then ∥x∥^{2} = d^{2} + ∑_{i=1}^{∞} \frac{ε^{2}}{2^{i}} = d^{2} + ε^{2} = 1. For n ≤ m_{0} we have

Re⟨T^{n}x, x⟩ = Re⟨T^{n}du_{0}, du_{0}⟩ + ∑_{i=1}^{∞}Re⟨\frac{εT^{n}u_{i}}{2^{i/2}}, \frac{εu_{i}}{2^{i/2}} ⟩ ≥ d^{2}Re⟨T^{n}u_{0}, u_{0}⟩ ≥ a_{0} ≥ a_{n}.

For m_{0} < n ≤ m_{1} we have

\begin{matrix}Re⟨T^{n}x, x⟩ = Re⟨T^{n}x_{0}, x_{0}⟩ + ∑_{i=1}^{∞}Re⟨\frac{εT^{n}u_{i}}{2^{i/2}}, \frac{εu_{i}}{2^{i/2}} ⟩ \\ ≥ \frac{−d^{2}ε^{2}}{4} + \frac{1}{2} ∑_{i=1}^{∞} \frac{ε^{2}}{2^{i}} ≥ \frac{ε^{2}}{4} ≥ a_{m_{0}} ≥ a_{n}. \end{matrix}

Let k ≥ 1 and m_{k} < n ≤ m_{k+1} . Then

Re⟨T^{n}x, x⟩ = Re⟨T^{n}x_{k−1}, x⟩ + Re⟨T^{n}(x − x_{k−1}), x_{k−1}⟩ + ∑_{i,j=k}^{∞} Re⟨\frac{εT^{n}u_{i}}{2^{i/2}}, \frac{εu_{j}}{2^{j/2}} ⟩

≥ −∥T^{n}x_{k−1}∥ − ∥T^{∗n}x_{k−1}∥ + ∑_{i=k}^{∞} \frac{ε^{2}}{2^{i}} Re⟨T^{n}u_{i}, u_{i}⟩

≥ −\frac{2ε^{2}}{2^{k+3}} + \frac{ε^{2}}{2^{k}}Re⟨T^{n}u_{k}, u_{k}⟩ + ∑_{i=k+1}^{∞} \frac{ε^{2}}{2^{i}} Re⟨T^{n}u_{i}, u_{i}⟩

≥ −\frac{3ε^{2}}{2^{k+3}} + \frac{1}{2} ∑_{i=k+1}^{∞} \frac{ε^{2}}{2^{i}} = \frac{ε^{2}}{2^{k+3}} ≥ a_{n}.

The next result, which is something between the statement of Theorem 3.1 and Problem 3.2, is a generalization of [12]. We need the following lemma:

LEMMA 3.5. Let X be a real or complex Banach space, T ∈ L(X), r_{e}(T) = 1, n_{0} ∈ N, ε > 0, m ∈ N. Then there are numbers n_{0} < n_{1} < ⋯ < n_{m} such that, in each subspace M ⊂ X of finite codimension, there exists a vector x ∈ M of norm one with ∥T^{n_{j}}x − x∥ ≤ ε (j = 1, 2, …, m).

PROOF. Suppose first that X is a complex Banach space and T ∈ L(X). Let λ ∈ σ_{e}(T), |λ| = 1. Then λ ∈ σ_{πe}(T). Find s ∈ N such that s > n_{0} and |λ^{s}−1| ≤ ε/2m. Then

|λ^{sj} − 1| = |λ^{s} − 1| · |λ^{s(j−1)} + λ^{s(j−2)} + ⋯ + 1| ≤ ε/2

for j = 1, 2, …, m. Let x ∈ M be a vector of norm one satisfying ∥T^{sj}x − λ^{sj}x∥ ≤ ε/2 (j = 1, …, m) so that ∥T^{sj}x − x∥ ≤ ∥T^{sj}x − λ^{sj}x∥ + ∥λ^{sj}x − x∥ ≤ ε.

This finishes the proof in the complex case.

In the real case consider the complexification X_{c} of X and T_{c} ∈ L(X_{c}). As in the complex case find a vector x ∈ M_{c} of norm one and s > n_{0} such that ∥T_{c}^{sj}x − x∥ ≤ ε/2 (j = 1, …, m). Let x = u + iv where u, v ∈ X. Then ∥u∥ + ∥v∥ ≥ ∥x∥ = 1 so that max{∥u∥, ∥v∥} ≥ 1/2; without loss of generality we can assume that ∥u∥ ≥ 1/2. Set y = \frac{u}{∥u∥}. Then y ∈ M, ∥y∥ = 1 and ∥T^{sj}y − y∥ = \frac{1}{∥u∥}∥T^{sj}u − u∥ ≤ 2∥T^{sj}x − x∥ ≤ ε (j = 1, …, m).

THEOREM 3.6. Let X be a real or complex Banach space, T ∈ L(X) and let (a_{j})_{j≥1} be a sequence of positive numbers with a_{j} → 0. Then there exist x ∈ X, x^{∗} ∈ X^{∗} and an increasing sequence (n_{j}) of positive integers such that

Re⟨T^{n_{j}}x, x^{∗}⟩ ≥ a_{j} · r(T)^{n_{j}}

for all j ≥ 1.

PROOF. Without loss of generality we can assume that r(T) = 1 and that 1/16 ≥ a_{0} ≥ a_{1} ≥ · · ·. We distinguish two cases:

A. Suppose that T^{n} does not tend to 0 weakly, so that there are x ∈ X, x^{∗} ∈ X^{∗} and c > 0 such that |⟨T^{n}x, x^{∗}⟩| ≥ c for infinitely many n’s.

In the real case |⟨T^{n}x, x^{∗}⟩| ≤ max{Re⟨T^{n}x, x^{∗}⟩, Re⟨T^{n}x, −x^{∗}⟩}; in the complex case

|⟨T^{n}x, x^{∗}⟩| ≤ \sqrt{2} · max\{Re⟨T^{n}x, x^{∗}⟩, Re⟨T^{n}x, ix^{∗}⟩, Re⟨T^{n}x, −x^{∗}⟩, Re⟨T^{n}x, −ix^{∗}⟩\}.

In both cases there are c_{1} > 0 and x_{1}^{∗} ∈ X^{∗} such that Re⟨T^{n}x, x_{1}^{∗}⟩ ≥ c_{1} for infinitely many powers n. By considering a suitable multiple of x_{1}^{∗} we get the statement of the Theorem.

B. Suppose that T^{n} → 0 weakly.

Using the uniform boundedness theorem twice it is easy to show that sup{∥T^{n}∥ : n = 0, 1, …} = M < ∞. Further T^{n} → 0 weakly implies that there are no eigenvalues of modulus 1, i.e., r_{e}(T) = 1. Let s = 8M. Find numbers m_{k} ∈ N such that 0 = m_{0} < m_{1} < m_{2} < ⋯ and

a_{j} ≤ \frac{1}{16s^{2k}} (k ≥ 0, j > m_{k}).

We construct inductively sequences (u_{k})_{k≥0} ⊂ X, (u_{k}^{∗})_{k≥0} ⊂ X^{∗} and an increasing sequence of positive integers (n_{j}) in the following way:

Set u_{0} = 0 and u_{0}^{∗} = 0. Let k ≥ 0 and suppose that u_{0}, …, u_{k} ∈ X, u_{0}^{∗}, …, u_{k}^{∗} ∈ X^{∗} and numbers n_{1}, …, n_{m_{k}} have already been constructed. Write x_{k} = ∑_{i=1}^{k} \frac{u_{i}}{s^{i−1}} and x_{k}^{∗} = ∑_{i=1}^{k} \frac{u_{i}^{∗}}{s^{i−1}} . Find q_{k} such that |⟨T^{j}x_{k}, x_{k}^{∗}⟩| ≤ \frac{1}{16s^{2k}} (j ≥ q_{k}). Find numbers n_{m_{k}+1} , …, n_{m_{k+1}} satisfying the properties of Lemma 3.5 such that

max{n_{m_{k}}, q_{k}} < n_{m_{k}+1} < n_{m_{k}+2} < ⋯ < n_{m_{k+1}} .

Let E_{k} = ∨{T^{n_{j}}u_{i} : 0 ≤ j ≤ m_{k+1} , 0 ≤ i ≤ k}. By Lemma 1.6 there exists a subspace Y_{k} of finite codimension such that

∥e + y∥ ≥ max{∥e∥/2, ∥y∥/4} (e ∈ E_{k}, y ∈ Y_{k}).

Let u_{k+1} ∈ Y_{k} ∩ (∨{T^{∗n_{j}}u_{i}^{∗} : 0 ≤ j ≤ m_{k+1} , 0 ≤ i ≤ k})^{⊥} be a vector of norm one such that

∥T^{n_{j}}u_{k+1} − u_{k+1} ∥ < 1/16 (m_{k} < j ≤ m_{k+1} ).

Find u_{k+1}^{∗} ∈ E_{k}^{⊥} such that ∥u_{k+1}^{∗} ∥ = 1 and

⟨u_{k+1} , u_{k+1}^{∗} ⟩ = dist{u_{k+1} , E_{k}} ≥ 1/4.

Note that ⟨T^{n_{j}}u_{i}, u_{k+1}^{∗} ⟩ = 0 and ⟨T^{n_{j}}u_{k+1} , u_{i}^{∗}⟩ = 0 for all i ≤ k and j ≤ m_{k+1} . Continue the inductive construction and set x = ∑_{i=1}^{∞} \frac{u_{i}}{s^{i−1}} and x^{∗} = ∑_{i=1}^{∞} \frac{u_{i}^{∗}}{s^{i−1}} . To show that x, x^{∗} and the sequence (n_{j}) satisfy the required properties, let

k ≥ 0 and m_{k} < j ≤ m_{k+1} . We have

Re⟨T^{n_{j}}x, x^{∗}⟩ = Re⟨T^{n_{j}}(x_{k} + ∑_{i=k}^{∞} \frac{u_{i+1}}{s^{i}} ), x_{k}^{∗} + ∑_{i=k}^{∞} \frac{u_{i+1}^{∗}}{s^{i}} ⟩

= Re⟨T^{n_{j}}x_{k}, x_{k}^{∗}⟩ + \frac{1}{s^{2k}} Re⟨T^{n_{j}}u_{k+1} , u_{k+1}^{∗} ⟩ + ∑_{i=k+1}^{∞} \frac{1}{s^{2i}} Re⟨T^{n_{j}}u_{i+1} , u_{i+1}^{∗} ⟩

≥ −\frac{1}{16s^{2k}} + \frac{1}{s^{2k}} (Re⟨u_{k+1} , u_{k+1}^{∗} ⟩ − Re⟨u_{k+1} − T^{n_{j}}u_{k+1} , u_{k+1}^{∗} ⟩) − ∑_{i=k+1}^{∞} \frac{M}{s^{2i}}

≥ \frac{1}{s^{2k}} (−\frac{1}{16} + \frac{1}{4} − \frac{1}{16} − \frac{2M}{s^{2}} ) ≥ \frac{1}{16s^{2k}} ≥ a_{j}.

COROLLARY 3.7. (cf. [16]) Let X be a real or complex Banach space, let T ∈ L(X), 1 ≤ p < ∞, r(T) ≠ 0. Then the set

\{(x, x^{∗}) ∈ X × X^{∗} : ∑_{n=0}^{∞}(\frac{|⟨T^{n}x, x^{∗}⟩|}{r(T^{n})})^{p} = ∞\}

is residual in X × X^{∗}. Consequently (see [17]), the set

\{x ∈ X : ∑_{n=0}^{∞}(\frac{∥T^{n}x∥}{r(T^{n})})^{p} = ∞\}

is residual in X.

PROOF. For k ∈ N set

M_{k} = \{(x, x^{∗}) ∈ X × X^{∗} : ∑_{n=0}^{∞}(\frac{|⟨T^{n}x, x^{∗}⟩|}{r(T^{n})})^{p} > k\}.

Clearly M_{k} is open in X × X^{∗}. To show that M_{k} is dense, let x ∈ X, x^{∗} ∈ X^{∗} and ε > 0. By the previous lemma for a suitable sequence (a_{n}) there are u ∈ X and u^{∗} ∈ X^{∗} such that ∑_{n=0}^{∞} (\frac{|⟨T^{n}u,u^{∗}⟩|}{r(T^{n})})^{p} = ∞. Clearly we can assume that ∥u∥ < ε and ∥u^{∗}∥ < ε. Since

(\frac{|⟨T^{n}u, u^{∗}⟩|}{r(T^{n})})^{p} = |\frac{⟨T^{n}(u + x), u^{∗} + x^{∗}⟩}{4r(T^{n})} + \frac{⟨T^{n}(u + x), u^{∗} − x^{∗}⟩}{4r(T^{n})}

+ \frac{⟨T^{n}(u − x), u^{∗} + x^{∗}⟩}{4r(T^{n})} + \frac{⟨T^{n}(u − x), u^{∗} − x^{∗}⟩}{4r(T^{n})}|^{p}

≤ max\{\frac{|⟨T^{n}(u + x), u^{∗} + x^{∗}⟩|}{r(T^{n})} , \frac{|⟨T^{n}(u + x), u^{∗} − x^{∗}⟩|}{r(T^{n})} ,

\frac{|⟨T^{n}(u − x), u^{∗} + x^{∗}⟩|}{r(T^{n})} , \frac{|⟨T^{n}(u − x), u^{∗} − x^{∗}⟩|}{r(T^{n})}\}^{p},

we have that ∑_{n=0}^{∞} (\frac{|⟨T^{n}y,y^{∗}⟩|}{r(T^{n})})^{p} = ∞ for at least one pair

(y, y^{∗}) ∈ {(x + u, x^{∗} + u^{∗}), (x + u, x^{∗} − u^{∗}), (x − u, x^{∗} + u^{∗}), (x − u, x^{∗} − u^{∗})}.

Thus M_{k} is dense in X × X^{∗} and

M = ⋂_{k} M_{k} = \{(x, x^{∗}) ∈ X × X^{∗} : ∑_{n=0}^{∞}(\frac{|⟨T^{n}x, x^{∗}⟩|}{r(T)^{n}})^{p} = ∞\}

is residual in X × X^{∗}.

The second statement can be proved similarly (or we can use the fact that the canonical projection X × X^{∗} → X maps residual subsets of X × X^{∗} onto residual subsets of X).

THEOREM 3.8. Let X, Y be complex Banach spaces, T_{n} ∈ L(X, Y) (n = 1, 2, …). Let (a_{n}) be a sequence of positive numbers satisfying ∑_{n=1}^{∞} a_{n}^{1/3} < 1/4. Let B ⊂ X, B^{∗} ⊂ Y^{∗} be balls of radii equal to 1/4. Then there exist x ∈ B and y^{∗} ∈ B^{∗} such that

|⟨T_{n}x, y^{∗}⟩| ≥ a_{n}∥T_{n}∥

for all n = 1, 2, . . ..

PROOF. By Lemma 1.4 there exists x ∈ B such that ∥T_{n}x∥ ≥ a_{n}^{1/2} ∥T_{n}∥ for all

n.

Consider operators S_{n} : Y^{∗} → C defined by S_{n}y^{∗} = ⟨T_{n}x, y^{∗}⟩ (y^{∗} ∈ Y^{∗}).

Clearly ∥S_{n}∥ = ∥T_{n}x∥ ≥ a_{n}^{1/2} ∥T_{n}∥ for all n. Using Lemma 1.4 again there exists y^{∗} ∈ B^{∗} such that

|⟨T_{n}x, y^{∗}⟩| = ∥S_{n}y^{∗}∥ ≥ a_{n}^{1/2} ∥S_{n}∥ ≥ a_{n}∥T_{n}∥

for all n = 1, 2, . . ..

COROLLARY 3.9. Let X be a complex Banach space, T ∈ L(X), and let (a_{n})_{n≥0} be a sequence of positive numbers satisfying ∑_{n=0}^{∞} a_{n}^{1/3} < ∞. Then there is a dense subset L ⊂ X × X^{∗} with the following property: for all pairs (x, x^{∗}) ∈ L there is k ∈ N such that

|⟨T^{n}x, x^{∗}⟩| ≥ a_{n}∥T^{n}∥ (n ≥ k).

If X is a real Banach space and T ∈ L(X) then the condition ∑ a_{n}^{1/3} < ∞ must be replaced by ∑ a_{n}^{1/4} < ∞ (cf. Theorem 2.1).

COROLLARY 3.10. Let X be a real or complex Banach space, let T ∈ L(X). Then the set {(x, x^{∗}) ∈ X ×X^{∗} : lim inf_{n→∞} |⟨T^{n}x, x^{∗}⟩|^{1/n} = r(T)} is dense in X ×X^{∗}.

EXAMPLE 3.11. There is a Hilbert space H and a non-nilpotent operator T ∈ L(H) such that

∑_{n=0}^{∞} \frac{|⟨T^{n}x, y⟩|}{∥T^{n}∥} < ∞

for all x, y ∈ H.

PROOF. For k ≥ 1 let H_{k} be the (k + 1)-dimensional Hilbert space with an orthonormal basis e_{k0}, e_{k1}, …, e_{kk}. Let S_{k} ∈ L(H_{k}) be the shift operator defined by S_{k}e_{k0} = 0, S_{k}e_{kj} = e_{k,j−1} (j ≥ 1). Set H = ⨁_{k=1}^{∞} H_{k} and T = ⨁_{k=1}^{∞} \frac{1}{2^{k}} S_{k}. Then ∥T^{n}∥ = 2^{−n^{2}} for all n.

Let k ∈ N and x_{k}, y_{k} ∈ H_{k}, x_{k} = ∑_{j=0}^{k} α_{j}e_{kj}, y_{k} = ∑_{j=0}^{k} β_{j}e_{kj}. Then

\begin{matrix}∑_{n=0}^{∞} \frac{|⟨T^{n}x_{k}, y_{k}⟩|}{∥T^{n}∥} = ∑_{n=0}^{k} \frac{2^{n^{2}}}{2^{kn}} |∑_{j=0}^{k−n}α_{j+n} β_{j}| ≤ ∑_{n=0}^{k} \frac{1}{2^{n(k−n)}} (∑_{j=0}^{k−n}|α_{j+n} |^{2})^{1/2}(∑_{j=0}^{k−n}|β_{j}|^{2})^{1/2} \\ ≤ ∥x_{k}∥ · ∥y_{k}∥ ∑_{n=0}^{k} \frac{1}{2^{n(k−n)}} ≤ ∥x_{k}∥ · ∥y_{k}∥(1 + ∑_{n=0}^{k−1} \frac{1}{2^{n}}) ≤ 3∥x_{k}∥ · ∥y_{k}∥. \end{matrix}

Let x, y ∈ H, x = ∑_{k=1}^{∞} x_{k}, y = ∑_{k=1}^{∞} y_{k} where x_{k}, y_{k} ∈ H_{k}. Then

\begin{matrix}∑_{n=0}^{∞} \frac{|⟨T^{n}x, y⟩|}{∥T^{n}∥} ≤ ∑_{n=0}^{∞} ∑_{k=1}^{∞} \frac{|⟨T^{n}x_{k}, y_{k}⟩|}{∥T^{n}∥} \\ ≤ ∑_{k=1}^{∞}3∥x_{k}∥ · ∥y_{k}∥ ≤ 3(∑_{k=1}^{∞}∥x_{k}∥^{2})^{1/2} · (∑_{k=1}^{∞}∥y_{k}∥^{2})^{1/2} = 3∥x∥ · ∥y∥ < ∞. \end{matrix}

REMARK 3.12. In the previous example we have ∑_{n=0}^{∞} (\frac{∥T^{n}x∥}{∥T^{n}∥})^{2} < ∞ for all x ∈ H.

Indeed, using the notation of the previous example, for x_{k} ∈ H_{k} we have

∑_{n=0}^{∞}(\frac{∥T^{n}x_{k}∥}{∥T^{n}∥})^{2} = ∑_{n=0}^{k}(\frac{2^{n^{2}}}{2^{nk}}(∑_{j=0}^{k−n}|α_{j+n} |^{2})^{1/2})^{2} ≤ ∑_{n=0}^{k} \frac{1}{2^{2n(k−n)}} ∥x_{k}∥^{2} ≤ 3∥x_{k}∥^{2}

and for x = ⨁_{k=1}^{∞} x_{k} ∈ H we have

∑_{n=0}^{∞}(\frac{∥T^{n}x∥}{∥T^{n}∥})^{2} = ∑_{k=1}^{∞} ∑_{n=0}^{∞}(\frac{∥T^{n}x_{k}∥}{∥T^{n}∥})^{2} ≤ ∑_{k=1}^{∞}3∥x_{k}∥^{2} = 3∥x∥^{2} < ∞.

REMARK 3.13. Using the method of Lemma 1.9 it is possible to show that for each Hilbert space operator T ∈ L(H) and all c < 1 there are vectors x, y ∈ H with

∑_{n=0}^{∞}(\frac{|⟨T^{n}x, y⟩|}{∥T^{n}∥})^{c} = ∞

(one can even get y = x).

For Banach space operators this is true for all c < 1/2.

IV. LOCAL CAPACITY

In this section we replace the powers of an operator by the set of all polynomials. All Banach spaces are supposed to be complex.

Denote by P_{n}^{1} the set of all monic polynomials of degree n (by monic we mean that the leading coefficient is equal to 1). For T ∈ L(X) we write cap_{n} T = inf{∥p(T)∥ : p ∈ P_{n}^{1}} and the capacity of T is defined by

capT = lim_{n→∞}(cap_{n} T)^{1/n} = inf_{n}(cap_{n} T)^{1/n},

see [6]. Clearly the capacity is related to the spectral radius, cap_{n} T ≤ ∥T^{n}∥ for all n and capT ≤ r(T).

For T ∈ L(X), x ∈ X and n ∈ N write cap_{n}(T, x) = inf{∥p(T)x∥ : p ∈ P_{n}^{1}}. In general the limit lim_{n→∞}(cap_{n}(T, x))^{1/n} does not exist; the local capacity cap(T, x) is defined by

cap(T, x) = lim sup_{n→∞}(cap_{n}(T, x))^{1/n}.

By [6], capT = capσ(T); recall that the classical capacity of a nonempty compact subset K ⊂ C is defined by capK = lim_{n→∞}(cap_{n} K)^{1/n}, where cap_{n} K = inf{∥p∥_{K} : p ∈ P_{n}^{1}} and ∥p∥_{K} = sup{|p(z)| : z ∈ K}.

Since σ(T) \ σ_{e}(T) consists of some bounded components of C \ σ_{e}(T) and at most countably many isolated points, it is easy to see that capσ_{e}(T) = capσ(T).

A nonempty compact subset K ⊂ C is called algebraic if p(K) = {0} for some nonzero polynomial p.

The basic results for local capacities are similar to those for the spectral radius.

THEOREM 4.1. Let X be a complex Banach space, T ∈ L(X) and let ε > 0. Then:

(i) the set of all x ∈ X with the property that

cap_{n}(T, x) ≥ (n + 1)^{−(2+ε)} (capT)^{n} for infinitely many n’s

is residual.

In particular, the set {x ∈ X : cap(T, x) = lim sup_{n→∞}(cap_{n}(T, x))^{1/n} = capT } is residual.

(ii) there is a dense subset R ⊂ X with the property that, for each x ∈ R, there exists k ∈ N with

cap_{n}(T, x) ≥ (n + 1)^{−(2+ε)} (capT)^{n} (n ≥ k).

In particular, the set {x ∈ X : lim_{n→∞}(cap_{n}(T, x))^{1/n} = capT } is dense.

PROOF. (ii) is a reformulation of [10]. To show (i), we need the following lemma:

LEMMA 4.2. Let T ∈ L(X), x ∈ X, η > 0 and k ∈ N. Suppose that the set K = σ_{e}(T) is not algebraic. Then there exists n ≥ k such that the set

L ={y ∈ X : ∥y − x∥ < η, ∥p(T)y∥ > (n + 1)^{−(2+η)} ∥p∥_{K} for all p with degp ≤ n}

is a non-empty open subset of X.

PROOF. Choose n ≥ k such that (n + 1)^{−η} < \frac{η(1−η)}{4} .

For a polynomial p(z) = ∑_{i=0}^{n} α_{i}z^{i} set |p| = ∑_{i=0}^{n} |α_{i}|. Since K is not an algebraic set, the norms |·| and ∥·∥_{K} are equivalent on the set of polynomials of degree ≤ n so that there exists a constant c > 0 such that |p| ≤ c · ∥p∥_{K} for all polynomials p with degp ≤ n.

We prove that the set L is open. Let y ∈ L. By a compactness argument, there exists δ > 0 such that ∥p(T)y∥ > δ + (n + 1)^{−(2+η)} for all polynomials p with degp ≤ n and ∥p∥_{K} = 1.

Let y′ ∈ X, ∥y′ − x∥ < η and ∥y′ − y∥ < \frac{δ}{c·max \{1,∥T ∥^{n}\}}. Let p(z) = ∑_{i=0}^{n} α_{i}z^{i} be a polynomial with ∥p∥_{K} = 1. Then

∥p(T)(y′ − y)∥ ≤ ∑_{i=0}^{n}|α_{i}| · ∥T ∥^{i}∥y′ − y∥ ≤ |p| max{1, ∥T ∥, …, ∥T ∥^{n}} · ∥y − y′∥ < δ

and

∥p(T)y′∥ ≥ ∥p(T)y∥ − ∥p(T)(y′ − y)∥ > δ + \frac{1}{(n + 1)^{2+η}} − δ = \frac{1}{(n + 1)^{2+η}} .

Thus y′ ∈ L.

It is sufficient to show that L is non-empty. Set E = ⋁{x, T x, …, T^{n}x}. By [10], Lemma 3 there exists u ∈ X of norm one such that

∥p(T)(e + u)∥ ≥ \frac{1 − η}{2(n + 1)^{2}} r_{e}(p(T))

for all e ∈ E and all polynomials p, degp ≤ n. Note that

r_{e}(p(T)) = max{|μ| : μ ∈ σ_{e}(p(T))} = max{|p(λ)| : λ ∈ σ_{e}(T)} = ∥p∥_{K}.

Set y = x + \frac{ηu}{2}. Then ∥y − x∥ = η/2 < η and, for all polynomials p of degree ≤ n,

∥p(T)y∥ = \frac{η}{2} ∥p(T)(\frac{2x}{η} + u)∥ ≥ \frac{η}{2} · \frac{1 − η}{2(n + 1)^{2}} ∥p∥_{K} > \frac{1}{(n + 1)^{2+η}} ∥p∥_{K}.

Hence y ∈ L.

Proof of Theorem 4.1 (i). If σ_{e}(T) is an algebraic set then capT = capσ_{e}(T) = 0, so that the statement is trivial.

Suppose that the set K = σ_{e}(T) is not algebraic. Fix ε > 0. For each k ∈ N let

M_{k} = {y ∈ X : there exists n ≥ k such that cap_{n}(T, y) > (n + 1)^{−(2+ε)} cap_{n} K}.

For each x ∈ X and η ∈ (0, ε) let n = n(x, η, k) ≥ k and L = L(x, η, k) ⊂ X be as constructed in Lemma 4.2. For y ∈ L(x, η, k) we have

cap_{n}(T, y) = inf{∥p(T)y∥ : p ∈ P_{n}^{1}} ≥ inf{(n + 1)^{−(2+η)} ∥p∥_{K} : p ∈ P_{n}^{1}}
≥(n + 1)^{−(2+η)} cap_{n} K > (n + 1)^{−(2+ε)} cap_{n} K.

Thus ⋃_{x,η} L(x, η, k) ⊂ M_{k} and ⋃_{x,η} L(x, η, k) is an open dense subset of X. Hence M_{k} is residual for all k and so is the intersection ⋂_{k} M_{k}.

For k ∈ N and y ∈ M there is n ≥ k with

cap_{n}(T, y) > (n + 1)^{−(2+ε)} cap_{n} K.

In particular, for y ∈ ⋂_{k=1}^{∞} M_{k} we have

cap(T, y) = lim sup_{n→∞}(cap_{n}(T, y))^{1/n} ≥ lim sup_{n→∞}(cap_{n} K)^{1/n} = capK = capσ_{e}(T) = capT.

The opposite inequality cap(T, y) ≤ capT is always true.

Example 1.3 shows that in general we can not expect to have

lim inf_{n→∞}(cap(T, y))^{1/n} = capT

for all y in a residual set. Thus (ii) cannot be improved.

REMARK 4.3. The statement of Theorem 4.1 can be generalized to commuting n-tuples of operators. The analogue of the second statement was proved in [11]; the analogue of part (i) can be proved as above.

References

[1] A. Atzmon, Power regular operators, Trans. Amer. Math. Soc. 347 (1995), 3101–3109.

[2] B. Beauzamy, Introduction to operator theory and invariant subspaces, North-Holland Mathematical Library Vol. 42, North-Holland, Amsterdam, 1988.

[3] S. Brown, B. Chevreau, C. Pearcy, Contractions with rich spectrum have invariant subspaces, J. Operator Theory 1 (1979), 123–136.

[4] S. Brown, B. Chevreau, C. Pearcy, On the structure of contraction operators II, J. Funct. Anal. 76 (1988), 30–55.

[5] J. Daneš, On local spectral radius, Čas. Pěst. Mat. 112 (1987), 177-188.

[6] P.R. Halmos, Capacity in Banach algebras, Indiana Univ. Math. J. 20 (1971), 855–863.

[7] R. Harte, A. Wickstead, Upper and lower Fredholm spectra II, Math. Z. 154 (1977), 253–256.

[8] A. Lebow, M. Schechter, Semigroups of operators and measures of noncompactness, J. Funct. Anal. 7 (1971), 1–26.

[9] V. Müller, Local spectral radius formula for operators in Banach spaces, Czechoslovak Math. J. 38 (1988), 726–729.

[10] V. Müller, Local behaviour of the polynomial calculus of operators, J. Reine Angew. Math. 430 (1992), 61–68.

[11] V. Müller and A. So ltysiak, On local joint capacities of operators, Czechoslovak Math. J. 43 (1993), 743–751.

[12] J.M.A.M. van Neerven, On the orbits of an operator with spectral radius one, Czechoslovak Math. J. 45 (1995), 495–502.

[13] J.M.A.M. van Neerven, Exponential stability of operators and operator semigroups, J. Funct. Anal. 130 (1995), 293–309.

[14] J.M.A.N. van Neerven, The asymptotic behaviour of semigroups of operators, Operator Theory: Advances and Applications, vol. 88, Birkhäuser, Basel, 1996.

[15] P. Vrbová, On local spectral properties of operators in Banach spaces, Czechoslovak Math. J. 25 (1973), 483–492.

[16] G. Weiss, Weakly ℓ^{p}-stable linear operators are power stable, Int. System Sci. 20 (1989), 2323–2328.

[17] J. Zabczyk, Remarks on the control of discrete-time distributed parameter systems, SIAM J. Control 12 (1974), 721–735.

Institute of Mathematics AV ČR Žitná 25, 115 67 Praha 1 Czech Republic e-mail: muller@math.cas.cz