SUBSCALAR OPERATORS AND GROWTH OF RESOLVENT

CĂTĂLIN BADEA AND VLADIMÍR MÜLLER

Abstract. We construct a Banach space bounded linear operator T which is not E(𝕋)-subscalar but ∥(T −z)^{−1}∥ ≤ (|z|−1)^{−1} for |z| > 1 and m(T − z) ≥ const · (1 − |z|)^{3} for |z| < 1 (here m denotes the minimum modulus). This gives a negative answer to a variant of a problem of K.B. Laursen and M.M. Neumann. We also give a sufficient condition (in terms of growth of resolvent and of an analytic left inverse of T − z) implying that T is an E(𝕋)-subscalar operator. This condition is also necessary for Hilbert space operators.

1. Introduction

Generalized scalar operators are those Banach spaces operators possessing a C^{∞}-functional calculus. To be more specific, let E(ℂ) denote the usual Fréchet algebra of all C^{∞}-functions on ℂ with the topology of uniform convergence of derivatives of all orders on compact subsets of ℂ. Let X be a complex Banach space. A bounded linear operator S ∈ B(X) is said ([CF]) to be an E(ℂ)-scalar (or generalized scalar) operator if there is a continuous algebra homomorphism Φ : E(ℂ) → B(X) for which Φ(1) = I and Φ(z) = S. Here z denotes the identity function on ℂ. A bounded linear operator is E(ℂ)-subscalar if it is similar to the restriction of an E(ℂ)-scalar operator to one of its closed invariant subspaces. We refer to three books [CF], [EP] and [LN] for more information on E(ℂ)-scalar and E(ℂ)-subscalar operators.

The following statements are known to be equivalent (see [CF, LN]) :

(1) S is E(𝕋)-scalar, i.e., it has a continuous functional calculus on the Fréchet algebra E(𝕋) of C^{∞} functions on the unit circle 𝕋;

(2) S is E(ℂ)-scalar with spectrum σ(S) in the unit circle 𝕋;

(3) S is invertible, and there exist constants C > 0, p ≥ 0 and q ≥ 0

such that

∥S^{n}∥ ≤ Cn^{p} (n ∈ ℕ)

2000 Mathematics Subject Classification. 47B40, 47A20, 47A10.

Key words and phrases. subscalar operators, growth conditions, resolvents.

2 CĂTĂLIN BADEA AND VLADIMÍR MÜLLER

and

∥S^{−n}∥ ≤ Cn^{q} (n ∈ ℕ);

(4) σ(S) ⊂ 𝕋 and there exist constants C > 0, p ≥ 0 and q ≥ 0 such

that

∥(S − z)^{−1}∥ ≤ C (|z| − 1)^{−p} (|z| > 1)

and

∥(S − z)^{−1}∥ ≤ C (1 − |z|)^{−q} (|z| < 1).

The distinction between the growth of norms of positive and negative powers (and the resolvent growth inside and outside unit disc) will become apparent later on.

For T ∈ B(X) we denote

m(T) = inf{∥T x∥ : x ∈ X, ∥x∥ = 1}.

This quantity is called the minimum modulus of T ([GT]) or the lower bound of T ([LN]). It is easy to see that m(T) > 0 if and only if T ∈ B(X) is one-to-one and with closed range. For invertible operators S we have m(S) = ∥S^{−1}∥^{−1}.

The main question we consider in this note is the problem of intrinsic characterizations of E(𝕋)-subscalar operators (i.e. operators similar to a restriction of an E(𝕋)-scalar operator to an invariant subspace). Compressions of E(𝕋)-scalar operators to invariant subspaces have been studied in [BP].

Let T ∈ B(X) be an E(𝕋)-subscalar operator. Using (3) for the invertible extension of T we obtain the existence of constants C > 0, p ≥ 0 and q ≥ 0 such that :

(P) ∥T^{n}∥ ≤ Cn^{p} and m(T^{n})^{−1} ≤ Cn^{q}.

It is natural to ask if the polynomial growth condition (P) above (in terms of norms and minimum moduli of iterates) characterizes E(𝕋)-subscalars operators (cf. K.B. Laursen and M.M. Neumann [LN, Problem 6.1.15] and M. Didas [Di]). This problem was also discussed in [MMN1, MMN2, MMN3, MMN4]. It was recently proved by the authors [BM2, BM1] that E(𝕋)-subscalar operators are indeed characterized by the polynomial growth condition (P).

Using the resolvent condition (4), it can be proved similarly that if T ∈ B(X) is an E(𝕋)-subscalar operator then there exist constants C > 0, p ≥ 0 and q ≥ 0 such that

(R) ∥(T −z)^{−1}∥ ≤ \frac{C}{(|z| − 1)^{p}} (|z| > 1) and m(T −z) ≥ C(1 − |z|)^{q} (|z| < 1).

Note that if T is E(𝕋)-subscalar then σ_{ap}(T), the approximate point spectrum of T given by

σ_{ap}(T) = {λ ∈ ℂ : inf{∥(T − λ)x∥ : ∥x∥ = 1} = 0},

is included in the unit circle. Moreover, either σ(T) is included in the unit circle (and so T is E(𝕋)-scalar) or σ(T) = \overline{𝔻}, the closed unit disc.

Again it is natural to ask if the condition (R) implies the E(𝕋)-subscalarity of T. This is a variant of the open Problem 6.1.14 in [LN]. A characterization of E(𝕋)-subscalar operators in terms of the growth of the local resolvent of the adjoint has been given by Didas [Di].

The aim of this note is to show that the answer to the above problem is negative : there is a Banach space operator T satisfying condition (R) (with suitable p and q) which is not E(𝕋)-subscalar. We also give a sufficient condition (in terms of growth of resolvent and of an analytic left inverse of T −z) implying that T is an E(𝕋)-subscalar operator. This condition is also necessary for Hilbert space operators.

2. A Counterexample

Recall that an equivalent definition of decomposable operators is the following : T ∈ B(X) is decomposable if for every open cover ℂ = U ∪V, there are closed invariant (for T) subspaces Y and Z of X such that X = Y + Z and σ(T | Y) ⊂ U, σ(T | Z) ⊂ V. We refer for instance to [CF] and [LN]. An operator T ∈ B(X) has Bishop’s property (β) if, for every open set U ⊂ ℂ, the operator T_{U} defined by T_{U}(f)(z) = (T − z)f(z) on the set O(U, X) of holomorphic functions from U into X is injective and has closed range. According to a result by E. Albrecht and J. Eschmeier [AE], T ∈ B(X) is subdecomposable (i.e., T is similar to the restriction of a decomposable operator) if and only if T has Bishop’s property (β).

Example 2.1. On the Banach space X = c_{0}, there exists an operator T ∈ B(X) such that

(i) ∥T ∥ ≤ 1, σ_{ap}(T) = 𝕋 and σ(T) = \overline{𝔻} ; (ii) ∥(T − z)^{−1}∥ ≤ (|z| − 1)^{−1} (|z| > 1), (iii) there is a constant C > 0 such that

m(T − z) ≥ C(1 − |z|)^{3} (z ∈ 𝔻);

(iv) T is not E(𝕋)-subscalar ;

(v) T has Bishop’s property (β).

Proof. Let X = c_{0} be the Banach space of all complex sequences converging to zero endowed with the supremum norm. We denote its standard basis by e_{1}, e_{2}, . . .. For n ≥ 1 let

w_{n} = e^{ln^{2}(n+2)−ln^{2}(n+3)}.

Let T ∈ B(X) be the weighted shift defined by T e_{n} = w_{n}e_{n+1} (n ≥ 1).

The proof of the properties of Example 2.1 will be obtained in several steps.

We first remark that 0 < w_{n} < 1 for all n.

Claim 1. (w_{n}) is an increasing sequence and lim_{n→∞} w_{n} = 1.

Proof. For each n ≥ 1 there exists x = x(n) such that n + 2 ≤ x ≤ n + 3 and

ln^{2}(n + 2) − ln^{2}(n + 3) = −2\frac{ln x}{x}.

The function g(x) = −2\frac{ln x}{x} is increasing since

g′(x) = −2 · \frac{1 − ln x}{x^{2}} > 0 (x > e).

Therefore (ln^{2}(n + 2) − ln^{2}(n + 3)) is an increasing sequence for n ≥ 1 and

lim_{n→∞}(ln^{2}(n + 2) − ln^{2}(n + 3)) = 0.

Hence (w_{n}) is an increasing sequence and lim_{n→∞} w_{n} = 1. □

The previous Claim implies that ∥T ∥ ≤ 1. Therefore, for |z| > 1, we have

∥(T − z)^{−1}∥ = ∥−\frac{1}{z} ∑_{n≥0} \frac{1}{z^{n}} T^{n}∥ ≤ \frac{1}{|z| − 1}.

This proves (ii).

For n ≥ 1 we have T^{n}e_{k} = w_{k}w_{k+1} ⋯ w_{k+n−1}e_{k+n} (k ≥ 1), and so

m(T^{n}) = inf_{k} w_{k} ⋯ w_{k+n−1}
= w_{1} ⋯ w_{n}
= e^{ln^{2} 3−ln^{2} 4}e^{ln^{2} 4−ln^{2} 5} ⋯ e^{ln^{2}(n+2)−ln^{2}(n+3)}
= e^{ln^{2} 3−ln^{2}(n+3)} = \frac{3^{ln 3}}{(n + 3)^{ln(n+3)}} .

SUBSCALAR OPERATORS AND RESOLVENT 5

Therefore T does not satisfy condition (P), and so T is not E(𝕋)-subscalar. This proves (iv).

We also have lim_{n→∞} m(T^{n})^{1/n} = 1. Therefore (see [MZ])

σ_{ap}(T) ⊂ {z : |z| = 1}.

Since the spectrum of a weighted shift is circularly symmetric, we have in fact σ_{ap}(T) = {z : |z| = 1}. But ∂σ(T) ⊂ σ_{ap}(T) ⊂ σ(T) and thus σ(T) is either equal to \overline{𝔻} or contained in 𝕋. Since T is not invertible we have σ(T) = \overline{𝔻}. This completes the proof of (i).

Note also that

∑_{n≥1} \frac{| ln m(T^{n})|}{n^{2}} < ∞,

so T satisfies the Beurling-type condition (B) (cf. [BM2, Section 4]). Consequently, T has Bishop’s property (β) (see [BM2, Theorem 4.5]).

We prove now (iii).

Claim 2. lim_{n→∞} \frac{(1−w_{n})^{3}}{w_{n+1}−w_{n}} = 0.

Proof. Let n ∈ ℕ. Then there is an x = x(n), n + 2 ≤ x ≤ n + 3, such that

w_{n+1} − w_{n} = e^{ln^{2}(n+3)−ln^{2}(n+4)} − e^{ln^{2}(n+2)−ln^{2}(n+3)}
= e^{ln^{2} x−ln^{2}(x+1)} (\frac{2 ln x}{x} − \frac{2 ln(x + 1)}{x + 1})

and there is a y = y(n), x ≤ y ≤ x + 1 (i.e., n + 2 ≤ y ≤ n + 4) such that

w_{n+1} − w_{n} = −2e^{ln^{2} x−ln^{2}(x+1)} · \frac{1 − ln y}{y^{2}} .

Similarly, there is an x′ = x′(n), n + 2 ≤ x′ ≤ n + 3, such that

ln^{2}(n + 2) − ln^{2}(n + 3) = −\frac{2 ln x′}{x′} .

We have

lim_{n→∞} \frac{(1 − w_{n})^{3}}{w_{n+1} − w_{n}}

= lim_{n→∞} \frac{(\frac{1−e^{ln^{2}(n+2)−ln^{2}(n+3)}}{ln^{2}(n+2)−ln^{2}(n+3)} )^{3} · (ln^{2}(n + 2) − ln^{2}(n + 3))^{3}}{−2e^{ln^{2} x−ln^{2}(x+1)} \frac{1−ln y}{y^{2}}}

= (−1)^{3} (−\frac{1}{2}) lim_{n→∞} \frac{(ln^{2}(n + 2) − ln^{2}(n + 3))^{3}}{\frac{1−ln y}{y^{2}}} = \frac{1}{2} lim_{n→∞} \frac{(\frac{−2 ln x′}{x′} )3}{\frac{1−ln y}{y^{2}}}

= −4 lim_{n→∞} \frac{y^{2}}{x^{′2}} · lim_{n→∞} \frac{ln^{3} x′}{x′(1 − ln y)} = 0.

□

Claim 3. There is an r > 0 such that m(T − z) ≥ (1 − |z|)^{3} for all z ∈ 𝔻, |z| ≥ r.

Proof. Find n_{0} such that

\frac{(1 − w_{n})^{3}}{w_{n+1} − w_{n}} < \frac{1}{16}

for all n ≥ n_{0}. Find r, 1/2 ≤ r < 1, such that r − (1 − r)^{3} > w_{n_{0}}. Suppose on the contrary that there is a λ ∈ 𝔻, |λ| ≥ r such that

m(T − λ) < (1 − |λ|)^{3}.

Thus there exists x = (x_{i}) ∈ X with ∥x∥ = max_{i} |x_{i}| = 1 and

∥(T − λ)x∥ < (1 − |λ|)^{3}.

Since

(T − λ)x = (−λx_{1}, w_{1}x_{1} − λx_{2}, w_{2}x_{2} − λx_{3}, ⋯),

we have

|λ| · |x_{1}| < (1 − |λ|)^{3}

and

sup_{i} |w_{i}x_{i} − λx_{i+1}| < (1 − |λ|)^{3}.

Without loss of generality we may assume that λ > 0 and x_{i} > 0 for all i ≥ 1. Indeed, replace λ by |λ| and x_{i} by |x_{i}| (i ≥ 1). We have

sup_{i}|w_{i}|x_{i}| − |λ| · |x_{i+1}|| ≤ sup_{i} |w_{i}x_{i} − λx_{i+1}| < (1 − |λ|)^{3}.

Thus we may assume that there are r, μ with 1/2 ≤ r < μ < 1 and u = (u_{i}) ∈ X with u_{i} ≥ 0 (i ∈ ℕ), ∥u∥ = max_{i} u_{i} = 1 and

(2.1)  μ · u_{1} < (1 − μ)^{3},  sup_{i} |w_{i}u_{i} − μu_{i+1}| < (1 − μ)^{3}.

We show that this is not possible. Write for short a = (1 − μ)^{3}.

Let m ∈ ℕ satisfy u_{m} = 1 and u_{j} < 1 for all j < m.

We have u_{1} < \frac{(1−μ)^{3}}{μ} < 1. Thus m ≥ 2.

We show that w_{m−1} ≥ μ−a. Suppose on the contrary that w_{m−1} < μ−a. By (2.1), we have

a > |w_{m−1}u_{m−1} − μu_{m}| ≥ μu_{m} − w_{m−1}u_{m−1}
≥ μ − (μ − a)u_{m−1}
= (μ − a)(1 − u_{m−1}) + a ≥ a,

SUBSCALAR OPERATORS AND RESOLVENT 7

a contradiction. Hence

(2.2)  w_{m−1} ≥ μ − a.

We show now that w_{m} ≥ μ+a. Suppose on the contrary that w_{m} < μ+a. Then w_{m} − w_{m−1} ≤ 2a and 1 − w_{m−1} ≥ 1 − w_{m} ≥ 1 − μ − a. Therefore we have

\frac{(1 − w_{m})^{3}}{w_{m} − w_{m−1}} ≥ \frac{(1 − μ − a)^{3}}{2a} = \frac{(1 − μ − (1 − μ)^{3})^{3}}{2(1 − μ)^{3}} ≥ 1/16,

since μ ≥ 1/2 and

(1 − μ) − (1 − μ)^{3} = (1 − μ)μ(2 − μ) ≥ \frac{1}{2}(1 − μ).

Thus m − 1 < n_{0}, and so

μ − a ≥ r − (1 − r)^{3} > w_{n_{0}} ≥ w_{m−1},

a contradiction with (2.2). Hence

(2.3)  w_{m} ≥ μ + a.

Since |w_{m}u_{m} − μu_{m+1}| < a, we have μu_{m+1} > w_{m} − a, and so

u_{m+1} > \frac{w_{m} − a}{μ} ≥ 1,

a contradiction with the assumption that ∥u∥ = 1.

Hence m(T − z) ≥ (1 − |z|)^{3} for all z ∈ 𝔻 with |z| ≥ r. □

Since m(T − z) > 0 for all z ∈ 𝔻 and the function

z ↦ \frac{m(T − z)}{(1 − |z|)^{3}}

is continuous on 𝔻, there is a constant C > 0 such that m(T − z) ≥ C · (1 − |z|)^{3} for all z ∈ 𝔻.

The proof of Example 2.1 is now complete. □

Remarks 2.2. (a) Another proof of Bishop’s property (β) for T can be given using [LN, 1.7.1].

(b) The fact that T has Beurling-type property (B) implies by [BM2, Theorem 4.5] that there exists a Banach space Y containing c_{0} and an invertible operator S ∈ B(Y) such that T = S_{|X} and S satisfies

∑_{n=−∞}^{∞} \frac{log ∥S^{n}∥}{1 + n^{2}} < ∞.

Note that this condition implies ([CF]) that S is decomposable.

8 CĂTĂLIN BADEA AND VLADIMÍR MÜLLER

(c) We don’t know if the weighted shift T on the Hilbert space ℓ_{2} = ℓ_{2}(ℕ) given by

T e_{n} = exp(ln^{2}(n + 2) − ln^{2}(n + 3))e_{n+1} (n ≥ 1)

is a hilbertian counterexample to the variant of Laursen-Neumann problem.

Theorem 2.3. Let X be a separable Banach space containing (an isomorphic copy of) c_{0}. Then there exist R ∈ B(X) and a constant C > 0 such that

(i) σ(R) = \overline{𝔻} ; (ii) ∥(R − z)^{−1}∥ ≤ C (|z| − 1)^{−1} (|z| > 1); (iii) m(R − z) ≥ C(1 − |z|)^{3} (z ∈ 𝔻) ; (iv) R is not E(𝕋)-subscalar ;

(v) R has Bishop’s property (β).

Proof. According to a result due to A. Sobczyk (see [BePe]), if X is a separable Banach space containing an isomorphic copy of c_{0}, then X contains a subspace Y, isomorphic to c_{0}, which is complemented in X. We consider the operator R on X equal to the operator of Example 2.1 on Y and equal to the identity on its complement. Then R satisfies all the requirements because of the properties of T. □

3. Sufficient conditions

We begin with the following sufficient condition.

Proposition 3.1. Let T ∈ B(X) be a Banach space operator satisfying

∥(T − z)^{−1}∥ ≤ C(|z| − 1)^{−p} (|z| > 1),

for some fixed constants C > 0 and p ≥ 0. Suppose that there are q ≥ 0 and an analytically dependant left inverse function L : 𝔻 → B(X) such that L(z)(T − z) = I and

∥L(z)∥ ≤ C(1 − |z|)^{−q} (z ∈ 𝔻).

Then T is E(𝕋)-subscalar.

We note that the growth condition on the analytically dependant left inverse function L implies that

∥x∥ = ∥L(z)(T − z)x∥ ≤ C(1 − |z|)^{−q}∥(T − z)x∥;

hence

m(T − z) ≥ C^{−1}(1 − |z|)^{q}.

SUBSCALAR OPERATORS AND RESOLVENT 9

Proof. A proof of this result can be given using Didas’ criterion [Di] in terms of local resolvent of the adjoint of T. We give here a different proof.

It is a classical result (see [LN, Th. 1.5.12]) that the resolvent growth condition outside the closed unit disc implies a polynomial growth condition for the powers of T : there is a constant c > 0 such that

∥T^{n}∥ ≤ cn^{p} (n ∈ ℕ).

Write

L(z) = ∑_{i=0}^{∞} L_{i}z^{i} (z ∈ 𝔻),

with L_{i} ∈ B(X). Let 0 < r < 1. By the Cauchy formula, for each n ∈ ℕ we have

∥L_{n}∥ ≤ \frac{max\{∥L(z)∥ : |z| ≤ r\}}{r^{n}} ≤ \frac{C}{r^{n}(1 − r)^{q}} .

In particular, for r = \frac{n}{n+q} (where the function r ↦ r^{−n}(1 − r)^{−q} attains the minimum) we obtain

∥L_{n}∥ ≤ C · (\frac{n}{n + q} )^{−n} (1 − \frac{n}{n + q} )^{−q} .

We have

lim_{n→∞} (\frac{n}{n + q} )^{−n} = lim_{n→∞} (1 + \frac{q}{n})^{n} = e^{q}.

Further, for n ≥ q we have

(1 − \frac{n}{n + q} )^{−q} = (\frac{n + q}{q} )^{q} ≤ (\frac{2n}{q})^{q} .

Thus there is a constant K > 0 such that ∥L_{n}∥ ≤ K · n^{q} for all n. We have

I = L(z)(T − z) = ∑_{i=0}^{∞} L_{i}z^{i}(T − z) = L_{0}T + ∑_{i=1}^{∞} z^{i}(L_{i}T − L_{i−1})

for all z ∈ 𝔻. Thus L_{0}T = I and L_{i}T = L_{i−1} for all i ≥ 1. Hence

L_{n}T^{n+1} = L_{n−1}T^{n} = ⋯ = L_{0}T = I.

Let x ∈ X, ∥x∥ = 1. Then

1 = ∥x∥ = ∥L_{n−1}T^{n}x∥ ≤ ∥L_{n−1}∥ · ∥T^{n}x∥.

Thus ∥T^{n}x∥ ≥ ∥L_{n−1}∥^{−1}, and so for some constant K′ we have m(T^{n}) ≥ K′n^{−q} for all n. Hence T is E(𝕋)-subscalar by [BM2, Theorem 4.1]. □

The next result gives an intrinsic characterization of E(𝕋)-subscalar operators on Hilbert spaces.

10 CĂTĂLIN BADEA AND VLADIMÍR MÜLLER

Theorem 3.2. Let H be a Hilbert space and T ∈ B(H). Then T is E(𝕋)- subscalar if and only if there are constants C > 0, p ≥ 0, q ≥ 0 and an analytic operator-valued function L : 𝔻 → B(H) such that

(i) ∥(T − z)^{−1}∥ ≤ C(|z| − 1)^{−p} (|z| > 1); (ii) L(z)(T − z) = I (|z| < 1); (iii) ∥L(z)∥ ≤ C(1 − |z|)^{−q} (|z| < 1).

Proof. Suppose that T is a Hilbert space E(𝕋)-subscalar operator. According to [BM2, Th. 4.1], there are a Hilbert space K, constants C′ > 0, s ≥ 0 and an E(𝕋)-scalar extension S ∈ B(K) such that σ(S) = σ_{ap}(T) ⊂ 𝕋 and

∥S^{m}∥ ≤ C′|m|^{s} (m ∈ ℤ, m ≠ 0).

It is known that the power growth estimate ∥S^{m}∥ ≤ C′|m|^{s} implies that [LN, 1.5.12]

∥(S − z)^{−1}∥ ≤ C||z| − 1|^{−s−1} (|z| ≠ 1)

for a suitable constant C > 0. This implies

∥(T − z)^{−1}∥ ≤ C(|z| − 1)^{−s−1} (|z| > 1).

We define L : 𝔻 ↦ B(H) by

L(z)x = P_{H}(S − z)^{−1}x (z ∈ 𝔻, x ∈ H),

where P_{H} ∈ B(K) is the orthogonal projection onto H.

Then L is analytic and we have

∥L(z)∥ ≤ ∥(S − z)^{−1}∥ ≤ C(1 − |z|)^{−s−1} (|z| < 1).

The equality L(z)(T −z) = I on 𝔻 follows from the equalities (S −z)^{−1}(S − z) = I and S_{|H} = T.

The second implication follows from Proposition 3.1. □

Acknowledgment. The first author was supported by CNRS (France) through a ’délégation CNRS’ at Institut Henri Poincaré, Paris. The second author was supported by grant No. 201/03/0041 of GA ČR. This paper was written during the second author’s stay at the University of Lille. He would like to thank for warm hospitality and perfect working conditions there.

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Département de Mathématiques, UMR CNRS no. 8524, Université Lille I, F–59655 Villeneuve d’Ascq, France

E-mail address: badea@math.univ-lille1.fr

Institut of Mathematics AV CR, Zitna 25, 115 67 Prague 1, Czech Republic

E-mail address: muller@math.cas.cz