SPECTRAL RADII OF BOUNDED OPERATORS ON TOPOLOGICAL VECTOR SPACES

VLADIMIR G. TROITSKY

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF TEXAS, AUSTIN, TX 78712. USA.

Abstract. In this paper we develop a version of spectral theory for bounded linear operators on topological vector spaces. We show that Gelfand’s formula for the spectral radius and Neumann series can be naturally interpreted for operators on topological vector spaces. Of course, the resulting theory has many similarities to the conventional spectral theory of bounded operators on Banach spaces, though there are several important differences. The main difference is that an operator on a topological vector space has several spectra and several spectral radii, which fit a well-organized pattern.

Key words: spectrum, spectral radius, bounded operator, topological vector space, locally convex space, closed operator.

AMS Subject Classification: 46A03, 47A10, 47A25.

Contents

0. Introduction 2 1. Preliminaries and notation 3 2. Bounded operators 4 3. Spectra of an operator 10 4. Spectral radii of an operator 11 5. Spectra and spectral radii 18 6. nb-bounded operators 25 7. Compact operators 29 8. Closed operators 32 References 33

PanAmerican Mathematical Journal, 11 (2001), no. 3, 1-35

1

0. Introduction

The spectral radius of a bounded linear operator T on a Banach space is defined by Gelfand’s formula r(T) = lim_{n} \sqrt[n]{∥T^{n}∥}. It is well known that r(T) equals the actual radius of the spectrum |σ(T)| = sup{|λ| : λ ∈ σ(T)}. Furthermore, it is known that the resolvent R_{λ} = (λI − T)^{−1} is given by the Neumann series ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} whenever |λ| > r(T). It is natural to ask if similar results are valid in a more general setting, e.g., for a bounded linear operator on an arbitrary topological vector space. The author arrived at these questions when generalizing some results on the Invariant Subspace Problem from Banach lattices to ordered topological vector spaces. A major difficulty is the lack of a readily available developed spectral theory. The spectral theory of bounded operators on Banach spaces has been thoroughly studied for a long time, and is extensively used. Unfortunately, little has been known about the spectral theory of bounded operators on general topological vector spaces, and many techniques used in Banach spaces cannot be applied for operators on topological vector spaces. In particular, the spectrum, the spectral radius, and the Neumann series are the tools which are widely used in the study of the Invariant Subspace Problem in Banach spaces, but which have not been sufficiently studied for general topological vector spaces. To overcome this obstacle we have developed a version of the spectral theory of bounded operators on general topological vector spaces and on locally convex spaces. Some results in this direction have also been obtained by B. Gramsch [10], and by F. Garibay and R. Vera [8, 9, 20].

We consider the following classification of bounded operators on a topological vector space. We call a linear operator T

- nb-bounded if T maps some neighborhood of zero into a bounded set,

- nn-bounded if there is a base of neighborhoods of zero such that T maps every neighborhood in this base into a multiple of itself, and

- bb-bounded if T maps bounded sets into bounded sets.

The classes of all linear operators, of all bb-bounded operators, of all continuous operators, of all nn-bounded operators, and of all nb-bounded operators form nested algebras. The spectrum of an operator T in each of these algebras is defined as usual, i.e., the set of λ’s for which λI − T is not invertible in this algebra. We show in Section 4 that the well known Gelfand formula for the spectral radius of an operator on a Banach space, r(T) = lim_{n} \sqrt[n]{∥T^{n}∥}, can be generalized to each of these five classes of operators on topological vector spaces, and then we use this formula to define the spectral radius of an operator with respect to each of the classes. Then in Section 5 we show that if T is a continuous operator on a sequentially complete locally convex space and |λ| is greater than the spectral radius of T, in any of the five classes, then the Neumann series ∑_{n=0}^{∞} \frac{T^{n}}{λ^{n+1}} converges in the topology of the class, and λ does not belong to the corresponding spectrum of T. In particular, the spectral radius of T with respect to each class is greater than or equal to the geometrical radius of the spectrum of T in this class. In Sections 6 and 7 we show that the radii are equal for nb-bounded and compact operators.

This paper is based on a part of the author’s Ph.D. thesis [19]. I would like to thank Yuri Abramovich for his encouragement and for numerous suggestions and improvements. Robert Kaufman read parts of the draft and asked interesting questions, which resulted in new developments. Thanks are also due to Heinrich Lotz, Michael Neumann, Ted Odell, and Joseph Rosenblatt for their support and interest in my work.

1. Preliminaries and notation

The symbols X and Y shall always denote topological vector spaces. A neighborhood of a point x ∈ X is any subset of X containing an open set which contains x. Neighborhoods of zero will often be referred to as zero neighborhoods. Every zero neighborhood V is absorbing, i.e., ⋃_{n=1}^{∞} nV = X. In every topological vector space (over ℝ or ℂ) there exists a base N_{0} of zero neighborhoods with the following properties:

(i) Every V ∈ N_{0} is balanced, i.e., λV ⊆ V whenever |λ| ⩽ 1; (ii) For every V_{1}, V_{2} ∈ N_{0} there exists V ∈ N_{0} such that V ⊆ V_{1} ∩ V_{2}; (iii) For every V ∈ N_{0} there exists U ∈ N_{0} such that U + U ⊆ V; (iv) For every V ∈ N_{0} and every scalar λ the set λV is in N_{0}.

Whenever we mention a basic neighborhood of zero, we always assume that it belongs to a base which satisfies these properties.

A topological vector space is called normed if the topology is given by a norm. In this case the collection of all balls centered at zero is a base of zero neighborhoods. A complete normed space is referred to as a Banach space. See [6] for a detailed study of normed and Banach spaces.

A set A in a topological vector space is called bounded if it is absorbed by every zero neighborhood, i.e., for every zero neighborhood V one can find α > 0 such that A ⊆ αV. A set A in a topological vector space is said to be pseudo-convex or semi-convex if A+A ⊆ αA for some number α (for convex sets α = 2). If U is a zero neighborhood, (x_{γ}) is a net in X, and x ∈ X, we write x_{γ} ⟶^{U} x if for every ε > 0 there exists an index γ_{0} such that x_{γ} − x ∈ εU whenever γ ⩾ γ_{0}. It is easy to see that when U is pseudo-convex, this convergence determines a topology on X, and the set of all scalar multiples of U forms a base of the topology. We denote X when equipped with this topology by (X, U). Clearly, (X, U) is Hausdorff if and only if ⋂_{n=1}^{∞} \frac{1}{n}U = \{0\}.

A topological vector space is said to be locally bounded if there exists a bounded zero neighborhood. Notice that if U is a bounded zero neighborhood then it is pseudo-convex. Conversely, if U is a pseudo-convex zero neighborhood, then (X, U) is locally bounded. Recall that a quasinorm is a real-valued function on a vector space which satisfies all the axioms of norm except the triangle inequality, which is replaced by ∥x+y∥ ⩽ k (∥x∥+∥y∥) for some fixed positive constant k. It is known (see, e.g., [13]) that a topological vector space is quasinormable if and only if it is locally bounded and Hausdorff. A complete quasinormed space is called quasi-Banach.

If the topology of a topological vector space X is given by a seminorm p, we say that X = (X, p) is a seminormed space. Clearly, in this case X = (X, U) where the convex set U is the unit ball of p and, conversely, p is the Minkowski functional of U. A Hausdorff topological vector space is called locally convex if there is a base of convex zero neighborhoods or, equivalently, if the topology is generated by a family of seminorms (namely, the Minkowski functionals of the convex zero neighborhoods). When dealing with a base of zero neighborhoods in a locally convex space we will always assume that it consists of convex sets. Similarly, a Hausdorff topological vector space is said to be locally pseudo-convex if it has a base of pseudo-convex zero neighborhoods. A complete metrizable topological vector space is usually referred to as a Fréchet space.

By an operator we always mean a linear operator between vector spaces. We will usually use the symbols S and T to denote operators. Recall that an operator T between normed spaces is said to be bounded if its operator norm defined by ∥T ∥ = sup{∥T x∥ : ∥x∥ ⩽ 1} is finite. It is well known that an operator between normed spaces is bounded if and only if it is continuous. An operator between two vector spaces is said to be of finite rank if the range of T is finite dimensional.

If A is a unital algebra and a ∈ A, then the resolvent set of a is the set ρ(a) of all λ ∈ ℂ such that e − λa is invertible in A. The resolvent set of an element a in a non-unital algebra A is defined as the set of all λ ∈ ℂ for which e − λa is invertible in the unitalization A_{×} of A. The spectrum of an element of an algebra is defined via σ(a) = ℂ \ ρ(a). It is well-known that whenever A is a unital Banach algebra then σ(a) is compact and nonempty for every a ∈ A. In this case the spectral radius r(a) is defined via Gelfand’s formula: r(a) = lim_{n} \sqrt[n]{∥a^{n}∥}. It is also well known that r(a) = |σ(a)|, where |σ(a)| is the geometric radius of σ(a), i.e., |σ(a)| = sup{|λ| : λ ∈ σ(a)}. An element a ∈ A is said to be quasinilpotent if σ(a) = {0}.

If T is a bounded operator on a Banach space X then we consider the spectrum σ(T) and the resolvent set ρ(T) in the sense of the Banach algebra of bounded operators on X. If λ ∈ ρ(T) then the inverse (I − λT)^{−1} is called the resolvent operator and is denoted by R(T; λ) or just R_{λ}. If λ ∈ ℂ satisfies |λ| > r(T) then the Neumann series ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} converges to R_{λ} in operator norm. Following [1, 2] we say that T is locally quasinilpotent at x ∈ X if lim_{n} \sqrt[n]{∥T^{n}x∥} = 0.

Further details on topological vector spaces can be found in [6, 13, 17, 7, 18, 12]. For details on locally bounded and quasinormed topological vector spaces we refer the reader to [13, 14, 16].

2. Bounded operators

Operator Theory on Banach spaces deals with the class of bounded operators, which coincides with the class of continuous operators. The operator zoo in general topological vector spaces is much richer. In this paper we will deal with the following types of bounded operators. To avoid confusion, we give different names to different types of boundedness.

Definition 2.1. Let X and Y be topological vector spaces. An operator T: X → Y is said to be

(i) bb-bounded if it maps every bounded set into a bounded set; (ii) nb-bounded if it maps some neighborhood into a bounded set;

Further, if X = Y we will say that T: X → X is nn-bounded if there exists a base N_{0} of zero neighborhoods such that for every U ∈ N_{0} there is a positive scalar α such that T (U) ⊆ αU.

Remark 2.2. [7] and [12] present (i) as the definition of a bounded operator on a topological vector space, while [17] and [18] use (ii) for the same purpose. As we will see, these definitions are far from being equivalent.

Proposition 2.3. Let X and Y be topological vector spaces. For an operator T: X → Y consider the following statements:

(i) T is bb-bounded; (ii) T is continuous; (iii) T is nn-bounded; (iv) T is nb-bounded.

Then (iv) ⇒ (ii) ⇒ (i). Furthermore, if X = Y then (iv) ⇒ (iii) ⇒ (ii) ⇒ (i).

Proof. The implications (iv) ⇒ (ii) ⇒ (i) are trivial. To show (iv) ⇒ (iii) ⇒ (ii) assume that X = Y and fix a base N_{0} of zero neighborhoods. If T is nb-bounded then T (U) is bounded for some U ∈ N_{0}. Note that N′_{0} = { λU ∩ V : λ > 0, V ∈ N_{0} } is another base of zero neighborhoods. For each W = λU ∩ V in N′_{0} we have T (W) ⊆ λT (U). But T (U) is bounded and so T (W) ⊆ λT (U) ⊆ λαW for some positive α, i.e., T is nn-bounded.^{1}

Finally, if T is nn-bounded, then there is a base N_{0} such that for every zero neighborhood U ∈ N_{0} there is a positive scalar α such that T (U) ⊆ αU. Let V be an arbitrary zero neighborhood. Then there exists U ∈ N_{0} such that U ⊆ V, so that T (U) ⊆ αU ⊆ αV for some α > 0. Taking W = \frac{1}{α}U we get T (W) ⊆ V, hence T is continuous. □

2.4. It can be easily verified that all the statements in Proposition 2.3 are equivalent if X is locally bounded. In general, however, these notions are not equivalent. Obviously, the identity operator I is always nn-bounded, continuous, and bb-bounded, but I is nbbounded if and only if the space is locally bounded. Every bb-bounded operator between two locally convex spaces is continuous if and only if the domain space is bornological. (Recall that a locally convex space is bornological if every balanced convex set which absorbs every bounded set is a zero neighborhood; for details see [18, 17].)

Example 2.5. A continuous but not nn-bounded operator. Let T be the left shift on the space of all real sequences ℝ^{ℕ} with the topology of coordinate-wise convergence, i.e., T: (x_{0}, x_{1}, x_{2}, …) ↦ (x_{1}, x_{2}, x_{3}, …). Clearly T is continuous. We will show that T is not nn-bounded. Assume that for every zero neighborhood U in some base N_{0} there is a positive scalar α such that T (U) ⊆ αU. Since the set {x = (x_{k}) : |x_{0}| < 1} is a zero neighborhood, there must be a neighborhood U ∈ N_{0} such that U ⊆ {x : |x_{0}| < 1}. Since T (U) ⊆ αU for some positive α then T^{n}(U) ⊆ α^{n}U, so that if x = (x_{k}) ∈ U then T^{n}x ∈ α^{n}U, so that |x_{n}| = |(T^{n}x)_{0}| < α^{n}. Hence U ⊆ {x : |x_{n}| < α^{n} for each n > 0}. But this set is bounded, while the space is not locally bounded, a contradiction.

1Note that if the topology is locally convex, then we can assume that U is convex and N_{0} consists of convex neighborhoods. In this case N′_{0} also consists of convex neighborhoods.

2.6. Algebraic properties of bounded operators. The sum of two bb-bounded operators is bb-bounded because the sum of two bounded sets in a topological vector space is bounded. Clearly the product of two bb-bounded operators is bb-bounded. It is well known that sums and products of continuous operators are continuous. Obviously, the product of two nn-bounded operators is nn-bounded, and it can be easily verified that the sum of two nn-bounded operators on a locally convex (or locally pseudo-convex) space is again nn-bounded. It is not difficult to see that the sum of two nb-bounded operator is nbbounded. Indeed, suppose that T_{1} and T_{2} are two nb-bounded operators, then the sets T_{1}(U_{1}) and T_{2}(U_{2}) are bounded for some zero neighborhoods U_{1} and U_{2}. There exists another zero neighborhood U so that U ⊆ U_{1} ∩ U_{2}. Then the sets T_{1}(U) and T_{2}(U) are bounded, so that (T_{1} + T_{2})(U) ⊆ T_{1}(U) + T_{2}(U) is bounded. Finally, it is not difficult to see that the product of two nb-bounded operators is again nb-bounded. This follows immediately from Proposition 2.3 and the following simple observation: if we multiply an nb-bounded operator by a bb-bounded operator on the left or by an nn-bounded operator on the right, the product is nb-bounded.

Thus, the class of all bb-bounded operators, the class of all continuous operators, and the class of all nb-bounded operators are subalgebras of the algebra of all linear operators. The class of nn-bounded operators is an algebra provided that the space is locally convex or locally pseudo-convex.

Boundedness in terms of convergence. Suppose T: X → Y is an operator between two topological vector spaces. It is well known that T is continuous if and only if it maps convergent nets to convergent nets.

Notice that a subset of a topological vector space is unbounded if and only if it contains an unbounded sequence. Therefore, an operator is bb-bounded if and only if it maps bounded sequences (nets) to bounded sequences (respectively, nets).

It is easy to see that T is nn-bounded if and only if T maps U-bounded (U-convergent to zero) sequences to U-bounded (respectively, U-convergent to zero) sequences for every U in some base of zero neighborhoods. If U is a zero neighborhood, we say that a net (x_{γ}) is U-bounded if it is contained in αU for some α > 0, we say that x_{γ} ⟶^{U} 0 if for every α > 0 there exits γ_{0} such that x_{γ} ∈ αU whenever γ > γ_{0}.

2.7. Suppose T is nb-bounded, then T (U) is bounded for some zero neighborhood U. Obviously x_{γ} ⟶^{U} 0 implies T x_{γ} → 0. The converse implication is also valid: if T maps U-convergent sequences to convergent sequences, then T (U) is bounded, hence T is nbbounded. Indeed, suppose that T (U) is unbounded, then there is a zero neighborhood V in Y such that V does not absorb T (U). Then for every n ⩾ 1 there exists y_{n} ∈ T (U)\nV. Further, if y_{n} = T x_{n} for some x_{n} ∈ U, then \frac{x_{n}}{n} ⟶^{U} 0, but T (\frac{x_{n}}{n}) = \frac{y_{n}}{n} ∉ V, so that T (\frac{x_{n}}{n}) does not converge to zero.

Normed, quasinormed, and seminormed spaces. Next, we discuss bounded operators in some particular topologies. Notice that every normed, seminormed, or quasinormed vector space is locally bounded. Therefore bb-boundedness, continuity, nn-boundedness and nb-boundedness coincide for operators on such spaces.

Locally convex topology. Similarly to the norm of an operator on a Banach space, we introduce the seminorm of an operator on a seminormed space.

Definition 2.8. Let T be an operator on a seminormed vector space (X, p). As in the case with normed spaces, p generates an operator seminorm p(T) defined by

p(T) = sup_{p(x)≠0} \frac{p(T x)}{p(x)}.

More generally, let S: X → Y be a linear operator between two seminormed spaces (X, p) and (Y, q). Then we define a mixed operator seminorm associated with p and q via

𝔪_{pq}(S) = sup_{p(x)≠0} \frac{q(Sx)}{p(x)}.

The seminorm 𝔪_{pq}(S) is a measure of how far in the seminorm q the points of the p-unit ball can go under S. Notice, that p(T) and 𝔪_{pq}(S) may be infinite. Clearly, if T is an operator on a seminormed space (X, p), then 𝔪_{pp}(T) = p(T).

Lemma 2.9. If S: X → Y is an operator between two seminormed spaces (X, p) and (Y, q), then

(i) 𝔪_{pq}(S) = sup_{p(x)=1} q(Sx) = sup_{p(x)⩽1}q(Sx);

(ii) q(Sx) ⩽ 𝔪_{pq}(S)p(x) whenever 𝔪_{pq}(S) < ∞.

Proof. The first equality in (i) follows immediately from the definition of p(T). We obviously have

sup_{p(x)=1} q(Sx) ⩽ sup_{p(x)⩽1}q(Sx).

In order to prove the opposite inequality, notice that if 0 < p(x) ⩽ 1, then q(Sx) ⩽ \frac{q(Sx)}{p(x)} ⩽ 𝔪_{pq}(S). Thus, it is left to show that p(x) = 0 implies q(Sx) ⩽ 𝔪_{pq}(S). Pick any z with p(z) > 0, then

p(\frac{z}{n}) = p(x + \frac{z}{n} − x) ⩽ p(x + \frac{z}{n}) + p(x) = p(x + \frac{z}{n}) ⩽ p(x) + p(\frac{z}{n}) = \frac{p(z)}{n},

so that p(x + \frac{z}{n}) = p(\frac{z}{n}) ∈ (0, 1) for n > p(z). Further, since Sx + \frac{Sz}{n} converges to Sx we have

q(Sx) = lim_{n→∞} q(Sx + \frac{Sz}{n}) ⩽ lim_{n→∞} \frac{q (S(x + \frac{z}{n}))}{p(x + \frac{z}{n})} ⩽ 𝔪_{pq}(S).

Finally, (ii) follows directly from the definition if p(x) ≠ 0. In the case when p(x) = 0, again pick any z with p(z) > 0. Then p(x + \frac{z}{n}) ≠ 0 and

q(Sx) = lim_{n→∞} q(Sx + \frac{Sz}{n}) = lim_{n→∞} q (S(x + \frac{z}{n})) ⩽ lim_{n→∞} 𝔪_{pq}(S)p(x + \frac{z}{n}) = 0.

□

Corollary 2.10. If T is an operator on a seminormed space (X, p), then

(i) p(T) = sup_{p(x)=1} p(T x) = sup_{p(x)⩽1}p(T x);

(ii) p(T x) ⩽ p(T)p(x) whenever p(T) < ∞.

The following propositions characterize continuity and boundedness of an operator on a locally convex space in terms of operator seminorms. We assume that X and Y are locally convex spaces with generating families of seminorms P and Q respectively.

Proposition 2.11. Let S be an operator from X to Y, then S is continuous if and only if for every q ∈ Q there exists p ∈ P such that 𝔪_{pq}(S) is finite.

Proposition 2.12. An operator T on X is nn-bounded if and only if p(T) is finite for every p ∈ P, or, equivalently, if T maps p-bounded sets to p-bounded sets for every p in some generating family P of seminorms.

Proposition 2.13. Let S: X → Y be a linear operator, then the following are equivalent: (i) S is nb-bounded; (ii) S maps p-bounded sets into bounded sets for some p ∈ P; (iii) There exists p ∈ P such that 𝔪_{pq}(S) < ∞ for every q ∈ Q.

Since the balanced convex hull of a bounded set in a locally convex space is again bounded, we also have the following characterization.

Proposition 2.14. An operator S: X → Y is bb-bounded if and only if 𝔪_{pq}(S) < ∞ whenever q ∈ Q and p is the Minkowski functional of a convex balanced bounded set.

Operator topologies. For each of the five classes of operators, we introduce an appropriate natural operator topology.

2.15. The class of all linear operators between two topological vector spaces will be usually equipped with the strong operator topology. Recall, that a sequence (S_{n}) of operators from X to Y is said to converge strongly or pointwise to a map S if S_{n}x → Sx for every x ∈ X. Clearly, S will also be a linear operator.

2.16. The class of all bb-bounded operators will usually be equipped with the topology of uniform convergence on bounded sets. Recall, that a sequence (S_{n}) of operators is said to converge to zero uniformly on A if for each zero neighborhood V in Y there exists an index n_{0} such that S_{n}(A) ⊆ V for all n > n_{0}. We say that (S_{n}) converges to S uniformly on bounded sets if (S_{n} − S) converges to zero uniformly on bounded sets.

Recall also that a family G of operators is called uniformly bounded on a set A ⊆ X if the set ⋃_{S∈G} S(A) is bounded in Y.

Lemma 2.17. If a sequence (S_{n}) of bb-bounded operators converges uniformly on bounded sets to an operator S, then S is also bb-bounded.

Proof. Fix a bounded set A and a zero neighborhood V. There exists a zero neighborhood U such that U + U ⊆ V. Since S − S_{n} converges to zero uniformly on bounded sets then there exists an index n_{0} such that (S_{n} − S)(A) ⊆ U whenever n ⩾ n_{0}. This yields S(A) ⊆ S_{n}(A) + U. The set S_{n}(A) is bounded, so that S_{n}(A) ⊆ γU for some γ ⩾ 1. Thus, S(A) ⊆ γU + U ⊆ γV. Hence S(A) is bounded for every bounded set A, and so S is bb-bounded. □

2.18. The class of all continuous operators will be usually equipped with the topology of equicontinuous convergence. Recall, that a family G of operators from X to Y is called equicontinuous if for each zero neighborhood V in Y there is a zero neighborhood U in X such that S(U) ⊆ V for every S ∈ G. We say that a sequence (S_{n}) converges to zero equicontinuously if for each zero neighborhood V in Y there is a zero neighborhood U in X such that for every ε > 0 there exists an index n_{0} such that S_{n}(U) ⊆ εV for all n > n_{0}.

Lemma 2.19. If a sequence S_{n} of continuous operators converges equicontinuously to S, then S is also continuous.

Proof. Fix a zero neighborhood V. There exist zero neighborhoods V_{1} and U and an index n_{0} such that V_{1} + V_{1} ⊆ V and (S_{n} − S)(U) ⊆ V_{1} whenever n > n_{0}. Fix n > n_{0}. The continuity of S_{n} guarantees that there exists a zero neighborhood W ⊆ U such that S_{n}(W) ⊆ V_{1}. Since (S_{n} − S)(W) ⊆ V_{1}, we get S(W) ⊆ S_{n}(W) + V_{1} ⊆ V_{1} + V_{1} ⊆ V, which shows that S is continuous. □

2.20. The class of all nn-bounded operators will be usually equipped with the topology of nn-convergence, defined as follows. We will call a collection G of operators uniformly nn-bounded if there exists a base N_{0} of zero neighborhoods such that for every U ∈ N_{0} there exists a positive real β such that S(U) ⊆ βU for each S ∈ G. We say that a sequence (S_{n}) nn-converges to zero if there is a base N_{0} of zero neighborhoods such that for every U ∈ N_{0} and every ε > 0 we have S_{n}(U) ⊆ εU for all sufficiently large n.

Question. Is the class of all nn-bounded operators closed relative to nn-convergence?

2.21. Finally, the class of all nb-bounded operators will be usually equipped with the topology of uniform convergence on a zero neighborhood.

Example 2.22. The class of nb-bounded operators is not closed in the topology of uniform convergence on a zero neighborhood. Let X = ℝ^{ℕ}, the space of all real sequences with the topology of coordinate-wise convergence. Let T_{n} be the projection on the first n components. Clearly, every T_{n} is nb-bounded because it maps the zero neighborhood U_{n} = {(x_{i})_{i=1}^{∞} : |x_{i}| < 1 for i = 1, …, n} to a bounded set. On the other hand, (T_{n}) converges uniformly on X to the identity operator, while the identity operator on X is not nb-bounded.

3. Spectra of an operator

Recall that if T is a continuous operator on a Banach space, then its resolvent set ρ(T) is the set of all λ ∈ ℂ such that the resolvent operator R_{λ} = (λI −T)^{−1} exists, while the spectrum of T is defined by σ(T) = ℂ \ ρ(T). The Open Mapping Theorem guarantees that if R_{λ} exists then it is automatically continuous. Now, if T is an operator on an arbitrary topological vector space and λ ∈ ℂ then the algebraic inverse R_{λ} = (λI − T)^{−1} may exist but it need not be continuous, or it may be continuous but not nb-bounded, etc. In order to treat all these cases properly we introduce the following definitions.

Definition 3.1. Let T be a linear operator on a topological vector space. We denote the set of all scalars λ ∈ ℂ for which λI − T is invertible in the algebra of linear operators by ρ^{l}(T). We say that λ ∈ ρ^{bb}(T) (respectively ρ^{c}(T) or ρ^{nn}(T)) if the inverse of λI − T is bb-bounded (respectively continuous or nn-bounded). Finally, we say that λ ∈ ρ^{nb}(T) if the inverse of λI − T belongs to the unitalization of the algebra of nb-bounded operators, i.e., when (λI − T)^{−1} = αI + S for a scalar α and an nb-bounded operator S.

The spectral sets σ^{l}(T), σ^{bb}(T), σ^{c}(T), σ^{nn}(T), and σ^{nb}(T) are defined to be the complements of the resolvent sets ρ^{l}(T), ρ^{bb}(T), ρ^{c}(T), ρ^{nn}(T), and ρ^{nb}(T), respectively.^{2} We will denote the (left and right) inverse of λI − T whenever it exists by R_{λ}.

3.2. It follows immediately from Proposition 2.3 that

σ^{l}(T) ⊆ σ^{bb}(T) ⊆ σ^{c}(T) ⊆ σ^{nn}(T) ⊆ σ^{nb}(T).

It follows from the Open Mapping Theorem that for a continuous operator T on a Banach space all the introduced spectra coincide with the usual spectrum σ(T). Since the Open Mapping Theorem is still valid on Fréchet spaces, we have σ^{l}(T) = σ^{bb}(T) = σ^{c}(T) for a continuous operator T on a Fréchet space.

3.3. If T is an operator on a locally bounded space (X, U), then by 2.4, bb-boundedness of T is equivalent to nb-boundedness, so that σ^{bb}(T) = σ^{c}(T) = σ^{nn}(T) = σ^{nb}(T). We will denote this set by σ_{U}(T) to avoid ambiguity. A spectral theory of continuous operators on quasi-Banach spaces was developed in [10].

3.4. There are several reasons why we define σ^{nb} in a slightly different fashion than the other spectra. Namely, for λ to be in ρ^{nb}(T) we require (λI −T)^{−1} be not just nb-bounded, but be nb-bounded up to a multiple of the identity operator. On one hand, this is the standard way to define the spectrum of an element in a non-unital algebra, and we know that the algebra of nb-bounded operators is unital only when the space is locally bounded.

2We use superscripts in order to avoid confusion with σ_{c}(T), which is commonly used for continuous spectrum.

On the other hand, if we defined ρ^{nb}(T) as the set of all λ ∈ ℂ for which (λI − T)^{−1} is nb-bounded, then we wouldn’t have obtained a reasonable theory because (λI − T)^{−1} is almost never nb-bounded unless the space is locally bounded.

Indeed, suppose that X is not locally bounded, T is a bb-bounded operator on X and λ ∈ ℂ. Then R_{λ} = (λI −T)^{−1} cannot be nb-bounded, because in this case I = (λI −T)R_{λ} would be nb-bounded by 2.6 as a product of a bb-bounded and an nb-bounded operator. But we know that I is not nb-bounded because X is not locally bounded.

We will see in Proposition 6.2 that in a locally convex but non locally bounded space nb-bounded operators are never invertible, which implies that in such spaces (λI − T)^{−1} is not nb-bounded for any linear operator T.

3.5. Next, let T be a (norm) continuous operator on a Banach space, σ(T) the usual spectrum of T, and let σ^{l}(T), σ^{bb}(T), σ^{c}(T) be computed with respect to the weak topology. It is known that an operator on a Banach space is weak-to-weak continuous if and only if it is norm-to-norm continuous; therefore it follows that σ^{c}(T) = σ(T). Furthermore, σ^{l}(T) does not depend on the topology, so that it also coincides with σ(T). Thus σ^{l}(T) = σ^{bb}(T) = σ^{c}(T) = σ(T).

4. Spectral radii of an operator

The spectral radius of a bounded linear operator T on a Banach space is usually defined via Gelfand’s formula, r(T) = lim_{n→∞} \sqrt[n]{∥T^{n}∥}. The formula involves a norm and so makes no sense in a general topological vector space. Fortunately, this formula can be rewritten without using a norm, and then generalized to topological vector spaces. Similarly to the situation with spectra, this generalization can be done in several ways, so that we will obtain various types of spectral radii for an operator on a topological vector space. We will show later that, as in the Banach space case, there are certain relations between the spectral radii, the radii of the spectra, and the convergence of the Neumann series of an operator on a locally convex topological vector space. This section may appear technical at the beginning, but later on the reader will see that all the facts lead to a simple and natural classification. We start with an almost obvious numerical lemma.

Lemma 4.1. 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}} < ∞ \}.

Proof. Suppose lim sup_{n} \sqrt[n]{t_{n}} = r. If 0 < ν < r, then \sqrt[n_{k}]{t_{n_{k}}} > μ > ν for some μ and some subsequence (t_{n_{k}}), so that \frac{t_{n_{k}}}{ν^{n_{k}}} > \frac{μ^{n_{k}}}{ν^{n_{k}}} → ∞ as k goes to infinity. It follows that lim sup_{n} \frac{t_{n}}{ν^{n}} = ∞. On the other hand, if r is finite and ν > r then \sqrt[n]{t_{n}} < μ < ν for some μ and for all sufficiently large n. Then lim_{n→∞} \frac{t_{n}}{ν^{n}} ⩽ lim_{n→∞} \frac{μ^{n}}{ν^{n}} = 0. □

This lemma implies that the spectral radius r(T) of a (norm) continuous operator T on a Banach space equals the infimum of all positive real scalars ν such that the sequence (\frac{T^{n}}{ν^{n}} ) converges to zero (or just is bounded) in operator norm topology. This can be considered as an alternative definition of the spectral radius, and can be generalized to any topological vector space with an operator topology. In view of 2.15-2.21, we arrive to the following definition.

Definition 4.2. Given a linear operator T on a topological vector space X, we define the following spectral radii of T:

r_{l}(T) = inf\{ ν > 0 : the sequence (\frac{T^{n}}{ν^{n}} ) converges strongly to zero \};
r_{bb}(T) = inf\{ ν > 0 : \frac{T^{n}}{ν^{n}} → 0 uniformly on every bounded set \};
r_{c}(T) = inf\{ ν > 0 : \frac{T^{n}}{ν^{n}} → 0 equicontinuously \};
r_{nn}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) nn-converges to zero\};
r_{nb}(T) = inf\{ ν > 0 : \frac{T^{n}}{ν^{n}} → 0 uniformly on some 0-neighborhood \}.

The following proposition explains the relations between the introduced radii.

Proposition 4.3. If T is a linear operator on a topological vector space X, then

r_{l}(T) ⩽ r_{bb}(T) ⩽ r_{c}(T) ⩽ r_{nn}(T) ⩽ r_{nb}(T).

Proof. Let T be a linear operator on a topological vector space X. Since every singleton is bounded, r_{l}(T) ⩽ r_{bb}(T). Next, assume ν > r_{c}(T) and fix μ such that r_{c}(T) < μ < ν. Then the sequence (\frac{T^{n}}{μ^{n}}) converges to zero equicontinuously. Take a bounded set A and a zero neighborhood U. There exists a zero neighborhood V and a positive integer N such that \frac{T^{n}}{μ^{n}} (V) ⊆ U whenever n ⩾ N. Also, A ⊆ αV for some α > 0, so that \frac{T^{n}}{ν^{n}} (A) ⊆ \frac{μ^{n}}{ν^{n}} \frac{T^{n}}{μ^{n}} (αV) ⊆ \frac{μ^{n}α}{ν^{n}}U ⊆ U for all sufficiently large n. It follows that the sequence (\frac{T^{n}}{ν^{n}}) converges to zero uniformly on A and, therefore, ν ⩾ r_{bb}(T). Thus, r_{bb}(T) ⩽ r_{c}(T).

To prove the inequality r_{c}(T) ⩽ r_{nn}(T) we let ν > r_{nn}(T). Then for some base N_{0} of zero neighborhoods and for every V ∈ N_{0} and ε > 0 there exists a positive integer N such that \frac{T^{n}}{ν^{n}} (V) ⊆ εV for every n ⩾ N. Given a zero neighborhood U, we can find V ∈ N_{0} such that V ⊆ U. Then \frac{T^{n}}{ν^{n}} (V) ⊆ εV ⊆ εU for every n ⩾ N, so that the sequence (\frac{T^{n}}{ν^{n}}) converges to zero equicontinuously, and therefore ν ⩾ r_{c}(T).

Finally, we must show that r_{nn}(T) ⩽ r_{nb}(T). Suppose that ν > r_{nb}(T). We claim that ν ⩾ r_{nn}(T). Take μ so that ν > μ > r_{nb}(T). One can find a zero neighborhood U such that for every zero neighborhood V there is a positive integer N such that \frac{T^{n}}{μ^{n}} (U) ⊆ V for every n ⩾ N. Fix a base N_{0} of zero neighborhoods, and define a new base N′_{0} of zero neighborhoods via N′_{0} = {mU ∩ W : m ∈ ℕ, W ∈ N_{0}}. Let V ∈ N′_{0} and ε > 0. Then V = mU ∩ W for some positive integer m and W ∈ N_{0}. Then \frac{T^{n}}{μ^{n}} (V) ⊆ m\frac{T^{n}}{μ^{n}} (U) ⊆ mV for every sufficiently large n. Thus \frac{T^{n}}{ν^{n}} (V) ⊆ \frac{μ^{n}}{ν^{n}} mV ⊆ εV, for each sufficiently large n, which implies ν ⩾ r_{nn}(T). □

The following lemma shows that, similarly to the case of Banach spaces, one can use boundedness instead of convergence when defining the spectral radii of an operator on a topological vector space.

Lemma 4.4. Let T be a linear operator on a topological vector space, then (i) r_{l}(T) = inf\{ ν > 0 : (\frac{T^{n}x}{ν^{n}}) is bounded for every x \}; (ii) if T is bb-bounded then

r_{bb}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) is uniformly bounded on every bounded set \}; (iii) if T is continuous then

r_{c}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) is equicontinuous \}; (iv) if T is nn-bounded then

r_{nn}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) is uniformly nn-bounded \};

(v) if T is nb-bounded then

r_{nb}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) is uniformly bounded on some 0-neighborhood \}. Moreover, in each of these cases it suffices to consider any tail of the sequence (\frac{T^{n}}{ν^{n}} ) instead of the whole sequence.

Proof. To prove (i) let

r′_{l}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} x) is bounded for every x \}.

Since every convergent sequence is bounded, we certainly have r_{l}(T) ⩾ r′_{l}(T). Conversely, suppose ν > r′_{l}(T), and take any positive scalar μ such that ν > μ > r′_{l}(T). Then for every x ∈ X the sequence \frac{T^{n}}{μ^{n}} x is bounded, and it follows that the sequence \frac{T^{n}x}{ν^{n}} = \frac{μ^{n}}{ν^{n}} \frac{T^{n}x}{μ^{n}} converges to zero, so that ν ⩾ r_{l}(T) and, therefore r′_{l}(T) ⩾ r_{l}(T).

To prove (ii), suppose T is bb-bounded and let

r′_{bb}(T) = inf\{ ν > 0 : (\frac{T^{n}}{ν^{n}} ) is uniformly bounded on every bounded set \}.

We’ll show that r′_{bb}(T) = r_{bb}(T). If (\frac{T^{n}}{ν^{n}} ) converges to zero uniformly on every bounded set, then for each bounded set A and for each zero neighborhood U there exists a positive integer N such that \frac{T^{n}}{ν^{n}} (A) ⊆ U whenever n ⩾ N. Also, since T is bb-bounded, for every n < N we have \frac{T^{n}}{ν^{n}} (A) ⊆ α_{n}U for some α_{n} > 0. Therefore, if α = max\{α_{1}, …, α_{N−1}, 1\}, then \frac{T^{n}}{ν^{n}} (A) ⊆ αU for every n, so that the sequence \frac{T^{n}}{ν^{n}} is uniformly bounded on A. Thus ν ⩾ r′_{bb}(T), so that r′_{bb}(T) ⩽ r_{bb}(T).

Now suppose ν > r′_{bb}(T). There exists μ such that ν > μ > r′_{bb}(T). The set ⋃_{n=1}^{∞} \frac{T^{n}}{μ^{n}} (A) is bounded for every bounded set A, so that for every zero neighborhood U there exists a scalar α such that \frac{T^{n}}{μ^{n}} (A) ⊆ αU for every n ∈ ℕ. Then \frac{T^{n}}{ν^{n}} (A) ⊆ \frac{μ^{n}α}{ν^{n}}U ⊆ U for all sufficiently large n. This means that the sequence (\frac{T^{n}}{ν^{n}} ) converges to zero uniformly on A, and it follows that ν ⩾ r_{bb}(T).

Further, if T is bb-bounded, then any finite initial segment (\frac{T^{n}}{ν^{n}})_{n=0}^{N} is always uniformly bounded on every bounded set, so that a tail (\frac{T^{n}}{ν^{n}})_{n=N}^{∞} is uniformly bounded on every bounded set if and only if the whole sequence (\frac{T^{n}}{ν^{n}})_{n=0}^{∞} is uniformly bounded on every bounded set.

The statements (iii), (iv), and (v) can be proved in a similar way. □

4.5. Locally bounded spaces. If T is a linear operator on a locally bounded topological vector space (X, U), then it follows directly from Definition 4.2 that r_{bb}(T) = r_{c}(T) = r_{nn}(T) = r_{nb}(T), because the corresponding convergences are equivalent. In this case we would denote each of these radii by r_{U}(T).

Spectral radii via seminorms. The following proposition provides formulas for computing spectral radii of an operator on a locally convex space in terms of seminorms.

Proposition 4.6. If T is an operator on a locally convex space X with a generating family of seminorms P, then

(i) r_{l}(T) = sup_{p∈P, x∈X} lim sup_{n→∞} \sqrt[n]{p(T^{n}x)};

(ii) r_{bb}(T) = sup_{p∈B, q∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})}, where B is the collection of Minkowski func-

tionals of all balanced convex bounded sets in X; (iii) r_{c}(T) = sup_{q∈P} inf_{p∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})};

(iv) r_{nn}(T) = inf_{Q} sup_{p∈Q} lim sup_{n→∞} \sqrt[n]{p(T^{n})}, where the infimum is taken over all generating

families of seminorms;

(v) r_{nb}(T) = inf_{p∈P} sup_{q∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})};

Proof. It follows from the definition of r_{l}(T) and Lemma 4.1 that

r_{l}(T) = inf\{ ν > 0 : lim_{n→∞} p (\frac{T^{n}x}{ν^{n}}) = 0 for every x ∈ X, p ∈ P \}

= sup_{x∈X, p∈P} inf\{ ν > 0 : lim_{n→∞} \frac{p(T^{n}x)}{ν^{n}} = 0 \} = sup_{x∈X, p∈P} lim sup_{n→∞} \sqrt[n]{p(T^{n}x)}.

Similarly, since the balanced convex hull of a bounded set is bounded,

r_{bb}(T) = inf\{ ν > 0 : ∀ bounded A ∀V ∈ N_{0} ∃N ∈ ℕ ∀n ⩾ N \frac{T^{n}}{ν^{n}} (A) ⊆ V \}

= inf\{ ν > 0 : ∀p ∈ B ∀q ∈ P ∃N ∈ ℕ ∀n ⩾ N 𝔪_{pq}(\frac{T^{n}}{ν^{n}} ) ⩽ 1 \}

= sup_{p∈B, q∈P} inf\{ ν > 0 : lim_{n→∞} \frac{𝔪_{pq}(T^{n})}{ν^{n}} ⩽ 1 \} = sup_{p∈B, q∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})}.

Let U_{p} = { x ∈ X : p(x) < 1 } for every p ∈ P. Then, rephrasing the definition of r_{c}(T) and applying Lemma 4.1, we have

r_{c}(T) = inf\{ ν > 0 : ∀q ∈ P ∃p ∈ P ∀ε > 0 ∃N ∈ ℕ ∀n ⩾ N \frac{T^{n}}{ν^{n}} (U_{p}) ⊆ εU_{q} \}

= sup_{q∈P} inf_{p∈P} inf\{ ν > 0 : ∀ε > 0 ∃N ∈ ℕ ∀n ⩾ N 𝔪_{pq}(\frac{T^{n}}{ν^{n}} ) < ε \}

= sup_{q∈P} inf_{p∈P} inf\{ ν > 0 : lim_{n→∞} \frac{𝔪_{pq}(T^{n})}{ν^{n}} = 0 \} = sup_{q∈P} inf_{p∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})}.

Similarly,

r_{nn}(T) = inf\{ ν > 0 : ∃Q ∀p ∈ Q ∀ε > 0 ∃N ∈ ℕ ∀n > N \frac{T^{n}}{ν^{n}} (U_{p}) ⊆ εU_{p} \}

= inf_{Q} sup_{p∈Q} inf\{ ν > 0 : lim_{n→∞} \frac{p(T^{n})}{ν^{n}} = 0 \} = inf_{Q} sup_{p∈Q} lim sup_{n→∞} \sqrt[n]{p(T^{n})}.

Finally,

r_{nb}(T) = inf\{ ν > 0 : ∃p ∈ P ∀q ∈ P ∃N ∈ ℕ ∀n > N \frac{T^{n}}{ν^{n}} (U_{p}) ⊆ U_{q} \}

= inf_{p∈P} sup_{q∈P} inf\{ ν > 0 : lim sup_{n→∞} \frac{𝔪_{pq}(T^{n})}{ν^{n}} ⩽ 1 \} = inf_{p∈P} sup_{q∈P} lim sup_{n→∞} \sqrt[n]{𝔪_{pq}(T^{n})}.

□

Some special properties of r_{c}(T). Continuity of an operator can be characterized in terms of neighborhoods (the pre-image of every neighborhood contains a neighborhood) or, alternatively, in terms of convergence (every convergent net is mapped to a convergent net). Analogously, though defined in terms of neighborhoods, r_{c}(T) can also be characterized in terms of convergent nets. This approach was used by F. Garibay and R. Vera in a series of papers [8, 9, 20]. Recall that a net (x_{α}) in a topological vector space is said to be ultimately bounded if every zero neighborhood absorbs some tail of the net, i.e., for every zero neighborhood V one can find an index α_{0} and a positive real δ > 0 such that x_{α} ∈ δV whenever α > α_{0}. As far as we know, ultimately bounded sequences were first studied in [5] for certain locally-convex topologies. The relationship between ultimately bounded nets and convergence of sequences of operators on locally convex spaces was studied in [8, 9, 20]. The following proposition (which is, in fact, a version of [20, Corollary 2.14]) shows how r_{c}(T) can be characterized in terms of the action of powers of T on ultimately bounded sequences. It also implies that r_{c}(T) coincides with the number γ(T) which was introduced in [8, 9, 20] for a continuous operator on locally convex spaces.

Proposition 4.7. Let T be a linear operator on a topological vector space X, then

r_{c}(T) = inf\{ν > 0 : lim_{n,α} \frac{T^{n}}{ν^{n}} x_{α} = 0 whenever (x_{α}) is ultimately bounded \}

= inf\{ν > 0 : (\frac{T^{n}}{ν^{n}} x_{α})_{n,α} is ultimately bounded whenever (x_{α}) is ultimately bounded \}.

Proof. To prove the first equality it suffices to show that r_{c}(T) < 1 if and only if lim_{n,α} T^{n}x_{α} = 0 whenever (x_{α}) is an ultimately bounded net. Suppose that r_{c}(T) < 1,

and let V be a zero neighborhood. One can find a zero neighborhood U such that for every ε > 0 there exists n_{0} ∈ ℕ such that T^{n}(U) ⊆ εV for each n > n_{0}. Let (x_{α}) be an ultimately bounded net. There exists an index α_{0} and a number δ > 0 such that x_{α} ∈ δU whenever α > α_{0}. Then for ε = δ^{−1} one can find n_{0} such that T^{n}(U) ⊆ δ^{−1}V for each n > n_{0}, so that T^{n}x_{α} ∈ δT^{n}(U) ⊆ V whenever α > α_{0} and n > n_{0}. This means that

lim_{n,α} T^{n}x_{α} = 0.

Conversely, suppose that lim_{n,α} T^{n}x_{α} = 0 for each ultimately bounded net (x_{α}), and assume that T^{n} does not converge equicontinuously to zero. Then there exists a zero neighborhood V such that for every zero neighborhood U one can find ε_{U} such that for every m ∈ ℕ there exists n_{U,m} > m with T^{n_{U,m}}(U) ⊈ ε_{U}V. Then there exists x_{U,m} ∈ U such that

T^{n_{U,m}}x_{U,m} ∉ ε_{U}V.  (1)

The collection of all zero neighborhoods ordered by inclusion is a directed set, so that (x_{U,n}) is an ultimately bounded net. Indeed, if W is a zero neighborhood then x_{U,n} ∈ W for each zero neighborhood U ⊆ W and every n ∈ ℕ. But it follows from (1) that the net (T^{n}x_{U,m})_{n,m,U} does not converge to zero.

To prove the second equality, let

γ_{1} = inf\{ν > 0 : lim_{n,α} \frac{T^{n}}{ν^{n}} x_{α} = 0 whenever (x_{α}) is ultimately bounded \} and
γ_{2} = inf\{ν > 0 : (\frac{T^{n}}{ν^{n}} x_{α})_{n,α} is ultimately bounded if (x_{α}) is ultimately bounded \}.

Since every net which converges to zero is necessarily ultimately bounded, it follows that γ_{1} ⩾ γ_{2}. Now let ν > γ_{2}, and let (x_{α}) be an ultimately bounded sequence. Suppose that γ_{2} < μ < ν. Then (\frac{T^{n}}{μ^{n}} x_{α})_{n,α} is ultimately bounded; that is, for each zero neighborhood V there exists an indices α_{0} and n_{0} and a positive ε such that \frac{T^{n}}{μ^{n}} x_{α} ∈ εV whenever α > α_{0} and n > n_{0}. It follows that \frac{T^{n}}{ν^{n}} x_{α} ∈ \frac{μ^{n}ε}{ν^{n}}V ⊆ V for α > α_{0} and all sufficiently large n. This implies that lim_{n,α} \frac{T^{n}}{ν^{n}} x_{α} = 0 so that ν ⩾ γ_{1}. □

Question. Are there similar ways for computing r_{l}(T), r_{bb}(T), r_{nn}(T), and r_{nb}(T) in terms of nets?

Proposition 4.7 enables us to prove some important properties of r_{c}. The following lemma is analogous to Lemma 3.13 of [20].

Lemma 4.8. If S and T are two commuting linear operators on a topological vector space X such that r_{c}(S) and r_{c}(T) are finite, then r_{c}(ST) ⩽ r_{c}(S)r_{c}(T).

Proof. Suppose μ > r_{c}(S) and ν > r_{c}(T) and let (x_{α}) be an ultimately bounded net in X. Then the net (\frac{T^{n}x_{α}}{ν^{n}})_{n,α} is ultimately bounded by Proposition 4.7. By applying Proposition 4.7 again we conclude that (\frac{S^{m}T^{n}x_{α}}{μ^{m}ν^{n}})_{m,n,α} converges to zero. In particular, lim_{n,α} \frac{(ST)^{n}x_{α}}{(μν)^{n}} = lim_{n,α} \frac{S^{n}T^{n}x_{α}}{μ^{n}ν^{n}} = 0, and applying Proposition 4.7 one more time we get μν > r_{c}(ST). □

Theorem 4.9. If S and T are two commuting continuous operators on a locally convex space X then r_{c}(S + T) ⩽ r_{c}(S) + r_{c}(T).

Proof. Assume without loss of generality that both r_{c}(S) and r_{c}(T) are finite. Suppose that η > r_{c}(S) + r_{c}(T) and take μ > r_{c}(S) and ν > r_{c}(T) such that η > μ + ν. Let (x_{α}) be an ultimately bounded net in X. By Proposition 4.7 it suffices to show that lim_{n,α} \frac{1}{η^{n}} (S + T)^{n}x_{α} = 0. Notice that the net (\frac{T^{n}}{ν^{n}} x_{α})_{n,α} is ultimately bounded. This implies that the net (\frac{S^{m}}{μ^{m}} \frac{T^{n}}{ν^{n}} x_{α})_{m,n,α} converges to zero. Fix a seminorm p, then there exist indices n_{0} and α_{0} such that p(S^{m}T^{n}x_{α}) < μ^{m}ν^{n} whenever m, n ⩾ n_{0} and α ⩾ α_{0}. Also, notice that we can split η into a product of two terms η = η_{1}η_{2} such that η_{1} > 1 while still η_{2} > μ + ν. Further, if n > 2n_{0} and α ⩾ α_{0} then we have

p (\frac{1}{η^{n}} (S + T)^{n}x_{α}) ⩽

\frac{1}{η^{n}} ∑_{k=0}^{n_{0}} \binom{n}{k}p(S^{k}T^{n−k}x_{α}) + \frac{1}{η^{n}} ∑_{k=n_{0}+1}^{n−n_{0}} \binom{n}{k}p(S^{k}T^{n−k}x_{α}) + \frac{1}{η^{n}} ∑_{k=n−n_{0}+1}^{n}\binom{n}{k}p(S^{k}T^{n−k}x_{α}).

Since \binom{n}{k} = \frac{(n−k+1)···(n−1)·n}{1·2···(k−1)·k} ⩽ n^{k} and ∑_{k=0}^{n} \binom{n}{k}μ^{k}ν^{n−k} = (μ + ν)^{n}, we have

p (\frac{1}{η^{n}} (S + T)^{n}x_{α}) ⩽

\frac{n^{n_{0}}}{η^{n}} ∑_{k=0}^{n_{0}} p(S^{k}T^{n−k}x_{α}) + \frac{1}{η^{n}} ∑_{k=n_{0}+1}^{n−n_{0}} \binom{n}{k}μ^{k}ν^{n−k} + \frac{n^{n_{0}}}{η^{n}} ∑_{k=n−n_{0}+1}^{n} p(S^{k}T^{n−k}x_{α})

⩽ \frac{n^{n_{0}}}{η_{1}^{n}} · \frac{1}{η_{2}^{n}} ∑_{k=0}^{n_{0}}(p(T^{n−k}S^{k}x_{α}) + p(S^{n−k}T^{k}x_{α})) + \frac{(μ + ν)^{n}}{η^{n}} .

Notice that lim_{n} \frac{(μ+ν)^{n}}{η^{n}} = 0 and that lim_{n} \frac{n^{n_{0}}}{η_{1}^{n}} = 0. Since T is continuous, the net (T^{k}x_{α})_{α} is ultimately bounded for every fixed k, so that lim_{n,α} \frac{1}{η_{2}^{n−k}} S^{n−k}T^{k}x_{α} = 0. It follows that

for every k between 0 and n_{0} the expression \frac{1}{η_{2}^{n}} p(S^{n−k}T^{k}x_{α}) is uniformly bounded for all sufficiently large n and α. Similarly, for every k between 0 and n_{0} the expression \frac{1}{η_{2}^{n}} p(T^{n−k}S^{k}x_{α}) is uniformly bounded for all sufficiently large n and α. Therefore there exist indices n_{1} and α_{1} such that the finite sum

\frac{1}{η_{2}^{n}} ∑_{k=0}^{n_{0}}(p(T^{n−k}S^{k}x_{α}) + p(S^{n−k}T^{k}x_{α}))

is uniformly bounded for all n ⩾ n_{1} and α ⩾ α_{1}. It follows that lim_{n,α} p(\frac{1}{η^{n}} (S +T)^{n}x_{α}) = 0,

so that η > r_{c}(S + T). □

Corollary 4.10. If T is a continuous operator on a locally convex space with r_{c}(T) < ∞ then r_{c}(P (T)) < ∞ for every polynomial P (z).

Definition 4.11. We say that a sequence (x_{n}) in a topological vector space is fast null if lim_{n→∞} α^{n}x_{n} = 0 for every positive real α.

Lemma 4.12. If T is a linear operator on a topological vector space with r_{c}(T) < ∞ then (T^{n}x_{n}) is fast null whenever (x_{n}) is fast null.

Proof. Suppose (x_{n}) is a fast null sequence in a topological vector space and r_{c}(T) < ∞. Let ν > r_{c}(T). The sequence ν^{n}α^{n}x_{n} converges to zero, hence is ultimately bounded. Thus by Proposition 4.7 we have

lim_{n→∞} α^{n}T^{n}x_{n} = lim_{n→∞} \frac{T^{n}}{ν^{n}} ν^{n}α^{n}x_{n} = 0.

□

5. Spectra and spectral radii

It is well known that for a continuous operator T on a Banach space its spectral radius r(T) equals the geometric radius of the spectrum |σ(T)| = sup{|λ| : λ ∈ σ(T)}. Furthermore, whenever |λ| > r(T), the resolvent operator R_{λ} = (λI − T)^{−1} is given by the Neumann series ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} . We are going to show in the following five theorems that the spectral radii that we introduced are upper bounds for the actual radii of the correspondent spectra, and that when |λ| is greater than or equal to any of these spectral radii, then the Neumann series converges in the corresponding operator topology to the resolvent operator.

In the following Theorems 5.1–5.5 we assume that T is a linear operator on a sequentially complete locally convex space, λ is a complex number and R_{λ} is the resolvent of T at λ in the sense of Definition 3.1.

Theorem 5.1. If |λ| > r_{l}(T) then the Neumann series converges pointwise to a linear operator R_{λ}^{0}, and R_{λ}^{0}(λI − T) = I. Moreover, if T is continuous, then R_{λ}^{0} = R_{λ} and |σ^{l}(T)| ⩽ r_{l}(T).

Proof. For any λ ∈ ℂ with |λ| > r_{l}(T) one can find z ∈ ℂ such that 0 < |z| < 1 and λz > r_{l}(T). Consider a point x ∈ X and a basic neighborhood of zero U. Since by the definition of r_{l}(T) the sequence (\frac{T^{n}x}{(λz)^{n}} ) converges to zero, there exist a positive integer n_{0} such that \frac{T^{n}x}{(λz)^{n}} ∈ U whenever n ⩾ n_{0}. Therefore, \frac{T^{n}x}{λ^{n}} ∈ z^{n}U ⊆ |z|^{n}U because U is balanced. Thus, if n ⩾ m ⩾ n_{0}, then ∑_{i=n}^{m} \frac{T^{i}x}{λ^{i}} ∈ ∑_{i=n}^{m}|z|^{i}U ⊆ (∑_{i=n}^{m}|z|^{i})U because U is convex. Since |z| < 1, we have ∑_{i=n}^{m}|z|^{i} < 1 for sufficiently large m and n, and so ∑_{i=n}^{m} \frac{T^{i}x}{λ^{i}} ∈ U because U is balanced. Therefore R_{λ,n}x = \frac{1}{λ} ∑_{i=0}^{n} \frac{T^{i}x}{λ^{i}} is a Cauchy sequence and hence it converges to some R_{λ}^{0}x because X is sequentially complete.

Clearly, R_{λ}^{0} is a linear operator. Notice that R_{λ,n}(λx − T x) = x − \frac{T^{n+1}x}{λ^{n+1}} for every x. As n goes to infinity, the left hand side of this identity converges to R_{λ}^{0}(λx − T x), while the right hand side converges to x. Thus it follows that R_{λ}^{0}(λI − T) = I.

Finally, notice that R_{λ,n} commutes with T for every n. Therefore, if T is continuous, then

R_{λ}^{0}T x = lim_{n→∞} R_{λ,n}T x = lim_{n→∞} T R_{λ,n}x = T ( lim_{n→∞} R_{λ,n}x) = T R_{λ}^{0}x

for every x. This implies that (λI − T)R_{λ}^{0} = R_{λ}^{0}(λI − T) = I, so that R_{λ}^{0} is the (left and right) inverse of λI − T. This means that R_{λ}^{0} = R_{λ} and λ ∈ ρ^{l}(T). Thus, |σ^{l}(T)| ⩽ r_{l}(T). □

Theorem 5.2. If T is bb-bounded and |λ| > r_{bb}(T), then the Neumann series converges uniformly on bounded sets, and its sum R_{λ}^{0} is bb-bounded. Moreover, if T is continuous, then R_{λ}^{0} = R_{λ} and |σ^{bb}(T)| ⩽ r_{bb}(T).

Proof. Suppose that |λ| > r_{bb}(T). The sum R_{λ}^{0} of the Neumann series exists by Theorem 5.1. As in the proof of Theorem 5.1 we denote the partial sums of the Neumann series by R_{λ,n}. Fix z ∈ ℂ such that 0 < |z| < 1 and λz > r_{bb}(T), and consider a bounded set A. Let U be a basic neighborhood of zero, without loss of generality U we assume that U is closed. Since \frac{T^{n}}{(λz)^{n}} converges to zero uniformly on A, there exits n_{0} ∈ ℕ such that \frac{T^{n}}{λ^{n}z^{n}} (A) ⊆ U for all n > n_{0}. Also, since |z| < 1, we can assume without loss of generality that ∑_{i=n_{0}}^{∞}|z|^{i} < |λ|. Then

\frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ \frac{1}{λ}( ∑_{i=n+1}^{m}|z|^{i})U ⊆ U

whenever x ∈ A and m > n > n_{0}. Since U is closed, we have

R_{λ}^{0}x − R_{λ,n}x = lim_{m→∞} \frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ U

for each x ∈ A and n > n_{0}, so that (R_{λ}^{0} − R_{λ,n})(A) ⊆ U whenever n > n_{0}. This shows that R_{λ,n} converges to R_{λ}^{0} uniformly on bounded sets. By Lemma 2.17 this implies that R_{λ}^{0} is bb-bounded.

Furthermore, if T is continuous, then by Theorem 5.1 we have R_{λ} = R_{λ}^{0}, so that λ ∈ ρ^{bb}(T), whence it follows that |σ^{bb}(T)| ⩽ r_{bb}(T). □

The next theorem is similar to Theorem 2.18 of [20].

Theorem 5.3. If T is a continuous and |λ| > r_{c}(T), then the Neumann series converges equicontinuously to R_{λ} and R_{λ} is continuous. In particular, |σ^{c}(T)| ⩽ r_{c}(T) holds.

Proof. Let |λ| > r_{c}(T). It follows from Theorem 5.1 that the Neumann series converges to R_{λ}. Again, we denote the partial sums of the Neumann series by R_{λ,n}. Let z ∈ ℂ be such that 0 < |z| < 1 and λz > r_{c}(T). Let U be a zero neighborhood, without loss of generality we can assume that U is closed. There exists a zero neighborhood V such that \frac{T^{n}}{λ^{n}z^{n}} (V) ⊆ U for every n ⩾ 0. Let ε > 0. Then ∑_{i=n_{0}}^{∞}|z|^{i} < ε|λ| for some n_{0}. Thus

\frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ \frac{1}{λ}( ∑_{i=n+1}^{m}|z|^{i})εU ⊆ U

whenever x ∈ V and m > n > n_{0}. Since U is closed, we have

R_{λ}x − R_{λ,n}x = lim_{m→∞} \frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ εU

for each x ∈ V and n > n_{0}, so that (R_{λ} − R_{λ,n})(V) ⊆ εU whenever n > n_{0}. This shows that R_{λ,n} converges to R_{λ} equicontinuously, and Lemma 2.19 yields that R_{λ} is continuous. □

Theorem 5.4. If T is nn-bounded and |λ| > r_{nn}(T), then the Neumann series nnconverges to R_{λ} and R_{λ} is nn-bounded. In particular, |σ^{nn}(T)| ⩽ r_{nn}(T).

Proof. Let |λ| > r_{nn}(T). By Theorem 5.1 the Neumann series ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} converges to R_{λ}. Again, we denote the partial sums of the Neumann series by R_{λ,n}. Fix some z such that 0 < |z| < 1 and λz > r_{nn}(T). There exists a base N_{0} of closed convex zero neighborhoods such that for every U ∈ N_{0} there is a scalar β > 0 such that \frac{T^{n}}{(λz)^{n}} (U) ⊆ βU for all n ⩾ 0. Fix U ∈ N_{0}. Then for each n ⩾ 0 we have \frac{T^{n}}{λ^{n}z^{n}} (U) ⊆ βU for some β > 0, so that \frac{T^{n}x}{λ^{n}} ∈ |z|^{n}βU whenever x ∈ U. It follows that

R_{λ,n}x = \frac{1}{λ} ∑_{i=0}^{n} \frac{T^{i}x}{λ^{i}} ∈ \frac{β}{λ}(∑_{i=0}^{n}|z|^{i})U.

Then R_{λ}x ∈ \frac{β}{λ(1−|z|)}U, so that R_{λ}(U) ⊆ \frac{β}{λ(1−|z|)}U, which implies that R_{λ} is nn-bounded, and, therefore, |σ^{nn}(T)| ⩽ r_{nn}(T).

Fix ε > 0. Then ∑_{i=N}^{∞}|z|^{i} < |λ| for some N. Then for every U ∈ N_{0} we have

\frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ \frac{1}{λ}( ∑_{i=n+1}^{m}|z|^{i})εU ⊆ U

whenever x ∈ U and N < n < m. Since U is closed, we have

R_{λ}x − R_{λ,n}x = lim_{m→∞} \frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ εU

for each x ∈ U and n > N, so that (R_{λ} − R_{λ,n})(U) ⊆ εU whenever N < n. This shows that R_{λ,n} nn-converges to R_{λ}. □

Theorem 5.5. If T is nb-bounded and |λ| > r_{nb}(T), then the Neumann series converges to R_{λ} uniformly on a zero neighborhood. Further, |σ^{nb}(T)| ⩽ r_{nb}(T).

Proof. Let |λ| > r_{nb}(T). By Theorem 5.1 the Neumann series ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} converges to R_{λ}. Since r_{bb}(T) ⩽ r_{nb}(T) then R_{λ} is bb-bounded by Theorem 5.2. But then ∑_{i=0}^{∞} \frac{T^{i}}{λ^{i+1}} = \frac{1}{λ}I + \frac{1}{λ}R_{λ}T. Notice that R_{λ}T is nb-bounded as a product of a bb-bounded and an nbbounded operator (see 2.6).

Suppose that |λ| > r_{nb}(T). Fix z ∈ ℂ such that 0 < |z| < 1 and λz > r_{nb}(T). Then the sequence (\frac{T^{n}}{λ^{n}z^{n}} ) converges to zero uniformly on U, where U is a basic neighborhood of zero. We will show that the Neumann series converges uniformly on U. As in the proof of Theorem 5.1, we denote the partial sums of the Neumann series by R_{λ,n}. Let V be a basic neighborhood of zero, without loss of generalityt we assume that V is closed. Since (\frac{T^{n}}{λ^{n}z^{n}} ) converges to zero uniformly on U, there exits n_{0} ∈ ℕ such that \frac{T^{n}}{λ^{n}z^{n}} (U) ⊆ V for all n > n_{0}. Also, since |z| < 1, we can assume without loss of generality that ∑_{i=n_{0}}^{∞}|z|^{i} < |λ|. Then

\frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ \frac{1}{λ}( ∑_{i=n+1}^{m}|z|^{i})V ⊆ V

whenever x ∈ A and m > n > n_{0}. Since V is closed, we have

R_{λ}x − R_{λ,n}x = lim_{m→∞} \frac{1}{λ} ∑_{i=n+1}^{m} \frac{T^{i}x}{λ^{i}} ∈ V

for each x ∈ U and n > n_{0}, so that (R_{λ} − R_{λ,n})(U) ⊆ V whenever n > n_{0}. □

In rest of this section we present some remarks on Theorems 5.1–5.5. In particular, we discuss the necessity of sequential completeness and local convexity, and consider several examples and special cases.

5.6. It is easy to see that each spectral radius is exactly the radius of convergence of the Neumann series in the corresponding operator convergence. Indeed, in each of Theorems 5.1–5.5 the convergence of the Neumann series implies that the terms of the series tend to zero. It follows that |λ| is greater than or equal to the corresponding spectral radius.

5.7. Sequential completeness. Recall that if X is a Banach space, then the norm topology on X and the weak^{∗} topology on X^{∗} are sequentially complete. The weak topology of X is sequentially complete when X is reflexive. Also, it is known that the weak topologies of ℓ_{1} and of L_{1}[0, 1] are sequentially complete. Since all these topologies are locally convex, Theorems 5.1–5.5 are applicable to each of them.

It can be easily verified that C([0, 1]) with the topology of pointwise convergence is not sequentially complete (consider x_{n}(t) = t^{n}). Similarly, the sequence spaces ℓ_{p} for 0 < p ⩽ ∞, c, c_{0}, and c_{00} (the space of eventually vanishing sequences) are not sequentially complete in the topology of coordinate-wise convergence: the sequence x_{n} = ∑_{i=1}^{n} ie_{i} serves as a counterexample.

Example 5.8. Theorems 5.1–5.5 fail without sequential completeness. Consider the space c_{0} with the topology of coordinate-wise convergence. Let T be the forward shift operator on c_{0}, that is, T e_{k} = e_{k+1}, where e_{k} is the k-th unit vector of c_{0}. Let V be a zero neighborhood, we can assume without loss of generality that V = {x ∈ c_{0} : |x_{i_{1}}| < 1, …, |x_{i_{k}}| < 1} where i_{1} < i_{2} < ⋯ < i_{k} are positive integers. If x ∈ c_{0} then T^{n}x has zero components 1 through n. In particular, for every positive ν we have \frac{T^{n}x}{ν^{n}} ∈ V whenever n > i_{k}. Therefore (\frac{T^{n}}{ν^{n}} ) converges uniformly on c_{0} for every ν > 0, so that r_{nb}(T) = 0. It follows from Proposition 4.3 that r_{l}(T) = r_{bb}(T) = r_{c}(T) = r_{nn}(T) = 0. On the other hand, ∑_{n=1}^{∞} T^{n}e_{1} diverges in c_{0}. Since T is obviously continuous, this shows that Theorems 5.1–5.5, do not hold in c_{0}. Thus, sequential completeness condition is essential in the theorems.

5.9. Monotone convergence property. Notice that if T is a positive operator on a locally convex-solid vector lattice (i.e., a locally convex space which is also a vector lattice such that |x| ⩽ |y| implies p(x) ⩽ p(y) for every generating seminorm p) then we can substitute the sequential completeness condition in Theorems 5.1–5.5 by a weaker condition called the sequential monotone completeness property: a locally convex-solid vector lattice is said to satisfy the sequential monotone completeness property if every monotone Cauchy sequence converges in the topology of X. For details, see [3]. Indeed, we used the sequential completeness at just one point — we used it in the proof of Theorem 5.1 to claim that since R_{λ,n}x = \frac{1}{λ} ∑_{i=0}^{n} \frac{T^{i}x}{λ^{i}} is a Cauchy sequence, then it converges to some R_{λ}x. But if T is positive, then R_{λ,n}x^{+} and R_{λ,n}x^{−} are increasing sequences, and the sequential monotone completeness property ensures the convergence.

5.10. Banach spaces. If T is a (norm) continuous operator on a Banach space, then it follows from 3.2 and 4.5 that σ^{l}(T) = σ^{bb}(T) = σ^{c}(T) = σ^{nn}(T) = σ^{nb}(T) = σ(T) and r_{bb}(T) = r_{c}(T) = r_{nn}(T) = r_{nb}(T) = r(T), where σ(T) and r(T) are the usual spectrum and the spectral radius of T. Further, it follows from Lemma 4.3 that r_{l}(T) ⩽ r(T). On the other hand, since r(T) = |σ(T)|, then r(T) ⩽ r_{l}(T) by Theorem 5.1, so that r_{l}(T) = r(T).

5.11. The following argument is a counterpart to 3.5. Let T be a (norm) continuous operator on a Banach space X, r(T) the usual spectral radius of T. Further, let r_{l}(T) and r_{bb}(T) be computed with respect to the weak topology of X. We claim that if the weak topology of X is sequentially complete, then r_{l}(T) = r_{bb}(T) = r(T). Indeed, r(T) ⩽ r_{l}(T) by 3.5 and Theorem 5.1 because σ(T) = σ^{l}(T). In view of Proposition 4.3 it suffices to show that r_{bb}(T) ⩽ r(T). Let ν > r(T), and let A be a weakly bounded subset of X. Then A is norm bounded, so that the sequence \frac{T^{n}}{ν^{n}} converges to zero uniformly on A in the norm topology. In particular, the set ⋃_{n=0}^{∞} \frac{T^{n}}{ν^{n}} (A) is norm bounded, hence weakly bounded, so that ν > r_{bb}(T).

Quasinilpotence. Recall that a norm continuous operator T on a Banach space X is said to be quasinilpotent if r(T) = 0 or, equivalently, if σ(T) = {0}. Quasinilpotent operators on Banach spaces have some nice properties. Therefore in the framework of topological vector spaces it is interesting to study operators having some of their spectra trivial or some of their spectral radii being zero. Notice, for example, that it follows from Proposition 4.6 that if T is an operator on a locally convex topological vector space, then r_{l}(T) = 0 if and only if lim_{n} \sqrt[n]{p(T^{n}x)} = 0 for every seminorm p in a generating family of seminorms and for every x ∈ X. Further, if the space is in addition sequentially complete, then for such an operator we would have σ^{l}(T) = {0} by Theorem 5.1.

Recall also that a norm continuous operator T on a Banach space X is said to be locally quasinilpotent at a point x ∈ X if lim_{n} \sqrt[n]{∥T^{n}x∥} = 0. Using Lemma 4.1, the concept of local quasinilpotence can be naturally generalized to topological vector spaces: an operator T on a topological vector space X is said to be locally quasinilpotent at a point x ∈ X if lim_{n} \frac{T^{n}x}{ν^{n}} = 0 for every ν > 0. It follows immediately from the definition of r_{l}(T) that r_{l}(T) = 0 if and only if T is locally quasinilpotent at every x ∈ X. It is known that a continuous operator on a Banach space is quasinilpotent if and only if it is locally quasinilpotent at every point. We see now that this is just a corollary of 5.10. The following example shows that a similar result for general topological vector spaces is not valid, that is, r_{l}(T) may be equal to zero without the other radii be equal to zero.

Example 5.12. A continuous operator with r_{l}(T) = 0 but r_{bb}(T) = r_{c}(T) = r_{nn}(T) = r_{nb}(T) = ∞. Consider the space of all bounded real sequences ℓ_{∞} = {x = (x_{1}, x_{2}, …) : sup |x_{k}| < ∞} with the topology of coordinate-wise convergence. This topology can be generated by the family of coordinate seminorms {p_{m}}_{m=1}^{∞} where p_{m}(x) = |x_{m}|. Let e_{k} denote the k-th unit vector in ℓ_{∞}.

Define an operator T: ℓ_{∞} → ℓ_{∞} via T: ∑_{k=1}^{∞} a_{k}e_{k} ↦ ∑_{k=2}^{∞} \frac{(k−1)^{k−1}}{k^{k}} a_{k}e_{k−1}. Then T^{n}e_{k} = \frac{(k−n)^{k−n}}{k^{k}} e_{k−n} if n < k and zero otherwise. Clearly T is continuous. In order to show that r_{l}(T) = 0 fix a positive real number ν and x ∈ ℓ_{∞}. Then

|(\frac{T^{n}x}{ν^{n}})_{m}| = |\frac{m^{m}}{(m+n)^{m+n}ν^{n}} x_{n+m}| ⩽ sup_{n} \frac{m^{m}}{(m+n)^{m+n}ν^{n}} · sup_{n}|x_{n}| < ∞

It follows from Lemma 4.4(i) that r_{l}(T) = 0.

Now we show that r_{bb}(T) = ∞ by presenting a bounded set A in ℓ_{∞} such that the sequence (\frac{T^{n}}{ν^{n}} ) in not uniformly bounded on A for every positive ν. Let

A = { x ∈ ℓ_{∞} : x_{n} ⩽ (2n)^{2n} for all n ⩾ 0 }.

Then (2n)^{2n}e_{n} ∈ A for each n > 0 and (\frac{T^{n−1}}{ν^{n−1}} (2n)^{2n}e_{n})_{1} = \frac{(2n)^{2n}}{n^{n}ν^{n}} is unbounded. Then by Lemma 4.4(ii) we have r_{bb}(T) = ∞, and it follows from Proposition 4.3 that r_{c}(T) = r_{nn}(T) = r_{nb}(T) = ∞.

It is not difficult to show that σ^{l}(T) = {0}, while σ^{c}(T) = σ^{nn}(T) = σ^{nb}(T) = ℂ.

Non-locally convex spaces. We proved the key Theorems 5.1–5.5 for locally convex spaces, but they are still valid for locally pseudo-convex spaces. The local convexity of X was used only once in the proof of Theorem 5.1, while Theorems 5.2–5.5 used Theorem 5.1. Hence it would suffice to modify the proof of Theorem 5.1 in such a way that it would work for locally pseudo-convex spaces instead of locally convex. Local convexity was used in the proof of Theorem 5.1 to show that if \frac{T^{n}x}{(λz)^{n}} ∈ U for all n > n_{0} and some n_{0} ∈ ℕ, then there exists m_{0} ∈ ℕ such that ∑_{i=n}^{m} \frac{T^{i}x}{λ^{i}} ∈ U for all m, n > m_{0}. (Recall that T is a linear operator, λ, z ∈ ℂ such that 0 < |z| < 1 and λz > r_{l}(T), x ∈ X, and U is a basic neighborhood of zero.) If X is locally pseudo-convex, then we can assume that U + U ⊆ αU for some α > 0, so that (X, U) is a locally bounded space. Let ∥·∥ be the Minkowski functional of U, then (see [14, pages 3 and 6]) for any x_{1}, x_{2}, …, x_{n} in X we have

∥x_{1} + ... + x_{k}∥ ⩽ 4^{\frac{1}{p}} (∥x_{1}∥^{p} + ⋯ + ∥x_{k}∥^{p})^{\frac{1}{p}} ,

where 2^{\frac{1}{p}} = α. Notice that ∥\frac{T^{n}x}{λ^{n}}∥ ⩽ |z|^{n} for all n > n_{0}. Since |z| < 1, then there exists m_{0} such that ∑_{i=n}^{m}|z|^{ip} < \frac{1}{4} whenever n, m > m_{0}. But then

∥∑_{i=n}^{m} \frac{T^{i}x}{λ^{i}} ∥ ⩽ 4^{\frac{1}{p}} (∑_{i=n}^{m}∥\frac{T^{i}x}{λ^{i}}∥^{p})^{\frac{1}{p}} ⩽ 4^{\frac{1}{p}} (∑_{i=n}^{m}|z|^{ip})^{\frac{1}{p}} < 1,

so that ∑_{i=n}^{m} \frac{T^{i}x}{λ^{i}} ∈ U.

The following example shows that Theorems 5.1-5.5 fail if we assume no convexity conditions at all.

Example 5.13. An operator on a complete non locally pseudo-convex space, whose spectral radii are 1, and whose Neumann series nevertheless diverges at λ = 2. Let X be the space of all measurable functions on [0, 1] with the topology of convergence in measure (which is not pseudo-convex). We identify the endpoints 0 and 1 and consider the interval as a circle. Fix an irrational α and define a linear operator T on X as the translation by α, i.e., (T f)(t) = f(t − α). It is easy to see that \frac{T^{n}f}{ν^{n}} converges in measure to zero for every f ∈ X if and only if ν > 1. We conclude, therefore, that r_{l}(T) = 1. Moreover, since the sets of the form W_{ε,δ} = {f ∈ X : μ(f > ε) < δ} form a base of zero neighborhoods for the topology of convergence in measure, and T (W_{ε,δ}) ⊆ W_{ε,δ}, it follows that r_{nn}(T) ⩽ 1. Then by Proposition 4.3 we have r_{l}(T) = r_{bb}(T) = r_{c}(T) = r_{nn}(T) = 1. Nevertheless, we are going to construct a function h ∈ X such that the Neumann series ∑_{n=0}^{∞} \frac{T^{n}h}{2^{n}} does not converge in measure, which means that the conclusions of Theorems 5.1–5.5 do not hold for this space

For each n = 1, 2, 3, … one can find a positive integer M_{n} such that the intervals [kα, kα + \frac{1}{n}] (mod 1) for k = 1, …, M_{n} cover the circle. Let s_{n} = ∑_{i=1}^{n} M_{i}, and let h be the step function taking value 2^{s_{n}} on the interval (\frac{1}{n+1}, \frac{1}{n}]. If s_{n−1} < k ⩽ s_{n} for some positive integers n and k, then on [0, \frac{1}{n}] we have h ⩾ 2^{s_{n}} ⩾ 2^{k}, so that \frac{h}{2^{k}} ⩾ 1 on [0, \frac{1}{n}], and it follows that \frac{T^{k}h}{2^{k}} ⩾ 1 on [kα, kα + \frac{1}{n}].

Now, given any positive integer N, we have N ⩽ s_{n−1} for some n. Then for each k = s_{n−1} + 1, …, s_{n} we have \frac{T^{k}h}{2^{k}} ⩾ 1 on the interval [kα, kα + \frac{1}{n}]. It follows that

∑_{k=s_{n−1}+1}^{s_{n}} \frac{T^{k}h}{2^{k}} ⩾ 1 on the set ⋃_{k=s_{n−1}+1}^{s_{n}}[kα, kα + \frac{1}{n}] = s_{n−1}α + ⋃_{k=1}^{M_{n}}[kα, kα + \frac{1}{n}] = [0, 1],

so that the series ∑_{n=0}^{∞} \frac{T^{n}h}{2^{n}} does not converge in measure.

Other approaches to spectral theory. There are different approaches to a spectral theory of operators on topological vector spaces, e.g., [21] and [4]. For example, Allan [4] defines the spectrum of an element x in a locally-convex algebra B as the set of all λ ∈ ℂ such that λe − x is not invertible or the inverse is not bounded, where y ∈ B is said to be bounded if {c^{n}y^{n}}_{n=1}^{∞} is a bounded set for some real c > 0. In our terms, this means that R_{λ} exists and has finite spectral radius. Allan’s spectrum is, therefore, bigger than ours. Allan defines the radius of boundedness of β(x), which in our terms is exactly the spectral radius, and he shows that β(x) is less than or equal to the geometric radius of his spectrum. This result nicely complements our Theorems 5.1–5.5 where we showed that a spectral radius is greater than or equal to the geometric radius of the corresponding spectrum.

For example, if X is a locally-convex space then it can be easily verified that the collection of all continuous operators on X equipped with the topology of uniform convergence on bounded sets is a locally convex algebra. For a base of convex neighborhoods of zero in this algebra one can take the sets V_{A,U} of all continuous operators T such that T (A) ⊆ U, where A ⊆ X is bounded and U is a convex basic neighborhood of zero neighborhood in X. Therefore, the result of Allan is applicable in the setup of Theorem 5.2.

6. nb-bounded operators

Since nb-boundedness is the strongest of the boundedness conditions we have introduced, it is natural to expect that stronger results can be obtained for nb-bounded operators.

6.1. The following argument is often useful when dealing with nb-bounded operators. Suppose that X and Y are topological vector spaces and T: X → Y is nb-bounded. Then T (U) is a bounded set in Y for some basic neighborhood of zero U. We claim that if Y is Hausdorff, then ⋂_{n=1}^{∞} \frac{1}{n}U ⊆ Null T. Indeed, it suffices to show that if x ∈ \frac{1}{n}U for every n ⩾ 1 then T x belongs to every zero neighborhood V of Y. But T (U) ⊆ αV for some positive α (depending on V), and hence T x ∈ \frac{1}{n}T (U) ⊆ \frac{α}{n}V ⊆ V whenever n ⩾ α.

It follows that if T is one-to-one, then U cannot contain any nontrivial linear subspaces. In particular, if U is convex then the locally bounded space (X, U) is Hausdorff, hence quasinormable. In this case T is a continuous operator from (X, U) to Y, and, moreover, if X = Y, then T is continuous as an operator from (X, U) to (X, U).

In fact, many “classical” topological vector spaces have the property that every zero neighborhood contains a nontrivial linear subspace, e.g., topologies of pointwise or coordinate-wise convergence, weak topologies. On the other hand, there exist spaces with a base of zero neighborhoods which consists of neighborhoods not containing any nontrivial linear subspaces. For example, let X be the space of all analytic functions on ℂ equipped with the topology of uniform convergence on compact subsets of ℂ. The sets U_{n,ε} = {f ∈ X : |f(z)| < ε whenever |z| ⩽ n}, (n ⩾ 0 and ε > 0) form a base of zero neighborhoods of this topology. It is easy to see that no U_{n,ε} contains a non-trivial linear subspace. Note that this topology is generated by the countable sequence of norms ∥f∥_{n} = sup_{|z|⩽n}|f(z)|; clearly ∥·∥_{n} is the Minkowski functional of U_{n,1}.

Theorem 6.2. If X is a complete locally convex space then X is locally bounded if and only if X admits an nb-bounded bijection.

Proof. If X is locally bounded then the identity map is an nb-bounded bijection. Suppose that T is an nb-bounded bijection on X. Then there exists a basic neighborhood of zero U such that T (U) is bounded. Without loss of generality we may assume that U is closed. Let A = \overline{T (U)}. Then A is convex, bounded, balanced, and absorbing. It follows that the space (X, A) is locally convex and locally bounded, — denote it by X_{A}. Notice also that the topology of X_{A} is finer than the original topology on X because A is bounded. In particular, X_{A} is Hausdorff.

We claim that X_{A} is complete. Indeed, if (x_{α}) is a Cauchy net in X_{A}, then it is also Cauchy in the original topology of X, which is complete, so that x_{α} converges to some x. Fix ε > 0. Then there exists α_{0} such that x_{α} − x_{β} ∈ εA whenever α, β ⩾ α_{0}. Letting β → ∞, since A is closed we have x_{α} −x ∈ εA, i.e., x_{α} → x in X_{A}. Thus, X_{A} is complete, hence Banach.

Since A is bounded, we can find m such that A ⊆ mU. Then T (A) ⊆ T (mU) ⊆ mA, so that T is bounded in X_{A}. Then T^{−1} is also bounded in X_{A} by the Open Mapping Theorem, so that U = T^{−1}(T (U)) ⊆ T^{−1}(A) ⊆ nA for some n > 0. Hence U is bounded. □

Proposition 6.3. Let T: X → Y be an nb-bounded operator between Hausdorff topological vector spaces such that X is not locally bounded. If

(i) every zero neighborhood in X contains a non-trivial linear subspace, or (ii) both X and Y are Fréchet spaces,

then T is not a bijection.

Proof. If every zero neighborhood of X contains a non-trivial linear subspace, then T cannot be one-to-one by 6.1. Suppose now that X and Y are Fréchet and assume that T is a bijection. Let S: Y → X be the linear inverse of T. The Open Mapping Theorem implies that S is continuous and hence bb-bounded. It follows that the identity operator of X is nb-bounded being the composition of the nb-bounded operator T and the bbbounded operator S. But the identity operator is nb-bounded if and only if the space is locally bounded, a contradiction. □

Weak topologies. We are going to show that an operator which is nb-bounded relative to a weak topology has to be of finite rank. In order to prove this we need the following well-known lemma. For completeness we provide a simple proof of it.

Lemma 6.4. Let T be a linear operator on a vector space L, and let f_{1}, …, f_{n} be linear functionals on L such that T x = 0 whenever f_{i}(x) = 0 for every i = 1, …, n. Then T is a finite rank operator of rank at most n.

Proof. Define a linear map π from L to ℝ^{n} via π(x) = (f_{1}(x), …, f_{n}(x)). Then the dimension of the range π(L) is at most n. Define also a linear map φ from π(L) to L via φ(π(x)) = T x. It can be easily verified that φ is well-defined. Then the range of T coincides with the range φ(π(L)), which is of dimension at most n. □

Proposition 6.5. Let X be a locally convex space, and T an operator on X such that T is nb-bounded with respect to the weak topology of X. Then T is of finite rank.

Proof. Suppose T maps some weak zero neighborhood U = {x ∈ X : |f_{i}(x)| < 1, i = 1, …, n}, (f_{1}, …, f_{n} ∈ X′), to a weakly bounded set. Since the weak topology is Hausdorff, it follows from 6.1 that ⋂_{n=1}^{∞} \frac{1}{n}U ⊆ ker T. In particular, T x = 0 whenever f_{i}(x) = 0 for every i = 1, …, n. Then Lemma 6.4 implies that T is a finite rank operator.

□

Spectra and spectral radii of nb-bounded operators.

Proposition 6.6. If T is an nb-bounded operator on a topological vector space then σ^{bb}(T) = σ^{c}(T) = σ^{nn}(T) = σ^{nb}(T).

Proof. If X is locally bounded then the result is trivial by 3.3. Suppose that X is not locally bounded. Then, in view of 3.2, it suffices to show that ρ^{bb}(T) ⊆ ρ^{nb}(T). Let λ ∈ ρ^{bb}(T), so that R_{λ} is bb-bounded. If λ ≠ 0, then it follows from R_{λ}(λI − T) = I that R_{λ} = \frac{1}{λ}R_{λ}T + \frac{1}{λ}I. Thus, R_{λ} is a sum of an nb-bounded operator and a multiple of the identity operator, which yields λ ∈ ρ^{nb}(T). To finish the proof, we will show that if X is not locally bounded then λ = 0 always belongs to σ^{bb}(T) and, therefore, to σ^{c}(T), σ^{nn}(T), and σ^{nb}(T). Indeed, if the resolvent R_{λ} = T^{−1} were bb-bounded, then I = T^{−1}T would be nb-bounded, which is impossible in a non-locally bounded space, a contradiction. □

Proposition 6.7. If T is an nb-bounded operator on a topological vector space, then r_{bb}(T) = r_{c}(T) = r_{nn}(T) = r_{nb}(T).

Proof. By Proposition 4.3 it suffices to show that r_{bb}(T) ⩾ r_{nb}(T). Since T is nb-bounded, then T (U) is a bounded set for some zero neighborhood U. Let ν > r_{bb}(T) and fix a zero neighborhood V. Then νV is again a zero neighborhood. In particular, since the sequence \frac{T^{n}}{ν^{n}} converges to zero uniformly on bounded sets, we have \frac{T^{n}}{ν^{n}} (T (U)) ⊆ νV for all sufficiently large n. Then \frac{T^{n+1}}{ν^{n+1}} (U) ⊆ V, so that \frac{T^{n}}{ν^{n}} converges to zero uniformly on U. Therefore ν ⩾ r_{nb}(T), so that r_{bb}(T) ⩾ r_{nb}(T). □

6.8. In view of Propositions 6.6 and 6.7 we can write σ(T) instead of σ^{bb}(T), σ^{c}(T), σ^{nn}(T), and σ^{nb}(T) and r(T) instead of r_{bb}(T), r_{c}(T), r_{nn}(T), and r_{nb}(T).

We have established in Theorems 5.1–5.5 that under certain conditions the spectral radii of a linear operator are greater than or equal to the geometric radii of the corresponding spectra. Of course we would like to know when the equalities hold. It is well known that the equality |σ(T)| = r(T) holds for every continuous operator on a Banach space. Moreover, it was shown in [10] that this equality also holds for every continuous operator on a quasi-Banach space (a complete quasinormed space). Further, by means of Proposition 4.7 the main result of [9] is equivalent to the following statement: r(T) = |σ(T)| for every nb-bounded operator T on a complete locally convex space. Here we present a direct proof of this. Our proof is a simplified version of the proof of [9].

Theorem 6.9. If T is an nb-bounded linear operator on a sequentially complete locally convex space, then |σ(T)| = r(T).

Proof. Suppose T (U) is bounded for some basic neighborhood of zero U. It follows from Propositions 6.6, 6.7, and 4.3, and Theorem 5.5 that it suffices to show that |σ^{nn}(T)| ⩾ r_{nb}(T). We are going to show that T induces a continuous operator T̃ on some Banach space such that σ(T̃) ⊆ σ^{nn}(T) ∪ {0} while r(T̃) ⩾ r_{nb}(T), and then appeal to the fact that the spectral radius of a continuous operator on a Banach space equals the geometric radius of the spectrum.

Consider T as a continuous operator on the locally bounded space X_{U} = (X, U). Then σ_{U}(T) is defined by 3.3 and r_{U}(T) is defined by 4.5. We claim that r_{U}(T) ⩾ r_{nb}(T). To see this, suppose r_{U}(T) < ν. Then \frac{T^{n}}{ν^{n}} (U) ⊆ U for all sufficiently large n. Let V be another zero neighborhood, then T (U) ⊆ αV for some α > 0, so that \frac{T^{n}}{ν^{n}} (U) = \frac{T}{ν} \frac{T^{n−1}}{ν^{n−1}} (U) ⊆ \frac{1}{ν}T (U) ⊆ \frac{α}{ν}V for sufficiently large n. This implies that ν ⩾ r_{nb}(T), and it follows that r_{U}(T) ⩾ r_{nb}(T).

On the other hand, we claim that σ_{U}(T) ⊆ σ^{nn}(T). Suppose λ ∈ ρ^{nn}(T). Then R_{λ} is nn-bounded with respect to some base N_{0} of zero neighborhoods. We can assume without loss of generality that U ∈ N_{0}, so that R_{λ}(U) ⊆ βU for some β > 0. It follows that λ ∈ ρ_{U}(T).

Since U is convex, the space X_{U} is, in fact, a seminormed space. We can assume without loss of generality that it is a normed space, because otherwise we can consider the quotient space X_{U}/(Null T) and the quotient operator T̂ on this quotient space instead of T. Indeed, since ⋂_{n=1}^{∞} \frac{1}{n}U ⊆ Null T by 6.1, we conclude that the quotient space X_{U}/(Null T) is Hausdorff. It follows then that X_{U}/(Null T) is a normed space, and T̂ is norm bounded. The spectrum σ_{U}(T) becomes even smaller when we substitute T with T̂. Indeed, suppose λ ∈ ρ_{U}(T). Then the resolvent R_{λ} exists in X_{U} and is continuous. If x ∈ ker T, then x = R_{λ}(λI − T)x = λR_{λ}x, so that R_{λ} leaves ker T invariant, and, therefore, induces a quotient operator R̂_{λ} on X_{U}/ ker T via R̂_{λ}([x]) = [R_{λ}x]. Clearly, R̂_{λ} is continuous: if [x_{n}] → [x] in X_{U}/ ker T then x_{n} − z_{n} → x in X_{U} for some (z_{n})_{n=1}^{∞} in ker T, so that [R_{λ}x_{n}] = [R_{λ}(x_{n} − z_{n})] → [R_{λ}x]. On the other hand, r_{U}(T̂) ⩾ r_{U}(T), because if ν > r_{U}(T̂) then \frac{T̂^{n}}{ν^{n}} ([U]) ⊆ [U] for all sufficiently large n, \frac{T^{n}}{ν^{n}} (U) ⊆ U + ker T, and so \frac{T^{n+1}}{ν^{n+1}} (U) ⊆ \frac{1}{ν}T (U) ⊆ \frac{α}{ν}U for some α > 0. It follows that ν ⩾ r_{U}(T) and, therefore, r_{U}(T̂) ⩾ r_{U}(T).

Finally, we consider the completion X̃_{U} of X_{U}, and extend T to a continuous linear operator T̃ on the completion. The spectrum of T̃ is smaller that the spectrum of T, because if λ ∈ ρ_{U}(T) then the resolvent R_{λ} can be extended by continuity to R̃_{λ} on X̃, and R̃_{λ} is a continuous inverse to λI − T̃, so that λ ∈ ρ(T̃). On the other hand, r(T̃) ⩾ r_{U}(T) because if ν > r(T̃) then \frac{T̃^{n}}{ν^{n}} (Ũ) ⊆ Ũ for all sufficiently large n, which implies \frac{T^{n}}{ν^{n}} (U) ⊆ U since T is a restriction of T̃ on X. □

7. Compact operators

As with bounded operators, there is more than one way to define a compact operator on an arbitrary topological vector space. A subset of a topological vector space is called precompact if its closure is compact. Given a linear operator T on a topological vector space, T is called Montel if it maps every bounded set into a precompact set and compact if it maps some neighborhood into a precompact set. To be consistent, we should have probably called these operators “b-compact” and “n-compact” respectively, but the names “Montel” and “compact” are commonly accepted. Obviously, every compact operator is Montel and nb-bounded (hence continuous); every Montel operator is bb-bounded.

7.1. If T is compact or Montel, then sequential completeness is not needed in Theorems 5.1–5.5. Indeed, we used sequential completeness just once, namely, in the proof of Theorem 5.1 to justify the convergence of the sequence R_{λ,n}x = \frac{1}{λ} ∑_{i=0}^{n} \frac{T^{i}x}{λ^{i}}. But since the sequence (R_{λ,n}x)_{n} is Cauchy and, therefore, bounded, the sequence (T R_{λ,n}x)_{n} has a convergent subsequence whenever T is compact or Montel. Furthermore, it follows from R_{λ,n+1}x = \frac{1}{λ}(I + T R_{λ,n})x that (R_{λ,n}x)_{n} has a convergent subsequence hence converges.

Let K be a compact operator on an arbitrary topological vector space, and let σ(K) and r(K) be as in 6.8. It was proved in [15] that σ(K) = {0} implies r_{l}(K) = 0. In the following theorem we use the technique of [15] to improve this result by showing that in general r(K) ⩽ |σ(K)|.

Theorem 7.2. If K is a compact operator on a Hausdorff topological vector space X, then r(K) ⩽ |σ(K)|.

Proof. Assume that |σ(K)| < r(K). Without loss of generality (by scaling K) we can assume that |σ(K)| < 1 < r(K). Since K is compact, there is a basic neighborhood of zero U such that \overline{K(U)} is compact. Without loss of generality we can assume that U is closed. In particular \overline{K(U)} is bounded, so that \overline{K(U)} ⊆ ηU for some η > 0. We can assume without loss of generality that η > 1. We define the following subsets of U:

U_{1} = \overline{K(U)} ∩ U, U_{n+1} = K(U_{n}) ∩ U (n = 1, 2, …), and U_{0} = ⋂_{n=1}^{∞} U_{n}.

Notice that U_{1} is compact because \overline{K(U)} is compact and U is closed. Also, if U_{n} is compact, then K(U_{n}) is compact being the image of a compact set under a continuous operator. Therefore, every U_{n} for n ⩾ 1 is compact. Using induction, we can show that the sequence (U_{n}) is decreasing. Indeed, U_{1} ⊆ U by definition, U_{2} = K(U_{1}) ∩ U ⊆ K(U) ∩ U ⊆ U_{1}, and if U_{n} ⊆ U_{n−1}, then U_{n+1} = K(U_{n}) ∩ U ⊆ K(U_{n−1}) ∩ U = U_{n}. It follows also that U_{0} is compact and contains zero.

Notice that K maps every balanced set to a balanced set. Since U is balanced, U_{n} is balanced for each n ⩾ 0. If A is a balanced subset of U, then obviously A ⊆ (ηA)∩U, and when we apply the same reasoning to \frac{1}{η}K(A) instead of A (which is also a balanced subset of U), we get \frac{1}{η}K(A) ⊆ K(A) ∩ U. We use this to show by induction that \frac{1}{η^{n}} K^{n}(U) ⊆ U_{n} for every n ⩾ 1. Indeed, for n = 1 we have \frac{1}{η}K(U) ⊂ K(U) ∩ U ⊆ U_{1}. Suppose \frac{1}{η^{n}} K^{n}(U) ⊆ U_{n} for some n ⩾ 1. Then

\frac{1}{η^{n+1}} K^{n+1}(U) ⊆ \frac{1}{η}K(U_{n}) ⊆ K(U_{n}) ∩ U = U_{n+1},

which proves the induction step.

Next, we claim that there exists an open zero neighborhood V and an increasing sequence of positive integers (n_{j}) such that U_{n_{j}} \V is nonempty for every j ⩾ 1. Assume for the sake of contradiction that for every open zero neighborhood V we have U_{n} ⊆ V for all sufficiently large n. Since \frac{1}{2}U contains an open zero neighborhood, then there exists a positive integer N such that U_{n} ⊆ \frac{1}{2}U whenever n ⩾ N. This implies that U_{N+m} = K^{m}(U_{N}) for all m ⩾ 0. Indeed, this holds trivially for m = 0. Suppose that U_{N+m} = K^{m}(U_{N}) for some m ⩾ 0. Then U_{N+m+1} = K(U_{N+m}) ∩ U = K^{m+1}(U_{N}) ∩ U, and this implies that U_{N+m+1} = K^{m+1}(U_{N}) because U_{N+m+1} ⊆ \frac{1}{2}U. Now take any open zero neighborhood V, then \frac{1}{η^{N}} V is again a zero neighborhood, and by assumption there exists a positive integer M such that U_{n} ⊆ \frac{1}{η^{N}} V whenever n ⩾ M. Let n ⩾ max\{M, N\}, then

V ⊇ η^{N}U_{n} = η^{N}K^{n−N}(U_{N}) ⊇ η^{N}K^{n−N} (\frac{1}{η^{N}} K^{N}(U)) = K^{n}(U),

which contradicts the hypothesis r_{nb}(K) = r(K) > 1.

It follows from U_{n_{j}} \ V ≠ ∅ for every j ⩾ 1 that U_{n} \ V ≠ ∅ for all sufficiently large n because U_{n} is a decreasing sequence. Since U_{n} \ V is a decreasing sequence of nonempty compact sets, then U_{0} \ V = ⋂_{n=1}^{∞}(U_{n} \ V) ≠ ∅, so that U_{0} ≠ {0}.

For every n ⩾ 1 we have U_{0} ⊆ U_{n}. It follows that K(U_{0}) ⊆ K(U_{n}) and, therefore, K(U_{0}) ⊆ ⋂_{n=1}^{∞} K(U_{n}). Actually, the reverse inclusion also holds. To see this, let y ∈ ⋂_{n=1}^{∞} K(U_{n}). Then y = Kx_{n}, where x_{n} ∈ U_{n} ⊆ U_{1}. Since U_{1} is compact, the sequence (x_{n}) has a cluster point, i.e., x_{n_{j}} → x for some subsequence (x_{n_{j}}) and some x. Since K is continuous we have y = Kx. On the other hand, since every U_{n_{j}} is closed we have x ∈ U_{n_{j}}, so that x ∈ ⋂_{n=1}^{∞} U_{n_{j}} = U_{0}. Thus K(U_{0}) = ⋂_{n=1}^{∞} K(U_{n}).

Next, we claim that U_{0} ⊆ K(U_{0}) ⊆ ηU_{0}. Indeed,

U_{0} = ⋂_{n=2}^{∞} U_{n} = ⋂_{n=2}^{∞}[K(U_{n−1}) ∩ U] ⊆ ⋂_{n=2}^{∞} K(U_{n−1}) = K(U_{0}).

On the other hand, since U_{n} are decreasing and η > 1, we have K(U_{n}) ⊆ K(U_{n−1}) ⊆ ηK(U_{n−1}) and K(U_{n}) ⊆ K(U) ⊆ ηU, so that K(U_{n}) ⊆ ηK(U_{n−1}) ∩ ηU = ηU_{n}. This implies K(U_{0}) ⊆ K(U_{n}) ⊆ ηU_{n} for every n. Thus K(U_{0}) ⊆ ηU_{0}.

Since \overline{K(U)} is compact, hence bounded, then \overline{K(U)} + \overline{K(U)} is also bounded. Thus there is a positive constant γ such that \overline{K(U)} + \overline{K(U)} ⊆ γU. Without loss of generality we can assume γ ⩾ 2. It follows that

U_{1} + U_{1} = \overline{K(U)} ∩ U + \overline{K(U)} ∩ U ⊆ \overline{K(U)} + \overline{K(U)} ⊆ γU.

We use induction to show that U_{n} + U_{n} ⊆ γU_{n−1}. Indeed, since A ∩ B + C ∩ D ⊆ (A + C) ∩ (B + D) for any four sets A, B, C, and D, then

U_{n+1} + U_{n+1} = K(U_{n}) ∩ U_{n} + K(U_{n}) ∩ U_{n}

⊆ [K(U_{n}) + K(U_{n})] ∩ (U_{n} + U_{n}) ⊆ K(U_{n} + U_{n}) ∩ (U_{n} + U_{n})

⊆ K(γU_{n−1}) ∩ γU_{n−1} = γ[K(U_{n−1}) ∩ U_{n−1}] = γU_{n}.

Finally, U_{0} + U_{0} ⊆ ⋂_{n=1}^{∞}(U_{n} + U_{n}) ⊆ ⋂_{n=1}^{∞} γU_{n} = γU_{0}.

Next, consider the set F = ⋃_{n=1}^{∞} nU_{0}. This set is closed under multiplication by a scalar, and U_{0} + U_{0} ⊆ γU_{0} implies that F is a linear subspace of X. We consider the locally bounded topological vector space (F, U_{0}) with multiples of U_{0} as the base of zero neighborhoods. Since U_{0} is balanced by definition, this topology is linear, and it is Hausdorff because U_{0} is compact. Also, it is finer than the topology on F inherited from X because U_{0} is compact and, therefore, bounded in X.

We claim that (F, U_{0}) is complete. Indeed, if (x_{n}) is a Cauchy sequence in (F, U_{0}) then there exists k > 0 such that x_{n} ∈ kU_{0} for each n > 0. Since U_{0} is compact, the sequence (x_{n}) has a subsequence which converges to some x ∈ kU_{0} in the topology of X. Moreover, lim_{n→∞} x_{n} = x because the sequence (x_{n}) is Cauchy in X. Fix ε > 0, then there exists n_{0} such that x_{n} − x_{m} ∈ εU_{0} whenever n, m ⩾ n_{0}. Let m → ∞, since U_{0} is is closed we have x_{n} − x ∈ εU_{0}, i.e., x_{n} → x in (F, U_{0}). Thus, (F, U_{0}) is complete and, therefore, quasi-Banach.

It follows from U_{0} ⊆ K(U_{0}) ⊆ ηU_{0} that F is invariant under K and the restriction K̃ = K|_{F} is continuous. We claim that σ(K̃) ⊆ σ(K) ∪ {0}. If λ ∈ ρ(K) and λ ≠ 0, then (λI − K) is a homeomorphism, so that (λI − K)(U) is a closed zero neighborhood, and αU_{1} ⊆ (λI − K)(U) for some positive real α because U_{1} is bounded. Further, αK(U_{1}) ⊆ K(λI − K)(U) ⊆ (λI − K)K(U). Therefore

αU_{2} ⊆ αK(U_{1}) ∩ αU_{1} ⊆ (λI − K)K(U) ∩ (λI − K)(U),

and since λI − K is one-to-one we get αU_{2} ⊆ (λI − K)(K(U) ∩ U) ⊆ (λI − K)(U_{1}). Similarly, we obtain αU_{n+1} ⊆ (λI − K)(U_{n}) for all n ⩾ 1, and then αU_{0} ⊆ (λI − K)(U_{0}). This implies that the restriction of λI − K to F is onto, invertible, and the inverse is continuous. Thus, λ ∈ ρ(K̃).

In particular this implies that |σ(K̃)| ⩽ |σ(K)| < 1. On the other hand, it follows from U_{0} ⊆ K(U_{0}) that U_{0} ⊆ K̃^{n}(U_{0}) for all n ⩾ 0, so that K̃^{n} does not converge to zero uniformly on U_{0}, whence r(K̃) = r_{bb}(K̃) ⩾ 1. This produces a contradiction because it was proved in [10] that the spectral radius of a continuous operator on a quasi-Banach space equals the radius of the spectrum. □

Corollary 7.3. If K is a compact operator on a locally convex (or pseudo-convex) space, then r(K) = |σ(K)|.

8. Closed operators

In certain situations one has to deal with unbounded linear operators on Banach spaces. For example, the generator of a strongly continuous operator semigroup is generally a closed operator with dense domain (see e.g. [11, 6]). Throughout this section T will be a closed operator on a Banach space X with domain D(T). As usually, we define D(T^{n+1}) = {x ∈ D(T^{n}) : T^{n}x ∈ D(T)} and D = ⋂_{n=0}^{∞} D(T^{n}). In case when T is the infinitesimal generator of an operator semigroup, D is dense in the range of the semigroup, which is usually assumed to be all of X. The set D with the locally-convex topology τ given by the sequence of norms ∥x∥_{n} = ∑_{k=0}^{n}∥T^{k}x∥ is a Fréchet space. Clearly, D is invariant under T, and the restriction operator T_{|D} is continuous because x_{α} ⟶^{τ} 0 in D implies ∥T x_{α}∥_{n} ⩽ ∥x_{α}∥_{n+1} → 0 for each n.

We investigate the relation between the spectral properties of the original operator T on X and of the restriction T_{|D} on D.^{3} Recall that λ ∈ ρ(T) if R(λ; T) = (λI −T)^{−1} : X → D(T) exists (it is automatically bounded by [11, Theorem 2.16.3]), and σ(T) = ℂ \ ρ(T).

Lemma 8.1. If λ ∈ ρ(T) then R(λ; T)_{|D} is a bijection from D to D commuting with T on D(T). Furthermore, for each n ⩾ 0 there is a constant C_{n} such that ∥R(λ; T)x∥_{n} ⩽ C_{n}∥x∥_{n−1} for each x ∈ D.

Proof. It can be easily verified that R(λ; T) is a bijection from D(T^{n}) onto D(T^{n+1}) and, therefore, the restriction R(λ; T)_{|D} is a bijection. Notice that since R(λ; T)(λI − T)x = x for each x ∈ D(T) and (λI − T)R(λ; T)x = x for each x ∈ X, then

T R(λ; T)x = λR(λ; T)x − x for each x ∈ D(T) and
R(λ; T)T x = λR(λ; T)x − x for each x ∈ X,

so that T and R(λ; T) commute on D(T). It also follows that for each x ∈ D we have

T^{2}R(λ; T)x = λT R(λ; T)x − T x = λ^{2}R(λ; T)x − λx − T x,
T^{3}R(λ; T)x = λ^{2}T R(λ; T)x − λT x − T^{2}x = λ^{3}R(λ; T)x − λ^{2}x − λT x − T^{2}x,
⋮
T^{k}R(λ; T)x = λ^{k}R(λ; T)x − λ^{k−1}x − λ^{k−2}T x − ⋯ − λT^{k−2}x − T^{k−1}x.

It follows that

∥T^{k}R(λ; T)x∥ ⩽ |λ|^{k}∥R(λ; T)x∥ + |λ|^{k−1}∥x∥ + |λ|^{k−2}∥T x∥ + ⋯ + ∥T^{k−1}x∥

for each x ∈ D, so that

∥R(λ; T)x∥_{n} = ∑_{k=0}^{n}∥T^{k}R(λ; T)x∥ ⩽

μ_{n}∥R(λ; T)x∥ + μ_{n−1}∥x∥ + μ_{n−2}∥T x∥ + ⋯ + μ_{0}∥T^{n−1}x∥, where μ_{k} = 1 + |λ| + ⋯ + |λ|^{k}. Since ∥R(λ; T)x∥ ⩽ ∥R(λ; T)∥∥x∥ it follows that ∥R(λ; T)x∥_{n} ⩽ C_{n}(∥x∥ + ∥T x∥ + ⋯ + ∥T^{n−1}x∥) for some C_{n}, so that ∥R(λ; T)x∥_{n} ⩽ C_{n}∥x∥_{n−1}. □

3Another approach to this question can be found in [22].

Proposition 8.2. The inclusion ρ(T) ⊆ ρ^{nn}(T_{|D}) holds. Moreover, if D is dense in X and T is the smallest closed extension of T_{|D}, then ρ(T) = ρ^{nn}(T_{|D}).

Proof. Suppose that λ ∈ ρ(T) and consider the resolvent operator R(λ; T) on X. Then ∥R(λ; T)x∥_{n} ⩽ C_{n}∥x∥_{n−1} ⩽ C_{n}∥x∥_{n} hence R(λ; T)_{|D} is nn-bounded and λ ∈ ρ^{nn}(T_{|D}).

Suppose now that D is dense in X, T is the smallest closed extension of T_{|D}, and λ ∈ ρ^{nn}(T_{|D}). Then there exists an nn-bounded operator R(λ; T_{|D}): D → D such that R(λ; T_{|D})(λI − T)x = (λI − T)R(λ; T_{|D})x = x for each x ∈ D. Then there is a constant C > 0 such that ∥R(λ; T_{|D})x∥ = ∥R(λ; T_{|D})x∥_{0} ⩽ C∥x∥_{0} = C∥x∥ for each x ∈ D. It follows that R(λ; T_{|D}) can be extended to a bounded operator R on X. Fix x ∈ X and pick (x_{n}) in D such that x_{n} → x. Then R(λ; T_{|D})x_{n} → Rx and (λI −T)R(λ; T_{|D})x_{n} = x_{n} → x. Since λI − T is closed we have (λI − T)Rx = x. It follows, in particular, that λI − T is onto.

Since ∥(λI − T)x∥ ⩾ \frac{1}{C} ∥x∥ for each x ∈ D, it follows that for every nonzero y ∈ X the pair (y, 0) doesn’t belong to the closure of the graph of λI − T_{|D}. But the closure of the graph of λI − T_{|D} is the graph of λI − T because λI − T is the smallest closed extension of λI − T_{|D}. It follows that λI − T is one-to-one, hence λ ∈ ρ(T). □

Suppose now that S is a bounded operator on X such that D is invariant under S and ST x = T Sx for each x ∈ D. Then

∥Sx∥_{m} = ∑_{k=0}^{m}∥T^{k}Sx∥ ⩽ ∥S∥ ∑_{k=0}^{m}∥T^{k}x∥ = ∥S∥ · ∥x∥_{m},

so that ∥S_{|D}∥_{n} ⩽ ∥S∥. Moreover, if m ⩽ k then ∥x∥_{m} ⩽ ∥x∥_{k}, so that the mixed seminorm 𝔪_{km}(S_{|D}) ⩽ ∥S∥. It also follows from Proposition 4.6 that r_{nn}(S_{|D}) ⩽ r(S).

Further, we claim that if R = R(λ; T) for some λ ∈ ρ(T), then r_{nb}(R_{|D}) ⩽ r(R). Indeed, recursive application of Lemma 8.1 yields ∥R^{n}x∥_{k} ⩽ M_{k}∥R^{n−k}x∥ for each x ∈ D and k ⩾ n, where M_{k} = Π_{i=1}^{k}C_{i}. It follows that the mixed seminorm

𝔪_{mk}(R_{|D}^{n}) = sup{∥R^{n}x∥_{k} : x ∈ D, ∥x∥_{m} ⩽ 1}

⩽ sup{M_{k}∥R^{n−k}x∥ : ∥x∥ ⩽ 1} = M_{k}∥R^{n−k}∥.

Therefore lim_{n} \sqrt[n]{𝔪_{mk}(R_{|D}^{n})} ⩽ lim_{n} \sqrt[n]{∥R^{n}∥} = r(T) for any m, k ⩾ 0. Now Proposition 4.6 yields r_{nb}(R_{|D}) ⩽ r(R).

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