Abstract and Applied Analysis

Volume 2014 (2014), Article ID 179027, 16 pages

http://dx.doi.org/10.1155/2014/179027

Research Article

Uniform Convergence and Spectra of Operators in a Class of Fréchet Spaces

Angela A. Albanese,¹ José Bonet,² and Werner J. Ricker³

¹Dipartimento di Matematica e Fisica “E. De Giorgi”, Università del Salento, C.P.193, 73100 Lecce, Italy

²Instituto Universitario de Matemática Pura y Aplicada IUMPA, Universitat Politècnica de València, 46071 Valencia, Spain

³Math.-Geogr. Fakultät, Katholische Universität Eichstätt-Ingolstadt, 85072 Eichstätt, Germany

Received 22 October 2013; Accepted 20 December 2013; Published 30 January 2014

Academic Editor: Alfredo Peris

Copyright © 2014 Angela A. Albanese et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Well-known Banach space results (e.g., due to J. Koliha and Y. Katznelson/L. Tzafriri), which relate conditions on the spectrum of a bounded operator to the operator norm convergence of certain sequences of operators generated by , are extended to the class of quojection Fréchet spaces. These results are then applied to establish various mean ergodic theorems for continuous operators acting in such Fréchet spaces and which belong to certain operator ideals, for example, compact, weakly compact, and Montel.

1. Introduction

Given a Banach space and a continuous linear operator on , there are various classical results which relate conditions on the spectrum of with the operator norm convergence of certain sequences of operators generated by . For instance, if , with denoting the operator norm, (even in the weak operator topology suffices), then necessarily , where , [1, p. 709, Lemma 1]. The stronger condition is equivalent to the requirement that both and hold [2]. An alternate condition, namely, that is a convergent sequence relative to the operator norm, is equivalent to the requirement that the three conditions , the range is closed in for some , and are satisfied [3]. Here with being the boundary of . Such results as above are often related to the uniform mean ergodicity of , meaning that the sequence of averages of is operator norm convergent. For instance, if and , then is uniformly mean ergodic [4, p. 90, Theorem 2.7]. Or if , then is uniformly mean ergodic if and only if is closed [5].

Our first aim is to extend results of the above kind to the class of all Fréchet spaces referred to as prequojections; this is achieved in Section 3. The extension to the class of all Fréchet spaces is not possible; see Proposition 17 below and [6, Example 3.11], for instance. We point out that a classical result of Katznelson and Tzafriri stating, for any Banach-space-operator satisfying , that if and only if [7], is also extended to prequojection Fréchet spaces; see Theorem 20.

Our second aim is inspired by well-known applications of the above mentioned Banach space results to determine the uniform mean ergodicity of operators which satisfy and belong to certain operator ideals, such as the compact or weakly compact operators; see, for example, [1, Ch. VIII, 8], [4, Ch. 2, 2.2], and [8, Theorem 6.1], where can even be quasi-compact. An extension of such a mean ergodic result to the class of quasi-precompact operators acting in various locally convex Hausdorff spaces is presented in [9]. For prequojection Fréchet spaces, this result is further extended to the (genuinely) larger class of quasi-Montel operators; see Proposition 32, Remark 33, and Theorem 35. A mean ergodic theorem for Cesàro bounded, weakly compact operators (and also reflexive operators) in a certain class of locally convex spaces (which includes all Fréchet spaces), is also presented; see Proposition 23 and Remark 24(ii).

2. Preliminaries and Spectra of Operators

Let be a lcHs and a system of continuous seminorms determining the topology of . The strong operator topology in the space of all continuous linear operators from into itself (from into another lcHs we write ) is determined by the family of seminorms , for , for each and , in which case we write . Denote by the collection of all bounded subsets of . The topology of uniform convergence on bounded sets is defined in via the seminorms , for , for each and ; in this case we write . For a Banach space, is the operator norm topology in . If is countable and is complete, then is called a Fréchet space. The identity operator on a lcHs is denoted by .

By we denote equipped with its weak topology , where is the topological dual space of . The strong topology in (resp. ) is denoted by (resp. ) and we write (resp. ); see [10, IV, Ch. 23] for the definition. The strong dual space of is denoted simply by . By we denote equipped with its weak-star topology . Given , its dual operator   is defined by for all , . It is known that and , [11, p. 134].

For a Fréchet space and , the resolvent set   of consists of all such that exists in . Then is called the spectrum of . The point spectrum   consists of all such that is not injective. Unlike for Banach spaces, it may happen that . For example, let be the Fréchet space equipped with the lc-topology determined via the seminorms , where , for . Then the unit left shift operator , for , belongs to and, for every , the element is an eigenvector corresponding to .

For a Fréchet space , the natural imbedding is an isomorphism of onto the closed subspace of . Moreover, we always have

that is, is an extension of .

The following result will be required in the sequel. Since the proof is standard we omit it. The polar of a set is denoted by .

Lemma 1. Let be a Fréchet space.

(i)Let be a fundamental, increasing sequence which determines the lc-topology of . For each define on via . Then is a fundamental, increasing sequence which determines the lc-topology of .(ii)Let be a fundamental, increasing sequence which determines the lc-topology of . For each , let denote the Minkowski functional (in ) of the bipolar of . Then is a fundamental, increasing sequence which determines the lc-topology of . Moreover, for each , we have

for each and . In particular, ; that is, the restriction of to coincides with , for each .

For Banach spaces the following fact is well-known.

Lemma 2. Let be a lcHs and be an equicontinuous sequence. Then also is equicontinuous.

Proof. Let . Then as is equicontinuous. So, for all and , we have with

As the seminorms generate the lc-topology of , the previous inequality shows that is equicontinuous.

Since is equicontinuous and the lc-topology of is generated by the polars of bounded subsets of , the same argument as above yields that is equicontinuous.

Lemma 3. Let be a Fréchet space and . Then is an isomorphism of onto itself if and only if is an isomorphism of onto itself.

Proof. If is an isomorphism of onto itself, then there exists with . It follows that and so . Accordingly, and . Thus, exists in and ; that is, is an isomorphism of onto itself.

Conversely, suppose that is an isomorphism of onto itself. Since is an extension of (i.e., ), we see that is one-to-one. Moreover, since is a closed subspace of (as is a complete, barrelled lcHs), it follows that is closed. It remains to show that . But, if , then there is such that for all . Hence, ; this is a contradiction because the surjectivity of implies that is necessarily one-to-one.

We remark that Lemma 3 remains valid for a complete barrelled lcHs.

The next result is an immediate consequence of (1) and Lemma 3.

Corollary 4. Let be a Fréchet space and . Then and . Moreover,

that is, the restriction of to the closed subspace of coincides with . Briefly, .

A Fréchet space is always a projective limit of continuous linear operators , for , with each a Banach space. If and can be chosen such that each is surjective and is isomorphic to the projective limit , then is called a quojection [12, Section 5]. Banach spaces and countable products of Banach spaces are quojections. Actually, every quojection is the quotient of a countable product of Banach spaces [13]. In [14] Moscatelli gave the first examples of quojections which are not isomorphic to countable products of Banach spaces. Concrete examples of quojection Fréchet spaces are , the spaces , with , and for , with any open set, all of which are isomorphic to countable products of Banach spaces. The spaces of continuous functions , with a -compact, completely regular topological space, endowed with the compact open topology, are also quojections. Domański exhibited a completely regular topological space such that the Fréchet space is a quojection which is not isomorphic to a complemented subspace of a product of Banach spaces, [15, Theorem]. A Fréchet space admits a continuous norm if and only if contains no isomorphic copy of [16, Theorem ]. On the other hand, a quojection admits a continuous norm if and only if it is a Banach space [12, Proposition 3]. So, a quojection is either a Banach space or contains an isomorphic copy of , necessarily complemented, [16, Theorem ]. Also [17] is relevant.

Lemma 5. Let be a quojection Fréchet space and . Suppose that , with a Banach space (having norm ) and linking maps which are surjective for all , and suppose, for each , that there exists satisfying

where , , denotes the canonical projection of onto (i.e., ). Then

Moreover,

If, in addition, for every , the resolvent operator satisfies

then .

Proof. For the containments (6) and (7) we refer to [18, Lemma 5.1].

Suppose now that (8) holds for each . To establish the desired equality, let . Then is surjective. Fix . Since is surjective, it is routine to check from the identity that also is surjective (with the identity operator). To verify is injective suppose that for some , in which case for some . Accordingly,

shows that . It then follows from (8) that ; that is, . Since , we have . Hence, is injective. This establishes that . Accordingly, as desired.

The following result occurs in [18, Lemma 5.2].

Lemma 6. Let be a quojection Fréchet space and . Suppose that , with a Banach space (having norm ) and linking maps which are surjective for all , and suppose, for each , that there exists satisfying

where , , denotes the canonical projection of onto (i.e., ). Then the following statements are equivalent.

(i)The limit - exists in .(ii)For each , the limit - exists in .

In this case, the operators and , for , satisfy

Moreover, (i) and (ii) remain equivalent if is replaced by .

Given any lcHs and , let us introduce the notation:

for the Cesàro means of . Then is called mean ergodic precisely when is a convergent sequence in . If happens to be convergent in , then will be called uniformly mean ergodic.

We always have the identities

and also (setting ) that

Some authors prefer to use in place of ; since

this leads to identical results.

Recall that is called power bounded if is an equicontinuous subset of .

The final result that we require (i.e., [18, Lemma 5.4]) is as follows.

Lemma 7. Let be a quojection Fréchet space and let operators and , for , be given which satisfy the assumptions of Lemma 5 (with , , denoting the canonical projection of onto and being the norm in the Banach space ).

(i) is power bounded if and only if each , , is power bounded.(ii) is mean ergodic (resp., uniformly mean ergodic) if and only if each , , is mean ergodic (resp., uniformly mean ergodic).

3. Spectrum, Uniform Convergence, and Mean Ergodicity

A prequojection is a Fréchet space such that is a quojection. Every quojection is a prequojection. A prequojection is called nontrivial if it is not itself a quojection. It is known that is a prequojection if and only if is a strict (LB) space. An alternative characterization is that is a prequojecton if and only if has no Köthe nuclear quotient which admits a continuous norm; see [12, 19–21]. This implies that a quotient of a prequojection is again a prequojection. In particular, every complemented subspace of a prequojection is again a prequojection. The problem of the existence of nontrivial prequojections arose in a natural way in [12]; it has been solved, in the positive sense, in various papers [19, 22, 23]. All of these papers employ the same method, which consists in the construction of the dual of a prequojection, rather than the prequojection itself, which is often difficult to describe (see the survey paper [24] for further information). However, in [25] an alternative method for constructing prequojections is presented which has the advantage of being direct. For an example of a concrete space (i.e., a space of continuous functions on a suitable topological space), which is a nontrivial prequojection, see [26].

In this section we extend to prequojection Fréchet spaces some well-known results from the Banach setting which connect various conditions on the spectrum , of a continuous linear operator , to the operator norm convergence of certain sequences of operators generated by . Such results have well-known consequences for the uniform mean ergodicity of .

We begin with a construction for quojection Fréchet spaces which is needed in the sequel.

Let be a quojection Fréchet space and be any fundamental, increasing sequence of seminorms generating the lc-topology of . For each , set and endow with the quotient lc-topology. Denote by the corresponding canonical (surjective) quotient map and define the quotient topology on via the increasing sequence of seminorms on by

for each . Then

Moreover,

which implies that is a norm on . As noted above, since is a quojection Fréchet space and every quotient space (of such a Fréchet space) with a continuous norm is necessarily Banach [12, Proposition 3], it follows that for each there exists such that the norm generates the lc-topology of . Moreover, it is possible to choose for all . Thus, is isomorphic to the projective limit of the sequence of Banach spaces with respect to the continuous, surjective linking maps defined by

This particular construction will be used on various occasions in the sequel, where will always denote the closed unit ball of , for . The so-constructed Banach space norm of will always be denoted by , for .

The following result is classical in Banach spaces [1, p. 709, Lemma 1].

Proposition 8. Let be a quojection Fréchet space and satisfy -. Then .

In case is a prequojection Fréchet space and -, the inclusion is again valid.

Proof. We have the following two cases.

Case (I)  ( is a quojection). Let be a fundamental, increasing sequence of seminorms generating the lc-topology of . Since in as and is a Fréchet space, the sequence is equicontinuous. So, for each there exists such that

there is no loss in generality by assuming that can be chosen.

Define on by , for . Then (20) ensures that is also a fundamental, increasing sequence of seminorms generating the lc-topology of . Moreover,

We now apply the construction (16)–(19) to the sequence of seminorms to yield the corresponding sequence of Banach spaces and the quotient maps , for ; recall that , for .

Fix . Define the operator via

Then   is a well-defined, continuous linear operator from into . Indeed, suppose for some ; that is, , so that . This, together with (21), yields . Since , it follows that , and hence, by (22) that . Therefore, . This means that is well defined. Clearly, is also linear. Moreover, (17), (21), and (22) imply that

for all and with . Taking the infimum with respect to , it follows that

Since generates the quotient topology of , (24) ensures the continuity of . Moreover, it follows from (22) that

The surjectivity and the continuity of together with (25) imply that -. Indeed, fix any . By the surjectivity of there exists such that . By (25) it follows that , for . Moreover, as by assumption. So, the continuity of yields that in the Banach space . We can then apply Lemma 1 in [1, p. 709] to obtain that .

We have just shown that . Moreover, the operators and satisfy (22). So, we can apply Lemma 5 which yields ; that is, .

Case (II).  ( is a prequojection and -). Observe that and are barrelled and, hence, quasi-barrelled as is a Fréchet space and is the strong dual of a prequojection Fréchet space. Since and , the condition - implies that - (see [27, Lemma 2.6] or [28, Lemma 2.1]). On the other hand, is a quojection Fréchet space. So, it follows from Case (I) that . Finally, Corollary 4 ensures that and so .

Remark 9. For a power-bounded operator it is always the case that - and so, whenever is a prequojection Fréchet space, it follows from Proposition 8 that .

For operators in Banach spaces, the following result is due to Koliha [2].

Theorem 10. Let be a prequojection Fréchet space and . The following assertions are equivalent.

(i)-.(ii)The series converges in .(iii)- and .

Moreover, if one (hence, all) of the above conditions holds, then is an isomorphism of onto with inverse and the series converging in .

Proof. We have the following two cases.

Case (I) ( is a quojection). (i)(ii). The assumption - implies that -. So, we can proceed as in the proof of Proposition 8 to obtain that in such a way that, for every , there exists in satisfying . Then also , for every . So, Lemma 6 implies that - for all . Thus, by [2, Theorem 2.1] the series converges in , for each . With , for , it follows again from Lemma 6 that the series converges in .

(ii)(iii). The assumption clearly implies -. So, as in the proof of (i)(ii), we may assume that in such a way that, for every , there exists in satisfying . Then also , for every . Since converges in and is a quojection, the series also converges in for all ; see Lemma 6. By [2, Theorem 2.1] we have that and so , for all . Accordingly, since for all , Lemma 5 yields ; that is, .

(iii)(i). Since , for every , the operator is invertible, that is, bijective with . On the other hand, - for every as - and . So, by Theorem 4.1 in [29] (see also Theorem 3.5 of [6]) we can conclude that

Let be a fundamental, increasing sequence of seminorms generating the lc-topology of . Arguing as in the proof of Proposition 8 (and adopting the notation from there) we conclude that (20) is satisfied. Define on by , for . Then again (21) is satisfied and, for each , there exists a continuous linear operator satisfying both (22) and (24). Moreover, it follows from (22) that

Fix and consider the sequences and in given by and , for . Then the operator satisfies for all . Moreover, (26) implies that in . Since all the assumptions of Lemma 3.4 in [6] are satisfied with , , and , we can proceed as in the proof of that result to conclude, for every , that the operator is invertible in (hence, also is invertible); that is, .

By the arbitrariness of , we have that , for all . So, there exists such that . It follows that

and, hence, that . Because of (27), with , it follows from Lemma 6 (with ) that -.

Case (II)  ( is a prequojection). As noted before and are barrelled with and .

(i)(ii). If in for , then an argument as for Case (II) in the proof of Proposition 8 shows that in for . Since is a quojection Fréchet space, we can apply (i)(ii) of Case (I) above to conclude that the series converges in . Then also converges in as and is a closed subspace of .

(ii)(iii). If converges in , then converges in ; see [27, Lemma 2.6] or [28, Lemma 2.1]. Since is a quojection Fréchet space, we can apply (ii)(iii) of Case (I) above to conclude that (the condition - clearly follows from the assumption). So, by Corollary 4.

(iii)(i). As already noted (cf. proof of Case (II) in Proposition 8) and are barrelled (hence, quasi-barrelled) and -. By Corollary 4, and so by assumption. Since is a quojection Fréchet space, we can apply Case (I) to conclude that -. So, also - as and is a closed subspace of .

Finally, suppose that one (hence, all) of the above conditions holds. Then the series converges in and so in for . But, for every , we have

and so, for , we can conclude that with convergence of the series in . In a similar way one shows that , with the series again converging in .

Remark 11. In the proof of (iii)(i) in Case (I) above, if , then it follows that . But, this is not the case in general as the following example shows.

Let be a Banach space and let be an increasing sequence with . Consider the quojection Fréchet space (endowed with the product topology) and the operator on defined by , for . It is easy to show that and that is even power bounded. Moreover, . Indeed, for a fixed , if , then ; that is, for all . Since , it follows that for all and so . On the other hand, if , then belongs to and . Hence, is bijective and so . Moreover, fix any and set for every . Then for every . Thus, each is an eigenvalue of .

Now, suppose that for some . Then . But for , and hence, there is such that . This contradiction as is an eigenvalue for .

If is uniformly mean ergodic, then (14) implies that -. With an extra condition the converse is also valid.

Corollary 12. Let be a prequojection Fréchet space and . If - and , then is uniformly mean ergodic.

Proof. Since , the operator is bijective and so the space is closed in . By [6, Theorem 3.5], is uniformly mean ergodic. In particular, as , we have that in for .

Remark 13. Let be a prequojection Fréchet space and let satisfy -. If , then the proof of Corollary 12 shows that is uniformly mean ergodic with -. On the other hand, if (a stronger condition than ), then Theorem 10 implies that - and hence again - follows [30, Remark 3.1]. However, the stronger conclusion that - does not follow from Corollary 12 in general. Indeed, let be any Banach space (even finite dimensional). Then every power of belongs to the set and so is power bounded. This implies that -. Since , surely and so, by Corollary 12, it follows that -. However, for every we have and so does not converge to zero. This does not contradict Theorem 10 as is not included in .

Remark 14. Let be a prequojection Fréchet space and . We observe the following.

(i)Proposition 8 and (14) yield that if is uniformly mean ergodic, then - and .(ii)Suppose that -. If , then is uniformly mean ergodic and - (cf. Remark 13).

For Banach spaces the next result is due to Mbekhta and Zemànek [3]. Recall that .

Theorem 15. Let be a prequojection Fréchet space and . The following statements are equivalent.

(i) is convergent in .(ii)-, the linear space is closed in for some and .(iii)- and is closed for some .

Proof. (i)(ii). If converges in to , say, then is uniformly mean ergodic with ergodic projection equal to [30, Remark 3.1]. Moreover, as is necessarily equicontinuous, it follows that -. Hence, by Theorem 3.5 and Remark 3.6 of [6] the space is closed for every . Moreover, by Proposition 8 we have . To establish the remaining condition we distinguish two cases.

(a) is a quojection. Let be any fundamental, increasing sequence of seminorms generating the lc-topology of . By equicontinuity of , for each , there exists such that

Define , for each , by , for . Then (30) ensures that is also a fundamental, increasing sequence of seminorms generating the lc-topology of . Moreover, it is routine to check (using also that for each ) that

With (31) in place of (21), we can argue as in the proof of Proposition 8 to deduce that and that, for every , there exist operators and in satisfying and . Hence, for every . Since also -, it follows from Lemma 6 (with and ) that -, for each . By [3, Corollaire 3] we have that for every . This implies that . Indeed, if , then for every we have and so ; that is, . As for every , an appeal to Lemma 5 yields that .

(b) is a prequojection. As noted before, and are barrelled (hence, quasi-barrelled) with and . Hence, - implies that -; see [27, Lemma 2.6] or [28, Lemma 2.1]. Since is a quojection Fréchet space, we can apply the result from case (a) to conclude that and so ; see Corollary 4.

(ii)(i). The assumptions - and the space being closed for some imply that is uniformly mean ergodic [6, Theorem 3.4 and Remark 3.6]. In particular, is closed and

[6, Theorem 3.4]. Moreover, Proposition 8 implies that . It then follows from the assumption that either or .

If , then necessarily and so, by (iii)(i) of Theorem 10, we have -.

In the event that we have that and so (otherwise, is injective and from also surjective; that is, ). Define and . Then is a prequojection Fréchet space (being a quotient space of the prequojection ) which is -invariant and so . The claim is that

It follows from (32) that . Fix (so that ). If for some (i.e., ), then as . Hence, is injective. Next, let . Then there exists such that . Since with and (cf. (32)), it follows that ; that is, , with and . As and , this implies that and so with ; that is, is surjective. These facts show that . This establishes .

Fix . Suppose that for some . Then with and (cf. (32)). It follows that with and . Arguing as in the previous paragraph, this implies that and . Since and , we can conclude that ; that is, is injective. Next, let . Then with and (cf. (32)). Since , the element exists. Moreover, with implies the existence of such that . It follows that satisfies . Hence, is also surjective and so . Accordingly, is proved. This establishes (33).

Since and (33) is equivalent to , it follows that . Moreover, is a prequojection Fréchet space and in as (because - and on ). So, we can apply Theorem 10 to conclude that in as . On the other hand, on . These facts ensure that in because and on .

(i)(iii). If converges to some in , then is uniformly mean ergodic with ergodic projection equal to [30, Remark 3.1]. Hence, by [6, Theorem 3.5 and Remark 3.6] the space is closed for every . Moreover, in as .

(iii)(i). We first observe that

This identity (together with the fact that - implies for the averages that - [30, Remark 3.1]) yields -. But, - and so we can conclude that -. As also is closed for some , we can apply [6, Theorem 3.4 and Remark 3.6] to conclude that is uniformly mean ergodic and, in particular, that (32) is valid with being closed. We claim that this fact, together with the assumption that -, implies that converges in . To see this, note that on and so in as . On the other hand, the surjective operator lifts bounded sets via [10, Lemma 26.13] because and , both being prequojections, are quasinormable Fréchet spaces [24, Proposition 2.1], [21]; that is, for every there exists such that . So, for fixed (with corresponding set ) and (every is the restriction of some ), we have

where as by assumption. Set . The arbitrariness of and shows that in (after observing that is -invariant and so ). These facts ensure that in as .

Remark 16. In assertion (ii) of Theorem 15 the condition that “ is closed in for some ” can be replaced with the condition that “   is uniformly mean ergodic”; see [6, Theorem 3.5 and Remark 3.6].

Theorems 10 and 15 do not necessarily hold for operators acting in general Fréchet spaces.

Proposition 17. Let or and let be a Köthe matrix on such that is a Montel space with . Then there exists an operator such that in as and but is not closed for every .

Proof. By the proof of Proposition 3.1 in [6] there exists with for all such that the diagonal operator given by , for , is power bounded, uniformly mean ergodic and is dense but, not closed in . So, for every , also is dense but not closed in . To see this, note that the arguments in the proof of [6, Remark 3.6, (5)(4)] are valid for any operator satisfying - and acting in any Fréchet space. So, in the case that was closed for some , we could apply [6, Remark 3.6, (5)(4)] to conclude that is also closed; a contradiction. So .

We claim that in as . Indeed, since is equicontinuous and convergence of a sequence in is equivalent to its convergence in (as is Montel), it suffices to show that in for each , where . But, this is immediate because , for all .

It remains to show that . Set . Then . Let . Then . It is routine to check that, for a fixed , the element belongs to and satisfies . This means that the operator is surjective. On the other hand which follows from . Therefore, as is a Fréchet space, ; that is, . Since , it follows that .

Concerning the example in Proposition 17 we note that (i) of Theorem 10 holds but (iii) of Theorem 10 fails (as implies that ). Moreover, (i) of Theorem 15 holds (as -) but (ii) and (iii) of Theorem 15 fail (because is not closed in for every ). Of course, is not a prequojection.

A well-known result of Katznelson and Tzafriri states that a power bounded operator on a Banach space satisfies if and only if , [7, Theorem 1 and p. 317 Remark]. In order to extend this result to prequojection Fréchet spaces (see Theorem 20 below) we require the following notion.

Let be a Fréchet space and . A fundamental, increasing sequence which generates the lc-topology of is called   contractively admissible if, for each , we have

Lemma 18. Let be a Fréchet space and . Then there exists a contractively admissible sequence of seminorms which generates the lc-topology of if and only if is power bounded.

Proof. If is contractively admissible, then it is clear from (36) that , for and every , . This means precisely that is equicontinuous in ; that is, is power bounded.

Conversely, suppose that is power bounded. Let be a fundamental, increasing sequence in which generates the lc-topology of . Via the equicontinuity of for every there exist and such that

Define , for and each . Then the previous inequality implies that

and so is a fundamental, increasing sequence determining the lc topology of , which clearly satisfies (36). That is, is contractively admissible.

Remark 19. (i) For a Banach space , Lemma 18 simply states that is power bounded if and only if it is a contraction for some equivalent norm in .

(ii) Let be a Fréchet space and let be an isomorphism which is bipower bounded; that is, is equicontinuous in . An examination of the proof of Lemma 18 shows that there exists a sequence , again called   contractively admissible, which generates the lc-topology of and satisfies, for each ,

Theorem 20. Let be a prequojection Fréchet space and let be power bounded. The following assertions are equivalent.

(i)-.(ii) and there exists a contractively admissible sequence such that, for each and , there exists satisfying

Remark 21. (i) If , then necessarily and so the resolvent family is defined.

(ii) If , then (i) of Theorem 20 follows without any further conditions. Indeed, by Remark 9 we actually have . Then Theorem 10 implies that - and, hence, also -.

(iii) If is a Banach space and is any norm in for which is a contraction (i.e., is contractively admissible), then the requirement (40) automatically holds with . That is, condition (ii) in Theorem 20 simply reduces to and we recover the result of Katznelson and Tzafriri.

Proof of Theorem 20 (i)(ii). As usual we distinguish two cases.

Case (I) ( is a quojection). According to Lemma 18 there is a contractively admissible sequence satisfying (36) and, hence, also , for and all . We proceed as in the proof of Proposition 8 (now using (36) in place of (21) so that (24) becomes , for and ) to obtain that in such a way that, for every , there exists a contraction   satisfying . Then also for all . For each , define for . By the properties of projective limits is a fundamental sequence generating the lc-topology of . Moreover,

shows that is also contractively admissible. According to Lemma 6 (applied to the norms and with , , and , for ), the assumption - implies that , for each . By [7, Theorem 1] we can conclude that . On the other hand, as is a contraction and so ; that is, , for . According to Lemma 5 also ; that is, .

Concerning (40), fix and . By the previous paragraph . It follows from that . Hence, for , we have

which establishes (40).

Case (II) ( is a prequojection). As noted before, and are barrelled (hence, quasi-barrelled) with and . So, the assumption - implies that -. Moreover, is a quojection Fréchet space and is power bounded; see Lemma 2. So, the result of Case (I) yields . But, (see Corollary 4) and so .

By (i)(ii) for quojections there exists a contractively admissible sequence such that, for every and , there exists satisfying

By Lemma 1 and Corollary 4 the seminorms , , satisfy (40).

(ii)(i). Case (I). ( is a quojection).

Let be as in the statement of (ii), in which case (36) holds. Proceed as in Case (I) of the proof of (i)(ii) to obtain that in such a way that, for every , there exists a contraction  , satisfying .

Claim  1. , for every .

To establish this, let . Since , it follows that , and hence, is surjective. But, also is surjective. It is then routine to check from the identity that is surjective. To verify that is injective suppose that for some , in which case for some . Accordingly,

shows that . It then follows from (40) that ; that is, . Since , we have . Hence, is injective. This establishes that , and hence, Claim 1 follows as was arbitrary.

Fix . From Claim 1 and the fact that is a contraction, it follows from [7, Theorem 1] that . According to Lemma 6 (with , ) we can conclude that -.

Case (II) ( is a prequojection). By Corollary 4 we have from that . Moreover, Lemma 2 implies that is power bounded.

Let be as stated in part (ii). Apply Lemma 1 to construct the seminorms given there. We first verify that is contractively admissible. Since is contractively admissible, we have with the closed unit ball of ; that is, , for . By the Bi-polar Theorem, [10, Theorem 22.13] applied twice we have

where denotes the closure for the weak topology of a subset (or, of ). Then (45) implies that for each and ; that is, is contractively admissible.

It follows from (40) that , for all and . Using (c.f. Corollary 4) one can repeat the argument via the Bi-polar Theorem to conclude that , which then implies that

So, the conditions in part (ii) are satisfied for the power bounded operator with respect to . Applying (ii)(i) for the quojection Fréchet space we conclude that -. But, with closed in . So, -; that is, (i) holds.

Let be a prequojection Fréchet space and be power bounded. By Remark 9 we have . Suppose that is actually bipower bounded. Then also . Clearly, . Moreover, if , then and so , that is, . It is routine to check that satisfies and hence, is invertible in with . This shows that . Accordingly, ; for a Banach space, see [31, Proposition 1.31], for example. Suppose now, in addition, that in which case ; that is, is quasinilpotent. For a Banach space, a classical result of Gelfand-Hille then states that necessarily ; see the survey article [32] for a complete discussion of this topic. The following fact is an extension of this result.

Corollary 22. Let be a prequojection Fréchet space and be an isomorphism which is bipower bounded. Suppose that and there exists a contractively admissible sequence such that, for each , the inequalities (40) are satisfied. Then .

Proof. According to Theorem 20 we can conclude that -. Fix . For each , it follows that

for every . Since , it follows that with arbitrary; that is, . So, .

4. Operator Ideals and Uniform Mean Ergodicity

Let , be lcHs'. An operator is called Montel (resp. reflexive) if maps bounded subsets of into relatively compact (resp. relatively weakly compact) subsets of [33] (resp., [34]). According to Grothendieck, [35, Chapter 5, Part 2], is called compact (resp., weakly compact) if there exists a 0-neighbourhood such that is relatively compact (resp., relatively weakly compact) in . Clearly, the 2-sided ideal (resp., ) of all Montel (resp., reflexive) operators coincides with the 2-sided ideal (resp., ) of all compact (resp., weakly compact) operators whenever , are Banach spaces. For general lcHs' we always have but the containment may be proper; consider the identity operator on an infinite dimensional Montel lcHs. Clearly, and . Criteria for membership of (resp. ) occur in Theorem (resp. Corollary ) of [36], for example.

In this section we present various connections between the uniform convergence of sequences of operators generated by an operator and the uniform mean ergodicity of , where stands for one of the operator ideals , , , .

Every compact operator acting in a Banach space has the property that has closed range. Hence, if , then is uniformly mean ergodic [1, p. 711, Corollary 4], [4, p. 87, Theorem 2.1]. For any lcHs and , it is also the case that is a closed subspace of [36, Theorem ]. Hence, if is a prequojection Fréchet space, then Theorem 3.5 of [6] implies that is uniformly mean ergodic whenever - (equivalently, - because ; see Remark 26(ii)). Since , the question arises of whether the same is true for ? This is indeed so; see Theorem 27 below.

In a lcHs all relatively -compact sets and all relatively sequentially -compact sets are necessarily relatively countably compact. These are the only implications between these three notions which hold in general. All three notions coincide whenever is angelic [37, p. 31]. Such spaces include all Fréchet spaces (actually, all (LF)-spaces), all (DF)-spaces and many more, [37, Section 3.10], [38, Theorem 11, Examples 1.2].

Operators for which is equicontinuous will be called Cesàro bounded; see [4, p. 72] for a Banach space.

Proposition 23. Let be a lcHs such that is angelic and .

(i)If    is Cesàro bounded and satisfies  - , then    is mean ergodic. (ii)If    is Cesàro bounded and satisfies  - ,   then    is uniformly mean ergodic.

Proof. (i) Fix . It follows from (13) that

The equicontinuity of ensures that . Since , the set is relatively weakly compact in . Moreover, in because of -. These facts, together with being angelic and (48), show that is relatively weakly (hence, relatively weakly sequentially) compact in . Since is arbitrary, we can apply Theorem 2.4 of [39] (an examination of its proof shows that it is not necessary to assume the barrelledness of stated there because of the equicontinuity of assumed here) to conclude that is mean ergodic.

(ii) By part (i) the operator is mean ergodic, that is, - exists in . In particular, (which follows from (13)) and so , for .

To establish the uniform mean ergodicity of , fix , , and . By the equicontinuity of there exist and such that

On the other hand, is a relatively compact subset of and so there exist such that, for every , we have for some . Hence, via (49) we obtain, for every and , that

It follows that

with . The arbitrariness of implies that . So, -.

Finally, the arbitrariness of and of together with the assumption - implies, via (48), that is uniformly mean ergodic.

Remark 24. (i) Let be a lcHs and let be mean ergodic with -. Then it follows from that whenever (here, stands for the operator ideal , , , or ). In particular, if , then is finite dimensional, [36, Theorem (1)].

(ii) Let be a lcHs such that is angelic. Then the class of all weakly completely continuous operators in in the sense of Definition 2 in [40] is precisely . Moreover, if is additionally barrelled, then, for any , the boundedness of the set in is equivalent to being power bounded. In particular, is necessarily Cesàro bounded and satisfies -. Accordingly, the containment shows that Proposition 23(i) is an extension of the following result of Altman [40, Theorem].

Fact 1. Let be a barrelled lcHs with being angelic. Then every power bounded operator is mean ergodic.

The following technical result should be compared with [33, Proposition 3.1].

Lemma 25. Let be a quojection Fréchet space, and let be a Fréchet space and   resp. . Suppose that , with a Banach space (having norm ) and surjective linking maps , for all , and that , with a Banach space (having norm ) and linking maps for all . Then, for every , there exist and   resp. such that

where , , is the canonical projection of into (i.e., ).

Proof. If we define for and and for and , then and are fundamental sequences of seminorms generating the lc-topology of and of , respectively.

Fix . The continuity of implies that there exist and satisfying

or equivalently, that

As noted before such an inequality ensures that there exists defined via .

Denote by the closed unit ball of . Since is a quojection Fréchet space, there exists such that [17, Proposition 1]. Since is Montel (resp. reflexive) and is continuous, it follows from , with a relatively compact subset (resp. relatively weakly compact subset) of , that is a relatively compact (resp. relatively weakly compact) subset of . That is, (resp. ).

Remark 26. (i) Let be a quojection Fréchet space and . Suppose, for every , that there exists such that for (here, the notation is according to Lemma 25 and its proof with ). Then, for every , there exists satisfying . So, if (resp., ), then each (resp., ).

(ii) Let be a Fréchet space and . Then - if and only if -.

As , it suffices to show - implies -.

Since is a Fréchet space and -, the set is equicontinuous in ; that is, for every there exist and such that

Now, fix , , and . Choose and according to (55). Since is a Montel operator, is a relatively compact subset of and so there exist such that

with . Let . By (56) there exist and such that . Then, by (55), we have for every that

But, as . So, there exists (depending only on ) such that for every . Since is arbitrary and the set is finite, we can conclude that for . By the arbitrariness of and we have -.

The following result should be compared with Proposition 23(ii). We point out (even if ) that a Cesàro bounded operator need not satisfy in [4, p. 85].

Theorem 27. Let be a prequojection Fréchet space and . If -, then is uniformly mean ergodic.

Proof. We have the following two cases.

Case (I)  ( is a quojection). The condition - ensures that both - (see Remark 26(ii)) and that we can represent such that, for every , there exists satisfying ; see the proof of Proposition 8. According to Lemma 25 and Remark 26(i) we have for all . Moreover, in for ; see Remark 26(ii) and Lemma 6 with , for .

Since and in for , for every , each is uniformly mean ergodic [1, p. 711, Corollary 4], which implies that is also uniformly mean ergodic; see Lemma 7.

Case (II) ( is a prequojection). As noted before and are barrelled (hence, quasi-barrelled) with and . So, the condition - (see Remark 26(ii)) implies that -. Moreover, is a quojection Fréchet space. Also, Corollaries 2.3 and 2.4 of [33] yield that . We can then apply Case (I) to conclude that is uniformly mean ergodic. So, is also uniformly mean ergodic as and is a closed subspace of .

It was noted prior to Proposition 23, for a prequojection Fréchet space and , that is uniformly mean ergodic whenever -. Since in general, Theorem 27 can be viewed as an extension of this fact.

Corollary 28. Let be a prequojection Fréchet space and let be power bounded. Then if and only if -.

Proof. If -, then Theorem 20 yields .

Conversely, suppose that . Since is power bounded, in for and so is uniformly mean ergodic by Theorem 27. By Theorem 3.5 of [6] this is equivalent to the fact that is closed in . So, by Theorem 15(ii)(iii) we can conclude that -.

In a Banach space , an operator is called quasi-compact if there exist and such that [8, 6], [4, p. 88]. For example, if some power of is compact or if some power of has norm less than one, then is quasi-compact. For a quasi-compact operator it is known that - suffices for to be uniformly mean ergodic [1, Ch.VIII, Corollary 8.4]. For non-normable, the question arises of how to extend the notion of a quasi-compact operator.

According to [9, Definition 1], for a lcHs an operator is called quasi-precompact if there exists a 0-neighbourhood such that for every 0-neighbourhood in there exist and a finite set (both depending on ) with the property that . For a Banach space, this notion coincides precisely with being quasi-compact [9, Theorem 3]. In [41] an operator is called -compact if is a relatively compact subset of , where is some 0-neighbourhood in . More generally, is called -quasicompact [41, Definition 2.1], if there exist , a -compact operator and such that and .

Lemma 29. Let be a lcHs and let be any 0-neighbourhood in . Then every -quasicompact operator is quasi-precompact.

Proof. Let be -quasicompact. Choose , a -compact operator and such that the set is bounded and . Then

Proceeding inductively yields

Fix . Note that and need not commute. By expanding it can be seen that , where is a finite sum of operators all of the form or with and . The claim is that is a -compact operator. Indeed, since is always -compact and the finite sum of -compact operators is clearly -compact, it suffices to show that (hence, also ) is -compact for all .

For , observe that yields

which is a relatively compact subset of . For , we then have

and, hence, that

Since both and are relatively compact, it follows that is also relatively compact. This argument can be continued to yield the above stated claim for all .

Define now and let be any convex, balanced 0-neighbourhood of . Since is bounded, there is such that . Choose large enough to ensure that . It follows from (59) that

and so

But, is relatively compact and so there is a finite set such that . Accordingly,

which establishes that is quasi-precompact.

Returning to mean ergodicity, we have the following result of Pietsch [9, Theorem 7].

Fact 2. Let be a complete, barrelled lcHs and let be a quasi-precompact operator satisfying -. Then is uniformly mean ergodic and is finite dimensional.

In order to be able to extend this result to a larger class of operators we recall, for a Banach space , that is quasi-compact if and only if there exists a sequence such that [4, p. 88, Lemma 2.4].

Definition 30. Let be a lcHs. An operator is called quasi-Montel (resp., quasi reflexive) if there exists a sequence (resp., ) such that in as .

Remark 31. (i) Let be a Fréchet space and be quasi-Montel. Then is also quasi-Montel. Indeed, in the notation of Definition 30, we have [33, Corollaries 2.3 and 2.4], with in as ; see [27, Lemma 2.6] or [28, Lemma 2.1].

(ii) Let be a Fréchet space and be quasi-Montel. Then - if and only if -.

Again it suffices to show that - implies -.

Arguing as in Remark 26(ii), for every there exist and such that (55) holds. Fix , , and . Choose and according to (55). Since is a quasi-Montel operator, there is with in as . So there exists such that

But, and so is a relatively compact subset of . It follows that there exist such that

where . From (66) and (67) it follows that

Fix . By (68) there exist and such that . Then, by (55), for every we have that

But, as . So, there exists (depending only on ) such that , for every . Since is arbitrary and the set is finite, we can conclude that for . By the arbitrariness of and we have -.

Proposition 32. Let be a prequojection Fréchet space and let satisfy -. If is quasi-precompact, then there exists a sequence such that -. In particular, is quasi-Montel as .

Proof. The completeness of ensures that every precompact subset of is also relatively compact. By Fact 2 the operator is uniformly mean ergodic and so -. By [9, Theorems 1, 2 and Satz 10] there exist and a projection commuting with such that and satisfying

Since , also for each . Moreover, (70) yields , for , and so -. Since (71) is equivalent to , it then follows from Theorem 10 applied to that -. It is then clear (see (70)) that in as .

Remark 33. There exist quasi-Montel operators, even in quojection Fréchet spaces, which fail to be quasi-precompact.

(i) For , define the projection via

Since is a Montel space, all of its bounded subsets are relatively compact. It is then clear that , and hence, is surely quasi-Montel. Of course, . On the other hand, since is infinite-dimensional, cannot be quasi-precompact [9, Satz 3].

(ii) Let be as in (i) and define the diagonal operator by

The same argument as in (i) shows that . In this case, in contrast to (i), the space is finite-dimensional. However, still fails to be quasi-precompact [9, p. 24].

Remark 34. The converse of Proposition 32 is not valid. Indeed, let and let be as Remark 33(ii), in which case is a quojection Fréchet space. For each , let be the finite rank operator given by

Then is a 0-neighbourhood in . Since has finite-dimensional range, it follows that is a relatively compact subset of ; that is, for each . Direct calculations show that the sequence of operators

converges to 0 in as . Since is a Montel space, also -. However, as noted in Remark 33(ii), the diagonal operator is not quasi-compact.

In view of Remark 33 the following result is an extension of Fact 2 above for prequojection Fréchet spaces (without the condition ).

Theorem 35. Let be a prequojection Fréchet space and . If   is a quasi-Montel operator and -, then is uniformly mean ergodic.

Proof. We have the following two cases.

Case (I) ( is a quojection). The assumption - ensures that we can proceed as in the proof of Proposition 8 to obtain in such a way that, for every , there exists in satisfying . Then also and , for every . So, Lemma 6 (with , for ) implies that - for all .

Since is quasi-Montel, there exists a sequence such that -. From this it follows that the operator , for any fixed , is quasi-precompact. To see this, let denote the norm of and . Since , there exists such that

with chosen such that . Since

it follows that

Hence, by the relative compactness (hence, precompactness) of in , due to and the continuity of , there exist such that

By the arbitrariness of it follows that is quasi-precompact. As is a Banach space, is quasi-compact [9, Theorem 3] and satisfies in for . By Fact 2, each operator , for , is uniformly mean ergodic. Then Lemma 7 implies that is also uniformly mean ergodic.

Case (II) ( is a prequojection). The condition - actually means that - because is quasi-Montel (see Remark 31(ii)). So, arguing as for Case (II) in the proof of Theorem 27, it follows that also -. Moreover, by Remark 31(i) the operator is quasi-Montel. Since is a quojection Fréchet space, we can apply Case (I) to conclude that is uniformly mean ergodic. Then is also uniformly mean ergodic as with a closed subspace of .

Since the only Fréchet-Montel spaces which are normable are the finite-dimensional ones, the following result may be viewed as an analogue of the fact that is finite dimensional whenever is quasi-precompact; see Definition 3 and Theorem 1 of [9].

Proposition 36. Let be a Fréchet space and let be a quasi-Montel operator. Then is a Fréchet-Montel space, for every .

Proof. It suffices to show that is a Fréchet-Montel space. Indeed, for every , the operator is quasi-Montel if and only if is quasi-Montel, with .

Let be any fundamental, increasing sequence of seminorms generating the lc-topology of . Let be a bounded sequence. Since is quasi-Montel, there exists such that - and so, for every , we have as .

Since is bounded and each operator , for , is Montel, we may construct recursively subsequences of such that each is a subsequence of and converges in for all . Consider the diagonal sequence . Clearly, converges in for each (by observing that ). Fix and . Then, for every and , we have

with as . So, there is such that for every . But, converges in and, hence, there is also such that for all . It follows that whenever . By the arbitrariness of and this means that is a Cauchy sequence in and so it converges in . Since is a Fréchet space, this shows that is a Fréchet-Montel space.

Proposition 37. Let be a prequojection Fréchet space and be a quasi-Montel operator. If -, then is closed.

Proof. By Theorem 35 the operator is uniformly mean ergodic. Also -. By [6, Theorem 3.5] this is equivalent to being closed in .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The research of the first two authors was partially supported by the projects MTM2010-15200 and GVA Prometeo II/2013/013 (Spain). The second author gratefully acknowledges the support of the Alexander von Humboldt Foundation.

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