# Does a typical $\ell_{p}\,$\- space contraction have a non-trivial invariant
subspace?

Sophie Grivaux  CNRS, Laboratoire Paul Painlevé, UMR 8524
Université de Lille
Cité Scientifique, Bâtiment M2
59655 Villeneuve d’Ascq Cedex (France<EMAIL_ADDRESS>Étienne
Matheron  Laboratoire de Mathématiques de Lens
Université d’Artois
Rue Jean Souvraz SP 18
62307 Lens (France<EMAIL_ADDRESS>Quentin Menet  Service de
Probabilité et Statistique, Département de Mathématique
Université de Mons
Place du Parc 20
7000 Mons (Belgium<EMAIL_ADDRESS>
###### Abstract.

Given a Polish topology $\tau$ on ${\mathcal{B}}_{1}(X)$, the set of all
contraction operators on $X=\ell_{p}$, $1\leq p<\infty$ or $X=c_{0}$, we prove
several results related to the following question: does a typical
$T\in{\mathcal{B}}_{1}(X)$ in the Baire Category sense has a non-trivial
invariant subspace? In other words, is there a dense $G_{\delta}$ set
$\mathcal{G}\subseteq({\mathcal{B}}_{1}(X),\tau)$ such that every
$T\in\mathcal{G}$ has a non-trivial invariant subspace? We mostly focus on the
Strong Operator Topology and the Strong∗ Operator Topology.

###### keywords:

Polish topologies, $\ell_{p}\,$\- spaces, typical properties of operators,
invariant subspaces, supercyclic vectors, Lomonosov Theorem

###### 1991 Mathematics Subject Classification:

47A15, 47A16, 54E52

This work was supported in part by the project FRONT of the French National
Research Agency (grant ANR-17-CE40-0021) and by the Labex CEMPI
(ANR-11-LABX-0007-01). The third author is a Research Associate of the Fonds
de la Recherche Scientifique - FNRS

##  1\. Introduction

Unless otherwise specified, all the Banach spaces – and hence all the Hilbert
spaces – considered in this paper are complex, infinite-dimensional, and
_separable_. If $X$ is a Banach space, we denote by ${\mathcal{B}}(X)$ the
space of all bounded linear operators on $X$ endowed with its usual norm. For
any $M>0$,

${\mathcal{B}}_{M}(X):=\\{T\in{\mathcal{B}}(X)\;;\;\|T\|\leq M\\}$

is the closed ball of radius $M$ in ${\mathcal{B}}(X)$. In particular,
${\mathcal{B}}_{1}(X)$ is the set of all _contractions_ of the Banach space
$X$.

In this paper, we will be interested in _typical_ properties of Banach space
contractions. The word “typical” is to be understood in the Baire category
sense: given a Baire topological space $\mathcal{Z}$ and any property (P) of
elements of $\mathcal{Z}$, we say that _a typical $z\in\mathcal{Z}$ satisfies
_(P)__ if the set $\\{z\in\mathcal{Z}\;;\;z\ \textrm{satisfies (P)}\\}$ is
comeager in $\mathcal{Z}$, i.e. this set contains a dense $G_{\delta}$ subset
of $\mathcal{Z}$. So, given a Banach space $X$, we need topologies on
${\mathcal{B}}_{1}(X)$ which turn ${\mathcal{B}}_{1}(X)$ into a Baire space.
The operator norm topology does so, but it appears to be too strong to get
interesting results by Baire category arguments; in particular, the fact that
$({\mathcal{B}}_{1}(X),\|\,\cdot\,\|)$ is “usually” non-separable is a real
disadvantage. However, there are quite a few weaker topologies on
${\mathcal{B}}(X)$ whose restriction to each ball ${\mathcal{B}}_{M}(X)$ is
Polish, i.e. completely metrizable and separable. We will mostly focus on two
of them: the _Strong Operator Topology_ (SOT) and the _Strong ${}^{\,*}$
Operator Topology_ (SOT∗). Recall that SOT is just the topology of pointwise
convergence, and that SOT∗ is the topology of pointwise convergence for
operators and their adjoints: a net $(T_{i})$ in ${\mathcal{B}}(X)$ converges
to $T\in{\mathcal{B}}(X)$ with respect to SOT if and only if $T_{i}x\to Tx$ in
norm for every $x\in X$, and $T_{i}\to T$ with respect to SOT∗ if and only if
$T_{i}x\to Tx$ for every $x\in X$ and $T_{i}^{*}x^{*}\to T^{*}x^{*}$ for every
$x^{*}\in X^{*}$. The _Weak Operator Topology_ (WOT) will also make an
occasional appearance ($T_{i}\to T$ with respect to WOT if and only if
$T_{i}x\to Tx$ weakly for every $x\in X$). Even though these topologies behave
badly when considered on the whole space ${\mathcal{B}}(X)$, each closed ball
${\mathcal{B}}_{M}(X)$ is indeed Polish when endowed with SOT, and the same
holds true for SOT∗ if one assumes additionally that $X^{*}$ is separable.
(For WOT, it is safer to assume $X$ is reflexive; and then
$\mathcal{B}_{M}(X)$ is even compact and metrizable.)

The study of typical properties of contractions was initiated, in a Hilbertian
setting, by Eisner [E] and Eisner-Mátrai [EM]. Given a Hilbert space $H$
(complex, infinite-dimensional and separable), Eisner studied in [E]
properties of typical operators $T\in{\mathcal{B}}_{1}(H)$ for the Weak
Operator Topology and proved that a typical
$T\in({\mathcal{B}}_{1}(H),\texttt{WOT})$ is unitary. The situation turns out
to be completely different if one considers the Strong Operator Topology:
indeed, it was shown in [EM] that a typical
$T\in({\mathcal{B}}_{1}(H),\texttt{SOT})$ is unitarily equivalent to the
backward shift of infinite multiplicity acting on
$\ell_{2}\bigl{(}\mathbb{Z}_{+},\ell_{2}\bigr{)}$. So the behaviour of typical
contractions on the Hilbert space is essentially fully understood in the SOT
case, and rather well so in the WOT case. The general picture in the SOT∗ case
is more complicated, and this was studied in some detail in [GMM].

Let $X$ be a Banach space, and let $\tau$ be a topology on ${\mathcal{B}}(X)$
turning the ball ${\mathcal{B}}_{1}(X)$ into a Polish space. If
${\mathcal{G}}$ is a subset of ${\mathcal{B}}(X)$ such that
${\mathcal{G}}_{1}:={\mathcal{G}}\cap{\mathcal{B}}_{1}(X)$ is a $G_{\delta}$
subset of ${\mathcal{B}}_{1}(X)$ for the topology $\tau$, then
$({\mathcal{G}}_{1},\tau)$ is also a Polish space in its own right. Our
initial goal in this paper was to determine, for certain suitable subsets
${\mathcal{G}}\subseteq{\mathcal{B}}(X)$, whether a typical contraction $T$
from ${\mathcal{G}}$ has a non-trivial invariant subspace, i.e. a closed
linear subspace $E\subseteq X$ with $E\neq\\{0\\}$ and $E\neq X$ such that
$T(E)\subseteq E$.

This is of course motivated by the famous _Invariant Subspace Problem_ , which
asks, for a given Banach space $X$, whether _every_ operator
$T\in{\mathcal{B}}(X)$ has a non-trivial invariant subspace. The Invariant
Subspace Problem was solved in the negative by Enflo in the 70’s [En] for some
peculiar Banach space; and then by Read [R1], who subsequently exhibited in
several further papers examples of operators without invariant subspaces on
some classical Banach spaces like $\ell_{1}$ and $c_{0}$. See [R2], [R3], and
also [GR2] for a unified approach to these constructions. On the other hand,
there are also lots of positive results. To name a few: the classical
Lomonosov Theorem from [L] (any operator whose commutant contains a non-scalar
operator commuting with a non-zero compact operator has a non-trivial
invariant subspace); the Brown-Chevreau-Pearcy Theorem from [BCP2] (any
Hilbert space contraction whose spectrum contains the whole unit circle has a
non-trivial invariant subspace); and more recently, the construction by
Argyros and Haydon [AH] of Banach spaces $X$ for which the Invariant Subspace
Problem has a _positive_ answer. We refer to the books [RR] and [CP], as well
as to the survey [CE], for a comprehensive overview of the known methods
yielding the existence of non-trivial invariant subspaces for various classes
of operators on Banach spaces.

A problem closely related to the Invariant Subspace Problem is the Invariant
_Subset_ Problem, which asks (for a given Banach space $X$) whether every
operator on $X$ admits a non-trivial invariant closed set; equivalently, if
there exists at least one $x\neq 0$ in $X$ whose orbit under the action of $T$
is not dense in $X$. Again, this question was solved in the negative by Read
in [R4], who constructed an operator on the space $\ell_{1}$ with no non-
trivial invariant closed set. Further examples of such operators on a class of
Banach spaces including $\ell_{1}$ and $c_{0}$ were exhibited in [GR2]. One
may also add that the Invariant Subset Problem has been a strong motivation
for the study of _hypercyclic_ operators. Recall that an operator
$T\in\mathcal{L}(X)$ is said to be hypercyclic if there exists some vector
$x\in X$ whose orbit under the action of $T$ is dense in $X$; in which case
$x$ is said to be hypercyclic for $T$. With this terminology, the Invariant
Subset Problem for $X$ asks whether there exists an operator
$T\in\mathcal{L}(X)$ such that every $x\neq 0$ is hypercyclic for $T$. We
refer to the books [BM] and [GEP] for more information on hypercyclicity and
other dynamical properties of linear operators.

Despite considerable efforts, the Invariant Subspace Problem and the Invariant
Subset Problem remain stubbornly open in the reflexive setting, and in
particular for the Hilbert space. Therefore, it seems quite natural to address
the a priori more tractable “generic” version of the problem; and this is
meaningful even on spaces like $\ell_{1}$ or $c_{0}$, since the mere existence
of counterexamples says nothing about the typical character of the property of
having a non-trivial invariant subspace.

We will mostly focus on operators on $X=\ell_{p}$, $1\leq p<\infty$ or
$X=c_{0}$. This already offers a wide range of possibilities: the spaces
$\ell_{1}$ and $c_{0}$ are known to support operators without non-trivial
invariant subspaces (and even without non-trivial invariant closed sets),
whereas the spaces $\ell_{p}$ for $1<p<\infty$ are all reflexive – so it is
not known if they support operators without non-trivial invariant subspaces –
but have rather different geometric properties in the three cases $p=2$,
$1<p<2$ and $p>2$. The main part of our work will consist in the study of the
spaces $({\mathcal{B}}_{1}(X),\tau)$ for one of the topologies
$\tau=\texttt{SOT}$ or $\tau=\texttt{SOT}\mbox{${}^{*}$}$ (when
$\tau=\texttt{SOT}\mbox{${}^{*}$}$, we have to exclude $X=\ell_{1}$ since it
has a non-separable dual).

It follows from the topological $0\;$-$\,1$ law that on any of the spaces
$X=\ell_{p}$ or $c_{0}$, either a typical contraction has a non-trivial
invariant subspace, or a typical contraction does not have a non-trivial
invariant subspace (see Proposition 3.2). In view of the complexity of the
known counterexamples to the Invariant Subspace Problem on $\ell_{1}$ and
$c_{0}$, and of the very serious difficulty encountered when trying to
construct such a counterexample on a reflexive space, it seems reasonable to
expect that the correct alternative is the first one, i.e. a typical
contraction on $X$ does have a non-trivial invariant subspace; but our efforts
to prove that have remained unsuccesful, except in the “trivial” case
$X=\ell_{2}$ (see below) and for $X=\ell_{1}$. All is not lost, however, since
we do obtain a few significant results related to our initial goal. In the
next section, we describe in some detail the contents of the paper.

###  1a. Notations

The following notations will be used throughout the paper.

\- We denote by $\mathbb{D}$ the open unit disk in $\mathbb{C}$, and by
$\mathbb{T}$ the unit circle

\- If $(x_{i})_{i\in I}$ is any family of vectors in a Banach space $X$, we
denote by $[\,x_{i}\;;\;i\in I\,]$ its closed linear span in $X$.

\- When $X=\ell_{p}$, $1\leq p<\infty$ or $X=c_{0}$, we denote by
$(e_{j})_{j\geq 0}$ the canonical basis of $X$, and by $(e_{j}^{*})_{j\geq
0}\subseteq X^{*}$ the associated sequence of coordinate functionals. For any
set $I$ of nonnegative integers, we denote by $P_{I}:X\to[e_{i}\,;\,i\in I]$
the canonical projection of $X$ onto $[e_{i}\,;\,i\in I]$. When $I=[0,N]$ for
some $N\geq 0$, we may also write $P_{N}$ instead of $P_{[0,N]}$. For every
$N\geq 0$, we set $E_{N}:=[e_{0},\dots,e_{N}]$ and $F_{N}:=[e_{j};\;j>N]$.

\- If $T$ is a bounded operator on a (complex) Banach space $X$, we denote by
$\sigma(T)$ its spectrum, by $\sigma_{e}(T)$ its essential spectrum, by
$\sigma_{p}(T)$ its point spectrum – i.e. the eigenvalues of $T$ – and by
$\sigma_{ap}(T)$ its approximate point spectrum, i.e. the set of all
$\lambda\in\mathbb{C}$ such that $T-\lambda$ is not bounded below.

##  2\. Main results

In Section 3, we review some properties of typical contractions of $\ell_{2}$
for the topology SOT, and we prove some general properties of SOT \- typical
contractions of the spaces $\ell_{p}$ and $c_{0}$. We mention in particular
the following result (labeled as Proposition 3.9).

######  Proposition 2.1.

Let $X=\ell_{p}$, $1\leq p<\infty$, or $X=c_{0}$. A typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has the following spectral
properties: $T-\lambda$ has dense range for every $\lambda\in\mathbb{C}$, and
$\sigma(T)=\sigma_{{ap}}(T)=\overline{\,\mathbb{D}}$.

When $X=\ell_{2}$, the Eisner-Mátrai result from [EM] mentioned above implies
that a typical operator $T$ in $({\mathcal{B}}_{1}(\ell_{2}),\texttt{SOT})$
satisfies much stronger properties: $T^{*}$ is a non-surjective isometry,
$T-\lambda$ is surjective for every $\lambda\in\mathbb{D}$, and every
$\lambda\in\mathbb{D}$ is an eigenvalue of $T$ with infinite multiplicity. It
immediately follows that an SOT \- typical contraction on $\ell_{2}$ has an
enormous amount of invariant subspaces.

A natural step towards the understanding of the behaviour of SOT \- typical
contractions on other $\ell_{p}\,$\- spaces is to determine whether they enjoy
similar properties. Somewhat surprisingly, it turns out that SOT \- typical
contractions on $\ell_{1}$ behave very much like SOT \- typical contractions
on $\ell_{2}$ in this respect. Indeed, we prove in Section 4 the following
result (labeled as Theorem 4.1):

######  Theorem 2.2.

Let $X=\ell_{1}$. A typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$
satisfies the following properties: $T^{*}$ is a non-surjective isometry,
$T-\lambda$ is surjective for every $\lambda\in\mathbb{D}$, and every
$\lambda\in\mathbb{D}$ is an eigenvalue of $T$ with infinite multiplicity. In
particular, a typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has non-
trivial invariant subspaces.

The situation is radically different when $X=\ell_{p}$ for $1<p<\infty$ and
$p\neq 2$, or when $X=c_{0}$. It is not hard to show that in this case, the
set of all co-isometries is nowhere dense in
$({\mathcal{B}}_{1}(X),\texttt{SOT})$, so that a statement analogous to the
first part of Theorem 2.2 cannot be true (see Proposition 5.1). Moreover, we
prove in Section 5 that, at least when $X=\ell_{p}$ for $p>2$ or $X=c_{0}$,
the typical behaviour regarding the eigenvalues is in fact to have no
eigenvalue at all. More precisely, we obtain the following results (see
Theorem 5.2 and Theorem 5.3).

######  Theorem 2.3.

If $X=c_{0}$, then a typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has no
eigenvalue. If $X=\ell_{p}$ with $p>2$ then, for any $M>1$, a typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is such that $(MT)^{*}$ is
hypercyclic.

The conclusion in the $\ell_{p}\,$\- case is indeed stronger than asserting
that $T$ has no eigenvalue, since it is well known that the adjoint of a
hypercyclic operator cannot have eigenvalues.

Our proofs in the $c_{0}\,$\- case and in the $\ell_{p}\,$\- case are quite
different. In the $c_{0}\,$\- case, we make use of the so-called _Banach-
Mazur_ game. In the $\ell_{p}\,$\- case, we observe that the result is already
known _for the topology_ SOT∗ by [GMM], and then we prove the following rather
unexpected fact, which might be useful in other situations: if $p>2$, then any
SOT∗\- comeager subset of $\mathcal{B}_{1}(\ell_{p})$ is also SOT \- comeager
(see Theorem 5.12). Our proof of this latter result heavily relies on
properties of the $\ell_{p}\,$\- norm which are specific to the case $p>2$,
and so it does not extend to the case $1<p<2$. In fact, we have not been able
to determine whether an SOT \- typical contraction on $X=\ell_{p}$, $1<p<2$
has eigenvalues. We do not know either if hypercyclicity of $(MT)^{*}$ is
typical in the $c_{0}\,$\- case.

Theorem 2.3 might suggest that in fact, an SOT \- typical contraction on
$\ell_{p}$, $p>2$ does _not_ have a non-trivial invariant subspace. If it were
true, this would solve the Invariant Subspace Problem for $\ell_{p}$,
presumably with rather “soft” arguments; so we would not bet anything on it.
To support this feeling, a result of Müller [M2] can be brought to use to
prove that for $X=\ell_{p}$ with $1<p<\infty$, a typical
$T\in({\mathcal{B}}_{1}(X),{\texttt{SOT}})$ has a non-trivial invariant
_closed convex cone_ (this is Proposition 5.26). When $X=c_{0}$, we show that
a typical $T\in({\mathcal{B}}_{1}(X),{\texttt{SOT}})$ has a non-zero _non-
supercyclic_ vector – i.e. a vector $x\neq 0$ whose scaled orbit $\\{\lambda
T^{n}x;\;n\geq 0\,,\,\lambda\in\mathbb{C}\\}$ is not dense in $X$ – and hence
a non-trivial invariant star-shaped closed set (this is Proposition 5.27).

The following table summarizes most of our knowledge about SOT \- typical
contractions on $\ell_{p}$ or $c_{0}$.

| $p=1$ | $1<p<2$ | $p=2$ | $2<p<\infty$ | $c_{0}$
---|---|---|---|---|---
$\sigma(T)$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$
| (Prop 3.9) | (Prop 3.9) | (Eisner-Mátrai) | (Prop 3.9) | (Prop 3.9)
$T^{*}$ is an isometry | Yes | No | Yes | No | No
| (Thm 4.1) | (Prop 5.1) | (Eisner-Mátrai) | (Prop 5.1) | (Prop 5.1)
$T$ has a non-trivial | Yes | Yes | Yes | Yes | ?
invariant closed cone | (Thm 4.1) | (Prop 5.26) | (Eisner-Mátrai) | (Prop 5.26) |
$T$ has a non-zero | Yes | Yes | Yes | Yes | Yes
non-supercyclic vector | (Thm 4.1) | (Prop 5.26) | (Eisner-Mátrai) | (Prop 5.26) | (Prop 5.27)
$\sigma_{p}(T)$ | $\mathbb{D}$ | $?$ | $\mathbb{D}$ | $\emptyset$ | $\emptyset$
| (Thm 4.1) |  | (Eisner-Mátrai) | (Thm 5.3) | (Thm 5.2)
$T$ has a non-trivial | Yes | ? | Yes | ? | ?
invariant subspace | (Thm 4.1) |  | (Eisner-Mátrai) |  |

Table 1. Properties of SOT \- typical $T\in{\mathcal{B}}_{1}(X)$

In Section 6, we change the setting and consider a subclass
${\mathcal{T}}_{1}(X)$ of contractions on $X=\ell_{p}$, $1\leq p<\infty$ or
$X=c_{0}$ defined as follows: if $(e_{j})_{j\geq 0}$ denotes the canonical
basis of $X$, then ${\mathcal{T}}_{1}(X)$ consists of all operators
$T\in{\mathcal{B}}_{1}(X)$ which are “triangular plus 1” with respect to
$(e_{j})$, with positive entries on the first subdiagonal:

${\mathcal{T}}_{1}(X)=\bigl{\\{}T\in{\mathcal{B}}_{1}(X)\;;\;Te_{j}\in[\,e_{0},\dots,e_{j+1}\,]\
\textrm{and}\ {\langle Te_{j},e_{j+1}\rangle}>0\ \textrm{for every}\ j\geq
0\bigr{\\}}.$

The class ${\mathcal{T}}_{1}(X)$ is rather natural for at least two reasons.
Firstly, in the Hilbertian setting, any cyclic operator
$T\in{\mathcal{B}}_{1}\bigl{(}\ell_{2}\bigr{)}$ is unitarily equivalent to
some operator in ${\mathcal{T}}_{1}\bigl{(}\ell_{2}\bigr{)}$. Secondly, all
the operators without non-trivial invariant subspaces constructed by Read in
[R1], [R2], [R3] belong to the classes ${\mathcal{T}}_{1}(\ell_{1})$ or
${\mathcal{T}}_{1}\bigl{(}c_{0}\bigr{)}$. The extensions considered in [GR2]
are also of this form, as well as the operators with few non-trivial invariant
subspaces constructed in [GR1]. Thus, “Read’s type operators” on $X=\ell_{p}$
or $c_{0}$ belong to ${\mathcal{T}}_{1}(X)$, and hence it is quite natural to
investigate whether an SOT \- typical or SOT∗\- typical operator from
${\mathcal{T}}_{1}(X)$ has a non-trivial invariant subspace. Again, our
results here are rather partial. We prove the following “local” version of the
Eisner-Mátrai result from [EM] (see Corollary 6.6).

######  Theorem 2.4.

Let $X=\ell_{2}$. A typical $T\in({\mathcal{T}}_{1}(X),\emph{{SOT}})$ is
unitarily equivalent to the backward shift of infinite multiplicity acting on
$\ell_{2}\bigl{(}\mathbb{Z}_{+},\ell_{2}\bigr{)}$. In particular, it has
eigenvalues, and hence non-trivial invariant subspaces.

In view of the results of Section 4, it is not too surprising that we are also
able to settle the case $X=\ell_{1}$ (see Proposition 6.7):

######  Theorem 2.5.

Let $X=\ell_{1}$. A typical $T\in({\mathcal{T}}_{1}(X),\emph{{SOT}})$ has all
the properties listed in _Theorem 2.2_. In particular, it has a non-trivial
invariant subspace.

We move over in Section 7 to the study of typical contractions on
$X=\ell_{p}$, $1<p<\infty$ with respect to the topology SOT∗. The spectral
properties of SOT∗\- typical contractions are rather different from those of
SOT \- typical contractions, at least for $p=2$, since for example a typical
$T\in(\mathcal{B}(\ell_{2}),\texttt{SOT}\mbox{${}^{*}$})$ has no eigenvalue
(see [EM] or [GMM]). Regarding invariant subspaces, the situation is exactly
the same as in the SOT \- case: we can prove that an SOT∗\- typical
contraction on $X=\ell_{p}$ has a non-trivial invariant subspace only in the
Hilbertian case $p=2$ (recall that $p=1$ is not considered in the SOT∗
setting); and to do that, we have to use a quite non-trivial result, namely
the Brown-Chevreau-Pearcy Theorem from [BCP2]. We are also able to prove
(using the result of Müller [M2]) that for any $1<p<\infty$, a typical
$T\in(\mathcal{B}_{1}(\ell_{p}),{\texttt{SOT}\mbox{${}^{*}$}})$ has a non-
trivial invariant closed cone. These facts are summarized in the following
table.

| $1<p<2$ | $p=2$ | $2<p<\infty$
---|---|---|---
$\sigma(T)$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$ | $\overline{\mathbb{D}}$
| (Prop 7.2) | (Prop 7.2) | (Prop 7.2)
$T^{*}$ is an isometry | No | No | No
| (Prop 7.1) | (Prop 7.1) | (Prop 7.1)
$T$ has a non-trivial | Yes | Yes | Yes
invariant closed cone | (Cor 7.4) | (Cor 7.4) | (Cor 7.4)
$\sigma_{p}(T)$ | $\emptyset$ | $\emptyset$ | $\emptyset$
| (Prop 7.1) | (Prop 7.1) | (Prop 7.1)
$T$ has a non-trivial | ? | Yes | ?
invariant subspace |  | (Cor 7.3) |

Table 2. Properties of SOT∗\- typical $T\in{\mathcal{B}}_{1}(X)$

Since SOT∗\- typical contractions of a Hilbert space are by far not so well
understood as SOT \- typical contractions, it is interesting to investigate
whether they satisfy some of the standard criteria implying the existence of a
non-trivial invariant subspace. What about the Lomonosov Theorem, for
instance? In this direction, we obtain the following result (this is Theorem
7.5):

######  Theorem 2.6.

Let $X=\ell_{2}$. A typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}\mbox{${}^{*}$}})$ does not commute with
any non-zero compact operator.

It should be apparent that this paper rises more questions than it answers. We
collect some of them in Section 8.

##  3\. Some general properties of SOT \- typical contractions

For the convenience of the reader, we begin this section by giving a quick
proof of the fact that when $X$ is a (separable) Banach space, the closed
balls $({\mathcal{B}}_{M}(X),\texttt{SOT})$ are indeed Polish spaces. The
proof for SOT∗ when $X^{*}$ is separable would be essentially the same.

######  Lemma 3.1.

For any (separable) Banach space $X$ and any $M>0$, the ball
${\mathcal{B}}_{M}(X)$ is a Polish space when endowed with the topology _SOT_.

###### Proof.

Let $\mathbf{Q}$ be a countable dense subfield of $\mathbb{C}$, and let
$Z\subseteq X$ be countable dense subset of $X$ which is also a vector space
over $\mathbf{Q}$. Consider the set

$\mathcal{L}_{M}:=\bigl{\\{}L:Z\to X\;;\;L\;\hbox{is $\mathbf{Q}$-linear and
$\forall z\in Z\;:\;\|L(z)\|\leq M\|z\|$}\bigr{\\}}.$

This is a closed subset of $X^{Z}$ with respect to the product topology, hence
a Polish space. Moreover, it is easily checked that the map $T\mapsto T_{|Z}$
is a homeomorphism from $({\mathcal{B}}_{M}(X),\texttt{SOT})$ onto
$\mathcal{L}_{M}$. So $({\mathcal{B}}_{M}(X),\texttt{SOT})$ is Polish. ∎

Next, we prove a general result showing that all the properties we will be
considering in this paper are either typical or “atypical”. If $X$ is a
(separable) Banach space, let us denote by ${\rm Iso}(X)$ the group of all
surjective linear isometries $J:X\to X$. We say that a set
$\mathcal{A}\subseteq{\mathcal{B}}_{1}(X)$ is _${\rm Iso}(X)\,$ \- invariant_
if $J\mathcal{A}J^{-1}=\mathcal{A}$ for every $J\in{\rm Iso}(X)$.

######  Proposition 3.2.

Let $X=\ell_{p}$, $1\leq p<\infty$ or $X=c_{0}$. If
$\mathcal{A}\subseteq({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has the Baire
property and is ${\rm Iso}(X)\,$\- invariant, then $\mathcal{A}$ is either
meager or comeager in ${\mathcal{B}}_{1}(X)$.

###### Proof of Proposition 3.2.

Recall first that ${\rm Iso}(X)$ becomes a Polish group when endowed with SOT.
This is well known and true for any (separable) Banach space $X$; but we give
a quick sketch of proof for convenience of the reader. First, ${\rm Iso}(X)$
is a $G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$, because an
operator $J\in{\mathcal{B}}_{1}(X)$ belongs to ${\rm Iso}(X)$ if and only if
it is an isometry with dense range (the first condition obviously defines a
closed set, and the second one is easily seen to define a $G_{\delta}$ set).
So $({\rm Iso}(X),\texttt{SOT})$ is a Polish space. Next, multiplication is
continuous on ${\rm Iso}(X)\times{\rm Iso}(X)$ since it is continuous on
${\mathcal{B}}_{1}(X)\times{\mathcal{B}}_{1}(X)$. Finally, if $(J_{n})$ is a
sequence in ${\rm Iso}(X)$ such that $J_{n}\to J\in{\rm Iso}(X)$, then
$\|J_{n}^{-1}x-J^{-1}x\|=\|x-J_{n}J^{-1}x\|\to 0$ for every $x\in X$, so
$J_{n}^{-1}\to J^{-1}$; hence the map $J\mapsto J^{-1}$ is continuous on ${\rm
Iso}(X)$.

The Polish group ${\rm Iso}(X)$ acts continuously by conjugacy on
${\mathcal{B}}_{1}(X)$. By the _topological $0\,$-$\,1$ law_ (see [Ke, Theorem
8.46]), it is enough to show that the action is topologically transitive, i.e.
that for any pair $(\mathcal{U},\mathcal{V})$ of non-empty open subsets of
$({\mathcal{B}}_{1}(X),\texttt{SOT})$, one can find $J\in{\rm Iso}(X)$ such
that $(J\mathcal{U}J^{-1})\cap\mathcal{V}\neq\emptyset$.

Choose $A\in\mathcal{U}$ and $B\in\mathcal{V}$. For any integer $N\geq 0$, set
$A_{N}:=P_{N}AP_{N}$ and $B_{N}:=P_{N}BP_{N}$, considered as operators on
$E_{N}=[e_{0},\dots,e_{N}]$. Denote by $\widetilde{A}_{N}$ and
$\widetilde{B}_{N}$ the copies of $A_{N}$ and $B_{N}$ living on
$\widetilde{E}_{N}:=[e_{N+1},\dots,e_{2N+1}]$. If $N$ is large enough, then
the operator $T_{N}:=A_{N}\oplus\widetilde{B}_{N}$ – considered as an operator
on $X$ – belongs to $\mathcal{U}$, and $S_{N}:=B_{N}\oplus\widetilde{A}_{N}$
belongs to $\mathcal{V}$. Let us fix such an integer $N$, and let $J\in{\rm
Iso}(X)$ be the isometry exchanging $e_{j}$ and $e_{N+1+j}$ for $j=0,\dots,N$
and such that $Je_{k}=e_{k}$ for all $k>2N+1$. Then $JT_{N}J^{-1}=S_{N}$; so
we have indeed $(J\mathcal{U}J^{-1})\cap\mathcal{V}\neq\emptyset$. ∎

######  Corollary 3.3.

If $X=\ell_{p}$ or $c_{0}$ then, either a typical $T\in{\mathcal{B}}_{1}(X)$
has a non-trivial invariant subspace, or a typical $T\in{\mathcal{B}}_{1}(X)$
does not have a non-trivial invariant subspace.

###### Proof.

Let $\mathcal{A}$ be the set of all $T\in{\mathcal{B}}_{1}(X)$ having a non-
trivial invariant subspace. It is clear that $\mathcal{A}$ is ${\rm
Iso}(X)\,$\- invariant; so we just have to show that $\mathcal{A}$ has the
Baire property. Now, if $T\in{\mathcal{B}}_{1}(X)$ then

$T\in\mathcal{A}\iff\exists(x,x^{*})\in X\times B_{X^{*}}\;:\;\bigl{(}x\neq
0\;,\;x^{*}\neq 0\quad{\rm and}\quad\forall n\in\mathbb{Z}_{+}\;:\;{\langle
x^{*},T^{n}x\rangle}=0\bigl{)}.$

Since the relation under brackets defines a Borel (in fact, $F_{\sigma}$)
subset of the Polish space ${\mathcal{B}}_{1}(X)\times
X\times(B_{X^{*}},w^{*})$, this shows that $\mathcal{A}$ is an _analytic_
subset of ${\mathcal{B}}_{1}(X)$. In particular, $\mathcal{A}$ has the Baire
property. (See [Ke] for background on analytic sets.) ∎

As mentioned in the introduction, we know what is actually true only for
$X=\ell_{2}$ and $X=\ell_{1}$. Likewise, one can say that either a typical
$T\in{\mathcal{B}}_{1}(X)$ has eigenvalues, or a typical
$T\in{\mathcal{B}}_{1}(X)$ does not have eigenvalues; but we do not know what
is true for $X=\ell_{p}$, $1<p<2$.

When $X$ is a Hilbert space (so that we call it $H$), the SOT \- typical
properties of contractions are essentially fully understood, thanks to the
following result of Eisner and Mátrai [EM]. Let us denote by
$\ell_{2}(\mathbb{Z}_{+},\ell_{2})$ the infinite direct sum of countably many
copies of $\ell_{2}$, and let $B_{\infty}$ be the canonical backward shift
acting on $\ell_{2}(\mathbb{Z}_{+},\ell_{2})$, i.e. the operator defined by
$B_{\infty}(x_{0},x_{1},x_{2},\dots):=(x_{1},x_{2},x_{3},\dots)$ for every
$(x_{0},x_{1},x_{2},\dots)\in\ell_{2}(\mathbb{Z}_{+},\ell_{2})$.

######  Theorem 3.4.

A typical $T\in({\mathcal{B}}_{1}(H),\emph{{SOT}})$ is unitarily equivalent to
$B_{\infty}$.

This theorem is proved in [EM] by showing that a typical
$T\in({\mathcal{B}}_{1}(H),{\texttt{SOT}})$ has the following properties:
$T^{*}$ is an isometry, $\dim\ker(T)=\infty$, and $\|T^{n}x\|\to 0$ as
$n\to\infty$ for every $x\in H$. The result then follows from the Wold
decomposition theorem.

######  Corollary 3.5.

A typical $T\in({\mathcal{B}}_{1}(H),\emph{{SOT}})$ has the following
properties.

  1. (a)

$T^{*}$ is a non-surjective isometry.

  2. (b)

$T-\lambda$ is surjective for every $\lambda\in\mathbb{D}$.

  3. (c)

Every $\lambda\in\mathbb{D}$ is an eigenvalue of $T$ with infinite
multiplicity.

  4. (d)

$\sigma(T)=\sigma_{e}(T)=\sigma_{ap}(T)=\overline{\,\mathbb{D}}$.

  5. (e)

$\|T^{n}x\|\to 0$ for every $x\in H$.

In particular, it follows immediately from (c) that an SOT \- typical
contraction on $H$ has a wealth of invariant subspaces. So we may state

######  Corollary 3.6.

A typical $T\in({\mathcal{B}}_{1}(H),\emph{{SOT}})$ has a non-trivial
invariant subspace.

One may also note that property (a) by itself implies as well that $T$ has a
non-trivial invariant subspace. Indeed, by a classical result of Godement
[Go], any Banach space isometry has a non-trivial invariant subspace. Hence,
if $X$ is a reflexive Banach space, then any operator $T\in{\mathcal{B}}(X)$
such that $T^{*}$ is an isometry has a non-trivial invariant subspace.

In the next three propositions, we show that something remains of Corollary
3.5 in a non-Hilbertian setting.

######  Proposition 3.7.

Let $X$ be a Banach space. A typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$
is such that $\|T^{n}x\|\to 0$ as $n\to\infty$ for all $x\in X$.

###### Proof.

The key point is that if $T\in{\mathcal{B}}_{1}(X)$ is a contraction and $x\in
X$, then the sequence $\|T^{n}x\|$ is non-increasing, and hence $T^{n}x\to 0$
if and only if $\underline{\lim}\,\|T^{n}x\|=0$. Let us denote by
${\mathcal{C}}_{0}$ the set of all $T\in{\mathcal{B}}_{1}(X)$ such that
$\forall x\in X\;:\;\|T^{n}x\|\to 0$. Let also $Z$ be a countable dense subset
of $X$. By what we have just said, the following equivalence holds true for
any $T\in{\mathcal{B}}_{1}(X)$:

$T\in{\mathcal{C}}_{0}\iff\forall z\in Z\;\forall K\in\mathbb{N}\;\,\exists
n\;:\;\|T^{n}z\|<1/K.$

Since the map $T\mapsto T^{n}$ is SOT-continuous on ${\mathcal{B}}_{1}(X)$ for
each $n\in\mathbb{N}$, this shows that ${\mathcal{C}}_{0}$ is a $G_{\delta}$
subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$. Moreover, ${\mathcal{C}}_{0}$
is dense in ${\mathcal{B}}_{1}(X)$ because it contains every operator $T$ with
$\|T\|<1$. ∎

Recall that a Banach space $X$ is said to have the _Metric Approximation
Property_ (for short, the MAP) if, for any compact set $K\subseteq X$ and
every $\varepsilon>0$, one can find a finite rank operator
$R\in{\mathcal{B}}(X)$ with $\|R\|\leq 1$ such that $\|Rx-x\|<\varepsilon$ for
all $x\in K$. Equivalently, $X$ has the MAP if and only if there exists a net
$(R_{i})\subseteq{\mathcal{B}}_{1}(X)$ consisting of finite rank operators
such that $R_{i}\xrightarrow{\texttt{SOT}}Id$ (and since $X$ is assumed to be
separable, one can in fact take a _sequence_ $(R_{i})$ with this property).
See e.g. [LT] for more on this notion.

######  Proposition 3.8.

Let $X$ be a Banach space which has the _MAP_. Then, the set of all
$T\in{\mathcal{B}}_{1}(X)$ such that $0\in\sigma_{ap}(T)$ is a dense
$G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\emph{{SOT}})$. In particular, a
typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is non-invertible.

###### Proof.

Set ${{\mathcal{G}}}:=\\{T\in{\mathcal{B}}_{1}(X);\;0\in\sigma_{ap}(T)\\}$.
Then,

$T\in{{\mathcal{G}}}\iff\forall K\in\mathbb{N}\;\exists x\in
S_{X}\;:\;\|Tx\|<1/K,$

so ${{\mathcal{G}}}$ is a $G_{\delta}$ subset of
$({\mathcal{B}}_{1}(X),{\texttt{SOT}})$. Moreover, ${{\mathcal{G}}}$ contains
all finite rank operators in ${\mathcal{B}}_{1}(X)$, and the latter are SOT \-
dense in ${\mathcal{B}}_{1}(X)$ because $X$ has the MAP.∎

In the particular case of $\ell_{p}$ and $c_{0}$, we have the following
proposition, which provides some spectral properties of SOT \- typical
contractions on these spaces:

######  Proposition 3.9.

Assume that $X=\ell_{p}$, $1\leq p<\infty$ or that $X=c_{0}$. A typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has the following properties:
$T-\lambda$ has dense range for every $\lambda\in\mathbb{C}$, and
$\sigma(T)=\sigma_{{ap}}(T)=\overline{\,\mathbb{D}}$.

###### Proof.

Let us set

${{\mathcal{G}}}_{1}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\hbox{$T-\lambda$
has dense range for every $\lambda\in\mathbb{C}$}\bigr{\\}},$

and

${{\mathcal{G}}}_{2}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\sigma(T)=\sigma_{ap}(T)=\overline{\,\mathbb{D}}\bigr{\\}}.$

By the Baire Category Theorem, it is enough to show that ${{\mathcal{G}}}_{1}$
and ${{\mathcal{G}}}_{2}$ are both dense $G_{\delta}$ subsets of
$({\mathcal{B}}_{1}(X),\texttt{SOT})$.

Let $Z$ be a countable dense subset of $X$. If $T\in{\mathcal{B}}_{1}(X)$,
then

$T\not\in{{\mathcal{G}}}_{1}\iff\exists\lambda\in\mathbb{C}\;\exists
K\in\mathbb{N}\;\exists z\in Z\;:\;\forall z^{\prime}\in
Z\;:\;\|(T-\lambda)z^{\prime}-z\|\geq 1/K.$

This shows that ${\mathcal{B}}_{1}(X)\setminus{{\mathcal{G}}}_{1}$ is the
projection of an $F_{\sigma}$ subset of
$\mathbb{C}\times{\mathcal{B}}_{1}(X)$, where ${\mathcal{B}}_{1}(X)$ is
endowed with the topology SOT. Since $\mathbb{C}$ is $\sigma$-compact, it
follows that ${\mathcal{B}}_{1}(X)\setminus{{\mathcal{G}}}_{1}$ is
$F_{\sigma}$ in $({\mathcal{B}}_{1}(X),{\texttt{SOT}})$, i.e.
${{\mathcal{G}}}_{1}$ is $G_{\delta}$.

Similarly, if $T\in{\mathcal{B}}_{1}(X)$ then (since
$\sigma_{ap}(T)\subseteq\sigma(T)\subseteq\overline{\,\mathbb{D}}$), we have
that $T\not\in{{\mathcal{G}}}_{2}$ if and only if $T-\lambda$ is bounded below
for some $\lambda\in\overline{\,\mathbb{D}}$, i.e.

$T\not\in{{\mathcal{G}}}_{2}\iff\exists\lambda\in\overline{\,\mathbb{D}}\;\exists
K\in\mathbb{N}\;\forall z\in Z\;:\;\|(T-\lambda)z\|\geq 1/K\,\|z\|;$

and since $\overline{\,\mathbb{D}}$ is compact, it follows that
${{\mathcal{G}}}_{2}$ is a $G_{\delta}$ subset of
$({\mathcal{B}}_{1}(X),{\texttt{SOT}})$.

A straightforward adaptation of the proof of [GMM, Proposition 2.3] or [BM,
Proposition 2.23] shows that for any $M>1$, the set of all hypercyclic
operators in ${\mathcal{B}}_{M}(X)$ is a dense $G_{\delta}$ subset of
$({\mathcal{B}}_{M}(X),\texttt{SOT})$. In particular, an SOT \- typical
$R\in{\mathcal{B}}_{M}(X)$ is such that $aR+b$ has dense range for any
$a,b\in\mathbb{C}$ with $(a,b)\neq(0,0)$. Since the map $T\mapsto MT$ is a
homeomorphism from $({\mathcal{B}}_{1}(X),{\texttt{SOT}})$ onto
$({\mathcal{B}}_{M}(X),{\texttt{SOT}})$, it follows that ${{\mathcal{G}}}_{1}$
is dense in $({\mathcal{B}}_{1}(X),{\texttt{SOT}})$.

Now, let us show that ${{\mathcal{G}}}_{2}$ is dense in
${\mathcal{B}}_{1}(X)$. Recall that $(e_{j})_{j\geq 0}$ is the canonical basis
of $X$, that for any $N\geq 0$, $P_{N}:X\to[e_{0},\dots,e_{N}]$ is the
canonical projection map onto $E_{N}=[e_{0},\dots,e_{N}]$, and that
$F_{N}=[e_{j};\;j>N]$. Choose an operator $B_{N}\in{\mathcal{B}}_{1}(F_{N})$
such that $B_{N}-\lambda$ is not bounded below for any
$\lambda\in\overline{\,\mathbb{D}}$ (for example, the backward shift built on
$(e_{j})_{j\geq N+1}$ will do). If $A\in{\mathcal{B}}_{1}(X)$ is arbitrary,
then $T_{N}:=P_{N}A_{|E_{N}}\oplus B_{N}$ satisfies $\|T_{N}\|\leq 1$ and is
such that $T-\lambda$ is not bounded below for any
$\lambda\in\overline{\,\mathbb{D}}$; that is, $T_{N}\in{{\mathcal{G}}}_{2}$.
Since $T_{N}\xrightarrow{\texttt{SOT}}A$ as $N\to\infty$, this shows that
${{\mathcal{G}}}_{2}$ is indeed dense in
$({\mathcal{B}}_{1}(X),\texttt{SOT})$. ∎

Proposition 3.9 says in particular that a typical
$T\in\mathcal{B}_{1}(\ell_{p})$ has the largest possible spectrum. However, it
is worth mentioning that there are also lots of operators
$T\in\mathcal{B}_{1}(\ell_{p})$ whose spectrum is rather small:

######  Proposition 3.10.

Let $X=\ell_{p}$, $1\leq p<\infty$. The set of operators
$T\in{\mathcal{B}}_{1}(X)$ such that $\sigma(T)\subseteq\mathbb{T}$ is dense
in $({\mathcal{B}}_{1}(X),\emph{{SOT}})$.

Perhaps surprisingly, the proof of this result is not straightforward. For
technical reasons, it will be convenient to identify the space $X$ with
$Y:=\ell_{p}(\mathbb{Z})$ endowed with its canonical basis
$(f_{k})_{k\in\mathbb{Z}}$.

For each $N\geq 0$, let $Y_{N}:=[\,f_{k}\;;\;|k|\leq N\,]$. To any operator
$A\in{\mathcal{B}}(Y_{N})$ and any bounded sequence of positive numbers
$\omega=(\omega_{k})_{k\in\mathbb{Z}}$ – such a sequence $\omega$ will be
called a _weight sequence_ – we associate the operator $S_{A,\omega}$ on $Y$
defined by

$S_{A,\omega}f_{k}:=\begin{cases}Af_{k}+\omega_{k-(2N+1)}f_{k-(2N+1)}\quad&\textrm{if}\
|k|\leq N,\\\ \omega_{k-(2N+1)}f_{k-(2N+1)}&\textrm{if}\ |k|>N.\end{cases}$

Some of the properties of these operators, related to linear dynamics, can be
found in [GMM, Proposition 2.14 and Remark 2.15].

In the next two lemmas, the integer $N$ and the operator
$A\in{\mathcal{B}}(Y_{N})$ are fixed. The first lemma gives an estimate on the
norm of $S_{A,\omega}$.

######  Lemma 3.11.

For any $\varepsilon>0$, there exists $\delta>0$ depending only on
$\varepsilon$ and $\|A\|$ such that, for any weight sequence $\omega$
satisfying $\sup\limits_{-(3N+1)\leq k\leq N}\omega_{k}<\delta$, one has

$\|S_{A,\omega}\|\leq\max\,\biggl{(}(\|A\|^{p}+\varepsilon^{p})^{1/p},\sup_{k\not\in[-(3N+1),N]}\omega_{k}\biggr{)}.$

###### Proof of Lemma 3.11.

For any set of integers $I$ we denote by $P_{I}$ the canonical projection of
$Y=\ell_{p}(\mathbb{Z})$ onto $[f_{k}\,;\;k\in I\,]$. For every
$y=\sum_{k\in\mathbb{Z}}y_{k}f_{k}$, we have

$\displaystyle\|S_{A,\omega}\,y\|^{p}=\biggl{\|}A\,P_{[-N,N]}y+\sum_{k=N+1}^{3N+1}\omega_{k-(2N+1)}$
$\displaystyle y_{k}f_{k-(2N+1)}\biggr{\|}^{p}$
$\displaystyle+\sum_{k\not\in[N+1,3N+1]}\omega_{k-(2N+1)}^{p}|y_{k}|^{p}$

Let $c_{1},c_{2}\in\mathbb{R}_{+}$ be such that $(a+b)^{p}\leq
a^{p}+c_{1}b^{p}+c_{2}a^{p-1}b$ for every $a,b\geq 0$. Using Hölder’s
inequality, we se that for any $u,v\in\ell_{p}$, we have

$\|u+v\|^{p}\leq\|u\|^{p}+c_{1}\,\|v\|^{p}+c_{2}\,\|u\|^{p-1}\,\|v\|.$

So we get

$\displaystyle\bigl{\|}\,S_{A,\omega}\,y\,\bigr{\|}^{p}\leq\bigl{\|}\,$
$\displaystyle
A\,\bigr{\|}^{p}\,\bigl{\|}\,P_{[-N,N]}y\,\bigr{\|}^{p}+c_{1}\sup_{N+1\leq
k\leq 3N+1}\omega_{k-(2N+1)}^{p}\,\bigl{\|}\,P_{[N+1,3N+1]}y\,\bigr{\|}^{p}$
$\displaystyle+c_{2}\,\bigl{\|}\,A\,\bigr{\|}^{p-1}\bigl{\|}\,P_{[-N,N]}\,y\,\bigr{\|}^{p-1}\sup_{N+1\leq
k\leq 3N+1}\omega_{k-(2N+1)}\,\bigl{\|}\,P_{[N+1,3N+1]}\,y\,\bigr{\|}$
$\displaystyle+\sup_{k\in[-N,N]}\omega_{k-(2N+1)}^{p}\,\bigl{\|}\,P_{[-N,N]}\,y\,\bigr{\|}^{p}$
$\displaystyle+\sup_{k\not\in[-N,3N+1]}\omega_{k-(2N+1)}^{p}\,\bigl{\|}\,(I-P_{[-N,3N+1]})\,y\,\bigr{\|}^{p}.$

Since $\sup_{-N\leq k\leq 3N+1}\omega_{k-(2N+1)}=\sup_{-(3N+1)\leq k\leq
N}\omega_{k}<\delta$ and since

$\bigl{\|}\,P_{[-N,N]}\,y\,\bigr{\|}^{p-1}\,\bigl{\|}\,P_{[N+1,3N+1]}\,y\,\bigr{\|}\leq\bigl{\|}\,P_{[-N,3N+1]}\,y\,\bigr{\|}^{p},$

this yields that

$\displaystyle\bigl{\|}\,S_{A,\omega}\,y\,\bigr{\|}^{p}$
$\displaystyle\leq\Bigl{(}\bigl{\|}\,A\,\bigr{\|}^{p}+\delta^{p}\Bigr{)}\,\bigl{\|}\,P_{[-N,N]}\,y\,\bigr{\|}^{p}+\bigl{(}c_{1}\delta^{p}+c_{2}\,\bigl{\|}\,A\,\bigr{\|}^{p-1}\delta\bigr{)}\,\bigl{\|}\,P_{[-N,3N+1]}\,y\,\bigr{\|}^{p}$
$\displaystyle\quad\quad+\sup_{k\not\in[-(3N+1),N]}\omega_{k}^{p}\;\bigl{\|}\,(I-P_{[-N,3N+1]})\,y\,\bigr{\|}^{p}$
$\displaystyle\leq\Bigl{(}\bigl{\|}\,A\,\bigr{\|}^{p}+(c_{1}+1)\delta^{p}+c_{2}\,\bigl{\|}\,A\,\bigr{\|}^{p-1}\delta\Bigr{)}\,\bigl{\|}\,P_{[-N,3N+1]}y\,\bigr{\|}^{p}$
$\displaystyle\quad\quad+\sup_{k\not\in[-(3N+1),N]}\omega_{k}^{p}\;\bigl{\|}\,(I-P_{[-N,3N+1]})\,y\,\bigr{\|}^{p}.$

Choosing $\delta$ such that
$(c_{1}+1)\delta^{p}+c_{2}\,\bigl{\|}\,A\,\bigr{\|}^{p-1}\delta<\varepsilon^{p}$,
we obtain that

$\bigl{\|}\,S_{A,\omega}\,\bigr{\|}\leq\max\biggl{(}\bigl{(}\,\bigl{\|}\,A\,\bigr{\|}^{p}+\varepsilon^{p}\bigr{)}^{1/p},\sup_{k\not\in[-(3N+1),N]}\omega_{k}\biggr{)},$

as claimed. ∎

The second lemma gives a description of the point spectrum of these operators
$S_{A,\omega}$. For any weight sequence $\omega$ and any $\lambda$ in
$\mathbb{C}$, let us define two subsets $\Lambda_{\omega,\lambda}^{-}$ and
$\Lambda_{\omega,\lambda}^{+}$ of $[-N,N]$ as follows:

$\displaystyle\Lambda_{\omega,\lambda}^{-}$
$\displaystyle:=\biggl{\\{}k\in[-N,N]\;;\;\sum_{i\geq 1}\
\Bigl{|}\,\dfrac{\omega_{k}\cdots\omega_{k-(i-1)(2N+1)}}{\lambda^{i}}\,\Bigr{|}^{p}<\infty\biggr{\\}}\
\quad\text{with}\quad\Lambda_{\omega,0}^{-}=\emptyset,$
$\displaystyle\Lambda_{\omega,\lambda}^{+}$
$\displaystyle:=\biggl{\\{}k\in[-N,N]\;;\;\sum_{i\geq 1}\
\Bigl{|}\,\dfrac{\lambda^{i}}{\omega_{k+(i-1)(2N+1)}\cdots\omega_{k}}\,\Bigr{|}^{p}<\infty\biggr{\\}}.$

######  Lemma 3.12.

Let $\lambda\in\mathbb{C}$. Then $\lambda$ is an eigenvalue of $S_{A,\omega}$
if and only if there exists a non-zero vector $u\in Y_{N}$ such that
$\emph{supp}(u)\subseteq\Lambda_{\omega,\lambda}^{-}$ and
$\emph{supp}\bigl{(}(A-\lambda)u\bigr{)}\subseteq\Lambda_{\omega,\lambda}^{+}$.
In particular, if $\Lambda_{\omega,\lambda}^{-}=\emptyset$ then $\lambda$ is
not an eigenvalue of $S_{A,\omega}$.

Here, we denote by $\textrm{supp}(y)$ the _support_ of a vector
$y=\sum_{k\in\mathbb{Z}}y_{k}f_{k}$, i.e. the set of all $k\in\mathbb{Z}$ such
that $y_{k}\neq 0$.

###### Proof of Lemma 3.12.

The number $\lambda$ is an eigenvalue of $S_{A,\omega}$ if and only if there
exists a non-zero vector $y\in Y$ such that

$\displaystyle(A-\lambda)\,P_{[-N,N]}\,y+\sum_{|k|\leq
N}\omega_{k}y_{k+(2N+1)}f_{k}=0$ (1)
$\displaystyle\omega_{k+(2N+1)}y_{k+(2N+1)}=\lambda\,y_{k}\qquad\qquad\textrm{for
every}\ k\geq N+1$ (2)
$\displaystyle\omega_{-k+(2N+1)}y_{-k+(2N+1)}=\lambda\,y_{-k}\qquad\,\textrm{for
every}\ k\geq N+1$ (3)

It follows from these expressions that $\lambda=0$ is never an eigenvalue of
$S_{A,\omega}$. We suppose for the rest of the proof that $\lambda\neq 0$. We
deduce from equations (1), (2), and (3) that

$\displaystyle y_{k+(2N+1)}$ $\displaystyle=-\dfrac{1}{\omega_{k}}\ {\langle
f_{k}^{*},\,(A-\lambda)\,P_{[-N,N]}y\rangle}\qquad\textrm{for every}\ |k|\leq
N,$ (given $P_{[-N,N]}y$, these equations determine the values of $y_{j}$ for
$j\in[N+1,3N+1]$). $\displaystyle y_{k+i(2N+1)}$
$\displaystyle=\dfrac{\lambda^{i}}{\omega_{k+i(2N+1)}\dots\omega_{k+(2N+1)}}\,y_{k}\qquad\textrm{for
every}\ k\geq N+1\ \textrm{and every}\ i\geq 1,$ (the values of $y_{j}$ for
$j\in[N+1,3N+1]$ determine the values of $y_{j}$ for $j>3N+1$) and
$\displaystyle y_{k-i(2N+1)}$
$\displaystyle=\dfrac{\omega_{k}\dots\omega_{k-(i-1)(2N+1)}}{\lambda^{i}}\,y_{k}\quad\textrm{for
every}\ k\leq N\ \textrm{and every}\ i\geq 1$

(the values of $y_{j}$ for $j\in[-N,N]$ determine the values of $y_{j}$ for
$j<-N$).

Hence, if $u$ is a vector of $Y_{N}$, there exists a vector $y\in Y$ with
$P_{[-N,N]}\,y=u$ and $y\in\ker(S_{A,\omega}-\lambda)$ if and only if the
series

$\sum_{i\geq 1}\
\Bigl{|}\,\dfrac{\omega_{k}\cdots\omega_{k-(i-1)(2N+1)}}{\lambda^{i}}\,\Bigr{|}^{p}\cdot|u_{k}|^{p}\quad\textrm{and}\quad\sum_{i\geq
1}\
\Bigl{|}\,\dfrac{\lambda^{i}}{\omega_{k+(i-1)(2N+1)}\cdots\omega_{k}}\,\Bigr{|}^{p}\cdot\big{|}{\langle
f_{k}^{*},(A-\lambda)u\rangle}\bigr{|}^{p}$

are convergent for every $k\in[-N,N]$. This is equivalent to the conditions
that, for every $k\in[-N,N]$, either $u_{k}=0$ or
$k\in\Lambda_{\omega,\lambda}^{-}$, and either ${\langle
f_{k}^{*},(A-\lambda)u\rangle}=0$ or $k\in\Lambda_{\omega,\lambda}^{+}$. Lemma
3.12 follows. ∎

With the two lemmas above at our disposal, we can now proceed to the

###### Proof of Proposition 3.10..

Let ${\mathcal{U}}$ be a non-empty open set in
$({\mathcal{B}}_{1}(Y),\texttt{SOT})$. We want to show that ${\mathcal{U}}$
contains an operator $T$ such that $\sigma(T)\subseteq\mathbb{T}$.

Choose an integer $N$, an operator $A\in{\mathcal{B}}(Y_{N})$ with $\|A\|<1$,
and $\varepsilon>0$, such that any operator $T\in{\mathcal{B}}_{1}(Y)$
satisfying $\|(T-A)f_{k}\|<\varepsilon$ for all $k\in[-N,N]$ belongs to the
open set ${\mathcal{U}}$. Then, choose a weight sequence
$\omega=(\omega_{k})_{k\in\mathbb{Z}}$ such that $\omega_{k}$ is extremely
small if $-(3N+1)\leq k\leq N$, and otherwise $\omega_{k}=1$. By Lemma 3.11
and since $\|A\|<1$, the operator $T:=S_{A,\omega}$ satisfies $\|T\|\leq 1$,
provided that the weights $\omega_{k}$ are sufficiently small for $-(3N+1)\leq
k\leq N$. Moreover, $Tf_{k}$ is very close to $Af_{k}$ for $k\in[-N,N]$. Hence
$T\in{\mathcal{U}}$.

It remains to show that $\sigma(T)\subseteq\mathbb{T}$. First, observe that
since all but finitely many weights $\omega_{k}$ are equal to $1$, the set
$\Lambda_{\omega,\lambda}^{-}$ is obviously empty for any
$\lambda\in\mathbb{D}$. By Lemma 3.12, it follows that $T$ has no eigenvalue
in $\mathbb{D}$. Let us show that, in fact,
$\sigma(T)\cap\mathbb{D}=\emptyset$. Let $\lambda\in\mathbb{D}$ be arbitrary.
Since all but finitely many weights $\omega_{k}$ are equal to $1$, we see that
the operator $T$ is a finite rank perturbation of $S^{2N+1}$, where $S$ is the
unweighted bilateral shift on $Y=\ell_{p}(\mathbb{Z})$. Hence, $T-\lambda$ is
a finite rank perturbation of $S^{2N+1}-\lambda$. Now, $S^{2N+1}$ is a
surjective isometry, so $S^{2N+1}-\lambda$ is invertible because
$|\lambda|<1$. Therefore, $T-\lambda$ is a Fredholm operator with index $0$.
But $T-\lambda$ is also one-to-one since $T$ has no eigenvalue in
$\mathbb{D}$. So $T-\lambda$ is invertible, i.e. $\lambda\not\in\sigma(T)$.
This terminates the proof of Proposition 3.10. ∎

##  4\. SOT \- typical contractions on $\ell_{1}$

We consider in this section the case of the space $X=\ell_{1}$, which turns
out to behave much like $\ell_{2}$ as far as the properties considered in
Corollary 3.5 are concerned:

######  Theorem 4.1.

Let $X=\ell_{1}$. A typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has the
following properties.

  1. (a)

$T^{*}$ is a non-surjective isometry.

  2. (b)

$T-\lambda$ is surjective for every $\lambda\in\mathbb{D}$.

  3. (c)

Every $\lambda\in\mathbb{D}$ is an eigenvalue of $T$ with infinite
multiplicity.

  4. (d)

$\sigma(T)=\sigma_{ap}(T)=\overline{\,\mathbb{D}}$.

  5. (e)

$\|T^{n}x\|\to 0$ for every $x\in X$.

###### Proof.

We already know that (e) holds true on any Banach space $X$; and (d) follows
from (c) because $\sigma_{ap}(T)$ is a closed set containing $\sigma_{p}(T)$.
So we just need to prove (a), (b) and (c).

(a) Since $X=\ell_{1}$ has the MAP, we know by Proposition 3.8 that a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is such that $T^{*}$ is non-
invertible; so it is enough to show that a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is such that $T^{*}$ is an isometry.
Let us denote by ${\mathcal{I}}_{*}$ the set of all $T\in{\mathcal{B}}_{1}(X)$
such that $T^{*}$ is an isometry; we are going to show that
${\mathcal{I}}_{*}$ is a dense $G_{\delta}$ subset of
$({\mathcal{B}}_{1}(X),\texttt{SOT})$.

The proof that ${\mathcal{I}}_{*}$ is $G_{\delta}$ works on any separable
Banach space $X$. This relies on the following fact.

######  Fact 4.2.

The map $(T,x^{*})\mapsto\|T^{*}x^{*}\|$ is lower-semicontinuous on
$({\mathcal{B}}_{1}(X),\texttt{SOT})\times(B_{X^{*}},w^{*})$.

###### Proof of Fact 4.2.

This is clear since $\|T^{*}x^{*}\|=\sup_{x\in B_{X}}|\langle
x^{*},Tx\rangle|$ and each map

$(T,x^{*})\mapsto\langle x^{*},Tx\rangle,\quad x\in X$

is continuous on $({\mathcal{B}}_{1}(X),\texttt{SOT})\times(B_{X^{*}},w^{*})$.
∎

If $T\in{\mathcal{B}}_{1}(X)$, then

$T\not\in{\mathcal{I}}_{*}\iff\exists x^{*}\in
B_{X^{*}}\;\exists\alpha\in\mathbb{Q}\;:\;\|T^{*}x^{*}\|\leq\alpha<\|x^{*}\|.$

The condition $\|T^{*}x^{*}\|\leq\alpha$ defines a closed subset of
$({\mathcal{B}}_{1}(X),\texttt{SOT})\times(B_{X^{*}},w^{*})$ by Fact 4.2; and
the condition $\|x^{*}\|>\alpha$ defines an open subset of
$(B_{X^{*}},w^{*})$, and hence an $F_{\sigma}$ set in $(B_{X^{*}},w^{*})$
because $(B_{X^{*}},w^{*})$ is metrizable. So we see that
${\mathcal{B}}_{1}(X)\setminus{\mathcal{I}}_{*}$ is the projection along
$B_{X^{*}}$ of an $F_{\sigma}$ subset of
$({\mathcal{B}}_{1}(X),\texttt{SOT})\times(B_{X^{*}},w^{*})$. Since
$(B_{X^{*}},w^{*})$ is compact, this shows that ${\mathcal{I}}_{*}$ is
$G_{\delta}$.

Now, let us show that ${\mathcal{I}}_{*}$ is SOT \- dense in
${\mathcal{B}}_{1}(X)$ when $X=\ell_{1}$. Given an arbitrary
$A\in{\mathcal{B}}_{1}(X)$, set $T_{N}:=P_{N}AP_{N}+B_{N}(I-P_{N})$, where
$B_{N}:F_{N}\to X$ is the operator defined by $B_{N}e_{N+1+k}=e_{k}$ for every
$k\geq 0$. Then $\|T_{N}e_{j}\|\leq 1$ for every $j\in\mathbb{Z}_{+}$, and
hence $\|T_{N}\|\leq 1$ since we are working on $X=\ell_{1}$. Moreover, if
$x^{*}\in X^{*}$ is arbitrary, then $\|T_{N}^{*}x^{*}\|\geq|\langle
x^{*},T_{N}e_{N+1+k}\rangle|=|\langle x^{*},e_{k}\rangle|$ for every $k\geq
0$, and hence $\|T^{*}x^{*}\|\geq\|x^{*}\|$ since $X^{*}=\ell_{\infty}$. So we
have $T_{N}\in{\mathcal{I}}_{*}$ for all $N\geq 0$; and since
$T_{N}\xrightarrow{\texttt{SOT}}A$ as $N\to\infty$, we conclude that
${\mathcal{I}}_{*}$ is indeed dense in $({\mathcal{B}}_{1}(X),\texttt{SOT})$.

To prove (b) and (c), observe that if $T\in{\mathcal{B}}_{1}(X)$ is such that
$T^{*}$ is an isometry, then $T^{*}-\lambda$ is bounded below for every
$\lambda\in\mathbb{D}$. By (a) and since the adjoint of an operator $R$ is
bounded below if and only if $R$ is surjective, it follows that a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is such that $T-\lambda$ is
surjective for every $\lambda\in\mathbb{D}$; which is (b). Since a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is also such that $T-\lambda$ is
non-invertible for every $\lambda\in\mathbb{D}$ (by Proposition 3.9), it
follows that a typical $T\in{\mathcal{B}}_{1}(X)$ is such that $T-\lambda$ is
not one-to-one for every $\lambda\in\mathbb{D}$. However, in order to obtain
(c), we still have to prove that for a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$, the multiplicity of every
$\lambda\in\mathbb{D}$ as an eigenvalue of $T$ is infinite. The proof of this
statement relies on the following lemma:

######  Lemma 4.3.

Let $X=\ell_{1}$. Let $\overline{\alpha}=(\alpha_{i})_{i\geq 0}$ be a sequence
of positive numbers with $\alpha_{0}=\alpha_{1}=1$ and $\sum_{i\geq
0}\alpha_{i}<\infty$. Let also $\overline{N}=(N_{i})_{i\geq 0}$ be a strictly
increasing sequence of nonnegative integers with $N_{0}=0$. For every $q\geq
0$, let ${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$ be the subset of
${\mathcal{B}}_{1}(X)$ defined as follows:

$\displaystyle
T\in{\mathcal{D}}_{q}(\overline{\alpha},\overline{N})\iff\forall\,0\leq
i_{0}<$ $\displaystyle\dots<i_{l}\leq
q\quad\textrm{with}\quad\biggl{\|}T\biggl{(}\sum_{j=0}^{l}\alpha_{j}e_{N_{i_{j}}}\biggr{)}\biggr{\|}\leq\alpha_{l+1}$
$\displaystyle\exists\,i>q\;:\;\biggl{\|}T\biggl{(}\sum_{j=0}^{l}\alpha_{j}e_{N_{i_{j}}}+\alpha_{l+1}e_{N_{i}}\biggr{)}\biggr{\|}<\alpha_{l+2}.$

Then ${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$ is a dense open
subset of $({\mathcal{B}}_{1}(X),\emph{{SOT}})$.

###### Proof of Lemma 4.3.

The set ${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$ is clearly SOT \-
open in ${\mathcal{B}}_{1}(X)$. Let $T_{0}\in{\mathcal{B}}_{1}(X)$, and let
${\mathcal{U}}$ be an SOT \- open neighborhood of $T_{0}$ in
${\mathcal{B}}_{1}(X)$. We choose $r\geq q$ such that any operator
$T\in{\mathcal{B}}_{1}(X)$ satisfying $Te_{n}=T_{0}e_{n}$ for every $0\leq
n\leq N_{r}$ belongs to ${\mathcal{U}}$. Next, we enumerate the (finite) set
$\Sigma_{q}$ consisting of all sequences $(i_{0},\dots,i_{l})$ with

$i_{0}<\dots<i_{l}\leq
q\quad\textrm{and}\quad\biggl{\|}T_{0}\biggl{(}\sum_{j=0}^{l}\alpha_{j}e_{N_{i_{j}}}\biggr{)}\biggr{\|}\leq\alpha_{l+1}$

as $\Sigma_{q}=\\{\sigma_{1},\dots,\sigma_{s}\\}$, and write for each
$k=1,\dots,s$ the sequence $\sigma_{k}$ as $(i_{k,0},\dots,i_{k,l_{k}})$ for a
certain integer $l_{k}\geq 0$. We have thus

$\biggl{\|}T_{0}\biggl{(}\sum_{j=0}^{l_{k}}\alpha_{j}e_{N_{i_{k,j}}}\biggr{)}\biggr{\|}\leq\alpha_{l_{k}+1}\quad\textrm{for
every}\quad k=1,\dots,s.$

Let now $i_{1},\dots,i_{s}$ be integers with $r<i_{1}<\dots<i_{s}$, and define
an operator $T$ on $X$ by setting

$Te_{n}:=\begin{cases}T_{0}e_{n}&\textrm{for every}\ 0\leq n\leq N_{r},\\\
-\dfrac{1}{\alpha_{l_{k}+1}}\,T_{0}\displaystyle\biggl{(}\sum_{j=0}^{l_{k}}\alpha_{j}e_{N_{i_{k,j}}}\biggr{)}&\textrm{if}\
n=N_{i_{k}}\ \textrm{for some}\ 1\leq k\leq s,\\\ 0&\textrm{in all other
cases.}\end{cases}$

Since $\|Te_{n}\|\leq 1$ for every $n\geq 0$ and $X=\ell_{1}$, we see that
$\|T\|\leq 1$. Also,

$\textrm{for every}\ k=1,\dots,s,\quad\quad
T\biggl{(}\sum_{j=0}^{l_{k}}\alpha_{j}e_{N_{i_{k,j}}}\biggr{)}=T_{0}\biggl{(}\sum_{j=0}^{l_{k}}\alpha_{j}e_{N_{i_{k,j}}}\biggr{)}$

since $i_{k,j}\leq q\leq r$ and $T$ and $T_{0}$ coincide on
$[\,e_{n}\;;\;0\leq n\leq N_{r}\,]$. Hence

$T\biggl{(}\sum_{j=0}^{l_{k}}\alpha_{j}e_{N_{i_{k,j}}}+\alpha_{l_{k}+1}e_{N_{i_{k}}}\biggr{)}=0\quad\quad\textrm{for
every}\ k=1,\dots,s.$

This implies that the operator $T$ belongs to
${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$. We have thus proved that
${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$ is SOT \- dense in
${\mathcal{B}}_{1}(X)$. ∎

The reason why these sets ${\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$
are introduced is given in the next lemma:

######  Lemma 4.4.

Let $\overline{\alpha}$ and $\overline{N}$ satisfy the assumptions of _Lemma
4.3_. If $T\in{\mathcal{B}}_{1}(X)$ belongs to the set $\bigcap_{q\geq
0}{\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$, there exists a non-zero
vector $x\in[\,e_{N_{i}}\;;\;i\geq 0\,]$ such that $Tx=0$.

###### Proof of Lemma 4.4.

Since $\alpha_{0}=\alpha_{1}=1$ and $N_{0}=0$, we have
$\|T(\alpha_{0}e_{N_{0}})\|\leq\alpha_{1}$. Since
$T\in{\mathcal{D}}_{q}(\overline{\alpha},\overline{N})$, setting $i_{0}=0$, we
obtain that there exists $i_{1}>0$ such that

$\bigl{\|}T(\alpha_{0}e_{N_{i_{0}}}+\alpha_{1}e_{N_{i_{1}}})\bigr{\|}<\alpha_{2}.$

An induction argument along the same line then shows the existence of a
strictly increasing sequence of nonnegative integers $(i_{j})_{j\geq 0}$ such
that

$\biggl{\|}T\biggl{(}\sum_{j=0}^{l}\alpha_{j}e_{N_{i_{j}}}\biggr{)}\biggr{\|}<\alpha_{l+1}\quad\textrm{for
every}\quad l\geq 0.$

The vector $x=\sum_{j\geq 0}\alpha_{j}e_{N_{i_{j}}}$ belongs to
$[\,e_{N_{i}}\;;\;i\geq 0\,]$, is non-zero since $\alpha_{j}>0$ for every
$j\geq 0$, and satisfies $Tx=0$. ∎

Thanks to Lemma 4.3 and 4.4, it is now not difficult to prove that an SOT \-
typical operator $T\in{\mathcal{B}}_{1}(X)$ satisfies $\dim\ker T=\infty$.
Indeed, let $(\overline{N}_{n})_{n\geq 1}$ be an infinite sequence consisting
of strictly increasing sequences $\overline{N}_{n}=(N_{i,n})_{i\geq 0}$ of
nonnegative integers with $N_{0,n}=0$, with the property that the sets
$\\{N_{i,n}\;;\;i\geq 1\\}$, $n\geq 1$, are pairwise disjoint. Let also
$\overline{\alpha}=(\alpha_{i})_{i\geq 0}$ be any sequence of positive numbers
with $\alpha_{0}=\alpha_{1}=1$ and $\sum_{i\geq 0}\alpha_{i}<\infty$. By Lemma
4.3, the set

${\mathcal{G}}:=\bigcap_{n\geq 1}\bigcap_{q\geq
0}{\mathcal{D}}_{q}(\overline{\alpha},\overline{N}_{n})$

is a dense $G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$. The
set ${\mathcal{U}}_{0}\subseteq{\mathcal{B}}_{1}(X)$ consisting of operators
$T\in{\mathcal{B}}_{1}(X)$ such that $Te_{0}\neq 0$ being open and dense in
$({\mathcal{B}}_{1}(X),\texttt{SOT})$, ${\mathcal{G}}\cap{\mathcal{U}}_{0}$ is
also a dense $G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$. By
Lemma 4.4, if $T$ belongs to ${\mathcal{G}}\cap{\mathcal{U}}_{0}$, there
exists, for each $n\geq 1$, a non-zero vector
$x_{n}\in[\,e_{N_{i,n}}\;;\;i\geq 0\,]$ such that $Tx_{n}=0$. Since
$Te_{0}\neq 0$, $x_{n}$ is not colinear to $e_{0}$, and the assumption that
the sets $\\{N_{i,n},\;\,i\geq 1\\}$, $n\geq 1$, are pairwise disjoint implies
that the vectors $x_{n}$, $n\geq 1$, are linearly independent. So $\dim\ker
T=\infty$.

At this point, we know that a typical $T\in{\mathcal{B}}_{1}(X)$ is such that
$T-\lambda$ is surjective for every $\lambda\in\mathbb{D}$ and
$\dim\ker(T)=\infty$. These two properties imply that
$\dim\ker(T-\lambda)=\infty$ for every $\lambda\in\mathbb{D}$. Indeed,
$T-\lambda$ is semi-Fredholm for every $\lambda\in\mathbb{D}$ since it is
surjective. By the continuity of the Fredholm index, ${\rm ind}(T-\lambda)$
does not depend on $\lambda\in\mathbb{D}$. Now, we have ${\rm ind}(T)=\infty$
since $T$ is surjective and $\dim\ker(T)=\infty$. So ${\rm
ind}(T-\lambda)=\infty$ for every $\lambda\in\mathbb{D}$; and since
$T-\lambda$ is surjective, it follows that $\dim\ker(T-\lambda)=\infty$. (See
e.g. [K, p. 229-244] for a treatment of the Fredholm index for operators on
Banach spaces.) This terminates the proof of assertion (c), and hence the
proof of Theorem 4.1. ∎

As an immediate consequence of property (c) in Theorem 4.1, we obtain:

######  Corollary 4.5.

Let $X=\ell_{1}$. A typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has a
non-trivial invariant subspace.

##  5\. SOT \- typical contractions on $\ell_{p}$ and $c_{0}$

We have proved in the previous sections that a typical contraction
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ for $X=\ell_{2}$ or $X=\ell_{1}$ has
the property that $T^{*}$ is an isometry on $X^{*}$. We begin this section by
observing that if $X=\ell_{p}$ with $1<p<\infty$ and $p\neq 2$, or if
$X=c_{0}$, this is no longer true. Thus one cannot follow this route to show
that a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ has a non-trivial
invariant subspace.

######  Proposition 5.1.

Let $X=c_{0}$ or $\ell_{p}$, with $1<p<\infty$ and $p\neq 2$. The set of all
$T\in{\mathcal{B}}_{1}(X)$ such that $T^{*}$ is an isometry is nowhere dense
in $({\mathcal{B}}_{1}(X),\emph{{SOT}})$. In particular, a typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is such that $T^{*}$ is not an
isometry.

###### Proof.

The dual space of $X$ is equal to $\ell_{q}$ for some $1\leq q<\infty$ with
$q\neq 2$. Now, the isometries of $\ell_{q}$ (or more generally the isometries
of $L_{q}$ spaces) are completely described by a classical result of Lamperti
[L]. All we need to know is that if $S$ is an isometry of $\ell_{q}$, then $S$
maps functions with disjoint supports to functions with disjoint supports: if
$f,g\in\ell_{q}$ and $f\cdot g=0$, then $(Sf)\cdot(Sg)=0$ (see e.g. [Caro,
Corollary 8.4] for a proof of this fact).

So if $T\in{\mathcal{B}}_{1}(X)$ is such that $T^{*}$ is an isometry, then
$(T^{*}e_{0}^{*})\cdot(T^{*}e_{1}^{*})=0$. In other words, the set

$\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\hbox{$T^{*}$ is an isometry}\bigr{\\}}$

is contained in

${\mathcal{F}}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\forall
j\in\mathbb{Z}_{+}\;:\;\langle e_{0}^{*},Te_{j}\rangle=0\;{\rm or}\;\,\langle
e_{1}^{*},Te_{j}\rangle=0\bigr{\\}}.$

The set ${\mathcal{F}}$ is obviously SOT \- closed in ${\mathcal{B}}_{1}(X)$,
and it is easily seen that its complement is SOT \- dense in
${\mathcal{B}}_{1}(X)$. In fact, it is clear that for any fixed $j\geq 0$, the
open set

${\mathcal{O}}_{j}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\langle
e_{0}^{*},Te_{j}\rangle\neq 0\;{\rm and}\;\,\langle
e_{1}^{*},Te_{j}\rangle\neq 0\bigr{\\}}$

is already SOT \- dense in ${\mathcal{B}}_{1}(X)$. This concludes the proof. ∎

In view of the results from Sections 3 and 4 above, it is also a natural
question to ask whether, for $X=\ell_{p}$ with $1<p<\infty$ and $p\neq 2$ or
$X=c_{0}$, a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ admits
eigenvalues or not. This question turns out to be surprisingly difficult, and
we are only able to answer it for $X=c_{0}$ and $X=\ell_{p}$ with $p>2$.

######  Theorem 5.2.

Let $X=c_{0}$. A typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has no
eigenvalue.

In the $\ell_{p}\,$\- case, we obtain the following stronger result.

######  Theorem 5.3.

Let $X=\ell_{p}$ with $p>2$. For any $M>1$, a typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is such that $(MT)^{*}$ is
hypercyclic. In particular, a typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ has no eigenvalue.

Recall that the adjoint of a hypercyclic operator cannot have any eigenvalue.
Indeed, suppose that $S$ is a hypercyclic operator acting on a Banach space
$Y$, and suppose that $S^{*}y^{*}=\lambda y^{*}$ for some $y^{*}\in Y^{*}$ and
$\lambda\in\mathbb{C}$. Then ${\langle
y^{*},S^{n}y\rangle}=\lambda^{n}{\langle y^{*},y\rangle}$ for every $y\in Y$
and every $n\geq 0$. It follows that the set $\\{{\langle
y^{*},S^{n}y\rangle}\;;\;n\geq 0\\}$ is never dense in $\mathbb{C}$, and thus
the vector $y$ cannot be hypercyclic for the operator $S$.

The reason why we have to assume, in the statement of Theorem 5.3, that $p>2$,
lies in the well-known fact that the $\ell_{p}\,$\- norms have very different
convexity and smoothness properties when $p>2$ and when $1<p<2$. Many
classical inequalities on $\ell_{p}\,$\- norms, like the Clarkson inequalities
for instance, reverse when one moves over from the case $p>2$ to the case
$1<p<2$. Also, if $p>2$, the $p\,$-th power $\|\,\cdot\,\|_{p}^{p}$ of the
$\ell_{p}\,$\- norm is $\mathcal{C}^{2}\,$\- smooth, while it is only
$\mathcal{C}^{1}\,$\- smooth when $1<p<2$ (see [DGZ, Chapter 5] for more on
smoothness of $L^{p}\,$\- norms). More prosaically, we will also use our
assumption that $p>2$ through the following classical inequality (see e.g.
[Kan, Lemma 2.1]):

######  Lemma 5.4.

Let $u$ and $v$ be complex numbers with $v\neq 0$. If $2<p<\infty$, we have

$|u+v|^{p}+|u-v|^{p}>2\,|u|^{p}+p\,|u|^{p-2}\,|v|^{2}.$

(If $0<p<2$ and $u\neq 0$, the reverse inequality holds.)

The proofs of Theorems 5.2 and 5.3 use rather different ideas. However, it is
possible to give a direct proof of a weaker yet apparently non-trivial
statement that follows the same pattern in the $c_{0}\,$\- case and in the
$\ell_{p}\,$\- case. (The second part of the statement follows from
Proposition 3.8.)

######  Corollary 5.5.

Let $X=\ell_{p}$ with $p>2$ or $X=c_{0}$. A typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is one-to-one; and hence a typical
$T\in{\mathcal{B}}_{1}(X)$ is not surjective.

Accordingly we will first give this direct proof of Corollary 5.5, and then we
will prove Theorem 5.2 and Theorem 5.3.

###  5a. Direct proof of Corollary 5.5

To show that a typical $T\in{\mathcal{B}}_{1}(X)$ is one-to-one when
$X=\ell_{p}$ for some $p>2$ or $X=c_{0}$, we are going to write down
explicitely a $G_{\delta}$ subset $\mathcal{G}$ of
$({\mathcal{B}}_{1}(X),\texttt{SOT})$ such that every $T\in\mathcal{G}$ is
one-to-one and $\mathcal{G}$ is dense in ${\mathcal{B}}_{1}(X)$.

The definition of $\mathcal{G}$ is rather technical looking, but it does not
require $p>2$ in the $\ell_{p}$ case; so assume temporarily that $X=\ell_{p}$,
$1\leq p<\infty$ or $X=c_{0}$. For every integer $k\geq 1$, let
$\mathcal{A}_{k}\subseteq{\mathcal{B}}_{1}(X)$ be defined in the following
way:

$T\in{\mathcal{A}}_{k}\iff\exists\varepsilon>0,\
\exists\,0<\delta_{1}\leq\delta_{2},\ \exists\,N>k,\
\exists\,r_{1},r_{2}\in\mathbb{Z}_{+}\quad\textrm{such that}$

  1. (i)

$\left\|e_{r_{1}}^{*}TP_{(N,\infty)}\right\|<\varepsilon\cdot 10^{-(k+1)}$ and
$\left\|e_{r_{2}}^{*}TP_{(N,\infty)}\right\|<\varepsilon\cdot 10^{-(k+1)}$;

  2. (ii)

$\left|\langle e_{r_{1}}^{*},Tx\rangle\right|>\left|\varepsilon
x_{k}\right|-|\delta_{1}x_{N}|-\varepsilon\cdot 10^{-(k+1)}$ and
$\left|\langle
e_{r_{2}}^{*},Tx\rangle\right|>|\delta_{2}x_{N}|-\varepsilon\cdot 10^{-(k+1)}$
for every $x\in B_{E_{N}}$.

Now set

$\mathcal{G}:=\bigcap_{k\geq
1}\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{A}}}_{k},$

where
$\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{A}}}_{k}$ is
the SOT \- interior of $\mathcal{A}_{k}$ relative to ${\mathcal{B}}_{1}(X)$.

######  Lemma 5.6.

Any operator $T\in\bigcap_{k\geq 1}\mathcal{A}_{k}$ is one-to-one.

###### Proof.

Indeed, let $T\in\bigcap_{k\geq 1}\mathcal{A}_{k}$, and suppose that
$x=\sum_{j\geq 0}x_{j}e_{j}\in X$ satisfies $\|x\|=1$ and $Tx=0$. Since
$\|x\|=1$, there exists $k\geq 0$ such that $|x_{k}|>5\cdot 10^{-(k+1)}$.

Since $T\in\mathcal{A}_{k}$, we may choose $\varepsilon$, $\delta_{1}$,
$\delta_{2}$, $N$, $r_{1}$, and $r_{2}$ such that conditions (i) and (ii)
above are satisfied. On the one hand, we have

$\displaystyle 0=|{\langle e_{r_{1}}^{*},Tx\rangle}|$
$\displaystyle\geq|{\langle
e_{r_{1}}^{*},TP_{[0,N]}x\rangle}|-\|e_{r_{1}}^{*}TP_{(N,\infty)}\|$
$\displaystyle>\varepsilon\,|x_{k}|-\delta_{1}|x_{N}|-2\varepsilon\cdot
10^{-(k+1)}>3\varepsilon\cdot 10^{-(k+1)}-\delta_{1}|x_{N}|.$

On the other hand,

$\displaystyle 0=|{\langle e_{r_{2}}^{*},Tx\rangle}|$
$\displaystyle\geq|{\langle
e_{r_{2}}^{*},TP_{[0,N]}x\rangle}|-\|e_{r_{2}}^{*}TP_{(N,\infty)}\|$
$\displaystyle>\delta_{2}|x_{N}|-2\varepsilon\cdot 10^{-(k+1)}.$

Hence $3\varepsilon\cdot
10^{-(k+1)}<\delta_{1}|x_{N}|\leq\delta_{2}|x_{N}|<2\varepsilon\cdot
10^{-(k+1)}$, which is a contradiction. ∎

######  Lemma 5.7.

Assume that $X=\ell_{p}$ for some $p>2$ or $X=c_{0}$. Then, for every integer
$k\geq 1$, the _SOT_ \- interior of $\mathcal{A}_{k}$ relative to
${\mathcal{B}}_{1}(X)$ is dense in $({\mathcal{B}}_{1}(X),\emph{{SOT}})$.

###### Proof.

Fix $k\geq 1$. Let ${{\mathcal{U}}}$ be a non-empty open subset of
$({\mathcal{B}}_{1}(X),\texttt{SOT})$, and pick a finite rank operator
$B\in{{\mathcal{U}}}$. There exist $\varepsilon\in(0,1/4)$ and an integer
$N>k+1$ such that ${{\mathcal{U}}}$ contains all operators
$T\in{\mathcal{B}}_{1}(X)$ satisfying $\|(T-B)e_{j}\|<4\varepsilon$ for every
$j=0,\dots,N-1$. We also fix an integer $M>N$ such that
$\textrm{Ran}(B)\subseteq E_{M}$.

Our aim is to exhibit an operator $A\in{\mathcal{B}}_{1}(X)$ belonging to the
SOT \- interior
$\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{A}}}_{k}$ of
$\mathcal{A}_{k}$ in ${\mathcal{B}}_{1}(X)$, and satisfying
$\|(A-B)e_{j}\|<4\varepsilon$ for every $j=0,\dots,N-1$. We will have to treat
separately in our discussion the cases $X=\ell_{p}$ and $X=c_{0}$.

_Case 1._ Suppose that $X=\ell_{p}$ for some $p>2$, and let $\delta>0$. We
define an operator $A$ in the following way:

$Ae_{j}:=\begin{cases}(1-2\varepsilon)\,Be_{j}&\textrm{if}\ 0\leq j<N\
\textrm{and}\ j\neq k,\\\
(1-2\varepsilon)\,Be_{k}+\varepsilon\,e_{M+1}\quad&\textrm{if}\ j=k,\\\
\delta(e_{M+1}+e_{M+2})&\textrm{if}\ j=N,\\\ 0&\textrm{if}\ j>N.\end{cases}$

By the Intermediate Value Theorem, we can now choose $\delta>0$ in such a way
that we have $\|A\|=1$. Obviously, $\|(A-B)e_{j}\|<4\varepsilon$ for every
$0\leq j<N$. What remains to be shown is that if $T\in{\mathcal{B}}_{1}(X)$ is
sufficiently close to $A$ for the topology SOT, then $T\in\mathcal{A}_{k}$.

For every $x=\sum_{j\geq 0}x_{j}e_{j}\in X$, we have

${\langle e_{M+1}^{*},Ax\rangle}=\varepsilon x_{k}+\delta
x_{N}\quad\textrm{and}\quad{\langle e_{M+2}^{*},Ax\rangle}=\delta x_{N}.$

Moreover, $AP_{(N,\infty)}=0$, and so $A$ belongs to $\mathcal{A}_{k}$, the
data witnessing this being $\varepsilon$, $\delta_{1}=\delta_{2}=\delta$, $N$,
$r_{1}=M+1$, and $r_{2}=M+2$. If $T$ is sufficiently close to $A$ with respect
to SOT, it will satisfy property (ii) of the definition of $\mathcal{A}_{k}$
for the same choices of $\varepsilon$, $\delta_{1}$, $\delta_{2}$, $N$,
$r_{1}$, and $r_{2}$ (since the ball $B_{E_{N}}$ is compact, the condition
given by property (ii) is SOT-open). Let us now show that if $T$ is
sufficiently close to $A$ for SOT, it also satisfies property (i).

Let $x\in E_{N}$ be a norming vector for $A$ (i.e. $\|Ax\|=\|x\|=1$). Since
$\|A_{|E_{N-1}}\|\leq 1-\varepsilon<1$, $x$ does not belong to $E_{N-1}$, so
that we have $x_{N}\neq 0$. Hence ${\langle e_{r_{2}}^{*},Ax\rangle}\neq 0$.
Moreover, we also have ${\langle e_{r_{1}}^{*},Ax\rangle}\neq 0$. Indeed,
suppose that ${\langle e_{r_{1}}^{*},Ax\rangle}=0$, and consider the vector
$x^{\prime}=x-2x_{N}e_{N}$. It satisfies $\|x^{\prime}\|=\|x\|=1$, and

$Ax^{\prime}=Ax-2x_{N}\delta(e_{M+1}+e_{M+2})=P_{[0,M]}Ax-2x_{N}\delta
e_{M+1}-x_{N}\delta e_{M+2}.$

So

$\displaystyle\|Ax^{\prime}\|^{p}$ $\displaystyle=\|P_{[0,M]}Ax\|^{p}+|2\delta
x_{N}|^{p}+|\delta x_{N}|^{p},$ while $\displaystyle\|Ax\|^{p}$
$\displaystyle=\|P_{[0,M]}Ax\|^{p}+|\delta x_{N}|^{p}=1.$

Hence $\|Ax^{\prime}\|>1$, which is a contradiction. So ${\langle
e_{r_{1}}^{*},Ax\rangle}\neq 0$. Let

$c:=\min\,\bigl{(}|\,{\langle e_{r_{1}}^{*},Ax\rangle}\,|,|\,{\langle
e_{r_{2}}^{*},Ax\rangle}\,|\bigr{)},$

and let $\eta\in(0,1)$. Any operator $T\in{\mathcal{B}}_{1}(X)$ which is
sufficiently close to $A$ for the topology SOT satisfies the following
properties:

$|{\langle e_{r_{i}}^{*},Tx\rangle}|\geq c/2,\
i=1,2,\quad\textrm{and}\quad\|Tx\|^{p}\geq 1-\eta.$

The inequality from Lemma 5.4 now comes into play: for every $y\in
F_{N}=[\,e_{j}\;;\;j>N\,{]}$, for any $\lambda>0$, and for every $j\geq 0$, we
have

$|{\langle e_{j}^{*},Tx+\lambda Ty\rangle}|^{p}+|{\langle e_{j}^{*},Tx-\lambda
Ty\rangle}|^{p}\geq 2|{\langle
e_{j}^{*},Tx\rangle}|^{p}+p\lambda^{2}\,|{\langle
e_{j}^{*},Tx\rangle}|^{p-2}\,|{\langle e_{j}^{*},Ty\rangle}|^{2}.$

Summing over $j$, we obtain that

$\|Tx+\lambda Ty\|^{p}+\|Tx-\lambda Ty\|^{p}\geq
2\|Tx\|^{p}+p\lambda^{2}\,\sum_{j\geq 0}|{\langle
e_{j}^{*},Tx\rangle}|^{p-2}\,|{\langle e_{j}^{*},Ty\rangle}|^{2}.$

Hence we have for $i=1,2$,

$\|Tx+\lambda Ty\|^{p}+\|Tx-\lambda Ty\|^{p}\geq
2\|Tx\|^{p}+p\lambda^{2}\,|{\langle e_{r_{i}}^{*},Tx\rangle}|^{p-2}\,|{\langle
e_{r_{i}}^{*},Ty\rangle}|^{2}.$

So if $T$ is SOT \- close to $A$,

$\|Tx+\lambda Ty\|^{p}+\|Tx-\lambda Ty\|^{p}\geq
2(1-\eta)+p\,\lambda^{2}\,(c/2)^{p-2}\,|{\langle
e_{r_{i}}^{*},Ty\rangle}|^{2}$

for every $y\in F_{N}$ and every $\lambda>0$.

On the other hand, since $\|T\|\leq 1$, and since the vectors $x$ and $y$ have
disjoint supports, we also have

$\displaystyle\|Tx+\lambda Ty\|^{p}+\|Tx-\lambda Ty\|^{p}$
$\displaystyle\leq\|x+\lambda y\|^{p}+\|x-\lambda y\|^{p}$
$\displaystyle=2(\|x\|^{p}+\lambda^{p}\|y\|^{p})$
$\displaystyle=2+2\lambda^{p}\|y\|^{p}.$

Hence

$|{\langle
e_{r_{i}}^{*},Ty\rangle}|^{2}\leq\dfrac{2}{p\,(c/2)^{p-2}}\Bigl{(}\dfrac{\eta}{\lambda^{2}}+\lambda^{p-2}\Bigr{)}$

for every $\lambda>0$, and every $y\in F_{N}$ such that $\|y\|\leq 1$. If we
choose first $\lambda>0$ sufficiently small, and then $\eta\in(0,1)$
sufficiently small, we obtain that for all
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ sufficiently close to $A$,

$\|e_{r_{i}}^{*}TP_{(N,\infty)}\|<\varepsilon\cdot 10^{-k},\quad i=1,2$

which is exactly property (i) from the definition of $\mathcal{A}_{k}$. This
terminates the proof of Lemma 5.7 in the $\ell_{p}\,$\- case.

_Case 2._ Suppose that $X=c_{0}$. In this case, we define the operator $A$ in
the following way:

$Ae_{j}:=\begin{cases}Be_{j}&\textrm{if}\ 0\leq j<N\ \textrm{and}\ j\neq k,\\\
Be_{k}+\varepsilon\,e_{M+1}&\textrm{if}\ j=k,\\\
(1-\varepsilon)\,e_{M+1}+e_{M+2}\quad&\textrm{if}\ j=N,\\\ 0&\textrm{if}\
j>N.\end{cases}$

Then $\|A\|=1$, and clearly $\|(A-B)e_{j}\|<4\varepsilon$ for every
$j=0,\dots,N-1$. Moreover, for every $x=\sum_{j\geq 0}x_{j}e_{j}\in X$,

${\langle e_{M+1}^{*},Ax\rangle}=\varepsilon
x_{k}+(1-\varepsilon)x_{N}\quad\textrm{and}\quad{\langle
e_{M+2}^{*},Ax\rangle}=x_{N}.$

Also $AP_{(N,\infty)}=0$, and so $A$ belongs to $\mathcal{A}_{k}$ for the
choices of the associated data as $\varepsilon$, $\delta_{1}=1-\varepsilon$,
$\delta_{2}=1$, $N$, $r_{1}=M+1$, and $r_{2}=M+2$. If $T$ is sufficiently
close to $A$ with respect to SOT, (ii) is clearly satisfied with the same
choices of $\varepsilon$, $\delta_{1}$, $\delta_{2}$, $N$, $r_{1}$, and
$r_{2}$, and what remains to be shown is that (i) is also satisfied.

The argument for this is much more direct than in the previous case, because
if we set $u_{1}:=e_{k}+e_{N}$ and $u_{2}:=e_{N}$, we have ${\langle
e_{r_{i}}^{*},Au_{i}\rangle}=1=\|u_{i}\|$, $i=1,2$.

Let $\eta\in(0,1)$. If $T$ is sufficiently close to $A$ for SOT then
$|{\langle e_{r_{i}}^{*},Tu_{i}\rangle}|>1-\eta$, and hence
$\|e_{r_{i}}^{*}TP_{[0,N]}\|>1-\eta$, $i=1,2$. The definition of the
$c_{0}\,$\- norm implies that

$\|e_{r_{i}}^{*}T\|=\|e_{r_{i}}^{*}TP_{[0,N]}\|+\|e_{r_{i}}^{*}TP_{(N,\infty)}\|,$

and thus $\|e_{r_{i}}^{*}TP_{(N,\infty)}\|\leq\eta$, $i=1,2$. Choosing
$\eta=\varepsilon\cdot 10^{-k}$, we obtain that any
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ sufficiently close to $A$ satisfies
(i) with the choices of $\varepsilon$, $\delta_{1}$, $\delta_{2}$, $N$,
$r_{1}$, and $r_{2}$ as above, and so $T\in\mathcal{A}_{k}$. ∎

###### Proof of Corollary 5.5.

Assuming that $X=\ell_{p}$, $p>2$ or $X=c_{0}$, it follows from the above two
lemmas that the set $\mathcal{G}=\bigcap_{k\geq
1}\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{A}}}_{k}$ is
a dense $G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$
consisting of one-to-one operators. ∎

###  5b. Proof of Theorem 5.2

We will make use of the so-called _Banach-Mazur game_ , which is often a very
effective tool for showing that a given set is comeager.

Let $\mathcal{E}$ be a Polish space, and let
$\mathcal{A}\subseteq\mathcal{E}$. The Banach-Mazur game
$\mathbf{G}(\mathcal{A})$ is an infinite game with two players, denoted by I
and II. The two players play alternatively open sets
$\mathcal{U}_{0}\supseteq\mathcal{U}_{1}\supseteq\mathcal{U}_{2}\supseteq\cdots$;
so, player I plays the open sets of even indices $\mathcal{U}_{2k}$ and player
II plays the open sets of odd indices $\mathcal{U}_{2k+1}$. Then, player II
wins the run if $\bigcap_{n\geq 0}\mathcal{U}_{n}\subseteq\mathcal{A}$. The
key result concerning this game is the following (see e.g. [Ke, p. 51] for a
proof): the set $\mathcal{A}$ is comeager in $\mathcal{E}$ if and only if
player II has a winning strategy in $\mathbf{G}(\mathcal{A})$. Moreover,
nothing changes if the open sets
$\mathcal{U}_{0},\mathcal{U}_{1},\mathcal{U}_{2},\dots$ are required to be
picked from a given basis for the topology of $\mathcal{E}$.

In our setting, the space $\mathcal{E}$ is
$(\mathcal{B}_{1}(c_{0}),\texttt{SOT})$, and $\mathcal{A}$ is the set of all
operators $T\in\mathcal{B}_{1}(c_{0})$ having no eigenvalue. The basic open
sets will be of the form

$\mathcal{U}({N,A,\varepsilon}):=\bigl{\\{}T\in\mathcal{B}_{1}(c_{0});\;\|Te_{j}-Ae_{j}\|<\varepsilon\;\,\hbox{for
$j=0,\dots,N$}\bigr{\\}},$

where $N\in\mathbb{N}$, $A\in\mathcal{B}_{1}(E_{N})$ and $0<\varepsilon\leq
1$. So player I will play open sets of the form
$\mathcal{U}_{2k}=\mathcal{U}(N_{2k},A_{2k},\varepsilon_{2k})$, and player II
will play open sets of the form
$\mathcal{U}_{2k+1}=\mathcal{U}(N_{2k+1},A_{2k+1},\varepsilon_{2k+1})$.

Let $(\alpha_{k})_{k\geq 0}$ be a sequences of positive real numbers, with
$\alpha_{k}\leq 1$. Let also $0<c\leq 1/2$ and $C>1$. We are going to describe
a strategy for player II depending on $(\alpha_{k})$, $c$ and $C$, and then to
show that this strategy is winning if $(\alpha_{k})$, $c$ and $C$ are suitably
chosen.

Assume that player I has just played a basic open set
$\mathcal{U}_{2k}=\mathcal{U}(A_{2k},N_{2k},\varepsilon_{2k})$. Let us set

$\tau_{k}:=\alpha_{k}\varepsilon_{2k},$

and choose a $\tau_{k}\,$\- net
$\Lambda_{k}=\\{\lambda_{1},\dots,\lambda_{L_{k}}\\}$ for the unit circle
$\mathbb{T}$, with $\lambda_{1}:=1$. Let us also choose an integer $R_{k}\geq
2$ such that

$\left(\frac{1-\tau_{k}/2}{1-\tau_{k}}\right)^{R_{k}-2}>\frac{C}{\varepsilon_{2k}}\cdot$

Now, set

$N_{2k+1}:=N_{2k}+(L_{k}+1)R_{k}-1,$

and define an operator $A_{2k+1}\in\mathcal{B}(E_{N_{2k+1}})$ as follows :

$A_{2k+1}e_{j}:=A_{2k}e_{j}\qquad\hbox{if $j\leq N_{2k}$ and $j\neq k$};$

$A_{2k+1}e_{k}:=A_{2k}e_{k}+\sum_{i=1}^{L_{k}}\lambda_{i}\frac{\varepsilon_{2k}}{2}\,e_{N_{2k}+iR_{k}};$

$A_{2k+1}e_{N_{2k}+iR_{k}}:=\lambda_{i}\left(1-\frac{\varepsilon_{2k}}{2}\right)e_{N_{2k}+iR_{k}}\qquad\hbox{for
every $1<i\leq L_{k}$};$

$A_{2k+1}e_{N_{2k}+R_{k}}:=\lambda_{1}\left(1-\frac{\varepsilon_{2k}}{2}\right)e_{N_{2k}+R_{k}}+e_{N_{2k}+R_{k}+1};$

$A_{2k+1}e_{N_{2k}+R_{k}+s}:=e_{N_{2k}+R_{k}+s+1}\qquad\hbox{for every $1\leq
s<R_{k}-1$};$

$A_{2k+1}e_{j}:=0\qquad\hbox{otherwise}.$

Observe that in fact $A_{2k+1}\in\mathcal{B}_{1}(E_{N_{2k+1}})$, i.e.
$\|A_{2k+1}\|\leq 1$. Indeed, if $x=\sum_{j\geq 0}x_{j}e_{j}\in E_{N_{2k+1}}$,
then

$\displaystyle\|A_{2k+1}x\|=\max\left\\{\max_{0\leq j\leq N_{2k}}\right.$
$\displaystyle|{\langle e_{j}^{*},A_{2k}x\rangle}|\,,\,$
$\displaystyle\left.\max_{1\leq i\leq
L_{k}}\left|\frac{\varepsilon_{2k}}{2}\,x_{k}+\left(1-\frac{\varepsilon_{2k}}{2}\right)x_{N_{2k}+iR_{k}}\right|\,,\,\max_{1\leq
s<R_{k}-1}|x_{N_{2k}+R_{k}+s}|\right\\}$ $\displaystyle\leq\|x\|.$

Finally, set

$\varepsilon_{2k+1}:=c\,\tau_{k}\,\varepsilon_{2k}.$

Then, the strategy of player II is to play the open set
$\mathcal{U}(N_{2k+1},A_{2k+1},\varepsilon_{2k+1})$. Note that this is indeed
allowed: since we are assuming that $c\leq 1/2$ and since $\tau_{k}\leq 1$, we
have $\|A_{2k+1}e_{j}-A_{2k}e_{j}\|\leq\varepsilon_{2k}/2$ for
$j=0,\dots,N_{2k}$, and hence
$\mathcal{U}(N_{2k+1},A_{2k+1},\varepsilon_{2k+1})\subseteq\mathcal{U}(N_{2k},A_{2k},\varepsilon_{2k})$.

Before proving that this strategy is winning for II if $(\alpha_{k})$, $c$ and
$C$ are suitably chosen, we point out the following fact.

######  Fact 5.8.

Let $(\mathcal{U}_{n})_{n\geq 0}$ be a run in the Banach-Mazur game where
player II has followed the strategy described above. If $T\in\bigcap_{n\geq
0}\mathcal{U}_{n}$ then, for any $k\geq 0$, every $1\leq i\leq L_{k}$ and
every $1\leq s<R_{k}-1$, we have

$\|e_{N_{2k}+iR_{k}}^{*}TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}\|\leq
2\varepsilon_{2k+1}\quad{\rm
and}\quad\|e_{N_{2k}+R_{k}+s}^{*}TP_{\mathbb{Z}_{+}\setminus\\{N_{2k}+R_{k}+s-1\\}}\|\leq\varepsilon_{2k+1}.$

###### Proof of Fact 5.8.

If $x\in c_{0}$ and $\|x\|\leq 1$, then $\|e_{k}+e_{N_{2k}+iR_{k}}+\omega
P_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}x\|=1$ for every
$\omega\in\mathbb{T}$, by definition of the $c_{0}\,$\- norm; so
$\|T(e_{k}+e_{N_{2k}+iR_{k}}+\omega
P_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}x)\|\leq 1$ for every
$\omega\in\mathbb{T}$ and hence

$\left|{\langle
e_{N_{2k}+iR_{k}}^{*},T(e_{k}+e_{N_{2k}+iR_{k}})\rangle}\right|+\left|{\langle
e_{N_{2k}+iR_{k}}^{*},TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}x)\rangle}\right|\leq
1.$

On the other hand,

$\displaystyle\left|{\langle
e_{N_{2k}+iR_{k}}^{*},T(e_{k}+e_{N_{2k}+iR_{k}})\rangle}\right|$
$\displaystyle\geq\bigl{|}\langle
e_{N_{2k}+iR_{k}}^{*}A_{2k+1}(e_{k}+e_{N_{2k}+iR_{k}})\rangle\bigr{|}$
$\displaystyle\phantom{e_{N_{2k}+iR_{k}}^{*}}-\bigl{\|}(T-A_{2k+1})(e_{k}+e_{N_{2k}+iR_{k}})\bigr{\|}$
$\displaystyle=1-\bigl{\|}(T-A_{2k+1})(e_{k}+e_{N_{2k}+iR_{k}})\bigr{\|}$
$\displaystyle\geq 1-2\varepsilon_{2k+1}.$

Thus, we see that $|{\langle
e_{N_{2k}+iR_{k}}^{*},TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}x)\rangle}|\leq
2\varepsilon_{2k+1}$ for every $x\in B_{c_{0}}$, i.e.
$\|e_{N_{2k}+iR_{k}}^{*}TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}\|\leq
2\varepsilon_{2k+1}$.

Likewise, if $x\in B_{c_{0}}$ then, on the one hand,

$\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},Te_{N_{2k}+R_{k}+s-1}\rangle}\right|+\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},TP_{\mathbb{Z}_{+}\setminus\\{N_{2k}+R_{k}+s-1\\}}x\rangle}\right|\leq
1$

and, on the other hand,

$\displaystyle\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},Te_{N_{2k}+R_{k}+s-1}\rangle}\right|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},A_{2k+1}e_{N_{2k}+R_{k}+s-1}\rangle}\right|$
$\displaystyle\phantom{e_{N_{2k}+R_{k}+s}^{*}}-\left\|(T-A_{2k+1})e_{N_{2k}+R_{k}+s-1}\right\|$
$\displaystyle\geq 1-\varepsilon_{2k+1};$

and this shows that
$\|e_{N_{2k}+R_{k}+s}^{*}TP_{\mathbb{Z}_{+}\setminus\\{N_{2k}+R_{k}+s-1\\}}\|\leq\varepsilon_{2k+1}$.
∎

We can now prove

######  Lemma 5.9.

Assume that $c<1/24$, and choose $\eta\geq 16c$ such that $8c+\eta<1$. If
$C>4/\eta$ and $\prod_{k=0}^{\infty}\frac{1-\alpha_{k}}{1+4\alpha_{k}}\geq
8c+\eta$, then the strategy described above is winning for player II.

###### Proof.

Let $(\mathcal{U}_{n})_{n\geq 0}$ be a run in the game where player II has
followed the strategy above, and let $T\in\bigcap_{n\geq 0}\mathcal{U}_{n}$.
We have to show that $T$ has no eigenvalue. Towards a contradiction, assume
that $Tx=\lambda x$ for some $x\in c_{0}$ such that $\|x\|=1$ and some
$\lambda\in\mathbb{C}$. Note that $|\lambda|\leq 1$ since $\|T\|\leq 1$. In
what follows, we write $x=\sum\limits_{j\geq 0}x_{j}e_{j}$.

######  Claim 5.10.

If $k\geq 0$ is such that $|x_{k}|\geq 8c+\eta$, then
$1-\tau_{k}\leq|\lambda|$.

###### Proof of Claim 5.31.

Assume that $|\lambda|<1-\tau_{k}$ for some $k\geq 0$ such that $|x_{k}|\geq
8c+\eta$. Looking at the $(N_{2k}+R_{k})$-th coordinate of $Tx-\lambda x$, we
may write

$\displaystyle 0$ $\displaystyle=\left|{\langle
e_{N_{2k}+R_{k}}^{*},Tx\rangle}-\lambda\,x_{N_{2k}+R_{k}}\right|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+R_{k}}^{*},TP_{\\{k,N_{2k}+R_{k}\\}}x\rangle}-\lambda\,x_{N_{2k}+R_{k}}\right|-\left|{\langle
e_{N_{2k}+R_{k}}^{*},TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+R_{k}\\}}x\rangle}\right|$
$\displaystyle\geq\,\left|{\langle
e_{N_{2k}+R_{k}}^{*},A_{2k+1}(x_{k}e_{k}+x_{N_{2k}+R_{k}}e_{N_{2k}+R_{k}})\rangle}-\lambda\,x_{N_{2k}+R_{k}}\right|-\left\|(T-A_{2k+1})x_{k}e_{k}\right\|$
$\displaystyle\phantom{e_{N_{2k}+R_{k}}^{*}}-\left\|(T-A_{2k+1})x_{N_{2k}+R_{k}}e_{N_{2k}+R_{k}}\right\|-\left|{\langle
e_{N_{2k}+R_{k}}^{*},TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+R_{k}\\}}x\rangle}\right|$
$\displaystyle\geq\left|\frac{\varepsilon_{2k}}{2}\,x_{k}+\left(1-\frac{\varepsilon_{2k}}{2}-\lambda\right)\,x_{N_{2k}+R_{k}}\right|-4\varepsilon_{2k+1},$

where we have used Fact 5.8. So
$\left|\frac{\varepsilon_{2k}}{2}\,x_{k}+\left(1-\frac{\varepsilon_{2k}}{2}-\lambda\right)\,x_{N_{2k}+R_{k}}\right|\leq
4\varepsilon_{2k+1}$. Since $|x_{k}|\geq 8c+\eta$,
$\varepsilon_{2k+1}=c\tau_{k}\varepsilon_{2k}\leq c\,\varepsilon_{2k}$ and
$\left|1-\frac{\varepsilon_{2k}}{2}-\lambda\right|\leq 2$, it follows that

$|x_{N_{2k}+R_{k}}|\geq\frac{(8c+\eta)\varepsilon_{2k}/2-4c\,\varepsilon_{2k}}{2}\geq\frac{\eta}{4}\,\varepsilon_{2k}.$

Now, for any $1\leq s<R_{k}-1$, we have

$\displaystyle 0$ $\displaystyle=\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},Tx\rangle}-\lambda\,x_{N_{2k}+R_{k}+s}\right|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},Tx_{N_{2k}+R_{k}+s-1}e_{N_{2k}+R_{k}+s-1}\rangle}-\lambda\,x_{N_{2k}+R_{k}+s}\right|$
$\displaystyle\phantom{e_{N_{2k}+R_{k}+s}^{*}}-\left\|{e_{N_{2k}+R_{k}+s}^{*}}{TP_{\mathbb{Z}_{+}\setminus\\{N_{2k}+R_{k}+s-1\\}}}\right\|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+R_{k}+s}^{*},A_{2k+1}x_{N_{2k}+R_{k}+s-1}e_{N_{2k}+R_{k}+s-1}\rangle}-\lambda\,x_{N_{2k}+R_{k}+s}\right|$
$\displaystyle\phantom{e_{N_{2k}+R_{k}+s}^{*}}-\left\|(T-A_{2k+1})e_{N_{2k}+R_{k}+s-1}\right\|-\left\|{e_{N_{2k}+R_{k}+s}^{*}}{TP_{\mathbb{Z}_{+}\setminus\\{N_{2k}+R_{k}+s-1\\}}}\right\|$
$\displaystyle\geq\left|x_{N_{2k}+R_{k}+s-1}-\lambda\,x_{N_{2k}+R_{k}+s}\right|-2\varepsilon_{2k+1}$
$\displaystyle\geq\left|x_{N_{2k}+R_{k}+s-1}\right|-|\lambda|\,\left|x_{N_{2k}+R_{k}+s}\right|-2\varepsilon_{2k+1}.$

In other words:

$\left|x_{N_{2k}+R_{k}+s}\right|\geq\frac{\left|x_{N_{2k}+R_{k}+s-1}\right|-2\varepsilon_{2k+1}}{|\lambda|}\cdot$

Since
$\varepsilon_{2k+1}=c\,\tau_{k}\varepsilon_{2k}\leq\frac{\eta}{16}\,\tau_{k}\varepsilon_{2k}$
and $|\lambda|<1-\tau_{k}$, it follows that

$\left|x_{N_{2k}+R_{k}+s}\right|\geq\frac{\left|x_{N_{2k}+R_{k}+s-1}\right|-(\eta/8)\,\tau_{k}\varepsilon_{2k}}{1-\tau_{k}}\qquad\hbox{for
every $1\leq s<R_{k}-1$}.$

Recall that $|x_{N_{2k}+R_{k}}|\geq\frac{\eta\,\varepsilon_{2k}}{4}\cdot$ With
$s:=1$, we get

$|x_{N_{2k}+R_{k}+1}|\geq|x_{N_{2k}+R_{k}}|\,\frac{1-\tau_{k}/2}{1-\tau_{k}}\geq\frac{\eta\varepsilon_{2k}}{4}\cdot$

Iterating this $R_{k}-2$ times, we obtain (keeping in mind that $C>4/\eta$)

$|x_{N_{2k}+R_{k}+R_{k}-2}|\geq\frac{\eta\,\varepsilon_{2k}}{4}\,\left(\frac{1-\tau_{k}/2}{1-\tau_{k}}\right)^{R_{k}-2}\geq\frac{\eta\,\varepsilon_{2k}}{4}\times\frac{C}{\varepsilon_{2k}}>1,$

which is a contradiction since $\|x\|=1$. ∎

######  Claim 5.11.

If $k\geq 0$ is such that $|x_{k}|\geq 8c+\eta$, then one can find
$k^{\prime}>k$ such that

$|x_{k^{\prime}}|\geq\frac{1-\alpha_{k}}{1+4\alpha_{k}}\,|x_{k}|\cdot$

###### Proof of Claim 5.32.

By Claim 5.31, we know that $|\lambda|\geq 1-\tau_{k}$. Since
$\Lambda_{k}=\\{\lambda_{1},\dots,\lambda_{L_{k}}\\}$ is a $\tau_{k}\,$\- net
for $\mathbb{T}$, it follows that one can find $1\leq i\leq L_{k}$ such that
$|\lambda-\lambda_{i}|<2\tau_{k}$. Then, we may write

$\displaystyle 0$ $\displaystyle=\left|{\langle
e_{N_{2k}+iR_{k}}^{*},Tx\rangle}-\lambda\,x_{N_{2k}+iR_{k}}\right|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+iR_{k}}^{*},T(x_{k}e_{k}+x_{N_{2k}+iR_{k}}e_{N_{2k}+iR_{k}})\rangle}-\lambda\,x_{N_{2k}+iR_{k}}\right|-\left\|e_{N_{2k}+iR_{k}}^{*}TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}\right\|$
$\displaystyle\geq\left|{\langle
e_{N_{2k}+iR_{k}}^{*},A_{2k+1}(x_{k}e_{k}+x_{N_{2k}+iR_{k}}e_{N_{2k}+iR_{k}})\rangle}-\lambda\,x_{N_{2k}+iR_{k}}\right|-\left\|(T-A_{2k+1})x_{k}e_{k}\right\|$
$\displaystyle\phantom{e_{N_{2k}+iR_{k}}^{*}}-\left\|(T-A_{2k+1})x_{N_{2k}+iR_{k}}e_{N_{2k}+iR_{k}}\right\|-\left\|e_{N_{2k}+iR_{k}}^{*}TP_{\mathbb{Z}_{+}\setminus\\{k,N_{2k}+iR_{k}\\}}\right\|$
$\displaystyle\geq\left|\lambda_{i}\,\frac{\varepsilon_{2k}}{2}\,x_{k}+\left(\lambda_{i}\left(1-\frac{\varepsilon_{2k}}{2}\right)-\lambda\right)x_{N_{2k}+iR_{k}}\right|-4\varepsilon_{2k+1}.$

So we have
$\left|\lambda_{i}\,\frac{\varepsilon_{2k}}{2}\,x_{k}+\left(\lambda_{i}\left(1-\frac{\varepsilon_{2k}}{2}\right)-\lambda\right)x_{N_{2k}+iR_{k}}\right|\leq
4\varepsilon_{2k+1}=4c\tau_{k}\varepsilon_{2k}$. Since $|x_{k}|\geq 8c+\eta$
and
$\left|\lambda_{i}\left(1-\frac{\varepsilon_{2k}}{2}\right)-\lambda\right|\leq\frac{\varepsilon_{2k}}{2}+2\tau_{k}$,
it follows that

$|x_{N_{2k}+iR_{k}}|\geq\frac{\varepsilon_{2k}/2-4c\tau_{k}\varepsilon_{2k}/(8c+\eta)}{\varepsilon_{2k}/2+2\tau_{k}}\,|x_{k}|=\frac{1-8c\tau_{k}/(8c+\eta)}{1+4\tau_{k}/\varepsilon_{2k}}\,|x_{k}|.$

Since $\tau_{k}=\alpha_{k}\varepsilon_{2k}$ and $\varepsilon_{2k}\leq 1$, this
implies that

$|x_{N_{2k}+iR_{k}}|\geq\frac{1-\alpha_{k}}{1+4\alpha_{k}}\,|x_{k}|.$

So we may set $k^{\prime}:=N_{2k}+iR_{k}$. ∎

It is now easy to conclude the proof of Lemma 5.9. Since $\|x\|=1$, we can
choose $k_{0}\geq 0$ such that $|x_{k_{0}}|=1$. Then, since we are assuming
that $\prod_{k=0}^{\infty}\frac{1-\alpha_{k}}{1+4\alpha_{k}}\geq 8c+\eta$, we
may use Claim 5.32 to find an increasing sequence of integers $(k_{l})_{l\geq
0}$ such that $|x_{k_{l}}|\geq 8c+\eta$ for all $l\geq 0$. Since $x\in c_{0}$,
this is the required contradiction. ∎

By Lemma 5.9, the proof of Theorem 5.2 is now complete.

###  5c. Outline of the proof of Theorem 5.3

The proof of Theorem 5.3 will now occupy us for a while. The main step will be
to prove the following statement, which interrelates in a rather curious way
the topologies SOT and SOT∗ on $\mathcal{B}_{1}(\ell_{p})$ when $p>2$.

######  Theorem 5.12.

Let $X=\ell_{p}$, $p>2$. If $\mathcal{O}\subseteq{\mathcal{B}}_{1}(X)$ is _SOT
∗_\- open, then the relative _SOT_ \- interior of $\mathcal{O}$ in
${\mathcal{B}}_{1}(X)$ is _SOT ∗_\- dense in $\mathcal{O}$, and hence _SOT_ \-
dense as well.

From this, we immediately deduce

######  Corollary 5.13.

Let $X=\ell_{p}$, $p>2$. If $\mathcal{G}\subseteq{\mathcal{B}}_{1}(X)$ is _SOT
∗_-$\,G_{\delta}$ and _SOT_ \- dense in ${\mathcal{B}}_{1}(X)$, then
$\mathcal{G}$ is _SOT_ \- comeager in ${\mathcal{B}}_{1}(X)$. Therefore, any
_SOT ∗_\- comeager subset of ${\mathcal{B}}_{1}(X)$ is _SOT_ \- comeager as
well.

###### Proof.

Choose a sequence $(\mathcal{O}_{k})_{k\in\mathbb{N}}$ of SOT∗\- open subsets
of ${\mathcal{B}}_{1}(X)$ such that
$\bigcap_{k\in\mathbb{N}}\mathcal{O}_{k}=\mathcal{G}$. Then each set
$\mathcal{O}_{k}$ is SOT \- dense in ${\mathcal{B}}_{1}(X)$. Denoting by
$\stackrel{{\scriptstyle\scriptscriptstyle\circ}}{{\;{\mathcal{O}_{k}}}}$ the
relative SOT \- interior of $\mathcal{O}_{k}$ in ${\mathcal{B}}_{1}(X)$, it
follows from Theorem 5.12 that each
$\stackrel{{\scriptstyle\scriptscriptstyle\circ}}{{\;{\mathcal{O}_{k}}}}$ is
(open and) dense in $({\mathcal{B}}_{1}(X),\texttt{SOT})$. Hence $\mathcal{G}$
is SOT \- comeager since it contains
$\bigcap_{k\in\mathbb{N}}\stackrel{{\scriptstyle\scriptscriptstyle\circ}}{{\;{\mathcal{O}_{k}}}}$.
∎

The other ingredient is the following result. This was proved for $p=2$ in
[GMM], and essentially the same proof would give the result for an arbitrary
$p$.

######  Proposition 5.14.

Let $X=\ell_{p}$, $1<p<\infty$. For any real number $M>1$, the set
$\\{T\in{\mathcal{B}}_{1}(X);\;(MT)^{*}\;\hbox{is hypercyclic and
$\|T\|=1$}\\}$ is a dense $G_{\delta}$ subset of
$\bigl{(}{\mathcal{B}}_{1}(X),\emph{{SOT}\mbox{${}^{*}$}}\bigr{)}$.

###### Proof.

The proof of [GMM, Proposition 2.3], which is given for $p=2$ but works in
fact for any $1<p<\infty$, shows that the set of hypercyclic operators is
$G_{\delta}$ and dense in
$\bigl{(}\mathcal{B}_{M}(X),\texttt{SOT}\mbox{${}^{*}$}\bigr{)}$. Since the
map $T\mapsto(MT)^{*}$ is a homeomorphism from
$\bigl{(}{\mathcal{B}}_{1}(X),\texttt{SOT}\mbox{${}^{*}$}\bigr{)}$ onto
$\bigl{(}\mathcal{B}_{M}(X^{*}),\texttt{SOT}\mbox{${}^{*}$}\bigr{)}$, it
follows that the set $\\{T\in{\mathcal{B}}_{1}(X);\;(MT)^{*}\;\hbox{is
hypercyclic }\\}$ is $G_{\delta}$ and dense in
$\bigl{(}{\mathcal{B}}_{1}(X),\texttt{SOT}\mbox{${}^{*}$}\bigr{)}$. Moreover,
the set $\\{T\in{\mathcal{B}}_{1}(X);\;\|T\|=1\\}$ is also $G_{\delta}$ and
dense in $\bigl{(}{\mathcal{B}}_{1}(X),\texttt{SOT}\mbox{${}^{*}$}\bigr{)}$;
in fact, SOT -$\,G_{\delta}$ and SOT∗\- dense. ∎

###### Proof of Theorem 5.3.

The result follows immediatey from Corollary 5.13 and Proposition 5.14. ∎

###  5d. Outline of the proof of Theorem 5.12

Let us recall our notation: the canonical basis of $X=\ell_{p}$ is denoted by
$(e_{j})_{j\geq 0}$, its dual basis by $(e_{j}^{*})_{j\geq 0}$, and for each
$N\geq 0$,

$E_{N}=[\,e_{0},\dots,e_{N}\,]\qquad\textrm{and}\qquad
F_{N}=[\,e_{j}\;;\;j>N\,].$

The canonical projection of $X$ onto $E_{N}$ is denoted by $P_{N}$. For any
operator $T\in{\mathcal{B}}(X)$, we identify the operator
$P_{N}TP_{N}\in{\mathcal{B}}(X)$ with
$P_{N}T_{|E_{N}}\in{\mathcal{B}}(E_{N})$. So we consider $P_{N}TP_{N}$ both as
an operator on $X$ and as an operator on $E_{N}$. Likewise, we may consider
any operator on $A\in{\mathcal{B}}(E_{N})$ as an operator on $X$ by
identifying it with $P_{N}AP_{N}$.

In what follows, we will use the following rather ad hoc terminology.

Recall that a vector $x$ is said to be _norming_ for an operator $A$ if
$\|x\|=1$ and $\|Ax\|=\|A\|$. Given $N\geq 0$, we will say that an operator
$A\in{\mathcal{B}}(E_{N})$ is _absolutely exposing_ if the set of all norming
vectors for $A$ is as small as possible, i.e consists only of unimodular
multiples of a single vector $x_{0}\in S_{E_{N}}$. We denote by
${\mathcal{E}}_{1}(E_{N})$ the set of absolutely exposing operators
$A\in{\mathcal{B}}_{1}(E_{N})$. Moreover, we say that an absolutely exposing
operator $A\in{\mathcal{B}}(E_{N})$ is _evenly distributed_ if for any norming
vector $x\in S_{E_{N}}$ for $A$, we have ${\langle e_{j}^{*},x\rangle}\neq 0$
and ${\langle e_{j}^{*},Ax\rangle}\neq 0$ for every $j\in[0,N]$.

The key steps in the proof of Theorem 5.12 are the next two propositions,
whose proofs are postponed to a later section.

######  Proposition 5.15.

Let $X=\ell_{p}$, $1<p<\infty$. Let also $N$ be a non-negative integer, and
let $A\in{\mathcal{B}}_{1}(E_{N})$. For any $\varepsilon>0$, there exists an
operator $B\in{\mathcal{B}}_{1}(E_{2N+1})$ such that

  1. _(i)_

$\|B\|=1$;

  2. _(ii)_

$B$ is absolutely exposing;

  3. _(iii)_

$B$ is evenly distributed;

  4. _(iv)_

$\|BP_{N}-A\|<\varepsilon$.

Moreover, if $\|A\|$ is sufficiently close to $1$, then one may require that
in fact

  1. _(iv’)_

$\|B-A\|<\varepsilon$.

This first proposition is valid on any $\ell_{p}\,$\- space with $p>1$. Our
assumption that $p>2$ will be needed only in the next statement:

######  Proposition 5.16.

Let $X=\ell_{p}$, $p>2$. Fix an integer $M\geq 0$, and suppose that
$B\in{\mathcal{B}}_{1}(E_{M})$ is such that $\|B\|=1$, $B$ is absolutely
exposing, and $B$ is evenly distributed. Then, for any $\varepsilon>0$, there
exists $\delta>0$ such that the following holds true:

$\textrm{if}\ T\in{\mathcal{B}}_{1}(X)\ \textrm{satisfies}\
\|P_{M}TP_{M}-B\|<\delta,\ \textrm{then}\ \|P_{M}T(I-P_{M})\|<\varepsilon.$

Taking the above two propositions for granted, we can now give the

###### Proof of Theorem 5.12.

For any $\mathcal{A}\subseteq{\mathcal{B}}_{1}(X)$, we denote by
$\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{A}}}$ the
relative SOT \- interior of $\mathcal{A}$ in ${\mathcal{B}}_{1}(X)$. It is in
fact enough to show that if $\mathcal{O}\subseteq{\mathcal{B}}_{1}(X)$ is
SOT∗\- open and $\mathcal{O}\neq\emptyset$, then
$\stackrel{{\scriptstyle\;\;\scriptscriptstyle\circ}}{{\mathcal{O}}}\neq\emptyset$.

Since the unit sphere of $\mathcal{B}(X)$ is SOT∗\- dense in
${\mathcal{B}}_{1}(X)$, one can find $T_{0}\in\mathcal{O}$ with $\|T\|=1$.
Choose $K\in\mathbb{N}$ and $\varepsilon$ such that for any
$T\in{\mathcal{B}}_{1}(X)$, the following implication holds true:

$\bigl{(}\,\|(T-T_{0})P_{K}\|<\varepsilon\quad{\rm
and}\quad\|(T^{*}-T_{0}^{*})P_{K}^{*}\|<\varepsilon\,\bigr{)}\implies
T\in\mathcal{O}.$

Next, choose $L\geq K$ such that, setting $A_{N}:=P_{N}T_{0}P_{N}$, we have

$\forall N\geq L\;:\;\|(A_{N}-T_{0})P_{K}\|<\varepsilon/3\quad{\rm
and}\quad\|(A_{N}^{*}-T_{0}^{*})P_{K}^{*}\|<\varepsilon/3.$

Now, let $\alpha>0$ be very small, and fix $N\geq L$ such that
$\|A_{N}\|>1-\alpha$.

If $\alpha$ is small enough then, by Proposition 5.15, one can find $M>N$ and
$B\in\mathcal{B}_{1}(E_{M})$ such that $\|B\|=1$, $B$ is absolutely exposing
and evenly distributed, and $\|B-A_{N}\|<\varepsilon/3$.

By Proposition 5.16, one can find $\gamma>0$ such that for any
$T\in{\mathcal{B}}_{1}(X)$, the following implication holds true:

$\|P_{M}TP_{M}-B\|<\gamma\implies\|P_{M}T(I-P_{M})\|<\varepsilon/6.$

We assume without loss of generality that $\gamma<\varepsilon/6$. Then
$\|P_{M}T-B\|<\gamma+\varepsilon/6<\varepsilon/3$ if
$\|P_{M}TP_{M}-B\|<\gamma$; and since $P_{M}TP_{M}-B=P_{M}(TP_{M}-B)$, we see
that

$\|TP_{M}-B\|<\gamma\implies\|P_{M}T-B\|<\varepsilon/3.$

Now, set

$\mathcal{U}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;\|TP_{M}-B\|<\gamma\bigr{\\}}.$

The set $\mathcal{U}$ is SOT \- open, and $\mathcal{U}\neq\emptyset$ since
$B\in\mathcal{U}$. Let us show that $\mathcal{U}\subseteq\mathcal{O}$. It is
enough to check that if $T\in\mathcal{U}$, then
$\|(T-T_{0})P_{K}\|<\varepsilon$ and
$\|(T^{*}-T_{0}^{*})P_{K}^{*}\|<\varepsilon$.

On the one hand, we have

$\|(T-T_{0})P_{K}\|\leq\|(T-A_{N})P_{K}\|+\|(A_{N}-T_{0})P_{K}\|<\|(T-A_{N})P_{K}\|+\varepsilon/3.$

Moreover,
$\|(T-A_{N})P_{K}\|\leq\|(T-B)P_{K}\|+\|(B-A_{N})P_{K}\|<\|(T-B)P_{K}\|+\varepsilon/3$,
and $\|(T-B)P_{K}\|=\|(TP_{M}-B)P_{K}\|<\gamma<\varepsilon/3$. Hence
$\|(T-T_{0})P_{K}\|<\varepsilon$.

On the the other hand,
$\|(T^{*}-T_{0}^{*})P_{K}^{*}\|<\|(T^{*}-A_{N}^{*})P_{K}^{*}\|+\varepsilon/3<\|(T^{*}-B^{*})P_{K}^{*}\|+2\varepsilon/3$.
Moreover, since $\|TP_{M}-B\|<\gamma$, we have $\|P_{M}T-B\|<\varepsilon/3$
and hence $\|T^{*}P_{M}^{*}-B^{*}\|<\varepsilon/3$. So we get
$\|(T^{*}-B^{*})P_{K}^{*}\|=\|(T^{*}P_{M}^{*}-B^{*})P_{K}^{*}\|<\varepsilon/3$,
and hence $\|(T^{*}-T_{0}^{*})P_{K}^{*}\|<\varepsilon$. ∎

###  5e. Proofs of Propositions 5.15 and 5.16

Since it is the shortest, we start with the proof of Proposition 5.16.

###### Proof of Proposition 5.16.

Let $\delta>0$ (how small $\delta$ needs to be will be determined during the
proof) and let $T\in{{\mathcal{B}}}_{1}(X)$ be such that
$\|P_{M}TP_{M}-B\|<\delta$. Let $x_{1}\in S_{E_{M}}$ be a norming vector for
$B$. For every non-zero vector $y\in F_{M}=[\,e_{i}\,;\,i>M]$ and every scalar
$\mu>0$, we have by Lemma 5.4 that

$\displaystyle|{\langle e_{j}^{*},Tx_{1}+\mu Ty\rangle}|^{p}+|{\langle
e_{j}^{*},Tx_{1}-\mu Ty\rangle}|^{p}\geq 2\,|$ $\displaystyle{\langle
e_{j}^{*},Tx_{1}\rangle}|^{p}$ $\displaystyle+p\,\mu^{2}\,|{\langle
e_{j}^{*},Tx_{1}\rangle}\,|^{p-2}|{\langle e_{j}^{*},Ty\rangle}\,|^{2}$

for every $j\geq 0$, which can be rewritten as

$\displaystyle p\mu^{2}\,|{\langle e_{j}^{*},Tx_{1}\rangle}|^{p-2}\,|{\langle
e_{j}^{*},Ty\rangle}|^{2}\leq|\,{\langle e_{j}^{*},Tx_{1}+\mu
Ty\rangle}\,|^{p}+|\,$ $\displaystyle{\langle e_{j}^{*},Tx_{1}-\mu
Ty\rangle}\,|^{p}$ $\displaystyle-2\,|\,{\langle
e_{j}^{*},Tx_{1}\rangle}\,|^{p}\\!.$

Summing over $j\geq 0$ yields that

$\displaystyle p\mu^{2}\,\sum_{j\geq 0}\,|{\langle
e_{j}^{*},Tx_{1}\rangle}|^{p-2}\,|\,{\langle e_{j}^{*},Ty\rangle}\,|^{2}$
$\displaystyle\leq\|T(x_{1}+\mu y)\|^{p}+\|T(x_{1}-\mu
y)\|^{p}-2\,\|Tx_{1}\|^{p}$ $\displaystyle\leq\|x_{1}+\mu y\|^{p}+\|x_{1}-\mu
y\|^{p}-2\|Tx_{1}\|^{p}.$

Since $\|Tx_{1}\|>1-\delta$ and since the vectors $x_{1}$ and $y$ have
disjoint supports, it follows that

$p\,\mu^{2}\sum_{j\geq 0}\,|{\langle
e_{j}^{*},Tx_{1}\rangle}|^{p-2}\,|{\langle e_{j}^{*},Ty\rangle}|^{2}\leq
2\,\bigl{(}1+\mu^{p}\,\|y\|^{p}-(1-\delta)^{p}\bigr{)}.$

Let us temporarily write

$\alpha:=\sum_{j\geq 0}\,|{\langle e_{j}^{*},Tx_{1}\rangle}|^{p-2}\,|{\langle
e_{j}^{*},Ty\rangle}|^{2}.$

Then $\|y\|^{p}\mu^{p}-(p\alpha/2)\mu^{2}+1-(1-\delta)^{p}\geq 0$ for every
$\mu>0$. Optimizing this inequality with respect to $\mu$, i.e. taking
$\mu:=\left(\frac{\alpha}{\|y\|^{p}}\right)^{1/(p-2)}$, we obtain

$\dfrac{\alpha^{\,p/(p-2)}}{\|y\|^{\,2p/(p-2)}}\,\biggl{(}1-\dfrac{p}{2}\biggr{)}+\bigl{(}1-(1-\delta)^{p}\bigr{)}\geq
0;$

which can be rewritten as

$\dfrac{p-2}{2}\,\Bigl{(}\,\dfrac{\alpha}{\|y\|^{2}}\,\Bigr{)}^{p/(p-2)}\leq
1-(1-\delta)^{p}\leq p\delta$

i.e.

$0\leq\alpha\leq\|y\|^{2}\cdot
K_{p}\cdot\delta^{(p-2)/p},\quad\textrm{where}\quad
K_{p}:={\biggl{(}\dfrac{2p}{p-2}\biggr{)}^{(p-2)/p}}\cdot$ (5.1)

Our assumption on the operator $B$ implies that

$\gamma:=\min_{j\in[0,M]}|{\langle e_{j}^{*},Bx_{1}\rangle}|>0.$

Let $\delta$ be so small that every operator $T\in{{\mathcal{B}}}_{1}(E_{M})$
with $\|P_{M}TP_{M}-B\|<\delta$ satisfies $\min_{j\in[0,M]}|{\langle
e_{j}^{*},Tx_{1}\rangle}|>\gamma/2$. For every $j\in[0,M]$, we have by (5.1)
that

$|{\langle e_{j}^{*},Ty\rangle}|^{2}\leq\|y\|^{2}\cdot
K_{p}\cdot\biggl{(}\dfrac{2}{\gamma}\biggr{)}^{p-2}\cdot\,\delta^{(p-2)/p},$

i.e.

$|{\langle e_{j}^{*},Ty\rangle}|^{p}\leq\|y\|^{p}\cdot
K_{p}^{p/2}\cdot\biggl{(}\dfrac{2}{\gamma}\biggr{)}^{\frac{p(p-2)}{2}}\cdot\,\delta^{(p-2)/2}.$

Summing over $j\in[0,M]$, we obtain that

$\|P_{M}Ty\|\leq\|y\|\cdot K_{p}^{1/2}\cdot
M^{1/p}\cdot\biggl{(}\dfrac{2}{\gamma}\biggr{)}^{\frac{p-2}{2}}\cdot\,\delta^{\frac{p-2}{2p}}.$

If we choose $\delta>0$ so small that $\displaystyle K_{p}^{1/2}\cdot
M^{1/p}\cdot(2/\gamma)^{\frac{p-2}{2}}\cdot\,\delta^{\frac{p-2}{2p}}<\varepsilon$
(this condition depends on $p$, $M$, and $B$, but not on $T$), we obtain that
$\|P_{M}T(I-P_{M})\|<\varepsilon$, which is exactly the inequality we were
looking for. ∎

It remains to prove Proposition 5.15, and this will be a bit longer. Recall
that for each $N\geq 1$, we denote by ${\mathcal{E}}_{1}(E_{N})$ the set of
absolutely exposing operators of ${{\mathcal{B}}}_{1}(E_{N})$. We begin by a
series of rather elementary lemmas which shed some light on the behaviour of
absolutely exposing contractions.

######  Lemma 5.17.

The set ${\mathcal{E}}_{1}(E_{N})$ is dense in ${{\mathcal{B}}}_{1}(E_{N})$.

###### Proof.

Let $A\in{{\mathcal{B}}}_{1}(E_{N})$ with $A\neq 0$ and $\|A\|<1$. Let
$x_{0}\in S_{E_{N}}$ be such that $\|Ax_{0}\|=\|A\|$, and let $x_{0}^{*}\in
E_{N}$ be such that $\|x_{0}\|={\langle x_{0}^{*},x_{0}\rangle}=1$. Let
$R_{0}$ be the rank 1 operator on $E_{N}$ defined by $R_{0}(x):={\langle
x_{0}^{*},x\rangle}\,Ax_{0}$, $x\in E_{N}$; and for any $\delta>0$, let
$A_{\delta}:=A+\delta R_{0}$. Then $A_{\delta}x_{0}=(1+\delta)Ax_{0}$, and
$\|A_{\delta}x_{0}\|=(1+\delta)\|Ax_{0}\|=(1+\delta)\|A\|$. Since
$\|R_{0}\|=\|A\|$, $\|A_{\delta}\|\leq(1+\delta)\|A\|$, and hence
$\|A_{\delta}\|=(1+\delta)\|A\|$. So $\|A_{\delta}x_{0}\|=\|A_{\delta}\|$.
Moreover, if $x\in S_{E_{N}}$ is such that $\|A_{\delta}x\|=\|A_{\delta}\|$,
then $\|Ax+\delta\,{\langle x_{0}^{*},x\rangle}\,Ax_{0}\|=(1+\delta)\|A\|$. So
${\langle x_{0}^{*},x\rangle}\neq 0$ and

$\displaystyle(1+\delta)\|A\|=\|Ax+\delta{\langle
x_{0}^{*},x\rangle}\,Ax_{0}\|$ $\displaystyle\leq\|A\|\,\|x+\delta\,{\langle
x_{0}^{*},x\rangle}\,x_{0}\|$
$\displaystyle\leq\|A\|\,\bigl{(}\|x\|+\delta\,|{\langle
x_{0}^{*},x\rangle}|\,\|x_{0}\|\bigr{)}\leq\|A\|\,(1+\delta).$

Since $\|A\|>0$, it follows that $\|x+\delta\,{\langle
x_{0}^{*},x\rangle}\,x_{0}\|=\|x\|+\|\delta\,{\langle
x_{0}^{*},x\rangle}\,x_{0}\|$, and since ${\langle x_{0}^{*},x\rangle}\neq 0$,
this implies that $x$ is colinear to $x_{0}$, by strict convexity of the
$\ell_{p}\,$\- norm. So $A_{\delta}$ is absolutely exposing. Given
$\varepsilon>0$, one can choose $\delta>0$ so small that $\|A_{\delta}\|<1$
and $\|A-A_{\delta}\|<\varepsilon$, and this proves that
${\mathcal{E}}_{1}(E_{N})$ is dense in ${{\mathcal{B}}}_{1}(E_{N})$. ∎

######  Lemma 5.18.

Let $A\in{\mathcal{E}}_{1}(E_{N})$, and let $x_{0}\in S_{E_{N}}$ be a norming
vector for $A$. For every $\varepsilon>0$, there exists $\delta>0$ such that
any vector $x\in S_{E_{N}}$ such that $\|Ax\|>(1-\delta)\|A\|$ satisfies
$\emph{dist}(x,\mathbb{T}x_{0})<\varepsilon$.

###### Proof.

Fix $\varepsilon>0$ and, towards contradiction, suppose that for every $n\geq
1$, there exists a vector $x_{n}\in S_{E_{N}}$ such that
$\|Ax_{n}\|>(1-2^{-n})\|A\|$ and
$\textrm{dist}(x_{n},\mathbb{T}x_{0})\geq\varepsilon$. Without loss of
generality, we can suppose that the sequence $(x_{n})_{n\geq 1}$ converges to
a vector $x_{\infty}\in S_{E_{N}}$. Then $\|Ax_{\infty}\|=\|A\|$ and thus
$x_{\infty}\in\mathbb{T}x_{0}$, which is a contradiction. ∎

######  Lemma 5.19.

Let $A\in{\mathcal{E}}_{1}(E_{N})$, and let $x_{0}\in S_{E_{N}}$ be a norming
vector for $A$. For every $\varepsilon>0$, there exists $\gamma>0$ with the
following property: for every operator $B\in{{\mathcal{B}}}_{1}(E_{N})$ with
$\|A-B\|<\gamma$ and any norming vector $x\in S_{E_{N}}$ for $B$, we have
$\emph{dist}(x,\mathbb{T}x_{0})<\varepsilon$.

###### Proof.

Given $A\in{{\mathcal{B}}}_{1}(E_{N})$ and $\varepsilon>0$, let $\delta>0$ be
given by Lemma 5.18: if $x\in S_{E_{N}}$ is such that
$\|Ax\|>(1-\delta)\|A\|$, then $\textrm{dist}(x,\mathbb{T}x_{0})<\varepsilon$.
Let $\gamma>0$ be so small that $\|A\|-2\gamma\geq(1-\delta)\|A\|$. If
$B\in{{\mathcal{B}}}_{1}(E_{N})$ is such that $\|B-A\|<\gamma$, and if $x\in
S_{E_{N}}$ is a norming vector for $B$, then
$\bigl{|}\,\|B\|-\|Ax\|\,\bigr{|}<\gamma$ , and since
$\bigl{|}\,\|B\|-\|A\|\,\bigr{|}<\gamma$, one has
$\bigl{|}\,\|A\|-\|Ax\|\,\bigr{|}<2\gamma$. Hence
$\|Ax\|>\|A\|-2\gamma\geq(1-\delta)\|A\|$, and so
$\textrm{dist}(x,\mathbb{T}x_{0})<\varepsilon$. ∎

Some of the arguments in the proofs to come will involve a duality mapping
$\mathbf{J}:\ell_{p}\to\ell_{p^{\prime}}$, where $1/p+1/p^{\prime}=1$. Recall
that we are assuming that $1<p<\infty$. The map $\mathbf{J}$ is defined as
follows: if $x\in X=\ell_{p}$, then $\mathbf{J}(x)$ is the unique element of
$X^{*}=\ell_{p^{\prime}}$ such that
${\langle\mathbf{J}(x),x\rangle}=\|\mathbf{J}(x)\|\,\|x\|=\|x\|^{p}$ (there
are several natural ways of “normalizing” the duality mapping when defining it
on the whole space rather than just on the unit sphere). Explicitely, we have:

$\mathbf{J}(x)=\sum_{j\geq 1}\overline{{\langle
e_{j}^{*},x\rangle}}\cdot\bigl{|}{\langle
e_{j}^{*},x\rangle}\bigr{|}^{p-2}\,e_{j}^{*},\quad x\in X.$

Recall also that an absolutely exposing operator
$B\in{\mathcal{E}}_{1}(E_{N})$ is said to be evenly distributed if for every
norming vector $x_{1}\in S_{E_{N}}$ for $B$, we have ${\langle
e_{j}^{*},x_{1}\rangle}\neq 0$ and ${\langle e_{j}^{*},Bx_{1}\rangle}\neq 0$
for all $j\in[0,N]$.

######  Proposition 5.20.

Let $A\in{\mathcal{E}}_{1}(E_{N})$ be such that $A\neq 0$ and $\|A\|<1$. For
every $\varepsilon>0$, there exists an operator $B\in{\mathcal{E}}_{1}(E_{N})$
such that $\|B-A\|<\varepsilon$ and $B$ is evenly distributed.

###### Proof.

The proof will proceed in two steps.

Step 1. _We prove that given $A\in{\mathcal{E}}_{1}(E_{N})$ with $A\neq 0$ and
$\|A\|<1$, and given $\varepsilon>0$, there exists
$C\in{\mathcal{E}}_{1}(E_{N})$ with $\|C-A\|<\varepsilon$ such that for any
norming vector $x_{1}\in S_{E_{N}}$ for $C$, and for every $j\in[0,N]$, we
have ${\langle e_{j}^{*},x_{1}\rangle}\neq 0$._

Let $x_{0}\in S_{E_{N}}$ be any norming vector for $A$, and let
$J:=\\{j\in[0,N]\;;\;{\langle e_{j}^{*},x_{0}\rangle}\neq 0\\}$. If $J=[0,N]$,
there is nothing to prove, so suppose that $J\neq[0,N]$. Observe that since
$x_{0}$ is supported on $J$ and $\|Ax_{0}\|=\|A\|$, we have
$\|\,A_{|E_{J}}\,\|=\|A\|$, where $E_{J}=[\,e_{j}\;;\;j\in J]$. Consider now
the sets

$\displaystyle J^{\prime}$
$\displaystyle:=\bigl{\\{}\,j\in[0,N]\;;\;\forall\,l\in[0,N],\;\exists\,s\in\mathbb{C}\setminus\\{0\\}\
\textrm{such that}\ \|A+s\,e_{l}\otimes e_{j}^{*}\|\leq\|A\|\bigr{\\}}$ and
$\displaystyle K$ $\displaystyle:=[0,N]\setminus(J\cup J^{\prime}).$

We first observe that $J^{\prime}$ is actually included in $J$:

######  Lemma 5.21.

For every $j\in J^{\prime}$, we have ${\langle e_{j}^{*},x_{0}\rangle}\neq 0$.

###### Proof of Lemma 5.21.

Let $j\in J^{\prime}$, and suppose that ${\langle e_{j}^{*},x_{0}\rangle}=0$.
Since $x_{0}\in S_{E_{N}}$ is a norming vector for $A$, the function
$\phi:\mathbb{R}\to\mathbb{R}$ defined by

$\phi(t)=\dfrac{\|A(x_{0}+te_{j})\|^{p}}{\|x_{0}+te_{j}\|^{p}}$

attains its maximum at $t=0$. As ${\langle e_{j}^{*},x_{0}\rangle}=0$,
$\|x_{0}+te_{j}\|^{p}=1+t^{p}$, and so

$\phi(t)=\dfrac{\|A(x_{0}+te_{j})\|^{p}}{1+t^{p}},\quad t\in\mathbb{R}.$

Since $p>1$, $\phi$ is differentiable on $\mathbb{R}$, and thus
$\phi^{\prime}(0)=0$. Using the fact that for any complex numbers $a$ and $b$,
the function $\varphi_{a,b}$ defined by $\varphi_{a,b}(t)=|a+tb|^{p}$,
$t\in\mathbb{R}$, is differentiable on $\mathbb{R}$ with
$\varphi^{\prime}_{a,b}(0)=p\cdot|a|^{p-2}\cdot\,\mathfrak{Re}(a\,\overline{b})$,
we obtain that

$\phi(0)=p\cdot\sum_{k=0}^{N}|\,{\langle
e_{k}^{*},Ax_{0}\rangle}\,|^{p-2}\cdot\,\mathfrak{Re}\bigl{(}{\langle
e_{k}^{*},Ax_{0}\rangle}\ \overline{{\langle
e_{k}^{*},Ae_{j}\rangle}}\,\bigr{)}=p\cdot\mathfrak{Re}\bigl{(}{\langle\mathbf{J}(Ax_{0}),Ae_{j}\rangle}\bigr{)}=0.$

Applying the same argument to the function $\psi:\mathbb{R}\to\mathbb{R}$
defined by

$\psi(t)=\dfrac{\|A(x_{0}+ite_{j})\|^{p}}{1+t^{p}},\quad t\in\mathbb{R},$

we obtain that
$\psi^{\prime}(0)=p\cdot\mathfrak{Im}\,\bigl{(}{\langle\mathbf{J}(Ax_{0}),Ae_{j}\rangle}\bigr{)}=0$.
Hence ${\langle\mathbf{J}(A{x_{0}}),Ae_{j}\rangle}=0$. In other words, the
vector $Ae_{j}$ belongs to the kernel of the functional
$\mathbf{J}(Ax_{0})_{|E_{N}}$ , which we denote by $H$. Since $Ax_{0}\neq 0$,
$H$ is a hyperplane of $E_{N}$.

Fix now $l\in[0,N]$. Since $j\in J^{\prime}$, there exists $s\neq 0$ such that
$\|A+s\,e_{l}\otimes e_{j}^{*}\|\leq\|A\|$. Setting $B=A+s\,e_{l}\otimes
e_{j}^{*}$, we have $\|B\|\leq\|A\|$ and $Bx_{0}=Ax_{0}+s\,{\langle
e_{j}^{*},x_{0}\rangle}\,e_{l}=Ax_{0}$ since we have supposed that ${\langle
e_{j}^{*},x_{0}\rangle}=0$. Hence $\|B\|=\|A\|$ and $x_{0}$ is a norming
vector for the operator $B$. Reasoning as above, we obtain that the vector
$Be_{j}$ belongs to the kernel of the functional
$\mathbf{J}(Bx_{0})_{|E_{N}}$, which is $H$. So $Be_{j}=Ae_{j}+s\,e_{l}\in H$.
Since $Ae_{j}\in H$ and $s\neq 0$, it follows that $e_{l}$ belongs to $H$.
This being true for every $l\in[0,N]$, we have $H=E_{N}$, which is a
contradiction. So ${\langle e_{j}^{*},x_{0}\rangle}\neq 0$, and Lemma 5.21 is
proved. ∎

We have thus shown that $J^{\prime}\subseteq J$, and hence we can write the
set $[0,N]$ as the disjoint union of the sets $J$ and $K$.

######  Lemma 5.22.

Suppose that $K\neq\emptyset$. Let $k\in K$ and $\gamma>0$. There exists an
operator $D\in{\mathcal{E}}_{1}(E_{N})$ with $D\neq 0$, $\|D\|<1$ and
$\|D-A\|<\gamma$ such that, for any norming vector $x_{2}\in S_{E_{N}}$ for
$D$, we have ${\langle e_{j}^{*},x_{2}\rangle}\neq 0$ for all $j\in
J\cup\\{k\\}$.

###### Proof of Lemma 5.22.

Since $k$ belongs to $K$, it does not belong to $J^{\prime}$, and hence there
exists $l\in[0,N]$ such that $\|A+s\,e_{l}\otimes e_{k}^{*}\|>\|A\|$ for every
$s>0$. For each $s>0$, set $B_{s}:=A+s\,e_{l}\otimes e_{k}^{*}$. Then
$\|B_{s}-A\|=s$, and the operator induced by $B_{s}$ on
$E_{\,[0,N]\setminus\\{k\\}}$ coincides with $A$ on
$E_{\,[0,N]\setminus\\{k\\}}$. Since $\|B_{s}\|>\|A\|$, it follows that if
$x\in S_{E_{N}}$ is a norming vector for $B_{s}$, then ${\langle
e_{k}^{*},x\rangle}\neq 0$.

On the other hand, since ${\langle e_{j}^{*},x_{0}\rangle}\neq 0$ for every
$j\in J$, Lemma 5.19 implies that if $s$ is sufficiently small, any norming
vector $x\in S_{E_{N}}$ for $B_{s}$ is such that ${\langle
e_{j}^{*},x\rangle}\neq 0$ for every $j\in J$. Combining this with the
observation above, we deduce that if $s\neq 0$ is sufficiently small, then any
norming vector $x\in S_{E_{N}}$ for $B_{s}$ is such that ${\langle
e_{j}^{*},x\rangle}\neq 0$ for every $j\in J\cup\\{k\\}$. By Lemmas 5.17 and
5.19, we can now choose $B\in{\mathcal{E}}_{1}(E_{N})$ so close to $B_{s}$
(with $s$ small enough chosen first) that any norming vector $x_{2}\in
S_{E_{N}}$ for $B$ satisfies ${\langle e_{j}^{*},x_{2}\rangle}\neq 0$ for
every $j\in J\cup\\{k\\}$. ∎

We now have all the tools handy to finish the proof of Step 1 of Proposition
5.20. Suppose that $K$ has cardinality $r\geq 1$, and write
$K=\\{k_{1},\dots,k_{r}\\}$. Choose $0<\gamma<\varepsilon/r$, and apply Lemma
5.22 $r$ times successively, to obtain operators
$D_{i}\in{\mathcal{E}}_{1}(E_{N})$, $i\in[1,r]$, with $D_{i}\neq 0$,
$\|D_{i}\|=1$, $\|D_{i+1}-D_{i}\|<\gamma$ for every $i\in[1,r-1]$, and such
that for every norming vector $y_{i}\in S_{E_{N}}$ for the operator $D_{i}$,
one has that ${\langle e_{j}^{*},y_{i}\rangle}\neq 0$ for every $j\in
J\cup\\{k_{1},\dots,k_{i}\\}$. Then, the operator $C:=D_{r}$ belongs to
${\mathcal{E}}_{1}(E_{N})$, satisfies $\|C-A\|<\varepsilon$, and whenever
$x\in S_{E_{N}}$ is a norming vector for $C$, we have ${\langle
e_{j}^{*},x\rangle}\neq 0$ for every $j\in J\cup K$. Since $J\cup K=[0,N]$ by
Lemma 5.21, this proves the statement we were looking for.

Step 2. _We now prove the statement of Proposition 5.20, namely that given
$A\in{\mathcal{E}}_{1}(E_{N})$ with $A\neq 0$ and $\|A\|<1$, and given
$\varepsilon>0$, there exists $B\in{\mathcal{E}}_{1}(E_{N})$ with
$\|A-B\|<\varepsilon$ such that for any norming vector $x_{2}\in S_{E_{N}}$
for $B$, and for every $j\in[0,N]$, we have ${\langle
e_{j}^{*},x_{2}\rangle}\neq 0$ and ${\langle e_{j}^{*},Bx_{2}\rangle}\neq 0$._

Let $C\in{\mathcal{E}}_{1}(E_{N})$ be given by Step 1, with $C\neq 0$,
$\|C\|<1$, $\|C-A\|<\varepsilon/2$, and ${\langle e_{j}^{*},x\rangle}\neq 0$
for every norming vector $x\in S_{E_{N}}$ for $C$ and all $j\in[0,N]$. We also
fix a norming vector $x_{1}\in S_{E_{N}}$ for $C$.

Our strategy is now to apply the result proved in Step 1 to the operator
$C^{*}$ acting on
$E_{N}^{*}=[e_{0}^{*},\dots,e_{N}^{*}]\subseteq\ell_{p^{\prime}}(\mathbb{Z}_{+})$,
where $1/p+1/p^{\prime}=1$. Since $1<p<\infty$, we have $1<p^{\prime}<\infty$.
We first observe that the operator $C^{*}$ belongs to
${\mathcal{E}}_{1}(E_{N}^{*})$, and that its norming vectors are the
unimodular multiples of the vector
$y_{1}^{*}:=\mathbf{J}(Cx_{1})\,/\,\|C\|^{p/p^{\prime}}$. Indeed, if $x^{*}\in
S_{E_{N}^{*}}$ is such that $\|C^{*}x^{*}\|=\|C^{*}\|$, there exists $x\in
S_{E_{N}}$ such that ${\langle C^{*}x^{*},x\rangle}=\|C^{*}\|$, i.e. ${\langle
x^{*},Cx\rangle}=\|Cx\|$. Hence $x$ is a norming vector for $C$, and thus $x$
belongs to $\mathbb{T}x_{1}$, and $|{\langle
x^{*},Cx_{1}\rangle}|=\|Cx_{1}\|=\|C\|$. It follows that $x^{*}$ is a
unimodular multiple of the vector
$\mathbf{J}(Cx_{1})\,/\,\|C\|^{p/p^{\prime}}=y_{1}^{*}$, and so $C^{*}$ is
absolutely exposing. Since moreover $\|C^{*}\|=1$ and $1<p^{\prime}<\infty$,
it is legitimate to apply the result proved in Step 1: there exists an
operator $D\in{\mathcal{E}}_{1}(E_{N}^{*})$ with $\|D-C^{*}\|<\varepsilon/2$
such that ${\langle y^{*},e_{j}\rangle}\neq 0$ for every norming vector
$y^{*}\in S_{E_{N}^{*}}$ for the operator $D$ and for every $j\in[0,N]$. Let
us fix a norming vector $y_{0}^{*}$ for $D$.

Set $B:=D^{*}\in{{\mathcal{B}}}_{1}(E_{N})$. Then $\|B-A\|<\varepsilon$, and
the same argument as above shows that $B\in{\mathcal{E}}_{1}(E_{N})$, and that
for any norming vector $x_{2}\in S_{E_{N}}$ for $B$, the vector
$\mathbf{J}(Bx_{2})\,/\,\|B\|^{p/p^{\prime}}$ is a unimodular multiple of
$y_{0}^{*}$. So $\mathbf{J}(Bx_{2})$ is a non-zero multiple of $y_{0}^{*}$,
and ${\langle\mathbf{J}(Bx_{2}),e_{j}\rangle}\neq 0$ for every $j\in[0,N]$. It
follows from the expression of $\mathbf{J}(Bx_{2})$ that ${\langle
e_{j}^{*},\mathbf{J}(Bx_{2})\rangle}\neq 0$ for all $j\in[0,N]$. Moreover,
Lemma 5.19 implies that if $\varepsilon>0$ is sufficiently small, the operator
$B$ is so close to $C$ that ${\langle e_{j}^{*},x_{2}\rangle}\neq 0$ for all
$j\in[0,N]$. Hence $B$ has all the required properties, and Proposition 5.20
is proved. ∎

We are now ready to prove Proposition 5.15.

###### Proof of Proposition 5.15.

Let $A\in{{\mathcal{B}}}_{1}(E_{N})$. By Lemma 5.17 and Proposition 5.20, we
can find an operator $A^{\prime}$ extremely close to it which is absolutely
exposing, evenly distributed, and is such that $\|A^{\prime}\|<1$. So we may
and do assume that $A$ satisfies these additional properties.

Write $E_{2N+1}$ as $E_{2N+1}=E_{N}\oplus G_{N}$, where
$G_{N}:=[\,e_{j}\;;\;N+1\leq j\leq 2N+1\,]$. Define $S_{N}:E_{N}\to G_{N}$ by
setting

$S_{N}x:=\sum_{i=0}^{N}e_{i}^{*}(x)e_{N+1+i}\quad\textrm{for every}\ x\in
E_{N}.$

For every parameters $\eta,\delta>0$, we define an operator
$B_{\eta,\delta}\in{{\mathcal{B}}}(E_{2N+1})$ in the following way:

$B_{\eta,\delta}(x+S_{N}y):=A(x+\delta y)+\eta\,S_{N}\,A(x+\delta y),\quad
x,y\in E_{N}.$

In matrix form with respect to the decomposition $E_{2N+1}=E_{N}\oplus G_{N}$,

$B_{\eta,\delta}=\begin{pmatrix}A&\delta A\\\ \eta A&\eta\delta
A\end{pmatrix}\cdot$

For every $x,y\in E_{N}$, we have

$\displaystyle\|B_{\eta,\delta}(x+S_{N}y)\|$
$\displaystyle=(1+\eta^{p})^{\,1/p}\,\|A(x+\delta y)\|$
$\displaystyle\leq(1+\eta^{p})^{\,1/p}\,\|A\|\,\bigl{(}\|x\|+\delta\,\|y\|\bigr{)}$
$\displaystyle\leq(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,\|A\|\cdot\Bigl{(}\,\|x\|^{p}+\|y\|^{p}\,\Bigr{)}^{\,1/p}$
$\displaystyle=(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,\|A\|\cdot\|x+S_{N}y\|.$

Hence
$\|B_{\eta,\delta}\|\leq(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,\|A\|$.
We now claim that $\|B_{\eta,\delta}\|$ is exactly equal to
$(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,\|A\|$.
Indeed, let $x_{0}\in S_{E_{N}}$ be a norming vector for $A$, and let $u$ be
the vector of $E_{2N+1}$ defined as
$u:=x_{0}+\delta^{p^{\prime}-1}S_{N}x_{0}$. We set also $u_{0}:=u\,/\,\|u\|$.
We have

$\displaystyle\|B_{\eta,\delta}(u)\|$
$\displaystyle=(1+\eta^{p})^{\,1/p}\,\|A(x_{0}+\delta^{p^{\prime}}x_{0})\|=(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})\,\|Ax_{0}\|$
$\displaystyle=(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})\,\|A\|$

since $x_{0}\in S_{E_{N}}$ is norming for $A$. Also
$\|u\|=(1+\delta^{(p^{\prime}-1)p})^{\,1/p}=(1+\delta^{p^{\prime}})^{\,1/p}$,
and hence

$\|B_{\eta,\delta}(u_{0})\|=(1+\eta^{p})^{\,1/p}\,(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,\|A\|.$

This proves our claim. Moreover, we have also shown that

$u_{0}=\frac{x_{0}+\delta^{p^{\prime}-1}S_{N}x_{0}}{(1+\delta^{p^{\prime}})^{\,1/p}}$

is a norming vector for $B_{\eta,\delta}$.

Let us now check that $B_{\eta,\delta}$ is absolutely exposing. Let $v\in
S_{E_{2N+1}}$ be such that $\|B_{\eta,\delta}v\|=\|B_{\eta,\delta}\|$. Write
$v=x+S_{N}y$ with $x,y\in E_{N}$ and $\|x\|^{\,p}+\|y\|^{\,p}=1$. Then
$\|B_{\eta,\delta}v\|=(1+\eta^{p})^{\,1/p}\,\|A(x+\delta y)\|$, so we have

$\|A(x+\delta
y)\|=\bigl{(}\,1+\delta^{p^{\prime}}\,\bigr{)}^{1/p^{\prime}}\,\|A\|.$

It follows from this equality that $x$ and $y$ are necessarily non-zero. Since

$\|x+\delta
y\|\leq\|x\|+\delta\|y\|\leq\bigl{(}\,1+\delta^{p^{\prime}}\,\bigr{)}^{1/p^{\prime}}\,\bigl{(}\,\|x\|^{p}+\|y\|^{p}\,\bigr{)}^{1/p}=\bigl{(}\,1+\delta^{p^{\prime}}\,\bigr{)}^{1/p^{\prime}},$

it also follows that $\|x+\delta
y\|=\|x\|+\delta\|y\|=(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}$. So there
exists $\alpha>0$ such that $y=\alpha x$. Finally, observe that
$\frac{x+\delta y}{\|x+\delta y\|}$ is a norming vector for $A$, and hence is
a unimodular multiple of $x_{0}$. We thus have
$(1+\alpha\delta)x=(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\lambda x_{0}$ for
some $\lambda\in\mathbb{T}$, so that

$v=\dfrac{(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}}{1+\alpha\delta}\,\lambda\,(x_{0}+\alpha
S_{N}x_{0}).$

Since $\|v\|=1$, we have
$1+\alpha\delta=(1+\delta^{p^{\prime}})^{\,1/p^{\prime}}\,(1+\alpha^{p})^{\,1/p}$.
By the equality case in Hölder’s inequality, it follows that the vectors
$\binom{1}{\delta^{p^{\prime}}}$ and $\binom{1}{\alpha^{p}}$ are colinear in
$\mathbb{R}^{2}$, which implies that $\alpha=\delta^{p^{\prime}-1}$. Looking
back at the definition of $u_{0}$, we thus see that $v$ is a unimodular
multiple of $u_{0}$. Hence $B_{\eta,\delta}$ is absolutely exposing, whatever
the choices of the parameters $\eta$ and $\delta$.

The next step is to verify that $B_{\eta,\delta}$ is evenly distributed. With
$u=x_{0}+\delta^{p^{\prime}-1}S_{N}x_{0}$ as above, it is enough to show that
${\langle e_{j}^{*},u\rangle}\neq 0$ and ${\langle
e_{j}^{*},B_{\eta,\delta}u\rangle}\neq 0$ for all $j\in[0,2N+1]$. We have
${\langle e_{j}^{*},u\rangle}={\langle e_{j}^{*},x_{0}\rangle}$ if $j\in[0,N]$
and ${\langle e_{j}^{*},u\rangle}=\delta^{p-1}{\langle
e_{j-N-1}^{*},x_{0}\rangle}$ if $j\in[N+1,2N+1]$; so ${\langle
e_{j}^{*},u\rangle}\neq 0$ in both cases since $x_{0}$ is a norming vector for
$A$ and $A$ is assumed to be evenly distributed. Likewise, since
$B_{\eta,\delta}(u)=A(x_{0}+\delta^{p^{\prime}}x_{0})+\eta\,S_{N}A(x_{0}+\delta^{p^{\prime}}x_{0})$
we have ${\langle
e_{j}^{*},B_{\eta,\delta}u\rangle}=(1+\delta^{p^{\prime}})\,{\langle
e_{i}^{*},Ax_{0}\rangle}$ if $j\in[0,N]$ and ${\langle
e_{j}^{*},B_{\eta,\delta}u\rangle}=\eta\,(1+\delta^{p^{\prime}})\,{\langle
e_{j-N-1}^{*},Ax_{0}\rangle}$ if $j\in[N+1,2N+1]$; so ${\langle
e_{j}^{*},B_{\eta,\delta}u\rangle}\neq 0$.

It remains to choose the parameters $\eta$ and $\delta$ in such a way that
$\|B_{\eta,\delta}P_{N}-A\|<\varepsilon$ and $\|B_{\eta,\delta}\|=1$; and,
moreover, to show that $\|B_{\eta,\delta}-A\|<\varepsilon$ if $\|A\|$ is close
enough to $1$. If $\eta>0$ is chosen sufficiently small, then it follows from
the definition of $B_{\eta,\delta}$ that
$\|B_{\eta,\delta}P_{N}-A\|<\varepsilon$ whatever the choice of $\delta$.
Moreover, once $\eta$ is chosen and is small enough, we can fix $\delta>0$ in
such a way that $\|B_{\eta,\delta}\|=1$. This is possible since $\|A\|<1$.
Indeed, if $\eta>0$ is such that $(1+\eta^{p})^{1/p}\|A\|<1$, then $\delta>0$
is uniquely determined by the equation

$(1+\eta^{p})^{1/p}\,(1+\delta^{p^{\prime}})^{1/p^{\prime}}\,\|A\|=1,$

namely

$\delta=\left[\biggl{(}\dfrac{1}{\|A\|\,(1+\eta^{p})^{1/p}}\biggr{)}^{p^{\prime}}-1\right]^{1/p^{\prime}}\leq\biggl{(}\dfrac{1}{\|A\|^{p^{\prime}}}-1\biggr{)}^{1/p^{\prime}}=:\delta_{A}.$

Finally, if $\|A\|$ is very close to $1$ then $\delta_{A}$ is very small; so,
looking at the definition of $B_{\eta,\delta}$, we see that
$\|B_{\eta,\delta}-A\|<\varepsilon$ provided that $\|A\|$ is close enough to
$1$. ∎

######  Remark 5.23.

The proof of Theorem 5.3 does not extend in a straightforward manner to the
case where $X=c_{0}$. The fact that we are working on an $\ell_{p}\,$\- space
with $p>1$ is used in an important manner in the proof of Proposition 5.20
(both in the differentiability argument underlying the proof of Lemma 5.21,
and in the duality argument used in Step 2 of the proof of Proposition 5.20),
as well as in the proof of Proposition 5.16. Although it seems likely, in view
of Proposition 5.5, that the statement of Theorem 5.3 also holds for
$X=c_{0}$, we are not able to provide a proof of it.

###  5f. Additional results

Theorem 5.3 rules out one trivial reason why a typical operator
$T\in{{\mathcal{B}}}_{1}(\ell_{p},\texttt{SOT})$, $p>2$ should have a non-
trivial invariant subspace, but it does not bring us any nearer to an answer
to this question.

In the Hilbertian case, here is a highly non-elementary proof of the fact that
a typical $T\in({\mathcal{B}}_{1}(H),\texttt{SOT})$ has non-trivial invariant
subspaces: by Theorem 3.4, we know that a typical
$T\in({\mathcal{B}}_{1}(H),\texttt{SOT})$ is such that
$\sigma(T)=\overline{\,\mathbb{D}}$; and by the Brown-Chevreau-Pearcy Theorem
from [BCP2], any $T\in{\mathcal{B}}_{1}(H)$ whose spectrum contains the unit
circle $\mathbb{T}$ has a non-trivial invariant subspace.

It is tempting to try to generalize this argument on more general Banach
spaces. The trouble is that there is no full analogue of the Brown-Chevreau-
Pearcy Theorem in a Banach space setting. A deep result due to Ambrozie and
Müller [AM] states that if $T$ is a _polynomially bounded_ operator on $X$
such that $\sigma(T)$ contains $\mathbb{T}$ and $\|T^{n}x\|\to 0$ for all
$x\in X$, then $T$ has a non-trivial invariant subspace. When $X=\ell_{p}$,
$1<p<\infty$, or $X=c_{0}$, we do know, thanks to Propositions 3.7 and 3.9,
that a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is such that
$\sigma(T)$ contains $\mathbb{T}$ and that $\|T^{n}x\|\to 0$ for all $x\in X$.
However, whereas any Hilbert space contraction is polynomially bounded by von
Neumann’s inequality, the next proposition shows that on $X=\ell_{p}$ with
$1\leq p<\infty$, $p\neq 2$, as well as on $c_{0}$, polynomial boundedness is
_not_ typical.

######  Proposition 5.24.

Let $X=c_{0}$ or $X=\ell_{p}$ with $1\leq p<\infty$ and $p\neq 2$. Then, the
set of all polynomially bounded contractions is $F_{\sigma}$ and meager in
$({\mathcal{B}}_{1}(X),\emph{{SOT}})$. In particular, a typical
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is not polynomially bounded.

###### Proof.

The key point of the proof is the following well known fact, of which we give
a proof for convenience of the reader.

######  Claim 5.25.

Let $S$ be the unweighted forward shift on $X$. Then $S$ is not polynomially
bounded.

###### Proof of Claim 5.25.

We first consider the case $X=\ell_{p}$, $1\leq p<2$. For any polynomial
$p(z)=\sum_{j=0}^{d}a_{j}z^{j}$, we have

$\|p(S)e_{0}\|=\left(\sum_{j=0}^{d}|a_{j}|^{p}\right)^{1/p}.$

For every $d\geq 0$, let $p_{d}$ be the classical _Rudin-Shapiro polynomial_
of degree $d$ (see [R]). This polynomial has the following properties :

$p_{d}(z)=\sum_{j=0}^{d}a_{j,d}z^{j},\quad\textrm{with}\ a_{j,d}\in\\{-1,1\\}\
\textrm{for every}\ 0\leq j\leq
d,\quad\hbox{and}\quad\|p_{d}\|_{\infty,\mathbb{D}}\leq\sqrt{2(d+1)}.$

So we have

$\|p_{d}(S)\|\geq\|p_{d}(S)e_{0}\|=(d+1)^{1/p}\geq\frac{1}{\sqrt{2}}\,(d+1)^{\frac{1}{p}-\frac{1}{2}}\|p_{d}\|_{\infty,\mathbb{D}},$

which shows that $S$ is not polynomially bounded since $p<2$.

Essentially the same proof (looking at the vector $e_{d+1}$ rather than at
$e_{0}$) shows that the backward shift $B$ is not polynomially bounded on
$\ell_{p}$, $1\leq p<2$. So the cases where $X=c_{0}$ and $X=\ell_{p}$, with
$2<p<\infty$ can be handled by a duality argument. ∎

Let us denote by ${{\mathcal{G}}}$ the set of all $T\in{\mathcal{B}}_{1}(X)$
which are not polynomially bounded. We have to show that ${{\mathcal{G}}}$ is
a dense $G_{\delta}$ subset of $({\mathcal{B}}_{1}(X),\texttt{SOT})$.

The fact that ${{\mathcal{G}}}$ is SOT -$\,G_{\delta}$ is straightforward:
indeed, if $T\in{\mathcal{B}}_{1}(X)$ then

$T\in{{\mathcal{G}}}\iff\forall K\in\mathbb{N}\;\exists
p\in\mathbb{C}[z]\;\exists x\in
S_{X}\;;\;\|p(T)x\|>K\|p\|_{\infty,\mathbb{D}}.$

Let us now show that ${{\mathcal{G}}}$ is SOT \- dense in
${\mathcal{B}}_{1}(X)$. Let $A\in{\mathcal{B}}_{1}(X)$ be arbitrary, and set
$T_{N}:=P_{N}A_{|E_{N}}\oplus S_{N}$, where $S_{N}$ is the forward shift
acting on $F_{N}$. Then $\|T_{N}\|\leq 1$, $T_{N}\xrightarrow{\texttt{SOT}}A$,
and $T_{N}\in{{\mathcal{G}}}$ because $S_{N}$ is not polynomially bounded on
$X$ by Claim 5.25. ∎

In the same circle of ideas, we mention the work of V. Müller, who proved in
[M2] the following result: Assume that $X$ is a Banach space not containing an
isomorphic copy of $c_{0}$, and let $T\in{\mathcal{B}}(X)$ be a power-bounded
operator on $X$ such that $1\in\sigma(T)$. Then, there exist $x_{0}\neq 0$ in
X and $x_{0}^{*}\neq 0$ in $X^{*}$ such that ${\rm Re}\,\langle
x_{0}^{*},T^{n}x_{0}\rangle\geq 0$ for all $n\in\mathbb{Z}_{+}$. Thus, $T$
admits a non-trivial invariant closed convex cone, namely the set

$L:=\\{x\in X;\;\forall n\geq 0\;:\;{\rm Re}\,\langle
x_{0}^{*},T^{n}x\rangle\geq 0\\}.$

It follows that $T$ has a (non-zero) non-supercyclic vector. Indeed, by a
result of [LM], any vector $x\in L$ is non-supercyclic if
$\sigma_{p}(T^{*})=\emptyset$; whereas if $T^{*}$ has an eigenvalue, then $T$
has a non-trivial invariant subspace and hence non-cyclic (non-zero) vectors.
Since a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is such that
$1\in\sigma(T)$ when $X=\ell_{p}$, $1\leq p<\infty$ or $X=c_{0}$ (by
Proposition 3.9), we may therefore state

######  Proposition 5.26.

Assume that $X=\ell_{p}$, $1<p<\infty$. Then a typical operator
$T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ admits a non-trivial invariant
closed cone and (hence) a non-zero non-supercyclic vector.

Müller’s result from [M2] cannot be applied when $X=c_{0}$. However, we are
able to prove the existence of non-supercyclic vectors in the $c_{0}\,$\- case
as well.

######  Proposition 5.27.

A typical $T\in(\mathcal{B}_{1}(c_{0}),\emph{{SOT}})$ admits a non-zero non-
supercyclic vector.

###### Proof.

As in the proof of Theorem 5.2, we use the Banach-Mazur game. Let us denote by
$\mathcal{A}\subseteq(\mathcal{B}_{1}(c_{0}),\texttt{SOT})$ the set of all
$T\in\mathcal{B}_{1}(c_{0})$ admitting a non-zero non-supercyclic vector. We
will describe a strategy for player II in the Banach-Mazur game
$\mathbf{G}(\mathcal{A})$, and show that this strategy is winning under a
suitable assumption.

With the notations of the proof of Theorem 5.2, assume that player I has just
played an open set
$\mathcal{U}_{2k}=\mathcal{U}(N_{2k},A_{2k},\varepsilon_{2k})$. Then, player
II plays the open set