Operators with a non-trivial closed invariant affine subspace

Abstract

We are concerned with the question of the existence of an invariant proper affine subspace for an operator A on a complex Banach space. It turns out that the presence of the number 1 in the spectrum of A or in the spectrum of its adjoint operator $A^*$ is crucial. For instance, an algebraic operator has an invariant proper affine subspace if and only if 1 is its eigenvalue. For an arbitrary operator A, we show that it has an invariant proper hyperplane if and only if 1 is an eigenvalue of $A^*$. If A is a power bounded operator, then every invariant proper affine subspace is contained in an invariant proper hyperplane, moreover, A has a non-trivial invariant cone.

1 Existence of non-trivial invariant affine subspaces

This paper is motivated by [8], where Wójcik is concerned with the question of which operator on a real Banach space has a non-trivial invariant affine subspace. We improve several results from [8], moreover, most of our proofs work for real and complex Banach spaces. However, to avoid unnecessary complications we confine ourselves to the complex case.

Let $\mathscr {X}$ be a complex Banach space of dimension at least 2. By $\mathcal {B}(\mathscr {X})$ we will denote the Banach algebra of all operators on $\mathscr {X}$; the identity operator will be denoted by I. A non-empty subset $\mathscr {S}\subseteq \mathscr {X}$ is invariant for $A\in \mathcal {B}(\mathscr {X})$ if $A\mathscr {S}\subseteq \mathscr {S}$. We are interested in closed invariant subsets. Note that $A\mathscr {S}\subseteq \mathscr {S}$ always implies $A\overline{\mathscr {S}}\subseteq \overline{\mathscr {S}}$ since A is continuous.

A subspace of $\mathscr {X}$ means a closed linear manifold. A subspace $\mathscr {M}\subseteq \mathscr {X}$ is non-trivial if $\mathscr {M}\ne \{ 0\}$ and $\mathscr {M}\ne \mathscr {X}$. A non-empty subset $\mathscr {U}\subseteq \mathscr {X}$ is an affine subspace if there exists a subspace $\mathscr {M}$ and a vector w such that $\mathscr {U}=w+\mathscr {M}$, that is, vectors in $\mathscr {U}$ are of the form $w+x$ with $x\in \mathscr {M}$. If $\mathscr {M}$ is a non-trivial subspace, then $\mathscr {U}$ is said to be a non-trivial affine subspace. Note that every subspace is an affine subspace, as well. We will say that $w+\mathscr {M}$ is a proper affine subspace if $\mathscr {M}\ne \{ 0\}$ and $w\not \in \mathscr {M}$. The dimension of an affine subspace $\mathscr {U}=w+\mathscr {M}$ is equal to the dimension of $\mathscr {M}$.

Let $\mathscr {S}\subseteq \mathscr {X}$ be a non-empty set. By $\textrm{span}(\mathscr {S})$ we denote the closed linear span of $\mathscr {S}$ and by $\textrm{aff}(\mathscr {S})$ the closed affine hull of $\mathscr {S}$. Hence, $\textrm{span}(\mathscr {S})$ is the closure of the set of all linear combinations $\lambda _1 s_1+\cdots +\lambda _n s_n$, where $s_j\in \mathscr {S}$ and $\lambda _j\in \mathbb {C}$ for all j, and $\textrm{aff}(\mathscr {S})$ is the closure of the set of all affine combinations $\lambda _1 s_1+\cdots +\lambda _n s_n$, where $s_j\in \mathscr {S}$ and $\lambda _j\in \mathbb {C}$ are such that $\lambda _1+\cdots +\lambda _n=1$. It is not hard to see that $\textrm{aff}(\mathscr {S})$ is the smallest affine subspace containing $\mathscr {S}$.

Let us begin with the following simple lemma.

Lemma 1.1

Let $\mathscr {U}=w+\mathscr {M}$ be an affine subspace and let $\mathscr {N}=\textrm{span}\{ w, \mathscr {M}\}$. For an operator $A\in \mathcal {B}(\mathscr {X})$, the following assertions are equivalent:

• (a) $\mathscr {U}$ is invariant for A;

• (b) $\mathscr {M}$ is invariant for A and $Aw-w\in \mathscr {M}$;

• (c) $(A-I)\mathscr {N}\subseteq \mathscr {M}$.

Proof

(a)$\iff$(b). Assume that $A\mathscr {U}\subseteq \mathscr {U}$. Then $Aw\in w+\mathscr {M}$ and therefore $Aw-w\in \mathscr {M}$. Let $x\in \mathscr {M}$ be arbitrary. Since $A(w+x)\in w+\mathscr {M}$, there exists $y\in \mathscr {M}$ such that $Aw+Ax=w+y$, which gives $Ax=w-Aw+y\in \mathscr {M}$. To prove the opposite implication, let $w+x\in w+\mathscr {M}$ be arbitrary. Then $A(w+x)=w+Aw-w+Ax$. By the assumption, $Aw-w$ and Ax are in $\mathscr {M}$. Hence $A(w+x)\in w+\mathscr {M}$.

(b)$\iff$(c). Suppose that (b) holds. It is clear that then $\mathscr {M}$ is invariant for $A-I$. Hence, $(A-I)(\lambda w+m)=\lambda (Aw-w)+(A-I)m\in \mathscr {M}$, for all $\lambda w+m\in \mathscr {N}$. For the opposite implication, note that $\mathscr {M}\subseteq \mathscr {N}$ implies that $\mathscr {M}$ is invariant for $A-I$ and therefore for A. Since $w\in \mathscr {N}$ we have $Aw-w\in \mathscr {M}$, as well. $\square$

By Lemma 1.1, $\mathscr {N}$ is invariant for A if $\mathscr {U}$ is invariant.

Corollary 1.2

An operator $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace if and only if it has invariant subspaces $\mathscr {M}$ and $\mathscr {N}$ such that $\{ 0\}\ne \mathscr {M}\subsetneq \mathscr {N}$ and $(A-I)\mathscr {N}\subseteq \mathscr {M}$.

Proof

If $\mathscr {U}=w+\mathscr {M}$ is a proper invariant affine subspace, then $\mathscr {M}\ne \{ 0\}$ and $\mathscr {N}=\textrm{span}\{ w,\mathscr {M}\}$ are invariant subspaces such that $(A-I)\mathscr {N}\subseteq \mathscr {M}$, by Lemma 1.1. On the other hand, if $\{ 0\}\ne \mathscr {M}\subsetneq \mathscr {N}$ are invariant subspaces such that $(A-I)\mathscr {N}\subseteq \mathscr {M}$, let $w\in \mathscr {N}\setminus \mathscr {M}$ and $\mathscr {U}=w+\mathscr {M}$. Since $(A-I)w\in \mathscr {M}$ we see, by Lemma 1.1, that $\mathscr {U}$ is a proper invariant affine subspace. $\square$

For an operator $A\in \mathcal {B}(\mathscr {X})$, let $\sigma (A)$ denote its spectrum and $\rho (A)=\mathbb {C}\setminus \sigma (A)$ its resolvent set. The unbounded component of $\rho (A)$ is denoted by $\rho _\infty (A)$ and its complement is $\widehat{\sigma }(A)$, the polynomial convex hull of the spectrum. The point spectrum of A, that is, the set of all eigenvalues of A, is denoted by $\sigma _p(A)$ and the spectral radius of A is $r(A)=\sup \{ |z|;\; z\in \sigma (A)\}$. The following proposition indicates that the presence of the number 1 in the spectrum is somehow crucial for the existence of a proper invariant affine subspace.

Proposition 1.3

Let $A\in \mathcal {B}(\mathscr {X})$. If $1\in \rho _\infty (A)$, then A does not have a proper invariant affine subspace.

Proof

Assume that $\mathscr {U}=w+\mathscr {M}$ is an invariant affine subspace for A. By Lemma 1.1, $\mathscr {M}$ is invariant for A and therefore it is invariant for $I-A$. By the assumption, $0\in \rho _\infty (I-A)$. It follows, by a result of Sarason [4, Theorem 11 (ii)], that $\mathscr {M}$ is invariant for $(I-A)^{-1}$. Since, by Lemma 1.1, $(I-A)w\in \mathscr {M}$ we conclude that $w\in (I-A)^{-1}\mathscr {M}\subseteq \mathscr {M}$. Hence, $\mathscr {U}$ is not a proper affine subspace. $\square$

The following corollary is an obvious consequence of Proposition 1.3.

Corollary 1.4

If $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace, then $1\in \widehat{\sigma }(A)$. In particular, if $r(A)<1$, then A does not have a proper invariant affine subspace.

Let $\mathscr {X}^*$ be the dual space of $\mathscr {X}$. The pairing between $\mathscr {X}$ and $\mathscr {X}^*$ is given by $\langle x,\xi \rangle =\xi (x)$, for all $x\in \mathscr {X}$, $\xi \in \mathscr {X}^*$. The adjoint operator of $A\in \mathcal {B}(\mathscr {X})$ is the operator $A^*\in \mathcal {B}(\mathscr {X}^*)$ that is given by $\langle x,A^*\xi \rangle =\langle Ax,\xi \rangle$ $(x\in \mathscr {X},\xi \in \mathscr {X}^*)$.

For $e\in \mathscr {X}$ and $\xi \in \mathscr {X}^*$, let $e\otimes \xi$ be given by $(e\otimes \xi )x=\langle x,\xi \rangle e$, for all $x\in \mathscr {X}$. It is easily seen that $e\otimes \xi$ is a bounded linear operator with norm $\Vert e\Vert \Vert \xi \Vert$. Thus, if $e\ne 0$ and $\xi \ne 0$, then $e\otimes \xi$ is a rank-one operator. Its range is $\mathscr {R}(e\otimes \xi )=\{ \lambda e; \lambda \in \mathbb {C}\}$ and its kernel $\mathscr {N}(e\otimes \xi )$ is equal to the kernel of $\xi$. An operator $F\in \mathcal {B}(\mathscr {X})$ is of finite rank if there exist vectors $e_1,\ldots , e_n\in \mathscr {X}$ and functionals $\xi _1,\ldots ,\xi _n\in \mathscr {X}^*$ such that $F=e_1\otimes \xi _1+\cdots +e_n\otimes \xi _n$.

An operator $A\in \mathcal {B}(\mathscr {X})$ is algebraic if there exists a non-zero polynomial $p\in \mathbb {C}[z]$ such that $p(A)=0$. It is well-known that in this case there exists a unique monic polynomial $m_A\in \mathbb {C}[z]$ of the lowest possible degree such that $m_A(A)=0$. This is the minimal polynomial of A. The spectrum of an algebraic operator consists of eigenvalues that are precisely all the zeros of the minimal polynomial. Thus, $\sigma (A)=\sigma _p(A)$. Every operator with finite rank is algebraic, in particular, all operators on a finite-dimensional Banach space are algebraic. For instance, it is not hard to see that the minimal polynomial of $e\otimes \xi \ne 0$ is $m_{e\otimes \xi }(z)=z\bigl (z-\langle e,\xi \rangle \bigr )$. Hence, $\sigma _p(e\otimes \xi )=\{ 0,\langle e,\xi \rangle \}$.

Theorem 1.5

An algebraic operator has a proper invariant affine subspace if and only if $1\in \sigma _p(A)$.

Proof

Let $A\in \mathcal {B}(\mathscr {X})$ be an algebraic operator. If $1\not \in \sigma _p(A)$, then A does not have a proper invariant affine subspace, by Proposition 1.3.

To prove the opposite implication assume that $1\in \sigma _p(A)$ and consider the minimal polynomial $m_A(z)$ of A. It is clear that $m_A(z)=(z-1)^k q(z)$, where $k\in \mathbb {N}$ and q is a monic polynomial such that $q(1)\ne 0$. We will distinguish two cases.

Suppose first, that $q(z)\not \equiv 1$. We claim that q(A) is not invertible. Indeed, if q(A) were invertible, then $0=m_A(A)=(A-I)^k q(A)$ would give $(A-I)^k=0$, which contradicts to the minimality of $m_A$. Thus, the kernel $\mathscr {N}\bigl ( q(A)\bigr )$ is a non-trivial invariant subspace for A. Let $0\ne w\in \mathscr {N}(A-I)$. Then $q(A)w=q(1)w\ne 0$ since $q(1)\ne 0$, that is, $w\not \in \mathscr {N}\bigl ( q(A)\bigr )$. It is straightforward to check that $A\bigl (w+\mathscr {N}\bigl ( q(A)\bigr )\bigr )\subseteq w+\mathscr {N}\bigl ( q(A)\bigr )$, which means that $w+\mathscr {N}\bigl ( q(A)\bigr )$ is a proper invariant affine subspace for A.

Now we suppose that $q(z)\equiv 1$. Hence $m_A(z)=(z-1)^k$. If $k=1$, then $A=I$, which obviously has a proper invariant affine subspace. Assume therefore that $k\ge 2$. It follows that $\mathscr {N}\bigl ( (A-I)^{k-1}\bigr )$ is a non-trivial invariant subspace for A. Let $w\in \mathscr {X}$ be such that $w\not \in \mathscr {N}\bigl ( (A-I)^{k-1}\bigr )$. Then $w+\mathscr {N}\bigl ( (A-I)^{k-1}\bigr )$ is a proper affine subspace. Since $(A-I)^{k-1}\bigr (Aw-w\bigr )=0$ we see, by Lemma 1.1, that $w+ \mathscr {N}\bigl ( (A-I)^{k-1}\bigr )$ is invariant for A. $\square$

Example

Let $0\ne e\in \mathscr {X}$ and $0\ne \xi \in \mathscr {X}^*$ be arbitrary. Then $e\otimes \xi$ is a rank-one operator. By Theorem 1.5, $e\otimes \xi$ has a proper invariant affine subspace if and only if $\langle e,\xi \rangle =1$. Let us show that in this case, $w+\mathscr {M}$ is a proper invariant affine subspace for $e\otimes \xi$ if and only if $\{ 0\}\ne \mathscr {M}\subseteq \mathscr {N}(\xi )$ and $w=\alpha e+m$, for a non-zero number $\alpha$ and a vector $m\in \mathscr {M}$.

Assume that $w+\mathscr {M}$ is a proper invariant affine subspace for $e\otimes \xi$. By Lemma 1.1, $\mathscr {M}$ is invariant for $e\otimes \xi$ and $w-\langle w,\xi \rangle e\in \mathscr {M}$. For a contradiction suppose that $\mathscr {M}\not \subset \mathscr {N}(\xi )$. Then there exists $x_0\in \mathscr {M}$ such that $\langle x_0,\xi \rangle =1$. Hence, for $x=-\langle w,\xi \rangle x_0\in \mathscr {M}$ we have $\langle x,\xi \rangle =-\langle w,\xi \rangle$. It follows that $(e\otimes \xi )(w+x)=0$. Since $w+x\in w+\mathscr {M}$ and $w+\mathscr {M}$ is assumed to be invariant for $e\otimes \xi$ we have the contradiction $0\in w+\mathscr {M}$. We conclude that $\mathscr {M}\subseteq \mathscr {N}(\xi )$. It follows from $w-\langle w,\xi \rangle e\in \mathscr {M}$ that there exists $m\in \mathscr {M}$ such that $w=\langle w,\xi \rangle e+m$. Since $w\not \in \mathscr {M}$ we see that $\langle w,\xi \rangle \ne 0$. For the opposite implication, let $\{ 0\}\ne \mathscr {M}\subseteq \mathscr {N}(\xi )$ and $w=\alpha e+m$, for a non-zero number $\alpha$ and a vector $m\in \mathscr {M}$. Then, for every $x\in \mathscr {M}$, we have $(e\otimes \xi )(w+x)=\alpha e=w+(-m)\in w+\mathscr {M}$, that is, $w+\mathscr {M}$ is invariant for $e\otimes \xi$. $\square$

A hyperplane in $\mathscr {X}$ is an affine subspace $\mathscr {H}\subseteq \mathscr {X}$ such that $\mathscr {H}=w+\mathscr {N}(\xi )$ for a non-zero functional $\xi \in \mathscr {X}^*$ and a vector $w\in \mathscr {X}$. Thus, $\mathscr {H}=\{ x\in \mathscr {X};\; \langle x,\xi \rangle =\langle w,\xi \rangle \}$. Clearly, $\mathscr {H}$ is a proper affine subspace if and only if $\langle w,\xi \rangle \ne 0$; in this case, we will say that $\mathscr {H}$ is a proper hyperplane.

If $\mathscr {Y}\subseteq \mathscr {X}$ is an invariant subspace for $A\in \mathcal {B}(\mathscr {X})$, then let $A|_\mathscr {Y}$ be the restriction of A to $\mathscr {Y}$. Thus $A|_\mathscr {Y}\in \mathcal {B}(\mathscr {Y})$ such that $A|_\mathscr {Y} y=Ay$, for all $y\in \mathscr {Y}$.

Theorem 1.6

An operator $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace if and only if there exists an invariant subspace $\mathscr {Y}\subseteq \mathscr {X}$ such that $1\in \sigma _p\bigl ( (A|_\mathscr {Y})^*\bigr )$. In particular, A has an invariant proper hyperplane if and only if $1\in \sigma _p(A^*)$.

Proof

Suppose that A has a proper invariant affine subspace $\mathscr {U}=w+\mathscr {M}$. Then $\mathscr {Y}=\textrm{span}\{ w,\mathscr {M}\}$ is an invariant subspace for A, by Lemma 1.1. Note that $\mathscr {U}$ is a proper hyperplane in $\mathscr {Y}$ and it is invariant for $A|_\mathscr {Y}$. Hence, there exists $\xi \in \mathscr {Y}^*$ such that $\mathscr {M}=\mathscr {N}(\xi )$ and $\langle w,\xi \rangle \ne 0$. By Lemma 1.1, $\mathscr {N}(\xi )$ is invariant for $A|_\mathscr {Y}$ and $w-A|_\mathscr {Y} w\in \mathscr {N}(\xi )$, that is, $\langle A|_\mathscr {Y} w,\xi \rangle =\langle w,\xi \rangle \ne 0$. Since every $y\in \mathscr {Y}$ is of the form $y=\lambda w+m$, where $\lambda \in \mathbb {C}$ and $m\in \mathscr {N}(\xi )$, we have $\langle A|_\mathscr {Y} y,\xi \rangle =\lambda \langle A|_\mathscr {Y} w,\xi \rangle =\lambda \langle w,\xi \rangle =\langle y,\xi \rangle$, which gives $\langle y,(A|_\mathscr {Y})^*\xi \rangle =\langle y,\xi \rangle$, for all $y\in \mathscr {Y}$. Hence, $(A|_\mathscr {Y})^*\xi =\xi$ and therefore $1\in \sigma _p((A|_\mathscr {Y})^*)$. If $\mathscr {U}$ is a hyperplane, then $\mathscr {Y}=\mathscr {X}$ and therefore $1\in \sigma _p(A^*)$.

Assume now that $\mathscr {Y}\subseteq \mathscr {X}$ is an invariant subspace for A such that $1\in \sigma _p((A|_\mathscr {Y})^*)$. Then $\mathscr {Y}$ cannot be equal to $\{0\}$. Let $0\ne \xi \in \mathscr {Y}^*$ be an eigenvector of $(A|_\mathscr {Y})^*$ at eigenvalue 1 and let $w\in \mathscr {X}$ be such that $\langle w,\xi \rangle =1$. Then $\mathscr {U}=w+\mathscr {N}(\xi )$ is a proper hyperplane in $\mathscr {Y}$. Hence, $y\in \mathscr {U}$ if and only if $\langle y,\xi \rangle =\langle w,\xi \rangle =1$. Let $y=w+m\in \mathscr {U}$, be an arbitrary vector. Then $\langle A|\mathscr {Y} y,\xi \rangle =\langle w+m,(A|_\mathscr {Y})^*\xi \rangle = \langle w,\xi \rangle =1$, that is, $\mathscr {U}$ is invariant for $A|_\mathscr {Y}$. Since $Ay=A|\mathscr {Y} y$, for all $y\in \mathscr {Y}$, we see that $\mathscr {U}$ is a proper non-trivial invariant affine subspace for A. If $\mathscr {Y}=\mathscr {X}$, that is, $1\in \sigma _p(A^*)$, then $\mathscr {U}=w+\mathscr {N}(\xi )$ is a proper hyperplane in $\mathscr {X}$. $\square$

The following corollary is a variant of Corollary 1.2.

Corollary 1.7

An operator $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace if and only if there exists an invariant subspace $\mathscr {Y}\subseteq \mathscr {X}$ of dimension at least 2 such that $\overline{(A-I)\mathscr {Y}}\subsetneq \mathscr {Y}$.

Proof

If A has a proper invariant affine subspace $\mathscr {U}=w+\mathscr {M}$, then $\mathscr {Y}=\textrm{span}\{ w,\mathscr {M}\}$ is an invariant subspace of dimension at least 2 and $\overline{(A-I)\mathscr {Y}}\subseteq \mathscr {M}\subsetneq \mathscr {Y}$. On the other hand, if $\mathscr {Y}\subseteq \mathscr {X}$ is a subspace of dimension at least 2 such that $\overline{(A-I)\mathscr {Y}}\subsetneq \mathscr {Y}$, then $\mathscr {Y}$ is invariant for A. Let $\mathscr {M}$ be a subspace such that $\{ 0\}\ne \mathscr {M}\subsetneq \mathscr {Y}$ and $\overline{(A-I)\mathscr {Y}}\subseteq \mathscr {M}$. Then $\mathscr {M}$ is invariant for A and, for every $w\in \mathscr {Y}\setminus \mathscr {M}$, we have $(A-I)w\in \mathscr {M}$. Hence,$w+\mathscr {M}$ is a proper invariant affine subspace. $\square$

Example

Let $1\le p<\infty$ and let $\ell ^p$ be the Banach space of all sequences $x=(x_j)_{j=1}^{\infty }\subseteq \mathbb {C}$ such that $\sum \nolimits _{j=1}^{\infty }|x_j|^p<\infty$. Let $e_n$ $(n\in \mathbb {N})$ be the standard basis of $\ell ^p$. For each $n\in \mathbb {N}$, let $\mathscr {M}_n=\textrm{span}\{ e_1,\ldots , e_n\}$. It is clear that $\mathscr {M}_1 \subsetneq \mathscr {M}_2 \subsetneq \cdots .$ It is well-known that all subspaces $\mathscr {M}_n$ $(n\in \mathbb {N})$ are invariant for an operator $A\in \mathcal {B}(\ell ^p)$ if A is represented by an upper triangular matrix $[ a_{ij}]_{i,j=1}^{\infty }$ with respect to the standard basis (hence, $a_{ij}=0$ if $i>j$). If A has a matrix $[ a_{ij}]_{i,j=1}^{\infty }$ with $a_{ij}=0$ whenever $i> j$ and $a_{nn}=1$ for an integer $n\ge 2$, then $(A-I)\mathscr {M}_n\subseteq \mathscr {M}_{n-1}$. Hence, A has a non-trivial invariant affine subspace $e_n+\mathscr {M}_{n-1}$. Compare this example with Corollary 2.9. Note that $\ell ^p$ is reflexive and our assumption on A actually says that $1\in \sigma _p(A)$. However, A is not necessarily a power bounded operator. $\square$

2 Invariant affine subspaces for power bounded operators

An operator $A\in \mathcal {B}(\mathscr {X})$ is power bounded if there exists a constant $M>0$ such that $\Vert A^n\Vert \le M$, for all $n\in \mathbb {N}$. It is clear that every contraction is power bounded. Since $\Vert A^*\Vert =\Vert A\Vert$ we see that an operator is power bounded if and only if its adjoint is power bounded. The following lemma is well-known.

Lemma 2.1

Let $A\in \mathcal {B}(\mathscr {X})$.

• (a) If A is power bounded, then $r(A)\le 1$.

• (b) If $r(A)<1$, then A is power bounded.

Proof

(a) Let $M>0$ be a number such that $\Vert A^n\Vert \le M$, for all $n\in \mathbb {N}$. Then $\Vert A^n\Vert ^{1/n}\le M^{1/n}$, for all $n\in \mathbb {N}$, and therefore, by Gelfand’s formula, $r(A)=\lim \limits _{n\rightarrow \infty } \Vert A^n\Vert ^{1/n}\le \lim \limits _{n\rightarrow \infty } M^{1/n}=1$.

(b) Suppose that $r(A)<1$. Then there exists $0<\epsilon <1$ such that $r(A)=\lim \limits _{n\rightarrow \infty } \Vert A^n\Vert ^{1/n}<1-\epsilon$. Hence, there exists $m\in \mathbb {N}$ such that $\Vert A^n\Vert ^{1/n}\le \lim \limits _{n\rightarrow \infty } \Vert A^n\Vert ^{1/n}+\epsilon <1$, for all $n\ge m$. It follows that $\Vert A^n\Vert \le \bigl (\lim \limits _{n\rightarrow \infty } \Vert A^n\Vert ^{1/n}+\epsilon \bigr )^n< 1$, for all $n\ge m$. Let $M=\max \{1, \Vert A\Vert ,\ldots ,\Vert A^{m-1}\Vert \}$. Then $\Vert A^n\Vert \le M$, for all $n\in \mathbb {N}$, that is, A is power bounded. $\square$

Hence, if a power bounded operator $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace, then $r(A)=1$, by Corollary 1.4.

As usual, let $\ell _{\mathbb {C}}^{\infty }$ be the Banach space of all bounded sequences $x=(x_j)_{j=1}^{\infty }\subseteq \mathbb {C}$ equipped with the norm $\Vert x\Vert _\infty =\sup \{ |x_j|;\; j\in \mathbb {N}\}$. Note that the subset $\ell _{\mathbb {R}}^{\infty }$ of all bounded real sequences is a real subspace of $\ell _{\mathbb {C}}^{\infty }$; moreover, $\ell _{\mathbb {C}}^{\infty }=\ell _{\mathbb {R}}^{\infty }+i\ell _{\mathbb {R}}^{\infty }$, that is, every bounded sequence of complex numbers $x=(x_j)_{j=1}^{\infty }$ is of the form $x=u+iv$, where $u=(u_j)_{j=1}^{\infty }$ and $v=(v_j)_{j=1}^{\infty }$ are bounded sequences of real numbers (clearly, $u_j$ is the real part and $v_j$ is the imaginary part of $x_j$, for all $j\in \mathbb {N}$).

Denote by S the backward shift operator on $\ell _{\mathbb {C}}^{\infty }$, that is,

$$\begin{aligned} S:(x_1, x_2,x_3,\ldots ) \mapsto (x_2,x_3,x_4,\ldots ),\quad \text {for all}\; (x_1, x_2,x_3,\ldots )\in \ell _{\mathbb {C}}^{\infty }. \end{aligned}$$

It is obvious that $S\in \mathcal {B}(\ell _{\mathbb {C}}^{\infty })$ and $\Vert S\Vert =1$. A functional $\Lambda \in \bigl (\ell _{\mathbb {C}}^{\infty }\bigr )^*$ is a Banach limit if it has the following properties:

• (i) $\Vert \Lambda \Vert =1$;

• (ii) if $x=(x_j)_{j=1}^{\infty }$ is a convergent sequence, then $\Lambda (x)=\lim \limits _{j\rightarrow \infty } x_j$;

• (iii) $\Lambda (Sx)=\Lambda (x)$, for all $x\in \ell _{\mathbb {C}}^{\infty }$;

• (iv) if $x, y\in \ell _{\mathbb {R}}^{\infty }$ are such that $x_j\ge y_j$, for all $j\in \mathbb {N}$, then $\Lambda (x), \Lambda (y)\in \mathbb {R}$ and $\Lambda (x)\ge \Lambda (y)$.

A Banach limit on $\ell _{\mathbb {R}}^{\infty }$ is defined similarly: it is a functional $L\in \bigl ( \ell _{\mathbb {R}}^{\infty }\bigr )^*$ satisfying conditions (i)–(iv), where in (iii) one applies only sequences $x\in \ell _{\mathbb {R}}^{\infty }$. By [1, Theorem III.7.1], Banach limits exist. A functional $\Lambda \in \bigl (\ell _{\mathbb {C}}^{\infty }\bigr )^*$ is a Banach limit if and only if there exists a Banach limit $L\in \bigl (\ell _{\mathbb {R}}^{\infty }\bigl )^*$ such that $\Lambda (x)=L(u)+i L(v)$, for all $x=u+iv\in \ell _{\mathbb {C}}^{\infty }$ (see the proof of [1, Theorem III.7.1]).

Let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator. Hence, for all $x\in \mathscr {X}$ and $\xi \in \mathscr {X}^*$, the sequence $s_A(x,\xi )=\bigl ( \langle x,\xi \rangle ,\langle A x,\xi \rangle ,\langle A^2 x,\xi \rangle ,\ldots \bigr )$ is bounded. Let $r_A(x,\xi )$ and $t_A(x,\xi )$ be the real and imaginary part of $s_A(x,\xi )$, respectively, that is $s_A(x,\xi )=r_A(x,\xi )+it_A(x,\xi )$. Hence, $r_A(x,\xi )=\bigl ( \textrm{Re}(\langle x,\xi \rangle ),$ $\textrm{Re}(\langle A x,\xi \rangle ),\ldots \bigr )$ and $t_A(x,\xi )=\bigl ( \textrm{Im}(\langle x,\xi \rangle ),\textrm{Im}(\langle A x,\xi \rangle ),\ldots \bigr )$. This notation will be used till the end of this section.

Lemma 2.2

Let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator and let $\Lambda \in \bigl (\ell _{\mathbb {C}}^{\infty }\bigr )^*$ be a Banach limit. Then there is a well-defined bounded bilinear form $\mu :\mathscr {X}\times \mathscr {X}^*\rightarrow \mathbb {C}$ given by $\mu (x,\xi )=\Lambda \bigl (s_A(x,\xi )\bigr )$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. There exists $Q\in \mathcal {B}(\mathscr {X}^*)$ such that $\mu (x,\xi )=\langle x,Q\xi \rangle$ and $\langle Ax,Q\xi \rangle =\langle x,Q\xi \rangle$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. If $\mathscr {X}$ is a reflexive Banach space, then there exists $R\in \mathcal {B}(\mathscr {X})$ such that $Q=R^*$.

Proof

Let $M>0$ be such that $\Vert A^n\Vert \le M$, for all $n\ge 0$. Then, for $x\in \mathscr {X}$ and $\xi \in \mathscr {X}^*$, we have $|\langle A^n x,\xi \rangle |\le \Vert A^n\Vert \Vert x\Vert \Vert \xi \Vert \le M\Vert x\Vert \Vert \xi \Vert$, for all $n\ge 0$. Thus, the sequence $s_A(x,\xi )=\bigl ( \langle x,\xi \rangle ,\langle A x,\xi \rangle ,\langle A^2 x,\xi \rangle ,\ldots \bigr )$ is bounded. It is clear that $\mu (x,\xi )=\Lambda \bigl (s_A(x,\xi )\bigr )$ defines a bilinear form on $\mathscr {X}\times \mathscr {X}^*$. Since $\Lambda$ is a bounded linear functional with norm 1 we have $|\mu (x,\xi )|\le \Vert s_A(x,\xi )\Vert _\infty$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. Hence, $\mu$ is a bounded bilinear form. It is known that the Banach space of bilinear forms on $\mathscr {X}\times \mathscr {X}^*$ can be identified with $\mathcal {B}(\mathscr {X}^*)$ and with $\mathcal {B}(\mathscr {X},\mathscr {X}^{**})$, where $\mathscr {X}^{**}$ is the second dual space of $\mathscr {X}$ (see [2, §1.1] or [3, p. 8]). More precisely, for a bounded bilinear form $\mu$ on $\mathscr {X}\times \mathscr {X}^*$, there exist operators $Q\in \mathcal {B}(\mathscr {X}^*)$ and $R\in \mathcal {B}(\mathscr {X},\mathscr {X}^{**})$ such that $\Vert Q\Vert =\Vert \mu \Vert =\Vert R\Vert$ and

$$\begin{aligned} \langle x,Q\xi \rangle =\mu (x,\xi )=\langle \xi , Rx\rangle ,\qquad \text {for all}\; (x,\xi )\in \mathscr {X}\times \mathscr {X}^*.\end{aligned}$$

If $\mathscr {X}$ is reflexive, then the second dual $\mathscr {X}^{**}$ may be identified with $\mathscr {X}$ and therefore one can consider R as an operator on $\mathscr {X}$. In this case, $\langle \xi , Rx\rangle =\langle Rx,\xi \rangle$ and therefore $\langle x, Q\xi \rangle =\langle Rx,\xi \rangle$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$, which gives $Q=R^*$. It follows from property (iii) of $\Lambda$ that $\mu (x,\xi )=\Lambda \bigl (s_A(x,\xi )\bigr )= \Lambda \bigl (s_A(Ax,\xi )\bigr )=\mu (Ax,\xi )$ and therefore $\langle Ax,Q\xi \rangle =\langle x,Q\xi \rangle$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. $\square$

Theorem 2.3

Let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator. If $w+\mathscr {M}$ is a proper invariant affine subspace for A, then there exists an invariant proper hyperplane $\mathscr {H}$ that contains $w+\mathscr {M}$.

Proof

Assume that $w+\mathscr {M}$ is a proper invariant affine subspace for A. Hence, $w\not \in \mathscr {M}$ and, by Lemma 1.1, $\mathscr {M}$ is invariant for A and $w-Aw\in \mathscr {M}$. It follows that $Aw\not \in \mathscr {M}$ and therefore $Aw=w+m_1$, for a vector $m_1\in \mathscr {M}$. By induction,

$$\begin{aligned} A^n w=w+m_n,\quad \text {for a vector}\; m_n\in \mathscr {M}, \end{aligned}$$

(2.1)

for all $n\ge 0$. By the Hahn-Banach theorem, there exists $\eta \in \mathscr {X}^*$ such that $\langle w,\eta \rangle =1$ and $\mathscr {M}\subseteq \mathscr {N}(\eta )$, that is, $w+\mathscr {M}\subseteq w+\mathscr {N}(\eta )$. It follows from (2.1) that

$$\begin{aligned} \langle A^n w,\eta \rangle =\langle w,\eta \rangle =1, \qquad \text {for all}\; n\ge 0. \end{aligned}$$

(2.2)

Let $\Lambda \in \bigl ( \ell _{\mathbb {C}}^{\infty }\bigr )^*$ be a Banach limit and let $\mu :\mathscr {X}\times \mathscr {X}^*\rightarrow \mathbb {C}$ be given by $\mu (x,\xi )=\Lambda \bigl ( s_A(x,\xi )\bigr )$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. By Lemma 2.2, $\mu$ is a bounded bilinear form and there exists $Q\in \mathcal {B}(\mathscr {X}^*)$ such that $\mu (x,\xi )=\langle x,Q\xi \rangle$ and $\langle Ax,Q\xi \rangle =\langle x,Q\xi \rangle$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. Note that $\mu (w,\eta )=1$, that is, $\langle w,Q\eta \rangle =1$, by (2.2). This shows that $Q\eta \ne 0$. On the other hand, if $m\in \mathscr {M}$, then $A^n m\in \mathscr {M}$ and therefore $\langle A^n m,\eta \rangle =0$, for all $n\ge 0$. Thus, $\langle m, Q\eta \rangle =0$, which means that $\mathscr {M}\subseteq \mathscr {N}(Q\eta )$. It is clear that $\mathscr {H}=\{ x\in \mathscr {X};\; \langle x, Q\eta \rangle =1\}$ is a proper hyperplane. Since $\langle w+m, Q \eta \rangle =\langle w,Q \eta \rangle =1$, for all $m\in \mathscr {M}$, we see that $w+\mathscr {M}\subseteq \mathscr {H}$. If $x\in \mathscr {H}$, then $\langle Ax, Q\eta \rangle =\langle x, Q\eta \rangle =1$. Hence, $A\mathscr {H}\subseteq \mathscr {H}$. $\square$

Corollary 2.4

A power bounded operator $A\in \mathcal {B}(\mathscr {X})$ has a proper invariant affine subspace if and only if $1\in \sigma _p(A^*)$.

Proof

By Theorem 2.3, A has a proper invariant affine subspace if and only if it has an invariant proper hyperplane. Now we apply Theorem 1.6. $\square$

A closed subset $\mathscr {C}\subseteq \mathscr {X}$ is a cone if $\mathscr {C}+\mathscr {C}\subseteq \mathscr {C}$ and $t\mathscr {C}\subseteq \mathscr {C}$, for all $t\ge 0$. A cone $\mathscr {C}$ is non-trivial if $\{ 0\} \ne \mathscr {C} \ne \mathscr {X}$.

Corollary 2.5

Let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator. If $1\in \sigma _p(A^*)$, then A has a non-trivial invariant cone.

Proof

By Corollary 2.4, A has an invariant proper affine subspace and therefore, by Theorem 2.3, A has an invariant proper hyperplane $\mathscr {H}$. In the proof of Theorem 2.3, we have seen that $\mathscr {H}=w+\mathscr {N}(\xi )$, where $0\ne \xi \in \mathscr {X}^*$ is such that $A^*\xi =\xi$ and $\langle w,\xi \rangle =1$. Let $\mathscr {C}=\{ x\in \mathscr {X};\; \langle x,\xi \rangle \ge 0\}$. It is easily seen that $\mathscr {C}$ is a non-trivial cone. If $x\in \mathscr {C}$, then $\langle Ax,\xi \rangle =\langle x,A^*\xi \rangle =\langle x,\xi \rangle \ge 0$. Thus, $\mathscr {C}$ is invariant for A. $\square$

In what follows, we will give an equivalent condition for 1 to be in the point spectrum of the adjoint operator of a power bounded operator. Let $u=(u_1,u_2,u_3,\ldots )\in \ell _{\mathbb {R}}^{\infty }$. Sucheston [7] proved that limits

$$\begin{aligned} q(u)=\lim _{n\rightarrow \infty } \inf \biggl \{\frac{1}{n}\sum _{k=m+1}^{m+n}u_k;\quad m\in \mathbb {N}\biggr \}\end{aligned}$$

and

$$\begin{aligned}p(u)=\lim _{n\rightarrow \infty } \sup \biggl \{\frac{1}{n}\sum _{k=m+1}^{m+n}u_k;\quad m\in \mathbb {N}\biggr \} \end{aligned}$$

exist. Of course, $q(u)\le p(u)$.

Lemma 2.6

Let $u\in \ell _{\mathbb {R}}^{\infty }$. For any number $\alpha$ in the interval [q(u), p(u)] there exists a Banach limit $L\in \bigl (\ell _{\mathbb {R}}^{\infty }\bigr )^*$ such that $L(u)=\alpha$.

Proof

The result follows from parts $(\beta )$ and $(\gamma )$ of Theorem in [7]; see page 727 in the translated version of [5, 6]. $\square$

Theorem 2.7

Let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator. Then the following assertions are equivalent:

• (a) $1\in \sigma _p(A^*)$;

• (b) there exist $u\in \mathscr {X}$ and $\eta \in \mathscr {X}^*$ such that at least one among the numbers

  $$\begin{aligned} q\big ( r_A(u,\eta )\bigr ),\quad p\big ( r_A(u,\eta )\bigr ),\quad q\big ( t_A(u,\eta )\bigr ),\quad \text {and}\quad p\big ( t_A(u,\eta )\bigr ) \end{aligned}$$

  (2.3)

  is not equal to 0.

Condition

• (c) $1\in \sigma _p(A)$

implies (a) and (b) and if $\mathscr {X}$ is reflexive, then all three conditions are equivalent.

Proof

(a)$\Rightarrow$(b) and (c)$\Rightarrow$(b). Suppose that $1\in \sigma _p(A^*)$. Then there exists $0\ne \eta \in \mathscr {X}^*$ such that $A^* \eta =\eta$. Let $u\in \mathscr {X}$ be such that $\langle u,\eta \rangle =1$. If $1\in \sigma _p(A)$, then there exists $0\ne u\in \mathscr {X}$ such that $Au=u$. Let $\eta \in \mathscr {X}^*$ be such that $\langle u,\eta \rangle =1$. In both cases we have $\langle A^n u,\eta \rangle =1$, for all $n\ge 0$, and therefore $r_A(u,\eta )=(1,1,1,\ldots )$. It follows that $q\big ( r_A(u,\eta )\bigr )=p\big ( r_A(u,\eta )\bigr )=1$.

(b)$\Rightarrow$(a). Assume that at least one among the numbers in (2.3) is not equal to 0. It is easily seen that $r_A(-iu,\eta )=t_A(u,\eta )$ and $t_A(-iu,\eta )=-r_A(u,\eta )$. Hence, there is no loss of generality if we assume that $q\big ( r_A(u,\eta )\bigr )\ne 0$ or $p\big ( r_A(u,\eta )\bigr )\ne 0$. Let $\alpha \in \bigl [q\big ( r_A(u,\eta )\bigr ),p\big ( r_A(u,\eta )\bigr )\bigr ]$ be a non-zero number. By Lemma 2.6, there exists a Banach limit L on $\ell _{\mathbb {R}}^{\infty }$ such that $L\big ( r_A(u,\eta )\bigr )=\alpha \ne 0$. Denote $L\big ( t_A(u,\eta )\bigr )$ by $\beta$. Let $\Lambda =L+i L$, that is, $\Lambda \in \bigl (\ell _{\mathbb {C}}^{\infty }\bigr )^*$ is a Banach limit. It follows that

$$\begin{aligned} \Lambda \bigl (s_A(u,\eta )\bigr )=\alpha +i\beta \ne 0. \end{aligned}$$

(2.4)

Let $\mu :\mathscr {X} \times \mathscr {X}^* \rightarrow \mathbb {C}$ be the bilinear form given by $\mu (x,\xi )=\Lambda \bigl ( s_A(x,\xi )\bigr )$, for $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$, and let $Q \in \mathcal {B}(\mathscr {X}^*)$ be such that $\mu (x,\xi )=\langle x, Q\xi \rangle$, for all $(x,\xi )\in \mathscr {X}\times \mathscr {X}^*$. It follows from (2.4) that $\langle u,Q\eta \rangle \ne 0$, which gives $Q\eta \ne 0$. By Lemma 2.2, $\langle x,A^*Q\eta \rangle =\langle Ax,Q\eta \rangle =\langle x,Q\eta \rangle$, for all $x\in \mathscr {X}$. Hence $A^*Q\eta =Q\eta$ and therefore $1\in \sigma _p(A^*)$.

To prove (b)$\Rightarrow$(c), assume that $\mathscr {X}$ is reflexive. We have already seen that (b) implies (a), that is $1\in \sigma _p(A^*)$. Hence, by the first paragraph of this proof (implication (c)$\Rightarrow$(b)), there exist $\eta \in \mathscr {X}^*$ and $u\in \mathscr {X}^{**}=\mathscr {X}$ such that at least one among the numbers $q\big ( r_{A^*}(\eta ,u)\bigr )$, $p\big ( r_{A^*}(\eta ,u)\bigr )$, $q\big ( t_{A^*}(\eta ,u)\bigr )$, and $p\big ( t_{A^*}(\eta ,u)\bigr )$ is not equal to 0. By the implication (b)$\Rightarrow$(a), we conclude that $1\in \sigma _p(A^{**})=\sigma _p(A)$. $\square$

Corollary 2.8

Let $\mathscr {X}$ be a reflexive Banach space and let $A\in \mathcal {B}(\mathscr {X})$ be a power bounded operator. Let $\lambda \in \mathbb {C}$ be such that $|\lambda |=1$. Then $\lambda \in \sigma _p(A)$ if and only if $\lambda \in \sigma _p(A^*)$.

Proof

Consider the operator $B=\frac{1}{\lambda }A$. It is clear that B is a power bounded operator. Since $\lambda \in \sigma _p(A)$ if and only if $1\in \sigma (B)$ and $\lambda \in \sigma _p(A^*)$ if and only if $1\in \sigma (B^*)$ we conclude, by Theorem 2.7, that $\lambda \in \sigma _p(A)$ if and only if $\lambda \in \sigma _p(A^*)$. $\square$

Corollary 2.9

Let $\mathscr {X}$ be a reflexive Banach space. A power bounded operator $A\in \mathcal {B}(\mathscr {X})$ has a proper affine subspace if and only if $1\in \sigma _p(A)$.

Proof

Combine Theorem 2.7 and Corollary 2.4. $\square$