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 nontrivial invariant cone.
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1 Existence of nontrivial 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 nontrivial 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 nonempty 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 nontrivial if \(\mathscr {M}\ne \{ 0\}\) and \(\mathscr {M}\ne \mathscr {X}\). A nonempty 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 nontrivial subspace, then \(\mathscr {U}\) is said to be a nontrivial 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 nonempty 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 \(Aww\in \mathscr {M}\);

(c)
\((AI)\mathscr {N}\subseteq \mathscr {M}\).
Proof
(a)\(\iff \)(b). Assume that \(A\mathscr {U}\subseteq \mathscr {U}\). Then \(Aw\in w+\mathscr {M}\) and therefore \(Aww\in \mathscr {M}\). 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 \(Ax=wAw+y\in \mathscr {M}\). To prove the opposite implication, let \(w+x\in w+\mathscr {M}\) be arbitrary. Then \(A(w+x)=w+Aww+Ax\). By the assumption, \(Aww\) 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 \(AI\). Hence, \((AI)(\lambda w+m)=\lambda (Aww)+(AI)m\in \mathscr {M}\), 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 \(AI\) and therefore for A. Since \(w\in \mathscr {N}\) we have \(Aww\in \mathscr {M}\), 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 \((AI)\mathscr {N}\subseteq \mathscr {M}\).
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 \((AI)\mathscr {N}\subseteq \mathscr {M}\), by Lemma 1.1. On the other hand, if \(\{ 0\}\ne \mathscr {M}\subsetneq \mathscr {N}\) are invariant subspaces such that \((AI)\mathscr {N}\subseteq \mathscr {M}\), let \(w\in \mathscr {N}\setminus \mathscr {M}\) and \(\mathscr {U}=w+\mathscr {M}\). Since \((AI)w\in \mathscr {M}\) 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 \(IA\). By the assumption, \(0\in \rho _\infty (IA)\). It follows, by a result of Sarason [4, Theorem 11 (ii)], that \(\mathscr {M}\) is invariant for \((IA)^{1}\). Since, by Lemma 1.1, \((IA)w\in \mathscr {M}\) we conclude that \(w\in (IA)^{1}\mathscr {M}\subseteq \mathscr {M}\). 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 rankone 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 nonzero polynomial \(p\in \mathbb {C}[z]\) such that \(p(A)=0\). It is wellknown 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 finitedimensional 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 \(m_A(z)=(z1)^k q(z)\), 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 \(0=m_A(A)=(AI)^k q(A)\) would give \((AI)^k=0\), which contradicts to the minimality of \(m_A\). Thus, the kernel \(\mathscr {N}\bigl ( q(A)\bigr )\) is a nontrivial invariant subspace for A. Let \(0\ne w\in \mathscr {N}(AI)\). 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 \(m_A(z)=(z1)^k\). If \(k=1\), then \(A=I\), which obviously has a proper invariant affine subspace. Assume therefore that \(k\ge 2\). It follows that \(\mathscr {N}\bigl ( (AI)^{k1}\bigr )\) is a nontrivial invariant subspace for A. Let \(w\in \mathscr {X}\) be such that \(w\not \in \mathscr {N}\bigl ( (AI)^{k1}\bigr )\). Then \(w+\mathscr {N}\bigl ( (AI)^{k1}\bigr )\) is a proper affine subspace. Since \((AI)^{k1}\bigr (Aww\bigr )=0\) we see, by Lemma 1.1, that \(w+ \mathscr {N}\bigl ( (AI)^{k1}\bigr )\) 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 rankone 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 nonzero 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 nonzero 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 nonzero 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 \(wA_\mathscr {Y} w\in \mathscr {N}(\xi )\), 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 nontrivial 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 \(\overline{(AI)\mathscr {Y}}\subsetneq \mathscr {Y}\).
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 \(\overline{(AI)\mathscr {Y}}\subseteq \mathscr {M}\subsetneq \mathscr {Y}\). On the other hand, if \(\mathscr {Y}\subseteq \mathscr {X}\) is a subspace of dimension at least 2 such that \(\overline{(AI)\mathscr {Y}}\subsetneq \mathscr {Y}\), then \(\mathscr {Y}\) is invariant for A. Let \(\mathscr {M}\) be a subspace such that \(\{ 0\}\ne \mathscr {M}\subsetneq \mathscr {Y}\) and \(\overline{(AI)\mathscr {Y}}\subseteq \mathscr {M}\). Then \(\mathscr {M}\) is invariant for A and, for every \(w\in \mathscr {Y}\setminus \mathscr {M}\), we have \((AI)w\in \mathscr {M}\). 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 wellknown 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 \((AI)\mathscr {M}_n\subseteq \mathscr {M}_{n1}\). Hence, A has a nontrivial invariant affine subspace \(e_n+\mathscr {M}_{n1}\). 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 wellknown.
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 \(M=\max \{1, \Vert A\Vert ,\ldots ,\Vert A^{m1}\Vert \}\). 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,
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 welldefined 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
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 \(wAw\in \mathscr {M}\). It follows that \(Aw\not \in \mathscr {M}\) and therefore \(Aw=w+m_1\), for a vector \(m_1\in \mathscr {M}\). By induction,
for all \(n\ge 0\). By the HahnBanach 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
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 nontrivial 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 nontrivial 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 nontrivial 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
and
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 nonzero 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
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
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Acknowledgements
The paper was partially supported by the Slovenian Research and Inovation Agency through the research program P20268.
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Bračič, J. Operators with a nontrivial closed invariant affine subspace. Aequat. Math. 98, 1305–1315 (2024). https://doi.org/10.1007/s00010024010900
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DOI: https://doi.org/10.1007/s00010024010900