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As already noted by several users in this comments, something seems to be a bit odd with the question due to the behaviour of the resolvent close to the spectrum. Here are a few details about what is …

There exists a Ritt operator $T$ on $\ell^\infty$ which is not mean ergodic. Indeed, let $T$ be the "multiplication operator" given by $Tx = \big((1-\frac{1}{n})x_n\big)_{n \in \mathbb{N}}$ for each …

Nik Weaver has already explained in his answer that this does not hold, in general. On the positive side, here is a sufficient condition for the implication to be true: Proposition. Let $\Omega_1, \Om …

My question is motivated by the following little proposition: Proposition. For a vector subspace $V$ of a Banach space $(X, \|\cdot\|_X)$ the following assertions are equivalent: (i) There exists a …

The characterisation of irreducible matrices as suggested in the question is not correct. The matrix $$ P = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} $$ is a counterexample: it has spectral ra …

Here are a few details which might be helpful to understand the argument in the paper better: (1) First note that $P$ is also the spectral projection for the spectral value $\frac{1}{\lambda}$ of the …

A few preliminary remarks: 1) Complexifications of Banach lattices are in fact a special case of the more general concept of complexifications of real Banach spaces. 2) Most books and articles about …

This answer consists of two parts: Part I. The answer is no, in general. Here is a counterexample: Example. Let $\Omega = \{1,2\}$ and let $P$ by $1/2$ times the counting measure. The matrix \begin{ …

The answer is no in general. For a counterexample, let $A$ be your favourite semigroup generator that has a sequence of eigenvalues $(\lambda_n) \subseteq \mathbb{R}$ such that $\lambda_n \to -\infty …

Here is one result that could, sometimes, be helpful in the setting of the question: Theorem. Let $(\Omega,\mu)$ be a finite measure space and let $0 \not= T: L^2(\Omega,\mu) \to L^2(\Omega,\mu)$ be a …

Convergence to $0$ is simple: Proposition 1. The following are equivalent: (i) The operator $e^{tQ}$ converges to $0$ with respect to the operator norm as $t \to \infty$. (ii) The spectrum of $Q$ is c …

If $V$ is merely an isomorphism from $D(A)$ to $D(B)$, the operator $V^{-1}BV$ is not well-defined (since $BV$ maps $D(A)$ to $Y$ rather than to $D(B)$). The "right" notion is a follows: Let's call th …

I quasi-nilpotent operator $T \in B(X)$ is nilpotent if and only if $0$ is a pole of its resolvent $(\cdot - T)^{-1}$, which means that there exists a number $M \ge 0$ and an integer $n \ge 1$ such th …

The answer is no in general (but it's not difficult to check that the answer is yes for injective operators). Counterexample. Let $H$ be a Hilbert space and $T_0: H \to H$ a bounded linear operator wi …

Preliminary remarks: In the comments the OP noted that he is primarly interested in the question whether the dual operator of an irreducible operator is irreducible, so I will focus on this aspect …