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Here's a way to construct such an example: For each integer $n \ge 1$ consider the matrices $A_n, M_n \in \mathbb{C}^{2 \times 2}$ given by $$ M_n = e_1 e_1^T = \begin{pmatrix} 1 & 0 \\ …

If $\lambda$ has modulus $1$, then both definitions are equivalent for power-bounded operators, i.e., for operators $A$ that satisfy $\sup_{n = 0,1,2,\dots} \|A^n\| < \infty$. Indeed, if $\lvert \lamb …

Yes, this is always true if $\lambda \not= 0$. The subsequent theorem shows a more general result. To formulate it, we need the following terminology: For an eigenvector $\lambda$ of a bounded linear …

From your assumption you can easily see that $T$ is unitarily similar to a multiplication operator on $\ell^2$ (and thus, $T$ is normal, by the way). This shows that the answer is "yes" (as it is easy …

General references. The references that you're probably looking for are books on the theory of $C_0$-semigroups. Some classics are: [1] Amnon Pazy: Semigroups of Linear Operators and Applications to P …

What you are looking for is actually true for every power-bounded operator, without any appeal to positivity: Theorem. Let $E$ be a Banach space and let $A: E \to E$ be a bounded linear operator such …

In the following answer I'll focus on the case for general $n$. Let $m: [0,1] \to \mathbb{C}^{n \times n}$ be measurable and bounded. Let $a \in \mathcal{L}(L^2([0,1]; \mathbb{C}^n))$ be the multiplic …

I'll try to answer with a few general remarks on what a pole is and what a meromorphic function is. Let $U \subseteq \mathbb{C}$ be non-empty and open and let $X$ be a complex Banach space (for instan …

Preliminary remark. As mentioned in the comments, I find the notion "resolvent formalism", as well as the description in the Wikipedia article, rather misleading - resolvents are not somekind of forma …

The major point here is that, for an operator $S$ on a Banach space (or Hilbert space) $X$, the number $\sup_{x \in X \setminus\{0\}} \frac{\|Sx\|}{\|x\|}$ is not the spectral radius of $S$ but the op …

I've looked it up now. The formula in question does indeed hold in the following sense: Theorem. Let $(e^{tA})_{t \in [0,\infty)}$ be a $C_0$-semigroup on a complex Banach space $X$. Let $\omega \in …

In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper): Theorem. Let $C$ and $D$ be bounded …

EDIT: I adjusted the answer to the new version of the question. Such an example does not exist. More precisely, for every compact Hausdorff space $K$ and every continuous mapping $T: K \to K$ the ass …

An additional remark (too long for a comment) that might be of interest: Given the finite-dimensional counterexamples in the answers by Iosif Pinelis and Pietro Majer it seems worthwhile to note that …

The essential spectrum (and even the spectrum) of the generator of a contractive $C_0$-semigroup on an $L^1$-space can be empty even if the generator does not have compact resolvent. Example. Endow $ …