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As pointed out by YCor in the comments, the following theorem is true: Theorem 1 Let $p \in [1,\infty] \setminus \{2\}$. If a matrix $A \in \mathbb{R}^{n \times n}$ is an isometry on $\mathbb{R}^n$ w …

In addition to Anthony Quas' answer, it might be worthwhile to mention the following general observations. A Banach space $X$ is said to have the Radon-Nikodym property if every Lipschitz mapping $f: …

I'll use the notion "ordered Banach space" to denote a Banach space $E$ that is ordered by a closed (and convex) cone $E_+$. Generally speaken, having non-empty interior is not a common property of co …

The answer is no, in general. In order to construct a counterexample, let $X = Y = \mathbb{R}^n$ for any $n \ge 2$ and endow this space with the $p$-norm for your favourite $p \in [1,\infty] \setminu …

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 …

Although András' comment already answers the question, I think it is worthwile to give a few more details explicitely here, in order to point out that the analyticity is in fact a consequence of the r …

The answer is no, in general. For a counterexample, consider the $\ell^p$-norm on $\mathbb{R}^2$ with $p=4$, and let $x = e_1 = (1,0)$. We first note that the vectors $e_2 = (0,1)$ and $-e_2$ do not …

The golden rule for conjectures in operator theory: Every ad-hoc conjecture is most likely false for $2 \times 2$-matrices. :-) So here's a $2 \times 2$-counterexample for the question: Let $A = \math …

The answer is "no", in general. An easy counterexample can be found as follows: Let $X = \mathbb{F}$ and let $Y$ be a non-reflexive Banach space. Then $\mathcal{B}(Y^*,X^*)$ is simply the bi-dual $Y^ …

(In the following I assume that the word "invertible" in the question means "bijective".) Your assumptions do not imply that $F$ is bijective (however, they imply that $F$ is injective and has closed …

Part 1 of the answer. In terms of $T$, the property you are looking for is characterized by the mean ergodic theorem: Theorem. Let $E$ be a Banach space and let $T$ be a bounded linear operator on $E …

The answer to the question is yes. Proof. Let $y$ be an extreme point of $T[S]$. Then $$ F :=T^{-1}(\{y\}) \cap S $$ is non-empty, compact and a face of $S$. By the Krein-Milman theorem, $F$ has an …

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 …

In addition to the information given by user bathalf15320, I think that a bit more information on the Banach space case could be useful: Here is a very general theorem about vector valued functions: T …

The answer is yes. Proof. Assume to the contrary that a sequence $(x_n)$ in $\ell^1$ converges, say to $0$, uniformly on convex weakly compact subsets of $c_0$, but is not norm convergent and hence no …