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It is known that the only Banach space that satisfies the von-Neumann inequality is the Hilbert space: Theorem (see e.g. Pisier, "Similarity Problems and Completely Bounded Maps", p 27) For a Banach ...

Let $A$ be a unital Banach algebra. An element $a \in A$ is hermitian if $\|\mathrm{exp}(ita)\|=1$ for every $t \in \mathbb{R}$. An element $a \in A$ is Moore-Penrose invertible if there exists $b \in ...

Recall that a Banach space $X$ has the approximation property (AP for short ) if for every compact subset $K$ of $X$ and every $\varepsilon > 0$, there exists a finite rank operator $S$ on $X$ such ...

Fix any $1\leq p\leq\infty$. If $X$ is a Banach space and $C\in(0,\infty)$, we say that $T\in\mathcal{A}_C(X)$ whenever, for each $(x_n)_{n=1}^\infty\subset B_X$ (where $B_X$ is the closed unit ball ...

In Rickart, page 50 Theorem 2.2.1, the statement is made: A linear subspace $\mathfrak{M}$ of the algebra $\mathfrak{A}-\mathfrak{L}$ is invariant with respect to the representation $a{\rightarrow}A_a^...

$X$, $Y$ are Banach spaces. Let $S \in L(X, Y)$, $T \in L(Y, X)$, where $L(X, Y)$ denotes the Banach algebra of bounded linear operators from $X$ to $Y$. If we have that $Id_Y - ST \in \mathbb{K}(Y)$ ...