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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
9
votes
0
answers
151
views
Moore-Penrose partial isometries and hermitian elements
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 …
28
votes
2
answers
1k
views
What is the Banach-Mazur distance between $\ell_\infty$ and $L_\infty$?
Given Banach spaces $X$ and $Y$, the Banach-Mazur distance between $X$ and $Y$ is defined as
$$ d(X,Y) = \inf\{ \|\varphi\|\|\varphi^{-1}\| : \varphi\colon X\to Y \text{ isomorphism} \}.
$$
We conside …
9
votes
1
answer
353
views
Scottish Book Problem 172
The problem is formulated using old terminology and I want to understand what it actually says.
The problem reads: "A space $E$ of type (B) has the property (a) if the weak closure of an arbitrary set …
8
votes
Which C*-algebras are complemented in their bidual?
(1) If $A$ is $1$-complemented in its bidual, then $A$ is an AW*-algebra.
Indeed, assume that $A$ is $1$-complemented in its bidual, via a contractive projection $p\colon A^{**}\to A$. By a theorem o …
15
votes
1
answer
435
views
Weak*-closure of finite rank operators on dual space
Given a Banach space $X$, we consider the space $B(X^*)$ of bounded, linear operators on $X^*$ with the weak*-topology from its canonical predual $B(X^*)_*=X^*\hat{\otimes}X$. What is $\overline{F(X^* …
4
votes
0
answers
114
views
Weakenings of the Bounded Approximation Property
Let $X$ be a Banach space, and let $F(X)$ be the finite-rank operators on $X$.
Then $X$ has the $\lambda$-BAP if there exists a net $(S_i)_i$ in $F(X)$ with $\sup_i\|S_i\|\leq\lambda$ such that $S_ …
16
votes
2
answers
681
views
Ultraweak topology on B(X): Is the map X\otimes X* -> B(X)* isometric?
Let $X$ be a Banach space. Consider the map
$$
\alpha\colon X\hat{\otimes} X^* \to B(X)^*,
$$
defined one simple tensors as
$$
\alpha(\xi\otimes\eta)(a) = \eta(a(\xi)).\quad (\xi\in X, \eta\in X^*, a\ …