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5 votes
1 answer
199 views

Is the unit ball of $B(H)$ a Baire space (with the SOT)?

Let $H$ be a Hilbert space, and let $B(H)$ be the set of bounded linear operators $t \colon H \to H$. Recall that we say $t_i \to t$ in the strong operator topology if $t_i \xi \to t \xi$ for every $\...
Diego Martinez's user avatar
4 votes
0 answers
152 views

Maximally fine topologies on $B(H)$ making the unit ball compact

Let $H$ be a Hilbert space, and $B(H)$ its algebra of bounded operators. One of the reasons the Ultraweak topology is (in a way) more useful than the weak operator topology is that the Ultraweak ...
Aareyan Manzoor's user avatar
2 votes
0 answers
198 views

A generalisation of closed and bounded subsets of non-Archimedean fields to topological spaces

The definition of compactness in topological spaces generalises the notion of a subset of $\mathbb{R}^n$ being closed and bounded, as expressed by the Heine-Borel Theorem. In finite-dimensional vector ...
Very Forgetful Functor's user avatar
9 votes
1 answer
551 views

Is the unit sphere of a Banach space dense in the unit sphere of its second dual with respect to the weak-$\ast$ topology

To be a bit more precise and fix notations, let $X$ be a Banach space (over $\mathbb{R}$ or $\mathbb{C}$), $X^{\ast\ast}$ its second dual (as a Banach space). Here and in the following we identify $X$ ...
Rick Sternbach's user avatar
6 votes
0 answers
124 views

Meagre sets of bounded operators

Let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathbb{B}(H)$ be the algebra of bounded operators on $H$. The norm topolology on $\mathbb{B}(H)$ is stricly finer, hence the ...
Matthias Ludewig's user avatar
6 votes
1 answer
212 views

Embedding of $C(X)$ into $B(H)$ where $H$ is separable

I would like to ask a question which may look strange at the first sight nevertheless I find it interesting. Let $H$ be a separable Hilbert space: for any separable $C^*$-algebra $A$ one can embed $A$ ...
truebaran's user avatar
  • 9,330
0 votes
1 answer
495 views

Separability of an algebra is equivalent to separability of its spectrum

Let $A$ be a commutative C*-algebra. I would like to show that $A$ is separable (i.e. has a countable dense subset) if and only if the spectrum of $A$ (denoted by $\Omega(A)$) is separable. Notes ...
ned grekerzberg's user avatar
0 votes
0 answers
147 views

Approximation of Inductive Tensor Product $C(X) \bar{\otimes} C(Y)$

The following question is from Banach Algebra Techniques in Operator Theory written by Ronald G. Douglas. Assume both $X, Y$ are Banach spaces and $X \otimes Y$ is the algebraic tensor product. Let ${...
Sanae Kochiya's user avatar
6 votes
1 answer
353 views

Equivalence of $\sigma$-weak topology to another topology

Let $\mathcal H$ be a Hilbert space. Define a topology $\tau_1$ on $B(\mathcal H)$ by the family of seminorms $x\mapsto |Tr(xa)|,$ $a\in L^1(B(\mathcal H)).$ Here $B(\mathcal H)$ denotes the set of ...
A beginner mathmatician's user avatar
1 vote
0 answers
253 views

When is the weak topology generated by a family of functions Baire?

Suppose we are given a locally compact space $X$ with $C_b(X)$ denoting the continuous bounded complex or real functions on $X$. Now, if $A\subset C_b(X)$ is given, I am trying to figure out when the ...
Merry's user avatar
  • 173
3 votes
2 answers
265 views

Ultraweak topology in abelian von Neumann algebras

Let $A$ be an abelian von Neumann algebra acting on the (not necessarily separable) Hilbert space $\mathcal{H}$ (with identity $I$). From the Gelfand-Neumark theorem, there is a compact Hausdorff ...
condexp's user avatar
  • 159
9 votes
1 answer
384 views

Comparing two $\sigma$-algebras on $B(\ell^1)$

Let us consider $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on $B(\ell^1)$ is determined as follow $$w-\lim T_i=T \Longleftrightarrow \...
ABB's user avatar
  • 4,058