Tagged Questions

4
votes
0answers
64 views

C* algebra of bounded Borel functions

Let $T\in B(H)$ is normal, and $B(\sigma(T))$ denote the $C^*$ algebra of all bounded Borel functions on $\sigma(T)$. Then is it true that $B(\sigma(T))$ is a closed $C^*$ algebra under the sup. norm ...
2
votes
1answer
57 views

Algebra (Not *)-Isomorphisms of von Neumann algebras

Let $A$ and $B$ be any two infinite-dimensional von Neumann algebras, they are operator algebras with operator composition as the multiplication and as infinite dimensional vector spaces they're ...
1
vote
0answers
55 views

Proving properties of exponential map on a Banach algebra

$$\exp(a) := \sum\frac {a^k}{k!}$$ Can you help me prove that: $\exp$ is well defined (i.e. converges for all $a$ in $A$) $\exp$ is continuous $\exp(A)$ is a subset of $A_0$ (where $A_0$ is the ...
3
votes
1answer
85 views

Polar decomposition of invertible elements in a unital C$ ^{*} $-algebra.

If $ A $ is a unital C$ ^{*} $-algebra and $ a $ is invertible, then $ a = u|a| $ for a unique unitary element $ u $ of $ A $. If $ \| a \| = \| a^{-1} \| = 1 $, what can you say about $ |a| $? I ...
3
votes
0answers
35 views

Biduals generated by projections

This question is motivated by a similar question recently posed at MO: http://mathoverflow.net/questions/122091/masas-in-second-duals-of-banach-algebras In this setting, let $B$ be a Banach algebra ...
3
votes
1answer
63 views

Double centralizers in the Murphy book

I've been into this for days and days and I still can't see why, given the definition of $L^\ast$ as $L^\ast =(L(a^\ast))^\ast$ we get that $(LM)^\ast =L^\ast M^\ast$. Where is my mistake: ...
6
votes
1answer
344 views

Some examples in C* algebras and Banach * algebras

I would like an example of the following things. A Banach * algebra that is not a C* algebra for which there exists a positive linear functional (it takes $x^*x$ to numbers $ \geq 0$) that is not ...
5
votes
2answers
137 views

Why are compact operators 'small'?

I have been hearing different people saying this in different contexts for quite some time but I still don't quite get it. I know that compact operators map bounded sets to totally bounded ones, that ...
1
vote
1answer
61 views

References on Algebraic Operators

Let $\mathcal{H}$ be a Hilbert space and $d$ is an inner derivation on $\mathcal{L}(\mathcal{H})$. An operator $T\in\mathcal{L}(\mathcal{H})$ is algebraic if $p(T)=0$ for some polynomial $p$. In ...
3
votes
2answers
207 views

Non-$C^{*}$ Banach algebras?

It suddenly occurred to me almost every Banach algebra I know is actually a $C^{*}$ algebra. Several kinds of function algebras are definitely $C^{*}$ algebras. So is the matrix algebra. Although one ...
2
votes
1answer
119 views

Subadditive, submultiplicative and scalar-multiplication-invariant functions

Let $\mathcal{A}$ be an algebra. $d: \mathcal{A}\to \mathbb{N}$ is a function satisfying 1) $d(S+T)\le d(S)+d(T)$, 2)$d(ST)\le d(S)+d(T)$ and 3) $d(aS)=d(S)$ for all $a\in \mathbb{C}, ...
11
votes
2answers
446 views

Is there an algebraic homomorphism between two Banach algebras which is not continuous?

According to wikipedia, you need the Axiom of Choice to find a discontinuous map between two Banach spaces. Does this procedure also apply for Banach algebras yielding a discontinuous multiplicative ...