7
votes
Can homomorphisms be Borel in some weaker topology?
It doesn't look so. There is a separable reflexive space $X$, so that $L(X)$ has $2^{2^{\aleph_0}}$ homomorphisms to $A = \mathbb C$ (see this thread).
As the space is reflexive, the unit ball of $L(X)...
6
votes
Accepted
Spectra of products variously permutated
This is false for other permutations. For example, call $(e_1,e_2,e_3)$ the canonical basis of $\mathbb{R}^3$ and let A,B,C be the $3 \times 3$ real matrices such that $Ae_1=e_2$, $Be_2=e_3$, $Ce_3=...
6
votes
Accepted
Closed prime ideal in $C[0, 1]$
No. Note that an ideal in a commutative ring with identity is prime if and only if the quotient ring is an integral domain.
Now consider $C[0,1]$. It is known that the closed ideals in this Banach ...
6
votes
Comparison between the operator norm and the $L^1$ norm on group algebras
This is only a partial answer, but it shows that if G has an element of infinite order then the weakest form of Q2 has a negative answer.
We use the so-called Rudin-Shapiro polynomials (really due to ...
5
votes
Arens regularity of $\mathrm{BV}(\mathbb{R})$
Writing this in a hurry on my way to a ritual English social cohesion exercise, so apologies for the lack of formatting and the sketchy nature of claims and references.
I think it might be "...
5
votes
Accepted
Do completely bounded maps on an operator space have a completely contractive Banach algebra structure?
Yes, this is true. More generally:$\newcommand{\CB}{\mathop{\sf{CB}}}\newcommand{\cbnorm}[1]{{\lVert#1\rVert}_{\sf cb}}$
Proposition 1. Let $X$, $Y$, $Z$ be operator spaces. Then composition of ...
4
votes
Complemented C*-algebras
No. The classical example is $A$ being the functions on $[0,1]$ that are right continuous, have left limits everywhere, and are left continuous except at dyadic rationals. $B$ is defined similarly but ...
4
votes
Accepted
Approximating continuous functions from $K\times L$ into $[0,1]$
The following provides a construction of $g_i, h_i$ satisfying the required conditions (the ranges of these functions can even be in $[0, 1]$):
Claim: For each $k \in K$, there exists an open ...
4
votes
Accepted
Comparison between the operator norm and the $L^1$ norm on group algebras
This is an alternative, more ad hoc proof, very closely mimicking a part of section 3 of the paper https://doi.org/10.4007/annals.2013.178.1.4
The start of the argument is the same. Write $M=L(G)$. ...
4
votes
Comparison between the operator norm and the $L^1$ norm on group algebras
Also for an arbitrary infinite group $G$, the weakest form of Q2 has a negative answer.
Denote by $M=L(G)$ the group von Neumann algebra with its canonical trace $\tau$. For every element $a \in M$, ...
4
votes
Accepted
Completely contractive Banach algebra structure on the dual of a Hopf $C^*$-algebra
There's already the idea of this argument in the comments, but let me flesh it out.
For any $C^*$-algebra $A$ and any von Neumann algebra $M$ with $A\subseteq M$ non-degenerately we can identify $M(A)...
4
votes
Accepted
When is $W^{1,p}(\Omega)$ a Banach algebra?
I'll just complete the answers in the comments showing that for $p\le d$ it is not a Banach algebra. Let me take $\Omega=B_1(0) \subset \mathbb{R}^n$ and $p=d$.
Assume that $W^{1,d}(B_1) $ is a Banach ...
4
votes
positive functional on Banach *-algebra (with appro. identity) is continuous?
F. F. Bonsall & J. Duncan, Complete Normed Algebras, Springer 1973: ยง37, Star Representations and Positive Functionals, Theorem 15 (p. 201).
Let $A$ (a Banach star algebra) have a bounded two-...
2
votes
Bound for the product of Sobolev functions in $W^{s,1}$
I believe that $W^{s,1}(\mathbb{R}^d)$ is a Banach algebra when $s > d$.
This is a particular case of Theorem 7.3 of the paper Multiplication in Sobolev spaces by Ali Behzadan and Michael Holst.
1
vote
Kernel of a map of Tate algebras
No. Take $U \subset X$ a connected rational domain in a two dimensional affinoid ball, and take $C \subset U$ a Zariski closed affinoid curve which is Zariski dense in $X$. Then $\mathcal{O}(X) \to \...
1
vote
Accepted
What is the socle of the $2\times 2$ matrix algebra over a Banach algebra?
We have $soc(\mathcal B) =M_2(soc(\mathcal A))$.
For $I$ a minimal right ideal of $\mathcal A$, $\{\begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \in M_2(\mathcal A) \mid a,b\in I \}$ is a right ...
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