Tag Info

New answers tagged

0

"Quotient of $X$ by $Y$" means that $Y$ is somehow thought of as a subset of $X$; a lot depends on how one embeds $Y$ into $X$. Here's a more concrete way to phrase this problem: you are looking for a surjective homomorphism $f:C[-1,1]\to C_0[-1,1]$ such that the kernel of $f$ is isomorphic to $\mathbb{C}$. This isn't going to be an algebra homomorphism, ...


2

Every (non-unital) C*-algebra has a bounded approximate identity consisting of self-adjoint elements. Every amenable Banach algebra has a bounded approximate identity and this class of algebras is quite substantial. If $X$ is a Banach space with the bounded approximation property, then the algebra $\mathscr{K}(X)$ of compact operators on $X$ has a bounded ...


0

No, this doesn't exists. Consider the sequences $b^{(i)}i$ such that $b^{(i)}_n = 1$ if $i=n$ and $0$ elsewhere Take an arbitrary $N \in \Bbb N$. Then as $e^{(r)}$ are an approximation of the identity, there exists $n_0$ such that $\forall n > n_0$, $\forall 1 \leq i \leq N$, $\| e^{(n)} b^{(i)} - b^{(i)} \| \leq \epsilon$ This imply that $\forall i ...


2

Well, of course if $G$ is any locally compact abelian group then $L^1(G)$ is a Banach algebra under convolution. If $G$ is not discrete then $L^1(G)$ has no identity, while if $G$ is first-countable then there is a bounded approximate identity. (If $G$ is not first countable there's still a net that gives a bounded approximate identity, but perhaps not a ...


3

Concerning bounded approximate identities: $\mathcal S$ is a Montel spaces, that is bounded sets are relatively compact ($\mathcal S$ is even nuclear). If it had a bounded approximate identity compactness would give a limit which then would be an identity element which certainly does not exist.


4

There's certainly an approximate identity in $\mathcal S$. For $\phi\in\mathcal S$ and $t>0$ define $\phi_t(x)=t^{-1}\phi(x/t)$. Then if $\int\phi=1$ it follows that $\phi_t*f\to f$ in $\mathcal S$ for every $f\in\mathcal S$. Say $\psi=\hat\phi$. It's easiest to verify that you have an approximate identity if you choose $\phi$ so that $\psi=1$ in a ...


1

A late answer, but maybe still helpful. All you have to keep in mind are the natural identification of $\ell^1$ and $c_0^\ast$ resp. $\ell^\infty$ and $(\ell^1)^\ast$. I will write $\ast$ for the product in the three steps of the construction of the Arens product so that there is no confusion with pointwise multiplication. For $a,b\in c_0, \omega\in\ell^1$ ...


0

Lemma: Let $A$ be a unital Banach algebra and $\{a_n\} \subset A$ such that $a_n \to a\in A$. Suppose $\lambda_n \in \sigma(a_n)$ are such that $\lambda_n \to \lambda$ in $\mathbb{C}$, then $\lambda \in \sigma(a)$ Proof: Suppose $\lambda \notin \sigma(a)$, then $(a-\lambda 1) \in GL(A)$, which is open. So $\exists \epsilon > 0$ such that $$ \|y - ...


0

There is no separable commutative Banach algebra that contains isometric copies of all separable commutative Banach algebras. Indeed, if $p$ and $q$ are commuting projections in a Banach algebra then $\|p-q\|\geqslant 1$ so each set of commuting projections is discrete. Now, projections in a Banach algebra can have arbitrarily large norms. Consequently, a ...


2

The notation $\prod_{i\in I}S_i$ denotes a set of functions. By definition, $f\in\prod_{i\in I}S_i$ if (i) $f$ is a function with domain $I$ and (ii) $f(i)\in S_i$ for every $i\in I$. So $\phi\in\prod_{a\in A}sp(a)$. Because $\phi$ is a function with domain $A$ and $\phi(a)\in sp(a)$ for every $a\in A$. Come to think of it, that raises an obvious ...


1

The product $\prod_{i\in I}A_i$ of an indexed family of sets is, by definition, the set of all functions $f$ whose domain is the index set $I$ and which satisfy, for each index $i\in I$, the requirement that $f(i)\in A_i$. So the product in your question is the set of functions that assign, to each $a$ in your algebra, an element of its spectrum. The ...



Top 50 recent answers are included