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## New answers tagged banach-algebras

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"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, ...

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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 ...

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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 ...

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