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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 ...


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.


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 ...


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 ...


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 $a,b\in c_0, \omega\in\ell^1$ ...



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