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Let $\mathcal{A} \equiv (A, \|\cdot\|_A)$ be a unital (associative) normed algebra over the real or complex field, and assume that $\mathcal{A}$ is not complete. Provided $\mathcal{B}_\mathcal{A}$ is the open unit ball of $\mathcal{A}$, define $N$ to be the set of all $a \in \mathcal{B}_\mathcal{A}$ such that the Neumann series $\sum_{n=0}^\infty a^n$ does not converge in $\mathcal{A}$.

Questions. 1) Is $N$ dense in $\mathcal{B}_\mathcal{A}$? 2) And what about $\mathcal{B}_\mathcal{A}\setminus N$?

Edit (11 Dec 2011). Following Davide's comment below, let $\mathcal{C}^0([0,1],\mathbb{R})$ be the usual Banach algebra (over the real field) of all continuous functions $[0,1] \to \mathbb{R}$ endowed with the uniform norm $\|\cdot\|_\infty$. Define $A$ to be the subalgebra of $\mathcal{C}^0([0,1],\mathbb{R})$ of all polynomial functions. For each $\phi \in A$ such that $\|\phi\|_\infty < 1$, the Neumann series $\sum_{n=0}^\infty \phi^n$ converges in $\mathcal{C}^0([0,1],\mathbb{R})$ to $(1 - \phi)^{-1}$, but it does not in $\mathcal{A} \equiv (A,\|\cdot\|_\infty)$ so far as $\phi$ is not a constant. Thus, $N$ is dense in $\mathcal{B}_\mathcal{A}$, and indeed in the unit ball of $\mathcal{C}^0([0,1],\mathbb{R})$ (by the Stone-Weierstrass theorem).

Martin Sleziak
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Salvo Tringali
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  • As far as I can understand, there still exists no tag for normed algebras, and I'm not yet enabled to create a new one by myself. – Salvo Tringali Dec 10 '11 at 18:38
  • What example(s) of $\mathcal A$ make(s) you think that $N$ can be dense in $\mathcal B_{\mathcal A}$? – Davide Giraudo Dec 10 '11 at 23:10
  • I'm editing the OP to add an example. – Salvo Tringali Dec 11 '11 at 13:57
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    I've retagged question - in my opinion, it's better to think twice before creating a new tag whether it will be useful. But if you think that the new tag would be useful, fell free to retag the question again, of course. – Martin Sleziak Dec 11 '11 at 14:20
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    @Martin. I don't care much about labels but still think that normed ring/algebras would deserve their own tag. :) – Salvo Tringali Dec 11 '11 at 15:01
  • As I've written, you can use your best judgement and retag. (I did what I thought was optimal, but of course, I might be wrong.) I'll ping @Davide as well, since after I checked edit history, I've seen that he introduced the tag. BTW we already do have [tag:operator-algebras] and [tag:c-star-algebras]. This answer at meta might be interesting in this context: http://meta.math.stackexchange.com/a/2829/ – Martin Sleziak Dec 11 '11 at 15:05
  • Formally speaking normed algebras is not the same as operator algebras. For example banach algebra is an operator algebra iff its multiplication bioperator is completely bounded. I think normed algebras diserves their own tag. – Norbert Dec 11 '11 at 19:20
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    Speaking as a Banach algebraist, to act as if that area is subsumed by "operator-algebras" and "c-star-algebras" seems quite mistaken in my view. (Just try applying continuous functional calculus to self-adjoint elements in $\ell^1$-group algebras and watch your norm estimates blow up in your face...) –  Jan 06 '12 at 06:52

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