4

I'm not sure. (Edit: actually I believe that yes, those functions are dense. See below) Some comments on this $H^2(-a,a)$ thing that may be helpful: First, it's not at all clear tat the norm you define is, or is equivalent to, a Hilbert-space norm. But in fact if $f\in H^2(-a,a)$ then the boundary values $f(x\pm ia)$ exist almost everywhere, and the norm ...


3

I take it you're talking about the result that says $\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)$. Yes, we need to assume that the limit of $f$ exists. Regarding the word "tauberian" in the title: I really think this is an abelian theorem; the corresponding tauberian theorem, which is not true without an extra "tauberian" hypothesis, would assume that $\lim sF(...


1

As for your first question the answer is no, as there exist unital, commutative Banach algebras that are also local rings. For example, consider the unitisation of $$R = \{f\in L_0((0,\infty))\colon \int\limits_0^\infty |f(t)|e^{-t^2}\, {\rm d}t <\infty\},$$ endowed with the convolution product: $$(f\ast g)(t) = \int\limits_0^t f(t-s)g(s)\,{\rm d}s.$$ ...


Only top voted, non community-wiki answers of a minimum length are eligible