Definition talk:Schauder Basis

Normed vs Banach Space
I find that many sources (including my own) assume Banach space over NVS. Is it simply restating the convergence condition in $\struct {X, \norm {\cdot} }$, or is it a stronger condition?--Julius (talk) 20:55, 12 September 2022 (UTC)


 * Not an expert, so don't take my words as the truth.
 * I believe the Banach space to be a stronger condition. Consider the subspace $V$ of $R^{\infty}$ that consists of sequences with only finitely many elements different from $0$. Give $V$ the usual norm from $\R^n$. Now $(1, 0 ,0, 0, \ldots), (0,1,0,0,\ldots), \ldots$ is a Schauder basis, but $V$ is not Banach, as far as I know.
 * My source "Introduction to Functional Analysis", by Meise and Vogt, assume a locally convex space in their definition of a Schauder basis. This is even weaker as the definition, as such a space need not have a norm. --Anghel (talk) 22:30, 12 September 2022 (UTC)