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 $\mathsf{Pr} \infty \mathsf{fWiki}$ definition, as such a space need not have a norm. --Anghel (talk) 22:30, 12 September 2022 (UTC)

clarification

Is it fair to say that a Schauder Basis is the extension of the concept of a Definition:Basis of Vector Space to an infinitude of dimensions? --prime mover (talk) 11:22, 15 June 2023 (UTC)