Let $ (e_n)_{n \in \mathbb{N}} $ be a linearly independent sequence in a Banach space $X$ such that
$$ X = \overline{\operatorname{span}} \{e_n : n \in \mathbb{N}\} $$
In particular, $X$ is separable.
However, $(e_n)$ may not be a Schauder basis for $X$, unless $X$ is a Hilbert space. The easiest counter-example is the sequence $ (1, x, x^2, \dots)$ in $ C(\mathbb{R}) $. Even worse, $X$ may not have a Schauder basis at all.
Questions:
- What are the conditions for $(e_n)$ to be a Schauder basis for $X$?
- Even if $(e_n)$ is not a Schauder basis, what are the conditions for $X$ to have a Schauder basis?
Best Answer
A well known necessary and sufficient condition for a countable linearly independent system of eigenfunctions, like $(e_k)$, to be a Schauder basis for its closed linear span is $$ \exists K>0\quad \forall (a_n)\subset \mathbb{C}\quad \forall n\in\mathbb{N}\quad\forall m\leq n\quad\left\Vert\sum\limits_{k=1}^m a_k e_k\right\Vert\leq K\left\Vert \sum\limits_{k=1}^n a_k e_k\right\Vert $$