It is known that if a Banach space $X$ has a Schauder basis, then $X$ is separable. On the other hand P. Enflo showed that there exist a separable Banach space without Schauder basis.
If $X$ is a separable Banach space, then we can find a increasing sequence of finite dimensional subspaces $X_n\subset X$ such that $\cup X_n$ is dense in $X$, hence, we can find a sequence $(x_n)$, such that $x_1,…,x_k$ is a basis of $X_k$ for all $k$.
My question is: What is a necessary condition (and maybe sufficient) in order to show that $(x_k)$ is a Schauder basis for $X$?
Best Answer
One of the necessary and sufficient conditions for a countable linearly independent system of vectors $(e_n)$ to be a Shauder basis of its closed linear span is $$ \exists K>0\quad \forall (a_n)\subset \mathbb{C}\quad \forall n\in\mathbb{N}\quad\forall m\leq n\quad\left\Vert\sum\limits_{k=1}^m a_k e_k\right\Vert\leq K\left\Vert \sum\limits_{k=1}^n a_k e_k\right\Vert $$ See Proposition 1.1.9 in Topics in Banach space theory by F. Albiac, N. Kalton