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.


  1. What are the conditions for $(e_n)$ to be a Schauder basis for $X$?
  2. Even if $(e_n)$ is not a Schauder basis, what are the conditions for $X$ to have a Schauder basis?

1 Answer 1


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 $$ \exists K>0\quad \forall (a_n)\subset \mathbb{C}\quad \forall n\in\mathbb{N}\quad\forall m\leq n\quad\left\Vert\sum\limits_{k=1}^m a_k e_k\right\Vert\leq K\left\Vert \sum\limits_{k=1}^n a_k e_k\right\Vert $$

  • $\begingroup$ Are there any prominent examples where this criterion has been used 'in practice'? $\endgroup$
    – fsp-b
    Commented Jun 1, 2023 at 21:50
  • $\begingroup$ Any link for a proof of this criteria ? $\endgroup$
    – Balaji sb
    Commented Jun 2, 2023 at 2:33

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