Does all total sets have a "stable" property?

Question

In Kreyszig's Introductory Functional Analysis with Applications, a total set in a normed space $X$ is defined as a subset $M \subset X$ whose span is dense in $X$. That is, $\overline{\text{span } M} = X$.

Let $X$ be a separable normed space, and $M = \{f_1, f_2, \dotsc\}$ be a countable total set in $X$.
Prove or disprove the following: For each $x \in X$, there exist coefficients $c_1, c_2, \dotsc$, such that $\|x - \sum_{k=1}^n c_k f_k\| \to 0$ as $n \to \infty$.

Note that $c_k$ does not change for each $n$, which is what I try to imply by "stable".

Note: The question may be badly posed in the sense that there may still be assumptions I need to make, otherwise there will be a trivial answer. It's a personal question rather than a homework question.

Answer 1

Requiring $c_k$ to remain fixed during the approximation process is a strong restriction, as you can see by looking at $f_n(x)=x^n$ on $C[0,1]$ with the sup norm. (Here $n$ starts from $0$ of course.) Weierstrass' theorem says this is total, but given a sequence $c_k$, if $\sum_{k=0}^\infty c_k x^k$ converges uniformly then the limit is analytic.

Answer 2

If the countable set $M$ has the property you describe it is said to be a Schauder basis for $X$; this term appears in Kreyszig's book, so you can see it there as well.

Lots of separable Banach spaces do have a Schauder basis, including $\ell^p$ or $L^p[0,1]$ for $1 \le p < \infty$.

Exhibiting a separable Banach space which does not have a Schauder basis is possible, but difficult. Enflo constructed an example in this paper: Enflo, P., A counterexample to the approximation property in Banach spaces, Acta Math. 130, 309–317 (1973). Note that existence of a Schauder basis implies that a Banach space has the approximation property.