Countable set in a Banach space which spans densely? Let $\mathcal{C}(\mathbf{T})$ be the algebra of continuous complex functions on the unit circle $\mathbf{T}$. Consider the following two statements:


*

*The $*$-subalgebra generated by $1$ and $z$ spans densely (by the Stone--Weierstrass Theorem).

*Every $f\in \mathcal{C}(\mathbf{T})$ can be written as a uniform limit $f(z) = \sum_{n\in \mathbf{Z}} \lambda_n z^n$ for some appropriate coefficients $\lambda_n \in \mathbf{C}$.


Is the second obvious from the first? The first statement means that every $f$ can be written as $f = \lim_{n\rightarrow \infty} f_n$ where the $f_n$'s are Laurent polynomials, but should it be clear that the $f_n$'s have the specific form $f_n = \sum_{i=-n}^n \lambda_i z^i$?

Maybe more generally, suppose $X$ is a Banach space with $X=\overline{\mathrm{span}}\{x_n\}_{n=1}^\infty$ for some countable set $\{x_n\}$. Then can we write each $x\in X$ as $x=\sum_{n=1}^\infty \lambda_nx_n$ for some appropriate coefficients $\lambda_n$?
Frequently I see arguments passing from a dense subspace to all of $X$ using this kind of argument: do it for finite linear combinations, and then extend to "infinite sums" by density. But I'm not sure how that works unless each element is indeed a "series" of the form $\sum \lambda_n x_n$.
In a Hilbert space it's true if $\{x_n\}$ is an orthonormal basis ... in that case we can take $\lambda_n = \langle x,x_n\rangle$. But that uses the additional structure of a Hilbert space.

Here is some context for this question. The Cuntz algebra $\mathcal{O}_1$ is the universal C*-algebra generated by a single unitary $s$. I'm trying to prove that $\mathcal{O}_1$ is isomorphic to $\mathcal{C}(\mathbf{T})$ via $s\mapsto \iota$, where $\iota(z):=z$. I have a $*$-homomorphism $\varphi:\mathcal{O}_1\rightarrow \mathcal{C}(\mathbf{T})$ given by $s\mapsto \iota$, and I'm trying to show it's injective. It clearly hits all Laurent polynomials, which are dense in $\mathcal{C}(\mathbf{T})$. So if I can show it's injective, it's isometric and so its image is simultaneously closed and dense, hence this map is surjective.
I've managed to show $\mathcal{O}_1$ is spanned densely by $\{s^n : n\in \mathbf{Z}\}$, but that doesn't mean that each element of $\mathcal{O}_1$ has the form
$$x = \sum_{n\in \mathbf{Z}} \lambda_n s^n$$
does it? In this particular case, maybe it does. If I can get that to work I'm done.
(The wikipedia page uses the fact that $C^*(s) \simeq \mathcal{C}(\sigma(s))$ by the functional calculus. There is another argument using the gauge invariant uniqueness theorem from graph C*-algebras, but that sounds like a big hammer for a tiny bug ...)
 A: Here is a theorem (paraphrased) from Ivan Singer's Bases in Banach Spaces, Vol. 1 (Springer-Verlag 1970). In what follows, "basis" means "Schauder basis".
Proposition 4.3 (pg. 29):
Let $E$ be a Banach space with a bounded  basis $\{x_i\}$. Let $\{\alpha_i\}$ be a sequence of non-zero scalars and for each positive integer $n$, let $$y_n=\sum\limits_{i=1}^n \alpha_i x_i.$$ Then $\{y_i\}$ is a basis of $E$ if and only if $\sup\limits_{n} {\Vert y_n\Vert\over|\alpha_{n+1}|}<\infty$.
Note $(y_n,y_n^*)\subset X\times X^*$ is   a complete biorthogonal  system ("biorthogonal" means $y_i^*y_j=\delta_{ij}$; "complete" means the linear span of $\{y_n\}$ is dense) where $y_n^*={1\over\alpha_n }x_n^* -{1\over\alpha_{n+1}}x_{n+1}^*$. 
It then follows from the above  that in any infinite dimensional Banach space with a basis, there is   biorthogonal system $(y_i,y_i^*)\subset X\times X^*$with $\{ y_i\}$ complete 
that is not a basis of the space.  
For a complete biorthogonal system $(y_i,y_i^*)$, your condition would imply $\{y_i\}$ is basic; so the above furnishes   counterexamples to your conjecture for infinite dimensional spaces with a basis. (Such counterexamples exist in separable Banach spaces without bases as well: every Banach space has a complete biorthogonal system).

For a concrete example, take $X=\ell_1$ and for each $n$, $y_n=\sum\limits_{k=1}^n e_k$ where $e_k$ is the standard $k$th unit vector in $\ell_1$. Then $\{y_n\}$ is not a basis of $\ell_1$, though its linear span is dense in $\ell_1$ (the linear span contains the unit vectors).
The proof that this is so follows more easily from the following characterization of bases: 
If $\{x_i\}$ is a complete set of non-zero vectors in the Banach space $X$, then $\{x_i\}$  is a basis of $X$ if and only if there is a $K>0$ such that for any sequence of scalars $\{\alpha_i\}$ and any $m<n$ one has $\Bigl\Vert \sum\limits_{k=1}^m \alpha_k x_k\Bigr\Vert \le K \Bigl\Vert \sum\limits_{k=1}^n \alpha_k x_k\Bigr\Vert$. 
(Note $\Vert y_n\Vert_{\ell_1}=n$ and $\Vert y_n-y_{n+1}\Vert_{\ell_1}=1$.)

The proofs of the above statements can be found in Singer. 
Singer's book may be hard to find; but you should be able to find  most of these results, and other results which imply a negative answer to your general question,  in any text covering Schauder bases in detail. E.g., Kalton and Albiac's text Topics in Banach Space Theory or Megginson's text An Introduction to Banach Space theory.

For your first question, see this.
