Hi,


Suppose we have a (real, separable) Banach space $V$ and a (linear) set $A\subseteq V$. I presume in general it might not be possible to write every element of the closed span of $A$ as an infinite linear combination $\sum\_{i=1}^\infty\beta\_i a\_i$ of elements of $A$. Are there simple (non-trivial) conditions guaranteeing that the closed linear span of $A$ coincides with its infinite linear span (perhaps with unconditional/absolute convergence)?


My example of interest is the following: Let $X$ be a compact metric space and $F:X\to X$ a continuous map. My space $V$ is the space $C(X)$ of continuous (real-valued) functions on $X$, and $A$ is the subset of functions that can be written as $\varphi\circ F - \varphi$ for some $\varphi\in C(X)$.


Thank you for any help.



Infinite linear span vs closed linear span
In the case of linear $A$, which seems to be your case of interest, you can simply do it as follows. For $x$ in the closure of $A$ take a sequence $(x\_n)$ in $A$ converging to $x$ such that $\|x\_n-x\_{n+1}\|<2^{-n}$. Then set $a\_i=x\_i-x\_{i-1}$, $\beta\_i=1$ and the sum will converge absolutely, hence also unconditionally to $x$. Or did I miss something?


