If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?

If not, does anyone know of any counterexamples?


No, certainly not. The linearly independent set $\{1, x, x^2, x^3, \dots\}$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.

A Schauder basis is, in general, much harder to construct than a set with dense span.

Since Enflo we know that there are separable Banach spaces (hence they have countable dense subset) that have no Schauder basis at all.


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