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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?

Keshav Srinivasan
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1 Answers1

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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.

GEdgar
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  • I am confused by your first paragraph. If $\text{span}({1, x, x^2, x^3, \dots})$ is dense in $C[0,1]$ that means, we can represent every element of $C[0,1]$ as a limit of a sequence in $\text{span}({1, x, x^2, x^3, \dots})$, so a limit of a sequence of polynomials. Does your answer ("not every continuous function is given by a power series.") imply then, that there exist limits of sequences of polynomials, that are not representable as a power series? – Zaph Dec 19 '24 at 14:50
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    @Zaph Exactly right. For example, if the sequence of polynomials converges uniformly to $\sqrt{x}$ on $[0,1]$, then the coefficient of $x$ in these polynomials may not converge at all, so we cannot use these polynomials to get a power series in the limit. – GEdgar Dec 19 '24 at 15:26