Closure of the span in a Banach space

Question

Let $X$ be a Banach space, and $S$ a subset. Is it true that $\overline {\operatorname{span}(S)}$ is equal to the set of the elements of $X$ that are obtained as norm convergent infinite sums of the scalar multiples of the elements of $S$? I can see that the infinite sums are in the closure of the span, and also that it would suffice to see that the collection of infinite sums of elements of $S$ forms a norm-closed set. I just can't show that.

Answer 1

With this new question, I think the answer is no.

For example, let $X=l_2(\mathbb{N})$ and $S=\left\{s_m=\sum\limits_{n=1}^m \frac{1}{n}e_n:m\in\mathbb{N}\right\}$, where $e_n=(0,\cdots,0,1,0,\cdots)$, with $1$ in the coordinate $n$.
Clearly we have $\overline{\textit{span}(S)}=l_2$, but the element $x=\sum\limits_{n=1}^\infty\frac{1}{n}e_n\in l_2$ isn't the sum of scalar multiples of elements of $S$. Indeed, suppose $x=\sum\limits_{n=1}^\infty a_ns_n$. Then necessarily each coordinate converges, and in particular, $\sum\limits_{n=1}^\infty a_ne_1=e_1$, so, $\sum\limits_{n=1}^\infty a_n=1$.
Analogously, necessarily $\sum\limits_{n=2}^\infty a_n\frac{1}{2}e_2=\frac{1}{2}e_2$, so $\sum\limits_{n=2}^\infty a_n=1$, and then $a_1=0$.
And so on, necessarily we have $\sum\limits_{n=m}^\infty a_n\frac{1}{m}e_m=\frac{1}{m}e_m$, and hence $\sum\limits_{n=m}^\infty a_n=1$, then $a_{m-1}=0$. But this implies $0=a_1=a_2=\cdots$, and hence $x=0$, absurd.

Answer 2

I'll add an indirect argument from analysis. Let $X$ be the space of continuous functions on $[0,1]$ with the supremum norm, and take $S$ to be the set of monomials. Weierstrass tells us that the closed span of $S$ is the entire space. Yet, a function cannot be written as the sum of a uniformly convergent power series unless it extends to a complex analytic function in the open unit disk (which nonsmooth functions obviously don't),

[Added to address the concern about rearrangement] Power series can be rearranged with impunity. Indeed, suppose $f(x)=\sum_{j=1}^\infty c_j x^{n_j}$ in $C[0,1]$. Since the sum converges at $x=1$, the coefficients are bounded: $|c_j|\le M$. Therefore, on every disk $\{ x\in \mathbb C: |x|\le r\}$, $r<1$, the j-th term is bounded by $Mr^{n_j}$. Hence the series converges uniformly and absolutely by the M-test of the same Weierstrass guy. We can rearrange it now, but don't really have to, because the limit of any uniformly convergent series of polynomials in a complex domain is a holomorphic function.

The same argument works if we replace $C[0,1]$ with the Hilbert space $L^2[0,1]$. Indeed, the $L^2$ norm of $c_j x^{n_j}$ is $|c_j|/\sqrt{2n_j+1}$, and this must be bounded if the series converges. On any disk of radius less than 1 the supremum of $c_j x^{n_j}$ is bounded by $Mr^{n_j}\sqrt{2n_j+1}$; these form a convergent numerical series and the M-test applies again.