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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.

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Well, if $X=\mathbb R$ and $S=\{1\}$, the sums of the elements of $S$ don't give you that much... – user31373 Jul 11 '12 at 0:19
@Leonid : If you allow sums of the form $\sum_{n=1}^{\infty} a_n s_n$ where $a_n \in \mathbb R$ and $s_n \in S$, then your example just gives $\mathbb R$. I believe this is a pretty much standard definition of the span of $S$ : $\mathrm{span}(S) = \{ \sum_{finite} a_n s_n \}$, so I don't think your example helps a lot. I don't know the answer to the question though ; I think there are things to worry about when taking the closure. – Patrick Da Silva Jul 11 '12 at 0:26
Sorry, I did indeed mean infinite sums of scalar multiples of $S$> – Jeff Jul 11 '12 at 1:22
@Patrick But it helps the OP to be more precise in his writing, which is not a useless thing, I think. – Dylan Moreland Jul 11 '12 at 1:25
I agree. I usually try to be precise, but this was actually not me being lazy, but actually me asking the wrong question. I'd like to modify my question now to the correct one. In fact, it is meant to be a more general case of part of this question, which may also help you to understand why I made the mistake of forgetting scalar multiples. (I also really just care about the specific case, although if the general question here is true I want to learn about it): – Jeff Jul 11 '12 at 1:27
up vote 7 down vote accepted

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.

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Wow this is very clever. If you reverse engineered this example, I'm curious how you did it. Thanks! – Jeff Jul 11 '12 at 4:14
@ Yuki This one even shows that Hilbert Spaces are not good enough, and that orthogonality is truly required for the flagship case of when S is an orthonormal basis. Do you know of any other important cases where my original question has "yes" as its answer? – Jeff Jul 11 '12 at 4:24
@Jeff: thanks! =p this was just a lucky example! =p – Yuki Jul 11 '12 at 11:31
@Jeff: hmmm... I'm still thinking in this last question... – Yuki Jul 11 '12 at 11:37
Actually it suffices for S to contain a Schauder basis for its closed linear span, but this is more like a reformulation. – user31373 Jul 11 '12 at 12:36

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.

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@ Leonid kovalev However, I'm not convinced that being in the "infinite sum span" relative to uniform norm necessarily means that the function is a power series. One thing that is critically lacking is pointwise absolute convergence that lets you reorder the series into its power series form. I suppose after that's done, standard arguments from complex analysis will finish the rest of this? – Jeff Jul 11 '12 at 4:36
@Jeff That is a healthy suspicion. I expanded the answer to address it. As a bonus, I added a Hilbert space version of the same argument. – user31373 Jul 11 '12 at 11:12
Nice indirect argument! – wildildildlife Jul 11 '12 at 12:19

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