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Let $H$ be an infinite-dimensional (separable) Hilbert space. $E$ is a linear independent set which is countable. Let $A:=span(E)$. Consider the closure of $A$, can every element $x\in\overline{A}$ be represented as $x=\sum_{n=1}^{\infty}\sum_{i=1}^{n}a_{ni}e_{i}$, where $a_{ni}\in\mathbb{C}$, $e_{i}\in E$. In other words, each element of $\overline{A}$ can be represented as infinite sum of elements in $A$?

I notice similar question have been asked, like this. My question is somewhat different, I know that not every element in $\overline{A}$ can be represented as infinite sum of elements in $E$. So I use elements in $A$ instead, but I am not sure if the form of elements in $A$ are general enough.

Ken.Wong
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    Whenever $A$ is a dense subspace of a Banach space $X$, it is possible to write any element $x$ of $X$, as a convergent series of elements of $A$ . Start with $x=\lim_n a_n$, and then $$x=a_1+\sum_{n=1}^\infty a_{n+1}-a_n.$$ – Ruy Jul 03 '22 at 15:54

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Apply the Gram-Schmidt procedure to $E=\{e_n\}_{n=1}^\infty.$ In this way you obtain the sequence of elements $v_n$ such that $v_n\in {\rm span}\ \{e_1,e_2,\ldots, e_n\}\subset A$ and $\{v_n\}_{n=1}^\infty $ forms an orthonormal basis in $A$, hence also in $\bar{A}.$ Thus every element $x\in \bar{A}$ is of the form $$x=\sum_{n=1}^\infty \langle x,v_n\rangle v_n$$

Equivalent description could be as follows. Let $P_n$ denote the orthogonal projection onto the finite dimensional subspace ${\rm span}\,\{e_1,e_2,\ldots, e_n\}.$ Then for every $x\in A$ there is $n_0$ such that $$P_nx=x,\quad n\ge n_0$$ Moreover for $x\in \bar{A}$ we have $$x=\lim_nP_nx$$ Therefore $$x=\sum_{n=1}^\infty (P_n-P_{n-1})x,\quad P_0=0$$ and $$(P_n-P_{n-1})x=P_nx-P_{n-1}x\in {\rm span}\{e_1,e_2,\ldots, e_n\}$$ In general let $A\subset X,$ where $X$ is a Banach space. Denote $V_n={\rm span}\{e_1,e_2,\ldots, e_n\}.$ For every $x\in \bar{A}$ let $$d_n(x)=\inf \{\|x-y\|\,:\, y\in V_n\}$$ Since $V_n$ is finite dimensional there exists $y_n\in V_n$ such that $$\|x-y_n\|=d_n(x)$$ As $x\in \bar{A},$ we have $\|x-y_n\|\to 0$ and $$x=y_1+\sum_{n=2}^\infty (y_n-y_{n-1}),\qquad y_n-y_{n-1}\in V_n$$ But this time the summands are not of the form $\lambda v_n$ for a fixed sequence of elements $v_n\in A.$

Ryszard Szwarc
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  • In the problem I am facing, I can't really use Gram-Schmidt procedure because it is difficult to compute. So I will stick to the general formalism, what I want to ask is if my proposed form of elements is general enough for $\overline{A}$. I guess my real question is, if $v_{n}$ can be chosen to be the arbitrary linear combinations of elements in $E$, with increasing order(meaning $v_{n}$ contain $e_{i}$, with $i\leq n$). – Ken.Wong Jul 03 '22 at 09:39
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    Your question does not require calculation of the coefficients $a_{ni}$, just the existence. In my answer $v_n$ are given in terms of $e_1,e_2,\ldots e_n.$ The Gram-Schmidt procedure allows one to calculate the coefficients explicitly in terms of the determinants of the matrix $\langle e_i,e_j\rangle.$ I have added an equivalent explanation using projections. – Ryszard Szwarc Jul 03 '22 at 10:10
  • If I understand correctly, my proposed form of $x$ is equivalent to using projection formalism? – Ken.Wong Jul 03 '22 at 10:20
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    Something of that sort, but in general there can be many sums of that form convergent to $x.$ The one using projections is the most efficient. – Ryszard Szwarc Jul 03 '22 at 10:40