Converges in inner product space Assume $(f_i)_{i\in I}$ is an orthonormal/orthogonal system in an (complex) inner product space. Does $$\sum_{i\in I}\langle f_i,f\rangle f_i$$ always converges for any $f$ (may not to $f$)? Especially, when $I$ is uncountable.
And we define convergency of a sum $\sum_I x_i$ on an arbitrary set by: if there exists $g$ and for any $\epsilon>0$, there exists a finite set $F$ and for all other finite set $H\supseteq F,|g-\sum_H x_i|<\epsilon$ as an analogue to the limit of series in analysis.
 A: Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis]. 
Now  assume that $X$ is a Hilbert space. 
Hints for proof of convergence: for any finite subcollection $\{g_1,g_2,..,g_n\}$ of $\{f_i\}$ we have $\|f-\sum\limits_{k=1}^{n} \langle f, g_k\rangle g_k\|^{2} \geq 0 $. Expanding this we get $\sum\limits_{k=1}^{n} |\langle f, g_k\rangle |^{2} \leq \|f\|^{2}$. This implies that $\{i: \langle f, f_i\rangle \neq 0\}$ is at most countable. If this set is $i_1,i_2,...$ then $\sum\limits_{k=1}^{\infty} \langle f, f_{i_k}\rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $\sum\limits_{k=1}^{\infty} \langle f, f_{i_k}\rangle f_{i_k}$.
A: Suppose you let $X$ be the set of sequences $x_n \in \ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is $\{ e_{2n} \mid n \in \mathbb{N}^+ \}$.  Then for $f = (\frac{1}{n})_{n=1}^\infty \in X$, the sum of $\sum_{n=1}^\infty \langle f, e_{2n} \rangle e_{2n}$ would have to be $g$ where $g_n = \frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
