let $V$ be some inner product space and $\lbrace {e_i\rbrace }_{i\in\mathbb{N}} \subset V$ be some countable orthonormal set. I am wondering if for any $x\in V$ the series $$\sum\limits_{i=1}^{\infty} \langle x,e_i \rangle e_i $$ is convergent in $V$?
If $V$ is a Hilbert space, then the series always converge to some alement $z\in V$ du to Bessel inequality. But my proof make use of the completeness of the space in an essential way. Furthermore, if the series converges to some $z\in Z$, then obviously $\langle x,e_i\rangle=\langle z,e_i\rangle $ for all $i$.
I would like to ask you if you know a counterexample or a proof without using the completeness of $V$?
Best wishes
In general, this series will not be convergent.
Let's consider the case where $V$ is the subset of a Hilbert space, which is not closed: For the Hilbert space $L^2(0,2\pi)$, we have an orthonormal basis
$$E= \left\{e_n: x \mapsto \frac{1}{2\pi}e^{inx}\right\}_{ n \in \mathbb N}.$$
Let $V= span E,$ i.e. the set of (finite) linear combinations of elements of $E$.
Then, $V$ is a vector space of continuous functions which we can equip with the $L^2$ scalar product and $E$ is an orthonormal set in $(V,\langle,\cdot,\cdot\rangle)$.
However, since $E$ is a Hilbert Basis for $L^2(0,2\pi)$, we can express any element of $L^2(0,2\pi)$ as a series above, in particular discontinuous functions like the characteristic function $\chi_{(0,1)}$. Then we have
$$\chi_{(0,1)}= \sum_{i=1}^\infty\langle\chi_{(0,1)},e_i\rangle e_i,$$
a series which converges in $L^2(0,2\pi)$, but its limit is clearly not an element of $V$, so it cannot converge in $V$ (even though it is Cauchy).