Show that a set is complete in a generic Hilbert space $H$ Let $\{e_n\}$, $n\geq1$ be an orthonormal base of an hilbert space $H$. Let 
$$
f_1=e_1-e_2+e_3,\quad f_2=e_2-e_3+e_4,\quad \ldots\quad f_n=e_n-e_{n+1}+e_{n+2}.
$$
Show that $\{f_n\}$, $n\geq1$ is complete. I think I have to use the definition, that is
$$
(f_n, x)=0,\ \forall n\ \ \Rightarrow \ \ x = 0
$$
but I cannot understand how. Some help?
 A: Hint:
By using $f_1+f_2=e_1+e_4$
we get
$$
 0= (f_1,x) + (f_2,x)
 =
 (e_1,x)+(e_4,x)
$$
Doing the same with $f_4,f_5$ and combining with the above, yields
$$
 (e_1,x)=(e_7,x).
$$
This can be generalized, so you have statements about $(e_n,x)$ for $n\in \mathbb N$.
Then you have to use $(e_n,x)\to0$,
which follows from $e_n \rightharpoonup 0$
or Bessels inequality.
After that you will be much closer to the goal.
A: Assume $\langle x, f_n \rangle = 0$, for every $n \in \mathbb{N}$. Notice that for $n \ge 2$ we have:
$$\sum_{k=1}^n f_k = \sum_{k=1}^n e_k-e_{k+1}+e_{k+2} = e_1+e_{n+2}$$
So 
$$0 = \sum_{k=1}^n\left\langle x,  f_k\right\rangle = \left\langle x, \sum_{k=1}^n f_k\right\rangle = \langle x, e_1 \rangle + \langle x, e_{n+2} \rangle$$
which implies
$$\langle x, e_1 \rangle =- \langle x, e_{n} \rangle, \quad \forall n \ge 4$$
Now, since $(e_n)_{n=1}^\infty$ is an orthonormal basis we have:
$$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x,e_n\rangle\right|^2 =  \left|\langle x,e_1\rangle\right|^2 + \left|\langle x,e_2\rangle\right|^2 + \left|\langle x,e_3\rangle\right|^2 + \sum_{n=4}^\infty \left|\langle x,e_1\rangle\right|^2$$
For the sum to be finite, it must hold $\langle x, e_1 \rangle = 0$.
Now
$$0 = \langle x,f_3\rangle = \langle x,e_3-e_4+e_5\rangle = \langle x,e_3\rangle - \langle x,e_4\rangle + \langle x,e_5\rangle = \langle x,e_3\rangle$$
so we have $\langle x,e_3\rangle = 0$.
Finally:
$$0 = \langle x,f_1\rangle = \langle x,e_1-e_2+e_3\rangle = \langle x,e_1\rangle - \langle x,e_2\rangle + \langle x,e_3\rangle = -\langle x,e_2\rangle$$
So, $\langle x,e_2\rangle = 0$.
Therefore, we have $\langle x,e_n\rangle = 0$, $\forall n\in\mathbb{N}$.
Now, since $(e_n)_{n=1}^\infty$ is an orthonormal basis:
$$x = \sum_{n=1}^\infty \underbrace{\langle x,e_n}_{=0}\rangle e_n = 0$$
An orthonormal set $\{f_i\}_{i\in I}$ with this property is sometimes called a maximal orthonormal set, and in a Hilbert space this is equivalent to $\{f_i\}_{i\in I}$ being an orthonormal basis. It is maximal in the sense that there does not exist a vector $f \in H$, such that $f \perp f_i, \forall i\in I$ and that $\{f_i\}_{i\in I} \cup \{f\}$ is also an orthonormal set.
