I'm studying the orthonormal systems in Hilbert spaces and I should prove that:
An orthonormal system is complete iff it is maximal.
Ideas?
DEFs:
"complete" means that its span is dense in $H$, while "maximal" means that if $y$ is orthogonal to the system then $y=0$.
Let $E$ be the orthonormal system in question.
Assume $E$ is complete.
Let $x \perp E$. Then $x \perp \operatorname{span} E$ by linearity of the inner product. Also, then $x \perp \overline{\operatorname{span} E} = X$ by continuity of the inner product.
In particular $x \perp x$ so $x = 0$. Therefore, $E$ is maximal.
Conversely, assume that $E$ is maximal.
We will show that every $x \in X$ can be written as $$x = \sum_{e \in E} \langle x, e\rangle e$$
A result we need is that for any orthonormal system $F$ and $x \in X$ the sum $$\sum_{f \in F} \langle x, f\rangle f $$ exists.
Therefore $$x_0 = \sum_{e \in E} \langle x, e\rangle e$$ exists.
Now notice that for every $f \in E$ we have:
$$\langle x_0, f\rangle = \left\langle \sum_{e \in E} \langle x, e\rangle e, f\right\rangle = \sum_{e \in E} \langle x, e\rangle \underbrace{\langle e, f\rangle}_{=\delta_{e,f}} = \langle x, f\rangle$$
So $\langle x - x_0, f\rangle = 0$ for all $f \in F$. Therefore $x - x_0 \perp F$ so maximality implies $x - x_0 = 0$.
We conclude $x = x_0$.
In particular, $\overline{\operatorname{span} E} = X$ so $E$ is complete.