# Schauder basis are dense in any Hilbert space

Definition: A sequence $(x_n)$ in a Hilbert space H is called total iff for any $x\in H, \quad$ $<x,x_n>=0$ implies $x=0$.

I want to show that any schauder basis $(e_n)_{n=1}^{\infty}$ in a Hilbert space $H$ is total, linearly independent and $\operatorname{span} \{e_n : n=1,2,... \}$ is dense in $H$.

I was able to show totality and linear independency.But , I could not find a way to show it is dense. I consider using somewhat density of $\mathbb{Q}$ in $\mathbb{R}$ as follows:

Take an element $x\in H$. Write $x=\sum_{n=1}^{\infty }\alpha_n e_n,\quad \alpha_n \in \mathbb{R}.$ Since $\overline{\mathbb{Q}}=\mathbb{R}$, for each $\alpha_n$, we can find a sequence $\beta_{n_k}$ in rationals s.t $\lim_{k \rightarrow \infty } \beta_{n_k} = \alpha_n.$ So, $$x = \sum_{n=1}^{\infty }\lim_{k \rightarrow \infty } \beta_{n_k} e_n .$$ Can we say directly since summation is a continuous function we can put the limit operator outside the summation and conclude that we find a sequence $x_k = \sum_{n=1}^{\infty } \beta_{n_k} e_n$ s.t $x_k \rightarrow x$. So $x \in \overline{\operatorname{span}(e_n)}.$ So, the conclusion holds.?

• When you write $x = \Sigma \alpha_n e_n$, you are already assuming that the span of the $e_n$ is dense. Commented May 14, 2017 at 18:11

In general, if $S$ is a subspace which is not dense in $H$, then the closure $\bar{S}$ is a proper subspace. Since we are in a Hilbert space, there is a nontrivial orthogonal complement to $\bar{S}$, containing some vector $v \not = 0$. This vector $v$ has the property that it is orthogonal to $S$, but nonzero. This will contradict the assumption that $S$ is generated by a total sequence.