When an orthonormal set is a basis (proof verification)

Question

Problem:

Let $B$ an orthonormal set in a Hilbert space $\mathcal{H}$. Show that $B$ is a basis (i.e $\overline{span(B)} = \mathcal{H})$ if and only if $B^\perp = \{0\}$.

My attempt:
$(\Rightarrow)$ Suppose $B$ is a basis and let $x\in B^\perp$. There exists then a finite linear combination of elements $e_i$ in $B$ such that for a given $\varepsilon > 0$ $$ ||x-\sum_{i=1}^n c_ie_i || < \varepsilon$$ for some scalars $c_i$. This inequallity reaches its minimum when $$ ||x-\sum_{i=1}^n <x,e_i>e_i || \le ||x-\sum_{i=1}^n c_ie_i || < \varepsilon $$ but since $x \in B^\perp$ this leads to $||x|| < \varepsilon$ and then $x = 0$.
$(\Leftarrow)$ Since $$ (span(B)^\perp)^\perp = \overline{span(B)}$$ we have $\overline{span(B)} = \{0\}^\perp = \mathcal{H}$

Thanks a lot for your help!!

Answer 1

With the changes, this looks good now. I do think some small tweaks to the wording could make this even clearer.

I think it's worth rewording

This inequality reaches its minimum when $$ \left\|x-\sum_{i=1}^n \langle x,e_i \rangle e_i \right\| \le \left\|x-\sum_{i=1}^n c_ie_i \right\| < \varepsilon.$$

Instead, you might want to say,

Note that the quantity $$\left\|x-\sum_{i=1}^n d_ie_i \right\|$$ reaches its minimum when $d_i = \langle x, e_i \rangle$ for $i = 1, \ldots, n$. Therefore, $$ \left\|x-\sum_{i=1}^n \langle x,e_i \rangle e_i \right\| \le \left\|x-\sum_{i=1}^n c_ie_i \right\| < \varepsilon.$$

You may also want to reference a theorem for this fact!

The other thing I would rewords is this:

...since $x \in B^\perp$ this leads to $\|x\| < \varepsilon$ and then $x = 0$.

I would put it like this:

...since $x \in B^\perp$ this leads to $\|x\| < \varepsilon$. As $\varepsilon > 0$ was arbitrary, we have $x = 0$.

Other than that, I think it's perfectly clear.