I need help only with the converse of the proof. Suppose, $M^{\perp} = \{0\}$, and $V=span(M)$, then if $x \perp V$, this implies $x \perp M$ (why?) so that $x \in M^{\perp}$ and $x=0$. Hence $V^{\perp} = \{0\}$ (again why?). Noting that $V$ is a subspace of $H$, we obtain $\bar{V}=H.$
The last equality is obtained from $H= \bar{V}\oplus \bar{V}^{\perp}$, I think, but I don't understand this very well either. Please explain the logic behind the steps indicated with a (why?) that I'm having trouble with. Do we have that $V$ is closed? Also, Is this true that $V^{\perp \perp} = \bar{V}$? How do I you prove this? Thanks for all your help.
Edit: I've proven this in the following way and need to know if this is correct:
$V^{\perp}$ is closed, $\bar{V}$ is a closed subspace of $H$. Then
$H = \bar{V} \oplus \bar{V^{\perp}} = \bar{V} \oplus V^{\perp}=\bar{V} \oplus \{0\}$ implies $H = \bar{V}$.
Best Answer
Your edit is correct modulo the proof that $(\overline{V})^\perp=V^\perp$ (i.e. be careful on the order of applying bars and perps).
To prove this, note first that since $V\subset \overline{V}$, we trivially have $\overline{V}^\perp \subset V^\perp$. To prove the converse, let $f \in V^\perp$, and let $g \in \overline{V}$. We must show that $\langle f, g \rangle = 0$. But $g=\lim_{n \to \infty}g_n$ for some sequence $\{g_n\}$ in $V$, hence $0 = \lim \langle f, g_n \rangle = \langle f, \lim g_n \rangle = \langle f,g \rangle$, and we are done.
Also, you could have used that $V^\perp$ is closed and $(V^\perp)^\perp=\overline{V}$, so $\mathcal{H}=V^\perp \oplus (V^\perp)^\perp = \{0\} \oplus \overline{V}=\overline{V}$ (Of course, this is really the same proof).