I am trying to understand the proof that the space of simple processes is dense in $\Lambda^2$. The proof in my lecture notes starts by assuming that for $\phi \in \Lambda^2$ which is orthogonal to $\mathcal{E}$. Then the proof claims that it is sufficient to show that $\phi\equiv 0 \in \Lambda^2$.
I cant precisely show why does this imply the density of $\mathcal{E}$ in $\Lambda^2$. Clearly since $\phi$ is orthogonal to $\mathcal{E}$ we have that $\langle \phi, F\rangle=0$ for every $F \in \mathcal{E}$. Now if I knew that $\mathcal{E}$ was dense in $\Lambda^2$, then by the continuity of the inner product I can easily show that $\phi$ is orthogonal to $\Lambda^2$. Indeed given any element $H \in \Lambda^2$ , there exists a sequence $\{H_n\}_{n=1}^\infty \subseteq \mathcal{E}$ such that
$$ \lim_{n \to \infty}\vert \vert H_n-H \vert \vert_{\Lambda^2}=0 $$
and hence $ \lim_{n \to \infty} \vert\langle H_n-H,\phi \rangle \vert \le \lim_{n \to \infty}\vert \vert H_n-H\vert\vert_{\Lambda^2}^{1/2} \vert\vert \phi \vert \vert_{\Lambda^2}^{1/2}=0$ which would imply $$ \langle H,\phi \rangle= \lim_{n \to \infty}\langle H_n,\phi \rangle= 0 $$
But all this seems to be useless, since I assume $\mathcal{E}$ to be dense where as I want to show the density
If $H$ is a hilbert space and $X$ a subspace we have that $X$ is dense in $H$ if and only if $h\perp X\implies h=0$. The forward direction is fairly trivial utilising the continuity of the innerproduct.
For the converse let us assume that $h\perp X\implies h=0$. We first note that $H=\bar X\oplus \bar X^\perp$ (it is a basic theorem that should be covered in any first course on Hilbert spaces. Specifically closed subspaces are complemented by their orthogonal complement). Now suppose $x\in \bar X^\perp$. This implies $x\in X^\perp$, which by assumption means $x=0$. Thus $\bar X^\perp=\{0\}$, meaning $H=\bar X$.
All that is left is to note that the space of simple functions is a subspace of any "ordinary" function space.