I am stuck on a lemma:
Prove that if $ A $ is self-adjoint, then $ (Ker A)^{\perp} $ is an $ A $-invariant subspace, that is $ A: (Ker A)^{\perp} \cap D(A) \to (Ker A)^{\perp} $.
First of all, one minor issue I am having: the definition of an invariant subspace is that $ A(U) \subseteq U $. So, I would have thought that $ A $ mapping $ (Ker A)^{\perp} \cap D(A) $ to the larger space $ (Ker A)^{\perp} $ would mean it is not an invariant subspace? Or are we saying that it is just $(Ker A)^{\perp}$, not $(Ker A)^{\perp} \cap D(A)$, that is an invariant subspace. In this case, I don't think that claim would make much sense.
Secondly, here is my attempt at a proof, and I will try to be clear where I got stuck.
First of all, we have \begin{equation}\label{key} \mathcal{H} = \overline{Im A} \oplus Ker(A^{*}), \end{equation} in general. But $ A $ is self-adjoint, therefore $ (Ker A)^{\perp} = \overline{Im(A)} $, where $ Im A $ denotes the image of $ A $ (or equivalently, the range of $ A $).
Thus, it suffices to prove that $ Ax \in \overline{Im(A)} $ whenever $ x \in \overline{Im(A)} $.
To that end, assume that $ x \in \overline{Im(A)}$. Then, there is a sequence $ x_n \in Im(A) $ such that $ x_n \to x $ in $ \mathcal{H} $.
I want to show that $ Ax \in \overline{Im(A)} $, so it should suffice to prove that $ Ax_n \in Im(A)$ converges to $ Ax $. This is where I seem to get stuck.
Is my approach correct?