Equivalence to essentially self-adjoint

Question

In a book I have found a statement (without proof):

Let $A$ be a symmetric operator on a Hilbert space $\mathcal H$, the following are equivalent:

1. $A$ is essentially self-adjoint
2. $\nu(i)=\nu(-i)=0$
3. The ranges of $A-i \mathbb{1}$ and $A+i\mathbb 1$ are dense in $\mathcal H$

Here $\nu(\lambda)$ is defined as the dimension of the $\ker(A^*-\lambda \mathbb 1)$.

I don't really know where to begin with the proof. For 2. and 3. is it necessary to have both parts of the statement or are they already equivalent to each other?

How can the statement be proven? I think at this point I would appreciate hints more than the direct proof.

Answer 1

Hint: If $A : \mathcal{D}(A)\subseteq H \rightarrow H$ is a densely-defined symmetric linear operator on a complex Hilbert space $H$, then the following holds in $H\times H$: $$ \mathcal{G}(A^*)=\overline{\mathcal{G}(A)}\oplus\mathcal{N}(A^*-iI)\oplus\mathcal{N}(A^*+iI), $$ where $\mathcal{G}(A^*)$ is the graph of $A^*$ and $\overline{\mathcal{G}(A)}$ is the closure of the graph of $A$ in $H\times H$. (I'm being sloppy about the notation of $\mathcal{N}(A^*\pm iI)$ because it's really the graph of the restriction of $A^*$ to these subspaces.)