TITLE: Closure of numerical range contains spectrum
QUESTION [6 upvotes]: Let $A: D(A) \subset \mathcal{H} \to \mathcal{H}$ be a densely defined operator on a Hilbert space $\mathcal{H}$ with adjoint operator $A^{*}$. Given that $D(A) = D(A^{*})$ I'm trying to show that the spectrum $\sigma(A)$ is a subset of the closure of the numerical range of A: $N(A)=\{ \langle x, Ax \rangle \vert \|x\| = 1, x \in D(A) \}$.
If $A$ is closed, I think we can argue as follows:
$ \| (z - A)x \| \geq \vert \langle x, (z-A)x \rangle \vert \geq \inf_{\|x\|=1} \vert z - \langle x, Ax \rangle\vert$.
So if $z \notin \overline{N(A)}$ we have that $(z - A)$ is injective und the inverse $(z - A)^{-1}$ is bounded on the range of $A$. 
By contradiction Ran$(z-A) = \mathcal H$ and hence $z$ has to be in the resolvent set of $A$.
What I am not sure about is: 

Is the above proof right?
If $A$ is not closed, does $\sigma(A) \subset \overline{N(A)}$ hold?  

Can anyone help? 
Thanking you in advance!

REPLY [2 votes]: $T$ should be closed. Then, as you said,
$\| (\lambda-T)x \| \ge \langle (\lambda-T)x, \tfrac{ x }{\| x \| } \rangle \ge \| x \| \inf_{\|z\|=1} |\lambda - \langle Tz , z \rangle|$.
This amounts that $(\lambda-T)$ has closed image. But since the same inequality holds for $T^*$, and since  $\lambda-T^*$ is clearly injective, $\lambda-T$ has dense range. It is thus surjective. Now the inequality asserts that the resolvent in $\lambda$ is well defined, and bounded by $1/ \text{dist}(\lambda, \overline{N(T)})$.