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!
Too long to be a comment.
Spectrum of a non-closed operator is not defined. If A is not closed, then $A−z, z\in\mathbb C$ is never closed and $(A−z)D(A)$ cannot be equal to $H$.
So far you have shown that $z\notin \overline{N(A)}$ implies that $(A-z)^{-1}$ is bounded on the range of $(z-A)$ which is closed. But the $Ran(z-A)$ need not to be the whole $H.$ You should probably use $D(A)=D(A^*).$
Can you prove it for bounded closed operators?