Weyl sequence for closure of an operator I'm trying to solve following exercise and need some hints.
Let $A= \bar{ A_0 }$ be closure of $A_0$ - a densely defined operator. Suppose $f_n \in D(A)$ is Weyl sequence for $z \in \sigma (A)$. Show that there exists a Weyl sequence in $D(A_0)$.
For reference I include definition of Weyl sequence:
$\psi _n \in D(A)$ is Weyl sequence for $z \in \mathbb C$ if $ \| \psi _n \| = 1 $ and $(A-z) \psi _n \to 0$.
I tried approximating terms of Weyl sequence by elements of unit norm in $D(A_0)$ (since it's dense) but I don't know anything about continuity of $A$ so it doesn't lead anywhere.
 A: Without loss of generality we can assume that $z=0$ (otherwise apply the proof to the operator $A-z$ instead). The sequence $\{f_n\}$ is Weyl, i.e. $\|f_n\|=1$ and $\|Af_n\|\to 0$ when $n\to+\infty$.


*

*The vector $(f_n,Af_n)$ belongs to the graph $\Gamma(A)$. Since $A=\bar A_0$ we know that $\Gamma(A)=\bar\Gamma(A_0)$, so we can approximate $(f_n,Af_n)$ arbitrary well by $(g_n,A_0g_n)$ with $g_n\in D(A_0)$ in the graph metric, for example, as
$$
\|f_n-g_n\|^2+\|Af_n-A_0g_n\|^2\le\frac{1}{n^2}
$$
which implies, in particular, that $\|f_n-g_n\|\le\frac1n$ and $\|Af_n-A_0g_n\|\le\frac{1}{n}$.

*By the reverse triangle inequality we have 
$$
0\le\Big|1-\|g_n\|\Big|=\Big|\|f_n\|-\|g_n\|\Big|\le\|f_n-g_n\|\le\frac1n\to 0,
$$
therefore, $\|g_n\|\to 1$ as $n\to+\infty$. Moreover,
$$
0\le\|A_0g_n\|=\|A_0g_n-Af_n+Af_n\|\le \|A_0g_n-Af_n\|+\|Af_n\|\le\frac1n+\|Af_n\|\to 0.
$$

*Let us show that $\psi_n=\frac{g_n}{\|g_n\|}\in D(A_0)$ is the Weyl sequence. Indeed, $\|\psi_n\|=1$ and
$$
\|A_0\psi_n\|=\frac{\|A_0g_n\|}{\|g_n\|}\to\frac{0}{1}=0.
$$
