Let $H$ be a Hilbert space, and let $A$ be a densely defined, unbounded operator on $H$. Then there is a sequence $v_n$ of vectors in Dom$(A)$ such that $\frac{\|Av_n\|}{\|v_n\|}\to \infty$. My question is essentially whether the obstruction to $\lim_{n\to \infty} Av_n$ being defined is only in norm. More precisely,
Let $h\in H\setminus \text{Dom}(A)$. Since $A$ is densely defined, there exists a Cauchy sequence $v_n$ in $\text{Dom}(A)$ converging to $h$. Does the sequence $ \frac{Av_n}{\|Av_n\|}$ converge in $H$ to some unit vector? If so, is it independent of which sequence we pick converging to $h$? I'm having trouble making any progress since $A$ doesn't play well with the norm. I've tried considering the specific example of $\frac{d}{dx}$ acting on $L^2(\mathbb{R})$ without much success.
Fix an orthonormal basis $\{e_n\}$, $$ \operatorname{Dom}(A)=\Big\{\sum_n x_ne_n:\ \exists n_0:\ x_n=0\ \forall n\geq n_0\Big\}, $$ and $$ A\sum_n x_ne_n=\sum_n nx_ne_n. $$ Consider $h=\displaystyle\sum_n\tfrac1n\,e_n\in H\setminus\operatorname{Dom}(A)$ and put $h_k=\displaystyle\sum_{n=1}^k\tfrac1n\,e_n$. Then $h_k\to h$, and $$ \frac{Ah_k}{\|Ah_k\|}=\frac{\displaystyle\sum_{n=1}^ke_n}{\displaystyle\Big\|\sum_{n=1}^ke_n\Big\|}=\frac1{\sqrt k}\,\sum_{n=1}^ke_n, $$ which does not converge, since for $k>2j$ \begin{align} \Big\|\frac{Ah_k}{\|Ah_k\|}-\frac{Ah_j}{\|Ah_j\|}\Big\|^2 &=\Big\|\sum_{n=1}^j\Big(\frac1{\sqrt k}-\frac1{\sqrt j}\Big)\,e_n+\sum_{n=j+1}^k\frac1{\sqrt k}\,e_n\Big\|^2\\[0.3cm] &\geq\sum_{n=j+1}^k\frac1k =\frac{k-j}{k}=1-\frac jk\\[0.3cm] &\geq\frac12. \end{align}