Let $ X $ be a Hilbert space. Can we find unbounded self-adjoint operators $ A, A_n $ on $ X $ such that $ A_n \stackrel{s.r.s}{\longrightarrow} A, $ that is, \begin{equation*} \ R_\lambda(A_n) \stackrel{s}{\longrightarrow} R_\lambda(A), \ \forall\ \lambda \in \mathbb{C}\setminus \mathbb{R}, \ \ \text{where} \ R_\lambda(A) := (\lambda I -A)^{-1} \end{equation*} but $ (\bigcap_{n \in \mathbb{N} } \mathcal{D}(A_n)) \bigcap \mathcal{D}(A) = \{ 0 \} $?
Edit: This question is equivalent to Convergence of (unbounded) self-adjoint operators
Let $$ D(S)=\left\lbrace f\in L^2(\mathbb{R})\mid \int (e^{x^2}f(x))^2\,dx<\infty\right\rbrace,\,Sf(x)=e^{x^2}f(x)\\ D(T)=\{f\in L^2(\mathbb{R})\mid \hat f\in D(S)\},\,Tf=\mathcal{F}^{-1}(S\hat f).$$
In particular, if $f\in D(S)\cap D(T)$, then $e^{|\cdot|^2/2}f, e^{|\cdot|^2/2}\hat f\in S'(\mathbb{R})$. It follows from Hardy's incertainty principle* that $f(x)=e^{-|x|^2/2}p(x)$ for some polynomial $p$. Using once again $f\in D(S)$, we see that $f$ must be zero.
Now you can take $A_1=S$ and $A_n=T$ for $n>1$. Clearly $A_n\to T$ and $\bigcap_n D(A_n)\cap D(T)=D(S)\cap D(T)=\{0\}$.
*Depending on you definition of the Fourier transform, you may need different factors in the exponentials.