Suppose we have a smooth domain $\Omega \in \mathbb{R}^n$ equipped with the Euclidean norm and a self-adjoint operator $A: \mathcal H^1_0(\Omega) \to \mathcal H^1_0(\Omega)$ defined as $A = -\Delta + V(x)$ and a sequence $A_n = P_n A P_n, \ n\in \mathbb N$ where $P_n$ is the projection onto a space spanned by the orthonormal basis $\mathfrak D_n = \operatorname{span}(\phi_k)_{k=1}^n$. How can I show that $P_n A P_n \longrightarrow A$ in the strong resolvent sense? I have been reading Mathematical Methods in Quantum Mechanics by Gerald Teschl for reference. Strong convergence is defined there as: $$ \operatorname{s-lim}_{n \rightarrow \infty} A_{n}=A \quad: \Leftrightarrow \quad A_{n} \psi \rightarrow A \psi \quad \forall \psi \in \mathfrak{D}(A) $$
Here is what I have so far:
Note that $i \in \rho(A)\cap \rho(A_n)$, hence it is enough to show that: $$\underset{n \rightarrow \infty}{\text { s-lim }} (A_n - i)^{-1} \longrightarrow (A - i)^{-1}$$ We can rewrite this using Neumann series. So we want to show that:
$$\underset{n \rightarrow \infty}{\text { s-lim }} \sum_j (-1)^ji^{j+1} P_n(A P_n)^j = \sum_j (-1)^ji^{j+1} A^j$$
So we are ultimately interested in whether $P_n(AP_n)^j \psi \longrightarrow A^j \psi$ for all $j \in \mathbb N$, right?. I do not know how to proceed further from here. It seems it should be intuitive, but I am stuck.
Edit: Is it enough to prove this by induction? How is this?:
By the dominated convergence theorem, we are able to move the limit into the summation (what is the dominating function?) and for $j=1$, we can now consider for $\psi \in H^1_0(\Omega)$: $$\lim_{n \to \infty} \|( P_n A P_n - A) \psi \|_{H^1_0} = 0$$ Because the Fourier coefficients of $ A^j \psi$ are $\ell^2(\mathbb N)$ for all $j \in \mathbb N$. Furthermore, $$\lim_{n \to \infty} \|( P_n (A P_n)^j(A P_n) - A^j A) \psi \|_{H^1_0} = 0$$ for all $j \in \mathbb N$, and we are done.