Does this sequence of operators converge in norm or strongly? Let $H$ be a Hilbert space and $\mathcal{L}(H)$ the set of all bounded linear operators $L:H\to H$, equiped with the usual norm $\|\cdot\|_{\mathcal{L}}$.
Let $T:D(T)\subset H\to H$ be a densely-defined linear operator and $(\lambda_n)$ a sequence in $\rho(T)$ (the resolvent set of $T$).
Suppose that there exists $\lambda\in\rho(T)$ such that $\lambda_n\to \lambda$.
What is it possible to conclude about the sequence $(S_n)$ of bounded linear operators given by
$$S_n=(\lambda_nI-T)^{-1}?$$
Is it possible to prove that $\|S_n-(\lambda I-T)^{-1}\|_{\mathcal{L}}\to 0$?
Is it possible to prove that $\|S_nx-(\lambda I-T)^{-1}x\|_H\to 0$ for all $x\in H$?
Thanks.
 A: This sequence does converge in the operator norm (actually, $\lambda \mapsto (T-\lambda I)^{-1}$ is analytic! See this link for a short proof). Note that operator convergence implies strong convergence, so we only have to prove your first claim.
The proof uses the ever helpful resolvent identity:
$$
(A-\lambda I)^{-1} - (A-\mu I)^{-1} = (\lambda-\mu)(A-\lambda I)^{-1}(A-\mu I)^{-1}
$$
See this page for more details as well as the other resolvent identity.
Now, we can start proving. $(T-\lambda_n I)^{-1} \to (T-\lambda I)^{-1}$. Begin by noticing that by replacing $T$ with $T-\lambda I$, we can assume without loss that $\lambda = 0$. Recall that $\|AB\|\leq \|A\|\,\|B\|$ for any operators $A$ and $B$. By the resolvent identity above,
$$
\|S_n - (T)^{-1}\| = |\lambda_n|\,\|S_n(T)^{-1}\|\leq |\lambda_n|\, \|S_n\|\,\|T^{-1}\|
$$
So it suffices to prove that $\|S_n\|$ is bounded.
$$
\|S_n\| = \|T^{-1}T(T-\lambda_n I)^{-1}\| \leq \|T^{-1}\|\,\|(I-\lambda_nT)^{-1}\|
$$
We now come to the main trick. When $\lambda_n$ is small enough, $\|\lambda_nT\| <1/2$ so 
$$
(I-\lambda_nT)^{-1} = \sum_{i=0}^\infty (\lambda_nT)^i 
$$
analogously to the geometric series from 1 dimension. So, using that $\|AB\| \leq \|A\|\|B\|$,
$$
\|(I-\lambda_nT)^{-1}\| \leq \sum_{i=0}^\infty \|(\lambda_nT)^i\| \leq  \sum_{i=1}^\infty (1/2)^n = 1
$$
Now we can conclude with $\|S_n\| \leq \|T^{-1}\|$ for $n$ large enough.
