Isolated singularities of the resolvent Let $T$ be a bounded operator on $l_2$ such that there exists $\mu$ in the spectrum of $T$ which is an isolated point of the spectrum. We know that for any $x\in l_2$ the resolvent function $R(x):\rho(T)\to l_2$ given by $R(x)(\lambda)=(\lambda I-T)^{-1} x$ is analytic, and so $\mu$ is an isolated singularity of $R(x)$.
Is it possible that, for some $x\neq 0, $ $\mu$ is a removable singularity for $R(x)$?

Yes, this is generally possible, as multiplying with $x$ may not 'look in the right direction'. I'll give an example for matrices which you can adapt for $\ell^2$:
Consider $\begin{pmatrix}  1 & 0 \\ 0 & 2 \end{pmatrix}, \ \mu=1, \ x= \begin{pmatrix}0\\1\end{pmatrix}$. Then $(A-\lambda)^{-1}x= \begin{pmatrix}0\\\frac{1}{2-\lambda}\end{pmatrix}$, which has a removable singularity at $\mu=1.$
