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 $R(x)$?
 A: By assumption, the resolvent $R(\lambda)=(T-\lambda I)^{-1}$, $\lambda \in \rho(T)$, of the bounded linear operator $T$ has an isolated singularity at $\lambda=\mu$, which guarantees that $R$ has a Laurent series expansion for all $\lambda$ in some punctured disk centered at $\mu$ of radius $r > 0$:
$$
           R(\lambda)=\sum_{n=-\infty}^{\infty}(\lambda-\mu)^{n}R_{n},\\
                               0 < |\lambda-\mu| < r.
$$
Each $R_{n}$ is a bounded linear operator on $X$ which is given by a residue integral. So, the set $\mathcal{A}$ of $x \in X$ for which $R(\lambda)x$ has a removable singularity at $\mu$ is a closed linear subspace of $X$ given by an intersection of closed null spaces:
$$
                             \mathcal{A}=\bigcap_{n=1}^{\infty}\mathcal{N}(R_{-n}).
$$
The subspace $\mathcal{A}$ is invariant under $T$ and under every $R(\lambda)$ for $\lambda \in \rho(T)$. This is because
$$
              R(\lambda)Tx = x+\lambda R(\lambda)x,\;\;\; R(\lambda)R(\lambda')x=\frac{1}{\lambda-\lambda'}\{R(\lambda)-R(\lambda')\}x
$$
have removable singularities in $\lambda$ at $\mu$ if $R(\lambda)x$ does. Notice that $T$ is continuously invertible on the closed invariant subspace $A$ because
$$
         (T-\mu I)R(\lambda)x=x+(\lambda-\mu)R(\lambda)x=R(\lambda)(T-\mu I)x
$$
implies the following
$$
            (T-\mu I)R_{0}x = R_{0}(T-\mu I)x=x,\;\;\; x \in \mathcal{A}.
$$
(Here $R_{0}$ is the constant term in the Laurent series expansion.) Therefore, if $R(\lambda)x$ has a removable singularity at $\lambda=\mu$ for some $x\ne 0$, then there is a non-trivial closed invariant subspace $\mathcal{A}$ for $T$ containing $x$ on $T-\mu I$ is continuously invertible. It's not too hard to show that the converse is also true: if there is a non-trivial closed invariant subspace of $T$ on which $T-\mu I$ is continuously invertible, then $R(\lambda)x$ has a removable singularity at $\lambda=\mu$ for all $x \in \mathcal{A}$. This is because the restriction of $R(\lambda)$ to $\mathcal{A}$ will equal the resolvent of $T$ on $\mathcal{A}$ due to the invariance of $\mathcal{A}$ under $T$ and its resolvent.
A: 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 $\mu=1.$
