Suppose one has a bounded operator $T$ on a Banach space $X$ say, which admits a spectral gap, i.e. the spectrum may be decomposed into $\sigma(T)=\sigma_1 \cup \{1\}$, where $\sigma_1 \subset B_r(0)$ for some $r<1$, and the rank of the Riesz projection of $T$ at $1$ is exactly one. Suppose further that one has a sequence of operators $T_k$ which converge in SOT to $T$, and which each also admit a spectral gap, with radius $r_k$ say. Is it possible that the $r_k \to 1$?
By classical spectral pertubation arguments, this cannot happen if the $T_k $ were to converge in norm to $T$ (in this case we wouldn't even need to assume that the $T_k$ admit a spectral gap, since it would be automatic for $k$ sufficiently small). However, I am not aware of any nice stability properties of the spectrum under SOT convergence. I therefore suspect that in general its possible that the $r_k \to 1$, but I have been unable to find a counterexample.
In the likely event that its possible that $r_k \to 1$, are there any conditions that are known, short of norm convergence, for which one can ensure that $r_k$ remains bounded away from $1$?