It was shown in \cite{NB} that if $T$ is a contraction in a
Hilbert space with finite defect (i.e., $\|T\|\le 1$ and
$\operatorname{rank} (I- T^*T) <\infty$), and if the
spectrum $\sigma(T)$ does not coincide with the closed unit
disk $\overline{\mathbb{D}}$, then the Linear Resolvent Growth
condition $$ \|(\la I - T)^{-1}
\|\le\frac{C}{\operatorname{dist}(\la,\si(T))},\
\la\in\bc\backslash \si(T) $$ implies that $T$ is similar to a
normal operator. The condition $\operatorname{rank}(I -
T^*T)<\infty$ measures how close $T$ is to a unitary
operator. A natural question is whether this condition can be
relaxed. For example, it was conjectured in \cite{NB} that
this condition can be replaced by the condition $I - T^*T\in
\fS_1$, where $\fS_1$ denotes the trace class. In this note we
show that this conjecture is not true, and that, in fact, one
cannot replace the condition $\operatorname{rank}(I -
T^*T)<\infty$ by any reasonable condition of closeness to a
unitary operator.