$\nu(e^A,{\mathsf{D}}_{x_0}\Phi^1)$ is always finite.

\[rem:no-nonres-needed\] The statement regarding the above limit actually holds without any nonresonance assumptions if such an approximate linearizing factor $P$ exists; see Lemma \[lem-make-approx-exact\] in §\[sec:main-proof-exist\].

In the uniqueness portion of the above theorem and also later in this paper, the point of the condition that either $$\nu(e^{A},{\mathsf{D}}_{x_0} \Phi^1) < k + \alpha \qquad \textnormal{ or } \qquad \nu(e^{A},{\mathsf{D}}_{x_0} \Phi^1) \leq k$$ is to require that $\nu(e^{A},{\mathsf{D}}_{x_0} \Phi^1) < k + \alpha$ be strictly less than $\alpha$ except when $\alpha = 0$, in which case non-strict inequality is allowed to hold. This is relevant for, e.g., the case that $k= 1$ and $\alpha = 0$. Of course the “or” above is inclusive, i.e., we allow both inequalities to hold in the hypotheses of the above theorem and later results.

\[rem:point-cinf-case\] In the case that $k = \infty$, the hypothesis $\nu(e^A,{\mathsf{D}}_{x_0}\Phi^1) < k + \alpha$ becomes $\nu(e^A,{\mathsf{D}}_{x_0}\Phi^1) < \infty$ which is automatically satisfied since $\nu(e^A,{\mathsf{D}}_{x_0}\Phi^1)$ is always finite. Hence for the case $k = \infty$, no assumption is needed on the spectral spread in Theorem \[th:main-thm\]; we need only assume that $(e^A, {\mathsf{D}}_{x_0}\Phi^1$) is $\infty$-nonresonant. Similar remarks hold for all of the following results which include a 