and $\mathbf S_{\rm erg}$-continuous measures need not be continuous (a measure $\sigma$ is $\mathbf S_{\rm erg}$-continuous iff $\sigma(\{ \mathbf 1\})=0$).

Theorem \[abstract\] will be proved in section \[proofabstract1\]. The following much simpler converse result (which corresponds to (3) in Theorem \[WS\]) will be proved in section \[proofabstract2\].

\[converse\] Let $X$ be a separable complex Banach space, and assume that $X$ has cotype 2. Let also $\mathbf S$ be an arbitrary subset of $\ell^\infty(\ZZ_+)$. If $T\in\mathfrak L(X)$ is $\mathbf S$-mixing in the Gaussian sense, then the $\TT$-eigenvectors of $T$ are $\mathbf S$-perfectly spanning for analytic sets.

Two “trivial" cases are worth mentioning. If we take $\mathbf S=\mathbf S_{\rm erg}$, then $\mathcal F_+^{-1}(\mathbf S)=\{ \sigma\in\mathcal M(\TT);\; \sigma (\{\mathbf 1\})=0\}$ and hence a set $D\subset\TT$ is $\mathbf S$-small if and only if $D\subset\{\mathbf 1\}$. If we take $\mathbf S=\ell^\infty (\ZZ_+)$, then $\mathcal F_+^{-1}(\mathbf S)=\mathcal M(\TT)$ and hence the only $\mathbf S$-small set is $D=\emptyset$. So, assuming that $X$ has cotype $2$, we get the following (known) results: if $T$ admits an invariant Gaussian measure with full support, then the $\TT$-eigenvectors of $T$ span a dense subspace of $X$; and if $T$ is ergodic in the Gaussian sense, then ${\overline{\rm span}}\,\left(\bigcup_{\lambda\in\TT\setminus\{ \mathbf 1\}}\ker(T-\lambda)\right)=X$. As 