already pointed out, much more is true in the ergodic case: it is proved in [@BG2 Theorem 4.1] that if $T$ is ergodic in the Gaussian sense, then in fact the $\TT$-eigenvectors are perfectly spanning (provided that $X$ has cotype 2).

Our last result (to be proved also in section \[proofabstract2\]) says that when the space $X$ is well-behaved, the assumptions in Theorem \[abstract\] can be relaxed: it is no longer necessary to assume that the family $\mathbf S$ is $c_0$-like, nor that $\mathbf S$-continuous measures are continuous. However, the price to pay is that one has to use the strengthened version of $\mathbf S$-perfect spanning.

\[abstracteasy\] Let $X$ be a separable complex Banach space, and assume that $X$ has *type 2*. Let also $\mathbf S$ be a norm-closed, translation-invariant ideal of $\ell^\infty(\ZZ_+)$, and let $T\in\mathfrak L(X)$. If the $\TT$-eigenvectors of $T$ are $\mathbf S$-perfectly spanning for analytic sets, then $T$ is $\mathbf S$-mixing in the Gaussian sense.

Since Hilbert space has both type 2 and cotype 2, we immediately deduce

\[Smix=span\] Let $\mathbf S$ be a norm-closed, translation-invariant ideal of $\ell^\infty(\ZZ_+)$. For Hilbert space operators, $\mathbf S$-mixing in the Gaussian sense is equivalent to the $\mathbf S$-spanning property of 