if $\sup_{n\geq 1} \vert w_1\cdots w_n\vert=\infty$. Likewise, $B_{\mathbf w}$ is strongly mixing in the topological sense (on any $X_p$) iff $\vert w_1\cdots w_n\vert\to\infty$. Hence, we see that strong mixing in the topological sense turns out to be equivalent to strong mixing in the Gaussian sense for weighted shifts on $c_0(\NN)$.

Let $T$ be the operator defined by formula (\[Kal\]), but acting on $L^2(0,2\pi)$. It is straightforward to check that for any $t\in (0,2\pi)$, the function $f_t=\mathbf 1_{(0,t)}$ is an eigenvector for $T$ with associated eigenvalue $\lambda =e^{it}$. Moreover the map $E:\TT\setminus\{ \mathbf 1\}\to L^2(0,2\pi)$ defined by $E(e^{it})=f_t$ is clearly continuous. Now, let $\Lambda$ be an arbitrary compact subset of $\TT\setminus\{ \mathbf  1\}$. Let us denote by $H_\Lambda$ the closed linear span of $E(\Lambda)$, and let $T_\Lambda$ be the restriction of $T$ to $H_\Lambda$. By a result of G. Kalisch ([@Kal], see also [@BG1 Lemma 2.12]), the point spectrum of $T_\Lambda$ is exactly equal to $\Lambda$. By Corollary \[eigvectfield\] and Theorem \[WS\] (3), it follows that the operator $T_\Lambda$ is weakly mixing in the Gaussian sense if and only if $\Lambda$ is a perfect set, and strongly mixing iff $\Lambda$ is $\mathcal U_0$-perfect. Hence, any perfect $\mathcal U_0$-set $\Lambda$ gives rise 