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A weakly mixing system has no nonconstant Koopman operator eigenfunction. If $U_{T}f=\lambda f$, then $f(x)\overline{f(y)}$ is invariant under $T\times T$, contradicting the product characterization of weak mixing unless $f$ is constant.

The statement $D-lim{a}_n=a$ means convergence in density of a sequence: for every $\epsilon >0$, the proportion of $n\le N$ with $|a_n-a|\ge \epsilon$ tends to zero. The statement $C-lim{a}_n=a$ means Cesaro convergence of a sequence, $N^{-1}{\sum }_{n=1}^N{a}_n\to a$.

Suppose $|a_n-a|\le M$. Splitting into the indices where $|a_n-a|<\epsilon$ and its complement givesThus density convergence implies Cesaro convergence of the absolute deviations to zero. Conversely, the Markov inequality givesproving the reverse implication.

The product characterization of weak mixing says that if $T$ is weakly mixing, then $T\times S$ is ergodic exactly when $S$ is ergodic; in particular, ergodicity of $T\times T$ characterizes weak mixing.

Suppose $U_{T}f=\lambda f$. Unitarity of the Koopman operator gives $|\lambda |=1$. Weak mixing implies ergodicity, so $|f|$ is almost everywhere constant. On the product,is invariant under $T\times T$. Since the square is ergodic, $F$ is constant, which forces $f$ to be constant almost everywhere. Thus there are no nonconstant Koopman eigenfunctions.

Finally use the density-one correlation characterization of a weakly mixing measure-preserving transformation. For each positive-measure pair $(A,B)$, the integers $n$ for which $\mu (T^{-n}A\cap B)>0$ form a density-one set after discarding finitely many terms; the corresponding sets for $(A,B)$ and $(A,C)$ therefore intersect, proving simultaneous hitting. Conversely, the simultaneous-hitting property is precisely the simultaneous hitting characterization of weak mixing, so it implies weak mixing.