= Solution
The statement $D\!\operatorname{-lim}a_n=a$ means <convergence in density of a sequence>: for every $\varepsilon>0$, the proportion of $n\leq N$ with $|a_n-a|\geq\varepsilon$ tends to zero. The statement $C\!\operatorname{-lim}a_n=a$ means <Cesaro convergence of a sequence>, $N^{-1}\sum_{n=1}^Na_n\to a$.
Suppose $|a_n-a|\leq M$. Splitting into the indices where $|a_n-a|<\varepsilon$ and its complement gives
$$
\frac1N\sum_{n=1}^N|a_n-a|
\leq\varepsilon+M\frac{|\{n\leq N:|a_n-a|\geq\varepsilon\}|}{N}.
$$
Thus density convergence implies Cesaro convergence of the absolute deviations to zero. Conversely, the <Markov inequality> gives
$$
\frac{|\{n\leq N:|a_n-a|\geq\varepsilon\}|}{N}
\leq\frac1{\varepsilon N}\sum_{n=1}^N|a_n-a|,
$$
proving 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 $U_Tf=\lambda f$. Unitarity of the <Koopman operator> gives $|\lambda|=1$. Weak mixing implies ergodicity, so $|f|$ is almost everywhere constant. On the product,
$$
F(x,y)=f(x)\overline{f(y)}
$$
is invariant under $T\times T$. Since the square is ergodic, $F$ is constant, which forces $f$ to be constant almost everywhere. Thus \b[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.
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