[proofplan] We first show that the set definition of weak mixing is equivalent to Cesaro decay of all zero-mean correlations. Then we use the spectral theorem and Wiener's Fourier-coefficient criterion: Cesaro square decay of the diagonal correlations is equivalent to the absence of atoms in every zero-mean spectral measure, and atoms are exactly eigenvectors. Polarization and Cauchy-Schwarz on Cesaro averages connect the diagonal and mixed forms. [/proofplan] [step:Translate weak mixing of sets into zero-mean correlation decay] For $A,B\in\mathcal{B}$, define \begin{align*} a_A:=\mathbb{1}_A-\mu(A)\mathbb{1}_X, \qquad b_B:=\mathbb{1}_B-\mu(B)\mathbb{1}_X. \end{align*} Then $a_A,b_B\in H_0$ and \begin{align*} \langle U_T^n a_A,b_B\rangle = \mu(T^{-n}A\cap B)-\mu(A)\mu(B). \end{align*} Thus (i) is exactly (iii) for centered indicator functions. The centered simple functions are dense in $H_0$. If $f,g\in H_0$ and $s,t\in H_0$ are centered simple functions, then for every $n\geq0$, \begin{align*} |\langle U_T^n f,g\rangle| &\leq |\langle U_T^n s,t\rangle| +|\langle U_T^n(f-s),g\rangle| +|\langle U_T^n s,g-t\rangle|\\ &\leq |\langle U_T^n s,t\rangle| +\|f-s\|_2\|g\|_2+\|s\|_2\|g-t\|_2, \end{align*} using the [Cauchy-Schwarz Inequality](/theorems/432) and unitarity of $U_T$. Averaging and then choosing $s,t$ close to $f,g$ proves that (i) implies (iii). The converse follows by applying (iii) to $a_A$ and $b_B$. [/step] [step:Relate Cesaro correlation decay to atoms of spectral measures] Let $V:=U_T|_{H_0}$. For $f\in H_0$, let $\sigma_f$ be its spectral measure for the unitary operator $V$. Then \begin{align*} \langle V^n f,f\rangle=\widehat{\sigma_f}(n) \end{align*} for every $n\in\mathbb{Z}$. We use Wiener's criterion for finite positive Borel measures on $\mathbb{T}$: \begin{align*} \lim_{N\to\infty}\frac1N\sum_{n=0}^{N-1}|\widehat{\sigma}(n)|^2 = \sum_{\zeta\in\mathbb{T}}\sigma(\{\zeta\})^2. \end{align*} Hence \begin{align*} \frac1N\sum_{n=0}^{N-1}|\langle V^n f,f\rangle|^2\to0 \end{align*} for every $f\in H_0$ if and only if every $\sigma_f$ is atomless. Atoms are exactly eigenvectors. If $\sigma_f(\{\lambda\})>0$, then the spectral projection $E(\{\lambda\})f$ is a non-zero vector $h\in H_0$ satisfying $Vh=\lambda h$. Conversely, if $0\neq h\in H_0$ and $Vh=\lambda h$, then the spectral measure of $h$ is supported at $\lambda$, so \begin{align*} \sigma_h(\{\lambda\})=\|h\|_2^2>0. \end{align*} Therefore $V$ has no eigenvectors if and only if every zero-mean spectral measure is atomless. [/step] [step:Pass between diagonal square decay and mixed absolute decay] Assume first that (iii) holds. Taking $g=f$ gives \begin{align*} \frac1N\sum_{n=0}^{N-1}|\langle V^n f,f\rangle|\to0. \end{align*} Since $|\langle V^n f,f\rangle|\leq\|f\|_2^2$, the Cesaro averages of the squares also tend to $0$. By the preceding step, $V$ has no eigenvectors. Thus (iii) implies (ii). Conversely, assume (ii). Then every zero-mean spectral measure is atomless, so for each $h\in H_0$, \begin{align*} \frac1N\sum_{n=0}^{N-1}|\langle V^n h,h\rangle|^2\to0. \end{align*} By the [Cauchy-Schwarz Inequality](/theorems/432), \begin{align*} \frac1N\sum_{n=0}^{N-1}|\langle V^n h,h\rangle| \leq \left(\frac1N\sum_{n=0}^{N-1}|\langle V^n h,h\rangle|^2\right)^{1/2}, \end{align*} so diagonal Cesaro absolute averages tend to $0$. For $f,g\in H_0$, polarization gives \begin{align*} 4\langle V^n f,g\rangle &= \langle V^n(f+g),f+g\rangle -\langle V^n(f-g),f-g\rangle\\ &\quad +i\langle V^n(f+ig),f+ig\rangle -i\langle V^n(f-ig),f-ig\rangle . \end{align*} Averaging absolute values and applying the diagonal result to the four zero-mean vectors on the right proves (iii). Therefore (ii) and (iii) are equivalent. [/step] [step:Combine the equivalences] The first step proves (i) if and only if (iii). The last two steps prove (ii) if and only if (iii). Hence weak mixing, absence of non-constant Koopman eigenfunctions, and Cesaro decay of all zero-mean correlations are equivalent. [/step]