[proofplan] Strong mixing is first rewritten as decay of all zero-mean correlations. The spectral theorem identifies the diagonal zero-mean correlations with Fourier coefficients of the spectral measures. Polarization then shows that decay of all diagonal correlations is equivalent to decay of all mixed correlations, which is exactly strong mixing. [/proofplan]

[step:Translate strong mixing into zero-mean correlation decay] For measurable sets $A,B\in\mathcal{B}$, 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 $a_A,b_B\in L^2_0(X,\mu)$ and, because $\mu(X)=1$, \begin{align*} \langle U_T^n a_A,b_B\rangle =\mu(T^{-n}A\cap B)-\mu(A)\mu(B). \end{align*} Thus strong mixing is equivalent to \begin{align*} \langle U_T^n a_A,b_B\rangle\longrightarrow 0 \end{align*} for all centered indicators. Finite linear combinations of indicators are dense in $L^2(X,\mu)$, and subtracting the mean preserves density in $L^2_0(X,\mu)$. If $f,g\in L^2_0(X,\mu)$ and $s,t$ are centered simple functions, then the [Cauchy-Schwarz Inequality](/theorems/432) and unitarity of $U_T$ give, for all $n$, \begin{align*} |\langle U_T^n f,g\rangle-\langle U_T^n s,t\rangle| &\leq |\langle U_T^n(f-s),g\rangle| +|\langle U_T^n s,g-t\rangle|\\ &\leq \|f-s\|_2\|g\|_2+\|s\|_2\|g-t\|_2. \end{align*} Hence strong mixing is equivalent to \begin{align*} \langle U_T^n f,g\rangle\longrightarrow0 \end{align*} for all $f,g\in L^2_0(X,\mu)$. [/step]

[step:Reduce mixed correlations to diagonal spectral coefficients] For $f\in L^2_0(X,\mu)$, the spectral theorem for the unitary operator $U_T|_{L^2_0}$ gives \begin{align*} \langle U_T^n f,f\rangle = \int_{\mathbb{T}}z^n\,d\sigma_f(z) = \widehat{\sigma_f}(n). \end{align*} Therefore the Rajchman condition for every $\sigma_f$ is exactly the diagonal correlation condition \begin{align*} \langle U_T^n f,f\rangle\longrightarrow0 \qquad(f\in L^2_0(X,\mu)). \end{align*} The diagonal condition is equivalent to mixed zero-mean correlation decay by polarization. With the inner product linear in the first variable, \begin{align*} 4\langle U_T^n f,g\rangle &= \langle U_T^n(f+g),f+g\rangle -\langle U_T^n(f-g),f-g\rangle\\ &\quad +i\langle U_T^n(f+ig),f+ig\rangle -i\langle U_T^n(f-ig),f-ig\rangle . \end{align*} Since $L^2_0(X,\mu)$ is a complex linear subspace, all four diagonal terms involve zero-mean functions. Thus all diagonal correlations tend to $0$ if and only if all mixed correlations tend to $0$. [/step]

[step:Conclude the equivalence] By the first step, $T$ is strongly mixing if and only if every mixed zero-mean correlation $\langle U_T^n f,g\rangle$ tends to $0$. By the second step, this is equivalent to \begin{align*} \widehat{\sigma_f}(n)\to0 \end{align*} for every $f\in L^2_0(X,\mu)$. This is precisely the assertion that every zero-mean spectral measure is Rajchman. [/step]