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I'm trying to prove an equivalent statement about weak mixing transformations that relies on the spectral theorem, but I can't find a reference to fill in the last details. A hint for solving it or finding details will be greatly appreciated.

The statement; a transformation $T:(X,\lambda) \rightarrow (X,\lambda)$ is weak mixing, i.e. $\frac{1}{N}\sum | \lambda(T^{-n}A \cap B) - \lambda(A)\lambda(B)|^2 \rightarrow 0$, if and only if the associated operator $U_T : L^2(X) \rightarrow L^2(X)$ has only constant eigenfunctions.

By the spectral theorem, if $\sigma$ is the spectrum of $U_T$, for any $f \in L^2(X)$ there is a measure $\mu \in \mathcal{M}(\sigma)$ for which $\int_\sigma z^n d\mu = < U_T^n f, f>$

What I'm missingfrom spectral theory; in the converse, if $U_T$ has only constant eigenvalues and $\int_X f = 0$ then the corresponding function $\mu$ is non atomic.

3 added 1 characters in body

I'm trying to prove an equivalent statement about weak mixing transformations that relies on the spectral theorem, but I can't find a reference to fill in the last details. A hint for solving it or finding details will be greatly appreciated.

The statement; a transformation $T:(X,\lambda) \rightarrow (X,\lambda)$ is weak mixing, i.e. $\frac{1}{N}\sum | \lambda(T^{-n}A \cap B) - \lambda(A)\lambda(B)|^2 \rightarrow 0$, if and only if the associated operator $U_T : L^2(X) \rightarrow L^2(X)$ has only constant eigenfunctions.

By the spectral theorem, if $\sigma$ is the spectrum of $U_T$, for any $f \in L^2(X)$ there is a measure $\mu \in \mathcal{M}(\sigma)$ for which $\int_\sigma z^n d\mu = < U_T^n f, f>$

What I'm missing from spectral theory; in the converse, if $U_T$ has only constant eigenvalues and $\int_X f = 0$ then the corresponding function $\mu$ is non atomic.

The statement; a transformation $T:X T:(X,\lambda) \rightarrow X$ (X,\lambda)$is weak mixing, i.e.$\frac{1}{N}\sum | \lambda(T^{-n}A \cap B) - \lambda(A)\lambda(B)|^2 \rightarrow 0$if and only if the associated operator$U_T : L^2(X) \rightarrow L^2(X)$has only constant eigenfunctions. By the spectral theorem, if$\sigma$is the spectrum of$U_T$, for any$f \in L^2(X)$there is a measure$\mu \in \mathcal{M}(\sigma)$for which$\int_\sigma z^n d\mu = < U_T^n f, f> $What I'm missing from spectral theory; in the converse, if$U_T$has only constant eigenvalues and$\int_X f = 0$then the corresponding function$\mu\$ is non atomic.