Weak mixing is equivalent to absence of nonconstant eigenfunctions

If $f$ is a nonconstant eigenfunction, center it when its eigenvalue is one; when the eigenvalue differs from one its mean already vanishes. This gives a nonzero $f_0\in H_0$ with $U^n{f}_0={\lambda }^n{f}_0$ and $\mathrm{\mid }\lambda \mathrm{\mid }=1$. Thus $\mathrm{\mid }⟨U^n{f}_0,f_0⟩\mathrm{\mid }=\mathrm{\parallel }{f}_0{\mathrm{\parallel }}^2>0$ for every $n$. Its absolute-Cesaro average cannot tend to zero. F8–F9 therefore prove (i) implies (ii).

On every measurable rectangle $E\times F$, the product map $S=T\times T$ satisfies $(\mu \times \mu )(S^{-1}(E\times F))=\mu (T^{-1}E)\mu (T^{-1}F)=\mu (E)\mu (F)\mathrm{.}$ The inverse images of rectangles are measurable, so the sets whose inverse images are product measurable form a sigma-algebra; hence $S$ is measurable because rectangles generate the product sigma-algebra. Rectangles together with $X^2$ form a generating pi-system, and F12 makes $S$ measure preserving on the uncompleted product. If $A\cup N$ is completed measurable as in F14, with $N\subseteq Z$ and $(\mu \times \mu )(Z)=0$, then $S^{-1}N\subseteq S^{-1}Z$ and the latter is a base null set. Thus $S^{-1}(A\cup N)$ is completed measurable and has the same completed measure as $A\cup N$. Therefore $S$ preserves the completed product, and F13 makes its Koopman operator $W$ a complex $L^2$ isometry.

Assume (ii). Every invariant indicator is an eigenfunction of eigenvalue one unless zero, so it is constant a.e.; its set is null or conull. Thus $T$ is ergodic. If $Wk=k$ and $k$ is nonconstant, subtract its total mean to make $\iint k=0$, still with $k\mathrm{\ne }0$. F15 gives a nonzero compact kernel operator $K$, while F4 gives $KU=UK$ and $K^{\ast }U=U{K}^{\ast }$. Since $U\mathbf{1}=\mathbf{1}$, the images $K\mathbf{1}$ and $K^{\ast }\mathbf{1}$ are invariant $L^2$ functions. Ergodicity makes both constant. Their means are $\iint k=0$ and its conjugate, respectively, by Fubini; hence both are zero. These are exactly the two zero-marginal conditions. F5 and F6 would then produce a nonconstant eigenfunction, contradicting (ii). Every $W$-fixed $L^2$ class is consequently constant. In particular each product-invariant indicator is constant, so (iii) follows. The kernel is integrable because its $L^2$ norm is finite and the product mass is one.

Assume (iii). If a nonconstant eigenfunction existed, center it as in step 1.1 to obtain $f_0\mathrm{\ne }0$ of mean zero. Put $F(x,y)=f_0(x)\overline{f_0(y)}$. Tonelli gives $\mathrm{\parallel }F{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_2^4>0$, and Fubini gives $\iint F=\mathrm{\mid }\int f_0{\mathrm{\mid }}^2=0$. Its pullback is $WF=\mathrm{\mid }\lambda {\mathrm{\mid }}^{2}F=F$; factor-null exceptional sets pull back to null subsets of the square by the completed-product preservation in step 1.2. Product ergodicity and F7 force $F$ to be constant, and its zero mean forces that constant to be zero, a contradiction. Thus (iii) implies (ii).

Under (iii), F7 says the fixed space of $W$ consists exactly of constants. For arbitrary $f,g\in L^2$, let $f_0=f-(\int f)\mathbf{1}$, $F=f_0\otimes \overline{f_0}$ and $G=g\otimes \bar{g}$. As in step 2.2, these are in product $L^2$ and $F$ has integral zero. Its projection onto constants is therefore zero. Since step 1.2 proves that $W$ is an isometry, F2 yields $N^{-1}{\sum }_{n<N}{W}^{n}F\to 0$ in norm, so its pairing with $G$ tends to zero by Cauchy–Schwarz. Fubini computes $⟨W^{n}F,G⟩=\mathrm{\mid }⟨U^n{f}_0,g⟩{\mathrm{\mid }}^2$: each factor $f_0\circ T^n\bar{g}$ is integrable by Cauchy–Schwarz, and the absolute double integral is the square of its finite $L^1$ norm.

Finite Cauchy–Schwarz on the $N$ real nonnegative numbers $b_n=\mathrm{\mid }⟨U^n{f}_0,g⟩\mathrm{\mid }$ gives $N^{-1}{\sum }_{n<N}{b}_n\le (N^{-1}{\sum }_{n<N}{b}_n^2{)}^{1\mathrm{/}2}\to 0$. The correlation here equals $C_n(f,g)$ of F8 because $U\mathbf{1}=\mathbf{1}$. Thus F8 proves (i). Together with steps 1.1, 2.1 and 2.2 this closes all three implications. If $f_0=0$ or $g=0$, both sides of the estimate are zero. The assumed AC supplies every cited projection, compactness and completion result that requires it; no spectral-measure construction is used.

For completeness let $a_n=\mathrm{\mid }{C}_n(f,g)\mathrm{\mid }$, bounded by $M=\mathrm{\parallel }f-(\int f)\mathbf{1}{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$. If its Cesaro mean tends to zero, each $E_j=\{n:a_n>1\mathrm{/}j\}$ has density zero since $\mathrm{\mid }{E}_j\cap [0,N)\mathrm{\mid }\mathrm{/}N\le j{N}^{-1}{\sum }_{n<N}{a}_n$. Choose $N_j$ recursively as the least integer larger than $N_{j-1}$ such that this density is below $1\mathrm{/}j$ for every $N\ge N_j$; the convergence just proved ensures existence. Put $D={\bigcup }_{j\ge 1}(E_j\cap [N_j,N_{j+1}))$. For $N_j\le N<N_{j+1}$, the increasing property $E_i\subseteq E_j$ for $i\le j$ implies $D\cap [0,N)\subseteq E_j\cap [0,N)$, so $D$ has density zero. Off $D$ on its $j$th interval, $a_n\le 1\mathrm{/}j$, proving the claimed convergence. Conversely, if $D$ has density zero and $a_n\to 0$ off $D$, choose $N_0$ so that $a_n<\epsilon$ off $D$ for $n\ge N_0$. Then $N^{-1}{\sum }_{n<N}{a}_n\le M{N}_0\mathrm{/}N+M\mathrm{\mid }D\cap [0,N)\mathrm{\mid }\mathrm{/}N+\epsilon$. Its limsup is at most every positive $\epsilon$, hence zero. This proves the pairwise density-zero equivalent criterion without asserting one exceptional set for all uncountably many pairs. Least integer cutoffs add no choice use.