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An increasing sequence of integers has full density when its range has natural density one. Convergence in density of a sequence means that, for each fixed tolerance, the exceptional indices have natural density zero.

The displayed condition implies convergence in density of a sequence to $\ell$, since the fraction of indices with $|a_n-\ell |\ge \epsilon$ is at most ${\epsilon }^{-2}$ times the averaged squared error. A signed Cesaro limit alone is insufficient: $a_n=(-1{)}^n$ has zero signed average but no density limit.

For a probability measure-preserving system, weak mixing is the vanishing of averaged absolute correlation discrepancies:By approximation with simple functions and the Cauchy-Schwarz inequality, this is equivalent toHere $⟨f,g⟩=\int f\overline{g}d\mu$, and $U=U_T$. Absolute values are part of the definition: signed Cesaro convergence of a sequence of these correlations alone expresses ergodicity, not weak mixing.

Suppose the product system is ergodic. Fix a mean-zero $f\in L^2(\mu )$ and any $g\in L^2(\mu )$. On the product takeThese belong to $L^2(\mu \otimes \mu )$, and $\int F=|\int f{|}^2=0$. The mean ergodic theorem on the product, followed by pairing with $G$, givesThe Cauchy-Schwarz inequality in $n$ bounds the averaged absolute correlation by the square root of this quantity. Subtract the mean from a general $f$ to obtain the definition above. This proves product ergodicity implies weak mixing, through the square-correlation proof of weak mixing from product ergodicity.

For the multiple averages, write $m(f)=\int fd\mu$ and use the following Van der Corput lemma. If $(v_n)$ is bounded in a Hilbert space, every correlation averageexists, and $H^{-1}{\sum }_{h=1}^H|L_h|\to 0$, then $N^{-1}{\sum }_{n<N}{v}_n\to 0$ in norm. One way to see the estimate is to replace $v_n$ by $H^{-1}{\sum }_{j=0}^{H-1}{v}_{n+j}$; for fixed $H$ the change in its long average tends to zero. The Cauchy-Schwarz inequality and expansion of the squared block norm giveLet $H\to \infty$. This proves the auxiliary implication needed here.

A weak mixing system is ergodic: a mean-zero invariant $f$ would have the nonvanishing correlation $⟨U^{n}f,f⟩=\Vert f{\Vert }_2^2$. Every positive power $T^r$ is also weak mixing, since for nonnegative correlation discrepancies $b_n$,This part of the stability of weak mixing under powers and products will control the induction.

In fact the stronger arithmetic-progression multiple averages under weak mixing holds:For $k=1$ this is the mean ergodic theorem and ergodicity. Suppose it is known for $k-1$ and first assume $m(f_k)=0$. Put $v_n={\prod }_{j=1}^k{U}^{jn}{f}_j$. For fixed $h$, setUsing invariance of the integral to remove the common $U^n$ givesThis identity does not require an inverse of $T$. Apply the induction hypothesis to the $k-1$ factors and pair their $L^2$ limit with $g_{1,h}$. It follows thatThe average in $h$ of the right side tends to zero because $T^k$ is weak mixing and $m(f_k)=0$. The Van der Corput lemma proves the zero $L^2$ limit. For general $f_k$, split it into $f_k-m(f_k)$ and its constant mean; the first term has the zero limit just proved, and the other term is $m(f_k)$ times the induction average for $k-1$. This completes the induction.

Pair this result with the real bounded $f_0$. Forthe requested Cesaro limit is

To obtain convergence in density of a sequence, we also need the product system to be weak mixing. For simple tensors, its correlations are products of single-system correlations. If $c_n\to c$ and $d_n\to d$ in averaged absolute discrepancy and both sequences are bounded, thenso the product discrepancy has zero average. Finite sums of tensors are dense in $L^2(\mu \otimes \mu )$, and the Cauchy-Schwarz inequality extends the conclusion to arbitrary $L^2$ functions. Hence $T\times T$ is weak mixing.

Apply the multiple-average result to $F_j=f_j\otimes f_j$ on that product. Because the original $f_j$ are real,Together with the first-moment limit this givesThe mean-square criterion for convergence in density now yields, for every $\epsilon >0$,Therefore the density convergence of multiple weak-mixing correlations gives the same answer:The second-moment argument is essential; the signed Cesaro limit alone would not imply this conclusion.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

The statement $D-lim{a}_n=a$ means convergence in density of a sequence: for every $\epsilon >0$, the proportion of $n\le N$ with $|a_n-a|\ge \epsilon$ tends to zero. The statement $C-lim{a}_n=a$ means Cesaro convergence of a sequence, $N^{-1}{\sum }_{n=1}^N{a}_n\to a$.

Suppose $|a_n-a|\le M$. Splitting into the indices where $|a_n-a|<\epsilon$ and its complement givesThus density convergence implies Cesaro convergence of the absolute deviations to zero. Conversely, the Markov inequality givesproving 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 $U_{T}f=\lambda f$. Unitarity of the Koopman operator gives $|\lambda |=1$. Weak mixing implies ergodicity, so $|f|$ is almost everywhere constant. On the product,is invariant under $T\times T$. Since the square is ergodic, $F$ is constant, which forces $f$ to be constant almost everywhere. Thus 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.

For a bounded sequence $(u_n)$ in a Hilbert space, put $c_h={\mathrm{limsup}}_{N\to \infty }|N^{-1}{\sum }_{n=0}^{N-h-1}⟨u_{n+h},u_n⟩|$. If $H^{-1}{\sum }_{h=1}^H{c}_h\to 0$, then $N^{-1}{\sum }_{n=0}^{N-1}{u}_n\to 0$ in norm. Since the $c_h$ are bounded, it suffices that $(c_h)$ has convergence in density of a sequence to zero. The proof applies the Cauchy-Schwarz inequality to an average of shifted sums.

A probability measure-preserving transformation $T$ is weakly mixing when, for every pair of measurable sets $A,B$, the average absolute correlation discrepancy tends to zero:Equivalently, the correlations converge to their products in convergence in density of a sequence, its Cartesian product with itself is an ergodic transformation, or its product with every ergodic transformation is ergodic. The absolute values are essential: signed Cesaro convergence of a sequence of the correlations to their products characterizes ergodic transformations and does not imply weak mixing. For example, the transformation interchanging two equally weighted points has discrepancy $(-1{)}^n/4$ for a singleton $A=B$: its signed average tends to zero while its average absolute discrepancy stays $1/4$.