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If a circle-valued sequence is not an equidistributed sequence, some positive-shift difference is not equidistributed either. The contrapositive follows by applying the Van der Corput inequality for finite scalar sequences to every nonzero integer exponential sum. Thus cancellation for all fixed nonzero differences implies equidistribution of the original sequence.

If $a$ is irrational and $b,c$ are real, the fractional parts of $a{n}^2+bn+c$ form an equidistributed sequence. In the Van der Corput lemma, every nonzero-lag correlation of $e^{2\pi im(a{n}^2+bn+c)}$ is a geometric series with irrational frequency $2mah$ and tends to zero. The Weyl criterion then applies.

Let $p,q\ge 2$ be relatively prime integers. If $\nu$ is invariant and ergodic for the integer multiplication map on the circle $T_p$ and $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. For a non-ergodic $T_p$ invariant measure, the same conclusion holds if almost every measure in its ergodic decomposition has positive entropy. Positive entropy of the whole measure alone does not suffice.

A real number $x$ is a normal number in base $K$ exactly when its iterates under the integer multiplication map on the circle, $T_K(x)=\{Kx\}$, form an equidistributed sequence. Braces denote the fractional part. Multiplication by $K$ shifts the positional notation one digit to the left:A length-$r$ word with base-$K$ integer value $a$ specifies the half-open interval $[a/K^r,(a+1)/K^r)$, whose length is $K^{-r}$. Visiting that interval is exactly encountering that word at the next digit position. These digit intervals form arbitrarily fine grids, so their correct limiting visiting frequencies imply equidistribution on every interval. This converts a question about digits into one about a dynamical system. Bailey and Crandall develop this connection using equidistribution of the fractional parts of $K^{n}x$. ^[ref]

In base ten the word $12$ corresponds to $[0.12,0.13)$. For Champernowne constant, counting the visits $T_{10}^n(x)$ for $0\le n\le 99998$ gives precisely the $2088$ occurrences of $12$ in its first $100000$ digits. For $x=1/3$, the base-two integer multiplication map on the circle instead alternates between $1/3$ and $2/3$; it never visits $[0,1/4)$, the interval for $00$. This two-point orbit is not an equidistributed sequence.

Put $e(t)=e^{2\pi it}$ and let $P(n)=\theta n^k+\cdots$ be a real polynomial of degree $k\ge 2$. A standard form of Weyl inequality is: if $(a,q)=1$ and $|\theta -a/q|\le q^{-2}$, then for every $\eta >0$The implied constant is uniform in all lower coefficients. For degree one the finite geometric series bound is $|\sum e(\theta n+\beta )|\le \min (N,(2\Vert \theta \Vert {)}^{-1})$, where $\Vert t\Vert$ is distance to the nearest integer.

Here is the Weyl differencing proof. Extend $z_n=e(P(n))$ by zero outside $[1,N]$. Expanding $|\sum z_n{|}^2$ by differences of indices gives a sum of correlations ${\sum }_n{z}_{n+h}\overline{z_n}$. Repeated Cauchy-Schwarz inequality and this correlation identity yield, after $k-1$ steps, with $p=2^{k-1}$,The interval $I(\mathbf{h})$ ensures that every shifted argument remains in $[1,N]$. For completeness, the induction is the following: at stage $j$, the factor is $N^{2^j-j-1}$ and there is a sum over $j$ shifts. Squaring, Cauchy-Schwarz over those $O(N^j)$ shifts supplies $N^j$; expanding the square of each remaining interval sum introduces the next shift. The new factor is $N^{2^{j+1}-j-2}$, as asserted for stage $j+1$.

The final differenced phase is linear, of slope $k!\theta h_1\cdots h_{k-1}$. Zero products contribute $O_k(N^{k-1})$ to the sum. For nonzero products the finite geometric series bound applies. Grouping by $r=|h_1\cdots h_{k-1}|\le N^{k-1}$, the subpower bound for the divisor function bounds the number of representations by $O_{k,\rho }(N^{\rho })$ for any chosen $\rho >0$. The divisor input is elementary: for $t=k-1$, the number of ordered factorizations of $r$ is ${\prod }_{p^v\parallel r}\left(\genfrac{}{}{0ex}{}{v+t-1}{t-1}\right)$. For all sufficiently large primes this is bounded factorwise by $p^{ϵv}$, and for the finitely many smaller primes the polynomial in $v$ is bounded by a constant times $p^{ϵv}$. Multiplication gives $O_{t,ϵ}(r^{ϵ})$. ThusThe reduced denominator of $k!a/q$ is between $q/k!$ and $q$. Its approximation error is bounded by a constant depending on $k$ times the inverse square of that denominator. Split the $r$-range into blocks of length a sufficiently small fixed multiple of $q$. Within a block the values $k!\theta r$ are separated modulo one by a constant multiple of $1/q$: rational differences have spacing at least $1/q$, while the approximation error in a block difference is smaller than half that spacing. At most a bounded number of values occupy each distance interval of width $1/q$. Summing reciprocals of their ordered distances proves the reciprocal fractional-part sum near a rational estimateFor $q>N^k$, the proposed inequality follows from $|S|\le N$. Otherwise the last display and the differencing estimate giveChoose $\rho$ sufficiently small, absorb the logarithm into a further arbitrarily small power of $N$, and take $p$th roots. This proves the stated Weyl inequality.

For the recurrence assertion, a rational $a=u/v$ gives an integer $a{m}^2$ with $m=v$. For irrational $a$, Weyl differencing implies that $a{m}^2$ is an equidistributed sequence modulo one. One way to avoid a rate issue for unusually good rational approximations is to keep a fixed differencing range $H$: the Van der Corput inequality for finite scalar sequences gives, for each nonzero integer $j$, a limsup bound ${\mathrm{limsup}}_N|N^{-1}{\sum }_{m\le N}e(ja{m}^2){|}^2\ll 1/H$, because every fixed nonzero difference has irrational linear slope and its geometric average tends to zero. Let $H\to \infty$ and apply the Weyl criterion. Hence some $m$ always enters the open $ϵ$-neighborhood of zero.

Finally, $U_m=\{a\in \mathbb{R}/\mathbb{Z}:\Vert a{m}^2\Vert <ϵ\}$ are open sets covering the compact circle group. A finite subcover has a largest index $N$. That single $N$ works for every real $a$. This is uniform square recurrence on the circle; the compactness step gives existence, not an explicit quantitative bound.

Write $e(t)=\exp (2\pi it)$. A sequence in the circle group is an equidistributed sequence if, for every interval $I\subseteq \mathbb{R}/\mathbb{Z}$,where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. Trigonometric polynomials approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant additive characters have integral zero. These facts give the Weyl criterion:This is the link between equidistribution and cancellation in exponential sums.

Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne 0$ and set $z_n=e(m\varphi (n))$. For every fixed $h\ge 1$, the Weyl criterion would giveThe omitted final $h$ terms change a normalized average by at most $h/N$.

Here is the needed Van der Corput inequality for finite scalar sequences. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to $H{\sum }_n{z}_n={\sum }_t{\sum }_{j=0}^{H-1}{z}_{t-j}$ and expand the squared inner sum. Taking $N\to \infty$ first leaves a bound $1/H$; then let $H\to \infty$. Every nonzero Fourier average of $\varphi$ vanishes, so the Weyl criterion makes $\varphi$ equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive $h$, hence for some $h\ne 0$ as requested.

For $\varphi (n)=\sqrt{2}{n}^2$, the difference is ${\Delta }_h\varphi (n)=-2h\sqrt{2}n-h^2\sqrt{2}$. For any nonzero integer $m$, its exponential sum is a constant phase times a geometric progression with ratio $e(-2mh\sqrt{2})\ne 1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt{2})|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves

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 Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the circle group is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive Kolmogorov-Sinai entropy, thenEquivalently, a jointly ergodic common invariant measure other than Lebesgue measure has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\frac{1}{2}m+\frac{1}{2}{\delta }_0$, with ${\delta }_0$ a Dirac measure, is a common invariant measure of positive entropy and is not $m$.

The Host equidistribution theorem states that if $p,q\ge 2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous $g$ on the circle,The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the ergodic decomposition $\nu =\int {\nu }_{\omega }d\lambda (\omega )$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_{\nu }(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the Rudolph measure rigidity theorem, suppose $h_{\mu }(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ ergodic decomposition as $\mu =\int \nu d\tau (\nu )$. Since $T_3$ commutes with $T_2$, its pushforward measure sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3{)}_{\ast }\nu$. On each component, $T_3$ is a factor of a measure-preserving system with fibres of size at most three. We use the standard entropy preservation under a finite-to-one factor:The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log 3$, and division by the orbit length gives zero relative entropy.

The component at $T_{3}x$ is $(T_3{)}_{\ast }{\nu }_x$ almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{{\nu }_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as $h_{\mu }(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so $\mu =m$, proving the deduction.

For the normal-number example, let ${\omega }_1,{\omega }_2,\dots$ be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set $C$. If $S$ is the Bernoulli shift, then $T_3\circ \pi =\pi \circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit measurable partition $\eta =\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. HenceApply the Host equidistribution theorem with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^{n}x)$ equidistributes for Lebesgue measure, so $x$ is a normal number in base $2$ by normality and equidistribution under integer multiplication.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a normal number in base $3$. We have proved the stronger almost-everywhere existence statement

For rational $a$, a multiple of its denominator gives exact integer recurrence. For irrational $a$, Weyl differencing and the Weyl criterion give an equidistributed sequence $a{m}^2$ modulo one. Thus every parameter belongs to one of the open sets $\{a:\Vert a{m}^2\Vert <ϵ\}$. Compactness of the circle group gives a finite subcover, hence a bound $N$ uniform in $a$; no explicit rate follows from this compactness step.

A sequence $(x_n)$ in $\mathbb{R}/\mathbb{Z}$ is an equidistributed sequence if and only if $N^{-1}{\sum }_{n=0}^{N-1}{e}^{2\pi im{x}_n}\to 0$ for every nonzero integer $m$. Approximation of continuous functions by trigonometric polynomials proves the criterion.