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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.

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.

We give a spectral proof of square-difference recurrence for finite colourings, proving the auxiliary results it uses. Extend the colouring arbitrarily to nonpositive integers, and let $f_j$ be the indicator of colour $j$. By successively taking subsequences of the interval lengths, followed by a diagonal subsequence, we may arrange that all the limitsexist, for every integer $h$. This uses only sequential compactness of bounded real sequences: list the countably many quantities, choose a convergent subsequence for each in turn, and take the diagonal. All limits below are along this common subsequence. We have $c(0)=1$, $c(-h)=c(h)$, $c(h)\ge 0$ and ${\sum }_j{\delta }_j=1$. Translating a fixed interval endpoint changes a normalized average by $O(|h|/L)$.

The sequence is a positive-definite function on $\mathbb{Z}$. Indeed, for finitely many complex numbers $z_u$,We next prove the required Bochner–Herglotz theorem for this sequence. The trigonometric polynomialis nonnegative by positive definiteness, and its integral on the circle is $c(0)=1$. Thus $F_H(t)dt$ is a probability measure. Such measures on a compact interval have a weakly convergent subsequence. For completeness, take a diagonal subsequence on all rational points of their nondecreasing distribution functions. The right-continuous extension of the limiting rational values is a distribution function and defines a probability measure by its interval increments. At its continuity points the original distribution functions converge, by sandwiching between rational endpoints. Approximating continuous functions by step functions with such endpoints proves convergence of their integrals. Identify the interval endpoints to obtain a measure on the circle. This proves the compactness of probability measures on a compact metric space in the case needed here.

For a fixed $h$, the Fourier coefficient of $F_{H}dt$ is $(1-|h|/H)c(h)$ once $H>|h|$. Taking the weak limit therefore supplies a positive probability measure $\mu$ such thatNo spectral representation has been assumed without proof.

Its atom at zero has positive mass. For each $H$,The last inequality is Cauchy-Schwarz on each colour's interval average. The integrand is at most one and tends to zero for every nonzero point of the circle, by the geometric-series formula, while it is one at zero. Bounded convergence gives

We also prove that quadratic phases with an irrational coefficient average to zero. For $z_0,\dots ,z_{L-1}$ bounded by one, extended by zero, Cauchy-Schwarz applied to ${\sum }_x{\sum }_{u=0}^{H-1}{z}_{x+u}=H{\sum }_x{z}_x$ yieldsThis identity follows by expanding the square of the inner shifted sum and grouping pairs by their difference; it is the needed finite Van der Corput inequality for finite scalar sequences. If $z_x=e(\theta x^2)$ and $\theta$ is irrational, each fixed nonzero lag correlation is a geometric sum with frequency $2\theta h$, hence is bounded independently of $L$. The geometric bound follows from $(1-z^L)/(1-z)$ with $z=e(2\theta h)\ne 1$. Dividing by $H^2{L}^2$ and letting $L$ tend to infinity gives $\mathrm{limsup}|L^{-1}\sum z_x{|}^2\le 1/H$. Letting $H$ tend to infinity proves the asserted cancellation. Rational coefficients instead give periodic sequences and therefore have convergent averages.

Choose a finite set $T$ of rational points of the circle, including zero, such that the total mass of the remaining rational atoms is below $1/(2r)$. This is possible because the rational points are countable and $\mu$ is finite. Let $q$ be a common multiple of their denominators. For $t\in T$, $e(t{q}^2{m}^2)=1$ for every integer $m$. For irrational $t$ the preceding cancellation applies to $q^{2}t$; at every other rational point the average exists by periodicity and has modulus at most one. Bounded convergence consequently givesHere the inequality takes real parts; $c$ itself is real. Thus some $b=qm>0$ has $c(b^2)>0$. By its definition as a limit of nonnegative averages, at least one actual positive $a$ has ${\sum }_j{f}_j(a)f_j(a+b^2)>0$. The same colour occurs at both points, provingThe arbitrary extension to nonpositive integers played no role in this final pair.

Write $e(t)=e^{2\pi it}$. We first prove the needed cancellation of quadratic exponential sums, rather than invoke equidistribution without proof. The number $\sqrt{2}$ is irrational: writing it as a reduced fraction would make its numerator even from $u^2=2v^2$, and then its denominator even as well, a contradiction. Thus $2hr\sqrt{2}$ is irrational for nonzero integers $h,r$.

Fix $h\ne 0$ and set $z_n=e(h\sqrt{2}{n}^2)$ for $1\le n\le L$, extending it by zero otherwise. For integers $L\ge H\ge 1$, each term appears exactly $H$ times in the shifted sum, soFor any complex numbers $w_1,\dots ,w_J$, nonnegativity of ${\sum }_j|w_j-J^{-1}{\sum }_i{w}_i{|}^2$ gives $|{\sum }_j{w}_j{|}^2\le J{\sum }_j|w_j{|}^2$. Apply this special Cauchy-Schwarz inequality to the displayed outer sum, and then expand the inner squared modulus. This givesFor fixed $H$ and large $L$, each correlation is the geometric serieswhose absolute value is at most $2/|1-e(2hr\sqrt{2})|$, independently of $L$. This follows by multiplying the geometric sum by $1-e(2hr\sqrt{2})$; its denominator is nonzero by irrationality. Divide the inequality by $L^2$ and let $L\to \infty$, keeping $H$ fixed. It gives a limit superior at most $1/H$. Taking arbitrarily large $H$ provesThis is the quadratic instance of the Van der Corput inequality for finite scalar sequences, with its proof supplied here.

To turn cancellation into the particular approximation, put $\epsilon =10^{-8}$ and choose an integer $M>2/{\sin }^2(\pi \epsilon )$. The Fejér kernel is the explicit finite trigonometric polynomialThe expansion follows by counting pairs of indices with difference $h$. Its constant coefficient is one, so the cancellation just proved givesIf $\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \epsilon$, the geometric-series formula givesFor all sufficiently large $L$, the displayed average exceeds $1/2$. Consequently at least one $n\in \{1,\dots ,L\}$ satisfiesIt is a positive integer, as required. Every auxiliary cancellation and kernel identity used in this existence proof has been derived; a smooth cutoff or an unproved Weyl criterion is unnecessary.

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

There is a minor range issue in the printed bound: for $\epsilon >1$ and positive $C$, the upper bound is below one. We prove the intended result for $0<\epsilon \le 1$; a valid formulation for every $\epsilon >0$ replaces the upper bound by $\max \{1,{\epsilon }^{-C}\}$. In fact the argument below gives $C=32$. All auxiliary estimates are proved here.

For $0<\epsilon \le 1/2$, let $m=\lceil {\epsilon }^{-2}\rceil$, and use the Fejér kernelExpanding the square proves the identity and nonnegativity. The finite geometric series formula and $|\sin \pi t|\ge 2\Vert t\Vert$ give $F_m(t)\le (4m\Vert t{\Vert }^2{)}^{-1}$ off the integers. In particular $F_m(t)\le 1/4$ whenever $\Vert t\Vert \ge \epsilon$.

Suppose, for a contradiction, that no $1\le n\le N$ has $\Vert \alpha n^2\Vert <\epsilon$. Summing the Fejér kernel along the quadratic sequence and separating its constant term givesThe coefficients on the right sum to $m-1$. Also $|S_{-k}|=|S_k|$. Hence some $1\le k<m$ has $|S_k|\ge \eta N$, where $\eta =1/(2m)$.

We need only an elementary Van der Corput inequality for finite scalar sequences. For $|z_n|\le 1$, extended by zero outside $[N]$, each term of $S={\sum }_n{z}_n$ occurs in exactly $H$ windows of length $H$. Applying the Cauchy-Schwarz inequality to the window sums and expanding their squares gives, for $H\le N$,Consequently $|S{|}^2\le 2N^2/H+(4N/H){\sum }_{h=1}^{H-1}|C_h|$. Take $H=\lceil 8{\eta }^{-2}\rceil$. If $|S|\ge \eta N$, some $1\le h<H$ must have $|C_h|\ge {\eta }^{2}N/8$; otherwise the displayed upper bound is less than $3{\eta }^2{N}^2/4$.

For the large quadratic exponential sum just found, put $z_n=e(k\alpha n^2)$. Its quadratic exponential sum has multiplicative derivativeThus $C_h$ is a finite geometric series. Its absolute value is at most $1/(2\Vert 2kh\alpha \Vert )$, unless that distance is zero, in which case the desired estimate is automatic. It follows thatSet $n_{\ast }=2kh$. The distance to the nearest integer satisfies $\Vert \ell x\Vert \le \ell \Vert x\Vert$ for a positive integer $\ell$, by multiplying a nearest integer to $x$. ThereforeThis proves the needed quantitative quadratic recurrence once $N$ is chosen polynomially in ${\epsilon }^{-1}$.

For explicit bookkeeping, $m\le 2{\epsilon }^{-2}$ and $H\le 33m^2$. Choose $N=\lfloor {\epsilon }^{-32}\rfloor$. For $0<\epsilon \le 1/2$ we have $N>33792{\epsilon }^{-11}\ge 8mH/({\eta }^2\epsilon )$ and $N\ge H$, while $2mH\le 528{\epsilon }^{-6}\le N$. Hence $n_{\ast }\le N$ and $\Vert \alpha n_{\ast }^2\Vert <\epsilon$, contradicting our supposition. The estimates have substantial slack even at $\epsilon =1/2$.

For $1/2<\epsilon \le 1$, simply take $n=1$, since $\Vert \alpha \Vert \le 1/2<\epsilon$ and $1\le {\epsilon }^{-32}$. For the intended range, the conclusion isFor $\epsilon >1$, the same choice $n=1$ proves the corrected all-range version.

A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.

Finally, if $\nu$ is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus $\nu =m_2$, since continuous functions determine Borel probability measures on a compact metric space. We concludeThe uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.

A quadratic exponential sum has a polynomial phase of degree two. Its multiplicative derivative at lag $h$ has the linear phase $2\alpha hn$ up to a constant factor. This degree reduction lets the Van der Corput inequality for finite scalar sequences reduce its size to estimates for finite geometric series.

For a universal constant $C>0$ and $0<\epsilon \le 1$, every real $\alpha$ admits $1\le n\le {\epsilon }^{-C}$ with distance to the nearest integer of $\alpha n^2$ less than $\epsilon$. A Fejér kernel detects failure of recurrence as a large quadratic exponential sum. The Van der Corput inequality for finite scalar sequences then produces a short linear near-return, whose suitable multiple gives the quadratic return. To include $\epsilon >1$, use the bound $\max \{1,{\epsilon }^{-C}\}$.

For an irrational skew shift, the averages of every continuous function $g$ converge uniformly in the starting point to $\int gd{m}_2$, with $m_2$ normalized Lebesgue measure. Nonconstant Fourier basis characters have either linear or quadratic phases. Linear phases are bounded geometric series; for quadratic phases the Van der Corput inequality for finite scalar sequences reduces to linear correlations of irrational frequency, uniformly in the starting point. Approximation by trigonometric polynomials proves the assertion and identifies every invariant Borel probability measure as $m_2$.