Differencing obstruction to equidistribution 2026-10-07
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.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 11 2 Solution Created 2026-10-03 Updated 2026-10-07
Write . A sequence in the circle group is an equidistributed sequence if, for every interval ,where 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 and set . For every fixed , the Weyl criterion would giveThe omitted final terms change a normalized average by at most .
Here is the needed Van der Corput inequality for finite scalar sequences. Extend by zero outside and average consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to and expand the squared inner sum. Taking first leaves a bound ; then let . Every nonzero Fourier average of vanishes, so the Weyl criterion makes equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive , hence for some as requested.
For , the difference is . For any nonzero integer , its exponential sum is a constant phase times a geometric progression with ratio . Its normalized magnitude is at most , which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 79 2 Solution Created 2026-10-03 Updated 2026-10-07
There is a minor range issue in the printed bound: for and positive , the upper bound is below one. We prove the intended result for ; a valid formulation for every replaces the upper bound by . In fact the argument below gives . All auxiliary estimates are proved here.
For , let , and use the Fejér kernelExpanding the square proves the identity and nonnegativity. The finite geometric series formula and give off the integers. In particular whenever .
Suppose, for a contradiction, that no has . Summing the Fejér kernel along the quadratic sequence and separating its constant term givesThe coefficients on the right sum to . Also . Hence some has , where .
We need only an elementary Van der Corput inequality for finite scalar sequences. For , extended by zero outside , each term of occurs in exactly windows of length . Applying the Cauchy-Schwarz inequality to the window sums and expanding their squares gives, for ,Consequently . Take . If , some must have ; otherwise the displayed upper bound is less than .
For the large quadratic exponential sum just found, put . Its quadratic exponential sum has multiplicative derivativeThus is a finite geometric series. Its absolute value is at most , unless that distance is zero, in which case the desired estimate is automatic. It follows thatSet . The distance to the nearest integer satisfies for a positive integer , by multiplying a nearest integer to . ThereforeThis proves the needed quantitative quadratic recurrence once is chosen polynomially in .
For explicit bookkeeping, and . Choose . For we have and , while . Hence and , contradicting our supposition. The estimates have substantial slack even at .
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 108 3 Solution Created 2026-10-03 Updated 2026-10-06
A continuous map 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 on , and let be any invariant Borel probability measure. For each integer , set . Invariance givesFor , irrationality forces , so the corresponding Fourier coefficient is zero. For it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure . By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence and integrate every continuous function equally and are the same Borel probability measure. Since is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure .
For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with modulo one, and it preserves as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.
Let satisfy . Its expansion in the Fourier basis is in . Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor , the magnitudes of the Fourier coefficients along all distinct indices , , are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When , the relation becomes , which forces for . Only remains. Thus every invariant 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 , using the old first coordinate in the second coordinate of , yieldsIn particular, for a Fourier basis element,If and , this is a geometric series in , anduniformly in .
For , we give the finite Van der Corput inequality for finite scalar sequences and its proof. For , extend by zero outside . Fix an integer and set . Every original summand appears 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 with fixed, followed by .
For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed , is irrational, so another geometric series estimate givesuniformly in . With fixed , the Van der Corput inequality for finite scalar sequences therefore impliesLetting 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 . If approximates a continuous with , thenTaking and then proves uniform convergence to for every continuous .
Finally, if is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus , 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.
Quadratic exponential sum 2026-10-07
A quadratic exponential sum has a polynomial phase of degree two. Its multiplicative derivative at lag has the linear phase 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.
Quantitative quadratic recurrence 2026-10-07
For a universal constant and , every real admits with distance to the nearest integer of less than . 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 , use the bound .
For an irrational skew shift, the averages of every continuous function converge uniformly in the starting point to , with 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 .