If is irrational and are real, the fractional parts of form an equidistributed sequence. In the Van der Corput lemma, every nonzero-lag correlation of is a geometric series with irrational frequency and tends to zero. The Weyl criterion then applies.
Host equidistribution theorem 2026-10-05
Let be relatively prime integers. If is invariant and ergodic for the integer multiplication map on the circle and , then for -almost every the sequence is an equidistributed sequence for Lebesgue measure. For a non-ergodic 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 number is a normal number in base exactly when is an equidistributed sequence. A word of length corresponds to a half-open interval of length ; these intervals form arbitrarily fine grids, which approximate any interval from inside and outside.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 108 2 Solution Created 2026-10-03 Updated 2026-10-05
For an increasing sequence of nonnegative integers, full-density sequence means that its range has natural density one:For a sequence in a metric space, convergence in density of a sequence to means that, for every ,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most plus its bound times the proportion of indices above .
Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose , and putThenSince , it suffices for to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed .
To prove the lemma, extend the finite list by zero outside this range, and write . For ,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient counts the pairs of shifts separated by . Dividing by , bounding real parts by absolute values, and first taking the limit superior in for fixed yieldsThe right side tends to zero as under the stated hypothesis. This proves both the lemma and its density consequence.
For the polynomial application, fix a nonzero integer and let . For each fixed , the correlation phase is linear:Thus, with ,The coefficient is irrational, so . This geometric series is bounded in modulus by and tends to zero. All 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.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 108 4 Solution Created 2026-10-03 Updated 2026-10-05
The Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure on the circle group is invariant under both and , 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 under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: , with a Dirac measure, is a common invariant measure of positive entropy and is not .
The Host equidistribution theorem states that if are relatively prime integers and is invariant and ergodic under , with , then for -almost every the sequence is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous on the circle,The non-ergodic form assumes invariance and positive entropy for almost every component in the ergodic decomposition . Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for . More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of alone does not eliminate zero-entropy components.
To deduce the joint version of the Rudolph measure rigidity theorem, suppose ; if only has positive entropy, interchange the roles. Write the ergodic decomposition as . Since commutes with , its pushforward measure sends a ergodic component to a ergodic component . On each component, 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 , and division by the orbit length gives zero relative entropy.
The component at is almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function is therefore invariant under both and . Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as . Thus almost every component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that -almost every point equidistributes for under .
For any continuous , invariance under and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so , proving the deduction.
For the normal-number example, let be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set . If is the Bernoulli shift, then on the circle. Consequently is invariant and ergodic: a invariant event pulls back to an invariant event, which has probability zero or one.
Take the ternary digit measurable partition . Its block partition of length has, up to null endpoints, positive-measure atoms under , each of measure , and all other atoms have measure zero. HenceApply the Host equidistribution theorem with , . For -almost every , the sequence equidistributes for Lebesgue measure, so is a normal number in base by normality and equidistribution under integer multiplication.
The ternary expansion of -almost every such contains only and , so the frequency of digit is zero rather than . The ambiguous ternary endpoints form a countable null set and can be removed. Therefore is not a normal number in base . We have proved the stronger almost-everywhere existence statement
Weyl criterion 2026-10-05
A sequence in is an equidistributed sequence if and only if for every nonzero integer . Approximation of continuous functions by trigonometric polynomials proves the criterion.