Bochner–Herglotz theorem 2026-10-14
A positive-definite function on the discrete group is the sequence of Fourier coefficients of a unique finite positive Borel measure on the circle, with total mass . Positivity makes the finite sums nonnegative. Their measures converge weakly to the required measure; uniqueness follows from density of trigonometric polynomials in continuous functions.
Bochner's theorem 2026-10-14
For a locally compact Hausdorff Abelian group, a continuous positive-definite function is uniquely representable as for a finite positive Radon measure on the Pontryagin dual group. Its total mass is . Conversely every such finite positive measure defines a continuous positive-definite function.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 11 4 Solution Created 2026-10-03 Updated 2026-10-14
For a locally compact Hausdorff Abelian group , Bochner's theorem states that a continuous positive-definite function has a unique representationwhere is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.
First prove existence without presupposing the representation. The elementary consequences of positive definiteness give , and . If , then and take . Otherwise form finite formal linear combinations of symbols , with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space . Since is Abelian, translation preserves the form and extends to a unitary operator. Its inverse is , so is a unitary representation. With ,The translated copies of span a dense subspace. MoreoverFinite sums and density, together with , give strong continuity on all of . This constructs the cyclic unitary representation of a positive-definite function directly.
For define the integrated unitary representation , with the integral applied to each vector. Its norm is at most . Fubini gives , and inversion in the Abelian group gives . Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every . These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.
The Commutative Gelfand--Naimark theorem identifies with , where is its character space. The positive vector functional becomes a bounded positive functional on , so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.
For each , the functional is a nonzero algebra character of , since is dense in . By Question 3 it has the form . The resulting map is continuous: pointwise convergence of on gives convergence on all , and the topological character identification in Question 3 gives compact-open convergence of . Push forward under to obtain . This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.
To recover , note that . HenceThe left side tends to . On the right, and uniformly on compact subsets of . For completeness, the evaluation is jointly continuous: near restrict to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at , proving the stated uniform convergence. Continuity of gives uniform convergence on compact subsets of , and the finite Radon measure makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking also gives the required total mass .
Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.
Finally suppose two finite Radon measures have the same transform. Multiply by and integrate over . Fubini shows that the measures agree on every function . Question 3 showed that these functions are uniformly dense in . Boundedness of both measure functionals extends equality to all of ; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular corresponds exactly to a probability measure on the dual group.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 25 1 Solution Created 2026-10-03 Updated 2026-10-14
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 be the indicator of colour . By successively taking subsequences of the interval lengths, followed by a diagonal subsequence, we may arrange that all the limitsexist, for every integer . 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 , , and . Translating a fixed interval endpoint changes a normalized average by .
The sequence is a positive-definite function on . Indeed, for finitely many complex numbers ,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 . Thus 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 , the Fourier coefficient of is once . Taking the weak limit therefore supplies a positive probability measure such thatNo spectral representation has been assumed without proof.
Its atom at zero has positive mass. For each ,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 bounded by one, extended by zero, Cauchy-Schwarz applied to 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 and is irrational, each fixed nonzero lag correlation is a geometric sum with frequency , hence is bounded independently of . The geometric bound follows from with . Dividing by and letting tend to infinity gives . Letting tend to infinity proves the asserted cancellation. Rational coefficients instead give periodic sequences and therefore have convergent averages.
Choose a finite set of rational points of the circle, including zero, such that the total mass of the remaining rational atoms is below . This is possible because the rational points are countable and is finite. Let be a common multiple of their denominators. For , for every integer . For irrational the preceding cancellation applies to ; 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; itself is real. Thus some has . By its definition as a limit of nonnegative averages, at least one actual positive has . The same colour occurs at both points, provingThe arbitrary extension to nonpositive integers played no role in this final pair.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 30 3 Solution Created 2026-10-03 Updated 2026-10-14
The discrete Bochner–Herglotz theorem says that a sequence is a positive-definite function, meaningif and only if there is a unique finite positive Borel measure on such thatThe mass is . The necessity of positive definiteness follows by integrating .
For sufficiency, positive definiteness makes the relevant Gram matrices Hermitian positive semidefinite, so and . DefineTaking in the positive-definiteness condition gives , while . Moreover,Thus the integrals against every trigonometric polynomial converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by Fourier inversion for smooth functions gives this density. The uniform mass bound then makes the integrals against every continuous converge as well. Their limit is a positive linear functional with . The standard Riesz-Markov-Kakutani representation theorem gives a finite positive Borel measure with . The preceding coefficient computation proves its required Fourier coefficients. Finally, two such measures agree on all trigonometric polynomials, hence on all continuous functions by uniform density, and therefore are identical. This proves the Bochner–Herglotz theorem, including existence and uniqueness; if , the estimate gives the zero measure directly.
Now let be the Koopman operator on the complex L2 space of the probability measure-preserving system. We use , linear in the first entry. Since is an isometry, the sequence for , extended by , is positive-definite: its matrix is the Gram matrix of . The Bochner–Herglotz theorem gives its spectral measure of a Koopman observable .
We first prove exactly how an atom of a measure detects an eigenfunction. Fix and setThe mean ergodic theorem, applied to the isometry , says these averages converge in L2 space to the orthogonal projection onto . Its usual Hilbert-space proof works without invertibility: the orthogonal complement of is , and the averages of telescope to . The equality of the two kernels follows from being an isometry, by expanding when .
Expanding the averaged norm through the spectral measure of a Koopman observable givesThe integrand is at most one and converges to . By the dominated convergence theorem,Thus a nonzero atom corresponds to a nonzero Koopman operator eigenfunction, with eigenvalue , in the cyclic spectral data of .
We also need the Wiener Fourier coefficient lemma, which here has a short proof. For any finite positive circle measure with coefficients ,The bracket is bounded in absolute value by one and tends to the indicator of the diagonal . Another application of the dominated convergence theorem givesIn particular the limit is zero for an atomless measure.
Suppose there are no nonconstant Koopman operator eigenfunctions. For every in the mean-zero L2 space, all its projections vanish: a nonzero projection would be a nonconstant eigenfunction, since it has mean zero. Hence is atomless, and the Wiener Fourier coefficient lemma givesThe Cauchy-Schwarz inequality also gives vanishing averaged absolute autocorrelations. For two mean-zero observables , the sesquilinear polarization identitythen gives vanishing averaged absolute cross-correlations. Taking and proveswhich is weak mixing.
Conversely, this set formulation of weak mixing extends to all pairs of mean-zero L2 space observables by finite-simple-function approximation and the uniform estimate . If and , the isometry property gives . If , invariance of the integral gives . If and is nonconstant, replace it by , a nonzero mean-zero eigenfunction. But in either casefor every , contradicting weak mixing. We have proved weak mixing is equivalent to absence of nonconstant eigenfunctions, including for noninvertible transformations.
Finally, the doubling map preserves normalized Lebesgue measure, as substitution on its two inverse branches shows. Let be an eigenfunction, , with Fourier coefficients . The Fourier basis calculation givesSince , all odd coefficients vanish. Every nonzero integer is with odd, and repeated use of makes all nonzero coefficients vanish. Completeness of the Fourier basis then says is constant. Therefore the doubling map has no nonconstant eigenfunctions, and
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 9 4 Solution Created 2026-10-03 Updated 2026-10-14
Fix Haar measure and use . A positive-definite function satisfies for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.
First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of . All its elements lie in the closed unit ball of , and the equations define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.
Let and let be the finite sums of convolution squares , . Each such square is continuous with compact support, and is positive-definite becauseIts transform is , so every has . Also : convolution of two squares is , and finite sums distribute. Insert the Bochner representation of into and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares
For every compact there is with throughout . Indeed, at any choose , where is a nonzero member of . Then , and continuity supplies a neighbourhood where it stays nonzero. A finite subcover of and the sum of the corresponding squares give such a .
For , choose positive in transform on its support and defineThe quotient is set to zero away from the support; its denominator is bounded away from zero there. If is another choice, the consistency identity, multiplied by on that compact support, shows that the integral is unchanged. Choosing one for a union of supports proves linearity; positivity follows from positivity of . On each fixed compact support it is bounded by times the uniform norm. Moreover, for every and , choosing positive on the support of givesA nonzero square has ; a suitable nonnegative compactly supported consequently makes this last expression positive. Thus is nonzero.
To prove translation invariance, fix and modulate by . This again belongs to , since modulation takes to . Direct calculation and Bochner uniqueness giveHence, for , using as the denominator function and changing variables gives . The Riesz representation on compactly supported continuous functions now produces a nonzero translation-invariant positive Radon measure on the dual. It is finite on compact sets, so it is Haar measure. This is the compatible dual Haar measure, denoted .
The identity for says, by uniqueness of Radon measures,Thus and the Bochner representation immediately provesIt also proves the formula for all finite complex linear combinations of these functions. The measure scale is fixed by this identity: at , a nonzero square gives .
A wider inversion class is with . To prove this extension, choose nonnegative with integral one and support in a small identity neighbourhood , and put . These functions form an approximate identity: they are nonnegative, have integral one and support in . Also , and uniformly on compact subsets of the dual. For this uniform convergence use equicontinuity of compact families of characters: evaluation is jointly continuous, as follows from the translation quotient in question 3, and a finite cover of a compact family gives uniform control near the identity.
Convolving the already established inversion formula for with , the Fubini theorem givesThis is justified by . Compact-set uniform convergence and an integrable-tail split implyExplicitly, first choose a compact set outside which is small, use the bound on that tail, and use uniform convergence on the compact remainder. This argument works for neighbourhood-indexed nets, without assuming metrizability. Consequently converges uniformly to the continuous inverse integral. On the other hand, translation continuity in Lp on a locally compact group gives in ; testing these two limits against shows they agree almost everywhere. At every continuity point of the chosen representative of , the shrinking-support approximate identity also converges pointwise to . We obtain Fourier inversion on a locally compact abelian group:valid everywhere for continuous with integrable transform, and in general as equality with a continuous representative almost everywhere. This class includes all convolution squares above, and their linear span.
Finally, for arbitrary , apply the inversion formula at zero to :Thus the integral Fourier transform is linear and norm-preserving on . That space is dense in , so for choose with in and define as the limit of . The norm identity makes this limit exist and independent of the approximating sequence, and preserves linearity. This proves the Plancherel theorem for locally compact abelian groups in the requested form:On , choose the approximating sequence to converge in both norms; uniform convergence of the integral transforms and their convergence show that this extension agrees almost everywhere with the original transform. For general functions the extension is defined by norm limits, not by presuming that the pointwise integral exists.
Spectral measure of a Koopman observable 2026-10-14
The finite positive measure representing an observable's autocorrelation sequence by the Bochner–Herglotz theorem. For a probability measure-preserving system, the Koopman operator is an isometry, so its forward autocorrelations, extended by complex conjugation to negative indices, form a positive-definite function on . The measure has mass , and its atoms are squared norms of projections onto Koopman operator eigenspaces, including for noninvertible transformations.