Asymptotically minimal maxima of real Littlewood polynomials

Proof of Theorem 1.1. Fix 0<\delta<1/20 and choose the family from Proposition 2.1. For all sufficiently large N, (2) gives

Since 1-|x|\le1-x^2 for |x|\le1, it follows that \mu(X_N)\le4\delta N. The function u\mapsto u\log(80N/u) is increasing on 0<u\le N/2. Applying Lemma 2.2 and then (1), we obtain

First the limit in N was taken with every auxiliary choice depending on \delta fixed. Now let \delta\downarrow0. The right side tends to one, and the lower bound m_N\ge1 follows from Parseval. This proves the limit through all positive integer lengths.□

Proof of Lemma 3.1. All constants implicit in asymptotic estimates may depend on the fixed forms and the fixed slot lengths chosen next. The sole parameter tending to infinity in this proof will be a prime q.

Slot lengths and finite-field ranks. Choose positive proportions \alpha_i>2w_i with \sum_i\alpha_i=1. Approximating T\alpha_i by positive odd integers as T\to\infty gives fixed odd integers L_i such that, with L=\sum_iL_i,

Let b be the rank over \mathbb Q of the matrix whose rows are the a_i. Then b\ge2. For every sufficiently large prime q, the full matrix has rank b over \mathbb F_q, and each pair of rows has rank two. Indeed, it suffices to exclude the prime divisors of a nonzero b-by-b minor and of one nonzero two-by-two minor for each pair. Fix such a prime for the following construction.

Random slots and oriented centers. Put n=(q-1)/2, R=\lfloor\sqrt q\rfloor, and U_q=\{1,\ldots,n\}. For s\in\mathbb F_q, let |s|_q be the absolute value of its representative in [-n,n]. Partition U_q into consecutive chunks of R sites, with a possible final shorter chunk. There are K=\lceil n/R\rceil chunks.

Consider a periodic partition of the integers into consecutive slots: each period has one slot of type i, consisting of L_i consecutive sites, for each i=1,\ldots,r. Independently in each chunk, translate this pattern by a uniform residue modulo L, and keep only slots entirely contained in that chunk. A kept slot is a vertex of its type. Its middle site c is an integer because L_i is odd. Give it an oriented center s=c or s=-c in \mathbb F_q, with equal probabilities, independently for all kept slots. The random data in different chunks are independent.

A chunk of length u contains u/L+O(1) kept slots of each type, uniformly in its shift. Thus in every realization the number of vertices of each type is

This estimate includes the last short chunk. Call a site interior if its distance from both endpoints of its chunk is at least L. For any nonzero u\in\mathbb F_q whose folded site |u|_q is interior, the probability that u is the oriented center of a vertex of a specified type j is exactly

There is exactly one shift modulo L placing a type-j middle site there; the resulting slot is fully contained in the chunk, and its orientation has the required sign with probability 1/2.

Compatible tuples form a simple hypergraph. An edge is a tuple of one vertex of each type, with oriented centers s_1,\ldots,s_r, for which the equations

have a solution y\in\mathbb F_q^m. We keep one edge per tuple, regardless of the number of solutions. Every compatible tuple has exactly q^{m-b} solutions, the size of the kernel of the rank-b evaluation map. This constructs a simple r-uniform hypergraph.

Write C_j(u) for the indicator that u is the oriented center of a type-j vertex. For nonzero s, define the random variable

When the vertex of type i with center s exists, its degree is exactly

This follows by dividing the number q^{m-1}Z_{i,s} of generating vectors by q^{m-b}. Likewise, two fixed evaluations of different types leave q^{m-2} possible vectors, so their codegree is at most

Two vertices of the same type have codegree zero.

We next show that nearly all vertices have nearly the same degree. Although evaluations of several forms may be dependent, their folded sites usually belong to different chunks. The independent chunk choices then supply the first and second moments we need.

Exceptional centers and the moment estimates. For a fixed type i, let E_i contain those nonzero s\in\mathbb F_q for which there exist j<l, both different from i, and a relation

Any such relation has a unique nonzero c: if c=0, the two forms a_j,a_l would be parallel over \mathbb F_q. For each relation, multiplication by c is a bijection, so at most 2R+1 residues are excluded. There are at most 2\binom{r-1}{2} relations, giving

Fix a nonzero s\notin E_i, and sample y uniformly on the affine fiber a_i\cdot y=s. Each other evaluation a_j\cdot y is uniform on \mathbb F_q, by pairwise rank two. Let Q_s be the chunk containing |s|_q. At most 2LK positive sites are noninterior; this bound also covers a short final chunk with no interior sites. Including their negatives, zero, and the residues folding into Q_s, the probability that a specified evaluation fails to fold into an interior site outside Q_s is at most

If two residues fold into one chunk, the difference of their folded sites has absolute value at most R. Consequently at least one of their sum and difference has folded absolute value at most R. For a_j\cdot y and a_l\cdot y, each of these signed combinations is either uniform on \mathbb F_q under a_i\cdot y=s, or is the constant cs from a relation in (9). The latter case cannot cause a collision because s\notin E_i; the former has probability at most (2R+1)/q. Thus the r-1 evaluations fold into distinct interior chunks different from Q_s, except on an O(q^{-1/2}) fraction of the fiber.

For the second moment, take two independent uniform vectors y,y' from the same fiber. The preceding bounds apply within each copy. Across copies, a_j\cdot y and a_l\cdot y' are independent uniform residues for any j,l\ne i, including j=l; their sum and difference are therefore uniform. A union bound shows that the combined list of 2(r-1) evaluations also occupies distinct interior chunks outside Q_s, apart from a fraction O(q^{-1/2}). More explicitly, for either u=1 or u=2 independent fiber samples, the failure fraction is at most

This bound uses no joint independence among the forms in one copy.

Let V_{i,s} be the event C_i(s)=1, and condition on it whenever it has positive probability. The event depends only on random data in Q_s. Each tuple counted as good in (11) uses distinct other chunks, where the required center hits are independent and each has probability p. The good-tuple condition itself depends only on the fiber vectors, not on the slot choices. Averaging over those vectors, and bounding the contribution of bad tuples between zero and one, gives

The expectations are over the random slots and orientations. The second identity follows by expanding the square as the average over the independent pair y,y'. Both error bounds are uniform in nonexceptional s. The conditioned vertex itself need not be interior: avoiding its entire chunk makes that unnecessary.

Deleting the few vertices of atypical degree. Set D_q=p^{r-1}q^{b-1} and \varepsilon_q=q^{-1/8}. Equations (12)–(13) imply

By Chebyshev’s inequality and (7), a nonexceptional present vertex has degree outside [(1-\varepsilon_q)D_q,(1+\varepsilon_q)D_q] with conditional probability O(q^{-1/4}). Summing over the at most r(q-1) candidate centers gives an expected O(q^{3/4}) such vertices. The vertices with centers in some E_i number O(q^{1/2}) in every realization. Markov’s inequality therefore provides a realization with at most q^{7/8} vertices in these two classes, for all sufficiently large q.

Delete these vertices and all incident edges. By (8), any remaining vertex loses at most

edges. Hence every remaining degree is (1+O(q^{-1/8}))D_q, and the maximum codegree remains at most q^{b-2}=o(D_q). By (5), each type still has (1+o(1))q/(2L) vertices.

A matching supplies the centers. We have obtained a hypergraph to which Theorem 3.2 applies. If it has v_q vertices, degree summation gives

An edge coloring into (1+o(1))D_q matchings therefore has a color class with at least (1-o(1))v_q/r=(1-o(1))q/(2L) edges. A matching has no more edges than the number of vertices of any one type. We have consequently found a matching of size

For each of its edges choose a generating vector y_h\in\mathbb F_q^m. Using integer representatives of its coordinates, set \theta_h=y_h/q\pmod{\Z^m}. Then a_i\cdot\theta_h is the oriented center of the corresponding slot divided by q, modulo one.

It remains to check the geometry, including endpoints. A slot of type i with sites t,\ldots,t+L_i-1 corresponds to the real interval

Its length is L_i/q, and its center is the middle site divided by q. The positive slots lie in [1/(2q),1/2] and have disjoint interiors. Their reflections modulo one lie in [1/2,1-1/(2q)] and also have disjoint interiors. Neighboring slots may share an endpoint, and a slot may meet its reflection at 1/2, but none of these intersections reaches their interiors.

By (4) and (14), a sufficiently large prime gives w_i/H<L_i/q for every i. The desired closed interval at each signed center is therefore contained strictly inside its slot or reflected slot. Both signs use this same pair of slots, irrespective of the chosen orientation. The matching uses distinct vertices, so all these containing slot interiors are disjoint. This proves the required disjointness of the closed intervals, including at 1/2 and across zero on the circle.□

Proof. Use the real-line Fourier transform \widehat g(\xi)=\int_{\R}g(x)\e(-\xi x)\,dx, and put x_0=u/(T\beta). Completing the square reduces the integral to

Insert the damping factor \exp(-\pi\rho y^2) with \rho>0, and use Fourier inversion on g(x_0+y). The resulting double integral is absolutely integrable. The inner Gaussian integral is

Here the square root is the branch analytic in the right half-plane and positive on the positive real axis. As \rho\downarrow0,

The modulus of the exponential multiplier is at most one, and |(\rho-iT\beta)^{-1/2}|\le(T|\beta|)^{-1/2}. Since \widehat g is Schwartz, dominated convergence applies on the Fourier side. On the original side it applies with dominating function |g(x_0+y)|. We obtain the exact identity

The difference of the last integral from g(x_0) has absolute value at most

This bound is independent of x_0, proving the required uniformity.□

Proof of Proposition 4.1. Fix \delta. The choices below are successive: once a parameter has been chosen, it stays fixed while later parameters vary.

A polynomial of almost unit mean square. Starting with p_0=0, introduce one new torus coordinate at each step and define

For |z|\le1, (1-z^2)/2=(1-|z|)(1+|z|)/2\le1-|z|. Thus |p_j|\le1 pointwise. Integrating first in y_j shows that p_j has mean zero. If M_j=\int_{\T^j}p_j^2, integration of the square in the new variable gives

Hence M_j\uparrow1: a limit below one would make the increments bounded below by a positive constant. Choose d\ge2 so that M_d\ge1-\delta.

Let \mathcal S\subset\Z^d be the Fourier support of p_d, consisting of the frequencies with nonzero coefficients, and put c_s=\widehat p_d(s). Every vector in \mathcal S has last nonzero coordinate equal to 1 or -1. Indeed, the old term in (19) keeps the previous frequencies with new coordinate zero, whereas every frequency in the new term has new coordinate 1 or -1. The two terms cannot cancel each other. There is no constant frequency because the mean is zero.

If two such vectors are parallel, comparing their last nonzero coordinates shows that they are equal or negatives. Let s_1,\ldots,s_D be those with positive last nonzero coordinate. They are pairwise nonparallel and \mathcal S=\{\pm s_1,\ldots,\pm s_D\}. There are at least two: the coefficient at the first coordinate vector in p_1 is nonzero, and the coefficient at the second coordinate vector in p_2 is (1-M_1)/4=7/32>0; both persist in later steps.

We have obtained the mean-square mass needed in (15). It remains to distribute its Fourier coefficients so that the total width and the coefficient-to-width ratios in (16) are both controlled.

Choosing the quadratic curvatures. Choose \gamma\in\R^d outside the finite union of hyperplanes s^\perp, s\in\mathcal S, and define \rho_s=|s\cdot\gamma|>0. We will choose vectors b_1,\ldots,b_D\in\R^d so that the products

are approximately a common multiple of |c_s|^2/\rho_s. These products will determine both the number and the sizes of the spread coefficients.

For each j, choose b_j^0\in s_j^\perp which is not orthogonal to any s_l with l\ne j. This is possible because each such orthogonality condition cuts out a proper subspace of s_j^\perp, and a finite union of proper subspaces cannot cover that space. Choose b_j^1 satisfying

and set b_j=b_j^0+\tau b_j^1 for a positive parameter \tau. Every denominator is nonzero by the choices just made. For s=s_j,

The same limit holds for -s_j. Fix a sufficiently small \tau>0 so that, simultaneously for all s\in\mathcal S,

In particular, every curvature \beta_{s,j}:=s\cdot b_j is nonzero. All these curvatures are fixed from now on; no lower bound uniform in \delta is needed.

Spreading the coefficients. Choose g\in C_c^\infty((-1/2,1/2)) with 0\le g\le1 and

For a large positive real parameter B, and z\in[-1/2,1/2]^D, define

The cutoff makes this function vanish near every boundary face in z, so it extends to a smooth real function on \T^{d+D}. Translation invariance in y gives

Fourier integration first in y shows that the only possible frequencies are (s,u) with s\in\mathcal S and u\in\Z^D, and

Lemma 4.2 and 0\le g\le1 imply

uniformly for every real w. The constants may depend on the already fixed curvatures and cutoff. Since there are only finitely many pairs (s,j), for all sufficiently large B we have

The stationary points for these integrals suggest keeping the boxes

Their cardinalities satisfy

for all sufficiently large B. Thus the product \Delta_s controls the number of retained frequencies as well as their coefficient bound.

A summable bound outside the boxes. We need uniform approximation after truncation, so the additive error in Lemma 4.2 cannot simply be summed over the whole lattice. Instead, outside the boxes the phase has no stationary point near the amplitude support.

Choose \alpha<1/2 with \operatorname{supp}g\subset[-\alpha,\alpha]. For a fixed \beta\ne0, if |w|>B|\beta|/2, then

with c_{\alpha,\beta}>0. Indeed, the middle expression is at least (1-2\alpha)|w|, while B<2|w|/|\beta|. Integrate by parts l times using

There are no boundary terms. Each term after these integrations contains a bounded derivative of g and a factor bounded by C_l B^h(B+|w|)^{-l-h} for some 0\le h\le l. Hence

For integer w the sum of this bound outside the one-dimensional box is O(B^{1-l}) when l>1. The full sum over integer w is O(B): inside there are O(B) terms, each at most \norm{g}_{L^1(\R)}, and the outside sum is bounded by (28).

The complement of \mathcal U_s is contained in the union of the D coordinate complements. Consequently

Fix an integer l>D. Summing over the finite set \mathcal S shows that the total discarded coefficient mass tends to zero. The Fourier series is absolutely convergent, so this mass also bounds the uniform truncation error.

Truncation and the final budgets. Fix one B large enough for (25), (26), and a truncation error at most \delta in (29). Define

The boxes satisfy \mathcal U_{-s}=\mathcal U_s=-\mathcal U_s. Since G is real, the retained coefficients have conjugate symmetry, and F is real. By (23) and the uniform error,

Set m=d+D and take the containing support

Its vectors are nonzero. If two are parallel, projection onto the first d coordinates forces the proportionality factor to be 1 or -1; the full vectors must then be equal or negatives. Thus one representative from each signed pair gives nonparallel vectors a_1,\ldots,a_r. Every box contains zero, so r\ge D\ge2. Some of the corresponding Fourier coefficients can vanish; the containing support and all estimates remain valid in that case.

Finally choose

For a=(s,u)\in\mathcal A,

Using (26) and (21), with both signs already included in \mathcal S, gives

The coefficient bound (25), divided by 1+\delta, also gives

These are all the assertions of the proposition.□

Proof. For b\in J and \ell\in\Z, define

Poisson summation, with the change of variable k/N, gives

Here the factor N from Poisson summation combines with the original factor N^{-1/2} to give \sqrt N.

Choose a fixed integer L so large that, for |\ell|>L,

on \operatorname{supp}\chi, uniformly in b\in J. Since the amplitude is compactly supported, two integrations by parts have no boundary terms and yield

These integrals are O(N^{-2}|\ell|^{-2}). Their total contribution to (31) is therefore O(N^{-3/2}), uniformly in b. This summable estimate handles the infinite tail before any stationary-phase errors are added.

For the finitely many indices |\ell|\le L, expand the phase as

Apply Lemma 4.2 with T=N, curvature \lambda, and u=N(\ell-b+\lambda x_*). Its stationary point is x_{b,\ell}, and \phi_{b,\ell}(x_{b,\ell})=\omega_{b,\ell}. After multiplication by \sqrt N, it gives the corresponding term in (30) with error O(N^{-1}), uniformly in b. Increasing L if necessary ensures \chi(x_{b,\ell})=0 whenever |\ell|>L and b\in J. Summing the finitely many remaining errors proves the lemma.□

Proof of Proposition 2.1. Fix 0<\delta<1/20. Proposition 4.1 supplies a real trigonometric polynomial on \T^m,

and a vector v\in\R^m. The representatives a_i are nonzero and pairwise nonparallel. Write \lambda_a=a\cdot v. The properties needed here are

The set \mathcal A is a containing Fourier support; coefficients equal to zero are allowed.

Apply Lemma 3.1 with w_i=|\lambda_{a_i}|. Its strict width hypothesis holds because 2\sum_i w_i=\sum_{a\in\mathcal A}|\lambda_a|<1. We obtain an integer H\ge1 and points \theta_h\in\T^m such that all the closed intervals

are pairwise disjoint. Each has length less than one.

Put x_h^*=(h-1/2)/H. Choose smooth cutoffs

Such cutoffs can be chosen to equal one except in sufficiently small neighborhoods of the block endpoints. All data chosen so far, including the cutoffs, are now fixed independently of N.

For every integer N\ge1 and 0\le k<N, define

The expression does not depend on the choice of real lifts of \theta_h. It is real, and at most one summand is nonzero, so |X_{N,k}|\le1. As every cutoff is supported strictly inside (0,1), extending the sum over k to all integers introduces no additional nonzero terms.

The maximum on the whole circle. Expand F in (35) and apply Lemma 5.1 with

For fixed real lifts of the finitely many \theta_h, all such b lie in one compact interval as t ranges over [0,1]. Thus every error is uniform in t. The leading term indexed by (a,h,\ell) contains

Disjointness of all intervals in (34) allows at most one pair (a,h). For that pair, their length being less than one allows at most one integer \ell. The cutoffs vanish near block endpoints, so this argument also covers angles at the ends of the packed intervals.

By (33), the modulus of the sole possible leading contribution, including its Fourier coefficient, is at most K_\delta. Summing the errors over the fixed finite set of pairs (a,h) therefore gives

The error tends to zero uniformly in t, with all construction data held fixed. We have obtained the desired maximum bound; it remains to show that sampling retains the mean-square mass of F.

The mean-square mass. The block supports are disjoint, so products from different blocks vanish. Within block h, a pair a,a'\in\mathcal A contributes the coefficient c_a\overline{c_{a'}} times

If a=a', this is a Riemann sum with limit \int\chi_h^2. If \kappa\ne0, Lemma 5.1, now with amplitude \chi_h^2, shows that the sum with normalization N^{-1/2} is bounded. The additional factor N^{-1/2} in (38) therefore makes this contribution tend to zero.

The remaining case is a\ne a' and \kappa=0. It cannot be omitted: the coefficient-spreading construction deliberately produces repeated curvatures. The centers of I_{a,h} and I_{a',h} are distinct modulo one, so \beta\notin\Z. Geometric summation gives

Writing \psi=\chi_h^2, summation by parts consequently bounds \sum_k\psi(k/N)\e(k\beta) by a fixed constant times \norm{\psi}_\infty+\int|\psi'|. Indeed the discrete variation is at most \int|\psi'|. Division by N again makes (38) tend to zero.

Together with (37), this proves Proposition 2.1. Every integer N was allowed throughout. In particular, neither the packing prime nor the number of blocks imposes a divisibility condition on N.□

Proof. We first state the partial-coloring input in its vector form. Given v_1,\ldots,v_R\in\R^u, a starting point x_0\in[-1,1]^u, and nonnegative numbers c_1,\ldots,c_R such that

the theorem of Lovett and Meka gives, for every sufficiently small \eta>0, a point x\in[-1,1]^u with

Letting \eta tend to zero and taking a convergent subsequence gives a point with at least \lceil u/2\rceil coordinates equal to signs and the same discrepancy bounds. Indeed, there are only finitely many choices of these coordinates, and all the displayed inequalities are closed under limits.

Start at zero in [-1,1]^s. At a stage with u coordinates not yet equal to signs, restrict the rows of B to those coordinates, take their current values as x_0, and choose

The threshold sum is exactly u/16, and every restricted row has Euclidean norm at most \sqrt u. Thus this step changes each row sum by at most 4\sqrt{u\log(16R/u)} and fixes at least half the remaining coordinates. Leave the fixed coordinates unchanged and repeat. Zero rows impose no restriction. The process terminates with a sign vector.

If u_h is the number of remaining coordinates before step h, starting with h=0, then u_h\le2^{-h}s. The function u\log(16R/u) is increasing for 0<u\le R. With L=\log(16R/s)\ge\log16, the sum of the row errors is therefore at most

The last series converges, and \log(16R/s)\le4\log(2R/s) for s\le R. This proves the lemma with an absolute constant, including when s=R.□

Proof of Lemma 2.2. Let X=(X_0,\ldots,X_{N-1})\in[-1,1]^N and \mu=\frac12\sum_k(1-|X_k|). If \mu=0, every X_k is already a sign, and we take \varepsilon_k=X_k. Suppose henceforth that 0<\mu\le N/2. Set

Then X_k=\sigma_k(1-2p_k), 0\le p_k\le1/2, and \sum_kp_k=\mu. In particular, this definition includes X_k=0.

Put M=20N and R=2M=40N. For the grid points t_\ell=\ell/M, 0\le\ell<M, define the real R\times N matrix A by

Its entries belong to [-1,1], and every selection of its columns has at most N\le R columns.

Choose an integer J\ge1 with N2^{-J}\le1, and first round down:

This does not increase the mass, and

We next construct p^{(j-1)} from p^{(j)}, for j=J,J-1,\ldots,1, preserving the two properties

Let S_j be the set of coordinates for which 2^jp^{(j)}_k is odd, and let s_j=|S_j|. If s_j=0, no change is needed. Otherwise apply Lemma 6.1 to the columns indexed by S_j, obtaining signs \xi_k. Replace all these signs by their negatives if necessary so that \sum_{k\in S_j}\xi_k\le0, and define

An odd integer in [0,2^j] lies between 1 and 2^j-1; adding either sign therefore gives an even integer still in [0,2^j]. Thus the new vector belongs to ([0,1]\cap2^{-(j-1)}\Z)^N. Its mass does not increase, and the simultaneous sign reversal has left the discrepancy unchanged. The step error is consequently at most

Every coordinate in S_j is at least 2^{-j}, so the mass bound gives

This is where the total defect enters the estimate. Put U_j=2^j\mu and L_\mu=\log(80N/\mu). The function f(s)=s\log(80N/s) is increasing on (0,N]. If U_j\le N, then

If U_j>N, then, using \mu\le N/2,

These estimates apply at every nonempty stage. Summing (41) and (40) therefore gives, for p'=p^{(0)}\in\{0,1\}^N,

Empty stages contribute zero, without invoking the discrepancy bound with no columns.

Now set \varepsilon_k=\sigma_k(1-2p'_k)\in\{-1,1\} and define the error polynomial

At a grid point, the real and imaginary parts of -Q/2 are entries of A(p'-p). Hence

It remains to pass from this grid bound to the whole circle. We apply this step only to Q, the rounding error.

Write S=\max_{|z|=1}|Q(z)|. If Q=0, there is nothing to prove. Otherwise let d\le N-1 be its degree. Applying the maximum principle to Q on the unit disk and to the reversed polynomial z^dQ(1/z) gives

Cauchy’s estimate on the disk of radius 1/N centered at a point of the unit circle yields |Q'(z)|\le\exp(1)NS there. Consequently

Every point of \T has circular distance at most 1/(2M) from the grid. Integrating this derivative along the shorter arc gives

The denominator is positive. Combining this inequality with (43) and (42) proves Lemma 2.2, with an absolute constant. The same argument covers N=1, when the error polynomial is constant.□

Proof. Choose a minimizing polynomial at each length; the finite set of sign vectors ensures that the minimum is attained. Put f_N(t)=|P_N(\e(t))|/\sqrt N. Parseval and Theorem 1.1 give \int f_N^2=1 and \norm{f_N}_\infty=1+o(1). Hence

For u\ge0, |u-1|^4\le|u^2-1|^2, so \int|f_N-1|^4\to0. For p\le4, the probability-space integral inequality gives the assertion. For p>4, use the eventual uniform bound on |f_N-1| and the same fourth-moment convergence.□

Proof. Choose A_N so that P_{A_N} attains the minimum defining m_N. By (44), Parseval, and the elementary fourth-moment bound,

The last limit is Theorem 1.1. Thus F(A_N)\to\infty and \mathcal F_N\ge F(A_N) has the same conclusion. The equivalence with the autocorrelation formulation follows directly from the definition of F(A).□

Proof. Corollary 8.1 supplies nontrivial finite binary words with arbitrarily large merit factors. The forward construction in [7, Theorem 2 and its proof] therefore gives a binary Morse shift with \sum_{n=1}^{\infty}|\widehat{\sigma_f}(n)|^2<\infty. Its defining words have frequencies of both signs tending to 1/2; their Fact 1 gives unique ergodicity, and their Fact 2 gives simple spectrum for that invariant measure. Conjugate symmetry extends the square summability to \Z. Fourier Plancherel and uniqueness of Fourier coefficients then give \sigma_f=h\,dt with h\in L^2(\T). Positivity of \sigma_f gives h\ge0 almost everywhere, and \int h\,dt=\sigma_f(\T)=\norm{f}_2^2=1.□