A power saving for square-difference-free sets

Proof. We first show that the kernel b_p(x,y)=A_p(y-x)-1/2 has operator norm at most \delta_p. The characters x\mapsto e(kx/p), k\in\mathbb F_p, are an orthonormal eigenbasis. The constant eigenvalue is

For k\ne0, put G_p(k)=\sum_{u\in\mathbb F_p}e(ku^2/p). The corresponding eigenvalue is

Here the subtraction of 1 removes the zero square, and division by 2 removes the multiplicity of every nonzero square. Since (u,v)\mapsto(u-v,u+v) is a bijection on \mathbb F_p^2,

Thus all eigenvalues have absolute value at most \delta_p. In particular,

The same bound holds with x-y in place of y-x, since the transposed operator has the same norm.

Now replace the m edge indicators one at a time by 1/2. In a telescoping error for an edge (u,v), fix every variable except z_u,z_v. No other edge has both these endpoints. Consequently all the other factors separate as C f(z_u)g(z_v), where |C|,|f|,|g|\le1. Equation (2.1) bounds the conditional error by \delta_p. Averaging the fixed variables and summing over the edges proves the assertion.□

Proof. Put c=2^{-m} and B=K-c. With respect to the normalized point-mass bases, T_B has matrix entries B(x,y)/\sqrt{|\mathbb F_p^I|\,|\mathbb F_p^J|}. If s_r are its singular values, then

where the four lists are independent and uniform. The kernel is real, so no conjugates appear in the last expression.

Let \Omega=\{0,1\}^2. Expanding the last line gives

For each F, its product is the constraint indicator of a graph on

Every base edge between i\in I and j\in J is copied, in its original direction, between (i,a) and (j,b) for every (a,b)\in F. The two endpoints determine the base unordered pair and the corner (a,b). Hence this graph has no loops or repeated unordered edges, and has exactly m|F| edges. Lemma 2.1 therefore gives

The main terms sum to (c-c)^4=0. The remaining terms have absolute value at most

Combining this with (2.3) and taking fourth roots proves (2.2).□

Proof. The array H^S is invariant under simultaneous application of \theta to both endpoints. In the reflected block, conjugating a symbol transposition changes only the names of the swapped symbols, not their positional gap. Interactions in other blocks retain their gap as well. Hence relabeling indices by j_v\mapsto j_{\theta v} preserves their uniform law and weight.

If the reflected block is marked, u_S(\theta v)=u_S(v), so no relabeling of lists is needed. If it is unmarked, \theta acts on addresses. Left symbol permutations commute with the right position shift c; therefore it preserves the directed address relation (3.4), including its orientation. Relabeling lists by this address involution preserves their uniform law and I_S. In either case these bijections in (3.7) send the output label at v to the label at \theta v, proving invariance. At the cut this invariance exchanges the two matrix indices, which is symmetry.□

Proof. Index an n\times n matrix by x,y\in V_+ and set

The invariance of H^S used in Lemma 3.2, together with its symmetry, gives \mathsf A_{xy}=\mathsf A_{yx}.

Suppose the two cut symbols occupy positions i<j in the reflected block of x, and set g=j-i. The diagonal entry is \mathsf A_{xx}=\delta^g. A transposition taking x across the cut must act in this block and involve a cut symbol. Apart from swapping the two cut symbols themselves, it must either move the first symbol from i to a position k>j, or move the second from j to a position k<i. Its gap is therefore at least g+1. There are fewer than h^2 such transpositions, so

For every real vector a, using 2|a_xa_y|\le a_x^2+a_y^2 gives

Write m=m_s and \mathcal J=[m]^{V_+}. For j\in\mathcal J, define a vector with one nonzero coordinate for each vertex x\in V_+ by

Using the reflected indices on the minus side, the cross factor of (3.6) is the matrix

Each unordered cross pair is uniquely \{x,\theta y\} with x,y\in V_+, so no extra factor of two appears in this formula.

By (3.12), \mathsf D=(\mathsf A-\kappa I_n)\otimes I_m is positive semidefinite. The matrix

is positive semidefinite: after taking a square root of \mathsf D, the bilinear form is an inner product, and each term of the exponential series is a Gram matrix of tensor powers with a nonnegative coefficient. The convergent sum is positive semidefinite. Moreover \mathsf R(j,j)\ge1.

The remaining factor is

As a matrix this is

Every factor dominates (e^\kappa-1)I_m, also for m=1. It follows that \mathsf E dominates \lambda I_{\mathcal J}. The entrywise product of positive semidefinite matrices is positive semidefinite: if they are Gram matrices, the products of their entries are the Gram matrix of pairwise tensor products. Denoting entrywise multiplication by \circ, we therefore obtain

The internal weight on the plus half is

The reflected minus weight is the same function of its indices. Thus the full index matrix is \operatorname{diag}(a)\mathsf C\operatorname{diag}(a), which still dominates \lambda I_{\mathcal J} by (3.13).

Fix the lists. Since b\in S, the two reflected vertices v,\theta v have the same address. Consequently both sides of the quadratic form use the same vector

This vector may have repeated entries. No injectivity of the map from index assignments to field assignments is required for the matrix lower bound. Since |\mathcal J|=m^n, the full uniform index average is bounded below by

Multiply by I_S(Y) and average over the lists. This proves (3.11). The stated uniform coefficient follows from m_s\le m_{t_0+1}.□

Proof. Use \theta to identify the minus-address list space with a second copy of the plus-address space. Symmetry of the directed graph gives the factorization

where K_b is the product of the constraints crossing the cut. Its graph is bipartite, simple on unordered pairs, and may contain both orientations on different pairs. If it has e_b edges, then e_b\le d^2 by (3.3). Lemma 2.2 yields

Every nonzero index interaction changes a marked block, and so preserves the unmarked word in block b. No such interaction crosses the cut. The index weight therefore factors as W_+(j_+)W_+(j_-), with the same function on the reflected halves and 1\le W_+\le B. Set

Then Q_{p,S}^{b}(F)=\langle G,T_{K_b}G\rangle. The constant kernel contributes the nonnegative quantity 2^{-e_b}(\mathbb E G)^2. Cauchy–Schwarz in the uniform index average, followed by I_+^2=I_+ and W_+\le B, gives

The operator error proves (3.14). For s=t_0+1, no field constraints are present, so the same factorization gives exactly Q_{p,S}^{b}(F)=(\mathbb E G)^2\ge0.□

Proof. Let \Pi_+\subseteq\Pi consist of words in which the first cut symbol precedes the second. It has r=K/2 elements. Above each address in U_{S'}, the old plus addresses are indexed by w\in\Pi_+. Identify the new list of length m_{s+1}=rm_s with the entries indexed by (w,i)\in\Pi_+\times[m_s] in those old lists. Applied at every new address, this identifies the entire new list collection with the old plus list collection; both have the same uniform product law. Figure 2 separates this relabeling from the index restriction used below.

If s+1\le t_0, any new address edge already has at least t_0+1 matching blocks outside b. Every pair of old plus addresses above its endpoints therefore satisfies the old address edge condition, regardless of their block words in b. Old internal validity imposes the square constraint on all entries of every such pair of lists, and hence implies every new validity condition. If s=t_0, the new component is terminal and imposes no conditions, so the same implication holds. Thus under this identification

Apply Lemma 3.3 to S'. In the expectation on its right-hand side, restrict every new half index j'_v to the sublist labeled by the actual word v_b\in\Pi_+. There are n independent indices, each selecting that sublist with probability 1/r. The restriction therefore has probability exactly r^{-n}. Conditional on it, the residual indices in [m_s] remain independent uniform variables, and the emitted half labels are exactly the old half labels. Since the test function is squared, restricting the expectation and then using I_+\le I_{S'} gives

This is (3.15).

By Lemma 3.4, the sum of the two cut forms is at least

The bracket is nonnegative under (3.16). Symmetry follows from Lemma 3.2, completing the proof.□

Proof. Choose a fixed threshold P_0 beyond which the construction and its component mass bounds hold. By (3.20), the leading mass is at least c_0=2^{-e_0-1}. The affine invariance proved above gives uniform single-label marginals in every normalized positive-mass component. Thus a function with uniform mean zero is centered in each component separately, and its input L^2 norm is the same in each component.

Fix v. Write x=z_v and y=(z_w)_{w\ne v}, and use independent uniform measures in these variables until otherwise stated. Since m_0=1, the density of \nu_{p,\varnothing} is L(y)K(x,y), where K is the product of the incident edge indicators and L is the product of all other edge indicators. Both are indicator functions. If m_v is the number of incident edges, put

For m_v>0, Lemma 2.2, with its operator transposed if necessary, gives

The incident constraints form a cross-list graph between the single variable x and the other labels. Incoming and outgoing edges are both allowed by that lemma. The first inequality applies the centered kernel to the constant function, while the second uses \mathbb E f=0. All norms in (3.27) use the uniform base measures.

Put Z_0=\nu_{p,\varnothing}(1), and let T_0 be the conditional operator for the normalized leading component. Its other-label marginal is Z_0^{-1}L(y)a(y)\,dy, and therefore

The ratio is defined as zero where a(y)=0, a set of zero marginal mass. On \{a\ge c_v/2\}, its contribution to (3.28) is at most 2c_0^{-1}c_v^{-1}\Delta_v^2\|f\|_2^2. On the complementary set, Chebyshev’s inequality gives

Weighted Cauchy–Schwarz gives the pointwise bound

again with the zero-ratio convention. Thus no lower bound for a(y) is needed on this small set. Since L\le1, its contribution is at most 4c_0^{-1}c_v^{-2}\Delta_v^2\|f\|_2^2. We conclude that

where the fixed constant

works for every slot, since m_v\le e_0. If m_v=0, the label is independent of all others in this component, so T_0f=0 and the same bound holds.

Now let \rho be either \mu_p or \mu_p^\circ. Represent it as a mixture of the normalized positive-mass components:

The components with zero mass are omitted. In the nonexceptional law, the terminal components |S|=t_0+1 are also omitted. In either case \alpha_S is p^{-|S|/16}\nu_{p,S}(1) divided by the sum of these masses over the retained components. By (3.23),

Introduce the component index S as a latent random variable and condition on it as well as the other labels. Conditional Jensen and the tower property imply

For the nonleading components we used only conditional-expectation contraction. Its input norm is the same uniform L^2 norm in each component, by the marginal statement proved above. Set C_{\mathrm{cond}}=C_{\mathrm{lead}}+C_{\mathrm{mix}}. Enlarging the fixed prime threshold so that P\ge\max\{P_0,3,C_{\mathrm{cond}}^{32}\} ensures

Taking square roots proves (3.25). All constants involved depend only on the fixed finite construction. In particular, this is one threshold for both laws and all slots, with no residual multiplicative constant in the operator estimate.

It remains to identify the pair marginals. By (3.24), the \mu_p^\circ-pair is supported on \Omega_p. The square-affine group acts transitively on this set: the map taking (x,y) to (x',y') has

The affine invariance already proved forces the probability of every pair in \Omega_p to equal 2/[p(p-1)].

In a terminal component, there are still \ell-(t_0+1)=80 clean blocks. The two selected vertices have distinct clean addresses, and there are no field constraints. Conditional on every index choice, their labels are therefore selected from two independent uniform lists and form an independent uniform pair. The index weight is independent of the field entries, so this remains true after averaging the indices. The normalized exceptional mixture consequently has pair law \operatorname{Unif}(\mathbb F_p^2). Combining it with the nonexceptional part gives (3.26).□

Proof of Proposition 3.1. Choose one integer P at least P_{\mathrm{ref}}, P_{\mathrm{mass}}, C_{\mathrm{exc}}^{16/17}, and the threshold in Lemma 3.6. All these numbers depend only on the fixed parameters (3.1). Equations (3.18)–(3.21) give the normalized reflection-positive measure and its exceptional submeasure. The invariance and uniform marginals were proved above; (3.22) and (3.24) give the exceptional-mass bound and cycle support. Lemma 3.6 supplies the selected pair laws and (3.2). This proves all assertions.□

Proof. Across any reflection cut, the matrix of \mu_{\mathcal P} is the tensor product of the corresponding matrices of the \mu_p. It is therefore symmetric positive semidefinite. This also holds for the empty prime set. We first record precisely what a reflection does to an assignment of functions.

Fix a block and two distinct symbols. Let \theta exchange those symbols in that block, and let V_+ be the vertices in which the first symbol occurs before the second. Put V_-=\theta V_+. Define the two fold maps \phi_+,\phi_-:V\to V by

For an assignment a:V\to\mathbb R^{X_{\mathcal P}}, abbreviate \mathcal Z_{\mathcal P}((a(v))_{v\in V}) by \Lambda(a). The bilinear form of the cut matrix is evaluated on the two real functions

Its mixed value is \Lambda(a); its two quadratic values are \Lambda(a\circ\phi_+) and \Lambda(a\circ\phi_-). Cauchy–Schwarz for this positive semidefinite form gives

This remains valid when the form has a nontrivial null space. For a constant assignment, F=G; hence Z_{\mathcal P}(g)\ge0.

We next specify a finite sequence of folds whose composition is constant. For a word w in block b, write p_i=w^{-1}(i) for the position of symbol i. Let \phi_{b,i} be the fold onto p_i<p_{i+1}, where 1\le i<h. Swapping symbols i,i+1 on the left exchanges precisely the two entries p_i,p_{i+1}. Thus this fold compares these adjacent entries and puts them in increasing order.

In each block use the fixed sequence of comparisons

repeated h-1 times. One sweep moves the largest entry to the last position. Once the last k entries are the largest k entries in increasing order, the next sweep leaves them there and puts the largest remaining entry immediately before them. Induction shows that these sweeps sort every position array to (1,\ldots,h). Concatenating the sequences for all \ell blocks gives fold maps \phi_1,\ldots,\phi_T, with T=\ell(h-1)^2, such that

where v_* has the identity word in every block.

To use this sequence, fix any finite nonempty list of real functions f_1,\ldots,f_k and set

The maximum ranges over all assignments, with repetition, and exists because there are only k^d assignments. If M>0, choose a maximizer. Each of its folded assignments still takes values in the same finite list, so each folded integral is at most M in absolute value. Equation (4.3) says that those two integrals are nonnegative and their product is at least M^2. They must therefore both equal M. In particular, either fold of any maximizing assignment is again maximizing.

Apply the folds in reverse sorting order, \phi_T,\phi_{T-1},\ldots,\phi_1. Pullbacks compose on the right, so the resulting assignment is a\circ\phi_T\circ\cdots\circ\phi_1, which is constant by (4.4). Every intermediate integral is M, and the last is one of the diagonal values Z_{\mathcal P}(f_i). Conversely, every constant assignment was included in the maximum. It follows that

When M=0, all these diagonal values are zero, so the same identity holds. This is a finite argument and requires no limiting sequence of folds. The maximizing-assignment argument is analogous to the proof of the abstract chessboard estimate in [6, Theorem 4.1].

Finally, for \epsilon>0, apply (4.5) to the list

Every denominator is positive and Z_{\mathcal P}(f_v)\le1. Multilinearity consequently gives

Letting \epsilon\downarrow0 proves (4.2), including zero diagonal values.□

Proof. Nonnegativity is already proved. Since d is even, homogeneity of the diagonal form gives \|cg\|_{\mathcal P}=|c|\|g\|_{\mathcal P} for every real c. For real functions g_1,g_2, expand Z_{\mathcal P}(g_1+g_2) multilinearly and apply (4.2) to each term. With a=\|g_1\|_{\mathcal P} and b=\|g_2\|_{\mathcal P}, this gives

Taking the nonnegative d-th root proves the triangle inequality, also when one of the seminorms vanishes.

The probability normalization gives Z_{\mathcal P}(1)=1. Insert g in one slot and 1 in all others in (4.2). The uniform single-slot marginal identifies the left side with |\mathbb E g|, proving (4.6). The simultaneous affine invariance of each \mu_p proves (4.7); its parameters may be chosen independently at different primes.

Let G_B be the additive subgroup of X_{\mathcal P} consisting of vectors supported on B. Then

The triangle inequality and translation invariance show that this average has seminorm at most \|g\|_{\mathcal P}. Since the averaged function is independent of B, integration over the prime laws in B, each of mass one, gives

This proves (4.8). All statements include the empty prime set, for which the seminorm is absolute value.□

Proof. Fix the distinguished slot v. For each p, let Y_p=\mathbb F_p^{V\setminus\{v\}}, let \lambda_p be the marginal law of the other labels under \rho_p, and define

This is an operator from L^2(\mathbb F_p,\mathrm{unif}) to L^2(Y_p,\lambda_p); its values on marginal-null points are irrelevant. It preserves constants, and the tower property gives

For uniform-mean-zero f, Lemma 3.6 gives

Let H_p^0 be the mean-zero subspace of L^2(\mathbb F_p,\mathrm{unif}), and let K_p^0 be the mean-zero subspace of L^2(Y_p,\lambda_p). Thus

with orthogonal sums. Equation (4.10) says that T_pH_p^0\subseteq K_p^0. Notice that centering in the codomain refers to \lambda_p, which need not be uniform.

For U\subseteq\mathcal P, let H_U be the tensor subspace with factor H_p^0 when p\in U and factor \mathbb C1 otherwise; define K_U in the same way using K_p^0. The H_U form an orthogonal direct sum of L^2(X_{\mathcal P},\mathrm{unif}). Distinct K_U,K_W are also orthogonal: at a prime in the symmetric difference of U,W, one tensor factor is constant and the other has mean zero. Inner products of pure tensors therefore vanish, and linearity gives the assertion for the whole subspaces.

Under the product law \bigotimes_p\rho_p, the marginal of all the other labels is \lambda=\bigotimes_p\lambda_p. Conditional independence across primes shows that the operator from the distinguished slot to all other labels is

For completeness, on an input \prod_pf_p(z_{v,p}), the conditional expectation is \prod_pT_pf_p, since the conditional law is a product over primes. Such pure tensors span the finite-dimensional domain, proving the operator identity. Consequently

The norm bound follows by applying (4.11) one tensor factor at a time; on the constant factors the norm is one.

Nontrivial characters at p form an orthonormal basis for H_p^0. Thus H_U consists precisely of Fourier sums with exact denominator q_U. Decompose the distinguished input as g_v=\sum_{q_U>R}g_{v,U}, with g_{v,U}\in H_U. The images of distinct summands are orthogonal by (4.12) and the orthogonality of the K_U. Hence

Only distinct exact supports need have orthogonal images; no such assertion is made for different characters within a single support.

Finally condition the multilinear integral on all labels except those in slot v, and apply Cauchy–Schwarz. Its absolute value is at most

Hölder’s inequality with d-1 factors and the uniform marginal in each remaining slot give

This proves (4.9). If \mathcal P is empty, the support hypothesis forces g_v=0, so the assertion holds there too. No reflection positivity of the laws \rho_p was used.□

Proof. Conditional square-sum bound. We establish a conditional square-sum bound before estimating moments. For every B\subseteq\mathcal P with q_B\le Q, and every fixed value z_B, we claim that

The conditional expectation here integrates the coordinates outside B; those outside U are irrelevant to its summand.

To compute its coefficients, write U=B\sqcup A and uniquely decompose a frequency on U as \xi=\xi_B+\eta, with exact supports B and A, respectively. After fixing z_B, the coefficient of the outside character e(z_A\eta) is

Indeed, the sum over the nonzero frequencies at p is

Let \Omega_B denote the union of the outside-frequency sets arising from U\in\mathcal F with U\supseteq B. Different outside supports give disjoint frequency sets, and all their denominators are at most Q/q_B and coprime to q_B. Conditional Parseval therefore identifies the left side of (5.2) with \sum_{\eta\in\Omega_B}|c_B(\eta;z_B)|^2.

Expand the product in (5.3) by subsets J\subseteq B. Apart from its sign, the corresponding coefficient vector is

The congruence in this expression is the simultaneous congruence at the primes of J, or equivalently its unique class modulo q_J. For J=\varnothing it imposes no restriction.

Write its progression as n=a+q_Jt, where t ranges over a consecutive interval T of length L_J. Since q_J\le Q and L\ge10Q^6,

After removing the phase e(-a\eta), (5.4) is q_JL_J/L times the normalized Fourier analysis of t\mapsto f(a+q_Jt) at frequencies q_J\eta.

These frequencies remain distinct. In fact the denominator of \eta-\eta' is coprime to q_J, so q_J(\eta-\eta')\in\mathbb Z implies \eta=\eta'\pmod1. Multiplication by q_J also preserves each reduced outside denominator. Thus all the resulting frequencies have denominators at most Q, are pairwise separated in circle distance by at least Q^{-2}, and number at most \sum_{q\le Q}\varphi(q)\le Q^2, where \varphi(1)=1 accounts for the zero frequency.

This is a separated-frequency estimate of large-sieve type; compare [13, Theorem 1]. The length hypothesis permits the following elementary Gram-matrix bound. Consider the Gram matrix of these normalized exponential vectors in L^2(T). Its diagonal is 1. For distinct frequencies \theta,\theta', geometric summation gives

Its operator norm is at most its maximum absolute row sum, hence at most

The norm squared of the Fourier analysis map equals this Gram-matrix norm. Consequently

The Euclidean triangle inequality over the 2^{|B|} subsets J proves (5.2).

Square-function moments. We next use this bound to control

Put T_r=\mathbb E S^{2r} and T_0=1. Expand the first r-1 factors in this moment and fix their supports U_1,\ldots,U_{r-1}. Their product depends only on coordinates in W=U_1\cup\cdots\cup U_{r-1}, and |W|\le(r-1)k. For a last support U, let B=U\cap W. Integrating outside W gives exactly \mathbb E(|f_U|^2\mid z_B), because the coordinates of f_U outside B are fresh. For each fixed intersection B, its sum over U\cap W=B is bounded by (5.2). Only B with q_B\le Q can occur, and there are at most 2^{(r-1)k} possible intersections. It follows that

Induction gives

This conditioning argument permits arbitrary overlaps among supports. The remaining task is to recover their sum from this square function.

Recovering the sum. Fix a size j\ge1 and color the coordinates independently and uniformly with j colors. A support is called transversal if it contains exactly one coordinate of each color. Each support of size j is transversal with probability \pi_j=j!/j^j. Fix one coloring c, and let \mathcal T be its retained transversal supports in \mathcal F.

Take independent copies X_p^0,X_p^1 of every coordinate. Write \operatorname{swap}_p for interchanging that pair, and define

Coordinatewise centering gives \mathbb E_{X^1}D_U=f_U(X^0): in the expanded product, every term using at least one copied coordinate has zero conditional mean. Jensen’s inequality therefore gives

Swapping independent copies coordinate by coordinate is a standard symmetrization device; see [5, Section 4 of the arXiv version]. Independently swapping each coordinate pair preserves their joint law and multiplies D_U by the product of the associated signs. With independent random signs (\epsilon_p)_{p\in\mathcal P},

The signs here are indexed by individual coordinates, not merely by colors. Each monomial contains one coordinate sign from each color class.

For completeness, the scalar random-sign bound with the constant K_{2r} above is

For nonnegative coefficients b_i, expansion of the even moment retains only multiplicities 2m_i with \sum_i m_i=r. Since (2m)!\ge2^m m!, that moment is at most

For complex a_i, expand (\sum a_i\epsilon_i)^r(\sum\overline{a_i}\epsilon_i)^r and take absolute values termwise; its surviving terms are bounded by the expansion with b_i=|a_i|. This proves (5.7) in the required generality.

For fixed copies X^0,X^1, iterate (5.7) through the j color classes. At each stage one uses Minkowski in L^r, in the form

Induction on the number of color classes therefore gives

To estimate this difference square function, index the choices of copies by \eta\in\{0,1\}^j, one choice for each color. Since each U\in\mathcal T has one coordinate of every color,

For each fixed \eta, the selected full coordinate vector has the original product law. Minkowski in L^{2r}(\ell^2(\mathcal T)) consequently yields

Together with (5.6), this proves the transversal bound (2K_{2r})^j\|S\|_{2r}.

Average over colorings. Since every support is retained with probability \pi_j, another application of Minkowski gives

Here j!\ge(j/e)^j, which also follows by integrating \log x from 1 to j. The term of size zero is constant and bounded in absolute value by S. Summing over 0\le j\le k and using (5.5) proves (5.1).

Finally, the product of k distinct primes is at least (k+1)!, so (k+1)!\le Q. The bound \log(k!)\ge k\log k-k gives k=O(\log Q/\log\log Q). For fixed r, the logarithm of the right side of (5.1), after removing M, is therefore o(\log Q). Every bound used above was independent of the interval’s location, the ambient prime set, and the retained family. This proves the asserted uniformity.□

Proof. The specified fiber has uniform probability 1/q_B. Its contribution to \mathbb E|G|^p is

Discarding all other nonnegative contributions proves (5.8). Apply Lemma 5.2 to each displayed family. Every character of f_U, when specialized on B, retains exactly the coordinate support U\setminus B. Combining such characters can cancel coefficients but cannot create frequencies outside the asserted remaining-denominator ranges. This proves the last assertion.□

Proof. Put r=b-a modulo D, and denote the frequency set in (6.3) by \Xi. We first define the model and actual-square kernels. We then compare their Fourier transforms on minor and major arcs, and finish with the exact bilinear Fourier identity.

The two kernels. For every odd prime p\mid d_0, choose one root c_p of r modulo p. The conditions m\equiv c_p\pmod p for all such primes, together with m odd, specify one residue class modulo D'=2d_0=D/4. Let \rho_D be D' times the indicator of this class. Thus \rho_D has periodic mean one and supremum D'. For every odd prime p\nmid D, set

where u is squarefree and coprime to D, and w_1=1. The second equality follows by summing all characters modulo p. In particular |w_u|\le1.

The factor 1-w_p(m)=p(p-1)^{-1}1_{p\nmid m} is the density of the uniform unit law modulo p relative to the uniform law. Hence for any positive integer q coprime to D,

Here \omega(u) counts the distinct prime divisors of u, and \operatorname{rad}(q) is the product of the distinct primes dividing q. Thus the complete divisor sum converts a uniform residue into a uniform unit, whose square has Fourier multiplier \mathfrak g. We truncate this sum at u\le H in the actual-square kernel. At a rational frequency with denominator q\le H coprime to D, all divisors of \operatorname{rad}(q) remain present; the periodic mean of every term containing a prime outside q vanishes. This will give the exact low-denominator comparison. The other rational coefficients will be bounded by quadratic Gauss sums.

Define two finitely supported functions on \mathbb Z by

The first kernel represents the rational expression in the proposition; the second is supported on actual positive integer squares. The factor 2m/N is the discrete counterpart of the substitution dn=2m\,dm, matching the uniform weight 1/N in the model kernel. For these kernels use the positive-sign Fourier transforms

Our immediate goal is

Put V_N(t)=N^{-1}\sum_{n=1}^N e(tn). Expanding the congruence indicator in (6.6) gives

There are at most DH^2 terms, each of coefficient modulus at most one, and their centers have denominators at most DH. The centers are distinct. Indeed, an equality \xi-j/D=\xi'-j'/D would make \xi-\xi' have denominator both coprime to D and dividing D, so that \xi=\xi' and j=j' modulo D.

Every \alpha_0\in\mathbb Q/\mathbb Z has a unique decomposition \alpha_0=\alpha_D+\alpha_1 in which the denominator of \alpha_D has only prime factors dividing D and q_1=q(\alpha_1) is coprime to D. Thus the coefficient of the center \alpha_0 in (6.9), taken to be zero when it is absent, is

The sign is worth recording: in (6.9), \alpha_D=-j/D, and its coefficient contains e(-jr/D).

Minor arcs. We initially assume N is sufficiently large by an absolute amount. Take the major arcs to be the closed balls of radius 2N^{-9/10} about all reduced rationals of denominator at most N^{1/10}. They are disjoint for sufficiently large N, since two distinct centers are at least N^{-1/5} apart. The minor arcs are their complement. We write \|t\| for the distance of a real number t to the nearest integer.

For \alpha on the minor arcs, Dirichlet approximation with Q_0=\lfloor N^{9/10}\rfloor gives coprime integers a,q satisfying

where the difference is represented by a real number of least absolute value. For completeness, partitioning the circle into Q_0 intervals and placing the Q_0+1 points 0,\alpha,\ldots,Q_0\alpha in them gives integers 1\le h\le Q_0 and b with |h\alpha-b|\le Q_0^{-1}. Reducing b/h to a/q gives the displayed bounds. If q\le N^{1/10}, the first approximation bound would put \alpha within 2N^{-9/10} of a major-arc center. Consequently

Let M=\lfloor\sqrt N\rfloor. Uniformly in real \theta and integers 1\le L\le M, differencing and a geometric sum give

Indeed, pairs at distance j leave a linear phase with coefficient 2\alpha j in the inner sum. Its remaining factors, including e(\theta j), have modulus one. This also proves the asserted uniformity in the linear twist.

Partition the j indices into consecutive blocks of length \lfloor q/4\rfloor. Two distinct indices in a block differ by 0<h<q/4. The residue 2ah is nonzero modulo q: since (a,q)=1, its vanishing would force q\mid2h, whereas 0<2h<q/2. This argument includes even q. Hence

In each block the points 2\alpha j are therefore separated on the circle by at least 1/(2q). Ordering their distances to zero bounds the contribution of one block by O(M+q\log(2q)): only a bounded number can lie within distance 1/q of zero, and the others contribute a harmonic sum. There are O(M/q+1) blocks, so

By (6.11), (6.12)–(6.13) imply

Partial summation with the weight 2m/N gives

For a periodic function, use Fourier coefficients normalized by its period. The Fourier coefficient \ell^1 norm of \rho_D is D': its character expansion has D' coefficients of modulus one. By (6.4), the corresponding norm of each w_p is one, and that of w_u is at most one by the convolution inequality. The entire periodic weight multiplying the quadratic phase in S thus has Fourier coefficient \ell^1 norm at most D'\lfloor H\rfloor\le DH. Expansion into these characters and (6.14) yield

Every center in (6.9) has denominator at most DH\le N^{3\beta/2}<N^{1/10}, so it is a major-arc center. On the minor arcs its distance from \alpha exceeds 2N^{-9/10}. The elementary estimate

therefore gives

Reduction on a major arc to periodic means. Write \alpha=\alpha_0+y, where q=q(\alpha_0)\le N^{1/10} and |y|\le2N^{-9/10}. If \alpha_c\ne\alpha_0 is a center in (6.9), rational separation gives \|\alpha_c-\alpha_0\|\ge1/(qDH). Because qDH\le N^{1/10+3\beta/2}, for all sufficiently large N we have \|\alpha-\alpha_c\|\ge1/(2qDH). The contributions of these other centers sum to

For an admissible u, put

The sequence a_u has period T_u\le qDu and supremum at most D. Breaking a partial sum into complete periods and one remaining interval shows, for every real t\ge0, that

This includes the fractional part of t. For f_y(x)=(2x/N)e(yx^2) on [0,\sqrt N], direct differentiation gives

Stieltjes partial summation consequently gives

The integral equals N^{-1}\int_0^N e(yt)\,dt. Comparing the integral to its integer Riemann sum bounds its difference from V_N(y) by O((1+|y|N)/N). Since |A_u|\le D, that error is dominated by the preceding displayed error. Summing over u\le H proves

The use of \sqrt N as a real endpoint in this calculation accounts for rounding, while the sum itself contains only integers m with m^2\le N.

We have reduced the major-arc comparison to c(\alpha_0) and c_0(\alpha_0). We next compute the contribution at primes dividing D, and then the divisor sieve on the coprime denominator q_1.

The prescribed roots, including the prime two. At an odd prime p\mid D, the local factor of \rho_D is p\mathbf1_{m\equiv c_p\ (p)}. For every j\ge1, squaring maps the lifts of c_p modulo p^j bijectively to the residues congruent to r modulo p. Indeed, if x is such a lift, then

and 2x is invertible modulo p. This proves the bijection inductively over k. For a primitive additive character e(am^2/p^j), its expectation under this normalized root restriction is therefore e(ar/p) if j=1, and zero if j\ge2: in the latter case one averages a nontrivial character over all the lifts of a single residue modulo p. If no p-part is present in the character, the expectation is one.

At two the local factor of \rho_D is 2\mathbf1_{m\text{ odd}}. For j\le3, every odd square is r modulo 2^j, so the expectation of e(am^2/2^j) is e(ar/2^j). For j\ge3, the kernel of the squaring homomorphism on (\mathbb Z/2^j\mathbb Z)^\times has exactly four elements. To see this, if (x-1)(x+1) is divisible by 2^j, exactly one of these two consecutive even factors has 2-adic valuation one, and the other is divisible by 2^{j-1}. The four solutions modulo 2^j are 1,-1,1+2^{j-1},-1+2^{j-1}. The image of squaring thus has 2^{j-3} elements. Every element of this image is 1 modulo 8, and this residue class also has 2^{j-3} elements. The square of a uniform odd residue modulo 2^j is consequently uniform among all residues equal to 1 modulo 8. For j>3, a primitive character averages to zero on that class.

Apply these local facts to the decomposition \alpha_0=\alpha_D+\alpha_1 already used in (6.10). Chinese remaindering shows that the factor at primes dividing D is

The expectation here is the uniform average over any common period. The normalized odd-root restrictions, and the normalized oddness restriction at two, are precisely responsible for the factor D in the model kernel. In particular, choosing only one odd root introduces no additional factor in (6.19).

The coprime divisor sieve. Every prime of an admissible u is coprime to D. If such a prime does not divide q_1, its mean-zero factor w_p is independent of the exponential and all other factors by the Chinese remainder theorem. Thus A_u=0 unless u\mid\operatorname{rad}(q_1).

Suppose first that q_1\le H. All divisors of \operatorname{rad}(q_1) then occur in the sum. By (6.5), the divisor weight is the density of the uniform unit law modulo q_1 relative to the uniform law. If p^j exactly divides q_1 with j\ge2, its local primitive unit quadratic average vanishes. Indeed, conditional on a unit x modulo p^{j-1}, the p lifts x+tp^{j-1} have phase factors differing by e(2atx/p), where p\nmid 2ax. Their sum is zero. If q_1 is squarefree, the unit average is exactly \mathfrak g(\alpha_1). Together with (6.19), this proves

For q_1>H the model coefficient is zero, and we need a bound for the truncated divisor sum. We give the prime-power calculation explicitly. For an odd prime p, an integer j\ge1, and p\nmid a, write

Because two is invertible modulo p^j, the substitution (m,n)\mapsto(m-n,m+n) is a bijection. Consequently

Let T_j=\mathbb E_{m\bmod p^j}\mathbf1_{p\mid m}e(am^2/p^j). For j=1 one has T_1=1/p; for j\ge2, substituting m=pt gives T_j=p^{-1}G_{j-2}(a). It follows that

For j\ge2, the last bound follows from (p+1)/(p-1)\le2 and (6.21). For j=1, the bound obtained from the formula is (1+p^{-1/2})/(p-1); its ratio to p^{-1/2} is 1/(\sqrt p-1)\le2. Thus the exponent-one case is included.

For a fixed divisor u\mid\operatorname{rad}(q_1), the coprime local mean is a product of factors from (6.21) and (6.22). Its modulus is at most 2^{\omega(u)}q_1^{-1/2}. The factor at primes of D has modulus at most one by (6.19). Dropping the restriction u\le H after taking absolute values therefore gives

The penultimate implied constant is absolute. For example, one may use

since 3\le p^{1/6} for every prime p\ge729. Repeated prime powers only increase q_1^{1/6}. This argument covers all prime powers in q_1, and if q(\alpha_D)\nmid D, every coefficient under discussion is already zero by (6.19).

Completion of the uniform comparison. On the minor arcs, (6.15) and (6.16) apply. On a major arc, (6.17), (6.18), (6.20) and (6.23) apply, with |V_N(y)|\le1. For clarity, every approximation error is bounded here using the fixed value (6.1):

In the last two lines q\le N^{1/10} and |y|\le2N^{-9/10}. Each bound is O(N^{-\beta/6})=O(H^{-1/3}) with an absolute constant. Together with the coefficient bound this proves (6.8) for all sufficiently large N. All size and separation thresholds used above are absolute. For the bounded remaining range, the elementary bounds \|S\|_\infty\ll DH and \|S_0\|_\infty\le DH^2 absorb that range by increasing the absolute constant. Thus (6.8) is uniform in all the data of the proposition.

The directional bilinear identity. It remains to pass from the Fourier comparison to the supports in the statement. Extend F and J by zero to \mathbb Z. For any finitely supported kernel h define

Both sums use counting measure; the displayed 1/N normalizes only the base variable. Let

Expanding and integrating the characters imposes y=x+n, giving the bilinear identity

There are no conjugates in this identity. Cauchy–Schwarz and counting-measure Plancherel now give

because each of the last two norms is at most \sqrt N.

Every term in \mathcal B_k(F,J) contains F(x)J(x+m^2) with m\ge1, and this product is zero by hypothesis. Thus \mathcal B_k(F,J)=0 term by term, regardless of the signed weights in (6.7). For a pair x,y with F(x)J(y)\ne0, the ordering assumption gives 1\le y-x\le N-1, and the residue conditions give y-x\equiv r\pmod D. Consequently both indicators in (6.6) are automatic for every contributing pair. Its expansion is exactly

Combining this with (6.24) proves (6.3).□

Proof. Write \chi_\xi(z)=e(z\xi) for a product-space character. Each original character has a unique decomposition \xi+\eta, where \xi is supported on \mathcal S and \eta on B. Their denominators are coprime, so q(\xi+\eta)=q(\xi)q(\eta). The hypothesis ensures that every character \eta on X_B is retained in the lift. Orthogonality over this entire character group gives

Substituting (7.5) proves (7.6).□

Proof. We first record the parameter inequalities that make both lifts available. Since \tau<\beta/2,

Also

Because 1/2-2\tau>0, this lower bound is at least N^{1/2} for sufficiently large N. As D\ge8, we have N'<N for every integer N\ge2. Hence, with Q'=(N')^\beta,

once N is sufficiently large. The length conditions follow from 6\beta<1 and the uniform growth of N' in (7.10).

Removing large remaining conductors. Split g=g_{\le H}+g_{>H} according to the denominator on the remaining prime coordinates \mathcal S. Before specialization, these pieces retain exactly the denominator families

Thus Lemma 5.2 applies to both pieces. For a function u on X_B\times X_{\mathcal S}, uniform counting gives the specialization bound

this is also the estimate in Lemma 5.3. Since |f_s|\le M_0, it follows that, for g_*\in\{g,g_{\le H},g_{>H}\} and every fixed even r,

All subpower bounds in this proof are uniform over the indicated sets and fibers.

Telescope the d inputs in the diagonal integral from g to g_{\le H}. Each difference has one input g_{>H}. Lemma 4.3 and (7.12) imply, for every fixed \varepsilon>0,

Indeed the specialization powers in that lemma add to

The factors d and M_0^d are fixed and are included in C_\varepsilon. Applying the moment lemma with sufficiently small individual exponents makes the aggregate moment loss at most N^\varepsilon.

Splitting the exact fiber into square-step progressions. Let a\bmod D combine s and z_B. By Lemma 7.1, for every remaining \xi with q(\xi)\le H,

Let

There are exactly D elements of J_a. The map (n',j)\mapsto D^2(n'-1)+j bijects [N']\times[D^2] with [D^2N']. Restricting j to J_a selects exactly the residue class a\bmod D. The definition of A'_j assigns zero to all padded points greater than N, so this decomposition incurs no additive error.

Each A'_j is square-difference-free: a difference m^2>0 between its elements would give the difference D^2m^2=(Dm)^2 between the corresponding elements of A. This argument concerns actual integer squares.

For p\in\mathcal S, the integer D is invertible modulo p. Define

Each T_j is a translation followed by multiplication by the nonzero square (D^{-1})^2. Multiplication of a remaining frequency by D^2 preserves its exact denominator, so changing variables \eta=D^2\xi in the Fourier expansion of (7.14) gives the exact identity

For clarity, the coefficient of e(z\xi) on the right is

which is exactly (7.14).

The seminorm triangle inequality and affine invariance in Lemma 4.2 now imply

Consequently,

The averaging by D is essential: the number of progressions causes no further loss.

Recovering the smaller-scale functional. Set

The difference h_j^{Q'}-h_j^H has conductor greater than H. By (7.11), the moment lemma applies at scale N' to these lifts and their pieces. A second telescoping application of Lemma 4.3, this time without specialization, gives

uniformly in j.

The lift h_j^{Q'} depends only on coordinates in \mathcal S'=\mathcal P_{N'}\setminus B, because every prime in a denominator at most Q' is itself at most Q'. On the other hand, the full lift

becomes exactly h_j^{Q'} after uniform averaging over B\cap\mathcal P_{N'}. Character orthogonality removes precisely the denominators containing one of those primes. By Lemma 4.2, coordinate averaging contracts the seminorm; unused probability factors integrate to one. Therefore

This also covers primes of B larger than Q', which are absent from \mathcal P_{N'} already.

Combining (7.13), (7.16), (7.17), and (7.18) proves (7.7) with error

Uniform error bounds and mixed inputs. The definition of N' gives 1\le\lambda<1+D^2/N, and

Since d is fixed, these inequalities prove (7.9) and show that \lambda^d is bounded for large N. Choose \varepsilon=\beta\gamma/8 in (7.19). Its two terms, before the bounded factor \lambda^d, are at most

respectively. As \tau\le\beta\gamma/1000 and \sigma_0=\gamma\tau/4\le\tau/4, they are O(N^{-\sigma_0}). This establishes the diagonal bound with constants independent of any inductive estimate for Y.

Every g_v in the mixed statement is real and satisfies the same diagonal bound. Its right-hand side is nonnegative and independent of v. Proposition 4.1 therefore gives (7.8).□

Proof. We construct a length-24 sequence of admissible increments with sum zero at each factor of M_0. Modulo 8, take all the increments equal to 1; their sum is 24\equiv0\pmod8.

Let p<P be an odd prime. If -1 is a square modulo p, repeat the pair (1,-1) twelve times. Both entries are nonzero squares and each pair has sum zero. If -1 is not a square, put

The sets S and -1-S each have (p+1)/2 elements, so they intersect. Thus -1=a+b for some a,b\in S. Neither a nor b is zero, since otherwise -1 would be a square. The triple (a,b,1) consists of nonzero squares and has sum zero. Repeat this triple eight times. This also covers p=3, where the triple is (1,1,1).

The Chinese remainder theorem now gives increments \Delta_0,\ldots,\Delta_{23}\pmod{M_0}, each a strict square class, with \sum_{j=0}^{23}\Delta_j=0. For each initial residue a\pmod{M_0}, prescribe

The last edge has increment \Delta_{23} because the sum of all the increments is zero. The slots v_0,\ldots,v_{23} are distinct, since their block words are the distinct cyclic shifts of the identity word. Repetition of residue values at nonadjacent slots causes no difficulty. The other d-24 residues are unrestricted. We have therefore exhibited at least M_0\cdot M_0^{d-24} good assignments among the M_0^d possible assignments, which proves the bound.□

Proof. Write \eta_p for the total mass of \varepsilon_p, and set

It suffices to prove the assertion with \Lambda^\circ in place of the measure in (8.2), since the latter is \prod_p(1-\eta_p)\Lambda^\circ and 0\le\prod_p(1-\eta_p)\le1. The single-slot marginals of \Lambda^\circ are uniform. Moreover, at each prime its marginal on any directed cycle edge is uniform on strict square pairs. We use conductor decay for \Lambda^\circ, without any claim that this normalized measure is reflection positive.

Selecting an ordered pair. Put K=\lfloor N^\tau\rfloor, and partition [N] into ordered consecutive intervals I_1,\ldots,I_K of sizes differing by at most one. Writing w_i=|I_i|/N, we have the exact identity

Thus the integral is an average over independent interval choices in all slots, with probabilities w_i. For large N, every interval has length at least \tfrac12N^{1-\tau}. Since 1-\tau>6\beta, this length is at least 10Q^6. Lemma 5.2 therefore applies uniformly to every interval lift and every exact-denominator truncation used below. Its application to f_s, which is bounded by the fixed constant M_0, changes only the constants. All occurrences of N^{o(1)} in this proof are uniform over these interval choices, residues and sets.

First consider assignments for which all 24 cycle slots choose the same interval. Their total probability is

For every assignment, ordinary Hölder and the uniform single-slot marginals give

Here d is even, so the required moment is supplied directly by Lemma 5.2. The contribution of these assignments is therefore O(N^{-23\tau+o(1)}); there is no restriction on the interval choices in the other slots.

In every remaining assignment there is a directed cycle edge u\to v whose first interval precedes its second. Indeed, if the cyclic sequence of interval indices had no ascent, it would be nonincreasing all the way around the cycle, and hence constant. Fix one ascending edge, and denote its two intervals by I_1,I_2 for the rest of this argument. All integers in I_1 precede all integers in I_2.

Eliminating the other slots. Retain these two lifts in full. In each of the other d-2 slots, truncate the lift to exact conductors at most T=N^\tau. Lemma 4.3, applied during a telescoping replacement of these slots, bounds the total error by

Every norm used here is controlled by Lemma 5.2: the retained and discarded pieces are subfamilies of complete exact denominator pieces. This error is estimated before the Fourier expansion that follows.

Each Fourier coefficient in a truncated slot has modulus at most M_0, and there are at most \sum_{q\le T}\varphi(q)\ll T^2 frequencies. The product expansion of all other slots consequently has total coefficient absolute sum at most

For one term in this expansion, let B\subseteq\mathcal P_N be the union of the prime supports of its characters, and put

The conductors are squarefree, so their least common multiple satisfies

Condition on all slot labels at the primes in B. The expanded characters become constants of modulus one. Since \Lambda^\circ is a product over primes, the tuple laws at the remaining primes are unchanged and independent. The two selected labels now have fixed coordinates a_{1,B},a_{2,B} at B, whose difference is a nonzero square at each prime in B. At every remaining prime their pair marginal is still uniform on strict square pairs.

Let \mathcal S=\mathcal P_N\setminus B, let s_1=s_u and s_2=s_v, and define functions on X_{\mathcal S} by

With Fourier coefficients normalized for uniform measure on X_{\mathcal S}, their conditional pair integral is exactly

To see this, simultaneous translation kills character pairs whose frequencies do not sum to zero. For a matching pair, the multiplier is the character average of the difference of the labels. At every odd prime a uniform nonzero square is the square of a uniform nonzero unit. The Chinese remainder theorem therefore identifies this average with the unit-square multiplier \mathfrak g(\xi) used in Proposition 6.1.

Estimating the selected pair. Specialization on a uniform coordinate block of size R costs at most R^{1/2} in L^2: for every function G,

This inequality holds at every specified fiber, independently of the law under which that fiber was selected. Hence \|g_i\|_2\le R^{1/2}N^{o(1)}. By the bound |\mathfrak g(\xi)|\le q(\xi)^{-1/3}, Parseval and Cauchy–Schwarz, the part of (8.6) with conductor exceeding H has modulus at most

We next identify the remaining coefficients with actual integer fibers. Put D=M_0R. The parameter inequalities

imply RH\le Q and, for sufficiently large N, D\le N^\beta. For a remaining frequency \xi with q(\xi)\le H, all frequencies \xi+\eta, where \eta ranges over the character group on B, therefore occur in the original lift. Summing this entire character group gives

where a_i\pmod D combines s_i\pmod {M_0} and the coordinates a_{i,B} by the Chinese remainder theorem. Character orthogonality supplies the factor R, while the definition of f_{s_i} supplies M_0. Goodness of \boldsymbol s and the constraints at the conditioned primes show that a_2-a_1 is a strict square class modulo D.

The remaining frequencies of conductor at most H are precisely the frequencies with squarefree conductor at most H coprime to D. Indeed, D contains every prime less than P; every prime divisor of such a conductor is consequently an available prime at least P, at most H\le Q, and outside B. This includes the zero frequency under the convention q(0)=1.

Define functions on [N] by

They are bounded by one, their supports are ordered, and there is no positive square difference from the first support to the second. Moreover, D is eight times an odd squarefree number and D\le N^\beta. All hypotheses of Proposition 6.1 hold. If the Fourier coefficients of F_i use normalization N^{-1}, then (8.9) gives \widehat g_i(\xi)=DN\widehat F_i(\xi)/|I_i|. Thus the part of (8.6) of conductor at most H is bounded in modulus by

The factor D in Proposition 6.1 has canceled one of the two fiber-normalization factors.

Combining (8.7) and (8.10) bounds each conditional pair integral by O(N^{-\beta/6+(d+2)\tau+o(1)}). This is uniform over all conditionings of positive probability. Averaging them, and then summing the Fourier expansion using (8.4), bounds the truncated integral for each unequal-cycle interval assignment by

Averaging interval assignments incurs no further loss. Including the equal-cycle contribution and (8.3), the original integral has modulus at most

Finally, 23\tau=5888\sigma_0, \gamma\tau=4\sigma_0, and

The finitely many uniform moment losses in each term can be bounded by N^{\sigma_0}, after increasing the absolute threshold in N. Equation (8.11) is consequently O(N^{-\sigma_0}), as required. All estimates used absolute values of signed Fourier lifts; no pointwise positivity of a lift was needed.□

Proof. Fix a square-difference-free set A\subseteq[N]. Write

where \mathbf s=(s_v)_{v\in V} is a small-residue assignment. By (7.5), 1_A=\E_s f_s. The lift is linear, and the product-space integral is multilinear, so

For B\subseteq\mathcal P_N, define the nonnegative finite measure

Its mass is \eta_B. The finite product expansion gives the following identity of signed measures:

For any real integrand \Phi, integration followed by the triangle inequality therefore yields

This step makes no sign assumption on \Phi. Notice also that the measure outside B in each summand is the full measure \mu_p.

We first bound the summands with q_B>N^\tau. A summand of zero mass is zero, so assume \eta_B>0. Each normalized exceptional measure \varepsilon_p/\eta_p has uniform single-label marginals: its invariance under simultaneous translation implies that any one label is uniformly distributed on \F_p. The same uniformity holds for \mu_p. Consequently every single-slot marginal of \kappa_B/\eta_B is uniform on X_{\mathcal P_N}. No independence between different slots is needed. Ordinary Hölder’s inequality gives

The last inequality is Lemma 5.2, applied at the fixed even order d and with the loss in each factor chosen sufficiently small. The bound |f_s|\le M_0 only introduces a fixed factor. The constant C_\epsilon is independent of B, A, N, and \mathbf s.

The exceptional masses are summable with the extra factor q_B:

It follows from (9.9) that

Choose \epsilon=\tau/2. Since \sigma_0=\gamma\tau/4<\tau/2, we obtain

uniformly in the residue assignment. Thus the number of exceptional subsets has already been accounted for by the convergent product (9.10).

Now suppose q_B\le N^\tau and \eta_B>0. Condition on all the labels at primes in B under \kappa_B/\eta_B. The remaining prime law is exactly \bigotimes_{p\in\mathcal P_N\setminus B}\mu_p. For each realization, the function in slot v is G_{s_v} specialized at its prescribed labels in B. The mixed assertion of Proposition 7.2 applies to these arbitrary slotwise residues and labels. With

it gives

The same inequality is trivially true when \eta_B=0. Crucially, the error and its lower threshold for N are uniform in B and in every conditioned tuple. No induction hypothesis has been used.

The ceiling in the smaller scale produces a uniform multiplicative error. For q_B\le N^\tau,

and hence \lambda_B^d\le1+u_N. Moreover, \sum_{B\ne\varnothing}\eta_B\le1/4 by (9.10). Summing the errors in (9.12) is therefore harmless.

If \mathbf s is good, Proposition 8.2 bounds the first term on the right of (9.8) by K_3N^{-\sigma_0}. Combining this with (9.11) and (9.12), we obtain

For a nongood \mathbf s, apply the mixed scale comparison directly with B=\varnothing:

Averaging these two estimates in (9.6) gives (9.4), since the good assignments have probability \rho. The estimate is uniform in A, so we may take its maximum. Finally, 2\tau<1 by the fixed parameter choice, and M_0,d are fixed, so u_N\to0. All constants in the proof came from the preceding estimates and fixed parameters; none depended on an induction constant.□

Proof. We make all remaining parameter choices before starting the induction. The finite construction fixes h,t_0,\ell,d; the conditional and kernel estimates fix \gamma and \beta. The threshold P was then chosen, and with it M_0, \tau, and \sigma_0. In particular the probability \rho of Lemma 8.1 is already a fixed positive number. We shall not enlarge P after choosing the exponent below.

Put r=1-3\rho/4, so 1/4\le r<1, and choose

Table 1 summarizes the roles of the parameters; among the listed quantities, only those in the last row vary with the interval length.

Every quantity in (9.13) is fixed and positive. Thus 0<a_0<\min(1/2,\sigma_0), and

There is no requirement that a_0 be numerically useful.

For every finite \mathcal P_N, the bound \eta_p\le p^{-4} and 2a_0\le1 give

Let \kappa=(1-\theta)/2>0. Choose an integer N_*\ge N_1 so large that, for every N\ge N_*,

where K,N_1 are supplied by Proposition 9.1. This is possible because u_N\to0 and a_0<\sigma_0. The choice of N_* is independent of C_0. Choose C_0\ge1 large enough that

Only finitely many values are involved, so such a finite constant exists. For an elementary bound, the lift in Y(n,A) has at most 2n^{2\beta} frequencies and each coefficient has modulus at most 1. Since its tuple law is a probability measure, 0\le Y(n,A)\le(2n^{2\beta})^d. Thus, for example, C_0=\max\{1,2^dN_*^{2\beta d+a_0}\} suffices in (9.17). This also covers n=1 and the cases of an empty prime set.

Suppose now that N\ge N_* and \mathcal Y(n)\le C_0n^{-a_0} for every positive integer n<N. Every smaller scale appearing in (9.4) lies in this range. Indeed, M_0q_B\ge8 gives

The ceiling goes in the favorable direction for a negative exponent:

Apply the induction hypothesis at every such scale in (9.4), and then use (9.15). Extending the positive subset sum to all nonempty B can only increase the bound, and therefore

In the third line we used (9.16) and \theta+\kappa=1-\kappa; in the last line we used C_0\ge1. This closes strong induction. In particular, increasing C_0 to cover the finite base range does not change the validity of the large-scale step.□

Proof of Theorem 1.1. Let N\ge1 and let A\subseteq[N] have no positive integer-square difference. Put G=\mathcal L_{[N],\mathcal P_N,Q}1_A. Its constant Fourier coefficient is |A|/N, and every nonconstant character on X_{\mathcal P_N} has uniform mean zero. The mean inequality in Lemma 4.2 and Theorem 9.2 yield

Thus C=C_0^{1/d}<\infty and c=a_0/d>0 are absolute constants for which |A|\le CN^{1-c} for every such N and A.□