Lemma 9 ([Large-denominator estimate]). Fix l≥1 and a finite prime set P as above. At each prime choose either ρ_p=μ_ l,p or ρ_p=μ_ l,p ^ . Let g_v:X_ P → be arbitrary functions. If, in one distinguished slot v , every Fourier coefficient …

At one prime, let m_p denote the marginal law of the labels in all slots other than v , and let T_p f=𝔼_ ρ_p (f(z_v)(z_u)_ u≠v ). This is an operator from uniform L^2(_p) to L^2( m_p) . It preserves constants. It sends uniform-mean-zero inputs to m_p -mean-zero outputs, since 𝔼_ m_p T_pf=𝔼 f . By the conditional mixing estimate, its norm on the mean-zero subspace is at most p^ -γ . The product law makes the conditional operator for all primes equal to T=_pT_p : this identity holds first for products of one-prime functions, by independence across primes, and then for every function by linearity. The exact-support B Fourier space has a mean-zero factor at primes in B and a constant factor elsewhere. On this space the operator norm is at most q_B^ -γ . Furthermore, the images of two distinct exact-support spaces are orthogonal in L^2(_p m_p) . At a prime in the symmetric difference of the supports, one image factor is constant and the other has mean zero. Their inner product is zero, and tensoring preserves this orthogonality. Decompose g_v=∑_ q_B>R g_ v,B by exact supports. Both the domain summands and their images are orthogonal, so \|Tg_v\|_2^2 =∑_ q_B>R \|Tg_ v,B \|_2^2 ≤∑_ q_B>R q_B^ -2γ \|g_ v,B \|_2^2 ≤R^ -2γ \|g_v\|_2^2. Condition the original integral on the other slots and apply Cauchy--Schwarz. The remaining factor is bounded by \|∏_ u≠v g_u(z_u)\|_2 ≤∏_ u≠v (𝔼|g_u(z_u)|^ 2(d-1) )^ 1/(2(d-1)) =∏_ u≠v \|g_u\|_ 2(d-1) , using ordinary H\"older and the uniform single-slot marginals. This proves. If P is empty, the frequency hypothesis forces g_v=0 , and the conclusion is immediate.