证明
Lemma 9 ([Free-shift coordinates]). The change of coordinates H^k⟶H^ 2s ×H^…Lemma 8 ([Average cost of the unsplit projection]). Let f be a function of the…观点
All expectations over H and ^s below use uniform probability measure; the displayed sums defining w_q and M are unnormalized. Conditional on the J_g , orthogonality of the characters of L gives _LU(z)^2 =∑_qc(q,z+ J(q))^2. Summing over z proves. If M=0 , every vector c(q,z) is zero and is immediate. Henceforth assume M>0 . Set h=2^s and ρ=_q w_q/M . We will prove the quantitative bound below for every C,D>0 , and then choose its parameters. The main estimate bounds the maximum squared norm by the square root of the sum of fourth powers. After a Fourier change of variables, most terms in that fourth moment force the same character q in all four factors; their total is controlled by the dispersion of the character masses. Two truncations will control the remaining terms and the cost of this change to the array: _zU(z)^2 ≤2M[ √2ρ+(2C^2)/(D^2)+(3D^4)/(h) +(1 √C +1D)^2+1h ]. Fourier representation. For p∈^s and u∈H , define α_q(p)=∑_ z∈^s (-1)^ p·z c(q,z), A_p(u)=∑_ q∈H^* χ_q(u)α_q(p), S_p=L+∑_ g=1 ^s p_gJ_g. Fourier inversion in z gives U(z)=_p(-1)^ p·z A_p(S_p). Writing m_p=_uA_p(u)^2 , Parseval's identity in the two groups gives m_p=∑_qα_q(p)^2, _p m_p=M, _pα_q(p)^2=w_q. For distinct p,p' , the variables S_p,S_ p' are independent and uniform in H : the coefficient vectors (1,p),(1,p') are linearly independent over . Similarly, any three distinct p_1,p_2,p_3 give three independent uniform variables S_ p_1 ,S_ p_2 ,S_ p_3 . Indeed, any nonempty subcollection of their coefficient vectors whose sum is zero must have even cardinality, and hence would consist of two equal vectors. Two truncations. We first discard indices p with unusually large mean squared norm m_p , and then discard unusually large values of A_p(u) on the remaining indices. The first cutoff makes the change to each individual character coefficient uniformly small under the second cutoff. The second gives the pointwise bound needed when the fourth-moment indices coincide. We will control the discarded part using a first-moment bound and a variance bound from the pairwise independence of the samples S_p . Define A'_p(u)= A_p(u),, m_p≤CM and A_p(u)≤D√M,; 0,, otherwise, and put α'_q(p)=_uχ_q(u)A'_p(u), w'_q=_pα'_q(p)^2, U'(z)=_p(-1)^ p·z A'_p(S_p). Pointwise truncation does not increase the total squared norm, so Parseval gives ∑_qw'_q≤M . If m_p≤CM , then α_q(p)-α'_q(p) ≤_uA_p(u) _ A_p(u)>D√M ≤(m_p)/(D√M) ≤(C√M)/(D). If m_p>CM , then α'_q(p)=0 . Consequently, ∑_qw'_q≤M, w'_q≤2w_q+(2C^2M)/(D^2). Let X_p=(A_p-A'_p)(S_p), X=_pX_p. Equation and the triangle inequality imply _zU(z)-U'(z)≤X . The contribution to X from the indices with m_p>CM is at most _p_ m_p>CM √m_p ≤_p m_p √CM =√M/C. The contribution from the remaining indices is at most _p m_p/(D√M)=√M/D . Moreover, the X_p are pairwise independent, by the pairwise independence of the S_p . Since A_p-A'_p is either A_p or zero at each point, Var(X) =h^ -2 ∑_pVar(X_p) ≤h^ -2 ∑_p_uA_p(u)-A'_p(u)^2 ≤ Mh. We have proved _zU(z)-U'(z)^2 ≤M[(1 √C +1D)^2+1h]. The fourth moment. For brevity write B_p=A'_p(S_p) . Expanding the fourth power of the norm and summing the characters in z gives the exact identity ∑_zU'(z)^4 , =h^ -3 ∑_ p_1,p_2,p_3 [ B_ p_1 B_ p_2 B_ p_3 B_ p_1+p_2+p_3 ]; , =_ p_1,p_2,p_3 [ B_ p_1 B_ p_2 B_ p_3 B_ p_4 ], p_4=p_1+p_2+p_3. Here p_1,p_2,p_3 are independent uniform indices in the second line. The factor h^ -3 in the first line comes from four factors h^ -1 in Fourier inversion and the sum of h characters over z . If any two of p_1,p_2,p_3,p_4 coincide, then some two of p_1,p_2,p_3 coincide. This event has probability at most 3/h . Since B_p≤D√M , its contribution to is at most 3D^4M^2/h in absolute value. Now fix four distinct indices with p_4=p_1+p_2+p_3 . The first three S_ p_i are independent uniform elements of H , and S_ p_4 =S_ p_1 +S_ p_2 +S_ p_3 . Expanding the four A'_ p_i in characters and averaging over those three independent elements yields [ B_ p_1 B_ p_2 B_ p_3 B_ p_4 ] =∑_q α'_q(p_1) α'_q(p_2) α'_q(p_3) α'_q(p_4) . In detail, characters indexed by q_1,q_2,q_3,q_4 produce the factor ∏_ i=1 ^3χ_ q_i+q_4 (S_ p_i ), whose expectation is zero unless q_1=q_2=q_3=q_4 . The right side of need not be nonnegative. For an upper bound, replace each summand by ∏_ i=1 ^4α'_q(p_i) , which is nonnegative, and then drop the restriction that the four indices be distinct. For every fixed q , Cauchy--Schwarz gives _ p_1,p_2,p_3 ∏_ i=1 ^4α'_q(p_i) , ≤ (α'_q(p_1)^2 α'_q(p_2)^2)^ 1/2 (α'_q(p_3)^2 α'_q(p_4)^2)^ 1/2 ; , =(w'_q)^2. The last equality uses the unconditioned independent uniform law of each pair (p_1,p_2) and (p_3,p_4) ; it does not assert independence after conditioning on distinctness. Thus the nondegenerate fourth moment is bounded by ∑_q(w'_q)^2 : dispersion among characters now controls concentration among table entries. Combining these estimates with proves ∑_zU'(z)^4 , ≤∑_q(w'_q)^2+(3D^4M^2)/(h); , ≤M^2(2ρ+(2C^2)/(D^2)+(3D^4)/(h)). Therefore, using Cauchy--Schwarz once more, _zU'(z)^2 ≤(∑_zU'(z)^4)^ 1/2 ≤M√2ρ+(2C^2)/(D^2)+(3D^4)/(h). The inequality (a+b)^2≤2a^2+2b^2 , together with and , proves . Finally, set ε=ζ,1 and choose C=(64)/(ε), D=(8C)/(ε), γ=(ε^2)/(128), 2^ s_0 ≥ (192D^4)/(ε^2),(16)/(ε). If s≥s_0 and ρ≤γ , the square-root term in is at most ε/4 . Also D≥√C , so the other two terms in its brackets are at most ε/16 each. The resulting bound is 3εM/4≤ζM , establishing with constants depending only on ζ .