证明
Lemma 4 ([Two-color localization and diagonal tilting]). Let p^2+q^2=1 , and…Proposition 3 ([Particle-number free-energy bounds]). Uniformly in the…观点
For h≥0 take p_h=1 at distance at most hs from U , and interpolate by a cosine to zero between distances hs and (h+1)s . Set q_h=(1-p_h^2)^ 1/2 using the corresponding sine. Both gradients vanish off this transition and |∇p_h|^2+|∇q_h|^2≤Cs^ -2 there. Choose an open support region Ω_h inside the (h+2)s neighborhood; if that neighborhood covers the torus, use the torus. Its volume is at most C(h+4)^3|U| . Apply (7). Inside, partition the torus into cubes of side in [D_0^ -1/3 ,2D_0^ -1/3 ] . For small density this side is between b_0 and s . The cubes meeting Ω_h number at most CD_0(h+4)^3|U| . Applying (2) to part of the inside energy of zero-extended Dirichlet states gives c_vD_0j-CD_0^2(h+4)^3|U| . Another fixed fraction of kinetic energy gives (3). An additional fixed fraction may remain unused. The external potential costs at most ρj . Outside, (6) gives F_ J-j ^ (-W) ≥F_J^ (-W) -C_vρj . The classical entropy of the color number costs at most T(C+j) . Since T≤ρ^2 , all coefficients of j other than c_vD_0 are bounded by C_vρ . All volume errors, including CT[1+|Ω_h|(V^ -1 +T^ 3/2 )] , are at most C_vD_0^2(h+4)^3|U| after reducing the density: here |U|≥s^3 and V≥s^3 . Average in h with probability density e^ -h \,dh . At each x , ∫_0^∞p_h(x)^2e^ -h \,dh≥φ_U(x) . The averaged gradient cost is at most Cs^ -2 φ_U(x) : the transition can occur only for h∈[(0,d(x,U)/s-1),d(x,U)/s] . The moment of (h+4)^3 is finite. Taking H_0 large makes c_vD_0-C_vρ positive and yields (9). In equilibrium the left side is zero; because n_U≤n_ φ_U this proves (10). For (11), take G=λn_ φ_U in (8). The distance function is one-Lipschitz, so ∑_i|∇_iG|^2≤λ^2s^ -2 n_ φ_U . Also G≥0 , hence M_G≥0 and the entropy term is at most 2Tλ𝔼_g n_ φ_U . Comparison with (9), increasing the fixed H_0 if necessary, absorbs both terms and gives 𝔼_g n_ φ_U ≤C_vρ|U| uniformly in 0≤λ≤1 . Integrate ∂_λ M_G=2𝔼_g n_ φ_U from zero to obtain (11). For a single cell, exponential Markov gives P(n_Q≥t)≤(1,e^ Cm-2t ) . Integration of 2tP(n_Q>t) gives 𝔼 n_Q^2≤C(m+1)^2 , which is at most Cm^2 since m→∞ uniformly with dilution.