观点

Lemma 6 ([Transfer between temperatures]). Let j≥0 and let m>k≥2 be even. Suppose e_k(j)<1 and 0<q<1 with q≥ρ_ j+1 ^ m-k . Define, for 0≤e<1 and 0<t≤1 , F(q,e,t)=(q)/(1-q)((1)/((1-e)^2t)-1). Then e_m(j+1)≤F(q,e_k(j),A(m,j)). In particular, …

Lemma 6 ([Transfer between temperatures]). Let j≥0 and let m>k≥2 be even. Suppose ek(j)<1 and 0<q<1 with q≥ρj+1m-k . Define, for 0≤e<1 and 0<t≤1 , F(q,e,t)=q1-q(1(1-e)2t-1). Then em(j+1)≤F(q,ek(j),A(m,j)). In particular, an upper bound strictly below one on the right proves cj+1 >0 without presupposing it.

证明 1证否 0作为观点被引用 1链条图发起推演

发起推演1

证明
Proposition 5 ([Concentration at the doubling scales]). For every j≥0 , X_ b_j…Proposition 4 ([Finite periodic inputs from~\cite{periodic}]). Set ℓ=.999,…观点
At the new temperature, yields B_m(j+1)-c_ j+1 ≤q[B_k(j+1)-c_ j+1 ] . Subtract and rearrange to obtain e_m(j+1)≤(q)/(1-q) ((B_k(j+1))/(B_m(j+1))-1). Write r_j=λ_ j+1 /λ_j^2 . From the definitions, (B_k(j+1))/(B_m(j+1)) =(A(k,j))/(A(m,j)) ((B_k(j))/(B_m(j)))^2 r_j^ m-k . Section gives r_j≤1 , and physical purity gives A(k,j)≤1 . The hypothesis e_k(j)<1 implies c_j>0 , and B_m(j)≥c_j gives B_k(j)/B_m(j)≤(1-e_k(j))^ -1 . Substitution proves. Only the old overlap c_j has been divided by. The new overlap follows from e_m(j+1)<1 and B_m(j+1)>0 .
更多详情 →