V for all k 1 q if ϕ β 0 r 0 k r 0 k negationslash

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(v) for all k = 1 ,...,q , if ϕ β 0 ,R 0 k ( R 0 ,k ) negationslash = ϕ β 0 ,R 0 k +1 ( R 0 ,k ), then for all θ ′′ B ( θ 0 ,r ) B (( z 0 1 ,R 0 ) , qr ) and for all x such that || x | − R 0 ,k | <r , we have parenleftBig ϕ β ,R k ( | x | ) ϕ β ′′ ,R ′′ k +1 ( | x | ) parenrightBigparenleftBig ϕ β 0 ,R 0 k ( R 0 ,k ) ϕ β 0 ,R 0 k +1 ( R 0 ,k ) parenrightBig > 0 , which is possible by continuity of ( β,R,x ) mapsto→ ϕ β,R k ( | x | ) and since β 1 can be chosen as close as we want to β 0 . (vi) for all k = 1 ,...,q , if ϕ β 0 ,R 0 k ( R 0 ,k ) = ϕ β 0 ,R 0 k +1 ( R 0 ,k ), then for all θ B ( θ 0 ,r ), θ B (( z 0 1 ,R 0 ) , qr ) and for all x such that || x | − R 0 ,k | <r , ϕ ( β,R ) k ( | x | ) ϕ ( β ,R ) k +1 ( | x | ) and ϕ ( β,R ) k +1 ( | x | ) ϕ ( β ,R ) k ( | x | ) , which is possible for the same reasons as in (iv). Now we choose N ( θ 0 ) = { ( z 0 1 ,R ) } where R = ( R 1 ,R 2 ,...,R q ) is as in the proof of Theorem 2, see (56) and before. Note that ( z 0 1 ,R ) is in B (( z 0 1 ,R 0 ) , qr ) and that β 1 can be chosen sufficiently close to β 0 to ensure N ( θ 0 ) B ( θ 0 ). Following the same calculus as in (57), we obtain that for any θ = ( z ,R ) B ( θ 0 ,r ) φ β 1 ,R ( x ) φ β ,R ( x ) . We deduce that 25
H θ 0 Λ ( ω Λ ) H θ Λ ( ω Λ ) N Λ ( ω Λ ) >z 0 z > r and the second part of [Regularity] is proved. 5.5 Proof of Theorem 4 We check the assumptions of Theorem 1. For any β ∈ B the pair potential φ β is regular and non- integrably divergent at the origin. As it is well known (see [25] for instance), φ β is superstable and the existence of an associated Gibbs measures follows, i.e. [Existence] holds true. Superstability implies stability which shows assumption [Stability] . Since the map β mapsto→ φ β is differentiable and so continuous, assumption [Argmax] is obvious. The decomposition in [MeanEnergy] is done as in (49). In particular the mean energy (15) is H θ ( P ) = E P zN Δ 0 + summationdisplay { x,y }∈ ω Δ 0 φ β ( x y ) + 1 2 summationdisplay x ω Δ 0 ,y ω Δ c 0 φ β ( x y ) (59) where θ = ( z,β ). To check assumptions [MeanEnergy] and [Boundary] , we thus have to prove that (59) is finite and that for any compact set K ⊂ B , sup β ∈K 1 | Λ n | summationdisplay x ω Λ n ,y ω Λ c n φ β ( x y ) 0 . (60) For this purpose, we need the following Ruelle estimates. Proposition 3 ([26]) We define, for any i Z d , ψ i = ψ ( max ( | i | − 1 , 0) , where the function ψ comes from assumption (27) . Then i Z d ψ i < + and for any c> 0 E P parenleftBig e c i Z d ψ i N τ i 0 ) parenrightBig < + . Let us show (60). Note that sup β ∈K 1 | Λ n | summationdisplay x ω Λ n ,y ω Λ c n φ β ( x y ) A 1 + A 2 with A 1 = 1 | Λ n | summationdisplay x ω Λ n \ Λ n n 0 y ω,y negationslash = x | x y |≤ r 0 sup β ∈K | φ β | ( x y ) and A 2 = 1 | Λ n | summationdisplay x ω Λ n y ω Λ c n | x y | >r 0 sup β ∈K | φ β | ( x y ) , where r 0 comes from assumption (27) and n 0 is an integer greater than r 0 . By the spatial ergodic theorem, P -almost surely lim n →∞ 1 | Λ n | summationdisplay x ω Λ n y ω,y negationslash = x | x y |≤ r 0 sup β ∈K | φ β | ( x y ) = E parenleftBigg summationdisplay x ω Δ 0 y ω,y negationslash = x | x y |≤ r 0 sup β ∈K | φ β | ( x y ) parenrightBigg . 26
Therefore, A 1 goes P -almost surely to 0 if the expectation above is finite. By the GNZ equation, stationarity of P and assumption (27) E parenleftbigg summationdisplay x ω Δ 0 y ω,y negationslash = x | x y |≤ r 0 sup β ∈K | φ β | ( x y ) parenrightbigg = z E parenleftbigg e z ω φ β ( z ) summationdisplay y ω | y |≤ r 0 sup β ∈K φ β ( y ) parenrightbigg z E

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