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By Pearn W. L., Lin G. H.

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14. At this stage we give the proof only in the case where β < 1, because the ideas of this simple argument will be used √ again and again. 75) 24 1. 14). We use this for the function f (y) = th2 y, that satisﬁes f (y) = 2 thy 1 − 2sh2 y ; f ≤2. 75) that the function ψ(q) = Eth2 (βz q + h) satisﬁes ψ (q) < 1. This function maps the unit interval into itself, so that it has a unique ﬁxed point. 74). As in the case of pN (β, h) this is a number depending only on β and the law of h. 73) implies that pN (β, h) ≤ SK(β, h) .

Of course, this immediately raises the question as for which values of (β, h) this equality remains true. This is a diﬃcult question that will be investigated later. It suﬃces to say now that, given h, the equality fails for large enough β, but this statement itself is far from being obvious. We have observed that, as a consequence of H¨older’s inequality, the function β → pN (β, h) is convex. 108) that, when β < 1/2, the function β → SK(β, h) is also convex. Yet, this is not really obvious on the deﬁnition of this function.

U2M )) ≤ exp s2 A2 . 23) for the convex function exp(−sx) while taking expectation in uM +1 , . . , u2M , so that E exp s(F (u1 , . . , uM ) − EF (uM +1 , . . , u2M )) ≤ E exp s(F (u1 , . . , uM ) − F (uM +1 , . . , u2M )) ≤ exp s2 A2 . 4, and since EF (uM +1 , · · · , u2M ) = EF (g) we have E exp s(F (g) − EF (g)) ≤ exp s2 A2 . 7) we get that for s, t > 0 P(F (g) − EF (g) ≥ t) ≤ exp(s2 A2 − st) and, taking s = t/(2A2 ), P(F (g) − EF (g) ≥ t) ≤ exp − t2 4A2 . 18 1. The Sherrington-Kirkpatrick Model Applying the same inequality to −F completes the proof when F is inﬁnitely diﬀerentiable (or even twice continuously diﬀerentiable).