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Discrete Holomorphic Local Dynamical Systems

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216 Tien-Cuong Dinh and Nessim Sibony<br />

Theorem 1.94. Let g : (X,F ,ν) → (X,F ,ν) be a map with bounded Jacobian as<br />

above. Define Fn := g −n (F ).Letψ be a bounded real-valued measurable function.<br />

Assume there are constants δ > 1 and c > 0 such that<br />

� ν,e δ n |E(ψ|Fn)−〈ν,ψ〉| � ≤ c for every n ≥ 0.<br />

Then ψ satisfies a weak large deviations theorem. More precisely, for every ε > 0,<br />

there exists a constant hε > 0 such that<br />

� �<br />

�<br />

� 1<br />

ν x ∈ X : � ψ ◦ g<br />

� N<br />

n � �<br />

�<br />

�<br />

(x) −〈ν,ψ〉 � > ε ≤ e<br />

� −N(logN)−2 hε<br />

for all N large enough 6 .<br />

N−1<br />

∑<br />

n=0<br />

We first prove some preliminary lemmas. The following one is a version of the<br />

classical Bennett’s inequality see [DZ, Lemma 2.4.1].<br />

Lemma 1.95. Let ψ be an observable such that �ψ� L ∞ (ν) ≤ b for some constant<br />

b ≥ 0, and E(ψ)=0.Then<br />

for every λ ≥ 0.<br />

E(e λψ ) ≤ e−λ b + eλ b<br />

2<br />

Proof. We can assume λ = 1. Consider first the case where there is a measurable<br />

set A such that ν(A)=1/2. Let ψ0 be the function which is equal to −b on A and<br />

to b on X \ A. Wehaveψ 2 0 = b2 ≥ ψ 2 .Sinceν(A)=1/2, we have E(ψ0)=0. Let<br />

g(t)=a0t 2 + a1t + a2, be the unique quadratic function such that h(t) := g(t) − e t<br />

satisfies h(b)=0andh(−b)=h ′ (−b)=0. We have g(ψ0)=e ψ0.<br />

Since h ′′ (t)=2a0 − e t admits at most one zero, h ′ admits at most two zeros. The<br />

fact that h(−b) =h(b) =0 implies that h ′ vanishes in ] − b,b[. Hence h ′ admits<br />

exactly one zero at −b and another one in ] − b,b[. We deduce that h ′′ admits a<br />

zero. This implies that a0 > 0. Moreover, h vanishes only at −b, b and h ′ (b) �=0. It<br />

follows that h(t) ≥ 0on[−b,b] because h is negative near +∞. Thus, e t ≤ g(t) on<br />

[−b,b] and then e ψ ≤ g(ψ).<br />

Since a0 > 0, if an observable φ satisfies E(φ)=0, then E(g(φ)) is an increasing<br />

function of E(φ 2 ). Now, using the properties of ψ and ψ0, we obtain<br />

E(e ψ ) ≤ E(g(ψ)) ≤ E(g(ψ0)) = E(e ψ0<br />

e<br />

)= −b + eb .<br />

2<br />

This completes the proof under the assumption that ν(A)=1/2 for some measurable<br />

set A.<br />

The general case is deduced from the previous particular case. Indeed, it is<br />

enough to apply the first case to the disjoint union of (X,F ,ν) with a copy<br />

6 In the LDT for independent random variables, there is no factor (logN) −2 in the estimate.

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