I am looking at the textbook Analysis and Geometry of Markov Diffusion Operators by Bakry, Gentil, Ledoux. In section 4.4.2 on exponential integrability there is proposition 4.4.2 that states that if $\mu $ satisfies a Poincare Inequality with constant $C$ then for every $1$-Lipschitz function $f$ and $s<\sqrt{\frac{4}{C}}$ $$\int e^{sf}d\mu < \infty$$ The proof relies on this claim: $$\int e^{sf} d\mu \leq e^{s\int fd\mu} \prod_{n=0}^\infty \left(1- \frac{Cs^2}{4^{n+1}}\right)^{-2^n}$$ I am able to follow the proof to the point where we have by iteration for $Z(s) = \int e^{sf}d\mu$ $$Z(s) \leq Z\left(\frac{s}{2^k}\right)^{2^k} \prod_{n=0}^k \left[1-\frac{Cs^2}{4^{n+1}}\right]^{-2^n}$$ Then to get the claim we need to take the limit as $k\to \infty$. The product term obviously matches with the claim. However it is not at all clear to me why $$\lim_{k\to \infty} Z\left(\frac{s}{2^k}\right)^{2^k} = e^{s\int fd\mu}$$ I tried writing it out $$Z\left(\frac{s}{2^k}\right)^{2^k} = \left(\int e^{\frac{sf}{2^k}}d\mu\right)^{2^k}$$ But now I am not sure how to take the limit. I also tried using Taylor series for the exponential function $$e^{\frac{sf}{2^k}} = \sum_{n\geq0} \left(\frac{sf}{2^k}\right)^n \bigg/n! = 1 + \frac{sf}{2^k} + \frac{s^2f^2}{2^{2k} \cdot 2!} + \dots$$ Integrating term by term $$\int e^{\frac{sf}{2^k}} d\mu = 1 + \frac{s \int f d\mu}{2^k} + \frac{s^2 \int f^2d\mu}{2^{2k} \cdot 2!} + \dots $$ But this does not lead me anywhere. I would appreciate any help.
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Expressed $(Z(s/2^k))^{2^k}$ as an $\mathbb L^p$-"norm", whose limit is discussed here.
Davide Giraudo
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