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Let $f: M\to M$, where $(M,\rho)$ is a closed Riemann manifold, and $(\widetilde{M},\widetilde{\rho})$ is the universal covering of $(M,\rho)$, $D$ is a fundamental domain of $(\widetilde{M},\widetilde{\rho})$. $\widetilde{f}$ is the lift of $f$, we suppose there is $x\in D$ s.t. $\widetilde{f}(x)=x$ then I want to prove that: there exist a constant $C$, for all $n\in \mathbb{N}$, $$max\{max_{x\in M}|d_xf^n|,max_{x\in M}|d_xf^{-n}|\} \ge C sup_{z\in D}distance_{\widetilde{\rho}}(x,\widetilde{f}^nz).$$

Antoine
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Yes, $C=1$ and you do not even need to include derivatives of inverse iterations in the left hand side. To prove this just use the definition of the length of the image under f of a geodesic in M connecting projection of x to projection of z.

Moishe Kohan
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  • Do you mean like this: let $a=\pi(x),b=\pi(z)$,$\gamma(t)$ is a geodesic connecting a and b, then we calculate $\int_0^1 |(f \gamma)^{'}(t)|dt$, and how could we continue? – Antoine Mar 18 '14 at 15:23
  • Now, use definition of the norm of derivative of $f$ and the chain rule. – Moishe Kohan Mar 18 '14 at 15:45
  • $|d_xf|=\sum g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j}$, $|(f\gamma)^{'}|=|\sum \frac{\partial f}{\partial x^i}\gamma^{'}(t)|$, and then? – Antoine Mar 19 '14 at 07:00
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    Now, use definition of the norm of a linear operator: $|Av|\le ||A|| \cdot |v|$. If $||A||\le C$ then $|Av|\le C|v|$ for all vectors $v$. The linear operator in question is the differential of $f$ at the given point and you use it for vectors $v$ tangent to the geodesic curve. Lastly, integrate the inequality. – Moishe Kohan Mar 19 '14 at 10:40