Topological entropy of $C^1$ function on compact manifold is finite I am trying to solve Exercise 2.5.7 from Introduction to Dynamical Systems by Brin and Stuck:

Prove that the topological entropy of a continuously differentiable
  map of a compact manifold is finite.

Definitions needed: we have
$$
d_n(x,y) := \max_{0\leq k\leq n-1} d(f^k(x), f^k(y))
$$
The topological entropy of a map $f$ is defined as
$$
\lim_{\epsilon\downarrow 0} \limsup_{n\to\infty} \tfrac1n \log\text{cov}(n,\epsilon,f),
$$
where $\text{cov}(n,\epsilon,f)$ is the minimum cardinality of a covering of  by sets of $d_n$-diameter less than $\epsilon$.
My ideas for this exercise: since this manifold is compact we know there exist $N,M$ such that
$
\|f(x)\|\leq M
$
and
$
\|Df(x)\|\leq N.
$
Compactness also definitely will give that for every $\epsilon$ the expression
$$
 \limsup_{n\to\infty} \tfrac1n \log\text{cov}(n,\epsilon,f),
$$
exists and is finite.
 A: In fact you can prove a more general result: If $(X,d)$ is a metric space, and $f:X\to X$ is a continuous Lipschitz map, with Lipschitz constant $L(f)$. Let $D(X)$ be the ball dimension of $X$, which is defined by $$D(X):=\overline{\lim}_{\epsilon\to 0}\frac{\log b(\epsilon)}{|\log\epsilon|},$$
where $b(\epsilon)$ is the minimum of the number of elements of a covering of $X$ by balls of radius $\epsilon$. Then we have that $h_{top}(f)\leq D(X)\max(0,\log L(f)).$ In your setting, $L(f)$ is $\sup_x\|Df_x\|$, so we have the finite entropy property, since for a compact smooth manifold, the ball dimension is equal to its topological dimension. You can check that the ball dimension is finite for any set in a bounded subset of a finite dimensional Euclidean space. Then note that a smooth manifold is locally Euclidean, and the space is compact, and the ball dimension is invariant passing to a biLipchitz-equivalent metric. Here I omit some details, but I think it is not too hard to check. 
To prove this, let $L>\max(1,L(f))$. Then $d(f(x),f(y))\leq Ld(x,y)$ for all $x,y\in X$. So $f^m(B_d(x,L^{-n}\epsilon))\subset B_d(f(x),\epsilon)$ when $0\leq m\leq n$ and hence $B_d(x,L^{-n})\subset B_{d_n}(x,\epsilon)$ for all $x\in X$, $n\in\mathbb{N}$, and $\epsilon>0$. Thus $$S(f,\epsilon,n)\leq b(L^{-n}\epsilon).$$ Note that $|\log(L^{-n}\epsilon)|=n\log L-\log\epsilon$, so
$$n=\frac{|\log(L^{-n}\epsilon)|}{\log L}(1+\frac{\log\epsilon}{|\log(L^{-n}\epsilon)|})=\frac{1}{\log L}|\log(L^{-n}\epsilon)|(1+O(\frac{1}{n})),$$ so $$\overline{\lim}_{n\to\infty}\frac{\log S(f,\epsilon,n)}{n}\leq\overline{\lim}_{n\to\infty}\frac{\log b(L^{-n}\epsilon)}{n}=\log L\overline{\lim}_{n\to\infty}\frac{\log b(L^{-n}\epsilon)}{|\log(L^{-n}\epsilon)|}=D(X)\log L$$ and $h_{top}(f)\leq D(X)\log L<\infty$. And so the proof is completed.
