I'm having problems trying to show that there are no Anosov diffeomorfisms on $S^1$, i.e. that $S^1$ is not an hyperbolic set for any diffeomorfism of $S^1$. Can someone help me?
I know I can assume that $\left\| d_x f^nv\right\| \le c\lambda^n\left\| v\right\|$ ($\lambda <1$, $c>1$) for every $x\in S^1$ and $v\in\mathbb R$, but don't know how to follow from here...
Given a diffeomorphism $f : S^1 \to S^1$, you can choose a lift $F : \Bbb R \to \Bbb R$ to the universal cover $p : \Bbb R \to S^1$, $p(x) = \exp(2\pi i x)$. Since $f$ is a diffeomorphism of the circle, $F$ restricts to a diffeomorphism $g : [0, 1] \to [0, 1]$ such that for all $x \in [0, 1]$ and $n \in \Bbb Z$, $F(x + n) = g(x) + n$.
If $f$ is Anosov, then $df_p$ is either a contracting or an expanding linear transformation (the perks of having a single dimension) of $T_p S^1$ for all $p$. By compactness of the circle you can pick a global stretch factor, so let's assume that $df$ is globally $\lambda$-contracting for some $\lambda < 1$, i.e, $\|df_p(v)\| \leq \lambda \|v\|$ for all $p \in S^1$ and $v \in T_p S^1$.
Then $F'$ is also $\lambda$-contracting on $T_x \Bbb R$ for all $x$; in particular $g'$, where $F|_{[0, 1]} = g$, is also $\lambda$-contracting. But $g(0) = 0$ and $g(1) = 1$ implies by Lagrange's mean value theorem that for some $c \in (0, 1)$, $g'(c) = 1 > \lambda$, contradicting hypothesis.
Similar argument holds if you assume $df$ is globally $\lambda$-expanding for some $\lambda > 1$.