I'm using J.Meiss -Differential dynamical systems, and have some trouble to understand a proof about Lyapunov exponents. We have a dynamical system $$ \dot{x} = f(x), $$ with the corresponding flow $ \varphi_t $. Now suppose $ \varphi_t(x) $ is a bounded orbit, and that the flow $ \varphi$ is not forward asymptotic to an equilibrium. Want to show that then it has a zero Lyapunov exponent, i.e $ \mu(x,v) = \limsup_{t \rightarrow \infty} \frac{1}{t} \ln |D_x\varphi_t(x)v| = 0 $.
So this is what they do: Consider the vector $v(t) = f( \varphi_t(x)) $. Now $$ \frac{d}{dt}v(t) = \frac{d}{dt}f( \varphi_t(x)) = Df( \varphi_t(x))\frac{d}{dt} \varphi_t(x) = Df( \varphi_t(x)) f( \varphi_t(x)) = Df( \varphi_t(x))v(t), $$ hence $$ \dot{v} = Df( \varphi_t(x)) v.$$ Thus $v$ is a solution to the linearization of the system about this orbit with initial condition $f(x)$.
Now they do some argument that I do not uderstand:
First they say that since $ \varphi $ is bounded, $v$ is also bounded. Why is that? Further they say that since $ \varphi_t$ is not asymptotic to an equilibrium $ \limsup |v(t)| > 0$. Why is that? And Therefore $ \mu(x,v) = 0$.
Thank you for your help.