The growth bound $\omega_*$ of a semigroup is defined as $$\omega_* = \inf \{ \omega \ \ \rm{s.t.} \ ||S(t)|| \leq Me^{\omega t}, \ \forall t \geq 0 \}$$ where $M$ may depend on $\omega$. Alternatively, we have the characterization $$\omega_* = \lim_{t \rightarrow \infty} \frac{1}{t} \log(||S(t)||)$$
What I would like to prove is
Suppose $\omega_* > 0$. Then there exists an $x$ such that $||S(t)x||$ has exponential growth at infinity. In other terms, there exists $\alpha>0$ such that $$||S(t)x|| > e^{\alpha t} \ \ \forall t > \bar t$$
Is this true? I think it should be: since $\omega_* >0$, for every $\varepsilon >0$ there exists $\bar t$ such that $$||S(t)||>e^{(\omega_*-\varepsilon) t}$$ Hence there is an $x$ such that $$||S(t)x||>e^{(\omega_*-\varepsilon) t}||x||$$ However, I don't know how to continue.
This is answer was given to me by the reddit user u\harmath. I think it could be useful to report it here.
We will work by contradiction and suppose that there is no $x$ such that $||S(t)x||$ blows up exponentially. Then, for $0<\alpha<\omega_*$ we have $$\lim_{t \rightarrow \infty} \frac{||S(t)x||}{e^{\alpha t}} = 0, \ \ \forall x$$ But then, by the Uniform Boundedness Principle $$ \frac{||S(t)||}{e^{\alpha t}} \leq M$$ which is a contradiction by the definition of $\omega_*$.