Let $\mathcal{T}=\{T(t)\}_{t\geq0}$ be a $C_0$-semigroup of linear operators of Banach space $X$.The Adjoint of $T(t)$ is denoted by $T^*(t)$. It is known that $T^*(t)$ is bounded linear operator and $||T(t)||=||T^*(t)||$.
Assume that $\mathcal{T}^*=\{T^*(t)\}_{t\geq 0}$ is $C_0$-semigroup of bounded linear operators.
In my research, every vector $0\neq x\in X$ does have unbounded orbit hat evin $\mathcal{T}=\{T(t)\}_{t\geq0}$. What can say about $\mathcal{T^*}$.
Is it true that every vector does have unbounded orbit?
Let $S$ denote the shift operator on $\ell^2(\mathbb{N}_0)$ acting by $$S(v_0,v_1,\ldots, v_n,\ldots)=(0,v_0,v_1,\ldots, v_n,\ldots )$$ The elements of the standard basis in $\ell^2(\mathbb{N}_0)$ are denoted by $\{\delta_n\}_{n=0}^\infty.$ Then $S\delta_n=\delta_{n+1},$ $S^*\delta_n=\delta_{n-1}$ for $n\ge 0$ and $S^*\delta_0=0.$ Fix $0\neq v\in \ell^2(\mathbb{N}_0).$ Let $k$ be the smallest index for which $v_k\neq 0.$ Then $$\langle e^{tS}v,\delta_{k+1} \rangle =\sum_{n=0}^\infty {t^n\over n!}\langle S^nv,\delta_{k+1}\rangle =\sum_{n=0}^\infty {t^n\over n!}\langle v,(S^*)^n\delta_{k+1}\rangle\\ = \sum_{n=0}^{k+1} {t^n\over n!}\langle v,\delta_{k+1-n}\rangle=v_{k+1}+tv_k$$ Therefore the orbit of $v$ is unbounded. On the other hand $\delta_0\in \ker S^*$ hence $$e^{tS^*}\delta_0=\sum_{n=0}^\infty {t^n\over n!}(S^*)^n\delta_0=\delta_0$$
The example can be modified so that $\ker T^*$ is infinite dimensional. Then $e^{tT^*}v=v$ for $v\in \ker T^*,$ i.e. the subspace of vectors with bounded orbit for $T^*$ is infinite dimensional. For $X=\ell^2(\mathbb{N})$ let $T\delta_n=\delta_{2n},$ $n\in\mathbb{N}.$ Then $T^*\delta_{2n}=\delta_n$ and $T^*\delta_{2n-1}=0.$ Therefore $\ker T^*$ is infinite dimensional. For $v\neq 0,$ let $k$ denote the smallest index such that $v_k\neq 0.$Then $$\langle e^{tT}v, \delta_{2k}\rangle =\sum_{n=0}^\infty {t^n\over n!}\langle v,(S^*)^n\delta_{2k}\rangle=v_{2k}+tv_k $$ Hence the orbit $e^{tT}v$ is unbounded.