Density of positive orbits of $C_0$-semigroup

Question

Suppose that $(T(t))_{t\geq0}$ is a $C_0$-semigroup on a Banach space $X$ and assume that there exists $x\in X$ such that $\{T(t)x:\ t\geq0\}$ is dense in $X$. I wonder why the set $\{T(t)x:\ t\geq t_0\}$ is dense for every $t_0>0$. I do not see the reason why this should be true. I tried this with a sequence argument, but it doesn't work. Is there any idea? Maybe by contradiction?

Answer 1

Here is a simple solution which has nothing to do with orbits of semigroups but uses only a simple fact from general topology: If $M$ is dense in a topological space $X$ and $A$ is nowhere dense (that is, $\overline A$ does not have interior points) then $M\setminus A$ remains dense (in fact, if $U$ is any non-empty open set then $U\setminus \overline A$ is open and non-empty since otherwise $U\subseteq \overline A$. Hence $M\cap (U\setminus \overline A)$ is not empty which implies that $U\cap (M\setminus A)$ is not empty.)

To apply this, let $M$ be the dense orbit and $A=\lbrace T(t)x: 0\le t\le t_0\rbrace$ and observe that $A$ is compact (due to the continuity of $t\mapsto T(t)x$) and hence does not have interior points if $X$ is an infinite-dimensional Banach space (I guess that you are not interested in finite-dimensional spaces because then there aren't any hypercyclic semigroups).

Answer 2

If $\{T(t)x: t\ge 0\}$ is dense in $X$, then for any $\delta$, there is a $t>t_0$ with $\|T(t)x-x\|<\delta$. Now pick any $y\in X$. There exists $t_1$ with $\|T(t_1)x-y\|<\epsilon/2$. Moreover, there exists $\delta$ such that $\|T(t_1)x-T(t_1)z\|<\epsilon/2$ if $\|x-z\|<\delta$. Now pick $t_2>t_0$ such that $\|T(t_2)x-x\|<\delta$. It follows that $\|T(t_1+t_2)x-T(t_1)x\|<\epsilon/2$, hence $\|T(t_1+t_2)x-y\|<\epsilon$.