Eventually compact semigroup: if $T(t_0))$ is compact for some $t_0 > 0$, then $T(t)$ is compact for all $t \geq t_0$

Question

Problem: Let $(T(t))_{t \geq 0}$ be a strongly continuous semigroup on a Banach space $X$. Show that if $T(t_0)$ is compact for some $t_0 > 0$, then $T(t)$ is compact for all $t \geq t_0$.

It seems clear how to show that using the fact that $T(t)$ is compact if and only if for every sequence $(x_n)_n \subset U$ from the unit ball $U \subset X$, the sequence of images $(T(t)x_n)_n$ contains a Cauchy subsequence. If $(T(t) x_{n_k})_k$ is this Cauchy subsequence, then using $T(t) = T(t-t_0) T(t_0)$, \begin{equation} \|T(t)(x_{n_k}-x_{n_l})\| \leq \| T(t-t_0)\| \, \|T(t_0)(x_{n_k}-x_{n_l})\| \end{equation} and since $\| T(t-t_0)\| < \infty$ and $(T(t_0)x_{n_k})_k$ is Cauchy, we have $(T(t)x_{n_k})_k$ is Cauchy.

But there are other equivalent conditions for compact operators, e.g. requiring that the image of any bounded set is relatively compact (wikipedia). Is there an equally direct way to show the above using these alternative conditions?

Thanks in advance!

Chris

Answer 1

For a bounded set $B\subset X$, and $t\geq t_0$,

$\overline{T(t)(B)}=\overline{T(t-t_0)T(t_0)(B)}\subset T(t-t_0)\overline{T(t_0)(B)}$, as $T(t-t_0)$ is continuous.

Then $T(t-t_0)\overline{T(t_0)(B)}$ is compact since $T(t-t_0)$ is continuous and $\overline{T(t_0)(B)}$ is compact. So the closed subset $\overline{T(t)(B)}=\overline{T(t-t_0)T(t_0)(B)}$ is compact, that is $T(t)(B)$ is relatively compact.