Let $A$ be an unbounded linear operator of domain $D(A)$ defined on a Banach space $X$. Suppose that $A$ generates a $C\_0$-semigroup $T(t)$ which is uniformly bounded. I would like to know if there are additional assumptions under which the following conjecture is true:


For every compact subset $K$ of $X$, there exists a constant $C\_K$ such that
$$\mid{((T(t)-T(t'))x\mid \leq C\_K\mid t-t'\mid},\quad \forall t,t'\in \mathbb{R}\_+,\quad \forall x\in K.$$



A question about uniformly bounded semigroups
If I understand your question correctly, this would mean that the function $t\mapsto T(t)x$ is Lipschitz continuous, which is equivalent for $x$ to be in the Favard space $\text{Fav}(A)$$. See for example Defintion 8.2. in these [lecture notes](https://numerical-analysis.uibk.ac.at/isem-mathematik/images/1/10/ISEM15_Lecture8.pdf).


If your space $X$ is reflexive, then the Favard space is exactly $D(A)$, the domain of $A$, see [Corollary II.5.21](https://www.math.uni-tuebingen.de/de/forschung/agfa/members/engel-nagel_one-parameter-semigroups.pdf) in 


*Engel, Klaus-Jochen; Nagel, Rainer*, [**One-parameter semigroups for linear evolution equations**](http://dx.doi.org/10.1007/b97696), Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). [ZBL0952.47036](https://zbmath.org/?q=an:0952.47036).


Hence if $D(A)\neq X$ (which is the case for unbounded generators), this fails.


Unfortunately, in non-reflexive spaces, where $D(A)\subset \text{Fav}(A)\subset X$ still holds but the Favard space generally fails to be the whole space. Hence, your property cannot be fulfilled even for $K=\{x\}$ singletons if $x\notin \mathrm{Fav}(A)$.


