If is $P(S(4^{-k}))$ true so $P(S(t))$ is too true?

Question

i am reading an article about Besov-Morrey spaces an there has the following sentense: "The semigroup propety of $e^{t\Delta}$ allows us to reduce easily to $t = 4^{−k}$ for $k\in \mathbb{Z}$". I could explain the details but it would take a lot of time, so to be brief my question is, if I want to prove somthing of the type $$t^{\frac{s}{2}}\|S(t)f\|_X\leq C\|f\|_Y, \forall t>0$$ is suficiently to prove that $$(4^{-k})^{\frac{s}{2}}\|S(4^{-k})f\|_X\leq C\|f\|_Y, \forall k \in \mathbb{Z}$$ provided that $S:[0, \infty)\to L(X)$ be a semigroup? $X$ may be a Banach space, for example. If not, exist some situation that it is true? For example for $S(t)=e^{t\Delta}$, where $$e^{t\Delta}f(x)=\left[(4\pi t)^{\frac{-n}{2}}e^{-\frac{\|x\|^2}{4t}}*f\right](x), \ x \in \mathbb{R}^n$$ I am trying use that $S(t+s)=S(t)S(s)$ or that $\lim_{t \to 0}\|S(t)-I\|=0$, but i cant formulate the idea. I thank every help.