I don't really have an idea how to attack this problem, and I also wasn't able to find a proof.
Let $Z$ be the generator of a semigroup of contraction on a Banach space $X$. We define $$ D(Z^{2})=\{u \in D(Z); \quad Zu \in D(Z) $$ Show the following Identity: $$ T(t) u-u=t Z u+\int_0^t (t-s) T(s) Z^2 u \,ds $$ Show that if $u \in D(Z^2),$ then $$ \|Z u\|^2 \leq 4\|Z^2 u\|\|u\| $$
I don't have a strong background in Operator Semigroups so some of these questions might be quite basic.
Let $u\in D(Z^2)$.
1) For all $x\in D(Z)$ and all $\tau\ge 0$ we have $$T(\tau)x=x +\int_0^\tau T(s)Zx \,ds.$$ For $x=Zu$, we get $$T(\tau) Zu=Zu +\int_0^\tau T(s)Z^2u \,ds. \quad (*).$$ Integrating $(*)$ over $(0,t)$ with Fubini theorem yields the result.
2) It's just an application of the equality: \begin{align*} t \|Z u\|&=\|T(t) u-u-\int_0^t (t-s) T(s) Z^2 u \,ds\|\\ & \le \|T(t)\| \|u\|+\|u\| +\int_0^t (t-s) \|T(s)\| \|Z^2 u\| \,ds\\ & \le 2\|u\| + \left(\int_0^t (t-s) \,ds \right) \|Z^2 u\|\\ & \le 2 \|u\| + \frac{1}{2}t^2 \|Z^2 u\|. \end{align*} So, the polynomial $\frac{1}{2}t^2 \|Z^2 u\| - t \|Z u\| + 2 \|u\|$ is nonnegative. Hence its discriminant is nonpositive. That is, $$ \|Z u\|^2 \leq 4\|Z^2 u\|\|u\|. $$