= Solution
For $y\in D(A)$, the <semigroup property> gives
$$
\frac{U(h)U(t)y-U(t)y}{h}
=U(t)\frac{U(h)y-y}{h}\longrightarrow U(t)Ay.
$$
Hence $U(t)y\in D(A)$ and
$$
AU(t)y=U(t)Ay.
$$
The <generator domain> is therefore an <invariant subspace>, and the <operator norm> bound gives
$$
\begin{aligned}
\|\widetilde U(t)y\|_Y
&=\|U(t)y\|+\|AU(t)y\|\\
&=\|U(t)y\|+\|U(t)Ay\|\\
&\leq Me^{\omega t}\bigl(\|y\|+\|Ay\|\bigr).
\end{aligned}
$$
Thus
$$
\boxed{\|\widetilde U(t)y\|_Y\leq Me^{\omega t}\|y\|_Y}.
$$
Moreover, <strong continuity> applied separately to $y$ and $Ay$ gives
$$
\|\widetilde U(t)y-y\|_Y
=\|U(t)y-y\|+\|U(t)Ay-Ay\|\longrightarrow0.
$$
Therefore the restrictions form the <semigroup restricted to its generator domain>. Its derivative at zero exists in the <graph norm> exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is
$$
\boxed{A|_{D(A^2)},\qquad D(A^2)=\{y\in D(A):Ay\in D(A)\}}.
$$
Back to article page
