= Solution
The <infinitesimal generator of a semigroup> is the <linear operator>
$$
Au=\lim_{h\downarrow0}\frac{U(h)u-u}{h}
$$
with <generator domain>
$$
D(A)=\left\{u\in H:\lim_{h\downarrow0}\frac{U(h)u-u}{h}\text{ exists in }H\right\}.
$$
For example, the <Bochner integral>
$$
u_t=\int_0^tU(s)u\,ds
$$
belongs to $D(A)$ for every $u\in H$ and $t>0$, since $Au_t=U(t)u-u$.
To prove that $A$ is a <closed linear operator>, suppose $u_n\in D(A)$, $u_n\to u$, and $Au_n\to v$. For vectors in the <generator domain>,
$$
U(t)u_n-u_n=\int_0^tU(s)Au_n\,ds.
$$
Passing to the <limit> in the <Banach space> gives
$$
U(t)u-u=\int_0^tU(s)v\,ds.
$$
After division by $t$, <strong continuity> makes the right side converge to $v$ as $t\downarrow0$. Hence $u\in D(A)$ and $Au=v$, so $A$ is closed.
For $\operatorname{Re}z>\omega$, define the <Bochner integral>
$$
R(z)u=\int_0^\infty e^{-zt}U(t)u\,dt.
$$
It converges absolutely because
$$
\|e^{-zt}U(t)u\|\leq Me^{-(\operatorname{Re}z-\omega)t}\|u\|.
$$
Integrating the semigroup difference quotient shows that $R(z)u\in D(A)$ and $(zI-A)R(z)u=u$. The same computation for $u\in D(A)$ gives $R(z)(zI-A)u=u$. Thus the <Laplace-transform formula for a semigroup resolvent> proves
$$
\boxed{\{z\in\mathbb C:\operatorname{Re}z>\omega\}\subseteq\rho(A)},
\qquad
\|R(z)\|\leq\frac{M}{\operatorname{Re}z-\omega}.
$$
Back to article page
