The setting consists of a Banach space $X$, a possibly unbounded, densely defined operator $A:X \rightarrow X$ with domain $D(A)$ and an evolution equation \[ \frac{d}{dt}u(t) = Au(t) \] for all $t>0$ and initial condition $u(0)=x \in X$ (not only $D(A)$). A solution to this problem is denoted $u(t)=T(t)x$ with an evolution operator that forms an evolution semigroup. The operator $A$ is then called the generator.

Definition. A one-parameter family $T(t)$, $t \geq 0$, in $\mathcal{B}(X,X)$ is called a $\mathcal{C}^0$ semigroup (strongly continuous semigroup) if

1. $T(0)=\mathrm{id}$,
2. $T(s+t)=T(t)T(s)$ for all $t,s \geq 0$, and
3. $\lim_{t \rightarrow 0} \|T(t)x-x\| = 0$ for all $x\in X$.

Definition. The infinitesimal generator $A$ of a $\mathcal{C}^0$ semigroup $T(t)$ is defined by \[ Ax = \lim_{t \rightarrow 0} \frac{1}{t}(T(t)x-x) \] with domain $D(A)$, i.e. all $x\in X$ for which the limit exists.

Theorems that show the existence of such a semigroup by demanding specific properties from its generator are called generation theorems. An example is Stone's theorem in the Hilbert space case and the theorems of Hille-Yoshida and Lumer-Phillips in the more general setting of Banach spaces. The important relation between the semigroup and the Cauchy problem above is given by the following theorem with two different types of solutions (discussed later).

Theorem. Let $T(t)$ be a $\mathcal{C}^0$ semigroup and $A$ its infinitesimal generator. Then it holds for $t>0$ that

1. if $x \in X$ then $\int_0^t T(s) x \,ds \in D(A)$ and $A \int_0^t T(s) x \,ds = T(t) x - x$,
2. if $x \in D(A)$ then $T(t) x \in D(A)$ and $\frac{d}{dt} T(t)x = A T(t) x = T(t) A x$.

Proof. (1.) Take $h >0$, then \begin{align*} \frac{T(h)-\id}{h} \int_0^t T(s) x \,ds &= \frac{1}{h} \int_0^t (T(s+h)x-T(s)x) \,ds \\ &= \frac{1}{h} \int_t^{t+h} T(s) x \,ds - \frac{1}{h} \int_0^h T(s) x \,ds \end{align*} and with $h \searrow 0$ the right-hand side goes to $T(t)x-x$.

(2.) Now because of boundedness of $T(t)$ we have as $h \searrow 0$ \[ \frac{T(h)-\id}{h} T(t) x = T(t) \frac{T(h)-\id}{h} x \longrightarrow T(t) A x \] thus $T(t)x \in D(A)$ and $AT(t)x = T(t)Ax$ as well as the right derivative of $T(t)x$ fulfilling \[ \frac{d^+}{dt} T(t)x = A T(t) x = T(t) A x. \] To conclude we have to show the same for the left derivative. \begin{align*} &\lim_{h \searrow 0} \frac{T(t)x-T(t-h)x}{h} - T(t)Ax \\ &= \lim_{h \searrow 0} T(t-h) \left( \frac{T(h)x-x}{h} - Ax \right) + \lim_{h \searrow 0} (T(t-h)Ax - T(t)Ax) \end{align*} Both limit terms vanish, the first due to $x \in D(A)$ and boundedness of $T(t-h)$, the second by strong continuity of $T(t)$.