Generator of a semigroup

Remarks

The generator is closed and densely defined, and it determines the semigroup. For $x \in D(A)$, the map $u(t) = T(t)x$ is differentiable, stays in $D(A)$, and satisfies $u'(t) = Au(t)$. One writes $T(t) = e^{tA}$.

Example: let $B$ be a bounded operator on $X$ and $T(t) = e^{tB} = \sum_{k \ge 0} t^k B^k / k!$, the exponential of a bounded operator. Then $\norm{(T(t)x - x)/t - Bx} \le \sum_{k \ge 2} t^{k-1}\norm{B}^k \norm{x}/k! \to 0$, so the generator is $B$ with $D(B) = X$.

Example: on $X = C_0(\R)$, the continuous functions that tend to $0$ at infinity with the sup norm, let $(T(t)f)(s) = f(s + t)$. Its generator is $Af = f'$ with domain the $C^1$ functions $f$ such that $f$ and $f'$ lie in $C_0(\R)$.

Non-example: the operator $f \mapsto f'$ on smooth functions with compact support is not the generator of this shift semigroup. It is a proper restriction of $A$: the limit in the definition exists for more functions, and the generator is defined on all of them.

The Hille–Yosida theorem and the Lumer–Phillips theorem say which operators are generators.