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The infinitesimal generator of a semigroup is the linear operatorwith generator domainFor example, the Bochner integralbelongs to $D(A)$ for every $u\in H$ and $t>0$, since $A{u}_t=U(t)u-u$.

To prove that $A$ is a closed linear operator, suppose $u_n\in D(A)$, $u_n\to u$, and $A{u}_n\to v$. For vectors in the generator domain,Passing to the limit in the Banach space givesAfter division by $t$, strong continuity makes the right side converge to $v$ as $t\downarrow 0$. Hence $u\in D(A)$ and $Au=v$, so $A$ is closed.

For $\mathrm{Re}z>\omega$, define the Bochner integralIt converges absolutely becauseIntegrating 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

A C0-semigroup on a Banach space $X$ is a family $U(t)\in \mathcal{B}(X)$ such thatfor every $x\in X$. Its infinitesimal generator of a semigroup iswith generator domain

For $M\ge 1$ and $\omega \in \mathbb{R}$, write $A\in \mathcal{G}(M,\omega )$ when $A$ generates a $C_0$-semigroup satisfying $\Vert U(t)\Vert \le M{e}^{\omega t}$. The Hille-Yosida theorem states that this holds exactly when $A$ is closed and densely defined,and, for every real $\lambda >\omega$ and every integer $n\ge 1$,The estimates for every resolvent power, rather than only $n=1$, are essential when $M>1$.