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Let $X$ be a Banach space, $M\ge 1$, and $\omega \in \mathbb{R}$. A linear operator $A:D(A)\subseteq X\to X$ generates a C0-semigroup $S(t)$ with $\Vert S(t)\Vert \le M{e}^{\omega t}$ exactly when $A$ is closed, $D(A)$ is dense, every real $\lambda >\omega$ belongs to the resolvent set, andThe same constant $M$ must work for every power. When $M=1$, the first-power bound implies all the power bounds by submultiplicativity. For general $M$, multiplying first-power bounds gives $M^n$, which is insufficient.^[ref]

Necessity follows from the Laplace-transform formula for a semigroup resolvent:The Bochner integral is bounded by $M\Vert x\Vert /(\lambda -\omega {)}^n$. Sufficiency constructs the semigroup from bounded Yosida approximations, using all the resolvent-power estimates for a common growth bound. For $M=1$ and $\omega =0$, this characterizes generators of contraction semigroups, providing a route from an unbounded differential linear operator to a well-posed time evolution.

A strongly continuous semigroup on the complex Banach space $E$ is a family $T_t\in \mathcal{L}(E)$ with $T_0=I$, $T_{s+t}=T_s{T}_t$ and $\Vert T_{t}f-f\Vert \to 0$ as $t\downarrow 0$ for every $f\in E$. Its semigroup generator is the linear operatorFor $f\in D(Z)$, boundedness of $T_t$ and the semigroup identity giveThus semigroup commutation with its unbounded generator meansThere is no assertion that $Z$ is defined on every vector. The orbit is differentiable on this domain, with $\frac{d}{dt}{T}_{t}f=T_{t}Zf$.

A contraction semigroup additionally satisfies $\Vert T_t\Vert \le 1$ for every $t\ge 0$. For $\mathrm{Re}\lambda >0$, define the Bochner integralStrong continuity gives a continuous integrand, and its norm is bounded by $e^{-t\mathrm{Re}\lambda }\Vert f\Vert$, so the integral exists andChanging variables in $T_h{L}_{\lambda }f$ yieldsAfter subtracting $L_{\lambda }f$ and dividing by $h$, the limit is $\lambda L_{\lambda }f-f$. ConsequentlyFor $f\in D(Z)$, integrate the derivative of $e^{-\lambda t}{T}_{t}f$ over $[0,\infty )$. Its limit at infinity is zero by the contraction bound, and its value at zero is $f$. ThereforeThus this is also the Laplace-transform formula for a semigroup resolvent, with $L_{\lambda }=(\lambda I-Z{)}^{-1}$.

To prove that $Z$ is a closed operator, suppose $f_n\in D(Z)$, $f_n\to f$ and $Z{f}_n\to g$. Fix one $\lambda$ with positive real part. The identity just proved and boundedness of $L_{\lambda }$ giveIts range lies in $D(Z)$, and applying $\lambda I-Z$ shows $Zf=g$. Thus the graph is closed. Linearity follows directly from linearity of the defining difference quotient, so $Z$ is a closed linear operator.

The closedness conclusion also holds without the contraction assumption. For a general strongly continuous semigroup, the Uniform boundedness principle gives a uniform operator bound on each compact time interval. Integrating the generator-orbit identity gives $T_t{f}_n-f_n={\int }_0^t{T}_{s}Z{f}_{n}ds$. Under $f_n\to f$ and $Z{f}_n\to g$, pass to the limit to obtain $T_{t}f-f={\int }_0^t{T}_{s}gds$. Divide by $t$ and let $t\downarrow 0$; strong continuity makes the right side converge to $g$, so $f\in D(Z)$ and $Zf=g$.

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

The Laplace-transform formula for a semigroup resolvent is locally uniformly convergent in the half-plane $\mathrm{Re}z>\omega$, so it may be differentiated under the Bochner integral:The resolvent identity gives $R^{\prime }(z)=-R(z{)}^2$ and, by mathematical induction,ThereforeFor real $\lambda >\omega$, the integral triangle inequality and the gamma integral giveConsequentlyThe exponent $-n+1$ printed in the question is a typographical error: already at $n=0$ it contradicts the first-resolvent bound.

Let $E$ be the underlying Banach space. By the Laplace-transform formula for a semigroup resolvent,These are Bochner integrals, defined for every $f\in E$ and $\lambda >0$. The contraction semigroup bound and semigroup property giveInterchanging and regrouping these Bochner integrals is justified by the integrable bound $e^{-\lambda s-\mu t}\Vert f\Vert$, whose integral is $\Vert f\Vert /(\lambda \mu )$. For $\lambda \ne \mu$, set $v=s+t$ and evaluate the inner scalar integral:Since $f$ was arbitrary, the resolvent identity follows:For $\lambda =\mu$, both sides vanish, so that case holds as well.

For a contraction semigroup with semigroup generator $L$, the bounded operators $L_{\lambda }$ approximate $L$ on its generator domain. The Laplace-transform formula for a semigroup resolvent and strong continuity give $\lambda R_{\lambda }g\to g$ for every vector $g$: split ${\int }_0^{\infty }\lambda e^{-\lambda t}(P_{t}g-g)dt$ into a short interval and an exponentially small tail. For $f\in D(L)$, integration by parts gives $R_{\lambda }Lf=\lambda R_{\lambda }f-f$, so $L_{\lambda }f=\lambda R_{\lambda }Lf\to Lf$. No operator norm convergence of $\lambda R_{\lambda }$ is implied.