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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

For $y\in D(A)$, the semigroup property givesHence $U(t)y\in D(A)$ andThe generator domain is therefore an invariant subspace, and the operator norm bound givesThusMoreover, strong continuity applied separately to $y$ and $Ay$ givesTherefore the restrictions form the semigroup restricted to its generator domain. Its derivative at zero exists in the graph norm exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is

The familyinherits the identity, semigroup property, and strong continuity from $U$, whileIts difference quotient satisfiesfor $u\in D(A)$. Conversely, existence of this limit implies existence of the generator limit for $U$, so $D(\hat{A})=D(A)$ andThis is the exponentially shifted semigroup construction.

The Hille-Yosida theorem in the uniformly bounded case says that a linear operator $B$ on a Banach space generates a C0-semigroup with $\Vert e^{tB}\Vert \le M$ if and only if:

A solution operator for a nonautonomous evolution equation is an evolution family $U(t,s)$ satisfyingand, on a suitable common domain $\mathcal{D}$,

One applicable nonautonomous generation theorem is the following. Suppose $\mathcal{D}$ is a dense linear subspace of $H$, each $A(t)$ has domain $\mathcal{D}$, the family is a stable family of semigroup generators with constants $M,0$, and $t\mapsto A(t)$ is continuously differentiable as a map from $[0,T]$ to $\mathcal{B}(\mathcal{D},H)$, where $\mathcal{D}$ carries one of the uniformly equivalent graph norms. Then there is a unique evolution family such that:

For the uniform partition $t_j=s+j(t-s)/N$, the frozen-generator product approximation isAs $N\to \infty$, $U_N(t,s)u\to U(t,s)u$ in the norm of $H$ for every $u\in H$, uniformly for $(t,s)$ in the compact time triangle $0\le s\le t\le T$. This is convergence in the strong operator topology, rather than convergence in the operator norm.

It remains to verify the second differential equation. The evolution family law gives, for $h>0$,Divide by $h$. Since $u\in \mathcal{D}$,while strong continuity gives $U(t,s+h)A(s)u\to U(t,s)A(s)u$. ThereforeThe left derivative follows in the same way, so $s\mapsto U(t,s)u$ is differentiable.

Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.