Well-posedness of the abstract Cauchy problem is equivalent to generation

Internal lemma (reduction to $X_1$), part 1: (EU) gives a semigroup on $X_1$. Put $T_1(t)x:=u(t,x)$ for $x\in X_1$. Uniqueness makes $T_1(t)$ linear and gives $T_1(0)=I$, $T_1(t+s)=T_1(t)T_1(s)$; the classical equation $u^{\mathrm{\prime }}=Au$ shows that $t\mapsto T_1(t)x$ is continuous into $X_1$ for each $x\in X_1$.

(a)$\Rightarrow$(b),(c),(d). If $A$ generates a semigroup $T$ with $\mathrm{\parallel }T(t)\mathrm{\parallel }\le M{e}^{\omega t}$: for $x\in D(A)$ the orbit $T(\cdot )x$ is a classical solution and any classical solution $v$ satisfies $v\equiv T(\cdot )x$ by the rigidity computation $\frac{d}{ds}[T(t-s)v(s)]=0$, so (EU) holds; [F2] gives $(\omega ,\mathrm{\infty })\subseteq \rho (A)$ and $({\lambda }_n-A)D(A)=X$ for any ${\lambda }_n\uparrow \mathrm{\infty }$ in that half-line, so (c) holds and $\rho (A)\mathrm{\ne }\mathrm{\varnothing }$; density and the local bound ${\sup }_{0\le s\le t_0}\mathrm{\parallel }T(s)\mathrm{\parallel }$ give, for $x_n\to 0$ in $D(A)$, ${\sup }_{0\le t\le t_0}\mathrm{\parallel }u(t,x_n)\mathrm{\parallel }\le {\sup }_{0\le s\le t_0}\mathrm{\parallel }T(s)\mathrm{\parallel }\mathrm{\parallel }{x}_n\mathrm{\parallel }\to 0$, which is (d).

(d)$\Rightarrow$(a). Define $T(t)x:=u(t,x)$ for $x\in D(A)$; uniqueness makes each $T(t)$ linear, and the semigroup law holds on $D(A)$ by uniqueness of solutions. The continuous-dependence hypothesis (d) transfers to a local bound: if no $\delta >0$ had $\mathrm{\parallel }T(t)x\mathrm{\parallel }\le 1$ for all $x\in D(A)$ with $\mathrm{\parallel }x\mathrm{\parallel }\le \delta$ and all $t\in [0,1]$, then choosing $x_n$ with $\mathrm{\parallel }{x}_n\mathrm{\parallel }\le 1\mathrm{/}n$ and $\mathrm{\parallel }T(t_n)x_n\mathrm{\parallel }>1$ would give a sequence $x_n\to 0$ with $u(t_n,x_n)\to ̸0$ uniformly, contradicting (d); hence ${\sup }_{0\le t\le 1}\mathrm{\parallel }T(t)x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}\delta$ for all $x\in D(A)$. Since $D(A)$ is dense, each $T(t)$ extends uniquely to a bounded operator on $X$ with the same bound, and the semigroup law and strong continuity extend by density, using $\mathrm{\parallel }T(t)\mathrm{\parallel }\le (M_1{)}^{n+1}$ on $[0,n]$ with $M_1:=\max \{1,1\mathrm{/}\delta \}$ from the semigroup law. The generator $B$ of the extension satisfies $A\subseteq B$, because on $D(A)$ the difference quotients are those of the classical solutions and converge to $Ax$. The extension leaves $D(A)$ invariant, so $D(A)$ is a core of $B$: for $x\in D(B)$ choose $x_j\in D(A)$ with $x_j\to x$; then $\frac{1}{t}{\int }_0^{t}T(s)x_{j}ds$ lies in the graph-norm closure of $D(A)$ (the integrand is $D(A)$-valued and graph-norm continuous) and the integrated-orbits identity $\frac{1}{t}{\int }_0^{t}T(s)xds\to x$ in graph norm as $t\downarrow 0$ together with $\frac{1}{t}{\int }_0^{t}T(s)x_{j}ds\to \frac{1}{t}{\int }_0^{t}T(s)xds$ in graph norm shows $x\in {\overline{D(A)}}^{\mathrm{\parallel }\cdot {\mathrm{\parallel }}_B}$. Since $A$ is closed, $A\subseteq B$ and $D(A)$ is a core of $B$, every graph limit from $D(A)$ remains in $\mathrm{\Gamma }(A)$, so $A=B$.

Boundedness of $T_1(t)$. For any Banach space $Z$, $C([0,t],Z)$ is Banach in the supremum norm: a uniformly Cauchy sequence converges pointwise by completeness, uniformly by its common Cauchy estimates, and its uniform limit is continuous by the three-term increment estimate. Fix $t>0$ and consider $\mathrm{\Phi }:X_1\to C([0,t],X_1)$, $\mathrm{\Phi }(x):=T_1(\cdot )x$. Its graph is closed: if $x_n\to x$ in $X_1$ and $\mathrm{\Phi }(x_n)\to f$ uniformly in $X_1$, then the integral identity $T_1(s)x_n=x_n+{\int }_0^{s}A{T}_1(r)x_{n}dr$ passes to the limit in $X$ and gives $f(s)=x+{\int }_0^{s}Af(r)dr$ for $s\le t$; the extension $\tilde{f}(s):=T_1(s-t)f(t)$ for $s>t$, $=f(s)$ for $s\le t$, then solves (ACP) with initial value $x$, so $\tilde{f}=T_1(\cdot )x$ by uniqueness and $f=\mathrm{\Phi }(x)$. By [F4] $\mathrm{\Phi }$ is bounded on the Banach space $X_1$, hence $T_1(t)\in \mathcal{B}(X_1)$ and $T_1$ is a strongly continuous semigroup on $X_1$.

The generator of $T_1$ is $A_1$. First $A{T}_1(t)x=T_1(t)Ax$ for $x\in D(A_1)$: the curve $f(t):=x+{\int }_0^t{T}_1(s)Axds$ is differentiable with $f^{\mathrm{\prime }}=T_1(t)Ax$ and satisfies $Af=f^{\mathrm{\prime }}$ (move $A$ inside the integral by the closed-graph argument in $X$), so $f=T_1(\cdot )x$ by uniqueness and $A{T}_1(t)x=T_1(t)Ax$. Hence for $x\in D(A_1)$ the quotient $\frac{1}{t}(T_1(t)x-x)$ converges to $Ax$ in $X$ and its $A$-image converges to $A^{2}x$ in $X$; that is, the convergence holds in $X_1$, so $A_1\subseteq B$. Conversely, if $x\in D(B)$, then $A\frac{1}{t}(T_1(t)x-x)$ converges in $X$ and $\frac{1}{t}(T_1(t)x-x)\to Ax$ in $X$; closedness of $A$ gives $Ax\in D(A)$, that is $x\in D(A_1)$. Thus $B=A_1$.

(b)$\Rightarrow$(a). Let $\lambda \in \rho (A)$. For $x\in X$ one has $x\in D(A)$ iff $(\lambda -A{)}^{-1}x\in D(A_1)$, and $Ax=(\lambda -A)A_1(\lambda -A{)}^{-1}x$ for $x\in D(A)$: indeed $(\lambda -A{)}^{-1}x=R$ and $AR=\lambda R-x$, and $AR\in D(A)$ exactly when $x\in D(A)$. Thus $S:=(\lambda -A{)}^{-1}:X\to X_1$ is a bounded isomorphism with bounded inverse $\lambda -A:X_1\to X$, and $A=S^{-1}{A}_{1}S$ with $D(A)=S^{-1}D(A_1)$. By [step 3.1] and the internal lemma, $A_1$ generates $T_1$ on $X_1$; then $T(t):=S^{-1}{T}_1(t)S$ is a strongly continuous semigroup on $X$ whose generator is $S^{-1}{A}_{1}S=A$, because the difference quotients of $T$ are those of $T_1$ conjugated by the bounded isomorphism $S$.

(c)$\Rightarrow$(b). By [step 3.1] the operator $A_1$ generates $T_1$ on $X_1$ (via [step 1.1] and [step 2.1]), so by [F2] its resolvent set contains a half-line $({\omega }_1,\mathrm{\infty })$; choose $\lambda ={\lambda }_n>{\omega }_1$ with $(\lambda -A)D(A)=X$. If $Ax=\lambda x$ for some $x\in D(A)$, then $Ax\in D(A)$, so $x\in D(A_1)$ and $A_{1}x=\lambda x$; since $\lambda \in \rho (A_1)$ this forces $x=0$. Hence $\lambda -A$ is injective and, by hypothesis, surjective, so it is bijective; being closed it has bounded inverse by [F4], and $\lambda \in \rho (A)$.

All implications are established, so (a)-(d) are equivalent; and in each direction the solution is $u(t,x)=T(t)x$ for the generated semigroup, as asserted.