The generator is closed and densely defined

Density: for $x\in X$ and $t>0$, [F1] gives $J_{t}x\in D(A)$, and by [F2] $\frac{1}{t}{J}_{t}x\to x$ as $t\downarrow 0$; hence $x$ lies in the closure of $D(A)$. Since $x$ was arbitrary, $D(A)$ is dense in $X$.

Closedness: suppose $x_n\in D(A)$ with $x_n\to x$ and $A{x}_n\to y$ in $X$. For fixed $h>0$ and every $n$, the orbit of $x_n$ is differentiable with derivative $T(s)A{x}_n$, so [F3] gives $T(h)x_n-x_n={\int }_0^{h}T(s)A{x}_{n}ds$.

As $n\to \mathrm{\infty }$, the left-hand side tends to $T(h)x-x$, because $T(h)$ is bounded and the orbit of $x$ is continuous; the right-hand side tends to ${\int }_0^{h}T(s)yds$, because $\parallel {\int }_0^{h}T(s)(A{x}_n-y)ds\parallel \le h{\sup }_{0\le s\le h}\mathrm{\parallel }T(s)\mathrm{\parallel }\mathrm{\parallel }A{x}_n-y\mathrm{\parallel }\to 0$ by the local bound [F3]. Hence $T(h)x-x={\int }_0^{h}T(s)yds$ for every $h>0$.

Dividing by $h$ and using the average-convergence limit of [F2] for the continuous curve $s\mapsto T(s)y$ gives $\frac{T(h)x-x}{h}=\frac{1}{h}{\int }_0^{h}T(s)yds\to y$ as $h\downarrow 0$. By the definition of the generator, $x\in D(A)$ and $Ax=y$; hence the graph of $A$ contains the limits of all convergent graph sequences, and under DC (hence Countable Choice) the closure-sequence criterion in Infinitesimal generator of a C0-semigroup makes the graph closed.

Together with [step 1.1], the generator of a strongly continuous semigroup is a closed and densely defined linear operator, in the sense of [F4].