The right-translation semigroup on Lp has the weak derivative as generator

$T$ is a strongly continuous semigroup of isometries: $T(t)$ is linear, $T(t+s)f=T(t)T(s)f$ and $T(0)=I$ hold pointwise, $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$ because translation preserves the integral of $\mathrm{\mid }f{\mathrm{\mid }}^p$ [F4], and $T(t)f\to f$ in $L^p$ as $t\downarrow 0$ by [F1].

Let $f\in W^{1,p}(\mathbb{R})$ and $h>0$. The curve $u\mapsto {\tau }_{-u}{f}^{\mathrm{\prime }}$ is continuous from $[0,h]$ to $L^p$ by [F1], so $M_h:=\frac{1}{h}{\int }_0^h{\tau }_{-u}{f}^{\mathrm{\prime }}du\in L^p$ is defined by [F5], and $\mathrm{\parallel }{M}_h-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\le {\sup }_{0\le u\le h}\mathrm{\parallel }{\tau }_{-u}{f}^{\mathrm{\prime }}-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$ as $h\downarrow 0$.

For $f\in W^{1,p}$ and $h>0$ the difference quotient $q_h:=\frac{T(h)f-f}{h}\in L^p$ equals $M_h$ almost everywhere. Indeed, for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, translation invariance [F4] gives ${\int }_{\mathbb{R}}{q}_h\phi =\frac{1}{h}({\int }_{\mathbb{R}}f(s)\phi (s-h)ds-{\int }_{\mathbb{R}}f(s)\phi (s)ds)=\frac{1}{h}{\int }_{\mathbb{R}}f(s)(\phi (s-h)-\phi (s))ds$; writing $\phi (s-h)-\phi (s)=-{\int }_0^h{\phi }^{\mathrm{\prime }}(s-u)du$ and applying Fubini [F5] and the weak-derivative identity of [F2] with the test function $\phi (\cdot -u)$, ${\int }_{\mathbb{R}}{q}_h\phi =-\frac{1}{h}{\int }_0^h{\int }_{\mathbb{R}}f(s){\phi }^{\mathrm{\prime }}(s-u)dsdu=\frac{1}{h}{\int }_0^h{\int }_{\mathbb{R}}{f}^{\mathrm{\prime }}(s)\phi (s-u)dsdu=\frac{1}{h}{\int }_0^h{\mathrm{\Lambda }}_{\phi }({\tau }_{-u}{f}^{\mathrm{\prime }})du$; by [F5] this equals ${\mathrm{\Lambda }}_{\phi }(M_h)={\int }_{\mathbb{R}}{M}_h\phi$. Two $L^p$ functions with the same pairing with every test function coincide almost everywhere by [F6].

Therefore $\parallel \frac{T(h)f-f}{h}-f^{\mathrm{\prime }}{\parallel }_p=\mathrm{\parallel }{M}_h-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$ as $h\downarrow 0$ for every $f\in W^{1,p}(\mathbb{R})$; by the definition of the generator [F7], $W^{1,p}(\mathbb{R})\subseteq D(A)$ and $Af=f^{\mathrm{\prime }}$ for $f\in W^{1,p}(\mathbb{R})$.

Conversely, suppose $f\in D(A)$, so that $q_h\to g$ in $L^p$ for some $g$. For every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, $\mathrm{\mid }\int (q_h-g)\phi \mathrm{\mid }\le \mathrm{\parallel }{q}_h-g{\mathrm{\parallel }}_p\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\to 0$ by [F3], so $\int g\phi ={\lim }_h\int q_h\phi$. On the other hand the identity of [step 1.3] (which used only $f\in L^p$) gives $\int q_h\phi =\frac{1}{h}\int f(s)(\phi (s-h)-\phi (s))ds$, and for $0<h\le 1$ the integrand is supported in a fixed compact interval $K$ and bounded there by $\mathrm{\mid }f\mathrm{\mid }{\sup }_K\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$, whose integral over $K$ is finite because $f\in L^p(K)\subseteq L^1(K)$; since $\frac{\phi (s-h)-\phi (s)}{h}\to -{\phi }^{\mathrm{\prime }}(s)$ pointwise, dominated convergence [F4] gives $\int g\phi =-\int f{\phi }^{\mathrm{\prime }}$ for every test function $\phi$. By the definition of the weak derivative [F2], $g$ is the weak derivative of $f$, so $f\in W^{1,p}(\mathbb{R})$ and $g=f^{\mathrm{\prime }}$ almost everywhere.

Combining [step 2.1] and [step 2.2], the generator of the right-translation semigroup is $Af=f^{\mathrm{\prime }}$ with $D(A)=W^{1,p}(\mathbb{R})$, and the difference quotients converge to $f^{\mathrm{\prime }}$ in $L^p$ for every $f\in W^{1,p}(\mathbb{R})$; the semigroup is strongly continuous by [step 1.1]. The verification assumes Countable Choice, inherited from the translation-continuity and distribution-embedding inputs.