A multiplication semigroup with an unbounded generator

For every $t\ge 0$ the map $T(t)$ is linear and $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_p^p={\int }_0^{\mathrm{\infty }}{e}^{-tps}\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\le \mathrm{\parallel }f{\mathrm{\parallel }}_p^p$, so $T(t)$ is a contraction; the pointwise identities $e^{-(t+r)s}=e^{-ts}{e}^{-rs}$ and $e^0=1$ give $T(t+r)=T(t)T(r)$ and $T(0)=I$.

Strong continuity: for fixed $f\in X$, $\mathrm{\parallel }T(t)f-f{\mathrm{\parallel }}_p^p={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{e}^{-ts}-1{\mathrm{\mid }}^p\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\to 0$ as $t\downarrow 0$ by [F2], since $e^{-ts}\to 1$ pointwise and $\mathrm{\mid }{e}^{-ts}-1{\mathrm{\mid }}^p\le 2^p$ for $t\ge 0$, so the integrand is dominated by the $L^1$ function $2^p\mathrm{\mid }f{\mathrm{\mid }}^p$.

Inclusion $\{qf\in X\}\subseteq D(A)$: if $qf\in X$, then $\parallel \frac{T(h)f-f}{h}-qf{\parallel }_p^p={\int }_0^{\mathrm{\infty }}\mid \frac{e^{-hs}-1}{h}+s{\mid }^p\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\to 0$ by [F2], because $\frac{e^{-hs}-1}{h}\to -s$ pointwise and, by the inequality $1-e^{-x}\le x$ for $x\ge 0$, the bracket is at most $2s$, so the integrand is dominated by $(2s\mathrm{\mid }f\mathrm{\mid }{)}^p\in L^1$.

Converse: suppose the difference quotients converge in $X$ to some $g$, and let $\phi \in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$. By [F2] and the boundedness of $s$ on $\mathrm{supp}\phi$, ${\int }_0^{\mathrm{\infty }}\frac{e^{-hs}-1}{h}f\phi ds\to -{\int }_0^{\mathrm{\infty }}sf\phi ds$, while ${\int }_0^{\mathrm{\infty }}\frac{T(h)f-f}{h}\phi ds\to {\int }_0^{\mathrm{\infty }}g\phi ds$ because $\mathrm{\parallel }\frac{T(h)f-f}{h}-g{\mathrm{\parallel }}_p\to 0$ and $\phi \in L^{p^{\mathrm{\prime }}}$; hence ${\int }_0^{\mathrm{\infty }}(g-qf)\phi ds=0$ for every test function. The locally integrable function $g-qf$ has $sf\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(0,\mathrm{\infty })$, so [F3] gives $g=qf$ almost everywhere; in particular $qf=g\in X$ and $f\in D(A)$ with $Af=qf$.

Unboundedness and proper domain: for $n\ge 1$ let $f_n$ be the normalised nonnegative bump supported in $[n,n+1]$ with $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_p=1$. Then $\mathrm{\parallel }A{f}_n{\mathrm{\parallel }}_p^p={\int }_n^{n+1}{s}^p\mathrm{\mid }{f}_n(s){\mathrm{\mid }}^{p}ds\ge n^p$, so $\mathrm{\parallel }A{f}_n{\mathrm{\parallel }}_p\ge n\to \mathrm{\infty }$ while $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_p=1$, and no constant bounds $A$ on its domain. Moreover $D(A)\mathrm{\ne }X$: the function $f(s):=s^{-1-1\mathrm{/}p}{\mathbf{1}}_{(1,\mathrm{\infty })}$ lies in $L^p(0,\mathrm{\infty })$ because ${\int }_1^{\mathrm{\infty }}{s}^{-p-1}ds=1\mathrm{/}p$, while $sf(s)=s^{-1\mathrm{/}p}$ is not in $L^p$ because ${\int }_1^{\mathrm{\infty }}{s}^{-1}ds=\mathrm{\infty }$, so $f\in X\setminus D(A)$.

Together with [step 1.1] and [step 2.1], the displayed claims follow: $T$ is a strongly continuous contraction semigroup on $L^p(0,\mathrm{\infty })$ whose generator has domain $\{f:qf\in X\}=\{f:sf\in L^p\}$ and acts by $Af=qf=-sf$, and this operator is unbounded.