The translation semigroup is not analytic

Strong continuity and isometries. Each $T(t)$ is linear and, by translation invariance of the $L^p$ norm [L4], an isometry with $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$; the functional equation $T(t+s)=T(t)T(s)$ and $T(0)=I$ are immediate, and the orbit of $f\in C_c(\mathbb{R})$ is continuous at $0$ by [L1]; for general $f\in L^p$ and $\epsilon >0$ choose $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }f-\phi {\mathrm{\parallel }}_p<\epsilon$ by [L1] and estimate $\mathrm{\parallel }T(t)f-f{\mathrm{\parallel }}_p\le \mathrm{\parallel }T(t)(f-\phi ){\mathrm{\parallel }}_p+\mathrm{\parallel }T(t)\phi -\phi {\mathrm{\parallel }}_p+\mathrm{\parallel }\phi -f{\mathrm{\parallel }}_p\le 2\epsilon +\mathrm{\parallel }T(t)\phi -\phi {\mathrm{\parallel }}_p$, so strong continuity extends to all of $X$; hence $(T(t){)}_{t\ge 0}$ is a strongly continuous semigroup of isometries.

The generator is differentiation on $W^{1,p}$. If the difference quotient of $f$ converges in $L^p$ to $g$, then for every test function $\phi$ the substitution $y=x+h$ and translation invariance give $\int g\phi ={\lim }_h\int f(y)\frac{\phi (y-h)-\phi (y)}{h}dy=-\int f{\phi }^{\mathrm{\prime }}$ (uniform convergence of the difference quotients of $\phi$ with compactly supported domination), so $g=D^{1}f$ weakly and $f\in W^{1,p}(\mathbb{R})$ by [L3]; conversely for $f\in W^{1,p}$ and $\phi \in C_c^{\mathrm{\infty }}$ one has $\mathrm{\parallel }\frac{T(h)\phi -\phi }{h}-{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$ while the shift inequality $\mathrm{\parallel }(T(h)-I)u{\mathrm{\parallel }}_p\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_p$ for $u\in W^{1,p}$ follows from $u(x+h)-u(x)={\int }_0^h{u}^{\mathrm{\prime }}(x+s)ds$ for smooth $u$ and extends by the density [L2]; given $\delta >0$ choose $\phi \in C_c^{\mathrm{\infty }}$ with $\mathrm{\parallel }f-\phi {\mathrm{\parallel }}_{W^{1,p}}<\delta$, write $u=f-\phi$, and estimate $\mathrm{\parallel }\frac{T(h)f-f}{h}-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\le \mathrm{\parallel }\frac{T(h)\phi -\phi }{h}-{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_p+2\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_p\le \mathrm{\parallel }\frac{T(h)\phi -\phi }{h}-{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_p+2\delta$, so the limsup is $0$ and $W^{1,p}(\mathbb{R})\subseteq D(A)$ with $Af=f^{\mathrm{\prime }}$; hence $D(A)=W^{1,p}(\mathbb{R})$.

The range obstruction. Translation commutes with weak differentiation: if $v=D^{1}u$ and $h\in \mathbb{R}$, then for every test function $\phi$ the identities $\int u(x+h){\phi }^{\mathrm{\prime }}(x)dx=\int u(y){\phi }^{\mathrm{\prime }}(y-h)dy=-\int v(y)\phi (y-h)dy=-\int v(x+h)\phi (x)dx$ show $v(\cdot +h)=D^1(u(\cdot +h))$; hence if $T(t)f=f(\cdot +t)\in W^{1,p}(\mathbb{R})$ for some $t>0$, then $f=T(-t)(T(t)f)$ is a translate of a $W^{1,p}$ function and lies in $W^{1,p}(\mathbb{R})$, a contradiction whenever $f\in L^p\setminus W^{1,p}$; such $f$ exist, for instance $f={\mathbf{1}}_{[0,1]}$, whose would-be weak derivative must vanish a.e. off the two jump points by testing away from them, hence a.e. everywhere, which is incompatible with ${\int }_{\mathbb{R}}f{\phi }^{\mathrm{\prime }}=\phi (1)-\phi (0)$ for a test function with $\phi (1)\mathrm{\ne }\phi (0)$; therefore $T(t)f\mathrm{\notin }D(A)$ for every $t>0$.

The translation semigroup is not analytic. If $T$ had any analytic extension, norm differentiability at $t>0$ would give $h^{-1}(T(h)-I)T(t)y=h^{-1}(T(t+h)-T(t))y\to T^{\mathrm{\prime }}(t)y$ for every $y$. The generator definition [L5] would therefore put $T(t)y$ in $D(A)$, with no global sector bound required; [step 1.3] exhibits a vector $f\in X$ with $T(t)f\mathrm{\notin }D(A)$ for every $t>0$, while [step 1.2] identifies $D(A)=W^{1,p}$ as the domain of the generator, so $(T(t){)}_{t\ge 0}$ admits no analytic extension with generator $A$, bounded or not.

The half-planes lie in the resolvent set. For $\mathrm{Re}\lambda >0$ and $f\in X$ put $R_{\lambda }f:={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}T(t)fdt$ and for $\mathrm{Re}\lambda <0$ put $R_{\lambda }f:=-{\int }_0^{\mathrm{\infty }}{e}^{\lambda t}T(-t)fdt$; both are absolutely convergent Bochner integrals by [L8] because the isometries of [step 1.1] give $\mathrm{\parallel }{e}^{-\lambda t}T(t)f\mathrm{\parallel }\le e^{-\mathrm{Re}\lambda t}\mathrm{\parallel }f\mathrm{\parallel }$, and the substitution $y=x+t$ (respectively $y=x-t$) with [L4] gives for every test function $\phi$ the identity $\int (R_{\lambda }f){\phi }^{\mathrm{\prime }}=\int f\phi -\lambda \int (R_{\lambda }f)\phi$ in both cases, so $R_{\lambda }f\in W^{1,p}(\mathbb{R})$ with weak derivative $\lambda R_{\lambda }f-f$, that is $(\lambda I-A)R_{\lambda }f=f$ with $A$ as in [step 1.2]; conversely, for $g\in W^{1,p}(\mathbb{R})$ the same Fubini computation with the weak-derivative identity of [L3] gives $\int (R_{\lambda }(\lambda g-g^{\mathrm{\prime }}))\phi =\int g\phi$ for every test function $\phi$, so $R_{\lambda }(\lambda I-A)g=g$; hence $\lambda \in \rho (A)$ with $R(\lambda ,A)=R_{\lambda }$ for every $\lambda \mathrm{\notin }\mathrm{i}\mathbb{R}$.

The imaginary axis lies in the spectrum. Fix $\xi \in \mathbb{R}$, choose $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\phi \equiv ̸0$ and set $u_N(x):=\phi (x\mathrm{/}N)e^{\mathrm{i}\xi x}\in W^{1,p}(\mathbb{R})$ for $N\ge 1$; then $(A-\mathrm{i}\xi )u_N=u_N^{\mathrm{\prime }}-\mathrm{i}\xi u_N=N^{-1}{\phi }^{\mathrm{\prime }}(x\mathrm{/}N)e^{\mathrm{i}\xi x}$ by [step 1.2], and by the substitution $x=Ny$ in the Lebesgue integrals $\mathrm{\parallel }(A-\mathrm{i}\xi )u_N{\mathrm{\parallel }}_p=N^{1\mathrm{/}p-1}\mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_p$ while $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_p=N^{1\mathrm{/}p}\mathrm{\parallel }\phi {\mathrm{\parallel }}_p$, so the ratios $\mathrm{\parallel }(A-\mathrm{i}\xi )u_N{\mathrm{\parallel }}_p\mathrm{/}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_p=N^{-1}\mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_p\mathrm{/}\mathrm{\parallel }\phi {\mathrm{\parallel }}_p$ tend to $0$; if $\mathrm{i}\xi$ lay in $\rho (A)$ the bounded inverse would give the positive lower bound $\mathrm{\parallel }(A-\mathrm{i}\xi )u{\mathrm{\parallel }}_p\ge \mathrm{\parallel }u{\mathrm{\parallel }}_p\mathrm{/}\mathrm{\parallel }R(\mathrm{i}\xi ,A)\mathrm{\parallel }$ for all $u\in D(A)$, a contradiction for large $N$; hence $\mathrm{i}\xi \in \sigma (A)$ for every $\xi$, that is $\mathrm{i}\mathbb{R}\subseteq \sigma (A)$.

The spectral obstruction and the conclusion. By [step 2.2] and [step 2.3] one has $\sigma (A)=\mathrm{i}\mathbb{R}$; since the point $\mathrm{i}$ has argument $\pi \mathrm{/}2$, it lies in ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$ for every $\delta >0$, so no sector ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$ is contained in $\rho (A)$ and [L7] rules out sectoriality of every positive exponent; the characterisation theorem [L7] therefore rules out generation of a bounded analytic semigroup, independently of the range obstruction of [step 2.1], which already excludes every analytic extension; the restriction $p<\mathrm{\infty }$ is essential, since right translation is not strongly continuous on $L^{\mathrm{\infty }}$, and the argument uses no choice principle beyond Dependent Choice, which implies the Countable Choice required by the $L^p$ translation-continuity, density and Sobolev vocabulary.