Sectorial resolvent characterisation of bounded analytic semigroups

Complex Laplace representation. Let $(S(s){)}_{s\ge 0}$ be a strongly continuous semigroup with generator $B$ and $\mathrm{\parallel }S(s)\mathrm{\parallel }\le M{e}^{\omega s}$, and fix $\lambda \in \mathbb{C}$ with $\mathrm{Re}\lambda >\omega$; the integral $v:={\int }_0^{\mathrm{\infty }}{e}^{-\lambda s}S(s)xds$ converges absolutely with $\mathrm{\parallel }v\mathrm{\parallel }\le M(\mathrm{Re}\lambda -\omega {)}^{-1}\mathrm{\parallel }x\mathrm{\parallel }$, and $(S(h)-I)v\mathrm{/}h=((e^{\lambda h}-1)\mathrm{/}h){\int }_h^{\mathrm{\infty }}{e}^{-\lambda s}S(s)xds-h^{-1}{\int }_0^h{e}^{-\lambda s}S(s)xds\to \lambda v-x$ by absolute convergence and average convergence at $0$, so $v\in D(B)$ and $Bv=\lambda v-x$; if $y\in D(B)$ has $By=\lambda y$ then $s\mapsto e^{-\lambda s}S(s)y$ has derivative $e^{-\lambda s}S(s)(By-\lambda y)=0$ by the commutation lemma and the fundamental theorem, so $\mathrm{\parallel }y\mathrm{\parallel }=e^{-\mathrm{Re}\lambda s}\mathrm{\parallel }S(s)y\mathrm{\parallel }\le M{e}^{(\omega -\mathrm{Re}\lambda )s}\mathrm{\parallel }y\mathrm{\parallel }\to 0$; and for $y\in D(B)$ the same computation gives ${\int }_0^{\mathrm{\infty }}{e}^{-\lambda s}S(s)(\lambda y-By)ds=y$, so the bounded linear map $x\mapsto v$ is a two-sided inverse of $\lambda I-B$ and $\lambda \in \rho (B)$ with $R(\lambda ,B)x=v$. Its graph is closed by continuity; swapping coordinates shows that $\lambda I-B$ has closed graph, and the continuous coordinate change $(y,w)\mapsto (y,\lambda y-w)$ proves that $B$ is closed. The integrated-orbit identity of Time integrals of semigroup orbits lie in the generator domain puts $h^{-1}{\int }_0^{h}S(s)xds$ in $D(B)$; average convergence [L4] shows that these vectors tend to every $x$, so $D(B)$ is dense.

Resolvent scaling and agreement. For a closed linear operator $B$, $c\mathrm{\ne }0$ and $\lambda \in \rho (B)$ one has $c\lambda \in \rho (cB)$ with $R(c\lambda ,cB)=c^{-1}R(\lambda ,B)$, because $c\lambda I-cB=c(\lambda I-B)$ and $D(cB)=D(B)$, and conversely; if $S,T$ are closed operators and $R(\lambda ,S)=R(\lambda ,T)$ for one $\lambda$, then $(\lambda I-S{)}^{-1}=(\lambda I-T{)}^{-1}$ gives $\lambda I-S=\lambda I-T$ and $D(S)=R(\lambda ,S)X=R(\lambda ,T)X=D(T)$, hence $S=T$.

Rotations of an analytic semigroup. Assume (a), so $T$ is a bounded analytic semigroup on ${\mathrm{\Sigma }}_{\delta }\cup \{0\}$ with generator $A$; for $0<\vartheta <\delta$ and $s\ge 0$ put $T_{\vartheta }(s):=T(e^{i\vartheta }s)$ and $T_{-\vartheta }(s):=T(e^{-i\vartheta }s)$: then $T_{\pm \vartheta }(0)=I$, $T_{\pm \vartheta }(s+s^{\mathrm{\prime }})=T_{\pm \vartheta }(s)T_{\pm \vartheta }(s^{\mathrm{\prime }})$ because $e^{\pm i\vartheta }(s+s^{\mathrm{\prime }})=e^{\pm i\vartheta }s+e^{\pm i\vartheta }{s}^{\mathrm{\prime }}$, each orbit is continuous on $[0,\mathrm{\infty })$ by holomorphy on the sector and strong continuity at the vertex, and $\mathrm{\parallel }{T}_{\pm \vartheta }(s)\mathrm{\parallel }\le {\sup }_{z\in {\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}}\mathrm{\parallel }T(z)\mathrm{\parallel }<\mathrm{\infty }$ for any $\vartheta <{\delta }^{\mathrm{\prime }}<\delta$; thus both are bounded strongly continuous semigroups.

The (c) calculus and the local series. Assume (c), and put $M_0:={\sup }_{t\ge 0}\mathrm{\parallel }T(t)\mathrm{\parallel }$ and $M_1:={\sup }_{t>0}t\mathrm{\parallel }AT(t)\mathrm{\parallel }$. For $B_s:=AT(s)$, the domain inclusion in (c) and generator commutation [L3] give $B_{s}T(s)=AT(2s)=T(s)B_s$, hence $B_s$ commutes with every $T(ns)$ and $B_{(n+1)s}=B_{s}T(ns)$. Induction yields $T(t)X\subseteq D(A^n)$ and $A^{n}T(t)=B_{t\mathrm{/}n}^n$, with $\mathrm{\parallel }{A}^{n}T(t)\mathrm{\parallel }\le (M_{1}n\mathrm{/}t{)}^n$. First, $T$ is locally Lipschitz in operator norm on $(0,\mathrm{\infty })$. Fix $0<a<b$ and $0<h\le a\mathrm{/}2$. Since $T(s)X\subseteq D(A)$, the orbit formula and the fundamental theorem [L3, L4] give, for $a\le s\le b$, $T(s+h)x-T(s)x={\int }_0^{h}T(r)AT(s)xdr,$ so $\mathrm{\parallel }T(s+h)-T(s)\mathrm{\parallel }\le h{\sup }_{0\le r\le a\mathrm{/}2}\mathrm{\parallel }T(r)\mathrm{\parallel }{M}_1\mathrm{/}a$. The same estimate applied from $s+h$ to $s$ handles negative increments on compact subintervals of $(0,\mathrm{\infty })$. Thus $T$ is locally norm-continuous. For $n\ge 1$ and $0<s_1<s_2$, the semigroup law gives $A^{n}T(s_j)=A^{n}T(s_1\mathrm{/}2)T(s_j-s_1\mathrm{/}2),j=1,2.$ The power bound just proved and norm continuity of $T$ show that $s\mapsto A^{n}T(s)$ is norm-continuous on every compact subinterval of $(0,\mathrm{\infty })$; this is also true for $n=0$. For each $n\ge 0$ and $h>0$, the generic generator-orbit formula and commutation with $A^n$ give $A^{n}T(s+h)x-A^{n}T(s)x={\int }_s^{s+h}{A}^{n+1}T(\sigma )xd\sigma \mathrm{.}$ For $h<0$ the same formula follows by reversing the endpoints. Since $A^{n+1}T(\sigma )$ is operator-norm continuous, division by $h$ shows $\frac{d}{ds}(A^{n}T(s))=A^{n+1}T(s)$ in operator norm. Hence $T\in C^{\mathrm{\infty }}((0,\mathrm{\infty }),\mathcal{B}(X))$ and $T^{(n)}=A^{n}T$. If $M_1>0$, set $\rho :=\min \{1\mathrm{/}2,1\mathrm{/}(2e{M}_1)\}$. Taylor's formula on $[t-\mathrm{\mid }h\mathrm{\mid },t+\mathrm{\mid }h\mathrm{\mid }]$ then gives $\mathrm{\parallel }T(t+h)-{\sum }_{n<N}{h}^n{A}^{n}T(t)\mathrm{/}n!\mathrm{\parallel }\le (e{M}_1\mathrm{\mid }h\mathrm{\mid }\mathrm{/}(t-\mathrm{\mid }h\mathrm{\mid }){)}^N\to 0$ for $\mathrm{\mid }h\mathrm{\mid }<\rho t$. The zeroth series term has norm at most $M_0$, and for $n\ge 1$ the power bound gives $\mathrm{\mid }z-t{\mathrm{\mid }}^n\mathrm{\parallel }{A}^{n}T(t)\mathrm{\parallel }\mathrm{/}n!\le (\rho e{M}_1{)}^n$; hence ${\sum }_{n\ge 0}\mathrm{\mid }z-t{\mathrm{\mid }}^n\mathrm{\parallel }{A}^{n}T(t)\mathrm{\parallel }\mathrm{/}n!\le M_0+{\sum }_{n\ge 1}(\rho e{M}_1{)}^n\le M_0+1$ for $\mathrm{\mid }z-t\mathrm{\mid }\le \rho t$. If $M_1=0$, then $AT(s)=0$ for all $s>0$; generator commutation and the fundamental theorem [L3, L4] give $T(t)x=x$ for every $x\in D(A)$, and density gives $T(t)=I$ and $A=0$.

Banach-valued identity theorem. Let $F:U\to Y$ be holomorphic on a connected open set. If $F$ vanishes on a real interval, choose a real center and a disc whose real diameter lies in that interval. The power series in [L7] then has all coefficients zero, so $F$ vanishes on a disc. If $F$ vanishes on any nonempty open set, put $Z=\{z\in U:F\text{ vanishes on a neighborhood of }z\}$. This set is nonempty and open. At every point in its closure in $U$, continuity of every derivative from [L7] gives $F^{(n)}(z)=0$ for every $n$, since all derivatives vanish on $Z$. The Taylor expansion at $z$ therefore vanishes on a neighborhood, so $z\in Z$. Thus $Z$ is also closed; connectedness gives $Z=U$.

(d) gives a wedge at the imaginary axis. Assume (d) with constants $M,C$; for fixed $s\mathrm{\ne }0$ the resolvent identity gives $\mathrm{\parallel }R(r+is,A)-R(r^{\mathrm{\prime }}+is,A)\mathrm{\parallel }\le \mathrm{\mid }r-r^{\mathrm{\prime }}\mathrm{\mid }{C}^2\mathrm{/}{s}^2$ for $r,r^{\mathrm{\prime }}>0$, so $R(r+is,A)$ converges in norm as $r\downarrow 0$ to some $R_0\in \mathcal{B}(X)$; the vectors $y_r:=R(r+is,A)x$ satisfy $((r+is)I-A)y_r=x$, so $A{y}_r=(r+is)y_r-x\to is{R}_{0}x-x$ and closedness of $A$ gives $R_{0}x\in D(A)$ with $(isI-A)R_{0}x=x$; if $(isI-A)x=0$ then $x=R(r+is,A)(rx)=rR(r+is,A)x$ and $\mathrm{\parallel }x\mathrm{\parallel }\le Cr\mathrm{\mid }s{\mathrm{\mid }}^{-1}\mathrm{\parallel }x\mathrm{\parallel }$ for all $r>0$, so $x=0$; hence $is\in \rho (A)$ and $\mathrm{\parallel }R(is,A)\mathrm{\parallel }\le C\mathrm{/}\mathrm{\mid }s\mathrm{\mid }$, and the Neumann series $R(\lambda ,A)=[I+rR(is,A){]}^{-1}R(is,A)$ converges for $\lambda =r+is$ with $\mathrm{\mid }r\mathrm{\mid }<q\mathrm{\mid }s\mathrm{\mid }\mathrm{/}C$, $q\in (0,1)$, giving $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le C\mathrm{/}((1-q)\cos \theta \mathrm{\mid }\lambda \mathrm{\mid })$, $\theta :=\arctan (q\mathrm{/}C)$, on that wedge.

(e) gives the contour semigroup. Assume (e), so $A$ is sectorial of some angle ${\delta }_0>0$; putting $\delta :=\min \{{\delta }_0,\pi \mathrm{/}2\}$ (a smaller sector is contained in a larger one, so $A$ is sectorial of angle $\delta$) and applying [L6], the contour family is a bounded analytic semigroup of angle $\delta$ with generator $A$, unique among exponentially bounded semigroups with that generator, and satisfies $T(t)X\subseteq D(A^m)$, $\mathrm{\parallel }{A}^{m}T(t)\mathrm{\parallel }\le C_m{t}^{-m}$ for all $m\ge 1$, $t>0$; in particular this semigroup satisfies (c) and supplies the semigroup of (a).

(a) implies (b). Assume (a) and fix $0<\vartheta <\delta$; by [step 1.1] with $\lambda =1$ one has $R(1,A)x={\int }_0^{\mathrm{\infty }}{e}^{-t}T(t)xdt$ for every $x$, and the function $z\mapsto e^{-z}T(z)x$ is holomorphic on the star-shaped sector ${\mathrm{\Sigma }}_{\delta }$, so [L5] applied to the closed contours $\epsilon \to R\to R{e}^{i\vartheta }\to \epsilon e^{i\vartheta }\to \epsilon$ gives $0={\int }_{\epsilon }^R{e}^{-t}T(t)xdt+{\int }_{{\mathrm{a}\mathrm{r}\mathrm{c}}_R}+{\int }_{\mathrm{r}\mathrm{a}\mathrm{y}}+{\int }_{{\mathrm{a}\mathrm{r}\mathrm{c}}_{\epsilon }}$ with $\mathrm{\parallel }{e}^{-z}T(z)x\mathrm{\parallel }\le e^{-\mathrm{Re}z}{M}_{{\delta }^{\mathrm{\prime }}}\mathrm{\parallel }x\mathrm{\parallel }$; since $\mathrm{Re}(s{e}^{i\vartheta })\ge s\cos \vartheta$ and $\mathrm{Re}z\ge R\cos \vartheta$ on the outer arc and $\mathrm{\parallel }{e}^{-z}\mathrm{\parallel }\le e^{\epsilon }$ on the inner arc, the limits $\epsilon \downarrow 0$, $R\to \mathrm{\infty }$ give ${\int }_0^{\mathrm{\infty }}{e}^{-t}T(t)xdt=e^{i\vartheta }{\int }_0^{\mathrm{\infty }}{e}^{-s{e}^{i\vartheta }}{T}_{\vartheta }(s)xds$; by [step 1.1] applied to the bounded semigroup $T_{\vartheta }$ at $\lambda =e^{i\vartheta }$ the last integral equals $R(e^{i\vartheta },A_{\vartheta })x$, so $R(1,A)=e^{i\vartheta }R(e^{i\vartheta },A_{\vartheta })=R(1,e^{-i\vartheta }{A}_{\vartheta })$ by [step 1.2]; both $A$ and $e^{-i\vartheta }{A}_{\vartheta }$ are closed by the hypothesis and step 1.1, so [step 1.2] gives $A=e^{-i\vartheta }{A}_{\vartheta }$, that is $A_{\vartheta }=e^{i\vartheta }A$ with domain $D(A)$; the same argument with $-\vartheta$ gives $A_{-\vartheta }=e^{-i\vartheta }A$, so (b) holds for this $\vartheta$ and hence for every $\vartheta \in (0,\delta )$.

(b) implies (e). Assume (b) for some $\vartheta$, with the two bounded semigroups satisfying $\mathrm{\parallel }{S}_{\pm }(t)\mathrm{\parallel }\le M$; [step 1.1] applied to $S_{\pm }$ gives ${\mathbb{C}}_{+}\subseteq \rho (e^{\pm i\vartheta }A)$ with $\mathrm{\parallel }R(\lambda ,e^{\pm i\vartheta }A)\mathrm{\parallel }\le M\mathrm{/}\mathrm{Re}\lambda$, so by [step 1.2] the half-planes $\{\lambda :\mathrm{Re}(e^{\pm i\vartheta }\lambda )>0\}=e^{\mp i\vartheta }{\mathbb{C}}_{+}$ lie in $\rho (A)$; their union is ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\vartheta }$, and for $\lambda =r{e}^{i\alpha }$ with $\mathrm{\mid }\alpha \mathrm{\mid }\le \pi \mathrm{/}2+\vartheta -\epsilon$ the sign $\alpha \ge 0$ gives $\mathrm{Re}(e^{-i\vartheta }\lambda )=r\cos (\alpha -\vartheta )\ge r\min \{\cos \vartheta ,\sin \epsilon \}$ (the angle $\alpha -\vartheta$ ranges over $[-\vartheta ,\pi \mathrm{/}2-\epsilon ]$) and symmetrically for $\alpha <0$, so $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M\mathrm{/}(r\min \{\cos \vartheta ,\sin \epsilon \})=M_{\epsilon }\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid }$ there; hence $A$ is sectorial of angle $\vartheta$, which is (e).

(d) implies (e). Assume (d); [step 1.6] gives the wedge $\mathrm{\mid }r\mathrm{\mid }<q\mathrm{\mid }s\mathrm{\mid }\mathrm{/}C$ with $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le C\mathrm{/}((1-q)\cos \theta \mathrm{\mid }\lambda \mathrm{\mid })$, and [step 1.1] applied to the bounded semigroup generated by $A$ gives $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M\mathrm{/}\mathrm{Re}\lambda$ on the right half-plane; set $\eta :=\frac{1}{2}\arctan (q\mathrm{/}C)>0$: for $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi \mathrm{/}2+\eta -\epsilon$ either $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi \mathrm{/}2-\epsilon \mathrm{/}2$ and the Laplace estimate gives $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M\mathrm{/}(\mathrm{\mid }\lambda \mathrm{\mid }\sin (\epsilon \mathrm{/}2))$, or $\mathrm{\mid }\arg \lambda \mathrm{\mid }>\pi \mathrm{/}2-\epsilon \mathrm{/}2$ and then $\mathrm{\mid }r\mathrm{\mid }\mathrm{/}\mathrm{\mid }s\mathrm{\mid }\le \tan \eta <q\mathrm{/}C$, so the wedge bound applies; hence $\rho (A)\supseteq {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\eta }$ and the bound $M_{\epsilon }\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid }$ holds on each ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\eta -\epsilon }$, which is (e) with exponent $\eta$.

(c) builds the holomorphic extension. Assume (c) and $M_1>0$, put $\rho :=\min \{1\mathrm{/}2,1\mathrm{/}(2e{M}_1)\}$ and $\mathrm{\Omega }:={\bigcup }_{t>0}D(t,\rho t)$; by [step 1.4] each series $S_t(z):={\sum }_{n\ge 0}(z-t{)}^n{A}^{n}T(t)\mathrm{/}n!$ converges in $\mathcal{B}(X)$ for $\mathrm{\mid }z-t\mathrm{\mid }<t\mathrm{/}(e{M}_1)$, is bounded by $M_0+1$ on $D(t,\rho t)$, and equals $T$ on the real interval $(t-\rho t,t+\rho t)$; if $z\in D(t_1,\rho t_1)\cap D(t_2,\rho t_2)$ the two series are holomorphic on the convex intersection and agree on its nonempty real interval, hence agree on it by [step 1.5], so $\tilde{T}(z):=S_t(z)$ is a well-defined holomorphic map $\mathrm{\Omega }\to \mathcal{B}(X)$ with $\mathrm{\parallel }\tilde{T}\mathrm{\parallel }\le M_0+1$ extending $T$; for real $t>0$ the holomorphic difference $z\mapsto \tilde{T}(z)T(t)-\tilde{T}(z+t)$ vanishes on a real interval and hence, by [step 1.5], on every connected component of $\{z:z,z+t\in \mathrm{\Omega }\}$ meeting it, so $\tilde{T}(z)T(t)=\tilde{T}(z+t)$ for $z$ in the sector ${\mathrm{\Sigma }}_{\eta }$, $\eta :=\arctan \rho >0$, which lies in $\mathrm{\Omega }$ because $\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }<\rho \mathrm{Re}z$; if $M_1=0$ then $T(t)=I$ and $A=0$ by [step 1.4], and $\tilde{T}(z):=I$ is a bounded analytic semigroup of angle $\pi \mathrm{/}2$ with generator $A$.

(c) implies (a). Restrict the holomorphic extension of step 2.4 to the connected sector ${\mathrm{\Sigma }}_{\eta }$, where $\tilde{T}$ is bounded by $M_0+1$. For each fixed real $\tau >0$, the maps $z\mapsto T(\tau )\tilde{T}(z)$ and $z\mapsto \tilde{T}(z)T(\tau )$ are holomorphic on ${\mathrm{\Sigma }}_{\eta }$. They agree for every positive real $s$, because $\tilde{T}(s)=T(s)$ and the real semigroup law gives $T(\tau )T(s)=T(\tau +s)=T(s+\tau )=T(s)T(\tau )$. By the Banach-valued identity theorem [step 1.5], they agree throughout ${\mathrm{\Sigma }}_{\eta }$. Combining this commutation with step 2.4 yields, for every $z\in {\mathrm{\Sigma }}_{\eta }$ and real $\tau >0$, $T(\tau )\tilde{T}(z)=\tilde{T}(z)T(\tau )=\tilde{T}(z+\tau )$. Now fix $w\in {\mathrm{\Sigma }}_{\eta }$. Since the sector is closed under addition, $G(z):=\tilde{T}(z)\tilde{T}(w)-\tilde{T}(z+w)$ is holomorphic on ${\mathrm{\Sigma }}_{\eta }$; for real $z=t>0$ the mixed-product identity just proved gives $G(t)=T(t)\tilde{T}(w)-\tilde{T}(t+w)=0$. A second application of [step 1.5] gives $G\equiv 0$, so $\tilde{T}(z)\tilde{T}(w)=\tilde{T}(z+w)$. Set $\tilde{T}(0):=I$. To check strong continuity at the vertex, for fixed small real $h>0$ and $z$ near zero in ${\mathrm{\Sigma }}_{\eta }$ with $z+h\in {\mathrm{\Sigma }}_{\eta }$, use the semigroup law and uniform bound to get $\mathrm{\parallel }\tilde{T}(z)x-x\mathrm{\parallel }\le (M_0+1)\mathrm{\parallel }x-T(h)x\mathrm{\parallel }+\mathrm{\parallel }\tilde{T}(z+h)x-T(h)x\mathrm{\parallel }+\mathrm{\parallel }T(h)x-x\mathrm{\parallel }$. Continuity at the interior point $h$ makes the middle term tend to zero as $z\to 0$; strong continuity of $T$ at zero lets $\mathrm{\parallel }T(h)x-x\mathrm{\parallel }$ be arbitrarily small, so $\tilde{T}$ is strongly continuous at the vertex. Thus $\tilde{T}$ is a bounded analytic semigroup of angle $\eta >0$ extending $T$, and (a) holds.

(a) implies (c) and (d). Assume (a); [step 2.1] gives (b) and [step 2.2] gives (e), so the contour semigroup of [step 1.7] is a bounded analytic semigroup generated by $A$, and by the uniqueness clause of [L6] it equals the given $T$; the smoothing estimates of [L6] therefore give $T(t)X\subseteq D(A)$ and ${\sup }_{t>0}t\mathrm{\parallel }AT(t)\mathrm{\parallel }\le C_1<\mathrm{\infty }$, which is (c); moreover for $r>0$, $s\mathrm{\ne }0$ the point $\lambda =r+is$ lies in ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\vartheta -\epsilon }$ and [step 2.2] gives $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M_{\epsilon }\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid }\le M_{\epsilon }\mathrm{/}\mathrm{\mid }s\mathrm{\mid }$, which is (d).

The maximal angle. Let $\psi$ be the supremum of the $\delta \in (0,\pi \mathrm{/}2]$ for which $A$ has a bounded analytic semigroup extension to ${\mathrm{\Sigma }}_{\delta }$, let $B$ be the set of $\vartheta \in (0,\pi \mathrm{/}2)$ for which (b) holds, and let $E$ be the set of $\delta >0$ with $A$ sectorial of exponent $\delta$; [step 2.1] shows $(0,\psi )\subseteq B$, so $\sup B\ge \psi$, while $\vartheta \in B$ gives $\vartheta \in E$ by [step 2.2] and then an extension to ${\mathrm{\Sigma }}_{\vartheta }$ by [step 1.7], so $\vartheta \le \psi$ and $\sup B=\psi$; likewise $(0,\psi )\subseteq E$ by [step 2.2], so $\sup E\ge \psi$, and $\delta \in E$ gives an extension to ${\mathrm{\Sigma }}_{\delta }$ by [step 1.7], so $\delta \le \psi$ and $\sup E=\psi$; these are equalities of suprema only, with no attainment asserted at $\psi$.

Assembly. The implications (a)$\Rightarrow$(b) [step 2.1], (b)$\Rightarrow$(e) [step 2.2], (e)$\Rightarrow$(a),(c) [step 1.7], (c)$\Rightarrow$(a) [step 3.1], (a)$\Rightarrow$(c),(d) [step 3.2] and (d)$\Rightarrow$(e) [step 2.3] close the cycle, so (a)-(e) are equivalent; when they hold, the semigroup of (c) is the contour semigroup by the uniqueness argument of [step 3.2]; the angle equalities are [step 3.3]; and no choice principle beyond Dependent Choice is used in the argument: every inverse appearing is an explicit absolutely convergent Laplace integral, a norm limit of resolvents, or a Neumann series, boundedness of each inverse is proved explicitly, so the closed-graph implication in the resolvent vocabulary is not invoked.