The sector changes under the sign convention

The witness $A=1$ is not sectorial. In $X=\mathbb{C}$ the operator $\lambda I-A$ is multiplication by $\lambda -1$, which is invertible exactly for $\lambda \mathrm{\ne }1$, so $\rho (A)=\mathbb{C}\setminus \{1\}$ and $\sigma (A)=\{1\}$; since $1\in {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$ for every $\delta >0$, no sector with vertex $0$, ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$, is contained in $\rho (A)$ and [L1] rules out sectoriality at vertex $0$ of every positive exponent; moreover $A$ is bounded with $\mathrm{\parallel }A\mathrm{\parallel }=1$, so [L2] gives the generated semigroup $E(t)=e^t$ with $\mathrm{\parallel }E(t)\mathrm{\parallel }=e^t$, which is unbounded on $[0,\mathrm{\infty })$.

The reflected operator $-A=-1$ is sectorial of angle $\pi \mathrm{/}2$. Here $\lambda I+A$ is multiplication by $\lambda +1$, invertible for $\lambda \mathrm{\ne }-1$, so $\rho (-A)=\mathbb{C}\setminus \{-1\}$ and $R(\lambda ,-A)=(\lambda +1{)}^{-1}$; the point $-1$ has argument $\pi$ while every $\lambda \in {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$ has $\mathrm{\mid }\arg \lambda \mathrm{\mid }<\pi \mathrm{/}2+\delta \le \pi$ for $\delta \le \pi \mathrm{/}2$, so $-1\mathrm{\notin }{\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }$ and ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }\subseteq \rho (-A)$; for $\lambda =r{e}^{i\alpha }\in {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta -\epsilon }$ with $\epsilon <\delta \le \pi \mathrm{/}2$ one has $\mathrm{\mid }\alpha \mathrm{\mid }\le \pi \mathrm{/}2+\delta -\epsilon <\pi -\epsilon$, hence $\cos \mathrm{\mid }\alpha \mathrm{\mid }\ge \cos (\pi -\epsilon )=-\cos \epsilon$ and $\mathrm{\mid }\lambda +1{\mathrm{\mid }}^2=r^2+2r\cos \mathrm{\mid }\alpha \mathrm{\mid }+1\ge r^2-2r\cos \epsilon +1=(r\cos \epsilon -1{)}^2+r^2{\sin }^2\epsilon \ge r^2{\sin }^2\epsilon$; hence $\mathrm{\parallel }R(\lambda ,-A)\mathrm{\parallel }=1\mathrm{/}\mathrm{\mid }\lambda +1\mathrm{\mid }\le M_{\epsilon }\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid }$ with $M_{\epsilon }=1\mathrm{/}\sin \epsilon$, and [L1] makes $-A$ sectorial of angle $\pi \mathrm{/}2$.

The refutation. The two computations show that $A=1$ is not sectorial in the $e^{tA}$ convention while $-A=-1$ is sectorial of the endpoint angle $\pi \mathrm{/}2$, so replacing $A$ by $-A$ does not preserve sectoriality or the sector angle; by [L3] the sectorial operator $-A$ generates a bounded analytic semigroup, which by [L2] is $t\mapsto e^{-t}$ with norm $e^{-t}\le 1$, the inverse of the unbounded semigroup $e^t$ generated by $A$; the dictionary of [L1] therefore has to be applied to the operator that actually occurs, and the functions here are explicit, so no choice principle beyond Dependent Choice is used.