Self-adjoint nonpositive operators generate bounded analytic semigroups

The lower bound, injectivity and closed range. For $u\in D(A)\setminus \{0\}$ put $a_u:=⟨Au,u⟩\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_H^2\le 0$; by [L4], $\mathrm{\parallel }(\lambda I-A)u\mathrm{\parallel }\mathrm{\parallel }u{\mathrm{\parallel }}_H\ge \mathrm{\mid }⟨(\lambda I-A)u,u⟩\mathrm{\mid }=\mathrm{\mid }\lambda -a_u\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_H^2\ge \mathrm{dist}(\lambda ,(-\mathrm{\infty },0])\mathrm{\parallel }u{\mathrm{\parallel }}_H^2$, so for every $\lambda \mathrm{\notin }(-\mathrm{\infty },0]$ the operator $\lambda I-A$ is injective with the lower bound $\mathrm{\parallel }(\lambda I-A)u\mathrm{\parallel }\ge \mathrm{dist}(\lambda ,(-\mathrm{\infty },0])\mathrm{\parallel }u\mathrm{\parallel }$; moreover its range is closed, because if $(\lambda I-A)u_n\to w$ then $(u_n)$ is Cauchy, $u_n\to u$ and $A{u}_n=\lambda u_n-(\lambda I-A)u_n\to \lambda u-w$, and closedness of $A$ from [L3] and [L1] gives $u\in D(A)$ with $(\lambda I-A)u=w$.

The range is dense. If $v\perp \mathrm{Ran}(\lambda I-A)$ for some $\lambda \mathrm{\notin }(-\mathrm{\infty },0]$, then [L2] with $T=A$ and $z=\lambda$ gives $v\in \ker (A^{\ast }-\bar{\lambda })$; by self-adjointness [L1] this says $Av=\bar{\lambda }v$, that is $(\bar{\lambda }I-A)v=0$, and the lower bound of [step 1.1] at $\bar{\lambda }$ (which also lies outside $(-\mathrm{\infty },0]$) forces $v=0$; hence $\mathrm{Ran}(\lambda I-A{)}^{\perp }=\{0\}$ and, since the range is closed by [step 1.1], [L13] gives $\mathrm{Ran}(\lambda I-A)=\{0{\}}^{\perp }=H$.

The resolvent bounds and sectoriality. By [step 2.1] and [step 1.1] the map $\lambda I-A$ is bijective with inverse bounded by $1\mathrm{/}\mathrm{dist}(\lambda ,(-\mathrm{\infty },0])$, so $\lambda \in \rho (A)$ with $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le 1\mathrm{/}\mathrm{dist}(\lambda ,(-\mathrm{\infty },0])$ for every $\lambda \mathrm{\notin }(-\mathrm{\infty },0]$ by [L5]; for $\mathrm{Re}\lambda >0$ the distance to the smaller set $(-\mathrm{\infty },0]$ dominates the distance to $\{z:\mathrm{Re}z\le 0\}$, which equals $\mathrm{Re}\lambda$, and for $\lambda =a+ib$ with $b\mathrm{\ne }0$ the distance to the real set $(-\mathrm{\infty },0]$ is at least $\mathrm{\mid }b\mathrm{\mid }$; finally, for $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi -\epsilon$ the nearest point of $(-\mathrm{\infty },0]$ is the origin when $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi \mathrm{/}2$ and has distance $\mathrm{\mid }\lambda \mathrm{\mid }\sin (\pi -\mathrm{\mid }\arg \lambda \mathrm{\mid })\ge \mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon$ otherwise, so $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M_{\epsilon }\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid }$ with $M_{\epsilon }=1\mathrm{/}\sin \epsilon$ on ${\mathrm{\Sigma }}_{\pi -\epsilon }$; since ${\mathrm{\Sigma }}_{\pi }=\mathbb{C}\setminus (-\mathrm{\infty },0]\subseteq \rho (A)$, this is sectoriality of angle $\pi \mathrm{/}2$.

Generation and contractivity. By [step 3.1] $A$ satisfies the sectorial resolvent condition with exponent $\pi \mathrm{/}2$, so [L7] provides a bounded analytic semigroup $T$ of angle $\pi \mathrm{/}2$ generated by $A$; separately $A$ is dissipative by [L8] because $\mathrm{Re}⟨Au,u⟩\le 0$, and $\mathrm{Ran}(\lambda I-A)=H$ for every $\lambda >0$ by [step 3.1], so Lumer-Phillips [L8] makes $A$ generate a strongly continuous contraction semigroup; that semigroup is bounded, hence exponentially bounded, and has generator $A$, so by uniqueness [L9] it coincides with the analytic semigroup $T$, giving $\mathrm{\parallel }T(t)\mathrm{\parallel }\le 1$ for $t\ge 0$; the argument assumes Dependent Choice and inherits Countable Choice from the adjoint vocabulary [L1] and uses no further choice principle beyond Dependent Choice.