Quadratic spectral bounds control a self-adjoint parabolic semigroup

The shifted operator. $D(B)=D(A)$ is dense; for $x,y\in D(A)$ one has $⟨Bx,y⟩=⟨Ax,y⟩+{\lambda }_1⟨x,y⟩=⟨x,Ay⟩+{\lambda }_1⟨x,y⟩=⟨x,By⟩$, so $B$ is symmetric, and [L1] identifies $D(B^{\ast })$ with the $y$ for which $x\mapsto ⟨Ax,y⟩$ is bounded on $D(A)$, namely $D(A^{\ast })=D(A)$, with $B^{\ast }y=A^{\ast }y+{\lambda }_{1}y=By$; hence $B$ is self-adjoint, and $⟨Bu,u⟩=⟨Au,u⟩+{\lambda }_1\mathrm{\parallel }u{\mathrm{\parallel }}^2\le 0$ for every $u\in D(A)$; by [L2] $B$ is sectorial of angle $\pi \mathrm{/}2$ and generates a bounded analytic semigroup $S$ of angle $\pi \mathrm{/}2$ that is contractive on $[0,\mathrm{\infty })$.

The rotated generators. Fix $\alpha \in (-\pi \mathrm{/}2,\pi \mathrm{/}2)$ and put $V(t):=S(e^{i\alpha }t)$ for $t\ge 0$: the functional equation of $S$ makes $V$ a semigroup, strong continuity at $0$ holds along the ray $e^{i\alpha }[0,\mathrm{\infty })\subseteq {\mathrm{\Sigma }}_{\pi \mathrm{/}2}$ by [L8], and $\mathrm{\parallel }V(t)\mathrm{\parallel }\le C_{\alpha }$ for a finite constant because the ray lies in a strictly smaller sector; moreover for $f\in D(B)$ the identity $S(z)f-f={\int }_0^{z}S(w)Bfdw$ holds on ${\mathrm{\Sigma }}_{\pi \mathrm{/}2}$, because both sides are holomorphic by [L4] and [L5] (the primitive of $w\mapsto S(w)Bf$ is holomorphic with derivative $S(z)Bf$) and they agree on the real axis by the fundamental theorem and the orbit derivative $S^{\mathrm{\prime }}(t)f=S(t)Bf$ of [L3], so the Banach-valued identity theorem proved in Sectorial resolvent characterisation of bounded analytic semigroups extends the identity to the sector; dividing by $z=e^{i\alpha }t$ and letting $t\downarrow 0$ gives $(V(t)f-f)\mathrm{/}t\to e^{i\alpha }Bf$, so the generator $C_{\alpha }$ of $V$ contains the closed operator $e^{i\alpha }B$; for real $\lambda >0$ one has $\lambda \in \rho (C_{\alpha })$ by [L6] and $\lambda \in \rho (e^{i\alpha }B)$ with $R(\lambda ,e^{i\alpha }B)=e^{-i\alpha }R(e^{-i\alpha }\lambda ,B)$ because $e^{-i\alpha }\lambda \mathrm{\notin }(-\mathrm{\infty },0]$ and $B$ is self-adjoint nonpositive; both operators are closed and their resolvents at $\lambda$ agree (the identity $(\lambda -C_{\alpha })R(\lambda ,e^{i\alpha }B)=I$ holds on $H$ since the two operators agree on $D(B)$), so [L1] and the resolvent definition give $C_{\alpha }=e^{i\alpha }B$.

Contractivity on the sector. The operator $e^{i\alpha }B$ is dissipative by [L7], because for $v\in D(B)$ one has $\mathrm{Re}⟨e^{i\alpha }Bv,v⟩=\cos \alpha ⟨Bv,v⟩\le 0$ (the value $⟨Bv,v⟩=\overline{⟨v,Bv⟩}$ is real and nonpositive by self-adjointness and the quadratic bound); by [step 2.1] it is the generator of $V$, so [L3] gives $\frac{d}{dt}\mathrm{\parallel }V(t)f{\mathrm{\parallel }}^2=2\mathrm{Re}⟨V(t)f,e^{i\alpha }BV(t)f⟩\le 0$ for $f\in D(B)$ and hence $\mathrm{\parallel }V(t)f\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }$; since $D(B)$ is dense and $V(t)$ is bounded this extends to all $f\in H$, so $\mathrm{\parallel }S(z)\mathrm{\parallel }\le 1$ for every $z=e^{i\alpha }t\in {\mathrm{\Sigma }}_{\pi \mathrm{/}2}$.

The semigroup generated by $A$. Define $T(z):=e^{-{\lambda }_{1}z}S(z)$ on ${\mathrm{\Sigma }}_{\pi \mathrm{/}2}\cup \{0\}$: the functional equation, operator-norm holomorphy and strong continuity at the vertex are inherited, and $\mathrm{\parallel }T(z)\mathrm{\parallel }\le e^{-{\lambda }_1\mathrm{Re}z}\mathrm{\parallel }S(z)\mathrm{\parallel }\le e^{-{\lambda }_1\mathrm{Re}z}$ by [step 3.1]. For every real ${\lambda }_1$ this is a holomorphic semigroup family with generator $A$, since for $f\in D(A)=D(B)$ the real difference quotient satisfies $\frac{e^{-{\lambda }_{1}t}S(t)f-f}{t}=\frac{S(t)f-f}{t}+\frac{e^{-{\lambda }_{1}t}-1}{t}S(t)f\to Bf-{\lambda }_{1}f=Af$; conversely $S(t)=e^{{\lambda }_{1}t}T(t)$ shows that a vector with a convergent $T$ difference quotient belongs to $D(B)=D(A)$, so the generator is exactly $A$. When ${\lambda }_1\ge 0$ the bound is at most $1$ on the whole sector, so $T$ is a bounded analytic semigroup of angle $\pi \mathrm{/}2$; when ${\lambda }_1>0$ it gives the stated strict exponential decay on the real axis.

The spectral clause. If $H=\{0\}$, the conclusion is immediate. Otherwise assume AC and $\sigma (A)\subseteq (-\mathrm{\infty },-{\lambda }_1]$, and let $E$ be supplied by [L9]. To prove its carrier assertion, fix a real $\lambda \in \rho (A)$ and $0<r<\mathrm{\parallel }R(\lambda ,A){\mathrm{\parallel }}^{-1}$. For $J=(\lambda -r,\lambda +r)$, $y=E(J)x$ belongs to $D(A)$ because $J$ is bounded. By [L10], $\mathrm{\parallel }(A-\lambda )y\mathrm{\parallel }\le r\mathrm{\parallel }y\mathrm{\parallel }$, whereas the bounded inverse gives $\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\mathrm{\parallel }(A-\lambda )y\mathrm{\parallel }$. Hence $y=0$, and $E(J)=0$. Every real point outside $\sigma (A)$ has such a neighborhood; a countable rational-interval base gives a countable cover of this open set by subsets of these zero-projection neighborhoods. Countable additivity therefore gives $E(\mathbb{R}\setminus \sigma (A))=0$. Now [L9, L10] yield $⟨Au,u⟩=\int sd{E}_u(s)\le -{\lambda }_1{E}_u(\mathbb{R})=-{\lambda }_1\mathrm{\parallel }u{\mathrm{\parallel }}^2$ for $u\in D(A)$, so step 4.1 applies. AC is used in the spectral branch; the quadratic branch inherits Countable Choice from the self-adjoint generation supplier.