The sectorial multiplication operator

The operator is densely defined and self-adjoint. For $f\in H$ the truncations $f_n:=f{\mathbf{1}}_{\{\mathrm{\mid }q\mathrm{\mid }\le n\}}$ lie in $D(A)$ and $f_n\to f$ in $H$ by [L8], so $D(A)$ is dense; for $f,g\in D(A)$ the identity $⟨M_{q}f,g⟩=\int qf\bar{g}d\mu =⟨f,M_{q}g⟩$ shows symmetry. If $g\in D(A^{\ast })$, then for every $h\in D(A)$ the adjoint relation gives $\int h(q\bar{g}-\overline{A^{\ast }g})d\mu =0$. Put $r:=A^{\ast }g-qg$ and $E_n:=\{\mathrm{\mid }q\mathrm{\mid }\le n,\mathrm{\mid }r\mathrm{\mid }\le n\}$, and define $h_n:={\mathbf{1}}_{E_n}r$. Then $h_n$ is in $L^2$, since $\mathrm{\parallel }{h}_n{\mathrm{\parallel }}_2\le \mathrm{\parallel }{A}^{\ast }g{\mathrm{\parallel }}_2+n\mathrm{\parallel }g{\mathrm{\parallel }}_2$, while $q{h}_n\in L^2$ because $\mathrm{\mid }q\mathrm{\mid }\le n$ on its support, so $h_n\in D(A)$. The adjoint identity with $h=h_n$ yields $0=\int h_n(q\bar{g}-\overline{A^{\ast }g})d\mu =-{\int }_{E_n}\mathrm{\mid }r{\mathrm{\mid }}^{2}d\mu$. The sets $E_n$ increase to full measure because $q$ is finite-valued and $g,A^{\ast }g\in L^2$, so $r=0$ almost everywhere. Thus $qg=A^{\ast }g\in H$, and $g\in D(A)$ with $A^{\ast }g=Ag$; therefore $D(A^{\ast })=D(A)$ and $A=A^{\ast }$ by [L2].

The contraction semigroup. For $f\in H$ define $T(t)f:=e^{tq}f$ for $t\ge 0$; since $q\le 0$ one has $\mathrm{\mid }{e}^{tq}\mathrm{\mid }\le 1$ and $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$ with $T(t)$ linear and $T(0)=I$, the functional equation is immediate from $e^{(t+s)q}=e^{tq}{e}^{sq}$, and $\mathrm{\mid }{e}^{tq}-1{\mathrm{\mid }}^2\mathrm{\mid }f{\mathrm{\mid }}^2\le 4\mathrm{\mid }f{\mathrm{\mid }}^2$ together with pointwise convergence $e^{tq}\to 1$ gives $\mathrm{\parallel }T(t)f-f{\mathrm{\parallel }}_2\to 0$ as $t\downarrow 0$ by [L8]; hence $T$ is a strongly continuous semigroup of contractions by [L6].

The resolvent formula and its norm. Fix $\lambda \mathrm{\notin }(-\mathrm{\infty },0]$ and put $m_{\lambda }:=(\lambda -q{)}^{-1}$, bounded with $\mathrm{\mid }{m}_{\lambda }(x)\mathrm{\mid }\le 1\mathrm{/}\mathrm{dist}(\lambda ,(-\mathrm{\infty },0])$ because $q(x)\in (-\mathrm{\infty },0]$ a.e.; the multiplication operator $M_{m_{\lambda }}$ is bounded with $\mathrm{\parallel }{M}_{m_{\lambda }}\mathrm{\parallel }={\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\mid }{m}_{\lambda }(x)\mathrm{\mid }$ (the upper bound from [L4] and the lower bound by testing on the indicator of a finite-measure subset of $\{\mathrm{\mid }{m}_{\lambda }\mathrm{\mid }>s-\epsilon \}$); also $q{m}_{\lambda }=\lambda m_{\lambda }-1$ is bounded, so $M_{m_{\lambda }}H\subset D(A)$. The pointwise identities $(\lambda -q)m_{\lambda }=1=m_{\lambda }(\lambda -q)$ give $(\lambda I-A)M_{m_{\lambda }}f=f$ for $f\in H$ and $M_{m_{\lambda }}(\lambda I-A)g=g$ for $g\in D(A)$; hence $M_{m_{\lambda }}$ is the two-sided inverse of $\lambda I-A$, so $\lambda \in \rho (A)$ with $R(\lambda ,A)=M_{m_{\lambda }}$ and $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }={\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\mid }\lambda -q(x){\mathrm{\mid }}^{-1}\le 1\mathrm{/}\mathrm{dist}(\lambda ,(-\mathrm{\infty },0])$.

Sectoriality with exponent $\pi \mathrm{/}2$. By [step 2.1] every $\lambda \mathrm{\notin }(-\mathrm{\infty },0]$, in particular every $\lambda \in {\mathrm{\Sigma }}_{\pi }=\mathbb{C}\setminus (-\mathrm{\infty },0]$, lies in $\rho (A)$ with $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le 1\mathrm{/}\mathrm{dist}(\lambda ,(-\mathrm{\infty },0])$; for $\mathrm{Re}\lambda \ge 0$ the nearest point of $(-\mathrm{\infty },0]$ is the origin and the distance is $\mathrm{\mid }\lambda \mathrm{\mid }$, for $\mathrm{Re}\lambda <0$ it is $\mathrm{\mid }\mathrm{Im}\lambda \mathrm{\mid }$, and on ${\mathrm{\Sigma }}_{\pi -\epsilon }$ the distance is at least $\mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon$ (for $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi \mathrm{/}2$ the distance is $\mathrm{\mid }\lambda \mathrm{\mid }$, and for $\pi \mathrm{/}2\le \mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi -\epsilon$ it is $\mathrm{\mid }\lambda \mathrm{\mid }\sin (\pi -\mathrm{\mid }\arg \lambda \mathrm{\mid })\ge \mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon$); hence $A$ is sectorial of angle $\pi \mathrm{/}2$ by [L5], and as the sectorial exponent is capped at $\pi \mathrm{/}2$ by the definition this exponent is maximal.

The generator is $A$. For real $\lambda >0$ the Bochner integral $v={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}T(t)fdt$ exists by the contraction bound. For $h\in H$, Cauchy–Schwarz and $q\le 0$ give ${\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\Omega }}{e}^{-(\lambda -q(x))t}\mathrm{\mid }f(x)h(x)\mathrm{\mid }d\mu dt\le {\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }h{\mathrm{\parallel }}_2$. Scalar Fubini (Fubini's theorem for L^1 functions on a sigma-finite product) therefore gives $⟨v,h⟩={\int }_{\mathrm{\Omega }}(\lambda -q{)}^{-1}f\bar{h}d\mu$. A bounded linear functional commutes with the Bochner integral by its simple-function definition, so this equality identifies $v=M_{(\lambda -q{)}^{-1}}f$. By [L7], $R(\lambda ,G)=M_{(\lambda -q{)}^{-1}}=R(\lambda ,A)$. The common inverse has range $D(G)=D(A)$ and determines both operators, so $G=A$. Thus $T$ is the real-time restriction of the bounded analytic semigroup generated by $A$, by [L5] and uniqueness of real-time semigroups.