A sectorial nonselfadjoint multiplication generator

The resolvent bound. Since $A=M_q$ is bounded with $D(A)=H$, for $\lambda \mathrm{\ne }0$ the operator $\lambda I-A=M_{\lambda -q}$ has the two-sided inverse $M_{(\lambda -q{)}^{-1}}$ as soon as $(\lambda -q{)}^{-1}$ is essentially bounded; fix $\epsilon \in (0,\theta )$ and $\lambda \in {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\theta -\epsilon }$. The essential-range definition implies $q(x)\in \mathrm{essran}(q)$ almost everywhere: every value outside the essential range has a neighbourhood with null preimage; a countable rational-ball base covers that complement by countably many such neighbourhoods, so its preimage is null. Hence for almost every $x$ the value $q(x)$ lies in the closed sector $\{\zeta :\mathrm{\mid }\arg (-\zeta )\mathrm{\mid }\le \pi \mathrm{/}2-\theta \}\cup \{0\}$ whose boundary rays have arguments $\pm (\pi \mathrm{/}2+\theta )$, while $\mathrm{\mid }\arg \lambda \mathrm{\mid }\le \pi \mathrm{/}2+\theta -\epsilon$. For nonzero $\zeta =q(x)$ let $\gamma \in [0,\pi ]$ be the principal angle between $\lambda$ and $\zeta$; the sector geometry gives $\gamma \ge \epsilon$. If $\gamma \le \pi \mathrm{/}2$, then $\mathrm{\mid }\lambda -\zeta {\mathrm{\mid }}^2=\mathrm{\mid }\lambda {\mathrm{\mid }}^2{\sin }^2\gamma +(\mathrm{\mid }\lambda \mathrm{\mid }\cos \gamma -\mathrm{\mid }\zeta \mathrm{\mid }{)}^2\ge \mathrm{\mid }\lambda {\mathrm{\mid }}^2{\sin }^2\epsilon$. If $\gamma \ge \pi \mathrm{/}2$, then $\cos \gamma \le 0$ and $\mathrm{\mid }\lambda -\zeta {\mathrm{\mid }}^2=\mathrm{\mid }\lambda {\mathrm{\mid }}^2+\mathrm{\mid }\zeta {\mathrm{\mid }}^2-2\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }\zeta \mathrm{\mid }\cos \gamma \ge \mathrm{\mid }\lambda {\mathrm{\mid }}^2\ge \mathrm{\mid }\lambda {\mathrm{\mid }}^2{\sin }^2\epsilon$; for $\zeta =0$ the same lower bound follows from $\mathrm{\mid }\lambda -\zeta \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }$. Thus $\mathrm{\mid }\lambda -q(x)\mathrm{\mid }\ge \mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon$ almost everywhere; for each $f\in H$, $\mathrm{\parallel }{M}_{(\lambda -q{)}^{-1}}f{\mathrm{\parallel }}_2^2={\int }_{\mathrm{\Omega }}\mathrm{\mid }\lambda -q{\mathrm{\mid }}^{-2}\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu \le (\mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon {)}^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2,$ so $\mathrm{\parallel }{M}_{(\lambda -q{)}^{-1}}\mathrm{\parallel }\le 1\mathrm{/}(\mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon )$ and $R(\lambda ,A)=M_{(\lambda -q{)}^{-1}}$ with $\lambda \in \rho (A)$.

The spectrum is the essential range. If $\lambda \mathrm{\notin }\mathrm{essran}(q)$ then by definition of the essential range there is $\delta >0$ with $\mu (\{\mathrm{\mid }q-\lambda \mathrm{\mid }<\delta \})=0$, so $\mathrm{\mid }\lambda -q\mathrm{\mid }\ge \delta$ almost everywhere and $M_{(\lambda -q{)}^{-1}}$ is a bounded inverse of $\lambda I-A$, hence $\lambda \in \rho (A)$; conversely, if $\lambda \in \mathrm{essran}(q)$ then for every integer $n\ge 1$ the set $E_n:=\{\mathrm{\mid }q-\lambda \mathrm{\mid }<1\mathrm{/}n\}$ has positive measure, so by $\sigma$-finiteness it contains a measurable $F_n$ with $0<\mu (F_n)<\mathrm{\infty }$, and $f_n:=\mu (F_n{)}^{-1\mathrm{/}2}{\mathbf{1}}_{F_n}$ is a unit vector with $\mathrm{\parallel }(\lambda I-A)f_n{\mathrm{\parallel }}_2^2={\int }_{F_n}\mathrm{\mid }\lambda -q{\mathrm{\mid }}^{2}d\mu \mathrm{/}\mu (F_n)\le 1\mathrm{/}{n}^2$; a bounded inverse $R$ of $\lambda I-A$ would give $1=\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2\le \mathrm{\parallel }R\mathrm{\parallel }\mathrm{/}n$ for every $n$, impossible, so $\lambda \mathrm{\notin }\rho (A)$ and $\sigma (A)=\mathrm{essran}(q)$.

Nonselfadjointness. If $q$ is nonreal on a set of positive measure, [L6] supplies a finite-measure subset $F$ of that set with $\mu (F)>0$. Then ${\mathbf{1}}_F\in L^2(\mu )$ and $(M_q-M_{\bar{q}}){\mathbf{1}}_F=(q-\bar{q}){\mathbf{1}}_F$ is a nonzero $L^2$ class, so $M_q\mathrm{\ne }{M}_{\bar{q}}$. By [L5] $(M_q{)}^{\ast }=M_{\bar{q}}$; hence $A^{\ast }\mathrm{\ne }A$ and $A$ is not self-adjoint.

Generation and the explicit semigroup. By [step 1.1] the sectorial resolvent condition of [L1] holds with exponent $\theta$: for every $\epsilon \in (0,\theta )$ the bound $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le 1\mathrm{/}(\mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon )$ holds on ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\theta -\epsilon }$; hence by [L4] the exponential series $E(z)={\sum }_n{z}^n{A}^n\mathrm{/}n!$ extends boundedly analytically to ${\mathrm{\Sigma }}_{\theta }$ and is generated by $A$; moreover ${\sum }_n{z}^n{q}^n\mathrm{/}n!$ converges in essential supremum norm to $e^{zq}$ because $q$ is essentially bounded, so $E(z)=M_{e^{zq}}$, that is $E(z)f=e^{zq}f$; on ${\mathrm{\Sigma }}_{\theta }$ one has $\mathrm{Re}(zq)\le 0$ almost everywhere, since the angle between $z$ and $q(x)$ is at least $\pi \mathrm{/}2$, so $\mathrm{\mid }{e}^{zq}\mathrm{\mid }\le 1$ and the family is bounded on every ${\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$ with ${\delta }^{\mathrm{\prime }}<\theta$; therefore the maximal analytic angle of $A$ is at least $\theta$, by the definition of the angle as the supremum of the admissible exponents in [L1] and [L3].

Assembly. By [L8], self-adjoint nonpositive generation is a sufficient route to bounded analytic semigroups. Here [step 2.1] shows that $A=M_q$ generates the bounded analytic semigroup $T(z)f=e^{zq}f$ with maximal angle at least $\theta$, while [step 1.3] shows that $A$ is nonselfadjoint when $q$ is nonreal on a set of positive measure; thus self-adjointness is not necessary. By [step 1.2] its spectrum is $\mathrm{essran}(q)$, [L7] is the real multiplier special case, and [L9] supplies the complementary form-based generation context without a symmetry restriction.