Closed sectorial form and its associated operator

Let $H$ be a complex Hilbert space with inner product $(\cdot ,\cdot )$ linear in the first argument and conjugate-linear in the second (Hilbert space, Real and complex inner-product spaces and their induced length), and let $V\subseteq H$ be a dense linear subspace carrying a Hilbert norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$ whose inclusion $V\hookrightarrow H$ is continuous (The Sobolev space $H^1$ is a Hilbert space; the standard instance is $V=H_0^1(\mathrm{\Omega })\subseteq H=L^2(\mathrm{\Omega })$). A sesquilinear form $a:V\times V\to \mathbb{C}$, linear in the first argument and conjugate-linear in the second (Bounded, coercive and symmetric sesquilinear forms), is a closed sectorial form on $V\subseteq H$ if:

(i) $a$ is bounded on $V$: there is $C<\mathrm{\infty }$ with $\mathrm{\mid }a(u,v)\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$ for all $u,v\in V$;

(ii) there are $M\ge 0$ and $\theta \in [0,\pi \mathrm{/}2)$ with $\mathrm{Re}a(u,u)\ge -M\mathrm{\parallel }u{\mathrm{\parallel }}_H^2,\mathrm{\mid }\mathrm{Im}a(u,u)\mathrm{\mid }\le \tan \theta (\mathrm{Re}a(u,u)+M\mathrm{\parallel }u{\mathrm{\parallel }}_H^2)$ for every $u\in V$;

(iii) $V$ is complete for the shifted form norm $\mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}^2:=\mathrm{Re}a(u,u)+M\mathrm{\parallel }u{\mathrm{\parallel }}_H^2+\mathrm{\parallel }u{\mathrm{\parallel }}_H^2,$ By (ii) the square is nonnegative and $\mathrm{\parallel }u{\mathrm{\parallel }}_H\le \mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}$, so $\mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}=0$ only for $u=0$; condition (iii) is the closedness of the form. The Hermitian pairing $b(u,v)=(a(u,v)+\overline{a(v,u)})\mathrm{/}2+(M+1)(u,v{)}_H$ has $b(u,u)=\mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}^2>0$ for $u\mathrm{\ne }0$, so it is an inner product and this expression is its norm (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs).

Comparison with a prescribed form-domain norm. Under Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain), $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{a,M}$ is equivalent to $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$. The upper bound is $\mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}^2\le (C+(M+1)\mathrm{\parallel }\iota {\mathrm{\parallel }}^2)\mathrm{\parallel }u{\mathrm{\parallel }}_V^2$. The identity in the reverse direction has closed graph: convergence in either norm implies convergence in $H$, so the two limits agree. Both normed versions of $V$ are Banach, and Closed graph theorem makes this inverse identity bounded, giving a positive lower comparison constant. Generation results below may instead take this norm equivalence as an explicit hypothesis, retaining only Countable Choice for Lax–Milgram.

The associated operator $A$ of such a form, in the $e^{tA}$ sign convention, is $D(A):=\{u\in V:\mathrm{\exists }f\in H\text{ with }a(u,v)=-(f,v)\text{ for all }v\in V\},Au:=f\mathrm{.}$

The vector $f$ is unique, so $A$ is well defined: if $f,g$ both satisfy the relation, then $(f-g,v)=0$ for every $v\in V$; since $V$ is dense in $H$ and the inner product is continuous, $(f-g,w)=0$ for every $w\in H$, and testing $w=f-g$ gives $\mathrm{\parallel }f-g{\mathrm{\parallel }}^2=0$. The operator $A$ is linear (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores): if $u_1,u_2\in D(A)$ with associated vectors $f_1,f_2$ and $\lambda \in \mathbb{C}$, then $a(u_1+\lambda u_2,v)=a(u_1,v)+\lambda a(u_2,v)=-(f_1+\lambda f_2,v)$ for all $v\in V$, so $u_1+\lambda u_2\in D(A)$ with $A(u_1+\lambda u_2)=f_1+\lambda f_2$. The defining relation reads $⟨Au,v⟩=-a(u,v)(u\in D(A),v\in V),$ and is the only sign convention used on this page.

The form is coercive with constant $\alpha >0$ when $\mathrm{Re}a(u,u)\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}_V^2$ for all $u\in V$. A coercive form is closed: (iii) holds with explicit estimates, since (i) and coercivity give, with $\iota :V\hookrightarrow H$ the inclusion, $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}_V^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}^2\le (C+(M+1)\mathrm{\parallel }\iota {\mathrm{\parallel }}^2)\mathrm{\parallel }u{\mathrm{\parallel }}_V^2\mathrm{.}$

Sign convention

The dictionary $⟨Au,v⟩=-a(u,v)$ fixes the orientation: for $u\in D(A)$ the quadratic form of $A$ is the negative of $a$, $⟨Au,u⟩=-a(u,u),$ so the sector condition (ii) says that the shifted operator $M-A$ is accretive, $\mathrm{Re}⟨(M-A)u,u⟩=\mathrm{Re}a(u,u)+M\mathrm{\parallel }u{\mathrm{\parallel }}_H^2\ge 0,$ and gives $\mathrm{Re}⟨(A-M)u,u⟩\le 0$ for every $u\in D(A)$. This is the convention in which $e^{tA}$ solves $u^{\mathrm{\prime }}=Au$ and the resolvent sector of $A$ opens to the right; the opposite pairing $a(u,v)=⟨Au,v⟩$ is the Pazy-Lunardi convention for $-A$ and is not used here.

Coercivity versus sectoriality

A coercive form satisfies (ii) with $M=0$ and $\theta =\arctan (C\mathrm{/}\alpha )$ at most, because $\mathrm{Re}a(u,u)\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}_V^2\ge 0$ and $\mathrm{\mid }\mathrm{Im}a(u,u)\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_V^2$. The sectorial condition is a one-sided quantitative hypothesis on $\mathrm{Re}a$ and $\mathrm{Im}a$ along the diagonal; the phrase "elliptic operator" is never used as a hypothesis (Real and imaginary parts, complex conjugation, and modulus).