Closed semibounded symmetric forms and their associated operators

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $(V,⟨\cdot ,\cdot {⟩}_V)$ and $(H,⟨\cdot ,\cdot {⟩}_H)$ be complex Hilbert spaces (Hilbert space, Real and complex inner product spaces, with the inner product linear in the first argument) and let $i:V\to H$ be a linear map with dense image.

Symmetric forms. A sesquilinear form on $V$ is a map $a:V\times V\to \mathbb{C}$ that is linear in its first argument and conjugate-linear in its second; it is symmetric if $a(u,v)=\overline{a(v,u)}$ for all $u,v\in V$, in which case $a(u,u)\in \mathbb{R}$ and $\mathrm{Re}a(u,v)=\mathrm{Re}\overline{a(v,u)}$.

Bounded and semibounded. Such a form is bounded if there is $C>0$ with $\mathrm{\mid }a(u,v)\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$ for all $u,v\in V$, and semibounded below with bound $M\ge 0$ if $a(u,u)\ge -M\mathrm{\parallel }i(u){\mathrm{\parallel }}_H^2$ for every $u\in V$.

Form seminorm and closedness. Let $M\ge 0$ be a bound as above and define the form seminorm $\mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}:=(\mathrm{Re}a(u,u)+(M+1)\mathrm{\parallel }i(u){\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}(u\in V)\mathrm{.}$ It is a seminorm because $a(u,v)+(M+1)⟨i(u),i(v){⟩}_H$ is a positive semidefinite Hermitian form; its Cauchy–Schwarz inequality follows by expanding its value on $u+tv$ and minimizing over the scalar $t$, and gives the triangle inequality. It need not separate points for a general dense-image map $i$. The form $a$ is closed if there are such an $M$ and constants $c,C>0$ with $c\mathrm{\parallel }u{\mathrm{\parallel }}_V\le \mathrm{\parallel }u{\mathrm{\parallel }}_{a,M}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_V$ for every $u\in V$; then the form seminorm is a norm, and $(V,\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{a,M})$ is complete because $V$ is. If $i$ is injective, it is already a norm before closedness, since its square is at least $\mathrm{\parallel }i(u){\mathrm{\parallel }}_H^2$.

Associated operator. With $a$ and $i$ as above, define $D(A):=\{u\in V:\text{ there is }f\in H\text{ with }a(u,v)=⟨f,i(v){⟩}_H\text{ for every }v\in V\},Au:=f\mathrm{.}$ The element $f$ is unique when it exists, because two candidates differ by an element of $H$ orthogonal to the dense subspace $i(V)$, hence coincide; $A$ is called the operator associated with $a$. This definition asserts neither that $A$ is densely defined, nor that $A$ is symmetric, nor that $A$ is self-adjoint or bounded below; the specific closed semibounded form of Transported radial endpoint realization establishes those properties for its own associated operator by a direct computation. When $V$ is a subspace of $H$ and $i$ is the inclusion, $\mathrm{\parallel }i(u){\mathrm{\parallel }}_H$ is written $\mathrm{\parallel }u{\mathrm{\parallel }}_H$ and the pairing $⟨f,i(v){⟩}_H$ is written $⟨f,v⟩$.

For $V={\mathbb{C}}^2$, $H=\mathbb{C}$, $i(u,v)=u$ and $a=0$, the map has dense image but the form seminorm vanishes on $(0,1)$. Thus the general dense-image setting requires the seminorm qualification; the radial consumer uses the actual inclusion and proves closedness directly.