Adjoint of a densely defined operator

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a densely defined linear operator on $H$, with domain $D(T)$ dense in $H$ (Densely defined, closed and closable operators, and cores, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets). Since $D(T)$ is dense, and since $H$ is complete (Hilbert space), every bounded linear functional on the normed space $(D(T),\mathrm{\parallel }\cdot \mathrm{\parallel })$ (A bounded linear operator between normed spaces) extends uniquely to a bounded linear functional $\tilde{\phi }$ on $H$ of the same norm, by setting $\tilde{\phi }(z)={\lim }_n\phi (z_n)$ for any sequence $z_n\in D(T)$ with $z_n\to z$ in $H$; the limit exists because $\phi$ is bounded on a dense subspace and $H$ is complete, and it does not depend on the sequence because a bounded functional is uniformly continuous.

A vector $y\in H$ belongs to the adjoint domain $D(T^{\ast })$ when the linear functional ${\phi }_y:D(T)\to \mathbb{C}$, ${\phi }_y(x)=⟨Tx,y⟩$, is bounded on $(D(T),\mathrm{\parallel }\cdot \mathrm{\parallel })$. In that case Hilbert space Riesz representation (Riesz representation for Hilbert spaces) applied to the extension $\widetilde{{\phi }_y}$ produces a unique vector $T^{\ast }y\in H$ with $T^{\ast }y\text{ is the unique }w\in H\text{ such that }⟨Tx,y⟩=⟨x,w⟩\text{for all }x\in D(T),$ equivalently $\widetilde{{\phi }_y}(z)=⟨z,T^{\ast }y⟩$ for all $z\in H$. The map $T^{\ast }:D(T^{\ast })\to H$, $y\mapsto T^{\ast }y$, is the adjoint of $T$.

Well-definedness. The functional ${\phi }_y$ is linear in $x$ for each $y$, so its domain of boundedness $D(T^{\ast })$ is a linear subspace: if ${\phi }_y,{\phi }_{y^{\mathrm{\prime }}}$ are bounded so is ${\phi }_{ay+b{y}^{\mathrm{\prime }}}$ for scalars $a,b$, because the first-variable-linear convention gives ${\phi }_{ay+b{y}^{\mathrm{\prime }}}=\bar{a}{\phi }_y+\bar{b}{\phi }_{y^{\mathrm{\prime }}}$. On $D(T^{\ast })$ the map $T^{\ast }$ is linear: if $y,y^{\mathrm{\prime }}$ are represented by $w,w^{\mathrm{\prime }}$, then $ay+b{y}^{\mathrm{\prime }}$ is represented by $aw+b{w}^{\mathrm{\prime }}$, since $⟨x,aw+b{w}^{\mathrm{\prime }}⟩=\bar{a}⟨x,w⟩+\bar{b}⟨x,w^{\mathrm{\prime }}⟩$; the representing vector is unique. By Riesz representation for the Hilbert space $H$ a vector $w\in H$ is determined by the values $⟨z,w⟩$ with $z$ ranging over $H$, and those values are determined by the functional $\widetilde{{\phi }_y}$. Equivalently, if $w,w^{\mathrm{\prime }}$ both represent ${\phi }_y$, then $⟨x,w-w^{\mathrm{\prime }}⟩=0$ for every $x\in D(T)$; density of $D(T)$ and continuity of the inner product extend this equality to every $x\in H$, and taking $x=w-w^{\mathrm{\prime }}$ gives $w=w^{\mathrm{\prime }}$. Finally only ambient-norm boundedness of ${\phi }_y$ on $D(T)$ is required: no extension, no closure and no closedness of $T$ is presupposed.