Adjoint of a densely defined operator

Let $T:\mathcal D(T)\subseteq H\to K$ be a densely defined operator between complex Hilbert spaces, with inner products linear in the first variable. The adjoint $T^*$ has domain

\mathcal D(T^*)= \left\{y\in K:\exists z\in H\ \text{such that}\ \langle Tx,y\rangle_K=\langle x,z\rangle_H \ \text{for all }x\in\mathcal D(T)\right\},

and $T^*y=z$ . Density of $\mathcal D(T)$ makes $z$ unique. Equivalently, $y\in\mathcal D(T^*)$ exactly when the functional $x\mapsto\langle Tx,y\rangle_K$ is bounded in the $H$ -norm; the Riesz representation theorem then supplies $T^*y$ .

Closedness and closability

The adjoint $T^*$ is always closed, although its domain need not be dense. The original operator $T$ is closable exactly when $\mathcal D(T^*)$ is dense in $K$ . In that case its closure satisfies

\overline T=T^{**}.

These identities include equality of domains as well as equality of the operator actions.

Domain-sensitive algebra

Adjoints reverse products only with domain qualifications. For densely defined $S$ and $T$ such that $ST$ is densely defined, one generally has

(ST)^*\supseteq T^*S^*,

and equality requires additional hypotheses. Likewise, the formal integration-by-parts expression for a differential operator does not by itself determine the adjoint: boundary terms and the chosen domain determine $\mathcal D(T^*)$ .

Symmetry and self-adjointness

For $T:\mathcal D(T)\subseteq H\to H$ , symmetry means $T\subseteq T^*$ , including $\mathcal D(T)\subseteq\mathcal D(T^*)$ . Self-adjointness requires the stronger equality $T=T^*$ , especially $\mathcal D(T)=\mathcal D(T^*)$ . A symmetric differential operator on a small test domain can therefore fail to be self-adjoint even when its formal expression is real.

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