Adjoint of an unbounded operator

Remarks

The vector $z$ is unique because $D(T)$ is dense: two such vectors differ by a vector orthogonal to $D(T)$, hence to $H$. Equivalently, $y \in D(T^*)$ if and only if the functional $x \mapsto \ip{Tx}{y}$ is bounded on $D(T)$. If $T$ is bounded with $D(T) = H$, then $T^*$ is the Hilbert space adjoint.

The adjoint $T^*$ is always closed. It is densely defined if and only if $T$ is closable, and then $T^{**}$ is the closure of $T$. If $S \subseteq T$, then $T^* \subseteq S^*$.

Example: on $\ell^2$, let $D(T) = \set{x : \sum n^2 \abs{x_n}^2 < \infty}$ and $T(x_n) = (n x_n)$. For $y \in D(T^*)$ with $T^* y = z$, testing with the unit vector $x = e_n$ gives $n \overline{y_n} = \overline{z_n}$, so $z = (n y_n) \in \ell^2$ and $y \in D(T)$. Conversely each $y \in D(T)$ lies in $D(T^*)$ with $T^* y = Ty$, so $T^* = T$.

Non-example: if $D(T)$ is not dense, there is a nonzero $w \perp D(T)$, and then $z$ and $z + w$ both satisfy clause 1, so clause 2 does not define a map.