The adjoint is well defined, closed, and reverses inclusions

By [A1] the adjoint is well defined, $D(T^{\ast })$ is a linear subspace and $T^{\ast }$ is linear.

Let $y_n\in D(T^{\ast })$ with $y_n\to y$ and $T^{\ast }{y}_n\to w$. For every $x\in D(T)$ we have $⟨Tx,y⟩={\lim }_n⟨Tx,y_n⟩={\lim }_n⟨x,T^{\ast }{y}_n⟩=⟨x,w⟩$, both limits being scalar limits; hence $x\mapsto ⟨Tx,y⟩$ is bounded, with $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }=\mathrm{\mid }⟨x,w⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$. So $y\in D(T^{\ast })$ and $T^{\ast }y=w$ by [A1], and therefore $T^{\ast }$ is closed.

Let $y\in H$ and $z\in \mathbb{C}$. If $y\perp \mathrm{ran}(T-z)$, then $⟨(T-z)x,y⟩=0$, that is $⟨Tx,y⟩=z⟨x,y⟩=⟨x,\bar{z}y⟩$, for every $x\in D(T)$; this exhibits $x\mapsto ⟨Tx,y⟩$ as a bounded functional, so $y\in D(T^{\ast })$ and $T^{\ast }y=\bar{z}y$ by uniqueness in [A1]. Conversely if $T^{\ast }y=\bar{z}y$, then $⟨(T-z)x,y⟩=⟨x,\bar{z}y⟩-z⟨x,y⟩=0$ for every $x\in D(T)$, so $y\perp \mathrm{ran}(T-z)$. Hence $\mathrm{ran}(T-z{)}^{\perp }=\ker (T^{\ast }-\bar{z})$.

If $S\subseteq T$ are densely defined, then $⟨Sx,y⟩=⟨Tx,y⟩=⟨x,w⟩$ for every $x\in D(S)$, so every pair $(y,w)\in \mathrm{\Gamma }(T^{\ast })$ lies in $\mathrm{\Gamma }(S^{\ast })$ by [A3]; hence $T^{\ast }\subseteq S^{\ast }$.

Claims collected: well-definedness and linearity from step 1.1, closedness from step 1.2, the kernel-range identity from step 1.3, and the inclusion reversal from step 1.4. ∎