I'd like to prove that the adjoint of a densely defined linear operator is closed.
Let $H$ be a Hilbert space, and $T: D(T)\rightarrow H$ a densely defined linear operator ($D(T)$ is a dense subspace of $H$). The domain of the adjoint is defined as $D(T^{\ast}):=\left\{x \in H: \exists y\in H \mbox{ with } \langle x,Tz \rangle=\langle y, z \rangle\; \forall z \in H\right\}$ and $T^{\ast}x=y$.
Let $\{x_{n}\}_{n \in \mathbb{N}}\subset D(T^{\ast})$ be a convergent sequence with $\lim_{n \to \infty}x_{n}=x$ and $\lim_{n \to \infty}T^{\ast}x_{n}=y$. For $T^{\ast}$ to be closed I need to show:
- $x \in D(T^{\ast})$
- $T^{\ast}x=y$
1.) $$ \begin{align} \langle x, Tz \rangle &= \langle \lim_{n \to \infty}x_{n}, Tz \rangle\\ &= \lim_{n \to \infty}\langle x_{n}, Tz \rangle\\ &= \lim_{n \to \infty}\langle T^{\ast}x_{n}, z \rangle\\ &= \langle\lim_{n \to \infty} T^{\ast}x_{n}, z \rangle\\ &= \langle y, z \rangle\\ \end{align} $$ and therefore $x \in D(T^{\ast})$.
2.)
Given 1.), the continuity of $\langle \cdot, \cdot \rangle$ and the definition of the adjoint we can show $$ \begin{align} \langle y, z \rangle &= \langle \lim_{n \to \infty}T^{\ast}x_{n}, z \rangle\\ &= \lim_{n \to \infty} \langle T^{\ast}x_{n}, z \rangle\\ &= \lim_{n \to \infty} \langle x_{n}, Tz \rangle\\ &= \langle \lim_{n \to \infty}x_{n}, Tz \rangle\\ &= \langle x, Tz \rangle\\ &= \langle T^{\ast}x, z \rangle\\ \end{align} $$ for all $z \in H$ and hence $y=T^{\ast}x$.
Is this proof correct?