Closure of the adjoint operator

Question

Let $\mathcal{H}$ be a Hilbert space, $T:\mathfrak{D}_T\subseteq\mathcal{H}\rightarrow\mathcal{H}$ a densely defined linear operator that is closable. Consider its closure $\overline{T}$, and its adjoint $T^*$. Consider the following equality

$$\overline{T^*}=\overline{T}^*.$$

That is, the adjoint of the closure of $T$ is equal to the closure of the adjoint of $T$. A proof goes simply as;

$$T^*=\overline{T^*}=(T^*)^{**}=(T^{**})^*=\overline{T}^*.$$

However, in this derivation, I use the following result; a densely defined operator $T$ is closable if and only if $T^*$ is densely defined, in which case one has $\overline{T}=T^{**}$. I realised however that in my proof of this last statement, I am using the identity $\overline{T^*}=\overline{T}^*$, which has thus not yet been proven at that point, as it is a result of the statement.

My question; is there a proof of $\overline{T^*}=\overline{T}^*$ that does not rely on the proposition I mentioned above?

Answer 1

Let $T$ be a fixed densely defined linear operator on $\mathcal{H}$ throughout this answer. Your proposed proof is valid. The reasons behind each step are discussed here. It is worth mentioning that the result that $T$ is closable if and only if its adjoint is densely defined can be proved without using the equality $\overline{T^{*}} = (\overline{T})^{*}$. One such proof can be found in Proposition X.1.6(c) of A Course in Functional Analysis by Conway.

Finally, that $T$ is closable implies $\overline{T^{*}} = (\overline{T})^{*}$ can in fact be shown without using that $T$ is closable if and only if its adjoint is densely defined. Let $V\colon \mathcal{H}\times\mathcal{H} \to \mathcal{H}\times\mathcal{H}$ be defined by $V(x,y) := (-y,x)$. By a well-known result that only takes a few lines to show (see Lemma 1.10 of Unbounded Self-adjoint Operators on Hilbert Space by Schmüdgen if needed), for any densely defined linear operator $A$ on $\mathcal{H}$ we have \begin{equation} \mathcal{G}(A^{*}) = V(\mathcal{G}(A))^{\perp} = V(\mathcal{G}(A)^{\perp}), \end{equation} where for a linear operator $B$ on $\mathcal{H}$ we denote $\mathcal{G}(B)$ as the graph of $B$. It now follows that \begin{equation} \mathcal{G}((\overline{T})^{*}) = V(\mathcal{G}(\overline{T})^{\perp}) = V((\overline{\mathcal{G}(T)})^{\perp}) = V(\mathcal{G}(T)^{\perp}) = \mathcal{G}(T^{*}) . \end{equation} Hence we have $(\overline{T})^{*} = T^{*}$. This proof can be found in Theorem 1.8(ii) of Unbounded Self-adjoint Operators on Hilbert Space by Schmüdgen.