If $T$ is a closable densely defined operator then $T^* = \overline{(T^*)} = T^{***} = (\overline{T})^*$

Question

Let $T$ be a closable densely defined operator on a Hilbert space $H$. In Reed & Simon's book on functional analysis they state the result: $$T^* = \overline{(T^*)} = T^{***} = (\overline{T})^*$$ but they do not elaborate on how these equalities are proven. How can one prove them?

Answer 1

• First equal sign is because $T^*$ is a closed operator (its graph is the ‘negative transpose’ of the orthogonal complement of the graph of $T$, and recall that orthogonal complement of any set is always closed (since it is an intersection of kernels of continuous linear maps)).
• For any closable operator, the closure is equal to the double adjoint. We have just said $T^*$ is closed, so it is closable, so its closure is its double adjoint $(T^*)^{**}$.
• Finally, reorder the adjoints to get $(T^*)^{**}=(T^{**})^*$, and once again, since $T$ is closable by hypothesis, the inner bracket is $\overline{T}$.