Can we show that $T^{\ast\ast}\subseteq T^\ast$?

Question

Let $H$ be a $\mathbb C$-Hilbert space and $T$ be a densely-defined linear operator on $H$.

Can we show that $T^{\ast\ast}\subseteq T^\ast$?

It's clear to me that the claim is true when $T$ is symmetric (i.e. $T\subseteq T^\ast$), since then $T$ is closable, $\overline T$ is symmetric as well and hence $$T^{\ast\ast}=\overline T\subseteq\overline T^\ast=T^\ast\tag1.$$

Answer 1

Take $\newcommand{\iu}{\mathrm{i}}T = \iu I$, which is a bounded and everywhere defined operator (therefore also densely-defined). Then we have \begin{align*} T^{\ast} &= \overline{\iu}I= -\iu I \quad\text{and}\\ T^{\ast\ast} &= \overline{-\iu}I= \iu I. \end{align*} So $T^{\ast\ast} \cap T^{\ast} = \{0\}$.