Concerning unbounded linear operators on a Hilbert space Let $H$ be some Hilbert space and let $B:H\rightarrow H$ be a bounded linear operator and $T:D(T)\rightarrow H$ an unbounded linear operator. Furthermore we assume that $T$ is closed ,i.e. it's graph in $H\times H$ is closed and we also have that that the domains $D(T)$ and $D(BT)$ of operator $T$ and $BT$ are both dense sets in $H$.
The closure of a closable linear operator $S:D(S)\rightarrow H$ is written as $\overline{S}$. The adjoint of a densely defined linear operator $S:D(S)\rightarrow H$ is written as $S^*$. Operator $A^*$ is always closed for any densely defined operator $A$.
With the given assumptions and notations I would like to show the following equation holds
$ (TB)^*=\overline{B^* T^*}$.
This also means that the domains must coincide. So far I am only able to show that $TB$ is closed and that $(TB)^*$ is an extension of $\overline{B^* T^*}$. But I still need to show that $\overline{B^* T^*}$ is an extension of $(TB)^*$. The most difficult and crucial part seems to me to show that the following inclusion holds
$D((TB)^*)\subset D(\overline{B^*T^*})$.
But so far I got nothing.
Kindly apreciated,
Aris

Here is a solution based on the fact that $S^{**}=S$ for any closed, densely defined operator $S$.
Since $B$ is bounded, we have, for any $x\in \mathrm{Dom}((B^*T^*)^*)$ and $y\in \mathrm{Dom}(T^*)$, the identity
$$
\langle (B^*T^*)^* x,y\rangle = \langle x, B^*T^* y\rangle = \langle B x, T^* y\rangle,
$$
which implies that $Bx\in\mathrm{Dom}(T)$ and $TBx=(B^*T^*)^* x$. Thus, $(B^*T^*)^*\subseteq TB$, and therefore $(B^* T^*)^{**}\supseteq (TB)^*$. Additionally, the inclusion $B^*T^*\subseteq \overline{B^*T^*}$ implies $(B^*T^*)^{**}\subseteq \overline{B^*T^*}$. But then we have
$$
\overline{B^*T^*} \supseteq(B^*T^*)^{**}\supseteq(TB)^*.
$$
