Product unbounded operators Let $A : D(A) \subset H \rightarrow H$ be unbounded and $B$ be a bounded operator, both of them are self-adjoint, then 
$(AB)^* = B^*A^*$ and $(BA)^* = A^*B^*$, right?
I just wanted to be sure that I am correct about this.
 A: On the First Identity
The first identity does not seem right because $ A B $ may not be densely defined, which would mean that it has no well-defined adjoint.
Additional Information

Example. Suppose that $ A $ is a densely defined unbounded operator on $ \mathcal{H} $ whose domain $ D_{A} $ omits some $ v \in \mathcal{H} \setminus \{ 0_{\mathcal{H}} \} $. By the linearity of $ D_{A} $, we must have $ (\mathbb{C} \cdot v) \cap D_{A} = \{ 0_{\mathcal{H}} \} $.
Let $ B $ denote the bounded orthogonal projection operator onto $ \mathbb{C} \cdot v $. Then $ D_{A B} = (\mathbb{C} \cdot v)^{\perp} $, which cannot be dense otherwise $ \mathbb{C} \cdot v = (\mathbb{C} \cdot v)^{\perp \perp} = \{ 0_{\mathcal{H}} \} $, which is a contradiction. Hence, $ (A B)^{*} $ does not exist.


On the Second Identity
As $ B A $ is densely defined (because $ D_{B A} = D_{A} $), it has an adjoint and
\begin{align}
       v \in D_{(B A)^{*}}
& \iff \left\{
       \begin{matrix}
       D_{A} & \to     & \mathbb{C} \\
       x     & \mapsto & \langle v \mid B A x \rangle_{\mathcal{H}}
       \end{matrix}
       \right\} ~
       \text{is bounded} \qquad (\text{By definition.}) \\
& \iff \left\{
       \begin{matrix}
       D_{A} & \to     & \mathbb{C} \\
       x     & \mapsto & \langle B^{*} v \mid A x \rangle_{\mathcal{H}}
       \end{matrix}
       \right\} ~
       \text{is bounded} \qquad (\text{As $ D_{B^{*}} = \mathcal{H} $.}) \\
& \iff B^{*} v \in D_{A^{*}} \qquad (\text{By definition again.}) \\
& \iff v \in D_{A^{*} B^{*}}.
\end{align}
Therefore, $ D_{(B A)^{*}} = D_{A^{*} B^{*}} $, so we know that $ (B A)^{*} $ and $ A^{*} B^{*} $ have the same domain. However, we must still prove that they are equal as operators, so let $ v $ be a vector lying inside their common domain. Then for all $ x \in D_{A} $, we have
\begin{align}
    \langle (B A)^{*} v \mid x \rangle_{\mathcal{H}}
& = \langle v \mid B A x \rangle_{\mathcal{H}} \\
& = \langle B^{*} v \mid A x \rangle_{\mathcal{H}} \\
& = \langle A^{*} B^{*} v \mid x \rangle_{\mathcal{H}}.
\end{align}
As $ D_{A} $ is dense in $ \mathcal{H} $, it follows that $ (B A)^{*} v = A^{*} B^{*} v $. Therefore, $ (B A)^{*} = A^{*} B^{*} $.
No assumption is made about the self-adjointness of $ A $ or $ B $ whatsoever; this is a general proof.
A: Even if $AB$ is densely defined, its adjoint is not necessarily $B^* A^*$.
Consider an unbounded  self-adjoint operator $A$ such that $\sigma(A) \subseteq [1,\infty)$ and let $B = A^{-1}$ which is bounded.  Then $AB = I$, but $B^* A^* = B A$ is the restriction of $I$ to  $\mathcal D(A)$.
