Product unbounded operators

Question

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.

Answer 1

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.

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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.

Answer 2

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 $\mathcal D(A)$.