**Product unbounded operators**without wasting too much if your time.

The question is published on by Tutorial Guruji team.

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

**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.

**Product unbounded operators**- If you find the proper solution, please don't forgot to share this with your team members.