Product unbounded operators Code Answer

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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.

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