# 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?

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.

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.