Showing that if $A$ is closed, then $A^\ast A$ is self-adjoint Let $A$ be a closed linear operator on a Hilbert space $H$. Then I want to show that $B = A^\ast A$ is self-adjoint.
Now, $B$ is positive, i.e. $\langle f, B f \rangle \geq 0 \forall f \in D(B)$.
Therefore the associated quadratic form is real-valued and $B$ is symmetric (by polarization).
Another possible way to show that $B$ is symmetric is the following I think:
$(A^\ast A)^\ast \supseteq A^\ast A^{\ast \ast} = A^\ast A$,  since $A$ is closed.
is this so far correct reasoning? (I am a bit confused because the first alternative didn't use the closedness of $A$.)
To show that $B$ is self-adjoint, it remains to show that $D(B^\ast)$ is contained in $D(B)$.
How do I go about this?
Edit: I got this idea: If I can show that for every $\phi \in H$ there exists a $\psi$ such that
$\phi = \psi + B \psi$, then it follows that $B$ is self-adjoint.
Now I guess, the existence of such a $\psi$ follows from the closedness of $A$, right? How can I show this?
 A: $A$ needs to be closed and densely-defined. If that is the case, then the graph $\mathcal{G}(A)$ is a closed subspace of $H \times H$. Hence,
$$
                     \mathcal{G}(A)\oplus\mathcal{G}(A)^{\perp}=H\times H,
$$
where the orthogonal complement is taken in $H\times H$. You should know that
$$
\begin{align}
            \mathcal{G}(A)^{\perp} & = \{ (a,b)\in H\times H : \langle(a,b),(c,Ac)\rangle =0,\; \forall c \in \mathcal{D}(A) \} \\
       & = \{ (a,b) \in H\times H : \langle a,c\rangle+\langle b,Ac\rangle = 0,\; \forall c \in \mathcal{D}(A) \} \\
       & = \{ (-A^{\star}b,b) \in H\times H : b\in\mathcal{D}(A^{\star})\}.
\end{align}
$$
Therefore, every $(y,z) \in H\times H$ may be uniquely written as
$$
                (y,z) = (a,Aa)+(-A^{\star}b,b)
$$
for some $a \in \mathcal{D}(A)$ and $b\in\mathcal{D}(A^{\star})$. Useful special cases occur where $y=0$ or $z=0$. Start with $z=0$. Then
$$
                 y=a-A^{\star}b, 0=Aa+b \implies y = a+A^{\star}Aa.
$$
Therefore, $A^{\star}A+I$ is surjective on its natural domain,
$$
          \mathcal{D}(A^{\star}A)=\{ a \in \mathcal{D}(A) : Aa \in \mathcal{D}(A^{\star})\}.
$$
It is trivial to show that $A^{\star}A$ is symmetric on its natural domain. This domain must be dense; indeed, suppose $y \perp \mathcal{D}(A^{\star}A)$ and write $y=x+A^{\star}Ax$ for some $x\in\mathcal{D}(A^{\star}A)$ in order to obtain
$$
       0=(y,x) = (x+A^{\star}Ax,x)=\|x\|^{2}+\|Ax\|^{2} \implies x=0.
$$
So $A^{\star}A$ is symmetric, densely-defined and, hence, closable. Therefore the adjoint $(A^{\star}A)^{\star}$ is closed and densely-defined.
To show that $A^{\star}A$ is selfadjoint, suppose $y \in \mathcal{D}((A^{\star}A)^{\star})$. Then there exists $z \in \mathcal{D}(A^{\star}A)$ such that
$$
   (I+(A^{\star}A))^{\star}y=(I+A^{\star}A)z.
$$
Hence, for all $x \in \mathcal{D}(A^{\star}A)$,
$$
\begin{align}
       ((I+A^{\star}A)x,y) 
          & =(x,(I+A^{\star}A)^{\star}y) \\
          & =(x,(I+A^{\star}A)z) \\
          & = ((I+A^{\star}A)x,z).
\end{align}
$$
Because $I+A^{\star}A$ is surjective, then $y=z$, which implies $y\in\mathcal{D}(I+A^{\star}A)$. This proves that
$$
      (I+A^{\star}A)^{\star} \preceq (I+A^{\star}A)
$$
The opposite graph inclusion holds because $I+A^{\star}A$ is symmetric. Hence $I+A^{\star}A$ is selfadjoint.
