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?
$A$ 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.