Product of operator with its adjoint is self-adjoint 
Suppose $A$ is a densely defined closed operator. Show that $A^*A$ (with domain $D(A^*A)=\{\psi\in D(A)|A\psi\in D(A^*)\}$) is self-adjoint.

Let $\psi\in D(A^*A)$ and observe that $A^*A$ is nonnegative:
$$(\psi,A^*A\psi)=(A\psi,A\psi)=\|A\psi\|\ge0.$$ Then $A^*A$ is symmetric and
$$(A^*A\psi,\psi)=(\psi,A^*A\psi)=((A^*A)^*\psi,\psi)$$ for all $\psi\in D(A^*A)\cap D((A^*A)^*)$, implying $(A^*A)^*=A^*A$ in this domain. We must show that $D(A^*A)=D((A^*A)^*)$. Recall the definition of the domain of the adjoint:
$$D(B^*)=\{\psi\in\mathcal{H}|\exists\tilde\psi\in\mathcal{H}:(\psi,B\phi)=(\tilde\psi,\phi),\forall\phi\in D(B)\}.$$
Then for a symmetric operator, say $B$, we have $D(B)\subset D(B^*)$ since
$$(\psi,B\phi)=(B\psi,\phi)$$ for all $\phi\in D(B)$. This gives us the forward inclusion since $A^*A$ is symmetric. How can the reverse inclusion be achieved?
 A: The graph of $\mathcal{G}(A)$ is closed. The orthogonal complement of $\mathcal{G}(A)$ consists of all pairs $(y,z)\in\mathcal{H}\times\mathcal{H}$ such that
$$
                    \langle (x,Ax),(y,z)\rangle =0,\;\; x\in\mathcal{D}(A).
$$
That is the same as
$$
                      \langle x,y\rangle+\langle Ax,z\rangle=0,\;\; x\in\mathcal{D}(A),
$$
which is exactly the requirement that $z\in\mathcal{D}(A^*)$ and $A^*z=-y$. Therefore, every $(z,w)\in\mathcal{H}\times\mathcal{H}$ may be written as the following sum for some $x\in\mathcal{D}(A)$ and $y\in\mathcal{D}(A^*)$:
$$
               (z,w) = (x,Ax)+(-A^*y,y)
$$
Therefore,
$$
                    z=x-A^*y,\;\; w=Ax+y
$$
We can choose $w=0$ in order to obtain $x,y$ such that
$$
                          z=x-A^*y=x+A^*Ax.
$$
It follows that $A^*A+I$ is surjective because for every $z$ there exists $x$ such that $z=(I+A^*A)x$. It is easy to verify that $A^*A+I$ is symmetric on its natural domain.
The domain of $A^*A+I$ has to be dense because it is surjective; to see why, suppose $y\perp\mathcal{D}(A^*A)$. Then there exists $x\in\mathcal{D}(I+A^*A)$ such that $(I+A^*A)x=y$. Because $y\perp x$, it follows that
$$
    0= \langle y,x\rangle = \langle (I+A^*A)x,x\rangle=\|x\|^2+\|Ax\|^2\implies x=0 \implies y = 0.
$$
So $I+A^*A$ is densely-defined, symmetric and surjective.
That is enough to conclude that $I+A^*A$ is self-adjoint. Indeed, suppose $y\in\mathcal{D}((I+A^*A)^*)$. Then
$$
      \langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle,\;\; x\in\mathcal{D}(I+A^*A).
$$
Then there exists $w\in\mathcal{D}(I+A^*A)$ such that $(I+A^*A)^*y=(I+A^*A)w$. Hence,
$$
      \langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle= \langle x,(I+A^*A)w\rangle = \langle (I+A^*A)x,w\rangle
$$
Because $I+A^*A$ is surjective, then $y=w$, from which it follows that $y\in\mathcal{D}(I+A^*A)$ and $(I+A^*A)^*y=(I+A^*A)y$. Therefore, $(I+A^*A)^*=I+A^*A$.
