Show self-adjointness elementary

Question

Is anybody aware of an elementary proof that $T^*T$ is self-adjoint where $T$ is closed and densely-defined? All proofs I found so far use the Friedrich's extension or other more sophisticated theorems. Is it possible to show it by only using the characterisation that an operator $A$ is self-adjoint if and only if $ran(A \pm i)= H$ or $A$ is closed and $ker(A\pm i)=\{0\}$?

Answer 1

Theorem: Suppose $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a surjective linear operator on a complex Hilbert space $X$ for which $(Ax,x) \ge \|x\|^{2}$ holds for all $x\in\mathcal{D}(A)$. Then $A$ is densely-defined and selfadjoint.

Proof: To show that $A$ is densely-defined suppose otherwise. Then there exists $y \perp \mathcal{D}(A)$. However, $y= Ax$ for some $x\in\mathcal{D}(A)$ because $A$ is surjective, which gives $$ 0=(y,x)=(Ax,x) \ge \|x\|^{2} \implies x = 0 \implies y=Ax=0. $$ Therefore $A$ is densely-defined. So $A$ is a closable, densely-defined, symmetric linear operator, which gives the existence of a closed densely-defined adjoint $A^{\star}$, and $A^{\star}|_{\mathcal{D}(A)}=A$, meaning that one has the graph inclusion $\mathcal{G}(A)\subseteq\mathcal{G}(A^{\star})$.

To show that $A=A^{\star}$, it is enough to show that $\mathcal{D}(A^{\star})\subseteq\mathcal{D}(A)$. To prove this, let $y\in\mathcal{D}(A^{\star})$ and choose $z\in\mathcal{D}(A)$ such that $Az=A^{\star}y$, which is possible because $A$ is surjective. Then, $$ (Ax,y)=(x,A^{\star}y)=(x,Az)=(Ax,z),\;\;\; x\in\mathcal{D}(A). $$ Because $A$ is surjective, then $y=z \in\mathcal{D}(A)$. $\;\;\blacksquare$

Corollary: Let $T:\mathcal{D}(T)\subseteq X\rightarrow X$ be a closed, densely-defined, linear operator on a complex Hilbert space $X$. Then $I+T^{\star}T$ is densely-defined and selfadjoint on its natural domain $\mathcal{D}(T^{\star}T)$ consisting of all $x \in \mathcal{D}(T)$ for which $Tx\in\mathcal{D}(T^{\star})$.

Proof: By the theorem, it is enough to show that $A=I+T^{\star}T$ is surjective on its natural domain because $(Ax,x)=\|x\|^{2}+\|Tx\|^{2}\ge \|x\|^{2}$ for $x\in\mathcal{D}(T^{\star}T)$. Because $T$ is closed, then the graph $\mathcal{G}(T)\subseteq X\times X$ gives the orthogonal decomposition $$ \mathcal{G}(T)\oplus\mathcal{G}(T)^{\perp} = X\times X. $$ Clearly $\langle u,v\rangle \in \mathcal{G}(T)^{\perp}$ iff $$ 0=(\langle x,Tx\rangle,\langle u,v\rangle)_{X\times X}=(x,u)+(Tx,v), \;\;\; x \in \mathcal{D}(T) \\ \iff v\in\mathcal{D}(T^{\star}) \mbox{ and } T^{\star}v=-u. $$ Therefore, for every $z \in X$, there exists $x \in \mathcal{D}(T)$ and $y\in \mathcal{D}(T^{\star})$ such that $$ \langle z,0 \rangle = \langle x, Tx \rangle + \langle T^{\star}y,-y\rangle. $$ Thus $y=Tx$ and $z=x+T^{\star}y=x+T^{\star}Tx$, which proves that $I+T^{\star}T$ is surjective. Hence, by the theorem, $I+T^{\star}T$ is densely-defined and selfadjoint. $\;\;\blacksquare$.