If $T$ is closed and densely-defined, then so is $T^\ast T$?

Question

We suppose $H$ is a Hilbert space, and that $T : H \to H$ is a closed and densely-defined operator.

Under what conditions is $T^\ast T: H \rightarrow H$ a closed and densely-defined operator?

Answer 1

The product space $H\times H$ with natural inner product has an orthogonal decomposition $$ H \times H = \mathcal{G}(T)\oplus J\mathcal{G}(T^*) $$ where $J(x,y)=(-y,x)$. Recall that this is how the proof of the existence of adjoint proceeds for a closed densely-defined $T$. The above orthogonality gives $\langle (x,Tx),(-T^*y,y)\rangle=0$ which is the adjoint relation $(Tx,y)=(x,T^*y)$ for all $x\in\mathcal{D}(T)$, $y\in\mathcal{D}(T^*)$.

Surjectivity: Because of the orthogonal decomposition, for any $z\in H$ there exists $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^*)$ such that $$ (z,0) = (x,Tx)+(-T^*y,y) \\ \implies y=-Tx, \;\; z=x-T^*y=x+T^*Tx. $$ So $(I+T^*T)$ is surjective because every $z\in H$ is in the range of $I+T^*T$.

Density of Domain: $T^*T$ is densely-defined because, if $y\perp\mathcal{D}(T^*T)$, then $y=(I+T^*T)x$ for some $x$ and $$ 0=(y,x)=\langle (I+T^*T)x,x\rangle =\|x\|^2+\|Tx\|^2 \\ \implies x=0 \implies y=(I+T^*T)x=0. $$ Closed Graph: Because $\langle (I+T^*T)x,x\rangle \ge \|x\|^2$ for $x\in\mathcal{D}(T^*T)$, then $(I+T^*T)^{-1}$ is a bounded operator on $H$ and, hence, closed. It follows that $I+T^*T$ is also closed because its graph is the transpose of the graph of $(I+T^*T)^{-1}$. So $T^*T$ is closed because $(T^*T+I)-I$ is a bounded perturbation of a closed operator.