# If $T$ is self-adjoint then $T^2$ is also self-adjoint,

Let $$T:D(T)\subset H \to H$$ be selfadjoint unbounded linear operator on a complex Hilbertspace $$H$$. Show that $$T^2$$ is self-adjoint.

Since $$T$$ is selfadjoint it's spectrum must be real, so $$T\pm i Id$$ are both surjective. According to this answer I can conclude $$T$$ is densely defined. Furthermore it is $$T^2+1Id=(T+iId)(T-iId),$$ so $$T^2+1Id$$ is surjective aswell. Now I need to show only two things

• $$D(T^2)$$ is dense (which can be proven identicaly as in answer) and
• $$D((T^2)^*)\subset D(T^2)$$ (other inclusion is trivial since $$T^2$$ is symmetric).

I am stuck here, how do I proceed?

This is a theorem that might interest you, which gives you what you want when you apply with $$S=T^2+I$$.
Theorem: Let $$\mathcal{H}$$ be a complex Hilbert space. Let $$S : \mathcal{D}(S)\subseteq\mathcal{H}\rightarrow\mathcal{H}$$ be a positive and surjective linear operator defined on the linear subspace $$\mathcal{D}(S)$$. Then $$S$$ is densely-defined and self-adjoint.
Proof: Let $$S$$ be as stated. First we show that $$S$$ is densely-defined. If that were not the case, then there would exist $$y \perp\mathcal{D}(S)$$, and there would exist $$x\in\mathcal{D}(S)$$ such that $$Sx=y$$, which would contradict the positivity of $$S$$ because $$0=\langle x,y\rangle=\langle x,Sx\rangle.$$ So $$S$$ is densely-defined.
To show that $$S$$ is self-adjoint, suppose that $$y\in\mathcal{D}(S^*)$$. Then $$\langle Sx,y\rangle = \langle x,S^*y\rangle,\;\;\; x\in\mathcal{D}(S).$$ Then there exists $$z\in\mathcal{D}(S)$$ such that $$Sz=S^*y$$ because $$S$$ is surjective. Hence, $$\langle Sx,y\rangle =\langle x,S^*y\rangle = \langle x,Sz\rangle = \langle Sx,z\rangle,\;\;\; x\in\mathcal{D}(S).$$ Because $$S$$ is surjective, then $$y=z\in\mathcal{D}(S)$$, and $$S^*y=Sz=Sy$$. So $$S^*=S$$. $$\;\;\blacksquare$$
• For an unbounded Operator to be positive means being symmetric and $\langle x,Sx \rangle \geq 0$? Or for $x$ non zero it must be $\langle x,Sx \rangle > 0$? Commented Aug 27, 2019 at 0:26
• I have assumed $\langle Sx,x\rangle \ge 0$ with equality iff $x = 0$. Commented Aug 27, 2019 at 3:24