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?
 A: 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 $S^*=S$. $\;\;\blacksquare$
