Suppose that $T$ is a (possible unbounded) self-adjoint operator on a Hilbert space $H$, thus the domain $D(T)$ of $T$ is dense in $H$ and the graph of $T$ is closed in $H\times H$. I want to prove that $D(T^2):=\{x\in D(T):Tx\in D(T)\}$ is dense in $H$. That is what i have done until now: Let $y\in H$ and $\epsilon>0$ be arbitrary, then we can find $x\in D(T)$ such that $||x-y||<\epsilon$ since $D(T)$ is dense in $H$. The question is thus of $Tx\in D(T)$. I want to prove this but there is the problem, i don't know how. I think this is the case because $D(T)$ is dense in $G(T)$ is closed but i don't see how to use this. Someone an idea or help for me? Thank you very much.
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Maybe this is too simple such that i cannot see the answer ... – MuHo33 Sep 10 '14 at 15:32
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This may be helpful http://math.stackexchange.com/questions/137536/the-exponent-of-self-adjoint-operator – PhoemueX Sep 10 '14 at 15:38
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Thus if have to find a good book where i can find this theorem from van Neumann – MuHo33 Sep 10 '14 at 15:48
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The domain of $T^{2}$ includes the range of $P=(T-i I)^{-1}(T+iI)^{-1}$; and the range of this product $P$ is dense in $X$ because $P$ is selfadjoint with $\mathcal{R}(P)^{\perp}=\mathcal{N}(P^{\star})=\mathcal{N}(P)$, a subspace which must be $\{0\}$ because $$ (Px,x) = \|(T+iI)^{-1}x\|^{2} \implies \mathcal{N}(P)=\mathcal{N}((T+iI)^{-1})=\{0\}. $$
Disintegrating By Parts
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