prop:Let $(H,(,))$ be a Hilbert space and $T$ be a non-negative self-adjoint closed operator s.t., domain of $T$ is dense in $H$. Then, $\sqrt{T}$ is closed operator.
By spectral decomposition, We can represent $T=\int _0 ^\infty \lambda dE_{\lambda}$ , so we can define $\sqrt {T}=\int _0 ^\infty \sqrt{\lambda }dE_{\lambda}$. If $\sqrt{T}$ is bounded, I can prove it is closed. However, in general case, I cannot prove.