How can we prove that $‎(D(T),‎\Vert \cdot\Vert‎_{T})‎$‎is Hilbert space?

Question

If $T$ is a closed operator on a Hilbert space $H$ such that $‎T(D(T))‎\subseteq‎ D(T)‎$ , how can we prove that $‎(D(T),‎\Vert \cdot\Vert‎_{T})‎$‎ is a Hilbert space?

Answer 1

The formula $$ \|f\|_T=\sqrt{\|f|^2+\|Tf\|^2} $$ suggests that the inner product should be $$ \langle f,g\rangle_T=\langle f,g\rangle+\langle Tf,Tg\rangle. $$ Now suppose that $\{f_n\}$ is Cauchy in $D(T)$ for the $\|\cdot\|_T$ norm. This implies that $\{f_n\}$ and $\{Tf_n\}$ are Cauchy in the usual norm. Since $T$ is closed, and $H$ is complete, this implies that there exists $f=\lim f_n$ and that $Tf_n\to TF$. So $f\in D(T)$ and $D(T)$ is closed for $\|\cdot\|_T$.