Let $ T: D(T) \rightarrow \scr H $ be a densely defined isometric operator, i.e. $$ \langle T \phi, T \psi \rangle = \langle \phi, \psi \rangle \quad \forall \ \phi,\psi \in D(T) $$
notation:〈.,.〉is the hermitian inner product
Is there a sense in which $T$ can be bounded, even if it's not everywhere defined ($D(T) \subset \scr H$)?
Taking $\phi = \psi$, we have $||T\psi||^{2} = ||\psi||^{2}$ so that $T$ is bounded on $D(T)$, with $||T|| = 1$. Now, extend $T$ to $\mathscr{H}$.