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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:γ€ˆ.,.〉is the hermitian inner product

Is there a sense in which $T$ can be bounded, even if it's not everywhere defined ($D(T) \subset \scr H$)?

ric.san
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  • Can $T$ not be extended to all of $\mathscr{H}$ by density of $D(T)$? Or is there something I am missing here? – rubikscube09 May 28 '21 at 13:13
  • If you put $\phi=\psi$ you get $|T\phi|=|\phi|$. Therefore, $T$ is continuous at $0$ and then continuous. – plop May 28 '21 at 13:29

1 Answers1

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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 $\mathscr{H}$.

JustWannaKnow
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