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The Hellinger-Toeplitz theorem states that if $T$ is a linear map from a Hilbert space $H$ to itself, satisfying $ \langle Tx,y\rangle=\langle x,Ty\rangle$ for all $ x,y \in H$, then $T$ is bounded.

But can we not instead have the assumption that there is a linear map $ T^*$ such that $ \langle Tx ,y \rangle = \langle x, T^* y \rangle$ for all $ x,y \in H$? Then we can take $ x_n \to x$ and set $ z = \lim_{n\to \infty} Tx_n $, so that for all $ y \in H$

\begin{align} \langle z,y \rangle \leftarrow \langle Tx_n,y \rangle = \langle x_n, T^*y\rangle \to\langle x , T^*y\rangle = \langle Tx,y\rangle \end{align} so that $ z = Tx$, and so the graph of $T$ is closed, so by closed graph theorem $T$ is bounded.

It is the same proof as for the usual Hellinger-Toeplits, so I don't see why we need the operator to be symmetric?

Tom
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1 Answers1

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You are right. Your proof shows that an adjointable linear operator on a Hilbert space is bounded. I guess the reason Hellinger-Toeplitz has a name is due to Quantum Mechanics, where they only care about selfadjoint (or symmetric, when unbounded) operators.

Martin Argerami
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