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Let $H$ be a complex Hilbert space. This post shows that there exist unbounded (which I will use to mean “not bounded”) operators on $H$ whose domain is all of $H$, i.e., $\mathcal D(T) = H$ (although these operators cannot be defined constructively). Said differently, there are unbounded operators in $L(H)$. I’m curious whether any such operators also have their adjoint defined on all of $H$. In other words, do there exist operators $T,T^* \in L(H)$ such that (1) $T$ is not bounded, and (2) $(x,Ty) = (T^*x, y)$ for all $x,y \in H$?

There is a theorem saying that this is impossible if $T^* = T$, i.e., if $(x, Ty) = (Tx, y)$ for all $x, y \in H$, then $T$ is bounded. I’m basically asking whether this theorem extends to operators that are not self-adjoint in this sense.

WillG
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  • The same question https://math.stackexchange.com/questions/3662185/weakening-the-assumptions-of-the-hellinger-toeplitz-theorem. The same standard proof gives that if two operators satisfy $\langle Tx,y\rangle =\langle x,Sy\rangle $ then $T$ and $S$ are bounded – Ryszard Szwarc Jun 23 '24 at 05:59

1 Answers1

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This is not possible. Any adjointable operator is closed. Indeed, let $x_n \to x$, $Tx_n \to y$. Then for any $z \in H$, we have,

$$\langle z, y \rangle = \lim_n \langle z, Tx_n \rangle = \lim_n \langle T^\ast z, x_n \rangle = \langle T^\ast z, x \rangle = \langle z, Tx \rangle$$

Thus, $y = Tx$, so $T$ is a closed operator. Any closed operator defined on the entirety of $H$ is necessarily bounded, per the closed graph theorem.

David Gao
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  • Here is an alternative approach using uniform boundedness instead of closed graph: https://math.stackexchange.com/a/4857429/465145 – David Gao Jun 23 '24 at 08:13