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Please read all before giving an answer...

Suppose you're given a densely defined not necessarily bounded operator: $\overline{\mathcal{D}(T)}=\mathcal{H}$
Moreover, assume it is injective: $\mathcal{N}(T)=\{0\}$
Now, imagine it has the property: $T^*Tx=x,x\in\mathcal{D}(T^*T)$

My point is now, it might happen that: $\mathcal{D}(T^*)=\{0\}$
So in this case the above equation seems trivially true as: $\mathcal{D}(T^*T)=\mathcal{N}(T)=\{0\}$

It is not clear to me wether this situation in particular can occur, but is it possible that sth. seemingly similar can happen, so that the above property does not apply preservation of the scalar product?

...clearly, such an operator must be unbounded otherwise one could extend the domain to all of the Hilbert Space, the adjoint would then be defined everywhere as well and in return the property above would imply preservation of the scalar oroduct by construction...

freishahiri
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1 Answers1

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Yes, this can happen. In fact, it tells precisely where the operator is isometric:

If it is isometric somewhere the range will belong to the domain so the adjoint will act as a left inverse. Conversely, if the adjoint acts as a left inverse somewhere then the operator acts isometrically there.

Moreover, then on the isometric part the operator is bounded.

So the consequence is that the adjoint is a left inverse iff the operator is isometric and then therefore bounded

freishahiri
  • 17,045