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Let $S$ be a linear operator with dense domain $\mathcal{D}(S)$ in the Hilbert space $\mathcal{H}$. Assume that the domain $\mathcal{D}(S)$ belongs to a larger domain, namely $\mathcal{D}(S) \subset \mathcal{D}(T)$, where $T$ is a bounded linear operator.

Given the aforementioned setup is it true that:

1) The domain $\mathcal{D}(T)$ is dense in $\mathcal{H}$.

2) The linear operator $S$ is bounded on $\mathcal{D}(S)$.

The reason that I ask 1) and 2) is that I am trying to deduce information about the adjoint operators $S^*$ and $T^*$ given the data.

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If $T$ is bounded then $D(T)=H$ by definition. Therefore, $2$ is false as well.

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  • $\begingroup$ No why what by definition what?! $\endgroup$ – C-Star-W-Star Feb 15 '15 at 23:09
  • $\begingroup$ A bounded linear operator may even have trivial domain: $\mathcal{D}(T)=(0)$ $\endgroup$ – C-Star-W-Star Feb 15 '15 at 23:11
  • $\begingroup$ @Freeze_S: Really!! If $D(T)=0$, then $T$ is the $0$ operator, and hence $D(T)=H$. The definition of a bounded operator is that it's defined everywhere. See any text on operator algebras. $\endgroup$ – voldemort Feb 16 '15 at 0:21
  • $\begingroup$ Yes because operator algebra studies $\mathcal{B}(\mathcal{H})$ as a space in its own right while operator theory adresses an operator $T:\mathcal{D}(T)\to\mathcal{H}$ as an individual object. (Think of for example the BLT which is a very common tool.) $\endgroup$ – C-Star-W-Star Feb 16 '15 at 0:32
  • $\begingroup$ @Freeze_S: I am not familiar with BLT. Maybe we are talking at cross purposes here. $\endgroup$ – voldemort Feb 16 '15 at 3:04
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You might be looking for...

A bounded operator is closable with: $$\|T\|<\infty:\quad\mathcal{D}(\overline{T})=\overline{\mathcal{D}(T)}$$

(Or similarly, a bounded operator is closed iff its domain is so.)

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