2
$\begingroup$

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.

$\endgroup$

2 Answers 2

1
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ No why what by definition what?! $\endgroup$ Commented Feb 15, 2015 at 23:09
  • $\begingroup$ A bounded linear operator may even have trivial domain: $\mathcal{D}(T)=(0)$ $\endgroup$ Commented Feb 15, 2015 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
    Commented Feb 16, 2015 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.) $\endgroup$ Commented Feb 16, 2015 at 0:32
  • $\begingroup$ @Freeze_S: I am not familiar with BLT. Maybe we are talking at cross purposes here. $\endgroup$
    – voldemort
    Commented Feb 16, 2015 at 3:04
1
$\begingroup$

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.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .