I am reading the paper http://www.ams.org/mathscinet-getitem?mr=3246935 and there some notation that I have found a bit confusing on page 1503 between Lemma 4.2 and 4.3. I'll give as much context as I can.

Let $\mathcal{H}$ be a infinite dimensional separable Hilbert space and $(X, \nu)$ a metric space equipped with a positive locally finite Borel measure. Let $T$ be a bounded linear operator on $L^2(X, \nu) \otimes \mathcal{H}$. Let $\varphi \in L^2(X, \nu) \otimes \mathcal{H}$ and define the finite measure $$ d \mu = \|T \varphi\|_{\mathcal{H}}^2 d \nu $$ My questions are: 1) What does this notation mean? 2) What is the measure $\mu$? 3) How did $\mu$ become a finite measure when $\nu$ is locally finite?

This paper it has given more hints what this notation means. For a Borel subset $\Omega \subset X$ let $P_{\Omega}$ be the orthogonal projection $P_{\Omega} \colon L^2(X, \nu) \otimes \mathcal{H} \to L^2(\Omega, \nu) \otimes \mathcal{H}$. Then $\| P_{\Omega} T \varphi \|^2 = \mu(\Omega)$. This makes some sense but I am confused why there is no reference to $\nu$.

  • 3
    $\begingroup$ The notation $\mathrm d\mu =f\,\mathrm d\nu$ for some non-negative, measurable $f$, means that $\mu$ is the measure defined by having density $f$ with respect to $\nu$. $\endgroup$ – Stefan Hansen Oct 1 '14 at 12:23
  • $\begingroup$ And what does density mean here? $\endgroup$ – Chris Cave Oct 1 '14 at 12:25
  • 5
    $\begingroup$ It means that $\mu$ is defined by $\mu(A)=\int_A f\,\mathrm d\nu$. $\endgroup$ – Stefan Hansen Oct 1 '14 at 12:27
  • 1
    $\begingroup$ Stefan has answered your Question 1. As for Question 2 (in case the answer to Question 1 didn't also solve Question 2 for you), $\mu$ is finite because the measure of the whole space is the integral of $\Vert T\phi\Vert^2$, which is finite because $T\phi$ is in $L^2$. $\endgroup$ – Andreas Blass Oct 1 '14 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.