3
$\begingroup$

Setup:

If $\langle \Omega, \mathfrak{F},\mu \rangle$ is a measure space, $f:\Omega \rightarrow E$ is a weakly-measurable function to a Banach space $E$, $g: E \rightarrow E'$ is a diffeomorphism and for simplicity say $E$ is reflexive.

Recall that the $E$-valued Dunford-Pettis integral of $f$ over $\Omega$ is defined as the (unique) solution $e$ to: \begin{equation} (\forall x^* \in E^*)\, x^*(e) = \int_{\Omega} x^*(f(\omega)) d\mu(\omega) \end{equation} and we denote it by $\int_\Omega f d\mu$.

Question 1

My question is: is it true that: $g(\int_{\Omega} f d\mu) = \int_{\Omega} g \circ f d\mu $, where the integral here are the Pettis integrals?

Question 2 Since question $1$ was shown to be false, for what $g$ if is does the above equation hold true in general (if any)?

Improve this question
$\endgroup$
7
  • $\begingroup$ Can you define $\int_E f$ where $f:\Omega\to E$? $\endgroup$
    – Liviu Nicolaescu
    Oct 6, 2015 at 13:01
  • $\begingroup$ Its the unique $e\in E$ solving: $(x^* \in E^*)\, x^*(e) = \int x^*(f) $ (where the right is the Lebesgue integral). $\endgroup$
    – ABIM
    Oct 6, 2015 at 13:11
  • 1
    $\begingroup$ I meant, what do you mean by the integral of $f$ over $E$ when $f$ is defined on $\Omega$? $\endgroup$
    – Liviu Nicolaescu
    Oct 6, 2015 at 13:40
  • $\begingroup$ I edited my post to clarify my notation. $\endgroup$
    – ABIM
    Oct 6, 2015 at 13:53
  • $\begingroup$ Do you mean $x^*(f)$ instead of $x^*(e)$? And I still don't get what is denoted by $\int_E f d \mu$, because the expression above seems to be dependent on the choice of $x^* \in E$. Moreover, $x$ is an element of $\Omega$ and $x^*$ of $E$? I apologize for the annoying intrusion... $\endgroup$
    – Giovanni De Gaetano
    Oct 6, 2015 at 14:02

1 Answer 1

Reset to default
1
$\begingroup$

This is true for any linear map $g: E \rightarrow E'$.

The proof of this can be found in these notes.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Did you expect some nonlinear diffeomorphisms would qualify ?? $\endgroup$
    – Jean Duchon
    Oct 6, 2015 at 15:23
  • $\begingroup$ Only linear diffeos qualify $\endgroup$
    – Liviu Nicolaescu
    Oct 6, 2015 at 15:32
  • $\begingroup$ Yes, that's what I thought only linear involutions work.. $\endgroup$
    – ABIM
    Oct 6, 2015 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.