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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)?

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  • $\begingroup$ Can you define $\int_E f$ where $f:\Omega\to E$? $\endgroup$ Commented 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). $\endgroup$ Commented Oct 6, 2015 at 13:11
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    $\begingroup$ I meant, what do you mean by the integral of $f$ over $E$ when $f$ is defined on $\Omega$? $\endgroup$ Commented Oct 6, 2015 at 13:40
  • $\begingroup$ I edited my post to clarify my notation. $\endgroup$ Commented 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... $\endgroup$ Commented Oct 6, 2015 at 14:02

1 Answer 1

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

The proof of this can be found in these notes.

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  • $\begingroup$ Did you expect some nonlinear diffeomorphisms would qualify ?? $\endgroup$ Commented Oct 6, 2015 at 15:23
  • $\begingroup$ Only linear diffeos qualify $\endgroup$ Commented Oct 6, 2015 at 15:32
  • $\begingroup$ Yes, that's what I thought only linear involutions work.. $\endgroup$ Commented Oct 6, 2015 at 15:39

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