Skip to main content
MathOverflow

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

2
  • 1
    $\begingroup$ If you are referring to the proof in the big McDuff-Salamon book, they state at the beginning of that section (2.3) an assumption for J to have weakest possible smoothness, C^1. I think this assumption is meant to permeate through. $\endgroup$
    – Chris Gerig
    Commented Oct 15, 2016 at 4:45
  • $\begingroup$ Thanks for the comment. Yes, I am talking exactly about the McDuff-Salamon book you mentioned. However, it seems they don't wish to increase the regularity: they take $J(z) := J(u(z))$, where $u \in W^{1,p}$. So, I believe the statement should hold for $J$ as mentioned above. Anyhow, how about taking a section of the projection $\Psi \mapsto \Psi^{-1} J_0 \Psi $ and then precomposing it with $J$? $\endgroup$
    – Boggie Georgiev
    Commented Oct 15, 2016 at 10:06