2
$\begingroup$

This is a follow-up question of this one.

Let $\mathbb{D}^2$ be the closed two-dimensional unit disk (endowed with some smooth Riemannian metric), and let $\sigma \in \Omega^1(\mathbb{D}^2)$ be a smooth one-form.

Does there exist a sequence of smooth one-forms $\sigma_n \in \Omega^1(\mathbb{D}^2)$ such that

  1. $\sigma_n \to \sigma$ in $L^2$.
  2. $\sigma_n$ do not vanish on $\mathbb{D}^2$.
  3. $\sup_n \|\delta d \sigma_n \|_{L^p}< \infty, \sup_n \| d \delta \sigma_n \|_{L^p}< \infty$ for some $p>1$.

Note that I don't fix any boundary conditions on $\sigma_n$, so we don't have standard elliptic estimates. (In particular, $\sup_n \|\Delta \sigma_n \|_{L^p}< \infty$ does not imply $\sup_n \|\nabla^2 \sigma_n \|_{L^p}< \infty$).

Comment:

I think that one can make a perturbation to $\sigma$ which gives a form with only finitely many zeroes. After that, there is a procedure to push out the zeroes, one by one -- by composing with diffeomorphisms. I think that this procedure does not satisfy condition $3$.


If it matters, I am fine with assuming that $\delta \sigma=0, d\sigma=0$.

Note that we cannot approximate $\sigma$ by nowhere-vanishing $\sigma_n$ in $W^{1,1}$.

$\endgroup$
2
  • 1
    $\begingroup$ While you don't have a global elliptic estimate, there should be a local elliptic regularity estimate. I don't have a good reference but web.math.ucsb.edu/~grigoryan/246B/lecs/246B_ch5.pdf Theorem 5.3 is the sort of thing you could use. (Then with a $W^{2,p}$ bound you can use the Rellich–Kondrachov theorem.) $\endgroup$
    – Dap
    Commented Dec 9, 2019 at 19:53
  • $\begingroup$ Thank you, you are right. These local estimates are exactly what I needed. It seems then that my final goal is not really achievable. However, I certainly learned a lot from all your great answers, even if I won't "use" them in any tremendous way for some paper (which probably won't be written now...). Thank you very much again. $\endgroup$ Commented Dec 10, 2019 at 14:36

0

You must log in to answer this question.