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Let $B=B(r)$ denote a ball of radius $r$ in $\Omega \subset \mathbb R^d$ and $$ u_B := \frac1{|B|}\int_B u \, dx. $$ The standard Sobolev-Poincaré inequality states that if $u \in W^{1,p}(\Omega)$, then $$ \|u-u_B\|_{L^\kappa(B)} \le Cr \|\nabla u\|_{L^p(B)} $$ for $$ \kappa = \frac{dp}{d-p}, \quad 1 < p < d. $$

My question is whether something similar is true for merely curl-integrable functions. More precisely, suppose $u \in L_{\rm loc}^p$ is divergence-free and $\text{curl } u \in L_{\rm loc}^p, p>1$. As pointed out below, a standard Poincaré inequality with the gradient replaced by the curl, $$ \|u-u_B\|_{L^\kappa(B)} \le Cr \|\text{curl }u\|_{L^p(B)}, $$ cannot hold in general. But I suspect there should be a corresponding result with some extra assumptions or with some additional term on the right hand side. Any references in to that direction are appreciated.

I am mostly interested in the case $d=2$ and $1<p<2$.

Edit: If it is easier, I would also be interested just in the case, where $\kappa$ is replaced by $p \in (1,2)$ for $d=2$.

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    $\begingroup$ I'm not sure that it will contain the answer, but the following paper treats a very similar subject: Note on explicit proof of Poincare inequality for differential forms on manifolds by Leonid Shartser (arXiv:1010.3356). $\endgroup$ May 29, 2015 at 17:30
  • $\begingroup$ Looks interesting. But figuring out whether this gives what I want seems a non-trivial task with my geometry skills. I was hoping for a more definite answer for the question. $\endgroup$ Jun 2, 2015 at 7:48
  • $\begingroup$ Do you mean Friedrich's inequality? $\endgroup$
    – username
    Sep 23, 2015 at 14:30
  • $\begingroup$ To best of my knowledge Friedrich's inequality requires that the function is in some Sobolev space (i.e. all the partial derivatives, up to some order, are integrable). I was hoping to obtain something for merely curl-integrable functions (say, in 2D). $\endgroup$ Oct 2, 2015 at 8:12

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Counterexample $(x,-y)$, or am I misunderstanding something here.

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  • $\begingroup$ No, I think this is a valid counter-example to the curl-Poincaré inequality. However, then my question is that how should the result be modified to get something reasonable out of it. Somehow I feel there should be a corresponding result also in this setting. But maybe I am wrong. $\endgroup$ Jun 7, 2015 at 13:25

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