Let $v_k$ be a radial sequence of function that satisfies in $\Omega\subset\mathbb{R}^4$

  • $(-\Delta)^2 v_k=e^{v_k}$
  • $v_k(x)\leq v_k(0)=0$
  • $\left\Vert (-\Delta)v_k\right\Vert_{L^1(B_R(0))}=O(1)\qquad R>0$
  • $\left\Vert (-\Delta)v_k \right\Vert_{C^1(B_{R/2}(0))}=O(1).$

How can I prove that from those assumptions and Harnack's inequality and Elliptic theory follows that there exists $v\in C^{3}(\mathbb{R}^{4})$ such that \begin{equation} \lim_{k\to+\infty} v_k=v \end{equation} in $C^{3}_{loc}(\mathbb{R}^4)$?

  • $\begingroup$ Can you say something about context of your question? -- And why do you know that the assertion you mention is indeed true? $\endgroup$
    – Stefan Kohl
    Nov 20, 2013 at 17:52
  • $\begingroup$ The reference is a paper from F. Robert: "Concentration phenomena for a fourth order equation with exponential growth: the radial case." Exactly is the proof of Lemma 5.1, equation (32). $\endgroup$
    – gin111
    Nov 20, 2013 at 19:45
  • $\begingroup$ What does radial sequence mean? How is $\Omega$ related to $B_R(0)$? $\endgroup$
    – Deane Yang
    Mar 30, 2021 at 14:32

1 Answer 1


By the second assumption, $e^{v_k}$ are uniformly bounded in $L^\infty$. Standard $W^{2,p}$ estimate with the third assumption, implies that $\Delta v_k$ are uniformly bounded in $W^{2,p}_{loc}$ for any $p<\infty$. Then you can use Sobolev embedding and bootstrapping.

  • $\begingroup$ ok, this is a standard argument but where do I use Harnack? $\endgroup$
    – gin111
    Nov 21, 2013 at 10:04
  • $\begingroup$ I don't know where to use Harnack. Maybe the bound on $v_k$ (which I forgot)? But it seems unnecessary to use Harnack inequality. By the second assumption with uniform bound on $\Delta v_k$, I think we can get a uniform bound for $\|v_k\|_{L^1}$. $\endgroup$
    – Kelei Wang
    Nov 22, 2013 at 4:23
  • $\begingroup$ Harnack is for the convergence $\endgroup$
    – username
    Mar 30, 2021 at 9:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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