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I’m reading a note on higher regularity boundary Harnack inequality by D. DE SILVA AND O. SAVIN and I’m kind of confused of the case k=1.

In the paper they used the Hopf lemma to show that $u_\nu>c>0$, but, as the boundary regularity is just $C^{1, \alpha}$, I don’t think that we can directly use Hopf lemma.

I tried to use the transformation of coordinates to do make a better regularity of boundary, but it only works in the divergence form of equations. I have no clue to the non-divergence form. Is there any way to do this estimate?

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  • $\begingroup$ This is the paper, right? arxiv.org/pdf/1403.2588.pdf $\endgroup$ Commented Sep 9, 2021 at 8:34
  • $\begingroup$ @AlessandroDellaCorte Yes it is the paper I’m reading. The other conditions doesn’t seem that essential. Could you please tell that where the claim $u_{\nu}$ is bounded? It seems that I’ve missed the condition. $\endgroup$
    – Holden Lyu
    Commented Sep 9, 2021 at 8:34
  • $\begingroup$ @AlessandroDellaCorte yes I do want to know how to get this condition. Did they claim it? I have no idea ever reading it… Could you please tell where they claimed it? $\endgroup$
    – Holden Lyu
    Commented Sep 9, 2021 at 8:39
  • $\begingroup$ @AlessandroDellaCorte But, as the boundary is $C^{1, \alpha}$, we cannot get the interior ball condition. How to use Hopf lemma here? $\endgroup$
    – Holden Lyu
    Commented Sep 9, 2021 at 9:18
  • $\begingroup$ In the proof of Hopf Lemma, the smoothness of the boundary is used when claiming that there is a small ball contained in $\Omega$ whose closure is tangent to the boundary at $x$ and intersects the boundary only at $x$. But isn't this what the authors assume at the fourth line of Section 2? (although they should also have added that they keep the assumption $x_n>g(x')$ made in the fourth line of the Introduction). $\endgroup$ Commented Sep 9, 2021 at 9:35

1 Answer 1

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Edit: The result is fine: Hopf's lemma was proved in

  • G. Giraud, Problèmes de valeurs à la frontière relatifs à certaines donn ás discontinues, Bull. de la Soc. Math. de France, 61 (1933), 1–54

Below is my incorrect answer (which I keep for reference), where I mistakenly assumed that $C^{1,Dini}$ is stronger than $C^{1,\alpha}$.


Old answer:

Looks like you are right: the regularity assumption on the boundary is insufficient, although I did not check this very carefully.

You may have a look at the paper A counterexample to the Hopf-Oleinik lemma (elliptic case) by D. E. Apushkinskaya and A. I. Nazarov, DOI:10.2140/apde.2016.9.439, arXiv:1503.02179. Let me quote from p. 2 of this paper:

The reduction of the assumptions on the boundary of $\Omega$ up to $C^{1,Dini}$-regularity was realized for various elliptic operators in the papers [Wid67], [Him70] and [Lie85] (see also [Saf08]). A weakened form of the Hopf-Oleinik lemma (the existence of a boundary point $x_1$ in any neighborhood of $x_0$ and a direction $\ell$ such that $\frac{\partial u}{\partial \ell}(x_1) \ne 0$) was proved in [Nad83] for a much wider class of domains including all Lipschitz ones. We mention also the paper [Swe97] where the behavior of superharmonic functions near the boundary of 2-dimensional domains with corners is described in terms of the main eigenfunction of the Dirichlet Laplacian.

The sharpness of some requirements was confirmed by corresponding counterexamples constructed in [Wid67], [Him70], [KH75], [Saf08], [ABM$^+$11] and [Naz12]. In particular, the counterexamples from [Wid67], [Him70] and [Saf08] show that the Hopf-Oleinik result fails for domains lying entirely in non-Dini paraboloids.

Later, the paper shows that lack of "Dini condition" on the boundary invalidates the assertion of Hopf's lemma.

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  • $\begingroup$ Thank you for this reference. I think I would pay some time on it. $\endgroup$
    – Holden Lyu
    Commented Sep 9, 2021 at 15:56
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    $\begingroup$ I’ve just read some of the references and maybe we can use thm. 1.8. in arxiv.org/pdf/0810.0522v1.pdf to get the estimate? Holder estimate would allow us to get the interior Q-condition. $\endgroup$
    – Holden Lyu
    Commented Sep 9, 2021 at 16:43
  • $\begingroup$ Ah, that's right! I'll edit the answer momentarily. $\endgroup$ Commented Sep 9, 2021 at 18:26
  • $\begingroup$ Thank you very much for your help! It’s happy to discuss with you $\endgroup$
    – Holden Lyu
    Commented Sep 10, 2021 at 3:55

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