Suppose $\Omega \subset \mathbb{R}^d$ be a domain, and let $\rho(x) = \mathrm{dist} (x, \partial \Omega)$ be the distance function to the boundary of $\Omega$. I want to know for which domains $\rho$ satisfies a Harnack type inequality. Harnack inequality says that $\sup _{x \in B} \rho (x) \leq C \inf _{x \in B} \rho(x)$ on a ball $B= B(a,r), a \in \Omega$, and $C$ is a constant depend on $B$. It is known that harmonic functions satisfy Harnack inequality. Is it enough if $\Omega$ satisfy regularity property (e.g.,if it is a Lipschitz or a NTA-domain)? What about boundary Harnack inequality?

  • $\begingroup$ It looks like you are a bit confused about what the conditions for the classical Harnack inequality really are, forget about the boundary one. Look them up and ask the question again in a meaningful way. $\endgroup$ – fedja May 27 at 20:47
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    $\begingroup$ I know that Harnack inequality holds for harmonic functions. But a continuous function may also satisfy such an inequality. See for example the introduction of this paper: acadsci.fi/mathematica/Vol07/vol07pp259-277.pdf . I want to know for which domains, the distance function is a Harnack function. Please let me know if I am still confused. @fedja $\endgroup$ – Humed May 27 at 21:05
  • $\begingroup$ Of course, it can. But the formulation of the Harnack inequality in your post is a bit off. Once you have it right (and the paper you quote has it right), the answer becomes clear too ;-) $\endgroup$ – fedja May 27 at 21:13
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    $\begingroup$ I think you mean that the inequality should hold in "every" ball $B \Subset \Omega$, and not "a" ball. In this case, you are absolutely right. I will correct my mistake. @fedja $\endgroup$ – Humed May 27 at 21:23
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    $\begingroup$ It is a bit more than that.If $C$ depends on $B$ in an arbitrary way, then any fixed function $u$ that is strictly positive in $\Omega$ satisfies the Harnack inequality as you posted it with $C=\max_{\bar B}u/\min_{\bar B}u$. The Harnack inequality in this qualitative formulation is interesting not for a single function but for an infinite family of functions. So, the question, as asked, is still not very meaningful. $\endgroup$ – fedja May 27 at 21:43

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