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I read a statement online that Harnack's inequality is a more accurate and quantitative version of the Maximum principle. But it seems that it's impossible to deduce the maximum principle from Harnack's inequality. The difficulties (among a few others) are:

  1. maximum principle has information on the boundary, while in Harnack's inequality, the estimates occur in a compact subset that is positively bounded away from the boundary.

  2. even if 1. can be fixed somehow, Harnack's inequality gives you estimates up to a multiple of constant. But in maximum principle, there is no extra constant appearing in the estimate.

Any thought is appreciated.

henryforever14
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    Can you post a link to this online statement? Some context might be useful here. – Giuseppe Negro May 10 '14 at 01:20
  • @GiuseppeNegro It was merely a sentence saying that. I can't find it again... sorry. But as for my doubt, there is at least no easy way to prove maximum principle with harnack, right? – henryforever14 May 10 '14 at 06:41
  • I don't know of any "easy" way to do that, but this means absolutely nothing. I am not an expert. – Giuseppe Negro May 10 '14 at 11:20
  • @GiuseppeNegro just trying to see if somebody might have an idea. I am not "requiring" people to know or answer. That's the purpose of this website, right? People raise questions and people discuss. – henryforever14 May 10 '14 at 15:49

1 Answers1

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The Harnack inequality is a quantitative version of the following result:

$\textbf{Strong maximum principle:}$ If $u\in C(\bar{\Omega})$ is a non-constant solution of $-\Delta u=0$ in $\Omega$ then the maximum of $u$ can't be attained at an interior point of $\Omega$.

To see this assume that the Harnack inequality is known and pick an interior maximum $x_0\in \Omega$ with $u(x_0)=M=\max_{\Omega}u$. Consider, for $\varepsilon\geq0$, $v_{\varepsilon}(x)=(M+\varepsilon)-u(x)$ then $v_{\varepsilon}>0$ is a harmonic function and Harnack's inequality implies, for any compact set $x_0\in K \subset \Omega$,

$$ 0\leq\sup_K v_0 \leq \sup_K v_{\varepsilon} \leq C(K,\Omega) \inf_K v_{\varepsilon}= C(K,\Omega)\varepsilon, \qquad 0<\varepsilon. $$

Since $\varepsilon$ was arbitrary we get $v_0=0$, or $u=M$ a contradiction.

Two remarks.

$\textbf{1.}$ What I think you're thinking of as 'the' maximum principle is the following

$\textbf{Weak Maximum principle:}$ If $u\in C(\bar{\Omega})$ is a solution of $Lu=0$ in $\Omega$, then the maximum of $u$ is attained at the boundary $\partial \Omega$.

Notice that, as their names suggest, strong implies weak since by compactness the maximum is always attained at a point in $\bar{\Omega}$.

$\textbf{2.}$ Most of the difficulties you mention more or less dissapear as soon as you realize that both the maximum principle and Harnack's inequality are, in essence, pointwise estimates on $u$. The maximum principle gives then a qualitative estimate: a non-positive solution can have no zero; while Harnack's inequality gives that a non-positve non-constant solution must satisfy $-u\geq C >0$ on any compact set.

Jose27
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