Let $u$ be nonnegative harmonic function in $\Omega $ and $V$ an open connected subset of $\Omega $ s.t. $V\subset \subset \Omega $. Then, $$\sup_V u\leq C\inf_Vu,$$ where $C$ depending only on $V$.

remark Harnack's inequality assert that the value of a nonnegative harmonic function within $V$ are all comparable.

Question : What does it mean exactly ? (especially the remark). And in what is it specific to nonnegative harmonic functions ?


Evans describes in what sense this can be understood:

$u$ cannot be very small (or very large) at any point of $V$ unless $u$ is very small (or very large) everywhere.

It is crucial that the $C$ depends on $V$ only, not on $u$, implying that a region, that is of certain regularity, hosts only harmonic functions that are of certain comparability.

The inequality can obviously only be stated for nonnegative functions. You would run into problems if e.g. $\inf u < 0$ and $\sup u > 0$

However for a (bounded) harmonic $u$ you can always look at some $\tilde{u} = u + C > 0$, uses Harnack's inequality on $\tilde{u}$ and hope that this is useful for your stuff you want to do with your original $u$

  • $\begingroup$ A bounded harmonic function is necessarily constant. $\endgroup$ – Surb Nov 12 '17 at 9:50
  • 1
    $\begingroup$ That is true in $/mathbb{R}^n$, but not in a compact set $V$. Actually Harnack's inequality would not make much sense if that was true generally $\endgroup$ – don-joe Nov 12 '17 at 11:25

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