This is on page 33 of my edition, under the proof of Harnack's inequality. Let $V\subset \overline{V} \subset U$ with $\overline{V}$ compact. Let $r=\frac{1}{4}\text{dist}\left(V,\partial U\right)$. Let $x,y\in V$ s.t. $\left|x-y\right|\leq r$. It seems as though Evans uses $$\int_{B\left(x,2r\right)} u dz\geq \int_{B\left(y,r\right)} udz,$$ but this does not seem obvious to me. What if $u$ is negative in $B\left(x,2r\right)\setminus B\left(y,r\right)$?

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    $\begingroup$ $u$ should be non-negative and harmonic in $U$. $\endgroup$
    – Shuhao Cao
    Jun 13, 2013 at 2:20

1 Answer 1


Harnack's inequality concerns about the non-negative function which is harmonic in $U$, by the choice of $r$, $x$, and $y$, we know that $B(y,r)\subset B(x,2r) \subset U$, hence the inequality holds.

  • $\begingroup$ Thank you--- apparently I read the statement of the Theorem incorrectly. $\endgroup$
    – user5525
    Jun 13, 2013 at 2:42
  • $\begingroup$ @user5525 No problem. :) $\endgroup$
    – Shuhao Cao
    Jun 13, 2013 at 2:43

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