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This is in Section 2.8 of Gilbarg and Trudinger. I believe there are some inaccuracies in the proof supplied, and in any case I think there is a more straightforward proof.


A $C^0(\Omega)$ function $u$ will be called subharmonic in $\Omega$ if for every ball $B\subset\subset \Omega$ and every function $h$ harmonic in $B$ satisfying $u\leq h$ on $\partial B$, we also have $u \leq h$ in $B$.


(i) If $u$ is subharmonic in a domain $\Omega$, it satisfies the strong maximum principle; and if $v$ is superharmonic in a bounded domain $\Omega$ with $v\geq u$ on $\partial \Omega$, then either $v > u$ throughout $\Omega$ or $v\equiv u$.

The statement before the semicolon follows from $u\leq \bar{u}$ in any ball $B\subset\subset \Omega$, where $\bar{u}$ is the harmonic lifting of $u$ in $B$ (existence was proved as Theorem 2.6). In any case, it is a special case of the statement after the semicolon.

I won't comment on the author's proof, but instead I will present my own proof, which seems more straightforward and avoids the inaccuracies in the presented proof.


Notice that if $u_1,u_2$ are subharmonic in a domain $\Omega$, then (by using Theorem 2.6 in every ball $B\subset\subset\Omega$) $u_1+u_2$ is subharmonic as well.


Thus $w = u-v$ is subharmonic in the given bounded domain $\Omega$. Furthermore $w\leq 0 $ on $\partial \Omega$. By the weak maximum principle, $w\leq 0$ in $\Omega$ (here is the only place we use the boundedness of $\Omega$).

If $w < 0$ fails to hold in $\Omega$, then (by the preceeding) there is a point $y\in\Omega$ where $w(y) = 0$. So by the strong maximum principle, $w\equiv 0$ in $\bar{\Omega}$.

Thus $v > u$ throughout $\Omega$ or $v\equiv u$. QED

Note: I might not know what I am doing, so please let me know if I am missing out on subtleties. When I start thinking there are mistakes in a book, it is a sure sign I am misunderstanding something. So please post your feedback, if you know something about this topic.

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I do not quite understand what kind of inaccuracy you detected in the reasoning in GT. However, where do you get the weak maximum principle from, for $w$ in $\Omega$? It's, admittedly, really not hard to show, but, to my understanding, this is what you, e.g., get from the (slightly) higher complexity of the proof in GT. – user20266 Jul 20 '12 at 17:41
Section 2.2 (Page 15) contains a formulation of the strong/weak maximum principles. With minor changes, these formulations also hold for $C^0$ subharmonic functions. Keep in mind, $\Omega$ is bounded [at least in the second half of (i)]. This is the only additional requirement for the weak maximum principle to hold. The "complexity" is due to the fact we are mimicking the proof of the strong maximum principle in the middle of our proof. This is unnecessary, since we can just use the strong maximum principle directly. – pre-kidney Jul 20 '12 at 18:55

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