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Note: as discussed below, there is no mistake here. The notation and conventions have been cleared up for me. However what I have written is not incorrect, either. At least to me, it is just a clarification of the situation.

On page 25 of Gilbarg/Trudinger during the exposition of the Perron Method, barriers are introduced. The authors state that the existence of a barrier (i.e., regularity) is a local condition, and to support this claim they show how to construct an arbitrary barrier from a local barrier.

It is during this construction that I had a notational misunderstanding. At the end of the fourth paragraph, during the construction of $\bar{w}(x)$, they define $\displaystyle m = \underset{N-B}{\inf} w$ and make the crucial statement $m > 0$.

From my understanding, it should in fact be $$m = \underset{\Omega \cap \left(N-B\right)}{\inf} w$$

The simple reason - $w$ is only defined on $\Omega$. For instance, we can let $w$ tend to any negative number outside of $\Omega$ (but in $n$) to force $m < 0$.

After making the change, $$\begin{align*}m &= \underset{\Omega \cap \left(N-B\right)}{\inf} w\\ &= \underset{\overline{\Omega \cap \left(N-B\right)}}{\inf} w\\ &= \underset{\overline{\Omega \cap \left(N-B\right)}}{\min} w\\ \end{align*}$$ The last line follows by compactness.

Suppose $x\in \overline{\Omega \cap \left(N-B\right) } \subset \overline{\Omega \cap N} - \{\xi\}$. Then $w(x) > 0$, yielding $m > 0$.

So now my question: is my "counterexample" for $w$ given above correct? Am I missing a condition that the authors are using which makes their statement correct? And, most importantly, is my "hot fix" valid (replacing $N - B$ with $\Omega \cap (N-B)$? Again, I am new to PDEs, and the exposition of Gilbarg/Trudinger

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The part about adding in the symbol $\Omega$ is fine. Where you are incorrect is in the idea that any extension of $w$ outside of $\bar{\Omega}$ would be considered or, if accomplished, would have any influence on the construction. In short, it is implied that everything takes place in $\bar{\Omega}.$ – Will Jagy Jul 20 '12 at 20:04
Didn't understand what you said. Are you saying that it is implied we are not working in $\mathbb{R}^n$, we are working in a space, like the space $\Omega$ with the relative topology? In the book, no such claim is made, so I think this is an inaccuracy. If the text had specified something like this, then it would all be correct. Where are you getting this $\bar{\Omega}$ condition from? Is it a convention I don't know about? I looked over the text again and paid more attention to the fact that $w\in C^0(\bar{\Omega})$. Is this what you're talking about? – pre-kidney Jul 20 '12 at 21:39
Well, I don't understand you either, let's call it a day. – Will Jagy Jul 20 '12 at 21:51
More simply: Does $w\in C^0(\bar{\Omega})$ mean $w$ is defined on $\mathbb{R}^n$ and is also continuous on $\bar{\Omega}$, or does it mean $w$ is only defined on $\Omega$, and is continuous everywhere it is defined? – pre-kidney Jul 20 '12 at 23:21
It means $w$ is only defined on $\bar{\Omega}$ and is continuous there. – Will Jagy Jul 21 '12 at 0:22

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