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Let $u:\Omega\to\mathbb{R}$ be continuous and $\Omega\subset\mathbb{R}^n$. I've lately seen a definition of viscosity solutions where the two conditions for the test functions in a point $x_0\in\Omega$ was:

$\phi\in C^2(\mathbb{R}^n)$ such that $(u-\phi)(x) \leq (\geq)\, 0$ on $\overline{B_r(x_0)}\subset \Omega$ and $(u-\phi)(x_0)=0$.

I've never seen it before that the first condition has to hold in the closed set $\overline{B_r(x_0)}$. Why doesn't it matter if one takes the closed set $\overline{B_r(x_0)}$ or the open set $B_r(x_0)$ or maybe even the whole set $\Omega$ for that first condition? I mean is the reason for this that one is simply interested in a little neighbourhood of $x_0$?

I hope my question is clear! Thanks along and Br!

ACB
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HelloEveryone
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1 Answers1

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It doesn't matter. If you read the whole definiton od viscosity solutions you'll find that only local property od $\phi$ is important: this inequality ($\phi$ is sort of supporting function for $u$) and the derivative of $\phi$. If $\phi$ is a supporting function for $u$ in $x_0$ defined on closed ball then it's also on open ball and then it's also on a closed ball with a bit smaller radius. Each supporting function $\phi$ on a small ball $B(x_0,2r)$ can be modified to obtain supporting function $\psi$ defined on $\Omega$ such that $\phi|B(x_0,r)=\psi|B(x_0,r)$ (good exercise). Therefore it doesn't matter.

It's convenient to use in definition small balls, since only the behaviour of $\phi$ near $x_0$ is important and it's easier to find such functions without need to find it's extensions on the whole $\Omega$. I don't know why in the above definition they used closed balls. Maybe it's a part of some proof where they needed compactness or something similar.

Mateo
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