# Theorem 6.2 in Gilbarg-Trudinger (Clarification)

In Theorem 6.2 of G-T's book we have in the hypothesis of the theorem that $\Omega$ is an open subset of $\mathbb{R}^n$ and that $u\in C^{2, \alpha}(\Omega)$ (here $\alpha\in (0,1)$) is a bounded solution of the equation:

$$Lu=a^{i,j}D_{i,j}u+b^{i}D_{i}u+cu=f$$

with $f\in C^{\alpha}(\Omega)$ and certain bounds on the coefficients.

I am not sure why they needed to specify that $u$ is bounded on $\Omega$ and that it is in $C^{2, \alpha}(\Omega)$.

Is it possible to have $u$ not bounded but still in $C^{2, \alpha}(\Omega)$?

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Is $\Omega$ assumed bounded? – user53153 Feb 22 '13 at 5:07
$\Omega$ is not assumed to be bounded. – Nirav Feb 22 '13 at 5:28
Then $\Omega$ could be half-space, or even the entire space, and $u$ could be a linear function, for example. – user53153 Feb 22 '13 at 5:45

$\sqrt{x}$ is Hölder continuous on $(0,\infty)$
Somehow this is always confusing. For example for $f(x)=x$ one would say that $f\in C(\mathbb{R})$ if one inteprets $C(\mathbb{R})$ to be the vectorspace of continuous functions on $\mathbb{R}$. However if one interprets it to be the Banach-Space $C(\mathbb{R})$ with the sup norm this is false since $f$ is not bounded. – Quickbeam2k1 Feb 22 '13 at 10:52