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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:

\begin{equation} Lu=a^{i,j}D_{i,j}u+b^{i}D_{i}u+cu=f \end{equation}

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

1 Answer 1

up vote 1 down vote accepted

$\sqrt{x}$ is Hölder continuous on $(0,\infty)$

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Yes I just realised that I had the wrong definition of Holder spaces in mind. I thought that the seminorms of the highest order derivative and the sup norms of the derivatives (of all orders) had to be finite. When actually it's just the seminorms that need to be finite. Thanks. –  Nirav Feb 22 '13 at 7:59
2  
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

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