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What is $C^{2,1}(\Omega)$ if $\Omega$ is arbitrary (i.e. neither open nor closed in general)?

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2 Answers 2

You should check the definition in the book or paper you are reading, because $u\in C^{k,\alpha}(\Omega)$ can mean one of the following:

  • All $k$-th derivatives of $u$ are locally Hölder continuous on $\Omega$.
  • Derivatives of $u$ up to $k$-th order are all continuous and bounded, and the $k$-th derivatives are all globally Hölder on $\Omega$.

For bounded closed domains these two conditions are the same, but different in general.

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AFAIK, you got $u\in C^{2,1}(\Omega)$ iff $u\in C^2(\Omega)$ and each second partial derivative $u_{x_ix_j}$ is Lipschitz continuous in $\Omega$. Obviously, if $\Omega$ is not open, then you have to suitably extend the derivatives on those parts of $\partial \Omega$ which are actually "attached" to the set.

In general, $u\in C^{k,\theta}(\Omega)$ (with $k\in \mathbb{N}$ and $\theta \in ]0,1]$) iff $u\in C^k(\Omega)$ and each $k$-th partial derivative of $u$ is Hölder continuous in $\Omega$ with exponent $\theta$.

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