# Unit Open ball in $\mathbb{R}^3$ is the domain of $C^k$ for any k? [closed]

My guess is to transform into spherical coordinates, but I fail to prove that the mapping with it's inverse is uniformly bounded.

Any suggestions?

Sorry for the late update. I was off-internet in the last few days.

The domain $C^k$ is a domain similar with Lipschitz domain, but the bijective function is $C^k$ continuous instead of Lipschitz continuous.

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What does "is the domain of $C^k$ for any $k$" mean? –  Chris Eagle Jun 6 '12 at 10:20
It's not entirely clear what you're asking, but $\mathbb{B} = \{ \|x\|^2 - 1 < 0 \}$, and the function $\rho(x) = \|x\|^2 - 1$ is $C^\infty$ (and its gradient vanishes only at the origin), so $\mathbb{B}$ has $C^\infty$ boundary.