# 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?

----------updates------------

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
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## closed as not a real question by Chris Eagle, t.b., William, Nate Eldredge, copper.hatAug 29 '12 at 5:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

## 1 Answer

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.

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