Let $B$ be the unit open ball in $\mathbb R^n$ and $\phi$ a self diffeomorphism of $B$. Can we find a self diffeomorphism $f$ of $\mathbb R^n$ such that $$ f=\phi $$ on $B$?
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Certainly not, without making some further assumptions about $ f $. Consider, for example $ f(x) = \left( 1 - \sqrt{1 - |x|^2} \right) \ x $. Any extension cannot be differentiable on the boundary $ |x| = 1 $.
Jake Mirra
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