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Let $B$ be the open unit ball in $\mathbb{R}^n$. I need to find continuous bijective function $f: B \to \mathbb{R}^n$ such that $f^{-1}$ is also continuous.

I thought of $f(x)=\frac{1}{d(x,x_0)}$ for some $x_0 \in B$. Is it Ok regrading the conditions? I'm not really sure.. Do you have any idea for another suitable function?

Thanks a lot!

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The problem with your idea is that $f$ is undefined at $x_0$. What you want is a radial expansion of $B$ that gets more and more extreme as you move away from the origin. That is, you want your function to send $x$ to some scalar multiple $r(x)x$ of $x$, where $r(x)\to\infty$ as $\|x\|\to 1^-$. Since $1-\|x\|\to 0^+$ as $\|x\|\to 1^-$, you might try $r(x)=\left(1-\|x\|\right)^{-1}$, i.e., $f(x)=\frac{x}{1-\|x\|}\;.$ You still have to check that this $f$ really does have the desired properties, but it should at least be pretty clear intuitively.