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Let $U,V$ be two homeomorphic open sets in $\mathbb R^n$ with the homeomorphism $\Phi$. If $p$ is a fixed point in $U$, can we find a convex open neighborhood $U_p$ of $p$ such that $\Phi(U_p)$ is convex?

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1 Answer 1

For $n=1$, the answer is (trivially) yes.

For $n\ge 2$ the answer is negative. I give a two-dimensional counterexample that also works in higher dimensions. Namely, $\Phi(x,y)=(x,y+f(x))$ where $f$ is a continuous nowhere differentiable function. That this is a homeomorphism is evident from $\Phi^{-1}(x,y)=(x,y-f(x))$.

Let $U$ be a convex neighborhood of some point $p\in\mathbb R^2$. The upper part of $\partial U$ can be written as $y=g(x)$ where $g$ is a concave function. In particular, $g$ has bounded variation. The image of this curve under $\Phi$ is described by $y=f(x)+g(x)$, which is the graph of a function of infinite variation (unrectifiable, etc.) Such a curve can't be a piece of the boundary of a convex set in $\mathbb R^2$.

You can also argue via differentiability a.e. (which holds for concave functions) instead of BV property.

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