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Let $f:U\longrightarrow V$ be a $C^2$ diffeomorphism where $U,V\subset\mathbb{R}^n$ are open sets.

How can we prove that

$$\forall a\in U,\exists \epsilon>0:r\le \epsilon \Longrightarrow f(B[a,r])\text{ is convex }$$

Any hints would be appreciated.

felipeuni
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1 Answers1

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Without loss of generality let $a=0$ and $f(0)=0$.

Consider the function $u(x) = |f^{-1}(x)|^2$ on $V$. Its Hessian at $0$ is (using $f(0)=0$)

$$ Hu(0)(e_i,e_j) = \partial_i \partial_j u(0) = 2 \partial_i (f^{-1} \cdot \partial_j f^{-1})(0) = 2 \partial_i f^{-1}(0) \cdot\partial_j f^{-1}(0).$$

This is positive definite: for any $v$ we have $$Hu(0)(v,v) = 2v^T {df^{-1}}^T df^{-1} v = 2|df^{-1} v|^2 > 0.$$

Since $f$ is a $C^2$ diffeomorphism, $Hu$ is continuous and thus will be positive definite in some ball $B_R [0]$. This implies that $u$ is convex in this ball, and thus its sublevel sets $$S_r = \{ x \in B_R[0] : |f^{-1}(x)|^2 \le r\}$$ are convex. But

$$|f^{-1}(x)|^2 \le r \iff |f^{-1}(x)| \le \sqrt r \iff x \in f(B[0, \sqrt r]),$$

so this means the sets $B_R[0] \cap f(B[0,\sqrt r])$ are convex. By continuity of $f$ there is some $\delta > 0$ such that $f(B[0,\sqrt r]) \subset B_R[0]$ whenever $\sqrt r \le \delta$; so setting $\epsilon = \delta$ we have the desired result.

user2345678
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Anthony Carapetis
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  • How did you prove that $\partial_i \partial_j u(0) = 2 \partial_i f^{-1}(0) \partial_j f^{-1}(0)$? – rfloc Apr 07 '20 at 05:43
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    @rfloc: Apply the product rule to $u=f^{-1} \cdot f^{-1}$ and use the fact $f^{-1}(0)=0$ in the last step. – Anthony Carapetis Apr 07 '20 at 05:51
  • Why can we assume that $a=0$ and $f(0)=0$? Thanks for your help! – rfloc Apr 07 '20 at 06:11
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    @rfloc: you can always ensure this is the case by translating the domain and adding a constant to $f;$ and these operations preserve both the fact that $f$ is a diffeomorphism and convexity of sets. – Anthony Carapetis Apr 07 '20 at 06:17