0
$\begingroup$

Let $f:D\mapsto\mathbb{R}^n$, $D\subset\mathbb{R}^m$ open, $m<n$ be a continuously differentiable function. I define $g:\mathbb{R}^n\mapsto D$ by $$ g(y)=\operatorname{argmin}_{x\in D}\|f(x)-y\|, $$ i.e., $g(y)$ is maped to $x\in D$ such that $f(x)$ is the closest point to $y$.

Of course, if $f$ is injective then $g$ is a function.

  1. In case $f$ is ijective, is $g$ the inverse of $f$? If so, is it correct that the Jacobian of $g$ at $f(x)$ equals the inverse of the Jacobian of $f$ at $x$?

  2. What happen if $f$ is not injective that implies $g(y)$ is not unique? In this case, can we choose $x$ in such a way that $g$ is the inverse function of $f$?

$\endgroup$
1
$\begingroup$

There is something I do not understand in your question.

  1. Let $f\colon D=(0,\pi)\to\mathbb{R}^2$ be given by $f(x)=(\cos x,\sin x)$. $f$ is injective and $f(D)$ is a semicircle in the upper half plane. $$ \forall x\in D\quad\|f(x)-(0,0)\|=1\implies g(0,0)\text{ is not well defined.} $$ Also, $g(0,y)$ is not defined for $y<0$.
  2. $g\colon\mathbb{R}^n\to D$ cannot be the inverse of $f$, since $f(D)$ is a proper subset of $\mathbb{R}^n$.
$\endgroup$
  • $\begingroup$ Thanks a lot. So, my definition is not valid, is it? $\endgroup$ – Jlamprong Nov 18 '14 at 8:27
  • $\begingroup$ No, I do not think it is correct. $\endgroup$ – Julián Aguirre Nov 18 '14 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.