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Let $f:\Omega\subseteq\mathbb{R}^n\to\mathbb{R}^n$ be a Lipschitz function and $h$ be a vector in $\Omega$. Assume that $0\in\Omega$ and $f(0) = 0$. Also, let $\sigma:[0,1]\to\mathbb{R}^n$ be the straight line that connects zero to the vector $f(h)$, i.e., $$ \sigma(t) = tf(h). $$ Does there exist a rectifiable curve $\gamma:[0,1]\to\Omega$ such that $\sigma = f \circ \gamma$ and $\ell(\gamma) \geq \vert h \vert$? (Here, $\ell(\gamma)$ denotes the length of the curve $\gamma$.)

If this is not true in general, can you think of some extra conditions on $\Omega$ or $f$ under which we can find such a curve $\gamma$?

Any comments would be greatly appreciated. Thanks.

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    $\begingroup$ I think your question needs a lot more details. What about $f=\mathrm{identity}$ and $\Omega$ nonconvex? $\endgroup$
    – Alex Degtyarev
    Commented Mar 1, 2014 at 19:57
  • $\begingroup$ Is $\Omega$ open? You should assume at least that $f(\Omega)$ is connected and star-shaped with respect to 0. Do you want $\gamma$ to be continuous? Also, you probably need to assume the Lipschitz constant to be $\leq 1$. $\endgroup$
    – Daniele Zuddas
    Commented Mar 2, 2014 at 2:55

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