Let me first introduce a defitinition.

Definition (Smooth Curve) Let $\gamma: [a,b] \to \mathbb{R}^2$ a curve (continous function). We call $\gamma$ a $C^k$-smooth curve if $\gamma'(t) \neq 0$ for all $t \in [a,b]$ and if it is of class $C^k$.


Definition (Level curve) Let $f: \mathbb{R^2} \to \mathbb{R}$ be of class $C^k$ with $k > 0$. The set $\Gamma_{c} = \{(x,y) \mid f(x,y) = c\}$ is said to be a level curve.

Is there for every smooth $\gamma: [a,b] \to \mathbb{R}^2$ an $f: \mathbb{R^2} \to \mathbb{R}$ such that $\gamma([a,b]) = \Gamma_{0}$?

Any tips/help is greatly appreciated! Thanks in advance!

  • $\begingroup$ The image of $\gamma$ is homeomorphic to either the segment $[0,1]$ or the circle $S^1$. If you can show that you can find a homeomorphism of class $C^k$ the answer becomes rather trivial. $\endgroup$ – Andrea Mori Jun 7 '18 at 13:39
  • $\begingroup$ @AndreaMori Nowhere it states that $\gamma$ is injective, so it may parametrize an $\infty$ shape $\endgroup$ – Saucy O'Path Jun 7 '18 at 13:42
  • $\begingroup$ there is a theorem of Whitney that every closed subset of $\mathbb{R}^n$ is a level set of a smooth function. So the answer is yes. Maybe the OP is thinking about level sets with non-zero gradient, a more interesting question. $\endgroup$ – Orest Bucicovschi Jun 7 '18 at 13:58
  • $\begingroup$ Just let $f(x,y)$ be the distance from the point $(x,y)$ to the curve $\gamma$? (Should work for continuous functions. I haven't thought about what happens with smoothness, though.) $\endgroup$ – Hans Lundmark Jun 7 '18 at 14:09
  • $\begingroup$ (Cont.) Maybe distance squared is better. $\endgroup$ – Hans Lundmark Jun 7 '18 at 14:15

Your Answer

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

Browse other questions tagged or ask your own question.