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Im wondering if there is any result to guarantee that the level sets of a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ are closed curves?

Specifically if $f\in C^1(\mathbb{R^2},\mathbb{R})$ and the level set $\{(x,y)\mid f(x,y) = C\}$ is connected and contained in some ball, does this imply that the level set is a closed curve? If not what further conditions could one place to guarantee this, would convexity work?

I have tried to find some result concerning this in my textbooks and online but could not find anything concerning this.

OgvRubin
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  • In fact in general the level sets of a continuous function $f:\Bbb R^2\rightarrow\Bbb R$ do not need to be curves at all (just closed sets). Level sets will be bounded closed sets if the function $f$ is proper. – Andrea Mori Feb 23 '19 at 17:30
  • Is there some condition one could add? $C^1$, convexity? – OgvRubin Feb 23 '19 at 17:32
  • As others mentioned, $f$ being $C^1$ and $f^{-1}(y)$ not containing any singular point is sufficient for $f^{-1}(y)$ to be a closed smooth curve. If you relax these conditions ($y$ not being a regular value or $f$ not being $C^1$ at all), then the problem seems to be quite hard; I doubt there is a simple if-and-only-iff reformulation. – Peter Franek Feb 23 '19 at 20:09
  • As for the other question, "level set being connected and contained in some ball" -- note that any closed set can be realised as a level set of some ($C^1$, if you want) $f$. – Peter Franek Feb 23 '19 at 20:13

1 Answers1

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Let $\Omega\subset{\mathbb R}^2$ be the domain of $f$, and let $$K:=\bigl\{(x,y)\in\Omega\bigm| \nabla f(x,y)=(0,0)\bigr\}$$ be the set of critical points of $f$. In many cases the set $K$ is a finite set of isolated points in $\Omega$, see below. If $(x_0,y_0)\in\Omega\setminus K$ and $f(x_0,y_0)=C_0$ then $(x_0,y_0)$ lies on the level set $N_{C_0}$ of $f$. Furthermore the implicit function theorem guarantees that $N_{C_0}$ is a $C^1$-curve in the neighborhood of $(x_0,y_0)$. If you have control over $K$ and know that your level sets are bounded you can say that for $C\notin f(K)$ the level set $N_C$ is a closed curve.

But there are other examples: The function $f$ could be $\equiv1$ in the unit disc, then smoothly decrease to $0$ when $r:=\sqrt{x^2+y^2}$ goes from $1$ to $2$, and finally be $\equiv0$ for $r\geq2$. In this case the level set $N_1$ is a full disc, and the level set $N_0$ consists of all points $(x,y)$ with $r\geq2$.

The decisive assumption on $f$ therefore is that all critical points of $f$ should be nondegenerate, i.e., that the Hessian of $f$ at these points has rank $2$. This makes sure that $K$ consists of isolated points.

Christian Blatter
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  • Thank you for your reply however where can I find the result which says that $C\notin f(K)$ implies that the level set $N_c$ is a closed curve? – OgvRubin Feb 23 '19 at 20:01
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    @OlofRubin For instance, Milnor, Topology from differentiable viewpoint, p. 13. Note that level sets are always closed, but preimages of regular values are, moreover, smooth manifolds (1-dimensional in your case). – Peter Franek Feb 23 '19 at 20:06
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    $N_C$ then is a compact one-dimensional manifold: It is closed and bounded, and locally a curve. – Christian Blatter Feb 23 '19 at 20:07
  • @PeterFranek With closed I mean that the parametrization of the curve is periodic and not closed as in a closed subset of $\mathbb{R^2}$ which indeed is obvious. – OgvRubin Feb 25 '19 at 12:37