Are level surfaces of differentiable functions differentiable?

Question

Suppose I have a differentiable function $f: {\rm I\!R}^m \rightarrow {\rm I\!R}^n$. I obtain a level surface of $f$ by taking the set of points where $f$ is some constant vector: $S = \{\mathbf{x} | f(\mathbf{x}) = \mathbf{c}\}$.

Since $f$ is differentiable, is it the case that $S$ is itself "smooth" in a sense? Specifically, does there exist a differentiable $g: {\rm I\!R}^k \rightarrow {\rm I\!R}^m$ so that the image of $g$ is $S$? If so, is there a standard reference for this factoid? If not, what is a counterexample? Also if not, is there a condition on $f$ that will produce a "smooth" $S$ (like being infinitely differentiable, or some such)?

Answer 1

This is not true in general. Take $f(x,y,z)=x^2+y^2-z^2$. It’s zeroth level is a cone which is not smooth at its vertex. You have to assume that at the given point the gradient does not vanish to guarantee that the level surface is smooth (use the implicit function theorem)

Answer 2

This is not true. Consider the case where $m=n=1$, and choose $f(x)=\sin(x)$ and $c=0$. Then $S$ is disconnected, and the image of a continuous function with domain $\mathbb{R}^k$ is connected.