2

Consider a $C^1$ curve $\gamma:I\to\mathbb{R}^2$, where $I\subset \mathbb{R}$ denotes an open interval. What conditions on $\gamma$ are necessary and sufficient in order to guarantee that there is no $C^1$ function $f:U \to \mathbb{R}$, with $U\subset \mathbb{R}^2$ an open set and such that $f(\gamma(t))=0$, for every $t\in I$?

Edit: Here I also ask that $f$ is non constant on any open subset of $U$ (otherwise, as well pointed by @Moishe Kohan, we could simply take $f$ to be identically zero). Since I'm asking $f$ to be $C^1$, we can express this by saying that the gradient $\nabla f$ is nonzero, up to an empty interior subset of $U$.

Does $\gamma$ need to self intersect? Self accumulate? Does it need to have a fractal behavior?

This is possibly a duplicate to this question. However, I'm interested in the case that $I$ is an open (not closed) interval. Therefore, I believe the Whitney theorem mentioned in the comments to that question doesn't apply (please, tell me if I'm wrong).

Also, the question has only comments, but no answers. One of them by the way gives a very good suggestion, to take $f$ as the square distance to the curve. However, I'm really interested in the smoothness (at least $C^1$) of $f$.

Derso
  • 2,897
  • 1
    The title question is different from the question in the post (the latter does not require the image of $\gamma$ to equal the zero level set, only be contained in the level set). You also did not explain the role of the set $U$ or its nature. – Moishe Kohan Jul 21 '24 at 22:19
  • @MoisheKohan I edited the question. I hope it makes it clearer now. – Derso Jul 21 '24 at 23:27
  • 1
    You can always take $f$ which is identically zero. I do not think you wrote the question that you've meant to ask. – Moishe Kohan Jul 21 '24 at 23:31
  • What a distraction of me! I'm so sorry. I'll edit it again. – Derso Jul 21 '24 at 23:35
  • 1
    An obvious necessary condition is that the image of $\gamma$ is a nonclosed subset of the plane. More generally, the closure of the image of $\gamma$ has nonempty interior. I am not sure what else you'd like to know. – Moishe Kohan Jul 21 '24 at 23:58
  • 1
    With these hypotheses, a level curve will be a $C^1$ submanifold, so you only need to violate that. What about $y=|x|$? (This can be the image of a $C^k$ map.) – Ted Shifrin Jul 22 '24 at 00:00
  • @TedShifrin What about roses? https://en.wikipedia.org/wiki/Rose_(mathematics) For example, the zero leveled set of $f(x,y) = (x^2+y^2)^3-a^2(x^2-y^2)^2$ will be a quadrifolium, which is not a $C^1$ manifold at the origin. – Derso Jul 22 '24 at 00:08
  • @MoisheKohan I think I see your point. Because otherwise, if the closure of $\gamma$ has empty interior then, since it's closed, smooth Urysohn's lemma gives us a function that vanishes precisely on that closure of $\gamma$, right? If it had nonempty interior, then such a function would contradict our gradient condition. – Derso Jul 22 '24 at 00:23
  • 1
    Did you not say that the gradient is non-vanishing on the curve? Oh, you allowed some singular points. My error. – Ted Shifrin Jul 22 '24 at 00:41
  • Right........... – Moishe Kohan Jul 22 '24 at 01:34
  • 1
    Adding to @MoisheKohan's point: The image itself of a $C^1$ mapping, however, has area $0$ (stronger than measure $0$). – Ted Shifrin Jul 22 '24 at 17:15
  • 1
    Agh. Lacking compactness, we may just get the weaker measure $0$. – Ted Shifrin Jul 22 '24 at 17:59
  • @TedShifrin Yes, we don't have compactness. Thinking about the question and also coming upon space filling curves, it's interesting how little intuition we have. For example, if we think of a smooth curve as being drawn by a pencil, with no sudden changes of movements and without lifting it off from the paper, then the non existence of smooth space filling curves says we're not able to fully paint a small square on the paper (of course, our mathematical pencil has lead number 0.00, so no thickness). – Derso Jul 22 '24 at 18:05
  • 1
    Right. There are no $C^1$ space-filling curves. (I believe there are no differentiable ones, but I have forgotten.) It’s also counter-intuitive, but space-filling (continuous) curves are necessarily very much non-injective. – Ted Shifrin Jul 22 '24 at 18:23

0 Answers0