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I'm new on stackexchange community (this is my second question on these topics) and this one is linked to my first (About differentiability of economics functions). Indeed, I asked on the economic forum for this and I've found a proof of it, but I don't really understand the maths behind all of this, specifically the second part of the demonstration.

Question. For all $n \geq 1$. How can we prove the fact that if a continuous function have level set that are $C^n$ differentiable manifolds, the the function is $C^n$ regular with no critical point.

I've found a begining of proof in a book of economics (A. MAS-COLELL, The theory of general economic equilibrium : A differentiable approach, 1985).

Proof of $\Leftarrow $ : Let $ U : E \rightarrow \mathbb{R}^{+}$ where $E\subset \mathbb{R}^{l},\forall l\in\mathbb{N}$ is open, separable and connected and $y$ fixed such that $U(y)$ is constant. Suppose $V(x,y)=U(x)-U(y)$ such that $V(x,y)=0 \Leftrightarrow U(x)=U(y)$ all the $x \in E$ that make $V(x,y)=0$ are $\{x \in E | U(x)=U(y)\}$ so $V^{-1} (0) = \{x \in E | U(x)=U(y)\}$. Furthermore, we have $U(\bullet)$ without critical point, so $V(\bullet)$ doesn't have any critical point too, by construction. Therefore, by implicit function theorem : $V^{-1} (0)$ is a differentiable $C^n$ manifold. (I don't really understand why the fact that if $U$ have $n-$continuous derivatives and have no critical point, then IFT implies that level sets are differentiable $C^n$ manifolds).

And for the $\Rightarrow$ proof, for me, it just make no sense so I tried to find another proof, and when they were making sense to me, the proof was just : By IFT.

So I'm a little bit lost, first because I don't really understand the first part of the proof, and secondly, because the second part of the proof doesn't make any sense for me. So if someone can purpose an answer or an explanation, I could get interested. Before I end, apologize for my bad english.

I thank all people that answer or add explanation, have a good day !

PS : I am not a maths student, but an economics one, so I'm not up-to-date with all differential geometry or advanced calculus.

EDIT I. After looking at Moishe Kohan example (The function $f(x)=|x|$ with the level set for $f(x)=0$), I looked again to the purpose in my book and make the change in edit, n must be positive or equal to 1. Indeed the exact purpose is : "if a function is $C^r$ for $r\geq 1$ then any sublevel is a $C^r$ manifold". And the author show $\Leftarrow$ and $\Rightarrow$ after. Apologize for this, maybe it is just me that don't understand anything.

Economos
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    The fact you are trying to prove is simply false. Just think about case of functions of one variable (submanifolds are zero-dimensional). – Moishe Kohan Mar 05 '24 at 19:50
  • Consider $f(x)=|x|$ and the equation $f(x)=0$. – Moishe Kohan Mar 05 '24 at 20:06
  • What is the book that you are reading? (If you copied correctly, the book is a combination of "wrong" and "badly written".) My suggestion is to learn this material by reading books written by mathematicians, not by economists. As for my example: Do you understand that a finite subset is always an infinitely-differentiable submanifold? Do you understand that $f(x)=|x|$ is not even once differentiable? – Moishe Kohan Mar 06 '24 at 11:03
  • I mentioned the book where I found the proof it is Theory of general equilibrium: A differential approach by Andreu Mas-Colell. Thank you for your advices. And no I didn't knew this property ! Thank you. And I knew that absolute function is not everywhere differentiable. – Economos Mar 06 '24 at 11:12
  • This shows that the last claim of your edit is not true even for $r = \infty$. – Didier Mar 06 '24 at 11:12
  • Thank you for your answer I'll check this ! Have a nice day ! – Economos Mar 06 '24 at 11:14

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