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Suppose $F:\mathbb{R}^3\mapsto\mathbb{R}^3$ via $F(x)=(f_1(x),f_2(x),f_3(x))$. Assume $DF(a)$ is invertible and $F(a)=0$. Prove that there are infinitely many values $x\in\mathbb{R}^3$ such that $f_1(x)=f_2(x)=f_3(x)$.

I was just asked this question on a quiz and I don't know where to start , especially because there is no assumption about continuity. My first instinct is to use inverse function theorem but I know this isn't right.

nmasanta
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Jungleshrimp
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    What is $Df(a)$? Do you mean $DF(a)$ instead? Don't you have local surjectivity by Brouwer? And no, you don't get $C^1$ at the end of all these. – user10354138 May 31 '19 at 01:11
  • I think there is a solution using the inverse function theorem, what was your attempt using it? – none May 31 '19 at 15:37
  • If this is an exercise, you can surely assume that $F\in C^1$ in a neighborhood of the origin; they probably just forgot to write it. It is very difficult to establish an inverse function theorem in the class of differentiable functions, if you do not assume that they are $C^1$: see Terry Tao's blog. – Giuseppe Negro May 31 '19 at 15:51
  • I think that this is false if you do not assume that F is C1 in a neighborhood of the origin. The example I wrote in the comments to @none: answer can probably be modified to produce an example for this exercise. – Giuseppe Negro May 31 '19 at 19:26

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As $DF(a)$ is invertible, we can apply the inverse function theorem. Thus, $F$ is a diffeomorphism on some neighborhood $U$ of $a$. In particular, this means $F$ is surjective on $U$. Since $F(a) = 0$, $F(U)$ contains some ball around $0$. This ball contains a segment of the line $x=y=z$. By surjectivity, $F(U)$ contains that segment as well. Clearly the segment has infinitely many points, which correspond to values such that $f_1(x) = f_2(x) = f_3(x)$.

none
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    Yes, but you need that $F$ be $C^1$ in a neighborhood of $0$ to apply the inverse function theorem. The function $$f(x)=x+\begin{cases} x^2, & x\in \mathbb Q, \ 0, & x\notin \mathbb Q, \end{cases}$$ is differentiable at $0$ and $f'(0)=1$, but you cannot apply any inverse function theorem. However, I am sure that in this case we can safely assume that $F$ is smooth, they just forgot to write it in the exercise sheet; see my comment to the original question. – Giuseppe Negro May 31 '19 at 15:56
  • @GiuseppeNegro you are correct, I did assume $C^1$. It seems like a very natural assumption, especially if this is a calculus course of some sort. Maybe there's a way to do it without the inverse function theorem. – none May 31 '19 at 16:09
  • Please do not delete this answer. It is very instructive. – Giuseppe Negro May 31 '19 at 16:23