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NOTE. This is not really the question I wanted to ask. Somehow I forgot to mention that I am assuming $f$ is continuous. However, since Iosif's answer has been well-received I have left this question as it was and opened a new one with the right hypotheses. You can find it here A seemingly trivial property of continuous functions differentiable at the origin (PART 2).

Let $F:\mathbb{R}^n\to\mathbb{R}^n$ be a function such that $F(0)=0$, $F$ is differentiable at $0$ and $DF(0)$ is invertible. Is it true that for all $\epsilon>0$ there is $\delta>0$ such that $$ DF(0)(B_{(1-\epsilon)\delta})\subset F(B_\delta)\subset DF(0)(B_{(1+\epsilon)\delta}), $$ where $B_r$ is the ball of radius $r$ centered in $0$?

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  • $\begingroup$ Isn't this or something very like it proved as part of any proof of the change-of-variables theorem in measure theory? $\endgroup$
    – LSpice
    Sep 1, 2022 at 19:22
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    $\begingroup$ Yes, indeed the question came to my mind when I was trying to prove a version of the coarea formula for Lipschitz functions (a generalization of the change-of-variables theorem). $\endgroup$
    – No-one
    Sep 1, 2022 at 21:00

1 Answer 1

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A counterexample to your assertion: Let $n=1$ and let $$F(x):=x+4^{-j}$$ if $x\in(2^{-j},2^{1-j}]$ for any integer $j$, with $F(x):=x$ for real $x\le0$.


Indeed, then $F(x)=x+O(x^2)$ as $x\to0$, so that $F'(0)=1\ne0$. On the other hand, for any integer $j$, the function $F$ does not take any values in the interval $(2^{-j}+4^{-j-1},2^{-j}+4^{-j})$, whereas such intervals are arbitrarily close to $0$ if $j$ is large enough.


Here is the graph $\{(x,F(x))\colon-1<x<2\}$:

enter image description here


Of course, if we assume that $F$ is continuously differentiable in a neighborhood of $0$, then your desired conclusion will hold by the inverse function theorem.

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    $\begingroup$ Thank you very much for the answer. This is a clean counterexample. The result I want to prove is for continuous functions, somehow I forgot to mention it in the body of the question. Since your answer has been well-received I will leave this question as it is and open a new one with the right hypothesis. $\endgroup$
    – No-one
    Sep 1, 2022 at 20:42
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    $\begingroup$ @Titti : For continuous $F$ I think the result is true -- but what makes you believe that there exists an easy proof in that case? Of course, you are welcome to post the amended question anyway, and I would be pleasantly surprised if someone comes up with a more-or-less easy proof. $\endgroup$ Sep 1, 2022 at 20:49
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    $\begingroup$ Yes for continuous function the result is true. I don't know, I am just curious to know if there is any way to avoid using Brouwer. $\endgroup$
    – No-one
    Sep 1, 2022 at 20:56

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