A seemingly trivial property of differentiable functions

Question

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 $0$?

Answer 1

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 $x\le0$.

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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.

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

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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.