# A seemingly trivial property of differentiable functions

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 $$0$$?

• Isn't this or something very like it proved as part of any proof of the change-of-variables theorem in measure theory? Sep 1, 2022 at 19:22
• 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). Sep 1, 2022 at 21:00

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:

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.

• 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. Sep 1, 2022 at 20:42
• @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. Sep 1, 2022 at 20:49
• 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. Sep 1, 2022 at 20:56