@Moishe Kohan Sorry, I don't know. I read a paper, which claims that such a bi-Lipschitz mapping exists, but it doesn't say more. I am not familar with those concepts, so could you provide me with some references about what you mentioned?

Thank you, @TaQ!

Thank you, @TaQ! I understand $(*)$, since differentiability is a local property; hence, locally Lipschitz is enough for us to desire $(*)$. But I am not sure about your comment on 3. Could you please write the precise process of it? Thank you!

Sorry, I extracted one sentence from the paper, and it seemed confusing. Therefore, I deleted this sentence and added a full description at the end. @NateRiver

Thanks for your answer! Now I have some questions: 1. Strictly, I think "$=$" in (*) should be replaced by "$\leqslant$". 2. I am not very sure why the following inequality $$ \|\tilde{f}\|_{W^1_\infty(\mathbb{R}^n)} \leqslant M\|f\|_{W^1_\infty(\Omega)} $$ holds. Could you please show the exact process of how to use the open mapping theorem? 3. $W^1_\infty(\Omega)\subset W^1_\infty(\mathbb{R}^n)|_\Omega$ is a given condition. The whole proof of the original paper is based on this condition.

I understand. Thank @TaQ!