If $f:\mathbb{R}^n\to \mathbb{R}^n $ that has continuous derivative satisfies: $$(\forall x,y\in{\mathbb{R}^n})\|f(x)-f(y)\|\geqslant 2\|x-y\|,$$ then $f$ is surjective. Can someone help me with solving this problem?
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This question is similar to https://math.stackexchange.com/questions/203237/how-to-prove-f-is-onto-if-f-is-continuous-and-satisfying-fx-fy-≥ – mlc Jun 04 '17 at 17:39
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Possible duplicate of how to prove, $f$ is onto if $f$ is continuous and satisfying $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$ – ncmathsadist Jun 05 '17 at 13:07
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That is only case $n=1$ and can't help for $n>1$ – bluemath Jun 05 '17 at 18:26
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@bluemath: Let's use your comment ("can't help for $n \gt 1$") to supply some context for the Question, esp. if you can put that claim in sharper form. – hardmath Jun 05 '17 at 21:23
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First, notice that $f$ is a proper map, indeed one has: $$\lim_{\|x\|\to+\infty}\|f(x)\|=+\infty.$$ Therefore, $f$ is a closed map.
Now, notice that for all $x\in\mathbb{R}^n$, since $f$ is continuously differentiable, one has: $$\|\mathrm{d}_xf\cdot h\|=\lim_{t\to 0}\frac{\|f(x+th)-f(x)\|}{t}\geqslant 2\|h\|.$$ Therefore, $\mathrm{d}_xf$ is injective and thus invertible. Hence, using the inverse function theorem, $f$ is locally invertible everywhere and thus is an open map.
Finally, $f(\mathbb{R}^n)$ is clopen in $\mathbb{R}^n$ and non-empty. Whence, by connectedness of $\mathbb{R}^n$, one has $f(\mathbb{R}^n)=\mathbb{R}^n$.
Remark. One can prove that $f$ is actually a diffeomorphism of $\mathbb{R}^n$, this is the Hadarmard-Lévy theorem. However, in this precise case it is easy to see that $f$ is a diffeomorphism, since $f$ is injective: $$\|f(x)-f(y)\|=0\Rightarrow \|x-y\|=0.$$
C. Falcon
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How do you conclude that $f:\mathbb{R^n}\rightarrow\mathbb{R^n}$ is a closed map? – Ibrahim Islam Apr 20 '20 at 13:03
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@Ibs Since $|f(x)|$ tends to $+\infty$ as $|x|$ goes to $+\infty$, it is seen that $f$ is a proper map (the inverse image of compact is a compact, indeed it is closed by continuity of $f$ and bounded with the assumption), then $f$ is closed (use that the union of a sequence with its limit is compact). – C. Falcon Apr 20 '20 at 13:42
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Ok.....but I have some other approach to that, Wikipedia says that A bijective map is open if and only if it is closed ,if it is so then with help of inverse function theorem we can say that $f$ is an open map ,hence a closed map – Ibrahim Islam Apr 20 '20 at 14:01
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Can you provide me the proof the statement given in Wikipedia i.e A bijective map is open if and only if it is closed – Ibrahim Islam Apr 20 '20 at 14:09