$F:\mathbb R^n \to\mathbb R^n$ is continuous,and $\|F(\vec{x})-F(\vec{y})\| > \|\vec x-\vec y\| $ for $\vec x,\vec y \in \mathbb R^n$. How to prove $F$ is onto?
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Hi! Welcome to MathSE. You should provide some of your own thoughts and people might be more milling to help. – Rumplestillskin Jun 22 '17 at 08:11
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Hint. Note that $F$ is injective (why?). Case $n=1$: $F$ is strictly monotone and $$|F(x)|\geq |F(x)-F(0)|-|F(0)|>|x|$$ which implies that $\lim_{x\to\pm \infty}|F(x)|=+\infty$.
For the general statement try to show the non-empty set $F(\mathbb{R}^n)$ is open and closed in the connected set $\mathbb{R}^n$.
Robert Z
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Hi, robert. I was trying to think in the following direction. Assume that $y \notin F(\mathbb{R^n})$. Then if we look at the function $g:\mathbb{R^n} \to \mathbb{R}$ defined as $g(x) = \frac{1}{||F(x)-y ||}$. Then, because of continuity of $F$, $\exists M>0$ such that $\forall x \in \mathbb{R^n}, 0<g(x)<M$. I have been trying this line for about an hour. Can you kindly give me some input about how it may or may not work. – Jun 22 '17 at 09:18
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1Take a look here: https://math.stackexchange.com/questions/2309589/if-fx-fy-geqslant-2x-y-then-f-is-surjective?noredirect=1&lq=1 – Robert Z Jun 22 '17 at 09:25
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very helpful of your hint, but I have no idea to show $F(R^n)$ is open, can your give some advice? – lin Jun 22 '17 at 11:52
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Firstly, remark that $F$ is injective: $F(x)=F(y)$ implies $0=\|F(x)-F(y)\|=\geq \|x-y\|$.
Let $C$ be the complementary space of the image of $F$. We show that $C$ is open. Let $y\in C$, suppose that for every integer $n$, there exists $x_n$ such that $F(x_n)$ is an element of the ball $B(y, 1/n)$, we deduce that $\|F(x_n)-F(x_m)\|\leq \|F(x_n)-y\|+\|y-F(x_m)\|\leq 1/n+1/m$. Consider $n,m>N$ This implies that $\|x_n-x_m\|\leq 2/N$ and $(x_n)$ is a Cauchy sequence, so it converges towards $x$, we have $F(x)=lim_nF(x_n)=y$ contradiction. So $C$ is open and the image of $F$ is closed.
Let $Im(F)$ be the image of $F$ endowed with the norm induced bythe norm $\mathbb{R}^n$, consider $F^{-1}:Im(F)\rightarrow \mathbb{R}^n$, we have $\|x-y\|=\|F(F^{-1}(x))-F(F^{-1}(y))\|\geq \|F^{-1}(x)-F^{-1}(y)\|$. This implies that $F^{-1}$ is continue. We deduce that for every open subset $U$ of $\mathbb{R}^n$, $({F^{-1}})^{-1}(U)=F(U)$ is open. This implies that $Im(F)$ is closed and open so it is $\mathbb{R}^n$.
Tsemo Aristide
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