If $f\in C(\mathbb R^n,\mathbb R^n)$ satisfies $\lim_{\|x\|\to+\infty}\frac{(f(x),x)}{\|x\|}=+\infty$, then $f(\mathbb R^n)=\mathbb R^n$.

Question

Let $(\cdot,\cdot)$ be the standard inner product on $\mathbb R^n$. Suppose that $f\in C(\mathbb R^n,\mathbb R^n)$ satisfies $$\lim_{\|x\|\to+\infty}\frac{(f(x),x)}{\|x\|}=+\infty.\tag{$*$}$$ Prove: $f(\mathbb R^n)=\mathbb R^n$.

This is an exercise on the chapter about degrees, Leray's product formula and Jordan-Brouwder's separation theorem in a Differential Topology book (Chinese version).

First thought. First of all, I noticed that it suffices to show that there exists $x_0\in\mathbb R^n$ such that $f(x_0)=0$. Indeed, if we want to show $0\neq y\in f(\mathbb R^n)$, we just need to consider the map $x\mapsto f(x)-y$, for which we also have $(*)$. Next, it follows from $(*)$ that $\lim_{\|x\|\to+\infty}\|f(x)\|=+\infty$. Hence, for some $R>0$, we have $f(x)\neq 0$ for all $\|x\|\geq R$. It suffices to show that $\mathrm{deg}(f, B_R,0)\neq 0$, and then Kronecker's existence property will imply the existence of $x_0\in B_R$ with $f(x_0)=0$. But I don't know how to prove $\mathrm{deg}(f, B_R,0)\neq 0$.

Second thought. If we can prove that $f$ is locally injective, then we can use invariance of domain and $\lim_{\|x\|\to+\infty}\|f(x)\|=+\infty$ to prove the surjectivity of $f$: $f(\mathbb R^n)$ is open and closed. But I don't think we can get the local injectivity of $f$ quickly.

It seems to me that the first thought may be a little more feasible, although I don't know how to do that.

I'm not good at geometric stuffs. If my question is silly, forgive me please. Any help would be appreciated!

Proof using degrees. @Paresseux Nguyen gives a proof using Brouwder's fixed point theorem, which is very nice. Here is a proof using fundamental properties of degrees. Again, it suffices to show that $0\in f(\mathbb R^n)$. By $(*)$, we can find $R>0$ such that $(f(x),x)>0$ for all $x\in\partial B_R$. Define $$H(t,x)=tf(x)+(1-t)x,\qquad x\in \overline{B_R}, t\in I=[0,1].$$ Then $H\in C(I\times \overline{B_R}, \mathbb R^n)$ is a homotopy connecting $f$ and the identity map $\mathrm{id}(x)=x$. Also, for $x\in\partial B_R$ and $t\in I$, we have $(H(t,x),x)=t(f(x),x)+(1-t)\|x\|^2>0$, so $0\not\in H(I\times\partial B_R)$. By the homotopy invariance of degrees, we have $$\mathrm{deg}(f, B_R, 0)=\mathrm{deg}(\mathrm{id}, B_R, 0)=1\neq 0.$$ Finally, Kronecker's existence property implies that there exists $x_0\in B_R$ such that $f(x_0)=0$, and hence $0\in f(\mathbb R^n)$.

score 1 · Accepted Answer · edited Jul 20 '22 at 08:58

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It is sufficient to show that $0 \in \text{Im}f$. Suppose $0 \notin \text{Im}f$, for each $r>0$, by Brouwer fixed point theorem there is a $v_r \in \mathbb{S}^{n-1}$ such that $-rv_r = \frac{ f(rv_r)}{\| f(rv_r)\|}r.$ Thus, $\limsup_{r \rightarrow \infty} \frac{\langle f(rv_r),r v_r\rangle}{r} \le 0$, which is a contradiction. Hence, the conclusion.

edited Jul 20 '22 at 08:58

Feng

13,705

answered Jul 20 '22 at 00:42

Paresseux Nguyen

7,331

To which map do you apply the Brouwer fixed point theorem? – Kritiker der Elche Jul 20 '22 at 09:31
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@KritikerderElche If I understood it corretly, it is the map $f_r : \mathbb D^n\to \mathbb S^{n-1}\subset \mathbb D^n$ defined by $f_r(v)=-\frac{f(rv)}{|f(rv)|}$. – Feng Jul 20 '22 at 10:17
@Feng Yes, you are right! – Kritiker der Elche Jul 21 '22 at 22:18

If $f\in C(\mathbb R^n,\mathbb R^n)$ satisfies $\lim_{\|x\|\to+\infty}\frac{(f(x),x)}{\|x\|}=+\infty$, then $f(\mathbb R^n)=\mathbb R^n$.

1 Answers1