Surjective vector field on $\mathbb{R}^n$

analysisfixed-point-theoremsfunctionsreal-analysisVector Fields

Let $V: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous with the property
$$\frac{\langle V(x), \, x\rangle}{|x|} \, \to \infty \quad \text{as} \quad |x| \to \infty \qquad \qquad (1)$$
where $\langle \cdot \, , \cdot \rangle$ denote the standard inner product and $| \cdot | = \sqrt{\langle \cdot \, , \cdot \rangle}$ is the Euclidian norm on $\mathbb{R}^n$.

I have to show that $V$ is surjective.

My attempt:
Take $z \in \mathbb{R}^n$ and define $\varphi: \mathbb{R}^n \rightarrow \mathbb{R}^n, \, \varphi(x): = V(x) – z$. The aim is to show that $\varphi$ has a zero. So assume by contradiction that $\varphi(x)\neq 0 \, \, \, \forall \, x \in \mathbb{R}^n$. Let $R > 0$. We define an auxiliary function $\psi : \mathbb{R}^n \rightarrow \mathbb{R}^n, \, \psi(x): = R \cdot \frac{\varphi(x)}{|\varphi(x)|}$. Then, $\psi$ is a continuous self-mapping $\psi: \overline{B}_R(0) \rightarrow \overline{B}_R(0)$ with $\text{im}(\psi) \subset \partial B_R(0)$. Using Schauder's fixed point theorem, there exists $x_0 \in \overline{B}_R(0)$ s.t. $\psi(x_0) = x_0$. In particular, $x_0 \in \partial B_R(0)$.

Now, I tried to get a contradiction with the assumption $(1)$ taking $R \to \infty$ without success. (Maybe it's not the right thing to do)

Any suggestions? Thanks in advance!

Best Answer

Assume $\varphi(x)$ has no zeros.

It follows from the given property that there is $R$ such that $\forall x \in S_R(0)$ $\varphi(x)$ in not tangent to $S_R(0).$ Let's assume that $\varphi(x)$ is outward-pointing (consider $-\varphi(x)$ otherwise). Then it's outward-pointing on the whole sphere by continuity.

Having such a vector field, it's possible to construct a retraction $r$ of $B_R(0)$ onto $S_R(0):$ for $x \in B_R(0)$ define $r(x)$ to be the intersection of the ray $(x; \varphi(x))$ with the sphere. Note that $r(x) = x$ on the sphere and $r$ is continuous.

Using Schauder's fixed point theorem we can prove that there's no such retraction: consider $$F: B_R(0) \to B_R(0)$$ $$F(x) = - r(x). $$ This map cannot have a fixed point. Contradiction.

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