For $f:\mathbb R^n \rightarrow \mathbb R^n \quad \exists R \gt 0 $ that $f$ is continous on $\bar B_R(0):= \{ x \in \mathbb R^n : \Vert x\Vert_2 \leq R \} \subset \mathbb R^n$ .

Let also $\langle f(x),x \rangle \geq 0 \quad \forall x \in \partial\bar B_R(0)$

Show that it exists a $x^* \in \bar B_R(0) $ so that $f(x^*) = 0$.

I tried to check the function $F:\bar B_R(0) \rightarrow \mathbb R^n , \quad F(x):= -R \frac{f(x)}{\Vert f(x)\Vert_2}$ with the Brouwer fixed-point theorem but didnt get far.

Any help is appreciated thanks.


Suppose by contradiction that $f(x)\neq0$ for all $x\in \bar{B}_R(0)$. Then your function $F:\bar{B}_R(0)\to\mathbb{R}^n$ is well defined and we have that $|F(x)|=R$ for all $x\in \bar{B}_R(0)$, hence we can write that $F:\bar{B}_R(0)\to\partial \bar{B}_R(0)\subset\bar{B}_R(0)$. Also $F$ is continuous, so by the Brower fixed point theorem there exists $x_0\in\bar{B}_R(0)$ such that $F(x_0)=x_0$. Moreover $|F(x_0)|=|x_0|=R$ implies that $x_0\in \partial \bar{B}_R(0)$. Now we have:

$$0<R^2=\langle x_0,x_0\rangle=\langle F(x_0),x_0\rangle=-\frac{R}{|f(x_0)|}\langle f(x_0),x_0\rangle\le0,$$

that is a contradiction.

  • $\begingroup$ wow thats smooth thank you alot $\endgroup$ – user391105 Oct 26 '17 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy