How to deal with this $\limsup$?

I'm dealing with problems about fixed point and appeared to me the following:

I've a continuous map $f\colon\Bbb{R}^n\to\Bbb{R}^n$ such that $$\limsup_{\|x\|\to\infty} \dfrac{\|f(x)\|}{\|x\|} < 1$$ and I need prove that there exists a closed ball $B\subseteq\Bbb{R}^n$ such that $f(B)\subseteq B$ (then, by Brouwer fixed point theorem, $f (x_0)=x_0$ for some $x_0\in B$).

Being as transparent as possible, I'm not sure how to deal with this $\limsup$. Can anyone give me a hint or something?

By the given condition, there is some positive $r$ such that $|f(x)|<|x|$ whenever $|x|>r$. Letting $B_r$ be the closed ball of radius $r$ centered at $0$, we see that $f(B_r)$ is bounded (since compact) and so contained in some closed ball $B_s$. Putting $R=\max(r,s)$, we check that $B:=B_R$ works.
Let $x\in B$. If $|x|\leq r$, then $x\in B_r$ and so $f(x)\in B_s\subset B_R$ as required. Otherwise, if $r<|x|\leq s$, then $|f(x)|<|x|\leq s$.
• The last part is flawed. You should replace $r$ by $R$. Otherwise, very nice proof. I got one by contradiction, but yours is nicer. Commented Mar 18, 2017 at 3:03