Take a continuous mapping $f: \bar{B^{n}} \rightarrow \bar{B^{n}}$, where $\bar{B^{n}}$ is a closed unit ball in $\mathbb{R}^{n}$. Assume that $f(x) \neq x$ for every $x \in \bar{B^{n}}$. Define another function $r$ by following the directed line segment from $f(x)$ through $x$ to its intersection with $\partial B^{n}$, and let the intersection point be $r(x)$. Is it immediately evident that $r$ is a continuous function? Is so how does it follow? Thanks.
$\begingroup$
$\endgroup$
6
-
5$\begingroup$ That depends. With enough experience, it is. For the first years, it isn't. $\endgroup$– Daniel FischerCommented May 20, 2014 at 22:49
-
$\begingroup$ @angryavian Yes I forgot the continuity, will edit. $\endgroup$– user100431Commented May 20, 2014 at 23:09
-
$\begingroup$ Unless I'm missing something, your $f$ cannot exist. It contradicts Brouwer's Fixed Point Theorem. $\endgroup$– Martin ArgeramiCommented May 20, 2014 at 23:26
-
4$\begingroup$ @Martin This is the first step of a proof of the fixed point theorem. One constructs this $r$, which is a retract onto the boundary, and then demonstrates a contradiction by considering the homology or homotopy groups of the ball and its boundary. $\endgroup$– user98602Commented May 20, 2014 at 23:54
-
$\begingroup$ @MartinArgerami Yes this is an argument that is used in a proof of Brouwer's fixed point theorem. It is assumed $f$ is as described and then it follows that $r$ is continuous which then gives rise to a contradiction from the Negative Retract Principle. But the point is that the continuity of $r$ follows before the contradiction. So what I want to know is how the continuity of $r$ follows? $\endgroup$– user100431Commented May 20, 2014 at 23:56
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
2
Since translations are continuous, we may assume that $B^n$ is the unit ball centred at the origin. Then $$ r(x)=f(x)+\lambda_x\,(x-f(x)), $$ with $\lambda_x\geq0$ such that $$ \|f(x)+\lambda_x\,(x-f(x))\|=1. $$ This equality can be written as $\|f(x)+\lambda_x\,(x-f(x))\|^2=1$, and it expands to $$ \|x-f(x)\|^2\,\lambda_x^2+2\langle f(x),x-f(x)\rangle\,\lambda_x+\|f(x)\|^2-1=0. $$ This is quadratic in $\lambda_x$, and we want the non-negative solution, which is $$ \lambda_x=\frac{-\langle f(x),x-f(x)\rangle+\sqrt{\langle f(x),x-f(x)\rangle^2-\|x-f(x)\|^2(\|f(x)\|^2-1)}}{\|x-f(x)\|^2}. $$ The term inside the square root is always non-negative and bigger than $\langle f(x),x-f(x)\rangle^2$, so $\lambda_x\geq0$.
It is clear that $\lambda_x$ depends continuously on $x$, and so $r(x)$ is continuous.
answered May 21, 2014 at 0:42
-
$\begingroup$ Thanks for your answer, very helpful. One question to confirm something, is it clear that $\lambda_{x}$ depends continuously on $x$ simply because it is a composition and addition of continuous functions such as $f$, $\Vert \cdot \Vert$, $\sqrt{}$, $\langle \cdot \rangle$? And then $r$ is continuous as it is the addition of continuous functions? $\endgroup$– user100431Commented May 21, 2014 at 11:20
-
2$\begingroup$ Exactly. And the hypothesis guarantees that the denominator is never zero. $\endgroup$ Commented May 21, 2014 at 14:20