# Classical proof of Brouwer fixed point theorem: why the projection is continuous?

In the classical proof of the Brouwer fixed point theorem, we suppose for absurd that if $f$ is a continuous function from the closed unity ball to itself with no fixed point, then we show that it's possible to build a retraction $F$ of the unity ball (which is impossible). In all the proofs it's said that $f$ is clearly continuous, and intuitively I agree with that, but I've not found any proof of that (neither in books or on the web). The same question was already asked here, but the answer use the inverse function theorem, which could not be applied since $f$ isn't derivable. So, does anibody has some advice?

Thanks

• I remember seeing the proof of the continuity of $F$ somewhere here at MSE, but I can't find it. – Stefan Hamcke Dec 5 '14 at 19:25
• I don't know what you are referring to and have trouble to understand your question: $f$ is continuous by assumption. So what exactly is it you want do show and do not understand? Is there a type and you are talking about $F$? If yes, how is it defined? – Thomas Dec 5 '14 at 19:30

I assume you're talking about something like this: If $f: B \to B$ with no fixed point, let $F(x) = x + T(x) (x - f(x))$ where $T(x) \ge 0$ is chosen so that $\|F(x)\| = 1$, mapping the unit ball $B$ to the unit sphere. Geometrically, the line through $x$ and $f(x)$ intersects the sphere in two points, of which $F(x)$ is the one closer to $x$ than to $f(x)$. You want to know why $F$ is continuous. It suffices to show that $T$ is continuous.
Let $\epsilon > 0$ be given. I want to find $\delta > 0$ such that $y \in B$ with $\|y - x\| < \delta$ implies $|T(y) - T(x)| < \epsilon$. For any $s > 0$ and $y \in B$ we have $$\left|\|y + s (y - f(y))\| - \|x + s (x - f(x))\|\right| \le (1+s) \|y - x\| + s \|f(y) - f(x)\|$$ Now taking $s_1 = T(x) + \epsilon$, so $\|x + s_1(x - f(x))\| > 1$, and $T(x) > s_2 \ge T(x) - \epsilon$ so that $\|x + s_2(x - f(x))\| < 1$, take $\eta > 0$ small enough so that $(1+2s_1) \eta < \|x + s_1 (x - f(x))\| - 1$ and $(1 + 2 s_2) \eta < 1 - \|x + s_2 (x - f(x))\|$.
Taking $\delta > 0$ so that $\delta < \eta$ and so that $\|y - x\| < \delta$ implies $\|f(y) - f(x)\| < \eta$, we find that if $\|y - x\| < \delta$, $$\|y + s_1 (y - f(y))\| > \|x + s_1 (x - f(x))\| - (1 + 2 s_1) \eta > 1$$ and $$\|y + s_2 (y - f(y))\| < \|x + s_2 (y - f(y))\| + (1 + 2 s_2) \eta < 1$$ so that $s_2 < T(y) < s_1$, and in particular $|T(y) - T(x)| < \epsilon$.