Proof of Brouwer's fixed point theorem without direct computation Brouwer's Fixed Point theorem says - If $f:\Bbb{D}\to\Bbb{D}$ be a continuous map, then $f$ has a fixed point.
Suppose not, then we look at the line segment joining $f(x)$ and $x$ and extend it till the boundary of the disc and call the intersection point to be $r(x)$. Formally, I can write-
Let us define for $x\in\Bbb{D}$, define $\rho_x(t)=(1-t)f(x)+tx\ \forall t\ge0$. Let $t_x:=\text{inf}\{t\ge1|\ \lVert \rho_x(t)\rVert\ge1\}$ and finally define $r(x):=\rho_x(t_x)$.
We can directly compute what the $t_x$ is,by solving the equation $\lVert \rho_x(t)\rVert^2=1$ (as done here)
I can prove that $r$ is continuous if the map $x\mapsto t_x$ from $\Bbb{D}$ to $[1,\infty)$ is continuous. Can we use this definition (involving infimum) of $t_x$ to conclude the continuity of $x\mapsto t_x$ without computation?
Can anyone help me to complete the proof? Thanks for your help in advance.
 A: This is an interesting question. In fact we can avoid to work with the explicit solutions of
$$\lVert \rho_x(t)\rVert^2=1 . \tag{1}$$
Unfortunately the proof of Brouwer's fixed point theorem does not become easier if your alternative approach is used.
The explicit solutions of $(1)$ are given in Continuous function on closed unit ball and Proof of Brouwer's fixed point theorem in Hatcher and I think this approach is very easy.
As Thomas Andrews comments, it is a bit more general (and more transparent) to study the following situation:
Let $D^2 = \{ x \in \mathbb R^2 \mid \lVert x \rVert \le 1\}$ be the unit disk and $S^1=  \{ x \in \mathbb R^2 \mid \lVert x \rVert = 1\}$ be the unit circle which is the boundary of $D^2$. Define
$$l : D^2 \times D^2 \times \mathbb R \to \mathbb R^2, l(x,y,t) = (1-t)x + ty = x + t(y-x).$$
This is a continuous map. Given two distinct points $x, y$ of $D^2$, the continuous map $l_{(x,y)} : \mathbb R \to \mathbb R^2, l_{(x,y)}(t) = l(x,y,t)$, parameterizes the line through $x$ and $y$ having direction $d = y - x$. Our geometric intuition tells us that $l_{(x,y)}$ intersects the unit circle $S^1$ at exactly two times $t^-_{(x,y)}  \le 0$ and $t^+_{(x,y)}  \ge 1$. We can formally prove this by deriving an explicit formula for $t^\pm_{(x,y)}$ as the solutions of a quadratic equation (similarly as in the above two links). This gives us  a function
$$R: \Delta = D^2 \times D^2 \setminus \{(x,x)\mid x \in D^2\} \to S^1, R(x,y) = l(x,y,t^+_{(x,y)}) .$$
Continuity of $R$ is immediate from the above explicit formula.
Brouwer's fixed point theorem can now proved as follows: Assume to the contrary that there exists a map $f : D^2 \to D^2$ without fixed points. Then we get the continuous map
$$r : D^2 \stackrel{(f,id)}{\to} \Delta  \stackrel{R}{\to} S^1$$
which is easily seen to be a retraction. Here $(f,id)(x) = (f(x),x)$. But the non-existence of such a retraction is a well-known theorem from algebraic topology.
Let us now come to the approach suggested in your question.
If we accept the above geometric intuition without calculating something, we expect that $t^+_{(x,y)}  \ge 1$ and $\lVert l_{(x,y)}(t) \rVert \ge 1$ for $t \ge t^+_{(x,y)}$. Therefore we must indeed have $$t^+_{(x,y)}  = s(x,y) := \inf\{ t \ge 1 \mid \lVert l_{(x,y)}(t) \rVert \ge 1 \} . \tag{2}$$
So let us take $(2)$ as the definition of $t^+_{(x,y)}$. This does not depend on any calculation, but of course we must prove that $s(x,y)$ is well-defined, $l_{(x,y)}(s(x,y)) \in S^1$ and $(x,y) \mapsto s(x,y)$ is continuous.
For each $(x,y) \in \Delta$ we define
$$\nu_{(x,y)} : [1,\infty) \to [0,\infty), \nu_{(x,y)}(t) = \lVert l_{(x,y)}(t) \rVert  . \tag{3}$$
The $\nu_{(x,y)}$ are continuous maps and we have
$$s(x,y) = \inf \nu_{(x,y)}^{-1}([1,\infty)). \tag{4}$$

*

*$\nu_{(x,y)}^{-1}([1,\infty)) \ne \emptyset$. Hence $s(x,y)$ is a well-defined element of $[1,\infty)$. In other words, we have defined a function $s : \Delta \to [1,\infty)$.

Proof. We have
$$\nu_{(x,y)}(t)  = \lVert x + t(y-x)) \rVert \ge \lVert t(y-x)) \rVert - \lVert x \rVert = \lvert t \rvert \cdot \lVert y-x \rVert - \lVert x \rVert \\ \ge \lvert t \rvert \cdot \lVert y-x \rVert - 1 . \tag{5}$$
Since $y-x \ne 0$, we see that $\nu_{(x,y)}(t) \to \infty$ as $t \to \infty$. This proves $\nu_{(x,y)}^{-1}([1,\infty)) \ne \emptyset$. Since $\nu_{(x,y)}^{-1}([1,\infty))  \subset [1,\infty)$, we have $s(x,y) \in [1,\infty)$.


*$\nu_{(x,y)}(s(x,y)) = 1$, i.e. $l_{(x,y)}(s(x,y)) \in S^1$.

Proof. Since $\nu_{(x,y)}^{-1}([1,\infty))$ is a closed subset of $[1,\infty)$, we have $s(x,y) \in \nu_{(x,y)}^{-1}([1,\infty))$, thus $\nu_{(x,y)}(s(x,y)) \ge 1$. Assume that $\nu_{(x,y)}(s(x,y)) > 1$ (which is only possible if $s(x,y) > 1$ because $\nu_{(x,y)}(1) = \lVert y \rVert \le 1$). Then there exists $\epsilon > 0$ such that $\nu_{(x,y)}(t) > 1$ for $t \in (s(x,y)- \epsilon, s(x,y)+ \epsilon) \cap [1,\infty)$. In particular, $\nu_{(x,y)}(t) > 1$ for $\max(1,s(x,y) - \epsilon) < t \le s(x,y)$, thus $s(x,y) = \inf\{ t \in [1,\infty) \mid \nu_{(x,y)}(t) \ge 1 \} \le \max(1,s(x,y) - \epsilon) < s(x,y)$ which is a contradiction.
What about the continuity of $s$?
Let us prove two more properties of $s$.


*$\lVert l_{(x,y)}(t) \rVert < 1$ for $0 < t < s(x,y)$.

Proof. For $1 \le t < s(x,y)$ this follows from the definition of $s(x,y)$. So let us consider $0 < t < 1$. If $\lVert x \lVert < 1$ or $\lVert y \lVert < 1$, this follows from
$$\lVert l_{(x,y)}(t) \lVert \le (1-t)\lVert x \lVert + t\lVert y )\lVert .$$
If $\lVert x \lVert = \lVert y \lVert = 1$, we need the Cauchy-Schwartz-inequality $\lvert a \cdot b \rvert \le \lVert a \rVert \cdot \lVert b \rVert$ in which equality holds iff $a, b$ are linearly dependent. Note that here it is essential that we work with the Euclidean norm (or any norm induced by a scalar product). Let us take $a = (1-t)x, b = ty$. Since $\lVert a \rVert + \lVert b \rVert = (1-)t\lVert x \rVert + t\lVert h(x) \rVert = t + 1-t =1$, we get
$$\lVert l_{(x,y)}(t) \rVert^2 = (a + b) \cdot (a + b) = \lVert a \rVert^2 + 2 a \cdot b + \lVert b \rVert^2 \le \lVert a \rVert^2 + 2 \lVert a \rVert \cdot \lVert b \rVert + \lVert b \rVert^2 \\ =(\lVert a \rVert + \lVert b \rVert)^2 = 1$$
where $\lVert l_{(x,y)}(t) \rVert^2 = 1$ is possible only when $a,b$ are linearly dependent. By definition of $a,b$ linearl dependency means $y = \lambda x$ for some $\lambda$ since $t, 1-t \ne 0$. Clearly $\lvert \lambda \rvert = 1$, thus we must have  $y = -x$. But then $l_{(x,y)}(t) = (1-t)x - tx = (1-2t)x$. Since $-1 < 1-2t < 1$, we get $\lVert l_{(x,y)}(t) \rVert = \lvert 1-2t \rvert \cdot \lVert x \rVert =  \lvert 1-2t \rvert  < 1$, a contradiction. Thus equality does not occur.


*$\lVert l_{(x,y)}(t) \rVert > 1$ for $t > s(x,y)$.

Proof. Asume that $\lVert l_{(x,y)}(\tau) \rVert \le 1$ for some $\tau > s(x,y)$. Thus $y' = l_{(x,y)}(\tau) = (1-\tau)x + \tau y \in D^2$ and clearly $x \ne y'$ since $l_{(x,y)}(t) = x$ if and only if $t = 0$. By 3. the line $l_{(x,y')}$ has the property $\lVert l_{(x,y')}(t) \rVert < 1$ for $0 < t < 1$. In particular $\lVert l_{(x,y')}(s(x,y)/\tau) \rVert < 1$ since $\tau > s(x,y)$. But
$$l_{(x,y')}(s(x,y)/\tau) = (1 - s(x,y)/\tau)x + s(x,y)/\tau ((1- \tau)x +\tau y) = (1 - s(x,y)) x  + s(x,y) y \\ = l_{(x,y)}(s(x,y)) $$
which contradicts 2.
Properties 3. and 4. show that there exists a unique value $s \ge 1$ such that $\lVert l_{(x,y)}(s) \rVert = 1$; this value is $s(x,y)$. We shall see that this is the crucial property to prove that $s$ is continuous.
Let us prove it by contradiction and assume that $s$ is not continuous at some $(\xi,\eta)$.  In that case there exists a sequence $(x_n,y_n) \in \Delta$ such that $(x_n,y_n) \to (\xi,\eta)$ and $s_n = s(x_n,y_n)$ does not converge to $s = s(\xi,\eta)$.
$(s_n)$ is bounded:
Let $r = \lVert \eta - \xi \rVert$. Then $r_n = \lVert y_n - x_n \rVert \ge r/2$ for $n \ge n_0$. By $(5)$ we get $\lVert l(x_n,y_n,t) \ge \lvert t \rvert \cdot r/2 - 1$ for $n \ge n_0$. Thus $\lVert l(x_n,y_n,t) \rVert > 1$ for $t > 2/r$ and $n \ge n_0$. We conclude that $s_n \le 2/r$ for $n \ge n_0$.
Since $(s_n)$ is bounded and $(s_n)$ does not converge to $s$, we can find a subsequence of $(s_n)$ converging to some $s' \ge 1$ such that $s' \ne s$. W.l.o.g. we may assume that $s_n \to s'$. But then $l(\xi,\eta,s') = \lim_{n \to \infty} l(x_n,y_n,s_n) = \lim_{n \to \infty} 1 = 1$ which shows that $s' = s$, a contradiction.
A: Here’s a different way to show continuity.
Let, for each $x \in X$ ($X$ being a metric space), $f_x: \mathbb{R} \rightarrow [0,\infty)$ be a strictly convex proper function such that $(x,t) \longmapsto f_x(t)$ is continuous and $f_x(0) \leq 1$, $f_x(1) \leq 1$.
Define $t(x)=\inf\,\{t \geq 1,\, f_x(t) \geq 1\}$.
$t(x)$ is well-defined, because $f_x$ is proper. Because $f_x$ is continuous, and $f_x(1) \leq 1$, $f_x(t(x))=1$. Because $f_x$ is strictly convex, if $u > t(x)$ is such that $f_x(u)=1$, then $\{f_x \leq 1\}=(t(x),u)$. But $\{f_x \leq 1\}$ contains $1 \leq t(x)$ — we get a contradiction.
So $t(x)$ is the unique element in $[1,\infty)$ such that $f_x(t(x))=1$.
For the same reason, there is an element $t’(x) \leq 0$ such that $f_x(t’(x))=1$ and $(t’(x),t(x))=\{f_x < 1\}$.
To show that $t(x)$ is continuous, it is enough to show that its restriction at any compact is continuous, so we may assume that $X$ is compact metric.
It’s formal topology to show that $t$ is then continuous iff it is bounded and its graph is closed.
We just saw that the graph of $t$ was cut out by the conditions $f(x,t)=1$ and $t \geq 1$, so it’s closed.
Assume $t$ is unbounded: there are sequences $x_n \in X$, $z_n$ growing to $\infty$ with $f_{x_n}(z_n) = 1$.
Up to extracting, we can assume that $x_n$ converges to some $x \in X$. Now, for each $n$ and each $m \geq n$, $f_{x_m}(z_n) \leq 1$ (as $z_n \leq z_m=t(x_m)$), so it follows that $f_x(z_n) \leq 1$. So $\{f_x \leq 1\}$ isn’t bounded, a contradiction.

I cannot resist to put in here an alternative way to deduce Brouwer’s theorem from the non-retraction lemma – in a simple way, that works in any dimension $n \geq 2$ and without any continuity computation. The idea isn’t very different. Let $f: B^n \rightarrow B^n$ be continuous without any fixed point.
Define $N: x \in \mathbb{R}^n \backslash \{0\} \longmapsto \frac{x}{|x|}\in  S^{n-1}$.
For $x \in B^n$, define $g(x)=N(2x-f(2x)) \in S^{n-1}$ if $2|x| \leq 1$ and, if $x =ru$ with $u \in S^{n-1}$ and $r \geq 1/2$, $g(x)=N(u-2(1-r)f(u))$.
Then $g: B^n \rightarrow S^{n-1}$ is a retraction.
