Proof of the Brouwer fixed-point theorem for intervals in $\mathbb{R}$

Question

Proof of the Brouwer fixed-point theorem for intervals in $\mathbb{R}$. I'm looking for feedback and corrections.

Lemma. If $\overline{B}_1(0)=\{x\in\mathbb{R}:\left|x\right|\leq1\}=[-1,1]\subset \mathbb{R}$ and $f\colon \overline{B}_1(0)\to \overline{B}_1(0)$ is continuous, then there exists $x_0\in \overline{B}_1(0)$ such that $f(x_0)=x_0$.

Proof of lemma. Let $g\colon \overline{B}_1(0)\to \overline{B}_1(0)$ be defined as $g(x)=f(x)-x$. As $f(x)$ and $x\mapsto x$ are continuous functions, then $g$ is also continuous.

Because $\overline{B}_1(0)$ is closed, we define $g(-1)$ and $g(1)$, and without loss of generality let us say that $g(-1) < 0< g(1)$. From the intermediate value theorem, we know that for $0\in \overline{B}_1(0)$, there is a $x_0\in \overline{B}_1(0)$ such that $g(x_0)=0$, but as $g(x)=f(x)-x$, we know that $g(x_0)=0=f(x_0)-x_0 \implies f(x_0)=x_0$. $\blacksquare$

(Brouwer fixed-point theorem for intervals in $\mathbb{R}$.) If $f\colon [a,b]\to [a,b]$ is continuous, then there is a $x_0\in [a,b]$ such that $f(x_0)=x_0$.

Proof. Consider $h\colon [a,b]\to [-1,1]$ a continuous bijection with $h(a)=-1$ and $h(b)=1$.

Define $g(x)=h(f(x))-h(x)$ with $g\colon [a,b]\to [-1,1]$. By the lemma there's a $x_0\in [a,b]$ such that $h(f(x_0))=h(x_0)$, and because both $h$ and $f$ are bijections (and therefore admit inverses), $f(x_0)=x_0$. $\blacksquare$

Answer 1

Your idea is correct, but your proof of the lemma does not work as it is. In fact, you cannot be sure that $g(x) \in [-1,1]$. Take for example $f(x) = -x$. But that does not matter, for your IVT-argument you do not need $g([-1,1]) \subset [-1,1]$. Next, you cannot be sure that $g(1), g(-1)$ are non-zero with opposite sign. Look at $f(x) = x$.

I suggest to proceed as follows.

Consider $f : [a,b] \to [a,b]$ and define $g : [a,b] \to \mathbb R, g(x) = f(x) - x$. Then $g(a) = f(a) - a \ge a - a = 0$ and $g(b) = f(b) - b \le b - b = 0$. Now the IVT tells us that $g(x_0) = 0$ for some $x_0 \in [a,b]$.

Answer 2

Using the lemma is good, but not the way you're trying to.

Consider $g(y)=h(f(h^{-1}(y)))$; this is a continuous function $g\colon[-1,1]\to[1,1]$, so it has a fixed point $y_0\in[-1,1]$. Then $x_0=h^{-1}(y_0)$ is a fixed point for $f$, because $$ y_0=g(y_0)=h(f(x_0)) $$ and so $x_0=h^{-1}(y_0)=f(x_0)$.

_{Of course, using $g(x)=f(x)-x$ and applying the intermediate value theorem to $g$ is much easier.}