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

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), $$f(x_0)=x_0$$. $$\blacksquare$$

• Domain of $g$ is [a,b], not [-1,1].
– Mars
Commented Nov 19, 2020 at 5:07
• @Mars thank you! Commented Nov 19, 2020 at 5:09
• you're on the right track though, You just want to define $h$ in the other direction so that $h(-1)=a$ and $h(1)=b$
– Mars
Commented Nov 19, 2020 at 5:11

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 $$f(x) = x$$.
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 $$x_0 \in [a,b]$$.
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.