# Brouwer Fixed Point Theorem $f(S^1)\subset B$

I have a question about the Brouwer Fixed Point Theorem:

Theorem 1.(Brouwer Fixed Point Theorem) Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow B$ has a fixed point.

Theorem 2. Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow \mathbb{R}^2$ such that $f(\partial B)\subset B$ has a fixed point.

Note that Theorem $2$ implies Theorem $1$.

We can prove Theorem $2$ using Theorem $1$(or a similar proof)?

$\partial B=S^{1}$

Any hints would be appreciated.

## 1 Answer

Since Theorem $1$ obviously generalises to arbitrary finite radii, we can deduce Theorem $2$ from Theorem $1$ in a simple way:

Since $B$ is compact, $f(B)$ is contained in a ball $B_R$ of some finite radius $R > 0$. Extend $f$ to $B_R$ in the following way:

$$F(x) = \begin{cases}f(x) &, \lVert x\rVert \leqslant 1\\ f\left(\lVert x\rVert^{-1}\cdot x\right) &, 1 < \lVert x\rVert \leqslant R. \end{cases}$$

Then $F \colon B_R \to B_R$ is continuous, hence it has a fixed point by the generalisation of Theorem $1$. Since $\lVert F(x)\rVert \leqslant 1$ for $\lVert x\rVert > 1$, all fixed points of $F$ must lie in $B$, and therefore are fixed points of $f$.