How to prove the 1 Lipschitz function defined on the closed unit ball has a fixed point I need to prove that the 1 Lipschitz function has a fixed point:
$\|f(x)-f(y)\|≤ \|x-y\|$ for all $x,y\in B$, where $B$ is the closed unit ball in the $R^n$.
I want to apply the contraction mapping theorem. The contraction mapping theorem is not suitable for 1 Lipschitz function because it requires that $\lambda \in [0,1)$, so I want to modify this function to obtain a contraction. Anyone could please gives me some hints about how to get such contraction?
 A: Note that $(1-{1 \over n}) f$ is $(1-{1 \over n})$ Lipschitz and so has a fixed point $x_n$ that satisfies
$(1-{1 \over n}) f(x_n) = x_n$. Since $B$ is compact, $x_n$ has an accumulation point $x^*$ and continuity of $f$ gives $f(x^*) = x^*$.
A: This is Problem 35 in Section 10.3 of Real Analysis, fourth edition, by Royden and Fitzpatrick. This answer is for readers of that text.
Instead of modifying the function, I modify the proof of Banach's contraction principle.
Select a point in $B$ and label it $x_0$. Now define sequence $\{x_k\}$ inductively by putting $x_1$ to $f(x_0)$ and if $k$ is a natural number such that $x_k$ is defined, putting $x_{k+1}$ to $f(x_k)$.
By Theorem 20(i)$\to$(iii) of Section 9.5, $\{x_k\}$ has a subsequence that converges to a point $x_*$ in $B$. For notational convenience, assume the whole sequence $\{x_k\}$ converges to $x_*$. Because $f$ is Lipschitz, it is continuous. Therefore
$$f(x_*) =\lim_{k\to\infty} f(x_k) =\lim_{k\to\infty} x_{k+1} = x_*.$$
Thus, mapping $f:B\to B$ has a fixed point.
