Every 1-Lipschitz function in the closed unit ball has a fixed point

Question

I'm currently trying to solve the following exercise:

Let B be the closed unit ball in $\mathbb R^n$ together with the euclidean metric. Show that every 1-Lipschitz function $f:B\to B$ has a fixed point.

I think I am supposed to use the Banach Fixed Point Theorem, but I somehow have to show that I am allowed to use it, since in general you can only use it for a Lipschitz constant $L$ with $0 \le L \lt 1$. However, since we look at the closed unit ball, I think I have to show that $f$ is a contraction even for $L = 1$. Can you give me any ideas?

Answer 1

Since $B$ is compact, it suffices to prove that $$\inf_{x\in B} |f(x)-x|=0 \tag1$$ (The infimum must be attained by compactness.)

For every $\epsilon>0$ the map $x\mapsto (1-\epsilon)f(x)$ is $(1-\epsilon)$-Lipschitz, and therefore has a fixed point $x_\epsilon$. Since $|f(x_\epsilon)-x_\epsilon|=|\epsilon f(x_\epsilon)|\le \epsilon$, we have $(1)$.

Answer 2

Here's another argument I came up with, although I personally prefer the accepted answer.

Let $g(x) = ||f(x) - x||$. Since $B$ is compact and $f$ is continuous, there exists $x_0 \in B$ such that $g(x_0) \leq g(x)$ for all $x \in B$. Suppose $g(x_0) > 0$. Let $\epsilon = g(x_0)$, $z = (f(x_0) - x_0) / 2$ and $x_n = x_0 + n z$. It is possible to show via induction that $f(x_{n}) = f(x_0) + n z$ for all $n$. This means $f$ eventually maps a point inside $B$ to a point outside $B$, violating the assumption that $f$ maps $B$ to $B$. We must therefore conclude $g(x_0) = 0$, implying that $f$ has a fixed point at $x_0$.

Answer 3

This answer is pretty much the first one presented (Please let me know for mistakes, this is my first answer):

If we consider $$f_n(x) := (\Big (1-\frac{1}{n} \Big)f(x) \\ f_n(x)_{_n} := \Big (\Big (1-\frac{1}{n} \Big)f(x) \Big)_n$$ We can see that for any $ n>2, f_n(x) $ is $1-\frac{1}{n}$-Lipschitz. For Banach Fixed point, $f_n(x)$ has a fixed point.

$\exists a_n \in B[0,1]^n$ such that $ f_n(a_n) = a_n $ for every $f_n$ function. We can set a sequence of these points. $$f_n(x_n)_{_n} = (x_n)_{_n} $$ And notice that as closed subset of $\mathbb{R} ^n $ it is compact (Heine Borel). Therefore there is a subsequence $(x_{n_k})_k$ such that it converges to $x_o$. By applying the limit we get: $$ \lim_{k\to \infty} f_{n_k}(x_{n_k})_k = \lim_{k\to \infty}(1-\frac{1}{n_k})f(x_{n_k}) = x_o $$ By using that $x_{n_k}$ converges to $x_o$ and continuity (guaranteed by Lipschitz) $$ 1 \cdot f(x_o) = x_o$$.

Examples to show that is even possible: any constant function with $c$ in $[-1, 1]$ would satisfy in one point. (ie: $f(x) = 0$ in $x=0$) and $f(x) = x$ everywhere.