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

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?

Solution 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 $(1)$.