Showing that a function $f$ has a unique fixed point in a metric space.

Question

Let $(X, d)$ be a compact metric space, and suppose $f : X → X$ satisfies $$d(f(x), f(y)) < d(x, y)$$ for all $x \neq y \in X$. Show that f has a unique fixed point.

All I've gotten it so far is that we need to somehow use another function $g(x)=(x,f(x))$.

Thanks

Answer 1

Hint: Consider the function $$ g(x) = d(x,f(x)) $$ Note that this is a continuous function (why?) over a compact space, so it attains its minimum.

Suppose for contradiction that the minimimum of $g$ over $X$ is not $0$. That is, suppose that for every $x \in X$, $g(x) \geq \epsilon > 0$. By compactness, there is an $x^* \in X$ is such that $g(x^*) = \epsilon$. Reach a contradiction (how?) to conclude that $g$ must have a minimum of $0$.

Uniqueness is easy using the inequality.

Answer 2

1. Show that $f$ is continuous.

2. Take $x_0 \in X$ and define $x_{n+1}:=f(x_n)$ for $n \ge 1$

3. By compactness of $X$ ,$(x_n)$ contains a convergent subsequence.

4. Show that the limit of this convergent subsequence is a fixed point of $f$

5. Show that this fixed point is unique.