# Compact Metric Space Question

Let $(X,d)$ be nonempty compact metric space and $f : X\to X$ be a function satisfying $d(f (x), f (y)) < d(x, y)$ for all distinct pair of points $x, x \in X$. Show that $f$ have a fixed point and this fixed point is unique.

Hint: Define the function $g : X \to R$ by $g(x) = d(f (x), x)$. Assume that $f (x) \neq x$ for all $x \in X$. Use compactness to show the existence of a point $a \in X$ such that $g(a) \leq g (x)$ for all $x \in X$. Deduce a contradiction by considering $x=f (a)$.

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Have you tried applying the hint? –  Chris Eagle Dec 2 '12 at 22:03
That hint is a solution. Don't see a question here. –  Norbert Dec 2 '12 at 22:07
I also thought that the hint is solution. However, the question is this and i guess that the aim is this question to prove it which i could not. That's why i asked this question. –  user51524 Dec 2 '12 at 22:51
Have you tried applying the hint? –  Chris Eagle Dec 2 '12 at 23:08
I think i made a mistake because by applying the hint i could not prove anything. How can the hint make the question proved while it is the answer? –  user51524 Dec 2 '12 at 23:15

You can use the result that tells you that a continuous function on a compact set achieves a minimum (notice that $f$ is continuous since it is a strict contraction). This gives you the existence of $a$. You get a contradiction from choosing $x=f(a)$ because then $$g(a) = d(f(a),a) \leq d(f(f(a)), f(a)) = g(x)$$ which contradicts the fact that $f$ is a contraction (since $a \neq f(a)$).
Uniqueness follows easily by contradiction as well since if you assume $\exists x, y \in X$ fixed points you get (using the fact that $f$ is a contraction) $$d(f(x),f(y)) = d(x,y) < d(x,y).$$