Let $(M,d)$ be a compact metric space and $d(f(x),f(y)) < d(x,y) $ for all $ x\neq y$ Prove that if $f$ is a continuous fuction then there is a unique $x_0 \in M $ such that $f(x_0)=x_0$
I know that since $g(x) = d((f(x),f(y))$ is continuous then a continuous function in a compact space reaches its maximum and minimum. Then by proving that the minimum of $g(x)$ = 0 can I conclude that $x_0$ is a fixed point?
The map $h:M\to \mathbb R$ defined by $h(x)=d(x,f(x))$ is continuous. To see this, note $$ d(x,f(x))\le d(x,y)+d(y,f(y))+d(f(y),f(x))<d(x,y)+d(y,f(y))+d(x,y) $$ so $h(x)-h(y)< 2d(x,y)$. The same logic implies $h(y)-h(x)<2d(x,y)$, and continuity follows.
Let $x^*$ attain the minimum value of $h$. I claim that $h(x^*)=0$. Suppose not; then $d(x,f(x))\neq 0$, so $x\neq f(x)$, implying $$ h(f(x^*))=d(\,\,f(x^*),f(f(x^*))\,)<d(x^*,f(x^*))=h(x^*), $$ contradicting the minimality of $h$.
Therefore, $h(x^*)=0$, so $f(x^*)=x^*$. To prove uniqueness, let $x$ and $y$ be two distinct fixed points, and use the contraction property to get a contradiction.