Let $ f:(X, d)\mapsto (X, d ) $ be a mapping on compact metric space with $ d (f (x), f (y))<d (x,y) $for $ x\ne y $

Question

I prove that $ f $ has a fixed point. My question is whether the point is unique and the mapping $ f $ is continuous.

Answer 1

Suppose $x_0$, $x_1$ are fixed points of $f$, i.e. $f(x_0)=x_0$ and $f(x_1)=x_1$. If $x_0\neq x_1$, then we can use the the inequality $d(f(x),f(y))<d(x,y)$ with $x=x_0$ and $y=x_1$, we have $$d(x_0,x_1)=d(f(x_0),f(x_1))<d(x_0,x_1),$$ which is a contradiction. Therefore, we must have $x_0=x_1$, i.e. there exists a unique fixed point of $f$.

Answer 2

Paul has already shown uniqueness. Your mapping $f$ is indeed continuous. Theorem 10.1 in Munkres Topology: A First Course says that $f$ is continuous if, given any $x$ and a real $\epsilon > 0,$ there exists a $\delta > 0$ such that $$ d(x,y) < \delta \; \; \Longrightarrow \; \; d(f(x), f(y)) < \epsilon. $$ In this case, given $\epsilon$ we can simply take $\delta = \epsilon.$

Which leads to the shortest proof of the existence of a fixed point: the mapping $$ g(x) = d(x, f(x)) $$ is also continuous, and is real valued. As the space is compact, $g$ achieves its minimum, at some $x_0.$

If we assume $g(x_0) > 0,$ then calculate $$ g(f(x_0)) = d(f(x_0), f(f(x_0)) ) < d(x_0, f(x_0)) = g(x_0), $$ which is a contradiction of the minimality of $g(x_0).$ So, in fact, $g(x_0) = 0$ and so $f(x_0) = x_0$ shows the fixed point.