Proving the existence of a root

Question

Let $ (X,d) $ be a compact metric space and $ f:X\to X $ a function with $ d(f(x),f(y))<d(x,y) $ for all $ x,y\in X $ with $ x\neq y $. I have the function $ g:X\to \mathbb{R}, x\mapsto d(f(x),x) $ which is Lipschitz-continuous. So $ g $ has a minimum which means there exists $ x'\in X $ with $$ \min\limits_{x\in X}\{g(x)\}=g(x')=d(f(x'),x')\geq 0. $$

I assume $ g(x')>0 $. I want to strive for a contradiction disprove this. How can I do this?

Answer 1

Assume $g(x') = \varepsilon>0$. Or in other words, $d(f(x'),x') = \varepsilon$. Then applying what we know about $f$, we get $d(f(f(x')),f(x')) < d(f(x'),x') = \varepsilon$. So, $g(f(x'))<\varepsilon$, which contradicts the assumption that $g(x')$ is the minimum. So $f(x')=x'$, giving your result.