Bounded metric in compact metric space with continuous function

Question

Let $X$ be a compact metric space with metric $d$ and $f:X \to X$ a continuous map so that $f(x)$ never equals $x$.

How do I show that the function $g(x) :=d(f(x),x)$ is continuous?

Answer 1

The hard way : Let $\epsilon >0$. As $f$ is continuous there is a $\tau >0$ such that $d(x,y) < \tau$ implies $d(f(x),f(y)) <\epsilon/2$. Set $\delta =\min(\tau, \epsilon/2)$

Using the reverse triangle inequality, we get for $d(x,y) < \delta$ $$|d(x,f(x)) - d(y,f(y))| \\ =|d(x,f(x))-d(x,f(y))+ d(x,f(y))- d(y,f(y))| \\ \le |d(x,f(x))-d(x,f(y))|+ |d(x,f(y))- d(y,f(y))|\\ \le d(f(x),f(y)) + d(x,y)= \epsilon/2+ \epsilon/2= \epsilon$$

The elegant way : see Viktor Glombik's comment

Answer 2

Let $x_0 \in X$ and $(x_n)_{n \in \mathbb{N}}$ be a sequence in $X$ with $x_n \to x_0$. Since $f$ is continuous, we have $f(x_n) \to f(x_0).$ Since $d$ is continuous, it follows that $$ g(x_n) = d(f(x_n),x_n) \to d(f(x_0),x_0)=g(x_0). $$ This shows that $g$ is continuous in $x_0$. Since $x_0$ was arbitrary, $g$ is continuous on $X$.