Krasnoselskii contraction mapping

Question

Definition: Give metric space $(X,d)$. Mapping $f: X \to X$ is called Krasnoselskii contraction mapping if $$\forall a,b>0, a<b, \exists k \in (0,1): a \le d(x,y) \le b \Rightarrow d(f(x),f(y)) \le kd(x,y).$$ Problem: Prove mappings below is Krasnoselskii contraction mapping

1. Mapping $f: X \to X$ satisfy $$d(f(x),f(y)) \le \varphi(d(x,y)),$$ where $\varphi: [0,+\infty) \to [0,+\infty)$ is continuous, $\varphi(0)=0$ and $0 < \varphi(t) < t$ for all $t>0$.
2. $X$ is compact space and $f:X \to X$ satisfy $$d(f(x),f(y)) < d(x,y), \forall x \neq y$$

My attempt:

1. By setting $\psi(t)=\dfrac{\varphi(t)}{t}, t\in[a,b]$, I have proved that $f$ is Krasnoselskii contraction mapping.

2. I don't know how to use $X$ is compact to solve the problem.

Answer 1

Suppose this is false. There exist $x_k,y_k$ such that $a \leq d(x_k,y_k) \leq b$ and $d(f(x_k),f(y_k)) >(1-\frac 1 k) d(x_k,y_k)$. By compactness $(x_k)$ and $(y_k)$ converge to some $x$ and $y$ along a subsequence. Let the limits be $x$ and $y$ . Then $ d(f(x),f(y) \geq d(x,y)$. This contradicts the hypothesis since $d(x,y) \geq a >0$ so $x \neq y$.