# show that $d(a, f(a)) \leq d(x, f(x)) \leq d(b, f(b)$

Let $$(E,d)$$ be a compact metric space and $$g:(E, d) \rightarrow(\mathbb{R},|.|), x \mapsto d(x, f(x))$$, such that $$f:(E, d) \rightarrow(E, d)$$ verify: $$d(f(x), f(y))
show that there exists $$a,b\in E$$ $$d(a, f(a)) \leq d(x, f(x)) \leq d(b, f(b)),\forall x \in E$$
my attempt:
since E is compact then $$a\leq x \leq b$$ and since g is continuous so $$f(E)$$ is compact so $$f(a)\leq f(x) \leq f(b)$$ Hence $$d(a, f(a)) \leq d(x, f(x)) \leq d(b, f(b))$$

• You are using the function $f$ in the definition of $f$ so the question doesn't make much sense. Sep 15 at 0:45
• @GiorgosGiapitzakis sorry wait a minute Sep 15 at 0:47
• @GiorgosGiapitzakis this is the third question in the problem that's why i missed that Sep 15 at 0:53
• Actually, $g$ is also continuous, since the metric $d:E\times E\to \mathbb{R}_+$ is always continuous. Sep 15 at 0:59
• @Z.Zhu yes it's lipschitz function Sep 15 at 1:01