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: $d(f(x), f(y))<d(x, y) \text { if } x \neq y$
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 $d(a, f(a)) \leq d(x, f(x)) \leq d(b, f(b))$

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

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