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Let $f : [a,b] \to \mathbb{R}$ a function so that:

1) $f([a,b]) \subset [a,b]$;

2) $\forall x,y \in [a,b] : x \neq y \Rightarrow |f(x)-f(y)| < |x-y|$

I’d like to prove that $f$ is continuous. I can’t really get started however can anyone enlighten me please? Thanks

Gabrielek
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Pablito
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    Hint: to prove that $f(x)$ is continuous at some point $x_0$, let $\epsilon > 0$ be given. You need to find $\delta$ such that $| x - x_0 | < \delta \Rightarrow |f(x) - f(x_0)| < \epsilon$. The given inequality (2) is really nice and gives you a simple choice for $\delta$ in terms of $\epsilon$. – hunter Sep 01 '19 at 13:52
  • I tried with $\delta = \epsilon >0$ and it seems to work! Was it the right idea? Thus $\forall x \neq x_{0}, |f(x)-f(x_{0}|<|x-x_{0}|<\delta$ – Pablito Sep 01 '19 at 14:01
  • yup, that's exactly it! – hunter Sep 01 '19 at 14:03
  • Okay thanks a lot! – Pablito Sep 01 '19 at 14:03

1 Answers1

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1) says that $f$ is bounded

2) says that $f$ is Lipschitz of Lipschitz's costant less than one

You can now finish because every Lipschitz function is continuous (you can say more: uniformly continuous)

Gabrielek
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