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Let $f:\Bbb R \longrightarrow \Bbb R$ be a function, and $ \lambda \in (\frac {1}{2},1)$.

For all $ x,y \in \Bbb R$ we have

$$\lambda |x-f(x)| \leq |x-y| \Longrightarrow |f(x)-f(y)| \leq |x-y|.$$

Does any $\mu \geq 1$ exist such that: $$|x-f(y)| \leq \mu |x-f(x)|+|x-y|\text{ ?}$$

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Isn't this homework? What have you tried? –  tomasz Dec 20 '12 at 19:06

1 Answer 1

Hint:

$$|x-f(y)|=|(x-f(x))+(f(x)-f(y))|\leq |x-f(x)|+|f(x)-f(y)|.$$

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How can I use this equation:$$\lambda|x−f(x)|\leq|x−y| \Longrightarrow |f(x)−f(y)|\leq|x−y|.$$ –  sooshky Dec 21 '12 at 11:41

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