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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?2012-12-20

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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|.$2012-12-21