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Let (C, ||.||) be a closed subset of a Banach space, and let f: C -> C be a mapping such that

||f(x) - f(y)|| < ||x - y|| for all x, y in C. Must there exist a fixed point in C that f maps to itself?

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Hint $C=[2, \infty) \subset \mathbb R$ and $$f(x)=x+\frac{1}{x}$$

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  • $\begingroup$ :Your example is not true, $\left \| f(x)-f(y) \right \|=\left \| x-\frac{1}{x}-(y-\frac{1}{y}) \right \|=\left \| x-y \right \|(1+\frac{1}{xy})$ , I think $f(x)=x+\frac{1}{x}$ will be OK.. $\endgroup$ – Motaka May 1 at 10:58

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