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Let $(X,d)$ be nonempty compact metric space and $f : X\to X$ be a function satisfying $d(f (x), f (y)) < d(x, y)$ for all distinct pair of points $x, x \in X$. Show that $f$ have a fixed point and this fixed point is unique.

Hint: Define the function $g : X \to R$ by $g(x) = d(f (x), x)$. Assume that $f (x) \neq x$ for all $x \in X$. Use compactness to show the existence of a point $a \in X$ such that $g(a) \leq g (x)$ for all $x \in X$. Deduce a contradiction by considering $x=f (a)$.

  • 1
    Have you tried applying the hint?2012-12-02
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    That hint is a solution. Don't see a question here.2012-12-02
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    I also thought that the hint is solution. However, the question is this and i guess that the aim is this question to prove it which i could not. That's why i asked this question.2012-12-02
  • 1
    Have you tried applying the hint?2012-12-02
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    I think i made a mistake because by applying the hint i could not prove anything. How can the hint make the question proved while it is the answer?2012-12-02

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