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Assume $K$ is a compact metric space with metric $\rho$ and $A$ is a map from $K$ to $K$,$\rho (Ax,Ay) < \rho(x,y)$ for $x\neq y$.Prove A have a unique fixed point in K.

The uniqueness is easy.My problem is to show that there a exist fixed point.$K$ is compact so every sequence has convergent subsequence.Construct a sequance ${x_n}$ by $x_{n+1}=Ax_{n}$,${x_n}$ has a convergent subsequence ${ x_{n_k}}$but how to show there is a fixed point using $\rho (Ax,Ay) < \rho(x,y)$?

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(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ... –  Neal Mar 10 '12 at 13:09
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@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness. –  Brian M. Scott Mar 10 '12 at 13:17
    
Oh, I totally missed "compact" in the question. My bad. –  Neal Mar 11 '12 at 0:24
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up vote 5 down vote accepted

Define $f(x):=\rho(x,A(x))$; it's a continuous map. Let $\alpha:=\inf_{x\in K}f(x)$, then we can find $x_0\in K$ such that $\alpha=f(x_0)$, since $K$ is compact. If $\alpha>0$, then $x_0\neq Ax_0$ and $\rho(A(Ax_0),Ax_0)<\rho(Ax_0,x_0)=\alpha$, which is a contradiction. So $\alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.


Note that completeness wouldn't be enough in this case, for example consider $\mathbb R$ with the usual metric, and $A(x):=\sqrt{x^2+1}$. It's the major difference between $\rho(Ax,Ay)<\rho(x,y)$ for $x\neq y$ and the existence of $0<c<1$ such that for all $x,y,$: $\rho(Ax,Ay)\leq c\rho(x,y)$.

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You want $\rho(x,Ax)$ in the definition of $f$. –  Brian M. Scott Mar 10 '12 at 13:18
    
Yes, I will correct it. Thanks. –  Davide Giraudo Mar 10 '12 at 13:18
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Also, I think you mean "completeness would NOT be enough in this case" –  Geoff Robinson Mar 10 '12 at 13:20
    
@GeoffRobinson Yes, other typo. Thank you too. –  Davide Giraudo Mar 10 '12 at 13:21
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@Jacques: $\delta: x \mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) \mapsto (x,A(y))$ is continuous, and $d:(x,y) \mapsto d(x,y)$ is continuous, so $f(x) = (d\circ g \circ \delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times. –  t.b. Apr 2 '12 at 9:52
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