Assume $K$ is a compact metric space with metric $\rho$ and $A$ is a map from $K$ to $K$ such that $\rho (Ax,Ay) < \rho(x,y)$ 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 sequence ${x_n}$ 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 $\rho (Ax,Ay) < \rho(x,y)$?

  • 1
    $\begingroup$ (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 ... $\endgroup$
    – Neal
    Mar 10, 2012 at 13:09
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    $\begingroup$ @Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness. $\endgroup$ Mar 10, 2012 at 13:17
  • $\begingroup$ Oh, I totally missed "compact" in the question. My bad. $\endgroup$
    – Neal
    Mar 11, 2012 at 0:24

3 Answers 3


Define $f(x):=\rho(x,A(x))$; it's a continuous map. (Note $$\rho(x,Ax)\le\rho(x,y)+\rho(y,Ay)+\rho(Ay,Ax)\quad\forall x, y\in K$$ or $$\rho(x,Ax)-\rho(y,Ay)\le\rho(x,y)+\rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$\left|\rho(x,Ax)-\rho(y,Ay)\right|\le\rho(x,y)+\rho(Ax,Ay)<2\delta \quad \text{ whenever }\rho(x,y)<\delta.$$ That is, $f$ is actually uniformly continuous.)

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 $x,y,$: $\rho(Ax,Ay)\leq c\rho(x,y)$.

  • $\begingroup$ Nice proof!Thank you! :)) $\endgroup$
    – Proton
    Mar 10, 2012 at 13:53
  • $\begingroup$ How do we show that f(x):=ρ(x,A(x)) is indeed continuous? $\endgroup$
    – Jacques
    Apr 2, 2012 at 9:22
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    $\begingroup$ @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. $\endgroup$
    – t.b.
    Apr 2, 2012 at 9:52
  • $\begingroup$ Can someone clarify about uniqueness? $\endgroup$
    – Niebla
    Nov 9, 2015 at 2:57
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    $\begingroup$ @Niebla In general if we have $\rho(A(x), A(y))<\rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $\rho(A(a), B(b))<\rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal. $\endgroup$
    – Jack M
    Dec 27, 2015 at 16:53

I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:

Since we are told that $K$ is a compact set. $f:K\rightarrow K$ being continuous implies that the $\mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $\inf_{x\in K} f(x)$.

If it is possible to show that $f(K) \subseteq K$ is a closed set, then it is necessarily compact as well: A subset of a compact set is compact? However, I am not aware of how you would do this in this case without relying on continuity of $f$.


you don't need to prove completeness or define any sequence. Define a nonnegative real function $$ h(x) = \rho(x,f(x) ) $$ This is continuous, so its minimum is achieved at some point $x_0.$ If $h(x_0) >0,$ we see that $$ h(f(x_0) ) = \rho( f(x_0), f(f(x_0 )) < \rho( x_0, f(x_0)) = h(x_0) $$ Put together, $$ h(f(x_0) ) < h(x_0) $$ Thus the assumption of a nonzero minimum of $h$ leads to a contradiction. Therefore the minimum is actually $0,$ so $h(x_0) = 0,$ so $f(x_0) = x_0 $

  • $\begingroup$ Please do not post the same answer to multiple questions. If you believe that an answer is appropriate for more than one question, please post only one answer, and nominate the other question for closure as a duplicate. $\endgroup$
    – Xander Henderson
    Feb 17 at 23:22

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