# Prove the map has a fixed point

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) 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
• @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

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

• Nice proof!Thank you! :)) – Jiangnan Yu Mar 10 '12 at 13:53
• How do we show that f(x):=ρ(x,A(x)) is indeed continuous? – Jacques Apr 2 '12 at 9:22
• @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
• Can someone clarify about uniqueness? – Niebla Nov 9 '15 at 2:57
• @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. – Jack M Dec 27 '15 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 $$f$$.