Prove the map has a fixed point

Question

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 $\rho (Ax,Ay) < \rho(x,y)$?

Answer 1

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.

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

Answer 2

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 $f$.

Answer 3

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 $