Contractive Operators on Compact Spaces

Suppose that $T: M \to M$ is a compact contractive Operator on a nonempty compact subset $M$ of a complete metric space $X$. Show that $T$ has a unique fixed point. Further show that the sequence defined by $x_{n+1}=Tx_n$ converges to the fixed point from an arbitrary point $x_0 \in M$. By a contractive operator I mean there exists a $1 \gt k \ge 0$ such that $d(Tx,Ty) \le k d(x,y)$.

My try: For the first part Let $S=\{(x,y): 0 \lt a \le d(x,y) \le b\}$. Let $f: M \times M \to K$ such that $f(x,y)=\frac{d(Tx,Ty)}{d(x,y)}$. $f$ is continuous on $S$ and $S$ being compact, $f$ attains its maximum say $K(a,b) \lt 1$ . Then by generalized fixed point theorem (Generalized Fixed Point Theorem) $T$ has a unique fixed point.

I have trouble showing the second part. Since $M$ is compact, every such $x_n$ will have a convergent subsequence say $x_{n_{k}}$ which goes to say $x'$. I need to show that all convergent subsequences go to $x$ which is the fixed point and somewhere use $x_{n+1}=Tx_n$. I am unable to do so.

Thanks for the help!!

Solution 1:

If I understand correctly, the assumption is that $d(T(x),T(y)<d(x,y)$ whenever $x\neq y$. Incidentally, the results you want to prove constitute what is called Edelstein's theorem, of which you can find lots of proofs and generalizations by googling a bit.

For the existence of the fixed point, consider the function $f(x):=d(x,T(x))$. This is a continuous function (because $T$ is continuous), so it has a minimum at some point $a\in M$ because $M$ is compact. Then $T(a)=a$ since otherwise we would have $d(T(a), T(T(a))<d(a,T(a))$, contradicting the minimality of $f(a)$.

The uniqueness of the fixed point is easy: if $a,b$ are two fixed points, then we must have $b=a$ since otherwise we would get the contradiction $d(a,b)=d(T(a),T(b))<d(a,b)$.

For the last part, I hope the following proof is correct. Let $x_0\in M$ and $x_n:=T^n(x_0)$, $n\geq 0$.

Let us first show that $d(x_n,T(x_n))\to 0$. Note that $d(x_n,T(x_n))$ is non-increasing (and even strictly decreasing if all the $x_n$ are distinct) by the assumption on $T$; so $d(x_n,T(x_n))$ has a limit $\delta\geq 0$. Then $d(x_n,T(x_n))\geq \delta$ for all $n$ and $\delta=\inf_n d(x_n,T(x_n))$. Denote by $K$ the closure of the set $X:=\{ x_n;\; n\geq 0\}$. Then $d(x,T(x))\geq \delta$ for all $x\in K$, because this is true for all $x\in X$ and $T$ is continuous. Also, the set $K$ is invariant under $T$, i.e. $T(K)\subseteq K$ (because $T(X)\subseteq X$ and $T$ is continuous). Since $K$ is compact, we may apply what has already been proved there is a fixed point $a$ for $T$ in $K$. But $d(a,T(a))\geq\delta$, so $\delta=0$.

Once we know that $d(x_n,T(x_n))\to 0$, it is easy to show by a compactness argument that $x_n\to a$ (the unique fixed point of $T$). Indeed, any convergent subsequence $(x_{n_k})$ of $(x_n)$ must in fact converge to $a$, because if $b=\lim x_{n_k}$, then $d(b,T(b))=\lim d(x_{n_k},T(x_{n_k}))=0$ and hence $T(b)=b$. So, $a$ is the only cluster point of the sequence $(x_n)$, and hence $x_n\to a$ by compactness.

Solution 2:

It is the same proof as in the Banach fixed point theorem. The idea is to show that the sequence $x_n$ is Cauchy in $X$ and thus it is convergent, say to $x\in X$. Because all $x_n$ are in $M$ and $M$ is closed, then $x\in M$. Now you have to show that this $x$ is indeed a solution to $Tx=x$. For this write $d(Tx_n,Tx)\leq k d(x_n,x)\to 0$ and therefore $d(x_{n+1},Tx)\to 0$, but also $d(x_{n+1},x)\to 0\Rightarrow Tx=x$.

Showing that the sequence $\{x_n\}$ is Cauchy: $$d(x_{n+1},x_n)=d(Tx_n,Tx_{n-1})\leq kd(x_n,x_{n-1})\leq ...\leq k^nd(x_1,x_0)$$ Therefore, by the triangle inequality and the sum formula for geometric series you get $$d(x_n,x_{n+m})\leq d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+...+d(x_{n+m-1},x_{n+m})$$ $$\leq (k^n+k^{n+1}+...+k^{n+m-1})d(x_1,x_0)$$ $$\leq k^n(1+k+k^2+...)d(x_1,x_0)=\frac{k^n}{1-k}d(x_1,x_0)\to 0$$ as $n\to\infty$, because $k<1$