Let \((M,d)\) be a nonempty complete metric space and let \(S:M\to M\). Suppose \(S^2=S\circ S\) is a strict contraction: there exists \(\lambda<1\) such that \[ d(S^2(x),S^2(y))\le \lambda d(x,y) \] for all \(x,y\in M\). Show that \(S\) has a unique fixed point in \(M\).
Solution
1. The map \(S^2\) has a unique fixed point \(p\in M\).
If \(M\) consists of a single point, the conclusion is immediate. Otherwise there exist distinct \(x,y\in M\), and the contraction inequality forces \(\lambda\geq0\), since its left-hand side is nonnegative while \(d(x,y)>0\). Thus \[ 0\leq\lambda<1. \] By hypothesis, \(S^2:M\to M\) is therefore a contraction on the nonempty complete metric space \(M\). The Banach fixed-point theorem gives a unique point \(p\in M\) such that \[ S^2(p)=p. \]
2. The point \(S(p)\) is also fixed by \(S^2\).
Using \(S^2(p)=p\), \[ S^2(S(p)) = S^3(p) = S(S^2(p)) = S(p). \] Thus \(S(p)\) is a fixed point of \(S^2\).
3. The point \(p\) is fixed by \(S\).
By step 1, \(S^2\) has only one fixed point. Step 2 shows that \(S(p)\) is such a fixed point, while \(p\) itself is one. Hence \[ S(p)=p. \]
4. The fixed point of \(S\) is unique.
If \(q\in M\) satisfies \[ S(q)=q, \] then \[ S^2(q)=S(q)=q. \] Thus every fixed point of \(S\) is a fixed point of \(S^2\). Step 1 says the latter has only the fixed point \(p\), so \(q=p\).
5. Q.E.D.
Steps 3 and 4 give existence and uniqueness of the fixed point of \(S\).