Let \(K\) be a nonempty compact subset of a metric space with distance \(d\). Suppose \(\varphi:K\to K\) satisfies \[ d(\varphi(x),\varphi(y))<d(x,y) \] for all distinct \(x,y\in K\). Show that \(\varphi\) has exactly one fixed point in \(K\).
Solution
1. For all \(x,y\in K\),\[ d(\varphi(x),\varphi(y)) \leq d(x,y). \]
If \(x\neq y\), this follows from the strict inequality in the hypothesis. If \(x=y\), both sides are zero.
2. The map \(\varphi:K\to K\) is continuous.
Step 1 gives \[ d(\varphi(x),\varphi(y)) \leq d(x,y) \] for all \(x,y\in K\). Thus \(\varphi\) is \(1\)-Lipschitz and hence continuous.
3. The function\[ F:K\longrightarrow\RR, \qquad F(x)=d(x,\varphi(x)), \]is continuous.
The metric \[ d:K\times K\longrightarrow\RR \] is continuous, and step 2 gives continuity of \(\varphi\). Therefore \[ x\longmapsto d(x,\varphi(x)) \] is continuous.
4. There is a point \(x_0\in K\) at which \(F\) attains its minimum.
3.The extreme value theorem therefore gives\[ F(x_0) = \min_{x\in K}F(x) \]for some \(x_0\in K\).
5. One has\[ F(x_0)=0. \]
4. Hence\(F(x_0)=0\).
6. The point \(x_0\) is a fixed point of \(\varphi\).
By step 5, \[ d(x_0,\varphi(x_0))=0. \] A metric separates points, so \[ \varphi(x_0)=x_0. \]
7. The fixed point is unique.
Suppose \(x,y\in K\) are fixed points and \(x\neq y\). Then the hypothesis gives \[ d(x,y) = d(\varphi(x),\varphi(y)) < d(x,y), \] which is impossible. Thus no two distinct fixed points exist.
8. Therefore \(\varphi\) has exactly one fixed point:\[ \boxed{ \exists!\,x\in K \text{ such that } \varphi(x)=x. } \]
Existence is step 6 and uniqueness is step 7.
9. Q.E.D.
Step 8 is the required conclusion.