Let \((X, d)\) be a metric space. If \(f\) satisfies the condition
\[ d(f(x), f(y)) < d(x, y) \]
for all \(x, y \in X\) with \(x \neq y\), then \(f\) is called a shrinking map. If there is a number \(\alpha < 1\) such that
\[ d(f(x), f(y)) \leq \alpha d(x, y) \]
for all \(x, y \in X\), then \(f\) is called a contraction. A fixed point of \(f\) is a point \(x\) such that \(f(x) = x\).
If \(f\) is a contraction and \(X\) is compact, show \(f\) has a unique fixed point. [Hint: Define \(f^1 = f\) and \(f^{n+1} = f \circ f^n\). Consider the intersection \(A\) of the sets \(A_n = f^n(X)\).]
Show more generally that if \(f\) is a shrinking map and \(X\) is compact, then \(f\) has a unique fixed point. [Hint: Let \(A\) be as before. Given \(x \in A\), choose \(x_n\) so that \(x = f^{n+1}(x_n)\). If \(a\) is the limit of some subsequence of the sequence \(y_n = f^n(x_n)\), show that \(a \in A\) and \(f(a) = x\). Conclude that \(A = f(A)\), so that \(\operatorname{diam} A = 0\).]
Let \(X = [0, 1]\). Show that \(f(x) = x - x^2/2\) maps \(X\) into \(X\) and is a shrinking map that is not a contraction. [Hint: Use the mean-value theorem of calculus.]
The result in (a) holds if \(X\) is a complete metric space, such as \(\mathbb{R}\); see the exercises of §43. The result in (b) does not: show that the map \(f: \mathbb{R} \to \mathbb{R}\) given by \(f(x) = [x + (x^2 + 1)^{1/2}]/2\) is a shrinking map that is not a contraction and has no fixed point.
Solution
Goal: Prove the existence and uniqueness of fixed points for contractions and shrinking maps on compact metric spaces, and analyze explicit counterexamples on non-compact and non-contraction spaces.
1. Part (a): Contractions on compact metric spaces have a unique fixed point.Proof:
1.1. Define \(A_n = f^n(X)\) for each \(n \ge 1\). Since \(X\) is compact and \(f\) is continuous, each \(A_n\) is a non-empty compact set, and \(A_{n+1} = f(A_n) \subseteq A_n\).
1.2. The intersection \(A = \bigcap_{n=1}^\infty A_n\) is non-empty by the finite intersection property of compact spaces.
1.3. For any \(n \ge 1\), the contraction bound gives \(\operatorname{diam}(A_n) \le \alpha^n \operatorname{diam}(X)\). Since \(\alpha < 1\), \(\operatorname{diam}(A_n) \to 0\) as \(n \to \infty\).
1.4. Hence \(\operatorname{diam}(A) = 0\), so \(A = \{x_0\}\) is a singleton.
1.5. Since \(f(A) \subseteq \bigcap_{n=1}^\infty f(A_n) = \bigcap_{n=1}^\infty A_{n+1} = A\), \(f(x_0) \in A\), which forces \(f(x_0) = x_0\).
1.6. If \(x \neq y\) were two distinct fixed points, \(d(x, y) = d(f(x), f(y)) \le \alpha d(x, y) < d(x, y)\), a contradiction. Thus the fixed point is unique.
2. Part (b): Shrinking maps on compact metric spaces have a unique fixed point (Edelstein’s Theorem).Proof:
2.1. Consider the function \(g: X \to \mathbb{R}\) defined by \(g(x) = d(x, f(x))\).
2.2. Since \(f\) is continuous and the metric \(d\) is continuous, \(g\) is continuous.
2.3. Because \(X\) is compact, by the Extreme Value Theorem there exists a point \(x_0 \in X\) at which \(g\) attains its absolute minimum: \(g(x_0) \le g(x)\) for all \(x \in X\).
2.4. Suppose for contradiction that \(x_0 \neq f(x_0)\). Let \(y_0 = f(x_0)\).
2.5. Because \(f\) is a shrinking map and \(x_0 \neq y_0\):\[g(y_0) = d(y_0, f(y_0)) = d(f(x_0), f(y_0)) < d(x_0, y_0) = d(x_0, f(x_0)) = g(x_0).\]
2.6. This contradicts the minimality of \(g(x_0)\), so \(x_0 = f(x_0)\).
2.7. Uniqueness follows because if \(x \neq y\) were two fixed points, \(d(x, y) = d(f(x), f(y)) < d(x, y)\), impossible.
3. Part (c): \(f(x) = x - x^2/2\) on \([0, 1]\) is a shrinking map but not a contraction.Proof:
3.1. The derivative is \(f'(x) = 1 - x\). For \(x \in [0, 1]\), \(0 \le f'(x) \le 1\), and \(f(0) = 0, f(1) = 1/2\), so \(f([0, 1]) = [0, 1/2] \subseteq [0, 1]\).
3.2. For distinct \(x, y \in [0, 1]\) with \(x < y\), the Mean Value Theorem gives:\[|f(y) - f(x)| = f'(c)|y - x| = (1 - c)|y - x|\]for some \(c \in (x, y)\). Since \(c > x \ge 0\), \(1 - c < 1\), so \(|f(y) - f(x)| < |y - x|\). Thus \(f\) is a shrinking map.
3.3. Since \(\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = f'(0) = 1\), \(\sup_{x \neq y} \frac{|f(x) - f(y)|}{|x - y|} = 1\), so no constant \(\alpha < 1\) exists. Thus \(f\) is not a contraction.
4. Part (d): \(f(x) = \frac{x + \sqrt{x^2+1}}{2}\) on \(\mathbb{R}\) is a shrinking map without fixed points.Proof:
4.1. The derivative is \(f'(x) = \frac{1}{2}\left(1 + \frac{x}{\sqrt{x^2+1}}\right)\).
4.2. Since \(|x| < \sqrt{x^2+1}\), we have \(-1 < \frac{x}{\sqrt{x^2+1}} < 1\), which implies \(0 < f'(x) < 1\) for all \(x \in \mathbb{R}\).
4.3. By the Mean Value Theorem, \(|f(x) - f(y)| = f'(c)|x - y| < |x - y|\) for all \(x \neq y\), so \(f\) is a shrinking map.
4.4. Since \(\lim_{x \to \infty} f'(x) = 1\), \(f\) is not a contraction.
4.5. Setting \(f(x) = x\) gives \(\frac{x + \sqrt{x^2+1}}{2} = x \iff \sqrt{x^2+1} = x\), which has no real solutions (since \(x^2 + 1 > x^2 \ge 0\)). Thus \(f\) has no fixed point. Q.E.D.