鲁丁《数学分析原理》第四章第14题：证明连续映射不动点存在性

Problem Statement #

Let ( I = [0,1] ), and suppose ( f: I \to I ) is a continuous function. Prove there exists at least one ( x \in I ) such that ( f(x) = x ) (a fixed point).

Note: For any interval ( [A,B] \subseteq \mathbb{R} ), define ( \text{first}([A,B]) = A ) and ( \text{second}([A,B]) = B ).

Proof #

Let’s start by defining ( M = \sup f(I) ) (the maximum value of ( f ) on ( I )) and ( m = \inf f(I) ) (the minimum value). Since ( f ) maps ( I ) to itself, we know ( 0 \leq m \leq M \leq 1 ).

Case 1: ( M = m )

If ( f ) is constant (all outputs equal to ( c = m = M )), then ( c \in I ), so ( f(c) = c ). We’ve immediately found our fixed point.

Case 2: ( M \neq m ), Assume No Fixed Points (for Contradiction)

Suppose for contradiction that there is no ( x \in I ) with ( f(x) = x ).

First, since ( f ) is continuous and has no fixed points, it must be strictly monotonic. Why? If ( f ) weren’t strictly monotonic, the Intermediate Value Theorem would guarantee a point where ( f(x) ) crosses the line ( y=x )—which would be a fixed point, contradicting our assumption. So ( f ) is either strictly increasing or strictly decreasing.

Next, we construct a nested sequence of closed intervals:

• Let ( L_1 = [f(m), f(M)] )
• For ( n \geq 2 ), define ( L_n = [f(\text{first}(L_{n-1})), f(\text{second}(L_{n-1}))] )

We need to show these intervals are nested (each ( L_n \subseteq L_{n-1} )) and non-empty, so their infinite intersection ( V = \bigcap_{n=1}^\infty L_n ) is non-empty (by the Nested Interval Theorem).

Let’s verify nesting for strictly increasing ( f ) (the strictly decreasing case is similar, just reverse inequalities):

• Since ( m = \inf f(I) ), ( f(m) \geq m ) (if ( f(m) < m ), then ( f(M) \leq M ), and continuity would force a fixed point between ( m ) and ( M )). Similarly, ( f(M) \leq M ). Thus ( L_1 = [f(m), f(M)] \subseteq [m, M] ).
• For ( L_2 = [f(f(m)), f(f(M))] ): since ( f ) is increasing, ( f(f(m)) \geq f(m) ) (because ( f(m) \geq m ), so applying ( f ) preserves the inequality) and ( f(f(M)) \leq f(M) ) (since ( f(M) \leq M )). So ( L_2 \subseteq L_1 ).
• By induction, every ( L_n \subseteq L_{n-1} ): each interval’s endpoints are mapped by ( f ) to endpoints of the next interval, and strict monotonicity keeps the new interval inside the previous one.

By the Nested Interval Theorem, ( V ) is non-empty. Let ( c \in V ). Now, consider the sequences of endpoints of ( L_n ): let ( a_n = \text{first}(L_n) ) and ( b_n = \text{second}(L_n) ). For strictly increasing ( f ), ( a_n ) is an increasing sequence bounded above by ( b_1 ), so it converges to some ( a \in V ). Similarly, ( b_n ) is a decreasing sequence bounded below by ( a_1 ), converging to some ( b \in V ).

Since ( f ) is continuous, ( f(a) = f(\lim_{n \to \infty} a_n) = \lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} a_{n+1} = a ). Wait a minute—this means ( f(a) = a ), which directly contradicts our assumption that there are no fixed points!

This contradiction proves our initial assumption (no fixed points) is wrong. Therefore, there must exist at least one ( x \in I ) such that ( f(x) = x ).

内容的提问来源于stack exchange，提问作者HAT