Let f : [a, b]→[a,b] be an arbitrary function. (i) Prove that if f is continuous, then f has a fixed point, that is, there exists x 0 ∈

4.5.1. Let f : [a, b]→[a,b] be an arbitrary function. (i) Prove that if f is continuous, then f has a fixed point, that is, there exists x 0 ∈

[a, b] such that f (x 0 )=x 0 .

Fig. 4.6 Brouwer’s fixed-point theorem (one-dimensional case).

4.5 Fixed Points 155

Fig. 4.7 Fixed-point theorem (case of increasing functions).

(ii) Show that the result remains true if f is nondecreasing. (iii) Find a decreasing function f : [a, b]→[a,b] with no fixed points.

Solution. (i) Consider the continuous function g : [a, b]→R defined by g(x) =

f (x) −x. We have g(a) = f (a)−a ≥ 0 and g(b) = f (b)−b ≤ 0. Since g(a)g(b) ≤ 0 and g is continuous, there exists x 0 ∈ [a,b] such that g(x 0 ) = 0, that is, f (x 0 )=x 0 . We observe that in order to deduce that g vanishes, it is enough to show that g has the intermediate value property. In the proof, we have deduced this by using the continuity of g, as the difference of continuous functions. In the general case, by Sierpi´nski’s theorem, it is not true that the difference of two functions having the intermediate value property is a function with the same property.

(ii) Set

A = {a ≤ x ≤ b; f (x) ≥ x}

and x 0 = sup A. The following situations may occur: (a) x 0 ∈ A. By the definition of x 0 it follows that f (x 0 )≥x 0 . If f (x 0 )=x 0 , then

the proof is concluded. If not, we argue by contradiction and assume that

f (x 0 )>x 0 . By the definition of x 0 we obtain f (x) < x, ∀x > x 0 . On the other hand, for any x 0 < x < f (x 0 ) we have x > f (x) ≥ f (x 0 ), contradiction, since

Fig. 4.8 A fixed-point theorem that is true for increasing functions but that fails in the decreasing case.

156 4 Continuity x ∈ (x 0 , f (x 0 )), that is, x < f (x 0 ). It follows that the assumption f (x 0 )>x 0 is

false, so f has a fixed point. (b) x 0 0 0 ∈ A, which reduces the problem to the case (a). If x 0 uence x n →x 0 ,x n <x 0 , such that x n ∈ A. Since f is increasing, it follows that lim n → ∞ f (x n )=x 0 . On the other hand, from f (x 0 )<x 0 we deduce that there exists x n <x 0 such that f (x n ) > f (x 0 ), contradiction with the fact that f is increasing.

(iii) Consider the function

A related counterexample is depicted in Figure 4.8. We point out some simple facts regarding the hypotheses of the Brouwer fixed-

point theorem on the real axis. (i) While a fixed point in [a, b] exists for a continuous function f : [a, b]→[a,b], it

need not be unique. Indeed, any point x ∈ [a,b] is a fixed point of the function

f : [a, b]→[a,b] defined by f (x) = x. (ii) The condition that f be defined on a closed subset of R is essential for the existence of a fixed point. For example, if f : [0, 1)→R is defined by f (x) = (1 + x)/2, then f maps [0, 1) into itself, and f is continuous. However, f has no fixed point in [0, 1).

(iii) The condition that f be defined on a bounded subset of R is essential for the existence of a fixed point. For example, if f : [1, ∞)→R is defined by

f (x) = x + x −1 , then f maps [1, ∞) into itself, f is continuous, but f has no fixed point in [1, ∞). (iv) The condition that f be defined on an interval in R is essential for the existence of a fixed point. For example, if D = [−2,−1] ∪ [1,2] and f : D→R is defined by f (x) = −x, then f maps D into itself, f is continuous, but f has no fixed point in D.

We have just observed that if f : [a, b] → [a,b] is a continuous function then f must have at least one fixed point, that is, a point x ∈ [a,b] such that f (x) = x.

A natural question in applications is to provide an algorithm for finding (or approximating) this point. One method of finding such a fixed point is by suc- cessive approximation. This technique is due to the French mathematician Emile

Picard. More precisely, if x 1 ∈ [a,b] is chosen arbitrarily, define x n +1 = f (x n ), and the resulting sequence (x n ) n ≥1 is called the sequence of successive approximations of f (or a Picard sequence for the function f ). If the sequence (x n ) n ≥1 converges to some x, then a direct argument based on the continuity of f shows that x is a fixed point of f . Indeed,

4.5 Fixed Points 157

n −1 ) = lim x n →∞ n =x.⊓ ⊔ The usual method of showing that the sequence (x n ) n ≥1 of successive approx-

imations converges is to show that it satisfies the Cauchy convergence criterion: for every ε > 0 there is an integer N such that for all integers j, k ≥ N, we have |x j −x k |< ε . The next exercise asserts that it is enough to set j = k + 1 in the Cauchy criterion.

4.5.2. Let f : [a, b] → [a,b] be a continuous function. Let x 1 be a point in [a, b] and let (x n ) n ≥1 denote the resulting sequence of successive approximations. Then the sequence (x n ) n ≥1 converges to a fixed point of f if and only if lim n → ∞ (x n +1 −x n )=0 .

Solution. Clearly lim n → ∞ (x n +1 −x n ) = 0 if (x n ) n ≥1 converges to a fixed point. Suppose lim n → ∞ (x n +1 −x n ) = 0 and the sequence (x n ) n ≥1 does not converge. Since [a, b] is compact, there exist two subsequences of (x n ) n ≥1 that converge to ξ 1 and ξ 2 respectively. We may assume ξ 1 < ξ 2 . It suffices to show that f (x) = x for all x ∈( ξ 1 , ξ 2 ). Suppose this is not the case; hence there is some x ∗ ∈( ξ 1 , ξ 2 ) such that f (x ∗

∗ . Then a δ > 0 can be found such that [x ∗ − δ ,x ∗ + δ ]⊂( ξ 1 , ξ 2 ) and

f ∗ − δ ,x ∗ + δ ). Assume ˜x − f ( ˜x) > 0 (the proof in the other case being analogous) and choose N such that |f n (x) − f n +1 (x)| < δ for n > N. Since ξ 2 is a cluster point, there exists a positive integer n > N such that f n (x) > x ∗ . Let n 0 be the smallest such integer. Then, clearly,

f n 0 −1 (x) < x ∗ <f n 0 (x),

and since f n 0 (x) − f n 0 −1 (x) < δ , we must have

f n 0 −1 n (x) − f 0 (x) > 0, so that f n 0 (x) < f n 0 −1 (x) < x ∗ ,

a contradiction. ⊓ ⊔

4.5.3. Let f : [0, 1]→[0,1] be a continuous function such that f (0) = 0 , f (1) = 1 . Define f n : = f ◦ f ◦ ··· ◦ f ( n times) and assume that there exists a positive integer m such that f m (x) = x for all x ∈ [0,1] . Prove that f (x) = x for any x ∈ [0,1] .

Solution. Our hypothesis implies that f is one-to-one, so increasing (since f is continuous). Assume, by contradiction, that there exists x ∈ (0,1) such that f (x) > x. Then, for any n ∈ N, we have f n (x) > f n −1 (x) > ··· > f (x) > x. For n = m we obtain

a contradiction. A similar argument shows that the case f (x) < x (for some x) is not possible. ⊓ ⊔

4.5.4. Let a , b be real numbers, a <b , and consider a continuous function f : [a, b]→R .

(i) Prove that if [a, b] ⊂ f ([a,b]) then f has a fixed point. (ii) Assume that there exists a closed interval I ′ ⊂ f ([a,b]) . Prove that I ′ = f (J) , where J is a closed interval contained in [a, b] .

158 4 Continuity

(iii) Assume that there exist n closed intervals I 0 ,...,I n −1 contained in [a, b] such that for all 0 ≤k≤n−2 , I k +1 ⊂ f (I k ) and I 0 ⊂ f (I n −1 ) . Prove that f n has a fixed point ( f n = f ◦ ··· ◦ f ).

Ecole Normale Sup´erieure, Paris, 2003 ´ Solution. (i) Define f ([a, b]) = [m, M] and let x m ,x M ∈ [a,b] be such that

f (x m ) = m and f (x M ) = M. Since f (x m )−x m ≤ 0 and f (x M )−x m ≥ 0, it follows by the intermediate value property that f has at least a fixed point.

(ii) Set I ′ = [c, d] and consider u, v ∈ I such that f (u) = c and f (v) = d. Assume, without loss of generality, that u ≤ v.

The set A = {x ∈ [u,v]; f (x) = c} is compact and nonempty, so there exists α = max{x; x ∈ A} and, moreover, α ∈ A. Similarly, the set B = {x ∈ [ α , v]; f (x) =

d } has a minimum point β . Then f ( α ) = c, f ( β ) = d and for all x ∈ ( α , β ) we have

f α , β )) and f (( α , β )) is an interval that contains neither c nor d. It follows that I ′ = f (J), where J =[ α , β ].

(iii) Since I 0 ⊂ f (I n −1 ), it follows by (ii) that there exists a closed interval J n −1 ⊂I n −1 such that I 0 = f (J n −1 ). But J n −1 ⊂I n −1 ⊂ f (I n −2 ). So, by (ii), there exists a closed interval J n −2 ⊂I n −2 such that J n −1 = f (J n −2 ). Thus, we

obtain n closed intervals J 0 ,...,J n −1 such that

J k ⊂I k , for all 0 ≤ k ≤ n − 1,

and J k +1 = f (J k ), for all 0 ≤ k ≤ n − 2 and I 0 = f (J n −1 ).

Consequently, J 0 is included in the domain of the nth iterate f n and J 0 ⊂I 0 =f n (J 0 ). By (i) we deduce that f n has a unique fixed point in J 0 . ⊓ ⊔

We have already seen that if f : [a, b] → [a,b] is a continuous function, then f has at least one fixed point. One method of finding such a point is by successive approximation, that is, for a point x 0 in [a, b], define x n +1 = f (x n ), and the resulting sequence (x n ) n ≥0 is called the sequence of successive approximations of f . If the sequence (x n ) n ≥0 converges, then it converges to a fixed point of f .

The usual method of showing that the sequence (x n ) n ≥0 of successive approx- imations converges is to show that it satisfies the Cauchy convergence criterion. The next result establishes that this happens if and only if the difference of two consecutive terms in this iteration converges to zero. The American mathematician Felix Browder has called this condition asymptotic regularity.

be a sequence of real numbers such that the sequence (x n +1 −x n ) converges to zero. Prove that the set of cluster points of (x n ) n ≥0 is a closed interval in R ¯ , possibly degenerate.

4.5.5. (i) Let (x n ) n ≥0

Barone’s theorem, 1939

4.5 Fixed Points 159

be a continuous function. Consider the sequence (x n ) n ≥0 defined by x 0 ∈ [a,b] and, for any positive integer n , x n = f (x n −1 ) . Prove that (x n ) n ≥0 converges if and only if (x n +1 −x n ) converges to zero.

(ii) Let f : [a, b]→[a,b]

Hillam’s fixed-point theorem, 1976 Solution. (i) Set ℓ − : = lim inf n → ∞ x n , ℓ + : = lim inf n → ∞ x n and choose a ∈

(ℓ − ,ℓ + ). By the definition of ℓ − , there exists x n 1 < a. Let n 2 be the least integer greater than n 1 such that x n 2 > a (the existence of n 2 follows by the definition of ℓ + ). Thus, x n 2 −1 ≤a<x n 2 . Since ℓ − < a, there exists a positive integer n 3 >n 2 such that x n 3 < a. Next, by the definition of ℓ + , there exists an integer N 4 >n 3 such that x N 4 > a. If n 4 denotes the least integer with these properties, then x n 4 −1 ≤a<x n 4 . In this manner we construct an increasing sequence of positive numbers (n 2k ) k ≥1 such that for all k ≥ 1, x n 2k −1 ≤a<x n 2k . Using the hypothesis we deduce that the sequences n 2k −1 k ≥1 and n 2k k ≥1 converge to a, so a is a cluster point.

(ii) Assume that the sequence of successive approximations (x n ) n ≥0 satisfies x n +1 −x n →0, as n→∞. With the same notation as above, assume that ℓ − <ℓ + . The proof of (i) combined with the continuity of f implies a = f (a), for all

a ∈ (ℓ − ,ℓ + ). But this contradicts our assumption ℓ − <ℓ + . Indeed, choose ℓ − <c<d<ℓ + and 0 < ε < (d − c)/3. Since x n +1 −x n →0, there exists N ε such that for all n ≥N ε , − ε <x n +1 −x n < ε . Let N 2 >N 1 >N ε

be such that x N 1 <c<d<x N 2 . Our choice of ε implies that there exists an integer n ∈ (N 1 ,N 2 ) such that a := x n ∈ (c,d). Hence x n +1 = f (a) = a, x n +2 = a, and

so on. Therefore x N 2 = a, contradiction. ⊓ ⊔

The reverse assertion is obvious. The following result is a particular case of a fixed-point theorem due to

Krasnoselski (see [61]).

4.5.6. Let f : [a, b]→[a,b] be a function satisfying | f (x)− f (y)| ≤ |x−y| , for all x , y ∈ [a,b] . Define the sequence (x n ) n ≥1 by x 1 ∈ [a,b] and, for all n ≥1 , x n +1 = [x n + f (x n )] /2 . Prove that (x n ) n ≥1 converges to some fixed point of f .

Solution. We observe that it is enough to show that (x n ) n ≥1 converges. In this case, by the recurrence relation and the continuity of f , it follows that the limit of (x n ) n ≥1 is a fixed point of f . We argue by contradiction and denote byA the set of all limit points of (x n ) n ≥1 , that is,

A : = n k ) k ≥1 of (x n ) n ≥1 such that x n k →ℓ By our hypothesis and the compactness of [a, b], we deduce that A contains at least

two elements and is a closed set. We split the proof into several steps.

(i) For any ε > 0 and n k ∈ N such that |x n k − ℓ| ≤ ε . Then

160 4 Continuity ℓ + f (ℓ) x n k + f (x n k )

and so on. This shows that |x n − ℓ| ≤ ε , for all n ≥n k . Hence (x n ) n ≥1 converges to ℓ, contradiction.

(ii) There exists ℓ 0 ∈ A such that f (ℓ 0 )>ℓ 0 . Indeed, arguing by contradiction, set ℓ − = min ℓ∈A ℓ. Then ℓ − ∈ A and f (ℓ − )≤ℓ − . The variant f (ℓ − )=ℓ − is excluded, by (i). But f (ℓ − )<ℓ − implies that [ℓ − + f (ℓ − )]/2 ∈ A and [ℓ − +

f (ℓ − )]/2 < ℓ − , which contradicts the definition of ℓ − . (iii) There exists ε > 0 such that | f (ℓ) − ℓ| ≥ ε , for all ℓ ∈ A. For if not, let ℓ n ∈A

be such that | f (ℓ n )−ℓ n | < 1/n, for all n ≥ 1. This implies that any limit point of (ℓ n ) n ≥1 (which lies in A, too) is a fixed point of f . This contradicts (i). (iv) Conclusion. By (ii) and (iii), there exists a largest ℓ + ∈ A such that

f (ℓ + )>ℓ + . Let ℓ ′ = [ℓ + + f (ℓ + )]/2 and observe that f (ℓ + )>ℓ ′ >ℓ + and

f (ℓ ′ )<ℓ ′ . By (iii), there exists a smallest ℓ ′′ ∈ A such that ℓ ′′ >ℓ + and

f (ℓ ′′ )<ℓ ′′ . It follows that ℓ + <ℓ ′′ < f (ℓ + ). Next note that f (ℓ ′′ )<ℓ + ; for if not, ℓ ′′′ : = [ℓ ′′ + f (ℓ ′′ )]/2 satisfies ℓ + <ℓ ′′′ <ℓ ′′ and, by definitions of ℓ + and ℓ ′′ , it follows that f (ℓ ′′′ )=ℓ ′′′ , contrary to (i). Thus f (ℓ ′′ )<ℓ + <ℓ ′′ < f (ℓ + ). It then follows that | f (ℓ ′′ ) − f (ℓ + )| > |ℓ ′′ −ℓ + |. This contradicts the hypothesis and concludes the proof. ⊓ ⊔

Remark. The iteration scheme described in the above Krasnoselski property does not apply to arbitrary continuous mappings of a closed interval into itself. Indeed, consider the function f : [0, 1]→[0,1] defined by

if 1 2 ≤ x ≤ 1. Then the sequence defined in the above statement is defined by x 2n = 1/2 and

x 2n +1 = 1/4, for any n ≥ 1. So, (x n ) n ≥1 is a divergent sequence. The contraction mapping theorem states that if f : R →R (or f : [a,∞)→[a,∞))

is a map such that for some k in (0, 1) and all x and y in R, | f (x) − f (y)| ≤ k |x − y|, then the iterates f n = f ◦ ··· ◦ f (n terms) of f converge to a (unique) fixed point ξ of f . This theorem can be accompanied by an example to show that the inequality cannot be replaced by the weaker condition | f (x) − f (y)| < |x − y|. The most common example of this type is f (x) = x + 1/x acting on [1, ∞). Then

f (x) > x, so that f has no fixed points. Also, for every x, the sequence x, f (x),

f 2 (x), . . . is strictly increasing and so must converge in the space [−∞,+∞]. In fact,

f n (x) → +∞, for otherwise f n (x) → a for some real a, and then f ( f n (x)) → f (a) (because f is continuous), so that f (a) = a, which is not so. Thus we define f (+∞)

4.5 Fixed Points 161 to be +∞ and deduce that this example is no longer a counterexample. The fol-

lowing property clarifies these ideas and provides an elementary, but interesting, adjunct to the contraction mapping theorem. We just point out that a mapping

f :R function.

4.5.7. Suppose that f :R →R satisfies | f (x) − f (y)| < |x − y| whenever x . Then there is some ξ in [−∞,+∞] such that for any real x , f n (x) → ξ as n →∞ .