Prove that any continuous function f : [a, b]→R is uniformly continuous. Solution. We give the original proof from 1872, which is due to Heine [45].

4.2.7. Prove that any continuous function f : [a, b]→R is uniformly continuous. Solution. We give the original proof from 1872, which is due to Heine [45].

Arguing by contradiction, there exist ε > 0 and sequences (x n ) n ≥1 and (y n ) n ≥1 in [a, b] such that for all n ≥ 1, |x n −y n | < 1/n and | f (x n ) − f (y n )| ≥ ε . By the Bolzano–Weierstrass theorem, passing to subsequences if necessary, we can assume that x n →x and y n →y as n→∞. Passing to the limit in |x n −y n | < 1/n, we deduce that x = y. On the other hand, since | f (x n ) − f (y n )| ≥ ε , we find as n →∞ that 0 ≥ ε ,a contradiction.

Next, we give an alternative proof found in 1873 by L¨uroth [72] based on Weierstrass’s theorem. Fix ε > 0. For each x ∈ [a,b] let δ (x) > 0 be the length of the largest open interval I of center x such that | f (y) − f (z)| < ε for y ,z∈I (see Figure 4.5). More precisely,

δ (x) = sup{x > 0; | f (y) − f (z)| < ε for all y , z ∈ [x − δ /2, x + δ /2]}, where x, y, and z are supposed to lie in [a, b].

By continuity of f at x, the set of all δ > 0 in the definition of δ (x) is nonempty, for all x ∈ [a,b]. If δ (x) = ∞ for some x ∈ [a,b], then the estimate | f (y) − f (z)| < ε holds without any restriction; hence any δ > 0 satisfies the condition in the definition of uniform continuity.

Fig. 4.5 Illustration of L¨uroth’s proof of the Weierstrass theorem.

4.3 The Intermediate Value Property 147

If δ (x) < ∞ for all x ∈ [a,b], we move x to x ± η . The new interval I ′ cannot be longer than δ (x) + 2| η |; otherwise, I would be entirely in I ′ and could be extended,

a contradiction. At the same time, I ′ cannot be smaller than δ (x) − 2| η |. Thus, this δ (x) is a continuous function. By Weierstrass’s theorem, there exists x 0 ∈ [a,b] such that δ (x 0 )≤ δ (x) for all x ∈ [a,b]. This value δ (x 0 ) is positive by definition and can

be used to satisfy the condition in the definition of uniform continuity.

A direct consequence of the above property is that any continuous periodic func- tion f : R →R is uniformly continuous. ⊓ ⊔

4.2.8. Let f , g : [a, b]→(0,∞) be continuous functions such that g (x 0 < f (x) for all x ∈ [a,b] . Prove that there exists λ >1 such that f ≥ λ g in [a, b] . Does the conclusion remains true if we replace [a, b] with (a, b) ?

Solution. We argue by contradiction. Thus, for any n ≥ 1, there exists x n ∈ [a,b] such that f (x n ) < (1 + 1/n)g(x n ). By the Bolzano–Weierstrass theorem, passing eventually to a subsequence, we can assume that x n →x 0 . Moreover, since [a, b] is compact, we have x 0 ∈ [a,b]. Passing to the limit and using the continuity of f and

g , we deduce that f (x 0 ) ≤ g(x 0 ), a contradiction.

For the second part it is enough to consider the functions f (x) = x and g(x) = x 2 , x ∈ (0,1). ⊓ ⊔

4.2.9. Let f : [0, 1]→[0,1]

be a continuous function such that f ◦f=f . Define

E f = {x ∈ [0,1]; f (x) = x} . (i) Prove that E f .

(ii) Show that E f is an interval. (iii) Find all functions with the above properties.

Solution. (i) We prove that f ([0, 1]) ⊂ E f . Indeed, take y = f (x) ∈ f ([0,1]), where x ∈ [0,1]. Then f ( f (x)) = f (x); hence f (y) = y, which shows that y ∈ E f .

(ii) We show that E f ⊂ f ([0,1]). Since f ([0,1]) is an interval, as the image of an interval by the continuous function f , it follows that E f is an interval, too. Choose x ∈E f , that is, x ∈ [0,1] and f (x) = x. This implies that x ∈ f ([0,1]). (iii) Continuous functions f : [0, 1]→[0,1] satisfying f ◦ f = f are those functions such that E f = f ([0, 1]). ⊓ ⊔