Proof. We indicate by α, β (finite or not) the end-points of I and call

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1 C.6 Continuous functions Pag. 111 Proof of Corollary 4.25 Corollary 4.25 Let f be continuous on the interval I and suppose it admits non-zero its (finite or infinite) that are different in sign for x tending to the end-points of I. Then f has a zero in I, which is unique if f is strictly monotone on I. Proof. We indicate by α, β (finite or not) the end-points of I and call f(x) = l α and f(x) = l β. x α + x β Should one end-point, or both, be infinite, these writings denote the usual its at infinity. We suppose l α < 0 < l β, for otherwise we can swap the roles of l α and l β. By Theorem 4.2 there exist a right neighbourhood I + (α) of α and a left neighbourhood I (β) of β such that x I + (α), f(x) < 0 and x I (β), f(x) > 0. Let us fix points a I + (α) and b I (β) with α < a, b < β. The interval [a, b] is contained in I, hence f is continuous on it, and by construction f(a) < 0 < f(b). Therefore f will vanish somewhere, in virtue of Theorem 4.23 (existence of zeroes) applied to [a, b]. If f is strictly monotone, uniqueness follows from Proposition 2.8 on the interval I. Pag. 114 Proof of the Theorem of Weierstrass Theorem 4.31 (Weierstrass) A continuous map f on a closed and bounded interval [a, b] is bounded and admits minimum and maximum

2 2 C.6 Continuous functions m = min f(x) and M = max f(x). x [a,b] x [a,b] Consequently, f([a, b]) = [m, M]. Proof. We will show first that f admits a maximum in [a, b], in other words there exists ξ [a, b] such that f(x) f(ξ), x [a, b]. For this, let M = sup f([a, b]), which is allowed to be both real or +. In the former case the characterisation of the supremum, (1.7) ii), tells that for any n 1 there is x n [a, b] with M 1 n < f(x n) M. Letting n go to +, from the Second comparison theorem (Theorem 5. on p. 137) we infer n f(x n) = M. In the other case, by definition of unbounded set (from above) we deduce that for any n 1 there is x n [a, b] such that The Second comparison theorem implies f(x n ) n. f(x n) = + = M. n In either situation, the sequence {x n } n 1 thus defined is bounded (it is contained in [a, b]). We are then entitled to use Theorem C.5.3 of Bolzano-Weierstrass and call {x nk } k 0 a convergent subsequence. Let ξ be its it; since all x nk belong in [a, b], necessarily ξ [a, b]. But {f(x nk )} k 0 is a subsequence of {f(x n )} n 0, so by Proposition C.5.2 f(x n k ) = M. k The continuity of f at ξ implies f(ξ) = f( k x n k ) = k f(x n k ) = M, which tells us that M cannot be +. Moreover, M belongs to the range of f, hence M = max f([a, b]). Arguing in a similar fashion one proves that the number m = min f([a, b])

3 C.6 Continuous functions 3 exists and is finite. The final claim is a consequence of Corollary Pag. 114 Proof of Theorems 4.32 and 4.33 Let us prove a preinary result before proceeding. Lemma C.6.1 Let f be continuous and invertible on an interval I. For any chosen points x 1 < x 2 < x 3 in I, then one, and only one, of (i) f(x 1 ) < f(x 2 ) < f(x 3 ) or (ii) f(x 1 ) > f(x 2 ) > f(x 3 ) holds. Proof. As f is invertible, hence one-to-one, the images f(x 1 ) and f(x 3 ) cannot coincide. Then either f(x 1 ) < f(x 3 ) or f(x 1 ) > f(x 3 ), and we claim that these cases imply (i) or (ii), respectively. Suppose f(x 1 ) < f(x 3 ), and assume by contradiction that (i) is false, so f(x 2 ) does not lie strictly between f(x 1 ) and f(x 3 ). For instance, f(x 1 ) < f(x 3 ) < f(x 2 ) (if f(x 2 ) < f(x 1 ) < f(x 3 ) the argument is the same). As f is continuous on the closed interval [x 1, x 2 ] I, the Intermediate value theorem 4.29 prescribes that it will assume every value between f(x 1 ) and f(x 2 ) on [x 1, x 2 ]. In particular, there will be a point x (x 1, x 2 ) such that f( x) = f(x 3 ), in contradiction to injectivity: x and x 3 are in fact distinct, because separated by x 2. Theorem 4.32 A continuous function f on an interval I is one-to-one if and only if it is strictly monotone. Proof. Thanks to Proposition 2.8 we only need to prove the implication f invertible oni f strictly monotone on I. Letting x 1 < x 2 be arbitrary points of I, we claim that if f(x 1 ) < f(x 2 ) then f is strictly increasing on I (f(x 1 ) > f(x 2 ) will similarly imply f strictly decreases on I). Let z 1 < z 2 be points in I, and suppose both lie within (x 1, x 2 ); the other possibilities are dealt with in the same way. Hence we have

4 4 C.6 Continuous functions x 1 < z 1 < z 2 < x 2. Let us use Lemma C.6.1 on the triple x 1, z 1, x 2 : since we have assumed f(x 1 ) < f(x 2 ), it follows f(x 1 ) < f(z 1 ) < f(x 2 ). Now we employ the triple z 1, z 2, x 2, to the effect that f(z 1 ) < f(z 2 ) < f(x 2 ). The first inequality in the above line tells f is strictly increasing, proving Theorem Theorem 4.33 Let f be continuous and invertible on an interval I. Then the inverse f 1 is continuous on the interval J = f(i). Proof. The first remark is that J is indeed an interval, by Corollary Using Theorem 4.32 we deduce f is strictly monotone on I: to fix ideas, suppose it is strictly increasing (having f strictly decreasing would not change the proof). By definition of a monotone map we have that f 1 is strictly increasing on J as well. But it is known that a monotone map admits at most discontinuities of the first kind (Corollary 3.28). We will show that f 1 cannot have this type either. By contradiction, suppose there is a jump point y 0 = f(x 0 ) J = f(i) for f 1. Equivalently, let z0 = sup f 1 (y) = f 1 (y), y<y 0 y y 0 z + 0 = inf f 1 (y) = f 1 (y), y>y 0 y y + 0 and suppose z0 < z+ 0. Then inside (z 0, z+ 0 ) there will be at most one element x 0 = f 1 (y 0 ) of the range f 1 (J). Thus f 1 (J) is not an interval. By definition of J, on the other hand, f 1 (J) = I is an interval by hypothesis. In conclusion, f 1 must be continuous at each point of J. C.6.1 Uniform continuity Let the map f be defined on the real interval I. Recall f is called continuous on I if it is continuous at each point x 0 I, i.e., for any x 0 I and any ε > 0 there exists δ > 0 such that x I, x x 0 < δ f(x) f(x 0 ) < ε. In general δ = δ(ε, x 0 ), meaning that δ depends on x 0, too. But if, for fixed ε > 0, we find δ = δ(ε) independent of x 0 I, we say f is uniformly continuous on I. More precisely,

5 C.6.1 Uniform continuity 5 Definition C.6.2 A function f is called uniformly continuous on I if, for any ε > 0, there is a δ > 0 satisfying x, x I, x x < δ f(x ) f(x ) < ε. (C.6.1) Examples C.6.3 i) Let f(x) = x 2, defined on I = [0, 1]. Then f(x ) f(x ) = (x ) 2 (x ) 2 = x + x x x 2 x x. If x x < ε 2 we see f(x ) f(x ) < ε, hence δ = ε 2 fulfills (C.6.1) on I, rendering f uniformly continuous on I. ii) Take f(x) = x 2 on I = [0, + ). We want to prove by contradiction that f is not uniformly continuous on I. If it were, with ε = 1 for example, there would be a δ > 0 satisfying (C.6.1). Choose x I and let x = x + δ 2, so that x x = δ 2 < δ; then f(x ) f(x ) = x + x x x < 1, or ( 2x + δ 2) δ 2 < 1. Now letting x tend to + we obtain a contradiction. iii) Consider f(x) = sin x on I = R. From we have sin x sin x = 2 sin x x 2 cos x + x 2 sin x sin x x x, x, x R. With a fixed ε > 0, δ = ε satisfies the requirement for uniform continuity. iv) Let f(x) = 1 x on I = (0, + ). Note that f(x ) f(x ) = 1 x 1 x = x x x x. By letting x, x tend to 0, one easily verifies that f cannot be uniformly continuous on I. But if we consider only I a = [a, + ), where a > 0 is fixed, then f(x ) f(x ) x x a 2, so δ = a 2 ε satisfies the requirement on I a, for any given ε > 0. Are there conditions guaranteeing uniform continuity? One answer is provided by the following result.

6 6 C.6 Continuous functions Theorem C.6.4 (Heine-Cantor) Let f be a continuous map on the closed and bounded interval I = [a, b]. Then f is uniformly continuous on I. Proof. Let us suppose f is not uniformly continuous on I. This means that there exists an ε > 0 such that, for any δ > 0, there are x, x I with x x < δ and f(x ) f(x ) ε. Choosing δ = 1 n, n 1, we find two sequences of points {x n} n 1, {x n} n 1 inside I such that x n x n < 1 n and f(x n) f(x n) ε. The condition on the left implies n (x n x n) = 0, while the one on the right implies that f(x n) f(x n) will not tend to 0 for n. On the other hand the sequence {x n} n 1 is bounded, a x n b for all n, so Theorem C.5.3, of Bolzano-Weierstrass, will give a subsequence {x n k } k 0 converging to a certain x I: k x n k = x. Also the subsequence {x n k } k 0 converges to x, for k x n k = k [x n k + (x n k x n k )] = k x n k + k (x n k x n k ) = x + 0 = x. Now, f being continuous at x, we have k f(x n k ) = f ( ) k x n k = f( x) and k f(x n k ) = f ( ) k x n k = f( x). Then contradicting the fact that ( f(x nk ) f(x n k k ) ) = f( x) f( x) = 0, f(x n k ) f(x n k ) ε > 0, k 0.