81 apply the intermediate value theorem to the function f on the interval [−1,0] with γ = 0 to conclude that there is c ∈ [−1,0] such that f (c)= 0. The point c is the desired solution to the original equation. ■ Example 3.4.2 We show now that, given n ∈ N, every positive real number has a positive nth root. Let n ∈ N and let a ∈ R with a > 0. First observe that (1 + a)n ≥ 1 + na > a (see Exercise 1.3.6). Now consider the function f : [0,∞) → R given by f (x)= xn . Since f (0)= 0 and f (1+ a) > a, it follows from the intermediate value theorem that there is x ∈ (0,1 + a) such that f (x)= a. That is, xn = a, as desired. (We show later in Example 4.3.1 that such an x is unique.) Now we are going to discuss the continuity of the inverse function. Theorem 3.4.7 Let f : [a,b] → R be strictly increasing and continuous on [a,b]. Let c = f (a) and d = f (b). Then f is one-to-one, f ([a,b]) = [c,d], and the inverse function f −1 defned on [c,d] by f −1( f (x)) = x where x ∈ [a,b], is a continuous function from [c,d] onto [a,b]. Proof: The frst two assertions follow from the monotonicity of f and the intermediate value theorem (see also Corollary 3.4.6). We will prove that f −1 is continuous on [c,d]. Fix any y 0 ∈ [c,d] and fx any sequence {yn} in [c,d] that converges to y0. Let x0 ∈ [a,b] and xn ∈ [a,b] be such that f (x0)= y0 and f (xn)= yn for every n. Then f −1(y 0)= x0 and f − 1(y n)= xn for every n. Suppose by contradiction that {xn} does not converge to x0. Then there exist ε0 > 0 and a subsequence {xnk } of {xn} such that |xnk − x0|≥ ε0 for every k. (3.5) Since the sequence {xnk } is bounded, it has a further subsequence that converges to some x1 ∈ [a,b]. To simplify the notation, we will again call the new subsequence {xnk }. Taking limits in (3.5), we get |x1 − x0|≥ ε0 > 0. (3.6) On the other hand, by the continuity of f , { f (xnk )} converges to f (x1). Since f (xnk )= ynk → y0 as k → ∞, it follows that f (x1)= y0 = f (x0). This implies x1 = x0, which contradicts (3.6). □ An analogous theorem can be proved for strictly decreasing functions. Remark 3.4.2 A similar result holds if the domain of f is the open interval (a,b) with some additional considerations. If f : (a,b) → R is increasing and bounded, following the argument in Theorem 3.2.4 we can show that both limx →a+ f (x)= c and limx →b− f (x)= d exist in R (see Exercise 3.2.11). Using the intermediate value theorem we obtain that f ((a,b)) = (c,d). We can now proceed as in the previous theorem to show that f has a continuous inverse from (c,d) to (a,b). If f : (a,b) → R is increasing, continuous, bounded below, but not bounded above, then limx →a+ f (x) = c ∈ R, but limx →b− f (x) = ∞ (again see Exercise 3.2.11). In this case we can show using the intermediate value theorem that f ((a,b)) = (c,∞) and we can proceed as above to prove that f has a continuous inverse from (c,∞) to (a,b). The other possibilities lead to similar results.
RkJQdWJsaXNoZXIy NTc4NTAz