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Unformatted text preview: SOLUTIONS FOR HOMEWORK 8 5.4.2. (a) Note that, for x, y ∈ R\{0}, f (x) − f (y ) = 1 y 2 − x2 (y − x)(y + x) 1 1 1 − 2= = = (y − x) 2 + 2 . x2 y x2 y 2 x2 y 2 x y xy If x, y ∈ A, then x, y ≥ 1, hence 1/(x2 y ) + 1/(xy 2) ≤ 2, and f (x) − f (y ) ≤ 2x − y . In particular, for a given ε > 0, f (x) − f (y ) < ε when x − y  < ε/2. Thus, f is uniformly continuous on A. (b) For n ∈ N, let xn = 2−n (clearly, xn ∈ B ). Then xn − xn+1 = 2−(n+1) . However, f (xn+1 ) − f (xn ) = 22(n+1) − 22n = 3 · 4n . Suppose, for the same of contradiction, that f is uniformly continuous on B . Then there exists δ > 0 s.t. f (x) − f (y ) < 1 whenever x, y ∈ B satisfy x − y  < δ . Pick n ∈ N so large that 2−n < δ . Then we must have f (xn ) − f (xn+1 ) < 1, which is false. 5.4.4. Note that, for x, y ∈ R, f (x) − f (y ) = Furthermore, 1 + x2 1 + y 2 + ≥ x + y  (1 + x )(1 + y ) = 1 + x + y + x y ≥ 2 2 (the last inequality follows from the ArithmeticGeometric Mean Inequality). Therefore, y − x · y + x f (x) − f (y ) ≤ ≤ y − x. x + y  Thus, f is Lipschitz, hence uniformly continuous.
2 2 2 2 22 1 1 (y − x)(y + x) − = . 2 2 1+x 1+y (1 + x2 )(1 + y 2 ) 5.4.7. Both f and g are Lipschitz functions. Indeed, for f it is obvious. As far as g is concerned, recall that, for any t,  sin(t) ≤ t. Furthermore, sin α − sin β = 2 sin hence α+β α − β  α−β · cos ≤2 = α − β . 2 2 2 Therefore, both f and g are uniformly continuous. However, h = f g is not uniformly continuous. Indeed, if it were, then there would exist δ > 0 s.t. h(x) − h(y ) < 1 whenever x − y  < δ . Pick N ∈ N s.t. π/N < δ . Fix M > N . For 0 ≤ k ≤ N , let xk = 2πM − π/2 + kπ/N . Then x0 = 2πM − π/2, and sin α − sin β = 2 sin
1 α−β α+β cos , 2 2 2 SOLUTIONS FOR HOMEWORK 8 xN = 2πM + π/2, hence f (xN ) − f (x0 ) = 4πM . On the other hand, xk − xk−1  < δ for 1 ≤ k ≤ N , hence f (xk ) − f (xk−1 ) < 1, and, by the triangle inequality,
N f (xN ) − f (x0 ) ≤
k =1 f (xk ) − f (xk−1 ) < N < M, a contradiction. 5.4.8. Let h = f ◦ g . We have to show that for every ε > 0 there exists δ > 0 s.t. h(x) − h(y ) < ε whenever x − y  < δ . As f is uniformly continuous, for any ε like this there exists σ > 0 s.t. f (s) − f (t) < ε is s − t < σ . Moreover, g is uniformly continuous, hence there exists δ > 0 s.t. g (x) − g (y ) < σ whenever x − y  < δ . Then f (g (x)) − f (g (y )) < ε if x − y  < δ , which is what we need. 5.4.9. Let g = 1/f . Then g (x) − g (y ) = (f (y ) − f (x))/(f (x)f (y )). For ε > 0, ﬁnd δ > 0 s.t. f (x) − f (y ) < k 2 ε whenever x − y  < δ . For such x and y , g (x) − g (y ) ≤ f (y ) − f (x) k2 ε < 2 = ε, f (x) · f (y ) k which establishes the uniform continuity of g . 5.4.13. This is a bonus problem – very little partial credit is given. Fixing ε > 0, we show that there exists δ > 0 s.t. f (x) − f (y ) < ε for any x, y ∈ A s.t. x − y  < δ . We know that there exists a uniformly continuous function g : A → R s.t. f (t) − g (t) < ε/3 for any t ∈ S . Find δ > 0 s.t. g (x) − g (y ) < ε/3 for any x, y ∈ A s.t. x − y  < δ . By the triangle inequality, for such x and y we have: εεε f (x) − f (y ) ≤ f (x) − g (x) + g (x) − g (y ) + g (y ) − f (y ) < + + = ε. 333 As ε > 0 is arbitrary, the uniform continuity of f is established. 5.6.12. Suppose f is not strictly increasing on [0, 1]. Then there exist 0 ≤ a < b ≤ 1 s.t. f (a) ≥ f (b). However, no value is attained twice, so f (a) = f (b), hence f (a) > f (b). Moreover, f (0) < f (1), hence either a > 0, or b < 1. We henceforth assume that a > 0 (the other case is tackled similarly). (1) f (a) < f (0). Applying the Intermediate Value Theorem to [0, a] and [a, 1], we conclude that there exist c1 ∈ (0, a) and c2 ∈ (a, 1) s.t. f (c1 ) = f (c2 ) = C , where C = (f (0) + f (a))/2. Thus, f takes the value C twice, which is impossible. (2) f (a) = f (0). Impossible, too. (3) f (a) > f (0). Let A = f (a), and B = max{f (0), f (b)}. Note that A > B . Let C = (A + B )/2. Applying the Intermediate Value Theorem to [0, a] and [a, b], we conclude that there exist c1 ∈ (0, a) and c2 ∈ (a, b) s.t. f (c1 ) = f (c2 ) = C . Thus, f takes the value C twice, which is impossible. 5.6.13. Suppose, for the sake of contradiction, that f is continuous on [0, 1]. Then it attains its maximum and minimum, which we denote by M and m, respectively. Therefore, there exist a1 < a2 and b1 < b2 s.t. f (a1 ) = f (a2 ) = M , and f (b1 ) < f (b2 ) < M . Note that m < M (otherwise, f (x) = m = M for each x ∈ [0, 1]). SOLUTIONS FOR HOMEWORK 8 3 We show that some value is attained by f at least three times. To this end, consider the following three (mutually exclusive) cases: (1) The intervals [a1 , a2 ] and [b1 , b2 ] partially overlap: a1 < b1 < a2 < b2 , or b1 < a1 < b2 < a2 . (2) One of the intervals [a1 , a2 ] and [b1 , b2 ] contains the other: a1 < b1 < b2 < a2 , or b1 < a1 < a2 < b2 . (3) The intervals [a1 , a2 ] and [b1 , b2 ] are disjoint: a1 < a2 < b1 < b2 , or b1 < b2 < a1 < a2 . In case (1), suppose a1 < b1 < a2 < b2 . Applying the Intermediate Value Theorem to the intervals [a1 , b1 ], [b1 , a2 ], and [a2 , b2 ], we show the existence of the points c1 ∈ [a1 , b1 ], c2 ∈ [b1 , a2 ], and c3 ∈ [a2 , b2 ], s.t. f (c1 ) = f (c2 ) = f (c3 ) = (m + M )/2. Thus, the value (m + M )/2 is attained at least three times, which is impossible. In case (2), suppose a1 < b1 < b2 < a2 . As f is continuous on [b1 , b2 ], there exists a point c ∈ (b1 , b2 ) where f attains its maximum on [b1 , b2 ]. Note that m < f (c) ≤ M . Applying the Intermediate Value Theorem to the intervals [a1 , b1 ], [b1 , c], and [c, b2 ], we show the existence of the points c1 ∈ [a1 , b1 ], c2 ∈ [b1 , c], and c3 ∈ [c, b2 ], s.t. f (c1 ) = f (c2 ) = f (c3 ) = y , where y = (m + f (c))/2. Indeed, M = f (a1 ) > y > f (b1 ) = m, hence the existence of c1 . The existence of c2 and c3 is established in the same way. In case (3), suppose a1 < a2 < b1 < b2 . By the continuity of f , there exists c ∈ (b1 , b2 ) where f attains its maximum on [b1 , b2 ]. As in (2), m < f (c) ≤ M . As before, an application of the Intermediate Value Theorem yields c1 ∈ [a2 , b1 ], c2 ∈ [b1 , c], and c3 ∈ [c, b2 ], s.t. f (c1 ) = f (c2 ) = f (c3 ) = y , where y = (m + f (c))/2. 5.6.15. (a) We have to show that xr xs = xr+s when r, s ∈ Q (we already know this for r, s ∈ Z). Write r = a/b and s = c/d, with a, c ∈ Z and b, d ∈ N. We proved in class (see also Exercise 5.6.14) that xr (or xs ) is welldeﬁned – that is, the particular choice of a and b satisfying a/b = r doesn’t matter. Recalling that tm tn = tm+n for m, n ∈ Z, and letting t = x1/bd , m = ad, and n = bc, we have: xr xs = xad/bd xbc/bd = (x1/bd )ad (x1/bd )bc = (x1/bd )ad+bc = x(ad+bc)/bd = xr+s . (b) We have to show that (xr )s = xrs when r, s ∈ Q (this is known for r, s ∈ Z). Note ﬁrst that, for m, n ∈ N and t ≥ 0, (1)
1/n mn t1/m 1/n = t1/mn .
1/n n m m Indeed, s = t1/mn is the unique nonnegative number such that smn = t. However, t1/m = t1/m = t1/m = t, which establishes (1). The equality (xr )s = xrs is trivial if r = 0 or s = 0. Suppose ﬁrst that both r and s are positive. As in (a), write r = a/b and s = c/d, with a, b, c, d ∈ N. Then xr
s = x1/b a 1/d c . 4 SOLUTIONS FOR HOMEWORK 8 Recall that (tm )1/n = (t1/n )m for m, n ∈ N (Theorem 5.6.7), hence, by (1), x1/b Thus, xr
s a 1/d = x1/b 1/d a = x1/bd a .
ac = x1/b a 1/d c = 1 x1/bd 1 ac = x1/bd = xrs . Now suppose r > 0, s < 0. Then xr
s = (xr )−s = x−rs = xrs . The case of r < 0 (and s either positive or negative) is tackled similarly. Back to: the syllabus, the main page of the course, the assignment. ...
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.
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