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Analysis Problems #6 Solutions 1. Suppose that f is diﬀerentiable with f ′ (x) = 1 1+x2 for all x. Show that (a) f (x) − f (y) ≤ x − y whenever x ≥ y and that (b) f (x) + f (1/x) = 2f (1) for all x > 0. • When x = y, the inequality in part (a) is obviously true. When x > y, on the other hand, we can use the mean value theorem to find some c ∈ (y, x) such that f (x) − f (y) 1 = f ′ (c) = ≤ 1. x−y 1 + c2 To prove the identity in part (b), we let g(x) = f (x) + f (1/x) and compute ( ) 1 1 1 ′ ′ ′ ′ g (x) = f (x) + f (1/x)(1/x) = + · − 2 = 0. 1 + x2 1 + 1/x2 x This shows that g(x) is actually constant, so g(x) = g(1) = 2f (1). 2. Compute the limit x3 + x2 − 5x + 3 . x→1 x3 − x2 − x + 1 L = lim • Substituting x = 1 gives 0/0, so L’Hôpital’s rule is applicable and we have x3 + x2 − 5x + 3 3x2 + 2x − 5 = lim . x→1 x3 − x2 − x + 1 x→1 3x2 − 2x − 1 L = lim The last limit still gives 0/0, so we may use L’Hôpital’s rule once again, namely 3x2 + 2x − 5 6x + 2 6+2 = lim = = 2. 2 x→1 3x − 2x − 1 x→1 6x − 2 6−2 L = lim 3. Show that f is integrable on [0, 1] when f is the function deﬁned by } { 0 if x ̸= 0 f (x) = . 1 if x = 0 • Given a partition P = {x0 , x1 , . . . , xn } of [0, 1], one clearly has inf [xk ,xk+1 ] f (x) = 0 for each 0 ≤ k ≤ n − 1 and this implies that − S (f, P ) = n−1 ∑ k=0 inf [xk ,xk+1 ] f (x) · (xk+1 − xk ) = 0, so sup S − (f, P ) = 0 as well. For the upper Darboux sums, we have sup f (x) = 0 for each 1 ≤ k ≤ n − 1 sup f (x) = 1, [x0 ,x1 ] [xk ,xk+1 ] and this similarly gives + S (f, P ) = n−1 ∑ sup f (x) · (xk+1 − xk ) = x1 − x0 = x1 . k=0 [xk ,xk+1 ] Taking the infimum over all possible partitions, we now get inf S + (f, P ) = inf x1 = 0 0<x1 <1 so that inf S + (f, P ) = sup S − (f, P ) and f is integrable on [0, 1], indeed. 4. Suppose f, g are integrable on [a, b] with f (x) ≤ g(x) for all x ∈ [a, b]. Show that ∫ ∫ b b f (x) dx ≤ a g(x) dx. a • Let P = {x0 , x1 , . . . , xn } be a partition of [a, b]. Starting with the inequality f (x) ≤ g(x) for all x ∈ [xk , xk+1 ], we take the infimum of both sides to get inf [xk ,xk+1 ] f (x) ≤ inf g(x). [xk ,xk+1 ] Multiplying through by xk+1 − xk and then adding, we conclude that n−1 ∑ k=0 inf [xk ,xk+1 ] f (x) · (xk+1 − xk ) ≤ n−1 ∑ k=0 inf [xk ,xk+1 ] g(x) · (xk+1 − xk ). In view of the definition of lower Darboux sums, this actually gives S − (f, P ) ≤ S − (g, P ) for all partitions P , so we may take the supremum of both sides to finally get ∫ b ∫ b − − f (x) dx = sup S (f, P ) ≤ sup S (g, P ) = g(x) dx. a a