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Homework 8 Math 501 Due October 31, 2014 Exercise 1 Let f : R → R be a function that satisfies the following: 1. f is continuous for x ≥ 0 2. f 0 (x) exists for x > 0 3. f (0) = 0 4. f 0 is increasing. Let g : (0, ∞) → R be given by g(x) = f (x) . x To show that g is increasing it suffices to show that g 0 (x) ≥ 0 for all x > 0. We begin by observing that g 0 (x) = xf 0 (x) − f (x) . x2 Thus it suffices to show that f (x) ≤ xf 0 (x) for all x > 0. Note that by the mean value theorem, for every x > 0 there exists θx ∈ (0, x) such that f 0 (θx ) = f (x) − f (0) f (x) = , x x which implies f (x) = xf 0 (θx ). Since θx < x and f 0 is increasing, it follows that f 0 (θx ) ≤ f 0 (x). The result now follows. 1 Exercise 2 Let f : [a, b] → R be a bounded function that is continuous at a point t ∈ (a, b). Let c ∈ R and let α : [a, b] → R be given by c x≥t α(x) = . 0 x<t Let > 0. By the continuity of f at t, there exists δ > 0 such that |f (x) − f (t)| < /(2c) whenever x ∈ [a, b] satisfies |x − t| < δ. Let P = {x0 , . . . , xn } be a partition of [a, b] such that ∆xi < δ for all i = 1, . . . , n. Note that ∆αi = c for the unique 1 ≤ i ≤ n such that xi−1 ≤ t ≤ xi and ∆αj = 0 for all j 6= i. Therefore, L(P, f, α) = cmi , U (P, f, α) = cMi . where mi = inf{f (x) : x ∈ [xi−1 , xi ]}, Mi = sup{f (x) : x ∈ [xi−1 , xi ]}. Note that |f (u)−f (v)| ≤ |f (u)−f (t)|+|f (t)−f (v)| < /c for all u, v ∈ [xi−1 , xi ] by continuity. Since u, v were arbitrary, it follows that Mi − mi < /c. Thus U (P, f, α) − L(P, f, α) = c(Mi − mi ) < . It follows that f is Riemann-Stieltjes integrable with respect to α. Note that Z L(P, f, α) ≤ cf (t) ≤ U (P, f, α), L(P, f, α) ≤ f dα ≤ U (P, f, α), a from which it follows that Z b f dα < . cf (t) − a Since was arbitrary Z b f dα = cf (t). a 2 b Exercise 3 Let f : [a, b] → R be Riemann integrable and continuous at t ∈ (a, b). Let c1 , c2 ∈ R and let α, β : [a, b] → R be given by c1 x x ∈ [a, t) c1 x x ∈ [a, t) α(x) = , β(x) = . c2 x x ∈ [t, b] c2 (x − t) + c1 t x ∈ [t, b] Let I1 = Rt a f dx and I2 = Rb t f dx. Let > 0 and let η= . 3(c2 − c1 )t By the continuity of f at t, there exists δ > 0 such that |f (x) − f (t)| < η whenever x ∈ [a, b] satisfies |x − t| < δ. By the Riemann integrability of f there exists a partition P = {x0 , . . . , xn } of [a, b] such that , U (P, f ) − L(P, f ) < min 3c1 3c2 Without loss of generality, we may assume ∆xi < δ for all i = 1, . . . , n. Indeed, if the above P does not satisfy this condition, we may refine P and it is clear that the refinement will still satisfy the above inequality. Moreover, arguing in the same fashion, we may assume that t ∈ P . Suppose xi = t Let P 0 = {x0 , . . . , xi } and P 00 = {xi , . . . , xn }. Note that U (P 0 , f ) − L(P 0 , f ) ≤ U (P, f ) − L(P, f ) < , 3c1 U (P 00 , f ) − L(P 00 , f ) ≤ U (P, f ) − L(P, f ) < 3c2 and Note then that ∆αi = c2 xi − c1 xi−1 = c1 ∆xi + (c2 − c1 )t and ∆βi = c1 ∆xi Moreover, ∆αj = ∆βj = 3 c1 ∆xj c2 ∆xj j<i j > i. Observe that U (P, f, α) = n X Mi ∆αi = c1 U (P 0 , f ) + c2 U (P 00 , f ) + Mi (c2 − c1 )t i=1 and L(P, f, α) = n X mi ∆αi = c1 L(P 0 , f ) + c2 L(P 00 , f ) + mi (c2 − c1 )t. i=1 It follows that U (P, f, α) − L(P, f, α) < 2/3 + (Mi − mi )(c2 − c1 )t < 2/3 + η(c2 − c1 )t < Thus f is Riemann-Stieltjes integrable with respect to α. Moreover, note that Z t c1 L(P 0 , f ) ≤ c1 f dx ≤ c1 U (P 0 , f ) a and similarly, c2 L(P 00 , f ) ≤ c2 b Z f dx ≤ c2 U (P 00 , f ), t from which it follows that L(P, f, α) ≤ c1 I1 + c2 I2 + (c2 − c1 )tf (t) ≤ U (P, f, α) Therefore, since Z L(P, f, α) ≤ b f dα ≤ U (P, f, α) a it follows that Z b f dα < . c1 I1 + c2 I2 + (c2 − c1 )tf (t) − a Therefore Z b f dα = c1 I1 + c2 I2 + (c2 − c1 )tf (t). a Furthermore, U (P, f, β) = n X Mi ∆βi = c1 U (P 0 , f ) + c2 U (P 00 , f ) i=1 and L(P, f, β) = n X mi ∆βi = c1 L(P 0 , f ) + c2 L(P 00 , f ) i=1 4 Therefore, U (P, f, β) − L(P, f, β) < 2/3 < We conclude that f is Riemann-Stieltjes integrable with respect to β. A similar argument to that above shows that Z b f dβ = c1 I1 + c2 I2 . a 5