Homework 1 Solutions. 1. (Theorem 6.12(c)) Let f R(α) on [a, b] and a < c < b. Show f R(α) on [a, c] and [c, b] and that

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1 Homework Solutions. (Theorem 6.(c)) Let f R(α) on [, b] nd < c < b. Show f R(α) on [, c] nd [c, b] nd tht fdα = ˆ c fdα + c fdα Solution. Let ɛ > 0 nd let be rtition of [, b] such tht U(, f, α) L(, f, α) < ɛ. Let = {c}; tht is, the rtition formed from ll the oints of nd the oint c. Then, U(, f, α) L(, f, α) U(, f, α) L(, f, α) < ɛ s is renement of. Write = Q R where Q = {x : x c} nd R = {x : x c}. Then, Q is rtition of [, c], R is rtition of [c, b], nd we hve U(, f, α) L(, f, α) = (U(Q, f, α) L(Q, f, α)) + (U(R, f, α) L(R, f, α)) from which it follows tht U(Q, f, α) L(Q, f, α) < ɛ nd U(R, f, α) L(R, f, α) < ɛ s both of these untities re nonnegtive. Hence, f R(α) on [, c] nd [c, b]. To comute the integrl, we note tht fdα = (U(, f, α)) where rnges over ll rtitions of [, b]. This cn be relced by (U(, f, α)) where rnges over ll rtitions of [, b] contining c s to every we cn ssocite = {c} with U(, f, α) U(, f, α). Now, if is such rtition, let Q, R be s bove (so = Q R) nd note U(, f, α) = U(Q, f, α)+u(r, f, α). Hence, U(, f, α) = where we hve identied = Q R s bove. To rove tht (U(Q, f, α) + U(R, f, α)) (U(Q, f, α)) + (U(R, f, α)) (U(Q, f, α) + U(R, f, α)) (U(Q, f, α)) + (U(R, f, α)) let ɛ > 0 nd let Q, R be rtitions of [, c], [c, b] rising from nd (ie j = Q j R j for j =, ) such tht U(Q, f, α) (U(Q, f, α)) < ɛ nd Then, if = Q R, we hve so U(R, f, α) (U(R, f, α)) < ɛ U(, f, α) U(, f, α) = U(Q, f, α) + U(R, f, α) < s ɛ > 0 ws rbitrry. Lstly, nd U(, f, α) (U(Q, f, α)) + (U(R, f, α)) (U(Q, f, α)) = (U(Q, f, α)) = Q (U(R, f, α)) = (U(R, f, α)) = R (which I will not rove here) from which the result follows. (U(Q, f, α)) + (U(R, f, α)) + ɛ ˆ c c fdα fdα

2 . Let f, α : [, b] R be two incresing functions. Suose f R(α). Show tht α R(f) nd tht fdα + αdf = f(b)α(b) f()α() Solution. If is rtition of [, b] given by = x 0 < x < < x n = b, then see nd similrly Thus, U(, f, α) L(, f, α) = f(x) = f(x j ) nd x [x j,x j] α(x) = α(x j ) nd x [x j,x j] x [x j,x j] f(x) = f(x j ) x [x j,x j] α(x) = α(x j ) (f(x j ) f(x j ))(α(x j ) α(x j )) = U(, α, f) L(, α, f) j= Let ɛ > 0. As f R(α), we cn nd such tht U(, f, α) L(, f, α) < ɛ. By the bove, for this we lso hve U(, α, f) L(, α, f) < ɛ, so α R(f). Now, fdα + We see, if is notted s bove, tht U(, f, α) + L(, α, f) = = αdf = U(, f, α) + L(, α, f) f(x j )(α(x j ) α(x j )) + j= α(x j )(f(x j ) f(x j )) j= f(x j )α(x j ) f(x j )α(x j ) j= = f(b)α(b) f()α() As we my tke common renements in evluting U(, f, α)+ L(, α, f) (ie if Q m is minimizing seuence of rtitions for the, R m mximizing seuence of rtitions for the, then we my evlute this exression by tking the limit s m with the rtitions m = Q m R m ), we see this exression is s climed. 3. (Exercise 6.3) Dene functions β j for j =,, 3 s follows: set β j (x) = 0 if x < 0 nd β j (x) = if x > 0 for ll j. Then, set β (0) = 0, β (0) =, nd β 3 (0) =. Let f be bounded function on [, ]. () Clim. f R(β ) if nd only if f(0+) = f(0) nd, in tht cse, fdβ = f(0). roof. Let be rtition of [, ] given by = x 0 < x < < x n = nd ose 0 (sy 0 = x ). Then, U(, f, β ) L(, f, β ) = ( f(x) f(x)) x [x,x +] s β (x j ) β (x j ) is 0 if j + nd if j = +. x [x,x +] Suose f R(β ). Let ɛ > 0. Then, there is rtition with 0 (eg tke renement if necessry) nd U(, f, β ) L(, f, β ) < ɛ. ut δ = x + x. Then, if 0 x < δ, we hve f(x) f(0) ( x [x,x +] f(x) f(x)) = U(, f, β ) L(, f, β ) < ɛ x [x,x +]

3 so f(0+) = f(0). Now, ose f(0+) = f(0). Let ɛ > 0 nd let δ (0, ) be such tht 0 x < δ = f(x) f(0) < ɛ. Then, for given by, 0, δ,, we hve U(, f, β ) L(, f, β ) = ( x [x,x +] so f R(β ). In this cse we see f(x) f(x)) = ( x [x,x +] U(, f, β ) = ( x [x,x +] x [x,x +] f(x)) = f(0+) = f(0) f(x) f(0)+f(0) x [x,x +] f(x)) ɛ (b) Clim. f R(β ) if nd only if f(0 ) = f(0) nd, in tht cse, fdβ = f(0). Let be rtition of [, ] given by = x 0 < x < < x n = nd ose 0 (sy 0 = x ). Then, U(, f, β ) L(, f, β ) = ( f(x) f(x)) x [x,x ] s β (x j ) β (x j ) is 0 if j nd if j =. x [x,x ] Suose f R(β ). Let ɛ > 0. Then, there is rtition with 0 (eg tke renement if necessry) nd U(, f, β ) L(, f, β ) < ɛ. ut δ = x x. Then, if δ < x 0, we hve so f(0 ) = f(0). f(x) f(0) ( x [x,x ] f(x) f(x)) = U(, f, β ) L(, f, β ) < ɛ x [x,x ] Now, ose f(0 ) = f(0). Let ɛ > 0 nd let δ (0, ) be such tht δ < x 0 = f(x) f(0) < ɛ. Then, for given by, δ, 0,, we hve U(, f, β ) L(, f, β ) = ( x [x,x ] so f R(β ). In this cse we see f(x) f(x)) = ( x [x,x ] U(, f, β ) = ( x [x,x ] x [x,x ] f(x)) = f(0 ) = f(0) f(x) f(0)+f(0) x [x,x ] f(x)) ɛ (c) Clim. f R(β 3 ) if nd only if f is continuous t 0. roof. Let be rtition of [, ] given by = x 0 < x < < x n = nd ose 0 (sy 0 = x ). Then, U(, f, β 3 ) L(, f, β 3 ) = ( x [x,x ] f(x) f(x)) + x [x,x ] ( x [x,x +] f(x) f(x)) x [x,x +] s β (x j ) β (x j ) is 0 if j, + nd if j =, +. Suose f R(β 3 ). Let ɛ > 0. Then, there is rtition with 0 (eg tke renement if necessry) nd U(, f, β 3 ) L(, f, β 3 ) < ɛ. ut δ = min(x x, x + x ). Then, if x < δ, we hve f(x) f(0) ( x [x,x ] f(x) f(x))+( x [x,x ] x [x,x +] f(x) f(x)) = (U(, f, β 3) L(, f, β 3 )) < ɛ x [x,x +] 3

4 so lim x 0 f(x) = f(0). Now, ose lim x 0 f(x) = f(0). Let ɛ > 0 nd let δ (0, ) be such tht x < δ = f(x) f(0) < ɛ. Then, for given by, δ, 0, δ,, we hve U(, f, β 3 ) L(, f, β 3 ) = ( so f R(β 3 ). In this cse we see x [x,x ] U(, f, β 3) = ( ( f(x) f(x))+ x [x,x ] ( x [x,x ] f(x)) + ( x [x,x +] x [x,x +] f(x) f(x)) x [x,x +] ɛ+ ɛ = ɛ f(x)) = f(0) (d) This ws roven bove. 4. (Exercise 6.8) Suose f : [, ) [0, ) is monotoniclly decresing. Clim. f(x)dx converges if nd only if n= f(n) converges. roof. We note by denition tht A f(x)dx converges if nd only if lim A f(x)dx converges. Now, A f(x)dx is n incresing function of A s f 0. Thus, f(x)dx converges if nd only if A f(x)dx is bounded bove uniformly in A. Suose n= f(n) converges. Let A > nd ose m A < m+ where m Z. Then, A f(x)dx m+ f(x)dx U(, f, x) for ny rtition of [, m + ]. Consider given by x k = k + for k = 0,..., m. Then, s f is decresing, U(, f, x) = f(x k )(x k x k ) = f(x k ) = f(k) f(k) k= Thus, k= f(k) is n uer bound for A f(x)dx indeendent of A, so f(x)dx converges. Now, ose n= f(n) diverges. Let M R. Chose m Z such tht m n= f(n) > M. Let be the rtition of [, m + ] given by x k = k + for k = 0,..., m. Then, s f is decresing, ˆ m+ k= k= k= k= f(x)dx L(, f, x) = f(x k )(x k x k ) = f(k) > M Thus, s M ws rbitrry, A f(x)dx is not bounded bove indeendent of A nd so does not converge. 5. (Exercise 6.0 -c) Let, > 0 be such tht + =. () Clim. If u 0, v 0, then uv u k= + v. Moveover eulity occurs if nd only if u = v. + v uv dened on [0, ). For u = 0, we see roof. Fix v 0 nd consider the function f(u) = u 0. Dierentiting with resect to u, we hve f (u) = u v. From + this is v =, we see >, so this is incresing in u. Thus, f is decresing in u until the criticl oint f (u 0 ) = 0 nd then incresing for u u 0, so to show the result we evlute f t u 0 where f (u 0 ) = 0. This is u 0 = v, which hs uniue solution for u 0 0, nmely u 0 = v. We hve f(u 0 ) = u 0 + v u 0v = v 4 + v v v

5 Solving + = for gives =, so f(u 0 ) = v + v v + = v v = 0 Thus, f hs uniue minimum (s f is strictly incresing) t u = v, which is the sme s u = v, nd t this oint it ttins the vlue 0. (b) Suose f, g R(α), f, g 0, nd b f dα = = b g dα. Clim. b fgdα roof. For x [, b], we hve 0 f(x), g(x), so by () we see f(x)g(x) f(x) [, b] this becomes f fgdα + g dα = + = s desired. (c) Suose f, g R(α) re comlex functions. + g(x). Integrting over Clim. We hve fgdα ( f dα) / ( g dα) / b roof. We use the nottion f = ( f dα) / b nd g = ( g dα) /. First, ose f, g > 0. Consider F = f f nd G = g g. Then, We see f g fgdα f g F dα = fg dα = f b f dα = f dα b f dα = F Gdα nd likewise b G dα =. Moreover, F, G 0, so by (b) we hve b F Gdα. With the bove, this gives so b fgdα f g s climed. f g fgdα Now, ose f = 0. We show b fgdα = 0. Indeed, g is bounded, so g is bounded, sy 0 g M. Then, b fgdα b f Mdα, so it suces to show b f dα = 0. Now, if f = 0, then f = b f dα = 0, so L(, f, α) = 0. Hence, for ny xed we must hve L(, f, α) = 0 (s f 0), so this shows tht for every [x j, x j ] given by, we either hve [x ( f(x) ) = 0 or α(x j ) α(x j ) = 0 j,x j] If is rtition of [, b] given by x 0 < < x n, we see ( [x j,x j] f(x) ) = [x j,x j] ( f(x) ) 5

6 Hence, on [x j, x j ] we hve so we hve L(, f, α) = 0 for ll. Thus, f(x) = 0 or α(x j) α(x j ) = 0 [x j,x j] f dα = L(, f, α) = 0 Likewise, if g = 0, then b fgdα = 0, so the clim holds. 6. (Exercise 6.) For u R(α), let u = ( b u dα). Suose f, g, h R(α). Show f h f g + g h roof. Let F = f h nd G = g h, so we wish to show F + G F + G. For this, it suces to show F + G ( F + G ). From the left hnd side, we hve F + G dα F + F G + G dα = F + F G dα + G Now, from exercise 6.0, we hve b F G dα F G, so we hve s climed. F + G F + F G + G = ( F + G ) 7. (Exercise 6.3) Dene f(x) = x+ sin(t )dt x () Clim. f(x) < x for x > 0 roof. Let u = t, so du = tdt nd Integrting by rts, this is ˆ x+ x sin(t )dt = ˆ (x+) x sin(u) du u Hence, cos(u) ˆ (x+) (x+) u u=x cos(u) du x 4u 3 = cos(x ) x cos((x + ) ) (x + ) ˆ (x+) cos(u) du x 4u 3 f(x) < ˆ x + (x+) (x + ) + x du 4u 3 Note the ineulity is strict here s both cos(u) re not the sme (see exercise 6.). = x + +( ) (x+) (x + ) u u=x = x + (x + ) + x (x + ) = x 4u 3 nd 4u 3 re continuous on [x, x + ] nd these functions (b) Clim. xf(x) = cos(x ) cos((x + ) ) + r(x) where r(x) < c x for some constnt c. roof. ut r(x) = xf(x) cos(x ) + cos((x + ) ). By the bove, r(x) = x cos((x + ) ) (x + ) ˆ (x+) + cos((x + ) ) x cos(u) du x 4u 3 6 = cos((x + ) ) (x + ) ˆ (x+) x cos(u) du x 4u 3

7 nd so r(x) ˆ (x+) x + + x x du 4u 3 = x + + x( x (x + ) ) = x + + x(x + x x(x + ) ) = x + so c = works. (c) Clim. The uer nd lower limits of xf(x) re nd resectively. roof. We know x f(x) < by (). From (b), we hve xf(x) = (cos(x ) cos((x + ) ) + r(x)) where r(x) 0 s x, so it suces to show the uer limit of cos(x ) cos((x + ) ) is nd the lower limit of cos(x ) cos((x + ) ) is. Consider the seuence formed by tking x n = πn for n Z +. We hve cos(x n) cos((x n + ) ) = cos(πn) cos(πn + πn + ) = cos( πn + ) where we hve used cos(a + B) = cos(a) cos(b) sin(a) sin(b). Hence, we need only show cos( πn + ) hs lower limit. We know tht this hs lower limit. Let ɛ > 0. Let δ > 0 be such tht x π < δ = cos(x) + < ɛ. Then, if x (m + )π < δ, we hve cos(x) + < ɛ by eriodicity. Let M > 0 nd nd N > M such tht π(n + ) πn < δ for n N. Then, s πn s n, we see there is n x M with x = πn for some n N nd x (m + )π < δ for some m Z. Thus, cos( πn) + < ɛ, from which it follows tht the lower limit of cos( πn + ) is. The roof tht the lower limit of xf(x) = is similr with π(n + ) in lce of πn. (d) Clim. 0 sin(t )dt converges. roof. It suces to show sin(t )dt converges. As bove, we hve nd A cos(u) u 3 ˆ A sin(t )dt = cos() du is bsolutely convergent, cos(a ) A cos(a ) A ˆ A cos(u) du u 3 0 s A, so this converges. 7