Math_Analysis_HW9_Solutions.pdf - Real Analysis Homework IX solutions November 22nd 2019 Exercise 5.1.1 Let f[0 1 \u2192 R be defined by f(x = x3 and let P

Math_Analysis_HW9_Solutions.pdf - Real Analysis Homework IX...

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Unformatted text preview: Real Analysis Homework IX solutions November 22nd, 2019 Exercise 5.1.1 Let f : [0, 1] → R be defined by f (x) = x3 and let P = {0, 0.1, 0.4, 1}. Compute L(P, f ) and U (P, f ). We must compute the data mi , Mi , and ∆xi . Since f is monotone increasing , mi is achieved at the left endpoint of [xi−1 , xi ] and Mi is achieved at the right endpoint . Hence, m1 = 0 m2 = 0.001 m3 = 0.064 M1 = 0.001 M2 = 0.064 M3 = 1 ∆x1 = 0.1 − 0 = 0.1 ∆x2 = 0.4 − 0.1 − 0.3 ∆x3 = 1 − 0.4 = 0.6 Finally, we compute the upper and lower sums: L(P, f ) = U (P, f ) = 3 X i=1 3 X m1 ∆xi = 0.0387 M1 ∆xi = 0.6193 i=1 Exercise 5.1.2 R1 Let f : [0, 1] → R be defined by f (x) = x. Show that f ∈ R[0, 1] and compute 0 f using the definition of the Riemann integral (but feel free to use any of the propositions from this section). R1 R1 R1 Proof. We will show that f ∈ R[0, 1] with 0 f = 0 f = 0 f = 12 . Fix n ∈ N, and let Pn consist of n + 1 evenly spaced points in [0, 1]:   1 2 n−1 Pn = 0, , · · · , ,1 n n n 1 We will compute L(Pn , f ) and U (Pn , f ). Then,   i−1 i i−i mi = inf x : ≤x≤ = n n n   i−1 i i Mi = sup x : ≤x≤ = n n n and so L(Pn , f ) = n X mi ∆xi = i=1 U (Pn , f ) = n X i=1 n 1 X n(n − 1) (i − 1) = 2 n i=1 2n2 n 1 X n(n + 1) Mi ∆xi = 2 i= n i=1 2n2 Pn n(n+1) (using the result that ). Thus, by proposition 5.1.8 and the definitions of k=1 k = 2 the upper and lower Darboux sums: 1 Z L(Pn , f ) ≤ f≤ f ≤ U (Pn , f ) 0 n(n − 1) ≤ 2n2 1 Z Z 0 1 Z f≤ 1 f≤ 0 0 n(n + 1) 2n2 Taking n → ∞ and using the fact that limits preserve inequalities, we find that Z 1 Z f= 0 Hence, f ∈ R[0, 1] and R1 0 1 f= 0 1 2 f = 12 . Exercise 5.1.3 Let f : [a, b] → R be a bounded function, and suppose there exists a sequence of partitions {Pk }∞ k=1 such that lim U (Pk , f ) − L(Pk , f ) = 0 k→∞ Show that f is Riemann integrable with Z b f = lim U (Pk , f ) = lim L(Pk , f ). a k→∞ k→∞ Proof. We first show f is Riemann integrable. Proposition 5.1.13 says that it is enough to show that for every  > 0 there exists a k such that U (Pk , f ) − L(Pk , f ) <  . But, since limk→∞ U (Pk , f ) − L(Pk , f ) = 0, there exists an N such that for every n ≥ N , 2 |U (Pk , f ) − L(Pk , f )| < . In particular, for k = N , U (Pk , f ) − L(Pk , f ) <  and so f is Riemann integrable. Rb We now need to show the second assertion. Notice that since a f is both an upper bound of all of the lower sums and a lower bound to the upper sums, we have Z b 0≤ f − L(Pk , f ) ≤ U (Pk , f ) − L(Pk , f ) a Rb The Squeeze theorem, implies that limk→∞ a f − L(Pk , f ) = 0, so, Z b f = lim L(Pk , f ) a k→∞ It follows by limit arithmetic that limk→∞ U (Pk , f ) exists as well, and Z b f = lim L(Pk , f ) = lim U (Pk , f ) a k→∞ k→∞ Exercise 5.1.4 Finish the proof of 5.1.7. That is, prove that if P˜ is a refinement of P , then U (P˜ , f ) ≤ U (P, f ). The proof can be obtained in much that same ways at the argument for lower sums (up to changing infs to sups and reversing inequalities). However, it is less work to use the following argument to get it ‘for free’ from the argument for lower sums: Proof. It has already been shown that L(P˜ , f ) ≥ L(P, f ). We claim that U (Q, f ) = −L(Q, −f ) for any function f and partition Q. Assuming this is true, then we have that L(P˜ , −f ) ≥ L(P, −f ) from the book, so multiplying by −1 gives the result. We now prove the claim. Let Q = {x0 , x1 , · · · , xn } be a partition. Then, mi (−f ) = inf{−f (x) : x ∈ [xi−1 , xi ]} = − sup{f (x) : x ∈ [xi−1 , xi ]} = −Mi (f ) To see why, notice if y = sup{f (x) : x ∈ [xi−1 , xi ]}, then −y ≤ f (x) for all x ∈ [xi−1 , xi ], then certainly −y is a lower bound of {−f (x) : x ∈ [xi−1 , xi ]}. Moreover, if z is another lower bound for {−f (x) : x ∈ [xi−1 , xi ]}, then −z is an upper bound for {f (x) : x ∈ [xi−1 , xi ]}, so −z ≥ y by the definition of sup. Thus, −y ≥ z for all lower bounds, so −y = inf{−f (x) : x ∈ [xi−1 , xi ]}. Hence, −L(Q, −f ) = − n X i=1 mi (−f )∆xi = n X −mi (−f )∆xi = i=1 n X i=1 3 Mi (f )∆xi = U (Q, f ) Exercise 5.1.5 Suppose f : [−1, 1] → R is defined as ( 1 x>0 f (x) = 0 x≤0 Prove that f ∈ R[−1, 1] and compute R1 −1 f using the definition of the integral. Proof. For each  > 0, let P be the partition P = { −1, 0, , 1 }. Then, U (P , f ) = 0(1) + 1() + 1(1 − ) = 1 L(P , f ) = 0(1) + 0() + 1(1 − ) = 1 − . Thus, Proposition 5.1.8 and the definition of the lower and upper Darboux integrals imply that Z b Z b 1 −  = L(P , f ) ≤ f≤ f ≤ U (P , f ) = 1. a a Taking  → 0, the Squeeze theorem implies that Rb Riemann integrable with a f = 1. Rb a f = Rb a f = 1 so, by definition, f is Exercise 5.1.6 Let c ∈ (a, b) and d ∈ R. Define f : [a, b] → R by ( d x=c f (x) = 0 else Prove that f ∈ R[a, b] and compute Rb a f using the definition of the integral. Proof. We will prove that f ∈ R[a, b] with Pn = {a, c − Rb a f = 0. Let Pn be a partition given by c−a b−c ,c + , b} n+1 n+1 4 For this partition, mi,n = inf{f (x) : x ∈ [xi , xi+1 ] = 0 c−a M1,n = sup{f (x) : x ∈ [a, c −=0 n+1 c−a b−c M2,n = sup{f (x) : x ∈ [c − ,c +=d n+1 n+1 b−c M3,n = sup{f (x) : x ∈ [c + , b] = 0 n+1   c−a n ∆x1 = c − −a= (c − a) n+1 n+1     b−c c−a b−a ∆x2 = c + − c− = n+1 n+1 n+1   b−c n ∆x1 = b − c − = (b − c) n+1 n+1 In particular, using Proposition 5.1.8 and the definition of the Darboux sums, Z b 0 = L(Pn , f ) ≤ f≤ f ≤ U (Pn , f ) = a a Taking n → ∞, we find that Rb a f= b Z d(b − a) n+1 f = 0 , so f ∈ R[a, b] with a Rb Rb a f = 0. Exercise 5.1.7 Suppose f : [a, b] → R is Riemann integrable. Let  > 0 be given. Then show that there exists a partition P = {x0 , x1 , · · · , xn } such that for any choice of numbers {c1 , c2 , · · · , cn } with ck ∈ [xk−1 , xk ] for all k = 1, 2, · · · , n, then Z n b X f − f (c )∆x k k <  a k=1 Proof. To begin, we find a partition ‘fine enough’ that L(P, f ) and U (P, f ) are good apRb Rb proximations of a f . Let  > 0 be given. Since f is Riemann integrable, a f is the sup Rb of the set of lower sums. Thus, there exists a partition PL such that a f − L(PL , f ) <  . Rb Similarly, a f is the inf of the set of upper sums, so there exists a partition PU such that Rb U (PU , f ) − a f <  . Let P be the common refinement of PL and PU , P = PL ∪ PU . Then, Z b f − L(P, f ) <  and a Z U (P, f ) − f < a 5 b by proposition 5.1.7. Write P as an ordered list of points P = {x0 , x1 , · · · , xn }, and suppose we have points {c1 , c2 , · · · , ck } with each ck ∈ [xk−1 , xk ]. Then, mk = inf x∈[xk−1 ,xk ] f (x) ≤ f (ck ) ≤ sup f (x) = Mk . x∈[xk−1 ,xk ] Multiplying by ∆xk and summing gives Z b f −  < L(P, f ) = a ∞ X k=1 so mk ∆xk ≤ ∞ X f (ck )∆xk ≤ k=1 ∞ X Z Mk ∆xk = U (P, f ) < k=1 b f + . a Z n b X f− f (ck )∆xk < . a k=1 Exercise 5.1.8 Let f : [a, b] → R be a Riemann integrable  a−β b−β function. Let α > 0 and β ∈ R. Define g(x) = f (αx + β) on the interval I = α , α . Show that g is Riemann integrable on I. Proof. We will make use of proposition 5.1.13. Let  > 0 be arbitrary. Since f is Riemann integrable, there exists a partition P = {x0 , x1 , · · · , xn } such that U (P, f ) − L(P, f ) < α  Consider the partition Q = {y0 , y1 , · · · , yn } = x0α−β , x1α−β , · · · , xnα−β . Then, Q is a partition of I. Moreover, we have that we can express the ‘partition data’ for (Q, g) in terms of the data for (P, f ): mi (g) = inf{g(y) : y ∈ [yi−1 , yi ]}    xi−1 − β xi − β = inf f (αy + β) : y ∈ , α α = inf{f (x) : x ∈ [xi−1 , xi ]} = mi (f ) Mi (g) = sup{g(y) : y ∈ [yi−1 , yi ]}    xi−1 − β xi − β = sup f (αy + β) : y ∈ , α α = sup{f (x) : x ∈ [xi−1 , xi ]} = Mi (f ) xi − β xi−1 − β xi − xi−1 1 ∆yi = yi − yi−1 = − = = ∆xi α α α α 6 Thus, L(Q, g) = U (Q, g) = n X n X mi (g)∆yi = i=1 i=1 n X n X Mi (g)∆yi = i=1 i=1 so U (Q, g) − L(Q, g) = and g is Riemann integrable on I. 1 1 mi (f ) ∆xi = L(P, f ) α α 1 1 Mi (f ) ∆xi = U (P, f ) α α 1 (U (P, f ) − L(P, f )) <  α Exercise 5.1.9 Suppose f, g : [0, 1] → R are such that for all x ∈ (0, 1], we R 1 haveR f1 (x) = g(x). Suppose f is Riemann integrable. Prove g is Riemann integrable and 0 f = 0 g. Proof. We will show Z 1 1 Z g≥ 0 and f, 1 Z g≤ 0 0 from which it will follow that Z 1 1 Z g= 0 Z g= 0 Z 1 f 0 1 f 0 R1 R1 so g is Riemann integrable with 0 g = 0 f . Since g(x) = f (x) on (0, 1], we have max |g(x)| ≤ max |f (x)| + |g(0)| ≤ B x∈[0,1] x∈[0,1] for some B ∈ R. Let  > 0 be arbitrary, and let P be a partition such that Z b Z b  f − L(P, f ) <  and U (P, f ) − f< 2 a a (This can be done as in problem 5.1.7 by choosing different partitions PL and PU for the   lower and upper sum and taking the common refinement P ). Now, let P˜ = P ∪ 4B . Then, Z b f − L(P˜ , g) = a Z b    ˜ ˜ ˜ f − L(P , f ) + L(P , f ) − L(P , g) a Since P˜ is a refinement of P , the first term in parentheses is less than /2. The second term can be rewritten as L(P˜ , f ) − L(P˜ , g) = n X    (mk (f ) − mk (g))∆xk = (m1 (f ) − m1 (g)) ≤ 2B ≤ 4B 4B 2 k=1 7 since g(x) = f (x) in [xk−1 , xk ] for k ≥ 2, so mk (g) = mk (f ) for k ≥ 2. Thus, Z b f − L(P˜ , f ) <  a Since  > 0 was arbitrary, this shows that Z b Z b g = sup{L(P, g) : P is a partition} ≥ f. a a Similar reasoning with the upper sums shows that Z b g = inf{U (P, g) : P is a partition} ≤ a Z b f a Rb Rb Rb Rb Rb Since Proposition 5.1.8 states that a g ≤ a g, this shows that a g = a g = a f so g is Rb Rb Riemann integrable with a g = a f . Exercise 5.1.10 Let f : [0, 1] → R be a bounded function. Let Pn = {x0 , x1 , · · · , xn } be the uniform partition; that is, xj = nj . Is {L(Pn , f )}∞ n=1 always monotone? No. Consider ( 1 13 ≤ x ≤ 23 f (x) = 0 else Then, whenever Pn contains the points 13 and 23 (which happens when n is a multiple of 3), R1 L(Pn , f ) = 13 = 0 f . However, for n not a multiple of three, L(P, f ) < 13 , since over the subintervals containing 13 and 23 , mi = 0. In particular, L(P1 , f ) = 0 L(P2 , f ) = 0 1 L(P3 , f ) = 3 L(P4 , f ) = 0 so L(Pn , f ) cannot be monotone. 8 ...
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