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] andthat ˆ b

a

fdα =

ˆ c

a

fdα+

ˆ b

c

fdα

Solution. Let ε > 0 and let P be a partition of [a, b] such that U(P, f, α) − L(P, f, α) < ε. Let P ′ =P∪{c}; that is, the partition formed from all the points of P and the point c. Then, U(P ′, f, α)−L(P ′, f, α) ≤U(P, f, α) − L(P, f, α) < ε as P ′ is a re�nement of P . Write P ′ = Q ∪ R where Q = {x ∈ P ′ : x ≤ c} andR = {x ∈ P ′ : x ≥ c}. Then, Q is a partition of [a, c], R is a partition of [c, b], and we have

U(P ′, f, α)− L(P ′, f, α) = (U(Q, f, α)− L(Q, f, α)) + (U(R, f, α)− L(R, f, α))

from which it follows that U(Q, f, α) − L(Q, f, α) < ε and U(R, f, α) − L(R, f, α) < ε as both of thesequantities are nonnegative. Hence, f ∈ R(α) on [a, c] and [c, b]. To compute the integral, we note that

ˆ b

a

fdα = inf(U(P, f, α))

where P ranges over all partitions of [a, b]. This can be replaced by inf(U(P ′, f, α)) where P ′ ranges over allpartitions of [a, b] containing c as to every P we can associate P ′ = P∪{c} with U(P ′, f, α) ≤ U(P, f, α). Now,if P ′ is such a partition, let Q, R be as above (so P ′ = Q∪R) and note U(P ′, f, α) = U(Q, f, α)+U(R, f, α).Hence,

infP ′U(P ′, f, α) = inf

P ′(U(Q, f, α) + U(R, f, α)) ≥ inf

P ′(U(Q, f, α)) + inf

P ′(U(R, f, α))

where we have identi�ed P ′ = Q ∪R as above. To prove that

infP ′

(U(Q, f, α) + U(R, f, α)) ≤ infP ′

(U(Q, f, α)) + infP ′

(U(R, f, α))

let ε > 0 and let Q1, R2 be partitions of [a, c], [c, b] arising from P ′1 and P ′2 (ie P ′j = Qj ∪ Rj for j = 1, 2)such that

U(Q1, f, α)− infP ′

(U(Q, f, α)) < ε

andU(R2, f, α)− inf

P ′(U(R, f, α)) < ε

Then, if P = Q1 ∪R2, we have

infP ′U(P ′, f, α) ≤ U(P ′, f, α) = U(Q1, f, α) + U(R2, f, α) < inf

P ′(U(Q, f, α)) + inf

P ′(U(R, f, α)) + 2ε

soinfP ′U(P ′, f, α) ≤ inf

P ′(U(Q, f, α)) + inf

P ′(U(R, f, α))

as ε > 0 was arbitrary. Lastly,

infP ′

(U(Q, f, α)) = infQ(U(Q, f, α)) =

ˆ c

a

fdα

and

infP ′

(U(R, f, α)) = infR(U(R, f, α)) =

ˆ b

c

fdα

(which I will not prove here) from which the result follows.

1

2. Let f, α : [a, b]→ R be two increasing functions. Suppose f ∈ R(α). Show that α ∈ R(f) and that

ˆ b

a

fdα+

ˆ b

a

αdf = f(b)α(b)− f(a)α(a)

Solution. If P is a partition of [a, b] given by a = x0 < x1 < · · · < xn = b, then see

supx∈[xj−1,xj ]

f(x) = f(xj) and infx∈[xj−1,xj ]

f(x) = f(xj−1)

and similarlysup

x∈[xj−1,xj ]

α(x) = α(xj) and infx∈[xj−1,xj ]

α(x) = α(xj−1)

Thus,

U(P, f, α)− L(P, f, α) =n∑j=1

(f(xj)− f(xj−1))(α(xj)− α(xj−1)) = U(P, α, f)− L(P, α, f)

Let ε > 0. As f ∈ R(α), we can �nd P such that U(P, f, α) − L(P, f, α) < ε. By the above, for this P wealso have U(P, α, f)− L(P, α, f) < ε, so α ∈ R(f).

Now, ˆ b

a

fdα+

ˆ b

a

αdf = infPU(P, f, α) + sup

PL(P, α, f)

We see, if P is notated as above, that

U(P, f, α) + L(P, α, f) =

n∑j=1

f(xj)(α(xj)− α(xj−1)) +n∑j=1

α(xj−1)(f(xj)− f(xj−1))

=

n∑j=1

f(xj)α(xj)− f(xj−1)α(xj−1)

= f(b)α(b)− f(a)α(a)

As we may take common re�nements in evaluating infP U(P, f, α)+supP L(P, α, f) (ie if Qm is a minimizingsequence of partitions for the inf, Rm a maximizing sequence of partitions for the sup, then we may evaluatethis expression by taking the limit as m→∞ with the partitions Pm = Qm ∪Rm), we see this expression isas claimed.

3. (Exercise 6.3) De�ne functions βj for j = 1, 2, 3 as follows: set βj(x) = 0 if x < 0 and βj(x) = 1 ifx > 0 for all j. Then, set β1(0) = 0, β2(0) = 1, and β3(0) =

12 . Let f be a bounded function on [−1, 1].

(a) Claim. f ∈ R(β1) if and only if f(0+) = f(0) and, in that case,´ 1−1 fdβ1 = f(0).

Proof. Let P be a partition of [−1, 1] given by −1 = x0 < x1 < · · · < xn = 1 and suppose 0 ∈ P (say0 = xp). Then,

U(P, f, β1)− L(P, f, β1) = ( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x))

as β1(xj)− β1(xj−1) is 0 if j 6= p+ 1 and 1 if j = p+ 1.

Suppose f ∈ R(β1). Let ε > 0. Then, there is a partition P with 0 ∈ P (eg take a re�nement if necessary)and U(P, f, β1)− L(P, f, β1) < ε. Put δ = xp+1 − xp. Then, if 0 ≤ x < δ, we have

|f(x)− f(0)| ≤ ( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x)) = U(P, f, β1)− L(P, f, β1) < ε

2

so f(0+) = f(0).

Now, suppose f(0+) = f(0). Let ε > 0 and let δ ∈ (0, 1) be such that 0 ≤ x < δ =⇒ |f(x)− f(0)| < ε.Then, for P given by −1, 0, δ2 , 1, we have

U(P, f, β1)−L(P, f, β1) = ( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x)) = ( supx∈[xp,xp+1]

f(x)−f(0)+f(0)− infx∈[xp,xp+1]

f(x)) ≤ 2ε

so f ∈ R(β1). In this case we see

infPU(P, f, β1) = inf

P( supx∈[xp,xp+1]

f(x)) = f(0+) = f(0)

(b) Claim. f ∈ R(β2) if and only if f(0−) = f(0) and, in that case,´ 1−1 fdβ2 = f(0).

Let P be a partition of [−1, 1] given by −1 = x0 < x1 < · · · < xn = 1 and suppose 0 ∈ P (say 0 = xp).Then,

U(P, f, β2)− L(P, f, β2) = ( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x))

as β1(xj)− β1(xj−1) is 0 if j 6= p and 1 if j = p.

Suppose f ∈ R(β2). Let ε > 0. Then, there is a partition P with 0 ∈ P (eg take a re�nement if necessary)and U(P, f, β2)− L(P, f, β2) < ε. Put δ = xp − xp−1. Then, if −δ < x ≤ 0, we have

|f(x)− f(0)| ≤ ( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x)) = U(P, f, β2)− L(P, f, β2) < ε

so f(0−) = f(0).

Now, suppose f(0−) = f(0). Let ε > 0 and let δ ∈ (0, 1) be such that −δ < x ≤ 0 =⇒ |f(x)−f(0)| < ε.Then, for P given by −1, −δ2 , 0, 1, we have

U(P, f, β2)−L(P, f, β2) = ( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x)) = ( supx∈[xp−1,xp]

f(x)−f(0)+f(0)− infx∈[xp−1,xp]

f(x)) ≤ 2ε

so f ∈ R(β2). In this case we see

infPU(P, f, β2) = inf

P( supx∈[xp−1,xp]

f(x)) = f(0−) = f(0)

(c) Claim. f ∈ R(β3) if and only if f is continuous at 0.

Proof. Let P be a partition of [−1, 1] given by −1 = x0 < x1 < · · · < xn = 1 and suppose 0 ∈ P (say0 = xp). Then,

U(P, f, β3)− L(P, f, β3) =1

2( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x)) +1

2( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x))

as β1(xj)− β1(xj−1) is 0 if j 6= p, p+ 1 and 12 if j = p, p+ 1.

Suppose f ∈ R(β3). Let ε > 0. Then, there is a partition P with 0 ∈ P (eg take a re�nement if necessary)and U(P, f, β3)− L(P, f, β3) < ε. Put δ = min(xp − xp−1, xp+1 − xp). Then, if |x| < δ, we have

|f(x)−f(0)| ≤ ( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x))+( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x)) = 2(U(P, f, β3)−L(P, f, β3)) < 2ε

3

so limx→0 f(x) = f(0).

Now, suppose limx→0 f(x) = f(0). Let ε > 0 and let δ ∈ (0, 1) be such that |x| < δ =⇒ |f(x)−f(0)| < ε.Then, for P given by −1, −δ2 , 0,

δ2 , 1, we have

U(P, f, β3)−L(P, f, β3) =1

2( supx∈[xp−1,xp]

f(x)− infx∈[xp−1,xp]

f(x))+1

2( supx∈[xp,xp+1]

f(x)− infx∈[xp,xp+1]

f(x)) ≤ 1

22ε+

1

22ε = 2ε

so f ∈ R(β3). In this case we see

infPU(P, f, β3) = inf

P(1

2( supx∈[xp−1,xp]

f(x)) +1

2( supx∈[xp,xp+1]

f(x)) = f(0)

(d) This was proven above.

4. (Exercise 6.8) Suppose f : [1,∞)→ [0,∞) is monotonically decreasing.

Claim.´∞1f(x)dx converges if and only if

∑∞n=1 f(n) converges.

Proof. We note by de�nition that´∞1f(x)dx converges if and only if limA→∞

´ A1f(x)dx converges.

Now,´ A1f(x)dx is an increasing function of A as f ≥ 0. Thus,

´∞1f(x)dx converges if and only if

´ A1f(x)dx

is bounded above uniformly in A.

Suppose∑∞n=1 f(n) converges. Let A > 1 and suppose m ≤ A < m+1 where m ∈ Z. Then,

´ A1f(x)dx ≤´m+1

1f(x)dx ≤ U(P, f, x) for any partition P of [1,m+1]. Consider P given by xk = k+1 for k = 0, . . . ,m.

Then, as f is decreasing,

U(P, f, x) =

m∑k=1

f(xk−1)(xk − xk−1) =m∑k=1

f(xk−1) =

m∑k=1

f(k) ≤∞∑k=1

f(k)

Thus,∑∞k=1 f(k) is an upper bound for

´ A1f(x)dx independent of A, so

´∞1f(x)dx converges.

Now, suppose∑∞n=1 f(n) diverges. Let M ∈ R. Chose m ∈ Z such that

∑mn=2 f(n) > M . Let P be the

partition of [1,m+ 1] given by xk = k + 1 for k = 0, . . . ,m. Then, as f is decreasing,

ˆ m+1

1

f(x)dx ≥ L(P, f, x) =m∑k=1

f(xk)(xk − xk−1) =m∑k=2

f(k) > M

Thus, as M was arbitrary,´ A1f(x)dx is not bounded above independent of A and so does not converge.

5. (Exercise 6.10 a-c) Let p, q > 0 be such that 1p +

1q = 1.

(a) Claim. If u ≥ 0, v ≥ 0, then uv ≤ up

p + vq

q . Moveover equality occurs if and only if up = vq.

Proof. Fix v ≥ 0 and consider the function f(u) = up

p + vq

q − uv de�ned on [0,∞). For u = 0, we see

this is vq

q ≥ 0. Di�erentiating with respect to u, we have f ′(u) = up−1− v. From 1p +

1q = 1, we see p > 1, so

this is increasing in u. Thus, f is decreasing in u until the critical point f ′(u0) = 0 and then increasing foru ≥ u0, so to show the result we evaluate f at u0 where f ′(u0) = 0. This is up−10 = v, which has a unique

solution for u0 ≥ 0, namely u0 = v1

p−1 . We have

f(u0) =up0p

+vq

q− u0v =

vp

p−1

p+vq

q− v

1p−1 v

4

Solving 1p +

1q = 1 for q gives q = p

p−1 , so

f(u0) =vq

p+vq

q− v

1+p−1p−1 = vq − vq = 0

Thus, f has a unique minimum (as f ′ is strictly increasing) at u = v1

p−1 , which is the same as up = vq, andat this point it attains the value 0.

(b) Suppose f, g ∈ R(α), f, g ≥ 0, and´ bafpdα = 1 =

´ bagqdα.

Claim.´ bafgdα ≤ 1

Proof. For x ∈ [a, b], we have 0 ≤ f(x), g(x), so by (a) we see f(x)g(x) ≤ f(x)p

p + g(x)q

q . Integrating over

[a, b] this becomes ˆ b

a

fgdα ≤ˆ b

a

fp

p+gq

qdα =

1

p+

1

q= 1

as desired.

(c) Suppose f, g ∈ R(α) are complex functions.

Claim. We have

|ˆ b

a

fgdα| ≤ (

ˆ b

a

|f |pdα)1/p(ˆ b

a

|g|qdα)1/q

Proof. We use the notation ‖f‖p = (´ ba|f |pdα)1/p and ‖g‖q = (

´ ba|g|qdα)1/q. First, suppose ‖f‖p , ‖g‖q >

0. Consider F = |f |‖f‖p

and G = |g|‖g‖q

. Then,

1

‖f‖p ‖g‖q|ˆ b

a

fgdα| ≤ 1

‖f‖p ‖g‖q

ˆ b

a

|fg|dα =

ˆ b

a

FGdα

We see ˆ b

a

F pdα =

ˆ b

a

|f |p

‖f‖ppdα =

´ ba|f |pdα´ b

a|f |pdα

= 1

and likewise´ baGqdα = 1. Moreover, F,G ≥ 0, so by (b) we have

´ baFGdα ≤ 1. With the above, this gives

1

‖f‖p ‖g‖q|ˆ b

a

fgdα| ≤ 1

so |´ bafgdα| ≤ ‖f‖p ‖g‖q as claimed.

Now, suppose ‖f‖p = 0. We show |´ bafgdα| = 0. Indeed, g is bounded, so |g| is bounded, say 0 ≤

|g| ≤ M . Then, |´ bafgdα| ≤ |

´ ba|f |Mdα, so it su�ces to show

´ ba|f |dα = 0. Now, if ‖f‖p = 0, then

‖f‖pp =´ ba|f |pdα = 0, so supP L(P, |f |p, α) = 0. Hence, for any �xed P we must have L(P, |f |p, α) = 0 (as

|f |p ≥ 0), so this shows that for every [xj−1, xj ] given by P , we either have

inf[xj−1,xj ]

(|f(x)|p) = 0 or α(xj)− α(xj−1) = 0

If P is a partition of [a, b] given by x0 < · · · < xn, we see

( inf[xj−1,xj ]

|f(x)|)p = inf[xj−1,xj ]

(|f(x)|p)

5

Hence, on [xj−1, xj ] we have

inf[xj−1,xj ]

|f(x)| = 0 or α(xj)− α(xj−1) = 0

so we have L(P, |f |, α) = 0 for all P . Thus,

ˆ b

a

|f |dα = suppL(P, |f |, α)P = 0

Likewise, if ‖g‖q = 0, then |´ bafgdα| = 0, so the claim holds.

6. (Exercise 6.11) For u ∈ R(α), let ‖u‖2 = (´ ba|u|2dα) 1

2 . Suppose f, g, h ∈ R(α). Show

‖f − h‖2 ≤ ‖f − g‖2 + ‖g − h‖2

Proof. Let F = f − h and G = g− h, so we wish to show ‖F +G‖2 ≤ ‖F‖2 + ‖G‖2. For this, it su�ces

to show ‖F +G‖22 ≤ (‖F‖2 + ‖G‖2)2. From the left hand side, we have

ˆ b

a

|F +G|2dα ≤ˆ b

a

|F |2 + 2|F ||G|+ |G|2dα = ‖F‖22 + 2

ˆ b

a

|F ||G|dα+ ‖G‖22

Now, from exercise 6.10, we have 2´ ba|F ||G|dα ≤ 2 ‖F‖2 ‖G‖2, so we have

‖F +G‖22 ≤ ‖F‖22 + 2 ‖F‖2 ‖G‖2 + ‖G‖

22 = (‖F‖2 + ‖G‖2)

2

as claimed.

7. (Exercise 6.13) De�ne f(x) =´ x+1

xsin(t2)dt

(a) Claim. |f(x)| < 1x for x > 0

Proof. Let u = t2, so du = 2tdt and

ˆ x+1

x

sin(t2)dt =

ˆ (x+1)2

x2

sin(u)du

2u12

Integrating by parts, this is

−cos(u)

2u12

|(x+1)2

u=x2 −ˆ (x+1)2

x2

cos(u)du

4u32

=cos(x2)

2x− cos((x+ 1)2)

2(x+ 1)−ˆ (x+1)2

x2

cos(u)du

4u32

Hence,

|f(x)| < 1

2x+

1

2(x+ 1)+

ˆ (x+1)2

x2

du

4u32

=1

2x+

1

2(x+ 1)+(−12u

12

)|(x+1)2

u=x2 =1

2x+

1

2(x+ 1)+

1

2x− 1

2(x+ 1)=

1

x

Note the inequality is strict here as both cos(u) 1

4u32and 1

4u32are continuous on [x, x+1] and these functions

are not the same (see exercise 6.2).

(b) Claim. 2xf(x) = cos(x2)− cos((x+ 1)2) + r(x) where |r(x)| < cx for some constant c.

Proof. Put r(x) = 2xf(x)− cos(x2) + cos((x+ 1)2). By the above,

r(x) =−2x cos((x+ 1)2)

2(x+ 1)+ cos((x+ 1)2)− 2x

ˆ (x+1)2

x2

cos(u)du

4u32

=cos((x+ 1)2)

(x+ 1)− 2x

ˆ (x+1)2

x2

cos(u)du

4u32

6

and so

|r(x)| ≤ 1

x+ 1+ 2x

ˆ (x+1)2

x2

du

4u32

=1

x+ 1+ 2x(

1

2x− 1

2(x+ 1)) =

1

x+ 1+ x(

x+ 1− xx(x+ 1)

) =2

x+ 1

so c = 2 works.

(c) Claim. The upper and lower limits of xf(x) are 1 and −1 respectively.

Proof. We know x|f(x)| < 1 by (a). From (b), we have xf(x) = 12 (cos(x

2)− cos((x+1)2)+ r(x)) wherer(x)→ 0 as x→∞, so it su�ces to show the upper limit of cos(x2)− cos((x+ 1)2) is 2 and the lower limitof cos(x2)− cos((x+ 1)2) is −2. Consider the sequence formed by taking xn =

√2πn for n ∈ Z+. We have

cos(x2n)− cos((xn + 1)2) = cos(2πn)− cos(2πn+ 2√2πn+ 1) = 1− cos(2

√2πn+ 1)

where we have used cos(A+B) = cos(A) cos(B)− sin(A) sin(B). Hence, we need only show cos(2√2πn+1)

has lower limit −1. We know that this has lower limit ≥ −1. Let ε > 0. Let δ > 0 be such that|x− π| < δ =⇒ | cos(x) + 1| < ε. Then, if |x− (2m+ 1)π| < δ, we have | cos(x) + 1| < ε by periodicity. LetM > 0 and �nd N > M such that 2

√2π(n+ 1)− 2

√2πn < δ for n ≥ N . Then, as 2

√2πn→∞as n→∞,

we see there is an x ≥M with x = 2√2πn for some n ≥ N and |x− (2m+ 1)π| < δ for some m ∈ Z. Thus,

| cos(2√2πn) + 1| < ε, from which it follows that the lower limit of cos(2

√2πn + 1) is −1. The proof that

the lower limit of xf(x) = −1 is similar with√π(2n+ 1) in place of

√2πn.

(d) Claim.´∞0

sin(t2)dt converges.

Proof. It su�ces to show´∞1

sin(t2)dt converges. As above, we have

ˆ A

1

sin(t2)dt =cos(1)

2− cos(A2)

2A−ˆ A2

1

cos(u)

u32

du

and´ A2

1cos(u)

u32du is absolutely convergent, cos(A2)

2A → 0 as A→∞, so this converges.

7