Homework 1 Solutions
• Problem 13.10
Solution. We know that for any x1 , x2 ∈ [ti−1 , ti ],
m0i ≤ f (x1 ),
m00i ≤ g(x2 ).
and
Consequently,
m0i + m00i ≤ f (x1 ) + g(x2 )
for all x1 , x2 ∈ [ti−1 , ti ], hence m0i + m00i is a lower bound for the set
{f (x1 ) + g(x2 ) : x1 , x2 ∈ [ti−1 , ti ]},
so we must have that
m0i + m00i ≤ inf{f (x1 ) + g(x2 ) : x1 , x2 ∈ [ti−1 , ti ]}
from the definition of infimum as the greatest lower bound. Now we note that
inf{f (x1 ) + g(x2 ) : x1 , x2 ∈ [ti−1 , ti ]} ≤ inf{f (x) + g(x) : x ∈ [ti−1 , ti ]} = mi
since the set of numbers on the left contains the one on the right (hence the infimum is smaller).
• Problem 13.11
Solution.
a) Constant functions (seen in class).
b) Piecewise constant functions. For upper and lower sums to coincide,
we need Mi = mi
P
(M
for each subinterval in the partition, since U (f, P ) − L(f, P ) = N
i − mi )(ti − ti−1 ).
i=1
Thus, the upper and lower sums will coincide provided the function is constant on each
interval in the partition.
c) Constant functions. Let m be the minimum value for f on [a, b]. For the partition
P = {a, b}, we have L(f, P ) = m(b − a). Suppose that f (x) > m for x ∈ [a, b], then there
exists a partion with ti−1 < x < ti such that mi > m. Hence the lower sum with respect
to this partition satisfies L(f, P ) > m(b − a).
• Problem 13.16
Solution. We assume that c > 0. For every partition P = {t1 , . . . , tn } of [a, b], we observe
cP = {ct1 , . . . , ctn } is a partition of [ca, cb]. Furthermore,
mi = sup{f (cx) : ti−1 ≤ x ≤ ti } = sup{f (x) : cti−1 ≤ x ≤ cti } = m0i
and similarly for the infimum. Hence
cL(f (c ·), P ) = c
n
X
i=1
mi (ti − ti−1 ) =
n
X
i=1
m0i (cti − cti−1 ) = L(f, cP ).
and similarly for the upper sums. Thus, recalling that inf cA = c inf A and similarly for
suprema, we obtain
Z cb
Z b
f (t)dt
f (ct)dt =
c
ca
a
• Problem 13.18
Solution. By Problem 16,
a) We have
a
Z
1
Z
n
n
(ax) dx = a
x dx = a
n+1
Z
xn dx = an+1 cn .
0
0
0
1
b) We have
2a
Z
a
Z
n
(x + a)n dx.
x dx =
−a
0
By part (a), the left-hand side is equal to
(2a)n+1 cn = 2n+1 cn an+1 ,
while the binomial theorem implies that the right-hand side is equal to
Z
a
n X
n
k
−a k=1
Z
xk an−k dx =
a
X n
xk an−k dx
k
−a k even
since
Z
a
xk = 0
−a
for odd k. Hence, we obtain
Z a
Z
n
(x + a) dx = 2
−a
a
X n
X n
k n−k
x a
dx = 2
ck ak+1 an−k
k
k
0 k even
k even
where we have used part (a) in the last equality. This yields the result since ak+1 an−k =
an+1
c) Finally, we examine the equality
n+1
2
n+1
cn a
n+1
= 2a
X n
ck .
k
k even
We will prove the claim by induction. The case n = 0 is clear (by integrating), so suppose
we know the result for n − 1. Then, we obtain
X n 1
n+1
n+1
n+1
2
cn a
= 2a
.
k k+1
k even
and since
n
1
n n+1 1
n+1
1
=
=
k k+1
k k+1n+1
k+1 n+1
we have
X n 1
X n + 1 1 X n+1
2(n+1)−1
1
2n
=
=
=
=
k k+1
n+1
k+1
n+1
k
n+1
n+1
k even
k even
k odd
where we have used problem 2 − 3 to evaluate the sum over odd k. Thus
2n+1 cn an+1 = 2an+1
2n
n+1
and rearranging this expression we obtain the result.
• Problem 13.19
Solution. Fix ε > 0. Suppose that |f (x)| ≤ M for all x ∈ [a, b]. We note that f is continuous
on any intervals [a, x0 − δ] and [x0 + δ, b] for and δ > 0. Hence by the theorem from class, f
is integrable on these intervals and there exist partitions P1 of [a, x0 − δ] and P2 of [x0 + δ, b]
such that
ε
ε
U (f, P1 ) − L(f, P1 ) <
and
U (f, P2 ) − L(f, P2 ) < .
3
3
Consider the partiaion P = P1 ∪ {x0 − δ, x0 + δ} ∪ P2 . Let tj := x0 − δ and tj+1 := x0 + δ Then
U (f, P )−L(f, P ) = U (f, P1 )−L(f, P1 )+U (f, P2 )−L(f, P2 )+(Mj −mj )(tj −tj−1 ) <
2ε
+2δM.
3
Hence, choosing δ = ε/6M yields a partition P such that
U (f, P ) − L(f, P ) < ε,
as required.
• Problem 13.22
Solution. We will treat the case b ≤ f (a). Since f (0) = 0, we note that f −1 (0) = 0. First
suppose that we can prove the identity
Z
a
Z
f (a)
f (t)dt +
0
f −1 (t) = af (a).
0
Since f is increasing, for any c ≤ a,
Z a
Z c
f (t)dt ≥
f (t)dt + f (c)(a − c).
0
0
Now taking c = f −1 (b), we have
Z
a
Z
f (t)dt ≥
0
f −1 (b)
f (t)dt + f (f
−1
(b))(a − f
−1
Z
f −1 (b)
(b)) =
0
0
By the identity above (with f replacing f −1 ), we obtain that
Z
f −1 (b)
f (t)dt − bf
0
−1
Z
(b) =
0
b
f −1 (t)dt
f (t)dt + ba − bf −1 (b).
hence
Z
a
Z
f (t)dt ≥ ba −
b
f −1 (t)dt.
0
0
which yields the desired result. To prove the claim above, we note that af (a) is the area of the
rectangle formed by connecting the point (a, f (a)) with the x- and y-axes. Since f is integrable
take a partition such that
U (f, P ) − L(f, P ) < ε,
and letting P1 = {f (t0 ), . . . , f (tn )}, we note that
U (f, P ) − L(f, P ) = U (f −1 , P1 ) − L(f −1 , P1 ) =
n
X
(ti − ti−1 )(f (ti ) − f (ti−1 )).
i=1
(A picture makes this more clear, the errors are simply the sum of the areas of the same small
rectangles). Since
! n
!
n
X
X
af (a) =
(ti − ti−1 )
(f (ti ) − f (ti−1 )) = U (f, P )+L(f −1 , P1 ) = L(f, P )+U (f −1 , P1 ).
i=1
i=1
(again, a picture is helpful here) we conclude that for any ε > 0
Z a
Z f (a)
f (t)dt −
f −1 (t) < ε.
af (a) −
0
0
• Problem 13.26
Solution.
a) Fix ε > 0. Let P be a partition such that
U (f, P ) − L(f, P ) < ε.
Let s1 be the step function which takes values mi on the interval [ti−1 , ti ) and similarly
let s2 be the step function which takes values Mi on the interval [ti−1 , ti ). Then
Z b
Z b
L(f, P ) =
s1 ,
and
U (f, P ) =
s2 .
a
Since
a
Z
L(f, P ) <
f < U (f, P )
the result follows.
b) Fix ε > 0 and let s1 and s2 be the described step-functions. Let P1 be the partition
corresponding to s1 and P2 the partition corresponding to s2 . Since
Z b
Z b
s1 ≤ L(f, P1 ),
and
U (f, P2 ) ≤
s2 ,
a
a
hence
U (f, P2 ) − L(f, P1 ) < ε.
Taking P = P1 ∪ P2 yields
U (f, P ) − L(f, P ) < ε.
hence f is integrable.
• Problem 13.32
Solution.
a) Take g = f , then 0 =
Rb
fg =
a
Z
Rb
a
f 2 . Now, we note that
b
f 2 = sup{L(f, P ) : P partition of [a, b]},
a
but since f 2 (x) ≥ 0 for all x ∈ [a, b], this implies that
n
X
m2i (ti − ti−1 ) = 0
i=1
for any partition P = {t1 , . . . , tn }. Thus mi = 0 for all 1 ≤ i ≤ n, and since f is
continuous, this implies that f = 0 by the argument in 13.11 c).
b) Suppose that f (x0 ) > 0 for x0 ∈ (a, b). Since f is continuous, we can find a δ > 0 such
that f (x) > 0 for x ∈ (x0 − δ, x0 + δ). Let g be a continuous functions with g > 0 on
(x0 − δ, x0 + δ) and g = 0 elsewhere. In particular, g(a) = g(b) = 0. Then
Z b
Z x0 +δ
fg =
f g > 0,
x0 −δ
a
which is a contradiction.
• Problem 13.40
Solution. Fix ε > 0. Let N > 0 be such that for x > N ,
|f (x) − a| < ε.
Let C > 0 be such that |f (t)| ≤ C on [0, N ]. Now note that
Z
x−N
x−N
1 x
(a − ε) <
(a + ε),
f (t)dt <
x
x N
x
thus,
Z x
1
< x − N − 1 |a| + x − N ε.
f
(t)dt
−
a
x
x
x
N
Furthermore, we also have
Z N
1
CN
≤
f
(t)dt
.
x
x
0
Thus, let M > 0 be sufficiently large so that for x > M ,
CN
< ε,
x
1−
x−N
ε
≤
,
x
2(|a| + 1)
then
Z x
1
CN
x−N
x−N
ε|a|
≤
f
(t)dt
−
a
+
−
1
|a| +
ε<ε+
+ ε < 3ε,
x
x
x
x
2(|a|
+ 1)
0
which concludes the proof.