HOMEWORK #2 – SOLUTIONS
Solution to Problem 6.2.2. Assume that P1 = {x0 , x1 , . . . , xn } and P2 = {y0 , y1 , . . . , ym }. Then, there exist
xi−1 and xi such that xi − xi−1 = gap P1 . Since P1 is a refinement of P2 , so there exists yj−1 and yj such that
[xi−1 , xi ] ⊂ [yj−1 , yj ], but this means that
gap P1 = xi − xi−1 ≤ yj − yj−1 ≤ gap P2
as desired.
Solution to Problem 6.2.8. Since f is integrable, there exist an Archimedean sequence of partitions {Pn∗ }, that is
lim U (f, Pn∗ ) − L(f, Pn∗ ) = 0
(1)
n→∞
∗
The problem is that Pn+1
is not necessarily a refinement of Pn∗ , so we define a new sequence of partitions {Pn } as
follows:
P1 = P1∗
P2 = P2∗ ∪ P1∗
..
.
∗
∗
Pn+1 = Pn+1
∪ Pn∗ ∪ . . . ∪ P1∗ = Pn+1
∪ Pn
∗ . Therefore,
We can see that Pn+1 is a refinement of both Pn and Pn+1
U (f, Pn ) − L(f, Pn ) ≤ U (f, Pn∗ ) − L(f, Pn∗ )
From (1), the right hand side converges to 0, so by the comparison lemma, the left hand side also converges to 0.
Therefore, {Pn } is an Archimedean sequence. The statement that L(f, Pn ) increases and U (f, Pn ) decreases as n
increases follows directly from the Refinement Lemma (6.2).
Solution
to Problem 6.2.10. Consider a sequence {Pn } of partitions of the interval [2, 4] defined by
Pn = x0 = 2, x1 , . . . , xn = 3, 3 + n1 , 4 , where xi = 2 + ni for 1 ≤ i ≤ n. Note that xi − xi−1 = n1 for 1 ≤ i ≤ n.
Since f is increasing on the interval [xi−1 , xi ], so
Mi =
sup
f (x) = f (xi )
and
mi =
x∈[xi−1 ,xi ]
inf
x∈[xi−1 ,xi ]
f (x) = f (xi−1 )
On the interval [3, 3 + n1 ], we have Mi − mi = 3 − 2 = 1, and on [3 + n1 , 4], we have Mi − mi = 0 since f is constant
on this interval. Then,
n
X
1
1
U (f, Pn ) − L(f, Pn ) =
(xi − xi−1 )(Mi − mi ) + (3 + − 3) · 1 + (4 − (3 + )) · 0
n
n
=
i=1
n
X
i=1
1
1
(f (xi ) − f (xi−1 ) +
n
n
1
1
(f (xn ) − f (x0 )) +
n
n
1
1
= (f (3) − f (2)) +
n
n
2
=
n
Therefore, limn→∞ [U (f, Pn ) − L(f, Pn )] = 0, which means f is integrable.
=
Solution to Problem 6.2.11. Indeed, since x0 = a and xn = b,
n
n
n
X
X
X
[xi − xi−1 ]2 =
[(xi − xi−1 ) · (xi − xi−1 )] ≤
(xi − xi−1 ) · gap P = (b − a) · gap P.
i=1
i=1
i=1
1
2
HOMEWORK #2 – SOLUTIONS
Solution to Problem 6.2.12. Since f is Lipschitz, so f is continuous. By the Extreme Value Theorem, there exists ai , bi ∈ [xi−1 , xi ] such that f (ai ) =
sup f (x) and f (bi ) =
inf
f (x). Hence, from the Lipschitz
x∈[xi−1 ,xi ]
x∈[xi−1 ,xi ]
condition,
0 ≤ U (f, P ) − L(f, P ) =
n
X
(f (ai ) − f (bi ))(xi − xi−1 )
i=1
≤
n
X
c|ai − bi |(xi − xi−1 )
i=1
Since ai , bi ∈ [xi−1 , xi ], so |ai − bi | ≤ xi − xi−1 . Consequently,
n
X
c|ai − bi |(xi − xi−1 ) ≤ c
i=1
n
X
(xi − xi−1 )2
i=1
From problem (6.2.11), we have
c
n
X
(xi − xi−1 )2 ≤ c(b − a) · gap P.
i=1
Solution to Problem 6.2.13. Let f to be a Lipschitz function as defined in (6.2.12). Taking Pn to be the regular
partition of [a, b] into n intervals (Thus, gap Pn = b−a
n ), we have from (6.2.12) that
c
U (f, Pn ) − L(f, Pn ) ≤ c(b − a) · gap P = (b − a)2 .
n
Since the right hand side converges to 0 as n → ∞, by the comparison lemma, the left hand side also converges to 0
as n → ∞. Consequently, f is integrable.
Solution to Problem 6.3.1. Since f 2 , f g and g 2 are integrable, we have that [f − g]2 = f 2 − 2f g + g 2 is integrable.
Rb
Rb
Since [f (x) − g(x)]2 ≥ 0 for all x, by the monotonicity of the integral, a [f − g]2 ≥ a 0 = 0. This is equivalent to
Z b
Z b
Z
1 b 2
2
2
2
fg
[f − 2f g + g ] ≥ 0 or
[f + g ] ≥
2 a
a
a
Solution to Problem 6.3.2. Write the polynomial p as
Z b
Z b
Z b
Z b
2
2
2
g − 2λ
fg +
f 2 = Aλ2 + Bλ + C.
p(λ) =
[f − λg] = λ
a
a
Rb
g2, B
Rb
a
Rb
a
f 2.
We have that A = a
= −2 a f g and C = a
Since p(λ) is always nonnegative, its graph (whis is a parabola) intersects x-axis at most one point. This means that
R
2 R
R
b
b
b 2
the discrimant B 2 − 4AC ≤ 0, but this is equivalent to a f g ≤ a f 2
g
.
a
Solution to Problem 6.3.3. Since bn ≥ 0 for all n, so an ≤ an + bn . The right hand side converges to 0, so by the
comparison lemma, an → 0. Similarly, from bn ≤ an + bn we have that bn → 0.
Solution to Problem 6.3.4. We will only prove that sup αS = α sup S for α ≥ 0 and inf αS = α sup S for α ≤ 0,
since we can apply similar arguments to the two remaining cases.
First, we will show that sup αS = α sup S for α ≥ 0.
For any x ∈ S, we have
x ≤ sup S,
and so
αx ≤ α sup S.
Hence, α sup S is an upper bound of αS. Consequently,
sup αS ≤ α sup S.
On the other hand, for any x ∈ S
αx ≤ sup αS.
HOMEWORK #2 – SOLUTIONS
3
This implies
1
sup αS.
α
(if α = 0 the problem becomes trivial) Hence, α1 sup αS is an upper bound of S. Consequently,
x≤
1
sup αS,
α
and so α sup S ≤ sup αS. We can now conclude that sup αS = α sup S.
sup S ≤
Now we will show that inf αS = α sup S for α ≤ 0. The argument will almost be the same. However, since
α ≤ 0, some of the inequalities will be reversed.
For any x ∈ S, we have x ≤ sup S, and so αx ≥ α sup S. Hence, α sup S is a lower bound of αS. Consequently,
inf αS ≥ α sup S.
On the other hand, αx ≥ inf αS for any x ∈ S. This implies x ≤ α1 inf αS. (if α = 0 the problem becomes trivial)
Hence, α1 inf αS is an upper bound of S. Consequently, sup S ≤ α1 inf αS, and so α sup S ≥ inf αS. We can now
conclude that sup αS = α sup S.
Solution to Problem 6.4.1. a) False. The function
(
−1, if a ≤ x ≤ a+b
2
f (x) =
1, if a+b
<
x
≤
b
2
Rb
satisfies a f = 0.
b) False, as we can see that any step functions are integrable.
Rb
Rb
c) True, by using Theorem 6.13: f ≥ 0 implies a f ≥ a 0 = 0.
d) False. The function f (x) = x1 defined on (0, 1) is not bounded.
e) True. This is an application of Extreme Value Theorem.
Solution to Problem 6.4.2. Consider f (x) on [xi−1 , xi ]. Since xi is an upper bounded of
Si = {f (x) : x ∈ [xi−1 , xi ]},
we have that sup Si ≤ xi .
For any x < xi , there is a rational number r such that x < r < xi and that f (r) = r > x. Therefore, any x < xi is
not an upper bound of Si , which implies that sup Si = xi . In other words,
Mi =
sup
f (x) = xi .
x∈[xi−1 ,xi ]
With similar argument, we get
mi =
We observe that xi >
xi +xi−1
,
2
inf
x∈[xi−1 ,xi ]
f (x) = −xi .
which implies
U (f, P ) =
n
X
i=1
xi (xi − xi−1 ) >
n
X
xi + xi−1
(
)(xi − xi−1 )
2
i=1
n
1X 2
=
(xi − x2i−1 )
2
i=1
1
= (x2n − x20 )
2
1
= .
2
i−1
Similarly, from −xi < − xi +x
, we have
2
1
L(f, P ) < − .
2
4
HOMEWORK #2 – SOLUTIONS
Z
b
1
Therefore,
f = inf L(f, P ) ≤ − and
P
2
a
f is not integrable.
Z
b
1
f = sup U (f, P ) ≥ , so these two quantities cannot be equal. Hence,
2
P
a