Solutions to Analysis - Homework13
51.4
Let c ∈ (a, b), let P be a partition of [a, b] and P ∗ = P ∪ {c}. Then P1 = P ∗ ∩ [a, c] and P2 = P ∗ ∩ [c, b]
are partitions of [a, c] and [c, b], respectively. Now
Z c
Z b
∗
U (f, P ) ≥ U (f, P ) ≥ U (f, P1 ) + U (f, P2 ) ≥
f dα +
f dα
a
It follows that
Z
b
Z
c
Z
f dα ≥
b
f dα +
a
c
f dα.
a
c
Otherwise, let P1 be a partition of [a, c] and P2 be a partition of [c, b]. P = P1 ∪ P2 is also a partition of
[a, b]. Now
Z b
U (f, P1 ) + U (f, P2 ) = U (f, P ) ≥
f dα
a
It follows that
Z
c
Z
b
a
Z
b
Hence,
Z
c
f dα =
a
Z
b
f dα +
a
Z
f dα.
c
Z
a
b
Z
c
f dα =
f dα. The proof of
c
b
f dα ≥
f dα +
a
Z
b
f dα +
a
f dα is similar.
c
51.6
R1
R1
(a) Since f (x) = x is continuous, 0 xdx = 0 xdx. So we have to show that glbU = 21 . For every > 0,
Pn
1
define partition Pn = {xk = nk |k ∈ {0, 1, . . . , n}}. Then U (f, Pn ) = n1 k=1 nk = 12 + 2n
≤ 21 + n1 . There
1
exist N ∈ N, such that if n > N , n < . Hence for its corresponding partition Pn , U (f, Pn ) < 12 + ; 12 is the
R1
greatest lower boundary of U . It follows that 0 xdx = 12 .
R1
R1
(b) Since f (x) = x2 is continuous, 0 x2 dx = 0 x2 dx. So we have to show that glbU = 31 . For every
2
Pn
Pn
> 0, define partition Pn = {xk = nk |k ∈ {0, 1, . . . , n}}. Then U (f, Pn ) = n1 k=1 nk = n13 k=1 k 2 =
1 (2n+1)(n+1)n
1
= 13 + 2n
+ 6n1 2 ≤ 31 + n1 . There exist N ∈ N, such that if n > N , n1 < . Hence for
n3
6
its corresponding partition Pn , U (f, Pn ) < 31 + ; 13 is the greatest lower boundary of U . It follows that
R1 2
x dx = 13 .
0
51.7
R1
R2
R2
Since f (x) = x is continuous and α(x) is increasing, then f ∈ Rα [0, 2]. 0 xdα(x) = 0 xdα + 1 xdα =
R2
xdx + 1 xdα(x). The first integral is 12 from exercise 51.6, now we compute the second one. Define
0
partition Pn = {xk = 1 + nk |k ∈ {0, 1, . . . , n}}, then ∆αk = αk − αk−1 = α(xk ) − α(xk−1 ) = (3 + nk ) − (3 +
R2
k−1
1
2 + n1 . Thus, 1 xd(x + 2) ≤ limn→∞ U (f, Pn ) =
n ) = n for k > 1. For k = 1, ∆α1 = α(x1 ) − α(x0 ) = Pn
Pn
limn→∞ k=1 ∆αk 1+ nk = (∆α1 )(1+ n1 )+ k=2 n1 1+ nk = limn→∞ (2+ n1 )(1+ n1 )+ n1 · (n−1)+ (n+2)n
=
2n
R
R
2
2
1
7
2 + 1 + 2 = 2 . It implies that 0 xdα(x) ≤ 4. The lower bound works similarly, such that 0 xdα(x) = 4.
R1
1
51.10
(a) It is easy to conclude from Mean-value theorem of the integral that there exists c ∈ [a, b] such that
f (x)dx = f (c)(b − a). We only need to show that when c is endpoint, there also exists some point d on
a
interval (a, b) such that f (c) = f (d). We only show when c = a, and the other case c = b is similar. Then
Rb
Rb
Rb
from Mean Value Theorem, a f (x)dx = f (a)(b − a) = a dx, it leads to a [f (x) − f (a)]dx = 0.
Since g(x) = f (x) − f (a) is also continuous on [a, b]. Assume for all x ∈ (a, b) that g(x) 6= 0, then g(x)
Rb
will not change sign on (a, b). W.l.o.g. assume that g(x) > 0 on (a, b). Then by 51.11 a g(x)dx 6= 0, which
is a contradiction.
Rb
(b) Let
α(x) =
0, x ∈ [0, 21 );
1, x ∈ [ 12 , 1]
R1
α(x)dx = 12 , but we cannot find a c ∈ (0, 1) such that f (c) = 21 . Also for α from 51.7
x,
x ∈ [0, 12 );
α(x) =
2 + x, x ∈ [ 12 , 1]
R2
we have 0 xdα(x) = 4. However, the integrand f (x) = x never takes the value 4 on the interval (0, 2).
Then
0
51.11
Suppose f (x0 ) 6= 0 for some x0 ∈ [a, b]. Since f (x) is continuous on [a, b] and f (x2 0 ) > 0, there exists δ > 0
such that |f (x) − f (x0 )| < f (x2 0 ) for all x ∈ [a, b] such that |x − x0 | < δ. Let η = min(δ, max(x0 − a, b − x0 )),
so that η > 0. Let I be the interval [x0 − η, x0 ] if it is contained in [a, b]; otherwise let I = [x0 , x0 + η].
Whichever is the case, I ⊂ [a, b] and f (x) = f (x0 ) + (f (x) − f (x0 )) ≥ f (x0 ) − |f (x) − f (x0 )| > f (x2 0 ) for all
x ∈ I. The functions f1 (x) and f2 (x) defined as
f (x), x ∈ I,
f (x), x ∈
/ I,
f1 (x) =
f2 (x) =
0,
x∈
/ I,
0,
x ∈ I,
and both nonnegative, bounded, and continuous except possibly at the two endpoints of the interval I. They
are therefore both Riemann-integrable. Consideration of Riemann sums shows that
b
Z
a
f1 (x)dx ≥ η ,
2
Z
b
and
f2 (x)dx ≥ 0,
a
It therefore follows that
Z
b
Z
b
f (x)dx =
a
Z
a
Z
contradicting the hypothesis that
b
f2 (x)dx ≥ η
f1 (x)dx +
a
> 0,
2
b
f (x)dx.
a
51.12
R2
R1
R2
R1
For a, we get by 51.4 that ¯ 0 f (x)dα(x) = ¯ 0 f (x)dα(x) + ¯ 1 f (x)dα(x) = ¯ 0 f (x)dα(x), since the
integrator α is constant of the interval [1,2]. However, it is easy to see that for any partition P of [0,1], we
R1
get U (f |[0,1] , P ) = 1, thus ¯ 0 f (x)dα(x) = 1.
2
R1
R2
R1
f (x)dα(x) = f (x)dα(x) + f (x)dα(x) = f (x)dα(x). For any partition P of [0,1], we
0
1
0
R¯ 1
, P ) = 0, thus f (x)dα(x) = 0. Therefore f 6∈ R [a, b].
Similarly,
R2
0
get L(f |[0,1]
For b, we get
α
0
Z¯ 2
Z¯ 1
g(x)dα(x) =
0
and
Z
Z¯ 2
g(x)dα(x) +
g(x)dα(x) = 0
0
2
Z
1
g(x)dα(x) =
0
1
Z
2
g(x)dα(x) +
0
g(x)dα(x) = 0,
1
since on the first interval [0,1] the function g is identically zero and on the interval [1,2], the integrator
α is constant. R
R1
R2
R2
R2
2
For c, we get ¯ 0 g(x)dx = ¯ 0 g(x)dx+ ¯ 1 g(x)dx = ¯ 1 g(x)dx ≤ ¯ 1 dx = 1, since g ≡ 0 on [0,1] and g ≤ 1 on
R1
R2
R2
R2
R2
R2
g(x)dx =
dx = 2 − (1 + ) = 1 − .
[1, 2]. Similarly g(x)dx = g(x)dx + g(x)dx = g(x)dx ≥
0
0
1
1
1+
1+
R2
Since g ≥ 0 and g = 1 on (1, 2]. This holds for any > 0, therefore g ∈ R[a, b] and 0 g(x)dx = 1.
R2
Similar reasoning shows that f ∈ R[a, b] and 0 f (x)dx = 1.
51.15
|f −g|
and f, g ∈ Rα [a, b], then f /2, g/2, |f − g|/2 ∈ Rα [a, b], so max{f, g} ∈
Since max{f, g} = f +g
2 +
2
Rα [a, b].
|f −g|
similarly, min{f, g} = f +g
and f, g ∈ Rα [a, b], then f /2, g/2, −|f −g|/2 ∈ Rα [a, b], so min{f, g} ∈
2 − 2
Rα [a, b].
51.16
Let > 0, since f ∈ Rα [a, b], then there exists a partition P = x0 , x1 , . . . , xn such that U (f, P )−L(f, P ) <
M 2 .
Therefore, for arbitrary x, y ∈ [xi−1 , xi ],
1
1 |f (x) − f (y)|
|f (x) − f (y)|
Mi (f ) − mi (f )
−
≤
.
f (x) f (y) = |f (x)f (y)| ≤
2
M
M2
)−mi (f )
It implies that Mi ( f1 ) − mi ( f1 ) ≤ Mi (f M
. It leads to
2
1
1
1
1
,P − L
, P ≤ 2 (U (f, P ) − L(f, P )) < 2 M 2 = .
U
f
f
M
M
Hence
1
f
∈ Rα [0, 2].
51.17
For given > 0, let x0 ∈ [a, b] such that f (x0 ) = M . There exists δ > 0 such that if |x − x0 | ≤ δ and
x ∈ [a, b] then |f (x) − M | < .
Define I = [a, b] ∩ (x0 − δ, x0 + δ); f (x) > M − on I. Hence by mean value theorem
Z b
Z
M n (b − a) ≥
f n (x)dx ≥ f n (x)dx ≥ δ(M − )n .
a
Taking the power
1
n,
I
this gives that
1
Z
M (b − a) n ≥
b
f n (x)dx
a
3
n1
1
≥ δ n (M − ).
The first in inequality gives that
b
Z
lim sup
n→∞
n1
f (x)dx
≤ M,
n
a
and the second one that for each > 0,
Z
lim inf
n→∞
Thus limn→∞
Rb
a
b
n1
f n (x)dx
≥ M − .
a
1
f n (x)dx n = M .
51.19
First note that f is increasing then f (a) ≤ f (x) ≤ f (b) for all x ∈ [a, b] so f is bounded on [a, b].
Denote Pn = {xk = a + k b−a
n |k ∈ {0, 1, . . . , n}} as the partition of [a, b] which divides [a, b] into n equal
intervals. Since f is increasing,
mk = inf{f (x)|xk−1 ≤ x ≤ xk } = f (xk−1 );
Mk = sup{f (x)|xk−1 ≤ x ≤ xk } = f (xk ).
Therefore,
L(f, Pn ) =
n
X
n
mk (xk − xk−1 ) =
k=1
U (f, Pn ) =
n
X
k=1
n
Mk (xk − xk−1 ) =
k=1
Thus, U (f, Pn ) − L(f, Pn ) =
1
n
n
X
1X
f (xk−1 );
n
1X
f (xk ).
n
k=1
[f (xk ) − f (xk−1 )] =
k=1
1
1
(f (xn ) − f (x0 )) = (f (b) − f (a)).
n
n
For any > 0, we can find N ∈ N such that if n ≥ N , n1 (f (b) − f (a)) < and therefore exists partition
Pn such that |U (f, Pn ) − L(f, Pn )| < . Then f ∈ R[a, b].
4