MA244 Analysis III
Solutions. Sheet 2.
NB. THESE ARE SKELETON SOLUTIONS, USE WISELY !
0.1 Properties of regulated functions and their
Integrals.
1. (Q.1) Pick any > 0. As f, g are regulated, there exist φ, ψ ∈ S[a, b]: ||f −
φ||∞ < /2, ||g − ψ||∞ < /2. Notice that ψ + φ ∈ S[a, b] (Assignment 1). But
||f + g − φ − ψ||∞ ≤ ||f − φ||∞ + ||g − ψ||∞ < .
Therefore f + g ∈ R[a, b].
To show that f g ∈ R[a, b], notice that regulated functions are bounded. (Assignment 1.) Therefore there exist M, N > 0: ||f ||∞ < M , ||g||∞ < N . Let
φ, ψ ∈ S[a, b]: ||f − φ||∞ < δ/M , ||g − ψ||∞ < δ/N , Notice that φψ ∈ S[a, b].
(Assignment 1.)
Then
||f g − φψ||∞ = ||f (g − ψ) + g(f − φ) + (f − φ)(ψ − g)||∞
≤ ||f (g − ψ)||∞ + ||(f − φ)g||∞ + ||(f − φ)(ψ − g)||∞
≤ 2δ + δ 2 /(M N ).
For any > 0 the equation 2δ +δ 2 /(M N ) = has a positive solution and we are
done. The last inequality above uses the following property of the sup-norm:
||ab||∞ = sup |a(x)||b(x)| ≤ sup |a(x)| sup |b(y)| = ||a||∞ ||b||∞
x
x
y
2. (Q.2) sin2 (t·) is continuous hence regulated on [a, b]. By Question 1, sin2 (t·)φ ∈
. By the
R[a, b], so the question is well posed. Recall that sin2 (tx) = 1−cos(2tx)
2
Rb
RL lemma, a cos(2tx)φ(x)dx → 0 as t → ∞. Therefore,
Z
lim
t→∞
a
b
1
sin (tx)φ(x)dx =
2
2
Z
b
φ(x)dx.
a
3. (Q.3) Let h := g◦f . Then h(0) = 0 and h(x) = sign(x sin(1/x)) = sign(sin(1/x))
1
1
for x > 0. So, h(x) = 1 for x ∈ ( 2π(n+1/2)
, 2πn
) and h(x) = −1 for x ∈
1
1
( 2πn , 2π(n+1/2) ). Such a function is not regulated as for any step function φ in
I, ||h − φ||∞ > minc∈R (|1 − c|, |1 + c|) ≥ 1.
4. (Q. 4) Let f be defined as follows: f (a + (b − a)/n) = 1/n, n ∈ N; f (x) = 0 if
x∈
/ a+(b−a)/N. Note that f ∈ R[a, b]: consider φk ∈ S[a, b] : φk (x) = f (x) for
x ≥ a+(b−a)/k and φk (x) = 0 for x < a+(b−a)/k. Then ||f −φk ||∞ ≤ 1/k → 0
as k → ∞. Therefore, the sequence of step functions {φk } converges to f
uniformly and f is regulated by definition. Of course, f itself is not a step
function as the cardinality of its range is infinite. The integral:
Z b
Z b
f = lim
φk = 0.
a
k→∞
a
5. (Q.5) Let f (x0 ±) = limy→x0 ± f (y). According to the Theorem cited in the
hint, if f ∈ R[a, b], then f (x±) exists for any x ∈ (a, b) and the corresponding
one-sided limits exist at the boundary points.
The first step is to show that the set of points
S = {x ∈ [a, b] : |f (x+) − f (x−)| > }
is finite. Notice that S is the set of all points of discontinuity of f with the
jump size greater than . As f is regulated, there exists φ ∈ S[a, b] such that
for any x ∈ [a, b],
|f (x) − φ(x)| < /4.
Taking one-sided limits of the above double inequality, we find after a simple
manipulation:
f (x+) − f (x−) − /2 ≤ φ(x+) − φ(x−) ≤ f (x+) − f (x−) + /2.
Take x ∈ S . As |f (x+) − f (x−)| > , the above inequality implies that
φ(x+) − φ(x−) 6= 0. But the set of discontinuities of a step function is finite.
As we have just showed, that S is contained in the set of discontinuities of φ,
we can conclude that |S | < ∞.
Finally, let us consider the sequence of sets S1/n , n ∈ N . As it is easy to check,
any point of discontinuity of f is in S1/n for some n. Note that
S1 ⊂ S1/2 ⊂ S1/3 ⊂ . . . .
The enumeration of the set of the set of all discontinuities can be done inductively: the points of S1 can be enumerated as S1 is finite. Assume that all
points in S1/n have been enumerated by 1, 2, . . . Nn . As the set S1/(n+1) \ S1/n
is finite, we can enumerate its points starting with Nn + 1. The result is the
enumeration of all points in S1/(n+1) . As any discontinuity of f is contained in
S1/n for some n, we have established a one-to-one correspondence between the
set of all discontinuities of f and N, which means that this set is countable by
definition.
0.2
Integration.
6. (Q.6) Let PN = N5 {0, 1, 2, . . . N } be a sequence of partitions of [0, 5]. Let φN :
φN (0) = 0, φN |(5/N )(k−1,k] = ((5/N ) · k)2 , k = 1, 2, . . . , N be a sequence of step
functions. Then ||f − φN ||∞ ≤ max1≤k≤N (((5/N ) · k)2 − ((2/N ) · (k − 1)2 )) ≤
50/N → 0 as N → ∞. Therefore, φN ’s converge to f uniformly. By definition,
Z
5
Z
f = lim
0
N →∞
0
5
3 X
N
N
5 X
5
2
φN = lim
(5/N · k) == lim
k2
N →∞ N
N →∞
N
k=1
k=1
PN 2
PN kx
(N +1)x
2
But
|x=0 = ∂x2 e ex −1−1 |x=0 = (1/3)N 3 + (1/2)N 2 +
k=1 k = ∂x
k=1 e
(1/6)N . Substituting this into the previous formula and computing the limit,
we get
Z 5
53
f= .
3
0
Rb 1
Ra 1
dx + 2 log(x)
dx. Each of the terms is differ7. (Q.7) By additivity, I = − 2 log(x)
entiable by FTC2. As ∂a F (b) = 0, ∂a I = −1/ log(a), ∂b I = 1/ log(b).
Ra
8. (Q.8) I(x) = −J(a(x)) + J(b(x)), where J(a) = 2 1/ log(t)dt. As J is differentiable on the domain of a, b and a, b are differentiable on R, the function I
is differentiable on R. So, we can apply the chain rule:
I 0 (x) = −J 0 (a(x))a0 (x) + J 0 (b(x))b0 (x) = −
a0 (x)
b0 (x)
+
.
log a(x) log b(x)
9. (Q.9)
(a) F 0 (x) = log10 (ex )ex = x10 ex
p
√
(b) G0 (x) = 2x 1 + (1 + x2 )4 − 2x 1 + x8
10
10
(c) H 0 (x) = 2x(e−x + e−x ) = 4xe−x
(d) I 0 (x) =
1
√
2 x
cos(x2 ) +
1
x2
10
cos(1/x4 )
10. (Q.10) For all the integrals below the integrands are continuous on the respective intervals of integration and we can apply the fundamental theorem of
calculus for their evaluation.
Re
R1
(a) Substitution: t = log(x). Then 1 sin(log(x))/xdx = 0 sin(t)dt = 1 −
cos(1).
R log(3)
R log(3) sinh(x) 0
2
(b) log(2) 1/ cosh (x)dx = log(2) cosh(x) dx = tanh(log(3))−tanh(log(2)) =
1/5.
R 1/√2 1
R π/4
√
(c) Substitution: x = sin(t). Then 0
dx
=
sign(cos(t))dt =
2
0
1−x
π/4.
R e2 1
R2
(d) Substitution: t = log(x). Then e x log(x)
dx = 1 1t dt = log(2).
R1
100
(e) −1 x1001 e−x dx = 0 as the integrand is odd and the limits of integration
are symmetric.
0.3
Improper integrals.
Rn
Rn
Rn
11. (Q.11) Let bn = n−1 F , an = n−1 f . Note that |an | ≤ n−1 |f | ≤ bn . The fact
R∞
that 0 F converges implies that the series
∞
X
bn < ∞
n=1
P
As |an | < bn , the series ∞
n=1 an convergesRabsolutely by the comparison test
N
for series (Analysis I). Therefore, limN →∞ 0 f exists and is finite. To finish
R Rn
the proof, we need to show that 0 f converges for any choice of sequence
Rn : limn→∞ Rn = ∞. Let dxe be the smallest integer greater or equal than x.
Then
Z
Z
Z
Z
dRe
R
|
f−
0
dRe
f| ≤
0
dRe
|f | ≤
R
R→∞
|f | ≤ bdRe −→ 0,
dRe−1
as
P∞
n=1 bn
converges. Therefore,
Z
Z R
f = lim
lim
R→∞
R→∞
0
dRe
f,
0
where the right hand side has already been shown to converge.
R∞
R∞
3
3
12. (Q.12) The integrals 0 e−x and 1 e−x dx diverge and converge simultane3
ously. But e−x ≤ e−x for any x ≥ 1. Therefore,
Z ∞
Z ∞
−x3
e
≤
e−x dx = 1.
1
We conclude that
11.
R∞
0
−x3
e
1
dx converges by the comparison principle of Question
13. (Q.13) (1 − x4 ) = (1 − x2 )(1 + x2 ) = (1 − x)(1 + x)(1 + x2 ) ≥ (1 − x) for
x ∈ [0, 1). Therefore,
Z 1
Z 1
1
1
√
√
dx ≤
< ∞.
1−x
1 − x4
0
0
Therefore the elliptic integral converges by the comparison principle.
R∞
RR
4
14. (Q.14) −R x3 dx = x4 |R
−R = 0. However, −∞ xdx is divergent, as
Z
0
lim
R1 →∞
Z
3
x dx + lim
−R1
R2 →∞
R2
x3 dx
0
does not exist.
15. (Q.15)
R 100
R 100 1
R 100 1
<
(a) 0 x1/3 +2x11/4 +x3 dx ≤ 0 x1/3
dx = lim↓0 x1/3
dx = lim↓0 23 x2/3 |100
∞. So the integral converges by the comparison principle.
R∞
(b) The integral 1 sin(x)
converges. (Use the comparison principle with
x2
R ∞ sin(x)
R1
2
F (x) = 1/x .) Therefore, 0 sin(x)
and
dx diverge and converge
2
x
x2
0
simultaneously. But
Z 1
Z 1
sin(x)
sin(x) 1
cos(x)
lim
= − lim
| −
dx
↓0 ↓0
x2
x
x
Z 1
sin(x) 1
1
= − lim
| − cos(x) log(x) | −
sin(x) log(x)dx ,
↓0
x
R1
which diverges logarithmically (verify that 0 log(x) sin(x)dx converges!).
R∞
Therefore, 0 sin(x)
dx diverges.
x2
0.4
Uniform convergence.
Pn−1 k
1
16. (Q.16) For any x ∈ (−1, 1), limn→∞ k=0
x = 1−x
:= f (x) pointwise. Let
Pn−1 k 1−xn
fn (x) = k=1 x = 1−x . Choose xn = 1 − 1/n ∈ (−1, 1). Then
lim |fn (xn ) −
n→∞
1
xnn
| = lim |
| = lim n(1 − 1/n)n = +∞.
n→∞ 1 − x
n→∞
1−x
So choosing = 1 in the negation of the definition of the uniform convergence,
we
always find n and x ∈ (−1, 1) such that |f (x) − fn (x)| > 1. Therefore,
P can
k
x
does
not converge uniformly in (−1, 1). On the other hand, for any
k
x ∈ [−R, R], where 0 ≤ R < 1,
|fn (x) − f (x)| ≤
Rn
→ 0 for n → 0,
1−R
which immediately implies the uniform convergence on [−R, R].
17. (Q.17) Without restricting generality, we can consider symmetric intervals
[−R, R]. For any x ∈ [−R, R],
|
n−1 k
X
x
k=1
k!
x
−e | ≤
∞
X
Rk
k=n
≤
∞
Rk X
Rk−n
=
k!
k! k=n (n + 1)(n + 2) . . . (n + (k − n))
∞
Rk X Rk
Rk R
=
e → 0 as k → ∞,
k! k=0 k!
k!
where the last step can be justified using Stirling’s formula. Therefore, the exponential sequence converges uniformly on [−R, R], hence on any finite subinterval of R.
18. (Q.18) ⇒. If fn → f uniformly on A, then for any > 0 there is N ∈ N: for any
n > N and any x ∈ A, |f (x)−fn (x)| < /2. Therefore, supx∈A |fn (x)−f (x)| < and we get the definition of limn→∞ supx∈A |fn (x) − f (x)| = 0.
⇐. If supx∈A |fn (x) − f (x)| < , then for any x ∈ A |fn (x) − f (x)| < and
the the uniform convergence follows immediately from limn→∞ supx∈A |fn (x) −
f (x)| = 0.
19. (Q.19) Let Sn = x2 − x4 + x4 − x6 + . . . + x2n − x2n+2 = x2 − x2n+2 . Clearly,
limn→∞ Sn (x) = x2 for |x| < 1 and limn→∞ Sn (x) = 0 for |x| = 1. Consider
xn = 1 − 1/n ∈ (−1, 1). Then for any δ > 0, there is nδ ∈ N such that for any
n > nδ
x2n − Sn = x2n+2
= (1 − 1/n)2n+2 ≥ (1 − 1/n)n = en log(1−1/n) ≥ e−1−δ ..
n
Therefore, limn→∞ supx∈(−1,1) |Sn (x) − x2 | =
6 0 and the convergence cannot be
uniform by Question 18.
20. (Q.20) Suppose the convergence Fn → F were not uniform. Negating the
definition of the uniform convergence, we conclude that there must exist c > 0
and a strictly increasing sequence {pn }n≥1 ⊂ N such that
sup (Fpn (x) − F (x)) > c, n = 1, 2, . . .
x∈[a,b]
As Fpn − F is continuous on the closed interval [a, b], it reaches its maximal
value at some xpn ∈ [a, b]. The sequence {xpn }n≥1 is bounded. By the BoltzanoWeierstrass theorem it contains a subsequence {xpnk }k≥1 , which converges to a
point x0 ∈ [a, b]. By construction,
Fpnk (xpnk ) − F (xpnk ) > c, k = 1, 2, . . .
As Fn (x) > Fm (x) for any x and m > n, we conclude that
Fpn (xpnk ) − F (xpnk ) > c.
for each n < nk . Taking the limit k → ∞ in the above inequality and using
the continuity, we find that
Fpn (x0 ) − Fp (x0 ) ≥ c > 0, n = 1, 2, 3, . . . ,
which contradicts the pointwise convergence Fn → F at x0 . Therefore, Fn
must converge to F uniformly.
November the 9th, 2015
Sergey Nazarenko and Oleg Zaboronski.