MA244 Analysis III
Solution Sheet 1.
NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!
0.1
Step functions. Integration of step functions.
1. Let us define f as follows: f |( 1 , 1 ) = (−1)n , n ∈ N, f (0) = 0. A. f ∈
/ S[0, 1]:
n+1 n
1
1
1
take any p0 ∈ (0, 1). Choose a natural n > p0 . Then n+1 , n ⊂ (0, p0 ) and
1
1
,
⊂ (0, p0 ) This implies that f |(0,p0 ) is not constant. As p0 is arbitrary,
n+2 n+1
we proved that there exists no partition of [0, 1] compatible with f . Therefore,
f∈
/ S[0, 1]. B. For any ∈ (0, 1), f ∈ S[, 1]: the compatible partition is given
by {} ∪ ((, 1] ∩ {1, 1/2, 1/3, 1/4, . . .}).
2. Let f, g ∈ S[a, b]. Let Pf and Pg are partitions of [a, b] compatible with f and
g correspondingly. The the refined partition Pf ∪ Pg is compatible with both
f and g.
3. The result follows by induction from the fact that for any φ, ψ ∈ S[a, b] and
λ, µ ∈ R, λφ + µψ ∈ S[a, b]. This fact is proved as follows: if P is a partition of
[a, b] compatible with φ and Q is a partition of [a, b] compatible with ψ, then
the refined partition P ∪ Q is compatible with λφ + µψ.
4. No, it is impossible. Assume that we found such a set. Applying the conclusion
of question 2 inductively, we can construct a partition of [a, b] compatible with
each member of this set. Therefore, there exists p < q: [p, q] ∈ [a, b] and φk |[p,q]
is constant for each k = 1, 2, . . . , n. Define the following function f on [a, b]:
f (x) = 1 for x ∈ [a, (p + q)/2), f (x) = −1 for x ∈ [(p + q)/2, b]. Clearly,
f ∈ S[a, b], but f cannot be exressed as a linear combination of φ’s as it is not
constant on [p, q].
5. Let P be a partition of [a, b] compatible with both φ and ψ. Existence of such
a partition was established in question 2. The function φψ is compatible with
P : if ψ|(pi ,pi+1 ) = ψi and φ|(pi ,pi+1 ) = φi are constants, then ψφ|(pi ,pi+1 ) = ψi φi
is also a constant.
6. The sketch is ommited. ψ is a step function in [0, x] due to the fact that
ψ |(n−1,n) is constant for any n ∈ N. (i) Using the additivity of the integral and
for n ∈ N,
Z
n
Ψ(n) :=
ψ=
0
n
X
k=1
ψ|(k−1,k) =
n
X
k=1
(k − 1) =
n(n − 1)
.
2
Any x ∈ [0, ∞) can be represented in the form x = ψ(x) + r, where r =
x − ψ(x) ∈ [0, 1). Therefore,
Z
Ψ(x) = Ψ(ψ(x))+
ψ(x)+r
ψ(x)
ψ=
ψ(x)(ψ(x) + 1)
ψ(x)(ψ(x) − 1)
+ψ(x)r = xψ(x)−
,
2
2
where the penultimate equality is due to our result for Ψ(n), and the last
equality used the definition of r. ψ is continuous on R \ {0, 1, 2, . . .} and right
continuous at {0, 1, 2, . . .}. Therefore, Ψ is right continuous on [0, ∞) and left
continuous on R \ {0, 1, 2, . . .} (Analysis II). But
ψ(n−)(ψ(n−) − 1)
2
(n − 1)(n − 2)
n(n − 1)
= n−1+
=
= Ψ(n).
2
2
lim Ψ(x) = ψ(n−) +
x→n−
Therefore, Ψ is also left continuous at {1, 2, 3, . . .}. Conclusion: C = [0, ∞).
(ii) It clear from the representation of Ψ derived above that D = R \ {1, 2, . . .}
(Ψ is right-differentiable at 0) and that for any y ∈ D, Ψ0 (y) = ψ(y).
7. If {a = p0 < p1 < . . . < pn < b = pn+1 } is a partition of [a, b] compatible with
ϕ, then {Ka < Kp1 < . . . < Kpn < Kb} is a partition of [Ka, Kb] compatible
with ψ = ϕ(K −1 ·). Therefore, ψ ∈ S[Ka, Kb]. Let ψi = ϕi be the value of
ψ or ϕ on the i-th interval of the appropriate partition. By definition of the
integral,
Z
Kb
ψ=
n+1
X
Ka
ϕi (Kpi − Kpi−1 ) = K
i=1
n+1
X
b
Z
ϕi (pi − pi−1 ) =
ϕ
a
i=1
8. Let P = {a = p0 < p1 < . . . < pn < b = pn+1 } be a partition compatible
with both ψ, ϕ (see Question 2.) If ϕ|(pi−1 ,pi ) is a constant equal to ϕi , then
|ϕ|(pi−1 ,pi ) is also a constant equal to |ϕi |. Therefore, P is compatible with |ϕ|
and |ϕ| is a step function. Then
Z
|
b
ϕ| = |
a
n+1
X
n+1
X
ϕi (pi − pi−1 )| ≤
i=1
Z
|ϕi |(pi − pi−1 ) =
b
|ϕ|.
a
i=1
Similarly,
Z
|
b
(ϕ+ψ)| = |
a
n+1
X
(ϕi +ψi )(pi −pi−1 )| ≤
i=1
n+1
X
b
Z
(|ϕi |+|ψi |)(pi −pi−1 ) =
Z b
|ϕ|+
|ψ|.
a
i=1
a
Here ψi = ψ|(pi−1 ,pi ) .
9. Let f be our continuous step function on [a, b], P = {a = p0 < p1 < . . . < pn <
b = pn+1 } - a partition of [a, b] compatible with f and let fi = f |(pi−1 ,pi ) . The
continuity requires that at every point c of [a, b], limx→c− f (x) = limx→c+ f (x) =
f (c). Applying this to c = pi gives fi−1 = fi = f (pi ) for i = 1, . . . , n + 1. Finally, using one-sided continuity at the boundary points a and b we conclude
that f ≡ f1 , i. e. constant.
10. Let f, g ∈ S[a, b], let P = {a = p0 < p1 < . . . < pn < b = pn+1 } - a partition
of [a, b] compatible with both f and g. Let fi = f |(pi−1 ,pi ) , gi = g|(pi−1 ,pi ) ,
i = 1, 2, . . . , n + 1.
(a) Linearity:
Z
b
n+1
X
(αf + βg) =
(αfi + βgi )(pi − pi−1 )
a
=α
n+1
X
i=1
fi (pi − pi−1 ) + β
i=1
n+1
X
i=1
Z
gi (pi − pi−1 ) = α
b
Z
f +β
a
b
g.
a
(b) Monotonicity: if f ≥ g, then fi ≥ gi for every i. Therefore,
b
Z
Z
b
f−
g=
a
a
n+1
X
(fi − gi )(pi − pi−1 ) ≥ 0.
i=1
(c) Additivity: restrictions of step functions to subintervals are also step functions. Let {q = u0 < u1 < . . . < un < p = un+1 } be a partition of [q, p]
compatible with f |[q,p] and let {p = v0 < v1 < . . . < vm < r = vm+1 }
be a partition of [p, r] compatible with f |[p,r] . Let f |[ui−1 ,ui ] = αi and
f |[vj−1 ,vj ] = βj . Then the set {q = u0 < u1 < . . . < un < v0 < v1 < . . . <
vm < r = vm+1 } is a partition of [q, r] compatible with f |[q,r] and
Z
p
Z
f=
f+
q
r
p
n+1
X
αi (ui − ui−1 ) +
i=1
m+1
X
Z
βi (vi − vi−1 ) =
i=1
r
f.
q
(d) Compatibility with integrals of constants: if f is equal to constant c on
[a, b], then {a, b} is a partition of [a, b] compatible with f . Then by defiRb
nition of the integral, a f = c(b − a).
(e) Insensitivity to values on finite sets: if f is zero on [a, b] \ {p1 , p2 , . . . pn },
then f ∈ S[a, b] with a compatible partition {a = p0 < p1 < p2 < . . . <
pn < pn+1 = b}. (Here we assume that pi ’s are internal points of [a, b],
all other cases can be studied in the similar vein.) But f |(pi−1 ,pi ) = 0.
Therefore,
Z b
n+1
X
f=
0 · (pi − pi−1 ) = 0.
a
i=1
0.2 Regulated functions. Integration of regulated functions.
11. (Q11) Let ψ be any step function on [0, 1]. Let P = {0 < p1 < p2 < . . .}
be a partition of [0, 1] compatible with ψ. Let ψ1 = ψ |(0,p1 ) . Let n ∈ N be
1
1
< p1 and a−1 = 3π/2+2πn
< p1 . Therefore,
large enough so that a1 := π/2+2πn
a1 , a2 ∈ (0, p1 ). But f (a1 ) = 1 and f (a2 ) = −1. Therefore, ||f − ψ||∞ ≥
max(|ψ1 − 1|, |ψ1 + 1|) ≥ 1. Thus there exists no sequence of step functions
converging to f uniformly meaning that f is not regulated.
12. (Q12). R[a, b] contains all constant functions as S[a, b] ⊂ R[a, b] and a constant
function is a step function. (1) R[a, b] is closed w. r. t. multiplication by
numbers: let r ∈ R[a, b], α ∈ R. For any > 0 there exists a step function
ψ : ||r − ψ ||∞ < . Therefore, ||αr − αψ ||∞ < |α|. As αψ ∈ S[a, b] and
is arbitrary positive, we conclude that there is a sequence of step functions
converging uniformly to αr. Therefore, αr is regulated. (2) R[a, b] is closed
under addition: let p, q be regulated. Then for any > 0 there exist φ , ψ ∈
S[a, b]: supx∈[a,b] |p(x) − φ (x)| < /2, supx∈[a,b] |q(x) − ψ (x)| < /2. So,
supx∈[a,b] |p(x) + q(x) − φ (x) − ψ (x)|
≤ supx∈[a,b] (|p(x) − φ (x)| + |q(x) − ψ (x)|)
≤ supx∈[a,b] |p(x) − φ (x)| + supy∈[a,b] |q(y) − ψ (y)| < .
As ψ + φ ∈ S[a, b], we conclude that p + q is regulated.
13. (Q13) For any n ∈ N, let φn (x) = n1 integer(nx), where integer is the integer
part function on [0, ∞). For any x ≥ 0, 0 ≤ nx − integer(nx) ≤ 1. Therefore,
||f − φn ||∞ ≤ n1 → 0 as n → ∞.
14. (Q14) Example: f (x) = 1/x: f is continuous on (0, 1] (Analysis II). Let ψ
be any step function on (0, 1] and let ψ1 = ψ |(0,p1 ) , where {p1 < p2 < . . . <
pn = 1} is a partition of (0, 1] compatible with ψ. Let ∈ (0, p1 ). Then
||f − ψ||∞ ≥ |1/ − ψ1 | → ∞ as → 0. Therefore, there exists no sequence of
step functions on (0, 1] converging to f uniformly.
15. (Q15) As f ∈ R[a, b], there is ψ ∈ S[a, b]: such that for any x ∈ [a, b], ψ(x)−1 ≤
f (x) ≤ ψ(x) + 1. Any step function is bounded: the upper (the lower) bound is
given by the maximum M (the minimum m) of the finite set of values of the step
function over [a, b]. Therefore, for any x ∈ [a, b], |f (x)| ≤ max(|M |+1, |m|+1),
i. e. f ∈ B[a, b].
16. (Q16) A monotonic function is either non-decreasing or non-increasing. Let
f : [a, b] → R be non-decreasing (the second case can be analysed in the
same way). To prove that f is regulated we will construct, for any n ∈ N,
a function ϕ ∈ S[a.b]: ||f − φ||∞ < 1/n. f (a) ≤ f (b) so pick r ∈ Z, k ∈ N
with r/n ≤ f (a) < (r + 1)/n and (r + k − 1)/n < f (b) ≤ (r + k)/n. For
0 < j ≤ k put pj := sup{x ∈ [a, b] : f (x) ≤ (r + j)/n} and put p0 := a.
Then x ∈ (pj−1 , pj ) ⇒ (r + j − 1)/n < f (x) ≤ (r + j)/n. (If pj−1 = pj
then ((r + j − 1)/n, (r + j)/n) ∩ f ([a, b]) = ∅ and we should renumber P =
{p0 , . . . , pk }.) Put ϕ = (r + j)/n on (pj−1 , pj ) and ϕ(pj ) := f (pj ). Then
∀j ∈ {1, . . . , k} ∀x ∈ (pj−1 , pj ) |ϕ(x) − f (x)| = |(r + j)/n − f (x)| ≤ 1/n (and
also for x = pj ) so kϕ − f k∞ ≤ 1/n. Thus f is regulated.
17. (Q17) Let {φn }n≥1 be a sequence of step functions on [a, b] converging to
f uniformly (a < b). Then {φn (K −1 ·)}n ≥1 is a sequence of step functions
on [min(Ka, Kb), max(Ka, Kb)] converging to f (K −1 ·) uniformly. Therefore,
f (K −1 ·) is regulated. Let P = {a = p0 < p1 < . . . < pn < b = pn+1 }
be a partition of [a, b] compatible with φn . Then K · P is a partition of
[min(Ka, Kb), max(Ka, Kb)] compatible with φn (K·). It is an immediate consequence of the definition of the step integral that
max(Ka,Kb)
Z
φn (K
−1
b
Z
·) = |K|
φn .
min(Ka,Kb)
a
Taking the large-n limit of both sides we find
max(Ka,Kb)
Z
f (K
−1
Z
f.
min(Ka,Kb)
Using the convention
Rd
Rc
a
c
h=−
Z
max(Ka,Kb)
d
h we find that
Z
Kb
h = sign(K)
min(Ka,Kb)
h.
Ka
Combining the last two formulae, we find that
Z Kb
Z b
−1
f (K ·) = K
f.
Ka
b
·) = |K|
a
18. (Q18) The formula of Question 17 specified to the case K = −1 states
Z −π
Z π
ϕ(−x)dx = −
ϕ(x)dx.
−π
π
Using φ(−x) = −φ(x) and swapping
the
Rπ
R π order of limits in the l. h. s. of the
above formula, we find that −π φ = − −π φ, which implies the desired result.
19. (Q19) Let ϕn : ϕn |(r/n,(r+1)/n] = n/(r + 1), r = n, n + 1, . . . , 2n − 1. ϕn (1) = 1.
Then ||f − ϕn ||∞ ≤ maxn≤r≤2n |n/r − n/(r + 1)| ≤ 1/n → 0 as n → ∞.
Therefore,
R 2 {φn }n∈N
R 2 is a sequence of step functions converging to f uniformly.
Thefore 1 φn → 1 f by definition of the regulated integral. So,
lim
n→∞
n
X
k=1
1
=
n+k
Z
1
2
1
dx = log(2).
x
Rq
20. (Q20) (i) | p cos(tx+α)dx| = |t−1 [sin(tx+α)]qp | = t−1 | sin(tp+α)−sin(tq+α)| ≤
2t−1 → 0 as t → ∞.
(ii) Let ϕ ∈ S[a, b] and let {p0 < p1 . . . < pk } be a compatible partition
R pj
Rb
P
cos(tx + α) dx| ≤
of [a,b]. Then | a ϕ(x) cos(tx + α) dx| = | kj=1 cj pj−1
Pk
−1
→ 0 as t → ∞.
j=1 |cj | · 2t
(iii) Given f ∈ R[a, b] and ε > 0 choose ϕ ∈ S[a, b] with kf − ϕk∞ ≤ ε. Then
supx {|f (x) · cos(tx + α) − ϕ(x) · cos(tx + α)|} ≤ kf − ϕk∞ ≤ ε.
Rb
By the last part ∃ T such that t ≥ T ⇒ | a ϕ(x) · cos(tx + α) dx| < ε, so
Rb
Rb
t ≥ T ⇒ | a f (x) · cos(tx + α) dx| ≤ | a (f (x) · cos(tx + α) − ϕ(x) · cos(tx +
Rb
α)) dx| + | a ϕ(x) · cos(tx + α) dx| < ε(b − a) + ε = ε(b − a + 1). Hence
Rb
f (x). cos(tx + α) dx → 0 as t → ∞.
a
October the 12th, 2015
Sergey Nazarenko and Oleg Zaboronski.