MA244 Analysis III
Solutions. Sheet 1.
Questions for credit: 5 (6 points), 14 (6 points), 15 (6 points) and 18 (7 points)
0.1
Step functions. Integration of step functions.
1. (i) Let P = {a, b}, k = 1 and a = p0 < p1 = b. By definition of squelch
functions, ψ|(a,b) = c, where c ∈ R is a constant. Therefore, Image(ψ) =
{ψ(a), ψ(b), c}. The maximal cardinality of this set is 3. (ii) P is a partition
of [a, b] such that ψ is constant on the only open subinterval this partition. So
ψ is a step function. (iii) No, f (x) = sign(x), x ∈ [−1, 1] is a step function on
[−1, 1] but not a squelch function.
2. (i) The result is proved by induction using the fact that for any φ, ψ ∈ S[a, b]
and λ, µ ∈ R, λφ + µψ ∈ S[a, b]. This fact was proved in the lecture: 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. (ii)
False: let φn (x) = 1/n for x ∈ (1 − 1/n, 1 −
+ 1)), n = 1, 2, 3, . . . and zero
P1/(n
∞
otherwise. Clearly, φn ∈ S[0, 1]. But φ := k=1 φn is not a step function as for
any partition 0 = p0 < p1 < . . . < pk−1 < pk = 1, φ |(pk−1 ,pk ) is not constant.
3. Define φ(x) = 1 if x is rational, φ(x) = −1 if x is irrational. Clearly, φ is not a
step function, as it is not constant on any non-empty open subinterval of [0, 1].
(Students should still present a full argument.) But |φ| ≡ 1 - a constant, hence
a step function.
4. (i) If h |(zi−1R,zi ) = ci - constant, then h(wi ) = ci for any wi ∈ (zi−1 , zi ). By
definition of h,
Z b
n
n
X
X
h=
ci (zi − zi−1 ) =
h(wi )(zi − zi−1 ).
a
i=1
i=1
(ii) Changing the value
R of h at z1 , z2 , . . . zn does not change h(wi ), where wi ∈
(zi−1 , zi ). Therefore, h does not change by part (i).
5. Let P, Q be partitions of [a, b] compatible with step functions h1 and h2 correspondingly. Then the refinement R = P ∪ Q ∪ {c} is compatible with
both h1 and h2 . Also, c ∈ R. Let R = {z0 , z1 , . . . zk }. By construction,
h1 (w) = h2 (w) for any w ∈ (zi−1 , zi ), 1 ≤ i ≤ k. By exercise 4 this implies
Rb
Rb
that a h1 = a h2 . (ii) The proof is identical to part (i), if one considers the
partition R = P ∪ Q ∪ {c1 , c2 , . . . , cN }
6. Let a = z0 < z1 < z2 < . . . < zk = b be a partition of [a, b] compatible with
both h1 and h2 . By exercise 4,
Z
b
h1 =
a
Z
h1 (wi )(zi − zi−1 ),
i=1
b
h2 =
a
k
X
k
X
h2 (wi )(zi − zi−1 ),
i=1
where wi ∈ (zi−1 , zi ). As h1 (wi ) ≥ h2 (wi ) for any i between 1 and k, we are
done.
7. (i) If φ ∈ S[a, b], then |φ| ∈ S[a, b] (use any partition compatible with φ).
Therefore, all integrals we are asked to analyze are defined. As |φ| ≥ ±φ, the
Rb
Rb
previous exercise and the linearity of the integral implies that a |φ| ≥ ± a φ,
Rb
Rb
which implies of course that a |φ| ≥ | a φ|. (ii) As |φ + ψ| ≤ |φ| + |ψ| (triangle
Rb
inequality) and |φ + ψ|, |φ|, |ψ| ∈ S[a, b], we have by exercise 6: a |φ + ψ| ≤
Rb
Rb
Rb
(|φ|
+
|ψ|)
=
|φ|
+
|ψ|, where the last equality is due to the linearity of
a
a
a
the integral.
8. Note that φ |[−5,x] is a step function for any x ∈ [−5, 5]. To calculate integral
we can use exercise 4 and the following partitions: {−5, x} for x ≤ 0 and
{−5, 0, x} for x > 0. The answer is Φ(x) = −5 + |x|. Therefore (Analysis
II), Φ is differentiable on [−5, 0) ∪ (0, 5] and Φ0 (x) = −1 = φ(x) for x < 0,
Φ0 (x) = 1 = φ(x) for x > 0.
9. By definition of the integral of the step function,
Z b
N
N
X
X
Dφ |(pi−1 ,pi ) (pi −pi−1 ) =
(φi −φi−1 ) = φN −φ0 = φ(b)−φ(a).
Dφ(x)dx =
a
k=1
k=1
10. The refinement of partitions compatible with ϕ and ψ is a partition compatible
with the product function ϕψ. This shows that ϕψ is a step function.
Then tϕ + ψ and (tϕ + ψ)2 are also step functions. So, ∀t ∈ R,
Z b
k
X
2
(tϕ + ψ) =
(tcj + c0j )2 (pj − pj−1 )
0 ≤
a
= t
2
1
k
X
c2j (pj
− pj−1 ) + 2t
k
X
1
= t2
Z
b
ϕ2 + 2t
Z
b
Z
ϕψ +
− pj−1 ) +
k
X
c02
j (pj − pj−1 )
1
1
b
ψ 2 = At2 + 2Bt + C,
a
a
a
cj c0j (pj
say. Therefore the quadratic polynomial At2 + 2Bt + C has at most one real
root. Thus
Z b
2
Z b Z b 2
ϕψ = B ≤ AC =
ϕ2
ψ2 ,
a
a
a
as required.
0.2 Regulated functions. Integration of regulated functions.
11. f is continuous at x > 0 (as the product of two continuous functions). Moreover,
lim f (x) = 0 = f (0),
x↓0
so f is continuous at 0. Therefore, f is continuous on [0, 1], hence it is regulated.
12. Let us consider an arbitrary step function φ ∈ S[0, 1]. Let P = {0, p1 , . . . , pk−1 , 1}
be a partition of [0, 1] compatible with φ. Let c = φ |(0,p1 ) . As f is unbounded
on (0, p1 ), it is always possible to find x ∈ (0, p1 ) such that |f (x) − c| > 1.
Therefore, there exists no step function φ ∈ S[0, 1] such that ||f − φ||∞ < 1.
Therefore, f is not regulated.
13. f is regulated. So for any > 0 there is φ ∈ S[0, a] such that for any x ∈ [0, a]
|f (x) − φ| < 0 .
As f ≥ 0 we can always choose φ ≥ 0. (Why?) Let t P = {0, p1 , . . . , pk−1 , a}
be a partition of [0, a] compatible with φ. Let ci = φ |( pi−1 , pi ), ci ≥ 0. For
x ∈ (pi−1 , pi ),
|f (x) − ci | < 0
If ci = 0, the above inequality implies that
p
√
f (x) < 0 .
(1)
If ci > 0, it gives
p
√
0
0
| f − ci | < √
√ <√ .
ci
f + ci
(2)
Now, for any > 0 we can choose 0 > 0 such that
0 √ 0
max max √ , < i:ci >0
ci
Then the inequalities (1, 2) show that
p
p
|| f − φ||∞ < .
√
As φ is a step function, the proof is complete. (It is worth verifying that
φ can be chosen in such a way that nothing goes wrong at the points of the
partition.)
14. Let Pn = {0, a/n, 2a/n, . . . a(n − 1)/n} be the sequence of partitions of [0, a].
Define a sequence of step functions φn on [0, a]: φn |(a (k−1) ,a k ) = a nk . Clearly,
n
n
limn→∞ ||x − φn ||∞ = 0. But
Z
0
Therefore,
Ra
0
dxx =
a
n
aX k
a2 n(n + 1) n→∞ a2
φn =
a = 2
→
.
n k=1 n
n
2
2
a2
2
by definition.
15. The set up is exactly the same as for the previous exercise, but φn |( (k−1) , k ) =
n
e
(k−1)
n
. Then
Z
0
1
n
n
1 X (k−1)
e−1
dxe = lim
e n = lim
= e − 1,
n→∞ n
n→∞ n(e1/n − 1)
k=1
x
where we had to sum a geometric series.
16. For any > 0 there is φ ∈ S[a, b]: such that for any x ∈ [a, b],
φ(x) − ≤ f (x) ≤ φ(x) + .
Define φ̃ ∈ S[a, b]:
φ̃(x) =
φ(x)
x 6= x1 , x2 , . . . , xk ,
gk x = xk , k = 1, 2, . . . , xk .
(3)
Then, for any x ∈ [a, b],
φ̃(x) − ≤ f (x) ≤ φ̃(x) + .
(4)
We know (Exercise 5(ii)), that
b
Z
Z
b
φ=
a
φ̃.
(5)
a
Using (3, 4, 5), we can establish the following two bounds on
b
Z
b
Z
g≤
b
Z
a
a
g:
b
Z
φ + (b − a) ≤
(φ̃ + ) =
R
f + 2(b − a)
a
a
and
Z
b
b
Z
g≥
a
Z
b
(φ̃ − ) =
a
b
Z
φ − (b − a) ≥
f − 2(b − a).
a
Taking the limit ↓ 0 we find that
a
Rb
a
f=
Rb
a
g.
17. f is regulated. Therefore, the restriction f |[0,π] is regulated. For any > 0,
there is φ ∈ S[0, π] such that ||f |[0,π] −φ||∞ < . Let 0 < p1 < p2 < . . . <
pk−1 < π be a partition of [0, π] compatible with φ. Let ci = φ(pi−1 ,pi ) . Let
ψ be a function on [−π, π] defined as follows: ψ(x) = φ(x) for x ≥ 0 and
ψ(x) = φ(−x) for x < 0. Please check that ψ ∈ S[−π, π]. By construction,
−π < −pk−1 < −pk−2 < −p−1 < 0 < p1 < . . . < pk−1 < π is a partition of
[−π, π] compatible with ψ and ψ |(−pi ,−pi−1 ) = ci . Moreover, as f (x) = f (−x),
||f − ψ||∞ < .
By definition,
Z
k
X
π
ψ=
−π
ψ |(−pi ,−pi−1 ) (pi − pi−1 ) +
i=1
k X
k
X
ψ |(pi−1 ,pi ) (pi − pi−1 )
i=1 i=1
=2
k
X
Z
φ |(pi−1 ,pi ) (pi − pi−1 ) = 2
π
φ.
0
i=1
As > 0 is arbitrary, and ψ and φ are -uniformly close to f , the derived
equality implies that
Z π
Z π
f =2
f.
−π
0
18. This question is very simple yet its importance for asymptotic analysis of
Fourier series is hard to overestimate. (i)
Z q
cos(tx) q
|p → 0
sint(x) == −
t
p
as t → ∞. (ii) Using a partition compatible with φ,
Z
b
sin(tx)φ(x) =
a
k
X
i=1
Z
pi
φ |(pi−1 ,pi )
sin(tx)dx → 0,
pi−1
as t → ∞ as each integral under the sign of summation goes to zero by part
(i). (iii) Fix > 0. Let φ be a step function on [a, b] such that ||f − φ||∞ < .
Then
Z b
Z b
|
sin(tx)(f (x) − φ(x))dx| ≤
|sin(tx)(f (x) − φ(x))|dx ≤ (b − a).
a
a
Therefore, uisng (ii),
Z
t→∞
b
sin(tx)f (x) ≤ (b − a).
−(b − a) ≤ lim
a
As is arbitrary, the above inequality leads to
Z b
sin(tx)f (x) = 0.
lim
t→∞
a
19. By definition,
D(
d
X
k=0
k
pk x ) =
d
X
kpk xk−1 ,
(6)
k=1
d
d
X
X
pk k+1
k
pk x ) =
x .
I(
k+1
k=0
k=0
(7)
It is clear from these formulae that I, D are linear maps, alternatively one could
quote the linearity of operations of differentiation and integration.
We can see from (6), that D is not injective: constants (polynomials of zero
degree d) are mapped to zero. In fact, only constants are mapped to zero.
Therefore Ker(D) consists of the set of polynomials of zero degree which is
isomophic
to P
R. D is surjective: the pre-image of an arbitrary polynomial
Pd
d+1
xk
k=1 qk−1 k .
k=0 qk xk is
It follows from (7) that I is not surjective: constants are not in the image of
I. However, I is injective: if I(p) = 0, it is easy to see from (7), that p = 0.
Finally, D ◦ I 6= I ◦ D, as D ◦ I(1) = 1, but I ◦ D(1) = 0. In fact, as it is easy
to check from (6, 7), D ◦ I = id, whereas I ◦ D is a projection onto a linear
subspace
of polynomials which are equal to zero at x = 0,
Pd of R[x]k consisting
Pd
I ◦ D( k=0 pk x ) = k=1 pk xk .
Rx
20. By the FTC and the continuity of f , the function F (x) = a f , x ∈ [a, b] is
continuous on [a, b] and differentiable on (a, b). Therefore, by the Mean Value
Theorem, there is c ∈ (a, b):
F (b) − F (a) = F 0 (c)(b − a) = f (c)(b − a).
We are done as F (a) = 0.
October the 6th, 2014
Daniel Ueltschi and Oleg Zaboronski.