MA244 Analysis III
Solutions. Sheet 2.
Questions for credit: 2 (4 points), 4 (5 points), 6 (8 points) and 9 (8 points)
0.1
Integration.
1. Let f ((a + b)/2) = 1, f (x) = 0, x ∈ [a, b] \ {(a + b)/2}. f is a step function,
therefore it is regulated; for any x ∈ [a, b] f (x) ≥ 0, f ((a + b)/2) > 0 and
Rb
f = 0.
a
2. If f (c) > 0 for some c ∈ [a, b] then 12 f (c) > 0 so there is δ > 0 such that
x ∈ (c − δ, c + δ) ∩ [a, b] ⇒ |f (x) − f (c)| < 21 f (c) ⇒ f (x) > 21 f (c) ⇒ ∀x ∈
[a, b] f (x) ≥ ϕ(x) where the step function ϕ : [a, b] → R is given by ϕ(x) :=
1
f (c) if x ∈ (c − δ, c + δ) ∩ [a, b] and ϕ(x) = 0 otherwise. Then, by the bounds
2
Rb
Rb
on the integral of regulated functions, a f ≥ a ϕ = δf (c) > 0 (but adjust
Rb
ϕ if min{c − a, b − c} < δ). Thus ∀c ∈ [a, b] f (c) = 0.
a
3. The integrand is continuous in [1, x]. Therefore the integral defines a differentiable function. The derivative at the extremum points is zero. Applying
FTC,
sin(x)
f 0 (x) =
= 0.
x
Therefore, x = π · k, k = 1, 2, . . ..
4. The hint leads to the following formula:
Z b(x)
d
f = b0 (x)f (b(x)) − a0 (x)f (a(x)).
dx a(x)
Its application gives:
(a) F 0 (x) = log(x).
√
(b) G0 (x) = − 1 + x4 .
4
(c) H 0 (x) = 2xe−x − e−x
2
(d) I 0 (x) = 12 x−1/2 cos(x) +
1
cos(1/x2 ).
x2
5. All the integrals are straightforward, the first one requires integration by parts,
the fourth - a substitution y = t2 .
R1
(a) 0 log(1 + x)dx = (xlog(x) − x) |21 = 2log(2) − 1;
R −1
3
(b) −2 x13 dx = −1/2x−2 |−1
−2 = − 8 ;
Rx t
(c) −x e dt = ex − e−x = 2sinh(x);
Rx
R x2
(d) 0 t · cos(t2 )dt = 12 0 cos(y)dy = 12 sin(x2 ).
0.2
Properties of regulated functions.
6. (i) Take a sequence of irrationals xn → 1/3; then f (xn ) = 0√→ 0 6= 1/3 =
f (1/3) = f (limn→∞ xn ) so f is not continuous at 1/3. As c = 1/ 2 is irrational,
f (c) = 0. For any fixed ε > 0, only finitely many rational numbers p/q have
q ≤ 1/ε, so let δ be the distance from c to the nearest of these. Then for any
rational x ∈ (c − δ, c + δ) and x = p/q, 0 < f (x) = 1/q < ε; for any irrational
x ∈ (c − δ, c + δ), f (x) = 0. The last two facts imply that f is continuous at c.
(ii) Every open interval in (0, 1) contains irrational points x where f (x) = 0
and rational x where f (x) 6= 0, so it is not constant there and f cannot be a
step function.
(iii) Given ε > 0, define ϕ : [0, 1] → R by ϕ(p/q) := 1/q if 0 ≤ p ≤ q ∈ N and
p, q are coprime and q ≤ 1/ε. For all other x ∈ [0, 1] set φ(x) = 0. Clearly,
ϕ ∈ S[0, 1]. Then f (t) = ϕ(t) unless t = p/q, q > 1/ε. Hence kf − ϕk∞ =
supp/q∈[0,1]∩Q:q>1/ε |f (p/q)| = supp/q∈[0,1]∩Q:q>1/ε 1q < ε.
7. Given ε > 0 take a step function ϕ[0, 2] → R such that kϕ − f k∞ < ε/2.
Take a partition P compatible with ϕ and include 1 in P . Then, for some j,
0 = p0 < p1 < . . . < pj = 1 < . . . < pk = 2 and ∀x ∈ (pj−1 , 1) ϕ(x) = cj .
Choose N such that ∀n ≥ N xn ∈ (pj−1 , 1). Then m ≥ n ≥ N ⇒ |f (xm ) −
f (xn )| = |f (xm )−ϕ(xm )+ϕ(xn )−f (xn )| ≤ |f (xm )−ϕ(xm )|+|ϕ(xn )−f (xn )| ≤
kf − ϕk∞ + kϕ − f k∞ < 2ε/2 = ε. Hence (f (xn )) is Cauchy and so converges.
0.3
8.
RR
Improper integrals.
xdx =
−R
x2
2
|R
−R = 0. However,
Z
R∞
−∞
0
lim
R1 →∞
xdx is divergent, as
Z
xdx + lim
−R1
R2 →∞
R2
xdx
0
does not exist.
9. In all cases, the relevant integrals are elementary, in (a) use integration by
parts to find the anti-derivative of log(t).
R1
R1
(a) 0 log(t)dt = lim↓0 log(t)dt = lim↓0 (t log(t) − t) |1 = −1. Here we used
lim→0 log() = 0, which follows from l’Hôpital’s rule.
(b) There
consider: 0 < p < 1, p = 1 and p > 1. If 0 < p <
R 1 to
R 1 1are three cases
1−p
1
1
. The improper integral
1, 0 tp dt = lim↓0 tp dt = lim↓0 t1−p |1 = 1−p
R1 1
R1
1
exists and equals to 1−p . If p = 1, 0 t dt = lim↓0 1t dt = lim↓0 log(t)|1 .
The
improper integral diverges. If p > 1,
R 1 1 limit does not
R 1 1exist, so thet1−p
1
dt
=
lim
dt
=
lim
↓0 tp
↓0 1−p | . The limit does not exist, so the
0 tp
improper integral diverges.
(c) There are three cases to consider: p > 1, p = 1 and 0 < p < 1. If
RR
R∞
1−p
1
p > 1, 1 t1p dt = limR→∞ 1 t1p dt = limR→∞ t1−p |R
1 = p−1 . The improper
R
RR
∞
1
integral exists and equals to p−1
. If p = 1, 0 1t dt = limR→∞ 1 1t dt =
limR→∞ log(t)|∞
so the improper integral di1 . The Rlimit does not exist,
RR 1
1−p
∞ 1
verges. If 0 < p < 1, 1 tp dt = limR→∞ 1 tp dt = limR→∞ t1−p |∞
1 . The
limit does not exist, so the improper integral diverges.
R∞
RR
(d) 0 cos(x)dx = limR→∞ 0 cos(x)dx = limR→∞ sin(R). The limit does not
exist, the improper integral is divergent.
0.4 Uniform convergence and uniform continuity.
10. As f, g are uniformly continuous on A, for any > 0, there exist δ1 , δ2 > 0 such
that ∀x, y ∈ A,
|x − y| < δ1 ⇒ |f (x) − f (y)| < ,
2
|x − y| < δ2 ⇒ |g(x) − g(y)| < .
2
Let δ = min(δ1 , δ2 ) > 0. Then for any x, y ∈ A : |x − y| < δ
|f (x) + g(x) − f (y) − g(y)| ≤ |f (x) − f (y)| + |g(x) − g(y)| < /2 + /2 = .
So, f + g is uniformly continuous on A.
11. Take δ > 0 such that x, y ∈ (a, b), |x − y| < δ ⇒ |f (x) − f (y)| < 1. Now
take n ∈ N such that n > (b − a)/δ and put pj := a + j(b − a)/n , 0 ≤ j ≤ n.
Then ∀x ∈ (a, b) ∃j ∈ {1, . . . , n} such that x ∈ [pj−1 , pj ]. As |pj − pj−1 | < δ,
it follows that |f (x)| < |f ( 12 (pj + pj−1 ))| + 1. Hence sup{|f (x)| : x ∈ (a, b)} ≤
max{|f ( 21 (pj + pj−1 ))| + 1 : 1 ≤ j ≤ n}, which shows that f is bounded.
f : R → R, f (x) := x is uniformly continuous (with δ = ε) but not bounded.
g : R → R, g := sin is uniformly continuous (with δ = ε, use sin(x) − sin(y) =
2cos( x+y
)sin( x−y
)) and bounded. A constant function g is a simpler example.
2
2
12. Suppose that f is the uniform limit of weakly increasing step functions and yet
not itself weakly increasing. Then there are u < v in [a, b] with f (u) > f (v).
Put ε = (f (u) − f (v))/3 > 0 and pick a weakly increasing step function ϕ :
[a, b] → R with kϕ − f k∞ ≤ ε. Then f (u) − ε ≤ ϕ(u) ≤ ϕ(v) ≤ f (v) + ε so
3ε = f (u) − f (v) ≤ 2ε contradicting ε > 0.
October the 20th, 2014
Daniel Ueltschi and Oleg Zaboronski.