1. No category

MA3209
Mathematical Analysis III
AY 2008/2009 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Chan Yu Ming, Poh Wei Shan Charlotte
MA3209
Mathematical Analysis III
AY 2008/2009 Sem 1
SECTION A
Question 1
(a) ρ(z, w) is well-defined, since ρ(z, w) =
Note that ρ(z, w) =
∞
X
∞
X
|zk − wk | ≤
∞
X
k=1
|zk | +
k=1
∞
X
|wk | < ∞.
k=1
|zk − wk | ≥ 0.
k=1
Suppose ρ(z, w) = 0, then
∞
X
|zk − wk | = 0 ⇒ ∀k ∈ N, zk = wk . So z = w.
k=1
Conversely, suppose z = w. Then ∀k ∈ N, zk = wk . So ρ(z, w) =
We also have symmetry: ρ(z, w) =
∞
X
|zk − wk | =
∞
X
k=1
∞
X
|zk − wk | = 0.
k=1
|wk − zk | = ρ(w, z).
k=1
To show triangle inequality, take z = (zk ), w = (wk ), v = (vk ) ∈ `∞ .
∞
∞
∞
X
X
X
Then ρ(z, w) =
|zk − wk | ≤
|zk − vk | +
|vk − wk | = ρ(z, v) + ρ(v, w).
k=1
k=1
k=1
Therefore, ρ is a metric on `∞ .
(b) Since x1 6= x2 , so d(x1 , x2 ) > 0. Take 1 =
d(x1 ,x2 )
, 2
2
=
d(x1 ,x2 )
2
> 0.
Suppose D(x1 , 1 ) ∩ D(x2 , 2 ) 6= φ. Take an element p ∈ D(x1 , 1 ) ∩ D(x2 , 2 ).
Then d(p, x1 ) < 1 and d(p, x2 ) < 2 .
So d(x1 , x2 ) ≤ d(x1 , p) + d(x2 , p) < 1 + 2 =
d(x1 ,x2 )
2
+ d(x12,x2 ) = d(x1 , x2 ), which is a contradiction.
Question 2
(i) We know that A = {y ∈ Rn : d(w, y) ≤ 1} = {y ∈ Rn : kw − yk2 ≤ 1} is closed and bounded in Rn .
By the Heine-Borel Theorem, A is compact.
(ii) It suffices to show that A = {y ∈ `∞ : d(w, y) ≤ 1} is not sequentially compact.
Given w = (w1 , w2 , w3 , ....) where w is a bounded sequence in C, define the following sequences:
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AY 2008/2009 Sem 1
z (1) = (w1 + 1, w2 , w3 , ...)
z (2) = (w1 , w2 + 1, w3 , ...)
z (3) = (w1 , w2 , w3 + 1, ...)
..
.
Since w is bounded, so all the z (k) ’s are bounded as well, i.e. ∀k ∈ N, z (k) ∈ `∞ .
Note that for all k, d(z (k) , w) = 1, so z (k) ∈ A. So {z (1) , z (2) , z (3) , ...} is a sequence in A.
Furthermore, note that if m 6= n, then d(z (m) , z (n) ) = 1, so any subsequence of {z (1) , z (2) , z (3) , ...}
cannot be Cauchy, and hence cannot be convergent.
Thus, {z (1) , z (2) , z (3) , ...} is a sequence in A that has no convergent subsequence, so A is not
sequentially compact, and hence not compact.
Question 3
(a) Write lim xk = x, lim yk = y. Note that x and y exist as (M, d) is complete.
k→∞
k→∞
(⇒) Assume x = y. So given any ε > 0, there exists K1 , K2 ∈ N such that ∀k ≥ K1 , d(xk , x) < 2ε ,
and ∀k ≥ K2 , d(yk , x) < 2ε . Then ∀k ≥ max{K1 , K2 }, d(xk , yk ) ≤ d(xk , x) + d(yk , x) <
ε
2
+
ε
2
= ε.
∴ d(xk , yk ) → 0 as k → ∞.
(⇐) Assume that d(xk , yk ) → 0 as k → ∞.
Then given any ε > 0, there exists K3 ∈ N such that ∀k ≥ K3 , d(xk , yk ) < 3ε .
There exists K4 ∈ N such that ∀k ≥ K4 , d(xk , x) < 3ε .
Similarly, there exists K5 ∈ N such that ∀k ≥ K5 , d(yk , y) < 3ε .
Let K0 = max{K3 , K4 , K5 }. Then d(x, y) ≤ d(x, xK0 ) + d(xK0 , yK0 ) + d(yK0 , y) < ε. Since ε is
arbitrary, d(x, y) = 0. Hence x = y.
(b) Since A is closed in N , so its complement N \ A is open in N . Note that:
• A ∩ N = A 6= φ (given)
• (N \ A) ∩ N = (N \ A) 6= φ (since A 6= N )
• A ∩ (N \ A) ∩ N = φ (since A ∩ (N \ A) = φ)
• N = A ∪ (N \ A).
Therefore, N is disconnected.
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Question 4
n
(i) Given any ε > 0, choose δ = ε. Then
v ∀x = (x1 , x2 , ..., xn ), y = (y1 , y2 , ..., yn ) ∈ R ,
u n
uX
whenever kx − yk2 < δ, we have t
|xk − yk |2 < δ.
k=1
v
v
u n
u s
X
uX
u
2
t
|xk − yk | ≤ t
|xk − yk |2 < ε.
Hence, kf (x) − f (y)k2 =
k=1
k=1
So f is uniformly continuous on Rn .
(ii) Since f is continuous from part (i), and A is closed in Rs , so B = f −1 (A) is closed in Rn .
Question 5
(i) For all k ∈ N, define gk : R2 → R, gk (x, y) =
(x, y) ∈ R2 , |gk (x, y)| =
2
2
1
e−k(x +y )
(k!)2
≤
1
k! .
(−1)k −k(x2 +y 2 )
e
.
(k!)2
∞
X
Since
k=1
Note that for each k ∈ N and for all
∞
X
1
converges by ratio test, so
gk (x, y)
k!
k=1
converges uniformly on R2 by Weierstrass M-test.
(ii) Let g : R2 → R, g(x, y) =
all k, and by (i),
∞
X
(−1)k
k=1
∞
X
(k!)2
e−k(x
2 +y 2 )
=
∞
X
gk (x, y). Note that since gk is continuous for
k=1
gk (x, y) converges uniformly on R2 , so g is continuous on R2 .
k=1
Define f : [0, 1] → R2 , f (t) = (t, cos t). Note that f is continuous on [0,1], since it is continuous at
each of its components. Hence h = g ◦ f is continuous on [0,1].
By the Weierstrass Approximation Theorem, given any > 0, there exists a polynomial p on [0,1]
such that |h(t) − p(t)| < for all t ∈ [0, 1].
SECTION B
Question 6
(a)(i) True.
∞
\
Take any p ∈ int
!
Ak . Then there exists ε > 0 such that D(p, ε) ⊆
k=1
for all k ∈ N, so p ∈ int(Ak ) for all k ∈ N. Thus, p ∈
∞
\
∞
\
Ak . Since D(p, ε) ⊆ Ak
k=1
int(Ak ).
k=1
Hence, int
∞
\
k=1
!
Ak
⊆
∞
\
int(Ak ).
k=1
(a)(ii) False. Let M = {3, 4}, x = 3, r = 1. Consider the discrete metric d(x, y) =
0 if x = y
.
1 if x 6= y
Then cl({y ∈ M : d(3, y) < 1}) = cl({3}) = {3} since any singleton set is closed.
However, {y ∈ M : d(3, y) ≤ 1} = {3, 4}.
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(b) Suppose Q can be expressed as a countable intersection of open subsets of R.
For simplicity of notation, ∀B ⊆ R, denote R \ B as B c .
∞
\
Write Q =
Ak , where each Ak is an open subset of R.
k=1
!c
∞
∞
\
[
So R − Q =
Ak
=
(Ak )c .
k=1
k=1
Recall the following theorem:
Let (M, d) be a metric space, and A ⊆ M . Then A is nowhere dense in M if and only if M \ [cl(A)]
is dense in M .
Since each Ak contains Q, and Q is dense in R, so Ak is dense in R.
Furthermore, each (Ak )c is closed in R, so cl((Ak )c ) = (Ak )c .
Using the above theorem with M = R, A = (Ak )c , and the fact that [cl((Ak )c )]c = [(Ak )c ]c = Ak
is dense in R, we conclude that (Ak )c is nowhere dense in R.
Let r1 , r2 , ... be an enumeration of Q.
Note that ∀k ∈ N, (Ak )c ∪ {rk } is nowhere dense. Otherwise, if ∃ x ∈ int[cl((Ak )c ∪ {rk })] =
int((Ak )c ∪ {rk }), then ∃ ε > 0 such that (x − ε, x + ε) ⊆ (Ak )c ∪ {rk } which is a contradiction as
the interval contains more than 2 rational numbers.
∞
∞
∞
[
[
[
Then R = (R − Q) ∪ Q =
(Ak )c ∪
{rk } =
[(Ak )c ∪ {rk }] is a countable union of nowhere
k=1
k=1
k=1
dense subsets of R. This is impossible because of Baire Category Theorem and that R is complete.
Therefore, Q cannot be expressed as a countable intersection of open subsets of R.
Question 7
(a) Substitute y = 0 into the given inequality. Then for each fixed x0 ∈ [−1, 1], we obtain for all k ∈ N,
|fk (x0 ) − fk (0)| ≤ C|x0 |.
|fk (x0 ) − 1| ≤ C|x0 |.
|fk (x0 )| ≤ 1 + C|x0 |.
Hence, for each fixed x0 ∈ [−1, 1], the sequence {fk (x0 )}∞
k=1 is bounded.
∴ {fk } is pointwise bounded.
Furthermore, given any ε > 0, choose δ = Cε . Then for all k ∈ N, and for all x, y ∈ [−1, 1],
whenever |x − y| < δ, we have |fk (x) − fk (y)| ≤ C|x − y| ≤ C( Cε ) < ε.
∴ {fk } is equicontinuous on [−1, 1].
Lastly, [−1, 1] is a compact subset of R, so by the Arzel`a-Ascoli Theorem, {fk } has a uniformly
convergent subsequence.
(b) Note that if (x, y) ∈ A, then (−x, −y) ∈ A. So define h : A → R, h(x, y) = f (x, y) − f (−x, −y).
Since f is continuous on A, so h is also continuous on A. Suppose there is no (x0 , y0 ) ∈ A
such that f (x0 , y0 ) = f (−x0 , −y0 ). Take (0, 1), (0, −1) ∈ A. Then either f (0, 1) > f (0, −1)
or f (0, 1) < f (0, −1). Without loss of generality, assume f (0, 1) > f (0, −1). Then h(0, 1) =
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f (0, 1) − f (0, −1) > 0, and h(0, −1) = f (0, −1) − f (0, 1) < 0. By the intermediate value theorem,
there exists a point (x1 , y1 ) ∈ A such that h(x1 , y1 ) = 0. This implies that f (x1 , y1 ) = f (−x1 , −y1 ),
which is a contradiction. Thus, there exists (x0 , y0 ) ∈ A such that f (x0 , y0 ) = f (−x0 , −y0 ).
Question 8
(a) Let ε > 0 be given. Since f is uniformly continuous on A, so there exists δ > 0 such that for
all x, y ∈ A, if d(x, y) < δ, then ρ(f (x), f (y)) < ε. Since A ⊆ M is totally bounded, so there
n
[
exists a finite subset {x1 , x2 , ..., xn } ⊆ M such that A ⊆
Dd (xi , δ), where Dd (xi , δ) = {y ∈ M :
i=1
d(y, xi ) < δ}.
Claim: f (A) ⊆
n
[
Dρ (f (xi ), ε), where Dρ (f (xi ), ε) = {z ∈ N : ρ(z, f (xi )) < ε}.
i=1
Proof: For any p ∈ f (A), there exists q ∈ A such that f (q) = p. But A ⊆
n
[
Dd (xi , δ) implies that
i=1
there exists i0 ∈ {1, 2, .., n} such that q ∈ Dd (xi0 , δ) ⇒ d(q, xi0 ) < δ. By the uniform continuity
n
[
of f , we have ρ(f (q), f (xi0 )) = ρ(p, f (xi0 )) < ε. So p ∈ Dρ (f (xi0 ), ε). Hence, p ∈
Dρ (f (xi ), ε).
i=1
S
Thus, f (A) ⊆ ni=1 Dρ (f (xi ), ε).
∴ f (A) ⊆ N is totally bounded.
(b)(i) Since Φr is a contraction, so there exists λ ∈ (0, 1) such that for all x, y ∈ M , we have
d(Φr (x), Φr (y)) ≤ λ d(x, y).
Suppose that for some positive integer m, we have d(Φrm (x), Φrm (y)) ≤ λm d(x, y) for all x, y ∈ M .
So for any x, y ∈ M , since Φr (x), Φr (y) ∈ M , we have:
d(Φr(m+1) (x), Φr(m+1) (y)) = d(Φrm (Φr (x)), Φrm (Φr (y)))
≤ λm d(Φr (x), Φr (y))
≤ λm+1 d(x, y).
So by induction, for any ` ≥ 1,
d(Φr` (x), Φr`) (y)) ≤ λ` d(x, y) for all x, y ∈ M .
(b)(ii) Lemma: Φ has a unique fixed point.
Proof: Since Φr is a contraction mapping on the complete metric space M , so by the contraction
mapping principle, Φr has a unique fixed point which we shall denote it by x0 .
From the definition of contraction of Φr , we have
d(Φr (Φ(x0 )), Φr (x0 )) ≤ λ d(Φ(x0 ), x0 )
d(Φ(Φr (x0 )), Φr (x0 )) ≤ λ d(Φ(x0 ), x0 )
d(Φ(x0 ), x0 ) ≤ λ d(Φ(x0 ), x0 )
Suppose Φ(x0 ) 6= x0 , then d(Φ(x0 ), x0 ) > 0. Dividing both sides by d(Φ(x0 ), x0 ), we obtain 1 ≤ λ
which is a contradiction. Therefore, Φ(x0 ) = x0 , i.e. x0 is also a fixed point of Φ.
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To prove uniqueness, suppose Φ has two fixed points x0 , y0 . From the definition of contraction of
Φr , we have
d(Φr (x0 ), Φr (y0 )) ≤ λ d(x0 , y0 )
d(x0 , y0 ) ≤ λ d(x0 , y0 )
Suppose x0 6= y0 , then d(x0 , y0 ). Dividing both sides by d(x0 , y0 ) > 0, we obtain 1 ≤ λ which is a
contradiction. Therefore, x0 = y0 , i.e. Φ has a unique fixed point.
Returning to the main problem, fix x ∈ M . If x is the fixed point x0 , then the sequence {Φk (x0 )}∞
k=1
is just the constant sequence {x0 , x0 , ...}, which converges in M . So we shall assume that x is not
a fixed point of M . Let B = max{d(x, x0 ), d(Φ(x), x0 ), d(Φ2 (x), x0 ), ..., d(Φr−1 (x), x0 )}. Note that
B is positive because x 6= x0 implies d(x, x0 ) > 0.
L
ε
Now, given any ε > 0, choose L large enough such that λb r c < , where b Lr c is the quotient when
B
L is divided by r. For any k, using the division algorithm, write k = rm1 + m2 , where m1 is the
quotient, and m2 ∈ {0, 1, 2, .., r − 1} is the remainder. Then for all k ≥ L,
d(Φk (x), x0 ) = d(Φrm1 +m2 (x), x0 )
= d(Φrm1 (Φm2 (x)), Φrm1 (x0 ))
≤ λm1 d(Φm2 (x), x0 )
L
≤ λb r c d(Φm2 (x), x0 )
ε
(B)
<
B
= ε
(1)
k
L
Note that (1) is derived from m1 = b kr c ≥ b Lr c and since λ < 1, λb r c ≤ λb r c .
Therefore, the sequence {Φk (x)}∞
k=1 converges to x0 for any fixed x ∈ M .
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