1. No category

MA3209
Mathematical Analysis III
AY 2007/2008 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 2007/2008 Sem 1
Throughout this document, let A denote the closure of A; Nr (x) be the open neighbourhood of x with
radius r. Assume that all the metric spaces stated are non-empty.
Question 1
(a) Note that for all x1 , x2 ∈ X, y1 , y2 ∈ Y , since dX and dY are metrics on X and Y respectively, so
1
dX (x1 , x2 ) < ∞ and dY (y1 , y2 ) < ∞. Hence, d((x1 , y1 ), (x2 , y2 )) = (dX (x1 , x2 )p + dY (y1 , y2 )p ) p <
∞.
1
Since dX (x1 , x2 ) ≥ 0 and dY (y1 , y2 ) ≥ 0, so d((x1 , y1 ), (x2 , y2 )) = (dX (x1 , x2 )p + dY (y1 , y2 )p ) p ≥ 0.
We have
1
d((x1 , y1 ), (x2 , y2 )) = 0 ⇔ (dX (x1 , x2 )p + dY (y1 , y2 )p ) p = 0
⇔ dX (x1 , x2 )p + dY (y1 , y2 )p = 0
⇔ dX (x1 , x2 ) = 0 and dY (y1 , y2 ) = 0
⇔ x1 = x2 and y1 = y2
⇔ (x1 , y1 ) = (x2 , y2 ).
We also have
1
d((x1 , y1 ), (x2 , y2 )) = (dX (x1 , x2 )p + dY (y1 , y2 )p ) p
1
= (dX (x2 , x1 )p + dY (y2 , y1 )p ) p
= d((x2 , y2 ), (x1 , y1 ))
It suffices to show that d satisfies the triangle inequality.
Take any (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) ∈ X × Y .
Since dX is a metric, so dX (x1 , x3 ) ≤ dX (x1 , x2 ) + dX (x2 , x3 ). Hence,
dX (x1 , x3 )p ≤ [dX (x1 , x2 ) + dX (x2 , x3 )]p
Similarly,
dY (y1 , y3 )p ≤ [dY (y1 , y2 ) + dY (y2 , y3 )]p
Recall the Mink¨
owski’s inequality:
" n
#1 " n
#1 " n
#1
p
p
p
X
X
X
|ai + bi |p
≤
|ai |p +
|bi |p
i=1
i=1
(1)
i=1
Putting n = 2, a1 = dX (x1 , x2 ), a2 = dY (y1 , y2 ), b1 = dX (x2 , x3 ) and b2 = dY (y2 , y3 ) into (1), we
obtain:
1
d((x1 , y1 ), (x3 , y3 )) = [dX (x1 , x2 )p + dY (y1 , y3 )p ] p
1
≤ [(dX (x1 , x2 ) + dX (x2 , x3 ))p + (dY (y1 , y2 ) + dY (y2 , y3 ))p ] p
1
1
≤ [dX (x1 , x2 )p + dY (y1 , y2 )p ] p + [dX (x2 , x3 )p + dY (y2 , y3 )p ] p
by (1)
= d((x1 , y1 ), (x2 , y2 )) + d((x2 , y2 ), (x3 , y3 ))
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∴ d is a metric on X × Y .
(b)(i) Let f ∈ C[0, 1] be a limit point of S. Then given any ε > 0, there exists a function gε ∈ S such
that d∞ (f, gε ) < ε. Hence,
|f (0)| = |f (0) − gε (0)|
since gε (0) = 0
≤ sup{|f (x) − gε (x)| : x ∈ [0, 1]}
= d∞ (f, gε )
< ε.
Since ε is arbitrary, so |f (0)| = 0, which implies f (0) = 0. So f ∈ S.
∴ S is closed in (C[0, 1], d∞ ).
(b)(ii) For any small positive ε, define the function fε (x) : [0, 1] → R,
(
2x
if 0 ≤ x < 2ε
fε (x) = ε
1
if 2ε ≤ x ≤ 1
Since fε is continuous on [0, 1] and fε (0) = 0, so fε ∈ S.
Consider the function g : [0, 1] → R, g(x) ≡ 1. So g ∈ C[0, 1], but g ∈
/ S.
Claim: g is a limit point of S.
Proof: Given any ε > 0,
Z
d1 (fε , g) =
1
|fε (x) − g(x)| dx
Z ε
2 2x
dx
−
1
ε
0
Z ε
2
2x
1−
dx
ε
0
ε
x2 2
x−
ε 0
ε
4
ε
0
=
=
=
=
<
In other words, for every ε > 0, there exists fε ∈ S such that d1 (fε , g) < ε. So g is a limit point of
S. Since g ∈
/ S, so S is not closed in (C[0, 1], d1 ).
Question 2
(i) Take any x, y ∈ X.
Since f (y) = inf{d(y, a) : a ∈ A}, so given any ε > 0, there exists z ∈ A such that d(y, z) ≤ f (y)+ε.
Then f (x) ≤ d(x, z) ≤ d(x, y) + d(y, z) ≤ d(x, y) + f (y) + ε. So f (x) − f (y) ≤ d(x, y) + ε.
Since ε is arbitrary, so f (x) − f (y) ≤ d(x, y). Similarly, f (y) − f (x) ≤ d(x, y).
∴ |f (x) − f (y)| ≤ d(x, y) for all x, y ∈ X.
Now, given any ε > 0, we let δ = ε. So whenever x, y ∈ X and d(x, y) < δ, we have |f (x) − f (y)| ≤
d(x, y) < δ = ε. So f is uniformly continuous on X.
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(ii)(a) Since d(K, A) = inf{d(x, a) : x ∈ K, a ∈ A}, so given any ε > 0, there exists k ∈ K and a ∈ A such
that d(k, a) ≤ d(K, A) + ε.
Thus, inf{f (x) : x ∈ K} ≤ f (k) = d(k, A) ≤ d(k, a) ≤ d(K, A) + ε.
Since ε is arbitrary, so
inf{f (x) : x ∈ K} ≤ d(K, A).
(2)
Furthermore, ∃k ∈ K such that f (k) ≤ inf{f (x) : x ∈ K} + ε, so
d(K, A) = inf{d(x, a) : a ∈ A, x ∈ K}
≤ inf{d(k, a) : a ∈ A}
= d(k, A)
= f (k)
≤ inf{f (x) : x ∈ K} + ε
Since ε is arbitrary, so
d(K, A) ≤ inf{f (x) : x ∈ K}
(3)
By (2) and (3), we have d(K, A) = inf{f (x) : x ∈ K}.
Since f is continuous on the compact set K, so f attains its minimum at some k1 ∈ K, i.e. ∃k1 ∈ K
such that f (k1 ) = inf{f (x) : x ∈ K} = d(K, A).
Note that d(K, A) ≥ 0. Suppose d(K, A) = 0. Then from above f (k1 ) = inf{d(k1 , a) : a ∈ A} = 0.
Therefore, for any ε > 0, there exists aε ∈ A such that d(k1 , aε ) < ε. So k1 is a limit point of A.
Since A is closed, so k1 ∈ A. However, K ∩ A = φ from the assumption given in the question, and
this contradicts k1 ∈ K ∩ A.
∴ d(K, A) > 0.
(ii)(b) From 2(ii)(a), d(K, A) > 0. Let m = d(K, A).
Let U := {x ∈ X : d(x, K) <
So K ⊆ U and A ⊆ V .
m
2 }.
Let V := {x ∈ X : d(x, A) <
m
2 }.
Claim: U ∩ V = φ.
Proof: Suppose U ∩ V 6= φ. Take any y ∈ U ∩ V . Since y ∈ U , so d(y, K) < m
2.
0 ∈ K such that d(y, k 0 ) < d(y, K) + [ m − d(y, K)] =
Since m
−
d(y,
K)
>
0,
so
there
exists
k
2
2
Similarly, there exists a0 ∈ A such that d(y, a0 ) < m
.
2
m
Thus, d(k 0 , a0 ) ≤ d(k 0 , y) + d(y, a0 ) < m
2 + 2 = m = d(K, A) = inf{d(k, a) : k ∈ K, a ∈ A}.
This is a contradiction. Therefore, U and V are disjoint.
m
2.
It remains to show that U and V are open in X.
Take any x0 ∈ U .
Case 1: x0 ∈ K.
For any p in the neighbourhood N m2 (x0 ), we have
d(p, K) = inf{d(p, k) : k ∈ K}
≤ d(p, x0 )
m
.
<
2
So p ∈ U . Therefore, N m2 (x0 ) ⊆ U . Thus, every x0 ∈ K has an open neighbourhood contained in
U.
Case 2: x0 ∈
/ K.
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Since K is closed, so d(x0 , K) > 0. Let n = d(x0 , K).
Consider the neighbourhood N m2 −n (x0 ) := {x ∈ X : d(x0 , x) <
Then for all q ∈ N m2 −n (x0 ),
d(q, K) =
AY 2007/2008 Sem 1
m
2
− n}.
inf{d(q, k) : k ∈ K}
≤
inf{d(q, x0 ) + d(x0 , k) : k ∈ K}
<
inf{( m
2 − n) + d(x0 , k) : k ∈ K}
= (m
2 − n) + inf{(d(x0 , k) : k ∈ K}
=
=
(m
2 − n) + d(x0 , K)
m
2
−n+n
m
2
=
Thus, for all q ∈ N m2 −n (x0 ), q ∈ U . Thus, every x0 ∈ U \ K has an open neighbourhood which is
contained in U .
Combining the two cases, we conclude that every x0 ∈ U is an interior point of U .
∴ U is open in X.
Note that in the above proof that U is open in X, we only made use of the assumption that K is
closed in X. Since A being compact implies that A is closed in X, so by the same argument as
above, we can conclude that V is open in X.
Question 3
(1) ⇒ (2):
Assume X is connected, and f is locally constant.
Fix a point p ∈ X. Define S1 := {x ∈ X : f (x) = f (p)}, S2 := {x ∈ X : f (x) 6= f (p)}
Since p ∈ S1 , so S1 is non-empty. Suppose f is not a constant function, then there exists x0 ∈ X
such that f (x0 ) 6= f (p). So S2 is non-empty.
For any x1 ∈ S1 , there exists an open neighbourhood Ux1 containing x1 such that f (Ux1 ) = {p}
since f is locally constant. Note that Ux1 ⊆ S1 . Therefore, S1 is open in X. Similarly S2 is open
in X.
Note that S1 ∩ S2 = φ and S1 ∪ S2 = X. So X can be written as the union of two non-empty
disjoint sets, so X is disconnected. This contradicts our assumption that X is connected.
(2) ⇒ (1):
Suppose X is not connected, then X is a disjoint union of two non-empty open sets A and B.
Define f : X → R,
(
0
if x ∈ A
f (x) =
1
if x ∈ B
For any x0 ∈ X. x0 ∈ A or x0 ∈ B, but not both. Without loss of generality, assume x0 ∈ A. Since
A is open, there exists an open neighbourhood Ux0 of x0 such that Ux0 ⊆ A. So f (Ux0 ) = {0}.
Thus, f is locally constant. By assumption, f must be a constant function. However, f is not a
constant function from definition. Therefore, X must be connected.
Question 4
(i) Since T N is a contraction mapping, so there exists c ∈ (0, 1) such that for all x, y ∈ X,
d(T N (x), T N (y)) ≤ c d(x, y)
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(4)
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Since X is complete, so by the contraction mapping principle, there exists a unique fixed point
x0 ∈ X such that T N (x0 ) = x0 . Using (4), we obtain:
d(T N (T (x0 )), T N (x0 )) ≤ c d(T (x0 ), x0 )
d(T (T N (x0 )), T N (x0 )) ≤ c d(T (x0 ), x0 )
d(T (x0 ), x0 ) ≤ c d(T (x0 ), x0 )
This can only happen if d(T (x0 ), x0 ) = 0. Thus, T (x0 ) = x0 . So x0 is a fixed point of T . It remains
to show that T has at most one fixed point.
Suppose T has at least two different fixed points. Denote two of these by x1 and x2 . Since x1 6= x2 ,
so d(x1 , x2 ) > 0. We have d(x1 , x2 ) = d(T N (x1 ), T N (x2 )) ≤ c d(x1 , x2 ). Dividing by d(x1 , x2 ), we
obtain 1 ≤ c, which is a contradiction.
∴ T has a unique fixed point, namely x0 .
(ii)(a) Suppose φ is a contraction mapping on C[0, 1], then there exists c ∈ (0, 1) such that for all
f1 , f2 ∈ C[0, 1],
d(φ(f1 ), φ(f2 )) ≤ c d(f1 , f2 )
where d stands for the uniform metric.
So for all f1 , f2 ∈ C[0, 1],
Z x
Z x
d sin x +
f1 (t) dt, sin x +
f2 (t) dt
≤ c d(f1 , f2 )
0
0
Z x
Z x
sup sin x +
f2 (t) dt − sin x +
f1 (t) dt ≤ c sup |f1 (x) − f2 (x)|
x∈[0,1]
0
0
Z
sup x∈[0,1]
x
x∈[0,1]
Z
x
f2 (t) dt −
0
0
f1 (t) dt ≤ c sup |f1 (x) − f2 (x)|
x∈[0,1]
In particular, let f1 ≡ 0 and f2 ≡ 1 on [0, 1]. Then
Z x
1 dt ≤ c sup 1
sup x∈[0,1]
0
x∈[0,1]
sup x ≤ c
x∈[0,1]
1 ≤ c
This is a contradiction. Therefore, φ cannot be a contraction mapping.
(ii)(b) Claim: φ2 is a contraction mapping on C[0, 1].
Proof: For all f ∈ C[0, 1] and for all x ∈ [0, 1],
Z x
2
(φ f )(x) = φ sin x +
(φf )(t) dt
0
Z x
Z t
= sin x +
sin t +
f (u) du dt
0
0
Z x
Z xZ t
= sin x +
sin t dt +
f (u) du dt
0
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0
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So for all f1 , f2 ∈ C[0, 1] and for all x ∈ [0, 1],
Z x Z t
2
2
(φ f2 )(x) − (φ f1 )(x) = f2 (u) − f1 (u) du dt
0
0
Z x Z t
≤
f2 (u) − f1 (u) du dt
Z0 x 0
t sup |f2 (u) − f1 (u)| dt
≤
0
u∈[0,1]
!Z
=
≤
t dt
0
u∈[0,1]
=
x
sup |f2 (u) − f1 (u)|
x2
d(f1 , f2 )
2
1
d(f1 , f2 )
2
Therefore,
sup (φ2 f2 )(x) − (φ2 f1 )(x) ≤
x∈[0,1]
d(φ2 f1 , φ2 f2 ) ≤
1
d(f1 , f2 )
2
1
d(f1 , f2 )
2
So φ2 is a contraction mapping on C[0, 1]. Since C[0, 1] is complete, so by part (i), φ has a unique
fixed point.
Question 5
(a)(i) Let (zi ) be a limit point of c. We want to show that (zi ) ∈ c.
Let ε > 0 be given.
ε
Then we can choose a convergent sequence (xi ) such that d((xi ), (zi )) = sup|xi − zi | < .
3
i∈N
Since (xi ) is Cauchy, so there exists N0 ∈ N such that for all m, n ≥ N0 , we have |xm − xn | < 3ε .
So for all m, n ≥ N0 ,
|zm − zn | = |(zm − xm ) + (xm − xn ) + (xn − zn )|
≤ |zm − xm | + |xm − xn | + |xn − zn |
ε
≤ sup |xi − zi | + + sup |xi − zi |
3 i∈N
i∈N
= ε
Thus, (zi ) is a Cauchy sequence. Since (zi ) is a real sequence, so it is also a convergent sequence,
i.e. (zi ) ∈ c.
∴ c is closed in (`∞ , d).
(a)(ii) Since a closed subset of a complete metric space is complete, so c being a closed subset of the
complete metric space (`∞ , d) is complete.
(b) We need to show that ∀ε > 0, ∃δ > 0 such that ∀f ∈ F and ∀x1 , x2 ∈ K, d(x1 , x2 ) < δ ⇒
|f (x1 ) − f (x2 )| < ε.
Let ε > 0 be given. Since F is totally bounded, there exists a finite subset S = {f1 , f2 , ..., fn } ⊆
C(K) such that
F⊆
n
[
Nε (fi )
(5)
i=1
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where Nε (fi ) = {f ∈ C(K) : kf − fi k < ε}.
Since continuous functions on compact sets are uniformly continuous, so f1 , f2 , ..., fn are all uniformly continuous on K. So for all i ∈ {1, 2, ..., n}, ∃δi > 0 such that for all x1 , x2 ∈ K with
d(x1 , x2 ) < δi , we have |fi (x1 ) − fi (x2 )| < ε.
Let δ = min{δ1 , δ2 , ..., δn }.
Given any f ∈ F, by (5), f ∈ Nε (fi ) for some i ∈ {1, 2, ..., n}, i.e. sup |f (x) − fi (x)| < ε.
x∈K
Then ∀x1 , x2 ∈ K with d(x1 , x2 ) < δ, we have
|f (x1 ) − f (x2 )| ≤ |f (x1 ) − fi (x1 )| + |fi (x1 ) − fi (x2 )| + |fi (x2 ) − f (x2 )|
< ε+ε+ε
= 3ε.
∴ F is equicontinuous on K.
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