Some Notes on Rudin’s book:
Principles of Mathematical Analysis, 3/e
Written by Meng-Gen, Tsai
email: [email protected]
Exercise 1.1. If r is rational (r 6= 0) and x is irrational, prove that r + x
and rx are irrational.
Proof. If r + x is rational, then x = (r + x) − r is also rational, a contradiction. Also, if rx is rational, then x = rx
is also rational since r 6= 0, a
r
contradiction. Exercise 1.2. Prove that there is no rational number whose square is 12.
Proof. Let f (x) = x2 − 12 ∈ Q[x]. Hence all possible rational roots are
1, −1, 2, −2, 3, −3, 4, −4, 6, −6, 12, −12.
But none of them satisfy f (x) = 0. Hence no rational number whose square
is 12. (If so, then my proof implies that f (x) = x2 −12 has a rational root.) Exercise 1.3. Prove Proposition 1.15: The axioms for multiplication imply
the following statements.
(a) If x 6= 0 and xy = xz then y = z.
(b) If x 6= 0 and xy = x then y = 1.
(c) If x 6= 0 and xy = 1 then y = 1/x.
(d) If x 6= 0 then 1/(1/x) = x.
1
Proof. We prove (a). By (M3), we have y
(M 5)
(M 4)
(1/x)xy = (1/x)xz = 1z = z.
(M 4)
(M 4)
=
1y
(M 5)
=
x(1/x)y
(M 2)
=
(M 2)
We prove (b). xy = x = 1x = x1. By part (a), y = 1.
(M 5)
We prove (c). xy = 1 = x(1/x). By part (a), y = 1/x.
(M 5)
We prove (d). x(1/x) = 1. By part (c), 1/(1/x) = x. Exercise 1.4. α is a lower bound of E ⇒ x ≥ α for all x ∈ E. Since E
is nonempty, we can pick an element x0 ∈ E. β is a upper bound of E ⇒
x0 ≤ β. Hence
α ≤ x0 ≤ β.
So α ≤ β. Exercise 1.5. If A is a set of real numbers which is bounded below, then
inf A is a lower bound, i.e.,
inf A ≤ x
for all x ∈ A and if inf A ≥ b for any other lower bound b of A. Now if A is
bounded below, then −A is bounded above. In fact, if b ≤ x for all x ∈ A,
then −b ≥ −x for all x ∈ A which means −b ≥ y for all y ∈ −A. Now if
sup(−A) is the least upper bound of −A it follows that − sup(−A) is a lower
bound of A. In fact,
x ∈ A ⇒ −x ∈ −A ⇒ sup(−A) ≥ −x ⇒ − sup(−A) ≤ x.
As noted above, if b is any lower bound for A then −b is an upper bound for
−A so −b ≥ sup(−A) and b ≤ − sup(−A). This is the definition of inf A so
inf A = − sup(−A).
Exercise 1.6. We prove (a). mq = np since m/n = p/q. Thus bmq = bnp .
By Theorem 1.21 we know that (bmq )1/(mn) = (bnp )1/(mn) , that is, (bm )1/n =
(bp )1/q , that is, br is well-defined.
2
We prove (b). Let r = m/n and s = p/q where m, n, p, q are integers, and
n > 0, q > 0. Hence (br+s )nq = (bm/n+p/q )nq = (b(mq+np)/(nq) )nq = bmq+np =
bmq bnp = (bm/n )nq (bp/q )nq = (bm/n bp/q )nq . By Theorem 1.21 we know that
((br+s )nq )1/(nq) = ((bm/n bp/q )nq )1/(nq) , that is br+s = bm/n bp/q = br bs .
We prove (c). Note that br ∈ B(r). For all bt ∈ B(r) where t is rational
and t ≤ r. Hence, br = bt br−t ≥ bt 1r−t since b > 1 and r − t ≥ 0. Hence br is
an upper bound of B(r). Hence br = sup B(r).
We prove (d). bx by = sup B(x) sup B(y) ≥ btx bty = btx +ty for all rational
tx ≤ x and ty ≤ y. Note that tx + ty ≤ x + y and tx + ty is rational.
Therefore, sup B(x) sup B(y) is a upper bound of B(x + y), that is, bx by ≥
sup B(x + y) = bx+y . Conversely, we claim that bx br = bx+r if x ∈ R and
r ∈ Q. The following is my proof.
bx+r = sup B(x + r) = sup{bs |s ≤ x + r, s ∈ Q}
= sup{bs−r br |s − r ≤ x, s − r ∈ Q}
= br sup{bs−r |s − r ≤ x, s − r ∈ Q}
= br sup B(x) = br bx .
And we also claim that bx+y ≥ bx if y ≥ 0. The following is my proof: (r ∈ Q)
B(x) = {br |r ≤ x} ⊂ {br |r ≤ x + y} = B(x + y),
Therefore, sup B(x + y) ≥ sup B(x), that is, bx+y ≥ bx . Hence,
bx+y = sup B(x + y)
= sup{br |r ≤ x + y, r ∈ Q}
= sup{bs br−s |r ≤ x + y, s ≤ x, r, s ∈ Q}
≥ sup{sup B(x)br−s |r ≤ x + y, s ≤ x, r, s ∈ Q}
= sup B(x) sup{br−s |r ≤ x + y, s ≤ x, r, s ∈ Q}
= sup B(x) sup{br−s |r − s ≤ x + y − s, s ≤ x, r − s ∈ Q}
= sup B(x) sup B(x + y − s)
≥ sup B(x) sup B(y) = bx by .
Therefore, bx+y = bx by . 3
Exercise 1.8. If we can define an other on the complex field, then one and
only one of the statements
0 < i,
0 = i,
0 > i.
is true. First, 0 = i is impossible. (0 = i ⇒ 02 = i2 , 0 = 1, a contradiction).
Second, suppose 0 < i. Then
0 < i, and 0 < i ⇒ 0 < i2 = −1,
a contradiction. Finally, suppose 0 > i. Hence 0 < −i. Then
0 < −i, and 0 < −i ⇒ 0 < (−i)2 = i2 = −1,
a contradiction again. Hence no order can be defined in the complex field
that turns it into an ordered field. Exercise 1.13. By Theorem 1.33, |x| + |x − y| ≥ |y| and |y| + |x − y| ≥ |x|.
Thus |x − y| ≥ ||x| − |y||. Exercise 1.14.
|1 + z|2 + |1 − z|2 = (1 + z)(1 + z) + (1 − z)(1 − z)
= (1 + z)(1 + z) + (1 − z)(1 − z)
= (1 + z + z + zz) + (1 − z − z + zz)
= 2 + 2zz = 2 + 2 = 4.
Exercise 1.15. In the proof of Theorem 1.35, equality holds if and only if
P
B = 0 or
|Baj − Cbj |2 = 0, that is,
b1 = · · · = bn = 0
or
aj = rbj
4
for all 1 ≤ j ≤ n, and for some r ∈ R. Exercise 1.17. |x + y|2 + |x − y|2 = (x + y) · (x + y) + (x − y) · (x − y) =
(x · x + 2x · y + y · y) + (x · x − 2x · y + y · y) = 2x · x + 2y · y = 2|x|2 + 2|y|2.
Exercise 1.18. Write x = (x1 , · · · , xk ) ∈ Rk . If the i-th coordinate xi of x
is zero for some 1 ≤ i ≤ k, we can pick
y = (0, · · · ,
1
|{z}
, · · · , 0) ∈ Rk .
i -th position
If all coordinates of x are nonzero, we can pick
y = (1, −x1 /x2 , 0, · · · 0).
It is possible since k ≥ 2. Thus we have x · y = 0. It is not true if k = 1
since 1 is not a zero-divisor in R1 . Exercise 2.1. For any element x of the empty set, x is also an element of every set since x does not exist. Hence, the empty set is a subset of every set. Exercise 2.2. For every positive integer N there are only finitely many
equations with
n + |a0 | + |a1 | + ... + |an | = N.
(since 1 ≤ n ≤ N and 0 ≤ |a0 | ≤ N). We collect those equations as CN .
S
Hence CN is countable. For each algebraic number, we can form an equa-
tion and this equation lies in CM for some M and thus the set of all algebraic
numbers is countable. Exercise 2.3. If not, R = { all algebraic numbers } is countable, a contradiction. Exercise 2.4. If R − Q is countable, then R = (R − Q)
contradiction. Thus R − Q is uncountable. 5
S
Q is countable, a
Exercise 2.5. Put
A = {1/n : n ∈ N}
[
[
{1 + 1/n : n ∈ N} {2 + 1/n : n ∈ N}.
A is bounded by 3, and A contains three limit points −0, 1, 2. Exercise 2.6. For any point p of X − E ′ , that is, p is not a limit point E,
there exists a neighborhood of p such that q is not in E with q 6= p for every
q in that neighborhood.
Hence, p is an interior point of X − E ′ , that is, X − E ′ is open, that is,
E ′ is closed.
Next, if p is a limit point of E, then p is also a limit point of E since
S
E = E E ′ . If p is a limit point of E, then every neighborhood Nr (p) of p
contains a point q 6= p such that q ∈ E. If q ∈ E, we completed the proof. So
we suppose that q ∈ E − E = E ′ − E. Then q is a limit point of E. Hence,
Nr′ (q)
where r ′ = 21 min(r − d(p, q), d(p, q)) is a neighborhood of q
and contains a point x 6= q such that x ∈ E. Note that Nr′ (q) contains
in Nr (p) − {p}. That is, x 6= p and x is in Nr (p). Hence, q also a limit point
of E. Hence, E and E have the same limit points.
Last, the answer of the final sub-problem is no. Put
E = {1/n : n ∈ N},
and E ′ = {0} and (E ′ )′ = ∅. Exercise 2.7. We proof (a). Method 1. Bn is the smallest closed subset of
S
X that contains Bn . Note that Ai is a closed subset of X that contains
Bn , thus
n
[
Bn ⊃
Ai .
i=1
If p ∈ Bn − Bn , then every neighborhood of p contained a point q 6= p
such that q ∈ Bn .
If p is not in Ai for all i, then there exists some
6
T
neighborhood Nri (p) of p such that (Nri (p) − p) Ai = φ for all i. Take
T
r = min{r1 , r2 , · · · , rn }, and we have Nr (p) Bn = φ, a contradiction. Thus
p ∈ Ai for some i. Hence
Bn ⊂
that is,
Bn =
n
[
Ai .
i=1
n
[
Ai .
i=1
S
S
Method 2. Since i=1 Ai is closed and Bn = ni=1 Ai ⊂ ni=1 Ai , Bn ⊂
Sn
i=1 Ai .
We prove (b). Since B is closed and B ⊃ B ⊃ Ai , B ⊃ Ai for all i. Hence
S
B ⊃ Ai .
Note: My example is Ai = (1/i, ∞) for all i. Thus, Ai = [1/i, ∞), and
Sn
B = (0, ∞), B = [0, ∞). Note that 0 is not in Ai for all i. Thus this inclusion
can be proper. Exercise 2.8. For the first part of this problem, the answer is yes.
Reason. For every point p of E, p is an interior point of E. That is, there
is a neighborhood Nr (p) of p such that Nr (p) is a subset of E. Then for every
real r ′ , we can choose a point q such that d(p, q) = 1/2 min(r, r ′ ). Note that
q 6= p, q ∈ Nr′ (p), and q ∈ Nr (p). Hence, every neighborhood Nr′ (p) contains
a point q 6= p such that q ∈ Nr (p) ⊂ E, that is, p is a limit points of E.
For the last part of this problem, the answer is no. Consider A = {(0, 0)}.
A′ = φ and thus (0, 0) is not a limit point of E. Exercise 2.9. We prove (a). If E is non-empty, take p ∈ E ◦ . We need to
show that p ∈ (E ◦ )◦ . Since p ∈ E ◦ , there is a neighborhood Nr of p such
that Nr is contained in E. For each q ∈ Nr , note that Ns (q) is contained in
Nr (p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interior point
of E, that is, Nr is contained in E ◦ . Hence E ◦ is always open.
We prove (b). (⇒) It is clear that E ◦ is contained in E. Since E is
open, every point of E is an interior point of E, that is, E is contained in
7
E ◦ . Therefore E ◦ = E. (⇐) Since every point of E is an interior point of E
(E ◦ (E) = E), E is open.
We prove (c). If p ∈ G, p is an interior point of G since G is open. Note
that E contains G, and thus p is also an interior point of E. Hence p ∈ E ◦ .
Therefore G is contained in E ◦ . (Thus E ◦ is the biggest open set contained
in E. Similarly, E is the smallest closed set containing E.)
We prove (d). Suppose p ∈ X − E ◦ . If p ∈ X − E, then p ∈ X − E
clearly. If p ∈ E, then N is not contained in E for any neighborhood N of p.
Thus N contains an point q ∈ X − E. Note that q 6= p, that is, p is a limit
point of X − E. Hence X − E ◦ is contained in X − E.
Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − E ◦ clearly.
If p ∈ E, then every neighborhood of p contains a point q 6= p such that
q ∈ X − E. Hence p is not an interior point of E. Hence X − E is contained
in X − E ◦ . Therefore X − E ◦ = X − E.
For (e), the answer is no. Take X = R and E = Q. Thus E ◦ = ∅ and
◦
E = (R)◦ = R 6= ∅.
For (f), the answer is no. Take X = R and E = Q. Thus E = R, and
E ◦ = ∅ = ∅. Exercise 2.10. (a) d(p, q) = 1 > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) = d(q, p)
since p = q implies q = p and p 6= q implies q 6= p. (c) d(p, q) ≤ d(p, r)+d(r, q)
for any r ∈ X if p = q. If p 6= q, then either r = p or r = q, that is, r 6= p or
r 6= q. Thus, d(p, q) = 1 ≤ d(p, r) + d(r, q). By (a)-(c) we know that d is a
metric.
Every subset of X is open and closed. We claim that for any p ∈ X,
p is not a limit point. Since d(p, q) = 1 > 1/2 if q 6= p, there exists an
neighborhood N1/2 (p) of p contains no points of q =
6 p such that q ∈ X.
Hence every subset of X contains no limit points and thus it is closed. Since
X − S is closed for every subset S of X, S = X − (X − S) is open. Hence
every subset of X is open.
Every finite subset of X is compact. Let S = {p1 , ..., pn } be finite. Consider an open cover {Gα } of S. Since S is covered by Gα , pi is covered by
Gαi , thus {Gα1 , ..., Gαn } is finite subcover of S. Hence S is compact. Next,
8
suppose S is infinite. Consider an open cover {Gp } of S, where
Gp = N 1 (p)
2
for every p ∈ S. Note that q is not in Gp if q 6= p. If S is compact, then S
can be covered by finite subcover, say
Gp1 , ..., Gpn .
Then there exists q such that q 6= pi for all i since S is infinite, a contradiction. Hence only every finite subset of X is compact. Exercise 2.11. (1) d1 (x, y) is not a metric. Since d1 (0, 2) = 4, d1 (0, 1) = 1,
and d1 (1, 2) = 1, d1 (0, 2) > d1 (0, 1) + d1 (1, 2). Thus d1 (x, y) is not a metric.
(2) d2 (x, y) is a metric. (a) d(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) =
p
p
p
|p − q| = |q − p| = d(q, p). (c) |p − q| ≤ |p − r| + |r − q|, |p − q| ≤
p
p
p
|p − r| + |r − q| ≤ |p − r| + |r − q|. That is, d(p, q) ≤ d(p, r) + d(r, q).
(3) d3 (x, y) is not a metric since d3 (1, −1) = 0.
(4) d4 (x, y) is not a metric since d4 (1, 1) = 1 6= 0.
(5) d5 (x, y) is a metric since |x − y| is a metric.
d(x,y)
Claim: d(x, y) is a metric, then d′ (x, y) = 1+d(x,y)
is also a metric.
(a) d′ (p, q) > 0 if p 6= q; d(p, p) = 0. (b) d′ (p, q) = d′ (q, p). (c) Let
x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y + z.
d′ (p, q) ≤ d′ (p, r) + d′ (r, q)
x
y
z
≤
+
⇔
1+x
1+y 1+z
⇔ x(1 + y)(1 + z) ≤ y(1 + z)(1 + x) + z(1 + x)(1 + y)
⇔ x + xy + xz + xyz ≤ (y + xy + yz + xyz) + (z + xz + yz + xyz)
⇔ x ≤ y + z + 2yz + xyz
⇐ x≤y+z
Thus, d′ is also a metric. 9
Exercise 2.12. Suppose that {Oα } is an arbitrary open covering of K. Let
E ∈ {Oα } consists 0. Since E is open and 0 ∈ E, 0 is an interior point of
E. Thus there is a neighborhood N = Nr (0) of 0 such that N ⊂ E. Thus N
contains
1
1
,
,···
[1/r] + 1 [1/r] + 2
Next, we take finitely many open sets En ∈ {Oα } such that 1/n ∈ En
for n = 1, 2, · · · , [1/r]. Hence {E, E1 , ..., E[1/r] } is a finite subcover of K.
Therefore, K is compact.
Note: The unique limit point of K is 0. Suppose p 6= 0 is a limit point of
K. Clearly, 0 < p < 1. Thus there exists n ∈ Z+ such that
1
1
<p< .
n+1
n
o
n
1
1
Hence Nr (p) where r = min n − p, p − n+1 contains no points of K, a
contradiction. Exercise 2.13. Let K be consist of 0 and the numbers 1/n for n = 1, 2, 3, · · ·
Let xK = {xk : k ∈ K} and x + K = {x + k|k ∈ K} for x ∈ R. We take
1
K
) + n+1 ,
n
2
2
∞
[
[
S =
Sn {1}.
Sn = (1 −
n=1
S
Claim: S is compact and the set of all limit points of S is hK {1}. Clearly,
i
1
S lies in [0, 1], that is, S is bounded in R. Note that Sn ⊂ 1 − 21n , 1 − 2n+1
.
T
By Exercise 12 and its note, we have that all limit points of S [0, 1) is
1
1
0, , · · · , n , · · ·
2
2
Clearly, 1 is also a limit point of S. Therefore, the set of all limit points of
S
S
S is K {1}. Note that K {1} ⊂ S, that is, K is compact. We completed
the proof of our claim. 10
Exercise 2.14. Take {On } = {(1/n, 1)} for n = 1, 2, 3, · · · . For every
x ∈ (0, 1),
1
x∈
, 0 ∈ {On }
[1/x] + 1
Hence {On } is an open covering of (0, 1). Suppose there exists a finite subcover
1
1
,1 ,··· ,
,1
n1
nk
where n1 < n2 < · · · < nk , respectively. Clearly
1
2np
∈ (0, 1) is not in any
elements of that subcover, a contradiction.
Note: By the above we know that (0, 1) is not compact. Exercise 2.15. For closed: [n, ∞). For bounded: (−1/n, 1/n) − {0}. √
√ √ S √
Exercise 2.16. Let S = ( 2, 3) (− 3, − 2). Then E = {p ∈ Q : p ∈
S}. Clearly, E is bounded in Q. Since Q is dense in R, every limit point of
Q is in Q. (We regard Q as a metric space). Hence, E is closed in Q.
To prove that E is not compact, we form a open covering of E as follows:
{Gα } = {Nr (p)|p ∈ E and (p − r, p + r) ⊂ S}.
Surely, {Gα } is a open covering of E. If E is compact, then there are finitely
many indices α1 , · · · , αn such that
E ⊂ Gα1
[
···
[
Gαn .
For every Gαi = Nri (pi ), take p = max1≤i≤n pi . Thus, p is the nearest point
√
√
to 3. But Nr (p) lies in E, thus [p + r, 3) cannot be covered since Q is
dense in R, a contradiction. Hence E is not compact.
Finally, the answer is yes. Take any p ∈ Q, then there exists a neighborhood N(p) of p contained in E. (Take r small enough where Nr (p) = N(p),
and Q is dense in R.) Thus every point in N(p) is also in Q. Hence E is also
open. 11
Exercise 2.17. Let
E=
∞
nX
o
an
|a
=
4
or
a
=
7
.
n
n n
10
n=1
Claim 1. E is uncountable. If not, we list E as follows:
x1 = 0.a11 a12 · · · a1n · · ·
x2 = 0.a21 a22 · · · a2n · · ·
···
···
xk = 0.ak1 ak2 · · · akn · · ·
···
···
(Prevent ending with all digits 9) Let x = 0.x1 x2 · · · xn · · · where
(
4 if ann = 7
xn =
7 if ann = 4
By my construction, x ∈
/ E, a contradiction. Thus E is uncountable.
T
Claim 2. E is not dense in [0, 1]. E (0.47, 0.74) = ∅.
Claim 3. E is compact. Clearly, E is bounded. For every limit point p
of E, we will show that p ∈ E. If not, write the decimal expansion of p as
follows
p = 0.p1 p2 · · · pn · · ·
Since p ∈
/ E, there exists the smallest k such that pk 6= 4 and pk 6= 7. When
pk = 0, 1, 2, 3, select the smallest l such that pl = 7 if possible. (If l does not
exist, then p < 0.4. Thus there is a neighborhood of p such that contains no
points of E, a contradiction.) Thus
0.p1 · · · pl−1 4pl+1 · · · pk−1 7 < p < 0.p1 · · · pk−1 4.
Thus there is a neighborhood of p such that contains no points of E, a
contradiction. When pk = 5, 6,
0.p1 · · · pk−147 < p < 0.p1 · · · pk−1 74.
12
Thus there is a neighborhood of p such that contains no points of E, a
contradiction. When pk = 8, 9, it is similar. Hence E is closed. Therefore E
is compact.
Claim. E is perfect. Take any p ∈ E, and I claim that p is a limit point
of E. Write p = 0.p1 p2 · · · pn · · · Let
xk = 0.y1y2 · · · yn · · ·
where


 pk if k 6= n
yn =
4 if pn = 7


7 if pn = 4
Thus, |xk − p| → 0 as k → ∞. Also, xk 6= p for all k. Hence p is a limit point
of E. Therefore E is perfect. Exercise 2.18. Yes. The following claim will show the reason.
Claim. Given a measure zero set S, we have a perfect set P contains no
elements in S.
Since S has measure zero, there exists a collection of open intervals {In }
such that
[
X
S⊂
In and
|In | < 1.
S
Consider E = R − In . E is nonempty since E has positive measure. Thus
E is uncountable and E is closed. Therefore there exists a nonempty perfect
T
set P contained in E by Exercise 28. P S = ∅. Thus P is our required
perfect set. Exercise 2.19. For (a), we recall the definition of separated: A and B
T
T
are separated if A B and A B are empty. Since A and B are closed sets,
T
T
T
A = A and B = B. Hence A B = A B = A B = ∅. Hence A and B
are separated.
T
T
For (b), Suppose A B is not empty. Thus we can pick p ∈ A B. For
p ∈ A, there exists a neighborhood Nr (p) of p contained in A since A is open.
S
T
For p ∈ B = B B ′ , if p ∈ B, then p ∈ A B. Note that A and B are
disjoint, and it’s a contradiction. If p ∈ B ′ , then p is a limit point of B. Thus
13
every neighborhood of p contains a point q 6= p such that q ∈ B. Take an
neighborhood Nr (p) of p containing a point q 6= p such that q ∈ B. Note that
Nr (p) ⊂ A, thus q ∈ A. With A and B are disjoint, we get a contradiction.
T
T
Hence A B is empty. Similarly, A B is also empty. Thus A and B are
separated.
T
For (c), suppose A B is not empty. Thus there exists x such that x ∈ A
S
and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = B B ′ , thus if x ∈ B,
then d(p, x) > δ, a contradiction. The only possible is x is a limit point of
B. Hence we take a neighborhood Nr (x) of x contains y with y ∈ B where
r=
δ−d(x,p)
.
2
Clearly, d(y, p) > δ. But,
d(y, p) ≤ d(y, x) + d(x, p) < r + d(x, p) =
δ − d(x, p)
+ d(x, p)
2
δ + d(x, p)
δ+δ
<
= δ,
2
2
T
T
and it is absurd. Hence A B is empty. Similarly, A B is also empty.
Thus A and B are separated.
=
Note: Take care of δ > 0. Think a while and you can prove the next
sub-exercise.
For (d), let X be a connected metric space. Take p ∈ X, q ∈ X with
p 6= q, thus d(p, q) > 0 is fixed. Let
A = {x ∈ X|d(x, p) < δ} and B = {x ∈ X|d(x, p) > δ}.
Take δ = δt = td(p, q) where t ∈ (0, 1). Thus 0 < δ < d(p, q). p ∈ A since
d(p, p) = 0 < δ, and q ∈ B since d(p, q) > δ. Thus A and B are non-empty.
S
By (c), A and B are separated. If X = A B, then X is not connected,
S
a contradiction. Thus there exists yt ∈ X such that y ∈
/ A B. Let
E = Et = {x ∈ X|d(x, p) = δt } ∋ yt .
For any real t ∈ (0, 1), Et is nonempty. Next, Et and Es are disjoint if
t 6= s (since a metric is well-defined). Thus X contains a uncountable set
{yt |t ∈ (0, 1)} since (0, 1) is uncountable. Therefore, X is uncountable. Note:
It is a good exercise. If that metric space contains only one point, then it
must be separated.
14
Supplement exercises given by Lee Sin-Yi. (a) Let A = {x|d(p, x) < r}
and B = {x|d(p, x) > r} for some p in a metric space X. Show that A, B
are separated.
(b) Show that a connected metric space with at least two points must be
uncountable. [Hint: Use (a)] For (a), by definition of separated sets, we want
T
T
to show A B = φ, and B A = φ. In order to do these, it is sufficient to
T
T
show A B = φ. Let x ∈ A B = φ, then we have:
(1) x ∈ A ⇒ d(x, p) ≤ r; (2) x ∈ B ⇒ d(x, p) > r
T
It is impossible. So, A B = φ.
For (b), suppose that C is countable, say C = {a, b, x3 , · · · }. We want to
show C is disconnected. So, if C is a connected metric space with at least
two points, it must be uncountable. Consider the set S = {d(a, xi )|xi ∈ C},
and thus let r ∈ R − S and inf S < r < sup S. And construct A and B
S
as in (a), we have C = A B, where A and B are separated. That is C
is disconnected. Another proof of (b). Let a ∈ C, b ∈ C, consider the continuous function f from C into R defined by f (x) = d(x, a). So, f (C) is
connected and f (a) = 0, f (b) > 0. That is, f (C) is an interval. Therefore,
C is uncountable. Exercise 2.20. Closures of connected sets is always connected, but interiors
of those is not. The counterexample is
[
[
S = N1 (2, 0) N1 (−2, 0) {x-axis} ⊂ R2 .
S
Since S is path-connected, S is connect. But S ◦ = N1 (2) N1 (−2) is disconnected clearly.
Claim. If S is a connected subset of a metric space, then S is connected.
If not, then S is a union of two nonempty separated set A and B. Thus
T
T
A B = A B = ∅. Note that
S = S − T = A ∪ B − T = (A ∪ B) ∩ T c = (A ∩ T c ) ∪ (B ∩ T c )
where T = S − S. Thus
(A ∩ T c ) ∩ B ∩ T c ⊂ (A ∩ T c ) ∩ B ∩ T c ⊂ A ∩ B = ∅.
15
Hence (A ∩ T c ) ∩ B ∩ T c = ∅. Similarly, A ∩ T c ∩ (B ∩ T c ) = ∅.
Now we claim that both A ∩ T c and B ∩ T c are nonempty. Suppose that
B ∩ T c = ∅. Thus
A ∩ T c = S ⇔ A ∩ (S − S)c = S
⇔ A ∩ (A ∪ B − S)c = S
⇔ A ∩ ((A ∪ B) ∩ S c )c = S
⇔ A ∩ ((Ac ∩ B c ) ∪ S) = S
⇔ (A ∩ S) ∪ (A ∩ Ac ∩ B c ) = S
⇔ A ∩ S = S.
Thus B is empty, and it is absurd. Thus B ∩ T c is nonempty. Similarly,
A ∩ T c nonempty. Therefore S is a union of two nonempty separated sets, a
contradiction. Hence S is connected. T
T
Exercise 2.21. For (a), we claim that A0 B0 is empty. (B0 A0 is simT
ilar). If not, take x ∈ A0 B0 . x ∈ A0 and x ∈ B0 . x ∈ B0 or x is a
T
T
limit point of B0 . x ∈ B0 will make x ∈ A0 B0 , that is, p(x) ∈ A B, a
contradiction since A and B are separated.
Claim: x is a limit point of B0 ⇒ p(x) is a limit point of B. Take
any neighborhood Nr of p(x), and p(t) lies in B for small enough t. More
precisely,
r
r
x−
<t<x+
.
|b − a|
|b − a|
Since x is a limit point of B0 , and (x−r/|b−a|, x+r/|b−a|) is a neighborhood
N of x, thus N contains a point y 6= x such that y ∈ B0 , that is, p(y) ∈ B.
T
Also, p(y) ∈ Nr . Therefore, p(x) is a limit point of B. Hence p(x) ∈ A B,
T
a contradiction since A and B are separated. Hence A0 B0 is empty, that
is, A0 and B0 are separated subsets of R.
We prove (b). Suppose not. For every t0 ∈ (0, 1), neither p(t0 ) ∈ A
S
nor p(t0 ) ∈ B (since A and B are separated.) Also, p(t0 ) ∈ A B for all
S
t0 ∈ (0, 1). Hence (0, 1) = A0 B0 , a contradiction since (0, 1) is connected.
For (c), let S be a convex subset of Rk . If S is not connected, then S
is a union of two nonempty separated sets A and B. By (b), there exists
16
t0 ∈ (0, 1) such that p(t0 ) ∈
/ A ∪ B. But S is convex, p(t0 ) must lie in A ∪ B,
a contradiction. Hence S is connected. Exercise 2.22. Consider S = the set of points which have only rational
coordinates. For any point x = (x1 , x2 , · · · , xk ) ∈ Rk , we can find a rational
sequence {rij } → xj for j = 1, · · · , k since Q is dense in R. Thus,
ri = (ri1 , ri2 , · · · , rik ) → x
and ri ∈ S for all i. Hence S is dense in Rk . Also, S is countable, that is, S
is a countable dense subset in Rk , Rk is separable. Exercise 2.23. Let X be a separable metric space, and S be a countable
dense subset of X. Let a collection {Vα } = { all neighborhoods with rational
radius and center in S }. We claim that {Vα } is a base for X.
For every x ∈ X and every open set G ⊂ X such that x ∈ G, there exists
a neighborhood Nr (p) of p such that Nr (p) ⊂ G since x is an interior point
of G. Since S is dense in X, there exists {sn } → x. Take a rational number
rn such that rn < 2r , and {Vα } ∋ Nrn (sn ) ⊂ Nr (p) for enough large n. Hence
we have x ∈ Vα ⊂ G for some α. Hence {Vα } is a base for X. Exercise 2.24. Fix δ > 0, and pick x1 ∈ X. Having chosen x1 , · · · , xj ∈ X,
choose xj+1 , if possible, so that d(xi , xj+1 ) ≥ δ for i = 1, · · · , j. If this process
cannot stop, then consider the set A = {x1 , x2 , · · · , xk }. If p is a limit point
of A, then a neighborhood Nδ/3 (p) of p contains a point q =
6 p such that
q ∈ A. q = xk for only one k ∈ N. If not, d(xi , xj ) ≤ d(xi , p) + d(xj , p) ≤
δ/3 + δ/3 < δ, and it contradicts the fact that d(xi , xj ) ≥ δ for i 6= j. Hence,
this process must stop after finite number of steps.
Suppose this process stop after k steps, and X is covered by Nδ (x1 ),
Nδ (x2 ), · · · , Nδ (xk ), that is, X can therefore be covered by finite many neigh-
borhoods of radius δ.
Take δ = 1/n(n = 1, 2, 3, · · · ), and consider the set A of the centers of
the corresponding neighborhoods.
Fix p ∈ X. Suppose that p is not in A, and every neighborhood Nr (p).
17
Note that Nr/2 (p) can be covered by finite many neighborhoods Ns (x1 ), · · · , Ns (xk )
of radius s = 1/n where n = [2/r] + 1 and xi ∈ A for i = 1, · · · , k. Hence,
d(x1 , p) ≤ d(x1 , q) + d(q, p) ≤ r/2 + s < r where q ∈ Nr/2 (p) ∩ Ns (x1 ). Therefore, x1 ∈ Nr (p) and x1 6= p since p is not in A. Hence, p is a limit point of A if
p is not in A, that is, A is a countable dense subset, that is, X is separable. Exercise 2.25. For every positive integer n, there are finitely many neighborhood of radius 1/n whose union covers K (since K is compact). Collect
all of them, say {Vα }, and it forms a countable collection. We claim that
{Vα } is a base.
For every x ∈ X and every open set G ⊂ X, there exists Nr (x) such that
Nr (x) ⊂ G since x is an interior point of G. Hence x ∈ Nm (p) ∈ {Vα } for
some p where m = [2/r] + 1. For every y ∈ Nm (p), we have
d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.
Hence Nm (p) ⊂ G, that is, Vα ⊂ G for some α, and therefore {Vα } is a
countable base of K. Next, collect all of the center of Vα , say D, and we
claim that D is dense in K. (D is countable since Vα is countable.) For all
p ∈ K and any ǫ > 0 we can find Nn (xn ) ∈ {Vα } where n = [1/ǫ] + 1. Note
that xn ∈ D for all n and d(p, xn ) → 0 as n → ∞. Hence D is dense in K.
Exercise 2.26. By Exercises 2.23 and 2.24, X has a countable base. It
follows that every open cover of X has a countable subcover {Gn }, n =
1, 2, 3, · · · . If no finite subcollection of {Gn } covers X, then the complement
S S
T
Fn of G1 · · · Gn is nonempty for each n, but Fn is empty. If E is a set
contains a point from each Fn , consider a limit point of E.
Note that Fk ⊃ Fk+1 ⊃ · · · and Fn is closed for all n, thus p lies in Fk for
T
T
all k. Hence p lies in Fn , but Fn is empty, a contradiction. Exercise 2.27. Let {Vn } be a countable base of Rk , let W be the union
T
of those Vn for which E Vn is at most countable, and we will show that
P = W c . Suppose x ∈ P . (x is a condensation point of E). If x ∈ Vn for
18
T
some n, then E Vn is uncountable since Vn is open. Thus x ∈ W c . (If
T
x ∈ W , then there exists Vn such that x ∈ Vn and E Vn is uncountable, a
contradiction.) Therefore P ⊂ W c .
T
Conversely, suppose x ∈ W c . x ∈
/ Vn for any n such that E Vn is countT
able. Take any neighborhood N(x) of x. Take x ∈ Vn ⊂ N(x), and E Vn is
T
uncountable. Thus E N(x) is also uncountable, x is a condensation point
of E. Thus W c ⊂ P . Therefore P = W c . Note that W is countable, and
T
T
thus W ⊂ W E = P c E is at most countable.
To show that P is perfect, it is enough to show that P contains no isolated
point. (since P is closed). If p is an isolated point of P , then there exists a
T
neighborhood N of p such that N E = φ. p is not a condensation point of
E, a contradiction. Therefore P is perfect. Exercise 2.28. Let X be a separable metric space, let E be a closed set on
X. Suppose E is uncountable. (If E is countable, there is nothing to prove.)
Let P be the set of all condensation points of E. Since X has a countable
T
base, P is perfect, and P c E is at most countable by Exercise 27. Since E
T
S
is closed, P ⊂ E. Also, P c E = E − P . Hence E = P (E − P ).
For corollary: if there is no isolated point in E, then E is perfect. Thus
E is uncountable, a contradiction.
Note: It’s also called Cauchy-Bendixon Theorem. Exercise 2.29. Since O is open, for each x in O, there is a y > x such that
(x, y) ⊂ O. Let b = sup{y | (x, y) ⊂ O}. Let a = inf{z | (z, x) ⊂ O}. Then
a < x < b, and Ix = (a, b) is an open interval containing x.
Now Ix ⊂ O, for if w ∈ Ix , say x < w < b, we have by the definition of b
a number y > w such that (x, y) ⊂ O, and so w ∈ O).
Moreover, b ∈
/ O, for if b ∈ O, then for some ǫ > 0 we have (b−ǫ, b+ǫ) ⊂ O,
whence (x, b + ǫ) ⊂ O, contradicting the definition of b. Similarly, a ∈
/ O.
Consider the collection of open intervals {Ix }, x ∈ O. Since each x ∈ O
S
is contained in Ix , and each Ix ⊂ O, we have O = Ix .
Let (a, b) and (c, d) be two intervals in this collection with a point in
common. Then we must have c < b and a < d. Since c ∈
/ O, it does not
19
belong to (a, b) and we have c ≤ a. Thus a = c. Similarly, b = d, and
(a, b) = (c, d). Thus two different intervals in the collection {Ix } must be
disjoint. Thus O is the union of the disjoint collection {Ix } of open intervals, and it remains only to show that this collection is countable. But each
open interval contains a rational number since Q is dense in R. Since we
have a collection of disjoint open intervals, each open interval contains a
different rational number, and the collection can be put in one-to-one correspondence with a subset of the rationals. Thus it is a countable collection. Exercise 2.30. I prove Baire’s theorem directly. Let Gn be a dense open
T
subset of Rk for n = 1, 2, 3, · · · . I need to prove that ∞
1 Gn intersects any
nonempty open subset of Rk is not empty.
Let G0 is a nonempty open subset of Rk . Since G1 is dense and G0 is
T
T
nonempty, G0 G1 6= ∅. Suppose x1 ∈ G0 G1 . Since G0 and G1 are
T
open, G0 G1 is also open, that is, there exist a neighborhood V1 such that
T
V1 ⊂ G0 G1 . Next, since G2 is a dense open set and V1 is a nonempty
T
open set, V1 G2 6= ∅. Thus, I can find a nonempty open set V2 such that
T
V2 ⊂ V1 G2 . Suppose I have get n nonempty open sets V1 , V2 , · · · , Vn
T
T
such that V1 ⊂ G0 G1 and Vi+1 ⊂ Vi Gn+1 for all i = 1, 2, · · · , n − 1.
T
Since Gn+1 is a dense open set and Vn is a nonempty open set, Vn Gn+1
is a nonempty open set. Thus we can find a nonempty open set Vn+1 such
T
that Vn+1 ⊂ Vn Gn+1 . By induction, I can form a sequence of open sets
T
T
{Vn |n ∈ Z+ } such that V1 ⊂ G0 G1 and Vi+1 ⊂ Vi Gi+1 for all n ∈ Z+ .
Since V1 is bounded and V1 ⊃ V2 ⊃ · · · ⊃ Vn ⊃ · · · , by Theorem 2.39 I know
that
∞
\
Vn 6= ∅.
n=1
T
TT
Since V1 ⊂ G0 G1 and Vn+1 ⊂ Gn+1 , G0 ( ∞
n=1 Gn ) 6= ∅.
Note: By Baire’s theorem, I’ve proved the equivalent statement. Next,
Fn has a empty interior if and only if Gn = Rk − Fn is dense in Rk . Hence
we completed all proof. Exercise 3.1. Since {sn } is convergent, for any ǫ > 0, there exists N
20
such that |sn − s| < ǫ whenever n ≥ N. By Exercise 1.13 I know that
||sn | − |s|| ≤ |sn − s|. Thus, ||sn | − |s|| < ǫ, that is, {sn } is convergent.
The converse is not true for sn = (−1)n . Exercise 3.2.
√
n2 + n − n = √
as n → ∞. n2
n
1
1
=p
→
2
+n+n
1/n + 1 + 1
Exercise 3.3. First,
that {sn } is strictly increasing. It is trivial that
q I show
p
√
√
p
√
2 > 2 = s1 . Suppose sk > sk−1 when k < n.
s2 = 2 + s1 = 2 +
By the induction hypothesis,
q
q
√
√
sn = 2 + sn−1 > 2 + sn−2 = sn−1
By the induction, {sn } is strictly increasing. Next, we will show that {sn }
is bounded by 2. Similarly, we apply the induction again. Hence {sn } is
strictly increasing and bounded, that is, {sn } converges. Exercise 3.4. By direct computing s1 = 0, s2 = 0, s3 =
3
, s6
4
=
3
, s7
8
=
7
,···
8
So lim inf sn =
1
, lim sup sn
2
1
, s4
2
=
1
, s5
4
=
= 1. Exercise 3.5. Let a = lim sup an , b = lim sup bn , and c = lim sup(an + bn ).
Suppose {an } and {bn } is bounded below. If a = +∞ or b = +∞, then the
conclusion always holds. Hence we assume that a and b are finite. Hence give
ǫ > 0, there is an integer N such that an + bn < a + b + ǫ whenever n ≥ N.
It follows that c is finite. Since c = lim sup(an + bn ), for the same ǫ there
is an integer m ≥ N such that c − ǫ < am + bm . Put n = m and combine
them, we have c − ǫ < a + b + ǫ, or c < a + b + 2ǫ. Since ǫ is arbitrary small,
c ≤ a + b, that is,
lim sup(an + bn ) ≤ lim sup an + lim sup bn .
n→∞
n→∞
n→∞
If a = −∞ and b = −∞. Then given any real M, an > M and bm > M
for at most a finite number of values of n and m, so (an + bn ) > 2M for at
21
most a finite number of values of n. Hence c = −∞. So we also have
lim sup(an + bn ) ≤ lim sup an + lim sup bn .
n→∞
n→∞
n→∞
Exercise 3.6. For (a). The partial sum sn =
sn =
n
X
k=1
P
Pn
k=1 ak
is
n
X
√
√
√
ak =
( k + 1 − k) = n + 1 − 1.
k=1
Hence
an = lim sn = ∞ is divergent.
For (b).
√
√
n+1− n
1
an =
= √
√ .
n
n( n + 1 + n)
So
1
.
2n3/2
P
By Theorem 3.25 and 3.28,
an converges.
an ≤
For (c).
lim sup
n→∞
p
n
q √
√
|an | = lim sup n ( n n − 1)n = lim ( n n − 1) = 0
n→∞
n→∞
by Theorem 3.20(d). The root test indicates that
P
an converges.
For (d). an → 0 implies that |z| > 1 (an → 0, ⇒ 1 + z n → ∞, ⇒
P
z n → ∞). Hence
an diverges when |z| ≤ 1. When |z| > 1, there is N such
that |z|n − 1 > 21 |z|n whenever n ≥ N. Hence
1
1
2
≤ n
< n
n
|1 + z |
|z| − 1
|z|
P
P
whenever n ≥ N. By Theorem 3.25 and 3.26,
|an | converges. So
an
converges absolutely if |z| > 1. |an | =
Exercise 3.7. By Cauchy’s inequality,
√
k
k
X
X
an
1
an
a
≥
n
n2
n
n=1
n=1
n=1
k
X
22
√
Pk
P
P 1
an
are
convergent;
thus
a
for all n ∈ N. Also, both
an and
n=1 n n
n2
√
√
P
a
an
is bounded. Besides, n n ≥ 0 for all n. Hence
is convergent. n
Exercise 3.8. We prove a more general exercise.
Dirichlet’s Convergence Test: Let {an } and {bn } be sequences of real
P
numbers such that { nk=0 ak } is bounded and {bn } decreases with 0 as its
P
limit. Then ∞
n=0 an bn converges.
P
Let An = nk=0 an and let M be an upper bound for {bn }. By partial
summation formula, we have
q
q−1
X
X
an bn ≤ |
An (bn − bn+1 )| + |Aq bq | + |Ap−1 bp |
n=p
n=p
≤ M
q−1
X
n=p
(bn − bn+1 ) + |Aq bq | + |Ap−1 bp |)
≤ M(bp − bq ) + Mbq + Mbp
= 2Mbp → 0
as p, q → ∞. Hence
P∞
n=0
an bn is a Cauchy sequence, that is,
P∞
n=0
an bn
converges.
Return to this exercise, suppose {bn } is decreasing and bn → b, then the
P
new sequence {bn − b} decreases with 0 as its limit. Also,
an converges
Pn
implies that { k=0 ak } is bounded. By Dirichlet’s Convergence Test,
∞
X
n=0
an (bn − b)
P
P
converges. Also,
an converges implies that
ban converges. Add them,
P
and we have
an bn converges. Finally, suppose {bn } is increasing and
bn → b. Thus the neq sequence {b − bn } decreases with 0 as its limit. Using
the same argument, we also get the same conclusion. Exercise 3.9. (a) limn→∞ αn = limn→∞ (n3 )1/n = 1. R = 1/α = 1.
(b) limn→∞ αn = limn→∞ (2n /n!)1/n = limn→∞ 2/(n!)1/n = 0. R = +∞.
(c) limn→∞ αn = limn→∞ (2n /n2 )1/n = 2/1 = 2. R = 1/α = 1/2.
23
(d) limn→∞ αn = limn→∞ (n3 /3n )1/n = 1/3. R = 1/α = 3. p
Exercise 3.10. Recall α = lim supn→∞ n |an | and R = α1 . Since an are
p
integers, infinitely many of which are distinct from zero, n |an | ≥ 1 for all
n. Hence
α = lim sup
n→∞
p
n
|an | ≥ 1,
1
≤ 1,
α
that is, the radius of convergence is at most 1. R=
an
→ 0 iff 1 1+1 → 0 iff a1n → ∞ iff
Exercise 3.11. For (a). Note that 1+a
n
an
P an
an → 0 as n → ∞. If
converges, then an → 0 as n → ∞. Thus for
1+an
P an
some ǫ′ = 1 there is an N1 such that an < 1 whenever n ≥ N1 . Since
1+an
am
an
converges, for any ǫ > 0 there is an N2 such that 1+a
+
·
·
·
+
<
ǫ
for
1+an
m
all n > m ≥ N2 . Take N = max(N1 , N2 ). Thus ǫ >
am
an
n
+ · · · + 1+1
= am +···+a
for all n > m ≥ N. Thus
1+1
2
am
1+am
+ ··· +
an
1+an
>
am + · · · + an < 2ǫ
aN+k
aN+1
+· · ·+ sN+k
sN+1
P an
converges,
sn
aN+1
sN+k
P
an
1+an
aN+k
+· · ·+ sN+k
for all n > m ≥ N. It is absurd. Hence
diverges.
a
+···+a
s
−s
N
For (b).
≥
= N+1 sN+k N+k = N+k
=
sN+k
sN
m
. If
for any ǫ > 0 there exists N such that asm
+ · · ·+
1 − sN+k
an
< ǫ for all m, n whenever n > m ≥ N. Fix m = N and let n = N + k.
sn
Thus
am
an
aN
aN +k
sN
+···+
=
+···+
≥1−
sm
sn
sN
sN +k
sN +k
P
for all k ∈ N. But sN +k → ∞ as k → ∞ since
an diverges and an > 0.
P an
Take ǫ = 1/2 and we obtain a contradiction. Hence
diverges.
sn
n−1
n
For (c). sn−1 ≤ sn iff s12 ≤ sn s1n−1 iff as2n ≤ snasn−1
= ssnn−s
iff as2n ≤
sn−1
ǫ>
1
sn−1
−
1
sn
n
k
X
an
n=2
Note that
n
n
for all n. Hence
1
sn
s2n
≤
k
X
n=2
(
1
sn−1
→ 0 as n → ∞ since
P
−
1
1
1
)=
− .
sn
s1 sn
an diverges. Hence
24
P an
s2n
converges.
P an
P an
For (d).
may converge or diverge, and
converges. To
2
1+nan
P an 1+n anP
an
1
see this, we put an = 1/n. 1+nan = 2n , that is,
= 2 1/n diverges.
1+nan
Besides, if we put
an =
1
n(log n)p
where p > 1 and n ≥ 2, then
1
1
an
=
<
2p
p
1 + nan
n(log n) ((log n) + 1)
2n(log n)3p
P an
for large enough n. By Theorem 3.25 and Theorem 3.29,
converges.
1+nan
Next,
X
X 1
X
1
an
=
<
.
1 + n2 an
1/an + n2
n2
P 1
P an
for all an . Note that
converges,
and
thus
converges. 2
n
1+n2 an
Exercise 3.12. For (a).
am
an
am + · · · + an
rm − rn
rn
+···+
>
=
= 1−
rm
rn
rm
rm
rm
P an
if m < n. If
converges, for any ǫ > 0 there exists N such that
rn
am
an
+···+
<ǫ
rm
rn
for all m, n whenever n > m ≥ N. Fix m = N. Thus
am
an
rn
rn
+···+
>1−
=1−
rm
rn
rm
rN
m
for all n > N. But rn → 0 as n → ∞; thus arm
+ · · · + arnn → 1 as n → ∞. If
we take ǫ = 1/2, we will get a contradiction.
√
√
√
√
√
For (b). Note that rn+1 < rn iff rn+1 < rn iff rn + rn+1 < 2 rn iff
√
√
√
√
√
√
√
√
rn + rn+1
r + r
−rn+1
√
< 2 iff ( rn − rn+1 ) n √rn n+1 < 2( rn − rn+1 ) iff rn√
<
rn
rn
√
√
√
√
2( rn − rn+1 ) iff √arnn < 2( rn − rn+1 ) since an > 0 for all n. Hence,
k
k
X
X
√
√
an
√
√
<
2( rn − rn+1 ) = 2( r1 − rk+1 ).
√
rn
n=1
n=1
P an
P an
√
√
Note that rn → 0 as n → ∞. Thus
is bounded. Hence
rn
rn
converges. 25
Note: We say that
P
an converges faster than
lim
n→∞
an
= 0.
bn
P
bn if
According the above exercise, we can construct a faster convergent series
from a known convergent one easily. It implies that there is no perfect tests
to test all convergences of the series from a known convergent one.
Exercise 3.13. Put An =
Pn
k=0
|ak |, Bn =
Pn
k=0
|bk |, Cn =
Pn
k=0
|ck |. Then
Cn = |a0 b0 | + |a0 b1 + a1 b0 | + · · · + |a0 bn + a1 bn−1 + · · · + an b0 |
≤ |a0 ||b0 | + (|a0 ||b1 | + |a1 ||b0 |) + · · ·
+(|a0 ||bn | + |a1 ||bn−1 | + · · · + |an ||b0 |)
= |a0 |Bn + |a1 |Bn−1 + · · · + |an |B0
≤ |a0 |Bn + |a1 |Bn + · · · + |an |Bn
= (|a0 | + |a1 | + · · · + |an |)Bn = An Bn ≤ AB
where A = lim An and B = lim Bn . Hence {Cn } is bounded. Note that
{Cn } is increasing, and thus Cn is a convergent sequence, that is, the Cauchy
product of two absolutely convergent series converges absolutely. Exercise 3.14. For (a). The proof is straightforward. Let tn = sn − s,
τn = σn − s. (Or you may suppose that s = 0.) Then
τn =
t0 + t1 + · · · + tn
.
n+1
Choose M > 0 such that |tn | ≤ M for all n. Given ǫ > 0, choose N so that
n > N implies |tn | < ǫ. Taking n > N in τn = (t0 + t1 + · · · + tn )/(n + 1),
and then
|τn | ≤
|t0 | + · · · + |tN | |tN +1 | + · · · + |tn |
(N + 1)M
+
<
+ ǫ.
n+1
n+1
n+1
Hence, lim supn→∞ |τn | ≤ ǫ. Since ǫ is arbitrary, it follows that limn→∞ |τn | =
0, that is, lim σn = s.
26
Note: If you know O.Stolz Theorem, the proof will be very simple. A
similar exercise is stated here: If lim an = a, lim bn = b, then
a1 bn + a2 bn−1 + · · · + an b1
= ab.
n→∞
n
lim
Another proof of (a). Given ǫ > 0, there exists N1 such that |sn − s| < ǫ
whenever n > N1 . Thus
|s0 − s| + |s1 − s| + · · · + |sn − s|
n+1
|s0 − s| + · · · + |sN1 − s| n + 1 − N1
≤
+
ǫ.
n+1
n+1
σn − s ≤
Note that |s0 − s| + · · · + |sN1 − s| is fixed. Take N2 > 0 such that
|s0 − s| + · · · + |sN1 − s|
<ǫ
n+1
whenever n > N2 . Let N = max{N1 , N2 }. Hence,
|σn − s| < ǫ +
n + 1 − N1
ǫ < 2ǫ.
n+1
For (b). Let sn = (−1)n . Hence |σn | ≤ 1/(n + 1), that is, lim σn = 0.
However, lim sn does not exists.
For (c). Let


, n = 0,
 1
sn =
n1/4 + n−1 , n = k 2 for some integer k,

 −1
n
, otherwise.
It is obvious that sn > 0 and lim sup sn = ∞. Also,
s0 + · · · + sn = 1 + nn−1 +
That is,
√ 1/4
√ 1/4
n n =2+
n n .
√
2
⌊ n⌋ n1/4
σn =
+
.
n+1
n+1
The first term 2/(n + 1) → 0 as n → ∞. Note that
√
⌊ n⌋ n1/4
0≤
< n1/2 n1/4 n−1 = n−1/4 .
n+1
27
It follows that the last term → 0. Hence, lim σn = 0.
For (d).
n
X
kak =
k=1
=
n
X
k=1
n
X
k=1
k(sk − sk−1 ) =
ksk −
= nsn +
= nsn −
n−1
X
k=1
n−1
X
k=1
n−1
X
n
X
k=1
ksk −
n
X
ksk−1
k=1
(k + 1)sk
k=0
ksk −
n−1
X
k=1
(k + 1)sk − s0
sk − s0 = (n + 1)sn −
= (n + 1)(sn − σn ).
That is,
n
X
sk
k=0
n
1 X
sn − σn =
kak .
n+1
k=1
Note that {nan } is a complex sequence. By (a),
n
1 X
lim
kak = lim nan = 0.
n→∞ n + 1
n→∞
k=1
Also, lim σn = σ. Hence by the previous equation, lim s = σ.
For (e). If m < n, then
n
X
i=m+1
(sn − si ) + (m + 1)(σn − σm )
= (n − m)sn −
= (n − m)sn −
n
X
i=m+1
X
n
i=0
si + (m + 1)(σn − σm )
m
X
si −
si + (m + 1)(σn − σm )
i=0
= (n − m)sn − (n + 1)σn + (m + 1)σm + (m + 1)(σn − σm )
= (n − m)sn − (n − m)σn .
Hence,
n
X
m+1
1
sn − σn =
(σn − σm ) +
(sn − si ).
n−m
n − m i=m+1
28
For these i, recall an = sn − sn−1 and |nan | ≤ M for all n,
n
n
n
X
X
X M
(n − i)M
|sn − si | = ak ≤
|ak | ≤
=
i+1
i+1
k=i+1
k=i+1
k=i+1
≤
(n − (m + 1))M
(n − m − 1)M
=
.
(m + 1) + 1
m+2
Fix ǫ > 0 and associate with each n the integer m that satisfies
n−ǫ
m≤
< m + 1.
1+ǫ
Thus
n−m
n−m−1
≥ ǫ and
< ǫ,
m+1
m+2
or
m+1
1
≤
and |sn − si | < Mǫ.
n−m
ǫ
Hence,
1
|sn − σ| ≤ |σn − σ| + (|σn − σ| + |σm − σ|) + Mǫ.
ǫ
Let n → ∞ and thus m → ∞ too, and thus
lim sup |sn − σ| ≤ Mǫ.
n→∞
Since ǫ was arbitrary, lim sn = σ. there is an integer N such that
P
if m ≥ n ≥ N.
X
m
≤ǫ
a
n
Exercise 3.15. Theorem 3.22.
an converges if and only if for every ǫ > 0
k=n
Theorem 3.23. If
P
an converges, then limn→∞ an = 0.
Theorem 3.25(a). If |an | ≤ cn for n ≥ N0 , where N0 is some fixed integer,
P
P
and if
cn converges, then
an converges.
Theorem 3.33 (Root Test). Given
Then
29
P
an , put α = lim supn→∞
p
n
|an |.
P
(a) if α < 1,
an converges;
P
(b) if α > 1,
an diverges;
(c) if α = 1, the test gives no information.
P
Theorem 3.34 (Ratio Test). The series
an
< 1,
(a) converges if lim supn→∞ aan+1
n
an+1 (b) diverges if an ≥ 1 for n ≥ n0 , where n0 is some fixed integer.
Theorem 3.42. Suppose
(a) the partial sums An of
(b) b0 ≥ b1 ≥ b2 ≥ · · · ;
(c) limn→∞ = 0.
Then
P
P
an form a bounded sequence;
bn an converges.
Theorem 3.45. If
P
an converges absolutely, then
P
an converges.
P
P
P
Theorem 3.47. If
an = A, and
bn = B, then (an + bn ) = A + B,
P
and
can = cA for any fixed c ∈ R.
P
Theorem 3.55. If
an is a series in Ck which converges absolutely, then
P
every arrangement of an converges, and they all converge to the same sum.
Proof of Theorem 3.22. Theorem 3.11(c) implies it.
Proof of Theorem 3.23. Put m = n + 1 in Theorem 3.22, we get the
conclusion.
Proof of Theorem 3.25(a). Given ǫ > 0, there exists N ≥ N0 such that
m ≥ n ≥ N implies
m
X
k=n
ck ≤ ǫ,
30
by the Cauchy criterion. Hence
m
m
m
X
X X
ak ≤
|ak | ≤
ck ≤ ǫ,
k=n
k=n
k=n
and (a) follows.
Exercise 3.16. For (a).
1
α
1
α
1 x2n − α
xn − xn+1 = xn −
xn +
=
xn −
=
>0
2
xn
2
xn
2
xn
since xn > α. Hence {xn } decreases monotonically. Also, {xn } is bounded
by 0; thus {xn } converges. Let lim xn = x. Hence lim xn+1 = lim 12 (xn + xαn )
√
iff x = 21 (x + αx ) iff x2 = α. Note that xn > 0 for all n. Thus x = α.
√
lim xn = α.
For (b).
1
α
xn+1 =
xn +
2
x
n
√
√
1
α
xn +
− α
⇒ xn+1 − α =
2
xn
√
√
1 x2n − 2xn α + α
⇒ xn+1 − α =
2
x
√ n2
√
(xn − α)
⇒ xn+1 − α =
2xn
ǫ2n
ǫ2n
⇒ ǫn+1 =
< √ .
2xn
2 α
√
n
Hence ǫn+1 < β( ǫβ1 )2 where β = 2 α by induction.
For (c).
√
ǫ1
2− 3
1
1
1
√ =
√ < .
= √ = √
β
10
2 3
2 3(2 + 3)
6+4 3
Thus
√
ǫ1 4
ǫ5 < β( )2 < 2 3 · 10−16 < 4 · 10−16 ,
β
√
ǫ1 25
ǫ6 < β( ) < 2 3 · 10−32 < 4 · 10−32 .
β
31
Note. It is an application of Newton’s method. Let f (x) = x2 − α in
Exercise 5.25.
Exercise 3.20. Since {pn } is a Cauchy sequence, ∀ǫ > 0 there exists N1
such that |pn − pm | < ǫ whenever m, n ≥ N1 . Also, since {pnj } converges to
p ∈ X, there exists N2 such that |pnj − p| < ǫ whenever j ≥ N2 . Thus
|pn − p| ≤ |pn − pnj | + |pnj − p|
for each n ≥ N1 . Choose j such that j ≥ N2 and nj ≥ N1 . Thus |pn −pnj | < ǫ
and |pnj − p| < ǫ. Hence |pn − p| < 2ǫ whenever n ≥ N1 . Therefore the full
sequence {pn } converges to p. Exercise 3.21. En is bounded ∀n ⇒ En is nonempty ∀n. So we pick pn ∈ En
for each n. With the relation En ⊃ En+1 and Page 53, we know that {pn }
is a Cauchy sequence in X. Since X is complete, pn converges to a point
p ∈ X. Since En is closed and En ⊃ En+1 , the limit point p lies in each En .
So
p∈
If there is q ∈
T∞
n=1 En
∞
\
En .
n=1
T
such that p 6= q, then diam( ∞
n=1 En ) ≥ d(p, q) > 0,
where d is the metric of X. But
∞
\
n=1
for all k. Hence
En ⊂ Ek
\
∞
diam
En ≤ diamEk
n=1
for all k. Take limit and limn→∞ diamEn ≥ d(p, q), where d(p, q) is a fixed
T
positive number, contradicts with limn→∞ diamEn = 0. Hence ∞
n=1 En consists of exactly one point. Exercise 3.22. See 2 Exercise 2.30. 32
Exercise 3.23. For any ǫ > 0, there exists N such that d(pn , pm ) < ǫ and
d(qm , qn ) < ǫ whenever m, n ≥ N. Note that
d(pn , qn ) ≤ d(pn , pm ) + d(pm , qm ) + d(qm , qn ).
It follows that
|d(pn , qn ) − d(pm , qm )| ≤ d(pn , pm ) + d(qm , qn ) < 2ǫ.
Thus {d(pn , qn )} is a Cauchy sequence in X. Hence {d(pn , qn )} converges. Exercise 3.24. For (a). Suppose there are three Cauchy sequences {pn },
{qn }, and {rn }. First, d(pn , pn ) = 0 for all n. Hence, d(pn , pn ) = 0 as n → ∞.
Thus it is reflexive. Next, d(qn , pn ) = d(pn , qn ) → 0 as n → ∞. Thus it is
symmetric. Finally, if d(pn , qn ) → 0 as n → ∞ and if d(qn , rn ) → 0 as
n → ∞, d(pn , rn ) ≤ d(pn , qn ) + d(qn , rn ) → 0 + 0 = 0 as n → ∞. Thus it is
transitive. Hence this is an equivalence relation.
For (b). To show △(P, Q) is well-defined. Suppose {pn } and {p′n } are
equivalent; {qn } and {qn′ } are equivalent. Let △′ (P, Q) = lim d(p′n , qn′ ). Since
d(pn , qn ) ≤ d(pn , p′n ) + d(p′n , qn′ ) + d(qn′ , qn ),
so we take the limit:
lim d(pn , qn ) ≤ lim d(pn , p′n ) + lim d(p′n , qn′ ) + lim d(qn′ , qn ).
n→∞
n→∞
n→∞
n→∞
Note that the equivalence of sequences, and we have
△(P, Q) ≤ △′(P, Q).
Similarly,
d(p′n , qn′ ) ≤ d(p′n , pn ) + d(pn , qn ) + d(qn , qn′ ),
we have
△′ (P, Q) ≤ △(P, Q).
Hence
△(P, Q) = △′ (P, Q).
33
So △(P, Q) is well-defined (unchanged). Hence △ is a distance function in
X ∗.
For (c). Let {Pn } be a Cauchy sequence in (X ∗ , △). We wish to show that
there is a point P ∈ X ∗ such that △(Pn , P ) → 0 as n → ∞. For each Pn ,
there is a Cauchy sequence in X, denoted {Qkn }, such that △(Pn , Qkn ) → 0
as k → ∞. Let ǫn > 0 be a sequence tending to 0 as n → ∞. From the double
sequence {Qkn } we can extract a subsequence Q′n such that △(Pn , Q′n ) < ǫn
for all n. From the triangle inequality, it follows that
△(Q′n , Q′m ) ≤ △(Q′n , Pn ) + △(Pn , Pm ) + △(Pm, Q′m ).
(*)
Since {Pn } is a Cauchy sequence, given ǫ > 0, there is an N > 0 such that
△(Pn , Pm ) < ǫ for m, n > N. We choose m and n so large that ǫm < ǫ,
ǫn < ǫ. Thus the inequality (*) shows that {Q′n } is a Cauchy sequence in X.
Let P be the corresponding equivalence class in S. Since
△(P, Pn ) ≤ △(P, Q′n ) + △(Q′n , Pn ) < 2ǫ
for n > N, we conclude that Pn → P as n → ∞. That is, X ∗ is complete.
For (d). {p} ∈ Pp and {q} ∈ Pq . By part (b),
△(Pp , Pq ) = lim d(p, q) = d(p, q).
n→∞
So the mapping ϕ is an isometry of X into X ∗ .
For (e). Take any element P ∈ X ∗ . Take {pn } ∈ P , where pn ∈ X. Then
Ppn converges to P . In fact,
lim △(Ppn , P ) = lim lim d(pn , pm ) = 0
n→∞
n→∞ m→∞
(since {pn } is a Cauchy sequence). Hence ϕ(X) is dense in X ∗ . If X is
complete, then we take any element P ∈ X ∗ . Take {pn } ∈ P , where pn ∈ X.
Since X is complete, pn converges to p ∈ X. So {p} ∈ P , P = Pp ∈ ϕ(X).
Hence ϕ(X) = X ∗ if X is complete. Exercise 3.25. The completion of X is the real number system with the
metric d(x, y) = |x − y|. 34
Exercise 4.1. No. Take f (x) = 1 if x ∈ Z; f (x) = 0 otherwise. Exercise 4.2. If f (E) is empty, the conclusion holds trivially. If f (E) is
non-empty, then we take an arbitrary point y ∈ f (E). Thus, there exists
p ∈ E such that y = f (p). Thus p ∈ E or p ∈ E ′ . Also, note that f (E) =
S
f (E) (f (E))′. Now we consider the following two cases:
Case 1: If p ∈ E, then y ∈ f (E) ⊂ f (E).
Case 2: Suppose p ∈ E ′ . Since f is continuous at x = p, given ǫ > 0,
there exists δ > 0 such that dY (f (x), f (p)) < ǫ whenever dX (x, p) < δ for all
x ∈ X. Since p is a limit point of E, then for some δ > 0 there exists x ∈ E.
Thus f (x) ∈ Nǫ (f (p)) for some f (x) ∈ f (E). Since ǫ is arbitrary, f (p) is an
adhere point of f (E) in Y . Thus f (p) ∈ f (E).
By case 1 and 2, we proved that f (E) ⊂ f (E).
Now we show that f (E) can be a proper subset of f (E). Define a function
f (x) from X = (0, +∞) to Y = R1 by
f (x) =
1
.
x
Let E = Z+ . Thus
f (E) = f (E) = {1/n : n ∈ Z+ },
[
f (E) = {1/n : n ∈ Z+ } = {0} {1/n : n ∈ Z+ }.
Exercise 4.3. To prove Z(f ) is closed, it suffice to show that every limit
point of Z(f ) is contained in Z(f ) itself. Let x ∈ X be a limit point of Z(f ).
Thus, there is a convergent sequence {xn } in Z(f ) converging to x. Since f
is continuous, limn→∞ f (xn ) = f (limn→∞ xn ). It follows that 0 = f (x) and
thus x ∈ Z(f ) exactly. Exercise 4.4. First we need to show f (E) is dense in f (X), that is, every
point of f (X) is a limit point of f (E), or a point of f (E) (or both).
Take any y ∈ f (X), and then there exists a point p ∈ X such that
y = f (p). Since E is dense in X, thus p is a limit point of E or p ∈ E. If
35
p ∈ E, then y = f (p) ∈ f (E). Thus y is a point of f (E), done. If p is a
limit point of E and p ∈
/ E. Since f is continuous on X, for every ǫ > 0
there exists δ > 0 such that dY (f (x), f (p)) < ǫ for all points x ∈ X for which
dX (x, p) < δ. Since p is a limit point of E, there exists q ∈ Nδ (p) such that
q 6= p and q ∈ E. Hence
f (q) ∈ Nǫ (f (p)) = Nǫ (y).
and f (q) ∈ f (E). Since p ∈
/ E, f (p) ∈
/ f (E), and f (q) 6= f (p). Hence f (p) is
a limit point of f (E). Thus f (E) is dense in f (X).
Suppose p ∈ X − E. Since E is dense in X, p is a limit point of E and
p∈
/ E. Hence we can take a sequence {qn } → p such that qn ∈ E and qn 6= p
for all n. (More precisely, since p is a limit point, every neighborhood Nr (p)
of p contains a point q 6= p such that q ∈ E. Take r = rn = 1/n, and thus
rn → 0 as n → ∞. At this time we can get q = qn → p as n → ∞.) Hence
g(p) = g lim qn = lim g(qn )
n→∞
n→∞
= lim f (qn ) = f lim qn = f (p).
n→∞
n→∞
Thus g(p) = f (p) for all p ∈ X. Exercise 4.5. Note that the following fact: Every open set of real numbers
is the union of a countable collection of disjoint open intervals.
S
Thus, consider E c = (ai , bi ), where i ∈ Z, and ai < bi < ai+1 < bi+1 .
We extend g on (ai , bi ) as following:
g(x) = f (ai ) + (x − ai )
f (bi ) − f (ai )
bi − ai
(g(x) = f (x) for x ∈ E). Thus g is well-defined on R1 , and g is continuous
on R1 clearly.
Next, consider f (x) = 1/x on a open set E = R−0. f is continuous on E,
but we cannot redefine f (0) = any real number to make new f (x) continue
at x = 0.
Next, consider a vector valued function
f (x) = (f1 (x), · · · , fn (x)),
36
where fi (x) is a real valued function. Since f is continuous on E, each
component of f , fi , is also continuous on E, thus we can extend fi , say gi ,
for each i = 1, · · · , n. Thus,
g(x) = (g1 (x), · · · , gn (x))
is a extension of f (x) since each component of g, gi , is continuous on R1
implies g is continuous on Rn . Exercise 4.6. (⇒) Let G = {(x, f (x))|x ∈ E}. Since f is a continuous
mapping of a compact set E into f (E), by Theorem 4.14 f (E) is also compact. We claim that the product of finitely many compact sets is compact.
Thus G = E × f (E) is also compact.
(⇐) Define
g(x) = (x, f (x))
from E to G for x ∈ E. We claim that g(x) is continuous on E. Consider
h(x, f (x)) = x from G to E. Thus h is injective, continuous on a compact set
G. Hence its inverse function g(x) is injective and continuous on a compact
set E.
Since g(x) is continuous on E, the component of g(x), say f (x), is continuous on a compact E.
Proof of claim: We prove that the product of two compact spaces is
compact; the claim follows by induction for any finite product.
Step 1. Suppose that we are given sets X and Y , with Y is compact.
Suppose that x0 ∈ X, and N is an open set of X × Y containing the “slice”
x0 × Y of X × Y . We prove the following: There is a neighborhood W of x0
in X such that N contains the entire set W × Y .
The set W × Y is often called a tube about x0 × Y .
First let us cover x0 × Y by basis elements U × V (for the topology of
X × Y ) lying in N. The set x0 × Y is compact, being homeomorphic to Y .
Therefore, we can cover x0 × Y by finitely many such basis elements
U1 × V1 , · · · , Un × Vn .
37
(We assume that each of the basis elements Ui ×Vi actually intersects x0 ×Y ,
since otherwise that basis element would be superfluous; we could discard it
from the finite collection and still have a covering of x0 × Y .) Define
W = U1
\
···
\
Un .
The set W is open, and it contains x0 because each set Ui × Vi intersects
x0 × Y . We assert that the sets Ui × Vi , which were chosen to cover the slice
x0 × Y , actually cover the tube W × Y . Let x × y ∈ W × Y . Consider the
point x0 × y of the slice x0 × Y having the same y-coordinate as this point.
Now x0 × y ∈ Ui × Vi for some i, so that y ∈ Vi . But x ∈ Uj for every j
(because x ∈ W ). Therefore, we have x × y ∈ Ui × Vi , as desired.
Since all the sets Ui × Vi ⊂ N, and since they cover W × Y , the tube
W × Y ⊂ N also.
Step 2. Now we prove the claim. Let X and Y be compact sets. Let A
be an open covering of X × Y . Given x0 ∈ X, the slice x0 × Y is compact
and may therefore be covered by finitely many elements A1 , · · · , Am of A.
S S
Their union N = A1 · · · Am is an open set containing x0 × Y ; by Step 1,
the open set N contains a tube W × Y about x0 × Y , where W is open in
X. Then W × Y is covered by finitely many elements A1 , · · · , Am of A.
Thus, for each x ∈ X, we can choose a neighborhood Wx of x such
that the tube Wx × Y can be covered by finitely many element of A. The
collection of all the neighborhoods Wx is an open covering of X; therefore by
compactness of X, there exists a finite subcollection
{W1 , · · · , Wk }
covering X. The union of the tubes
W1 × Y, · · · , Wk × Y
is all of X × Y ; since each may be covered by finitely many elements of A,
so may X × Y be covered. 38
Exercise 4.7. Since x2 + y 4 ≥ 2xy 2 , f (x, y) ≤ 2 for all (x, y) ∈ R2 . That is,
f is bounded (by 2). Next, select
1 1
(xn , yn ) =
,
.
n3 n
(xn , yn ) → (0, 0) as n → ∞, and g(xn , yn ) = n/2 → ∞ as n → ∞, that
is, g(x, y) is unbounded in every neighborhood of (0, 0) by choosing large
enough n.
Next, select
(xn , yn ) =
1 1
,
.
n2 n
(xn , yn ) → (0, 0) as n → ∞, and f (xn , yn ) = 1/2 for all n. Thus,
lim f (xn , yn ) =
n→∞
1
6= 0 = f (0, 0)
2
for some sequence {(xn , yn )} ⊂ R2 . Thus, f is not continuous at (0, 0).
Finally, we consider two cases of straight lines in R2 : (1) x = c and (2)
y = ax + b. (equation of straight lines).
(1) x = c: If c 6= 0, f (x, y) = cy 2 /(c2 + y 4 ) and g(x, y) = cy 2 /(c2 + y 6) are
continuous since cy 2, c2 + y 4 , and c2 + y 6 are continuous on R1 respect to y,
and c2 + y 4 , c2 + y 6 are nonzero. If c = 0, then f (x, y) = g(x, y) = 0, and it
is continuous trivially.
(2) y = ax+b: If b 6= 0, then this line dose not pass (0, 0). Then f (x, y) =
x(ax+b)2 /(x2 +(ax+b)4 ) and g(x, y) = x(ax+b)2 /(x2 +(ax+b)6 ). By previous
method we conclude that f (x, y) and g(x, y) are continuous. If b = 0, then
f (x, y) = 0 if (x, y) = (0, 0); f (x, y) = a2 x/(1 + a4 x2 ), and g(x, y) = 0 if
(x, y) = (0, 0); g(x, y) = a2 x/(1 + a6 x4 ). Thus, f (x, y) → 0/1 = 0 = f (0, 0)
and g(x, y) → 0/1 = 0 = f (0, 0) as x → 0. Thus, f and g are continuous.
Both of two cases imply that the restriction of both f and g to every
straight line in R2 are continuous. Exercise 4.8. Let E is bounded by M > 0, i.e., |x| ≤ M for all x ∈ E.
Since f is uniformly continuous, take ǫ = 1 there exists δ > 0 such that
|f (x) − f (y)| < ǫ
39
whenever |x−y| < δ where x, y ∈ E. For every x ∈ E, there exists an integer
n = nx such that
nδ ≤ x < (n + 1)δ.
Since E is bounded, the collection of S = {nx |x ∈ E} is finite. Suppose
x ∈ E and x is the only one element satisfying nδ ≤ x < (n + 1)δ for some
n. Let x = xn , and thus
|f (x)| ≤ |f (xn )|
T
for all x ∈ E [nδ, (n + 1)δ). If there are more than two or equal to two
element satisfying that condition, then take some one as xn . Since |x−xn | < δ
T
for all x ∈ E [nδ, (n + 1)δ),
|f (x) − f (xn )| < 1
for all x ∈ E
T
[nδ, (n + 1)δ), that is,
|f (x)| < 1 + |f (xn )|.
So
|f (x)| < max (1 + f (xn )).
n∈S
(The maximum is meaningful because S is finite.) Thus f (x) is bounded. Note: If boundedness of E is omitted from the hypothesis, then we define
f (x) = x for x ∈ R1 . Hence f is uniformly continuous on R1 , but f (R1 ) = R1
is unbounded.
Exercise 4.9. Recall the original definition of uniformly continuity: for
every ǫ > 0 there exists δ > 0 such that dY (f (p), f (q)) < ǫ for all p and q in
X for which dX (p, q) < δ.
(⇐) Given ǫ > 0. ∀p, q ∈ X for which dX (p, q) < δ. Take
E = {p, q},
and thus diamE = supp,q∈E d(p, q) = d(p, q) < δ. Hence diamf (E) < ǫ. Note
that diamf (E) ≥ d(f (p), f (q)) since p, q ∈ E. Hence d(f (p), f (q)) < ǫ. Thus
for every ǫ > 0 there exists δ > 0 such that dY (f (p), f (q)) < ǫ for all p and
q in X for which dX (p, q) < δ.
40
(⇒) ∀E ⊂ X with diamE < δ. ∀p, q ∈ R, d(p, q) ≤ diamE < δ. Thus we
have
ǫ
d(f (p), f (q)) <
2
for all p, q ∈ E. Hence diamf (E) ≤ ǫ/2 < ǫ. Thus to every ǫ > 0 there exists
a δ > 0 such that diamf (E) < ǫ for all E ⊂ X with diamE < δ. Exercise 4.10. If f is not uniformly continuous, then for some ǫ > 0
there are sequences {pn }, {qn } in X such that dX (pn , qn ) → 0 but dY (f (pn ),
f (qn )) > ǫ.
Since E = {pn : n ∈ Z+ } is an infinite subset of a compact set X, then
E has a limit point in X; that is, there is a subsequence of {pn }, which we
denote {pkn }, converges to some point x ∈ X. Then, since dX (pn , qn ) → 0,
we see that qkn → x as n → ∞. Using the fact that f is continuous on X,
we have
f (pkn ) → f (x), f (qkn ) → f (x).
That is, there is a positive integer N such that
ǫ
,
2
ǫ
dY (f (qkn ), f (x)) ≤
2
dY (f (pkn ), f (x)) ≤
for all n ≥ N. Hence for all n ≥ N, we find
dY (f (pkn ), f (qkn )) ≤ dY (f (pkn ), f (x)) + dY (f (x), f (qkn ))
ǫ
ǫ
+ = ǫ.
<
2 2
The last inequality contradicts our hypotheses, and the result is established.
Exercise 4.11. Let {xn } be a Cauchy sequence in X. ∀ǫ′ > 0, ∃N such
that dX (xn , xm ) < ǫ′ whenever m, n ≥ N. Since f is a uniformly continuous,
∀ǫ > 0, ∃δ > 0 such that dY (f (x), f (y)) < ǫ whenever d(x, y) < δ. Take
ǫ′ = δ. Thus
dY (f (xn ), f (xm )) < ǫ
41
whenever m, n ≥ N for some N; that is, {f (xn )} is a Cauchy sequence. Exercise 4.12. Theorem. Let f be a function defined on a set S in X and
assume that f (S) ⊂ Y . Let g be defined on f (S) with value in Z, and let h
denote the composite function defined by
h(x) = g(f (x))
if x ∈ S. If f is uniformly continuous on S and if g is uniformly continuous
on f (S), then h is uniformly continuous on S.
Proof. Since f is uniformly continuous on S, ∀ǫ′ > 0, ∃δ > 0 such that
dY (f (x), f (y)) < ǫ′
whenever for all x, y ∈ S for which dX (x, y) < δ. Also, since g is uniformly
continuous on f (S), ∀ǫ > 0, ∃δ ′ > 0 such that
dZ (g(x), g(y)) < ǫ
whenever for all x, y ∈ f (S) for which dY (x, y) < δ ′ . Take ǫ′ = δ ′ ; that is,
∀ǫ > 0, ∃δ > 0 such that
dZ (g(f (x)), g(f (y))) < ǫ
whenever for all x, y ∈ S for which dX (x, y) < δ. Thus h is uniformly continuous on S. Exercise 4.13. Here I give a natural proof. Take x ∈ X − E. E is dense in
X, ⇒ ∃{xn } in E such that xn → x, ⇒ {xn }: Cauchy. Since f is a uniformly
continuous mapping, by Exercise 4.11, {f (xn )} is a Cauchy sequence in E.
R: complete ⇒ f (xn ) → L ∈ R. Define
(
f (x)
if x ∈ E
g(x) =
lim f (xn ) if x ∈ X − E, xn → x.
n→∞
g(x) is well-defined. Suppose there is another sequence x′n → x, and f (x′n ) →
L′ . Then
{x1 , x′1 , x2 , x′2 , · · · , xn , x′n , · · · }
42
is still a Cauchy sequence. Hence
|L − L′ | ≤ |L − f (xn )| + |f (xn ) − f (x′n )| + |f (x′n ) − L′ | → 0,
that is, L = L′ . Hence g(x) is an extension of f (x). We prove that g(x) is also
uniformly continuous. Take x, y ∈ X, then there exist xn → x, and yn → y,
where {xn }, {yn } in E. Given ǫ > 0. Since f is uniformly continuous, there
is δ > 0 such that |f (xn ) − f (yn )| < ǫ whenever d(xn , yn ) < δ. If we take
n large enough (d(xn , x) < δ/3, d(yn , y) < δ/3 whenever n ≥ N ′ for some
integer N.) and d(x, y) small enough (d(x, y) < δ/3), we have
d(xn , yn ) ≤ d(xn , x) + d(x, y) + d(y, yn) < δ.
So
|f (xn ) − f (yn )| < ǫ.
Also, g(x) = lim f (xn ), g(y) = lim f (yn ), there is an integer N ′′ such that
n→∞
n→∞
|g(x) − f (xn )| < ǫ, and |g(y) − f (yn )| < ǫ
whenever n ≥ N ′′ . Take N = max{N ′ , N ′′ }. Hence
|g(x) − g(y)| ≤ |g(x) − f (xn )| + |f (xn ) − f (yn )| + |f (yn ) − g(y)| < 3ǫ
whenever d(x, y) < δ/3 and n ≥ N. Hence g is uniformly continuous. To
prove the uniqueness of g: Suppose h is another such function. On E, g = h.
On X − E,
h(x) = lim h(xn ) = lim f (xn ) = g(x).
n→∞
n→∞
Hence g = h on X. So the g is unique.
In the above proof, we know that the conclusion is still true if we replace
the range space R by any complete metric space (including Rk surely). Also,
a metric space is compact iff it is complete and totally bounded. So it is also
true if we replace R by any compact metric space.
But if we replace R by any metric space, the conclusion may be wrong.
Consider a subset S of Q ∩ [0, 1]:
n
+
S=
|n ≤ am , n ∈ Z ∪ {0}
am
43
P
3
67
k
where am = m
k=0 10 . For example, 11 , 111 , · · · ∈ S. Hence S is dense in
Q ∩ [0, 1]. Define f on S by
(
1 if 2n ≥ am
n
f
=
am
0 if 2n < am .
3
67
For example, f ( 11
) = 0 and f ( 111
) = 1. So f is a uniformly continuous
1
function. But f cannot
extend
to Q ∩ [0,
n
o
n 1]. Take
o 2 ∈ Q ∩ [0, 1], we have two
sequences {pn } = ⌊aann/2⌋ and {qn } = ⌈aann/2⌉ such that lim f (pn ) = 0 and
lim f (qn ) = 1. Exercise 4.14. Let g(x) = f (x) − x. If g(1) = 0 or g(0) = 0, then the
conclusion holds trivially. Now suppose g(1) 6= 0 and g(0) 6= 0. Since f is
from I to I, 0 6= f (x) 6= 1. Thus,
g(1) = f (1) − 1 < 0,
g(0) = f (0) − 0 > 0.
Since g is continuous on [0, 1], by Intermediate Value Theorem (Theorem
4.23)
g(c) = 0
for some c ∈ (0, 1). Hence f (c) = c for some c ∈ (0, 1). Exercise 4.15. If not, there exist three points x1 < x2 < x3 ∈ R1 such that
f (x2 ) > f (x1 ), f (x2 ) > f (x3 )
or
f (x2 ) < f (x1 ), f (x2 ) < f (x3 ).
WLOG, we only consider the case that f (x2 ) > f (x1 ), f (x2 ) > f (x3 ) for
(x1 )
some x1 < x2 < x3 . Since f is continuous on R1 , for ǫ = f (x2 )−f
> 0 there
2
exists δ1 > 0 such that
|f (x) − f (x1 )| < ǫ
whenever |x − x1 | < δ1 . That is,
f (x) <
f (x1 ) + f (x2 )
< f (x2 )
2
44
whenever x < x1 + δ1 . Note that δ1 < x2 − x1 . Hence we can take y1 ∈
(x3 )
(x1 , x1 + δ1 ). Similarly, for ǫ = f (x2 )−f
> 0 there exists δ2 > 0 such that
2
|f (x) − f (x3 )| < ǫ
whenever |x − x3 | < δ2 . That is,
f (x) <
f (x2 ) + f (x3 )
< f (x2 )
2
whenever x > x3 − δ2 . Note that δ2 < x3 − x2 . Hence we can take y2 ∈
(x3 − δ2 , x3 ). Note that y2 > y1 . Since f is continuous on a closed set [y1 , y2 ],
f take a maximum value at p ∈ [y1 , y2]. Note that
sup f (x) ≤ sup f (x)
x∈(x1 ,x3 )
x∈[y1 ,y2 ]
by previous inequations. Also,
sup f (x) ≥ sup f (x)
x∈(x1 ,x3 )
x∈[y1 ,y2 ]
Hence supx∈(x1 ,x3 ) f (x) = supx∈[y1 ,y2 ] f (x). Since (x1 , x3 ) is an open set,
f ((x1 , x3 )) is also open. Note that f (p) ∈ f ((x1 , x3 )) but f (p) is not an
interior point of f ((x1 , x3 )). (Otherwise f (p) + ǫ ∈ f ((x1 , x3 )) for some
ǫ > 0. That is, f (p) + ǫ > f (p) which is a contradiction with the maximum
of f (p).) Exercise 4.16. Since [x] + (x) = x is a continuous function, [x] and (x) have
the same discontinuities.
For [x], its discontinuities are integers. So (x)’s discontinuities are also
integers. Exercise 4.17. Recall the definition of simple discontinities: both f (x+)
and f (x−) exist. First, let E be the set on which f (x−) < f (x+). With
each point x ∈ E, associate a rational triple (p, q, r) such that
(a) f (x−) < p < f (x+),
(b) a < q < t < x implies f (t) < p,
(c) x < t < r < b implies f (t) > p.
45
This association is possible. Q is dense in R ⇒ the existence of p. f (x−)
exists ⇒ for any ǫ > 0, there exists x−a > δ > 0 such that |f (t)−f (x−)| < ǫ
whenever t ∈ (x − δ, x) ∩ (a, b). Choose ǫ = p−f2(x−) > 0, and take δ = q ∈ Q.
So a < q < t < x implies f (t) < p, that is, the existence of (b) is OK.
Similarly, the existence of (c) is OK too.
The set of all such triples is countable. Claim: each triple is associated
with at most one point of E. If not, then there are at least two points x and
y with x < y such that
(a) f (x−) < p < f (x+),
(b) a < q < t < x implies f (t) < p,
(c) x < t < r < b implies f (t) > p.
(a′ ) f (y−) < p < f (y+),
(b′ ) a < q < t < y implies f (t) < p,
(c′ ) y < t < r < b implies f (t) > p.
for some rational triple (p, q, r). Hence q < x < y < r. However, if we take
t0 ∈ (x, y), then f (t0 ) > p by (c) and f (t0 ) < p by (b′ ), a contradiction.
Hence our claim holds. So E is countable.
Similarly, let F be the set on which f (x−) = f (x+). With each point
x ∈ F , associate a rational triple (p, q) such that
(a) a < p < t < x implies f (t) < f (x−),
(b) x < t < q < b implies f (t) > f (x−).
As the above argument, we conclude that F is also countable. Combine E
and F . Hence E ∪ F , the set of point at which f has a simple discontinuity,
is at most conutable. Exercise 4.18. WLOG, we only consider f on [0, 1]. Given ǫ > 0, for any
x0 ∈ [0, 1], there are finitely many positive integers n satisfying that
1
n< ;
ǫ
that is, there are finitely many rational numbers m/n such that
m
f
≥ ǫ.
n
46
Thus we can take δ > 0 such that the neighborhood Nδ (x0 ) of x0 contains
no such rational numbers. (If x0 ∈ Q, we may remove x0 .) Hence for all x
the following
|f (x)| < ǫ
holds whenever 0 < |x − x0 | < δ. Thus we proved that
f (x0 +) = f (x0 −) = 0
for all x0 ∈ [0, 1].
If x0 is rational, then f (x0 ) = 0; that is, f (x) is continuous at x = x0 .
If x0 is not rational, then f (x0 ) 6= 0; that is, has a simple discontinuity at
x = x0 . Note. f is Riemann-integrable on [0, 1].
Exercise 4.19. Let S = {x|f (x) = r}. If xn → x0 , but f (xn ) > r > f (x0 )
for some r and all n since Q is dense in R, then f (tn ) = r for some tn
between x0 and xn ; thus tn → x0 . Hence x0 is a limit point of S. Since S is
closed, f (x0 ) = r, a contradiction. Hence, lim sup f (xn ) ≤ f (x0 ). Similarly,
lim inf f (xn ) ≥ f (x0 ). Hence, lim f (xn ) = f (x0 ), and f is continuous at x0 .
Note: Original problem is stated as follows: Let f be a function from the
reals to the reals, differentiable at every point. Suppose that, for every r, the
set of points x, where f ′ (x) = r, is closed. Prove that f ′ is continuous. If
we replace Q into any dense subsets of R, the conclusion also holds.
Exercise 4.20. We prove (a). (⇐) If x ∈ E ⊂ E, then
inf d(x, z) ≤ d(x, x) = 0
z∈E
since we take z = x ∈ E. Hence ρE (x) = 0 if x ∈ E. Suppose x ∈ E − E,
that is, x is a limit point of E. Thus for every neighborhood of x contains
a point y 6= x such that q ∈ E. It implies that d(x, y) → 0 for some y ∈ E,
that is, ρE (x) = inf z∈E d(x, z) = 0 exactly.
(⇒) Suppose ρE (x) = inf z∈E d(x, z) = 0. Fixed some x ∈ X. If d(x, z) =
0 for some z ∈ E, then x = z, that is x ∈ E ⊂ E. If d(x, z) > 0 for all z ∈ E,
47
then by inf z∈E d(x, z) = 0, for any ǫ > 0 there exists z ∈ E such that
d(x, z) < ǫ,
that is,
z ∈ Nǫ (x).
Since ǫ is arbitrary and z ∈ E, x is a limit point of E. Thus x ∈ E ′ ⊂ E.
We prove (b). For all x ∈ X, y ∈ X, z ∈ E,
ρE (x) ≤ d(x, z) ≤ d(x, y) + d(y, z).
Take infimum on both sides, and we get that
ρE (x) ≤ d(x, y) + ρE (y).
Similarly, we also have ρE (y) ≤ d(x, y) + ρE (x). Hence
|ρE (x) − ρE (y)| ≤ d(x, y)
for all x ∈ X, y ∈ X. Thus ρE is a uniformly continuous function on X. Exercise 4.21. Let
ρF (x) = inf d(x, z)
z∈F
for all x ∈ K. By Exercise 4.20(a), we know that
ρF (x) = 0 ⇔ x ∈ F = F
(since F is closed). That is, ρF (x) = 0 if and only if x ∈ F . Since K and
F are disjoint, ρF (x) is a positive function. Also ρF (x) is continuous by
Exercise 4.20(b). Thus ρF (x) is a continuous positive function. Since K is
compact, ρF (x) takes minimum m > 0 for some x0 ∈ K. Take δ = m/2 > 0
as desired.
Next, let X = R, A = Z+ − {2}, and B = {n + 1/n|n ∈ Z+ }. Hence
A and B are disjoint, and they are not compact. Suppose there exists such
δ > 0. Take
1
1
x=
+ 1 ∈ A, y = x + ∈ B.
δ
x
48
However,
d(x, y) =
1
1
<
= δ,
x
1/δ
a contradiction. Exercise 4.22. Note that ρA (p) and ρB (p) are (uniformly) continuous on
X, and ρA (p) + ρB (p) > 0. (Clearly, ρA (p) + ρB (p) ≥ 0 by the definition. If
T
ρA (p) + ρB (p) = 0, then p ∈ A B by Exercise 20, a contradiction). Thus
f (p) = ρA (p)/(ρA (p) + ρB (p)) is continuous on X. Next, f (p) ≥ 0, and
f (p) ≤ 1 since ρA (p) ≤ ρA (p) + ρB (p). Thus f (X) lies in [0, 1].
Next, f (p) = 0 ⇔ ρA (p) = 0 ⇔ p ∈ A, and f (p) = 1 ⇔ ρB (p) = 0 ⇔ p ∈
B by Exercise 20.
Now we prove a converse of Exercise 3: Every closed set A ⊂ X is Z(f )
for some continuous real f on X. If Z(f ) = φ, then f (x) = 1 for all x ∈ X
satisfies our requirement. If Z(f ) 6= φ, we consider two possible cases: (1)
Z(f ) = X; (2) Z(f ) 6= X. If Z(f ) = X, then f (x) = 0 for all x ∈ X. If
Z(f ) 6= X, we can choose p ∈ X such that f (p) =
6 0. Note that Z(f ) and
{p} are one pair of disjoint closed sets. Hence we let
f (x) =
ρZ(f ) (x)
.
ρZ(f ) (x) + ρ{p} (x)
By the previous result, we know that f (x) satisfies our requirement. Hence
we complete the whole proof.
Note that [0, 1/2) and (1/2, 1] are two open sets of f (X). Since f is
continuous, V = f −1 ([0, 1/2)) and W = f −1 ((1/2, 1]) are two open sets.
f −1 ({0}) ⊂ f −1 ([0, 1/2)), and f −1 ({1}) ⊂ f −1 ((1/2, 1]). Thus, A ⊂ V and
B ⊂ W . Thus a metric space X is normal. Exercise 4.23. First, we prove that if f is convex in (a, b) and if a < s <
t < u < b, then
f (t) − f (s)
f (u) − f (s)
f (u) − f (t)
≤
≤
.
t−s
u−s
u−t
Since s < t < u, ∃λ > 0 such that t = λs + (1 − λ)u. Since f is convex in
(a, b),
f (t) ≤ λf (s) + (1 − λ)f (u).
49
Let
f (u) − f (s)
(t − s) + f (s)
u−s
= (f (u) − f (s))(1 − λ) + f (s)
m =
= λf (s) + (1 − λ)f (u).
Thus
m − f (s)
f (u) − f (s)
=
.
t−s
u−s
Hence
(λf (s) + (1 − λ)f (u)) − f (s)
f (t) − f (s)
≤
t−s
t−s
m − f (s)
f (u) − f (s)
=
=
.
t−s
u−s
Similarly,
f (u)−f (s)
u−s
≤
f (u)−f (t)
.
u−t
Lemma. If f is convex on (a, b) and if x, y, x′ , y ′ are points of (a, b) with
x ≤ x′ < y ′ and x < y ≤ y ′, then the chord over (x′ , y ′) has larger slope than
the chord over (x, y); that is,
f (y) − f (x)
f (y ′) − f (x′ )
≤
.
y−x
y ′ − x′
The above lemma is very easy by our previous result. Now we turn to
show that f is continuous on (a, b). Choose p ∈ (a, b). Since p is an interior
point of (a, b), there exists α > 0 such that (p − α, p + α) ⊂ (a, b). Hence
[c, d] ⊂ (a, b) where c = p − α/2, d = p + α/2. Then by lemma, we have
f (c) − f (a)
f (y) − f (x)
f (b) − f (d)
≤
≤
c−a
y−x
b−d
for x, y ∈ [c, d]. Thus |f (y)−f (x)| ≤ M|x−y| in [c, d]. Hence f is continuous
at x = p.
Finally, we show that every increasing convex function of a convex function is convex. Let f (x) be a convex function on (a, b), g(x) be an increasing
convex function on f ((a, b)). For any x, y ∈ (a, b), 0 < λ < 1,
g(f (λx + (1 − λ)y)) ≤ g(λf (x) + (1 − λ)f (y))
≤ λg(f (x)) + (1 − λ)g(f (y))
50
since g is increasing. Hence g(f (x)) is convex. Exercise 4.24. Claim 1.
x1 + · · · + x2n
f (x1 ) + · · · + f (x2n )
f
≤
.
2n
2n
for all n.
Induction on n. When n = 1, it is OK. Assume n = k the conclusion
holds, then as n = k + 1,
x1 + · · · + x2k+1
f
2k+1
1 x1 + · · · + x2k
x2k +1 + · · · + x2k+1
=f
(
+
)
2
2k
2k
1
x1 + · · · + x2k
x2k +1 + · · · + x2k+1
≤
f
+f
2
2k
2k
1 f (x1 ) + · · · + f (x2k ) f (x2k +1 ) + · · · + f (x2k+1 )
+
≤
2
2k
2k
f (x1 ) + · · · + f (x2k+1 )
.
=
2k+1
by induction hypothesis. Hence by the induction, claim 1 holds.
Claim 2.
for all n.
x1 + · · · + xn
f
n
≤
f (x1 ) + · · · + f (xn )
.
n
Reverse induction on n. Put
xn =
x1 + · · · + xn−1
.
n−1
Then
x1 + · · · + xn−1 xn
f (xn ) = f
+
n
n
x1 + · · · + xn
= f
n
f (x1 ) + · · · + f (xn )
.
≤
n
51
So
x1 + · · · + xn−1
f
n−1
Hence claim 2 holds.
≤
f (x1 ) + · · · + f (xn−1 )
.
n−1
Back to the proof. Given x, y ∈ (a, b), λ ∈ (0, 1). irst suppose λ = p/q ∈
(0, 1) is rational, where p, q ∈ N. Let xi = x for 1 ≤ i ≤ p, and xj = y for
p + 1 ≤ j ≤ q. Hence
p
q−p
p
p
f
x+
y ≤ f (x) + 1 −
f (y).
q
q
q
q
Finally we take λ ∈ (0, 1) and choose a rational sequence λn → λ. Since
f is continuous,
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y).
Exercise 4.25. We prove (a). Take z ∈
/ K + C. (If we cannot take such z,
then K + C = Rk . K + C is closed trivially.) Put F = z − C, the set of all
z − y with y ∈ C. Then K and F are disjoint. Note that F is closed, and
we choose δ as in Exercise 21.
We will show that the open ball with center z and radius δ does not
intersect K + C. If not, there exists y ∈ K + C such that d(z, y) ≤ δ. Since
y ∈ K + C, ∃k ∈ K, c ∈ C such that y = k + c. Let f = z − c ∈ F . Thus
d(k, f) = d(y − c, z − c) = d(y, z) > δ,
a contradiction.
We prove (b). Let A = C1 + C2 . Claim: 0 is a limit point of A. Let ǫ > 0.
Take arbitrary n ∈ N such that 1/n < ǫ, and we partition the interval [0, 1)
into n subintervals:
1
1 2
n−1
0,
, ,
,··· ,
,1 .
n
n n
n
Since n + 1 numbers,
0 · α − [0 · α], 1 · α − [1 · α], · · · , nα − [nα],
52
are all in [0, 1), there exist two different integer k, l with 0 ≤ k, l ≤ n such
that kα − [kα] and lα − [lα] in the same subinterval. If the front number is
larger, then
0 < (kα − [kα]) − (lα − [lα]) <
1
< ǫ.
n
Let u = [lα] − [kα], v = k − l; thus
0 < u + vα < ǫ
T
where u + vα ∈ Nǫ (0) (A − {0}). Thus 0 is a limit point of A.
For any x ∈ R and a neighborhood N of x, we can take two reals a, b
such that x ∈ (a, b) ⊂ N. According the previous conclusion, we can find
two integers u, v such that
0 < u + vα < b − a.
Since
b
u+vα
−
a
u+vα
> 1, there exists an integer m such that
a
b
<m<
,
u + vα
u + vα
a < m(u + vα) < b.
T
T
Thus m(u + vα) ∈ N A, that is, N A is not empty. Thus x is a limit
point of A.
Note. Let C1 = {n + 1/3n |n ∈ N} and C2 = {−n + 1/3n |n ∈ N}. Then
C1 and C2 are closed but C1 + C2 isn’t. Exercise 4.26. Let Z ′ = g(Y ) be a metric space induced from Z. Then
g : Y → Z ′ is injective and surjective. Since Y is compact, so is Z ′ . Also, the
inverse of g, denoted by g −1 is also continuous. Since Y and Z ′ are compact,
by Theorem 4.19 both g and g −1 are uniformly continuous. By Exercise 4.12,
we know that
f (x) = g −1 (h(x))
is uniformly continuous if h is uniformly continuous. Also, by Theorem 4,9,
f (x) = g −1 (h(x))
53
is continuous if h is continuous.
For a counterexample, let X = Z = [0, 1], Y = [0, 1/2) ∪ [1, 3/2],
(
x
if x ∈ [0, 1/2),
f (x) =
x + 1/2 if x ∈ [1/2, 1],
g(x) =
(
x
if x ∈ [0, 1/2),
x − 1/2 if x ∈ [1, 3/2].
Observe that g is uniformly continuous on Y , and h(x) = x is uniformly
continuous on Z, but f is not (uniformly) continuous on X. (y)
Exercise 5.1. |f (x)−f (y)| ≤ (x−y)2 for all real x and y. Fix y, | f (x)−f
|≤
x−y
|x − y|. Let x → y, therefore,
f (x) − f (y)
≤ lim |x − y| = 0.
x→y
x→y
x−y
0 ≤ lim
It implies that (f (x) − f (y))/(x − y) → 0 as x → y. Hence f ′ (y) = 0, or
f = constant. Exercise 5.2. For every pair x > y in (a, b), f (x) − f (y) = f ′ (c)(x − y)
where y < c < x by Mean Value Theorem. Note that c ∈ (a, b) and f ′ (x) > 0
in (a, b), hence f ′ (c) > 0. f (x) − f (y) > 0 or f (x) > f (y) if x > y, i.e., f is
strictly increasing in (a, b).
Let ∆g = g(x0 + h) − g(x0 ). Note that x0 = f (g(x0 )). So
(x0 + h) − x0 = f (g(x0 + h)) − f (g(x0 )),
and
h = f (g(x0 ) + ∆g) − f (g(x0 )) = f (g + ∆g) − f (g).
Apply the fundamental lemma of differentiation,
h = [f ′ (g) + η(∆g)]∆g,
1
∆g
=
f ′ (g) + η(∆g)
h
54
Note that f ′ (g(x)) > 0 for all x ∈ (a, b) and η(∆g) → 0 as h → 0, thus,
lim ∆g/h = lim
h→0 f ′ (g)
h→0
Thus g ′ (x) =
1
,
f ′ (g(x))
g ′ (f (x)) =
1
.
f ′ (x)
1
1
= ′
.
+ η(∆g)
f (g(x))
Exercise 5.3. For every x < y with x, y ∈ R, we will show that f (x) 6= f (y).
By Mean Value Theorem, g(x) − g(y) = g ′ (c)(x − y) for some x < c < y.
(x − y) + ǫ((x) − g(y)) = (ǫg ′ (c) + 1)(x − y),
that is,
f (x) − f (y) = (ǫg ′ (c) + 1)(x − y).
(*)
Since |g ′(x)| ≤ M, −M ≤ g ′ (x) ≤ M for all x ∈ R. Thus 1 − ǫM ≤
1
ǫg ′ (c) + 1 ≤ 1 + ǫM, where x < c < y. Take c = 2M
, and ǫg ′ (c) + 1 > 0
where x < c < y for all x, y. Take into equation (*), and f (x) − f (y) < 0
since x − y < 0, that is, f (x) 6= f (y), that is, f is injective. Cn n+1
Exercise 5.4. Let f (x) = C0 x + · · · + n+1
x . f is differentiable in R and
′
f (0) = f (1) = 0. Thus, f (1) − f (0) = f (c) for some c ∈ (0, 1) by Mean
Value Theorem. Also,
f ′ (x) = C0 + C1 x + · · · + Cn−1 xn−1 + Cn xn .
Thus, c ∈ (0, 1) is one real root between 0 and 1 of that equation. Exercise 5.5. f (x + 1) − f (x) = f ′ (c)(x + 1 − x) where x < c < x + 1 by
Mean Value Theorem. Thus, g(x) = f ′ (c) where x < c < x + 1, that is,
lim g(x) = lim f ′ (c) = lim f ′ (c) = 0.
x→+∞
x→+∞
c→+∞
Exercise 5.6. Our goal is to show g ′(x) > 0 for all x > 0 ⇔ g ′(x) =
xf ′ (x)−f (x)
x2
> 0 ⇔ f ′ (x) >
f (x)
.
x
Since f ′ (x) exists, f (x) − f (0) = f ′ (c)(x − 0)
55
where 0 < c < x by Mean Value Theorem. ⇒ f ′ (c) = f (x)
where 0 < c < x.
x
Since f ′ is monotonically increasing, f ′ (x) > f ′ (c), that is, f ′ (x) > f (x)
for
x
all x > 0. Exercise 5.7.
(x)
limt→x f (t)−f
f ′ (t)
f (t)
limt→x f (t)
t−x
=
=
= lim
g(t)−g(x)
′
g (t)
limt→x g(t) t→x g(t)
limt→x t−x
Surely, this holds also for complex functions. Exercise 5.8. Since f ′ (x) is continuous on a compact set [a, b], f ′ (x) is
uniformly continuous on [a, b]. Hence, for any ǫ > 0 there exists δ > 0 such
that
|f ′ (t) − f ′ (x)| < ǫ
(x)
whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Thus, f ′ (c) = f (t)−f
where
t−x
c between t and x by Mean Value Theorem. Note that 0 < |c − x| < δ and
thus |f ′ (c) − f ′ (x)| < ǫ, thus,
|
f (t) − f (x)
− f ′ (x)| < ǫ
t−x
whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.
It does not hold for vector-valued functions. Consider f (x) = (cos x, sin x),
[a, b] = [0, 2π], and x = 0. Hence f ′ (x) = (− sin x, cos x). Take any 1 > ǫ > 0,
there exists δ > 0 such that
f (t) − f (0)
′
− f (0) < ǫ
t−0
whenever 0 < |t| < δ by our hypothesis. Thus
cos t − 1 sin t
< ǫ,
,
−
(0,
1)
t
t
cos t − 1 sin t
< ǫ,
,
−
1
t
t
2
cos t − 1 2
sin t
) +(
− 1 < ǫ2 < ǫ,
t
t
56
2
2(cos t + sin t)
+
1
−
<ǫ
t2
t
since 1 > ǫ > 0. Note that
4
2
2(cos t + sin t)
2
+1− < 2 +1−
2
t
t
t
t
But
2
t2
+1−
4
t
→ +∞ as t → 0. It is absurd. Exercise 5.8. Note. We prove a more general exercise as following. Suppose
that f is continuous on an open interval I containing x0 , suppose that f ′ is
defined on I except possibly at x0 , and suppose that f ′ (x) → L as x → x0 .
Prove that f ′ (x0 ) = L.
By L’Hospital’s rule,
f (x0 + h) − f (x0 )
= lim f ′ (x0 + h)
h→0
h→0
h
lim
By our hypothesis, f ′ (x) → L as x → x0 . Thus,
f (x0 + h) − f (x0 )
= L,
h→0
h
lim
Thus f ′ (x0 ) exists and f ′ (x0 ) = L. Exercise 5.10. Write f (x) = f1 (x) + if2 (x), where f1 (x), f2 (x) are realvalued functions. Thus,
df (x)
df1 (x)
df2 (x)
=
+i
,
dx
dx
dx
Apply L’Hospital’s rule to
f1 (x)
x
and
f2 (x)
,
x
we have
f1 (x)
f2 (x)
= lim f1′ (x) and lim
= lim f2′ (x)
x→0
x→0
x→0
x→0
x
x
lim
Combine f1 (x) and f2 (x), we have
f1 (x)
f2 (x)
f1 (x)
f2 (x)
f (x)
+ i lim
= lim
+i
= lim
,
x→0
x→0
x→0
x→0 x
x
x
x
x
lim
or
f1 (x)
f2 (x)
+ i lim
= lim f1′ (x) + i lim f2′ (x) = lim f ′ (x)
x→0
x→0
x→0
x→0
x→0
x
x
lim
57
Thus, limx→0 f (x)
= limx→0 f ′ (x). Similarly, limx→0 g(x)
= limx→0 g ′(x). Note
x
x
that B 6= 0. Thus,
f (x)
f (x)
x
x
lim
= lim
−A
+A
x→0 g(x)
x→0
x
g(x)
g(x)
1
A
A
= (A − A) + = .
B B
B
In Theorem 5.13, we know g(x) → +∞ as x → 0. (f (x) = x, and
i
g(x) = x + x2 e x2 ). Exercise 5.11. By L’Hospital’s rule with respect to h,
f (x + h) + f (x − h) − 2f (x)
f ′ (x + h) − f ′ (x − h)
=
lim
.
h→0
h→0
h2
2h
lim
Note that
1 ′′
(f (x) + f ′′ (x))
2
1
f ′ (x + h) − f ′ (x)
f ′ (x − h) − f ′ (x)
=
(lim
+ lim
)
h→0
2 h→0
h
−h
1
f ′ (x + h) − f ′ (x − h)
=
lim
2 h→0
h
f ′ (x + h) − f ′ (x − h)
.
= lim
h→0
2h
f ′′ (x) =
Thus,
f (x + h) + f (x − h) − 2f (x)
→ f ′′ (x)
2
h
as h → 0.
Counterexample: f (x) = x|x| on R. Exercise 5.12. f ′ (x) = 3|x|2 if x 6= 0. Consider
f (h) − f (0)
|h|3
=
.
h
h
Note that |h|/h is bounded and |h|2 → 0 as h → 0. Thus,
f (h) − f (0)
= 0.
h→0
h
f ′ (0) = lim
58
Hence, f ′ (x) = 3|x|2 for all x. Similarly,
f ′′ (x) = 6|x|.
Thus,
Since
f ′′ (h) − f (0)
|h|
=6
h
h
|h|
h
= 1 if h > 0 and = −1 if h < 0, f ′′′ (0) does not exist. Exercise 5.13. For (a). (⇒) f is continuous iff for any sequence {xn } → 0
with xn 6= 0, xan sin x−c
n → 0 as n → ∞. In particular, we take
xn =
1
2nπ + π/2
1c
>0
and thus xan → 0 as n → ∞. Hence a > 0. If not, then a = 0 or a < 0. When
a = 0, xan = 1. When a < 0, xan = 1/x−a
n → ∞ as n → ∞. It contradicts.
(⇐) f is continuous on [−1, 1] − {0} clearly. Note that
−|xa | ≤ xa sin (x−c ) ≤ |xa |,
and |xa | → 0 as x → 0 since a > 0. Thus f is continuous at x = 0. Hence f
is continuous.
For (b). f ′ (0) exists iff xa−1 sin (x−c ) → 0 as x → 0. In the previous proof
we know that f ′ (0) exists if and only if a − 1 > 0. Also, f ′ (0) = 0.
Exercise 5.15. Suppose h > 0. By using Taylor’s theorem, f (x + h) =
2
f (x) + hf ′ (x) + h2 f ′′ (ξ) for some x < ξ < x + 2h. Thus h|f ′ (x)| ≤ |f (x +
h)| + |f (x)| +
h2
|f ′′ (ξ)|,
2
h|f ′ (x)| ≤ 2M0 +
h2
M2 ,
2
or
h2 M2 − 2h|f ′ (x)| + 4M0 ≥ 0.
(*)
Since equation (*) holds for all h > 0, its determinant must be non-positive.
Thus 4|f ′ (x)|2 − 4M2 (4M0 ) ≤ 0, |f ′ (x)|2 ≤ 4M0 M2 , or
(M1 )2 ≤ 2M0 M2
59
Note: There is a similar exercise. Suppose f (x) defined on R is a twicedifferentiable real-valued function, and
Mk =
sup
−∞<x<+∞
|f (k) (x)| < +∞
for k = 0, 1, 2. Prove that M12 ≤ 2M0 M2 . Write
f (x + h) = f (x) + f ′ (x)h +
f ′′ (ξ1 ) 2
h
2
(*)
f ′′ (ξ2 ) 2
h
2
(**)
for x < ξ1 < x + h or x > ξ1 > x + h, and
f (x − h) = f (x) − f ′ (x)h +
for x − h < ξ2 < x or x − h > ξ2 > x. (*) minus (**) implies that f (x + h) −
2
f (x − h) = 2f ′(x)h + h2 (f ′′ (ξ1 ) − f ′′ (ξ2 )). 2h|f ′ (x)| ≤ |2hf ′(x)| implies that
2
2h|f ′ (x)| ≤ |f (x + h)| + |f (x − h)| + h2 (|f ′′ (ξ1 )| + |f ′′ (ξ2 )|), or 2h|f ′ (x)| ≤
2M0 + h2 M2 . Thus
M2 h2 − 2|f ′ (x)|h + 2M0 ≥ 0.
Since this equation holds for all h, its determinant must be non-positive.
Thus 4|f ′ (x)|2 − 4M2 (2M0 ) ≤ 0, |f ′ (x)|2 ≤ 2M0 M2 , or
M12 ≤ 2M0 M2
Exercise 5.16. Suppose a ∈ (0, ∞), and M0 , M1 , M2 are the least up-
per bounds of |f (x)|, |f ′(x)|, |f ′′(x)| on (a, ∞). Hence, M12 ≤ 4M0 M2 . Let
a → ∞, M0 = sup |f (x)| → 0. Since M2 is bounded, therefore M12 → 0 as
a → ∞. It follows that sup |f ′ (x)| → 0 as x → ∞. Exercise 5.17. We take α = 0andβ = 1 in Theorem 5.15, and we obtain
that
f ′′ (0) f (3) (s)
f (1) = f (0) + f ′ (0) +
+
2
6
60
where s ∈ (0, 1). Take α = 0andβ = −1, and
f (−1) = f (0) − f ′ (0) +
f ′′ (0) f (3) (t)
−
2
6
where t ∈ (−1, 0). Thus
1=
f ′′ (0) f (3) (s)
+
where s ∈ (0, 1),
2
6
f ′′ (0) f (3) (s)
−
where s ∈ (−1, 0).
2
6
(*) minus (**) and thus
0=
(*)
(**)
f (3) (s) f (3) (t)
+
where s ∈ (0, 1) and t ∈ (−1, 0),
6
6
or
f (3) (s) + f (3) (t) = 6 where s, t ∈ (−1, 1).
Hence
f (3) (x) ≥ 3 for some x ∈ (−1, 1).
(0) −αn
f (βn )−f (0) βn
+ f (αnα)−f
.
βn
βn −αn
βn −αn
n
f (αn )−f (0)
f (βn )−f (0)
limn→∞
= limn→∞
. Thus for
αn
βn
f (αn )−f (0)
such that L − ǫ <
< L + ǫ, and L −
αn
′
Exercise 5.19. For (a). Write Dn =
Note that f ′ (0) =
ǫ > 0, there exists N
f (βn )−f (0)
βn
any
ǫ <
< L + ǫ whenever n > N where L = f (0) respectively. Note that
βn /(βn − αn ) and −αn /(βn − αn ) are positive. Hence,
and
f (βn ) − f (0) βn
βn
βn
(L − ǫ) <
<
(L + ǫ),
βn − αn
βn
βn − αn
βn − αn
−αn
f (αn ) − f (0) −αn
−αn
(L − ǫ) <
<
(L + ǫ).
βn − αn
αn
βn − αn
βn − αn
Combine two inequations, and we get L − ǫ < Dn < L + ǫ. Hence, lim Dn =
L = f ′ (0).
For (b). We process as above prove. Note that −αn /(βn − αn ) < 0 only
implies the following two inequations
βn
f (βn ) − f (0) βn
βn
(L − ǫ) <
<
(L + ǫ),
βn − αn
βn
βn − αn
βn − αn
61
−αn
f (αn ) − f (0) −αn
−αn
(L + ǫ) <
<
(L − ǫ).
βn − αn
αn
βn − αn
βn − αn
Combine them, and we have
βn + αn
βn + αn
ǫ < Dn < L +
ǫ.
βn − αn
βn − αn
βn Note that {βn /(βn − αn )} is bounded, ie, βn −αn ≤ M for some constant M.
Thus
βn + αn 2βn
=
≤ 2M + 1.
−
1
βn − αn βn − αn
L−
Hence, L − (2M + 1)ǫ < Dn < L + (2M + 1)ǫ, or lim Dn = L = f ′ (0).
For (c). By Mean Value Theorem, Dn = f ′ (tn ) for some tn between
αn and βn . Note that min{αn , βn } < tn < max{αn , βn }, max{αn , βn } =
1
(αn + βn + |αn − βn |), and min{αn , βn } = 12 (αn + βn − |αn − βn |). Thus,
2
max{αn , βn } → 0 and min{αn , βn } → 0 as αn → 0 and βn → 0. By squeezing
principle for limits, tn → 0. With the continuity of f ′ , we have
lim Dn = lim f ′ (tn ) = f ′ (lim tn ) = f ′ (0).
Example: Let f be defined by
(
x2 sin(1/x) (x 6= 0),
f (x) =
0
(x = 0).
Thus f ′ (x) is not continuous at x = 0, and f ′ (0) = 0. Take αn =
1
and βn = 2nπ
. It is clear that αn → 0, and βn → 0 as n → ∞. Also,
Dn =
1
π/2+2nπ
−4nπ
2
→−
π(π/2 + 2nπ)
π
as n → ∞. Thus, lim Dn = −2/π exists and is different from 0 = f ′ (0). Exercise 5.22. For (a). If not, then there exists two distinct fixed points,
say x and y, of f . Thus f (x) = x and f (y) = y. Since f is differentiable,
by applying Mean Value Theorem we know that f (x) − f (y) = f ′ (t)(x − y)
where t is between x and y. Since x 6= y, f ′ (t) = 1 and it is absurd.
For (b). We show that 0 < f ′ (t) < 1 for all real t. First, f ′ (t) =
et
t
t 2
t
t
1 + (−1)(1 + et )−2 et = 1 − (1+e
t )2 . Since e > 0, (1 + e ) = (1 + e )(1 + e ) >
62
1(1 + et ) = 1 + et > et > 0 for all real t. Thus 1 > (1 + et )−2 et > 0 for all
real t. Hence 0 < f ′ (t) < 1 for all real t.
Next, since f (t) − t = (1 − et )−1 > 0 for all real t, f (t) has no fixed point.
For (c). Suppose xn+1 6= xn for all n. (If xn+1 = xn , then xn = xn+1 = · · ·
and xn is a fixed point of f ).
By Mean Value Theorem, f (xn+1 ) − f (xn ) = f ′ (tn )(xn+1 − xn ) where tn
is between xn and xn+1 . Thus,
|f (xn+1 ) − f (xn )| = |f ′ (tn )||(xn+1 − xn )|.
Note that |f ′ (tn )| is bounded by A < 1, f (xn ) = xn+1 , and f (xn+1 ) = xn+2 .
Thus |xn+2 − xn+1 | ≤ A|xn+1 − xn | or
|xn+1 − xn | ≤ CAn−1
where C = |x2 − x1 |. For two positive integers p > q, |xp − xq | ≤ |xp − xp−1 | +
q−1
· · · + |xq+1 − xq | = C(Aq−1 + Aq + · · · + Ap−2 ) ≤ CA
. Hence
1−A
|xp − xq | ≤
CAq−1
.
1−A
Therefore, for any ǫ > 0, there exists N = [logA ǫ(1−A)
] + 2 such that |xp −
C
xq | < ǫ whenever p > q ≥ N. By Cauchy criterion we know that {xn }
converges to x. Thus,
lim xn+1 = f ( lim xn )
n→∞
n→∞
since f is continuous. Thus, x = f (x), i.e., x is a fixed point of f .
For (d). Since xn+1 = f (xn ), it is trivial. Exercise 5.25. For (b). We show that xn ≥ xn+1 ≥ ξ by induction. By
Mean Value Theorem, f (xn )−f (ξ) = f ′ (cn )(xn −ξ) where cn ∈ (ξ, xn ). Since
f ′′ ≥ 0, f ′ is increasing and thus
f (xn )
xn −ξ
= f ′ (cn ) ≤ f ′ (xn ) =
f (xn )
,
xn −xn+1
or
f (xn )(xn − ξ) ≤ f (xn )(xn − xn+1 ).
Note that f (xn ) > f (ξ) = 0 since f ′ ≥ δ > 0 and f is strictly increasing.
Thus, xn − ξ ≤ xn − xn+1 or ξ ≤ xn+1 .
63
Note that f (xn ) > 0 and f ′ (xn ) > 0. Thus xn+1 < xn . Hence, xn >
xn+1 ≥ ξ. Thus, {xn } converges to a real number ζ. Suppose ζ 6= ξ, then
xn+1 = xn −
Note that
f (xn )
f ′ (xn )
>
f (ζ)
.
δ
Let α =
f (ζ)
δ
f (xn )
.
f ′ (xn )
> 0, be a constant. Thus,
xn+1 < xn − α
for all n. Thus, xn < x1 − (n − 1)α, that is, xn → −∞ as n → ∞. It
contradicts. Thus, {xn } converges to ξ.
′′
f (tn )
For (c). By Taylor’s theorem, f (ξ) = f (xn ) + f ′(xn )(ξ − xn ) + 2(x
2 , or
n −ξ)
′′
′′
f (xn )
f (tn )
f (tn )
2
0 = f (xn ) + f ′ (xn )(ξ − xn ) + 2(x
2 , or 0 = f ′ (x ) − xn + ξ + 2f ′ (x ) (xn − ξ) .
n −ξ)
n
n
Thus
f ′′ (tn )
xn+1 − ξ = ′
(xn − ξ)2
2f (xn )
where tn ∈ (ξ, xn ).
For (d). By (b) we know that 0 ≤ xn+1 − ξ for all n. Again by (c) we
′′
f (tn )
2
′′
know that xn+1 − ξ = 2f
≤ M and f ′ ≥ δ > 0.
′ (x ) (xn − ξ) . Note that f
n
n
Thus xn+1 − ξ ≤ A(xn − ξ)2 ≤ A1 (A(x1 − ξ))2 by the induction. Thus,
0 ≤ xn+1 − ξ ≤
1
n
[A(x1 − ξ)]2 .
A
0)
For (e). If x0 is a fixed point of g(x), then g(x0 ) = x0 , that is, x0 − ff′(x
=
(x0 )
x0 or f (x0 ) = 0. It implies that x0 = ξ and x0 is unique since f is strictly
increasing. Thus, we choose x1 ∈ (ξ, b) and apply Newton’s method, we can
find out ξ. Hence we can find out x0 . Next, by calculating g ′ (x) =
we have
M
0 ≤ g ′ (x) ≤ f (x) 2 .
δ
′
As x near ξ from right hand side, g (x) near f (ξ) = 0.
f (x)f ′′ (x)
,
f ′ (x)2
n)
For (f). xn+1 = xn − ff′(x
= −2xn by calculating. Thus, xn = (−2)n−1 x1
(xn )
for all n, thus {xn } does not converges for any choice of x1 , and we cannot
find ξ such that f (ξ) = 0 in this case. 64
Exercise 5.26. Suppose A > 0. (If not, then f = 0 on [a, b] clearly.) Fix
x0 ∈ [a, b], let M0 = sup |f (x)|, M1 = sup |f ′ (x)| for a ≤ x ≤ x0 . For any such
x, f (x)−f (a) = f ′ (c)(x−a) where c is between x and a by using Mean Value
Theorem. Thus |f (x)| ≤ M1 (x − a) ≤ M1 (x0 − a) ≤ A(x0 − a)M0 . Hence
1
M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0 ] by taking x0 = a + 2A
.
1
Repeat the above argument by replacing a with x0 , and note that 2A is a
constant. Hence, f = 0 on [a, b]. Exercise 5.27. Suppose y1 and y2 are solutions of that problem. Since
|φ(x, y2 ) − φ(x, y1 )| ≤ A|y2 − y1 |, y(a) = c, y1′ = φ(x, y1 ), and y2′ = φ(x, y2 ),
by Exercise 26 we know that y1 − y2 = 0, y1 = y2 . Hence, such a problem
has at most one solution.
Note: Suppose there is initial-value problem
y ′ = y 1/2 with y(0) = 0.
If y 1/2 6= 0, then y 1/2 dy = dx. By integrating each side and noting that
y(0) = 0, we know that f (x) = x2 /4. With y 1/2 = 0, or y = 0. All solutions
of that problem are
f (x) = 0 and f (x) = x2 /4.
Why the uniqueness theorem does not hold for this problem? One reason is
that there does not exist a constant A satisfying
|y1′ − y2′ | ≤ A|y1 − y2 |
if y1 and y2 are solutions of that problem. (since 2/x → ∞ as x → 0 and
thus A does not exist.) Theorem. Let φj (j = 1, · · · , k) be real functions defined on a rectangle Rj
in the plane given by a ≤ x ≤ b, αj ≤ yj ≤ βj .
A solution of the initial-value problem
yj′ = φ(x, yj ), yj (a) = cj (αj ≤ cj ≤ βj )
is, by definition, a differentiable function fj on [a, b] such that fj (a) = cj ,
αj ≤ fj (x) ≤ βj , and
fj′ (x) = φj (x, fj (x)). (a ≤ x ≤ b)
65
Then this problem has at most one solution if there is a constant A such that
|φj (x, yj2 ) − φj (x, yj1 )| ≤ A|yj2 − yj1 |
whenever (x, yj1 ) ∈ Rj and (x, yj2 ) ∈ Rj .
Proof. Suppose y1 and y2 are solutions of that problem. For each components of y1 and y2 , say y1j and y2j respectively, y1j = y2j by using Exercise
26. Thus, y1 = y2 Exercise 5.29. Theorem. Let Rj be a rectangle in the plain, given by
a ≤ x ≤ b, min yj ≤ yj ≤ max yj . (since yj is continuous on the compact
set, say [a, b], we know that yj attains minimal and maximal.) If there is a
constant A such that
(
|yj+1,1 − yj+1,2| ≤ A|yj,1 − yj,2|
(j < k)
Pk
| j=1 gj (x)(yj,1 − yj,2 )| ≤ A|yk,1 − yk,2|
whenever (x, yj,1) ∈ Rj and (x, yj,2) ∈ Rj .
Proof. Since the system y1′ , · · · , yk′ with initial conditions satisfies a fact
that there is a constant A such that |y1′ − y2′ | ≤ A|y1 − y2 |, that system has
at most one solution. Hence,
y (k) + gk (x)y (k−1) + · · · + g2 (x)y ′ + g1 (x)y = f (x),
with initial conditions has at most one solution. Exercise 6.1. Note that L(P, f, α) = 0 for all partition P of [a, b]. Thus
Z a
f dα = 0.
b
Take a partition P such that
1
k
n−1
P = a, a + (b − a), · · · , a + (b − a), · · · , a +
(b − a), b
n
n
n
for all positive integer n > 1. Thus
U(P, f, α) =
n
X
i=1
Mi ∆αi ≤
66
2(b − a)
n
for all positive integer n > 1. Thus
0 ≤ inf U(P, f, α) ≤
2(b − a)
n
for all positive integer n. Thus inf U(P, f, α) = 0. Hence
Ra
f dα = 0. b
Ra
b
f dα = 0; thus,
Exercise 6.2. If not, then there is p ∈ [a, b] such that f (p) > 0. Since
f is continuous at x = p, for ǫ = f (p)/2, there exist δ > 0 such that
T
|f (x) − f (p)| < ǫ whenever x ∈ (x − δ, x + δ) [a, b], that is,
1
3
0 < f (p) < f (x) < f (p)
2
2
for x ∈ Br (p) ⊂ [a, b] where r is small enough. Next, consider a partition P
of [a, b] such that
r
r
P = a, p − , p + , b .
2
2
So
1
rf (p)
L(P, f ) ≥ r · f (p) =
,
2
2
or
rf (p)
sup L(P, f ) ≥ L(P, f ) ≥
> 0.
2
Rb
It is absurd since a f (x)dx = sup L(P, f ) = 0. Hence f = 0 for all x ∈ [a, b].
Note: The above conclusion holds under the condition that f is continuous. If f is not necessary continuous, then we cannot get this conclusion. (A
counter-example is shown in Exercise 6.1).
Exercise 6.4. If f (x) = 0 for all irrational x, f (x) = 1 for all rational x,
prove that f ∈
/ R on [a, b] for any a < b.
Proof. Take any partition P of [a, b], say
a = x0 ≤ x1 ≤ · · · ≤ xn−1 ≤ xn = b.
By P we can construct the new partition P ′ without repeated points, and
U(P, f ) = U(P ′ f ), L(P, f ) = L(P ′ , f ). Say P ′
a = y0 < y1 < · · · < ym−1 < ym = b.
67
Note that Q is dense in R, and R \ Q is also dense in R. Hence,
Mi =
mi =
sup
f (x) = 1,
yi−1 ≤x≤yi
inf
yi−1 ≤x≤yi
f (x) = 0.
Hence,
′
U(P, f ) = U(P , f ) =
L(P, f ) = L(P ′ , f ) =
m
X
i=1
m
X
Mi ∆yi = b − a,
mi ∆yi = 0.
i=1
for any partition of [a, b]. So,
Z
Z
b
a
b
f dx = inf U(P, f ) = b − a,
f dx = sup L(P, f ) = 0.
a
Thus, f ∈
/ R on [a, b]. Exercise 6.5. The first answer is NO. Define f (x) = −1 for all irrational
x ∈ [a, b], f (x) = 1 for all rational x ∈ [a, b]. Similarly, f ∈
/ R by Exercise
6.4.
However, if we assume that f 3 ∈ R, the answer is YES. Let Φ = x1/3 ,
we apply Theorem 6.11 and thus get f ∈ R. Exercise 6.7. (a) Take any c > 0, and note that f is bounded, say |f | ≤ M
for some M. Hence,
Z c
Z c
Z c
−Mdx ≤
f (x)dx ≤
Mdx,
0
0
0
Z c
−cM ≤
f (x)dx ≤ cM,
0
Z 1
Z 1
−cM ≤
f (x)dx −
f (x)dx ≤ cM.
c
0
68
R1
R1
R1
Letting c → 0 and thus c f (x)dx − 0 f (x)dx → 0, that is, c f (x)dx →
R1
f (x)dx as c → 0. Thus this definition of the integral agrees with the old
0
one.
(b) Define
f (x) = n(−1)n , where
or
f (x) =
1
1
<x≤
n+1
n
sin(1/x)
.
x
Exercise 6.8. Since f is nonnegative and decreases monotonically on [1, ∞),
we have for n ≥ 2,
k
X
n=2
f (n) ≤
Z
k
1
We define
g(x) =
f (x)dx ≤
Z
k−1
X
f (n).
n=1
y
f (t)dt.
1
If this integral converges, then g(x) is a non-decreasing function which tends
to a limit, and so g(k) is a bounded non-decreasing sequence. Thus denote
P
sk = kn=2 f (n), we see that sk ≤ g(k), and sk tends to a limit. The fact that
R∞
P∞
f
(n)
converges
if
f (x)dx converges is now established. If the inten=1
1
gral diverges, then g(x) → +∞ as x → +∞. Therefore g(k) → +∞. Since
R∞
P
g(k) ≤ k−1
f
(n),
we
conclude
that
the
series
diverges.
Hence
f (x)dx
n=1
1
R∞
P∞
P
∞
converges if n=1 f (n). Thus, 1 f (x)dx converges if and only if n=1 f (n)
converges. Exercise 6.9. Integration by parts gives us
x=M Z M
Z M
sin x cos x
cos x
dx =
+
dx
1+x
1 + x x=0
(1 + x)2
0
0
Z M
sin M
sin x
=
+
dx
1+M
(1 + x)2
0
69
for M > 0. By letting M → ∞, we have
Z ∞
Z ∞
cos x
sin x
dx =
dx.
1+x
(1 + x)2
0
0
R∞ x
R∞
dx does not converges abolutely, but 0
Also, 0 cos
1+x
sin x
dx
(1+x)2
does.
Exercise 6.10. We prove (a). Claim: if a and b are positive and 0 < α < 1,
then
aα b1−α ≤ αa + (1 − α)b
and that the equality holds if and only if a = b. To prove it, we let f (x) =
αx + (1 − α) − xα , where x = a/b. Thus
f ′ (x) = α − αxα−1 = α(1 − xα−1 ).
Hence f (x) attains its minimum at x = 1 (, a = b,) and f (1) = 0. Hence
f (x) = αx + (1 − α) − xα ≥ 0. Put x = a/b,
α
a
a
α
− (1 − α) −
≥0
b
b
or
aα b1−α ≤ αa + (1 − α)b.
Next, let α = 1/p, a = u1/α , and b = v 1/(1−α) . Hence
uv ≤
up v q
+ .
p
q
Equality holds if and only if a = b, that is, up = v q .
We prove (b). If 0 ≤ f ∈ R(α) and 0 ≤ g ∈ R(α) then f p and g q ∈ R(α)
by Theorem 6.11. It also follows that f g ∈ R(α) and using the part (a):
Z b
Z
Z
1 b p
1 b q
f gdα ≤
f dα +
g dα = 1.
p a
q a
a
We prove (c). If f and g are complex-valued in R(α) then |f | and |g| are
nonnegative elements of R(α) and f g ∈ R(α). Moreover
Z b
Z b
f gdα ≤
|f ||g|dα.
a
a
70
If I =
Rb
a
|f |p 6= 0 and J =
Rb
a
|g|q 6= 0 then apply the conclusion of the part
(b) to |f |/c and |g|/d where cp = I and dq = J. Hence
Z b
Z b
1/p Z b
1/q
p
q
f gdα ≤ cd =
|f | dα
|g| dα
a
a
a
If I = 0 (J = 0 is similar), we can reverse the roles of p and q, then
Z
b
|f |(c|g|)dα ≤ c
a
q1
p
Z
b
a
|g|q dα
for any c > 0. Since c is arbitrary,
Z b
|f ||g|dα = 0.
a
Hence the inequality still holds.
Exercise 6.11. Put p = q = 2 in Exercise 4.10, we have
2 Z b
Z b
Z b
2
|u| dα +
|v|2 dα.
|uv|dα ≤
a
a
(*)
a
Hence,
ku +
vk22
Z
b
|u + v|2dα
a
Z b
Z b
Z b
2
=
|u| dα +
(uv + vu)dα +
|v|2dα
a
a
a
Z b
Z b
Z b
=
|u|2 dα +
|uv + vu|dα +
|v|2dα
a
a
a
Z b
Z b
Z b
≤
|u|2dα + 2
|uv|dα +
|v|2 dα
=
a
≤
Z
a
b
2
|u| dα
a
1/2
= (kuk2 + kvk2 )2 .
(by (*))
a
+
Z
b
2
a
|v| dα
1/2 2
Therefore, ku + vk2 ≤ kuk2 + kvk2 . Putting u = f − g and v = h − g gives
kf − hk2 ≤ kf − gk2 + kg − hk2 .
71
Exercise 6.13. (a) Put t2 = u,
Z x+1
f (x) =
sin(t2 )dt
x
Z
(x+1)2
sin u
du
2u1/2
x2
u=(x+1)2 Z
− cos u −
=
2u1/2 2
=
(x+1)2
cos u
du
4u3/2
x2
u=x
Z (x+1)2
cos(x2 ) cos[(x + 1)2 ]
cos u
=
−
−
du.
2x
2(x + 1)
4u3/2
x2
To get a bound of
Z
R (x+1)2
x2
(x+1)2
x2
cos u
du,
4u3/2
we replace cos u by −1:
Z
2
cos u (x+1) −1
1
du < du =
3/2
3/2
4u
4u
2x(x + 1)
x2
for x > 0. Hence, for x > 0,
Z
cos(x2 ) cos[(x + 1)2 ] (x+1)2 cos u +
+
|f (x)| ≤ du
2x 2(x + 1) x2
4u3/2 1
1
1
1
<
+
+
= .
2x 2(x + 1) 2x(x + 1)
x
(b) By (a),
cos(x2 ) cos[(x + 1)2 ]
−
−
f (x) =
2x
2(x + 1)
Z
(x+1)2
x2
cos u
du,
4u3/2
1
2xf (x) = cos(x2 ) − cos[(x + 1)2 ] +
cos[(x + 1)2 ]
x+1
Z (x+1)2
cos u
− 2x
du.
4u3/2
x2
Let
1
r(x) =
cos[(x + 1)2 ] − 2x
x+1
Note that
Z
(x+1)2
x2
cos u
du.
4u3/2
1
1
1
2
x + 1 cos[(x + 1) ] ≤ x + 1 < x ,
72
2x
Z
(x+1)2
x2
cos u
1
1
du < 2x
<
3/2
4u
2x(x + 1)
x
if x > 0. Thus |r(x)| < 1/x + 1/x = c/x and c = 2 is a constant if x > 0.
Exercise 7.1. Prove that every uniformly convergent sequence of bounded
functions is uniformly bounded.
Proof. Let fn → f uniformly on S in a metric space. fn is bounded for all
n. Given ǫ = 1, there is an integer N such that |fn (x) − fm (x)| < 1 whenever
n, m ≥ N and x ∈ S. Put m = N, and we have
|fn (x)| ≤ |fN (x)| + 1
whenever n ≥ N and x ∈ S. Let
M = max sup |f1 (x)|, · · · , sup |fN −1 (x)|, sup |fN (x)| + 1 .
x∈S
x∈S
x∈S
(fn is bounded ⇒ the existence of supx∈S |fn (x)|.) Therefore |fn (x)| ≤ M
for all n and x ∈ S. Hence {fn } is uniformly bounded on S. Exercise 7.2. If {fn } and {gn } converge uniformly on a set E, prove that
{fn + gn } converges uniformly on E. If, in addition, {fn } and {gn } are
sequences of bounded functions, prove that {fn gn } converges uniformly on
E.
Proof. Let hn (x) = fn (x) + gn (x) for each n. Since {fn } (resp. {gn })
converges uniformly on E, given ǫ > 0, there exists N such that |fn (x) −
fm (x)| < ǫ/2 (resp. |gn (x) − gm (x)| < ǫ/2) for all x ∈ E whenever n, m ≥ N.
Thus |hn (x) − hm (x)| ≤ |fn (x) − fm (x)| + |gn (x) − gm (x)| < ǫ for all x ∈ E,
i.e., {fn + gn } converges uniformly on E.
By exercise 7.1, {fn } and {gn } are uniformly bounded on E. Let M be the
bound of them, i.e., |fn (x)| ≤ M and |gn (x)| ≤ M for all x and all n. Again
we let hn (x) = fn (x)gn (x) for each n. Since {fn } (resp. {gn }) converges
uniformly on E, given ǫ > 0, there exists N such that |fn (x) − fm (x)| <
ǫ/(4M) (resp. |gn (x) − gm (x)| < ǫ/(4M)) for all x ∈ E whenever n, m ≥ N.
73
Thus,
|hn (x) − hm (x)| = |fn (x)(gn (x) − gm (x)) + gm (x)(fn (x) − fm (x))|
≤ |fn (x)||gn (x) − gm (x)| + |gm (x)||fn (x) − fm (x)|
< ǫ
whenever n, m ≥ N and x ∈ E, i.e, {fn gn } converges uniformly on E. Exercise 7.3. Construct sequences {fn }, {gn } which converge uniformly on
some set E, but such that {fn gn } does not converge uniformly on E. (of
course, {fn gn } must converge on E).
Proof. Let E = [1, 2] ⊂ R. Let
1
fn (x) = x(1 + ),
n
(
gn (x) =
1
n
b+
1
n
if x ∈ R \ Q
if x =
a
b
with (a, b) = 1 and b > 0
on E. ({gn (x)} is a sequence of unbounded functions clearly.)
Claim 1. {fn } and {gn } converge uniformly on E.
lim fn (x) = f (x) = x
(
n→∞
lim gn (x) = g(x) =
n→∞
0
b
if x ∈ R \ Q
if x =
a
b
∈ Q with (a, b) = 1 and b > 0.
Given ǫ > 0, there exists N = [ 2ǫ ] + 1 > 2ǫ such that |fn (x) − f (x)| =
| nx | ≤ n2 ≤ N2 < ǫ whenever n ≥ N, x ∈ E. Therefore, fn → f uniformly on
E. Next, given ǫ > 0, there exists N = [ 1ǫ ] + 1 > 1ǫ such that |gn (x) − g(x)| ≤
1
≤ N1 < ǫ whenever n ≥ N, x ∈ E. Hence, gn → g uniformly on E.
n
Claim 2. {fn gn } does not converge uniformly on E.
lim fn (x)gn (x) = f (x)g(x)
(
0 if x ∈ R \ Q
=
a if x = ab ∈ Q with (a, b) = 1 and b > 0.
n→∞
It suffices to show that {fn gn } does not converge uniformly on F = E ∩ Q.
If not, given ǫ = 1 and consequently there exists N such that |fn (x)gn (x) −
74
f (x)g(x)| < 1 whenever n ≥ N and x ∈ F . Now we let x =
n = N. Hence,
N +1
N
∈ F,
1 > |fn (x)gn (x) − f (x)g(x)|
N + 1
1
1
=
1+
N+
− (N + 1)
N
N
N
3
2
N + 2N + 2N + 1 = N3
> 1,
and thus we reach a contradiction. Exercise 7.4. Consider
f (x) =
∞
X
n=1
1
.
1 + n2 x
For what values of x does the series converge absolutely? On what intervals does it converge uniformly? On what intervals does it fail to converge
uniformly? Is f continuous whenever the series converges? Is f bounded?
Pn=m 1
Proof. Let sm (x) = n=1
be the partial sum of f .
1+n2 x
First, we will find values of x such that the series converge absoP
lutely. x ∈
/ {− k12 | k ∈ N} clearly. When x = 0, the series
1 diverges.
P
P 1
P −2
2 −1
−1
−1
When x > 0, the series (1+n x) = x
<x
n converges
x−1 +n2
(absolutely) by the comparison test. To complete our calculation, we consider
P
P 1
the rest case x ∈ R− \ {− k12 | k ∈ N}. Recall: (1 + n2 x)−1 = x−1
.
x−1 +n2
−1
2
For each fixed x, there exists N such that x + n > 0 whenever n ≥ N.
Hence, it suffices to determine whether the series
∞
X
n=N
∞
X
1
1
−1
=x
2
−1
1+n x
x + n2
n=N
converges absolutely. It is similar to the case x > 0, and consequently the
new series from N to infinity converges absolutely too. Therefore, the series
P
(1 + n2 x)−1 converges absolutely for all real x except 0 ∪ {− k12 | k ∈ N}.
Second, we will find intervals such that the series converge uniformly or does not converge uniformly. By the above deduction and
75
Weierstrass M-test (theorem 7.10), f (x) converges uniformly on [A, +∞)
with A > 0. Does f converge uniformly on (0, +∞)? The answer is no. If
not, sm → f uniformly on (0, +∞).
sm (x) =
n=m
X
n=1
n=m
X
1
<
1=m
1 + n2 x
n=1
is bounded (by m) for all x ∈ (0, +∞). By exercise 7.1, f (x) is uniformly
bounded. Since
−3
sm (m ) =
n=m
X
n=1
=
1
1 + n2 · m−3
>
n=m
X
n=1
2
1
1 + m2 · m−3
m
m
=
>m−1
2
−3
1+m ·m
1+m
for each m, we obtain a contradiction. It also implies that f is unbounded.
Next, we consider other possible intervals that f converge uniformly on. We
list two possible cases: (i) (−∞, −1) and (ii) Ek = (−k −2 , −(k + 1)−2 ).
Case (i). Note that 0 > (1 + n2 x)−1 > (1 − n2 )−1 for every x ∈ (−∞, −1).
Thus,
∞ ∞ X
1 X
1 0≤
1 + n2 x <
1 − n2 .
n=2
n=2
P∞
By Weierstrass M-test, n=2 (1 + n2 x)−1 (and so f ) converges uniformly on
(−∞, −1).
Case (ii). If x ∈ Ek (or −k −2 < x < −(k + 1)−2 ), then
1 1
<
1 + n2 x 1 − n2 /k 2 P
2 −1
for n ≥ k + 2. Anain by Weierstrass M-test, ∞
(and so
n=k+1 (1 + n x)
f ) converges uniformly on Ek . In sum, f converges uniformly on all real
numbers except (0, A) ∪ {−k −2 | k ∈ N}), and does not converges uniformly
on (0, A) where A > 0.
Third, we will show that f is continuous whenever f converges. It
is clear. For any x ∈ R \ (0 ∪ {−k −2 | k ∈ N}), x must lie in some closed
interval [a, b] that f converges uniformly on. Since sm are continuous for all
m, f is continuous whenever f converges. 76
Exercise 7.5. Let


 0
fn (x) =
sin2


0
x<
π
x
1
n+1
1
≤x
n+1
1
<x .
n
,
≤
1
n
,
Show that {fn } converges to a continuous function, but not uniformly. Use
P
the series
fn to show that absolute convergence, even for all x, does not
imply uniform convergence.
Proof. By the definition of fn (x), it suffices to consider that fn (x) lie on
(0, 1) only. We will show that the limit function
lim fn (x) = 0
n→∞
and thus {fn } converges to a continuous function 0. To show this, we fix
any x ∈ (0, 1). There exists N such that x > n1 whenever n ≥ N. Hence
limn→∞ fn (x) = 0. However, {fn } does not converge uniformly. If not, then
given ǫ = 1 > 0 and consequently there is N such that |fn (x) − 0| < 1
1
whenever n ≥ N and x ∈ R. In particular, |fN ( N +1/2
)| = 1 < 1, which is
absurd.
P
Pn=m
Let s(x) =
fn (x) and sm (x) =
n=1 fn (x) be the partial sums of
P
P
fn (x). To show the absolute absolute convergence of s(x) =
fn (x), we
fix x ∈ (0, 1) first. Thus, there exists an unique N such that N 1+1 < x ≤ N1 .
Hence,
π
,
x
and consequently s(x) converges absolutely on (0, 1) (and on R).
X
|fn (x)| = s(x) = fN (x) = sin2
To show that {sn } is not uniformly convergent, we suppose {sn } is uniformly convergent and get a contradiction. By our assumption, sn → s
uniformly on R. Given ǫ = 1 > 0, there is N such that |sn (x) − s(x)| < 1
whenever n ≥ N and x ∈ R. Hence, for n = N and x =
sN
1
2N +
1
2
1
−s
2N +
which is absurd. 77
1
,
2N + 12
1 = |0 − 1| = 1,
2
Exercise 7.6. Prove that the series
∞
X
(−1)n
n=1
x2 + n
n2
converges uniformly in every bounded interval, but does not converge absolutely for any value of x.
Proof. Let
x2 + n
fn (x) = (−1)n , gn (x) =
.
n2
P
Fixed a bounded interval [a, b]. Note: (i)
fn (x) is uniformly bouned, (ii)
gn (x) → 0 uniformly on [a, b], and (iii) gn+1(x) ≤ gn (x) on [a, b]. Hence,
P
2
by Dirichlet test for uniform convergences, the series (−1)n x n+n
converges
2
uniformly on every bounded interval [a, b].
Next, notice that
∞
X
n=1
∞
|(−1)
∞
X x2 + n X 1
+n
|=
>
n2
n2
n
n=1
n=1
2
nx
diverges for any x ∈ R, and thus
for any value of x. P
2
(−1)n x n+n
does not converge absolutely
2
Exercise 7.7. For n = 1, 2, 3, · · · , x real. Put
fn (x) =
x
.
1 + nx2
Show that {fn } converges uniformly to a function f , and that the equation
f ′ (x) = lim fn′ (x)
n→∞
is correct if x 6= 0, but false if x = 0.
Proof. For a fixed x ∈ R, limn→∞ fn (x) = 0. Also,
x ≤ √2
|fn (x) − f (x)| = 2
1 + nx n
for all x ∈ R. The last inequality is due to the inequality for arithmetic and
geometric means. Thus, fn → f uniformly on R.
78
2
1−nx
′
′
f ′ (x) = 0 and fn′ (x) = (1+nx
2 )2 . For x = 0, f (0) = 0 and limn→∞ fn (0) =
limn→∞ 1 = 1. For x 6= 0, f ′ (x) = 0 and
−nx2 + 1
= 0,
n→∞ n2 x4 + 2nx2 + 1
lim fn′ (x) = lim
n→∞
i.e., f ′ (x) = limn→∞ fn′ (x) if x 6= 0. Exercise 7.8. If
I(x) =
(
0 (x ≤ 0),
1 (x > 0),
if {xn } is a sequence of distinct points of (a, b), and if
that the series
∞
X
f (x) =
cn I(x − xn )
P
|cn | converges, prove
n=1
converges uniformly, and that f is continuous for every x 6= xn .
P
Proof. Note that |cn | = |cn I(x − xn )| and
|cn | converges. By Weierstrass
M-test, f (x) converges uniformly on (a, b).
Next, let
sm (x) =
n=m
X
n=1
for each m.
cn I(x − xn ) = c1 I(x − x1 ) + · · · cm I(x − xm )
sm (x) are continuous on (a, b) \ {x1 , · · · , xm } and thus on
(a, b) \ {xk | k ∈ N}. Hence, f is continuous on (a, b) \ {xk | k ∈ N}.
Exercise 7.9. Let {fn } be a sequence of continuous functions which converges uniformly to a function f on a set E. Prove that
lim fn (xn ) = f (x),
n→∞
for every sequence of points xn ∈ E such that xn → x, and x ∈ E. Is the
converse of this true?
Proof. Given ǫ > 0. By our assumption, there exists N1 such that
|fn (x) − f (x)| <
79
ǫ
2
whenever n ≥ N1 and all x ∈ E. Also, since f is continuous on E by theorem
7.12, limn→∞ f (xn ) = f (x), that is, there exists N2 such that
|f (xn ) − f (x)| <
ǫ
2
whenever n ≥ N2 . Now, if n ≥ N = max{N1 , N2 }, we have
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| <
ǫ
ǫ
+ = ǫ,
2 2
i.e.,
lim fn (xn ) = f (x).
n→∞
The converse is not true. Here is our example:
fn (x) = xn
(0 < x < 1).
limn→∞ fn (x) = 0 but {fn } is not convergent uniformly. Exercise 7.16. Suppose that {fn } is an equicontinuous sequence of functions
on a compact set K, and {fn } converges pointwise on K. Prove that {fn }
converges uniformly on K.
Proof. First, by the hypothesis that {fn } is equicontinuous on K, given
ǫ > 0 and thus there exists δ > 0 such that
|fn (x) − fn (y)| <
ǫ
3
(*)
for all n whenever d(x, y) < δ with x, y ∈ K. For such δ > 0, we form an
open covering {Ui } of K where Ui = {x ∈ K | d(x, xi ) < δ} for each xi ∈ K.
By the compactness of K, there are finitely many indices 1, · · · , r such that
{U1 , · · · , Ur } also covers K.
Second, since {fn } converges pointwise on K, for a fixed x ∈ K and given
ǫ > 0, there is N such that
|fn (x) − fm (x)| <
whenever n, m ≥ N.
80
ǫ
3
(**)
Back to the proof. Now we fix any x ∈ K. there exists Ui such that
x ∈ Ui , i.e., d(x, xi ) < δ. Hence,
ǫ
(by (*)) and
3
ǫ
|fn (xi ) − fm (xi )| < (by (**)) and
3
ǫ
|fm (xi ) − fm (x)| < (by (*)).
3
|fn (x) − fn (xi )| <
Three inequalities implies that
|fn (x) − fm (x)| <
ǫ
ǫ ǫ
+ + =ǫ
3 3 3
for all x ∈ K. Therefore, {fn } converges uniformly on K by Cauchy criterion. Exercise 8.1. Define
f (x) =
(
e−1/x
0
2
(x 6= 0),
(x = 0).
Prove that f has derivatives of all orders at x = 0, and that f (n) (0) = 0 for
n = 1, 2, 3, · · · .
Proof. Induction.
2
e−1/x − 0
1/x
y
f (0) = lim
= lim 1/x2 = lim y2 = 0
x→0
x→0 e
y→±∞ e
x−0
′
where y = 1/x. By induction hypothesis, f (n−1) (0) = 0. Clearly,
1 − 12
(n−1)
f
(x) = Pn−1
e x
x
where Pn−1 ( x1 ) is a polynomial of an indeterminate variable x1 . Hence,
2
f
(n)
Pn−1 (1/x)e−1/x − 0
yPn−1(y)
(0) = lim
= lim
=0
x→0
y→±∞
x−0
ey2
where y = 1/x. 81
Exercise 8.2. For a fixed i,
X
X
X
X
aij =
aij +
aij +
aij
j
j<i
=
X
j=i
2
j−i
j<i
j>i
− 1 = (1 − 21−i ) − 1 = −21−i .
For a fixed j,
X
aij =
i
Hence
P P
i
j
aij =
P
X
aij +
i<j
1−i
aij +
i=j
= −1 +
i (−2
X
X
2
) = −2 and
aij
i>j
j−i
i>j
X
= −1 + 1 = 0.
P P
j
i
aij =
P
j
0 = 0. Exercise 8.4. (a) By L’Hospital’s rule,
bx − 1
log(b)bx
= lim
= log b
x→0
x→0
x
1
lim
if b > 0.
(b) By L’Hospital’s rule,
log(1 + x)
(1 + x)−1
= lim
= 1.
x→0
x→0
x
1
lim
(c) By part (b),
1
log(1 + x)
log(1 + x)
e = exp lim
= lim exp
= lim (1 + x) x .
x→0
x→0
x→0
x
x
(d) By part (c),
x
n
1
x
x
x
y
y
e = lim (1 + y)
= lim (1 + y) = lim 1 +
.
y→0
y→0
n→∞
n
Exercise 8.7. Let f (x) =
sin x
x
for all x > 0. Thus f ′ (x) =
′
x cos x−sin x
.
x2
′
Let
g(x) = x cos x − sin x for all real x. Thus g (x) = −x sin x. g (x) < 0 on
(0, π/2), and g(x) is decreasing strictly on (0, π/2). Also, g(π/2) = −1 < 0,
82
and g(x) < 0 for all 0 < x < π/2. f ′ (x) < 1 on (0, π/2), and thus f (x) is
decreasing strictly on (0, π/2).
Note that limx→0 sinx x = 1 and f (π/2) = 2/π. Hence
is,
2
π
< f (x) < 1, that
sin x
2
<
< 1.
π
x
Exercise 8.8. For n = 0, 1, we have | sin 0x| ≤ 0| sin x| and | sin x| ≤ | sin x|.
Assume that | sin kx| ≤ k| sin x|. We have
| sin(k + 1)x| = | sin(kx + x)| = | sin kx cos x + sin x cos kx|
≤ | sin kx|| cos x| + | sin x|| cos kx| ≤ k| sin x| + | sin x|
= (k + 1)| sin x|.
Exercise 8.9. (a) Let γn = sn − log n = 1 + 21 + · · · +
1
n
− log n. Thus
1
− log(n + 1) + log n
n+1
Z n+1
1
1
−
dx
=
n+1
x
n
Z n+1 1
1
=
−
dx
n+1 x
n
≤ 0
γn+1 − γn =
for all n ≥ 0. Hence {γn } is a decreasing sequence. Also,
γn =
n
X
1
k=1
k
1
=
+
n
≥ 0,
−
n−1 Z
X
k=1
n−1 Z
X
k=1
k
k+1
k
k+1
1
dx
x
1 1
−
dx
k x
that is, {γn } is bounded. Therefore, γ = limn→∞ γn exists.
83
(b) Note that sN − log N ≥ 0 for all N, and thus we can choose m >
100/ log 10. Exercise 8.10. Given N, let p1 , · · · , pk be those primes that divide at least
one integer ≤ N. Then
Y
−1
X
N
k k k
X
1 Y
1
1
1
2
≤
1+ + 2 +··· =
1−
≤ exp
.
n j=1
pj pj
pj
p
n=1
j=1
j=1 j
The last inequality holds because (1 − x)−1 ≤ e2x for 0 ≤ x ≤ 12 . Now letting
P
P 2
P∞ 1
1
N → ∞, and thus ∞
≤
exp
n=1 n
p p . Since the left hand side,
n=1 n ,
P
diverges, p 1/p is divergent. Exercise 8.11. Given ǫ > 0. Let
Z ∞
Z
−tx
g(t) = t
e f (x)dx − 1 = t
0
0
∞
e−tx [f (x) − 1]dx
on [0, 1]. Since f (x) → 1 as x → +∞, there exists M > 0 such that
|f (x) − 1| < ǫ/2 whenever x ≥ M. Thus
Z ∞
|g(t)| ≤ t
e−tx |f (x) − 1|dx
0
Z M
Z ∞
−tx
= t
e |f (x) − 1|dx + t
e−tx |f (x) − 1|dx
0
M
Z M
Z ∞
ǫ
≤ t
e−tx Ldx + t
e−tx dx
2
0
M
Z M
ǫ
≤ t
Ldx + e−tM
2
0
ǫ
≤ tML +
2
where L > 0 is a bound of |f (x) − 1| on [0, M]. (f ∈ R([0, M]) and thus f
is bounded.) Choose t = min(1, 2Mǫ L ) and thus |g(t)| < ǫ, i.e.,
Z ∞
lim t
e−tx f (x)dx = 1.
t→0+
0
84
Exercise 8.12. (a) c0 = δ/π. For m 6= 0,
Z π
Z δ
1
1
−imx
cm =
f (x)e
dx =
e−imx dx
2π −π
2π −δ
x=δ
sin(mδ)
1 e−imx =
·
=
.
2π −im
mπ
x=−δ
(b) Thus,
∞
X
sin(nδ)
δ
f (x) ∼ + 2
cos(nx).
π
nπ
n=1
For x = 0, we can apply theorem 8.14 to get
∞
X
sin(nδ)
δ
f (0) = + 2
,
π
nπ
n=1
that is,
∞
X
sin(nδ)
n
n=1
=
π−δ
.
2
(c) By Parseval’s theorem,
Z π
∞
∞
X
X
1
δ2
sin2 (nδ)
2
2
|cn | = 2 + 2
.
|f (x)| dx =
2π −π
π
n2 π 2
−∞
n=1
1
2π
Hence
Z
π
1
|f (x)| dx =
2π
−π
2
∞
X
sin2 (nδ)
n=1
(d)
Z
∞
0
sin x
x
2
δ→0
π
2
=
δ
12 dx =
−δ
δ
.
π
π−δ
.
2
2
∞ ∞
X
sin2 Nn
1 X sin Nn
dx = lim
= lim
n
n2
N →∞ N
N →∞
N
N
n=1
n=1
= lim
(e) Put δ =
n2 δ
Z
∞
X
sin2 (nδ)
n2 δ
n=1
π−δ
π
= .
δ→0
2
2
= lim
in (c), we have
∞
X
sin2 ( nπ )
2
n=1
n2 π
2
85
=
π − π2
.
2
that is,
∞
∞
)
π
2 X sin2 ( nπ
2X
1
2
=
=
,
2
4
π n=1
n
π n=1 (2n − 1)2
∞
X
n=1
Exercise 8.13. c0 =
1
2π
R 2π
0
Z
1
π2
=
.
(2n − 1)2
8
xdx = π. For m 6= 0,
2π
1
xe−imx dx
2π 0
x=2π
1
1
1
1 −imx −imx
=
·
xe
+
·
e
2π −im
2π m2
x=0
1
=
.
−im
By Parseval’s theorem,
Z 2π
∞
X
1
2
|f (x)| dx =
|cn |2 .
2π 0
−∞
cm =
Thus,
or
∞
X
1
4π 2
= π2 + 2
3
n2
n=1
∞
X
1
π2
=
.
n2
6
n=1
Rπ
R π 2 −imx
1
1
−imx
Exercise 8.14. We deduce 2π
|x|e
dx
and
xe
dx with
−π
2π −π
m 6= 0 for further computing.
Z
Z
1
1 −imx
−imx
−imx
xe
dx =
xe
−
e
dx
−im
−im
1
1
xe−imx + 2 e−imx + constant;
=
−im
m
Z
Z
1
2
2 −imx
2 −imx
xe
dx =
xe
−
xe−imx dx
−im
−im
Z
1 2 −imx
2
=
xe
−
xe−imx dx.
−im
−im
86
Thus
Z
π
−π
−imx
|x|e
dx =
Z
0
−imx
−π
−xe
dx +
Z
π
xe−imx dx
0
−imx
0
−imx
π
xe
e−imx xe
e−imx =−
+
+
+
−im
m2 −π
−im
m2 0
(
π
π
+ −im
if m is even
im
=
π
π
(− m22 − im
) + (− m22 + im
) if m is odd
(
0
if m is even
=
− m42 if m is odd;
(
Z π
π
π
+ −im
if m is even
−im
xe−imx dx =
π
π
( 22 + im
) + (− m22 + im
) if m is odd
−π
( m
2π
− im
if m is even
=
2π
if m is odd
im
2π
= (−1)m+1 ,
im
and
Z
π
2 −imx
xe
−π
π
Z π
1 2 −imx 2
dx =
xe
xe−imx dx
− −im
−im
−π
−π
4π
= (−1)m 2 .
m
Hence,
Z π
1
π2
c0 =
(π − |x|)2 dx = ,
2π −π
3
Z π
1
cm =
(π − |x|)2 e−imx dx
2π −π
Z π
1
=
(π 2 − 2π|x| + x2 )e−imx dx
2π −π
Z π
Z π
1
−imx
=−
|x|e
dx +
x2 e−imx dx
2π −π
−π
(
0 + m22
if m is even
=
4
−2
+ m2 if m is odd
m2
2
= 2.
m
87
Therefore,
∞
π2 X 4
f (x) = (π − |x|) ∼
+
cos nx.
3
n2
n=1
2
Given δ = 1. Note that
|f (x + t) − f (x)| = (π − |x + t|)2 − (π − |x|)2 = 2π − |x + t| − |x| · |x + t| − |x|
≤ (2π + 1)|t|
for all t ∈ (−δ, δ) = (−1, 1). By theorem 8.14,
∞
π2 X 4
f (x) = (π − |x|) =
+
cos nx
3
n2
n=1
2
on [−π, π]. Let x = 0, π 2 =
π2
3
+
P∞
4
n=1 n2 ,
i.e.,
P∞
1
n=1 n2
=
π2
.
6
By Parseval’s theorem,
1
2π
1
2π
Z
Z
π
−π
|f (x)|2 dx =
∞
X
−∞
P∞
∞
π4 X 8
+
.
4
9
n
n=1
Z π
1
|f (x)| dx =
(π − |x|)4 dx
2π −π
−π
Z 0
Z π
1
1
4
=
(π + x) dx +
(π − x)4 dx
2π −π
2π 0
0
π
1 (π + x)5 1 −(π − x)5 =
+ 2π
2π
5
5
π
2
−π
π4
= .
5
Hence
|cn |2 =
1
n=1 n4
=
π4
.
90
0
Exercise 8.15. For KN (x) =
1
N +1
·
1−cos(N +1)x
.
1−cos x
Recall
sin(n + 12 )x
Dn (x) =
,
sin x2
88
or
x
1
sin Dn (x) = sin(n + )x.
2
2
Hence
N
sin
n
X
xX
1
Dn (x) =
sin(n + )x.
2 n=0
2
n=0
Multiply sin x2 on both sides,
N
N
X
x 2X
1
x
(sin )
Dn (x) =
sin(n + )x · sin
2 n=0
2
2
n=0
N
1X
=
(cos nx − cos(n + 1)x)
2 n=0
=
1
(1 − cos(N + 1)x).
2
Note that 2(sin x2 )2 = 1 − cos x, and thus
N
KN (x) =
1 X
1
1 − cos(N + 1)x
Dn (x) =
·
.
N + 1 n=0
N +1
1 − cos x
+1)x
(a) By KN (x) = N 1+1 · 1−cos(N
, and | cos x| ≤ 1, KN (x) ≥ 0.
1−cos x
(b) In Theorem 8.14, we know
Z π
1
Dn (x)dx = 1.
2π −π
for all 0 ≤ n ≤ N. Hence
1
2π
Z
π
1
KN (x)dx =
2π
−π
Z
π
N
1 X
Dn (x)dx
−π N + 1 n=0
Z π
N
1 X 1
=
Dn (x)dx
N + 1 n=0 2π −π
=
1
(N + 1) = 1.
N +1
(c) cos x < cos δ whenever 0 < δ ≤ |x| ≤ π. Thus
KN (x) =
1 − cos(N + 1)x
2
≤
.
(N + 1)(1 − cos x)
(N + 1)(1 − cos δ)
89
By equation (78) on page 189,
Hence
1
sN (f ; x) =
2π
Z
π
1
σN (f ; x) =
2π
Z
π
−π
−π
f (x − t)DN (t)dt.
f (x − t)KN (t)dt.
For Fejer’s theorem. By (a) and (b),
Z π
1
|σN (f ; x) − f (x)| = f (x − t)KN (t)dt − f (x)
2π −π
Z π
1
≤
|f (x − t) − f (x)|KN (t)dt.
2π −π
Given ǫ > 0, we choose π > δ > 0 such that |y−x| < δ implies |f (y)−f (x)| <
ǫ/2. Let M = sup |f (x)|. Hence,
Z −δ
Z −δ
1
1
2
|f (x − t) − f (x)|KN (t)dt ≤
2M ·
dt
2π −π
2π −π
(N + 1)(1 − cos δ)
4M(π − δ)
=
2π(N + 1)(1 − cos δ)
2M
<
,
(N + 1)(1 − cos δ)
Z π
1
2M
,
|f (x − t) − f (x)|KN (t)dt <
2π δ
(N + 1)(1 − cos δ)
and
1
2π
Z
Z δ
ǫ
|f (x − t) − f (x)|KN (t)dt ≤
KN (t)dt
4π −δ
−δ
Z π
ǫ
ǫ
≤
KN (t)dt = .
4π −π
2
δ
Therefore,
1
|σN (f ; x) − f (x)| ≤
2π
≤
Z
π
−π
|f (x − t) − f (x)|KN (t)dt
4M
ǫ
+
(N + 1)(1 − cos δ) 2
<ǫ
90
for all large enough N, which proves the conclusion that σN (f ; x) → f (x)
uniformly on [−π, π]. Exercise 8.17. (a) Suppose f increases monotonically on [−π, π]. By Exercise 17 of Chapter 6,
Z π
1
ncn =
f (x)ne−inx dx
2π −π
Z π
ie−inπ
ie−inπ
1
=
f (π) −
f (−π) −
ie−inx df
2π
2π
2π −π
for n 6= 0. Hence,
−inπ
Z π
ie
ie−inπ
1
−inx f (π) −
f (−π) −
ie
df |ncn | = 2π
2π
2π −π
−inπ
Z π
inπ
ie
ie
1
−inx
f (π) + f (−π) + ie
df ≤ 2π
2π
2π
−π
1
1
≤
(|f (π)| + |f (−π)|) +
(f (π) − f (−π))
2π
2π
are bounded for n 6= 0, that is, {ncn } is a bounded sequence.
Exercise 8.29. Prove that every continuous mapping f of D into D has a
fixed point in D. (This is the 2-dimensional case of Brouwer’s fixed-point
theorem.) Hint: Assume that f (z) 6= z for every z ∈ D. Associate to each
z ∈ D the point g(z) ∈ T which lies on the ray that starts at f (z) and passes
through z. Then g maps D into T , g(z) = z if z ∈ T , and g is continuous,
because
g(z) = z − s(z)[f (z) − z],
where s(z) is the unique nonnegative root of a certain quadratic equation
whose coefficients are continuous function of f and z. Apply Exercise 28.
Exercise 8.30. Use Stirling’s formula to prove that
Γ(x + c)
=1
x→∞ xc Γ(x)
lim
for every real constant c.
91
Proof. By Stirling’s formula,
p
x+c−1
)
2π(x + c − 1)
( x+c−1
Γ(x + c)
e
p
lim c
= lim
x−1
x→∞ x Γ(x)
x→∞
)x−1 2π(x − 1)
xc (
r e
x+c−1
1
x+c−1 x+c−1
= c lim
.
e x→∞
x−1
x−1
Note that
x+c−1
x−1 x+c−1
x−1
c
x+c−1
= lim
1+
= ec .
lim
x→∞
x→∞
x−1
x−1
q
= 1. Hence
and limx→∞ x+c−1
x−1
Γ(x + c)
=1
x→∞ xc Γ(x)
lim
for every real c. Exercise 8.31. In the proof of Theorem 7.26 it was shown that
Z 1
4
(1 − x2 )n dx ≥ √
3 n
−1
for n = 1, 2, 3, · · · . Use Theorem 8.20 and Exercise 30 to show the more
precise result
Z 1
√
√
lim n
(1 − x2 )n dx = π.
n→∞
−1
Proof. By change of variable, letting t = x2 ,
Z 1
Z 1
Γ( 1 )Γ(n + 1)
1 −1
2 n
t 2 (1 − t)n dt = 2
.
(1 − x ) dx =
2Γ(n + 23 )
0
0 2
Note that
R0
(1 − x2 )n dx =
−1
lim
n→∞
√
R1
0
(1 − x2 )n dx. Hence
√ 1
√
Γ(
n 2 )Γ(n + 1)
1
2 n
n
(1 − x ) dx =
=
Γ
=
π.
2
Γ(n + 32 )
−1
Z
1
92
Exercise 9.1. Given x = c1 x1 + · · ·+ ck xk , y = d1 y1 + · · ·+ dh yh ∈ span(S),
where xi , yj ∈ S and ci , dj are scalars. Thus x + y = c1 x1 + · · · + ck xk +
d1 y1 + · · · + dh yh is a linear combinations of elements of S. Also, for every
scalar c, cx = c · (c1 x1 + · · · + ck xk ) = (cc1 )x1 + · · · + (cck )xk again is a linear
combinations of elements of S. Thus span(S) is a vector space. Exercise 9.2. To show BA is linear, it suffices to show that (BA)(x1 +x2 ) =
(BA)(x1 ) + (BA)(x2 ) and (BA)(cx) = c(BA)(x) for all x, x1 , x2 ∈ X and
all scalars c. In fact, (BA)(x1 + x2 ) = B(A(x1 + x2 )) = B(Ax1 + Ax2 ) =
B(Ax1 ) + B(Ax2 ) = (BA)(x1 ) + (BA)(x2 ), and (BA)(cx) = B(A(cx)) =
B(cAx) = cB(Ax) = c(BA)(x). Thus BA is linear.
To show A−1 is linear, it suffices to show that A−1 (x1 + x2 ) = A−1 (x1 ) +
A−1 (x2 ) and A−1 (cx) = cA−1 (x) for all x, x1 , x2 ∈ X and all scalars c.
Since A is invertible, A maps X onto X and thus x1 = Ay1 , x2 = Ay2 ,
x = Ay for some y1 , y2 , y ∈ X. Hence, A−1 (x1 + x2 ) = A−1 (Ay1 + Ay2 ) =
A−1 (A(y1 + y2 )) = y1 + y2 = A−1 (x1 ) + A−1 (x2 ); A−1 (cx) = A−1 (cAy) =
A−1 (A(cy)) = cy = cA−1 (x). Thus A−1 is linear.
To show A−1 is invertible, it needs to show that A−1 is a linear operator on X which (i) is one-to-one and (ii) maps X onto X. For (i), given
A−1 (x1 ) = A−1 (x2 ). Since A is invertible, A maps X onto X and thus
x1 = Ay1 , x2 = Ay2 for some y1 , y2 ∈ X. Thus, A−1 (Ay1 ) = A−1 (Ay2 ) or
y1 = y2 , that is, x1 = x2 . For (ii), given x ∈ X we have Ax ∈ X such that
A−1 (Ax) = x. Thus A−1 is invertible. Exercise 9.3. To show A is 1-1, it suffices to show that Ax1 = Ax2 implies
x1 = x2 . Since A ∈ L(X, Y ), 0 = Ax1 − Ax2 = A(x1 − x2 ). By assumption,
x1 − x2 = 0, that is, x1 = x2 . Exercise 9.4. For all x, y ∈ N (A), and for all scalars c, we need to show
that x + y ∈ N (A) and cx ∈ N (A). It follows by A(x + y) = A(x) + A(y) =
0 + 0 = 0 and A(cx) = cA(x) = c0 = 0. Hence N (A) is a vector space.
For all x, y ∈ R(A), and for all scalars c, we need to show that x + y ∈
R(A) and cx ∈ R(A). x and y are all in R(A) implies that there exist
93
x1 , y1 ∈ X such that Ax1 = x and Ay1 = y. Thus, x + y = Ax1 + Ay1 =
A(x1 + y1 ), i.e., x + y ∈ R(A). Also, cx = cAx1 = A(cx1 ), i.e., cx ∈ R(A).
Hence R(A) is a vector space. Exercise 9.5. By Schwarz inequality,
kAk = sup |Ax| = sup |x · y| ≤ sup |x||y| = |y|.
|x|≤1
|x|≤1
|x|≤1
Now we pick x = |y|−1y if y 6= 0. (The case y = 0 is trivial.) |x| = 1 and
|Ax| = |x · y| = |y| ≤ kAk. Hence, kAk = |y|. Exercise 9.6. For (x, y) 6= (0, 0), (D1 f )(x, y) =
3x2 y+y 3
(x2 +y 2 )2
and (D2 f )(x, y) =
3xy 2 +x3
.
(x2 +y 2 )2
For (x, y) = (0, 0), (D1 f )(0, 0) = (D2 f )(0, 0) = 0. But f is not continuous at (0, 0). In fact, to calculate the limit lim(x,y)→(0,0) f (x, y), we make
changes of variables by x = r cos θ and y = r sin θ. (x, y) → (0, 0) if and only
if r = 0, and f (x, y) = cos θ sin θ = 21 sin(2θ). Hence lim(x,y)→(0,0) f (x, y) =
limr→0 12 sin(2θ) =
1
2
sin(2θ) dose not exist. Exercise 9.7. Fix x ∈ E and ǫ > 0. Since E is open, there is an open ball
P
S ⊂ E, with center at x and radius r. Suppose h =
hj ej , h < r, put
v0 = 0, and vk = h1 e1 + · · · + hk ek , for 1 ≤ k ≤ n. Then
f (x + h) − f (x) =
n
X
j=1
(f (x + vj ) − f (x + vj−1 )).
(*)
Since |vk | < r for all k and since S is convex, the segments with end points
x + vj−1 and x + vj lie in S. Since vj = vj−1 + hj ej , the mean value theorem
shows that the j-th summand in (*) is equal to
hj (Dj f )(x + vj−1 + θj hj ej )
for some θj ∈ (0, 1). Since Dj f are bounded in E, there is M > 0 such that
94
|Dj f | ≤ M. Thus
|f (x + h) − f (x)| ≤
=
n
X
j=1
n
X
j=1
≤ M
|f (x + vj ) − f (x + vj−1 ))|
|hj (Dj f )(x + vj−1 + θj hj ej )|
n
X
j=1
|hj | ≤ Mnr < ǫ
if we choose r < ǫ/(Mn). Hence f is continuous in E. Exercise 9.8. By definition 5.7, there exists δ > 0 such that f (x+h) ≤ f (x)
for all h ∈ E with |h| < δ. To show f ′ (x) = 0, it suffices to show that
(Dj f )(x) = 0 for 1 ≤ j ≤ n. If −δ < t < 0, then |tej | < δ and
f (x + tej ) − f (x)
≥ 0.
t
Letting t → 0, we see that (Dj f )(x) ≥ 0. If 0 < t < δ, then |tej | < δ and
f (x + tej ) − f (x)
≤ 0,
t
which shows that (Dj f )(x) ≤ 0. Hence (Dj f )(x) = 0. (The idea of this
proof comes from theorem 5.8.)
Exercise 9.11. For each x ∈ Rn , the gradient of f g at x is
▽(f g)(x) =
=
=
n
X
i=1
n
X
i=1
n
X
(Di (f g))(x)ei
(gDi (f ) + f Di (g))(x)ei
(gDi (f ))(x)ei +
i=1
n
X
(f Di (g)(x)ei
i=1
= (g ▽ f + f ▽ g)(x).
Thus ▽(f g) = f ▽ g + g ▽ f . In particular, 0 = ▽1 = ▽(f · 1/f ) =
f ▽ (1/f ) + 1/f ▽ f , i.e., ▽(1/f ) = −f −2 ▽ f whenever f 6= 0. 95
Exercise 9.13. Let g(x, y, z) = (x, y, z) · (x, y, z) = x2 + y 2 + z 2 of R3 into
R, and F (t) = g(f(t)) = f(t) · f(t) = 1 of R into R. Note that g′ (x) = 2x
since (D1 g)(x, y, z) = 2x, (D2 g)(x, y, z) = 2y, and (D3 g)(x, y, z) = 2z. By
theorem 9.15,
F ′ (t) = g′ (f(t))f ′ (t) = 2f(t) · f ′ (t).
Since 1 = |f(t)|, 1 = f(t) · f(t) = F (t) and thus 0 = f(t) · f ′ (t). We interpret
this result geometrically. Given a particle on the unit sphere with a smooth
movement f(t) by the time t. The result f(t)·f ′ (t) = 0 says that the direction
of velocity of this particle point at the center of this unit sphere. Exercise 9.15. (a) (x4 + y 2)2 − 4x4 y 2 = (x4 − y 2) ≥ 0 implies that 4x4 y 2 ≤
(x4 + y 2 )2 . To show f is continuous on R2 , it suffices to show f is continuous
at (0, 0). Note that
f (x, y) = x2 + y 2 − 2x2 y − x2
4x4 y 2
(x4 + y 2 )2
and thus lim(x,y)→(0,0) f (x, y) = 0 = f (0, 0).
Exercise 9.16.
f (t) − f (0)
1
= lim 1 + 2t sin
f (0) = lim
= 1.
t→0
t→0
t
t
′
For t 6= 0,
1
1
f (t) = 1 + 4t sin
− cos
.
t
t
′
Hence, f ′ (0) = 1 and f ′ (t) is bounded (by 6). But f is not one-to-one in any
neighborhood of 0 since f ′ is not monotonic in any neighborhood of 0. R
Exercise 11.1. If f ≥ 0 and E f dµ = 0, prove that f (x) = 0 almost
everywhere on E. Hint: Let En be the subset of E on which f (x) > 1/n.
Write A = ∪En . Then µ(A) = 0 if and only if µ(En ) = 0 for every n.
Proof. Let En be the subset of E on which f (x) > 1/n. Write A = ∪En . In
fact, A = {x ∈ E | f (x) > 0}. Claim: µ(A) = 0 if and only if µ(En ) = 0 for
96
every n. One side is trivial. Conversely, if µ(En ) = 0 for all n, then we have
µ(En ) → µ(A) by theorem 11.3, that is, µ(A) = 0. Therefore, for every n
Z
Z
1
µ(En ) ≤
f dµ ≤
f dµ = 0
n
En
E
implies µ(En ) = 0. Since {x ∈ E | f (x) > 0} = A = ∪En , we have µ(A) = 0.
R
Exercise 11.2. If A f dµ = 0 for every measurable subset A of a measurable
set E, then f (x) = 0 almost everywhere on E.
Proof. Let A = {x ∈ E | f (x) ≥ 0} be a measurable subset of E. Thus,
R
R
f dµ = A f + dµ = 0. By exercise 11.1, f + = 0 almost everywhere on
A
A and thus on E. Similarly, f − = 0 almost everywhere on E. Hence,
f = f + − f − = 0 almost everywhere on E. Exercise 11.3. If {fn } is a sequence of measurable functions, prove that
the set of points x at which {fn (x)} converges is measurable.
Proof. Let E be the set of points x at which {fn (x)} converges. x ∈ E if
and only if for every ǫ > 0 there exists N such that n ≥ N implies
|fn (x) − f (x)| < ǫ
for some f (x), which can be restate as that
x∈
∞ \
∞
\
k=1 n=N
{x ∈ X | |fn (x) − f (x)| < 1/k}
for some integer N = N(x, k). For simplicity, we define f (x) = 0 if x ∈ X \E.
Hence,
∞ [
∞ \
∞
\
E=
{x ∈ X | |fn (x) − f (x)| < 1/k}
k=1 m=1 n=m
is measurable since every fn is a measurable function. 97
Exercise 11.5. Put
(
0
1
(0 ≤ x ≤ 12 ),
( 21 < x ≤ 1),
f2k (x) = g(x)
(0 ≤ x ≤ 1),
g(x) =
f2k+1 (x) = g(1 − x)
(0 ≤ x ≤ 1).
Show that
(0 ≤ x ≤ 1),
lim inf fn (x) = 0
n→∞
but
Z
1
0
1
fn (x)dx = .
2
f2k (x) =
(
0
1
f2k+1 (x) =
(
1
0
[Compare with (77).]
Proof. Since
lim inf n→∞ fn (x) = 0, but
R1
0
(0 ≤ x ≤ 12 ),
( 21 < x ≤ 1),
(0 ≤ x < 12 ),
( 21 ≤ x ≤ 1),
fn (x)dx =
Exercise 11.6. Let
fn (x) =
(
1
2
for all n. 1
n
(|x| ≤ n),
0 (|x| > n).
Then fn (x) → 0 uniformly on R1 , but
Z ∞
fn (x)dx = 2
(n = 1, 2, 3, · · · ).
(We write
R∞
−∞
−∞
in place of
R
R1
.) Thus uniform convergence does not imply
dominated convergence in the sense of Theorem 11.32. However, on sets
of finite measure, uniformly convergent sequences of bounded functions do
satisfy Theorem 11.32.
Proof. Clearly, limn→∞ fn (x) = 0. Put
Mn = sup |fn (x) − 0| =
x∈R1
98
1
.
n
By theorem 7.9, fn → 0 uniformly on R1 if and only if Mn =
R∞
n → ∞. But −∞ fn (x)dx = n1 · 2n = 2 for all n.
99
1
n
→ 0 as