## Sequences of Functions

### Uniform convergence

9.1 Assume that f_{n} → f uniformly on S and that each f_{n} is bounded on
S. Prove that {f_{n}} is uniformly bounded on S.

Proof : Since f_{n} → f uniformly on S, then given ε = 1, there exists a
positive integer n_{0} such that as n ≥ n_{0}, we have

|f_{n}(x) − f (x)| ≤ 1 for all x ∈ S. (*)
Hence, f (x) is bounded on S by the following

|f (x)| ≤ |f_{n}_{0}(x)| + 1 ≤ M (n_{0}) + 1 for all x ∈ S. (**)
where |f_{n}_{0}(x)| ≤ M (n_{0}) for all x ∈ S.

Let |f_{1}(x)| ≤ M (1) , ..., |f_{n}_{0}−1(x)| ≤ M (n_{0}− 1) for all x ∈ S, then by
(*) and (**),

|f_{n}(x)| ≤ 1 + |f (x)| ≤ M (n_{0}) + 2 for all n ≥ n_{0}.
So,

|f_{n}(x)| ≤ M for all x ∈ S and for all n
where M = max (M (1) , ..., M (n_{0}− 1) , M (n_{0}) + 2) .

Remark: (1) In the proof, we also shows that the limit function f is bounded on S.

(2) There is another proof. We give it as a reference.

Proof : Since Since f_{n}→ f uniformly on S, then given ε = 1, there exists
a positive integer n_{0} such that as n ≥ n_{0}, we have

|f_{n}(x) − f_{n+k}(x)| ≤ 1 for all x ∈ S and k = 1, 2, ...

So, for all x ∈ S, and k = 1, 2, ...

|f_{n}_{0}_{+k}(x)| ≤ 1 + |f_{n}_{0}(x)| ≤ M (n_{0}) + 1 (*)
where |f_{n}_{0}(x)| ≤ M (n_{0}) for all x ∈ S.

Let |f1(x)| ≤ M (1) , ..., |fn0−1(x)| ≤ M (n0− 1) for all x ∈ S, then by (*),

|f_{n}(x)| ≤ M for all x ∈ S and for all n

where M = max (M (1) , ..., M (n_{0}− 1) , M (n_{0}) + 1) .

### 9.2

Define two sequences {f_{n}} and {g

_{n}} as follows:

f_{n}(x) = x

1 + 1

n

if x ∈ R, n = 1, 2, ...,

g_{n}(x) =

_{1}

n if x = 0 or if x is irrational,

b + _{n}^{1} if x is rational, say x = ^{a}_{b}, b > 0.

Let h_{n}(x) = f_{n}(x) g_{n}(x) .

(a) Prove that both {f_{n}} and {g_{n}} converges uniformly on every bounded
interval.

Proof : Note that it is clear that

n→∞lim f_{n}(x) = f (x) = x, for all x ∈ R
and

n→∞lim g_{n}(x) = g (x) =

0 if x = 0 or if x is irrational,
b if x is ratonal, say x = ^{a}_{b}, b > 0.

In addition, in order to show that {f_{n}} and {g_{n}} converges uniformly
on every bounded interval, it suffices to consider the case of any compact
interval [−M, M ] , M > 0.

Given ε > 0, there exists a positive integer N such that as n ≥ N, we

have M

n < ε and 1 n < ε.

Hence, for this ε, we have as n ≥ N

|f_{n}(x) − f (x)| =
x
n
≤ M

n < ε for all x ∈ [−M, M ] and

|g_{n}(x) − g (x)| ≤ 1

n < ε for all x ∈ [−M, M ] .

That is, we have proved that {f_{n}} and {g_{n}} converges uniformly on every
bounded interval.

Remark: In the proof, we use the easy result directly from definition
of uniform convergence as follows. If f_{n} → f uniformly on S, then f_{n} → f
uniformly on T for every subset T of S.

(b) Prove that h_{n}(x) does not converges uniformly on any bounded in-
terval.

Proof : Write
h_{n}(x) =

_{x}

n 1 + ^{1}_{n}

if x = 0 or x is irrational
a + _{n}^{a} 1 + ^{1}_{b} + _{bn}^{1}

if x is rational, say x = ^{a}_{b} .
Then

n→∞lim h_{n}(x) = h (x) =

0 if x = 0 or x is irrational
a if x is rational, say x = ^{a}_{b} .

Hence, if h_{n}(x) converges uniformly on any bounded interval I, then h_{n}(x)
converges uniformly on [c, d] ⊆ I. So, given ε = max (|c| , |d|) > 0, there is a
positive integer N such that as n ≥ N, we have

max (|c| , |d|) > |h_{n}(x) − h (x)|

=

_{n}^{x} 1 + _{n}^{1}
= ^{|x|}_{n}

1 + ^{1}_{n}

if x ∈ Q^{c}∩ [c, d] or x = 0

^{a}_{n} 1 + ^{1}_{b} +_{bn}^{1}

if x ∈ Q ∩ [c, d] , x = ^{a}_{b}
which implies that (x ∈ [c, d] ∩ Q^{c} or x = 0)

max (|c| , |d|) > |x|

n

1 + 1 n

≥ |x|

n ≥ max (|c| , |d|) n

which is absurb. So, h_{n}(x) does not converges uniformly on any bounded
interval.

### 9.3

Assume that f_{n}→ f uniformly on S, g

_{n}→ f uniformly on S.

(a) Prove that fn+ gn→ f + g uniformly on S.

Proof : Since f_{n} → f uniformly on S, and g_{n} → f uniformly on S, then
given ε > 0, there is a positive integer N such that as n ≥ N, we have

|f_{n}(x) − f (x)| < ε

2 for all x ∈ S and

|g_{n}(x) − g (x)| < ε

2 for all x ∈ S.

Hence, for this ε, we have as n ≥ N,

|f_{n}(x) + g_{n}(x) − f (x) − g (x)| ≤ |f_{n}(x) − f (x)| + |g_{n}(x) − g (x)|

< ε for all x ∈ S.

That is, f_{n}+ g_{n}→ f + g uniformly on S.

Remark: There is a similar result. We write it as follows. If f_{n} → f
uniformly on S, then cf_{n}→ cf uniformly on S for any real c. Since the proof
is easy, we omit the proof.

(b) Let h_{n}(x) = f_{n}(x) g_{n}(x) , h (x) = f (x) g (x) , if x ∈ S. Exercise 9.2
shows that the assertion h_{n} → h uniformly on S is, in general, incorrect.

Prove that it is correct if each f_{n} and each g_{n} is bounded on S.

Proof : Since f_{n} → f uniformly on S and each f_{n} is bounded on S, then
f is bounded on S by Remark (1) in the Exercise 9.1. In addition, since
g_{n} → g uniformly on S and each g_{n} is bounded on S, then g_{n} is uniformly
bounded on S by Exercise 9.1.

Say |f (x)| ≤ M_{1} for all x ∈ S, and |g_{n}(x)| ≤ M_{2} for all x and all n. Then
given ε > 0, there exists a positive integer N such that as n ≥ N, we have

|f_{n}(x) − f (x)| < ε

2 (M_{2}+ 1) for all x ∈ S
and

|g_{n}(x) − g (x)| < ε

2 (M1+ 1) for all x ∈ S which implies that as n ≥ N, we have

|h_{n}(x) − h (x)| = |f_{n}(x) g_{n}(x) − f (x) g (x)|

= |[f_{n}(x) − f (x)] [g_{n}(x)] + [f (x)] [g_{n}(x) − g (x)]|

≤ |fn(x) − f (x)| |gn(x)| + |f (x)| |gn(x) − g (x)|

< ε

2 (M_{2}+ 1)M_{2}+ M_{1} ε
2 (M_{1} + 1)

< ε 2+ ε

2

= ε

for all x ∈ S. So, hn → h uniformly on S.

9.4 Assume that f_{n} → f uniformly on S and suppose there is a constant
M > 0 such that |f_{n}(x)| ≤ M for all x in S and all n. Let g be continuous
on the closure of the disk B (0; M ) and define hn(x) = g [fn(x)] , h (x) =
g [f (x)] , if x ∈ S. Prove that h_{n} → h uniformly on S.

Proof : Since g is continuous on a compact disk B (0; M ) , g is uniformly continuous on B (0; M ) . Given ε > 0, there exists a δ > 0 such that as

|x − y| < δ, where x, y ∈ S, we have

|g (x) − g (y)| < ε. (*)

For this δ > 0, since f_{n} → f uniformly on S, then there exists a positive
integer N such that as n ≥ N, we have

|f_{n}(x) − f (x)| < δ for all x ∈ S. (**)
Hence, by (*) and (**), we conclude that given ε > 0, there exists a positive
integer N such that as n ≥ N, we have

|g (f_{n}(x)) − g (f (x))| < ε for all x ∈ S.

Hence, h_{n} → h uniformly on S.

9.5 (a) Let f_{n}(x) = 1/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that {f_{n}}
converges pointwise but not uniformly on (0, 1) .

Proof : First, it is clear that lim_{n→∞}f_{n}(x) = 0 for all x ∈ (0, 1) . Supppos
that {f_{n}} converges uniformly on (0, 1) . Then given ε = 1/2, there exists a
positive integer N such that as n ≥ N, we have

|fn(x) − f (x)| =

1 1 + nx

< 1/2 for all x ∈ (0, 1) .

So, the inequality holds for all x ∈ (0, 1) . It leads us to get a contradiction since

1

1 + N x < 1

2 for all x ∈ (0, 1) ⇒ lim

x→0^{+}

1

1 + N x = 1 < 1/2.

That is, {f_{n}} converges NOT uniformly on (0, 1) .

(b) Let gn(x) = x/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that gn → 0 uniformly on (0, 1) .

Proof : First, it is clear that lim_{n→∞}g_{n}(x) = 0 for all x ∈ (0, 1) . Given
ε > 0, there exists a positive integer N such that as n ≥ N, we have

1/n < ε

which implies that

|g_{n}(x) − g| =

x 1 + nx

=

1

1 x + n

< 1 n < ε.

So, g_{n}→ 0 uniformly on (0, 1) .

9.6 Let fn(x) = x^{n}. The sequence {fn(x)} converges pointwise but not
uniformly on [0, 1] . Let g be continuous on [0, 1] with g (1) = 0. Prove that
the sequence {g (x) x^{n}} converges uniformly on [0, 1] .

Proof : It is clear that f_{n}(x) = x^{n} converges NOT uniformly on [0, 1]

since each term of {f_{n}(x)} is continuous on [0, 1] and its limit function
f = 0 if x ∈ [0, 1)

1 if x = 1.

is not a continuous function on [0, 1] by Theorem 9.2.

In order to show {g (x) x^{n}} converges uniformly on [0, 1] , it suffices to
shows that {g (x) x^{n}} converges uniformly on [0, 1). Note that

n→∞lim g (x) x^{n} = 0 for all x ∈ [0, 1).

We partition the interval [0, 1) into two subintervals: [0, 1 − δ] and (1 − δ, 1) .
As x ∈ [0, 1 − δ] : Let M = max_{x∈[0,1]}|g (x)| , then given ε > 0, there is a
positive integer N such that as n ≥ N, we have

M (1 − δ)^{n}< ε
which implies that for all x ∈ [0, 1 − δ] ,

|g (x) x^{n}− 0| ≤ M |x^{n}| ≤ M (1 − δ)^{n} < ε.

Hence, {g (x) x^{n}} converges uniformly on [0, 1 − δ] .

As x ∈ (1 − δ, 1) : Since g is continuous at 1, given ε > 0, there exists a δ > 0 such that as |x − 1| < δ, where x ∈ [0, 1] , we have

|g (x) − g (1)| = |g (x) − 0| = |g (x)| < ε which implies that for all x ∈ (1 − δ, 1) ,

|g (x) x^{n}− 0| ≤ |g (x)| < ε.

Hence, {g (x) x^{n}} converges uniformly on (1 − δ, 1) .

So, from above sayings, we have proved that the sequence of functions
{g (x) x^{n}} converges uniformly on [0, 1] .

Remark: It is easy to show the followings by definition. So, we omit the proof.

(1) Suppose that for all x ∈ S, the limit function f exists. If f_{n} → f
uniformly on S_{1}(⊆ S) , then f_{n} → f uniformly on S, where # (S − S_{1}) <

+∞.

(2) Suppose that f_{n} → f uniformly on S and on T. Then f_{n} → f uni-
formly on S ∪ T.

9.7 Assume that f_{n}→ f uniformly on S and each f_{n} is continuous on S.

If x ∈ S, let {x_{n}} be a sequence of points in S such that x_{n}→ x. Prove that
f_{n}(x_{n}) → f (x) .

Proof : Since f_{n} → f uniformly on S and each f_{n} is continuous on S, by
Theorem 9.2, the limit function f is also continuous on S. So, given ε > 0,
there is a δ > 0 such that as |y − x| < δ, where y ∈ S, we have

|f (y) − f (x)| < ε 2.

For this δ > 0, there exists a positive integer N_{1} such that as n ≥ N_{1}, we
have

|x_{n}− x| < δ.

Hence, as n ≥ N_{1}, we have

|f (x_{n}) − f (x)| < ε

2. (*)

In addition, since fn → f uniformly on S, given ε > 0, there exists a
positive integer N ≥ N_{1} such that as n ≥ N, we have

|f_{n}(x) − f (x)| < ε

2 for all x ∈ S which implies that

|f_{n}(x_{n}) − f (x_{n})| < ε

2. (**)

By (*) and (**), we obtain that given ε > 0, there exists a positie integer N such that as n ≥ N, we have

|fn(xn) − f (x)| = |fn(xn) − f (xn)| + |f (xn) − f (x)|

< ε 2+ ε

2

= ε.

That is, we have proved that f_{n}(x_{n}) → f (x) .

### 9.8

^{Let {f}n} be a seuqnece of continuous functions defined on a compact set S and assume that {f

_{n}} converges pointwise on S to a limit function f.

Prove that f_{n}→ f uniformly on S if, and only if, the following two conditions
hold.:

(i) The limit function f is continuous on S.

(ii) For every ε > 0, there exists an m > 0 and a δ > 0, such that n > m
and |f_{k}(x) − f (x)| < δ implies |f_{k+n}(x) − f (x)| < ε for all x in S and all
k = 1, 2, ...

Hint. To prove the sufficiency of (i) and (ii), show that for each x0 in S
there is a neighborhood of B (x_{0}) and an integer k (depending on x_{0}) such
that

|fk(x) − f (x)| < δ if x ∈ B (x0) .

By compactness, a finite set of integers, say A = {k_{1}, ..., k_{r}} , has the property
that, for each x in S, some k in A satisfies |f_{k}(x) − f (x)| < δ. Uniform
convergence is an easy consequences of this fact.

Proof : (⇒) Suppose that fn → f uniformly on S, then by Theorem 9.2, the limit function f is continuous on S. In addition, given ε > 0, there exists a positive integer N such that as n ≥ N, we have

|fn(x) − f (x)| < ε for all x ∈ S Let m = N, and δ = ε, then (ii) holds.

(⇐) Suppose that (i) and (ii) holds. We prove f_{k}→ f uniformly on S as
follows. By (ii), given ε > 0, there exists an m > 0 and a δ > 0, such that
n > m and |f_{k}(x) − f (x)| < δ implies |f_{k+n}(x) − f (x)| < ε for all x in S
and all k = 1, 2, ...

Consider

f_{k(x}_{0}_{)}(x0) − f (x0)

< δ, then there exists a B (x0) such that as
x ∈ B (x_{0}) ∩ S, we have

f_{k(x}_{0}_{)}(x) − f (x)
< δ

by continuity of f_{k(x}_{0}_{)}(x) − f (x) . Hence, by (ii) as n > m
f_{k(x}_{0}_{)+n}(x) − f (x)

< ε if x ∈ B (x_{0}) ∩ S. (*)
Note that S is compact and S = ∪_{x∈S}(B (x) ∩ S) , then S = ∪^{p}_{k=1}(B (x_{k}) ∩ S) .
So, let N = max^{p}_{i=1}(k (x_{p}) + m) , as n > N, we have

|fn(x) − f (x)| < ε for all x ∈ S
with help of (*). That is, f_{n}→ f uniformly on S.

### 9.9

(a) Use Exercise 9.8 to prove the following theorem of Dini: If {f_{n}} is a sequence of real-valued continuous functions converginf pointwise to a continuous limit function f on a compact set S, and if f

_{n}(x) ≥ f

_{n+1}(x) for each x in S and every n = 1, 2, ..., then f

_{n}→ f uniformly on S.

Proof : By Exercise 9.8, in order to show that f_{n}→ f uniformly on S,
it suffices to show that (ii) holds. Since f_{n}(x) → f (x) and f_{n+1}(x) ≤ f_{n}(x)
on S, then fixed x ∈ S, and given ε > 0, there exists a positive integer
N (x) = N such that as n ≥ N, we have

0 ≤ f_{n}(x) − f (x) < ε.

Choose m = 1 and δ = ε, then by f_{n+1}(x) ≤ f_{n}(x) , then (ii) holds. We
complete it.

Remark: (1) Dini’s Theorem is important in Analysis; we suggest the reader to keep it in mind.

(2) There is another proof by using Cantor Intersection Theorem.

We give it as follows.

Proof : Let gn = fn− f, then gnis continuous on S, gn→ 0 pointwise on
S, and g_{n}(x) ≥ g_{n+1}(x) on S. If we can show g_{n} → 0 uniformly on S, then
we have proved that f_{n} → f uniformly on S.

Given ε > 0, and consider Sn := {x : gn(x) ≥ ε} . Since each gn(x) is
continuous on a compact set S, we obtain that S_{n} is compact. In addition,
S_{n+1} ⊆ S_{n} since g_{n}(x) ≥ g_{n+1}(x) on S. Then

∩Sn6= φ (*)

if each S_{n} is non-empty by Cantor Intersection Theorem. However (*)
contradicts to g_{n} → 0 pointwise on S. Hence, we know that there exists a
positive integer N such that as n ≥ N,

S_{n} = φ.

That is, given ε > 0, there exists a positive integer N such that as n ≥ N, we have

|g_{n}(x) − 0| < ε.

So, g_{n}→ 0 uniformly on S.

(b) Use he sequence in Exercise 9.5(a) to show that compactness of S is essential in Dini’s Theorem.

Proof : Let f_{n}(x) = _{1+nx}^{1} , where x ∈ (0, 1) . Then it is clear that each
f_{n}(x) is continuous on (0, 1) , the limit function f (x) = 0 is continuous on
(0, 1) , and fn+1(x) ≤ fn(x) for all x ∈ (0, 1) . However, fn→ f not uniformly
on (0, 1) by Exercise 9.5 (a). Hence, compactness of S is essential in Dini’s
Theorem.

### 9.10

^{Let f}n(x) = n

^{c}x (1 − x

^{2})

^{n}for x real and n ≥ 1. Prove that {f

_{n}} converges pointwsie on [0, 1] for every real c. Determine those c for which the convergence is uniform on [0, 1] and those for which term-by-term integration on [0, 1] leads to a correct result.

Proof : It is clear that fn(0) → 0 and fn(1) → 0. Consider x ∈ (0, 1) ,
then |1 − x^{2}| := r < 1, then

n→∞lim f_{n}(x) = lim

n→∞n^{c}r^{n}x = 0 for any real c.

Hence, f_{n}→ 0 pointwise on [0, 1] .
Consider

f_{n}^{0} (x) = n^{c} 1 − x^{2}n−1

(2n − 1)

1

2n − 1 − x^{2}

,
then each f_{n} has the absolute maximum at x_{n}= ^{√} ^{1}

2n−1. As c < 1/2, we obtain that

|f_{n}(x)| ≤ |f_{n}(x_{n})|

= n^{c}

√2n − 1

1 − 1 2n − 1

n

= n^{c−}^{1}^{2}

r n

2n − 1

1 − 1 2n − 1

n

→ 0 as n → ∞. (*)

In addition, as c ≥ 1/2, if f_{n}→ 0 uniformly on [0, 1] , then given ε > 0, there
exists a positive integer N such that as n ≥ N, we have

|fn(x)| < ε for all x ∈ [0, 1]

which implies that as n ≥ N,

|f_{n}(x_{n})| < ε
which contradicts to

n→∞lim fn(xn) =

√1

2e if c = 1/2

∞ if c > 1/2 . (**)

From (*) and (**), we conclude that only as c < 1/2, the seqences of functions converges uniformly on [0, 1] .

In order to determine those c for which term-by-term integration on [0, 1] , we consider

Z 1 0

f_{n}(x) dx = n^{c}
2 (n + 1)
and

Z 1 0

f (x) dx = Z 1

0

0dx = 0.

Hence, only as c < 1, we can integrate it term-by-term.

9.11 Prove that P x^{n}(1 − x) converges pointwise but not uniformly on
[0, 1] , whereas P (−1)^{n}x^{n}(1 − x) converges uniformly on [0, 1] . This illus-
trates that uniform convergence of P fn(x) along with pointwise con-
vergence of P |fn(x)| does not necessarily imply uniform conver-
gence of P |f_{n}(x)| .

Proof : Let sn(x) =Pn

k=0x^{k}(1 − x) = 1 − x^{n+1}, then
s_{n}(x) → 1 if x ∈ [0, 1)

0 if x = 1 .

Hence, P x^{n}(1 − x) converges pointwise but not uniformly on [0, 1] by The-
orem 9.2 since each s_{n} is continuous on [0, 1] .

Let g_{n}(x) = x^{n}(1 − x) , then it is clear that g_{n}(x) ≥ g_{n+1}(x) for all x ∈
[0, 1] , and gn(x) → 0 uniformly on [0, 1] by Exercise 9.6. Hence, by Dirich-
let’s Test for uniform convergence, we have proved thatP (−1)^{n}x^{n}(1 − x)
converges uniformly on [0, 1] .

9.12 Assume that g_{n+1}(x) ≤ g_{n}(x) for each x in T and each n = 1, 2, ...,
and suppose that g_{n} → 0 uniformly on T. Prove thatP (−1)^{n+1}g_{n}(x) con-
verges uniformly on T.

Proof : It is clear by Dirichlet’s Test for uniform convergence.

### 9.13

Prove Abel’s test for uniform convergence: Let {g_{n}} be a sequence of real-valued functions such that g

_{n+1}(x) ≤ g

_{n}(x) for each x in T and for every n = 1, 2, ... If {g

_{n}} is uniformly bounded on T and ifP fn(x) converges uniformly on T, then P f

_{n}(x) g

_{n}(x) also converges uniformly on T.

Proof : Let F_{n}(x) = Pn

k=1f_{k}(x) . Then
s_{n}(x) =

n

X

k=1

f_{k}(x) g_{k}(x) = F_{n}g_{1}(x)+

n

X

k=1

(F_{n}(x) − F_{k}(x)) (g_{k+1}(x) − g_{k}(x))

and hence if n > m, we can write

s_{n}(x)−s_{m}(x) = (F_{n}(x) − F_{m}(x)) g_{m+1}(x)+

n

X

k=m+1

(F_{n}(x) − F_{k}(x)) (g_{k+1}(x) − g_{k}(x))

Hence, if M is an uniform bound for {g_{n}} , we have

|s_{n}(x) − s_{m}(x)| ≤ M |F_{n}(x) − F_{m}(x)| + 2M

n

X

k=m+1

|F_{n}(x) − F_{k}(x)| . (*)

Since P fn(x) converges uniformly on T, given ε > 0, there exists a positive integer N such that as n > m ≥ N, we have

|F_{n}(x) − F_{m}(x)| < ε

M + 1 for all x ∈ T (**)

By (*) and (**), we have proved that as n > m ≥ N,

|s_{n}(x) − s_{m}(x)| < ε for all x ∈ T.

Hence, P fn(x) g_{n}(x) also converges uniformly on T.

Remark: In the proof, we establish the lemma as follows. We write it as a reference.

(Lemma) If {a_{n}} and {b_{n}} are two sequences of complex numbers, define

A_{n}=

n

X

k=1

a_{k}.

Then we have the identity

n

X

k=1

akbk= Anbn+1−

n

X

k=1

Ak(bk+1− bk) (i)

= A_{n}b_{1}+

n

X

k=1

(A_{n}− A_{k}) (b_{k+1}− b_{k}) . (ii)

Proof : The identity (i) comes from Theorem 8.27. In order to show (ii), it suffices to consider

b_{n+1} = b_{1}+

n

X

k=1

b_{k+1}− b_{k}.

9.14 Let f_{n}(x) = x/ (1 + nx^{2}) if x ∈ R, n = 1, 2, ... Find the limit function
f of the sequence {f_{n}} and the limit function g of the sequence {f_{n}^{0}} .

(a) Prove that f^{0}(x) exists for every x but that f^{0}(0) 6= g (0) . For what
values of x is f^{0}(x) = g (x)?

Proof : It is easy to show that the limit function f = 0, and by f_{n}^{0} (x) =

1−nx^{2}

(1+nx^{2})^{2}, we have

n→∞lim f_{n}^{0} (x) = g (x) = 1 if x = 0
0 if x 6= 0 .

Hence, f^{0}(x) exists for every x and f^{0}(0) = 0 6= g (0) = 1. In addition, it is
clear that as x 6= 0, we have f^{0}(x) = g (x) .

(b) In what subintervals of R does f_{n} → f uniformly?

Proof : Note that

1 + nx^{2}

2 ≥√

n |x|

by A.P. ≥ G.P. for all real x. Hence,

x
1 + nx^{2}

≤ 1

2√
n
which implies that f_{n}→ f uniformly on R.

(c) In what subintervals of R does f_{n}^{0} → g uniformly?

Proof : Since each f_{n}^{0} = ^{1−nx}^{2}

(1+nx^{2})^{2} is continuous on R, and the limit function
g is continuous on R − {0} , then by Theorem 9.2, the interval I that we
consider does not contains 0. Claim that f_{n}^{0} → g uniformly on such interval
I = [a, b] which does not contain 0 as follows.

Consider

1 − nx^{2}
(1 + nx^{2})^{2}

≤ 1

1 + nx^{2} ≤ 1
na^{2},

so we know that f_{n}^{0} → g uniformly on such interval I = [a, b] which does not
contain 0.

9.15 Let f_{n}(x) = (1/n) e^{−n}^{2}^{x}^{2} if x ∈ R, n = 1, 2, ... Prove that f_{n} → 0
uniformly on R, that f_{n}^{0} → 0 pointwise on R, but that the convergence of
{f_{n}^{0}} is not uniform on any interval containing the origin.

Proof : It is clear that f_{n} → 0 uniformly on R, that f_{n}^{0} → 0 pointwise
on R. Assume that f_{n}^{0} → 0 uniformly on [a, b] that contains 0. We will prove
that it is impossible as follows.

We may assume that 0 ∈ (a, b) since other cases are similar. Given ε = ^{1}_{e},
then there exists a positive integer N^{0} such that as n ≥ max N^{0},^{1}_{b} := N
(⇒ _{N}^{1} ≤ b), we have

|f_{n}^{0} (x) − 0| < 1

e for all x ∈ [a, b]

which implies that

2 N x
e^{(N x)}^{2}

< 1

e for all x ∈ [a, b]

which implies that, let x = _{N}^{1},
2
e < 1

e

which is absurb. So, the convergence of {f_{n}^{0}} is not uniform on any interval
containing the origin.

9.16 Let {f_{n}} be a sequence of real-valued continuous functions defined
on [0, 1] and assume that f_{n} → f uniformly on [0, 1] . Prove or disprove

n→∞lim

Z 1−1/n 0

f_{n}(x) dx =
Z 1

0

f (x) dx.

Proof : By Theorem 9.8, we have

n→∞lim Z 1

0

f_{n}(x) dx =
Z 1

0

f (x) dx. (*)

Note that {f_{n}} is uniform bound, say |f_{n}(x)| ≤ M for all x ∈ [0, 1] and all
n by Exercise 9.1. Hence,

Z 1 1−1/n

f_{n}(x) dx

≤ M

n → 0. (**)

Hence, by (*) and (**), we have

n→∞lim

Z 1−1/n 0

fn(x) dx = Z 1

0

f (x) dx.

9.17 Mathematicinas from Slobbovia decided that the Riemann integral was too complicated so that they replaced it by Slobbovian integral, de- fined as follows: If f is a function defined on the set Q of rational numbers in [0, 1] , the Slobbovian integral of f, denoted by S (f ) , is defined to be the limit

S (f ) = lim

n→∞

1 n

n

X

k=1

f k n

,

whenever the limit exists. Let {f_{n}} be a sequence of functions such that
S (fn) exists for each n and such that fn → f uniformly on Q. Prove that
{S (f_{n})} converges, that S (f ) exists, and S (f_{n}) → S (f ) as n → ∞.

Proof : f_{n}→ f uniformly on Q, then given ε > 0, there exists a positive
integer N such that as n > m ≥ N, we have

|f_{n}(x) − f (x)| < ε/3 (1)

and

|f_{n}(x) − f_{m}(x)| < ε/2. (2)
So, if n > m ≥ N,

|S (fn) − S (fm)| =

k→∞lim 1 k

k

X

j=1

fn

j k

− fm

j k

= lim

k→∞

1 k

k

X

j=1

f_{n} j

k

− f_{m} j
k

≤ lim

k→∞

1 k

k

X

j=1

ε/2 by (2)

= ε/2

< ε

which implies that {S (f_{n})} converges since it is a Cauchy sequence. Say its
limit S.

Consider, by (1) as n ≥ N, 1

k

k

X

j=1

f_{n} j

k

− ε/3

≤ 1 k

k

X

j=1

f j k

≤ 1 k

k

X

j=1

f_{n} j

k

+ ε/3

which implies that

"

1 k

k

X

j=1

f_{n} j
k

#

− ε/3 ≤ 1 k

k

X

j=1

f j k

≤

"

1 k

k

X

j=1

f_{n} j
k

# + ε/3

which implies that, let k → ∞

S (f_{n}) − ε/3 ≤ lim

k→∞sup1 k

k

X

j=1

f j k

≤ S (f_{n}) + ε/3 (3)

and

S (f_{n}) − ε/3 ≤ lim

k→∞inf 1 k

k

X

j=1

f j k

≤ S (f_{n}) + ε/3 (4)

which implies that

k→∞lim sup 1 k

k

X

j=1

f j k

− lim

k→∞inf 1 k

k

X

j=1

f j k

≤

k→∞lim sup 1 k

k

X

j=1

f j k

− S (f_{n})

+

k→∞lim inf 1 k

k

X

j=1

f j k

− S (f_{n})

≤ 2ε

3 by (3) and (4)

< ε. (5)

Note that (3)-(5) imply that the existence of S (f ) . Also, (3) or (4) implies that S (f ) = S. So, we complete the proof.

9.18 Let f_{n}(x) = 1/ (1 + n^{2}x^{2}) if 0 ≤ x ≤ 1, n = 1, 2, ... Prove that {f_{n}}
converges pointwise but not uniformly on [0, 1] . Is term-by term integration
permissible?

Proof : It is clear that

n→∞lim f_{n}(x) = 0

for all x ∈ [0, 1] . If {fn} converges uniformly on [0, 1] , then given ε = 1/3, there exists a positive integer N such that as n ≥ N, we have

|f_{n}(x)| < 1/3 for all x ∈ [0, 1]

which implies that

f_{N} 1
N

= 1 2 < 1

3

which is impossible. So, {f_{n}} converges pointwise but not uniformly on [0, 1] .
Since {f_{n}(x)} is clearly uniformly bounded on [0, 1] , i.e., |f_{n}(x)| ≤ 1
for all x ∈ [0, 1] and n. Hence, by Arzela’s Theorem, we know that the
sequence of functions can be integrated term by term.

9.19 Prove that P∞

n=1x/n^{α}(1 + nx^{2}) converges uniformly on every finite
interval in R if α > 1/2. Is the convergence uniform on R?

Proof : By A.P. ≥ G.P., we have

x
n^{α}(1 + nx^{2})

≤ 1

2n^{α+}^{1}^{2} for all x.

So, by Weierstrass M-test, we have proved thatP∞

n=1x/n^{α}(1 + nx^{2}) con-
verges uniformly on R if α > 1/2. Hence, P∞

n=1x/n^{α}(1 + nx^{2}) converges
uniformly on every finite interval in R if α > 1/2.

9.20 Prove that the series P∞

n=1((−1)^{n}/√

n) sin (1 + (x/n)) converges uniformly on every compact subset of R.

Proof : It suffices to show that the seriesP∞

n=1((−1)^{n}/√

n) sin (1 + (x/n))
converges uniformly on [0, a] . Choose n large enough so that a/n ≤ 1/2, and
therefore sin 1 + _{n+1}^{x} ≤ sin 1 +^{x}_{n} for all x ∈ [0, a] . So, if we let f_{n}(x) =
(−1)^{n}/√

n and g_{n}(x) = sin 1 + ^{x}_{n} , then by Abel’s test for uniform con-
vergence, we have proved that the series P∞

n=1((−1)^{n}/√

n) sin (1 + (x/n)) converges uniformly on [0, a] .

Remark: In the proof, we metion something to make the reader get more. (1) since a compact set K is a bounded set, say K ⊆ [−a, a] , if we can show the series converges uniformly on [−a, a] , then we have proved it. (2) The interval that we consider is [0, a] since [−a, 0] is similar. (3) Abel’s test for uniform convergence holds for n ≥ N, where N is a fixed positive integer.

9.21 Prove that the series P∞

n=0(x^{2n+1}/ (2n + 1) − x^{n+1}/ (2n + 2)) con-
verges pointwise but not uniformly on [0, 1] .

Proof : We show that the series converges pointwise on [0, 1] by con- sidering two cases: (1) x ∈ [0, 1) and (2) x = 1. Hence, it is trivial. De- fine f (x) = P∞

n=0(x^{2n+1}/ (2n + 1) − x^{n+1}/ (2n + 2)) , if the series converges
uniformly on [0, 1] , then by Theorem 9.2, f (x) is continuous on [0, 1] .
However,

f (x) =

_{1}

2 log (1 + x) if x ∈ [0, 1) log 2 if x = 1 . Hence, the series converges not uniformly on [0, 1] .

Remark: The function f (x) is found by the following. Given x ∈ [0, 1), then both

∞

X

n=0

t^{2n} = 1

1 − t^{2} and 1
2

∞

X

n=0

t^{n}= 1
2 (1 − t)

converges uniformly on [0, x] by Theorem 9.14. So, by Theorem 9.8, we

have Z x

0

∞

X

n=0

t^{2n}− 1
2

∞

X

n=0

t^{n} =
Z x

0

1

1 − t^{2} − 1
2 (1 − t)dt

= Z x

0

1 2

1

1 − t+ 1 1 + t

−1 2

1 1 − t

dt

= 1

2log (1 + x) . And as x = 1,

∞

X

n=0

x^{2n+1}/ (2n + 1) − x^{n+1}/ (2n + 2)

=

∞

X

n=0

1

2n + 1− 1 2n

=

∞

X

n=0

(−1)^{n+1}

n + 1 by Theorem8.14.

= log 2 by Abel’s Limit Theorem.

9.22 Prove thatP a_{n}sin nx andP b_{n}cos nx are uniformly convergent on
R if P |an| converges.

Proof : It is trivial by Weierstrass M-test.

### 9.23

^{Let {a}n} be a decreasing sequence of positive terms. Prove that the series P a

_{n}sin nx converges uniformly on R if, and only if, na

_{n}→ 0 as n → ∞.

Proof : (⇒) Suppose that the seriesP a_{n}sin nx converges uniformly on
R, then given ε > 0, there exists a positive integer N such that as n ≥ N,
we have

2n−1

X

k=n

a_{k}sin kx

< ε. (*)

Choose x = _{2n}^{1} , then sin ^{1}_{2} ≤ sin kx ≤ sin 1. Hence, as n ≥ N, we always

have, by (*) (ε >)

2n−1

X

k=n

a_{k}sin kx

=

2n−1

X

k=n

a_{k}sin kx

≥

2n−1

X

k=n

a_{2n}sin1

2 since a_{k}> 0 and a_{k} &

= 1 2sin1

2

(2na_{2n}) .

That is, we have proved that 2na_{2n} → 0 as n → ∞. Similarly, we also have
(2n − 1) a_{2n−1} → 0 as n → ∞. So, we have proved that na_{n}→ 0 as n → ∞.

(⇐) Suppose that nan → 0 as n → ∞, then given ε > 0, there exists a
positive integer n_{0} such that as n ≥ n_{0}, we have

|na_{n}| = na_{n}< ε

2 (π + 1). (*)

In order to show the uniform convergence of P∞

n=1ansin nx on R, it suffices to show the uniform convergence of P∞

n=1a_{n}sin nx on [0, π] . So, if we can
show that as n ≥ n_{0}

n+p

X

k=n+1

a_{k}sin kx

< ε for all x ∈ [0, π] , and all p ∈ N
then we complete it. We consider two cases as follows. (n ≥ n_{0})

As x ∈h

0,_{n+p}^{π} i
, then

n+p

X

k=n+1

a_{k}sin kx

=

n+p

X

k=n+1

a_{k}sin kx

≤

n+p

X

k=n+1

a_{k}kx by sin kx ≤ kx if x ≥ 0

=

n+p

X

k=n+1

(ka_{k}) x

≤ ε

2 (π + 1) pπ

n + p by (*)

< ε.

And as x ∈h

π n+p, πi

, then

n+p

X

k=n+1

a_{k}sin kx

≤

m

X

k=n+1

a_{k}sin kx +

n+p

X

k=m+1

a_{k}sin kx

, where m = hπ x i

≤

m

X

k=n+1

akkx + 2am+1

sin^{x}_{2} by Summation by parts

≤ ε

2 (π + 1)(m − n) x + 2a_{m+1}
sin^{x}_{2}

≤ ε

2 (π + 1)mx + 2am+1

π

x by 2x

π ≤ sin x if x ∈ h 0,π

2 i

≤ ε

2 (π + 1)π + 2a_{m+1}(m + 1)

< ε

2+ 2 ε 2 (π + 1)

< ε.

Hence, P∞

n=1a_{n}sin nx converges uniformly on R.

Remark: (1) In the proof (⇐), if we can make sure that na_{n}& 0, then
we can use the supplement on the convergnce of series in Ch8, (C)-
(6) to show the uniform convergence ofP∞

n=1a_{n}sin nx =P∞

n=1(na_{n}) ^{sin nx}_{n}
by Dirichlet’s test for uniform convergence.

(2)There are similar results; we write it as references.

(a) Suppose a_{n} & 0, then for each α ∈ 0,^{π}_{2} , P^{∞}_{n=1}a_{n}cos nx and
P∞

n=1a_{n}sin nx converges uniformly on [α, 2π − α] .

Proof: The proof follows from (12) and (13) in Theorem 8.30 and Dirichlet’s test for uniform convergence. So, we omit it. The reader can see the textbook, example in pp 231.

(b) Let {an} be a decreasing sequence of positive terms. P∞

n=1ancos nx uniformly converges on R if and only if P∞

n=1a_{n} converges.

Proof: (⇒) Suppose thatP∞

n=1a_{n}cos nx uniformly converges on R, then
let x = 0, then we have P∞

n=1an converges.

(⇐) Suppose thatP∞

n=1a_{n} converges, then by Weierstrass M-test, we
have proved that P∞

n=1a_{n}cos nx uniformly converges on R.

9.24 Given a convergent series P∞

n=1a_{n}. Prove that the Dirichlet series
P∞

n=1a_{n}n^{−s} converges uniformly on the half-infinite interval 0 ≤ s < +∞.

Use this to prove that lim_{s→0}^{+}P∞

n=1ann^{−s} =P∞
n=1an.
Proof : Let f_{n}(s) =Pn

k=1a_{k} and g_{n}(s) = n^{−s}, then by Abel’s test for
uniform convergence, we have proved that the Dirichlet seriesP∞

n=1a_{n}n^{−s}
converges uniformly on the half-infinite interval 0 ≤ s < +∞. Then by
Theorem 9.2, we know that lim_{s→0}^{+}P∞

n=1a_{n}n^{−s}=P∞
n=1a_{n}.
9.25 Prove that the series ζ (s) =P∞

n=1n^{−s} converges uniformly on every
half-infinite interval 1 + h ≤ s < +∞, where h > 0. Show that the equation

ζ^{0}(s) = −

∞

X

n=1

log n
n^{s}

is valid for each s > 1 and obtain a similar formula for the kth derivative
ζ^{(k)}(s) .

Proof : Since n^{−s} ≤ n^{−(1+h)} for all s ∈ [1 + h, ∞), we know that ζ (s) =
P∞

n=1n^{−s} converges uniformly on every half-infinite interval 1 + h ≤ s < +∞

by Weierstrass M-test. Define T_{n}(s) =Pn

k=1k^{−s}, then it is clear that
1. For each n, T_{n}(s) is differentiable on [1 + h, ∞),

2. lim

n→∞T_{n}(2) = π^{2}
6 .
And

3. T_{n}^{0} (s) = −

n

X

k=1

log k

k^{s} converges uniformly on [1 + h, ∞)
by Weierstrass M-test. Hence, we have proved that

ζ^{0}(s) = −

∞

X

n=1

log n
n^{s}

by Theorem 9.13. By Mathematical Induction, we know that
ζ^{(k)}(s) = (−1)^{k}

∞

X

n=1

(log n)^{k}
n^{s} .

### 0.1 Supplement on some results on Weierstrass M- test.

1. In the textbook, pp 224-223, there is a surprising result called Space- filling curve. In addition, note the proof is related with Cantor set in exercise 7. 32 in the textbook.

2. There exists a continuous function defined on R which is nowhere differentiable. The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 154.

Remark: The first example comes from Bolzano in 1834, however, he
did NOT give a proof. In fact, he only found the function f : D → R that
he constructed is not differentiable on D^{0}(⊆ D) where D^{0} is countable and
dense in D. Although the function f is the example, but he did not find the
fact.

In 1861, Riemann gave g (x) =

∞

X

n=1

sin (n^{2}πx)
n^{2}

as an example. However, Reimann did NOT give a proof in his life until 1916, the proof is given by G. Hardy.

In 1860, Weierstrass gave

h (x) =

∞

X

n=1

a^{n}cos (b^{n}πx) , b is odd, 0 < a < 1, and ab > 1 + 3π
2 ,
until 1875, he gave the proof. The fact surprises the world of Math, and
produces many examples. There are many researches related with it until
now 2003.

### Mean Convergence

9.26 Let f_{n}(x) = n^{3/2}xe^{−n}^{2}^{x}^{2}. Prove that {f_{n}} converges pointwise to 0
on [−1, 1] but that l.i.m._{n→∞}f_{n}6= 0 on [−1, 1] .

Proof : It is clear that {f_{n}} converges pointwise to 0 on [−1, 1] , so it

remains to show that l.i.m._{n→∞}f_{n}6= 0 on [−1, 1] . Consider
Z 1

−1

f_{n}^{2}(x) dx = 2
Z 1

0

n^{3}x^{2}e^{−2n}^{2}^{x}^{2}dx since f_{n}^{2}(x) is an even function on [−1, 1]

= 1

√2 Z

√ 2n 0

y^{2}e^{−y}^{2}dy by Change of Variable, let y =√
2nx

= 1

−2√ 2

Z

√2n

0

yd
e^{−y}^{2}

= 1

−2√ 2

"

ye^{−y}^{2}

√2n

0 −

Z

√2n

0

e^{−y}^{2}dy

#

→

√π 4√

2 since Z ∞

0

e^{−x}^{2}dx =

√π

2 by Exercise 7. 19.

So, l.i.m._{n→∞}f_{n}6= 0 on [−1, 1] .

### 9.27

Assume that {f_{n}} converges pointwise to f on [a, b] and that l.i.m.

_{n→∞}f

_{n}= g on [a, b] . Prove that f = g if both f and g are continuous on [a, b] .

Proof : Since l.i.m._{n→∞}f_{n} = g on [a, b] , given ε_{k} = _{2}^{1}k, there exists a n_{k}
such that

Z b a

|f_{n}_{k}(x) − g (x)|^{p}dx ≤ 1

2^{k}, where p > 0
Define

h_{m}(x) =

m

X

k=1

Z x a

|f_{n}_{k}(t) − g (t)|^{p}dt,
then

a. h_{m}(x) % as x %
b. h_{m}(x) ≤ h_{m+1}(x)

c. hm(x) ≤ 1 for all m and all x.

So, we obtain h_{m}(x) → h (x) as m → ∞, h (x) % as x %, and

h (x) − h_{m}(x) =

∞

X

k=m+1

Z x a

|f_{n}_{k}(t) − g (t)|^{p}dt % as x %

which implies that

h (x + t) − h (x)

t ≥ h_{m}(x + t) − h_{m}(x)

t for all m. (*)

Since h and h_{m} are increasing, we have h^{0} and h^{0}_{m} exists a.e. on [a, b] . Hence,
by (*)

h^{0}_{m}(x) =

m

X

k=1

|f_{n}_{k}(t) − g (t)|^{p} ≤ h^{0}(x) a.e. on [a, b]

which implies that

∞

X

k=1

|f_{n}_{k}(t) − g (t)|^{p} exists a.e. on [a, b] .

So, f_{n}_{k}(t) → g (t) a.e. on [a, b] . In addition, f_{n} → f on [a, b] . Then we
conclude that f = g a.e. on [a, b] . Since f and g are continuous on [a, b] , we
have

Z b a

|f − g| dx = 0

which implies that f = g on [a, b] . In particular, as p = 2, we have f = g.

Remark: (1) A property is said to hold almost everywhere on a set S (written: a.e. on S) if it holds everywhere on S except for a set of measurer zero. Also, see the textbook, pp 254.

(2) In this proof, we use the theorem which states: A monotonic function h defined on [a, b] , then h is differentiable a.e. on [a, b] . The reader can see the book, The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 113.

(3) There is another proof by using Fatou’s lemma: Let {f_{k}} be a
measruable function defined on a measure set E. If f_{k} ≥ φ a.e. on E and
φ ∈ L (E) , then

Z

E

lim

k→∞inf f_{k}≤ lim

k→∞inf Z

E

f_{k}.

Proof : It suffices to show that f_{n}_{k}(t) → g (t) a.e. on [a, b] . Since
l.i.m.n→∞fn = g on [a, b] , and given ε > 0, there exists a nk such that

Z b a

|f_{n}_{k}− g|^{2}dx < 1
2^{k}

which implies that

Z b a

m

X

k=1

|fn_{k}− g|^{2}dx <

m

X

k=1

1
2^{k}
which implies that, by Fatou’s lemma,

Z b a

m→∞lim inf

m

X

k=1

|f_{n}_{k} − g|^{2}dx ≤ lim

m→∞inf Z b

a m

X

k=1

|f_{n}_{k}− g|^{2}dx

=

∞

X

k=1

Z b a

|f_{n}_{k}− g|^{2}dx < 1.

That is,

Z b a

∞

X

k=1

|f_{n}_{k}− g|^{2}dx < 1
which implies that

∞

X

k=1

|f_{n}_{k}− g|^{2} < ∞ a.e. on [a, b]

which implies that fn_{k} → g a.e. on [a, b] .

Note: The reader can see the book, Measure and Integral (An In- troduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 75.

(4) There is another proof by using Egorov’s Theorem: Let {f_{k}} be a
measurable functions defined on a finite measurable set E with finite limit
function f. Then given ε > 0, there exists a closed set F (⊆ E) , where

|E − F | < ε such that

f_{k} → f uniformly on F.

Proof : If f 6= g on [a, b] , then h := |f − g| 6= 0 on [a, b] . By continuity of h, there exists a compact subinterval [c, d] such that |f − g| 6= 0. So, there exists m > 0 such that h = |f − g| ≥ m > 0 on [c, d] . Since

Z b a

|fn− g|^{2}dx → 0 as n → ∞,

we have

Z d c

|fn− g|^{2}dx → 0 as n → ∞.

then by Egorov’s Theorem, given ε > 0, there exists a closed susbet F of [c, d] , where |[c, d] − F | < ε such that

f_{n}→ f uniformly on F
which implies that

0 = lim

n→∞

Z

F

|f_{n}− g|^{2}dx

= Z

F

n→∞lim |f_{n}− g|^{2}dx

= Z

F

|f − g|^{2}dx ≥ m^{2}|F |

which implies that |F | = 0. If we choose ε < d−c, then we get a contradiction.

Therefore, f = g on [a, b] .

Note: The reader can see the book, Measure and Integral (An In- troduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 57.

9.28 Let f_{n}(x) = cos^{n}x if 0 ≤ x ≤ π.

(a) Prove that l.i.m._{n→∞}f_{n} = 0 on [0, π] but that {f_{n}(π)} does not
converge.

Proof : It is clear that {f_{n}(π)} does not converge since f_{n}(π) = (−1)^{n}.
It remains to show that l.i.m._{n→∞}f_{n}= 0 on [0, π] . Consider cos^{2n}x := g_{n}(x)
on [0, π] , then it is clear that {g_{n}(x)} is boundedly convergent with limit
function

g =

0 if x ∈ (0, π) 1 if x = 0 or π . Hence, by Arzela’s Theorem,

n→∞lim Z π

0

cos^{2n}xdx =
Z π

0

g (x) dx = 0.

So, l.i.m._{n→∞}f_{n}= 0 on [0, π] .

(b) Prove that {f_{n}} converges pointwise but not uniformly on [0, π/2] .
Proof : Note that each f_{n}(x) is continuous on [0, π/2] , and the limit
function

f = 0 if x ∈ (0, π/2]

1 if x = 0 .

Hence, by Theorem9.2, we know that {f_{n}} converges pointwise but not
uniformly on [0, π/2] .

9.29 Let f_{n}(x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let f_{n}(x) = n if
1/n < x < 2/n. Prove that {f_{n}} converges pointwise to 0 on [0, 1] but that
l.i.m.n→∞fn 6= 0 on [0, 1] .

Proof : It is clear that {f_{n}} converges pointwise to 0 on [0, 1] . In order
to show that l.i.m._{n→∞}f_{n} 6= 0 on [0, 1] , it suffices to note that

Z 1 0

f_{n}(x) dx = 1 for all n.

Hence, l.i.m._{n→∞}f_{n} 6= 0 on [0, 1] .

### Power series

9.30 If r is the radius of convergence ifP an(z − z_{0})^{n}, where each a_{n} 6= 0,
show that

n→∞lim inf

a_{n}
a_{n+1}

≤ r ≤ lim

n→∞sup

a_{n}
a_{n+1}

.

Proof : By Exercise 8.4, we have 1

lim_{n→∞}sup

an+1

an

≤ r = 1

lim_{n→∞}sup |a_{n}|^{n}^{1} ≤ 1
lim_{n→∞}inf

an+1

an

.

Since

1
lim_{n→∞}sup

an+1

an

= lim

n→∞inf

a_{n}
a_{n+1}

and

1 limn→∞inf

an+1

an

= lim

n→∞sup

a_{n}
a_{n+1}

,

we complete it.

9.31 Given that two power series P a_{n}z^{n} has radius of convergence 2.

Find the radius convergence of each of the following series: In (a) and (b), k is a fixed positive integer.

(a) P∞
n=0a^{k}_{n}z^{n}
Proof : Since

2 = 1

lim_{n→∞}sup |a_{n}|^{1/n}, (*)

we know that the radius of P∞

n=0a^{k}_{n}z^{n} is
1

lim_{n→∞}sup |a^{k}_{n}|^{1/n} = 1

lim_{n→∞}sup |a_{n}|^{1/n}k = 2^{k}.

(b) P∞

n=0a_{n}z^{kn}
Proof : Consider

n→∞lim sup
a_{n}z^{kn}

1/n = lim

n→∞sup |a_{n}|^{1/n}|z|^{k} < 1
which implies that

|z| < 1

lim_{n→∞}sup |a_{n}|^{1/n}

!1/k

= 2^{1/k} by (*).

So, the radius of P∞

n=0anz^{kn} is 2^{1/k}.
(c) P∞

n=0a_{n}z^{n}^{2}
Proof : Consider

lim sup
a_{n}z^{n}^{2}

1/n

= lim

n→∞sup |a_{n}|^{1/n}|z|^{n}
and claim that the radius of P∞

n=0a_{n}z^{n}^{2} is 1 as follows.

If |z| < 1, it is clearly seen that the series converges. However, if |z| > 1,

n→∞lim sup |a_{n}|^{1/n} lim

n→∞inf |z|^{n} ≤ lim

n→∞sup |a_{n}|^{1/n}|z|^{n}