MAT 127B

HW 10 Solutions(6.2.3/6.2.5/6.2.13)

Exercise 1 (6.2.3)
For each n ∈ N and x ∈ [0, ∞), let
(
x 1, if x ≥ n1
gn (x) = and hn (x) = 1
1 + xn nx, if 0 ≤ x < n

Answer the following questions for the sequences (gn ) and (hn ):
a) Find the pointwise limit on [0, ∞).

b) Explain how we know that the convergence cannot be uniform on [0, ∞).

c) Choose a smaller set over which the convergence is uniform and supply an argument
to show that this is indeed the case.
Proof.

a) Pointwise limit of gn .

• If 0 ≤ x < 1,
x
lim xn = 0 =⇒ lim gn (x) = lim = x.
n→∞ n→∞ n→∞ 1 + xn

• If x = 1

1 1
lim 1n = 1 =⇒ lim gn (x) = lim n
= .
n→∞ n→∞ n→∞ 1 + 1 2
• If x > 1
x
lim xn = ∞ =⇒ lim gn (x) = lim = 0.
n→∞ n→∞ n→∞ 1 + xn

1
Hence let g(x) be the pointwise limit of gn (x)

x, if 0 ≤ x < 1

g(x) = 12 , if x = 1

0, if x > 1


Pointwise Convergence of hn (x)

• If x=0

lim hn (x) = lim 0 = 0.

n→∞ n→∞

• If x > 1,
Since n1 −→ 0 as n −→ ∞.
Hence for x > 1, there exists N ∈ N(depends on x) such that for all n ≥ N,
1
≤ x.
n
Hence
lim hn (x) = lim hn (x) = lim 1=1
n→∞ n→∞;n≥N n→∞;n≥N

Hence let h(x) be the pointwise limit of hn (x)

(
0, if x = 0
h(x) =
1, if x > 0

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b) We know by a Theorem that if fn (x) be a sequence of continuous functions defined on
A ⊂ R which converges uniformly to a function f (x), then f (x) is continuous in A.

We see that A = [0, ∞), and gn are continuous functions in A. If the sequence gn
converges uniformly to g, then the function g(x) should be continuous in [0∞).

Since that is not the case, hence the convergence is not uniform.

Similarly for hn (x).

We see that A = [0, ∞), and hn are continuous functions in A.
This is because

lim+ hn (x) = lim+ 1 = 1

1 1
x→ n x→ n

1
lim− hn (x) = lim+ nx = n = 1.
1
x→ n 1
x→ n n

If the sequence hn converges uniformly to h, then the function h(x) should be contin-
uous in [0, ∞).

Since that is not the case, hence the convergence is not uniform.

c) Let us choose the interval,  

1
A = 0, .
2
Given  > 0, choose N ∈ N such that
 
− ln()
N > max ,1 .
ln 2

Let n ≥ N.
For any x ∈ [0, 12 ]
   n+1 
 x   x 
|gn (x) − g(x)| =  − x =
  1 + xn  .

1 + xn
Since x ≥ 0,

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Hence

1 1
1 + xn ≥ 1 =⇒ ≤1 and xn+1 ≤
1 + xn 2n+1
 n+1 
 x
≤ 1 .

=⇒ 
1 + xn  2n+1

Now, for the choice of n,

− ln()
n≥N >
ln 2
1
=⇒ 2n+1 > 2n >

1
=⇒ < .
2n+1
Hence we have,
|gn (x) − g(x)| < .

for any n ≥ N and for any x ∈ 0, 21 .

 

Hence in A = [0, 12 ], gn (x) converges uniformly to g(x).

For hn (x) choose

A = [2, 3].

Since n ≥ 1, hence for x ∈ [2, 3], hn (x) = 1 for every x ∈ [0, 1].

Given  > 0, choose N = 1 ∈ N, such that for any n ≥ N,

|hn (x) − h(x)| = 0 < 

for any x ∈ [2, 3]
Hence in A = [2, 3], hn (x) converges uniformly to h(x).

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Exercise 2 (6.2.5)
Using the Cauchy Criterion for convergent sequences of real numbers, supply a proof for
Theorem 6.2,5 (First, define a candidate for f(x), and then argue that fn −→ f uniformly.)
Proof.

Theorem 6.2.5

A sequence of functions (fn ) defined on a set A ⊂ R converges uniformly on A if and only

if for every  > 0 there exists an N ∈ N such that |fn (x) − fm (x)| <  whenever m, n ≥ N
and x ∈ A.

=>
Let (fn ) be a sequence of functions which converge uniformly on A to f (say).

Given  > 0,
there exists N 0 ∈ N such that

|fn (x) − f (x)| <
2
for n ≥ N and for all x ∈ A.

Choose N = N 0 ∈ N.
For n, m ≥ N, and for all x ∈ A we have
 
|fn (x) − fm (x)| ≤ |fn (x) − f (x)| + |f (x) − fm (x)| < + = .
2 2

<=
Let for every  > 0, there exists N ∈ N such that |fn (x) − fm (x)| <  whenever m, n ≥ N
and x ∈ A.

Given  > 0,
there exists N ∈ N such that for every x ∈ A and for any m, n ≥ N, |fn (x) − fm (x)| < .
Let x ∈ A be any element.
Consider the sequence {fn (x)} ⊂ R.

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Since we know that the sequence {fn (x)} is a Cauchy sequence for every x ∈ A in R, hence
we know that this sequence converges to a real number. We define,

f (x) = lim fn (x).

n→∞

for every x ∈ A.

We will show that fn converges uniformly to f .

Given  > 0,
choose the N ∈ N such that |fn (x) − fm (x)| < 2
for every n, m ≥ N and for every x ∈ A. [From the assumption]

We will show,
|fn (x) − f (x)| < 
for all n ≥ N and for every x ∈ A.

Let x ∈ A be any element.

Since the sequence fn (x) converges to f (x), hence given the  > 0, choose Nx ∈ N(may
depend on x) such that |fm (x) − f (x)| < 2 . for all m ≥ Nx .

Let n ≥ N.
Choose m > max{N, Nx }.
Then
 
|fn (x) − f (x)| < |fn (x) − fm (x)| + |fm (x) − f (x)| < + = .
2 2

Given  > 0, there exists an N ∈ N such that for any n ≥ N and for any x ∈ A, we have

|fn (x) − f (x)| < .

This proves that fn converges uniformly to f on A.

Exercise 3 (6.2.13)
Recall that the Bolzano–Weierstrass Theorem (Theorem 2.5.5) states that every bounded
sequence of real numbers has a convergent subsequence. An analogous statement for bounded
sequences of functions is not true in general, but under stronger hypotheses several different

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conclusions are possible. One avenue is to assume the common domain for all of the functions
in the sequence is countable. (Another is explored in the next two exercises.)
Let A = x1 , x2 , x3 , . . . be a countable set. For each n ∈ N , let fn be defined on A and assume
there exists an M > 0 such that |fn (x)| ≤ M for all n ∈ N and x ∈ A. Follow these steps
to show that there exists a subsequence of (fn ) that converges pointwise on A.

a) Why does the sequence of real numbers fn (x1 ) necessarily contain a convergent sub-
sequence (fnk )? To indicate that the subsequence of functions (fnk ) is generated by
considering the values of the functions at x1 , we will use the notation fnk = f1,k .

b) Now, explain why the sequence f1,k (x2 ) contains a convergent subsequence.

c) Carefully construct a nested family of subsequences (fm,k ), and show how this can be
used to produce a single subsequence of (fn ) that converges at every point of A.

Proof.

a) Consider the sequence

{fn (x1 )}n∈N .

We know by Bolzano-Weierstrass Theorem states that every bounded sequence of real

numbers has a convergent subsequence.
|fn (x1 )| ≤ M for every n ∈ N implies that there is a convergent subsequence in
{fn (x1 )}.
Let {fnk (x1 )}k∈N be the subsequence which converges. We call fnk = f1,k .

{f1,k }k∈N such that f1,k (x1 ) is a convergent sequence.

b) Consider the sequence in R,

{f1,k (x2 )}.

Since |f1,k (x2 )| ≤ M is a bounded sequence implies there is a convergent subsequence

in R, call it f2,k (x2 ).

We have

{fn } ⊃ {f1,k } ⊃ {f2,k } such that {f2,k (x2 )}, {f1,k (x1 )} converges in R.

c) We will continue the process. After we obtain {fi,k } such that fi,k (xi ) converges, we
consider the sequence {fi,k (xi+1 )}.
Since |fi,k (xi+1 )| ≤ M is a bounded sequence in R, hence there exists a convergent
subsequence {fi+1,k (xi+1 )}.
Hence we obtain

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{fn } ⊃ {f1,k } ⊃ {f2,k } ⊃ · · · ⊃ {fi,k } ⊃ {fi+1,k } ⊃ . . . such that {fj,k (xj )} converges.

Consider the subsequence

{gn } = {f1,1 , f2,1 , f2,2 , f3,1 , f3,2 , f3,3 , f4,1 . . . }.

OR

{gn } = {fi,j |i ≥ j}

with dictionary order.

If,
fi,j = gk and fi0 .j 0 = gk0 then
k < k 0 if (i, j) < (i0 , j 0 ) in dictionary order.

Need to show that the sequence {gn } converges pointwise on A.

Let xi be any point on A.
Consider the sequence {fi,k(xi ) }.
We know that this sequence {fi (xi )} converges hence every subsequence of it.

{gn } = {f1,1 , f2,1 . . . fi−1,i−1 , fi,1 , fi,2 , . . . fi,i , fi+1,1 . . . }

Since we have the nested family of subsequence, hence {fj,k } ⊂ {fi,k } for every j ≥ i,
hence the sequence

{fi,1 , fi,2 , . . . fi,i , fi+1,1 . . . fj,t . . . } ⊂ {gn }; i ≤ j; t ≤ j

is a subsequence {fi,k }, and we know fi,k (xi ) converges

hence

{fi,1 (xi ), fi,2 (xi ), . . . fi,i (xi ), fi+1,1 (xi ) . . . fj,t (xi )} . . . } ⊂ {fi,k (xi ); i ≤ j; t ≤ j

converges.

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Hence
lim gn (xi ) = lim fl,k (xi ) = lim fl,k (xi )
n→∞ l→∞;k≤l l→∞;k≤l;l≥i

which is exactly the limit of the sequence

{fi,1 (xi ), fi,2 (xi ), . . . fi,i (xi ), fi+1,1 (xi ) . . . fj,t (xi )} . . . } ⊂ {fi,k (xi ); i ≤ j; t ≤ j

which converges.
This implies
{gn (xi )} converges and since xi was an arbitrary element of A, hence the subsequence
{gn } ⊂ {fn } converges pointwise on A.