for any  &gt

1. Quizz 11

(1) Let (f_n) be a sequence in C[a, b] such that (f_n) is convergent to f in (C[a, b], k · k∞).

Prove that

n→∞lim Z b

a

f_n(x)dx = Z b

a

f (x)dx.

Proof. Let s =Rb

af (x)dx and sn=Rb

afn(x)dx for n ≥ 1. Since (fn) is convergent to f in (C[a, b], k · k∞), for any  > 0, there exists N_∈ N such that

kf_n− f k_∞< 

b − a, for any n ≥ N. For any n ≥ N,

|s_n− s| ≤ Z b

a

|f_n(x) − f (x)|dx

≤ kf_n− f k_∞· (b − a)

< 

b − a· (b − a)

= .

This proves that lim

n→∞sn= s.



(2) Let fn(x) = nx(1 − x²)ⁿ for x ∈ [0, 1].

(a) Let x₀ ∈ [0, 1]. Prove that

n→∞lim f_n(x₀) = 0.

Proof. When x₀ = 0 or 1, f_n(x₀) = 0 for all n ≥ 1. Thus lim

n→∞f_n(x₀) = 0. When 0 < x0 < 1, let r = (1 − x²₀). Then 0 < r < 1.

Lemma 1.1. For 0 < r < 1, lim

n→∞nrⁿ= 0.

Proof. Since 0 < r < 1, (1/r) > 1. Set (1/r) = 1 + a for a > 0. By binomial theorem and a > 0,

1

rⁿ = (1 + a)ⁿ= 1 +

n

X

k=0

n k



a^k≥ 1 + na + n(n − 1) 2 a². This implies that for any n ≥ 1,

0 < nrⁿ< n

1 + na +ⁿ⁽ⁿ⁻¹⁾₂ a² . Since lim

n→∞0 = lim

n→∞

n

1 + na + ⁿ⁽ⁿ⁻¹⁾₂ a² = 0, lim

n→∞nrⁿ = 0 by Sandwich princi-

ple. 

Since 0 < x₀< 1,

0 < fn(x0) = nx0(1 − x²₀)ⁿ< n(1 − x²₀)ⁿ. By Lemma (1.1) and the Sandwich principle, lim

n→∞f_n(x₀) = 0. 

1

2

(b) Is (fn) convergent to the zero function in (C[0, 1], k · k∞)? (Hint: consider (1)).

Proof. If (fn) is convergent to f in (C[0, 1], k·k∞), by Problem (1), lim

n→∞

Z 1 0

fn(x)dx = Z 1

0

f (x)dx. On the other hand, Z ₁

0

f_n(x)dx = Z ₁

0

nx(1 − x²)ⁿdx = n

2(n + 1), n ≥ 1.

Hence

n→∞lim Z 1

0

fn(x)dx = 1 2 6= 0 =

Z 1 0

f (x)dx.



(3) Let (f_n) be a sequence in (C[a, b], k · k∞). Suppose that (f_n) is convergent to f in (C[a, b], k · k∞). Prove or disprove that (f_n²) is convergent to f² in (C[a, b], k · k∞).

Proof. Since (f_n) is convergent to f in (C[a, b], k · k∞), there exists M > 0 such that kf_nk∞≤ M for any n ≥ 1. Hence kf k∞≤ M. For any x ∈ [a, b],

|f_n(x) + f (x)| ≤ |f_n(x)| + |f (x)| ≤ kf_nk_∞+ kf k∞≤ 2M.

Therefore kfn+ f k∞≤ 2M for any n ≥ 1. Since (f_n) is convergent to f in (C[a, b], k · k_∞), for any  > 0, there exists N_∈ N such that

kf_n− f k_∞< 

2M for n ≥ N_. For x ∈ [a, b] and any n ≥ 1,

|f_n(x)²− f (x)²| = |f_n(x) − f (x)||fn(x) + f (x)| ≤ 2M kfn− f k_∞. We find that for any n ≥ 1,

kf_n²− f²k_∞≤ 2M kf_n− f k_∞. Hence if n ≥ N_,

kf_n²− f²k_∞≤ 2M kf_n− f k_∞< 2M ·  2M = .

 (4) Let (an) be a sequence of real numbers such thatP∞

n=1|a_n| is convergent. Define f (x) =

∞

X

n=1

a_nsin nx, x ∈ [0, 2π].

(a) Show that f defines a real valued continuous function on [0, 2π], i.e. f ∈ C[0, 2π].

Proof. Let fn(x) = ansin nx for x ∈ [0, 2π]. If x ∈ [0, 2π],

|f_n(x)| = |an|| sin nx| ≤ |a_n|.

Hence kf_nk_∞≤ |a_n| for any n ≥ 1. SinceP∞

n=1|a_n| is convergent in R,P∞

n=1kf_nk_∞ is convergent in R. Since (C[0, 2π], k · k∞) is a Banach space,P∞

n=1fnconverges to f in (C[0, 2π], k · k∞). Hence f ∈ C[0, 2π].



3

(b) Let k be a natural number. Find Z 2π

0

f (x) sin kxdx in terms of a_n. Proof. Let s_n(x) = Pn

j=1f_j(x) for n ≥ 1 and t_n(x) = s_n(x) sin kx and t(x) = f (x) sin kx for x ∈ [0, 2π]. For x ∈ [0, 2π] and n ≥ 1,

|t_n(x) − t(x)| = |s_n(x) − f (x)|| sin kx| ≤ ks_n− f k_∞.

Therefore ktn− tk∞ ≤ ks_n− f k∞ for any n ≥ 1. Since (fn) converges to f in (C[0, 2π], k · k∞), (tn) converges to t in (C[0, 2π], k · k∞). By Problem (1),

n→∞lim Z 2π

0

tn(x)dx = Z 2π

0

t(x)dx.

For n ≥ k, Z 2π

0

tn(x)dx =

n

X

j=1

aj

Z 2π 0

sin jx sin kxdx = πa_k. We see that

Z 2π 0

f (x) sin kxdx = lim

n→∞

Z 2π 0

tn(x)dx = lim

n→∞πak= πak.

 (c) Find

Z 2π 0

|f (x)|²dx in terms of a_n. Proof. For any n ≥ 1,

ks_nf − f²k_∞≤ ks_n− f k_∞kf k_∞,

we can show that (snf ) is convergent to f in (C[0, 2π], k · k∞). By Problem (1),

n→∞lim Z 2π

0

s_n(x)f (x)dx = Z 2π

0

f (x)²dx.

For n ≥ 1, Z 2π

0

s_n(x)f (x)dx =

n

X

k=1

a_k Z 2π

0

f (x) sin kxdx =

n

X

k=1

a_k· πa_k. This shows that for n ≥ 1,

Z 2π 0

s_n(x)f (x)dx = π ·

n

X

k=1

a²_k. This implies that

Z 2π 0

f (x)²dx = lim

n→∞ π ·

n

X

k=1

a²_k

!

= π

∞

X

n=1

a²_n.