ASSIGNMENT I - HINTS TO SELECTED QUESTIONS
First let us observe the following elementary result from calculus:
∑∞ ∑∞
Let n=1 an be a convergent series of real numbers. Then the tail of the series n=N an
converges to zero as N → ∞. ∑∞
Let (sn ) be the∑sequence of partial sums of the series n=1 an and (s′n ) be the sequence of
∞
partial sums of n=N an . Let ϵ > 0. Since (sn ) is convergent, there exists M ∈ N such that
|sn − sm | < ϵ ∀ n, m > M.
Note that
∑
n ∑
N −1 ∑
n
sn = ak = ak + ak = sN −1 + s′n .
k=1 k=1 k=N
Similarly, sm = sN −1 + s′m . Consequently,
|sn − sm | < ϵ ⇒ |sn − sN −1 + sN −1 − sm | < ϵ ⇒ |s′n − s′m | < ϵ.
′
∑∞ (sn ) is a Cauchy sequence of real numbers
Therefore, ∑∞ and hence convergent.
∑∞ That is, the
series
∑∞ n=N a n is convergent. Note that l := n=1 a n = sN −1
∑∞ + n=N a n and hence,
n=N an = l − sN −1 . Since sN −1 → l as N → ∞, it follows that n=N an → 0 as N → ∞.
∑∞
(11) Let x ∈ lp , so that j=1 |x(j)|p := l < ∞. Then,
∞
∑ ∞
∑
∥xn − x∥pp = |xn (j) − x(j)|p = |x(j)|p → 0 as n → ∞.
j=1 j=n+1
p 1
Thus, (xn ) converges to x in l . Since l is a Banach space, and a Banach space cannot have a
denumerable Hamel basis, {e1 , e2 , . . . } cannot be a basis for l1 .
(15) Consider l2 and the bounded subset S = {e1 , e2 , . . . }. Since ∥en − em ∥2 = 1 for all
m, n ∈ N, m ̸= n, S cannot have a limit point and consequently, S is closed. However, S is not
compact, since the sequence (en ) in S has no convergent subsequence.
(16) Consider R. The sequence ( n1 ) is convergent in R with respect to the usual Euclidean
metric, but it is not convergent with respect to the discrete metric. Therefore, the Euclidean
metric and the discrete metric are not equivalent.
(17) Consider X = R2 and for x = (x(1), x(2)) ∈ X, consider the norm ∥x∥∞ = ∥(x(1), x(2))∥∞ =
max{|x(1)|, |x(2)|} and the subspace X0 = {(x(1), x(2)) ∈ R2 : x(2) = 0}. Note that for
x = (0, 1) ∈ R2 , the elements {(a, 0) : −1 ≤ a ≤ 1} are best approximants from X0 .
(21) When T is the set of natural numbers with usual Euclidean metric (equivalent to discrete
metric on N),
C(T ) = l∞ , C0 (T ) = c0 , Cc (T ) = c00 .
(22) Trivial if x = 0. Let 0 ̸= x ∈ lr . We have
( ∥x∥ )p ∑ ∞ (
|x(n)| )p ∑ ( |x(n)| )r ( ∥x∥r )r
∞
p
= ≤ = .
∥x∥∞ n=1
∥x∥∞ n=1
∥x∥∞ ∥x∥∞
That is,
( ∥x∥ ) pr
r
∥x∥p ≤ ∥x∥∞ . (∗)
∥x∥∞
1
First let us observe the following elementary result from calculus:
∑∞ ∑∞
Let n=1 an be a convergent series of real numbers. Then the tail of the series n=N an
converges to zero as N → ∞. ∑∞
Let (sn ) be the∑sequence of partial sums of the series n=1 an and (s′n ) be the sequence of
∞
partial sums of n=N an . Let ϵ > 0. Since (sn ) is convergent, there exists M ∈ N such that
|sn − sm | < ϵ ∀ n, m > M.
Note that
∑
n ∑
N −1 ∑
n
sn = ak = ak + ak = sN −1 + s′n .
k=1 k=1 k=N
Similarly, sm = sN −1 + s′m . Consequently,
|sn − sm | < ϵ ⇒ |sn − sN −1 + sN −1 − sm | < ϵ ⇒ |s′n − s′m | < ϵ.
′
∑∞ (sn ) is a Cauchy sequence of real numbers
Therefore, ∑∞ and hence convergent.
∑∞ That is, the
series
∑∞ n=N a n is convergent. Note that l := n=1 a n = sN −1
∑∞ + n=N a n and hence,
n=N an = l − sN −1 . Since sN −1 → l as N → ∞, it follows that n=N an → 0 as N → ∞.
∑∞
(11) Let x ∈ lp , so that j=1 |x(j)|p := l < ∞. Then,
∞
∑ ∞
∑
∥xn − x∥pp = |xn (j) − x(j)|p = |x(j)|p → 0 as n → ∞.
j=1 j=n+1
p 1
Thus, (xn ) converges to x in l . Since l is a Banach space, and a Banach space cannot have a
denumerable Hamel basis, {e1 , e2 , . . . } cannot be a basis for l1 .
(15) Consider l2 and the bounded subset S = {e1 , e2 , . . . }. Since ∥en − em ∥2 = 1 for all
m, n ∈ N, m ̸= n, S cannot have a limit point and consequently, S is closed. However, S is not
compact, since the sequence (en ) in S has no convergent subsequence.
(16) Consider R. The sequence ( n1 ) is convergent in R with respect to the usual Euclidean
metric, but it is not convergent with respect to the discrete metric. Therefore, the Euclidean
metric and the discrete metric are not equivalent.
(17) Consider X = R2 and for x = (x(1), x(2)) ∈ X, consider the norm ∥x∥∞ = ∥(x(1), x(2))∥∞ =
max{|x(1)|, |x(2)|} and the subspace X0 = {(x(1), x(2)) ∈ R2 : x(2) = 0}. Note that for
x = (0, 1) ∈ R2 , the elements {(a, 0) : −1 ≤ a ≤ 1} are best approximants from X0 .
(21) When T is the set of natural numbers with usual Euclidean metric (equivalent to discrete
metric on N),
C(T ) = l∞ , C0 (T ) = c0 , Cc (T ) = c00 .
(22) Trivial if x = 0. Let 0 ̸= x ∈ lr . We have
( ∥x∥ )p ∑ ∞ (
|x(n)| )p ∑ ( |x(n)| )r ( ∥x∥r )r
∞
p
= ≤ = .
∥x∥∞ n=1
∥x∥∞ n=1
∥x∥∞ ∥x∥∞
That is,
( ∥x∥ ) pr
r
∥x∥p ≤ ∥x∥∞ . (∗)
∥x∥∞
1
