Problem Solving - Normed Linear Space

 Problem Solving - Normed Linear Space  

M.Velrajan

Pure Mathematics learning will not be completed unless the learner solves related

problems. Only when the learner solves related problems, the learner has a better

understanding of pure mathematics concepts and can also develop the applying skill

and mathematical or critical thinking. 

To inculcate problem solving among young mathematics learners, how to develop

problem solving was illustrated by some of the problems in groups, linear algebra

and real analysis in the following posts : 

 

https://velrajanm.blogspot.com/2024/04/problem-solving-illustration-m.html

https://velrajanm.blogspot.com/2024/05/problem-solving-groups.html

https://velrajanm.blogspot.com/2024/05/problem-solving-linear-transformation.html

https://velrajanm.blogspot.com/2024/06/problem-solving-analysis.html

In this connection this post illustrates a way of developing problem solving skills, by

some of the problems in Normed Linear Space given in Introduction to Topology and

Modern Analysis by George F. Simmons.

The way of thinking and applying the known concepts are typed in

blue colour.

Learners are advised to use this to develop their mathematical thinking in this or in

any other ways. And with this experience the learner can solve other problems. 

If the learners change their way of learning they can enjoy  the beauties of Pure

Mathematics.

1. Let T be a continuous linear transformation of a normed linear space N into a normed

linear space N′. Then

sup {||T(x)|| / ||x|| ≤ 1} = sup {||T(x)|| / ||x|| = 1} 

                                    = inf {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}.

Let 𝛂 = sup {||T(x)|| / ||x|| ≤ 1}, 𝛃 = sup {||T(x)|| / ||x|| = 1} and

𝛄 = inf {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}.

First we observe and use that {||T(x)|| / ||x|| = 1} ⊆ {||T(x)|| / ||x|| ≤ 1}.

Since {||T(x)|| / ||x|| = 1} ⊆ {||T(x)|| / ||x|| ≤ 1},

𝛃 = sup {||T(x)|| / ||x|| = 1} ≤  𝛂 = sup {||T(x)|| / ||x|| ≤ 1}.

Now observe and use that for x ≠ 0 then ||x/||x|||| = 1.

Let x ∈ N.

If x ≠ 0 then ||x/||x|||| = 1, hence ||T(x/||x||)|| ≤ 𝛃 and hence ||T(x)|| ≤ 𝛃||x||.

If x = 0 then ||T(x)|| = 0 = 𝛃 ||x||.

Hence  ||T(x)|| ≤ 𝛃||x||, for all x ∈ N. Also 𝛃  ≥ 0. So,

𝛃 ∈ {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}. 

Therefore, 𝛄 = inf {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N} ≤ 𝛃.

So far we have 𝛄 ≤ 𝛃 ≤ 𝛂. To complete we have to show that 𝛂 ≤ 𝛄.

Since 𝛄 is the greatest lower bound of 

{K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N} enough to show that 

𝛂 is a lower bound of {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}.

Suppose K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N.

To connect K and 𝛂 we have to consider x ∈ N with ||x|| ≤ 1.

Then for all x ∈ N with ||x|| ≤ 1, ||T(x)|| ≤ K.

Hence K is an upper bound of {||T(x)|| / ||x|| ≤ 1}.

Since 𝛂 is the least upper bound of {||T(x)|| / ||x|| ≤ 1}, 𝛂 ≤ K.

Hence 𝛂 is a lower bound of {K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}.

Since 𝛄 is the greatest lower bound of 

{K / K ≥ 0 and ||T(x)|| ≤ K||x||, for all x ∈ N}, we get 𝛂 ≤ 𝛄.

Therefore, 𝛄 ≤ 𝛃 ≤ 𝛂 ≤ 𝛄.

i.e. 𝛂 = 𝛃 = 𝛄.

2. Let a linear space L be made into a normed linear space in two ways, and let the two

norms of a vector x be denoted by ||x|| and ||x||′. These norms are said to be equivalent if

they generate the same topology on L. 

Show that this the case if and only if there exist two positive real numbers K1 and K2

such that K1 ||x|| ≤ ||x||′ ≤ K2 ||x|| for all x.

A linear transformation T is continuous if and only if there exists K > 0

such that ||T(x)|| ≤ K||x|| for all x.

So we have to use the identity transformation I, so that I(x) = x.

Think how to connect the identity transformation and the two topologies on L. 

Yes,

Two topologies ℑ and ℑ′ on a set X are the same if and only if the identity function 

I : (X, ℑ) → (X, ℑ′) is a homeomorphism.

Hence the norms ||x|| and ||x||′ on L are equivalent 

if and only if the identity linear transformation I : (L, || ||) → (L, || ||′) is continuous 

                      and its inverse is continuous

if and only if there exist K, K′ > 0 such that ||x|| = ||I-1(x)|| ≤ K||x||′ for all x 

                      and ||x||′ = ||I(x)||′ ≤ K′ ||x|| for all x.

if and only if there exist K, K′ > 0 such that 1/K ||x|| ≤ ||x||′ ≤ K′ ||x|| for all x.

3. If n is a fixed positive integer, the spaces lpn (1 ≤ p < ∞) consist of a single underlying

linear space with different norms defined on it. 

Show that these norms are all  equivalent to one another. 

These norms are equivalent if and only if they generate the same topology on the space of all n-tuples of scalars.

Two metric topologies on a set are the same if and only if a sequence is convergent

in one metric ⇔ it is convergent in the other metric.

So, these norms are equivalent  

if and only if  for any 1 ≤ p ≤ q ≤ ∞, xk → x in lpn ⇔ xk → x in lqn.

We proceed with xk → x in lpn. 

Let 1 ≤ p < ∞. Suppose (xk) be a sequence of n-tuples of scalars, 

where xk = ( xk1, xk2, . . . ,xkn) and x = (x1, x2, . . . , xn) is a n-tuple of scalars. 

Then xk → x in lpn 

⇔ for any ε > 0, there exists k0 such that ||xk - x||p < ε, for all k ≥ k0

⇔ for any ε > 0, there exists k0 such that

⇒ for any ε > 0, there exists k0 such that |xki - xi| p ≤

                   and for all i = 1, 2, . . . , n

Note that here for the converse we have to do something more.

So we take it separately. 

⇒ for any ε > 0, there exists k0 such that 

          |xki - xi| < ε, for all k ≥ k0 and for all i = 1, 2, . . . , n                

⇒ xki → xi as k → ∞, for all i = 1, 2, . . . , n

What about the converse ?

And xki → xi as k → ∞, for all i = 1, 2, . . . , n

⇒ for any ε > 0, for each i = 1, 2, . . . , n there exists ki such that 

          |xki - xi| < ε/n1/p, for all k ≥ ki  

⇒ for any ε > 0, there exists k0 = max {k1, k2, . . . , kn} such that

         |xki - xi| < ε/n1/p, for all k ≥ k0 and for all i = 1, 2, . . . , n

⇒ for any ε > 0, there exists k0 such that

         |xki - xi| p < ε p/n, for all k ≥ k0 and for all i = 1, 2, . . . , n 

⇒ for any ε > 0, there exists k0 such that  

⇒ xk → x in lpn

Hence, for any 1 ≤ p < ∞, xk → x in lpn ⇔ xki → xi as k → ∞, for all i = 1, 2, . . . , n.

What about p = ∞ ?

And xk → x in l∞n

⇔ for any ε > 0, there exists k0 such that ||xk - x||∞ < ε, for all k ≥ k0

⇔ for any ε > 0, there exists k0 such that 

                            max {|xki - xi| / i = 1, 2, . . . , n} < ε, for all k ≥ k0

⇒ for any ε > 0, there exists k0 such that

|xki - xi| ≤ max {|xki - xi| / i = 1, 2, . . . , n} < ε, for all k ≥ k0 and

for all i = 1, 2, . . . , n                

⇒ xki → xi as k → ∞, for all i = 1, 2, . . . , n

And xki → xi as k → ∞, for all i = 1, 2, . . . , n

⇒ for any ε > 0, for each i = 1, 2, . . . , n there exists ki such that 

          |xki - xi| < ε, for all k ≥ ki  

⇒ for any ε > 0, there exists k0 = max {k1, k2, . . . , kn} such that

         |xki - xi| < ε, for all k ≥ k0 and for all i = 1, 2, . . . , n

⇒ for any ε > 0, there exists k0 such that 

                                        max {|xki - xi| / i = 1, 2, . . . , n} < ε, for all k ≥ k0

⇒ xk → x in l∞n

Hence, xk → x in l∞n ⇔ xki → xi as k → ∞, for all i = 1, 2, . . . , n.

Therefore, for any 1 ≤ p ≤ ∞, 

xk → x in lpn ⇔ xki → xi as k → ∞, for all i = 1, 2, . . . , n. — (A)

Hence, for any 1 ≤ p ≤ q ≤ ∞, 

xk → x in lpn ⇔ xki → xi as k → ∞, for all i = 1, 2, . . . , n.

                     ⇔ xk → x in lqn

4. If N is an arbitrary normed linear space, show that

any linear transformation T of lpn (1 ≤ p < ∞) into N is continuous.

Since N is a metric space T is continuous if and only if for any sequence 

(xk) in lpn,  xk → x in lpn implies T(xk) → T(x) as k → ∞.

Let 1 ≤ p ≤ ∞. 

Suppose (xk) be a sequence of n-tuples of scalars, where xk = ( xk1, xk2, . . . ,xkn), 

x = (x1, x2, . . . , xn) is a n-tuple of scalars and xk → x in lpn.

Then, by (A) of 3, xki → xi as k → ∞, for all i = 1, 2, . . . , n.

Think how to use this along with T is linear. 

Yes, 

Think how to prove T(xk) → T(x) as k → ∞ using xki → xi in the scalar space.

Yes,

Since the scalar multiplication of the normed linear space N is jointly continuous, 

xkiT(ei) → xiT(ei) as k → ∞, for all i = 1, 2, . . . , n.

Since the addition of the normed linear space N is jointly continuous, 

i.e. T(xk) → T(x) as k → ∞.

5. Let M be a closed linear subspace of a normed linear space N, and x0 be a vector not

in M. Then there exists a functional f0 in N* such that f0(M) = 0, f0(x0) = 1 and ||f0|| = 1/d,

where d is the distance from x0 to M. 

By Hahn Banach Theorem and its proof, we have to get a functional on the smallest

space containing M and x0 such that f(M) = 0, f(x0) = 1 and ||f|| = 1/d. 

Let M0 = (x0) + M be the linear subspace of N generated by x0 and M.

Then any vector y ∈ M0 can be uniquely expressed as y = 𝛂x0 + m, where m ∈ M and 𝛂 is a scalar. 

Think how to define a functional f on M0 with f(m) = 0 and f(x0) = 1.  Yes,

For each y = 𝛂x0 + m ∈ M0, define f(y) = 𝛂.

Then it can be verified that f is a scalar valued linear transformation of M0.

Clearly, f(m) = 0, for all m ∈ M and f(x0) = 1.

We have to show f is continuous and ||f|| = 1/d.

i.e. there exists K > 0 such that |f(𝛂x0 + m)| = |𝛂| ≤ K ||𝛂x0 + m||, for all m ∈ M and

for all scalar 𝛂 and 1/d is the smallest such K.

i.e. |𝛂| / ||𝛂x0 + m|| ≤ 1/d.

Think how to get d in terms of ||𝛂x0 + m|| and |𝛂|.  

Since x0 is a vector not in M, d = d(x0, M) ≠ 0. And

d = inf {||x0 - m|| / m ∈ M} = inf {||x0 + m|| / m ∈ M} (since m ∈ M ⇔ - m ∈ M).

And ||x0 + m|| =  ||𝛂x0 + 𝛂m|| / |𝛂|,  for all 𝛂 ≠ 0.

Hence {||x0 + m|| / m ∈ M} 

= {||𝛂x0 + 𝛂m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0}

= {||𝛂x0 + m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0} 

                [since for m ∈ M, 𝛂m ∈ M, for all scalar 𝛂 ≠ 0,

                 {||𝛂x0 + 𝛂m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0}

                 ⊆ {||𝛂x0 + m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0} and 

                 since for m ∈ M, (m/𝛂) ∈ M and m = 𝛂(m/𝛂), for all scalar 𝛂 ≠ 0

                 {||𝛂x0 + m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0}

                 ⊆ {||𝛂x0 + 𝛂m|| / |𝛂| / m ∈ M and for all scalar 𝛂 ≠ 0}]

= {||𝛂x0 + m|| / |f(𝛂x0 + m)| / m ∈ M and for all scalar 𝛂 ≠ 0}

And d = inf {||𝛂x0 + m|| / |f(𝛂x0 + m)| / m ∈ M and for all scalar 𝛂 ≠ 0}.

Hence d ≤ ||𝛂x0 + m|| / |f(𝛂x0 + m)|, for all m ∈ M and for all scalar 𝛂 ≠ 0.

|f(𝛂x0 + m)| ≤ (1/d) ||𝛂x0 + m||, for all m ∈ M and for all scalar 𝛂 ≠ 0.

(since d > 0 and |f(𝛂x0 + m)| = |𝛂| > 0 for all scalar 𝛂 ≠ 0)

For 𝛂 = 0, |f(𝛂x0 +m)| = 0 ≤ (1/d) ||𝛂x0 + m||, for all m ∈ M.

Hence |f(𝛂x0 + m)| ≤ (1/d) ||𝛂x0 + m||, for all m ∈ M and for all scalar 𝛂.

i.e. |f(y)| ≤ (1/d) ||y||, for all y ∈ M0. — (1)

Hence f is a continuous functional of M0 and ||f|| ≤ (1/d).

Suppose K > 0 and |f(y)| ≤ K ||y||, for all y ∈ M0.

Then |f(𝛂x0 + m)| ≤ K ||𝛂x0 + m||, for all m ∈ M and for all scalar 𝛂.

Hence |f(𝛂x0 + m)| / ||𝛂x0 + m|| ≤ K, for all m ∈ M and for all scalar 𝛂 ≠ 0.

And hence sup {|f(𝛂x0 + m)| / ||𝛂x0 + m|| / m ∈ M and for all scalar 𝛂 ≠ 0} ≤ K.

Since |f(𝛂x0 + m)| / ||𝛂x0 + m|| > 0, for all m ∈ M and for all scalar 𝛂 ≠ 0, 

sup {|f(𝛂x0 + m)| / ||𝛂x0 + m|| / m ∈ M and for all scalar 𝛂 ≠ 0} 

= 1/inf {||𝛂x0 + m|| / |f(𝛂x0 + m)| / m ∈ M and for all scalar 𝛂 ≠ 0} = 1/d.

So, 1/d ≤ K.

Hence K > 0 and |f(y)| ≤ K ||y||, for all y ∈ M0 ⇒ 1/d ≤ K. — (2)

From (1) and (2), ||f|| = inf {K / K > 0 and |f(y)| ≤ K ||y||, for all y ∈ M0} = 1/d.

Therefore, f ∈ M*, ||f|| = 1/d, f(m) = 0, for all m ∈ M and f(x0) = 1.

Hence, by Hahn-Banach theorem, the functional f ∈ M* can be extended to a functional f0 ∈ N* such that ||f0|| = ||f|| = 1/d. 

And f0(m) = f(m) = 0, for all m ∈ M and f0(x0) = f(x0) = 1.

6. A linear subspace of a normed linear space is closed if and only if it is weakly closed.

Suppose M is a closed linear subspace of a normed linear space N.

If M = N then M = N is closed in the weak topology on N.

Since M = N is closed in any topology on N.

i.e. M is weakly closed.

Suppose M ≠ N. 

We know the weak topology on N is the smallest topology on N in which all 

f ∈ N* are continuous. To show M is closed in the weak topology on N we have to

connect M and f ∈ N*. Yes,

Since M is a closed linear subspace of N, for each x ∈ N - M, 

there exists fx ∈ N* such that fx(M) = 0 and fx(x) ≠ 0.

Hence M ⊆ fx -1({0}), for all x ∈ N - M and hence M ⊆ ∩x ∈ N - M fx -1({0}).

Let y ∈ N - M.

Since fy(y) ≠  0, y ∉ fy -1({0}), hence y ∉ ∩x ∈ N - M fx -1({0}).

Hence ∩x ∈ N - M fx -1({0}) ⊆ M.

Therefore, M = ∩x ∈ N - M fx -1({0}).

Since fx is continuous in the weak topology on N, fx -1({0}) is closed in the weak topology on N.

Hence M = ∩x ∈ N - M fx -1({0}) is closed in the weak topology on N.

i.e. M is weakly closed.

Conversely, if M is weakly closed in N then M is closed in N,

since weak topology on N ⊆ topology on N.