Problem Solving - Banach Space

Problem Solving - Banach Space  

M.Velrajan

 

Problem Solving - Banach 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,

real analysis and normed linear space 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

https://velrajanm.blogspot.com/2025/02/problem-solving-normed-linear-space.html

In this connection this post illustrates how to develop problem solving skills, by

some of the problems in Banach 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 similar ways. And with this experience the  learner can solve other

problems.

1. Let N be a non-zero normed linear space. Then 

N is a Banach space ⇔ {x ∈ N / ||x|| = 1} is complete.

Suppose N is a Banach space. 

Since N is a Banach space, N is complete.

We have to prove that the subset {x ∈ N / ||x|| = 1} of N is complete.

Think when a subset of a complete space is complete.

Yes a closed subset of a complete space is complete. So, we have to prove that 

the subset {x ∈ N / ||x|| = 1} is a closed subset of N.

Clearly, {x ∈ N / ||x|| = 1} is the preimage of {1} of the norm function on N.  

Since norm is a continuous function and {1} is closed in the scalar space,

{x ∈ N / ||x|| = 1} is a closed subset of N.

Since a closed subset of a complete space is complete, 

{x ∈ N / ||x|| = 1} is complete.

Conversely, suppose B = {x ∈ N / ||x|| = 1} is complete.

To prove N is a Banach space, we have to prove that N is complete

i.e. every Cauchy sequence in N is convergent.

So, we consider a Cauchy sequence (xn) in N. 

And from the sequence we have to get a Cauchy sequence in B. 

The natural sequence in B is the sequence (xn / ||xn||). 

( xm / ||xm|| ) -  ( xn / ||xn|| ) = [||xn||xm - ||xm||xn] / ||xm|| ||xn||

= [||xn||(xm - xn) - (||xm|| - ||xn|| )xn ] / ||xm|| ||xn||

|| xm / ||xm|| ) -  ( xn / ||xn|| || ≤ ( ||xm - xn|| / ||xm|| ) + ( | ||xm|| - ||xn || | ) / ||xm||

Since (xn) is a Cauchy sequence there exists N0 such that 

| ||xm|| - ||xn|| | ≤ ||xm - xn|| < ε/2, for all m, n ≥ N0.

So, to show || xm / ||xm|| ) -  ( xn / ||xn|| || < ε, we have to show that there exists N1 such

that ||xn|| > 0 for all n ≥ N1 and (1/||xn||) is bounded or (1/||xn||) is convergent or

(||xn||) is convergent. 

Let (xn) be a Cauchy sequence in N.

Since | ||xm|| - ||xn|| | ≤ ||xm - xn||, (||xn||) is a Cauchy sequence of real numbers and

hence (||xn||) is a convergent sequence.

Let ||xn|| → a. 

Since ||xn|| ≥ 0, for all n, a ≥ 0.

Suppose a = 0. 

Since ||xn - 0|| = ||xn|| = | ||xn|| - 0|, xn → 0 in N.

Suppose a ＞0. 

Since ||xn|| → a, there exists N1 such that | ||xn|| - a | ≤ a/2, for all n ≥ N1. 

Hence -a/2  ≤ ||xn|| - a, for all n ≥ N1.

i.e. ||xn|| ≥ a/2 > 0, for all n ≥ N1. 

Hence (1 / ||xn||) → 1/a and (1 / ||xn||) ≤ 2/a, for all n ≥ N1. 

( xm / ||xm|| ) -  ( xn / ||xn|| ) = [||xn||xm - ||xm||xn] / ||xm|| ||xn||

= [||xn||(xm - xn) - (||xm|| - ||xn|| )xn ] / ||xm|| ||xn||

|| xm / ||xm|| ) -  ( xn / ||xn|| || ≤ ( ||xm - xn|| / ||xm|| ) + ( | ||xm|| - ||xn || | ) / ||xm||

Since (xn) is a Cauchy sequence  there exists N2 such that 

| ||xm|| - ||xn|| | ≤ ||xm - xn|| < aε/4, for all m, n ≥ N2.

Let N = max {N1, N2}. Then

|| xm / ||xm|| ) -  ( xn / ||xn|| || < (2/a) ( aε/4 + aε/4) = ε, for all m, n ≥ N.

(xn / ||xn||) is a Cauchy sequence in B.

Since B is complete, (xn / ||xn||) is a convergent sequence.

Suppose (xn / ||xn||) → y in B.

Then, since the scalar multiplication of a normed linear space is jointly continuous

and xn = ||xn|| (xn / ||xn||), xn → ay in N.

Hence N is complete and hence N is a Banach space. 

2. Let a Banach space B be made into a Banach space B′ by means of a new norm.

Then the topologies generated by these norms are the same if either is stronger than

the other.

Let ℑ and ℑ′ be the topologies on B and B′ generated by their norms.

Suppose ℑ is stronger than ℑ′.

Then the identity transformation I : B → B′ is continuous

(since ℑ′ ⊆ ℑ, if V ∈ ℑ′ then I-1(V) = V ∈ ℑ).

We have to prove that ℑ ⊆ ℑ′ i.e. I-1 : B′ → B is continuous. But I : B → B′ is a

one-to-one continuous linear transformation of B onto B′. Yes,

By an application of open mapping theorem one-to-one continuous linear

transformation of one Banach space onto another  is a homeomorphism.

Hence I is a homeomorphism.

i.e. I-1 : B′ → B is continuous.

So, if V ∈ ℑ then (I-1) -1(V) = I(V) = V ∈ ℑ′.

i.e. ℑ ⊆ ℑ′.

Hence ℑ = ℑ′.

Similarly, if ℑ′ is stronger than ℑ by using the identity transformation I : B′ → B

we can prove that ℑ = ℑ′.

  

3. A one-to-one continuous linear transformation of one Banach space onto another

is a homeomorphism – Show that it follows from the closed graph theorem.    

Let T be a one-to-one continuous linear transformation of a Banach space B onto a

Banach space B′.

We have to use the closed graph theorem.

By closed graph theorem, the graph {(x, T(x)) / x ∈ B} of T is closed in B × B′. 

To prove T is a homeomorphism, we have to prove T-1 : B′ → B is continuous 

i.e. the graph of T-1 = {(y, T-1(y)) / y ∈ B′}is closed in B′ × B.

Since T is one-to-one and onto, for each y ∈ B′ there exists a unique x ∈ B such

that y = T(x) or x = T-1(y). 

Hence {(T-1(y), y) / y ∈ B′} = {(x, T(x)) / x ∈ B} and hence {(T-1(y), y) / y ∈ B′} is

closed in B × B′.

But we want {(y, T-1(y)) / y ∈ B′} is closed in B′ × B. 

Oh, (x, y) has to become (y, x).

Since (x, y) → (y, x) is a homeomorphism from B × B′ onto  B′ × B, 

{(y, T-1(y)) / y ∈ B′} is closed in B′ × B.

i.e. the graph of T-1 is closed in B′ × B.

Hence by closed graph theorem, T-1 : B′ → B is continuous and hence T is a

homeomorphism.

4. Let T be a linear transformation of a Banach space B into a Banach space B′. 

If {fi} is a set of functionals in B′ *  which separates the vectors in B′, 

and if fiT is continuous for each fi then T is continuous.

To prove T is continuous, by closed graph theorem, we have to show that the

graph G of T is closed. We can prove G is closed by proving G = G or by

proving Gc is open.

To prove (x, y) ∈ G ⇒ (x, y) ∈ G we have to use the continuity of fi and fiT.

Think.

Yes, B × B′ is a metric space,  hence there exists a sequence (xn, T(xn)) in G

such that (xn, T(xn)) → (x, y) in B × B′. 

Let G = {(x, T(x)) / x ∈ B} be the graph of T.

Let (x, y) ∈ G ⊆ B × B′.

Then, since B × B′ is a metric space,  there exists a sequence (xn, T(xn)) in G 

such that (xn, T(xn)) → (x, y) in B × B′.

Since the convergence in B × B′ is equivalent to coordinate wise convergence, 

xn → x in B and T(xn) → y in B′.

Since fiT is continuous for each fi and xn → x in B, fiT(xn) → fiT(x) in B′.

i.e. fi(T(xn)) → fi(T(x)) in B′, for all fi 

Since T(xn) → y in B′ and each fi is continuous, fi(T(xn)) → fi(y) in B′. 

Hence fi(T(x)) = fi(y), for all fi.

We have to use that {fi} separates the vectors in B′

Since {fi} separates the vectors in B′, T(x) = y.

(For, if T(x) ≠ y then there exists a fi ∈ {fi} such that fi(T(x)) ≠ fi(y), a contradiction).

Therefore, (x, y) = (x, T(x)) ∈ G.

i.e. G ⊆ G.

Hence G = G.

i.e. G is closed in B × B′.

Therefore, by closed graph theorem, T is continuous.

Aliter : 

Let (x, y) ∈ Gc = B × B′ - G. Then y ≠ T(x).

To prove Gc is open, we have to get a neighbourhood of (x, y) contained in Gc.

We use the given conditions. Since y ≠ T(x) first we use that {fi} separates the

vectors in B′.

Since {fi} separates the vectors in B′, 

there exists a fi ∈ {fi} such that fi(T(x)) ≠ fi(y).

Yes, fi(T(x)) and fi(y) are scalars and the scalar space is Hausdorff.

Since the scalar spaces ℝ or ℂ is Hausdorff there exist open sets U and V such that

fi(T(x)) ∈ U and fi(y) ∈ V and U ∩ V = Ø.

Since fiT and fi are continuous, 

(fiT)- 1 (U) and  fi- 1 (V) are neighbourhoods of x and y.

Hence (fiT)- 1 (U) ✖  fi- 1 (V) is open  in B × B′ and (x, y) ∈ (fiT)- 1 (U) ✖  fi- 1 (V) .

If (u, v) ∈ (fiT)- 1 (U) ✖  fi- 1 (V) then fi(T(u)) ∈ U, fi(v) ∈ V. 

Since U ∩ V = Ø, v ≠ T(u). 

Hence (u, v) ∈ Gc and hence (fiT)- 1 (U) ✖  fi- 1 (V) ⊆ Gc.

So, Gc is open in B × B′.

i.e. G is closed in B × B′.

Therefore, by closed graph theorem, T is continuous.

5. A Banach space B is reflexive if and only if B* is reflexive.

Suppose B is reflexive. Then B = B**, hence B* = B*** = (B*)**. 

i.e. B* is reflexive.

Suppose B* is reflexive. Then B* = B***, hence B** = B****. 

i.e. B** is reflexive.

Since B is closed in B** and a closed subspace of reflexive space is reflexive, 

B is reflexive.

6. The closed unit sphere S in a reflexive Banach space B is weakly compact.

Let B be a reflexive Banach space.

Then the closed unit sphere S* in B* is compact in the weak* topology.

Since B is reflexive, the set {Fx / x ∈ B} = B**.

Hence the weak* topology on B* is same as the weak topology on B*.

Therefore, the closed unit sphere S* in B* is weakly compact.

Since B is reflexive B = B**, hence B* = B***. i.e. B* is reflexive.

Hence, the closed unit sphere S** in B** is weakly compact.

Since B = B**,  in B the closed unit sphere S = S** is weakly compact.

7. Let T be an operator on a Banach space B. 

Then T has an inverse T -1 ⇔ T* has an inverse (T*) -1 

and in this case (T*) -1 = (T -1)*.

Suppose T has an inverse, T -1.

Then TT -1 = I = T -1T, where I is the identity transformation on B.

Hence (TT -1)* = I* = (T -1T)*.

i.e. (T -1)*T* =  I* = T*(T -1)*, where I* is the identity transformation on B*

(since T T* reverses the products and preserves the identity transformation). 

Hence T* has an inverse and (T*) -1 = (T -1)*.

Conversely, suppose T* has an inverse. Then T* is one-one and onto.

Claim : T is one-one.

For, Suppose T is not one-one.

Then there exists x ≠  0 ∈ B such that T(x) = 0. 

And there exists f0 ∈ B* such that f0(x) = ||x|| and ||f0|| = 1.

Since T* is onto, there exists g ∈ B* such that T*(g) = f0. 

Hence ||x|| = f0(x) = T*(g)(x) = g(T(x)) = g(0) = 0 and hence x = 0, a contradiction.

Hence T is one-one.

Let M = T(B). Then M is closed in B.

Suppose M ≠ B. Then there exists y ∈ B and y ∉ M. 

Hence there exists f ∈ B* such that f(M) = 0 and f(y) ≠ 0. 

For all x ∈ B, since T(x) ∈ M, T*(f)(x) = f(T(x)) = f(0) = 0.

Hence T*(f) = 0. But, since f(y) ≠ 0, f ≠ 0, a contradiction to T* is one-one.

Therefore T(B) = M = B.

Hence T is onto.

Therefore T is a one-one continuous linear transformation of the Banach space B

onto B.

Hence T -1 is continuous and T -1 is an operator on B.

i.e. T has an inverse.