Problem Solving - Hilbert Space

                                             Problem Solving - Hilbert  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, normed linear space and Banach 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

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

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

some of the problems in Hilbert 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. In a Hilbert space, if ||x|| = ||y|| = 1 and ϵ > 0 is given then there exists 

0 < δ < 1 such that ||(x + y)/2|| > 1- δ ⇒ ||x - y|| < ϵ. A Banach space with this

property is said to be uniformly convex.

Think what is the relation between ||x - y|| and ||x + y|| ? 

Yes, parallelogram law.

By parallelogram law,

||x - y||2 = 2 ||x||2 + 2||y||2 - ||x + y||2

            = 4 - ||x + y||2.

||x - y||2 =  4 - ||x + y||2 < ϵ2 if ||x + y||2 > 4 - ϵ2.

So we can consider two cases ϵ ≥ 2 and ϵ < 2 and get δ accordingly. 

If ϵ ≥ 2, for all 0 < δ < 1,

||(x + y) /2|| > 1- δ ⇒ ||x + y|| >  2(1- δ)  > 0 

                             ⇒ ||x - y||2 = 4 - ||x + y||2 < 4 ≤ ϵ2 

                             ⇒ ||x - y|| < ϵ

If 0 < ϵ < 2, let δ = 1- ½  (4 - ϵ2)½ .

Since (4 - ϵ2)½  < 2, 0 < δ < 1. And 

||(x + y)/2|| > 1- δ ⇒ ||(x + y)/2||2 > (1- δ)2 = ¼ (4 - ϵ2)

                                 ⇒ ||x + y||2 > 4 - ϵ2

                                 ⇒ ||x - y||2 =  4 - ||x + y||2 < ϵ2

                                 ⇒ ||x - y|| < ϵ.

Therefore for any ϵ > 0 there exists 0 < δ < 1 such that 

||(x + y)/2|| > 1- δ ⇒ ||x - y|| < ϵ.

3. Show that an orthonormal set in a Hilbert space is linearly independent. And use

this to prove that a Hilbert space is finite dimensional ⇔ every complete

orthonormal set is a basis.  

Let S be an orthonormal set in a Hilbert space H.

Let x1, x2, . . . , xn ∈ S and 𝛂1x1 + 𝛂2x2 + . . . + 𝛂nxn = 0.

To show 𝛂i  = 0 for each i, we have to use that x1, x2, . . . , xn  are orthonormal

vectors. So, we consider the inner product (𝛂1x1 + 𝛂2x2 + . . . + 𝛂nxn, xi).

Then for each i, 

0 = (0, xi) = (𝛂1x1 + 𝛂2x2 + . . . + 𝛂nxn, xi)

                 = 𝛂1(x1, xi) + 𝛂2(x2, xi) + . . . + 𝛂n(xn, xi)

                 = 𝛂i (since (xj, xi) = 0, if j ≠ i and (xj, xi) = 1, if j = i).

Therefore S is linearly independent.

Suppose H is finite dimensional.

Let S be a complete orthonormal set in H.

Then S is linearly independent. 

Since H is finite dimensional, S has only a finite number of elements.

Suppose S = {x1, x2, . . . , xn}.

Since S is a complete orthonormal set in H,

Hence S = {x1, x2, . . . , xn} generates H.

Therefore S is a basis of H.

Conversely, suppose every complete orthonormal set in H is a basis of H.

Since any Hilbert space has a complete orthonormal set, 

H has a complete orthonormal set S.

By hypothesis, S is a basis.

To prove H is finite dimensional we have to show S is a finite set. For that we

cannot proceed with S is an orthonormal basis. So we prove it contra-positively. 

Suppose S is infinite.

Think how to proceed.

Since S is a basis for any x ∈ H there exist finite x1, x2, . . . , xn ∈ S such that 

x = 𝛂1x1 + 𝛂2x2 + . . . + 𝛂nxn.

Since S is infinite there exists an infinite number of e ∈ S – {x1, x2, . . . , xn}.

For such e’s, 

(x, e) = (𝛂1x1 + 𝛂2x2 + . . . + 𝛂nxn, e)

         = 𝛂1(x1, e) + 𝛂2(x2, e) + . . . + 𝛂n(xn, e)

         = 𝛂10 + 𝛂20 + . . . + 𝛂n0

         = 0 (since e ∉ {x1, x2, . . . , xn}, (xj, e) = 0, for all j = 1, 2, . . . , n).

So we can get a contradiction if we get a x ∈ H such that there are infinite

e ∈ S – {x1, x2, . . . , xn} with (x, e) ≠ 0.

Since S is orthonormal for any x in H, {e ∈ S / (x, e) ≠ 0} is either empty or countable. So we have to get a x with {e ∈ S / (x, e) ≠ 0} is countably infinite.

How to get such x?

Yes, we can get x in H from a countably infinite subset of S, as in the proof of theorems on complete orthonormal sets.

Let {en  /  n = 1, 2, . . . } be a countably infinite subset of S.

Let (an) be a sequence of nonzero scalars.