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Lecture 01 - Introduction to Functional Analysis

We start the lecture with a problem from financial mathematics, posed by Dr. Yerkin to Dr. Mokhtar.

2t e(t,x)=ln(xx e(t,x))

Taking exponential both sides we obtain

xxe=e2te

This is simply a differential equations problem, and DE is apart of functional analysis.


We take the following spring problem:
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By Hooke's Law we know that the restoring force of the spring g=kf with k>0.

By Newton's 2nd law of motion

g=mx¨

Where x is the displacement. Thus the net force acting on the body is

mf¨=kf+g(t)

By setting m=k=1 we obtain:

(1)f(t)+f(t)=g(t)(2)f(a)=1f(a)=0 (Initial Conditions)

This is a non-homogenous equation. Thus we first solve the homogenous case (that is g=0).

f(t)+f(t)=0λ2+1=0λ=±i

Thus we obtain the solution that

(3)f(t)=Asin(t)+Bcos(t)

with A,BR. Now let gC([a,b])]. We can solve the inhomogenous case using variation of the parameters method, where

A=A(t)B=B(t)

Leading us to the solution of the problem:

f(t)=cos(ta)+atsin(tξ)g(ξ)dξ

Till this point we havent used anything from functional analysis. Let us modify the problem to:

f(t)+f(t)=f(t)σ(t)

with the same initial conditions. Using our solution above we get:

f(t)=cos(ta)+atsin(tξ)f(ξ)σ(ξ)dξ

That is we tranformed our problem from a differential equation into an integral equation. Let us define the operator and function:

(Kf)(t)=atsin(tξ)σ(ξ)f(ξ)dξu(t)=cos(ta)

Our problem transforms to:

(4)f=u+Kf

This is simply a fixed point problem. We can try to solve this using an iterative scheme of maps (by Picard AlKashi).

fn+1=u+Kfn

This gives us a sequence of functions that (ideally) converges to a solution to our problem. We want our solution to be continuous so we are interested in uniform convergence since (recall) that the uniform convergence of continuous functions is continuous.

Definition

We say a sequence fn uniformly converges iff ϵ>0 NN s.t m,n>N

Max |fn(x)fm(x)|<ϵ

The notion of taking the distance between two functions (Max|fn(x)fm(x)|) is called a norm on the functions which is of particular interest in functional analysis.


Let us return back to our fixed point problem and iterate through with f0=u. We obtain, by induction, that

fn=u+Ku+K2u+...+Knu

where Kn is defined to be K composed with itself n times. Thus, w.l.o.g, let n>m,

fnfm=Km+1u+Km+2u+...+Knu

We want ||fnfm||0 as n,m. Taking the triangle inequality both sides we obtain:

||fnfm||||Km+1u||+||Km+2u||+...+||Knu||

This sequence goes to zero provided that 0||Knu||<
Note that since f(t)=σ(t)f(t)f(t) and our solution f(t) is continuous thus fC2 (that the second derivative exists and is continuous). Now we take notice of the following:

(Kf)(t)=atsin(tξ)σ(ξ)f(ξ)dξ|(Kf)(t)|at|σ(ξ)|f(ξ)|dξ||Kf||M||f||

where M=||σ||(ba). By induction one can show:

||Knu||||u|| ||σ||n(ba)nn!

Thus our series:

n=0||Knu||||u||01n!||σ||n(ba)n=||u||e||σ||(ba)<

That is fn is a Cauchy sequence, hence our fixed point iteration will indeed converge and is a solution to our differential equation.

Here are some questions to ponder:

  1. Why did the method work?
  2. Could it work in other situations?
  3. What properties of continuous functions are used?

We used the fact that Ξ=C([a,b]) is a normed vector space, and Ξ is a complete space. A normed vector space which is complete is called a Banach space.

Theorem

Let Ξ be a Banach space, and K:ΞΞ be an operator such that:

  • K(u+v)=K(u)+K(v)
  • K(v)=K(v)
  • ||Kv||<M||v|| for some MR+
  • n=0||Knv||< n,vΞ

Then for each uΞ !fΞ such that f=u+Kf.

To prove uniqueness:

Proof

Let f=f1f2 where both fi are solutions. Then f=Kf=Knf thus:

||f||=||Knf||n0f=0