# Metric Spaces, Contraction mapping and the fredholm integral equation.

Greetings to all who choose to look at this question. i'm about half way through a 20 credit unit named metric spaces and have run into a problem which im hoping (And fairly sure) that someone can help clarify.

the section in particular is about contraction mapping and it's applications. so some basic definitions first

$$\text{Def: }x \in X \text{ is a fixxed point of the mapping }T, \text{if } \\Tx=x$$

$$\text{Def: }T:x \rightarrow x \text{ is a contraction if } (\exists \lambda \in [0,1))(\forall x,y \in X)\left[ d(Tx,Ty) \leq \lambda d(x,y) \right]$$

we've also proven the uniqueness of a fixed point in complete metric space $(X,d)$

the notes then provides several examples using the euclidean and supreme metric then we come to a new example. where the notes shows the existance of a unique solution to the Fredholm integral Equation of the second kind for $\lambda$ sufficiently small.

-tl;dr- skip to the bottom of the yellow section

quickly:

let $K:[a,b] \times [a,b] \rightarrow \mathbb{R} \text{ be a continous function and let, } \theta:[a,b] \rightarrow \mathbb{R} \text{ be continous.}$ consider the integral

$$\psi(x) = \lambda \int_{a}^{b}K(x,y)\psi(y)dy+\theta(x), \psi \in C[a,b]$$ (the Fredholm integral equation of the second kind) where $\psi$ is in the set of all continous functions on [a,b] and $\lambda$ constant.

we now prove that for $|\lambda|$ small enough there exists a unique $\psi \in C[a,b]$ which solves the above equation. In a later chapter we shall prove that continous functions on closed bounded subsets of $\mathbb{R^m}$ are bounded ie, $\exists M ~s.t: ~|K(x,y)|\leq M.$ we'll also show that continous functions on closed bounded subsets of $\mathbb{R^m}$ are uniformly continous, ie $$\forall \epsilon > 0~\exists \delta > 0,~s.t: ~ |x_1-x_2| < \delta \Longrightarrow |\theta(x_1)-\theta(x_2)| \leq \epsilon$$ and $|K(x_1,y)-K(x_2,y)| < \epsilon$. also for $\psi \in C[a,b] \exists~N_{\psi} ~ s.t:~ |\psi(y)| \leq N_{\psi}, ~ \forall (a\leq y \leq b)$

(students note...so yeah i've got that to look forward to!)

Define T on C[a,b] by

$$(T\psi)(x) = \lambda \int_{a}^{b}K(x,y)\psi(y)dy+\theta(x), ~a \leq x \leq b$$

then

$$|T\psi(x_1)-T\psi(x_2)| \leq |\lambda||\int_{a}^{b}(K(x_1,y)-K(x_2,y))\psi(y)dy|+|\theta(x_1)-\theta(x_2)|\leq |\lambda| \epsilon N_{\psi}(b-a)+\epsilon$$

for $|x_1 - x_2| \leq \delta$

Hence $T\psi \in C[a,b]. ~\forall \psi_1, \psi_2 \in C[a,b]$ we have

$$d(T\psi_1,T\psi_2)= \max_{a \leq x \leq b}|T\psi_1(x)-T\psi_2(x)| \leq |\lambda| \int_{a}^{b}|K(x,y)||\psi_{1}(y)-\psi_{2}(y)|dy \leq |\lambda|M(b-a)d(\psi_1,\psi_2)$$

hence T is a contraction mapping on a complete metric space for $|\lambda| \leq (M(b-a))^{-1}$ and $T\psi=\psi$ has a unique solution. then $\psi(x) = \lambda \int_{a}^{b}K(x,y)\psi(y)dy+\theta(x)$ has a unique solution in C[a,b] for $|\lambda| \leq (M(b-a))^{-1}$

now i chose to add in the proof so that the reader may have a better understanding of the current level i'm at (being a second year undergrad) now im no stranger to uniform convergence and the epsilon proof of continous functions so given that [a,b] is closed i can accept the assumption that $K(x,y)$ is closed, i've also come to understand that $K(x,y)$ is called the kernel function though i have no experience in such things yet (though i'll (for now) assume that this will be somewhat linked to the kernel as described in linear algebra)

now reading on the internet leads me to a variety of ways of solving these problems including the resolvant formalism in the effect of the Liouville-Neumann series which i admit at this point escapes me and in all honesty i dont believe im expected to know at this point. but none the less i understand that the goal here is to find $\psi(x)$

-tl;dr-

my issue stems stems from the next example:

Consider

$$\psi(x)=\lambda \int_{-1}^{1}(x-y)\psi(y)dy+1,~ \psi \in C[-1,1]$$ then $b-a = 2$ and $|x-y| \leq 2$ on $[-1,1] \times [-1,1].$ so there exists a unique solution for $|\lambda| \leq \frac{1}{4}$

(student note: i understand that this restriction is because of $M = 2$ (in this case and as said above $b-a = 2$)). we see the solution is of the form $\psi(x) = c+dx$ (student note: and this is the leap i'm unable to see) which im sure has simple reasoning which im just missing entirely.

Hence

$$c+dx = \lambda \int_{-1}^{1}(x-y)(c+dy)dy+1 = 1 - \frac{2d\lambda}{3}+2c\lambda x$$

so $c = 1 - \frac{2d \lambda}{3}$, $d= 2c\lambda$ with solutions $c = \frac{3}{3+4\lambda^2}$ and $d = \frac{6\lambda}{3+4\lambda^2}$ and we see that this is also the unique solution for $|\lambda| \geq \frac{1}{4}$

so there you have it... any one willing to help please do so. although i may not be expected to know the liouville-neumann method i feel that the example here displays a level i should be familiar with. so if anyone could explain the form $\psi(x) = c+dx$ and the end results that would be great and i garentee you a shiny new upvote.

• Use linearity of the integral: expand the bracket to get a difference of integrals, one of which you can factor x out of. Now neither integral is dependent on x, i.e. the RHS is a constant plus constant times x. – Max Freiburghaus Nov 3 '17 at 19:17
• i....really didnt see that. thank you. – Vaas Nov 3 '17 at 19:20
• even with $\int_{-1}^{1} x(c+dy)-y(c+dy) dy$ i admit i dont actually know how to integrate with infinitismals, should i be considering them purely as a constant then? – Vaas Nov 3 '17 at 21:07
• **** notes. dx is a constant times x and not an infinitismal. – Vaas Nov 3 '17 at 22:23
• ok i understand why its the end result but not our choice of c+d•x – Vaas Nov 3 '17 at 22:25