# Homework Help: Existence of solution to integral equation

1. Feb 2, 2008

### quasar987

[SOLVED] Existence of solution to integral equation

1. The problem statement, all variables and given/known data
There's k:[0,1]²-->R square integrable and the operator T from L²([0,1],R) to L²([0,1],R) defined by

$$T(u)(x)=\int_{0}^{1}k(x,y)u(y)dy$$

(a) Show that T is linear and continuous.

(b) If ||k||_2 < 1, show that for any f in L²([0,1],R) , there exists a u in L²([0,1],R) solution of the integral equation

$$u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy$$

(almost everywhere on [0,1])

3. The attempt at a solution

I have done (a), and in demonstrating that T is continuous, I have shown that $$||T(u)||_2\leq ||k||_2||u||_2$$

Therefor, the condition ||k||_2 < 1 in (b) + the fact that T is linear is equivalent to saying that T is a contraction. So by Banach's fixed point theorem, there exists a (unique) fixed point to T; call it u*:

$$u^*(x)=\int_{0}^{1}k(x,y)u^*(y)dy$$

Fine, but what about the equation

$$u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy$$

??? If we define an operator T_f by

$$T_f(u)(x)=f(x)+\int_{0}^{1}k(x,y)u(y)dy$$

this is not a contraction.

How do the dots connect?!?!?

2. Feb 3, 2008

### morphism

Why is T_f not a contraction?

3. Feb 3, 2008

### quasar987

You're right, I spoke too fast!

I wanted to look at ||T_f(u) - T_f(0)||=||T_(u)||, but I looked at ||T_f(u)|| instead.

Thank you morphism!

4. Feb 3, 2008