Existence of solution to integral equation

[quasar987]

[SOLVED] Existence of solution to integral equation

Homework Statement

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])

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?

 

[morphism]

Why is T_f not a contraction?

 

[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!

 

[Defennder]

Wow, morphism's one line reply was enough to answer your lengthy post.

 

[HallsofIvy]

'Tis often so.

And often, the lengthy (and, hopefully, complete) first post results in the response being brief.