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**[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

[tex]T(u)(x)=\int_{0}^{1}k(x,y)u(y)dy[/tex]

(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

[tex]u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

(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 [tex]||T(u)||_2\leq ||k||_2||u||_2[/tex]

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*:

[tex]u^*(x)=\int_{0}^{1}k(x,y)u^*(y)dy[/tex]

Fine, but what about the equation

[tex]u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

? If we define an operator T_f by

[tex]T_f(u)(x)=f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

this is not a contraction.

How do the dots connect?