# Analysis. Function spaces/Contraction mappings

1. Feb 5, 2009

### datenshinoai

1. The problem statement, all variables and given/known data

Let K: [0,1] x [0,1] -> R be a continuous function such that K(x,y) > 0 and the integral K(x,y) dy <= C < 1 (from 0 to 1) for all x within [0,1].

Let g:[0,1] -> R be any continuous Function. Show that there is a unique continuous function f:[0,1] -> R such that f(x)= g(x) + integral K(x,y) dy

3. The attempt at a solution

I know that we are given 2 functions and that we need to input those into the formula. We want to see the output is closer than the input by a factor of k, but I'm not sure what it is that I should be doing.

To show uniqueness, I say to let g(f(x)) = h(x). Which means we are sending [0,1] -> R to [0,1] -> R, but I don't think that's right...

2. Feb 5, 2009

### quasar987

Did you make a typo in the question? Because the unique continuous function f such that f(x)= g(x) + integral K(x,y) dy is the function f defined by f(x):=f(x)= g(x) + integral K(x,y) dy

3. Feb 6, 2009

### HallsofIvy

Staff Emeritus
I suspect you mean
[tex]f(x)= g(x)+ \int K(x,y) f(y) dy[/itex]
That's a variation of the Poisson proof of the existance and uniqueness of solutions to first order differential equations. Are you allowed to use the Banach fixed point theorem? (If f(x) is a contraction map then there is a unique x such that f(x)= x.)

4. Feb 6, 2009

### quasar987

Probably he is allowed to use Banach's fixed point theorem because of the title of the thread.

The question as to whether Banach's fixed point theorem can be applied to this situation amounts to finding
(1) a complete metric space inhabited by continuous maps from [0,1] to R
(2) a contraction F on the above space such that [F(f)](x)=g(x) + integral K(x,y)f(y) dy