X=C[0,1/2] , F:X---->X defined by
F(f)(x)=1+f(x)/3+integral of f(s)ds from 0 to x
how to prove F is a contraction?
how to find the unique fixed point?
Thanx
To prove it's a contraction you proceed as follows:
$\displaystyle \vert F(f)-F(g) \vert = \vert \frac{f(x)-g(x)}{3} + \int_{0}^{x} f(t)-g(t)dt \vert \leq \frac{ \vert f(x)-g(x) \vert }{3} + \int_{0}^{x} \vert f(t)-g(t) \vert dt$ now taking the supremum over $\displaystyle x$ we get that this last is $\displaystyle \leq \frac{\Vert f-g \Vert }{3} + \frac{1}{2} \Vert f-g \Vert = \frac{5}{6} \Vert f-g \Vert$
To find the fixed point you could follow th proof of the fixed point theorem and iterate $\displaystyle F$ given a simple initial function (say $\displaystyle f=x$ or some such) or you could try by inspection see if you can find it.
Edit: For simplicity you could assume $\displaystyle f$ is differentiable, derive the expression $\displaystyle F(f)=f$ and obtain an easy diff. equation, solve, substitute and determine the constant.
Spoiler: