How do we find the fixed point of the map F:X-->X
for F(f)(x)=1+ integration of f(x-y)dy from 0 to x
X=C[0,1/2]
Thank you
You want $\displaystyle f$ such that $\displaystyle f(x)=1+\int_0^x f(x-y)dy$. Put $\displaystyle z=x-y$ in the integral, then differentiate the equation you get, this gives you a differential equation satisfied by $\displaystyle f$.