let X=C[0,1/2] with standard sup-norm and F:X---->X defined by
F(f)(x)=1+integral of f(x-t)dt from 0 to x
how to prove F is a contraction with factor k=1/2?
and how to find the unique fixed point?
let X=C[0,1/2] with standard sup-norm and F:X---->X defined by
F(f)(x)=1+integral of f(x-t)dt from 0 to x
how to prove F is a contraction with factor k=1/2?
and how to find the unique fixed point?
What is $\displaystyle t$? a constant? if it is then $\displaystyle F(f)= 1+xf(x-t)$ which is clearly a contraction. The fixed point of this function however is not so easy to find (to me at least).
If however you meant $\displaystyle \int_{0}^{x} f(x-t)dt$ then, arguing as in your last question, one gets it's indeed a contraction and using induction we prove that $\displaystyle F^n(1)= \sum_{j=0}^{n} \frac{x^j}{j!} \rightarrow e^x$ (where $\displaystyle F^n$ means composition $\displaystyle n$-times)