1. ## [SOLVED] unique point

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

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 $f$ such that $f(x)=1+\int_0^x f(x-y)dy$. Put $z=x-y$ in the integral, then differentiate the equation you get, this gives you a differential equation satisfied by $f$.

3. thank you

I will try now

4. Hi LAURENT

IS THAT GIVING f '(x)= -f(x)?

and the unique fixed point is e^(-x)

am I right?

Hi LAURENT

IS THAT GIVING f '(x)= -f(x)?

and the unique fixed point is e^(-x)

am I right?
No, it is f'(x)=f(x). The integral becomes $\int_x^0 f(z)(-dz)=\int_0^x f(z) dz$, there's no minus sign.

6. why is the limits from x to 0?

why is the limits from x to 0?
It comes from the change of variable: if y=0, then z=x, and vice-versa.

8. oh yes. I substitute wrong. Thank you very much.

And how do we show F is a contraction on X?

$| Ff(x)-Fg(x) | \leq \int_{0}^{x} |f(x-y)-g(x-y) |dy \leq \frac{1}{2} \sup_{z\in [0, \frac{1}{2} ]} \{ |f(z)-g(z) |\}$