Consider the DE $y' = 1 + xy$ and $y(0) = 0$ on $[-1,1]$.
Show that the associated integral operator is a contraction mapping.
Attempt:
In order to do this one has to rewrite the DE into a fixed point problem. Performing those steps I got the following fixed point problem:
$$TF(x) = \Gamma + \int_{0}^{x}\Phi(t,F(t))dt = 0 + \int_{0}^{x}1 + ty(t) dt = x + \int_{0}^{x}ty(t) dt$$, where $T$ is defined as the integral operator.
Now I have to show that this is a contraction mapping. I did the following:
$$\|TF(x) - TF(z)\| = \|x + \int_{0}^{x}ty(t) dt - z + \int_{0}^{z}ty(t) dt\| = \| x - z + \int_{z}^{x}ty(t) dt\| \leq \| x - z \| + \|\int_{z}^{x}ty(t) dt\| \leq \| x - z \| + \int_{z}^{x}\|ty(t)\| dt = \| x - z \| + \|ty(t)\| \int_{z}^{x}1 dt $$
This is where I am stuck. I tried a few things after this particularly:
$$\| x - z \| + \|ty(t)\| \|x-z\| = (1 + \|ty(t)\|) \|x-z\|$$.
This would not be a contraction mapping because it is greater than or equal to $1$. I know I am on the right path, but I am making a small mistake somewhere along this chain of inequalities.
Could I get some assistance?
We use the max norm $$ || T(y) - T(z) ||_{\infty} = \max_{x \in [-1, 1]} | T(y)(x) - T(z)(x) | $$ For any $x \in [-1, 1]$, we write $$ | T(y)(x) - T(z)(x) | = \left| \int_{0}^{x} t( y(t) - z(t) )dt \right| $$ and $$ | T(y)(x) - T(z)(x) | \leq \left| \int_{0}^{x} t dt \right| ||y-z ||_{\infty} \leq \frac{1}{2} ||y-z ||_{\infty} $$