Suppose $g \in C[a,b]$, and $M(b-a) < 1$, and let $y$, $y_N$ $\in C[a,b]$ be the unique solutions of the equations,
$y = g + Ky$ and $y_N = g + K_N y_N$.
$y_N$ has been previously defined as an approximation to $y$. $K$ and $K_N$ are linear operators.
We are asked to show that, $$|| y - y_N||_\infty \leq \dfrac{M_N \|g\|_\infty}{(1 - M(b-a))^2}$$
where $M =\max | k(x,t)|$ ( $k$ is the kernel applied by the operator $K$) and $M_N = \max | k(x,t) - k_N(x,t)|$ ( $k_N$ is the kernel applied by $K_N$).
Some previous results we have to use are,
$\|K\|, \|K_N\|$ are both $\leq M(b-a)$.
$\|K - K_N\| \leq M_N (b-a)$.
Also, I believe
$\|Ky\| \leq M\|y\|$.
Any help would be much appreciated! I'm failing to even work the inequality backwards at the moment! I believe the first step should be to write $\|y - y_N\|$ as,
$\| K(y - y_N) + (K - K_N) y_N \|$, but beyond that I'm stumped!
$K$ is defined as, $K\phi = \int_a^b k(x,t) \phi(t) dt$
$K_N$ is defined as, $K_N \phi = \int_a^b k_N(x,t) \phi(t) dt$ where,
$k_N(x,t)$ is an approximation to $k$ by picking a step size $h$ to cut up $[a,b] $into $a=t_0 < t_1 < ... <t_N = b$. and evaluating $t$ at the midpoint of the interval it's in. Although I don't think the definition of $k_N$ will be needed in showing the desired result.
$k \in $ $(C[a,b])^2$.
$$\|y-y_n\|_\infty=\| K(y - y_N) + (K - K_N) y_N \|\leq \|K\|\|y-y_n\|_\infty+\|K-K_N\|\|y_N\|$$ $$\Leftrightarrow (1-\|K\|)\|y-y_N\|_\infty\leq \|K-K_N\|\|y_N\|$$ $$\|y-y_N\|_\infty\leq \frac{\|K-K_N\|\|y_N\|_\infty}{1-\|K\|}\quad\quad (*)$$ For $\|y_N\|$ you have: $$y_N=g+K_Ny_N\Rightarrow \|y_N\|_\infty\leq \|g\|_\infty+\|K_N\|\|y_N\|_\infty\Rightarrow \|y_N\|_\infty\leq \frac{\|g\|_\infty}{1-\|K_N\|}$$ Plug this in $(*)$: $$\|y-y_N\|_\infty\leq \frac{\|K-K_N\|\|y_N\|_\infty}{1-\|K\|}\leq \frac{\|K-K_N\|\|g\|_\infty}{(1-\|K\|)(1-\|K_N\|)}\leq \frac{M_N(b-a)\|g\|_\infty}{(1-M(b-a))^2}$$