the solution of Fredholm´s integral equation Be  $\lambda \in \mathbb{R}$ such that $\left | \lambda  \right |> \left \| \kappa  \right \|_{\infty }(b-a)$.
Prove that the solution $f^*$ of the integral equation of Fredholm $$\lambda f -\int_{a}^{b}\kappa (x,y)f(y)dy=g(x)$$  for all $x\in [a,b]$ satisfies $$\left \| f^*-\sum_{m=1}^{k}\frac{1}{\lambda ^m}\Im ^{m-1}g \right \|_{\infty }\leq \frac{\alpha ^{k}}{(1-\alpha) \left | \lambda  \right |} \left \| g \right \|_{\infty }$$ for all $k\in \mathbb{N}$ where $$\alpha :=\frac{\left \| \kappa  \right \|(b-a)}{\left |\lambda  \right | }$$
i´m really stuck in this problem, I know that I can use that $$\Im :C_{\infty }^{0}[a,b]\rightarrow C_{\infty }^{0}[a,b]$$ is Lipschitz continuous and that linear Fredholm´s operator is linear, can anybody just give a hint please?
thanks!

Since $\lambda \ne 0$, we may write
\begin{align*}
f(x) - \frac{1}{\lambda}\int_{a}^{b} \kappa(x,y) f(y)\, \mathrm{d}y = \frac{g(x)}{\lambda}
\end{align*}
Let $\Im\colon C^{\infty}_{0} \to C^{\infty}_{0}$ be defined by
\begin{align*}
\Im[f](x):=\int_{a}^{b} \kappa(x,y) f(y) \, \mathrm{d}y
\end{align*}
Then by standard integral inequalities
\begin{align*}
\frac{\left|\Im[f](x)  \right|}{\lambda}
\le \frac{\left\|\kappa\right\|_{\infty}}{\lambda}(b-a)\left\|f\right\|_{\infty} = \alpha\left\|f\right\|_{\infty} <\left\|f\right\|_{\infty}, 
\end{align*}
which shows that $\left\|\Im \right\|/|\lambda|\le \alpha < 1$, i.e, that $\Im/\lambda$ is a contractive linear operator on the Banach space $C_{0}^{\infty}$. 
Symbolically, we can now write the equation as 
\begin{align*}
(I-\Im/\lambda)f = g/\lambda.
\end{align*}
In elementary algebra, we would write $(1-x)^{-1} = \sum_{k=0}^{\infty} x^{k}$ if $|x|<1$. A similar identity holds here:
\begin{align*}
(I-\Im/\lambda)\sum_{m=0}^{k} \Im^{m}/\lambda^{m}
&=
I-(\Im/\lambda)^{k+1} \to I \text{ as } k\to \infty
\end{align*} 
as $\left\|\Im\right\|_{\infty}/|\lambda| <1$. 
Hence we may write
\begin{align*}
(I-\Im/\lambda)^{-1} = \sum_{m=0}^{\infty} \frac{\Im^{m}}{\lambda^{m}}
\end{align*}
Applying this operator gives
\begin{align*}
f 
&= (1-\Im/\lambda)^{-1}g/\lambda 
= \sum_{m=0}^{\infty} \frac{\Im^{m}}{\lambda^{m+1}}g \\
&= \sum_{m=1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m}}g
\end{align*}
Now the problem is easy:
\begin{align*}
\left\|f^{*} -\sum_{m=1}^{k} \frac{\Im^{m-1}}{\lambda^{m}}g
  \right\|_{\infty}
&=
\left\| \sum_{m=1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m}}g -\sum_{m=1}^{k} \frac{\Im^{m-1}}{\lambda^{m}}g
  \right\|_{\infty} \\
&\le
\left\|\sum_{m=k+1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m-1}}\right\| \frac{\left\|g\right\|_{\infty}}{|\lambda|} \\
&\le
\sum_{m=0}^{\infty} \left(\frac{\left\|\Im \right\| }{|\lambda|}\right)^{m+k}
\frac{ \left\|g\right\|_{\infty} }{ |\lambda| } \\
&\le
\sum_{m=0}^{\infty} \alpha^{m}\alpha^{k}\frac{\left\|g\right\|_{\infty}}{|\lambda|} \\
&=
\frac{\alpha^{k}\left\|g\right\|_{\infty}}{(1-\alpha)|\lambda|}
\end{align*}
