solution of an integral equation of $u(x)=f(x)+\lambda \int_\Omega K(x,y)u(y)dy$ Can someone help me with this problem? 
I have shown that $Tu(x)$ is a contraction. Then if I understand one of the theorems correctly, there is a unique $u(x)$ such that $Tu(x) = u(x)$. I am not sure what to do after this or how to come up with $\lambda_o$
Any help is appreciated. Thanks
Here is the full statement of the problem:
Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ and let $K(x,y)$ be a Lebesgue measurable function on the product space $\Omega \times \Omega$ such that $\int_\Omega \int_\Omega |K(x,y)|^2dxdy <\infty$. Let $f(x)$ be a function in $L^2(\Omega)$. Consider the integral equation $u(x)=f(x)+\lambda \int_\Omega K(x,y)u(y)dy$, where $\lambda$ is a complex number. Prove that there exists a small positive number $\lambda_o$ such that, for any complex number $\lambda$, with |$\lambda$|<$\lambda_o$ there exists unique solution $u(x)$ of the integral equation. (It gives hint to look at the operator $(Tu)(x) = f(x)+\lambda \int_\Omega K(x,y)u(y)dy$ on $L^2(\Omega)$)
Edit:
Here is the work for showing T is a contraction: 
$||Tu(x)-Tv(x)||_2 = (\int_\Omega {|Tu(x)-Tv(x)|^2}dx)^{1/2} = (\int_\Omega |\lambda \int_\Omega K(x,y)(u(y)-v(y))dy|^2dx)^{1/2} \leq (\int_\Omega \lambda ^2 \int_\Omega |K(x,y)|^2dy \int_\Omega |u(y)-v(y)|^2dy dx)^{1/2} = |\lambda| ||u(x)-v(x)||_2 (\int \int _{\Omega \times \Omega} |K(x,y)|^2dydx)^{1/2} $ so T is a contraction

Just to summarize the comments, you have shown that
$$
\|T(u)-T(v)\|_2\le|\lambda|\|u-v\|_2\left(\int_\Omega\int_\Omega|K(x,y)|^2\,\mathrm{d}x\,\mathrm{d}y\right)^{1/2}
$$
This implies that $T$ is a contraction when
$$
\lambda\lt\left(\int_\Omega\int_\Omega|K(x,y)|^2\,\mathrm{d}x\,\mathrm{d}y\right)^{-1/2}
$$
Thus, for
$$
\lambda_0=\left(\int_\Omega\int_\Omega|K(x,y)|^2\,\mathrm{d}x\,\mathrm{d}y\right)^{-1/2}
$$
we get that $T$ is a contraction for $|\lambda|\lt\lambda_0$. Therefore, $T$ has a unique fixed point $u_0$ so that
$$
u_0=T(u_0)
$$
