Resolving Integral equations with Neumann's series I need to resolve the integral equation, $x(s)-\int_o^1 2stx(t)dt=sin(\pi s)$ ,where $  s\in [0,1], y\in C[0,1], k\in C([0,1]^2)$, using the method of Neumann series. Following a proposition, if $ X $ is Banach, $T\in L(X)$ and $\lVert T\rVert < 1$, then the operator $Id-T$ is invertible such that $ (Id-T)^{-1}=\sum_{n=0}^\infty T^n.$ After introducing the integral operator, $T_k:C[0,1]\rightarrow C[0,1],T_k x=\int_0^1 k(.,t)x(t)dt$ and $\lVert T_k\rVert=sup_s\int_o^1 |k(s,t)|dt$ the integral equation can be written as an operator equation, $(Id-T_k)x=y$. Unfortunately one concludes that $\lVert T_k\rVert=1.$ 
So the question is, how to make the proposition work under weaker conditions ? Can somebody make a proposal for a solution of the integral equation ?
Many thanks.

Well, $T^2$ is an integral operator, too, with kernel $k_2(s,t)=\int^1_0k(s,u)\,k(u,t)\,du$. If you calculate that, you'll see $T^2=\frac23T$ in this simple case, so $\|T^2\|=\frac23$. Now, you could use the elementary identity $(Id-T)^{-1}=(Id-T^2)^{-1}(Id+T)$.
