Define the operator $K: L^2[0,1] \to L^2[0,1]$ by $$K f(x) = \int _0 ^1 k(x,y) f(y) \, dy .$$ where $k(x,y)$ is a continuous complex valued function on the unit square. $[0,1]\times[0,1]$ Also assume that $$sup_{x\in[0,1]}\int_{0}^{1}|k(x,y)|dy<1$$ How to show that given $g\in (C[0,1],\mathbb{C})$, the integral equation $$f(x)=g(x)+\int _0 ^1 k(x,y) f(y) \, dy$$ has exactly one continuous solution $f\in(C[0,1],\mathbb{C})$?
Note: I only know introductory measure theory and functional analysis, so I am suppose to do this problem without heavy integral function theory.
One way to get started is to show that K is compact and its operator norm $||K||<1$ because we are given that $sup_{x\in[0,1]}\int_{0}^{1}|k(x,y)|dy<1$ Then we know that $(I-K)$ is invertible and its inverse is given by the Neumann series. At this point, I'm not sure how to proceed.
The equation
$$f(x)=g(x)+\int _0 ^1 k(x,y) f(y) \, dy$$ is equivalent to
$$f=g +Kf.$$
Hence $(I-K)f=g$. Since $I-K$ is invertble, we get
$$f=(I-K)^{-1}g.$$