So let $X = C([0, 1])$ the space of continuous functions and let $\|f\|_{\infty} := \sup_{s \in[0,1]}|f(s)|$. Now we have the map $A:X \to X$ defined by $$(Af)(x) = \int_0^x k(t)f(t)\,\mathrm{d}t$$ for $x \in [0, 1]$ and $k \in X$ such that $0 \leq k(x) \leq 1/2$.
We know that $A$ is linear, continuous and a contraction. Therefore by the Banach fixed-point theorem it has an unique fixpoint. Now I want to find $f^*$ with $(Af^*)(x) = f^*(x)$. How do I do that?
I tried $f^* = e^{\int k(x)}$. Therefore $$(Af^*)(x) = e^{\int k(x)} - e^{\int k(0)}$$
It looks close, but still not it..
Some times it's easier than you think. The fixed point is the zero function. The big hint is that $A$ is linear, which makes $0$ a fixed point regardless of contractivity.