While I can't answer your question on a mathematical basis, I'll try on an intuitive one:
fixpoint techniques need a "flat" function graph around their ..well.. fixpoint. This means: if you picture your fixpoint function on an X-Y chart, you'll see that the function crosses the diagonal (+x,+y) exactly at the true result. In one step of your fixpoint algorithm you are guessing an X value which needs to be within the interval around the intersection point where the first derivative is between (-1..+1) and take the Y value. The Y that you took will be closer to the intersection point because starting from the intersection it is reachable by following a path which has a smaller slope than +/-1 , in contrast to the previous X value that you utilized, which has in this sense, the exact slope -1. It is immediately clear now that the smaller the slope, the more way you make towards the intersection point (the true function value) when using the Y as new X. The best interpolation function is trivially a constant, which has slope 0, giving you the true value in the first step.
Sorry to all mathematicians.