Find an interval about 1/A for which fixed-point iteration converges, provided p0 is in that interval.
Our interval will be of the form
[1/A−δ,1/A+δ] for some small
δ.
One thing we need to ensure is that
∣g′(x)∣≤k<1 over all of the interval, for our choice of
k. Know that
g′(x)=2−2Ax
Since this
function is
linear it suffices to ensure
∣g′(x)∣≤k on endpoints only. Know
g′(1/A−δ)=2−2A(1/A−δ)=2−2A/A+2Aδ=2Aδ and likewise
g′(1/A+δ)=−2Aδ.
Hence need
∣±2Aδ∣≤k<1 for our choice of
k. This is satisfied by choosing
k=0.5 and then ensuring
δ≤4A1.
Another thing we need to ensure is that
g maps
[1/A−δ,1/A+δ] into itself. As per the extreme value theorem, it suffices to verify that this holds on endpoints of the interval and points where
g′ is
zero, which is at
1/A exactly. So we compute:
g:(1/A−δ)1/A(1/A+δ)↦1/A−Aδ2↦1/A↦1/A−Aδ2
for all three of these values to be in
[1/A−δ,1/A+δ] we need that
Aδ2≤δ, which is satisfied when
δ≤1/A.
Hence overall it suffices to choose
δ so that
δ≤4A1 and
δ≤A1. One such
δ is
δ=4A1. Then performing
fix-point iteration with an
initial p0 in
[1/A−δ,1/A+δ] will converge.