The polar inequality, up to the branch point #
The polar identity says
(∏ q) ∏ⱼ (1 - a zⱼ) = n ∫₀¹ ∏ⱼ (t(1-a²)qⱼ + a) dt, (∏ q) ∏ⱼ (a - zⱼ) = n.
The Möbius estimate |a - zⱼ| ≤ |1 - a zⱼ|, valid because zⱼ lies in the closed unit disk
and a is real with |a| ≤ 1, therefore gives
n = |(∏ q) ∏ (a - zⱼ)| ≤ |(∏ q) ∏ (1 - a zⱼ)| = n |∫₀¹ ∏ⱼ (t(1-a²)qⱼ + a) dt|,
and after the integral triangle inequality
1 ≤ ∫₀¹ ∏ⱼ |a + t(1-a²)qⱼ| dt. (⋆)
(⋆) is the branch point of the whole argument. The high-degree route relaxes it by
AM–GM into the raw polar inequality (1Q) of Sendov.Reduction.Setup; the low-degree route
for 2 ≤ n ≤ 5 (see docs/plan-low-degrees.md) instead bounds each factor by a + (1-a²)t
and integrates a quartic. Neither may be folded into the other, so (⋆) is stated on its own.
Note that no division occurs anywhere: splitting the blog post's quotient
∏(1-azⱼ)/(a-zⱼ) into its numerator and denominator identities is what makes that possible,
and p'(a) ≠ 0 is never needed.
Main statements #
Sendov.norm_sub_le_norm_one_sub_mul: the Möbius estimate;Sendov.one_le_integral_prod_norm: the branch point(⋆).
The Möbius estimate #
|a - z| ≤ |1 - a z| for real a with |a| ≤ 1 and z in the closed unit disk. The
difference of squares is (1-a²)(1-|z|²), which is why a must be real: for complex a the
cross terms contribute (conj a - a)(z - conj z), which need not vanish.
Products of norms over a multiset #
The branch point #
The branch point (⋆). 1 ≤ ∫₀¹ ∏ⱼ |a + t(1-a²)qⱼ| dt.
The high-degree argument relaxes this by AM–GM to the raw polar inequality; the low-degree
argument bounds each factor by a + (1-a²)t. It is stated separately so that neither route
has to reproduce the other's work.
The AM–GM relaxation to (1Q) #
Splitting ∑ⱼ ‖a + s qⱼ‖² into its three pieces.
The pointwise quadratic-mean bound. ∏ⱼ |a + t(1-a²)qⱼ| ≤ P(t)^{(n-1)/2)}, by AM–GM
on the squares followed by ∑ ‖qⱼ‖² ≤ n-1.
The raw polar inequality (1Q).