Rubinstein's boundary theorem, at the zero a = 1 #
If p has degree n ≥ 2, all zeroes in the closed unit disk, and p(1) = 0, then p' has a
zero strictly inside D(1,1) unless p = c(Xⁿ - 1).
The polar identity is useless here: at a = 1 the reflected point 1/a coincides with a,
1 - a² = 0, and the identity degenerates. One elementary identity replaces it, obtained from
p''(1)/p'(1). Writing p = c(X-1)Q and p' = ncR, evaluation at 1 gives
Q(1) = n R(1) and 2 Q'(1) = n R'(1)
— the factor 2 coming from p''(1) = 2Q'(1) — whence the boundary reciprocal identity
∑ⱼ qⱼ = 2 ∑ⱼ 1/(1 - zⱼ), qⱼ = 1/(1 - wⱼ).
Both sides are then sandwiched. On the left Re qⱼ ≤ ‖qⱼ‖ ≤ 1, so the real part is at most
n-1; on the right Re 1/(1-z) ≥ 1/2 for ‖z‖ ≤ 1, because
Re 1/(1-z) - 1/2 = (1-‖z‖²)/(2‖1-z‖²), so the real part is at least n-1. Equality forces
Re qⱼ = 1 for every j, hence qⱼ = 1, hence every critical point is 0, hence
p' = n c Xⁿ⁻¹ and p = c(Xⁿ - 1).
The repeated-root case is handled before any division: if 1 is a multiple zero then p'(1) = 0
and ζ = 1 is a strict witness, so under the contradiction hypothesis p'(1) ≠ 0 and every
zⱼ ≠ 1.
No polar integral, origin identity, defect lemma, numerical certificate or Gauss–Lucas theorem
is used. Only the two factorizations of Sendov.Counterexample.Factor are shared with the
interior argument.
Main statements #
Sendov.rubinstein_one: the strict-or-extremal alternative;Sendov.sendov_boundary_one: the closed-disk form,‖ζ - 1‖ ≤ 1;Sendov.boundary_reciprocal:(BR), in division-free form.
Evaluating ∏ⱼ (X - zⱼ) and its derivative at a point #
The versions in Sendov.Counterexample.Identities are specialized to the origin; the boundary
argument needs them at 1.
Scalar facts #
The boundary reciprocal identity #
(BR), in division-free form. From ∏ Zs = n ∏ Qi and 2 ∑ₑ Zs = n ∑ₑ Qi — the two
readings of p'(1) and p''(1) — one gets ∑ⱼ 1/Qiⱼ = 2 ∑ⱼ 1/Zsⱼ. Applied with
Zsⱼ = 1 - zⱼ and Qiⱼ = 1/qⱼ this is ∑ⱼ qⱼ = 2 ∑ⱼ 1/(1-zⱼ).
The sandwich #
Both sides of (BR) are pinned: the left has real part at most N, the right at least N,
so every qⱼ has real part 1 and hence, lying in the closed unit disk, equals 1.
Rubinstein's theorem at a = 1 #
Rubinstein's boundary theorem at a = 1, with its equality case.
The closed-disk form. In the extremal case p = c(Xⁿ - 1) the only critical point is 0,
at distance exactly 1 from the zero 1.