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import FormalConjecturesUtil
import FormalConjectures.ErdosProblems.«1043»Erdős Problem 1048
References:
[EHP58] Erdős, P. and Herzog, F. and Piranian, G., Metric properties of polynomials. J. Analyse Math. (1958), 125-148.
[Po61] Pommerenke, Ch., On metric properties of complex polynomials. Michigan Math. J. (1961), 97-115.
namespace Erdos1048open Polynomial Erdos1043The set ${ z \in \mathbb{C} : \lvert f(z)\rvert < 1}$
def openLevelSet (f : Polynomial ℂ) : Set ℂ :=
{z : ℂ | ‖f.eval z‖ < 1}If $f\in \mathbb{C}[x]$ is a monic polynomial with all roots satisfying $\lvert z\rvert \leq r$ for some $r<2$, then must $${ z: \lvert f(z)\rvert <1}$$ have a connected component with diameter $>2-r$?
A problem of Erdős, Herzog, and Piranian [EHP58].
Pommerenke [Po61] proved the answer is no for $r>1$, showing that if $f(z)=z^n-r^n$ then ${ z: \lvert f(z)\rvert \leq 1}$ has $n$ connected components, all with diameter $\to 0$ as $n\to \infty$.
@[category research solved, AMS 30, formal_proof using lean4 at "https://github.com/plby/lean-proofs/blob/main/src/v4.29.1/ErdosProblems/Erdos1048.lean"]
theorem erdos_1048 : answer(False) ↔
∀ (r : ℝ) (f : ℂ[X]), r < 2 → f.Monic → f.degree ≥ 1 → (∀ z ∈ f.roots, ‖z‖ ≤ r) →
∃ z ∈ openLevelSet f, ENNReal.ofReal (2 - r) <
Metric.ediam (connectedComponentIn (openLevelSet f) z) := ⊢ False ↔
∀ (r : ℝ) (f : ℂ[X]),
r < 2 →
f.Monic →
f.degree ≥ 1 →
(∀ z ∈ f.roots, ‖z‖ ≤ r) →
∃ z ∈ openLevelSet f, ENNReal.ofReal (2 - r) < Metric.ediam (connectedComponentIn (openLevelSet f) z)
All goals completed! 🐙Pommerenke [Po61] proved the answer is no for $r>1$, showing that if $f(z)=z^n-r^n$ then ${ z: \lvert f(z)\rvert \leq 1}$ has $n$ connected components, all with diameter $\to 0$ as $n\to \infty$.
@[category research solved, AMS 30]
theorem erdos_1048.variants.pommerenke_ncard_components (r : ℝ) (hr : 1 < r) (n : ℕ) (hn : 1 ≤ n)
(f : ℂ[X]) (hf : f = (X : ℂ[X]) ^ n - C ((r : ℂ) ^ n)) :
{C : Set ℂ | ∃ z ∈ levelSet f, C = connectedComponentIn (levelSet f) z}.ncard = n := r:ℝhr:1 < rn:ℕhn:1 ≤ nf:ℂ[X]hf:f = X ^ n - C (↑r ^ n)⊢ {C | ∃ z ∈ levelSet f, C = connectedComponentIn (levelSet f) z}.ncard = n
All goals completed! 🐙Pommerenke [Po61] proved the answer is no for $r>1$, showing that if $f(z)=z^n-r^n$ then ${ z: \lvert f(z)\rvert \leq 1}$ has $n$ connected components, all with diameter $\to 0$ as $n\to \infty$.
@[category research solved, AMS 30]
theorem erdos_1048.variants.pommerenke_diam_tendsto_zero (r : ℝ) (hr : 1 < r) (ε : ℝ)
(hε : 0 < ε) :
∀ᶠ n : ℕ in Filter.atTop, ∀ f : ℂ[X], f = (X : ℂ[X]) ^ n - C ((r : ℂ) ^ n) →
∀ z ∈ levelSet f,
Metric.ediam (connectedComponentIn (levelSet f) z) < ENNReal.ofReal ε := r:ℝhr:1 < rε:ℝhε:0 < ε⊢ ∀ᶠ (n : ℕ) in Filter.atTop,
∀ (f : ℂ[X]),
f = X ^ n - C (↑r ^ n) → ∀ z ∈ levelSet f, Metric.ediam (connectedComponentIn (levelSet f) z) < ENNReal.ofReal ε
All goals completed! 🐙On the other hand, if $0<r\leq 1$, then the answer is yes, as also shown by Pommerenke [Po61].
@[category research solved, AMS 30]
theorem erdos_1048.variants.r_le_one (r : ℝ) (hr₀ : 0 < r) (hr₁ : r ≤ 1) (f : ℂ[X])
(hmonic : f.Monic) (hdeg : f.degree ≥ 1) (hroots : ∀ z ∈ f.roots, ‖z‖ ≤ r) :
∃ z ∈ openLevelSet f, ENNReal.ofReal (2 - r) <
Metric.ediam (connectedComponentIn (openLevelSet f) z) := r:ℝhr₀:0 < rhr₁:r ≤ 1f:ℂ[X]hmonic:f.Monichdeg:f.degree ≥ 1hroots:∀ z ∈ f.roots, ‖z‖ ≤ r⊢ ∃ z ∈ openLevelSet f, ENNReal.ofReal (2 - r) < Metric.ediam (connectedComponentIn (openLevelSet f) z)
All goals completed! 🐙If $0\leq r\leq 1/2$ then the component which contains $0$ must have diameter $\geq 2$, which $f(z)=z^n$ shows is best possible.
@[category research solved, AMS 30]
theorem erdos_1048.variants.diam_ge_two (r : ℝ) (hr₀ : 0 ≤ r) (hr₁ : r ≤ 1 / 2) (f : ℂ[X])
(hmonic : f.Monic) (hdeg : f.degree ≥ 1) (hroots : ∀ z ∈ f.roots, ‖z‖ ≤ r) :
2 ≤ Metric.ediam (connectedComponentIn (levelSet f) 0) := r:ℝhr₀:0 ≤ rhr₁:r ≤ 1 / 2f:ℂ[X]hmonic:f.Monichdeg:f.degree ≥ 1hroots:∀ z ∈ f.roots, ‖z‖ ≤ r⊢ 2 ≤ Metric.ediam (connectedComponentIn (levelSet f) 0)
All goals completed! 🐙If $0\leq r\leq 1/2$ then the component which contains $0$ must have diameter $\geq 2$, which $f(z)=z^n$ shows is best possible.
@[category research solved, AMS 30]
theorem erdos_1048.variants.diam_ge_two_is_best (n : ℕ) (hn : 1 ≤ n) :
Metric.ediam (connectedComponentIn (levelSet ((X : ℂ[X]) ^ n)) 0) = 2 := n:ℕhn:1 ≤ n⊢ Metric.ediam (connectedComponentIn (levelSet (X ^ n)) 0) = 2
All goals completed! 🐙If $1/2<r\leq \frac{\sqrt{5}-1}{2}$ then the component which contains $0$ must have diameter $>1/r$.
@[category research solved, AMS 30]
theorem erdos_1048.variants.diam_gt_inv_r (r : ℝ) (hr₀ : 1 / 2 < r)
(hr₁ : r ≤ (Real.sqrt 5 - 1) / 2) (f : ℂ[X]) (hmonic : f.Monic) (hdeg : f.degree ≥ 1)
(hroots : ∀ z ∈ f.roots, ‖z‖ ≤ r) :
ENNReal.ofReal (1 / r) < Metric.ediam (connectedComponentIn (levelSet f) 0) := r:ℝhr₀:1 / 2 < rhr₁:r ≤ (√5 - 1) / 2f:ℂ[X]hmonic:f.Monichdeg:f.degree ≥ 1hroots:∀ z ∈ f.roots, ‖z‖ ≤ r⊢ ENNReal.ofReal (1 / r) < Metric.ediam (connectedComponentIn (levelSet f) 0)
All goals completed! 🐙If $\frac{\sqrt{5}-1}{2}\leq r\leq 1$ then the component which contains $0$ must have diameter $>2-r^2$.
@[category research solved, AMS 30]
theorem erdos_1048.variants.diam_gt_two_sub_sq (r : ℝ) (hr₀ : (Real.sqrt 5 - 1) / 2 ≤ r)
(hr₁ : r ≤ 1) (f : ℂ[X]) (hmonic : f.Monic) (hdeg : f.degree ≥ 1)
(hroots : ∀ z ∈ f.roots, ‖z‖ ≤ r) :
ENNReal.ofReal (2 - r ^ 2) < Metric.ediam (connectedComponentIn (levelSet f) 0) := r:ℝhr₀:(√5 - 1) / 2 ≤ rhr₁:r ≤ 1f:ℂ[X]hmonic:f.Monichdeg:f.degree ≥ 1hroots:∀ z ∈ f.roots, ‖z‖ ≤ r⊢ ENNReal.ofReal (2 - r ^ 2) < Metric.ediam (connectedComponentIn (levelSet f) 0)
All goals completed! 🐙end Erdos1048