/- Copyright 2026 The Formal Conjectures Authors. Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at https://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. -/ import FormalConjecturesUtil import FormalConjectures.ErdosProblems.«1043»

Erdős Problem 1048

References:

    erdosproblems.com/1048

    [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 Erdos1043

The set ${ z \in \mathbb{C} : \lvert f(z)\rvert < 1}$

def openLevelSet (f : Polynomial ) : Set := {z : | 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 : ) (f : [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 : ) (f : [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 : ) (hr : 1 < r) (n : ) (hn : 1 n) (f : [X]) (hf : f = (X : [X]) ^ n - C ((r : ) ^ n)) : {C : Set | z levelSet f, C = connectedComponentIn (levelSet f) z}.ncard = n := r:hr:1 < rn:hn:1 nf:[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 : ) (hr : 1 < r) (ε : ) ( : 0 < ε) : ∀ᶠ n : in Filter.atTop, f : [X], f = (X : [X]) ^ n - C ((r : ) ^ n) z levelSet f, Metric.ediam (connectedComponentIn (levelSet f) z) < ENNReal.ofReal ε := r:hr:1 < rε::0 < ε∀ᶠ (n : ) in Filter.atTop, (f : [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 : ) (hr₀ : 0 < r) (hr₁ : r 1) (f : [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:hr₀:0 < rhr₁:r 1f:[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 : ) (hr₀ : 0 r) (hr₁ : r 1 / 2) (f : [X]) (hmonic : f.Monic) (hdeg : f.degree 1) (hroots : z f.roots, z r) : 2 Metric.ediam (connectedComponentIn (levelSet f) 0) := r:hr₀:0 rhr₁:r 1 / 2f:[X]hmonic:f.Monichdeg:f.degree 1hroots: z f.roots, z r2 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 : ) (hn : 1 n) : Metric.ediam (connectedComponentIn (levelSet ((X : [X]) ^ n)) 0) = 2 := n:hn:1 nMetric.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 : ) (hr₀ : 1 / 2 < r) (hr₁ : r (Real.sqrt 5 - 1) / 2) (f : [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:hr₀:1 / 2 < rhr₁:r (5 - 1) / 2f:[X]hmonic:f.Monichdeg:f.degree 1hroots: z f.roots, z rENNReal.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 : ) (hr₀ : (Real.sqrt 5 - 1) / 2 r) (hr₁ : r 1) (f : [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:hr₀:(5 - 1) / 2 rhr₁:r 1f:[X]hmonic:f.Monichdeg:f.degree 1hroots: z f.roots, z rENNReal.ofReal (2 - r ^ 2) < Metric.ediam (connectedComponentIn (levelSet f) 0) All goals completed! 🐙end Erdos1048