Erdős Problem #1120

Erdős Problem 1120 --- collection record

PROVENANCE AND STATUS

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1120.tex.

REUSED SOURCE 1

1) FORMAL RESTATEMENT

Fix an integer $n\ge 1$. Let $f\in\mathbb{C}[z]$ be a monic polynomial of degree $n$ whose roots $\alpha_1,\dots,\alpha_n$ satisfy $|\alpha_j|\le 1$. Define the lemniscate sublevel set \[ E(f):=\{z\in\mathbb{C}: |f(z)|\le 1\}. \] For such $f$, define \[ \ell(f):=\inf\{\operatorname{length}(\gamma): \gamma\text{ is a rectifiable path in }E(f),\ \gamma(0)=0,\ |\gamma(1)|=1\}. \] (We interpret "a path joining $0$ to $|z|=1$" as a continuous curve with one endpoint at $0$ and the other on the unit circle; rectifiability is imposed to make "length" well-defined.)

The natural extremal quantity is \[ L_n:=\sup\{\ell(f): f\text{ monic degree }n\text{ with all roots }|\alpha_j|\le 1\}. \] The question asks for the behavior of $L_n$, in particular how large $\ell(f)$ can be as a function of $n$.

2) QUICK LITERATURE/CONTEXT CHECK

I did not browse the web; I only record what is stated in the problem text:

- Clunie and Netanyahu (via personal communication reported in Hayman) showed that such a path always exists. - The trivial lower bound is $\ell(f)\ge 1$, achieved by $f(z)=z^n$. - Erdős suggested $L_n\to\infty$ with $n$ but not too fast.

3) ATTACK PLAN

- Prove basic necessary properties of $E(f)$: e.g. show $E(f)$ always meets the unit circle $|z|=1$, and record easy lower bounds on $\ell(f)$. - Try to get any nontrivial upper bound on $\ell(f)$ (or $L_n$) in terms of $n$. (I do not succeed beyond trivialities.)

4) WORK

Lemma 1120.1 (Existence of a point on the unit circle with $|f|\le 1$). Let $f$ be monic of degree $n$ with all roots $|\alpha_j|\le 1$. Then there exists $\zeta$ with $|\zeta|=1$ such that $|f(\zeta)|\le 1$. Equivalently, $E(f)\cap\{z:|z|=1\}\ne\emptyset$.

Proof. Write $f(z)=\prod_{j=1}^n (z-\alpha_j)$ with $|\alpha_j|\le 1$. A standard Jensen formula for polynomials states that \[ \frac{1}{2\pi}\int_0^{2\pi} \log|f(e^{i\theta})|\,d\theta = \log|a_n| + \sum_{|\alpha_j|>1} \log|\alpha_j|, \] where $a_n$ is the leading coefficient. Here $a_n=1$ and there are no roots with $|\alpha_j|>1$, so the right-hand side equals $0$. Hence \[ \frac{1}{2\pi}\int_0^{2\pi} \log|f(e^{i\theta})|\,d\theta = 0. \] If $|f(e^{i\theta})|>1$ for all $\theta$, then $\log|f(e^{i\theta})|>0$ for all $\theta$, forcing the integral to be positive, contradiction. Therefore there exists some $\theta$ with $\log|f(e^{i\theta})|\le 0$, i.e. $|f(e^{i\theta})|\le 1$. Taking $\zeta=e^{i\theta}$ completes the proof. $\square$

Lemma 1120.2 (Universal lower bound $\ell(f)\ge 1$). For any path $\gamma$ with $\gamma(0)=0$ and $|\gamma(1)|=1$, its Euclidean length satisfies $\operatorname{length}(\gamma)\ge 1$. Consequently $\ell(f)\ge 1$ for every admissible $f$, and thus $L_n\ge 1$ for all $n$.

Proof. For any rectifiable curve $\gamma$, the length dominates the straight-line distance between endpoints: $\operatorname{length}(\gamma)\ge |\gamma(1)-\gamma(0)|$ (triangle inequality applied to polygonal approximations, then take infimum). Here $\gamma(0)=0$ and $|\gamma(1)|=1$, so $|\gamma(1)-\gamma(0)|=1$. $\square$

Lemma 1120.3 (The case $n=1$: $L_1=1$). If $n=1$, every admissible polynomial has the form $f(z)=z-\alpha$ with $|\alpha|\le 1$, and for every such $f$ we have $\ell(f)=1$. Hence $L_1=1$.

Proof. For $f(z)=z-\alpha$, $E(f)=\{z:|z-\alpha|\le 1\}$ is the closed disk of radius $1$ centered at $\alpha$. Since $|\alpha|\le 1$, this disk contains $0$. By Lemma 1120.1 there exists $\zeta$ with $|\zeta|=1$ and $|\zeta-\alpha|\le 1$, i.e. $\zeta\in E(f)\cap\{|z|=1\}$. Because $E(f)$ is convex, the straight segment from $0$ to $\zeta$ lies in $E(f)$. Its length is $|\zeta-0|=1$, so $\ell(f)\le 1$. Lemma 1120.2 gives $\ell(f)\ge 1$, so $\ell(f)=1$ for all such $f$. $\square$

FAST REALITY CHECK (non-rigorous numerical exploration for small $n$). To sanity-check that paths exist and to get a feel for the geometry, I discretized the square $[-1.2,1.2]^2$ with mesh size $h=0.02$, marked grid points satisfying $|f(z)|\le 1$, and ran a shortest-path search on the resulting grid graph from the grid point nearest $0$ to grid points lying within distance $\approx h/2$ of the unit circle. For random degree-4 polynomials built from 4 random roots in the unit disk, the algorithm typically found paths of length very close to $1$ (often within a few percent), consistent with the trivial lower bound. This computation is only a coarse heuristic because: (a) the endpoint is only approximately on $|z|=1$, (b) the grid introduces discretization artifacts, and (c) the path is constrained to grid edges.

5) VERIFICATION

- Lemma 1120.1: the only nontrivial input is Jensen's formula for polynomials; the specialization to monic polynomials with all roots $|\alpha_j|\le 1$ gives an average log-modulus of $0$, forcing some point with $|f|\le 1$ on $|z|=1$. - Lemma 1120.2 is purely metric and applies to any rectifiable path. - Lemma 1120.3 checks the base case $n=1$ completely.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: $E(f)$ always meets the unit circle (Lemma 1120.1), and every admissible $f$ satisfies the universal lower bound $\ell(f)\ge 1$ (Lemma 1120.2). In the base case $n=1$, the extremal value is exactly $L_1=1$ (Lemma 1120.3).

(ii) First gap (crisp): Prove any nontrivial upper bound on $L_n$ in terms of $n$ (or prove $L_n\to\infty$), i.e. control how long the shortest path inside $E(f)$ might have to be in the worst case.

(iii) Top 3 next moves: 1. Use the branched-covering viewpoint: $E(f)=f^{-1}(\overline{\mathbb{D}})$. Try to relate existence/length of a path from $0$ to $|z|=1$ inside $E(f)$ to the location of critical values of $f$ relative to the unit disk, potentially via a "connected component" criterion. 2. Search for explicit candidate extremizers for $L_n$ by arranging roots near $|z|=1$ so that the lemniscate $|f|=1$ develops narrow "necks" and forces long detours; then try to bound neck widths using product structure $\prod|z-\alpha_j|$. 3. Attempt to bound $\ell(f)$ in terms of geometric quantities of $E(f)$ (e.g. diameter, conformal modulus of an annulus separating $0$ from $|z|=1$ inside $E(f)$), and then bound those quantities using potential theory and the root constraint $|\alpha_j|\le 1$.

(iv) Minimal counterexample structure (if $L_n$ grows quickly): one would expect a polynomial whose sublevel set $E(f)$ contains $0$ but connects to the unit circle only through a long, thin corridor (high conformal modulus), with the corridor formed by the interaction of many root factors $|z-\alpha_j|$ being slightly larger than 1 except along a winding route.

COMPLETION ESTIMATE

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

1) FORMAL RESTATEMENT

Fix an integer $n\ge 1$. Let $f\in\mathbb{C}[z]$ be a monic polynomial of degree $n$ whose roots $\alpha_1,\dots,\alpha_n$ satisfy $|\alpha_j|\le 1$. Define the lemniscate sublevel set \[ E(f):=\{z\in\mathbb{C}: |f(z)|\le 1\}. \] For such $f$, define \[ \ell(f):=\inf\{\operatorname{length}(\gamma): \gamma\text{ is a rectifiable path in }E(f),\ \gamma(0)=0,\ |\gamma(1)|=1\}. \] (We interpret "a path joining $0$ to $|z|=1$" as a continuous curve with one endpoint at $0$ and the other on the unit circle; rectifiability is imposed to make "length" well-defined.)

The natural extremal quantity is \[ L_n:=\sup\{\ell(f): f\text{ monic degree }n\text{ with all roots }|\alpha_j|\le 1\}. \] The question asks for the behavior of $L_n$, in particular how large $\ell(f)$ can be as a function of $n$.

2) QUICK LITERATURE/CONTEXT CHECK

I did not browse the web; I only record what is stated in the problem text:

- Clunie and Netanyahu (via personal communication reported in Hayman) showed that such a path always exists. - The trivial lower bound is $\ell(f)\ge 1$, achieved by $f(z)=z^n$. - Erdős suggested $L_n\to\infty$ with $n$ but not too fast.

3) ATTACK PLAN

- Prove basic necessary properties of $E(f)$: e.g. show $E(f)$ always meets the unit circle $|z|=1$, and record easy lower bounds on $\ell(f)$. - Try to get any nontrivial upper bound on $\ell(f)$ (or $L_n$) in terms of $n$. (I do not succeed beyond trivialities.)

4) WORK

Lemma 1120.1 (Existence of a point on the unit circle with $|f|\le 1$). Let $f$ be monic of degree $n$ with all roots $|\alpha_j|\le 1$. Then there exists $\zeta$ with $|\zeta|=1$ such that $|f(\zeta)|\le 1$. Equivalently, $E(f)\cap\{z:|z|=1\}\ne\emptyset$.

Proof. Write $f(z)=\prod_{j=1}^n (z-\alpha_j)$ with $|\alpha_j|\le 1$. A standard Jensen formula for polynomials states that \[ \frac{1}{2\pi}\int_0^{2\pi} \log|f(e^{i\theta})|\,d\theta = \log|a_n| + \sum_{|\alpha_j|>1} \log|\alpha_j|, \] where $a_n$ is the leading coefficient. Here $a_n=1$ and there are no roots with $|\alpha_j|>1$, so the right-hand side equals $0$. Hence \[ \frac{1}{2\pi}\int_0^{2\pi} \log|f(e^{i\theta})|\,d\theta = 0. \] If $|f(e^{i\theta})|>1$ for all $\theta$, then $\log|f(e^{i\theta})|>0$ for all $\theta$, forcing the integral to be positive, contradiction. Therefore there exists some $\theta$ with $\log|f(e^{i\theta})|\le 0$, i.e. $|f(e^{i\theta})|\le 1$. Taking $\zeta=e^{i\theta}$ completes the proof. $\square$

Lemma 1120.2 (Universal lower bound $\ell(f)\ge 1$). For any path $\gamma$ with $\gamma(0)=0$ and $|\gamma(1)|=1$, its Euclidean length satisfies $\operatorname{length}(\gamma)\ge 1$. Consequently $\ell(f)\ge 1$ for every admissible $f$, and thus $L_n\ge 1$ for all $n$.

Proof. For any rectifiable curve $\gamma$, the length dominates the straight-line distance between endpoints: $\operatorname{length}(\gamma)\ge |\gamma(1)-\gamma(0)|$ (triangle inequality applied to polygonal approximations, then take infimum). Here $\gamma(0)=0$ and $|\gamma(1)|=1$, so $|\gamma(1)-\gamma(0)|=1$. $\square$

Lemma 1120.3 (The case $n=1$: $L_1=1$). If $n=1$, every admissible polynomial has the form $f(z)=z-\alpha$ with $|\alpha|\le 1$, and for every such $f$ we have $\ell(f)=1$. Hence $L_1=1$.

Proof. For $f(z)=z-\alpha$, $E(f)=\{z:|z-\alpha|\le 1\}$ is the closed disk of radius $1$ centered at $\alpha$. Since $|\alpha|\le 1$, this disk contains $0$. By Lemma 1120.1 there exists $\zeta$ with $|\zeta|=1$ and $|\zeta-\alpha|\le 1$, i.e. $\zeta\in E(f)\cap\{|z|=1\}$. Because $E(f)$ is convex, the straight segment from $0$ to $\zeta$ lies in $E(f)$. Its length is $|\zeta-0|=1$, so $\ell(f)\le 1$. Lemma 1120.2 gives $\ell(f)\ge 1$, so $\ell(f)=1$ for all such $f$. $\square$

FAST REALITY CHECK (non-rigorous numerical exploration for small $n$). To sanity-check that paths exist and to get a feel for the geometry, I discretized the square $[-1.2,1.2]^2$ with mesh size $h=0.02$, marked grid points satisfying $|f(z)|\le 1$, and ran a shortest-path search on the resulting grid graph from the grid point nearest $0$ to grid points lying within distance $\approx h/2$ of the unit circle. For random degree-4 polynomials built from 4 random roots in the unit disk, the algorithm typically found paths of length very close to $1$ (often within a few percent), consistent with the trivial lower bound. This computation is only a coarse heuristic because: (a) the endpoint is only approximately on $|z|=1$, (b) the grid introduces discretization artifacts, and (c) the path is constrained to grid edges.

5) VERIFICATION

- Lemma 1120.1: the only nontrivial input is Jensen's formula for polynomials; the specialization to monic polynomials with all roots $|\alpha_j|\le 1$ gives an average log-modulus of $0$, forcing some point with $|f|\le 1$ on $|z|=1$. - Lemma 1120.2 is purely metric and applies to any rectifiable path. - Lemma 1120.3 checks the base case $n=1$ completely.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: $E(f)$ always meets the unit circle (Lemma 1120.1), and every admissible $f$ satisfies the universal lower bound $\ell(f)\ge 1$ (Lemma 1120.2). In the base case $n=1$, the extremal value is exactly $L_1=1$ (Lemma 1120.3).

(ii) First gap (crisp): Prove any nontrivial upper bound on $L_n$ in terms of $n$ (or prove $L_n\to\infty$), i.e. control how long the shortest path inside $E(f)$ might have to be in the worst case.

(iii) Top 3 next moves: 1. Use the branched-covering viewpoint: $E(f)=f^{-1}(\overline{\mathbb{D}})$. Try to relate existence/length of a path from $0$ to $|z|=1$ inside $E(f)$ to the location of critical values of $f$ relative to the unit disk, potentially via a "connected component" criterion. 2. Search for explicit candidate extremizers for $L_n$ by arranging roots near $|z|=1$ so that the lemniscate $|f|=1$ develops narrow "necks" and forces long detours; then try to bound neck widths using product structure $\prod|z-\alpha_j|$. 3. Attempt to bound $\ell(f)$ in terms of geometric quantities of $E(f)$ (e.g. diameter, conformal modulus of an annulus separating $0$ from $|z|=1$ inside $E(f)$), and then bound those quantities using potential theory and the root constraint $|\alpha_j|\le 1$.

(iv) Minimal counterexample structure (if $L_n$ grows quickly): one would expect a polynomial whose sublevel set $E(f)$ contains $0$ but connects to the unit circle only through a long, thin corridor (high conformal modulus), with the corridor formed by the interaction of many root factors $|z-\alpha_j|$ being slightly larger than 1 except along a winding route.