Erdős Problem #1046

Erdős Problem 1046: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Separate the questions on the current page

The radius-two disk assertion is true, with centre equal to the arithmetic mean of the roots. The least possible diameter of a connected closed unit lemniscate is also two. The conjectured greatest width two is false. Thus the page's negative overall status must not be read as a counterexample to its opening disk-containment question.

We reconstruct the conformal-map proof of containment and give an explicit width counterexample. The historical source is Pommerenke, On some problems by Erdős, Herzog and Piranian, Michigan Math. J. 6 $1959$, 221--225, as cited at https://www.erdosproblems.com/1046. Our analytic inputs are the Riemann mapping theorem, continuous boundary extension for locally connected boundaries, and elementary complex analysis. The coefficient estimate needed below is proved rather than imported.

Normalize the exterior map

Let $f$ be monic of degree $n\ge1$, with roots $z_1,\ldots,z_n$, and let $K=\{|f|\le1\}$ be connected. This includes the hypothesis that the strict sublevel set is connected. The complement of $K$ has no bounded component: on such a component $\log|f|$ would be positive harmonic with boundary value zero, contradicting the maximum principle. Thus $K$ is a full continuum and its exterior is simply connected on the sphere.

The normalized exterior conformal map has expansion \[ \psi(w)=a w+b+O(w^{-1}),\qquad |w|>1,\quad a>0. \] The boundary of a polynomial lemniscate is locally connected, being a finite union of algebraic arcs, so the map extends continuously to $|w|=1$ and maps that circle onto $\partial K$. The nonvanishing holomorphic function \[ Q(w)=\frac{f(\psi(w))}{w^n} \] extends at infinity, with $Q(\infty)=a^n$. Its modulus is one on the unit circle. After inversion, $\log|Q|$ is harmonic on the disk, including zero, with zero boundary values. Thus $Q$ has constant modulus one, and is constant. Its value at infinity is positive, so $Q=1$ and $a=1$. Comparing the coefficient of $w^{n-1}$ in $f(\psi(w))=w^n$ gives \[ b=\frac{z_1+\cdots+z_n}{n}.\tag{1} \]

The coefficient bound needed for containment

We prove that every univalent holomorphic function \[ F(\zeta)=\zeta+A_2\zeta^2+\cdots\qquad(|\zeta|<1) \] satisfies $|A_2|\le2$. Its only zero is zero, so \[ h(\zeta)=\sqrt{F(\zeta^2)} =\zeta+\tfrac12 A_2\zeta^3+\cdots \] has an analytic choice of square root with derivative one at zero, and is odd. It is univalent: equality of two values implies equality of their squares, then equality of the squared arguments by injectivity of $F$; opposite arguments cannot give equal values of the odd function except at its sole zero. Therefore \[ H(w)=\frac1{h(1/w)}=w-\frac{A_2}{2w}+\cdots \] is univalent for $|w|>1$.

For completeness, if an exterior univalent map is $w+b_0+\sum_{j\ge1}b_jw^{-j}$, the area enclosed by its image of the circle $|w|=R>1$ is \[ \pi\left(R^2-\sum_{j\ge1}j|b_j|^2R^{-2j}\right)\ge0. \] This follows by inserting the Laurent series in the signed-area integral $\frac1{2i}\int\overline z\,dz$; distinct Fourier modes integrate to zero. The image is a positively oriented Jordan curve. Letting $R\downarrow1$ yields $\sum j|b_j|^2\le1$, and in particular $|A_2/2|\le1$. This proves the coefficient bound.

For each fixed $z\in K$, apply it to \[ F_z(\zeta)=\frac1{\psi(1/\zeta)-z} =\zeta+(z-b)\zeta^2+O(\zeta^3). \] The denominator does not vanish, and the extension at zero is univalent with derivative one. We conclude $|z-b|\le2$ for every $z\in K$. Together with $1$, this proves the disk assertion with its specified centre.

The minimum diameter

For $R>1$, the Laurent expansion gives \[ \frac1{2\pi}\int_0^{2\pi} [\psi(Re^{it})-\psi(-Re^{it})]e^{-it}\,dt=2R. \] Hence the diameter of the image of that circle is at least $2R$. Continuous extension implies uniform convergence of these curves as $R\downarrow1$, so $\operatorname{diam}(K)\ge2$. Equality is attained by $f(z)=z^n$, whose closed unit lemniscate is the unit disk. This proves the diameter assertion also when the closed set is connected but its strict interior has several components touching at critical boundary points.

The width assertion fails even for connected strict sublevels

First put $K_1=\{|z^3-1|\le1\}$. In polar coordinates its three petals satisfy $0\le\rho^3\le2\cos(3\theta)$ where the right side is nonnegative. The set is connected through zero, so every line projection is an interval. Its width in every direction is at least \[ W=2^{1/3}+2^{-1/3}>2.\tag{2} \] Here is a direct bound. Rotational symmetry and reversal of a projection direction reduce its angle to $0\le\phi\le\pi/3$. Put $x=3\phi/4\in[0,\pi/4]$. The boundary points at angles $\phi/4$ and $5\pi/4+\phi/4$ have projections respectively \[ 2^{1/3}\cos^{4/3}x,\qquad -2^{1/3}\cos^{4/3}(x-\pi/4). \] Their separation bounds the width from below. The function $h(t)=\cos^{4/3}t$ is concave on $[-\pi/4,\pi/4]$, since \[ h''(t)=\tfrac43\cos^{-2/3}t (\tfrac13\sin^2t-\cos^2t)<0. \] Thus $h(x)+h(x-\pi/4)$ has its minimum at an endpoint, where it is $1+2^{-2/3}$. This proves $2$.

Choose $a$ with $2/W<a<1$ and let $f_a(z)=z^3-a^3$. Its strict unit sublevel set is star-shaped about zero: the disk $\{w:|w-a^3|<1\}$ contains zero and is convex, so $z^3$ in it implies $t^3z^3$ in it for $0\le t\le1$. In particular the strict sublevel set is connected. Moreover it contains $aK_1$, because \[ |f_a(az)|=a^3|z^3-1|\le a^3<1\qquad(z\in K_1). \] Every projection therefore has width at least $aW>2$. This disproves the width conjecture while satisfying even the stronger connectedness condition for the strict sublevel set.

Assessment

The disk containment, sharp minimum diameter, and failure of the proposed width bound are all proved. The conformal existence and boundary-extension theorems are explicit standard inputs; the coefficient estimate and counterexample calculation are included in full.

Completion rate estimate: 100%.