Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

From the root count to an area contradiction.

At a direction avoiding the finitely many critical-value arguments, lift the value radius from $0$ to $t{e}^{i\theta }$ from each of the $k$ roots in $C_t$ and split each lift at level $\mu$. Put $A_0=\mathrm{Area}(C_t\cap \{|f|<\mu \})$. This is the sum of the areas of the $k$ one-root components inside $C_t$, not of all the root components of $f$; in particular, $0\le A_0\le \pi a$. On each of these components $f/\mu$ is conformal, since it is a proper map with no critical point onto a simply connected disc. Take $\psi (z)=\sum _{\ell \ge 0}{\gamma }_{\ell }{z}^{\ell }$ to be its inverse, so $f(\psi (z))=\mu z$ and $\psi (0)$ is the selected root. The radial curves in this coordinate therefore trace exactly the required value-ray lifts. Cauchy–Schwarz in the radial variable and Parseval give

First stop the radial integrals short of the unit circle, then pass to the endpoint by monotone convergence. Thus no smoothness of the boundary at level $\mu$ is assumed. Summing the $k$ square-root area bounds gives a mean total low lift length at most $\sqrt{k{A}_0/\pi }$. At a regular intermediate level $u$, let $A_C(u)$ be the area of $C_t\cap \{|f|<u\}$ and $P_C(u)$ its total boundary length. The argument principle and the coarea formula give

because $|dz|=ud(\arg f)/|f^{\prime }|$ on the level curve, its total argument variation is $2\pi k$, and Cauchy–Schwarz applies. Integrating $P_C(u)/(2\pi u)$ from $\mu$ to $t$ bounds the mean high lift length by $\sqrt{kx(\pi a-A_0)/(2\pi )}$. Cauchy–Schwarz combines the low and high bounds as

Choose a direction whose total lift length is at most $\sqrt{ka(x+2)/2}$. Order its $k$ boundary endpoints cyclically and join each adjacent pair using the two lifts and the intervening boundary arc. Each path lies in $\overline{C_t}\subset \{|f|<1\}$ and, under failure, has length at least $2$. Each lift occurs twice in the sum of these lengths, while the boundary occurs once. Therefore

Now let $t=\mu e^x$ vary, and write $a(x)=\mathrm{Area}(C_{\mu e^x})/\pi$ for the component containing the chosen first pair; its root count $k$ may increase at later mergers. On a regular interval, $a^{\prime }(x)=t{A}_C^{\prime }(t)/\pi$. Applying (9) at the outer level and rearranging therefore gives

on each regular interval for the component containing the selected pair, using the largest of the root-count bounds just derived. At merger levels this component only gains area, which strengthens the integrated comparison. Pólya’s inequality gives $a\le 1$ until the level reaches $1$. Thus an integrated lower bound greater than $1$ before $t=1$ contradicts the assumption.

No positive initial area has to be assumed. While $a\le 1$, for $0<x\le 3/{10}^5$ the bound $k\ge 2$ alone gives

at regular levels. For example, the last inequality follows using $\pi <22/7$, $\sqrt{2}>140/99$ and $\sqrt{2(2+3/{10}^5)}<20001/10000$. Integration, including the nonnegative merger jumps, gives $a(3/{10}^5)>{10}^{-6}$. This is the universal starting bound used by the certificate.

The comparison is integrated cell by cell without assuming that the independently certified lower bounds increase with the table index. On a cell $[x_{\ell },x_r]$, suppose $a(x_{\ell })\ge a_{\iota }$ and choose a trial upper bound $m$ for the area. Let $g$ be a certified lower bound for the right side of (10) throughout that cell under the trial assumption $a\le m$. If $m<a_{\iota }+(x_r-x_{\ell })g$, then $a(x_r)>m$. Indeed, the contrary assumption $a(x_r)\le m$ would imply $a\le m$ throughout the cell by monotonicity. Integrating the differential inequality, with the nonnegative merger jumps, would then give $a(x_r)>m$, a contradiction.