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

Two further root-count bounds.

Order the distances as $d_1\le \cdots \le d_k$. The triangle inequality gives $d_j\ge D-d_1$ for $j\ge 2$. The function $\lambda$ is decreasing and convex, since ${\lambda }^{\prime }(d)=-1/\sinh d$ and ${\lambda }^{″}(d)=\cosh d/{\sinh }^{2}d>0$. On $[d_{\mathrm{low}},D/2]$, the convex function $\lambda (d)+(k-1)\lambda (D-d)$ takes its maximum at an endpoint. If $d_1\ge D/2$, each summand is at most $\lambda (D/2)$ instead. (The definitions give $d_{\mathrm{low}}<D/2$.) Thus

and consequently

The minimum is essential: the two expressions arise from alternative positions of the nearest root, not from simultaneous restrictions.

For the second bound, integrate over the disjoint hyperbolic balls of radius $D/2$. Put $E=e^{2/a}-1$, so each ball has hyperbolic area $2\pi E$. At most one contains the origin; its contribution to $\sum _j\lambda (d_j)$ is at most $\delta (a)/2$. On every other ball, $-\log |z|$ is harmonic. To average it, first move the ball’s centre to $0$ by a disc automorphism. Hyperbolic area is invariant under this change of coordinates and has radial density $4/(1-|z{|}^2{)}^2$ there. The ordinary mean-value property on each centred Euclidean circle therefore gives hyperbolic area mean $\lambda (d_j)$. If the original ball has $0$ on its boundary, use increasing smaller concentric balls and monotone convergence.

There are $m=k-1$ nonexceptional balls if one contains the origin, and $m=k$ otherwise. Their union has area $2\pi mE$. Each superlevel set of $-\log |z|$ is a centred ball, so its intersection with any set of this area has area at most the smaller of the two areas. The centred ball of area $2\pi mE$ attains that upper bound at every superlevel. Integrating these intersection areas over the superlevel parameter therefore bounds the integral over the union by the integral over that centred ball. Its radius $R$ satisfies $\cosh R=1+mE$. With $v=\cosh r-1$, the identity $-\log \tanh (r/2)=\frac{1}{2}\log (1+2/v)$ reduces the radial integral to $(2E{)}^{-1}{\int }_0^{mE}\log (1+2/v)dv$. Evaluating it gives

where the improper integral is finite because $v\log v\to 0$ as $v\downarrow 0$, and the last inequality uses $\log (1+u)\le u$ for $u>0$. Adding the possible exceptional contribution and inverting each case yields

Again the minimum accounts for the two possible cases. This estimate uses area rearrangement, whereas the preceding one uses only ordered distances; neither replaces the circle-slice inequalities.

The comparison takes the maximum of $2$, $2x/\delta (a)$, these two bounds, and the certified circle-slice bounds. Each may first be rounded up because $k$ is an integer. Any of these estimates derived with an upper bound for $a$ remains valid at the actual smaller area: the root configuration still satisfies the weaker separation and radius conditions at that upper bound.