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

The weighted Poisson inequality and its equality case.

For one point $c$ of weight $1$, the quantity to bound is $(1-|c{|}^2{)}^2\le 1$; equality holds only at $c=0$. For two points $c,-c$ of equal weight, it is $1-|c{|}^4\le 1$. These examples suggest why a nonzero configuration should lose from equality. Now let $w_j\ge 0$, $\sum _j{w}_j=1$, and $|c_j|\le 1$, with repetitions allowed. Write $G(z)=\prod _k|1-\overline{c_k}z{|}^{w_k}$, taking a factor of exponent zero to be one. We seek an analytic function with modulus $G$: its logarithmic derivative will express the weighted Poisson kernel, and its Taylor coefficients will measure the loss from equality. For $|c|<1$ and $|\zeta |=1$, let $P_c(\zeta )=(1-|c{|}^2)/|\zeta -c{|}^2$, and write $dm=dt/(2\pi )$ on $\zeta =e^{it}$. When every centre is strictly inside the disc, choose analytic logarithms zero at the origin and put

On the unit circle $P=1-2\mathrm{\Re }(\zeta g^{\prime }/g)$, by logarithmic differentiation. The Poisson formula reproduces $g(c)$ from its boundary values and gives $\int P_{c}dm=1$ for $|c|<1$. Expanding a square therefore yields

This explains why the same kernel bounds the value at each centre. Take $c=c_j$, multiply by $w_j$ and sum to obtain

To see the last equality, expand $|g{|}^2$ and $\zeta g^{\prime }\bar{g}$ and integrate on the circle. Orthogonality leaves $1+\sum _{\nu \ge 1}|a_{\nu }{|}^2$ and $\sum _{\nu \ge 1}\nu |a_{\nu }{|}^2$, respectively. The analytic function $g$ extends to a neighbourhood of the closed disc in this interior case, which justifies the termwise integrations. For closed-disc centres replace $c_j$ by $r{c}_j$, $0<r<1$. The actual coefficient of degree $\nu$ becomes $r^{\nu }{a}_{\nu }$, where the fixed coefficients are defined from the analytic function near zero. Pass to $r\uparrow 1$ first with an arbitrary finite coefficient sum. Nonnegative finite deficits then give summability and the infinite inequality; no boundary holomorphic logarithm is required. Lean checks the exact radial transport of each Cauchy coefficient, the finite boundary-deficit inequality, and the summable full boundary deficit.

If equality holds, all positive-degree coefficients vanish, so $g=1$ near zero and $\sum _j{w}_j\overline{c_j}/(1-\overline{c_j}z)=0$ there. Group equal centres before clearing denominators. A nonzero centre of positive total weight gives a nonzero pole coefficient, which is impossible. Thus equality is equivalent to $c_j=0$ on positive support. In particular, strictly positive weights force all centres zero. A point of weight zero is unconstrained: weight one at $0$ and weight zero at $1$ still give equality. Grouping equal centres is necessary: an argument that assumed they were distinct would not prove the stated equality case.

Equal weights in $\sum _j{w}_{j}G(c_j{)}^2\le 1$ give the quadratic finite inequality above for every $m\ge 1$, with equality only at the zero configuration. The linear inequality follows separately. For $M=\sum _j{w}_{j}G(c_j)$, the variance identity

gives $M\le 1$. Equality in this Cauchy–Schwarz step means that the values $G(c_j)$ are constant at all indices of positive weight; equality in the final bound also requires equality in the quadratic inequality. The three-point and four-point specialisations follow with their equality cases. Lean checks the quadratic weighted inequality. The hypotheses permit repeated centres and zero weights; the equality condition refers only to centres of positive weight.

This equality condition also identifies all equality cases of Theorem 10.2 for a fixed containing disc $\bar{D}(h,R)$ with $R>0$. Equality in the critical-value bound forces equality in the quadratic finite inequality after normalisation, so every critical point is $0$. Thus the normalised derivative is $n{z}^{n-1}$ and the polynomial is $z^n-\lambda$, with $|\lambda |=1$ forced by equality. Restoring the centre and radius gives exactly $f(z)=(z-h{)}^n-\lambda$ with $|\lambda |=R^n$. Equality in either lower-power consequence also forces equality in the quadratic bound, so it has the same family. For $R=0$ the only possible polynomial is $(z-h{)}^n$. Lean does not check this classification.

The earlier three-point Hölder argument and four-point matching argument are not used here. Their scalar identities do not justify the omitted inequalities; the proof is the weighted Poisson calculation above.