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

Will Cook

Long paper

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

Erdős #1041 By Will Cook 29 September 2026 Authorship, AI use and citation

PDFOpen the PDF, 86 pages TeXLaTeX source Source on GitHub Problem page

Summary. The record gives the rational estimates and geometric proof for ani's degree-seven polynomial: every connected set in its strict unit lemniscate joining two roots has Hausdorff measure greater than two. It develops contained paths for trinomials, collinear roots and two sparse families. Ordinary analytic arguments for critical-value criteria retain their unformalised inputs beside the statements. Critical-value moments, counterexamples to prescribed paths and questions about flow-strip geometry complete the account. The one-polynomial formal statement and correspondence with the historical wording remain distinct.

This paper covers the exact geometry of ani's one-polynomial counterexample, proved positive families and critical-value arguments with their unformalised inputs identified; formal-statement status and historical correspondence remain owned by the claims registry.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or the current status of Erdős #1041, including the later one-polynomial Formal Conjectures path-image Hausdorff-length check and its unreviewed historical curve-length correspondence; those distinctions belong to docs/claims.json in the mathematics repository.

Start here, by the editorial route

In this paper

A path estimate from area and boundary length

We can also average inverse-ray lengths and boundary arcs directly. This ordinary construction keeps both the number of roots and the capacity of the selected component in the estimate. Its formal source assumes the construction in degrees above two, so the resulting path bounds below are unformalised remarks.

The scaled packing criterion, Remark 4.5, has smaller constants: $5 / 2 < 71 / 10$ for length and $25 / 13 < 2$ for the containment level. The advantage of the direct argument is its explicit dependence on component geometry. Its $μ \leq 1 / 2$ specialisations are already within the range of Remark 4.1. For a monic degree- $n$ polynomial put

K_{t} = {z : | f (z) | \leq t}, μ = min_{f^{'} (c) = 0} | f (c) |, ρ = μ^{1 / n} .

Remark 5.1 (unformalised path bound at level $2 μ$ ). For every monic polynomial $f$ of degree $n \geq 2$ , two zero occurrences are joined by a possibly degenerate path of length at most

\frac{71}{10} ρ

inside $K_{2 μ}$ . If $f$ is squarefree, their locations are distinct. If $μ \leq 1 / 2$ , the construction may be chosen inside ${| f | < 1}$ with length at most $5.7$ .

The constant is the rationally certified specialization of a two-parameter bound. If the selected component contains $k \geq 2$ roots, then for every $r \in (0, 1)$ and $λ > 1$ the proof constructs a path in $K_{λ μ}$ whose length is at most

\begin{matrix} (CF) & \sqrt{\frac{2}{k}} (\frac{\sqrt{2} r}{(1 - r)^{2}} + λ^{1 / n} (\sqrt{\log (λ / r)} + \frac{π}{\sqrt{\log λ}})) ρ . \end{matrix}

For $n \geq 3$ , the choice $λ = 2$ , $r = 3 / 20$ makes the bracket less than $71 / 10$ : the largest case is $n = 3$ , where exact rational bounds give $66517563 / 9392500 < 71 / 10$ . Degree two has the exact root-segment bound $2 ρ$ .

Proof. A repeated zero gives the constant path, so assume $f$ is squarefree and $μ > 0$ . Degree two is the segment of length $2 ρ$ in $K_{μ}$ ; assume $n \geq 3$ . Put $T = λ μ$ , let $C_{T}$ be the component of $K_{T}$ containing a critical point at level $μ$ , and write $A (σ) = Area (K_{σ} \cap C_{T})$ . We will first choose a regular level $t$ , then a direction in the value plane, and finally an adjacent pair of roots. This order ensures that the three estimates below apply to the same path.

At a regular level $σ$ , a component $C^{'}$ containing $k^{'}$ roots has $| d z | = σ d φ / | f^{'} |$ on its boundary and total argument variation $2 π k^{'}$ . Cauchy–Schwarz, followed by the coarea formula, gives

\begin{matrix} (CF1) & H^{1} (\partial C^{'})^{2} \leq 2 π k^{'} σ A^{'} (σ) . \end{matrix}

Here $A^{'}$ sums the area derivatives of all the components in $C_{T}$ , so it bounds the contribution from $C^{'}$ . Pólya’s inequality $A (T) \leq π T^{2 / n}$ [2] and averaging with respect to $d σ / σ$ on $(μ, T)$ give a regular level $t$ such that $t A^{'} (t) \leq π T^{2 / n} / \log λ$ . The component $C_{t}$ containing the chosen first merger has $k \geq 2$ roots and

H^{1} (\partial C_{t}) \leq π \sqrt{2 k / \log λ} T^{1 / n} .

Averaging over levels avoids estimating the area derivative at the critical level itself.

Choose a direction avoiding all critical-value arguments. Lift its radial segment from each of the $k$ roots to $\partial C_{t}$ , splitting each lift at value modulus $r μ$ . Below $μ$ , the inverse branches of $f / μ$ are univalent. Their derivatives at $0$ are $μ / f^{'} (z_{i})$ , where $z_{i}$ ranges over the roots. The area formula for their disjoint images gives

\begin{matrix} (CF2) & \sum_{i} \frac{μ^{2}}{| f^{'} (z_{i}) |^{2}} \leq ρ^{2} . \end{matrix}

Indeed, the squared derivative at the centre is at most the image area divided by $π$ , and the total image area is at most $π μ^{2 / n}$ . Koebe distortion and Cauchy–Schwarz therefore bound the sum of the lengths below $r μ$ , uniformly in the direction. Write $ℓ_{i}^{h i g h} (θ)$ for the length of the $i$ th piece above $r μ$ in direction $θ$ . At an intermediate regular level, apply (CF1) to all the lower components contained in $C_{t}$ ; their root counts sum to $k$ . Integrating their total perimeter with measure $d σ / (2 π σ)$ gives the mean of $\sum_{i} ℓ_{i}^{h i g h}$ . Cauchy–Schwarz in $σ$ , using an area increment at most $Area (C_{t}) \leq π T^{2 / n}$ , gives

\frac{1}{2 π} \int_{0}^{2 π} \sum_{i} ℓ_{i}^{h i g h} (θ) d θ \leq \sqrt{\frac{k Area (C_{t})}{2 π} \log \frac{t}{r μ}} \leq \sqrt{\frac{k}{2}} T^{1 / n} \sqrt{\log (λ / r)} .

Choose a direction avoiding the critical-value arguments at which the sum is at most this bound. The excluded directions form a finite set and hence have measure zero.

Order its $k$ endpoints cyclically on $\partial C_{t}$ . Each adjacent pair of endpoints gives a path consisting of two lifted segments and the intervening boundary arc. In the sum over these $k$ paths, every lifted segment occurs twice and the boundary arcs partition $\partial C_{t}$ . Thus some pair has length at most the sum of the following three bounds:

\begin{aligned} \frac{2}{k} \sum_{i} ℓ_{i}^{l o w} & \leq \frac{2 r ρ}{\sqrt{k} (1 - r)^{2}}, \\ \frac{2}{k} \sum_{i} ℓ_{i}^{h i g h} & \leq \sqrt{\frac{2}{k}} \sqrt{\log (λ / r)} T^{1 / n}, \\ \frac{H^{1} (\partial C_{t})}{k} & \leq \sqrt{\frac{2}{k}} \frac{π T^{1 / n}}{\sqrt{\log λ}} . \end{aligned}

The first holds in every direction, the second in the direction just chosen, and the third at the previously selected level. Substituting $T^{1 / n} = λ^{1 / n} ρ$ proves (CF). For the final assertion, the repeated-zero case is already a constant path, and in degree two the root segment has length $2 ρ = 2 \sqrt{μ} \leq \sqrt{2}$ . For $n \geq 3$ , take $λ = 2$ and $r = 3 / 20$ . Since $k \geq 2$ and $μ \leq 1 / 2$ , (CF) gives

\begin{aligned} length & \leq \frac{60 \sqrt{2}}{289} μ^{1 / n} + (2 μ)^{1 / n} (\sqrt{\log (40 / 3)} + \frac{π}{\sqrt{\log 2}}) \\ \leq \frac{60 \sqrt{2}}{289} + \sqrt{\log (40 / 3)} + \frac{π}{\sqrt{\log 2}} \\ = 5.676476 \dots < 5.7 . \end{aligned}

Here $2 μ \leq 1$ , and the mean-value choice may be taken at a regular level strictly below the top of its positive-measure window. Thus the entire path lies in ${| f | < 1}$ . ◻

Before replacing the right side of (CF) by the constant $71 / 10$ , we can retain its dependence on the selected component. Its number $k$ of roots occurs in the factor $\sqrt{2 / k}$ , and its logarithmic capacity bounds the area used in the estimate. Each gives sufficient conditions for the length bound to be less than $2$ .

The number of roots in the first merged component can be much smaller than the degree. At a first merger caused by one simple critical point it is $2$ , so the following thresholds do not apply. For $z^{n} - b$ , however, all $n$ roots merge at the same level, giving $k_{0} = n$ . For example, $n \geq 17$ and $0 < | b | \leq 1 / 2$ satisfy the first pair of hypotheses. These conditions illustrate what the area argument gains from simultaneous mergers; the preceding unformalised $13 / 25$ criterion already covers these three ranges without the additional lower bound on the number of roots.

Remark 5.2 (unformalised root-count refinement). Let $f$ be monic with every root in the open unit disc, let $c_{*}$ be a critical point with $| f (c_{*}) | = μ$ , and let $k_{0}$ be the number of roots, counted with multiplicity, in the component of $K_{μ}$ containing $c_{*}$ . Then Erdős #1041 holds for $f$ in each of the three cases

μ \leq \frac{1}{2} and k_{0} \geq 17, μ \leq \frac{1}{4} and k_{0} \geq 12, μ \leq \frac{1}{8} and k_{0} \geq 10 .

Proof. If $μ = 0$ , a repeated zero gives the constant path. Assume henceforth $μ > 0$ . Every selected component $C_{t}$ in the proof of Remark 5.1 contains the first-merge component, so $k \geq k_{0}$ . For the first case take $λ = 2$ , $r = 13 / 100$ ; then $(2 μ)^{1 / n} \leq 1$ and $ρ \leq 1$ , and the exact bounds $\sqrt{2} < 283 / 200$ , $\sqrt{\log (200 / 13)} < 5 / 3$ and $π / \sqrt{\log 2} < (22 / 7) / (104 / 125) = 1375 / 364$ make the bracket in (CF) at most $15668813 / 2755116$ , whose square is $34 - 12570881068535 / 7590664173456 < 34$ . Hence $k_{0} \geq 17$ gives squared length below $(2 / 17) \cdot 34 = 4$ . For the second case take $λ = 4$ , $r = 3 / 25$ : the bracket is below $6075221 / 1273888$ , whose square is $24 - 2038665078215 / 1622790636544 < 24$ , and $2 / k_{0} \leq 1 / 6$ gives squared length below $4$ . For the third take $λ = 8$ , $r = 11 / 100$ : the bracket is below $55629121 / 12475575$ , whose square is $20 - 18200328379859 / 155639971580625 < 20$ , and $2 / k_{0} \leq 1 / 5$ again gives squared length below $4$ . In each case $λ μ \leq 1$ , so the freedom in the choice of the regular level $t$ keeps $t < 1$ and the containment strict. ◻

The next criterion uses logarithmic capacity, a measure of the size of a compact set that scales linearly under dilation. It compares the size of one component with that of the entire polynomial sublevel set. If that sublevel set is connected, the ratio $κ$ below is $1$ ; a cutoff such as $κ \leq 1 / 3$ therefore requires the component to have smaller capacity than the full sublevel set. A small number of roots in the component does not imply this capacity bound. The ratio requires $μ > 0$ . When $μ = 0$ , $f$ has a repeated zero and the constant path already gives the conclusion; there is no capacity ratio to evaluate.

Remark 5.3 (unformalised capacity refinement). Keep the hypotheses of Remark 5.2 with $0 < μ \leq 1 / 2$ , let $C$ be the component of ${| f | < 2 μ}$ containing $c_{*}$ , and put $κ = cap (\overset{―}{C}) / (2 μ)^{1 / n}$ . If $κ \leq τ_{k_{0}}$ , where

τ_{k} = \frac{\sqrt{2 k} - A}{B}, A = \frac{283}{3610}, B = \frac{52029}{9100},

then Erdős #1041 holds for $f$ . In particular $κ \leq 1 / 3$ suffices for every root count $k_{0} \geq 2$ , and the rational cutoffs $2 / 5, 12 / 25, 1 / 2, 7 / 12, 16 / 25, 2 / 3, 7 / 10$ suffice at $k_{0} = 3, \dots, 9$ , rising to $39 / 40$ at $k_{0} = 16$ .

Proof. Repeat the averaging proof of Remark 5.1 with $A (σ) = Area (K_{σ} \cap C)$ for $σ < 2 μ$ . This keeps the chosen paths in the open component $C$ , even if distinct components have touching boundaries at level $2 μ$ . Replace the global area input by the component form $Area (C) \leq π cap (\overset{―}{C})^{2} = π κ^{2} (2 μ)^{2 / n}$ of the area–capacity inequality [2]. Only the high-lift and boundary terms acquire the factor $κ$ ; the low-lift term (CF2) is unchanged. Taking $λ = 2$ , $r = 1 / 20$ , and the exact bounds above together with $\log 40 < 12641 / 3402 < (97 / 50)^{2}$ , gives $length < \sqrt{2 / k_{0}} (A + B κ)$ . The definition of $τ_{k_{0}}$ makes the right side at most $2$ , and for each displayed rational $q_{k}$ integer arithmetic gives $(A + B q_{k})^{2} < 2 k$ . Discarding $\sqrt{2 / k_{0}} \leq 1$ altogether gives the uniform cutoff $κ \leq 1 / 3$ . ◻

By the component-capacity formula, $κ = e^{- Σ / n}$ with $Σ$ the sum of exterior Green function values at the roots excluded from $C$ . Indeed, let $Ω$ be the unbounded component of $\hat{C} \ \overset{―}{C}$ , let $G$ be its Green function with pole at infinity, extended by zero off $Ω$ , and let $g (\cdot, a)$ be its Green function with pole at $a \in Ω$ . The function $\frac{1}{n} \log \frac{| f |}{2 μ} - G + \frac{1}{n} \sum_{a} g (\cdot, a)$ , summed over the roots $a \in Ω$ with multiplicity, extends harmonically to $Ω$ , is bounded, and vanishes on $\partial Ω \subseteq \partial C$ , so it vanishes identically; its value at infinity is $\log κ + \frac{1}{n} \sum_{a} G (a)$ by the symmetry $g (\infty, a) = G (a)$ . Failure of these component-sensitive criteria alone, in the range $μ \leq 1 / 2$ , forces $k_{0} \leq 16$ and exterior Green sum $Σ < n \log (40 / 39)$ when $k_{0} = 16$ . This is a limitation of the area-based criteria. The low-critical theorem already excludes an actual counterexample throughout that range.

The proof of Remark 5.1 averages over levels in $[μ, λ μ]$ . To keep the path at the first critical level instead, one would need a perimeter estimate of the following kind. The regular-level argument (CF1) gives no finite pointwise bound at the critical level: even at a simple critical point, the area derivative diverges logarithmically.

The addendum [26] reports a Runge-approximation construction of monic level-one components with one simple zero and arbitrarily long analytic boundaries. It imposes neither a unit root disc nor a global lower bound on critical-value moduli. Its proof is not available, and the report is not an input to the path estimates here. The question below requires a stronger hypothesis than the existence of one unramified component: the level must lie below every critical-value modulus of the polynomial. A construction without that global condition cannot by itself answer the question.

Problem 5.4 (perimeter below the smallest critical-value modulus). Let $f$ be squarefree and monic of degree $n \geq 2$ , and put $μ = min_{f^{'} (c) = 0} | f (c) |$ . Is there a constant $β$ , independent of $n, f$ and $0 < σ < μ$ , such that every component $C$ of ${| f | \leq σ}$ satisfies

H^{1} (\partial C) \leq β σ^{1 / n} ?

Equivalently, for each univalent inverse branch defined on $D$ by $f (ϕ_{j} (w)) = σ w$ , is

\int_{0}^{2 π} | ϕ_{j}^{'} (e^{i t}) | d t \leq β σ^{1 / n} (1 \leq j \leq n) ?

There is no root-location hypothesis. If such constants exist, write $β_{*}$ for their infimum.

All these components contain one zero. The condition $σ < μ$ controls all inverse branches at once; it is stronger than requiring only the selected component to contain one zero. The examples below compare particular constants, not the necessity of this global hypothesis for every possible perimeter bound. For $1 < R < μ / σ$ , the area estimate gives

\frac{1}{2 π} \int_{0}^{2 π} | ϕ_{j}^{'} (e^{i t}) | d t \leq \frac{(σ R)^{1 / n}}{\sqrt{R^{2} - 1}} .

Indeed, circular means of $| ϕ_{j}^{'} |^{2}$ increase, the image of the annulus $1 < | w | < R$ has area at least $π (R^{2} - 1)$ times the mean at $1$ , and Pólya’s preimage-area bound [2] bounds that area by $π (σ R)^{2 / n}$ . The loss as $R ↓ 1$ belongs to this argument and does not prove an endpoint divergence.

For $d > 0$ and $f (z) = z^{2} - d^{2}$ , one has $μ = d^{2}$ and $ρ = d$ . For $σ < d^{2}$ the component of ${| f | \leq σ}$ around $d$ contains one root. Its parametrisations $d \sqrt{1 + (σ / d^{2}) e^{i θ}}$ converge uniformly, as $σ ↑ d^{2}$ , to one loop of Bernoulli’s lemniscate $| z^{2} - d^{2} | = d^{2}$ , of length $d Γ (1 / 4)^{2} / (2 \sqrt{π})$ by the substitution $φ (w) = d \sqrt{1 + w}$ and $Γ (1 / 4) Γ (3 / 4) = π \sqrt{2}$ . At $σ = d^{2}$ the two loops meet at the critical point, and the closed sublevel set is a single component containing both roots. Lower semicontinuity of length under the stated uniform convergence gives $β_{*} \geq Γ (1 / 4)^{2} / (2 \sqrt{π}) = 3.70814935 \dots$ ; convergence of the lengths themselves is not required. For $z^{n} - r^{n}$ , polar parametrisation of one limiting loop gives its length divided by $r$ as

\frac{2^{1 / n}}{n} \int_{- π / 2}^{π / 2} (\cos θ)^{1 / n - 1} d θ = \frac{2^{1 / n}}{n} \sqrt{π} \frac{Γ (1 / (2 n))}{Γ (1 / (2 n) + 1 / 2)} .

The evaluation is Euler’s beta integral (5.12.1), in its trigonometric form (5.12.2). To verify the claimed decrease with $n$ , put $p = 1 / n$ . The expression is $2^{p + 1} \sqrt{π} Γ (1 + p / 2) / Γ ((1 + p) / 2)$ ; its logarithmic derivative is

\log 2 + \frac{1}{2} (\frac{Γ^{'} (1 + p / 2)}{Γ (1 + p / 2)} - \frac{Γ^{'} ((1 + p) / 2)}{Γ ((1 + p) / 2)}) > 0.

Here $Γ^{'} / Γ$ is increasing on the positive real axis, by the positive trigamma series (5.15.1). The expression therefore decreases with $n$ and tends to $2$ .

The comparison with the next example also has an elementary bound, independent of rounded gamma values. One loop of $z^{2} - d^{2}$ has parametrisation

z (s) = \sqrt{2} d \frac{\cos s (1 + i \sin s)}{1 + \sin^{2} s}, - π / 2 \leq s \leq π / 2,

whose speed is $\sqrt{2} d / \sqrt{1 + \sin^{2} s}$ . Convexity puts $(1 + x)^{- 1 / 2}$ below its secant on $[0, 1]$ . Integrating that secant at $x = \sin^{2} s$ bounds the loop length divided by $d$ by $(π / 2) (1 + \sqrt{2})$ , as used below.

Proposition 5.5 (a counterexample to the proposed one-root perimeter bound). Let $p (z) = z^{8} - (3 / 2) z$ and let $C$ be the component of ${| p | \leq 1}$ containing the origin. Then $C$ contains exactly one zero and a neighbourhood of the closed disc of radius $5 / 8$ , so $H^{1} (\partial C) > 5 π / 4$ . The constant $Γ (1 / 4)^{2} / (2 \sqrt{π})$ is at most $(π / 2) (1 + \sqrt{2}) < 5 π / 4$ .

The quadratic loop is exceeded in degree eight, even with the global subcritical hypothesis. For $p (z) = z^{8} - (3 / 2) z$ , every critical point $c$ satisfies $c^{7} = 3 / 16$ and $p (c) = - (21 / 16) c$ . Hence

μ = \frac{21}{16} {(\frac{3}{16})}^{1 / 7} > 1, μ^{7} = \frac{3 \cdot 21^{7}}{16^{8}} > 1.

Thus $σ = 1$ is below every critical-value modulus, not merely a regular level for the selected component. The component $C$ of ${| p | \leq 1}$ containing the origin lies in ${| z | < 4 / 5}$ , because $| p (z) | \geq 6 / 5 - (4 / 5)^{8} > 1$ on $| z | = 4 / 5$ , and it contains a neighbourhood of the closed disc of radius $5 / 8$ , because $| p (z) | \leq (5 / 8)^{8} + 15 / 16 < 1$ on that disc. The other roots satisfy $| z |^{7} = 3 / 2$ , so $C$ contains exactly one root, and $H^{1} (\partial C) > 5 π / 4 > (π / 2) (1 + \sqrt{2}) \geq Γ (1 / 4)^{2} / (2 \sqrt{π})$ . Moreover, for fixed $a > 1$ and all sufficiently large $N$ , the one-root component of ${| z^{N} - a z | \leq 1}$ contains every disc of radius $r < 1 / a$ , and a circle of radius between $1 / a$ and $1$ isolates it by Rouché’s theorem; letting $a ↓ 1$ shows that every admissible $β$ is at least $2 π$ . For the latter family all critical values have modulus $(1 - 1 / N) a (a / N)^{1 / (N - 1)} \to a > 1$ , so it also gives the necessary bound $β_{*} \geq 2 π$ under the corrected global subcritical hypothesis. Kuznetsova–Tkachev [24] establish Laplace-transform and log-convexity properties for regular level-length functions. Those results do not provide a uniform bound at the first critical level. Total lemniscate length and one-branch endpoint length are different extremal quantities.

The preceding examples do not establish a universal perimeter bound. The implication below applies to an individual polynomial; a constant $β$ independent of the polynomial and degree would give a uniform path bound. The construction uses two components meeting at a first critical point and joins each root to that point at a cost of at most half its component’s perimeter. It does not give the constant $2$ of the historical question.

Remark 5.6 (unformalised consequence of a perimeter bound). Let $f$ be squarefree and monic of degree $n \geq 2$ , and put $μ = min_{f^{'} (c) = 0} | f (c) | > 0$ . Suppose $β > 0$ satisfies

H^{1} (\partial C) \leq β σ^{1 / n}

for every $0 < σ < μ$ and every component $C$ of ${| f | \leq σ}$ . Then two distinct roots of $f$ are joined inside $K_{μ} = {| f | \leq μ}$ by a rectifiable path of length at most $β μ^{1 / n}$ .

Proof.

Put $ρ = μ^{1 / n}$ . On each component of ${| f | < μ}$ , $f$ is a proper unramified covering of the disc ${| w | < μ}$ , hence univalent. Equivalently, exhaust by regular levels below $μ$ and apply the component count [8]. At a critical point $c_{*}$ with $| f (c_{*}) | = μ$ , the local degree $d \geq 2$ gives $d$ nearby preimages of an inward radial value. Univalence places them in different one-root components. Choose two such components $U_{a}, U_{b}$ , containing roots $a, b$ and with $c_{*}$ in both closures; simplicity of $c_{*}$ is unnecessary. For the boundary passage, take $f (ϕ (w)) = μ w$ on $D$ . Fix $| w_{0} | = 1$ . The finitely many preimages of $μ w_{0}$ have disjoint neighbourhoods. Properness of $f$ and connectedness of a sufficiently small disc cap force $ϕ$ to stay in one of them as $w \to w_{0}$ . At its preimage $c$ , write

f (z) - f (c) = (z - c)^{d} h (z), h (c) \neq 0.

A local analytic $d$ th root of $h$ makes $(z - c) h (z)^{1 / d}$ a conformal coordinate. In each inverse sector, $ϕ$ is therefore an analytic function of $(w - w_{0})^{1 / d}$ and extends continuously to $c$ . The finitely many critical boundary values are the only exceptional points; at every other boundary value the ordinary inverse function theorem applies. Moreover, near $e^{i θ_{0}} = w_{0}$ the boundary speed is $O (| θ - θ_{0} |^{1 / d - 1})$ , which is integrable even when $d > 2$ . Thus the extension is continuous on the closed disc and its boundary curve is rectifiable. The identity $f (ϕ (e^{i θ})) = μ e^{i θ}$ makes the boundary values injective, so the curve is Jordan.

As $r ↑ 1$ , the curves $ϕ (r e^{i θ})$ converge uniformly to that boundary and have length at most $β (μ r)^{1 / n}$ by the hypothesis. Lower semicontinuity of length gives boundary length at most $β ρ$ for each component.

A bounded rectifiable Jordan domain $U$ has the following elementary property: any $a \in U$ can be joined to any $c \in \partial U$ in $\overset{―}{U}$ with length at most $H^{1} (\partial U) / 2$ . Choose a line through $a$ but not $c$ , and let $[u, v]$ be the closure of the connected interval in its intersection with $U$ containing $a$ . Of the two boundary arcs from $u$ to $v$ , write $A$ for the one containing $c$ and $A^{'}$ for the other. The paths from $a$ through $u$ or $v$ and then along $A$ to $c$ have total length

| a - u | + | a - v | + length (A) = | u - v | + length (A) \leq H^{1} (\partial U),

because $| u - v | \leq length (A^{'})$ . One path is at most half this sum. The connected interval ensures containment of the straight parts even when $U$ is not convex.

Apply this property in $U_{a}, U_{b}$ with common boundary point $c_{*}$ . Concatenating the two paths gives length at most $\frac{1}{2} (H^{1} (\partial U_{a}) + H^{1} (\partial U_{b})) \leq β ρ$ , as required. Only these two component perimeters are used; the subcritical hypothesis bounds them by the limiting argument above. Boundary arcs are allowed, so containment is in $K_{μ}$ , not the open sublevel. ◻

Sources and scope.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

Prefer the manuscript?

PDFOpen the PDF, 86 pages TeXOpen the LaTeX source Problem page

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:4bee9a881dbbe940. Its file fingerprint matched the published source record, so the PDF above, the LaTeX, and this page are one manuscript. Redeploying the site regenerates this page from whatever the public repository holds at that moment.