Brennan’s conjecture and sharp inverse-square integral means

Proof. The coefficient and derivative bounds are the classical estimates just cited. The classical lower growth bound r/(1+r)^2 implies the weaker bound r/4 recorded here. A univalent holomorphic map has nonzero derivative, so for the last assertion the map

belongs to \SSS. Applying the Koebe quarter theorem to this normalized map gives (2.2).□

Proof. The upper growth bound and (2.3) give local uniform boundedness. Montel’s theorem therefore gives a locally uniformly convergent subsequence from every sequence in \KK; see [Dur83, Sections 1.1 and 1.3] for this normal-family argument and the univalent-limit form of Hurwitz’s theorem. Cauchy’s formula preserves the derivative normalization, so a limit F has F'(i)=1 and is nonconstant. For any fixed z_0\in\HH, apply Hurwitz’s theorem to F_n(z)-F_n(z_0) on the connected domain \HH\setminus\{z_0\}. The limit is not identically zero and has no zero there. Thus F is injective and belongs to \KK. The compact-open topology is metrizable using a countable compact exhaustion of \HH, so this sequential compactness proves compactness.

Cauchy’s formula gives locally uniform convergence of all derivatives. The limiting derivative is nonzero, so reciprocals also converge locally uniformly, as do their derivatives. These observations give the asserted joint continuity. Compactness of \KK\times E then gives the uniform bounds.

For the explicit estimate, put D=|x+i(y+1)|^2 and r=|m(x+iy)|. Then

The lower derivative bound in (2.1) gives

When |x|\le C_0 and y\le1, this proves (2.4).

Finally |m(z)|<1/7 on B(i,1/4), while (2.2) gives B(0,1/16)\subset F(B(i,1/4)). Injectivity prevents any point outside B(i,1/4) from having an image of modulus less than 1/16. This proves (2.5).□

Proof. Differentiating (2.7) proves the first identity in (2.9). Applying the normalization twice proves the second. In the operator product the weights telescope:

This gives the third identity, with its displayed order of multiplication. Joint continuity of (z,F)\mapsto T_zF follows from locally uniform convergence on affine images of compact source sets and the nonzero normalizing denominator. Together with Lemma 2.2, this proves boundedness of A_z. For fixed \Phi, joint continuity on \KK times a compact neighborhood of z gives the asserted supremum-norm continuity.

A word of length n in the affine maps -1/2+i/2 and 1/2+i/2 has height 2^{-n} and real part \sum_{l=1}^n\sigma_l2^{-l}, where \sigma_l\in\{-1,1\}. Writing \sigma_l=2\epsilon_l-1 and j=\sum_{l=1}^n\epsilon_l2^{n-l} gives real part -1+(2j+1)2^{-n}. Each j occurs once and each word has coefficient 2^{-n}. Expansion of L^n proves the grid formula; at n=0 it reduces to A_i=I.

For the positive sums defining L^n,

Testing with \Phi=\mathbf1 proves the norm identity. At the map F(z)=z-i every summand in L^n\mathbf1(F) equals one, so \|L^n\|\ge1. Since |u_{n,j}|\le1, (2.4) gives

The logarithms of these positive norms are subadditive. More explicitly, for fixed N\ge1 write n=qN+r, 0\le r<N, and use \|L^n\|\le\|L^N\|^q\|L^r\|. Taking the upper limit of the nth roots and then the infimum over N proves (2.11); the displayed lower and upper bounds give 1\le\rho\le4. If \lambda>\rho, the root limit bounds \|L^n\| by \lambda^n for all sufficiently large n. Increasing a constant to cover the finitely many remaining n proves the last claim.□

Proof. Fix f\in\SSS and 1/2\le r<1. Put \delta=1-r and choose n\ge1 with v=2^{-n}\le\delta<2v. On the fixed angular interval I=[3\pi/4,5\pi/4], write

The inverse formula gives

On I we have \cos\theta\le0 and hence D\ge1+r^2\ge2r; therefore |x|\le|\sin\theta|<1. Also

The height bounds use D\le(1+r)^2, D\ge1, and 1+r\le2. For the derivative bound use 4r^2\ge1 and D^2\le16.

The cells [u_{n,j}-v,u_{n,j}+v] cover [-1,1] with disjoint interiors. If x(\theta) belongs to cell j, then

Let B_E=\max_{G\in\KK,\,w\in E}|Q_G(w)|, which is finite by Lemma 2.2. The first identity in (2.9) gives

Thus the comparison covers every point of the varying-height arc in each cell.

The exact derivative conversion in (2.3) is

Since x(\theta) is increasing and d\theta\le16\,dx, (2.14) and the cell lengths 2v give

Here the bars around the intersection denote interval length.

For \alpha\in\{1,i,-1,-i\}, the map f_\alpha(z)=\alpha^{-1}f(\alpha z) belongs to \SSS and has derivative f'(\alpha z). The four rotated copies of I cover the full circle, with overlap only at endpoints. Applying the preceding estimate to these four maps therefore proves

with an absolute constant C. If \rho\le2, apply Lemma 2.3 with \lambda=2^{1+\varepsilon}. It gives

The operator L and the compact set E were fixed before f and r were chosen, so this constant depends only on \varepsilon.□

Proof. We first obtain a probability eigenmeasure for L. The dyadic formula for L^n will then give one horizontal period. Averaging over that period and over one logarithmic scale interval produces (3.1).

The eigenmeasure. For \tau>\rho, the root limit defining \rho implies that

converges in operator norm. Set

Positivity gives 1\leq g_\tau\leq M_\tau, and the maximum is attained because g_\tau is continuous and \KK is compact. We claim that

Indeed,

Iteration, using positivity and \mathbf 1\leq g_\tau, gives

Since \|L^n\|=\|L^n\mathbf 1\|_\infty by Lemma 2.3, it follows that

If M_{\tau_j}\leq M<\infty along a sequence \tau_j\downarrow\rho, taking the nth-root limit for each fixed j would give

Letting j\to\infty contradicts \rho>0, proving (3.2).

Choose a maximizer F_\tau and define

This is a positive functional of mass one, hence a Borel probability measure. The norm-convergent series gives R_\tau L=\tau(R_\tau-I), where I is the identity operator, so

Choose a sequence \tau\downarrow\rho. Since \KK is compact metric, C(\KK) has a countable uniformly dense subset. Diagonal selection gives a subsequence along which the integrals of those functions converge. The bound \|\mu_\tau\|=1 extends convergence to every function in C(\KK). The limiting functional is positive and has mass one, hence is a probability measure \mu by Riesz representation. For each \Phi\in C(\KK), the function L\Phi is continuous, and the final term above has absolute value at most \tau\|\Phi\|_\infty/M_\tau. Consequently

A horizontal period. For v=2^{-n} and u_j=-1+(2j+1)v, the formula from Lemma 2.3 reads

Right composition with A_{2+i} replaces u_j by u_j+2v. Thus (3.3) gives the exact cancellation

Both remaining abscissae have absolute value at most 2. The bound |Q_F(u+iv)|\leq C/v from Lemma 2.2 therefore implies

The left side is independent of n, so \mu A_{2+i}=\mu. Composition with A_{-2+i} gives the negative period as well.

Horizontal and scale averages. Define, by integration on continuous test functions,

For each continuous test, the integrand is continuous in x, by joint continuity of the normalized transformations and their weights. Compactness of the parameter interval makes this a finite positive functional, and its mass is strictly positive because the weights |Q_F(x+i)|^2 are positive. The period just proved shows that

We next compute the change under a factor of two in height. Using (3.3) before expanding L, we obtain

Here the substitutions s=\pm1/2+x/2 give two adjacent intervals. The last integrand has period two, since

Moving its integration interval to [0,2] yields

In particular

Since 2^\beta=\rho, the functional

satisfies H(2v)=H(v). Moreover, each \bar\mu A_{iv} remains horizontally invariant: the affine product gives

and (3.5) applies to the first operator on the right.

We have obtained all horizontal translations and a dyadic scaling relation. To obtain every positive scale, set

As before, this is a finite positive functional of strictly positive mass. For every y>0, substitution u=vy gives

The last equality follows because t\mapsto H(e^t) has period \log2; the integral over one full period is independent of its starting point. Horizontal invariance also persists under this average. Normalize to the probability P=\nu/\nu(\mathbf 1). Since A_{x+iy}=A_{x+i}A_{iy}, we conclude that

on continuous tests.

Borel tests. For each fixed z=x+iy, the expression

defines a finite Borel measure on \KK. Indeed, its weight and transformation are continuous in F, and the weight is bounded on \KK. The identity on continuous functions identifies this measure with y^{-\beta}P. Equality on nonnegative simple functions, followed by monotone convergence, proves (3.1) in its stated generality.□

Proof. Taking \Phi=\mathbf 1 in (3.1) gives

On a compact neighborhood of any interior point, the functions Q_F,Q'_F,Q''_F are uniformly bounded over \KK, by Lemma 2.2. These bounds control the first two real derivatives of |Q_F|^2, and hence justify differentiation under expectation in (3.10).

Holomorphicity and (3.7) give the pointwise identities

At i we have Q_F(i)=1, so the y derivative, the x derivative, and the Euclidean Laplacian of (3.10) give, respectively,

This proves (3.8). Boundedness of q_F(i) follows from the same compact-family bounds and Q_F(i)=1.

Finally, differentiating Q_{T_zF}(w)=Q_F(z\cdot w)/Q_F(z) with respect to w introduces the factor y=\operatorname{Im}z. Since \operatorname{Im}(z\cdot w)=y\operatorname{Im}w, formula (3.9) follows directly from the definition of q.□

Proof. For z\ne i, normalization gives

Apply (4.1) with \Phi(H)=|H(z^\#)|^{-4}. Both evaluations avoid the unique zero, and we obtain

The transformation (x,y)\mapsto(u,Y)=(-x/y,1/y) is an involution, with

Consequently dx\,dy=Y^{-3}du\,dY. Moreover,

so the involution preserves S_0.

Lemma 2.2 gives |F|\ge1/16 on S_0. Injectivity and conformal change of variables therefore imply

All the integrands are nonnegative and jointly Borel, by continuity of interior evaluations. Integrating (4.4), using Tonelli and the involution, gives

The same joint measurability makes I Borel. The conclusion follows from (4.5).□

Proof. Write R(z)=|x|+y. If z'=x'+iy'\in B(z,y/2), then

Thus these disks lie in \{R(z')\ge R(z)/3\} and eventually in S_0. On them 1/F is holomorphic. Submean for |1/F|^4, followed by comparison of y with y', gives

The convergence follows from the tail of the single integrable function in (4.3), and is uniform in z as R(z)\to\infty. Since 4k=\beta+3, this proves (4.6).

At heights y\ge\epsilon>0, (4.6) also gives |F(z)|\to\infty as R(z)\to\infty. Hence the preimage of a compact K, restricted to these heights, is bounded. It is closed in the plane, because a limit stays at height at least \epsilon and F is continuous there. It is therefore compact in \HH. Taking \epsilon=1 and then including the points below height one proves the bound on H_{F,K}.

To prove (4.7), choose a nonnegative \psi\in C_c^\infty(\CC) with \psi\ge1 on K, and fix \epsilon>0. The function g=\psi\circ F has compact support when restricted to y\ge\epsilon/2, by the compactness just proved. Thus

is smooth for y\ge\epsilon, differentiation passes under the integral, and h,h' vanish at sufficiently large height. The horizontal second derivative of g integrates to zero. Therefore

Write \Omega=F(\HH) and Y(w)=\Im F^{-1}(w) for w\in\Omega. Since \Delta(\psi\circ F)=|F'|^2(\Delta\psi)\circ F, conformal change of variables gives

Taking the absolute value of this signed integral yields

This finite bound is independent of \epsilon, as required.□

Proof of Proposition 4.1. Fix F with I(F)<\infty, and put \Omega=F(\HH). For each fixed \xi\in\CC and \epsilon>0, Lemma 4.3 implies

Indeed |F|\to\infty in this region, so eventually |F-\xi|\ge|F|/2.

For small heights first consider omitted targets. Fix R,\delta>0 and a dyadic h\le1. Partition \mathbb R into the half-open cells I_j=[jh,(j+1)h), j\in\mathbb Z. Call a cell a witness cell if there exist x\in I_j, h\le y\le2h, and \xi\notin\Omega with |\xi|\le R such that V_\xi^F(x+iy)\ge\delta. At any such witness z=x+iy,

The disk B(z,y) lies in \HH, and its image omits \xi. Lemma 2.1 therefore gives

For every x'\in I_j, the relative point

belongs to a fixed compact subset of \HH. The values of T_zF\in\KK there are bounded by a common constant M, by Lemma 2.2. The definition of T_zF and (4.10) imply

In particular, each witness cell at height h maps entirely into the fixed target disk K_{R,\delta}=\{|w|\le R+C_\delta\}. By (4.7), the disjoint cells, each of length h, have number at most C_{F,R,\delta}/h. This count permits all the omitted targets with |\xi|\le R at once. For every witness cell place a closed target disk of radius C_\delta h^k at the image of its midpoint at height h. Estimate (4.11), applied to any witness in that cell, shows that these disks cover every target witnessed at scale h. Their union E_h has area at most

The sum over dyadic h\to0 is finite. Hence \limsup_{n\to\infty}E_{2^{-n}} has area zero: its area is bounded by each tail of that convergent sum.

Take the countable union of these null sets for positive integer R and \delta=1/n, n\ge1. Every omitted target outside the union satisfies

Indeed each positive threshold eventually has no witness in any smaller dyadic slab, and these slabs cover all sufficiently small heights. This argument only uses finite disk covers for the fixed map F.

For an interior target \xi\in\Omega, choose a>0 so that \overline B(\xi,a)\subset\Omega. Its image under the continuous inverse F^{-1} is compact in \HH, and has positive minimum height. Below that height, |F(z)-\xi|\ge a, so V_\xi^F(z)\le y^k/a. Thus (4.13) holds for every interior target as well.

For each target retained above and each \delta>0, (4.13) first excludes all sufficiently small heights from \{V_\xi^F\ge\delta\}. Equation (4.9) then excludes all sufficiently large |x|+y. Finally

is closed, includes the possible pole, and lies in a bounded box with positive minimum height. It is compact in \HH. The assertion follows from Lemma 4.2.□

Proof. Use the compact exhaustion C_m=\{x+iy:|x|\le m,\ 1/m\le y\le m\}, m\ge1. The stated property is equivalent to

Necessity follows by containing the closed 1/n-superlevel in a box. For sufficiency, given \delta>0 choose 1/n<\delta; the corresponding box contains the closed \delta-superlevel. For fixed n,m, the containment is equivalent to

Any violation occurs in a nonempty open subset of the open complement of C_m and is therefore detected at a rational point. Each displayed inequality is closed in the parameters by continuity of evaluation. The countable tests prove the claim.□

Proof. Since F''/F'=iq_F/y, the two real derivative columns are

For complex vectors u,v, their real determinant is \operatorname{Im}(\overline u v). Applying this identity to the displayed columns gives (5.3). Local derivative continuity and compactness of \KK give the asserted continuity and boundedness at i. Lemma 3.2 gives

Because 2k-1=(\beta+1)/2, the last two terms in the expectation of (5.3) cancel. This proves (5.4).□

Proof. Write r=F'/(F-\xi). Since \log|F-\xi| is harmonic off the pole,

and (5.5) follows. Thus the critical equation is r=-ik/y, or F-\xi=iyF'/k, which is precisely G_F(z)=\xi and gives (5.6). A pole cannot solve this equation because y>0 and F'\ne0.

At a critical point the holomorphic derivative of r is

Differentiating the two real gradient entries now gives (5.7), whose determinant is (5.8) by (5.3). Its trace is -k/y^2<0. A negative determinant therefore gives a saddle; a positive determinant gives two negative eigenvalues and hence a strict maximum. At a critical point, \operatorname{Hess}V^F_\xi=V^F_\xi\operatorname{Hess}\ell, so these classifications apply to V^F_\xi as well.□

Proof. The compact-superlevel condition is Borel by Lemma 4.4. For completeness, regularity can be tested on the compact source boxes C_m=\{|x|\le m,\ 1/m\le y\le m\}: failure means that for some m there is w\in C_m with

The set defined by these equalities is closed in \KK\times\CC\times C_m; its projection onto \KK\times\CC is closed because C_m is compact. A countable union gives a Borel exceptional set.

For each fixed F, the critical values of G_F have area zero. This is the plane case of Sard’s theorem [Sar42, Theorem 4.1]; we give the short covering argument for the smooth maps at hand. Fix a compact source box and a slightly larger compact box in \HH. For small h>0, cover the first box by O(h^{-2}) squares of side h. If a square meets the critical set, choose a critical point there. Taylor expansion at that point, using the bounded second derivative on the larger box, puts the image of the square in an O(h^2)-neighborhood of a line segment of length O(h): the linear part has rank at most one. This neighborhood has area O(h^3). The image of all critical points in the first box consequently has outer area O(h), and therefore area zero. Exhausting \HH by boxes proves the claim. Combine it with Proposition 4.1 to obtain the asserted full-area sections for almost every F.

The derivative identity (T_tF)'(w)=F'(t\cdot w)/F'(t) gives (5.9) by substitution and gives the q identity in (5.10). Formula (5.3) then gives the J identity. The definition of V gives (5.11). Its multiplying factor is positive and independent of w. The source affine map is a homeomorphism of \HH, while the target change is a nonsingular similarity. They preserve the two conditions defining \mathcal G in both directions.□

Proof. Put \ell=\log V^F_\xi away from the pole. The solutions of G_F(z)=\xi are isolated by regularity, and form a closed subset of \HH; isolation follows from the real inverse-function theorem [Lee13, Theorem C.34]. Those with V^F_\xi\ge h lie in a compact set and are therefore finite. This includes the possibility of a pole in that compact set without allowing accumulation there: if F(z_0)=\xi, then G_F(z_0)-\xi=-i\operatorname{Im}(z_0)F'(z_0)/k\ne0.

Choose \lambda>0 which is not a critical value of V^F_\xi and put

including the pole if it exists. This is open, with compact closure in \HH. Its boundary is exactly the level \{V^F_\xi=\lambda\}: the pole is an interior point of U_\lambda, and at every level point the nonzero gradient gives values on both sides of \lambda. The level is a compact smooth one-manifold without boundary. Its components are open in that manifold, and thus there are finitely many. Each component is a smooth regular simple closed curve. One way to see the last assertion is to flow a unit tangent field, obtained by rotating \nabla\ell/|\nabla\ell| along the level. The flow is complete by compactness. Its orbits are open in each connected component, so a component is a single orbit. If this orbit were not periodic, its parametrization would be an injective surjective local homeomorphism from \mathbb R to that compact component, which is impossible. The return times form a closed subgroup with a gap around zero by local injectivity; its least positive element gives a regular simple closed parametrization.

Consider the bounded Jordan interior D of one level curve \gamma. The Jordan curve theorem is used here in its plane separation form [Mun14, Theorem 63.4]. The closure of D lies in \HH: points below the positive minimum height of \gamma connect to infinity without crossing it and hence lie in the unbounded complementary component. In fact

If a point inside had smaller value, \ell would attain a minimum below its boundary value \log\lambda at an interior point. A neighborhood of any pole can be excluded when taking this minimum, since \ell tends to +\infty there. Such a minimum contradicts \Delta\ell=-k/y^2<0. Once smaller values are excluded, equality at an interior point is likewise a local minimum and is impossible.

The interiors of the level curves are disjoint and are exactly the components of U_\lambda. Indeed, nested curves are excluded by (5.13). Conversely, a nonempty component of U_\lambda has a boundary point because its closure is compact. This point is on the level: a boundary point cannot belong to a component of the open set U_\lambda. Near it, regularity and (5.13) identify the superlevel side with the interior side of its level curve. The component therefore meets and contains that connected Jordan interior, and cannot leave it without crossing the curve. This proves the assertion, including that no component has a hole.

The superlevel components are therefore bounded Jordan domains with smooth boundaries. It remains to turn this topological information into a count of their critical points, keeping the possible pole visible in the index sum. On the positive orientation of each boundary curve, \nabla\ell is an inward normal and has winding number +1. We recall the elementary tangent-turning argument of Hopf [Hop35, §2, pp. 53–55] to fix this sign. Parametrize a regular simple closed curve periodically by \gamma:[0,1]\to\mathbb R^2, positively oriented and starting at a leftmost point. Its initial tangent points down. The unit secant

extends continuously to the closed triangle: on the diagonal its value is the unit tangent, and at (0,1) it is the negative initial tangent. The latter extension follows by the endpoint expansion

The extended secant has a continuous angle lift A(s,t) on the triangle. For example, uniform continuity permits one common number of subdivisions of every segment from (0,0) to (s,t) so that successive unit-vector ratios have a unique small continuous argument; adding those arguments constructs the lift. Along (0,t), secants lie in the closed right semicircle, from the downward tangent to its upward negative. Thus A(0,1)-A(0,0)=\pi. The secant along (t,1) is the negative of that along (0,t), including their extensions, so continuity gives A(t,1)-A(0,t)=\pi throughout. Therefore A(1,1)-A(0,0)=2\pi. This is the turn of the tangent along the diagonal; rotation to the inward normal preserves its winding number.

Inside one component of U_\lambda, remove small disjoint disks about its finitely many critical points and its pole, if present. On the remaining compact smooth domain, X=\nabla\ell is nonzero and its angle form

is closed. Stokes’ theorem [Lee13, Theorem 16.11] shows that the winding on the outer boundary equals the sum of the windings on the positively oriented small circles. At a nondegenerate critical point, the gradient on a sufficiently small circle is homotopic through nonzero vectors to its invertible linear part. Its index is consequently the sign of \det\operatorname{Hess}\ell, namely +1 at a maximum and -1 at a saddle.

If z_0 is the pole, the simple zero of F-\xi gives

where vectors are identified with complex numbers in the second formula. The negative radial leading field has winding number +1. Thus the pole also contributes +1, although it is not a finite critical point.

Let C_\lambda be the number of components of U_\lambda, and let B\in\{0,1\} be the number of poles. Summing the index identity gives

If B=1, its pole belongs to U_\lambda, so C_\lambda\ge B. The same inequality is automatic when B=0. Hence maxima are at least as numerous as saddles above every regular positive level. An empty superlevel gives the same identity with all counts zero.

Finally, for a given h>0, finiteness of the critical points of value at least h/2 permits a regular \lambda\in(h/2,h) with no critical value in [\lambda,h). The counts above \lambda are exactly those at or above h, including all ties. This proves (5.12).□

Proof. Fix a good pair (F,\xi). Order its saddles by decreasing value of V^F_\xi, breaking ties by increasing lexicographic order of (x,y). Order its maxima separately by the same rule. Every point has finitely many predecessors, since all predecessors have value at least its own positive value. Give it the zero-based rank equal to the number of its predecessors. Distinct points have distinct ranks: if one precedes another, its predecessors and the point itself precede the latter. Moreover, if rank j occurs, the finite ordered set of its predecessors has ranks 0,\ldots,j-1. Thus there are no missing ranks below an attained one, whether the entire fiber is finite or infinite.

Suppose a saddle has rank j and value v. There are at least j+1 saddles of value at least v. Lemma 5.5 gives at least j+1 maxima of value at least v. These maxima form an initial segment of the maximum order, so the maximum of rank j exists and has value at least v. Assign the saddle to this maximum. This is injective within the fiber, and gives (5.15) by Lemma 5.2. Distinct fibers cannot share a point, so the assignment is injective for fixed F across its entire domain.

Figure 1 illustrates the two ordered lists.

We give the Borel construction explicitly. Countable Borel enumeration is the subject of the Lusin–Novikov theorem; see [Mos09, 4F.6 and 4F.17]. Here nonsingularity supplies continuous local branches, and the rank definition will also supply affine equivariance. Consider the nonsingular solution graph

It can be covered by countably many continuous branches over open parameter sets in \KK\times\CC. To prove this without differentiating in the parameter F, fix (F_0,\xi_0,w_0)\in\Gamma and the invertible real matrix D=DG_{F_0}(w_0). Joint continuity of derivatives on source compacts gives a closed ball \overline B(w_0,r)\subset\HH and an open parameter neighborhood U such that

for every (F,\xi)\in U. The map

is a contraction of the closed ball into its interior. Its unique fixed point exists by the contraction principle [Lee13, Lemma C.35]. This point solves G_F(w)=\xi, is nonsingular, and depends continuously on (F,\xi). Continuity follows, for instance, by taking any convergent parameter sequence and using compactness of the ball and uniqueness to identify every subsequential limit of its fixed points.

Consequently \Gamma\cap(U\times B(w_0,r)) is the graph of one continuous branch over U, and is an open neighborhood in \Gamma. The space \Gamma is second countable, being a subspace of \KK\times\CC\times\HH. A countable collection of these branch neighborhoods therefore covers it. Enumerate the resulting continuous partial branches as b_j:U_j\to\HH, j\ge1. Activate branch j at a parameter if the parameter is in U_j and its value differs from every earlier branch defined there. These activation sets are Borel: there are finitely many earlier indices for a given j, and equality of two continuous branches is closed relative to their open overlap. Each nonsingular root occurs in exactly one active branch.

On a branch, its critical value can be evaluated without division by F(w)-\xi, using the positive continuous function

At every root this equals V^F_\xi(w) by (5.6). The signs of J_F, comparisons of H, equalities of values, and lexicographic comparisons of coordinates are Borel. The rank of an active branch among branches of its sign is thus the countable sum of the Borel indicators of its active predecessors. This is an extended-integer-valued Borel function, and is finite on good fibers. Requiring a positive-sign and a negative-sign active branch to have the same finite rank gives a countable Borel test for the assigned pair.

The set \mathcal D is Borel by Lemma 5.4 and continuity of (F,z)\mapsto(F,G_F(z)). At this target, require additionally that the positive-sign branch equal z. Exactly one active negative-sign branch passes the equal-rank test, by the existence and uniqueness proved above. Selecting its value defines a Borel map on \mathcal D: the preimage of an open source set is the countable union of the Borel tests just described with the selected branch constrained to that set.

Finally, under the affine source change w\mapsto t\cdot w, the critical fibers and signs correspond by (5.9)–(5.10). All critical heights in a fiber are multiplied by the same positive constant in (5.11). The positive real source scale preserves lexicographic order, including ties of the first coordinate. Hence the two orders, their ranks, and the equal-rank assignment correspond exactly, giving (5.16). The enumeration was used only to prove measurability and does not enter the geometric definition of the assignment.□

Proof. For P-almost every F, the complement of \{\xi:(F,\xi)\in\mathcal G\} has planar area zero by Lemma 5.4. Apply (6.1) on the positive sign with

Its target integral is supported on that null set, so

for almost every F.

Define the nonnegative Borel function

The normalized Jacobian and goodness are affine-equivariant, so

The weighted covariance identity consequently gives

Taking expectations in (6.3) and using Tonelli yields 0=2\log2\,\EE\Phi, which proves the claim.□

Proof. Continue under the contrary assumption \rho>2. By weighted covariance and (6.5),

Likewise \EE[\omega_F(z)J_{F,-}(z)]=y^{-1}\EE J_{F,-}(i). Thus expectation in (6.7) yields

The root quantities are integrable: q_F(i) is uniformly bounded on the compact normalized class, and J_F(i) is a fixed quadratic expression in this jet.

Hold N fixed, set R=Y^2, and let Y\to\infty. The ratio of the right box factor to the left one is

Dividing (6.8) by 2R\log Y gives

Now let integer N\to\infty. Monotone convergence exhausts the good-target saddles, and Lemma 6.1 accounts for the remaining positive-J weight. Hence

For the same J_F and the same law P, Lemma 5.1 gives

since 2k-1=(\beta+1)/2 and k>1. This contradiction proves \rho\leq2.□

Proof of the integral-means assertion in Theorem 1.1. Proposition 6.2 gives \rho\leq2. Proposition 2.4 therefore implies that, for every \epsilon>0, there is C_\epsilon<\infty, independent of f\in\SSS, such that

In particular \beta_f(-2)\leq1 for every f\in\SSS.

For the reverse inequality consider the Koebe map

It belongs to \SSS: equality of its values at z,u\in\DD gives (z-u)(1-zu)=0, hence z=u. Write \delta=1-r and take r sufficiently close to 1. On the arc z=re^{i(\pi+\theta)} with |\theta|\leq\delta,

Consequently

No exponent smaller than 1 can bound these means near the boundary. Hence \beta_{f_{\mathrm K}}(-2)\geq1, and B_{\SSS}(-2)=1.□

Proof. Put a=|t|-2\geq0. The distortion lower bound (2.1) and (7.1) give

Since a+1=|t|-1, this proves (7.2) with C_{t,\epsilon}=8^{|t|-2}C_\epsilon. At t=-2 it is exactly (7.1). For each fixed f and t, letting \epsilon\downarrow0 in the definition of \beta_f(t) gives \beta_f(t)\leq|t|-1; taking the supremum over f gives the same upper bound for B_{\SSS}(t).

For the Koebe map and the arc used above, |f_{\mathrm K}'(z)|\leq2\delta. Thus, for \delta=1-r sufficiently small,

Hence \beta_{f_{\mathrm K}}(t)\geq|t|-1, proving the equality.□

Proof. Away from the simple zero of f at 0,

The right side is locally integrable also at 0, where it is O(|z|^{p-2}). Green’s radial mean identity gives

To justify it at the zero, apply Green’s theorem on an annulus [Lee13, Theorem 16.17] with inner radius \delta. The circular mean at that radius is O(\delta^p) and its derivative is O(\delta^{p-1}), because f(z)=z+O(z^2). The inner terms are consequently O(\delta^p(1+|\log\delta|)) and vanish as \delta\downarrow0.

Bound the logarithm in (7.7) by \log(1/|z|). On |z|\leq1/2, the classical bounds give |f(z)|\geq c|z| and |f'(z)|\leq C, uniformly in \SSS. Since p-2<0, this part is bounded by a constant times

On |z|>1/2, the growth bound |f(z)|\leq |z|/(1-|z|)^2 implies

The conformal change of variables w=f(z) therefore bounds the remaining integral by a constant times

This proves (7.5).

For (7.6), the case a=0 is immediate. For 0<a<1, choose

If a\leq p, the power-mean inequality and (7.5) bound the means of |f|^a. If a>p, the growth bound gives instead

Thus the means of (1+|f|)^a are bounded by C(1-r)^{-2(a-p)_+}, where (b)_+=\max\{b,0\}. Since 2(a-p)_+<1, radial integration proves the claim.□

Proof. Choose 0<\eta<2/t-3 and set

Injective conformal area change gives

Here the last constant is the polar integral over \CC. The exact factorization

and Hölder’s inequality with exponents 2/t and 2/(2-t) now yield

by Lemma 7.2.□

Proof of the area assertion in Theorem 1.1. Equations (7.4) and Lemma 7.3, together with t=0, prove the disk integral for f\in\SSS throughout -2<t<2/3. For an arbitrary univalent f:\DD\to\CC,

The multiplier is finite and positive for every real t.

If \varphi:W\to\DD is a conformal bijection and f=\varphi^{-1}, the identities \varphi'(f(w))=1/f'(w) and \dd A(f(w))=|f'(w)|^2\dd A(w) give

This change of variables is valid for nonnegative integrands, whether or not the integrals are finite. Since 4/3<s<4 is equivalent to -2<2-s<2/3, the desired domain integral is finite.□

Proof. This is the known Sobolev formulation of Brennan’s conjecture due to Gol’dshtein and Ukhlov [GU13, Equivalence Theorem and its following Remark 2]. Indeed, 1<q<p and (p-2)q/(p-q)=s, so Theorem 1.1 supplies the derivative integrability required by their equivalence. The cited remark gives the case p=\infty, with q=s.□

Proof of Corollary 1.2. Use the same Koebe map f_{\mathrm K}. For sufficiently small |w|,

The fixed sector

lies inside \DD under either substitution z=-1+w or z=1-w: in both cases |z|^2=1-2\Re w+|w|^2<1. The area integral at t=-2 near -1 and the one at t=2/3 near 1 therefore each dominate a positive multiple of

More generally these sectors diverge for t\leq-2 and t\geq2/3, respectively. The image of f_{\mathrm K} meets the domain hypotheses just verified, so its inverse, by (7.9), also gives divergence at s=4 and s=4/3. Thus both open endpoints are sharp.□