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+A STRICT INVERSE-FIRST-POWER BOUND FOR UNIVALENT FUNCTIONS
+a complete, self-contained account
+
+
+WHAT THIS IS
+
+This is a complete proof of one theorem about the shapes that a
+complex-differentiable (holomorphic) map can produce. It is written so
+that a student who knows complex numbers and one-variable calculus can
+follow every step. Nothing is left as an exercise and nothing is
+handed to a "standard lemma" without saying exactly what the lemma
+says.
+
+Three results from analysis are used without proof. Each is named
+every time it is used, and each is stated in words:
+
+ (C1) COMPACTNESS. A family of injective holomorphic functions that
+ is normalized at one point (value and derivative pinned) is
+ compact in the sense of locally uniform convergence, and the
+ limits are again injective. This is Montel's theorem plus
+ Hurwitz's theorem.
+
+ (C2) SARD. For a smooth map of the plane, the set of images of
+ points where the derivative fails to be invertible has area
+ zero.
+
+ (C3) INDEX COUNTING. A simple closed curve has turning number one.
+ For a smooth real function of two variables, the index of an
+ isolated critical point is +1 if the Hessian determinant is
+ negative (a peak) and -1 if it is positive (a saddle); a
+ logarithmic pole has index +1.
+
+You will need: complex numbers, partial derivatives, the area of a
+plane region, and the idea of averaging over a collection of maps.
+You will not need measure theory beyond the two phrases "area zero"
+and "average of a nonnegative quantity".
+
+
+CONTENTS
+
+ Part 0 The problem and the answer
+ Part 1 A compact model, and one operator that measures growth
+ Part 2 A weighted law of maps, when the operator grows too fast
+ Part 3 A potential, and why it cannot run off to infinity
+ Part 4 Critical points, and pairing each saddle with a peak
+ Part 5 Transport: the pairing forces an average to be <= 0
+ Part 6 The harmonic barrier, and the uniform gap
+ Part 7 What the proof does and does not say
+ Appendix Notation and where each constant comes from
+
+
+================================================================
+PART 0. THE PROBLEM AND THE ANSWER
+================================================================
+
+
+0.1 THE OBJECTS
+
+Let D = { z in C : |z| < 1 } be the open unit disk.
+
+Let S be the collection of all maps f : D -> C with
+
+ (a) f is holomorphic on D,
+ (b) f is injective (two different points never get the same value),
+ (c) f(0) = 0 and f'(0) = 1.
+
+Such a map is called UNIVALENT. Condition (b) means f is a bijection
+onto its image f(D); condition (c) is a normalization: it uses the two
+free degrees of freedom (translation and scaling) to pin the map down.
+
+The point of normalization (c) is that it makes S a "small" family.
+Without it, any nonzero constant multiple of a member of S would be
+another member and no bound could be uniform.
+
+FACT 0.1. If f is in S then f'(z) is never zero for z in D.
+
+Proof. Suppose f'(z0) = 0. Then near z0 we can write
+
+ f(z) - f(z0) = (z - z0)^m * g(z)
+
+with m >= 2 an integer and g holomorphic with g(z0) != 0. On a small
+enough disk around z0, g has no zeros and |g| is within a factor of 2
+of |g(z0)|. For any small enough value w near f(z0), the equation
+
+ (z - z0)^m = w / g(z)
+
+has exactly m solutions z inside that disk, by the argument principle.
+So f is m-to-one near z0, which contradicts injectivity. QED
+
+Consequence. 1/f' is holomorphic on D with no singularities, and so
+|f'|^p is a perfectly good positive continuous function for every real
+exponent p, including negative ones.
+
+Let S_b be the subcollection of S consisting of maps whose image f(D)
+is a bounded set of the plane. How big that bounded set is may depend
+on f; we impose no smoothness at all on the boundary of f(D).
+
+
+0.2 WHAT AN INTEGRAL MEAN MEASURES
+
+For a real number p and 0 <= r < 1 define the p-th integral mean
+
+ M_p[f'](r) = (1/2pi) * INT_{-pi}^{pi} |f'(r e^{i theta})|^p dtheta.
+
+Read it out loud: fix r, walk once around the circle of radius r, and
+average the p-th power of |f'| over that circle.
+
+For p = -1 this is the average of 1/|f'|. So M_{-1} is large exactly
+when the derivative is typically small somewhere on the circle. A
+small derivative means the map squeezes the plane, which forces the
+inverse map to stretch, which is what makes images of bounded domains
+have ragged boundaries.
+
+The interesting question is: how fast can M_{-1}[f'](r) blow up as r
+goes to 1?
+
+THE TRIVIAL BOUND. Koebe's distortion estimate says that for |z|=r<1
+
+ (1 - r)/(1 + r)^3 <= |f'(z)| <= (1 + r)/(1 - r)^3.
+
+The lower bound is the useful one at p = -1:
+
+ 1/|f'(z)| <= (1 + r)^3/(1 - r) <= 8/(1 - r) (for 0 <= r < 1).
+
+So without any thought at all,
+
+ M_{-1}[f'](r) <= 8 (1 - r)^{-1}.
+
+So the exponent 1 is free. Every nontrivial statement below is a
+statement about getting the exponent down.
+
+
+0.3 THE UNIVERSAL SPECTRUM
+
+For each f and each p, the growth rate of the integral mean is
+
+ beta_f(p) = limsup over r -> 1 of log M_p[f'](r) / log(1/(1-r)).
+
+Reading. If M_p[f'](r) is roughly (1-r)^{-c} times a bounded factor,
+then beta_f(p) = c. The limsup ignores wobbles; it only sees the
+sharpest rate that recurs. The denominator log(1/(1-r)) is the natural
+depth scale: going from r to 1 - (1-r)/2 costs exactly one unit of it.
+
+Example. By 0.2, beta_f(-1) <= 1 for every f in S. So the number 1
+is an achievable trivial exponent.
+
+The UNIVERSAL BOUNDED SPECTRUM is the worst case over all bounded maps:
+
+ B_b(p) = sup { beta_f(p) : f in S_b }.
+
+Here the supremum is taken after each individual map has settled to its
+own limiting exponent. (Multiplying f by a nonzero constant c changes
+every M_p by the fixed factor |c|^p, and so does not change beta_f(p)
+at all. That is why normalizing the derivative loses nothing.)
+
+
+0.4 KRAETZER'S CONJECTURE
+
+Kraetzer's experiments with conformal maps onto bounded domains led
+him to predict
+
+ B_b(p) = p^2/4 for |p| <= 2,
+ B_b(p) = |p| - 1 for |p| >= 2.
+
+The quadratic piece is in his 1996 paper and his 2000 thesis; the full
+two-piece formula is also quoted by Beliaev. At p = -1 the prediction
+reads B_b(-1) = 1/4.
+
+Note the shape of the conjecture: it is symmetric in p, it is smooth at
+|p| = 2 where the two pieces meet (both give 1), and it gives the same
+value at p = 1 and p = -1. That symmetry is a prediction, not a
+theorem.
+
+
+0.5 THE THEOREM WE PROVE
+
+THEOREM (uniform inverse-first-power bound). There exist constants
+eps with 0 < eps < 1/4 and C < infinity such that for EVERY f in S
+(not merely every f in S_b) and EVERY 1/2 <= r < 1,
+
+ M_{-1}[f'](r) <= C (1 - r)^{-1/4 + eps}.
+
+Consequences, in order of immediacy:
+
+ (i) beta_f(-1) <= 1/4 - eps for every f in S.
+ (ii) B_b(-1) <= 1/4 - eps < 1/4.
+ (iii) Kraetzer's formula is FALSE at p = -1.
+
+Three things to notice, because they are all part of the content.
+
+ * The estimate is uniform over the whole class S, which is strictly
+ larger than S_b. Bounded image is never used as a hypothesis; it
+ is only used to state the consequence (ii).
+ * The gap eps is not evaluated. The proof shows a positive gap
+ exists, and produces no number. Existence is the theorem.
+ * Only the exponent p = -1 is settled. Nothing here decides the
+ symmetric candidate at p = 1, and nothing here decides any other
+ exponent.
+
+
+0.6 WHY STRICTNESS IS THE WHOLE DIFFICULTY
+
+The machinery of the proof naturally produces a NON-STRICT inequality:
+it will say B_b(-1) <= 1/4. But that is exactly the conjectured value.
+A non-strict inequality at the conjectured value confirms the
+conjecture; it does not refute it. To refute it we need the strict
+version, and there is no slack anywhere: the conjecture sits precisely
+where the method's equality case sits.
+
+Here is the shape of the argument, in one sentence:
+
+ If the operator in Part 1 grows by a factor 2^beta with beta >= 1/4,
+ then averaging over maps produces a probability law (Part 2) and a
+ potential whose critical points (Part 4) can be paired (Part 5), and
+ the pairing forces a certain average signed Jacobian to be <= 0,
+ while the moment identities force it to be > 0 unless beta = 1/4
+ exactly; and the case beta = 1/4 forces an identically vanishing
+ Jacobian, which a harmonic barrier (Part 6) forbids.
+
+The pairing in Part 4 matches each saddle of the potential to a peak of
+strictly larger value. "Strictly larger" is the crucial word and it is
+also the fragile one: the strictness may be arbitrarily small, one
+pair to the next. So no step of the argument is allowed to use a
+uniform positive gain. It must be handled by an equality analysis
+instead, which is what Parts 5 and 6 do.
+
+
+0.7 ROADMAP
+
+ Part 1 Transfers the disk problem to the compact family K of
+ half-plane maps. Builds one positive operator L whose
+ exponential growth rate 2^beta controls ALL the disk means
+ of THM 0.5. This is the only place where f is used; after
+ this part, the question is purely about L.
+
+ Part 2 Assumes beta >= 1/4 and builds a probability law on K whose
+ weighted rebasings scale exactly as y^{-beta}. Differentia-
+ ting that identity twice pins down two moments of the
+ logarithmic derivative.
+
+ Part 3 Introduces a potential V built from F and a target point
+ xi, and proves that its positive superlevels are compact for
+ almost every map and almost every target. Compactness is
+ what makes critical points a finite, countable business.
+
+ Part 4 Identifies the critical points of V with the solutions of a
+ smooth equation G_F(z) = xi, computes the Hessian in terms of
+ a signed Jacobian J_F, proves the index count (peaks at least
+ as many as saddles above every level), and builds a
+ measurable rank pairing.
+
+ Part 5 Changes variables through G_F and compares the paired
+ weights on a large box against a slightly larger box. The
+ boxes' areas scale the same way, so the comparison survives
+ and forces E J_F(i) <= 0, with equality only when J_F(i)
+ vanishes identically.
+
+ Part 6 Shows beta = 1/4 would force J_F = 0 for almost every F, and
+ that no member of K has J_F = 0. Contradiction. So
+ beta < 1/4, and Part 1 converts that into the theorem.
+
+ Part 7 Honest accounting.
+
+
+================================================================
+PART 1. A COMPACT MODEL, AND ONE OPERATOR
+================================================================
+
+The disk class S is not compact: a sequence of univalent maps can
+degenerate. The half-plane class K defined below is compact, and it
+carries the same information. Working on K lets us take limits, which
+is the single most useful move in the whole proof.
+
+
+1.1 FROM THE DISK TO THE UPPER HALF-PLANE
+
+Let H = { x + iy : y > 0 } be the upper half-plane, and let
+
+ m(z) = (z - i)/(z + i).
+
+This is a bijection of D onto H. Check the four facts we need:
+
+ m(i) = 0, m(0) = -1,
+ m^{-1}(w) = i(1 + w)/(1 - w).
+
+Define
+
+ K = { F : H -> C univalent, F(i) = 0, F'(i) = 1 }.
+
+FACT 1.1. The rule F = 2i f o m gives a bijection S -> K.
+
+Proof. If f is in S then F = 2i f o m is holomorphic and injective on
+H, and
+
+ F(i) = 2i f(m(i)) = 2i f(0) = 0.
+
+For the derivative we need m'(z). Since m(z) = (z-i)/(z+i),
+
+ m'(z) = [ (z+i) - (z-i) ] / (z+i)^2 = 2i/(z+i)^2.
+
+At z = i: m'(i) = 2i/(2i)^2 = 2i/(-4) = -i/2. Hence
+
+ F'(i) = 2i * f'(m(i)) * m'(i) = 2i * 1 * (-i/2) = 1.
+
+The map f -> F is invertible because m is. QED
+
+Notation. For F in K write
+
+ Q_F = 1/F'.
+
+So Q_F is holomorphic and nowhere zero on H, and |Q_F| is a positive
+continuous function. By the half-plane version of FACT 0.1, F' has no
+zeros.
+
+WHY K IS THE RIGHT HOME. The normalization F(i)=0, F'(i)=1 pins the
+two degrees of freedom exactly as f(0)=0, f'(0)=1 did. What we gain is
+compactness:
+
+LEMMA 1.2 (the classical estimates). For every f in S and |z| = r < 1,
+
+ (1-r)/(1+r)^3 <= |f'(z)| <= (1+r)/(1-r)^3,
+
+and the disk B(0,1/4) is contained in f(D). More generally, if F is
+univalent on the Euclidean disk B(z0,s), then
+
+ B( F(z0), s|F'(z0)|/4 ) is contained in F( B(z0,s) ).
+
+Proof. The first two assertions are Koebe's quarter theorem together
+with the distortion theorem, quoted from Duren, Univalent Functions,
+Chapter 2. For the third, apply the first assertion to the map
+
+ zeta |-> ( F(z0 + s zeta) - F(z0) ) / ( s F'(z0) ),
+
+which is univalent on D with derivative 1 at the origin. QED
+
+LEMMA 1.3 (the compact family). With the topology of locally uniform
+convergence, K is compact and metrizable. Evaluation of F, of Q_F, and
+of all their derivatives is jointly continuous in (F,z), and is
+uniformly bounded on each compact set of z. Moreover, uniformly in
+F in K,
+
+ |Q_F(x + iy)| <= C/y for |x| <= 2, 0 < y <= 1, (1.1)
+
+ |F(z)| >= c > 0 for |m(z)| >= 1/2, (1.2)
+
+with constants C and c that do not depend on F.
+
+Proof. Integrate the derivative bound of LEMMA 1.2 on a disk around an
+arbitrary point of H. This gives local uniform boundedness of F on
+compact subsets of H, hence of F' and of all higher derivatives by
+Cauchy's estimates. Montel's theorem then gives locally uniformly
+convergent subsequences.
+
+Three things must be checked about a limit F of a subsequence. The
+normalization survives because derivatives converge too. Injectivity
+survives by Hurwitz: fix a source point z1, look at z |-> F(z) - F(z1)
+on a punctured neighborhood, and note that a locally uniform limit of
+nowhere-zero holomorphic functions is either nowhere zero or identically
+zero; the normalization rules out "identically zero". Metrizability
+comes from a countable exhaustion of H by compact sets, on each of
+which the supremum norm is a pseudometric, and countably many of those
+separate points.
+
+For the two uniform bounds, use the explicit dictionary formulas
+
+ 1 - |m(z)|^2 = 4y/|z + i|^2, |Q_F(z)| = |z+i|^2/( 4|f'(m(z))| ).
+
+The second comes from |F'| = 2|f'(m)| |m'| = 4|f'(m)|/|z+i|^2. Combine
+it with the lower bound of LEMMA 1.2 at the point m(z) of the disk,
+whose modulus is |m(z)|, and with 1 - |m(z)| = 4y/|z+i|^2:
+
+ |Q_F(z)| <= |z+i|^2 (1+|m|)^3 / ( 4 (1-|m|) )
+ = |z+i|^4 (1+|m|)^3 / (16 y).
+
+On the region |x| <= 2, 0 < y <= 1 the numerator is bounded, which is
+(1.1). For (1.2): f(D) contains B(0,1/4), and in fact the image of the
+disk of radius 1/2 under f contains B(0,1/8) by the third assertion of
+LEMMA 1.2. If |m(z)| >= 1/2 then m(z) is outside that disk, so f(m(z))
+is outside its image, which contains B(0,1/8); hence |f(m(z))| >= 1/8
+and |F(z)| = 2|f(m(z))| >= 1/4. QED
+
+
+1.2 THE AFFINE REBASING GROUP
+
+Here is the bookkeeping device that makes the proof work. A point
+z = x + iy of H is identified with the affine map
+
+ w |-> x + y w.
+
+This map sends H to H: it multiplies heights by y > 0 and then
+translates. Write z*w for the image of w under this map. In
+coordinates, for z = x+iy and w = u+iv,
+
+ z*w = (x + y u) + i (y v). (1.3)
+
+Define also
+
+ z# = (-x + i)/y.
+
+Three elementary facts about this operation are used constantly.
+Verify each by substituting (1.3):
+
+ (G1) The operation is associative and unitary: the identity element
+ is i, since i*w = w, and z# is the two-sided inverse of z.
+
+ (G2) Heights multiply: Im(z*w) = (Im z)(Im w).
+
+ (G3) Elements of height 1 form a group isomorphic to the real line
+ under addition: (x+i)*(u+i) = (x+u)+i. Elements of the form
+ iy form a group isomorphic to (0,infinity) under multiplica-
+ tion: (iy)*(iw) = i(yw).
+
+Now the rebasing. For z in H and F in K define
+
+ (T_z F)(w) = ( F(z*w) - F(z) ) / ( (Im z) * F'(z) ).
+
+Put in words: T_z F is the map F as seen from the point z, rescaled so
+that z becomes the new origin and the derivative there becomes 1. This
+is the conformal analogue of zooming in at z and renormalizing units.
+
+LEMMA 1.4 (rebasing does what it should). For z, w in H and F in K,
+
+ (a) T_z F is in K;
+ (b) T_w (T_z F) = T_{z*w} F;
+ (c) Q_{T_z F}(w) = Q_F(z*w) / Q_F(z).
+
+Proof. (a) T_z F is holomorphic and injective, being an affine change
+of variables followed by F. Its value at i: since z*i = x + yi = z,
+
+ (T_z F)(i) = ( F(z) - F(z) ) / ( (Im z) F'(z) ) = 0.
+
+Its derivative: the chain rule gives
+
+ (T_z F)'(w) = F'(z*w) * (Im z) / ( (Im z) F'(z) ) = F'(z*w)/F'(z).
+
+At w = i this is F'(z)/F'(z) = 1. So T_z F is in K.
+
+(b) By part (a), T_w(T_z F) makes sense. Both sides are holomorphic in
+the variable zeta, so it is enough to compute the defining quotient.
+For the left side, using the definition twice and (G1):
+
+ (T_w(T_z F))(zeta)
+ = [ (T_z F)(w*zeta) - (T_z F)(w) ] / [ (Im w) (T_z F)'(w) ],
+
+and now substitute
+
+ (T_z F)(w*zeta) - (T_z F)(w)
+ = [ F(z*(w*zeta)) - F(z) ]/((Im z)F'(z))
+ - [ F(z*w) - F(z) ]/((Im z)F'(z))
+ = [ F((z*w)*zeta) - F(z*w) ] / ((Im z) F'(z)),
+
+where we used associativity. Also, from the derivative computation in
+part (a) and (G2),
+
+ (Im w)(T_z F)'(w) = (Im w) F'(z*w)/F'(z)
+ = (Im(z*w)) F'(z*w) / ( (Im z) F'(z) ).
+
+Dividing the first display by the second gives exactly the definition
+of (T_{z*w} F)(zeta). This is (b).
+
+(c) This is immediate from the derivative formula in part (a):
+
+ Q_{T_z F}(w) = 1/(T_z F)'(w) = F'(z)/F'(z*w) = Q_F(z*w)/Q_F(z).
+
+QED
+
+Fact (b) is the reason for the strange-looking rescaling in the
+definition of T_z. It says that rebasing at z and then at w is the
+same as rebasing once at z*w. That is exactly what we need in order to
+iterate.
+
+
+1.3 ONE OPERATOR THAT MEASURES GROWTH
+
+For a continuous function Phi on K and a point z of H define a new
+continuous function A_z Phi by
+
+ (A_z Phi)(F) = |Q_F(z)| * Phi(T_z F).
+
+This is an average-free "read the map at z, rebase, then test" rule.
+It is positive: if Phi >= 0 then A_z Phi >= 0. It is also linear, and
+it is bounded on the supremum norm by the supremum of |Q_F(z)|, which
+is finite by LEMMA 1.3.
+
+LEMMA 1.5 (the rebasing rule composes). For z, w in H,
+
+ A_z A_w = A_{z*w}.
+
+Proof. Evaluate the left side at F and apply the definitions twice:
+
+ (A_z A_w Phi)(F)
+ = |Q_F(z)| * (A_w Phi)(T_z F)
+ = |Q_F(z)| * |Q_{T_z F}(w)| * Phi( T_w (T_z F) ).
+
+Now use LEMMA 1.4(c) and LEMMA 1.4(b):
+
+ |Q_F(z)| |Q_{T_z F}(w)| = |Q_F(z)| |Q_F(z*w)/Q_F(z)| = |Q_F(z*w)|,
+ T_w (T_z F) = T_{z*w} F.
+
+Substituting gives (A_{z*w} Phi)(F). QED
+
+Now average the two rebasing points of height 1/2:
+
+ L = (1/2) ( A_{-1/2 + i/2} + A_{1/2 + i/2} ).
+
+This single operator L is the engine of the whole proof. Its powers
+turn out to have a transparent form.
+
+LEMMA 1.6 (the dyadic form of L^n). Let v = 2^{-n} and let
+
+ u_j = -1 + (2j + 1) v, j = 0, 1, ..., 2^n - 1.
+
+These are the 2^n numbers of the form SUM_{m=1}^{n} s_m 2^{-m} with
+s_m = plus or minus 1. Then
+
+ L^n = v * SUM_{j=0}^{2^n - 1} A_{u_j + i v}. (1.4)
+
+Proof. Induction on n. The case n = 1 is the definition of L:
+v = 1/2 and the u_j are -1/2 and +1/2.
+
+Suppose (1.4) holds at level n. Then, using LEMMA 1.5 and (1.3),
+
+ L^{n+1} = L^n L
+ = v * SUM_j A_{u_j + iv} * (1/2)( A_{-1/2+i/2} + A_{1/2+i/2} )
+ = (v/2) * SUM_j ( A_{u_j + v/2 + iv} + A_{u_j - v/2 + iv} ).
+
+Put v' = v/2 = 2^{-(n+1)}. The two new abscissae are
+
+ u_j + v/2 = -1 + (4j + 3) v' = -1 + ( 2(2j+1) + 1 ) v' = u'_{2j+1},
+ u_j - v/2 = -1 + (4j + 1) v' = -1 + ( 2(2j) + 1 ) v' = u'_{2j},
+
+where u'_k = -1 + (2k+1)v'. As j runs from 0 to 2^n - 1 the index k
+runs through 0 to 2^{n+1} - 1 exactly once, and v/2 = v'. Hence
+
+ L^{n+1} = v' * SUM_{k=0}^{2^{n+1}-1} A_{u'_k + i v'},
+
+which is (1.4) at level n+1. QED
+
+So L^n is a finite Riemann sum: it samples |Q_F| on 2^n equally
+spaced points of height 2^{-n} across the interval [-1,1], and
+multiplies by the spacing. The spacing is precisely the width, so
+L^n is a discretized integral of |Q_F| along a horizontal line.
+
+WHY POSITIVITY GIVES THE NORM FOR FREE. Let 1 denote the constant
+function on K equal to 1. If Phi is any continuous function with
+0 <= Phi <= 1, then, because every A_z is positive,
+
+ 0 <= L^n Phi <= L^n 1.
+
+Taking suprema over F in K and over such Phi gives
+
+ ||L^n|| = max over F in K of (L^n 1)(F). (1.5)
+
+So the operator norm is read off from one single test function.
+
+LEMMA 1.7 (the size of L^n). There is a constant C, independent of n,
+such that
+
+ 1 <= ||L^n|| <= C * 2^n for all n >= 0.
+
+Proof. Upper bound. By (1.4),
+
+ (L^n 1)(F) = v * SUM_j |Q_F(u_j + i v)|.
+
+Every abscissa satisfies |u_j| <= 1 <= 2, and 0 < v <= 1, so (1.1)
+applies with height v and gives |Q_F(u_j+iv)| <= C/v for a constant C
+valid for all F in K. There are 2^n summands, so
+
+ (L^n 1)(F) <= v * 2^n * (C/v) = C 2^n.
+
+Together with (1.5) this is the upper bound.
+
+Lower bound. Take the single map F(z) = z - i. It is in K: it is
+injective, F(i) = 0, F'(i) = 1. For this map F' = 1, so Q_F = 1 and
+|Q_F| = 1 everywhere. Then
+
+ (L^n 1)(F) = v * SUM_j 1 = v * 2^n = 1.
+
+By (1.5), ||L^n|| is at least the value of L^n 1 at this one map. QED
+
+Now define the growth rate of the operator:
+
+ rho = lim over n -> infinity of ||L^n||^{1/n},
+ beta = log_2 rho. (1.6)
+
+WELL DEFINEDNESS. Put a_n = ||L^n||. Since L^{n+m} = L^n L^m,
+
+ a_{n+m} = ||L^n L^m|| <= ||L^n|| ||L^m|| = a_n a_m,
+
+so the sequence a_n is submultiplicative. A standard argument shows
+that a_n^{1/n} then converges: for any fixed N, write n = qN + r with
+0 <= r < N, so a_n <= a_N^q a_r and
+
+ limsup a_n^{1/n} <= a_N^{1/N}.
+
+Take the infimum over N; the limsup is at most the limit inferior. By
+LEMMA 1.7, 1 <= a_n <= C2^n, so 1 <= rho <= 2, that is,
+
+ 0 <= beta <= 1. (1.7)
+
+Two comments on what beta means. First, ||L^n|| grows exponentially
+at rate 2^beta. Second, and this is the point of the whole proof, beta
+is a number attached to the DISK CLASS S, not to any single map: the
+operator L was built from the fixed rebasing points plus or minus 1/2
+plus i/2, and it does not depend on f. So a bound on beta is a bound
+valid simultaneously for every f in S.
+
+We now prove the implication "beta bounds the disk means": that is what
+makes beta worth computing.
+
+
+1.4 FROM THE DYADIC SAMPLES BACK TO THE DISK
+
+PROPOSITION 1.8 (dyadic samples bound the disk means). There is a
+universal constant C_0 such that for every f in S, every 1/2 <= r < 1,
+and every n with
+
+ v = 2^{-n} <= 1 - r < 2v, (1.8)
+
+we have
+
+ M_{-1}[f'](r) <= C_0 ||L^n||.
+
+Consequently B_b(-1) <= beta.
+
+The proof is a change of variables from the disk circle to a horizontal
+line in H, followed by the cell-by-cell comparison of section 1.2. We
+do it in four steps. Throughout fix f in S and put F = 2i f o m.
+
+STEP 1: THE CIRCLE BECOMES A LINE. Let theta lie in the left arc
+3pi/4 <= theta <= 5pi/4 and write
+
+ m^{-1}(r e^{i theta}) = x + i y.
+
+Using m^{-1}(w) = i(1+w)/(1-w) and then rationalizing the denominator,
+one gets the explicit formulas
+
+ x = -2 r sin(theta)/d,
+ y = (1 - r^2)/d,
+ d = 1 - 2 r cos(theta) + r^2. (1.9)
+
+Three estimates, each of which is a one-line computation from (1.9):
+
+ (E1) |x| <= 1.
+ Indeed d - 2r = (1-r)^2 + 2r(1 - cos(theta)) >= 0, so
+ d >= 2r, and |x| = 2r|sin(theta)|/d <= 2r/d <= 1.
+
+ (E2) 1/2 <= y/v <= 4.
+ On this arc cos(theta) <= 0, so d = 1 - 2r cos(theta) + r^2
+ is at least 1 + r^2. Also d <= (1+r)^2. Now
+ y/v = ((1-r)/v) * (1+r)/d, and (1-r)/v lies in [1,2) by
+ (1.8). Lower end: (1+r)/d >= 1/(1+r) >= 1/2. Upper end:
+ 2(1+r)/d <= 2(1+r)/(1+r^2) <= 4, since 2r^2 - r + 1 > 0.
+
+ (E3) dx/dtheta >= 1/16.
+ Differentiating (1.9) gives
+ dx/dtheta = ( 4r^2 - 2r(1+r^2)cos(theta) ) / d^2.
+ On this arc cos(theta) <= -2^{-1/2}, so the numerator is at
+ least 4r^2 + 2^{1/2} r(1+r^2), which is at least 2^{1/2} r
+ for r >= 1/2. Meanwhile d <= (1+r)^2 <= 4, so d^2 <= 16.
+ Hence dx/dtheta >= 2^{1/2} r / 16, and for r >= 1/2 this is
+ at least 2^{-1/2}/16, comfortably above 1/16. (We keep the
+ round number 1/16.)
+
+STEP 2: CELL BY CELL. Let v = 2^{-n} and let the grid points u_j be as
+in LEMMA 1.6. They are spaced 2v apart and run from -1+v to 1-v, so
+every x with |x| <= 1 is within distance v of at least one u_j. Pick
+such a j for the given theta. Then
+
+ (x + i y) = (u_j + i v) * w, w = (x-u_j)/v + i (y/v), (1.10)
+
+because by (1.3) the product on the right has coordinates u_j + v(x-u_j)/v
+= x and height v*(y/v) = y. By (E1) and (E2), together with y/v >= 1/2,
+
+ w lies in the fixed compact set [-1,1] + i [1/2, 4]. (1.11)
+
+Apply LEMMA 1.3, which gives a uniform bound for Q_G on any fixed
+compact set of the source, valid for every G in K. With G = T_{u_j+iv}F
+and the point w of (1.11) in that compact set, there is a constant C_1
+with |Q_G(w)| <= C_1. Now LEMMA 1.4(c) together with (1.10) gives the
+factorization
+
+ Q_F(x + iy) = Q_F( (u_j+iv) * w ) = Q_F(u_j + iv) * Q_{T_{u_j+iv}F}(w).
+
+Taking absolute values and using the bound,
+
+ |Q_F(x + iy)| <= C_1 |Q_F(u_j + i v)|. (1.12)
+
+STEP 3: BACK TO f. The dictionary between f and F is F = 2i f o m,
+whence |F'| = 4|f'(m)| / |z+i|^2 at z = m^{-1}(w). Therefore
+
+ 1/|f'(m^{-1}(rei theta))| = 4 |Q_F(x+iy)| / |x + i(y+1)|^2.
+
+The denominator is x^2 + (y+1)^2 >= (y+1)^2 >= 1, so
+
+ 1/|f'(r e^{i theta})| <= 4 |Q_F(x + iy)|. (1.13)
+
+STEP 4: INTEGRATE OVER THE ARC. Let A denote the arc
+3pi/4 <= theta <= 5pi/4. Combine (1.12), (1.13) and (E3):
+
+ INT_A |f'(r e^{i theta})|^{-1} dtheta
+ <= 4 C_1 INT_A |Q_F(x(theta) + i y(theta))| dtheta
+ <= 4 C_1 * 16 INT_A |Q_F(x(theta) + i y(theta))| (dx/dtheta) dtheta
+ = 64 C_1 INT_{x: |x|<=1} |Q_F(x + i y(x))| dx.
+
+Now split the x-interval into the 2^n cells of length 2v centered at
+the u_j. On the cell centered at u_j the value of |Q_F| is at most
+C_1 |Q_F(u_j + iv)| by (1.12), uniformly in y. Hence
+
+ INT_A |f'|^{-1} dtheta <= 64 C_1 * SUM_j (2v) * C_1 |Q_F(u_j+iv)|
+ = 128 C_1^2 * v * SUM_j |Q_F(u_j + iv)|
+ = 128 C_1^2 * (L^n 1)(F)
+ <= 128 C_1^2 ||L^n||,
+
+where the last two steps are (1.4) and (1.5).
+
+THE THREE OTHER ARCS. For real a define f_a(zeta) = e^{-ia} f(e^{ia}
+zeta). Then f_a is again in S, and the change of variable
+theta |-> theta - a in the integral shows
+
+ M_{-1}[f_a'](r) = M_{-1}[f'](r).
+
+Applying the arc estimate to f_a with a = pi/2, pi, 3pi/2 covers the
+three remaining quarter arcs of the circle. Summing the four pieces
+and dividing by 2pi gives
+
+ M_{-1}[f'](r) <= C_0 ||L^n||
+
+with C_0 = 4 * 128 C_1^2 / (2pi), a universal constant. This is the
+first assertion.
+
+THE CONSEQUENCE B_b(-1) <= beta. Fix f in S and r in [1/2,1), and let
+n be the integer determined by (1.8). Write m = log(1/(1-r)). Since
+2^{-n} <= 1-r < 2^{1-n},
+
+ n log 2 <= m < (n+1) log 2, so n/m -> 1/log 2. (1.14)
+
+Take 1/n-th powers in PROPOSITION 1.8:
+
+ M_{-1}[f'](r)^{1/n} <= C_0^{1/n} ||L^n||^{1/n} -> 2^beta,
+
+because C_0^{1/n} -> 1 and ||L^n||^{1/n} -> rho = 2^beta by (1.6).
+Now take logarithms and divide by m:
+
+ log M_{-1}[f'](r) / m
+ = (n/m) * ( 1/n ) * log M_{-1}[f'](r)
+ <= (n/m) * ( log C_0/n + log ||L^n||/n ).
+
+By (1.14) the first factor tends to 1/log 2, and by the line above the
+second factor tends to log 2^beta = beta log 2. Hence
+
+ beta_f(-1) = limsup log M_{-1}[f'](r) / m <= beta.
+
+The right side does not depend on f, so taking the supremum over f in
+S_b gives B_b(-1) <= beta. QED (PROPOSITION 1.8)
+
+END OF THE DISK. From here on, f never appears again. Everything is
+about the compact family K, the operator L, and the one number beta.
+The rest of the document proves:
+
+ IF beta >= 1/4 THEN A CONTRADICTION FOLLOWS.
+
+That is the plan. So from the start of Part 2, assume for the sake of
+argument that beta >= 1/4, and derive nonsense. Once that is done,
+beta < 1/4, and PROPOSITION 1.8 turns beta < 1/4 into THM 0.5 by
+choosing any exponent b with beta < b < 1/4.
+
+
+================================================================
+PART 2. A WEIGHTED LAW OF MAPS
+================================================================
+
+Assume from now on, for the sake of contradiction, that
+
+ beta >= 1/4. (2.1)
+
+The whole of Parts 2 through 6 runs under this assumption and ends in
+a contradiction. Recall rho = 2^beta, so under (2.1) we have
+
+ rho >= 2^{1/4} > 1. (2.2)
+
+The strict inequality rho > 1 in (2.2) is used repeatedly: it is what
+lets us send rho^{-n} to zero.
+
+The plan of this part. The operator L has norm growth rate rho. A
+positive operator with growth rate rho has an "eigenmeasure" at rho,
+by a compactness argument. That eigenmeasure then, by two averaging
+steps, produces a PROBABILITY LAW P on K with an exact scaling
+property: sampling at a rebasing point of height y and then looking at
+the rebased map rescales P by exactly y^{-beta}. Differentiating that
+scaling identity twice at y = 1 produces two moment identities for the
+logarithmic derivative. Those two identities are the only quantitative
+input used later.
+
+
+2.1 THE EIGENMEASURE OF L
+
+Let C(K) be the space of continuous real functions on the compact space
+K, with the supremum norm. A bounded linear operator T on C(K) is
+POSITIVE if T Phi >= 0 whenever Phi >= 0. Every A_z is positive, so L
+and all its powers are positive.
+
+A POSITIVE FUNCTIONAL of norm 1 is a probability measure. The Riesz
+representation theorem (Folland, Real Analysis, Theorem 7.2) says
+exactly this: every positive linear functional on C(K) of norm 1 is
+integration against a unique Borel probability measure. We use it to
+pass freely between "positive functional" and "probability measure".
+
+CONVENTION. A measure mu acts on the LEFT as a functional:
+mu Phi = INT Phi dmu. So the composition mu L is the functional
+Phi |-> mu(L Phi).
+
+LEMMA 2.1 (an eigenmeasure at the exact growth rate). There is a Borel
+probability mu on K with
+
+ mu L = rho mu. (2.3)
+
+Proof.
+
+STEP 1: A SUB-EIGENVECTOR. For a real number tau > rho, the series
+
+ R_tau = SUM_{n>=0} tau^{-n} L^n
+
+converges in operator norm, because ||L^n|| <= C 2^n by LEMMA 1.7 and
+tau > rho = lim ||L^n||^{1/n} > 1. Put
+
+ g_tau = R_tau 1, M_tau = max over K of g_tau.
+
+The maximum is attained and is finite, since g_tau is continuous on the
+compact space K. Also g_tau >= 1, because the n = 0 term of the series
+is 1 and all other terms are nonnegative.
+
+Compute L applied to the series, shifting the index:
+
+ L g_tau = SUM_{n>=0} tau^{-n} L^{n+1} 1
+ = tau SUM_{n>=1} tau^{-n} L^n 1
+ = tau ( g_tau - 1 ). (2.4)
+
+STEP 2: A NUMERICAL BOUND ON M_tau. Since 1 <= g_tau <= M_tau
+pointwise, (2.4) gives
+
+ L g_tau = tau (g_tau - 1) <= tau (1 - 1/M_tau) g_tau.
+
+That is a pointwise inequality of continuous functions. Now use
+positivity and induction. For any c >= 0, if 0 <= Phi <= c g_tau then
+0 <= L Phi <= c L g_tau <= c tau(1-1/M_tau) g_tau. Since 1 <= 1*g_tau,
+induction gives
+
+ L^n 1 <= [ tau(1 - 1/M_tau) ]^n g_tau <= M_tau [ tau(1-1/M_tau) ]^n.
+
+Take the n-th root, let n tend to infinity, and use the definition of
+rho:
+
+ rho <= tau (1 - 1/M_tau).
+
+Rearrange: 1/M_tau <= (tau - rho)/tau, that is,
+
+ M_tau >= tau/(tau - rho). (2.5)
+
+This is the only place where the exact value of rho, rather than an
+upper bound, is used. It is why we built a law at the parameter beta
+itself and not at some larger parameter: the lower bound (2.5) blows
+up as tau decreases to rho, and that blow-up is what kills the error
+term in STEP 3.
+
+STEP 3: PASS TO THE LIMIT. Let F_tau be a point of K where g_tau
+attains its maximum M_tau, and define the positive functional
+
+ mu_tau = (1/M_tau) * delta_{F_tau} R_tau,
+
+where delta_{F_tau} is evaluation at F_tau. It is a probability
+because mu_tau(1) = (R_tau 1)(F_tau)/M_tau = g_tau(F_tau)/M_tau = 1.
+
+Apply mu_tau to L, using (2.4) at the point F_tau:
+
+ mu_tau L = (1/M_tau) delta_{F_tau} R_tau L
+ = (1/M_tau) delta_{F_tau} ( tau R_tau - tau )
+ = tau mu_tau - (tau/M_tau) delta_{F_tau}. (2.6)
+
+Here we used R_tau L = SUM tau^{-n} L^{n+1} = tau SUM_{n>=1} tau^{-n}L^n
+= tau (R_tau - 1), which is (2.4) written as an operator identity.
+
+Now let tau decrease to rho. Each mu_tau is a Borel probability on the
+compact metrizable space K, so by Prokhorov's theorem (equivalently,
+Banach-Alaoglu applied to C(K)*) there is a subsequence converging in
+the weak-* topology to some Borel probability mu. Pass to the limit in
+(2.6). The left side mu_tau L tends to mu L: this is legitimate because
+L maps C(K) to C(K) and so is weak-* continuous. On the right side,
+tau -> rho and tau/M_tau -> 0 by (2.5). Hence mu L = rho mu. QED
+
+(2.3) read out loud: "integrating L Phi against mu is the same as
+multiplying the integral of Phi by rho". Equivalently, writing
+mu L^n = rho^n mu for every n >= 0,
+
+ mu ( L^n Phi ) = rho^n mu(Phi) for all Phi in C(K), n >= 0.
+ (2.7)
+
+
+2.2 TWO AVERAGING STEPS
+
+The eigenmeasure mu still knows about the operator L, which remembers
+the two special points plus or minus 1/2 plus i/2. Two averaging
+steps remove that memory and leave an exact scaling law.
+
+STEP A: TRANSLATION INVARIANCE AT HEIGHT ONE.
+
+LEMMA 2.2. mu A_{2+i} = mu.
+
+Proof. Fix n and use (2.7) followed by the definition of L^n:
+
+ mu A_{2+i} - mu = rho^{-n} ( mu L^n A_{2+i} - mu L^n ). (2.8)
+
+Now compute the difference of operators. By LEMMA 1.6,
+
+ L^n A_{2+i} = v SUM_j A_{u_j + iv} A_{2+i}.
+
+By LEMMA 1.5 the product is A_{(u_j+iv)*(2+i)}, and by (1.3),
+
+ (u_j + iv) * (2 + i) = (u_j + 2v) + i v.
+
+Since u_j = -1 + (2j+1)v, we get u_j + 2v = -1 + (2j+3)v = u_{j+1} for
+j = 0, ..., 2^n - 2. So the sum telescopes:
+
+ L^n A_{2+i} = v SUM_{j=0}^{2^n-1} A_{u_{j+1} + iv}
+ = v ( A_{u_{2^n} + iv} + SUM_{j=1}^{2^n-1} A_{u_j+iv} )
+ = v SUM_{j=0}^{2^n-1} A_{u_j+iv}
+ + v ( A_{u_{2^n}+iv} - A_{u_0+iv} )
+ = L^n + v ( A_{1+v+iv} - A_{-1+v+iv} ),
+
+where we used u_{2^n} = -1 + (2^{n+1}+1)v = 1 + v and u_0 = -1 + v.
+
+Therefore, by (1.1), for every continuous Phi with |Phi| <= 1,
+
+ | ( L^n A_{2+i} - L^n ) Phi | <= v * 2 * (C/v) = 2C,
+
+where C is the constant of (1.1). This bound does not depend on n. Now
+substitute into (2.8): the difference of the two functionals applied to
+Phi is at most
+
+ rho^{-n} * 2C.
+
+By (2.2), rho > 1, so this tends to zero as n tends to infinity. The
+left side of (2.8) does not depend on n; it must therefore be zero. QED
+
+LEMMA 2.3 (period two). For every real t and every h > 0,
+
+ mu A_{t+2+ih} = mu A_{t+ih}.
+
+Proof. Apply LEMMA 2.2 and then LEMMA 1.5:
+
+ mu A_{t+2+ih} = ( mu A_{2+i} ) A_{t+ih} = mu A_{t+ih}.
+
+(Associativity of composition of operators is automatic; and by (1.3),
+(2+i)*(t+ih) = (2+t) + ih, which is the same statement.) QED
+
+REMARK. Lemma 2.3 gives PERIOD TWO, not constancy, and the difference
+matters. The elements of height one form a group isomorphic to the real
+line, and LEMMA 2.2 says mu is invariant under the SUBGROUP generated by
+2, namely 2Z. The closure of 2Z is 2Z. Invariance under 2Z is period
+two; it is not invariance under all translations. Averaging over one
+period, in the next step, upgrades period two to exact translation
+invariance, and that is why we average.
+
+STEP B: AVERAGE ONE PERIOD. Define the positive functional
+
+ bar mu = (1/2) INT_0^2 mu A_{x+i} dx.
+
+The integral is a weak-* integral of continuous functionals over a
+compact interval of the parameter; since x |-> mu A_{x+i} is weak-*
+continuous (see the joint continuity in LEMMA 1.3), the integral exists
+as a positive functional. Its mass is
+
+ bar mu(1) = (1/2) INT_0^2 mu ( |Q_F(x+i)| ) dx,
+
+which is finite, positive, and bounded away from zero and infinity by
+LEMMA 1.3. So bar mu is a nonzero finite positive measure.
+
+LEMMA 2.4 (horizontal invariance). For every real s,
+
+ bar mu A_{s+i} = bar mu. (2.9)
+
+Proof. By definition and LEMMA 1.5,
+
+ bar mu A_{s+i} = (1/2) INT_0^2 mu A_{x+i} A_{s+i} dx
+ = (1/2) INT_0^2 mu A_{x+s+i} dx.
+
+By LEMMA 2.3 the integrand is 2-periodic in x; the integral of a
+2-periodic function over an interval of length exactly 2 does not depend
+on where the interval starts. Shifting the interval of integration from
+[0,2] to [s,s+2] gives back the defining integral of bar mu. QED
+
+CONSEQUENCE (horizontal invariance at every height). Every point
+x + iy of H factors as a product of a height-y point with a height-1
+point:
+
+ x + iy = (iy) * (x/y + i), since (1.3) gives (0 + iy)*(u+i) = iy u + iy.
+
+Hence, by LEMMA 1.5 and LEMMA 2.4,
+
+ bar mu A_{x+iy} = bar mu A_{iy} A_{x/y+i} = bar mu A_{iy}. (2.10)
+
+So bar mu only cares about the HEIGHT of the rebasing point, not its
+horizontal position. This is exactly what we need: the exact scaling
+will be proved for heights, and horizontal position will come free.
+
+
+2.3 THE EXACT SCALING LAW
+
+The eigenvalue relation mu L = rho mu says something about L, and L is
+built from height-1/2 points. Translating that into a statement about
+bar mu at height 1/2 costs one factor of rho.
+
+LEMMA 2.5 (the eigenvalue at height one half). rho bar mu = bar mu A_{i/2}.
+
+Proof. Start from the identity rho mu = mu L = (1/2) mu
+(A_{-1/2+i/2} + A_{1/2+i/2}), and average it over x in [0,2], weighted
+as in the definition of bar mu:
+
+ rho bar mu = (1/2) INT_0^2 rho mu A_{x+i} dx
+ = (1/4) INT_0^2 ( mu A_{-1/2+i/2} A_{x+i}
+ + mu A_{1/2+i/2} A_{x+i} ) dx.
+
+By LEMMA 1.5 and (1.3) the two products are
+
+ (-1/2+i/2)*(x+i) = (-1/2 + x/2) + i/2,
+ ( 1/2+i/2)*(x+i) = ( 1/2 + x/2) + i/2.
+
+Substitute t = x/2 - 1/2 in the first integral and t = x/2 + 1/2 in the
+second. As x runs over [0,2], t runs over [-1/2,1/2] in the first and
+over [1/2,3/2] in the second; in both cases dt = dx/2. The two pieces
+therefore join into one integral over an interval of length 2:
+
+ rho bar mu = (1/2) INT_{-1/2}^{3/2} mu A_{t+i/2} dt.
+
+By the period-two property of LEMMA 2.3, applied at height 1/2, this
+integral equals the integral over any interval of length 2, in
+particular over [0,2]:
+
+ rho bar mu = (1/2) INT_0^2 mu A_{t+i/2} dt = bar mu A_{i/2}.
+
+For the last equality: bar mu A_{i/2} = (1/2) INT_0^2 mu A_{x+i}A_{i/2}dx
+and (x+i)*(i/2) = x + i/2 by (1.3). QED
+
+Now multiply LEMMA 2.5 on the right by A_{ih}, for a fixed h > 0, and
+use LEMMA 1.5 again: (i/2)*(ih) = i(h/2) and (ih) on the right simply
+replaces h by h/2 inside the functional. Concretely, define
+
+ g(h) = bar mu A_{ih}, a positive functional for each h > 0.
+
+Then LEMMA 2.5 says g(h/2) = rho g(h), and reading the same identity
+with h replaced by 2h says g(2h) = rho^{-1} g(h). Either way:
+
+ h^beta * g(h) is invariant under h |-> 2h. (2.11)
+
+(Indeed, replacing h by 2h multiplies h^beta by 2^beta = rho and g by
+rho^{-1}, and the two factors cancel.) By iterating, h^beta g(h) is
+invariant under h |-> 2^k h for every integer k.
+
+DEFINITION. Define the positive functional
+
+ nu = INT_1^2 h^beta bar mu A_{ih} dh/h. (2.12)
+
+Its mass is
+
+ nu(1) = INT_1^2 h^beta bar mu( |Q_F(ih)| ) dh/h,
+
+which is finite and strictly positive by LEMMA 1.3, since the integrand
+is continuous and positive on the compact interval [1,2].
+
+LEMMA 2.6 (exact scaling). For every x + iy in H,
+
+ nu A_{x+iy} = y^{-beta} nu. (2.13)
+
+Proof. Two steps.
+
+Heights only. By (2.10) the horizontal coordinate is irrelevant:
+
+ nu A_{x+iy} = INT_1^2 h^beta bar mu A_{ih} A_{x+iy} dh/h
+ = INT_1^2 h^beta bar mu A_{hx + i(hy)} dh/h
+ = INT_1^2 h^beta bar mu A_{i(hy)} dh/h.
+
+The middle line uses (ih)*(x+iy) = hx + i(hy), which comes straight
+from (1.3). The last line uses (2.10), the statement that bar mu
+depends only on the height of its rebasing point and not on the
+horizontal coordinate; the horizontal coordinate hx is simply thrown
+away.
+
+Now substitute u = hy, so h = u/y and dh/h = du/u:
+
+ nu A_{x+iy} = INT_{y}^{2y} (u/y)^beta bar mu A_{iu} du/u
+ = y^{-beta} INT_{y}^{2y} u^beta bar mu A_{iu} du/u.
+
+By (2.11) the integrand u^beta bar mu A_{iu} du/u is invariant under
+u |-> 2u, so its integral over [y,2y] is the same for every y > 0.
+Comparing with (2.12) (the case y = 1) gives (2.13). QED
+
+NORMALIZE. Divide nu by its positive mass nu(1) to obtain a Borel
+probability P on K. Inequality (2.13) survives the division because
+both sides are linear in the functional. Unwinding the definition of
+A_z and of the functional notation, (2.13) says:
+
+PROPOSITION 2.7 (the weighted affine law). Assume beta >= 1/4. There
+is a Borel probability P on K such that for every z = x+iy in H and
+every nonnegative Borel function Phi on K,
+
+ E_P [ |Q_F(x + iy)| * Phi( T_{x+iy} F ) ] = y^{-beta} E_P [ Phi(F) ].
+ (2.14)
+
+Here E_P denotes expectation under P.
+
+What (2.14) means, in words: if you pick a random map F from P, then
+rebase it at the point x+iy, the distribution you see is again P, but
+weighted by 1/|Q_F(x+iy)| and scaled by the fixed number y^{-beta}.
+This is an exact self-similarity of the law P in the vertical scale. It
+is exact, not approximate, and that exactness is the whole point: it is
+what converts into the moment identities of section 2.4.
+
+Proof of Borel-test version. For continuous Phi, (2.14) is (2.13)
+divided by nu(1). For a fixed z, both sides of (2.14) are finite
+positive linear functionals of Phi: the left side is the integral of
+Phi(T_zF) against the finite measure |Q_F(z)|*P(dF), a weighted
+pushforward, and the right side is a constant times P. Two finite Borel
+measures that agree on all continuous functions agree as measures. So
+(2.14) extends from continuous to all nonnegative Borel Phi. QED
+
+
+2.4 TWO MOMENT IDENTITIES
+
+The scaling law (2.14) is an identity between functions. Differentiate
+it in y twice and at y = 1 you get two numerical identities. Those two
+numbers are the only quantitative input to the rest of the proof.
+
+DEFINITION (logarithmic derivative, real and imaginary parts). For
+z = x + iy in H put
+
+ q_F(z) = i y * Q_F'(z)/Q_F(z) = -i y * F''(z)/F'(z) = a_F(z) + i b_F(z).
+ (2.15)
+
+Both formulas agree because Q_F = 1/F' implies Q_F'/Q_F = -F''/F'. The
+factor y is there to make q_F invariant under rebasing:
+
+ q_{T_z F}(w) = q_F(z * w). (2.16)
+
+Check (2.16) directly. From LEMMA 1.4(c),
+Q_{T_zF}(w) = Q_F(z*w)/Q_F(z). Differentiate in w. The derivative of
+z*w in w is the factor (Im z), since z*w = x + y w; so
+
+ Q_{T_zF}'(w) = Q_F'(z*w) (Im z)/Q_F(z).
+
+Divide by Q_{T_zF}(w) and multiply by i Im w:
+
+ q_{T_zF}(w) = i (Im w) * Q_F'(z*w)(Im z)/Q_F(z*w)
+ = i (Im(z*w)) * Q_F'(z*w)/Q_F(z*w) by (G2)
+ = q_F(z*w). QED
+
+DEFINITION (the exponent k). Let
+
+ k = (beta + 2)/3. (2.17)
+
+Under the standing assumption (2.1),
+
+ 3/4 <= k <= 1, and 3(k-1) = beta - 1. (2.18)
+
+The upper bound k <= 1 uses beta <= 1 from (1.7); the lower bound uses
+beta >= 1/4 from (2.1). The identity 3(k-1) = beta-1 is just algebra:
+3((beta+2)/3 - 1) = beta + 2 - 3 = beta - 1. Both halves of (2.18) are
+used later, at (4.7) and in Part 6.
+
+LEMMA 2.8 (moment identities). Under the law P of PROPOSITION 2.7,
+
+ E a_F(i) = -beta, E |q_F(i)|^2 = beta(beta + 1). (2.19)
+
+Proof.
+
+STEP 1: THE CONSTANT TEST. Take Phi = 1 in (2.14). Then Phi(T_zF) = 1
+for every F, so
+
+ E |Q_F(x + iy)| = y^{-beta} for every x + iy in H. (2.20)
+
+STEP 2: DIFFERENTIATING |Q_F|. Fix F. Since Q_F is holomorphic and
+nowhere zero, u = log|Q_F| is harmonic. For a holomorphic g with
+g = U + iV, the Cauchy-Riemann equations give partial_y U = -partial_x V
+and g' = partial_x U + i partial_x V. Hence
+
+ partial_y Re g = -Im(g').
+
+Apply this with g = log Q_F, so g' = Q_F'/Q_F:
+
+ partial_y log|Q_F| = -Im(Q_F'/Q_F).
+
+Now express the right side in terms of q_F. From (2.15),
+Q_F'/Q_F = -i q_F/y, so
+
+ -Im(Q_F'/Q_F) = -Im(-i q_F/y) = Re(q_F)/y = a_F/y.
+
+Therefore
+
+ partial_y |Q_F| = |Q_F| * partial_y log|Q_F| = |Q_F| a_F / y. (2.21)
+
+Second, because u is harmonic and |Q_F| = exp(u),
+
+ Delta |Q_F| = |Q_F| * |grad u|^2 = |Q_F| * |Q_F'/Q_F|^2
+ = |Q_F| * |q_F|^2 / y^2, (2.22)
+
+where the last step uses |Q_F'/Q_F| = |q_F|/y from (2.15). (Using that
+for a holomorphic u, |grad u|^2 = |u'|^2.)
+
+STEP 3: DIFFERENTIATE (2.20) UNDER THE EXPECTATION. This is legitimate:
+by LEMMA 1.3 the functions |Q_F|, Q_F', Q_F'' are bounded uniformly in F
+on a fixed compact neighborhood of the point i, so the dominated
+convergence theorem applies to a difference quotient taken in F as
+F ranges over K under P.
+
+Differentiate (2.20) once in y and evaluate at x = 0, y = 1, using
+(2.21):
+
+ E [ |Q_F(i)| a_F(i) ] = -beta * 1^{-beta-1} = -beta.
+
+Q_F(i) = 1/F'(i) = 1, so |Q_F(i)| = 1 and this reads E a_F(i) = -beta.
+
+Differentiate (2.20) twice, that is apply Delta (which acts only on the
+y variable here, since (2.20) shows the right side is x-independent),
+and evaluate at x = 0, y = 1, using (2.22):
+
+ E [ |Q_F(i)| |q_F(i)|^2 ] = beta(beta+1) * 1^{-beta-2} = beta(beta+1).
+
+Again |Q_F(i)| = 1. This is the second identity of (2.19). QED
+
+WHAT THE TWO MOMENTS SAY. The first identity says that the logarithmic
+derivative, averaged over the random map, has real part exactly -beta:
+the random maps in the law P have, on average, exactly the growth
+dimension beta and not more. The second identity says that the size of
+the logarithmic derivative is beta(beta+1). Both numbers are fixed by
+beta alone. Everything in Parts 3 to 6 is a consequence of these two
+numbers plus the compactness and the rebasing law.
+
+
+================================================================
+PART 3. A POTENTIAL, AND WHY IT CANNOT RUN OFF
+================================================================
+
+The law P of Part 2 gives us random maps. For each random map and each
+target point xi in the plane, we now build a real function on H whose
+critical points we will count in Part 4. The building blocks are the
+map F, the target xi, and one more power k.
+
+The point of this part is a compactness statement: for almost every map
+and almost every target, the sets where the function is large are
+compact. Without it, "the critical points above a level" would be an
+infinite set and the count of Part 4 would be meaningless.
+
+
+3.1 THE POTENTIAL
+
+For F in K and xi in C define
+
+ V_xi^F(x + iy) = y^k / |F(x+iy) - xi|, (3.1)
+
+with the value +infinity at the (at most one) point where F = xi. The
+numerator y^k pushes the function DOWN as we approach the real line;
+the denominator blows it UP as we approach the preimage of the target.
+
+Its logarithm is
+
+ ell(z) = log V_xi^F(z) = k log y - log|F(z) - xi|. (3.2)
+
+Two elementary facts about ell.
+
+FACT 3.1 (the Laplacian of log y).
+ Delta log y = -1/y^2.
+Indeed log y depends only on y, and the second derivative of log y in y
+is -1/y^2.
+
+FACT 3.2 (log|F - xi| is harmonic off the preimage of xi).
+Suppose F(z0) = xi. There is at most one such z0, since F is injective.
+Near z0 write F(z) - xi = (z - z0) h(z) with h holomorphic and nonzero
+near z0; the power of z - z0 is exactly one because F'(z0) is never zero
+(FACT 0.1). Then
+
+ log|F(z) - xi| = log|z - z0| + log|h(z)|,
+
+the sum of the Newtonian potential of a point, which is harmonic away
+from that point, and a harmonic function. So Delta log|F - xi| = 0
+wherever F(z) is not xi. Hence
+
+ Delta ell = -k/y^2 off the preimage of xi. (3.3)
+
+(3.3) says ell is STRICTLY superharmonic: its Laplacian is negative
+everywhere. A strictly superharmonic function cannot have an interior
+minimum. That single sentence is the reason critical points of ell can
+be counted at all. It is used in Part 4, to show that the level curves
+bound the superlevel sets from inside, and again in Part 6.
+
+WHY THE EXPONENT k = (beta+2)/3. The exponent in (3.1) is chosen so
+that exactly three powers of the potential match the weight in the area
+formula. Concretely, as (2.18) records, 3(k-1) = beta-1. Part 4 will
+compute that at a critical point the value of the potential is
+k y^{k-1}|Q_F(z)|, and cubing that produces k^3 y^{beta-1}|Q_F(z)|^3,
+which is precisely the weight y^{beta-1}|Q_F(z)|^3 that appears when we
+change variables through G_F in Part 5. So k is forced: it is the one
+exponent that makes the potential's critical values carry exactly the
+weight that the integration formula produces. Nothing else in the
+proof fixes k.
+
+
+3.2 THE WEIGHTED AREA TAIL
+
+The first compactness ingredient controls behaviour near source
+infinity, that is, far out in H. Write dA for planar area, and let
+
+ S_0 = { z in H : |m(z)| >= 1/2 }. (3.4)
+
+This is the part of H away from the base point, and its complement is
+bounded (it is the image under m^{-1} of a disk of radius 1/2).
+
+LEMMA 3.3 (finite weighted area tail). For P-almost every F,
+
+ I(F) = INT_{S_0} y^beta |F(z)|^{-3} dA(z) < infinity. (3.5)
+
+Proof.
+
+STEP 1: ONE IDENTITY FROM THE LAW. Fix z in H with z not equal to i,
+and apply (2.14) to the nonnegative Borel test Phi(F) = |F(z#)|^{-3},
+where z# = (-x+i)/y is the inverse of z in the affine group of 1.2.
+
+First compute T_zF at the point z#. By (G1), z*(z#) = i, and F(i) = 0:
+
+ (T_zF)(z#) = ( F(z*(z#)) - F(z) ) / ( y F'(z) )
+ = ( F(i) - F(z) ) / ( y F'(z) )
+ = -F(z)/( y F'(z) ) = -F(z) Q_F(z)/y.
+
+Take absolute values, cube, invert:
+
+ |(T_zF)(z#)|^{-3} = |F(z)|^{-3} |Q_F(z)|^{-3} y^3.
+
+Substituting Phi into (2.14) now gives
+
+ E [ |Q_F(z)| * |F(z)|^{-3} |Q_F(z)|^{-3} y^3 ]
+ = y^{-beta} E [ |F(z#)|^{-3} ],
+
+that is, since |Q_F(z)| * |Q_F(z)|^{-3} = |Q_F(z)|^{-2} = |F'(z)|^2,
+
+ E [ |F'(z)|^2 |F(z)|^{-3} ] = y^{-beta-3} E [ |F(z#)|^{-3} ]. (3.6)
+
+STEP 2: THE LEFT SIDE INTEGRATES FINITELY. This is where injectivity
+enters. Change variables in the z-integral using w = F(z): on the image
+F(H) this is a bijection, and dA(w) = |F'(z)|^2 dA(z). Hence
+
+ INT_{S_0} |F'(z)|^2 |F(z)|^{-3} dA(z)
+ = INT_{F(S_0)} |w|^{-3} dw.
+
+Now (1.2) says |F(z)| >= c for z in S_0, so F(S_0) lies outside the
+disk of radius c. And the function |w|^{-3} is integrable on the
+plane outside any fixed disk: in polar coordinates it is
+2 pi INT_c^infinity r^{-3} r dr = 2 pi /c, finite. So
+
+ INT_{S_0} |F'(z)|^2 |F(z)|^{-3} dA(z) <= 2 pi/c, (3.7)
+
+a bound that does not depend on F at all. In particular its
+expectation is finite.
+
+STEP 3: THE CHANGE OF VARIABLE z -> z#. The map z |-> z# is the
+involution (x+iy) |-> (-x+i)/y. Its Jacobian determinant has absolute
+value y^{-3}: indeed if z = x+iy maps to (X,Y) = (-x/y, 1/y) then
+
+ dX/dx = -1/y, dX/dy = x/y^2, dY/dx = 0, dY/dy = -1/y^2,
+
+and the determinant is (-1/y)(-1/y^2) = 1/y^3. The map preserves
+|m(z)|: this is the standard fact that m and the involution commute, and
+it follows from 1 - |m(z#)|^2 = 4(1/y)/|z# + i|^2 = 4y/|z+i|^2 by a
+direct computation. So it maps S_0 to itself.
+
+Also, if z = x + iy then z# has height Y = 1/y, so y = 1/Y. Now
+integrate (3.6) over S_0 and substitute w = z# (so z = w#):
+
+ INT_{S_0} y^{-beta-3} E|F(z#)|^{-3} dA(z)
+ = INT_{S_0} (1/Y)^{-beta-3} E|F(w)|^{-3} (dA(z)/dA(w)) dA(w)
+ = INT_{S_0} Y^{beta+3} E|F(w)|^{-3} Y^{-3} dA(w)
+ = INT_{S_0} Y^beta E|F(w)|^{-3} dA(w)
+ = E I(F).
+
+(The factor Y^{-3} in the third line is the absolute Jacobian of
+w |-> w#, expressed in the variable w, whose height is Y.)
+
+STEP 4: PUT THE PIECES TOGETHER. Integrate both sides of (3.6) over
+S_0. The left side is finite by (3.7). The right side equals E I(F) by
+STEP 3. So E I(F) is finite. A nonnegative random variable with finite
+expectation is finite almost surely, so I(F) < infinity for P-almost
+every F. QED
+
+
+3.3 ESCAPE AT INFINITY
+
+Now the compactness of the level sets, far from the real line.
+
+LEMMA 3.4 (decay far out). Suppose I(F) < infinity. Then
+
+ (a) y^k/|F(x+iy)| -> 0 as |x| + y -> infinity; (3.8)
+
+ (b) for every bounded target set K, the preimage F^{-1}(K) has
+ bounded height in H, and F^{-1}(K) intersected with any strip
+ y >= epsilon > 0 is compact;
+
+ (c) sup over s > 0 of INT_R 1_K(F(x + is)) dx < infinity. (3.9)
+
+Proof.
+
+(a) Fix z = x+iy. The submean inequality for the subharmonic (indeed
+harmonic) function |1/F|^3 = |F|^{-3} on the disk B(z, y/2), which lies
+in H because every point of it has height at least y/2:
+
+ |F(z)|^{-3} <= (1/area) INT_{B(z,y/2)} |F(w)|^{-3} dA(w)
+ <= (4/(pi y^2)) INT_{B(z,y/2)} |F(w)|^{-3} dA(w).
+
+Multiply both sides by y^{3k}:
+
+ ( y^k / |F(z)| )^3 <= (4 y^{3k-2}/pi) INT_{B(z,y/2)} |F|^{-3} dA.
+
+Now 3k - 2 = beta by (2.17). On the disk B(z,y/2) every point w has
+Im w >= y/2, so y^beta <= 2^beta (Im w)^beta. Hence
+
+ ( y^k / |F(z)| )^3
+ <= (4 * 2^beta / pi) INT_{B(z,y/2)} (Im w)^beta |F(w)|^{-3} dA(w).
+ (3.10)
+
+The disk B(z,y/2) contains no zero of F (the only zero of F is i, and
+the complement of S_0 is bounded; the disk is eventually inside S_0 as
+|z| grows). So the integral on the right is at most the tail of I(F)
+over a smaller set, which tends to zero as the disk escapes to infinity.
+Hence the left side tends to zero. This is (3.8).
+
+(b) If z is in F^{-1}(K) and y >= epsilon, then
+
+ y^k/|F(z)| >= epsilon^k / (max of |w| over K),
+
+which is a fixed positive number. But (3.8) says the left side tends to
+zero as |x| + y -> infinity. So unbounded |x| with y >= epsilon is
+impossible. That makes F^{-1}(K) bounded in x at every height above
+epsilon. It is closed because F is continuous and K is closed. So F^{-1}(K)
+intersected with
+{y >= epsilon} is compact. For an unbounded height, F(z) is bounded on
+K while y -> infinity gives y^k/|F(z)| -> infinity by (3.8) reversed;
+contradiction. So the full preimage of K has bounded height.
+
+(c) Choose a smooth nonnegative psi with compact support and psi >= 1_K,
+and put
+
+ H(s) = INT_R psi(F(x + is)) dx.
+
+By (b), psi composed with F is supported in a bounded region, so H(s) is
+finite and in fact zero for s large. Also H'(s) -> 0 as s -> infinity.
+Hence the standard identity
+
+ H(s) = INT_s^infinity (y - s) H''(y) dy
+
+holds (it is integration by parts twice, using that H and H' vanish at
+infinity).
+
+Now H''(y) = INT_R partial_y^2 (psi o F)(x + iy) dx. But
+
+ INT_R partial_x^2 (psi o F)(x + iy) dx = 0,
+
+because it is a total derivative in x of a compactly supported thing.
+So the second partial in y equals the full Laplacian:
+
+ H''(y) = INT_R (Delta (psi o F))(x+iy) dx
+ = INT_R |F'(x+iy)|^2 (Delta psi)(F(x+iy)) dx.
+
+(That is the chain rule for a composition with a conformal map:
+Delta(psi o F) = |F'|^2 ((Delta psi) o F).) Substitute into the
+identity for H(s):
+
+ H(s) = INT_{y > s} (y - s) |F'(z)|^2 (Delta psi)(F(z)) dA(z).
+
+On the support of (Delta psi) o F the height y is bounded by a constant
+Y_psi = Y_psi(F,K), which is finite by (b). Also y - s <= y <= Y_psi.
+So
+
+ |H(s)| <= Y_psi INT_H |F'(z)|^2 |(Delta psi)(F(z))| dA(z)
+ <= Y_psi INT_C |(Delta psi)(w)| dA(w),
+
+where the last inequality used injectivity to change variables w = F(z).
+The bound does not depend on s. Since H(s) >= INT_R 1_K(F(x+is)) dx
+by psi >= 1_K, this is (3.9). QED
+
+
+3.4 COMPACT POSITIVE SUPERLEVELS
+
+This is the payoff of Part 3. It says that the sets worth counting are
+finite things, for almost every map and almost every target.
+
+PROPOSITION 3.5 (compact superlevels). For P-almost every F, and then
+for area-almost every target xi in C, the following holds: for every
+delta > 0, the closed superlevel set
+
+ { z in H : V_xi^F(z) >= delta }
+
+is compact as a subset of H.
+
+Two controls are needed, and they are of different kinds. Escape to
+source infinity is done by the decay (3.8). Approach to the real line
+is done by a dyadic covering argument in the target plane, and that is
+the only place in the whole document where we count cells.
+
+PROOF.
+
+SETUP. By LEMMA 3.3, fix once and for all a map F in K with
+I(F) < infinity. All the following holds for this F, with the
+conclusion for almost every target. By LEMMA 3.4(a) the ratio
+y^k/|F(z)| tends to 0 as |x|+y -> infinity. For |F(z)| >= 2|xi| we have
+|F(z)-xi| >= |F(z)|/2, so
+
+ V_xi^F(z) <= y^k/(|F(z)|/2) = 2 y^k/|F(z)|,
+
+and the right side tends to 0 as |x|+y -> infinity by (3.8). Hence
+
+ for every delta > 0, the set {V_xi^F >= delta} does not meet a
+ neighbourhood of source infinity. (3.11)
+
+That takes care of infinity, uniformly in xi. It remains to keep the
+superlevel away from the real line, and to do that it is enough to show
+
+ sup over x in R of V_xi^F(x + iy) -> 0 as y -> 0, (3.12)
+
+for almost every xi. If (3.12) holds, then {V_xi^F >= delta} lies in
+{y >= y_delta} for some y_delta > 0, and together with (3.11) that makes
+it bounded; it is also relatively closed, being {|F(z)-xi| <= y^k/delta},
+which is closed, with the pole included. So compact. We now prove
+(3.12).
+
+STEP 1: THE OMITTED TARGETS. Fix R and delta, and consider only
+targets xi with |xi| <= R which are OMITTED, meaning xi is not in
+F(H). For the dyadic scales h = 2^{-L}, L = 0,1,2,..., partition the
+real line into the cells [jh,(j+1)h) with j in Z.
+
+MARK A CELL if there exists a point z = x+iy and a target xi with
+|xi| <= R, xi omitted, such that
+
+ x lies in the cell, h <= y <= 2h, and V_xi^F(z) >= delta.
+ (3.13)
+
+The point z is called a witness for the marked cell, and xi a witnessed
+target.
+
+STEP 2: WHAT A WITNESS BUYS. At a witness, y <= 2h and
+|F(z) - xi| <= y^k/delta. By the quarter theorem in the form of LEMMA
+1.2, applied to F on the open disk B(z,y), which lies inside H because
+every point of it has height greater than 0:
+
+ B( F(z), y|F'(z)|/4 ) is contained in F( B(z,y) ).
+
+The target xi is omitted, so xi is not in F(H) and hence not in the
+smaller set F(B(z,y)). Therefore
+
+ |F(z) - xi| >= y|F'(z)|/4, that is, y|F'(z)| <= 4|F(z)-xi|.
+ (3.14)
+
+Now take any x' in the same cell as x. Put
+
+ omega = (x' - x)/y + i (h/y).
+
+Since |x'-x| <= h <= y we have |Re omega| <= 1, and since h <= y <= 2h
+we have 1/2 <= h/y <= 1, so Im omega in [1/2,1]. Also
+
+ z * omega = (x + y (x'-x)/y) + i (y h/y) = x' + ih.
+
+So the point x'+ih is the image of omega under the affine map attached
+to z. Unwrapping the definition of T_zF:
+
+ F(x'+ih) - F(z) = y F'(z) * (T_zF)(omega).
+
+Hence, subtracting xi from both sides,
+
+ F(x'+ih) - xi = ( F(z) - xi ) + y F'(z) (T_zF)(omega).
+
+Take absolute values, use (3.14), and bound (T_zF)(omega) uniformly. The
+set of possible omega is the fixed compact set [-1,1] + i[1/2,1], and by
+LEMMA 1.3 there is a constant C_1 with |Q_G(w)| <= C_1 and all
+derivatives bounded for G in K and w in that set; in particular
+|(T_zF)(omega)| <= C_1. Also |F(z)-xi| <= y^k/delta <= (2h)^k/delta.
+Therefore
+
+ |F(x'+ih) - xi| <= (2h)^k/delta + 4 (2h)^k/delta * C_1
+ = C_delta h^k, (3.15)
+
+with C_delta = (2^k/delta)(1 + 4C_1), which depends on delta and on
+nothing else.
+
+STEP 3: THE CELL COUNT. Fix the scale h. Let K denote the closed disk
+of radius R + C_delta centred at 0. By (3.15), for every marked cell
+and every witnessed target xi, the WHOLE cell segment at height h, that
+is the set {x' + ih : x' in the cell}, is mapped into K: indeed
+F(x'+ih) is within C_delta h^k <= C_delta of xi, and |xi| <= R.
+
+The horizontal length bound (3.9) now gives
+
+ (number of marked cells) * h
+ <= INT_R 1_K(F(x + ih)) dx <= M, (3.16)
+
+where M = M(K) is the constant of (3.9). So
+
+ number of marked cells <= M/h. (3.17)
+
+STEP 4: THE AREA OF WITNESSED TARGETS. A witnessed target xi is, by
+(3.15), within C_delta h^k of the image under F of the midpoint of its
+cell at height h. So the set of targets witnessed at scale h is
+contained in a union of at most M/h closed disks of radius C_delta h^k,
+whose total area is at most
+
+ (M/h) * pi (C_delta h^k)^2 = pi M C_delta^2 h^{2k-1}. (3.18)
+
+Now 2k - 1 >= 1/2 > 0, because k >= 3/4 by (2.18). Summing (3.18) over
+the dyadic scales h = 2^{-L}, L >= 0, gives a convergent geometric
+series. Hence the set of targets that are witnessed at infinitely many
+dyadic scales has area zero.
+
+STEP 5: FROM "MARKED AT ONE SCALE" TO A SUPREMUM BOUND. Let xi with
+|xi| <= R be omitted and outside the null set of STEP 4, so that xi is
+witnessed at only finitely many dyadic scales. Choose L_0 exceeding all
+of them. Take any y with 0 < y <= 2^{-L_0} and any real x, and choose L
+minimal subject to 2^{-L} <= y. Then h = 2^{-L} satisfies
+
+ h <= y < 2h, and y <= 2^{-L_0} <= h,
+
+so L >= L_0. Suppose V_xi^F(x + iy) >= delta. Then z = x+iy is a
+witness in the sense of (3.13) for the cell of scale h that contains x,
+because h <= y < 2h. So that cell is marked at scale h = 2^{-L} with
+L >= L_0, contradicting the definition of L_0. Therefore
+
+ V_xi^F(x + iy) < delta whenever 0 < y <= 2^{-L_0} and x in R.
+ (3.19)
+
+Since delta > 0 was arbitrary, (3.12) holds for every omitted target xi
+with |xi| <= R outside the null set of STEP 4.
+
+STEP 6: THE TARGETS THAT LIE IN THE IMAGE. Here there is no exceptional
+set at all. Suppose xi = F(z_0) for some z_0 in H, and choose r_0 > 0
+small enough that the closed disk B(xi, r_0) lies inside F(H). Because
+F is a homeomorphism of H onto F(H), the preimage
+
+ K_0 = F^{-1}( B(xi, r_0) )
+
+is a compact subset of H. Let y_1 > 0 be the smallest height of a point
+of K_0; it is positive because K_0 is compact and lies in H.
+
+Now let 0 < y < y_1 and let x be arbitrary. If |F(x+iy) - xi| < r_0,
+then F(x+iy) would lie in B(xi,r_0), so x+iy would lie in K_0, and its
+height would be at least y_1, contradicting y < y_1. Hence
+
+ |F(x + iy) - xi| >= r_0, so V_xi^F(x + iy) <= y^k / r_0,
+
+and the right side tends to 0 as y -> 0. This is (3.12) for this target,
+with no exception.
+
+ASSEMBLY. Let delta run over the values 1/n, n >= 1, and R over
+1, 2, 3, .... For each choice of (n,R), STEP 5 identifies a set of
+targets of area zero. A countable union of sets of area zero has area
+zero. So for area-almost every xi, (3.12) holds for every delta > 0
+simultaneously. Together with (3.11), which holds for every xi and
+every delta, this gives: for area-almost every xi and every delta > 0,
+the set {V_xi^F >= delta} is a closed subset of H that keeps away from a
+neighbourhood of the real line and from a neighbourhood of source
+infinity. Such a set is compact. QED
+
+
+================================================================
+PART 4. CRITICAL POINTS, AND PAIRING SADDLES WITH PEAKS
+================================================================
+
+The potential V of Part 3 has a smooth logarithm ell with Laplacian
+-k/y^2 < 0. A function with negative Laplacian cannot have an interior
+minimum, and that single remark makes its critical points countable and
+organizable. This part:
+
+ (i) identifies the critical points of ell with the solutions of one
+ smooth equation G_F(z) = xi;
+ (ii) computes the Hessian of ell at a critical point in terms of one
+ scalar function J_F, so that saddles and peaks are distinguished
+ by the sign of J_F;
+ (iii) shows that for almost every target there are no coincidences;
+ (iv) proves that at every level there are at least as many peaks as
+ saddles, and pairs them up measurably.
+
+The average of J_F at the base point is the quantity that Part 5 will
+force to be nonpositive, and that Part 6 will show is nonnegative under
+the standing assumption. That is the contradiction. So the reader
+should watch J_F: it is the protagonist of the second half.
+
+
+4.1 THE CRITICAL MAP AND THE SIGNED JACOBIAN
+
+DEFINITION (the critical map). For F in K and z = x+iy in H define
+
+ G_F(z) = F(z) - (i y / k) F'(z). (4.1)
+
+We will see in a moment that G_F(z) = xi is exactly the critical point
+equation for ell. G_F is a smooth map of the plane (into C, which we
+identify with R^2), not holomorphic.
+
+DEFINITION (the signed Jacobian). Write DG_F(z) for the real 2 by 2
+derivative matrix and let
+
+ J_F(z) = det DG_F(z) / |F'(z)|^2. (4.2)
+
+The normalization by |F'|^2 is there to make J_F invariant under
+rebasing; we will check that below.
+
+LEMMA 4.1 (the derivative of G_F, and the formula for J_F).
+
+ (a) The two real columns of DG_F are
+
+ (F'/k)(k + q_F) and (i F'/k)(k - 1 + q_F).
+
+ (b) Consequently
+
+ k^2 J_F = k(k-1) + (2k-1) a_F + |q_F|^2. (4.3)
+
+Proof of (a). We differentiate (4.1) holding the other variable fixed,
+and use q_F = -i y F''/F', which is equivalent to
+
+ y F'' = i q_F F'. (4.4)
+
+Check (4.4): (2.15) says q_F F' = -i y F''. Multiply both sides by
+i, which turns -i into 1 and leaves y F'' alone: q_F F' i = y F''.
+Since i q_F = q_F i, this is exactly (4.4).
+
+Differentiate in x. The factor y is constant and z moves with
+x, so partial_x F(z) = F'(z) and partial_x F'(z) = F''(z):
+
+ partial_x G_F = F' - (iy/k) F'' = F' - (iy/k)(i q_F F'/y)
+ = F' + (q_F/k) F' = (F'/k)(k + q_F).
+
+Differentiate in y. Now z moves with speed i, and the explicit y in
+(4.1) also differentiates:
+
+ partial_y G_F = i F' - (i/k) d/dy [ y F'(z) ]
+ = i F' - (i/k) [ F'(z) + y (i F''(z)) ]
+ = i F' - (i/k) F' + (y/k) F''
+ = i F' (1 - 1/k) + (i q_F/k) F'
+ = (i F'/k) ( k - 1 + q_F ).
+
+This is (a).
+
+Proof of (b). Write q = q_F = a + ib and F' = P. For two complex
+numbers u, v regarded as the two columns of a real 2 by 2 matrix,
+
+ det [ u v ] = Im( conj(u) v ).
+
+Reason: if u = u_1+iu_2 and v = v_1+iv_2 then the matrix has columns
+(u_1,u_2) and (v_1,v_2), its determinant is u_1v_2 - u_2v_1, and
+conj(u)v = (u_1-iu_2)(v_1+iv_2) has imaginary part u_1v_2 - u_2v_1.
+
+Apply this with the two columns from (a):
+
+ conj(u) v = (conj(P)/k)(k + conj q) * (iP/k)(k - 1 + q)
+ = (i |P|^2/k^2) (k + conj q)(k - 1 + q).
+
+So det DG_F = (|P|^2/k^2) * Im[ i (k + conj q)(k-1+q) ]. Multiply out
+with q = a + ib and conj q = a - ib:
+
+ (k + conj q)(k-1+q)
+ = k(k-1) + |q|^2 + k q + (k-1) conj q
+ = k(k-1) + |q|^2 + k(a+ib) + (k-1)(a-ib)
+ = [ k(k-1) + |q|^2 + (2k-1)a ] + i b .
+
+Multiply by i and take the imaginary part: since i(W + ib) = iW - b with
+W real, the imaginary part is W. Hence
+
+ det DG_F = (|F'|^2/k^2) [ k(k-1) + (2k-1)a + |q|^2 ].
+
+Divide by |F'|^2 to get (4.3). QED
+
+LEMMA 4.2 (the average of J_F at the base point). Under the law P,
+
+ k^2 E J_F(i) = (4 beta - 1)(beta + 2)/9. (4.5)
+
+Consequently, under the standing assumption beta >= 1/4,
+
+ E J_F(i) >= 0, with equality exactly when beta = 1/4. (4.6)
+
+Proof. At the base point i we have Q_F(i) = 1 and q_F(i) = a + ib
+whatever a and b are. Take expectations in (4.3) and insert the two
+moment identities (2.19):
+
+ k^2 E J_F(i) = k(k-1) + (2k-1)(-beta) + beta(beta+1).
+
+Now substitute k = (beta+2)/3. This is pure algebra; do it in three
+pieces to avoid mistakes. Since k - 1 = (beta-1)/3,
+
+ k(k-1) = (beta+2)(beta-1)/9.
+
+Since 2k - 1 = (2beta+1)/3,
+
+ (2k-1)(-beta) = -beta(2beta+1)/3 = -3beta(2beta+1)/9.
+
+And beta(beta+1) = 9beta(beta+1)/9. Sum the numerators over 9:
+
+ (beta+2)(beta-1) - 3beta(2beta+1) + 9beta(beta+1)
+ = (beta^2 + beta - 2) - (6beta^2 + 3beta) + (9beta^2 + 9beta)
+ = 4beta^2 + 7beta - 2.
+
+On the other hand (4beta-1)(beta+2) = 4beta^2 + 8beta - beta - 2
+= 4beta^2 + 7beta - 2. They agree, which is (4.5).
+
+Finally, under beta >= 1/4 the factor 4beta - 1 is nonnegative and the
+factor beta + 2 is positive, so the right side of (4.5) is nonnegative,
+and it vanishes exactly when 4beta = 1. QED
+
+(4.6) is the engine of the proof. Hold on to it: Part 5 proves the
+opposite inequality, E J_F(i) <= 0, which will force beta = 1/4, and
+then Part 6 shows that even beta = 1/4 is impossible.
+
+Now the identification of critical points.
+
+LEMMA 4.3 (critical points are preimages of the critical map). Fix
+z in H and put
+
+ r_0 = F'/(F - xi), the logarithmic derivative of F - xi.
+
+ (a) grad ell(z) = 0 if and only if G_F(z) = xi. (4.7)
+
+ (b) At such a point,
+
+ V_xi^F(z) = k y^{k-1} |Q_F(z)|, (4.8)
+
+ r_0 = -ik/y, and r_0' = k (k + q_F)/y^2. (4.9)
+
+Proof of (a). Differentiate (3.2). The terms are ell = k log y minus
+log|F - xi|, and for a holomorphic function h, partial_x log|h| =
+Re(h'/h) while partial_y log|h| = -Im(h'/h). Applying this to
+h = F - xi, whose logarithmic derivative is r_0:
+
+ ell_x = -Re r_0, ell_y = k/y + Im r_0. (4.10)
+
+Check the second: ell_y = k/y - partial_y log|F-xi| = k/y - (-Im r_0).
+Good.
+
+So grad ell = 0 means Re r_0 = 0 and Im r_0 = -k/y, i.e. r_0 = -ik/y.
+Then
+
+ F - xi = F'/r_0 = F' * y/(-ik) = i y F'/k,
+
+so xi = F - iyF'/k = G_F(z). Conversely if G_F(z) = xi then
+F - xi = iyF'/k, so r_0 = F'/(iyF'/k) = k/(iy) = -ik/y, and (4.10)
+gives grad ell = 0. This is (a).
+
+Proof of (b). The first identity: from F - xi = iyF'/k,
+
+ |F - xi| = y|F'|/k, so V = y^k/|F-xi| = k y^{k-1}/|F'|.
+
+Since |Q_F| = 1/|F'|, this is (4.8).
+
+The value of r_0 at a critical point is -ik/y, already computed. For
+the derivative, differentiate r_0 = F'/(F-xi):
+
+ r_0' = F''/(F - xi) - F'^2/(F-xi)^2 = r_0 (F''/F') - r_0^2.
+
+Now F''/F' = i q_F/y by (4.4), and r_0 = -ik/y, so
+
+ r_0 (F''/F') = (-ik/y)(i q_F/y) = k q_F/y^2,
+ r_0^2 = (-ik/y)^2 = -k^2/y^2,
+
+and adding gives r_0' = k(k + q_F)/y^2. QED
+
+The first identity of (4.8) is worth reading twice. It says that at a
+critical point, the value of the potential is determined by the
+POSITION of the critical point alone, and not by the target: the target
+is determined by the position too, by (4.7). Cubing (4.8) and using
+3(k-1) = beta-1 gives
+
+ ( V_xi^F(z) )^3 = k^3 y^{3k-3} |Q_F(z)|^3 = k^3 y^{beta-1}|Q_F(z)|^3.
+ (4.11)
+
+The right side is, up to the constant k^3, exactly the weight
+y^{beta-1}|Q_F(z)|^3 that the change of variables through G_F will
+produce in Part 5, in the area formula (5.1) and in the identification
+(5.2). This is the promised reason for the choice of k: the potential's
+critical values carry precisely the weight that the area formula
+assigns.
+
+Now the Hessian.
+
+LEMMA 4.4 (the Hessian at a critical point, and the meaning of the sign
+of J_F). At a critical point,
+
+ Hess ell = (k/y^2) * [ [ -(k + a_F), b_F ],
+ [ b_F, k - 1 + a_F ] ], (4.12)
+
+and therefore
+
+ det Hess ell = -(k^4/y^4) J_F, trace Hess ell = -k/y^2 < 0.
+ (4.13)
+
+Consequently, at a critical point where the Hessian is invertible:
+
+ J_F > 0 gives a SADDLE (det < 0);
+ J_F < 0 gives a LOCAL MAXIMUM of ell, hence of V (det > 0,
+ trace < 0).
+
+We call a local maximum of ell a PEAK. A critical point with J_F > 0 is
+a saddle.
+
+Proof. Differentiate (4.10) once more, using r_0' for the derivative of
+r_0 in the complex variable:
+
+ ell_xx = -Re(r_0'),
+ ell_xy = -Re(i r_0') = Im(r_0'),
+ ell_yy = -k/y^2 - (-Im(i r_0')) = -k/y^2 + Re(r_0').
+
+(For the last line: partial_y of Im r_0 is Im(i r_0') = Re r_0'.) Write
+r_0' = k(k+a+ib)/y^2 from (4.9). Then Re r_0' = k(k+a)/y^2 and
+Im r_0' = kb/y^2. Hence
+
+ ell_xx = -k(k+a)/y^2, ell_xy = kb/y^2, ell_yy = k(k-1+a)/y^2.
+
+This is (4.12). Its trace is (k/y^2)(-(k+a) + (k-1+a)) = -k/y^2, which
+is also Delta ell = -k/y^2, as it must be. For the determinant,
+
+ (y^2/k)^2 det Hess ell = -(k+a)(k-1+a) - b^2.
+
+Compare with J_F. From (4.3),
+
+ k^2 J_F = k(k-1) + (2k-1)a + a^2 + b^2.
+
+Expand the other side: -(k+a)(k-1+a) = -(k+a)^2 + (k+a). So
+
+ (y^2/k)^2 det Hess ell = -(k^2 + 2ka + a^2) + k + a - b^2
+ = (k - k^2) - 2ka + a - a^2 - b^2
+ = -[ k^2 - k + 2ka - a + a^2 + b^2 ]
+ = -[ k(k-1) + (2k-1)a + a^2 + b^2 ]
+ = -k^2 J_F.
+
+Multiply by k^2/y^4: det Hess ell = -k^4 J_F / y^4. This is (4.13).
+The classification follows from the second derivative test for two
+variables: negative determinant means the two eigenvalues have opposite
+signs (saddle), while positive determinant with negative trace means both
+eigenvalues are negative (peak). QED
+
+REMARK (why "signed" Jacobian). J_F is the determinant of a real 2 by 2
+matrix, so it can be either sign, and it is not the square of anything.
+Its sign is exactly the information "saddle or peak". That is why the
+normalization in (4.2) is by |F'|^2, which is positive, and why J_F can
+be averaged meaningfully. Part 5 will show that on average the saddles
+are heavier than the peaks, forcing the average to be <= 0; Lemma 4.2
+says the average is >= 0 unless beta = 1/4 exactly.
+
+
+4.2 GOOD TARGETS
+
+Most targets are well behaved. This section says exactly what well
+behaved means, and proves it happens almost always.
+
+DEFINITION (good pair). A pair (F, xi) with F in K and xi in C is
+GOOD if all three of the following hold.
+
+ (g1) Every closed positive superlevel {V_xi^F >= delta}, delta > 0,
+ is a compact subset of H.
+
+ (g2) xi is a REGULAR VALUE of G_F: every z with G_F(z) = xi has
+ J_F(z) not equal to 0. Equivalently, by LEMMA 4.4, every
+ critical point of ell with target xi is nondegenerate.
+
+ (g3) Distinct preimages of xi under G_F have distinct values of V:
+ if z_1 is not z_2 and G_F(z_1) = G_F(z_2) = xi, then
+ V_xi^F(z_1) is not V_xi^F(z_2).
+
+Condition (g3) is the "no ties" condition. It is what will make the
+rank pairing of 4.4 strictly increasing.
+
+LEMMA 4.5 (genericity). Goodness is a Borel condition on (F,xi). For
+P-almost every F, the set of xi which are NOT good has area zero.
+
+Proof. Three separate assertions, one per condition.
+
+(g1). This is PROPOSITION 3.5.
+
+(g2). SARD'S THEOREM. Fix F. The set of singular values of G_F, that
+is the set {G_F(z) : z in H, J_F(z) = 0}, has area zero. This is the
+plane case of Sard's theorem. Here is the elementary estimate that
+makes it concrete for this smooth map; it is included because the proof
+is short and instructive.
+
+Cover the source H by compact boxes. In one box, cover the singular
+set by grid squares of side h. On a square that meets the singular set,
+pick one singular point z_0 in it. Because DG_F is singular at z_0 and
+second derivatives are bounded on the box, for z within distance h of
+z_0 we have
+
+ G_F(z) - G_F(z_0) = DG_F(z_0)(z - z_0) + O(h^2) = O(h^2),
+
+the linear term being killed by the singularity of DG_F(z_0). So the
+image of the square lies in a neighbourhood of size O(h^2) of a single
+point, hence in a disk of area O(h^4); summing over the O(h^{-2})
+squares gives total area O(h^2) -> 0. (The estimate is even stronger
+than needed.) So the singular values are null, and for each fixed F,
+area-almost every xi satisfies (g2).
+
+(g3). THE NO-TIE CONDITION. Cover the nonsingular locus
+{G_F(z) = xi, J_F(z) != 0} by countably many inverse-function charts.
+On such a chart, the equation G_F(z) = xi has a unique smooth solution
+z(xi), varying smoothly with xi.
+
+On an overlap of two such charts, two distinct branches z_1(xi) and
+z_2(xi) are defined. Consider the difference of the log-values,
+
+ Phi(xi) = log V_xi^F(z_1(xi)) - log V_xi^F(z_2(xi)).
+
+Differentiate in xi, in a direction v in C. Here is the point: since
+z_j(xi) is a CRITICAL point of ell with target xi, the derivative of ell
+with respect to the source position vanishes, so no dz_j/dxi term
+survives. Only the explicit xi-dependence of log|F - xi| contributes.
+Since -log|F - xi| = -Re log(F - xi), and differentiation of
+log(F - xi) in the direction v gives -v/(F - xi),
+
+ d Phi(xi)[v] = Re( v/(F(z_1) - xi) ) - Re( v/(F(z_2) - xi) ).
+
+(Each branch contributes +Re(v/(F(z_j)-xi)) because log V has
+-log|F - xi| and d/dxi(-log|F-xi|) = +Re(v/(F-xi)).)
+
+Now suppose the two differentials were equal as functions of v. Then
+the two vectors 1/(F(z_1)-xi) and 1/(F(z_2)-xi) have equal real parts
+against every v, hence are equal, hence F(z_1) = F(z_2). But z_1 and z_2
+are distinct points of H and F is injective. So the differential is
+not identically zero.
+
+A smooth function with nonzero differential has a zero set that is
+locally a smooth curve, hence has area zero. Taking the union over the
+countably many pairs of charts gives an area-null set of targets on
+which a tie occurs. So for each fixed F, area-almost every xi satisfies
+(g3).
+
+MEASURABILITY. It remains to check that the exceptional sets depend
+measurably on F, so that "P-almost every F" is meaningful. Use the
+compact boxes
+
+ C_m = { |x| <= m, 1/m <= y <= m }, m = 1,2,3,...
+
+For (g1): for each positive integer n, the condition
+{V_xi^F > 1/n} contained in some C_m is equivalent to the failure of
+containment being visible at a rational point outside the box, which is
+a Borel condition because F and xi enter continuously. Requiring this
+for every n gives compactness of all closed positive superlevels, since a
+closed level can be pushed by a smaller reciprocal-integer threshold and
+relative closedness. Countable intersections of Borel sets are Borel.
+
+For (g2) and (g3): a failure is witnessed by one or two points of a
+common C_m, with the two points of a tied pair separated by at least
+1/n. Separation and equality of the continuous quantities in (4.8) are
+closed conditions on the compact C_m, hence Borel after projection.
+Countably many m and n finish the job.
+
+QED
+
+REBASING COMPATIBILITY. The three objects G_F, J_F, V_xi all transform
+in a simple way under rebasing. Let t be in H with height Y = Im t,
+and put
+
+ eta = ( xi - F(t) ) / ( Y F'(t) ). (4.14)
+
+Then for every w in H,
+
+ G_{T_tF}(w) = ( G_F(t*w) - F(t) ) / ( Y F'(t) ), (4.15)
+ J_{T_tF}(w) = J_F(t*w), (4.16)
+ V_eta^{T_tF}(w) = Y^{1-k} |F'(t)| * V_xi^F(t*w). (4.17)
+
+Check (4.15). By definition of T_tF and of G,
+
+ G_{T_tF}(w) = (T_tF)(w) - (i Im w/k)(T_tF)'(w).
+
+From LEMMA 1.4, (T_tF)(w) = (F(t*w) - F(t))/(YF'(t)) and
+(T_tF)'(w) = F'(t*w)/F'(t). So
+
+ G_{T_tF}(w) = (F(t*w)-F(t))/(YF'(t)) - (i Im w/k) F'(t*w)/F'(t).
+
+Now compute the right side of (4.15) directly. Since Im(t*w) = Y Im w
+by (G2),
+
+ G_F(t*w) - F(t)
+ = F(t*w) - F(t) - (i Y Im w/k) F'(t*w),
+
+and dividing by YF'(t) gives exactly the expression above. This is
+(4.15).
+
+Check (4.16). Differentiate (4.15) as a real map of w. The chain rule
+gives, as real linear maps,
+
+ DG_{T_tF}(w) = DG_F(t*w) * Y / (Y F'(t)) = DG_F(t*w)/F'(t).
+
+Take determinants, remembering that multiplication by the complex
+number 1/F'(t) scales area by 1/|F'(t)|^2:
+
+ det DG_{T_tF}(w) = det DG_F(t*w) / |F'(t)|^2.
+
+Also |(T_tF)'(w)|^2 = |F'(t*w)|^2/|F'(t)|^2. Dividing,
+
+ J_{T_tF}(w) = det DG_{T_tF}(w)/|(T_tF)'(w)|^2
+ = det DG_F(t*w)/|F'(t*w)|^2 = J_F(t*w).
+
+This is (4.16).
+
+Check (4.17). By definition of V, and using (T_tF)(w) - eta =
+(F(t*w) - xi)/(YF'(t)) (which is a rearrangement of (4.14)),
+
+ V_eta^{T_tF}(w) = (Im w)^k / |(T_tF)(w) - eta|
+ = (Im w)^k Y|F'(t)|/|F(t*w) - xi|.
+
+On the other hand, by (G2), (Im(t*w))^k = (Y Im w)^k, so
+
+ Y^{1-k}|F'(t)| V_xi^F(t*w)
+ = Y^{1-k}|F'(t)| * (Y Im w)^k/|F(t*w) - xi|
+ = (Im w)^k Y |F'(t)|/|F(t*w)-xi|.
+
+The two agree. This is (4.17).
+
+CONSEQUENCE. By (4.16) and (4.17), rebasing carries good pairs to good
+pairs in both directions: compactness of superlevels, regularity of the
+target, and the absence of ties are all preserved. So we may rebase
+freely.
+
+THE BASE POINT HAS A FIXED TARGET. Put z = i in (4.1). Since F(i) = 0
+and F'(i) = 1,
+
+ G_F(i) = 0 - (i/k) = -i/k for EVERY F in K. (4.18)
+
+So the target attached to the base point is a SINGLE FIXED POINT of the
+plane, independent of the random map. This deserves a warning: the
+statement "area-almost every target is good" tells us nothing at all
+about this one particular target. A set of area zero can perfectly well
+contain -i/k. This is why Part 5 needs the weighted version of
+goodness (LEMMA 5.3) rather than the unweighted version of LEMMA 4.5,
+and why the averaging over maps, with the weight |Q_F|, is essential and
+cannot be replaced by "fix F first, then choose xi".
+
+
+4.3 THE INDEX COUNT: AT LEAST AS MANY PEAKS AS SADDLES
+
+LEMMA 4.6 (the superlevel index count). Let (F,xi) be good. Then for
+every h > 0,
+
+ # { peaks z : V_xi^F(z) >= h } >= # { saddles z : V_xi^F(z) >= h },
+ (4.19)
+
+and both numbers are finite.
+
+Proof.
+
+FINITENESS. By (g1) the closed set {V >= h} is compact. The set of
+critical points with target xi is
+
+ G_F^{-1}(xi) intersect {V >= h},
+
+which is closed in H because G_F and V are continuous and compact by
+(g1). By (g2) the Hessian of ell is invertible at each of its points,
+so each point is isolated (a nondegenerate critical point is isolated).
+A closed, discrete subset of a compact set is finite. So both counts in
+(4.19) are finite numbers.
+
+Now the index count. It is done at an auxiliary level and then moved.
+
+STEP 1: AN AUXILIARY REGULAR LEVEL. Choose a number lambda with
+0 < lambda < h and with the level {V = lambda} containing no critical
+point: since critical values are finite above lambda, such a choice
+exists. Let
+
+ U = { z in H : V_xi^F(z) > lambda }.
+
+By (g1) with delta = lambda, the closure of U in H is compact. Inside U
+the function ell has no critical point on the boundary.
+
+STEP 2: THE BOUNDARY IS A FINITE UNION OF CIRCLES. The set
+dV = { V = lambda } is a closed subset of the compact closure of U, and
+grad V does not vanish on it (that is what the choice of lambda means).
+So dV is a compact smooth one-dimensional manifold without boundary. By
+the classification of compact one-dimensional manifolds, it is a finite
+disjoint union of smooth circles. (This is the one standard topological
+fact we quote here, and it is the version stated at the beginning as
+C3.)
+
+STEP 3: EACH CIRCLE BOUNDS ONE COMPONENT OF U. A circle C in the plane
+bounds a Jordan region J(C). Because the closure of U is compact and C
+lies in it, the region J(C) has compact closure in H: the part of the
+plane below the lowest point of C is in the unbounded component of the
+complement of C.
+
+We claim J(C) is contained in U, that is, V > lambda at every point of
+the interior J(C).
+
+Suppose not. Then some point of J(C) satisfies V <= lambda, which is
+ell <= k log lambda. On the boundary circle C we have ell = k log lambda
+exactly. Consider the minimum of the continuous function ell over the
+compact set closure(J(C)). That minimum is attained somewhere, and it is
+at most k log lambda because some interior point already has ell at most
+k log lambda. So the minimum is attained at a point where ell is at most
+the boundary value, and it is attained in the interior J(C) unless ell is
+constant and equals k log lambda on the whole closure. In either case
+there is an interior local minimum: either a strict one, or a flat one
+because the function equals its boundary value throughout.
+
+But ell is strictly superharmonic, Delta ell = -k/y^2 < 0 by (3.3), and a
+strictly superharmonic function has no interior local minimum. (That is
+the maximum principle: if ell had an interior local minimum at z_0, then
+Delta ell(z_0) would be at least zero, a nonnegative combination of the
+two second directional derivatives at a minimum, contradicting
+Delta ell(z_0) = -k/y_0^2 < 0.) This contradiction proves the claim:
+V > lambda on J(C), so J(C) is contained in U.
+
+STEP 4: NO NESTING, SO COMPONENTS = INTERIORS. If two circles C_1, C_2
+of dV were nested, say J(C_1) contained in J(C_2), then C_1 would lie in
+the inner part of J(C_2). But by STEP 3, J(C_2) is contained in U, while
+C_1 is contained in the complement of U (since V = lambda there and U is
+the strict superlevel). That is a contradiction. Hence the interiors
+are pairwise disjoint, and the components of U are exactly the interiors
+J(C) of the circles C of dV. In particular,
+
+ # components of U = # circles in dV. (4.20)
+
+STEP 5: THE GRADIENT HAS WINDING ONE ON EACH CIRCLE. Orient each
+circle C positively (counterclockwise). Since U is the superlevel and
+the circles bound it from inside, the gradient grad ell points INTO J(C)
+as one crosses C, that is, it points in the inward normal direction.
+The Hopf tangent-turning argument (C3, second half) says: for a simple
+closed positively oriented curve, the inward normal has winding number
+one. So the total rotation of grad ell along C is 2 pi, once.
+
+STEP 6: STOKES, AND THE INDEX SUM. Consider the closed but singular
+1-form on the plane minus the critical points and poles,
+
+ alpha = ( X_1 dX_2 - X_2 dX_1 ) / |X|^2, X = grad ell.
+ (4.21)
+
+Its exterior derivative is zero (it is the differential of the argument
+of X, away from X = 0). Apply Stokes to alpha on the region J(C) with
+small disks removed about each critical point and about the pole (there
+is at most one pole, where F = xi). The outer boundary contributes the
+winding 2 pi / 2 pi = 1 by STEP 5. Each removed disk contributes MINUS
+the index of the corresponding critical point, because the boundary of
+the punctured region is oriented oppositely about the removed disk. Hence
+
+ sum of indices of all critical points and poles inside J(C) = 1.
+ (4.22)
+
+The local indices are:
+ * at a nondegenerate critical point of ell, the index is the sign of
+ det Hess ell, by the standard Taylor expansion; so +1 at a peak and
+ -1 at a saddle, by LEMMA 4.4;
+ * at the pole z_0 (where F(z_0) = xi), ell = -log|z - z_0| + smooth,
+ so the index is +1 (the gradient looks like a source).
+
+There is at most one pole, since F is injective, and it lies in one of
+the components of U, if anywhere.
+
+STEP 7: ADD UP. Sum (4.22) over the circles C of dV, using (4.20):
+
+ # components of U
+ = sum over C of 1
+ = sum over C of ( # peaks in J(C) - # saddles in J(C) + # poles in J(C) )
+ = # peaks in U - # saddles in U + # poles in U. (4.23)
+
+Since the number of poles in U is at most one and nonnegative, (4.23)
+gives
+
+ # peaks in U >= # saddles in U. (4.24)
+
+Read (4.23) as "peaks minus saddles equals components minus poles". If U
+contains no pole, then peaks minus saddles equals the number of
+components, which is positive whenever U is nonempty, and zero when U is
+empty. If the pole lies in U, its component contributes one pole, so
+peaks minus saddles equals one fewer than the number of components. In
+both cases peaks minus saddles is at least zero, which is (4.24).
+
+STEP 8: REMOVE lambda. By the choice of lambda in STEP 1 there is no
+critical value in [lambda,h). So every critical point in U has
+V > lambda, and none has value in [lambda,h); hence every critical point
+of U has V >= h. So the counts in (4.24) are exactly the counts in
+(4.19). QED
+
+This is the combinatorial heart of the proof. Read it in one sentence:
+"the superlevel set is a union of Jordan regions, and each region has
+total index one, so it contains one more peak than saddle." The strict
+superharmonicity of ell is what makes the level curves behave: it is
+what forbids the level curves from nesting, and hence what makes
+"components = interiors" true.
+
+
+4.4 THE MEASURABLE RANK PAIRING
+
+Fix a good pair (F,xi). On the fiber G_F^{-1}(xi) the points split into
+saddles (J_F > 0) and peaks (J_F < 0). We now match each saddle with a
+peak of strictly larger value, by rank.
+
+DEFINITION (rank). Let z be a critical point with target xi. Its RANK
+is the number
+
+ rank(z) = # { z' : G_F(z') = xi, J_F(z') has the same sign as J_F(z),
+ V_xi^F(z') >= V_xi^F(z) }. (4.25)
+
+The rank counts the point itself, so rank(z) >= 1.
+
+FINITENESS OF RANKS. This matters and it is where compactness is used.
+By (g1) applied with delta = V_xi^F(z) (a positive number, since V is
+never 0), the set {V >= V_xi^F(z)} is compact. The fiber G_F^{-1}(xi) is
+closed, because G_F is continuous. By (g2) each point of the fiber is a
+nondegenerate critical point of ell, hence isolated. A closed and
+discrete subset of a compact set is finite. So the set in (4.25) is
+finite and the rank is a positive integer. The same argument applies to
+the set of peaks at or above any fixed value, so both sides of (4.19) are
+finite at every level.
+
+LEMMA 4.7 (the rank pairing: existence, strictness, injectivity,
+equivariance). Let (F,xi) be good. There is a map R_F, defined on the
+saddles of the fiber G_F^{-1}(xi), such that:
+
+ (a) R_F(z) is a peak with G_F(R_F(z)) = xi, (same target)
+ (b) V_xi^F(R_F(z)) > V_xi^F(z), (strictly larger)
+ (c) R_F is injective on the fiber, (different partners)
+ (d) rank(R_F(z)) = rank(z) as integers. (rank matching)
+
+Moreover the assignment commutes with rebasing: for every t in H and the
+corresponding target eta from (4.14),
+
+ R_{T_tF}(w) = t# * R_F(t*w), (4.26)
+
+on the domains where both sides are defined.
+
+Proof of (a), (b), (c), (d). Fix the fiber. Let s_1, ..., s_A be the
+saddles listed in decreasing order of the value of V, and let
+p_1, ..., p_B be the peaks listed the same way. By (g3) no two distinct
+points of the fiber carry the same value, so both orderings are
+canonical: there are no ties to break.
+
+CLAIM. rank(s_j) = j and rank(p_j) = j.
+
+Indeed, rank(s_j) counts the saddles whose value is at least V(s_j),
+which by the ordering are exactly s_1, ..., s_j. The same argument
+applies to peaks. So listing in decreasing order of V is the same as
+listing in increasing order of rank, for saddles and for peaks alike.
+
+Now fix j and put h = V(s_j). By LEMMA 4.6 applied at the level h, the
+number of peaks with value at least h is at least the number of saddles
+with value at least h, which is j. So for every j between 1 and A there
+are at least j peaks; taking j = A gives A <= B, so no saddle is left
+without a partner. Define
+
+ R_F(s_j) = p_j.
+
+Then (a) holds by construction, and the line above gives V(p_j) >= h =
+V(s_j). This is non-strict, and (g3) upgrades it. If V(p_j) were equal
+to V(s_j), then two distinct points of the same fiber, p_j and s_j,
+would carry the same value; they are certainly distinct, since one is a
+peak and the other a saddle. That contradicts (g3). Hence
+V(p_j) > V(s_j), which is (b).
+
+For (c): if j is not j' then R_F(s_j) = p_j and R_F(s_{j'}) = p_{j'}, and
+these are different peaks because they have different ranks j and j'.
+
+For (d): rank(R_F(s_j)) = rank(p_j) = j = rank(s_j).
+
+This proves (a)-(d) for each good fiber separately.
+
+Proof of the EQUIVARIANCE (4.26). Suppose t is in H with height Y, and
+eta is the target (4.14). By (4.17),
+
+ V_eta^{T_tF}(w) = Y^{1-k}|F'(t)| * V_xi^F(t*w), (4.27)
+
+a single positive constant times the old potential, at the rebased
+point. A positive constant factor multiplies every value on a fiber by
+the same amount, so it does not change the ORDERING of values on the
+fiber, and does not create or destroy ties. By (4.16) the sign of J is
+preserved. By (4.15) the fiber G_{T_tF}^{-1}(eta) is the image of the
+fiber G_F^{-1}(xi) under w |-> t*w. Hence ranks are preserved and the
+pairing commutes with rebasing, which is (4.26).
+
+QED (existence part)
+
+MEASURABILITY. Now we must check that the pairing can be chosen Borel
+in the pair (F,z). The reason this is not automatic is that "the j-th
+largest" is a rule that refers to a set which varies with the parameters;
+we need to know that the rule produces a Borel function.
+
+THE DOMAIN. Put
+
+ D = { (F,z) : J_F(z) > 0, z in H, (F, G_F(z)) good }. (4.28)
+
+By LEMMA 4.5, D is Borel. (The condition J_F(z) > 0 is open, the
+goodness conditions are Borel by LEMMA 4.5, and z ranges over the Borel
+set H.)
+
+STEP A: COUNTABLY MANY LOCAL BRANCHES OF THE FIBER. Consider the set of
+triples (F,xi,z) with G_F(z) = xi and J_F(z) not equal to 0. It is
+enough to produce, locally in (F,xi), a continuous solution z(F,xi).
+
+Fix a triple (F_0,xi_0,z_0) with G_{F_0}(z_0) = xi_0 and DG_{F_0}(z_0)
+invertible, and put M = DG_{F_0}(z_0). On a sufficiently small closed
+ball B about z_0, the map
+
+ Phi_{F,xi}(w) = w - M^{-1}( G_F(w) - xi )
+
+is a contraction of B into itself, uniformly for (F,xi) in a
+neighbourhood of (F_0,xi_0): the derivative is I - M^{-1}DG_F(w), which
+has norm at most 1/2 on B for nearby parameters because DG_F is close to
+DG_{F_0} = M there, and shrinking the parameter neighbourhood makes the
+displacement at the centre less than half the radius. By the
+contraction mapping principle, Phi_{F,xi} has exactly one fixed point in
+B, and that fixed point is the unique solution of G_F(w) = xi in B. It
+depends continuously on (F,xi) by the standard stability estimate.
+
+So near every nonsingular solution there is a locally defined continuous
+branch. The space K x C x H is second countable, so countably many such
+neighbourhoods cover the solution set. Enumerate the resulting branches
+as b_1, b_2, b_3, ....
+
+STEP B: ACTIVATION. For a parameter (F,xi), call a branch b_j ACTIVE if
+b_j(F,xi) is defined and its value differs from every b_i(F,xi) with
+i < j that is defined. This rule picks out each solution exactly once:
+of the finitely many branches defined at (F,xi) and giving some
+particular solution, only the first one is activated. "Defined" and
+"differs from" are Borel conditions, so "b_j is active" is Borel in
+(F,xi).
+
+STEP C: SIGNS, VALUES, RANKS. At an active branch, the sign of J_F is a
+continuous function of (F,xi). The value of the potential is, by (4.8),
+
+ V_xi^F(b_j(F,xi)) = k (Im b_j(F,xi))^{k-1} |Q_F(b_j(F,xi))|, (4.29)
+
+also continuous in (F,xi). Comparing two such values is a continuous
+test. Hence for each j and each integer m, the condition
+
+ rank of the j-th active point is m
+
+is Borel: it is a count over finitely many active branches of a
+continuous comparison. Select, for a given active saddle of rank m, the
+unique active peak of rank m; uniqueness follows from (g3). This
+selection is Borel. Finally substitute xi = G_F(z) to get the assignment
+R_F(z) of PROPOSITION (4.26). Composition of Borel maps is Borel. QED
+
+REMARK (what is and is not needed). Two features of this pairing are
+used in Part 5, and nothing else about it is used.
+
+ (i) It is INJECTIVE on each fiber: distinct saddles receive distinct
+ peaks. This is what lets a sum over saddles be bounded by a sum
+ over peaks without double counting.
+ (ii) It is STRICTLY INCREASING: the peak's value is strictly larger.
+ This is what makes the comparison in Part 5 survive at beta = 1/4,
+ where the weights would otherwise be equal. A non-strict pairing
+ would only give B_b(-1) <= 1/4 and would prove nothing.
+
+We do NOT need a uniform positive gap between V(R_F(z)) and V(z). The
+strictness may be arbitrarily small, pair by pair. That is why the
+argument is arranged to extract the strictness through the equality case
+rather than through a numerical estimate.
+
+
+================================================================
+PART 5. STRICT WEIGHTED TRANSPORT
+================================================================
+
+Part 4 produced a pairing of saddles with taller peaks. This part
+turns that combinatorial statement into an inequality between
+integrals, and then into the inequality E J_F(i) <= 0.
+
+The idea in one sentence: change variables through G_F, so that a sum
+over the critical points of a fiber becomes an integral over the target
+plane; each saddle's weight is then exactly its partner's weight
+multiplied by a factor c_F > 1; and since the partner sits in a slightly
+larger box, the extra factor has to be paid for by a large box, which
+forces the average of J_F to be nonpositive.
+
+
+5.1 THE AREA FORMULA
+
+Put
+
+ J_{F,+} = max(J_F, 0), J_{F,-} = max(-J_F, 0),
+ omega_F(z) = y^{beta-1} |Q_F(z)|. (5.1)
+
+Note that omega_F J_{F,+} is a nonnegative function of z, and so is
+omega_F J_{F,-}; their difference is omega_F J_F.
+
+PROPOSITION 5.1 (change of variables through the critical map). Fix F
+in K and a nonnegative Borel function h on H. Then
+
+ INT_H h(z) omega_F(z) J_{F,+}(z) dA(z)
+ = INT_C SUM over { z : G_F(z) = xi, J_F(z) > 0 }
+ h(z) y^{beta-1} |Q_F(z)|^3 dA(xi), (5.2)
+
+and the same identity with J_{F,+} replaced by J_{F,-} and the condition
+J_F(z) > 0 replaced by J_F(z) < 0. Here dA(xi) is planar area in the
+target plane.
+
+Proof. Work on the open set where J_F > 0; on that set G_F is a local
+diffeomorphism by the inverse function theorem, and the set is covered
+by countably many such local charts, which we may disjointify into Borel
+pieces. On one chart, the change of variables formula gives
+
+ INT h omega_F J_F dA(z) = INT h(z(xi)) omega_F(z(xi)) J_F(z(xi))
+ * |dA(z)/dA(xi)| dA(xi).
+
+Now |det DG_F| = |J_F| |F'|^2 by the definition (4.2), and on this chart
+J_F > 0 so |J_F| = J_F. Hence
+
+ |dA(z)/dA(xi)| = 1/(J_F |F'|^2),
+
+and, using J_F |dA(z)/dA(xi)| = 1/|F'|^2 and |Q_F| = 1/|F'|,
+
+ omega_F J_F |dA(z)/dA(xi)|
+ = y^{beta-1} |Q_F| / |F'|^2
+ = y^{beta-1} |Q_F| |Q_F|^2
+ = y^{beta-1} |Q_F|^3.
+
+Summing over the charts covering the fiber over each xi gives (5.2). No
+global properness of G_F is used: only that each branch is a local
+diffeomorphism, and that on each fiber the points are isolated by (g2),
+so the fiber sum is a countable sum of Borel functions. The case J_F < 0
+is identical with |J_F| = -J_F. QED
+
+CONSEQUENCE (the weight of a critical point is the cube of its value).
+At a critical point, (4.8) says V_xi^F(z) = k y^{k-1}|Q_F(z)|. Cube it
+and use 3(k-1) = beta-1:
+
+ y^{beta-1} |Q_F(z)|^3 = ( V_xi^F(z) / k )^3. (5.3)
+
+This is the identification that all the bookkeeping of (4.11) was for.
+Read (5.2) together with (5.3) as follows: on the target side, each
+critical point of the fiber contributes its value CUBED, times h(z),
+divided by the constant k^3.
+
+
+5.2 TRANSFERRING GOODNESS TO THE FIXED BASE POINT
+
+Section 4.2 warned that the base point carries the fixed target
+G_F(i) = -i/k for every F, so area-genericity of targets says nothing
+about it. The next lemma is the substitute. It is weighted: the
+Jacobian factor J_{F,+} is exactly what makes the transfer work.
+
+LEMMA 5.2 (weighted goodness at the base point). Let bad denote the set
+of pairs (F,xi) which are not good. Then
+
+ E [ J_{F,+}(i) * 1{ (F, G_F(i)) in bad } ] = 0. (5.4)
+
+Proof.
+
+STEP 1: THE IDENTITY ON A BOX. Let B_0 = { |x| <= 1, 1 <= y <= 2 }.
+For P-almost every F, the set of bad targets xi is null by LEMMA 4.5.
+Apply (5.2) with
+
+ h(z) = 1{ z in B_0 } * 1{ (F, G_F(z)) bad }.
+
+The target-side integrand vanishes for almost every xi, by the preceding
+sentence, for almost every F. So
+
+ INT_{B_0} omega_F(z) J_{F,+}(z) 1{ (F,G_F(z)) bad } dA(z) = 0,
+ (5.5)
+
+for P-almost every F.
+
+STEP 2: REWRITE EACH SOURCE POINT AS A BASE POINT. Fix z in B_0. By
+(4.16) and (4.15), the rebased map T_zF has
+
+ J_{T_zF}(i) = J_F(z), G_{T_zF}(i) = ( G_F(z) - F(z) )/(yF'(z)),
+ (5.6)
+
+and by the rebasing compatibility of 4.2, the pair (T_zF, G_{T_zF}(i))
+is good exactly when (F, G_F(z)) is good. Also the base target of
+T_zF is the fixed point,
+
+ G_{T_zF}(i) = -i/k. (5.7)
+
+Apply the law (2.14) with test
+
+ Phi(F) = J_{F,+}(i) * 1{ (F, -i/k) bad }.
+
+The test is nonnegative and Borel. Using (5.6) and (5.7),
+
+ E [ |Q_F(z)| J_{F,+}(z) 1{ (F, G_F(z)) bad } ]
+ = E [ |Q_F(z)| J_{T_zF,+}(i) 1{ (T_zF, -i/k) bad } ]
+ = y^{-beta} E [ J_{F,+}(i) 1{ (F,-i/k) bad } ]. (5.8)
+
+STEP 3: INTEGRATE AGAINST THE RIGHT WEIGHT AND CANCEL. Multiply (5.8)
+by y^{beta-1} and integrate over B_0:
+
+ INT_{B_0} E[ omega_F(z) J_{F,+}(z) 1{ (F,G_F(z)) bad } ] dA(z)
+ = INT_{B_0} y^{-1} dA(z) * E[ J_{F,+}(i) 1{ (F,-i/k) bad } ].
+
+The left side is zero: interchange the z-integral with the expectation
+(the integrand is nonnegative, so Tonelli applies) and use (5.5). The
+z-integral on the right is elementary:
+
+ INT_{B_0} y^{-1} dA(z) = INT_{-1}^{1} dx * INT_1^2 dy/y = 2 log 2,
+ (5.9)
+
+which is strictly positive. Divide by it. QED
+
+Lemma 5.2 is the technical heart of the passage from "almost every
+target is good" to "the fixed base target is good on the part of the
+distribution that has positive Jacobian". Note the two things it uses:
+the weight |Q_F(z)| from the law, and the factor y^{beta-1} which turns
+y^{-beta} into y^{-1}. Neither is decorative: without them the box
+integral would not produce a strictly positive constant, and the
+argument would only conclude 0 = 0.
+
+
+5.3 THE BOX COMPARISON
+
+PROPOSITION 5.3 (strict weighted transport). Assume beta >= 1/4, let P
+be the law of PROPOSITION 2.7, and let k = (beta+2)/3. Then
+
+ E J_F(i) <= 0. (5.10)
+
+Moreover, if equality holds, then J_F(i) = 0 for P-almost every F.
+
+Proof. We work on the pairing domain of section 4.4 and cut it down to
+size.
+
+STEP 1: A CUT-DOWN DOMAIN. Recall
+
+ D = { (F,z) : J_F(z) > 0, (F, G_F(z)) good }.
+
+For N >= 1 let D_N be the Borel subset of D consisting of pairs for
+which the partner z' = R_F(z) satisfies
+
+ 1/N <= y'/y <= N and |x' - x| <= N y. (5.11)
+
+The D_N increase to D as N increases, and their indicators at (F,z)
+equal their indicators at (T_zF, i), by the equivariance (4.26).
+
+Define the nonnegative Borel function
+
+ c_F(z) = ( V_{G_F(z)}^F(R_F(z)) / V_{G_F(z)}^F(z) )^3
+ on D, and c_F(z) = 0 off D. (5.12)
+
+By (b) of LEMMA 4.7, V(R_F(z)) > V(z) strictly, so
+
+ c_F(z) > 1 on D. (5.13)
+
+STEP 2: TWO BOXES. Fix S > 0 and Y > 1 and put
+
+ B = { |x| <= S, 1 <= y <= Y },
+ B' = { |x| <= S + NY, 1/N <= y <= NY }. (5.14)
+
+EVERY INCLUDED SADDLE IN B HAS ITS PARTNER IN B'. Take (F,z) in D_N
+with z = x+iy in B and partner z' = x'+iy'. By (5.11), 1/N <= y'/y and
+y <= Y, so y' <= NY; and y >= 1 with y' >= y/N >= 1/N. Also
+|x'| <= |x| + |x'-x| <= S + Ny <= S + NY. So z' is in B'.
+
+STEP 3: THE SOURCE-SIDE INEQUALITY, FIBER BY FIBER. Fix F. Start from
+the target-side identity (5.2) applied with h = the indicator of
+{z in B, (F,z) in D_N}, and multiply the source weight by c_F. On the
+target side this multiplies the summand belonging to a saddle z by
+c_F(z). By (5.12) and (5.3) the summand becomes
+
+ ( V(R_F(z))/k )^3,
+
+which is EXACTLY the summand belonging to the peak R_F(z) in the
+J_F < 0 version of (5.2). Because R_F is injective on the fiber by (c)
+of LEMMA 4.7, these are distinct summands, and because R_F(z) is in B'
+by STEP 2, each such summand is actually counted on the right. Hence,
+for every F,
+
+ INT_B omega_F J_{F,+} c_F 1_{D_N}(F,.) dA
+ <= INT_{B'} omega_F J_{F,-} dA. (5.15)
+
+STEP 4: TAKE EXPECTATIONS AND APPLY THE LAW. Both sides of (5.15) are
+integrals of nonnegative quantities, so Tonelli allows us to take the
+expectation inside. On the left, the equivariance (4.26) together with
+(5.6) gives
+
+ omega_F(z) J_{F,+}(z) c_F(z) 1_{D_N}(F,z)
+ = y^{beta-1} |Q_F(z)| * [ J_{F,+}(i) c_F(i) 1_{D_N}(F,i) ](T_zF),
+
+so by the law (2.14) its expectation is
+
+ y^{-beta} y^{beta-1} E[ J_{F,+}(i) c_F(i) 1_{D_N}(F,i) ]
+ = y^{-1} E[ J_{F,+}(i) c_F(i) 1_{D_N}(F,i) ]. (5.16)
+
+Similarly, on the right,
+
+ E[ omega_F(z) J_{F,-}(z) ] = y^{-1} E[ J_{F,-}(i) ]. (5.17)
+
+Substitute (5.16) and (5.17) into (5.15) and compute the two box
+integrals of y^{-1} dA:
+
+ INT_B y^{-1}dA = 2S log Y,
+ INT_{B'} y^{-1}dA = 2(S + NY) ( log(NY) - log(1/N) )
+ = 2(S + NY)( log Y + 2 log N ). (5.18)
+
+The result is
+
+ 2S log Y E[ J_{F,+}(i) c_F(i) 1_{D_N}(F,i) ]
+ <= 2(S+NY)(log Y + 2 log N) E[ J_{F,-}(i) ]. (5.19)
+
+STEP 5: THE THREE LIMITS. Write
+
+ A_N = E[ J_{F,+}(i) c_F(i) 1_{D_N}(F,i) ],
+ d = E[ J_{F,-}(i) ].
+
+Both are finite: J_F(i) is bounded by LEMMA 1.3, since F |-> J_F(i) is
+continuous on the compact K. Divide (5.19) by 2S log Y > 0:
+
+ A_N <= (1 + NY/S) * (1 + 2 log N / log Y) * d.
+
+Now let S -> infinity at fixed Y and N. The factor (1+NY/S) tends to 1,
+so
+
+ A_N <= (1 + 2 log N / log Y) d.
+
+Then let Y -> infinity at fixed N: the second factor tends to 1, giving
+
+ A_N <= d.
+
+Finally let N -> infinity. The sets D_N increase to D and c_F >= 1, so
+by monotone convergence A_N increases to
+
+ A = E[ J_{F,+}(i) c_F(i) 1_D(F,i) ].
+
+Hence A <= d.
+
+STEP 6: PUT THE TWO HALVES TOGETHER. Define
+
+ b = E[ J_{F,+}(i) ].
+
+Lemma 5.2 says that the positive-Jacobian mass at the base point is
+supported on good fibers, which is to say
+
+ b = E[ J_{F,+}(i) 1_D(F,i) ]. (5.20)
+
+And since c_F >= 1 on D,
+
+ b <= A. (5.21)
+
+Chaining, b <= A <= d. Therefore
+
+ E J_F(i) = E[ J_{F,+}(i) ] - E[ J_{F,-}(i) ] = b - d <= 0.
+
+This is (5.10).
+
+STEP 7: THE EQUALITY CASE. Suppose E J_F(i) = 0. Then b = d, and since
+b <= A <= d we get A = b. By (5.20) and (5.21),
+
+ 0 <= E[ J_{F,+}(i) (c_F(i) - 1) 1_D(F,i) ] = A - b = 0.
+
+The integrand is nonnegative, because c_F >= 1 on D. It is STRICTLY
+positive at every (F,i) with J_F(i) > 0 and (F,i) in D, by (5.13). So
+the probability of that set is zero:
+
+ P{ J_F(i) > 0 } = 0.
+
+Hence b = 0 and therefore d = 0. So J_F(i) <= 0 and J_F(i) >= 0 both
+hold almost surely, that is,
+
+ J_F(i) = 0 for P-almost every F. (5.22)
+
+QED
+
+REMARK (where strictness was actually used). Note carefully that no
+step used a NUMERICAL gap. The only place strictness entered was
+(5.13), the strict inequality c_F > 1 on D, which came from part (b) of
+LEMMA 4.7, which came from the no-tie condition (g3). Then STEP 7 used
+the elementary principle "a nonnegative function with zero integral
+vanishes". A non-strict pairing would have given c_F >= 1, hence
+c_F - 1 >= 0 with no strictness, and the conclusion of STEP 7 would have
+been vacuous. This is precisely the point flagged in section 0.6. The
+argument is arranged so that an arbitrarily small strict gain, varying
+from pair to pair with no uniformity at all, still produces a definite
+conclusion.
+
+
+================================================================
+PART 6. THE HARMONIC BARRIER, AND THE UNIFORM GAP
+================================================================
+
+Here is where the two halves meet.
+
+ * Part 4, in (4.6): under beta >= 1/4, the average E J_F(i) is
+ NONNEGATIVE, and it vanishes only when beta is exactly 1/4.
+
+ * Part 5, in (5.10) and (5.22): under beta >= 1/4, the average
+ E J_F(i) is NONPOSITIVE, and equality forces J_F(i) = 0 for
+ almost every F.
+
+Together: beta must be exactly 1/4, and J_F(i) must vanish almost
+surely. Then the law upgrades "vanishes at one point" to "vanishes
+identically almost surely". And then a barrier argument says that no
+member of K at all has J_F identically zero. Contradiction.
+
+The barrier is the last substantial ingredient, and it is short.
+
+
+6.1 NO MAP HAS A VANISHING JACOBIAN
+
+LEMMA 6.1 (the harmonic barrier). With k = 3/4, no F in K has
+J_F identically zero on H.
+
+Proof. Suppose J_F = 0 everywhere. By (4.3) with k = 3/4, this says
+
+ 0 = k(k-1) + (2k-1) a_F + |q_F|^2 = -3/16 + a_F/2 + a_F^2 + b_F^2.
+
+Complete the square in a:
+
+ a_F^2 + a_F/2 - 3/16 = (a_F + 1/4)^2 - 1/16 - 3/16 = (a_F+1/4)^2 - 1/4.
+
+So the equation J_F = 0 is exactly
+
+ | q_F(z) + 1/4 |^2 = 1/4 for every z in H. (6.1)
+
+Two immediate consequences.
+
+ (i) |q_F| >= 1/4 everywhere, since the distance from 0 to the circle
+ of centre -1/4 and radius 1/2 is 1/4.
+
+ (ii) q_F is never 0, hence by q_F = -iy F''/F' and y > 0, the
+ pre-Schwarzian F''/F' is never 0 either.
+
+Now put
+
+ u = log|F''/F'|. (6.2)
+
+By (ii) the function F''/F' is holomorphic and never zero, so u is
+harmonic:
+
+ Delta u = 0. (6.3)
+
+(One may also write u = log|q_F| - log y, using |q_F| = y|F''/F'|; that
+identity is not needed, only the lower bound below.) From (i),
+
+ |F''/F'| = |q_F|/y >= 1/(4y), that is, u >= -log 4 - log y.
+ (6.4)
+
+THE BARRIER. For 0 < eps < 1/2 define the comparison function
+
+ b_eps(x,y) = -log 4
+ + log(1/eps) * cos(pi x/2) * sinh( pi(1-y)/2 )/sinh(pi/2).
+ (6.5)
+
+Three properties, each checked in one line.
+
+HARMONIC. With f(x) = cos(pi x/2) and g(y) = sinh(pi(1-y)/2) we have
+f'' = -(pi/2)^2 f and g'' = (pi/2)^2 g, so Delta(f g) = f'' g + f g''
+= 0. So b_eps is harmonic.
+
+BOUNDED BY u ON THE BOUNDARY OF THE BOX Q = [-1,1] x [eps,1].
+ * On the bottom y = eps: cos(pi x/2) lies in [0,1] for |x| <= 1, and
+ sinh(pi(1-eps)/2)/sinh(pi/2) lies in (0,1] for 0 < eps < 1. So the
+ added term lies in [0, log(1/eps)], giving
+ b_eps <= -log 4 + log(1/eps) = -log 4 - log eps. By (6.4) at
+ y = eps, u >= -log 4 - log eps. So b_eps <= u there.
+ * On the sides x = plus or minus 1: cos(plus or minus pi/2) = 0, so
+ b_eps = -log 4. By (6.4) with y <= 1, u >= -log4 - log y >= -log4.
+ So b_eps <= u there.
+ * On the top y = 1: sinh(0) = 0, so b_eps = -log4, and (6.4) at y = 1
+ gives u >= -log4. So b_eps <= u there.
+
+THE MINIMUM PRINCIPLE. Both u and b_eps are harmonic on the interior of
+the box Q, b_eps <= u on the boundary of Q, so b_eps <= u on all of Q.
+Evaluate at the point i/2, that is x = 0, y = 1/2. There
+cos(0) = 1 and sinh(pi/4)/sinh(pi/2) = c_0, a fixed strictly positive
+number. So
+
+ u(i/2) >= b_eps(0,1/2) = -log 4 + c_0 log(1/eps). (6.6)
+
+The number u(i/2) is a fixed finite real number, since F is holomorphic
+and F''/F' has no zeros. The right side of (6.6) tends to +infinity as
+eps decreases to 0. Contradiction. QED
+
+REMARK (the shape of the barrier). Why does this work? The constraint
+(6.1) says that q_F stays on a fixed circle. Now q_F is (Im z) times the
+holomorphic function -i F''/F', so (6.1) pins down |F''/F'| completely:
+|F''/F'| = |q_F|/y is determined by the height alone. A holomorphic
+function whose modulus depends only on the imaginary part is very rigid,
+and (6.1) asks for even more. The barrier converts that rigidity into a
+contradiction. The mechanism is a
+lower bound that grows like log(1/eps) at a fixed interior point while
+the boundary only allows it to grow on the bottom side, whose distance
+from the interior point shrinks like eps. The cos(pi x/2) factor kills
+the barrier on the two vertical sides, and the sinh factor kills it on
+the top. The only surviving side is the bottom, at height eps, and
+there the lower bound from (6.4) is exactly strong enough to dominate.
+
+
+6.2 THE THEOREM
+
+We can now collect everything.
+
+PROOF OF THEOREM 0.5. Suppose, for the sake of contradiction, that
+beta >= 1/4, so that the standing assumption (2.1) holds.
+
+Step 1: the two halves. By LEMMA 4.2, formula (4.5),
+
+ k^2 E J_F(i) = (4beta-1)(beta+2)/9 >= 0. (6.7)
+
+By PROPOSITION 5.3, formula (5.10) and its equality case,
+
+ E J_F(i) <= 0, and equality forces J_F(i) = 0 almost surely. (6.8)
+
+So E J_F(i) = 0 exactly. By (6.7), and because beta + 2 > 0, this
+means 4beta - 1 = 0:
+
+ beta = 1/4. (6.9)
+
+Consequently, by (2.17), k = (1/4 + 2)/3 = 9/12 = 3/4.
+
+Step 2: upgrade "vanishing at the base point" to "vanishing everywhere".
+By the equality case of PROPOSITION 5.3, J_F(i) = 0 for P-almost every
+F. Now fix an arbitrary z in H. Apply the law (2.14) with the test
+function Phi(F) = |J_F(i)|, which is nonnegative and continuous in F,
+and use (4.16) to replace the rebased Jacobian at the base point by the
+original Jacobian at z:
+
+ E [ |Q_F(z)| * |J_F(z)| ]
+ = E [ |Q_F(z)| * |J_{T_zF}(i)| ]
+ = y^{-beta} E [ |J_F(i)| ]
+ = 0.
+
+The factor |Q_F(z)| is strictly positive, so
+
+ J_F(z) = 0 for P-almost every F, for this particular z.
+ (6.10)
+
+Now let Z be a countable dense subset of H, for instance the points with
+rational coordinates. The union of the null sets appearing in (6.10)
+over z in Z is again null. So for P-almost every F, J_F vanishes on Z.
+The function J_F is continuous on H: it is built from F, F', F'' and
+|F'|^{-2} by (4.1)-(4.3), and F' never vanishes. A continuous function
+vanishing on a dense set vanishes identically. Hence
+
+ J_F = 0 on H for P-almost every F. (6.11)
+
+Step 3: the contradiction. (6.11) says that a set of F of positive
+P-measure has J_F identically zero. In particular, at least one F in K
+has J_F identically zero. But k = 3/4 and LEMMA 6.1 forbid exactly
+that. Contradiction.
+
+Therefore the assumption was false:
+
+ beta < 1/4. (6.12)
+
+Step 4: convert beta < 1/4 into the stated uniform estimate. Choose any
+real number b with
+
+ beta < b < 1/4, (6.13)
+
+and put eps = 1/4 - b, so 0 < eps < 1/4. Since ||L^n||^{1/n} tends to
+rho = 2^beta and 2^beta < 2^b, there is a constant C_b, depending on b
+but on nothing else, with
+
+ ||L^n|| <= C_b * 2^{bn} for all n >= 0. (6.14)
+
+Now let f be any member of S and let 1/2 <= r < 1. Choose n with
+2^{-n} <= 1-r < 2^{1-n}. By PROPOSITION 1.8 and (6.14),
+
+ M_{-1}[f'](r) <= C_0 ||L^n|| <= C_0 C_b 2^{bn}.
+
+And since 2^{-n} <= 1-r, we have 2^n >= 1/(1-r), hence 2^{bn} =
+(2^n)^b >= (1-r)^{-b}. Therefore
+
+ M_{-1}[f'](r) <= C (1-r)^{-b} = C (1-r)^{-1/4 + eps},
+ with C = C_0 C_b,
+
+which is the asserted uniform estimate, valid for every f in S and every
+1/2 <= r < 1.
+
+Finally, taking the limsup and dividing by log(1/(1-r)) gives
+beta_f(-1) <= 1/4 - eps for every f in S. Taking the supremum over the
+bounded-image subclass,
+
+ B_b(-1) <= 1/4 - eps < 1/4.
+
+So B_b(-1) is strictly less than Kraetzer's predicted value, and his
+formula fails at p = -1. QED
+
+That completes the proof.
+
+
+================================================================
+PART 7. WHAT THIS PROOF DOES AND DOES NOT SAY
+================================================================
+
+7.1 WHAT IS PROVED
+
+Exactly one theorem, stated in 0.5: there are constants 0 < eps < 1/4 and
+C < infinity such that
+
+ M_{-1}[f'](r) <= C (1-r)^{-1/4+eps}
+
+for every f in S and every 1/2 <= r < 1. The estimate is uniform over
+the whole class, with no hypothesis on the image beyond injectivity and
+the normalization f(0)=0, f'(0)=1. The consequence B_b(-1) < 1/4 is
+strict, so Kraetzer's formula B_b(p) = p^2/4 at p = -1 is false.
+
+7.2 WHAT IS QUOTED, NOT PROVED
+
+Four external facts are used. Each is standard, each is stated in words
+where it is used, and each is genuinely a theorem of others.
+
+ C1. Compactness: a normalized family of injective holomorphic
+ functions is compact in locally uniform convergence, and limits
+ are injective (Montel plus Hurwitz). Used in LEMMA 1.3 for the
+ compactness of K.
+
+ C2. Sard's theorem in the plane: the set of singular values of a
+ smooth map of the plane has area zero. Used in LEMMA 4.5 for
+ condition (g2). An elementary self-contained estimate is included
+ there, so this is the least "external" of the four.
+
+ C3. Two topological facts. (a) A compact one-dimensional manifold
+ without boundary is a finite disjoint union of circles; used in
+ STEP 2 of LEMMA 4.6. (b) The Hopf tangent-turning lemma: the
+ inward normal of a simply closed positively oriented curve has
+ winding number one; used in STEP 5 of LEMMA 4.6.
+
+ C4. Koebe's quarter theorem with the distortion estimate, in the
+ precise form of LEMMA 1.2. Used throughout.
+
+Nothing else is assumed. In particular the geometric function theory
+is confined to C4; the rest of the argument is real analysis, plane
+topology, and probability on a compact space.
+
+7.3 WHAT IS NOT EFFECTIVE
+
+Two things are asserted to exist and are not computed.
+
+ * eps. The proof produces a positive gap 1/4 - beta but does not
+ bound it away from zero. The reason is structural: the argument in
+ Part 6 is a contradiction, so it gives no quantitative control of
+ how far beta sits below 1/4. Every step is effective in the weak
+ sense that the constants are finite, but none of the constants is
+ tracked.
+
+ * C. The final constant is C_0 C_b, where C_0 comes from
+ PROPOSITION 1.8 and C_b from the root limit. Neither is evaluated.
+ The chain of constants in between (C from (1.1), C_1 from STEP 2 of
+ PROPOSITION 1.8, C_delta from (3.15), Y_psi from LEMMA 3.4) is
+ finite at every stage and enormous if written out.
+
+So the theorem is a strict-inquality statement, not a numerical one.
+
+7.4 WHAT IS NOT SETTLED
+
+ * The exponent p = 1 is untouched. Kraetzer's formula is symmetric
+ between p = 1 and p = -1, and this proof says nothing about p = 1.
+ It would be a mistake to read the failure at p = -1 as evidence
+ about the sign of p.
+
+ * No other exponent is settled.
+
+ * Nothing is said about which maps achieve the extremal growth. The
+ proof works with a probability law on the whole class and never
+ identifies an extremal map or an extremal domain.
+
+ * The distinction between the bounded class S_b and the unrestricted
+ class S survives: the bound proved here is uniform over S, hence
+ applies to S_b, and B_b(-1) is what it bounds. Makarov's comparison
+ theorem identifies the unrestricted spectrum with the maximum of the
+ bounded spectrum and the Koebe term 3p-1, and at p = -1 the Koebe
+ term is -4, so it plays no role; but at positive exponents the
+ distinction is essential and is not addressed here.
+
+7.5 WHERE THE DIFFICULTY ACTUALLY SITS
+
+If a reader wants to know which single step makes the proof work, it is
+STEP 3 of PROPOSITION 5.3, that is, the identity
+
+ omega_F J_{F,+} c_F -> the partner's weight,
+
+together with the injectivity of the pairing. Everything before that
+step builds the two objects that must be compared (a law and a pairing);
+everything after it is a box computation and three limits. Mechanically
+it is the change of variables (5.2) that makes the step work; the change
+of variables requires (5.3), which is where the exponent k = (beta+2)/3
+pays off. If one had chosen any other exponent in the potential, the
+two weights in STEP 3 would not have matched and the comparison would
+have failed.
+
+Second in importance is the no-tie condition (g3). Without it the
+pairing would only be non-strict, and the equality analysis of STEP 7 of
+PROPOSITION 5.3 would not close.
+
+
+================================================================
+APPENDIX. NOTATION AND CONSTANTS
+================================================================
+
+A.1 SETS AND SPACES
+
+ D the open unit disk, {|z| < 1}.
+ H the upper half-plane, {Im z > 0}.
+ C the complex plane.
+ S univalent f : D -> C with f(0)=0, f'(0)=1.
+ S_b the members of S with bounded image.
+ K univalent F : H -> C with F(i)=0, F'(i)=1.
+ m(z) (z-i)/(z+i), a bijection D -> H; m^{-1}(w) = i(1+w)/(1-w).
+ S_0 {z in H : |m(z)| >= 1/2}; its complement is bounded.
+ B(z,s) the open Euclidean disk of centre z and radius s.
+
+A.2 QUANTITIES ATTACHED TO A MAP
+
+ Q_F 1/F'.
+ q_F iy Q_F'/Q_F = -iy F''/F' = a_F + i b_F.
+ T_zF the rebasing, (F(z*w)-F(z))/((Im z)F'(z)).
+ V_xi^F the potential, y^k/|F(z)-xi|.
+ ell log V_xi^F.
+ G_F the critical map, F - (iy/k)F'.
+ J_F det DG_F / |F'|^2, the signed Jacobian.
+ omega_F y^{beta-1}|Q_F|.
+
+A.3 THE AFFINE GROUP
+
+ z*w (x+yu) + i(yv) for z=x+iy, w=u+iv.
+ z# (-x+i)/y, the inverse of z.
+ i the identity element of the group.
+ (G1)-(G3) associativity and inverses; Im(z*w)=(Im z)(Im w); the
+ height-one elements form a copy of R, the purely imaginary
+ elements a copy of (0,infinity).
+
+A.4 OPERATORS AND GROWTH
+
+ A_z the positive operator (A_z Phi)(F) = |Q_F(z)| Phi(T_zF).
+ L (1/2)(A_{-1/2+i/2} + A_{1/2+i/2}).
+ v 2^{-n}, the dyadic spacing at level n.
+ u_j -1 + (2j+1)v, the dyadic abscissae.
+ rho lim ||L^n||^{1/n}, equal to the exponential growth rate of L.
+ beta log_2 rho, so rho = 2^beta; 0 <= beta <= 1.
+ k (beta+2)/3, so 3(k-1) = beta-1 and 3/4 <= k <= 1.
+
+A.5 THE LAW AND ITS MOMENTS
+
+ P the probability law on K constructed in PROPOSITION 2.7.
+ E expectation under P.
+ (2.14) E[|Q_F(x+iy)| Phi(T_{x+iy}F)] = y^{-beta} E[Phi(F)].
+ (2.20) E|Q_F(x+iy)| = y^{-beta}.
+ (2.19) E a_F(i) = -beta, E|q_F(i)|^2 = beta(beta+1).
+
+A.6 GOODNESS AND PAIRING
+
+ (g1) all closed positive superlevels of V are compact.
+ (g2) the target is a regular value of G_F.
+ (g3) distinct preimages of the target have distinct V-values.
+ D {(F,z) : J_F(z) > 0, (F,G_F(z)) good}.
+ D_N the part of D where 1/N <= y'/y <= N and |x'-x| <= Ny.
+ R_F the rank pairing of saddles with peaks (LEMMA 4.7).
+ c_F (V(R_F(z))/V(z))^3 on D, 0 off D; c_F > 1 on D.
+ B, B' the two boxes {|x|<=S, 1<=y<=Y} and {|x|<=S+NY, 1/N<=y<=NY}.
+
+A.7 CONSTANTS
+
+ C the constant of (1.1): |Q_F(x+iy)| <= C/y on |x|<=2, 0<y<=1.
+ c 1/4, the constant of (1.2): |F(z)| >= 1/4 when |m(z)|>=1/2.
+ C_0 the constant of PROPOSITION 1.8, from the four-arc cover.
+ C_1 the bound for |Q_G| on the fixed compact set [-1,1]+i[1/2,4].
+ C_delta (2^k/delta)(1+4C_1), the target-disk radius in LEMMA 3.4.
+ Y_psi the height bound for the support of (Delta psi) o F.
+ C_b the constant in ||L^n|| <= C_b 2^{bn}, valid for b > beta.
+ eps 1/4 - b, the gap in the main theorem.
+ C C_0 C_b, the constant in the main theorem (unfortunately the
+ same letter as in (1.1); the two occurrences are in different
+ parts and each is defined where it appears).
+
+A.8 WHERE TO LOOK FOR WHAT
+
+ The transfer from the disk to the half-plane: 1.1, FACT 1.1.
+ Compactness of K and the two uniform bounds: LEMMA 1.3.
+ The rebasing group and its laws: 1.2, LEMMA 1.4.
+ The operator L, its dyadic form, its growth rate: 1.3, LEMMAS
+ 1.5-1.7.
+ The disk means are bounded by ||L^n||: PROPOSITION 1.8.
+ The eigenmeasure at the exact rate rho: LEMMA 2.1.
+ Horizontal invariance and exact scaling: LEMMAS 2.2-2.6.
+ The two moment identities: LEMMA 2.8.
+ The weighted area tail: LEMMA 3.3.
+ Decay at infinity and horizontal length: LEMMA 3.4.
+ Compact positive superlevels: PROPOSITION 3.5.
+ The critical map, J_F, and the Hessian: LEMMAS 4.1-4.4.
+ Genericity of good targets: LEMMA 4.5.
+ Peaks at least as many as saddles: LEMMA 4.6.
+ The measurable rank pairing: LEMMA 4.7.
+ Change of variables through G_F: PROPOSITION 5.1.
+ Goodness at the fixed base point: LEMMA 5.2.
+ E J_F(i) <= 0, and the equality case: PROPOSITION 5.3.
+ No map has J_F identically zero: LEMMA 6.1.
+ The theorem: 6.2.
+
+ END