A STRICT INVERSE-FIRST-POWER BOUND FOR UNIVALENT FUNCTIONS
a complete, self-contained account
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.
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