1e1ae8f c674361
STRICT HOT SPOTS ON SMOOTH SIMPLY CONNECTED PLANAR DOMAINS
A complete proof that every first Neumann eigenfunction has a nonzero
gradient inside the domain
This is a self-contained account of the theorem named in the title. It
is meant to be read line by line by someone who knows multivariable
calculus, linear algebra, and a little complex analysis.
Nothing is quoted except the six classical theorems listed in 1.6, each
stated there in the exact form in which it is used. Every other step is
proved, including every estimate on the Green kernel.
There are no pictures to decode; geometry is described in words.
CONTENTS
0. The claim, in plain words
1. Setting, notation, the quoted toolkit, and a map of the proof
2. The eigenvalue gap: mu < lambda_D
3. The conformal change of variables
4. The boundary weight beta
5. Sign sets of an eigenfunction are connected
6. Multiplicity at most two
7. A variational principle for tangent vector fields
8. The curvature identity
9. Disk kernels for a subcritical potential
10. The boundary energy and the multiplier identity
11. Reciprocal matrices and property (P)
12. Regularizing the Green kernel
13. Rank-one updates
14. The interior resolvent
15. The boundary limit
16. From conditional negativity to a Hilbert space
17. No interior critical point
A. The six quoted theorems, stated exactly
B. Where every hypothesis is used
C. References
0. THE CLAIM, IN PLAIN WORDS
Put a metal plate on a table and insulate its edge. Heat it unevenly and
watch. The heat sloshes about, then settles to a uniform temperature. The
interesting moment is the tail of the sloshing, when a single pattern of
hot and cold regions survives and decays at its own slow rate. That
pattern is a "first Neumann eigenfunction" u of the plate, and its hottest
and coldest points are the "hot spots".
Rauch asked in the mid 1970s whether the hot spots must lie on the
insulated edge. For a plate with no holes and a smooth edge the answer is
yes, and more: the temperature gradient can never vanish inside the
plate, so there is not even an interior saddle point. That sharper
statement is the theorem below.
Simple connectivity is essential. Plates with one or two holes can have
their hot spots strictly inside; see section 1.7 for the history.
1. SETTING, NOTATION, THE QUOTED TOOLKIT, A MAP
1.1 THE DOMAIN. Throughout, Omega is a nonempty bounded open subset of
the plane R^2, connected and simply connected (no holes), with
C-infinity boundary dOmega. No convexity and no symmetry is assumed.
Since Omega is bounded and smooth, the divergence theorem, the Poincare
inequality, and elliptic regularity all hold on it.
1.2 THE EIGENVALUE. For a function v let grad v be its gradient, let
Delta v = d_11 v + d_22 v be its Laplacian, and let nu be the outward
unit normal of dOmega. Define
mu = min { int_Omega |grad v|^2 / int_Omega v^2 :
0 != v in H^1(Omega), int_Omega v = 0 }.
The minimum is attained (1.6.1), and its minimizers are exactly the
nonzero solutions of
-Delta u = mu u in Omega, d_nu u = 0 on dOmega, int_Omega u = 0.
Let V be the real vector space of all such solutions. We count only
positive eigenvalues; books that count the constant function first call
this eigenvalue mu_2.
Why this is the slowest mode: solving the heat equation d_t w = Delta w
with d_nu w = 0 on dOmega gives a sum of terms c_k exp(-nu_k t) phi_k
over the Neumann eigenvalues 0 = nu_0 < nu_1 <= nu_2 <= ..., so the mode
with the smallest positive eigenvalue mu dominates for large time.
1.3 THE THEOREM.
THEOREM. Let Omega be as in 1.1 and let 0 != u in V. Then
grad u(x) != 0 for every x in Omega.
Hence for every x in Omega
min_{dOmega} u < u(x) < max_{dOmega} u,
so the hottest and coldest points of u lie on dOmega and only there.
The second sentence follows from the first: u is continuous on the
compact set cl(Omega), so both extreme values are attained; an interior
extreme value would be a critical point; and an interior point where u
equals a boundary extreme value is itself a global extreme value, hence
also a critical point.
1.4 NOTATION, ALL OF IT.
D = {z in C : |z| < 1} the unit disk, S1 = dD the unit circle
ds arclength on S1 or on dOmega
dA area
Phi : D -> Omega the conformal map of 3.1
a, q a(z) = mu |Phi'(z)|^2, dq(z) = a(z) dA(z)
d the constant mu/lambda_D < 1 of 2.2
lambda_D the first Dirichlet eigenvalue of Omega
G, P_s the Green function and Poisson kernel of D
T, A, W_h Green operator, harmonic extension, tilted
extension; all defined in 9.2
K, R, N, N_0 boundary kernels; defined in 9.4
Q(X) int (|div X|^2 + |curl X|^2), see 7.1
E(h) the reduced energy, see 10.1
tau, nu, kappa tangent, normal, curvature; see 8.1
beta the boundary weight of section 4
(P) "at most one positive eigenvalue", 11.1
D_p the kernel K(p,s)K(p,t)/N(s,t), see 11.3
H^1(Omega) is the space of functions with one square-integrable weak
derivative; H^1_0(Omega) is the subspace of functions with zero boundary
trace. "int" means int_Omega unless a subscript says otherwise. X . Y is
the dot product of vectors, X : Y the entrywise product of matrices,
|X| = sqrt(X . X).
1.5 WEAK SOLUTIONS. A function w in H^1 is a weak solution of
-Delta w = f with d_nu w = 0 when int grad w . grad phi = int f phi for
every phi in H^1(Omega). That is the only meaning of "solution" we need.
By 1.6.2, weak solutions with smooth data are smooth, so nothing is lost.
1.6 THE SIX QUOTED THEOREMS. These are the only outside facts. Each is
stated in exactly the form used, with the place of use.
(1.6.1) SPECTRAL THEOREM FOR THE NEUMANN LAPLACIAN. On a bounded
Lipschitz domain the operator -Delta with Neumann condition has discrete
spectrum 0 = nu_0 < nu_1 <= nu_2 <= ..., every eigenspace is finite
dimensional, and nu_1 is the minimum in 1.2. Used everywhere: mu is
attained, V is finite dimensional.
(1.6.2) ELLIPTIC REGULARITY. (a) If f is in L^2(Omega) then -Delta v = f
with v = 0 on dOmega, and -Delta v = f with d_nu v = 0 on dOmega and
int v = 0 (this one needs int f = 0), each have a unique solution in
H^2(Omega). (b) If f is C-infinity then the solutions are C-infinity on
cl(Omega). (c) A solution of Delta w + mu w = 0 in an open set is real
analytic there. Used in 2.1, 4.1, 7.2, 9.2, 10.1, 17.6. Reference:
Showalter, Hilbert Space Methods for Partial Differential Equations,
Chapter III, Theorems 6.1, 6.4, 6.5.
(1.6.3) RIEMANN MAPPING WITH KELLOGG REGULARITY. For Omega as in 1.1
there is a biholomorphic Phi : D -> Omega that extends to a C-infinity
diffeomorphism of the closed disks, with Phi'(z) != 0 for |z| <= 1.
Used from section 3 on. Reference: Pommerenke, Boundary Behaviour of
Conformal Maps, Chapter 3.
(1.6.4) HOPF'S BOUNDARY POINT LEMMA. Let w be C^2 on cl(Omega) with
-Delta w > 0 in Omega and w >= 0 in Omega. If w(x_0) = 0 at a point x_0
of dOmega, then d_nu w(x_0) < 0. Used once, in 4.1.
(1.6.5) CROSSCUT FORM OF THE JORDAN CURVE THEOREM. Let gamma be a simple
arc in cl(D) whose two endpoints a, c lie on S1 and whose interior lies
in D. Then D minus gamma has two components; the two open arcs of S1
from a to c lie in different components; and every arc in cl(D) that
joins a point of one open arc to a point of the other and meets S1 only
at its endpoints must intersect gamma. Used once, in 5.2.
(1.6.6) POINCARE INEQUALITY. There is a constant C with
int_Omega |v - avg v|^2 <= C int_Omega |grad v|^2 for all v in
H^1(Omega), where avg v is the mean value. Used in 2.1 and 7.2.
That is the whole list. In particular nothing about reciprocal kernels is
quoted: sections 11 to 15 build exactly what is needed, from scratch.
1.7 A SHORT HISTORY, so the reader knows why the theorem is news.
Burdzy and Werner (1999) found a plate with two holes whose hot spots are
not on the edge; Bass and Burdzy (2000), and then Burdzy (2005) with only
one hole, found examples where both extremes are strictly interior. So
simple connectivity is indispensable. In the plane, Nadirashvili (1988)
proved dim V <= 2, which is step 6 below. Symmetric and convex cases were
settled by Jerison and Nadirashvili (2000) and, by probabilistic
coupling, by Banuelos and Burdzy (1999) and Pascu (2002); lip domains by
Atar and Burdzy (2004); triangles by Judge and Mondal (2020, with a 2022
erratum). Rohleder (2021, 2024) introduced the vector-field variational
principle used in section 7. What was missing was a supply of boundary
multipliers on a domain with no symmetry at all; sections 11 to 15 supply
them. References are in appendix C.
1.8 MAP OF THE PROOF. Six steps, in order.
Preparation. 2: mu < lambda_D. 3: pull everything back to the disk by
the conformal map. 4: every eigenfunction is orthogonal to a positive
boundary weight. 5: its positive and negative regions are single
connected blobs. 6: dim V <= 2.
The energy. 7: for any tangent vector field X,
int (|div X|^2 + |curl X|^2) >= mu int |X|^2, with equality exactly
for X = grad v, v in V. 8: that energy equals int |grad X|^2 plus a
boundary curvature term.
The disk kernels. 9: the Helmholtz equation on the disk with potential
a = mu |Phi'|^2 gives positive kernels K, R, N and a formula for the
energy of a boundary field in terms of them. 10: multiplying the
boundary gradient g by a scalar b changes the energy by a double
integral with a factor (b(s)-b(t))^2.
The hard analysis. 11 to 15: the kernel
D_p(s,t) = K(p,s)K(p,t)/N(s,t) is conditionally negative semidefinite.
This is where the reciprocals of Green-kernel matrices enter.
The contradiction. 16: such a kernel is a squared distance in a
Hilbert space, so there are multipliers b_j with the differences
|b_j(s)-b_j(t)| summing to D_p(s,t). 17: if p is a critical point the
energies of all b_j g vanish, so each b_j g is the gradient trace of
an eigenfunction, and there are too many of them (case 1) or one of
them extends by zero across the boundary and dies by analyticity
(case 2).
Read 2 to 10 first; they are elementary. Sections 11 to 15 are the engine
and can be skipped on a first pass without losing the logic.
2. THE EIGENVALUE GAP: mu < lambda_D
2.1 THE DIRICHLET EIGENVALUE. Define
lambda_D = min { int_Omega |grad v|^2 / int_Omega v^2 :
0 != v in H^1_0(Omega) }.
By (1.6.6) it is positive, it is attained, and its minimizers are the
nonzero solutions of -Delta phi = lambda_D phi with phi = 0 on dOmega.
Replacing a minimizer by its absolute value keeps it a minimizer, so we
may take phi real and nonnegative; then int_Omega phi > 0. By (1.6.2)
phi is C-infinity on cl(Omega).
FIRST-VARIATION FACT, used here and several times later: if 0 != w in
H^1(Omega) has mean zero and int |grad w|^2 = mu int w^2, then
int grad w . grad psi = mu int w psi for every psi in H^1(Omega).
Reason: w + t psi has mean zero for every real t, so the function
t -> int |grad(w+t psi)|^2 / int (w+t psi)^2 has a minimum at t = 0 and
derivative zero there; a one-line computation of that derivative shows
it vanishes exactly when the displayed identity holds.
2.2 LEMMA (Filonov). mu < lambda_D.
Proof. STEP 1: mu <= lambda_D. Fix k in R^2 with |k|^2 = lambda_D and
put e_k(x) = exp(i k . x), so that -Delta e_k = lambda_D e_k. Define
w_k = e_k - c_k phi, c_k = (int_Omega e_k) / (int_Omega phi),
which is legitimate because int_Omega phi > 0. Then w_k has mean zero,
and w_k != 0: a multiple of phi would have to vanish on dOmega, but e_k
never vanishes. Let
B(u,v) = int_Omega (grad u . grad conj v - lambda_D u conj v)
be the Hermitian form whose diagonal is int (|grad v|^2 - lambda_D |v|^2).
Then B(e_k,e_k) = 0 and B(phi,phi) = 0, and the cross term vanishes:
integrating by parts and using -Delta e_k = lambda_D e_k together with
phi = 0 on dOmega gives int grad e_k . grad conj phi = lambda_D int
e_k conj phi. Hence
B(w_k,w_k) = B(e_k,e_k) - 2 Re(conj(c_k) B(e_k,phi)) + |c_k|^2 B(phi,phi)
= 0,
that is, int |grad w_k|^2 = lambda_D int |w_k|^2. Since w_k is nonzero
with mean zero, mu <= lambda_D.
STEP 2: THE INEQUALITY IS STRICT. Suppose mu = lambda_D. Then each w_k
has mean zero and Rayleigh quotient mu, so by the first-variation fact
it belongs to V. The w_k are linearly independent. Indeed, if a finite
combination sum a_k w_k vanished, then sum a_k e_k = c phi for some
constant c; the left side is a finite sum of plane waves, hence real
analytic in the whole plane, and it vanishes on dOmega because phi does.
A real analytic function vanishing on a nonempty open arc of dOmega
vanishes identically, so sum a_k e_k = 0, and distinct plane waves are
linearly independent; hence all a_k = 0. The set of k with
|k|^2 = lambda_D is a circle in R^2, so there are infinitely many such
k and V is infinite dimensional, contradicting (1.6.1). Hence
mu < lambda_D.
3. THE CONFORMAL CHANGE OF VARIABLES
3.1 THE MAP. By (1.6.3) there is a biholomorphic Phi : D -> Omega, of
class C-infinity up to the boundary, with Phi' != 0 on cl(D). Think of
D as the reference plate and Omega as the real plate: Phi carries one to
the other without distorting angles.
3.2 WHY WE DO THIS. On the disk the Green function and the Poisson
kernel are explicit; on Omega they are not. The price is that the
eigenfunction equation acquires the weight a = mu |Phi'|^2. The only
property of that weight we will need is the inequality of 3.5, which
says the weight is subcritical, that is, small enough that it never
reaches the Dirichlet threshold.
3.3 DIRICHLET ENERGY IS CONFORMALLY INVARIANT. For C^1 functions f,
int_Omega |grad f|^2 dA = int_D |grad (f o Phi)|^2 dA,
and the same holds component by component for vector-valued f. Reason:
Phi is conformal, so its Jacobian at z is a rotation followed by a
dilation of factor |Phi'(z)|, and the chain rule gives
grad(f o Phi)(z) = rotation x |Phi'(z)| x (grad f)(Phi(z)). The rotation
preserves lengths and areas of the images of small vector pairs scale by
|Phi'(z)|^2, so the two factors cancel. The identity extends to H^1
functions by approximation.
3.4 AREA AND THE MEASURE q. Put a(z) = mu |Phi'(z)|^2 and
dq(z) = a(z) dA(z). The density a is bounded, positive, and continuous
on cl(D). For a function f on Omega set psi = f o Phi. The change of
variables formula int_Omega g dA = int_D g(Phi(z)) |Phi'(z)|^2 dA, applied
to g = |f|^2, says
int_Omega |f|^2 dA = int_D |psi|^2 |Phi'|^2 dA = (1/mu) int_D |psi|^2 dq.
So we may write, for any function psi on D,
mu int_Omega |psi o Phi^{-1}|^2 dA = int_D |psi|^2 dq = ||psi||^2_q.
3.5 LEMMA (subcritical weight). For every psi in H^1_0(D),
||psi||^2_q <= d int_D |grad psi|^2, where d = mu/lambda_D < 1.
Proof. By 3.4, ||psi||^2_q = mu int_Omega |psi o Phi^{-1}|^2 dA. The
function psi o Phi^{-1} lies in H^1_0(Omega), so (1.6.6) bounds its
L^2 norm by (1/lambda_D) int_Omega |grad(psi o Phi^{-1})|^2, and by 3.3
the latter is (1/lambda_D) int_D |grad psi|^2. Multiply by mu and use
2.2.
3.6 WHAT HAPPENS TO THE EIGENFUNCTION. Let 0 != u in V and let
W = (grad u) o Phi, a vector-valued C-infinity function on cl(D) with
components W_1, W_2. Pulling the Laplacian back through a conformal map
multiplies it by |Phi'|^2, and each component of grad u solves
-Delta_Omega (grad u)_i = mu (grad u)_i. Hence
-Delta_D W_i = a W_i in D, W_i = g_i on S1, g = (grad u) o Phi on S1.
The Neumann condition says that g is tangent to dOmega at the point
Phi(s), because 0 = d_nu u = (grad u) . nu. Two facts about g will be
used: it is a Lipschitz (indeed smooth) function of s, and W is the
unique solution of the displayed Dirichlet problem with boundary trace g,
by 9.2.
4. THE BOUNDARY WEIGHT beta
4.1 THE AUXILIARY FUNCTION L. By 2.2 and (1.6.2) the problem
(-Delta - mu) L = 1 in Omega, L = 0 on dOmega
has a unique solution, and it is C-infinity on cl(Omega). Also L >= 0.
Indeed, test the equation against the negative part L^-, which lies in
H^1_0(Omega): with -Delta L = 1 + mu L,
-int |grad L^-|^2 = int grad L . grad L^- = int (1 + mu L) L^-
= int L^- + mu int (L^-)^2 >= 0,
so L^- = 0. Therefore -Delta L = 1 + mu L > 0 in Omega, and Hopf's
lemma (1.6.4) gives d_nu L < 0 at every boundary point.
4.2 LEMMA (boundary weight). Put beta(s) = |Phi'(s)| (-d_nu L)(Phi(s)).
Then beta is C-infinity and positive on S1, and for every v in V with
trace h_v = v o Phi,
int_S1 beta(s) h_v(s) ds = 0.
If v != 0 then h_v is not identically zero, and it takes both signs.
Proof. Positivity of beta is 4.1 together with |Phi'| > 0. For the
identity, use 0 = int_Omega v and the defining equation of L:
0 = int_Omega v = int_Omega v (-Delta L - mu L)
= int_Omega (-Delta v) L - int_dOmega v d_nu L - mu int_Omega v L
= mu int_Omega v L - int_dOmega v d_nu L - mu int_Omega v L.
The middle line is Green's identity, whose boundary term survives
because d_nu v = 0; the last line uses -Delta v = mu v. So
int_dOmega v d_nu L dS = 0. On the boundary dS = |Phi'(s)| ds and
v(Phi(s)) = h_v(s), hence
int_dOmega v d_nu L dS = int_S1 h_v (d_nu L)(Phi) |Phi'| ds
= -int_S1 beta h_v ds,
which gives the identity. If h_v = 0 then v lies in H^1_0(Omega), and
int |grad v|^2 = mu int v^2 contradicts the characterization of
lambda_D and 2.2 unless v = 0. So a nonzero v has a nonzero trace, and
a nonzero continuous function with zero integral against the positive
weight beta must take both signs.
4.3 COROLLARY (the trace map is injective). The map v -> h_v from V to
C(S1) is linear and injective, by the last paragraph of 4.2. Hence the
traces of V form a space of the same dimension as V.
5. THE TWO SIGN SETS OF AN EIGENFUNCTION
5.1 For a continuous function v on cl(Omega) write {v>0} for the open
subset {x in Omega : v(x) > 0} of Omega, and likewise {v<0}.
5.2 LEMMA. Let 0 != v in V and let h = h_v be its boundary trace.
(a) {v>0} and {v<0} are nonempty and connected.
(b) h has no four cyclically ordered points of S1 at which the signs
of h are +, -, +, -.
Proof of (a). Nonemptiness: by 4.2 the trace h takes both signs, and v
is continuous on cl(Omega), so v takes both signs in Omega.
Let U be a component of {v>0}. Then v 1_U lies in H^1(Omega). Reason: v
is C-infinity on cl(Omega) by (1.6.2), hence Lipschitz; v 1_U agrees
with the Lipschitz function v on cl(U) and with the Lipschitz function 0
on cl(Omega minus U), and these agree on the common boundary, where
v = 0. Gluing gives a Lipschitz function with weak gradient 1_U grad v.
Testing the eigenfunction equation against v 1_U gives
int |grad(v 1_U)|^2 = int grad v . grad(v 1_U) = mu int (v 1_U)^2.
Now suppose there are three or more components in total. Then two of
them, U1 and U2, have the same sign; say v > 0 on both. Put
y = v 1_{U1} - c v 1_{U2}, c = (int v 1_{U1}) / (int v 1_{U2}) > 0.
Then y != 0, y has mean zero, and since the supports of the two pieces
are disjoint, int |grad y|^2 = mu int y^2. By the first-variation fact
of 2.1, y belongs to V. But y vanishes on the third component, which is
a nonempty open subset of Omega, contradicting the analyticity of
eigenfunctions (1.6.2c). Hence there are exactly two components, one of
each sign, and (a) follows.
Proof of (b). Suppose s1, s2, s3, s4 occur in this cyclic order on S1
with h(s1) > 0, h(s2) < 0, h(s3) > 0, h(s4) < 0. Near s1 and s3 there
are points x1, x3 of Omega with v > 0, because v is continuous up to
the boundary and v(s1) = h(s1) > 0. Since {v>0} is open and connected,
hence path connected, there is a polygonal path in {v>0} from x1 to x3;
append the short straight segments from s1 to x1 and from x3 to s3,
which lie in {v>0} except for their outer endpoints, and remove loops.
The result is a simple arc gamma_+ in cl(Omega) whose interior lies in
{v>0} and whose endpoints are s1 and s3. The same construction with the
negative set gives a simple arc gamma_- from s2 to s4 with interior in
{v<0}.
Apply Phi^{-1}. We get two simple arcs in cl(D), with interiors in D,
one joining s1 to s3 and one joining s2 to s4 (we keep the same
letters). By the crosscut theorem (1.6.5) the first arc separates the
two open arcs of S1 from s1 to s3, and s2 and s4 lie on different ones,
so the second arc must meet the first. But the first arc minus its
endpoints lies in {v>0} and the second minus its endpoints lies in
{v<0}; they are disjoint. Contradiction.
5.3 LEMMA (the positive set is one piece). Let h be a continuous
function on S1 that takes both signs and has no four cyclically ordered
points with alternating strict signs. Choose s0 = exp(i theta_0) with
h(s0) < 0 and identify S1 minus {s0} with the interval
(theta_0, theta_0 + 2 pi). Let
P = { theta in (theta_0, theta_0 + 2 pi) : h(exp(i theta)) > 0 }.
Then P is nonempty and open, and with a = inf P, b = sup P one has
theta_0 < a < b < theta_0 + 2 pi, P is contained in (a,b),
h >= 0 on (a,b), h <= 0 outside [a,b], h(a) = h(b) = 0.
Proof. P is open because h is continuous, and nonempty because h takes
both signs. It does not accumulate at s0, since h < 0 in a neighbourhood
of s0, so a > theta_0 and b < theta_0 + 2 pi. By definition of infimum
and supremum, P is contained in (a,b) and h <= 0 outside [a,b]. Also
a is not in P (P is open), so h(a) <= 0, while h(a) >= 0 as a limit of
values on P; hence h(a) = 0, and likewise h(b) = 0. Finally, if some
theta in (a,b) had h(theta) < 0, choose p in P with a < p < theta and
q in P with theta < q < b; then s0, p, theta, q are four cyclically
ordered points with signs -, +, -, +, which is forbidden. Hence h >= 0
on (a,b).
6. MULTIPLICITY AT MOST TWO
6.1 PROPOSITION (Nadirashvili). dim V <= 2.
Proof. Suppose dim V >= 3. By 4.3 the trace map is injective, so the
traces of V form a real vector space W of dimension at least three. The
two linear conditions
int_S1 beta h cos(theta) ds = 0, int_S1 beta h sin(theta) ds = 0,
(s = exp(i theta))
leave a kernel of dimension at least one, so there is v in V with
nonzero trace h satisfying them. By 4.2 we also have
int_S1 beta h ds = 0.
So h is orthogonal, against the positive weight beta, to the three
functions 1, cos(theta), sin(theta).
By 4.2 choose s0 = exp(i theta_0) with h(s0) < 0, cut the circle there,
and let P, a, b be as in 5.3. Define
F(theta) = cos(theta - m) - cos(delta),
m = (a+b)/2, delta = (b-a)/2.
Since theta - m ranges over (-delta, delta) when theta is in (a,b), and
cos is larger there than at the endpoints, F > 0 on (a,b). When theta is
outside [a,b] but inside the cut interval, |theta - m| exceeds delta and
is less than 2 pi - delta, so F < 0 there. Also 0 <= delta < pi because
a < b < a + 2 pi.
Now h >= 0 on (a,b) and h <= 0 outside [a,b], by 5.3. So F h >= 0 on
the whole circle, and F h > 0 on the nonempty open set P. As beta > 0,
int_S1 beta F h ds > 0.
On the other hand F = cos(m) cos(theta) + sin(m) sin(theta) - cos(delta),
so the three orthogonality relations give
int_S1 beta F h ds = cos(m) 0 + sin(m) 0 - cos(delta) 0 = 0.
Contradiction. Hence dim V <= 2.
6.2 COROLLARY. Let 0 != v in V. Its trace h_v takes both signs. With s0,
a, b as in 5.3, the set where h_v > 0 is the open set P contained in the
single arc (a,b), and h_v <= 0 outside [a,b].
7. A VARIATIONAL PRINCIPLE FOR TANGENT VECTOR FIELDS
7.1 VECTOR FIELDS WITH TANGENT TRACE. Let X = (X_1, X_2) be a vector
field on Omega whose two components lie in H^1(Omega). Write
div X = d_1 X_1 + d_2 X_2, curl X = d_1 X_2 - d_2 X_1,
Q(X) = int_Omega (|div X|^2 + |curl X|^2).
Call X tangent if X(x) . nu(x) = 0 for every x on dOmega. In the weak
language of 1.5 this says that the boundary trace of X . nu is zero,
which makes sense for an H^1 field.
7.2 PROPOSITION (Rohleder). For every tangent X in H^1(Omega; R^2),
Q(X) >= mu int_Omega |X|^2,
with equality if and only if X = grad v for some v in V.
Proof. STEP 1: A DECOMPOSITION. Solve the two scalar problems
Delta v = div X in Omega, d_nu v = 0 on dOmega, int_Omega v = 0,
Delta j = curl X in Omega, j = 0 on dOmega.
The Neumann problem is compatible: int_Omega div X = int_dOmega X . nu dS
= 0 because X is tangent. Both solutions exist and lie in H^2(Omega) by
(1.6.2). Put
Z = X - grad v - grad^perp j, grad^perp j = (-d_2 j, d_1 j).
Then div grad^perp j = -d_1 d_2 j + d_2 d_1 j = 0 and
curl grad^perp j = d_1(d_1 j) - d_2(-d_2 j) = Delta j, so
div Z = div X - Delta v - 0 = 0, curl Z = curl X - 0 - Delta j = 0.
Also Z is tangent: X is tangent, grad v has normal trace d_nu v = 0, and
grad^perp j has normal trace
(grad^perp j) . nu = -d_2 j nu_1 + d_1 j nu_2 = -d_tau j = 0,
because j vanishes on dOmega.
STEP 2: Z = 0. Since div Z = 0 and curl Z = 0, each component of Z is
harmonic in the interior: for instance,
Delta Z_1 = d_1 d_1 Z_1 + d_2 d_2 Z_1 = d_1(-d_2 Z_2) + d_2(d_1 Z_2) = 0.
So Z is smooth in Omega. Omega is simply connected, so the closed 1-form
Z_1 dx_1 + Z_2 dx_2 is exact: there is a smooth r on Omega with
grad r = Z. For M > 0 put r_M = max(-M, min(r, M)). Then r_M lies in
H^1(Omega), its gradient is 1_{|r|<M} Z, and its normal trace is zero,
since (grad r_M) . nu = 1_{|r|<M} (Z . nu) = 0.
Apply the divergence theorem to the field r_M Z, whose normal component
r_M (Z . nu) vanishes on dOmega:
0 = int_dOmega r_M Z . nu dS = int_Omega div(r_M Z)
= int_Omega (grad r_M . Z + r_M div Z) = int_Omega 1_{|r|<M} |Z|^2.
The right side increases to int_Omega |Z|^2 as M grows, so by monotone
convergence int_Omega |Z|^2 = 0 and Z = 0. Hence
X = grad v + grad^perp j.
STEP 3: ORTHOGONALITY AND THE SPLIT OF Q. The two summands are
orthogonal in L^2: since grad^perp j is divergence-free and tangent,
int grad v . grad^perp j = -int v div grad^perp j + int_dOmega v (Z . nu)
= 0 + 0.
And div X = Delta v, curl X = Delta j, so
Q(X) = ||Delta v||^2 + ||Delta j||^2.
STEP 4: TWO POINCARE BOUNDS. Since v has mean zero, the definition of mu
in 1.2 bounds its L^2 norm: ||v|| <= mu^{-1/2} ||grad v||. Therefore
int |grad v|^2 = -int v Delta v <= ||v|| ||Delta v||
<= mu^{-1/2} ||grad v|| ||Delta v||,
so ||Delta v||^2 >= mu ||grad v||^2. Since j vanishes on dOmega, the
same argument with the definition of lambda_D in 2.1 gives
||Delta j||^2 >= lambda_D ||grad j||^2.
STEP 5: CONCLUSION. By step 3 and orthogonality of the summands,
Q(X) = ||Delta v||^2 + ||Delta j||^2
>= mu ||grad v||^2 + lambda_D ||grad j||^2
= mu ||X||^2 + (lambda_D - mu) ||grad j||^2
>= mu ||X||^2,
and the last inequality is 2.2. Equality forces ||grad j||^2 = 0, so
j = 0, and equality in the first bound, which by the first-variation
fact of 2.1 says v lies in V. Conversely, if X = grad v with v in V,
then div X = Delta v = -mu v and curl X = 0, so
Q(X) = mu^2 int v^2 = mu int |grad v|^2 = mu int |X|^2.
That proves the proposition.
8. THE CURVATURE IDENTITY
8.1 ORIENTATION AND CURVATURE. Traverse dOmega counterclockwise, and let
x(s) be the arclength parametrization. The unit tangent is
tau(s) = x'(s), and the outward unit normal is nu = (tau_2, -tau_1),
which is tau rotated clockwise by a right angle. The signed curvature is
kappa = det(tau, d_tau tau) = tau_1 (d_tau tau)_2 - tau_2 (d_tau tau)_1,
where d_tau is the derivative with respect to arclength along dOmega.
For a circle of radius R traversed counterclockwise, kappa = 1/R > 0; on
a part of dOmega that bulges outward kappa is positive, on an inward
dip it is negative.
8.2 LEMMA (curvature identity). For every tangent X in H^1(Omega; R^2),
Q(X) = int_Omega |grad X|^2 dA + int_dOmega kappa |X|^2 dS.
Proof. STEP 1: ALGEBRA. For any smooth field, with
|grad X|^2 = sum_{i,j} (d_j X_i)^2 and
det(grad X) = d_1 X_1 d_2 X_2 - d_2 X_1 d_1 X_2,
|div X|^2 + |curl X|^2 = |grad X|^2 + 2 det(grad X).
Indeed, expanding (X_1,1 + X_2,2)^2 + (X_1,2 - X_2,1)^2 and subtracting
X_1,1^2 + X_1,2^2 + X_2,1^2 + X_2,2^2 leaves 2 X_1,1 X_2,2 - 2 X_1,2 X_2,1,
which is 2 det(grad X).
STEP 2: DIVERGENCE FORM. A direct computation of mixed partials gives
2 det(grad X) = div F, F = (2 X_1 d_2 X_2, -2 X_1 d_1 X_2).
Therefore, by the divergence theorem,
int_Omega 2 det(grad X) dA = int_dOmega F . nu dS
= 2 int_dOmega X_1 (d_2 X_2 nu_1 - d_1 X_2 nu_2) dS.
STEP 3: TANGENTIAL DERIVATIVE. Since tau = (-nu_2, nu_1),
d_tau X_2 = tau . grad X_2 = -nu_2 d_1 X_2 + nu_1 d_2 X_2
= d_2 X_2 nu_1 - d_1 X_2 nu_2.
So the boundary term of step 2 is 2 int_dOmega X_1 d_tau X_2. The integral
of the tangential derivative d_tau(X_1 X_2) over the closed curve dOmega
is zero, so
2 int_dOmega X_1 d_tau X_2 = int_dOmega (X_1 d_tau X_2 - X_2 d_tau X_1) dS.
STEP 4: TANGENT TRACE. Write X = a tau on dOmega, with a real. Since
d_tau X = (d_tau a) tau + a d_tau tau,
X_1 d_tau X_2 - X_2 d_tau X_1 = a^2 (tau_1 (d_tau tau)_2 - tau_2 (d_tau tau)_1)
= a^2 kappa = kappa |X|^2.
So int_Omega 2 det(grad X) dA = int_dOmega kappa |X|^2 dS, and step 1
gives the identity for smooth fields. Both sides are continuous in the
H^1 topology (the boundary term through the trace map into L^2(dOmega)),
so the identity extends to H^1 fields by approximation.
8.3 A CHECK. On the unit disk take X = tau = (-x_2, x_1), the
counterclockwise unit tangent. Then div X = 0 and curl X = 2, so
Q(X) = 4 pi. On the other side |grad X|^2 = 2 so int_Omega |grad X|^2 =
2 pi, and kappa = 1, |X| = 1, so int_dOmega kappa |X|^2 = 2 pi. The two
sides agree.
9. DISK KERNELS FOR A SUBCRITICAL POTENTIAL
9.1 GREEN FUNCTION AND POISSON KERNEL OF THE DISK. Use complex notation
for points of D and S1. Put
G(x,y) = (1/(2 pi)) log | (1 - x conj(y)) / (x - y) |
= (1/(4 pi)) log (1 + (1-|x|^2)(1-|y|^2)/|x-y|^2),
P_s(x) = (1 - |x|^2) / (2 pi |s - x|^2).
The two formulas for G agree because
|1 - x conj(y)|^2 = |x - y|^2 + (1-|x|^2)(1-|y|^2). Facts we use:
(i) G >= 0 on D x D, G(x,x) = +infinity, G(x,.) = 0 on S1 for fixed x,
and -Delta_x G(x,.) = delta_x, the Dirac mass at x. So G is the Dirichlet
Green function of the disk.
(ii) Consequently, for f in L^2(D), the function
psi(x) = int_D G(x,y) f(y) dA(y) is the unique H^1_0 solution of
-Delta psi = f in D.
(iii) int_S1 P_s(x) ds = 1 for every x, and x -> int_S1 P_s(x) h(s) ds
is the harmonic function in D with boundary values h (Poisson formula).
Also int_D P_s(x) dA(x) = 1 for every s.
(iv) The elementary bounds, for x, y in D and s in S1:
P_s(y) <= 1/(pi |y - s|), 0 <= G(x,y) <= (1/(2 pi)) log (2/|x-y|).
First bound: 1 - |y|^2 = (1-|y|)(1+|y|) <= 2 |s-y| because
1 - |y| = | |s| - |y| | <= |s - y|. Second: |1 - x conj(y)| <= 2 and
G = (1/(2 pi)) log |1 - x conj(y)| - (1/(2 pi)) log |x - y|.
9.2 THE GREEN OPERATOR AND THE TWO EXTENSIONS. Keep a = mu |Phi'|^2 and
dq = a dA from 3.4.
(a) For f in L^2(q) define Tf(x) = int_D G(x,y) f(y) dq(y).
(b) For a Lipschitz function h on S1 define
Ah(x) = int_S1 P_s(x) h(s) ds, the harmonic extension of h.
(c) For a Lipschitz function h on S1 define W_h to be the solution of
-Delta W_h = a W_h in D, W_h = h on S1.
For vector-valued h both constructions are componentwise.
9.3 LEMMA (the operator T and the tilted extension).
(a) T is bounded and self-adjoint on L^2(q) with ||T|| <= d < 1. The
same holds when q is replaced by any nonnegative measure rho whose
density is bounded by that of q.
(b) W_h exists, is unique, and equals (I-T)^{-1} A h. In particular
W_h = Ah + T(I-T)^{-1}Ah.
(c) u = W_h - Ah has zero trace and satisfies -Delta u = a W_h.
Proof. (a) Write dq = a dA. For f in L^2(q), the function a f lies in
L^2(dA) because int |a f|^2 dA <= ||a||_infinity int |f|^2 dq. Put
psi = Tf; by 9.1(ii) applied to the source a f, psi has zero trace and
-Delta psi = a f. Multiplying by psi and integrating,
int_D |grad psi|^2 = int_D psi f dq.
By the subcritical inequality 3.5, ||psi||^2_q <= d int_D |grad psi|^2.
Combining,
||psi||^2_q <= d int_D psi f dq <= d ||psi||_q ||f||_q,
so ||Tf||_q <= d ||f||_q. Symmetry is G(x,y) = G(y,x) and Fubini. For
rho <= q, the subcritical inequality for rho follows from the one for q,
and the same proof applies.
(b) Uniqueness: two solutions differ by u with zero trace and
-Delta u = a u, so ||u||^2_q = int_D |grad u|^2 and 3.5 give
||u||^2_q <= d ||u||^2_q, hence u = 0. Existence: let
u = sum_{n>=1} T^n(Ah), a convergent series in L^2(q) because
||T|| <= d < 1. Then u = T(Ah) + T(u); reading this as
-Delta u = a(Ah + u), u = 0 on S1, shows that W_h = Ah + u is a weak
solution of the required problem. Hence W_h = (I-T)^{-1}Ah.
(c) Ah is harmonic, so -Delta(W_h - Ah) = a W_h, and the trace of the
difference is h - h = 0.
9.4 THE KERNELS. For p in D and s, t in S1 put
K(p,s) = P_s(p) + sum_{n>=1} (T^n P_s)(p),
R(s,t) = sum_{n>=0} int_D P_s(y) (T^n P_t)(y) dq(y),
N(s,t) = N_0(s,t) + R(s,t), N_0(s,t) = 1/(pi |s - t|^2).
Here T^0 P_t = P_t, and for n >= 1 the symbol T^n P_t means
T^{n-1}(T P_t): one Green integration of P_t produces a bounded
function, after which the powers of the operator T of 9.2 apply. All
kernels are defined by the absolutely convergent series of 9.5.
9.5 LEMMA (kernel estimates and representations). Fix p in D.
(a) K(p,.) is positive, bounded, and continuous on S1.
(b) R is symmetric and nonnegative, and
iint_{S1 x S1} R(s,t) ds dt <= mu |Omega| / (1-d),
where |Omega| is the area of Omega.
(c) For Lipschitz h, k on S1,
W_h(p) = int_S1 K(p,s) h(s) ds,
<Ah, (I-T)^{-1}Ak>_q = iint R(s,t) h(s) . k(t) ds dt.
Proof. (a) By 9.1(iv), changing to polar coordinates around s,
sup_{s in S1} ||P_s||_{L^{3/2}(D)}
<= (1/pi) (2 pi int_0^2 r^{-1/2} dr)^{2/3} < infinity,
and sup_{x in D} ||G(x,.)||_{L^3(D)} < infinity. Put F_s = T P_s, which is
a bounded function of its argument. Holder's inequality with the bounded
density a gives, uniformly in s,
||F_s||_infinity <= ||a||_infinity ||G||_{L^3} ||P_s||_{L^{3/2}} <= C.
In particular F_s lies in L^2(q), with ||F_s||_q <= C' uniformly. For
n >= 2, self-adjointness of T and -Delta_y G(p,y) = delta_p give
(T^n P_s)(p) = (T^{n-1}F_s)(p) = <G(p,.), T^{n-2}F_s>_q,
so |(T^n P_s)(p)| <= ||G(p,.)||_q d^{n-2} ||F_s||_q <= C_p d^{n-2}. The
series for K therefore converges uniformly in s. Each term is continuous
in s: P_s depends continuously on s in L^{3/2}(D), by dominated
convergence with the dominating function 2/(pi|y-s|), and the operators
applied afterwards are bounded. Positivity is clear: P_s > 0, G > 0,
a > 0.
(b) For distinct s, t the product P_s P_t is integrable on D: near y = s
the factor P_t is bounded and |y - s|^{-1} is integrable in the plane;
symmetrically near y = t. The terms with n >= 1 are dominated by
||F_t||_infinity int_D P_s dq <= C ||a||_infinity, using int_D P_s dA = 1
from 9.1(iii). So R is nonnegative and finite. Symmetry: reverse a chain
of Green functions and use G(x,y) = G(y,x). Finally, by Tonelli and
int_S1 P_s ds = 1,
iint R ds dt = sum_{n>=0} <1, T^n 1>_q
<= sum_{n>=0} d^n ||1||^2_q = mu |Omega| / (1-d).
(c) By 9.3(b), W_h = sum_{n>=0} T^n Ah, and
(T^n Ah)(p) = int_S1 (T^n P_s)(p) h(s) ds. Summing over n gives
W_h(p) = int_S1 K(p,s) h(s) ds.
For the second identity, expand (I-T)^{-1} as a series:
<Ah, (I-T)^{-1}Ak>_q = sum_{n>=0} int_D (int_S1 P_s h ds)
(T^n int_S1 P_t k dt) dq.
By Tonelli (with |h|, |k| for integrability and then Fubini for signed
data), this is sum_{n>=0} iint h(s) . k(t) int_D P_s (T^n P_t) dq ds dt,
which is iint R(s,t) h(s) . k(t) ds dt.
10. THE BOUNDARY ENERGY AND THE MULTIPLIER IDENTITY
10.1 THE ENERGY OF A BOUNDARY FIELD. A vector function h on S1 is called
tangent if h(s) is tangent to dOmega at the point Phi(s), that is, if
h(s) . nu(Phi(s)) = 0 for every s. For tangent Lipschitz h, solve
-Delta W_h = a W_h in D with W_h = h on S1 by 9.3, and put
X_h = W_h o Phi^{-1} on Omega, E(h) = Q(X_h) - mu int_Omega |X_h|^2,
and let E(h,k) be the associated symmetric bilinear form. Two remarks:
(i) X_h solves the Helmholtz equation on Omega. Indeed, pulling the
Laplacian through the conformal map multiplies it by |Phi'|^2, so
-Delta_D W_h = a W_h becomes -Delta_Omega X_h = mu X_h.
(ii) By 7.2, E >= 0 on tangent fields, and
E(h) = 0 if and only if X_h = grad v for some v in V.
10.2 LEMMA (boundary formula for the energy). For tangent Lipschitz h, k,
E(h,k) = (1/2) iint N_0(s,t) (h(s)-h(t)) . (k(s)-k(t)) ds dt
- iint R(s,t) h(s) . k(t) ds dt
+ int_S1 |Phi'|(s) (kappa o Phi)(s) h(s) . k(s) ds.
Proof. Write c = |Phi'| (kappa o Phi). By the curvature identity 8.2,
conformal invariance 3.3, and 3.4,
E(h,k) = int_D grad W_h : grad W_k - <W_h, W_k>_q
+ int_S1 c (s) h(s) . k(s) ds.
Now write W_h = Ah + u_h and W_k = Ak + u_k, so that u_h and u_k have
zero trace and satisfy -Delta u_h = a W_h and -Delta u_k = a W_k by
9.3(c). Three integrations by parts:
- int_D grad Ah : grad u_k = -int_D u_k Delta Ah + int_S1 u_k d_nu Ah = 0,
because Ah is harmonic and u_k has zero trace;
- int_D grad u_h : grad Ak = 0, by the same argument;
- int_D grad u_h : grad u_k = -int_D u_h Delta u_k + int_S1 u_h d_nu u_k
= int_D u_h a W_k dA = <u_h, W_k>_q, because u_h has zero trace.
Therefore
int_D grad W_h : grad W_k = int_D grad Ah : grad Ak + <u_h, W_k>_q,
and, subtracting <W_h, W_k>_q = <Ah + u_h, W_k>_q,
E(h,k) = int_D grad Ah : grad Ak - <Ah, W_k>_q
+ int_S1 c h . k ds.
By 9.3(b) and 9.5(c), <Ah, W_k>_q = <Ah, (I-T)^{-1}Ak>_q
= iint R(s,t) h(s) . k(t) ds dt. By 10.3 below,
int_D grad Ah : grad Ak = (1/2) iint N_0 (h(s)-h(t)) . (k(s)-k(t)).
Substituting gives the formula.
10.3 LEMMA (the harmonic energy). For Lipschitz h, k on S1,
int_D grad Ah : grad Ak
= (1/2) iint N_0(s,t) (h(s)-h(t)) . (k(s)-k(t)) ds dt.
Proof. Both sides are Hermitian forms, and both are invariant under
rotations of the circle, because N_0(s,t) depends only on |s-t|. An
invariant Hermitian form is diagonal in the Fourier basis s^m: if B is
such a form and s -> s e^{i phi} is a rotation, then
B(s^m,s^n) = e^{i(m-n)phi} B(s^m,s^n) for every phi, forcing
B(s^m,s^n) = 0 when m != n. So it suffices to compare the two sides for
h = k = s^m with m >= 0; the case m < 0 is the complex conjugate.
Left side. The harmonic extension of s^m is z^m. For a holomorphic f,
|grad Re f|^2 + |grad Im f|^2 = |grad f|^2 = 2 |f'|^2, so
int_D (|grad Re z^m|^2 + |grad Im z^m|^2) dA
= 2 int_D m^2 |z|^{2m-2} dA = 2 m^2 . 2 pi / (2m) = 2 pi m.
Right side. Since s^m - t^m = (s-t) sum_{j=0}^{m-1} s^j t^{m-1-j},
|s^m - t^m|^2 / |s-t|^2 = |sum_j s^j t^{m-1-j}|^2
= sum_{j,l} s^{j-l} t^{l-j}.
Integrating over S1 x S1 keeps exactly the m terms with j = l and gives
m (2 pi)^2. Hence
(1/2) iint N_0 |s^m - t^m|^2 = (1/(2 pi)) m (2 pi)^2 = 2 pi m.
This matches the left side. Both sides are continuous in the H^{1/2}
norm on S1, and trigonometric polynomials are dense there, so the
identity holds for all Lipschitz h and k.
10.4 THE GRADIENT DATA. Fix 0 != u in V and put
g(s) = (grad u)(Phi(s)), s in S1.
By 3.6 the field g is tangent, and by uniqueness in 9.3, together with
the fact that each component of grad u solves the Helmholtz equation,
W_g = (grad u) o Phi, hence X_g = grad u.
By 7.2 and the Neumann equation int_Omega |grad u|^2 = mu int_Omega u^2,
Q(grad u) = int_Omega |Delta u|^2 + 0 = mu^2 int_Omega u^2
= mu int_Omega |grad u|^2,
so by (10.1)(ii),
E(g) = 0.
10.5 PROPOSITION (multiplier identity). For every real Lipschitz
function b on S1,
E(b g) = (1/2) iint N(s,t) (b(s)-b(t))^2 g(s) . g(t) ds dt.
Proof. Because E >= 0 and E(g) = 0, the bilinear form annihilates g:
for every admissible phi and every real t,
0 <= E(g + t phi) = t^2 E(phi) + 2 t E(g,phi), which is possible for all
t only if E(g,phi) = 0. In particular E(g, b^2 g) = 0.
Apply 10.2 three times, with c = |Phi'|(kappa o Phi):
E(bg,bg) = (1/2) iint N_0 (b(s)-b(t))^2 g(s) . g(t)
- iint R b(s) b(t) g(s) . g(t) + int c b^2 g . g,
E(g,b^2 g) = (1/2) iint N_0 (g(s)-g(t)) . (b(s)^2 g(s) - b(t)^2 g(t))
- iint R b(t)^2 g(s) . g(t) + int c b^2 g . g.
The curvature terms cancel. For vectors A, B and real scalars alpha,
beta one has
(alpha A - beta B)^2 - (A-B) . (alpha^2 A - beta^2 B)
= (alpha - beta)^2 A . B,
so the N_0 terms combine into (1/2) iint N_0 (b(s)-b(t))^2 g . g. For
the R terms, use symmetry of R in s and t: iint R b(t)^2 g . g
= iint R b(s)^2 g . g, hence
iint R [b(t)^2 - b(s) b(t)] g . g
= (1/2) iint R (b(s)-b(t))^2 g . g.
Adding the two contributions and using N = N_0 + R gives the claim.
Integrability holds because |b(s)-b(t)| <= Lip(b) |s-t| makes the N_0
integrand bounded, R is integrable by 9.5(b), and g is bounded.
11. RECIPROCAL MATRICES AND PROPERTY (P)
11.1 PROPERTY (P). A real symmetric matrix has property (P) if it has at
most one positive eigenvalue, counted with multiplicity. Equivalent
formulation, used constantly: M has (P) if and only if every
two-dimensional subspace contains a nonzero vector x with x^T M x <= 0.
(If some two-dimensional subspace had x^T M x > 0 for all nonzero x in
it, the restriction of M to that subspace would be positive definite and
would have two positive eigenvalues; by interlacing M would too.
Conversely, the span of two positive eigenvectors is such a subspace.)
Consequences, each immediate from the formulation or from Sylvester's
law of inertia:
(a) If D is a positive diagonal matrix and M has (P), then D M D has
(P).
(b) If M has (P) and N is positive semidefinite, then M - N has (P):
the quadratic form of M - N never exceeds that of M.
(c) A principal submatrix of a matrix with (P) has (P): extend test
vectors by zero.
(d) A limit of matrices of fixed size with (P) has (P): the
eigenvalues depend continuously on the entries.
Throughout, the reciprocal of a matrix means the entrywise reciprocal,
never the matrix inverse.
11.2 LEMMA (the bordered algebra). Let B be a real symmetric m x m matrix
with zero diagonal, and let ell be a vector in R^m with ell^T x = 0 for
at least one x. Suppose the bordered matrix
M = [ B ell ]
[ ell^T 0 ]
has (P). Then x^T B x <= 0 for every x with ell^T x = 0.
Proof. Suppose x^T B x > 0 for some such x. Since ell != 0, there is y
with ell^T y != 0; replacing y by y - (x^T B y)/(x^T B x) x keeps
ell^T y != 0 and makes x^T B y = 0. Choose a real c with
y^T B y + 2 c ell^T y > 0; this is possible for either sign of ell^T y.
Now u = (x,0) and w = (y,c) satisfy
u^T M u = x^T B x > 0, w^T M w = y^T B y + 2 c ell^T y > 0,
u^T M w = x^T B y + c ell^T x = 0.
So M is positive definite on the two-dimensional span of u and w,
contradicting (P).
11.3 THE TARGET THEOREM. Let dq = a dA on D with a bounded and
nonnegative, and suppose the subcritical inequality of 3.5 holds with
some constant d < 1. Let K, R, N be the kernels of 9.4. Then for every
p in D the kernel
D_p(s,t) = K(p,s) K(p,t) / N(s,t) for s != t, D_p(s,s) = 0
is conditionally negative semidefinite: for every finite family
s_1,...,s_m in S1 and real coefficients a_1,...,a_m with sum a_i = 0,
sum_{i,j} a_i a_j D_p(s_i,s_j) <= 0.
The theorem uses only the measure hypotheses; the conformal origin of q
is irrelevant. That generality is needed, because the proof passes
through cutoff measures 0 <= rho <= q, which do not come from a
conformal map.
11.4 THE PLAN, in the order of sections 12 to 15.
(i) 12: a bounded regularization C_eps of G, increasing to G, whose
entrywise reciprocal has (P) on every finite set.
(ii) 13: adding a rank-one update c M_iz M_zj along a column of the
current matrix preserves (P) of the reciprocal.
(iii) 14: a finite quadrature of the resolvent, then eps -> 0.
(iv) 15: a radial boundary limit, then rho -> q.
At the very end, 11.2 converts (P) of a bordered matrix into 11.3.
12. THE REGULARIZED GREEN KERNEL
12.1 THE FUNCTIONS h, L, f. For x, y in D put
h(x,y) = |x-y|^2 / ((1-|x|^2)(1-|y|^2)),
L(h) = 1 / log(1 + 1/h) for h > 0, L(0) = 0,
f(h) = h + 1/2 - L(h).
By 9.1, G = (1/(4 pi)) log(1 + 1/h), so
1/G = 4 pi L(h).
12.2 LEMMA. For every finite set of points x_1,...,x_m in D, the matrix
(h(x_i,x_j) + 1/2)_{i,j} has (P).
Proof. Multiply row i by (1-|x_i|^2) and column j by (1-|x_j|^2). This is
a positive diagonal congruence, so (P) is preserved by 11.1(a). A direct
computation gives
(h + 1/2)(1-|x|^2)(1-|y|^2)
= |x-y|^2 + (1/2)(1-|x|^2)(1-|y|^2)
= (1/2)(1+|x|^2)(1+|y|^2) - 2 x . y,
using |x-y|^2 = |x|^2 + |y|^2 - 2 x.y. The first term is a positive
rank-one kernel; the second is minus a positive semidefinite kernel,
because x . y = sum_k x_k y_k is a Gram kernel. A rank-one positive
kernel minus a positive semidefinite kernel has (P). Indeed its quadratic
form is L(x)^2 - q(x) with q >= 0 and L linear; if it were positive on a
two-dimensional subspace V, then L(x)^2 > q(x) >= 0 for every nonzero x in
V, so L would have no zero on V minus {0}, which is impossible for a
nonzero linear functional on a two-dimensional space.
12.3 LEMMA. For all x, y in D one has f(h(x,y)) >= 0, and the matrix
(f(h(x_i,x_j)))_{i,j} is positive semidefinite.
Proof. STEP 1: AN INTEGRAL FORMULA FOR L. For h > 0,
L(h) = int_0^1 h^{1-alpha} (h+1)^alpha d alpha.
Indeed, substitute u = h/(h+1) and t = 1 - alpha: the integrand becomes
u^t/(1-u), and int_0^1 u^t dt = (u-1)/log u, so the integral is
-1/log u = 1/log(1/u) = L(h).
STEP 2: A MIXTURE FORMULA FOR f. For 0 < alpha < 1 let c_alpha normalize
the measure v^{alpha-1}(1+v)^{-1} dv on (0,infinity) to total mass one,
set r = 1/(1+v), and write E_alpha for expectation under the normalized
measure. Integration by parts gives
E_alpha (1 - r) = c_alpha int_0^inf v^alpha (1+v)^{-2} dv = alpha.
For h > 0, scaling v by (h+1)/h in the normalized integral gives
h^{1-alpha}(h+1)^alpha
= c_alpha int_0^inf h(h+1) v^{alpha-1} / ((h+1) + v h) dv
= E_alpha h(h+1)/(h+r)
= E_alpha [ h + (1-r) - r(1-r)/(h+r) ],
the last equality being an algebraic identity after multiplying by h+r.
Integrating in alpha and using int_0^1 alpha d alpha = 1/2,
L(h) = h + 1/2 - int_0^1 E_alpha r(1-r)/(h+r) d alpha,
so by 12.1,
f(h) = int_0^1 E_alpha r(1-r)/(h+r) d alpha >= 0.
STEP 3: 1/(h+r) IS A POSITIVE SEMIDEFINITE KERNEL. First,
k(x,y) = 1/(1 + h(x,y)) = (1-|x|^2)(1-|y|^2) / |1 - x conj(y)|^2
is positive semidefinite: the function (1 - x conj(y))^{-1} is a Gram
kernel (its power series), so its modulus squared is positive
semidefinite, and multiplying by the positive factors (1-|x|^2)(1-|y|^2)
keeps that property. For fixed 0 < r < 1, the identity h = (1-k)/k gives
1/(h+r) = k/(1 - (1-r)k) = sum_{n>=0} (1-r)^n k^{n+1},
with entrywise powers, a series of nonnegative terms. Each entrywise
power of a positive semidefinite kernel is positive semidefinite (Schur
product theorem), so 1/(h+r) is positive semidefinite. By step 2, f is a
positive mixture of such kernels, so f(h(x,y)) is positive semidefinite
and nonnegative.
12.4 LEMMA (the regularization). For 0 < eps <= 1 define C_eps by
1/C_eps(x,y) = 4 pi [ L(h) + eps f(h) ]
= 4 pi [ (h + 1/2) - (1-eps) f(h) ].
Then C_eps is bounded, continuous, and strictly positive on D x D;
C_eps <= G; C_eps increases to G as eps decreases to 0, including on the
diagonal; C_eps(x,x) = 1/(2 pi eps); and for every finite set of points
the matrix (1/C_eps(x_i,x_j))_{i,j} has (P).
Proof. Since L + eps f = eps(h + 1/2) + (1-eps) L >= eps/2, the
denominator is at least 2 pi eps, so C_eps is bounded, continuous, and
positive. As f >= 0, the denominator L + eps f decreases as eps
decreases, so C_eps increases; its limit is 1/(4 pi L) = G. On the
diagonal h = 0 = L(0), so the denominator is 4 pi eps f(0) = 2 pi eps,
since f(0) = 1/2.
For (P): the reciprocal matrix is 4 pi times (h + 1/2) - (1-eps) f, which
is a matrix with (P) by 12.2 minus a positive semidefinite matrix by
12.3, hence has (P) by 11.1(b).
13. RANK-ONE UPDATES
13.1 LEMMA. Let M be a real symmetric matrix with strictly positive
entries whose entrywise reciprocal has (P). Fix an index z and a real
c >= 0, and define
M~_{ij} = M_{ij} + c M_{iz} M_{zj}.
Then the entrywise reciprocal of M~ has (P).
Proof. STEP 1: A POSITIVE DIAGONAL CONGRUENCE. Let H be the reciprocal of
M. Congruence with the positive diagonal matrix diag(M_iz)_i turns H into
H'_{ij} = M_{iz} M_{zj} / M_{ij}.
By 11.1(a), H' has (P). Put a = M_zz > 0. The z-row and the z-column of
H' are constant equal to a: H'_{iz} = M_iz M_zz / M_iz = a and
H'_{zj} = a.
STEP 2: SUBTRACTING THE z ROW AND COLUMN. Relabel so that z is the last
index, and let H_0 be the top-left block of H'. Subtracting the last row
from every other row, then the last column from every other column, is a
congruence, so it preserves (P); it gives
H' ~ [ H_0 - a 1 1^T 0 ]
[ 0 a ].
Because a > 0 and H' has (P), the top-left block must be negative
semidefinite. Hence
U = a 1 1^T - H'
is positive semidefinite, where the expression is read blockwise with
U_iz = U_zj = 0 (they vanish automatically, since H'_{iz} = H'_{zj} = a
and H'_{zz} = a). Also H'_{ii} > 0 because M and hence H' has positive
entries, so 0 <= U_ii = a - H'_{ii} < a; by the Cauchy-Schwarz inequality
for a positive semidefinite form, |U_ij| < a.
STEP 3: THE NEW RECIPROCAL. The reciprocal of M~ has entries
H'_{ij}/(1 + c H'_{ij}), because M~_{ij} = M_ij (1 + c H'_{ij}). Compare
with the constant rank-one kernel a/(1+ca):
a/(1+ca) - H'_{ij}/(1+cH'_{ij})
= U_ij / ((1+ca)(1+ca-cU_ij))
= sum_{n>=0} c^n U_ij^{n+1} / (1+ca)^{n+2}.
The series converges absolutely, since |cU_ij| < ca < 1+ca. All
coefficients are nonnegative, and every entrywise power U^{n+1} of the
positive semidefinite matrix U is positive semidefinite (Schur product
theorem). So the normalized reciprocal of M~ is a positive rank-one
kernel minus a positive semidefinite matrix, and therefore has (P) by
11.1(b). For c = 0 the same computation reduces to U itself.
14. THE INTERIOR RESOLVENT
14.1 NOTATION. Let 0 <= rho <= q be a nonnegative measure supported in a
compact subset of D, and let T_rho be the operator of 9.2 with q replaced
by rho, that is,
T_rho f(x) = int_D G(x,y) f(y) d rho(y).
By 9.3(a) applied to rho, ||T_rho|| <= d < 1 on L^2(rho). Put
M^rho(x,y) = G(x,y) + <G(x,.), (I - T_rho)^{-1} G(y,.)>_rho
for distinct interior points x, y, the inner product being in L^2(rho),
and M^rho(x,x) = +infinity. This is finite off the diagonal, because
each Green section G(x,.) lies in L^2(rho): rho is compactly supported in
D and G(x,.) has only a logarithmic singularity.
14.2 CHAIN EXPANSION. Expanding the resolvent,
M^rho(x,y) = sum_{n>=0} int ... int G(x,z_1) G(z_1,z_2) ... G(z_n,y)
d rho(z_1) ... d rho(z_n).
All factors are nonnegative, so all sums and interchanges are legitimate
by Tonelli. Write "a chain" for a finite sequence of intermediate points.
14.3 PROPOSITION. For every finite set of distinct interior points
x_1,...,x_m, the entrywise reciprocal of M^rho, whose diagonal is zero,
has (P).
Proof. Fix eps in (0,1] and replace G by C = C_eps of 12.4 throughout
14.1; call the result M_eps. Since 0 < C <= G, the operator T_C on
L^2(rho) satisfies |T_C f| <= T_rho |f| pointwise, hence ||T_C|| <= d.
I will show that the reciprocal of M_eps has (P); then eps -> 0 gives the
proposition, because the reciprocal entries converge (the diagonal ones
to 0) and (P) is closed by 11.1(d).
STEP 1: QUADRATURE. Partition the compact support of rho into finitely
many measurable cells of uniformly small diameter. In each cell of
positive mass choose a representative z_alpha and take its mass w_alpha
as a weight. Since C is uniformly continuous on the union of the
evaluation points and the support, the step-function operator T_pi
converges to T_C in operator norm, and each vector
(C(x,z_alpha))_alpha converges to C(x,.) in L^2(rho). Hence for
sufficiently fine partitions ||T_pi|| < 1 and the quadrature resolvent
C(x,y) + <C(x, pi(.)), (I - T_pi)^{-1} C(y, pi(.))>_rho
is arbitrarily close to the regularized resolvent evaluation.
Concretely, put
T_{alpha beta} = sqrt(w_alpha) C(z_alpha, z_beta) sqrt(w_beta),
xi_x = (sqrt(w_alpha) C(x, z_alpha))_alpha.
On the span of the normalized cell indicators, T_pi is represented by the
matrix T, and T_pi vanishes on the orthogonal complement. So ||T|| < 1
and the quadrature correction equals xi_x^T (I - T)^{-1} xi_y.
STEP 2: (P) FOR A FIXED QUADRATURE. Consider the finite set consisting
of the evaluation points x_1,...,x_m and the nodes z_alpha. Start with
the matrix M_{ij} = C(x_i,x_j) on this set. By 12.4 its reciprocal has
(P). Now allow the nodes one at a time, and let M denote the chain sum
using the nodes allowed so far. If we allow one more node z of weight
w > 0, the chains that visit z exactly k >= 1 times have total weight
w^k M_{iz} M_{zz}^{k-1} M_{zj},
so summing over k replaces M by
M_{ij} + (w / (1 - w M_zz)) M_{iz} M_{zj}.
Finiteness of the full chain sum forces w M_zz < 1. This is the update of
13.1 with c = w/(1 - w M_zz) >= 0, so the reciprocal keeps (P) at each
step. An endpoint that happens to be a node receives no weight, since
only internal visits are weighted; coincident nodes may be merged by
adding their weights. Thus every sufficiently fine quadrature kernel has
a reciprocal with (P).
STEP 3: TAKE THE TWO LIMITS. The quadrature limit of step 1, applied to
matrices of the fixed size m + (number of nodes), gives (P) for the
reciprocal of the regularized continuum kernel M_eps, by 11.1(d).
Now let eps -> 0. Each chain integral in M_eps increases to its
G-counterpart by monotone convergence, since C_eps increases to G. Off
the diagonal the limit is M^rho, and the correction is finite because
|<G(x,.), (I-T_rho)^{-1} G(y,.)>_rho|
<= ||G(x,.)||_rho ||G(y,.)||_rho / (1-d).
The direct diagonal term tends to infinity, so its reciprocal tends to
zero. Thus the limit of the reciprocals of M_eps is the reciprocal of
M^rho with zero diagonal, and it has (P) by 11.1(d).
15. THE BOUNDARY LIMIT
15.1 SETUP. Fix a compactly supported measure 0 <= rho <= q and let
K^rho, R^rho, N^rho be the kernels of 9.4 with q replaced by rho. The
limit r -> 1 will move evaluation points radially to S1 while the
integration measure stays compactly supported; this is why the boundary
limit comes before the cutoffs are removed.
15.2 LEMMA (radial limits). For distinct s, t in S1 and x in D,
G(rs,x)/(1-r) -> P_s(x) as r -> 1, uniformly for x in a compact subset
of D and in s in S1;
G(rs,rt)/(1-r)^2 -> N_0(s,t) as r -> 1, for s != t.
Proof. By 9.1,
G(rs,x) = (1/(4 pi)) log(1 + (1-r^2)(1-|x|^2)/|rs - x|^2).
For r close to 1 and x in a compact subset of D, |rs - x| is bounded
away from zero, so log(1+u) = u + O(u^2) with u = O(1-r) uniformly.
Hence G(rs,x) = (1-r)(1-|x|^2)/(2 pi |s-x|^2) + O((1-r)^2), which is the
first claim since P_s(x) = (1-|x|^2)/(2 pi|s-x|^2). For the second, use
|rs - rt| = r |s-t| and 1-|rs|^2 = 1-r^2:
G(rs,rt) = (1/(4 pi)) log(1 + (1-r^2)^2/(r^2 |s-t|^2)),
so G(rs,rt)/(1-r)^2 = (1/(4 pi)) (1+r)^2/(r^2|s-t|^2) + O((1-r)^2), which
tends to 1/(4 pi) * 4/|s-t|^2 = N_0(s,t).
15.3 LEMMA (bordered matrix, cutoff version). For distinct s_1,...,s_m in
S1 and p in D, the matrix
M_rho = [ B^rho ell^rho ]
[ (ell^rho)^T 0 ]
with B^rho_{ij} = 1/N^rho(s_i,s_j) for i != j, B^rho_{ii} = 0, and
ell^rho_i = 1/K^rho(p,s_i), has (P).
Proof. Fix r with |p| < r < 1, and apply 14.3 to the m + 1 points
rs_1,...,rs_m, p. The reciprocal matrix of M^rho at those points has (P).
Scale it by the positive diagonal congruence with factors (1-r) at the
moving points and 1 at p; this preserves (P) by 11.1(a). As r -> 1, by
15.2 and the boundedness of the resolvent on the compact support,
M^rho(rs_i,rs_j)/(1-r)^2 -> N^rho(s_i,s_j) (i != j),
M^rho(p,rs_i)/(1-r) -> K^rho(p,s_i).
The first limit uses that the resolvent correction tends to
<P_{s_i}, (I-T_rho)^{-1}P_{s_j}>_rho = R^rho(s_i,s_j), and the second
that it tends to P_{s_i}(p) + <G(p,.), (I-T_rho)^{-1}P_{s_i}>_rho
= K^rho(p,s_i); both pairings are legitimate because rho is compactly
supported and the Poisson kernels are bounded there. Since the diagonal
entries of the reciprocal are zero and stay zero under the congruence,
the limit of the scaled reciprocal matrices has entries
(1-r)^2 / M^rho(rs_i,rs_j) -> 1/N^rho(s_i,s_j) (i != j),
(1-r) / M^rho(p,rs_i) -> 1/K^rho(p,s_i),
and the corner entry 0. So the limit is M_rho, which therefore has (P)
by 11.1(d).
15.4 PROOF OF THEOREM 11.3. Choose increasing disk cutoffs
d rho_n = 1_{|x| <= r_n} dq with r_n -> 1. Every term of K^{rho_n} and
R^{rho_n} is a nonnegative chain integral, so by monotone convergence
K^{rho_n}(p,s) -> K(p,s), N^{rho_n}(s,t) -> N(s,t) (s != t),
with finite limits by 9.5. Taking reciprocals and the limit of the
finite matrices of 15.3 gives (P) for the bordered matrix of 11.2 with
B_{ij} = 1/N(s_i,s_j), B_ii = 0, ell_i = 1/K(p,s_i). Again 11.1(d).
Now take real coefficients a_1,...,a_m with sum a_i = 0 and put
x_i = K(p,s_i)a_i. Then ell^T x = sum a_i = 0, so 11.2 gives x^T B x <= 0,
which is
sum_{i != j} K(p,s_i) K(p,s_j) a_i a_j / N(s_i,s_j) <= 0.
The left side is exactly sum_{i,j} a_i a_j D_p(s_i,s_j), because
D_p(s_i,s_i) = 0. This proves 11.3 for distinct points s_i; repeated
points are handled by adding their coefficients.
16. FROM CONDITIONAL NEGATIVITY TO A HILBERT SPACE
16.1 PROPOSITION (Schoenberg). Let X be a set and let D be a real
symmetric kernel on X with D(x,x) = 0 that is conditionally negative
semidefinite. Then there is a real Hilbert space H and a map
b : X -> H with
|b(x) - b(y)|^2 = D(x,y) for all x, y in X.
Proof. Fix x_0 in X and put
Gamma(x,y) = (1/2) ( D(x,x_0) + D(y,x_0) - D(x,y) ).
Gamma is positive semidefinite. Indeed, for real coefficients c_1,...,c_m
and x_1,...,x_m, put c_0 = -sum_i c_i and apply conditional negativity to
the family x_1,...,x_m,x_0 with coefficients c_1,...,c_m,c_0. Since
D(x_0,x_0) = 0 the resulting inequality reads
-sum_{i,j} c_i c_j D(x_i,x_j) + 2 sum_{i,j} c_i c_j D(x_i,x_0) <= 0,
which rearranges to sum_{i,j} c_i c_j Gamma(x_i,x_j) >= 0.
Now take the free real vector space on X, give it the symmetric bilinear
form with Gram kernel Gamma, divide by the nullspace of that form, and
complete. Let b(x) be the class of x. Then <b(x),b(y)> = Gamma(x,y), and
since Gamma(x,x) = D(x,x_0),
|b(x)-b(y)|^2 = Gamma(x,x) - 2 Gamma(x,y) + Gamma(y,y) = D(x,y).
16.2 THE EMBEDDING FOR OUR KERNEL. Apply 16.1 to X = S1 and D = D_p of
11.3, which is conditionally negative by 15.4. By 9.5(a) the function
K(p,.) is positive and bounded, and N >= N_0 > 0. Hence for s != t
0 < D_p(s,t) = K(p,s)K(p,t)/N(s,t)
<= pi ||K(p,.)||^2_infinity |s-t|^2.
So b is injective and Lipschitz. Replacing H by the closed span of b(S1)
makes it separable.
16.3 COORDINATES. In an orthonormal basis (e_j) of H put
b_j(s) = <b(s), e_j>. Each b_j is a real Lipschitz function on S1, and
sum_j (b_j(s) - b_j(t))^2 = |b(s)-b(t)|^2 = D_p(s,t) for all s, t.
17. NO INTERIOR CRITICAL POINT
17.1 ASSUME A CRITICAL POINT. Fix 0 != u in V and let g be the tangent
field of 10.4. Suppose that Phi(p) is a critical point of u, that is,
W_g(p) = (grad u)(Phi(p)) = 0. Take b as in 16.2 and its coordinates b_j
as in 16.3. Then, summing the multiplier identity 10.5 over j,
sum_j E(b_j g)
= (1/2) iint N(s,t) sum_j (b_j(s)-b_j(t))^2 g(s) . g(t) ds dt
= (1/2) iint K(p,s) K(p,t) g(s) . g(t) ds dt
= (1/2) | int_S1 K(p,s) g(s) ds |^2
= (1/2) |W_g(p)|^2 = 0.
The exchange of sum and integral is justified by absolute convergence:
the integrands are dominated by K(p,s) K(p,t) |g(s)| |g(t)|, and
K(p,.) is bounded while g is bounded, so the dominating function is
integrable. The third line is the identification of 9.5(c) with h = g,
and the fourth is W_g(p) = 0.
Since each summand E(b_j g) is nonnegative by 7.2 and (10.1)(ii), every
summand vanishes. By (10.1)(ii) again, for each j there is v_j in V with
X_{b_j g} = grad v_j,
and therefore, taking traces,
the boundary trace of grad v_j equals b_j g, for every j.
17.2 THE GRADIENT TRACE IS INJECTIVE. If v in V has grad v = 0 on dOmega,
then the tangential derivative of v vanishes there, so v is a constant c
on dOmega; the function v - c lies in H^1_0(Omega) and solves
-Delta(v-c) = mu (v-c), so by 2.2 and the characterization of lambda_D,
v - c = 0. Thus v = 0 (it has mean zero). Hence the map
v -> (trace of grad v) is injective on V.
17.3 THE SET S_g = {s in S1 : g(s) != 0} IS NONEMPTY AND OPEN. It is open
because g is continuous. If it were empty then g = 0, so each component
of grad u would solve the Helmholtz equation with zero Dirichlet trace;
by 2.2 and the characterization of lambda_D both components would vanish
in Omega, so grad u = 0, so u would be constant, so u = 0 (it has mean
zero). Contradiction.
17.4 CASE 1: S_g IS DENSE IN S1. Then b(S_g) is not contained in an
affine line. Otherwise, by continuity, b(S1) would be contained in the
closed line, and the image of the circle under a continuous injective
map is homeomorphic to the circle; but a compact connected subset of a
line is a point or a closed interval, and deleting an interior point
disconnects an interval while deleting a point does not disconnect a
circle.
So there are three points s_a, s_b, s_c of S_g whose images are affinely
independent: if every three points of b(S_g) were collinear, then b(S_g)
would lie in a line. The two difference vectors
u = b(s_a)-b(s_c) and w = b(s_b)-b(s_c) are then linearly independent.
Consequently some two coordinates j, k satisfy
det [ u_j w_j ] != 0.
[ u_k w_k ]
Otherwise all vectors (u_i,w_i) in R^2 would be proportional, and since
u != 0 that would make w a multiple of u.
Equivalently, the three vectors (1, b_j(s_i), b_k(s_i)) in R^3, for
i = a, b, c, are linearly independent. Since g is nonzero at these three
points, the vector functions
g, b_j g, b_k g
are linearly independent on S1: a linear combination that vanishes as a
function vanishes at s_a, s_b, s_c, and the corresponding 3 x 3 system is
invertible. They are the gradient traces of u, v_j, v_k in V, so by 17.2
those three elements are linearly independent and dim V >= 3. That
contradicts 6.1.
17.5 CASE 2: S_g IS NOT DENSE. Then there is a nonempty open arc I of S1
on which g = 0. Since b is injective, some coordinate b_j is not constant
on S1; since S_g is dense and b_j is continuous, b_j is not constant on
S_g. Hence g and b_j g are linearly independent as functions on S1: if
b_j g = c g then (b_j - c) g = 0, so b_j = c on S_g. By 17.2 the
elements u and v_j of V are linearly independent.
Both u and v_j have gradient zero along the boundary arc Phi(I), since
grad u = g = 0 and grad v_j = b_j g = 0 there. So both are constant on
that arc, say with values c_u and c_v. The map
(alpha,beta) -> alpha c_u + beta c_v has a nonzero kernel in R^2, and any
nonzero pair in that kernel gives w = alpha u + beta v_j != 0 (by
independence) with
w = 0 on the arc Phi(I) and d_nu w = 0 everywhere on dOmega,
so in particular d_nu w = 0 on the arc.
Take a disk B around an interior point of the arc, small enough that B
meets dOmega only in points of the arc. Extend w by zero on B minus
Omega, getting a function W on B. For every smooth phi supported in B,
integration by parts over Omega cap B shows
int_B W (Delta + mu) phi = int_{Omega cap B} w (Delta + mu) phi
= int_{Omega cap B} ((Delta + mu) w) phi
+ int_{Phi(I) cap B} (w d_nu phi - phi d_nu w) dS = 0,
because (Delta + mu)w = 0 in Omega and both w and d_nu w vanish on the
arc. So (Delta + mu)W = 0 in the sense of distributions on B, and by
(1.6.2)(c), W is real analytic in B. It vanishes on the nonempty open
set B minus cl(Omega), so W = 0 on all of B; in particular w vanishes on
the nonempty open set Omega cap B. But w is real analytic in Omega by
(1.6.2)(c) and is not identically zero. Contradiction.
17.6 CONCLUSION. Both cases are impossible, so no point of Omega is a
critical point of u: grad u != 0 on Omega.
Every interior local maximum or minimum of u would be a critical point,
so the maximum and the minimum of u on the compact set cl(Omega) are
attained on dOmega. Moreover, if x in Omega satisfied u(x) = max u or
u(x) = min u (the extremes over cl(Omega)), then x would be an interior
extremum, hence a critical point, which is impossible. So
min_{dOmega} u < u(x) < max_{dOmega} u for every x in Omega.
This completes the proof of the theorem of 1.3.
A. THE SIX QUOTED THEOREMS, IN SHORT
The full statements, with the place of use, are in 1.6. For reference:
1.6.1 -Delta with Neumann data has discrete spectrum
0 = nu_0 < nu_1 <= nu_2 <= ..., finite dimensional eigenspaces,
nu_1 = the minimum in 1.2.
1.6.2 (a) -Delta v = f with Dirichlet or Neumann data, f in L^2, has a
unique H^2 solution; (b) smooth f gives a solution smooth up to
the boundary; (c) solutions of Delta w + mu w = 0 are real
analytic. (Showalter, Chapter III.)
1.6.3 There is a biholomorphic Phi : D -> Omega, C-infinity up to the
boundary, with Phi' != 0. (Riemann mapping plus Kellogg;
Pommerenke, Chapter 3.)
1.6.4 Hopf: if -Delta w > 0 and w >= 0 in Omega with w(x_0) = 0 at a
boundary point, then d_nu w(x_0) < 0.
1.6.5 A crosscut of the disk separates the two boundary arcs between
its endpoints, and any arc joining those arcs must meet it.
1.6.6 int_Omega |v - avg v|^2 <= C int_Omega |grad v|^2.
Everything else in this document is proved from these.
B. WHERE EVERY HYPOTHESIS IS USED
B.1 SMOOTH BOUNDARY. It enters in four places: Kellogg's theorem
(1.6.3) produces the smooth conformal map used from section 3 on;
elliptic regularity up to the boundary (1.6.2) provides smooth
eigenfunctions, the H^2 solutions in 7.2, and the smooth auxiliary
function L in 4.1; Hopf's lemma (1.6.4) is applied in 4.1 to get
d_nu L < 0; and the divergence theorem for H^1 fields with a trace is
used in 7.2, 8.2, 10.2 and 17.5. A merely Lipschitz boundary would need
every one of these to be redone.
B.2 SIMPLE CONNECTIVITY. It enters exactly twice: the conformal map
Phi of 1.6.3 exists only because Omega has no hole, and step 2 of 7.2
uses "closed 1-form on a simply connected plane domain is exact" to
produce the potential r with grad r = Z. The counterexamples of Burdzy
and Werner, Bass and Burdzy, and Burdzy (see C) show that neither use
can be dropped.
B.3 CHAIN OF DEPENDENCE, in order.
2.1, 2.2 (mu < lambda_D) <- 1.6.1, 1.6.2, 1.6.6
3.1 <- 1.6.3
3.3, 3.4 <- chain rule for conformal maps
3.5 <- 3.3, 3.4, 1.6.6, 2.2
4.1 <- 2.2, 1.6.2, 1.6.4
4.2, 4.3 <- 4.1, Green's identity
5.2 <- 4.2, 1.6.2(c), 1.6.5
6.1 <- 4.2, 4.3, 5.2, 5.3
7.2 <- 2.1, 2.2, 1.6.2
8.2 <- divergence theorem
9.3 <- 3.5, 9.1
9.5 <- 9.1(iv), 9.3
10.2 <- 8.2, 9.3, 9.5, 10.3
10.3 <- Fourier decomposition of S1
10.5 <- 10.2, 10.4, 7.2
11.1, 11.2 <- linear algebra
12.2, 12.3, 12.4 <- Sylvester, Schur product theorem
13.1 <- 11.1, Schur product theorem
14.3 <- 12.4, 13.1, 11.1(d)
15.3, 15.4 <- 14.3, 15.2, 11.1(d), 11.2
16.1, 16.2 <- 15.4, 9.5(a)
17.1 to 17.6 <- 10.5, 16.3, 6.1, 7.2, 1.6.2(c)
B.4 WHAT EACH PART PROVES, in one line each.
2 a strict spectral gap between Neumann and Dirichlet.
3 the problem becomes a disk problem with a subcritical weight.
4 eigenfunctions are orthogonal to one positive boundary weight and
change sign on the boundary.
5 the two sign regions are connected and cannot interlace.
6 dim V <= 2.
7 the vector energy dominates mu times the L^2 norm, with equality
exactly on gradients of eigenfunctions.
8 the vector energy is a Dirichlet energy plus a curvature term.
9 positive kernels K, R, N represent the tilted extension and its
L^2 pairing.
10 the energy of b g is a double integral of (b(s)-b(t))^2 times g.g
against N.
11 to 15 the kernel D_p = K(p,s)K(p,t)/N(s,t) is conditionally
negative semidefinite.
16 such a kernel is a squared distance in a Hilbert space.
17 a critical point would produce either three independent
eigenfunctions or a nonzero eigenfunction vanishing on an open set.
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