Strict hot spots and absence of interior critical points on smooth simply connected planar domains

Proof. Choose a nonzero nonnegative Dirichlet minimizer \phi, using absolute values in the Rayleigh quotient. Then \int_\Omega\phi>0. Work temporarily over \C. For k\in\R^2 with |k|^2=\lambda_D, put

Each w_k has mean zero. The form \int(|\grad v|^2-\lambda_D|v|^2) vanishes on e_k and \phi; its cross term vanishes because -\Delta e_k=\lambda_De_k and \phi has zero trace. Thus every nonzero w_k has Rayleigh quotient \lambda_D, proving \mu\le\lambda_D.

If equality held, each such w_k would be a Neumann eigenfunction by variational equality. The Euler equation first holds against mean-zero tests and then against all tests, since w_k has mean zero. But distinct plane waves are linearly independent: restrict any finite family to a line segment in an interior ball whose direction gives distinct frequencies. Subtracting multiples of one fixed function leaves their span infinite-dimensional. This contradicts finite eigenspace dimension. The real Neumann inequality and its equality case apply in the complexified space by separating real and imaginary parts.□

Proof. Identify the plane with \C and fix a\in\Omega. Let l solve the harmonic Dirichlet problem with boundary values \log|z-a|. It is smooth up to the boundary. Simple connectivity gives a harmonic conjugate \widetilde l; its gradient extends smoothly, so the conjugate also extends smoothly in boundary charts. The holomorphic function

has modulus one on the boundary, modulus less than one inside, and precisely one zero, of order one. Near the boundary, \log|m| is harmonic, negative inside, and zero on the boundary. The boundary point lemma gives a nonzero normal derivative, so m' does not vanish there.

For any w\in\D, take a smooth inner approximation of \Omega so close to its boundary that |m|>|w| throughout the omitted collar. Rouché’s theorem gives exactly one zero of m-w in that approximation, counting multiplicity, because m has exactly one. There are none in the collar. Thus m is a biholomorphism in the interior. It maps the boundary onto the circle by compactness. Its local boundary inverses imply injectivity there: two boundary preimages would give two nearby interior preimages. The inverse map is the required \Phi.□

Proof. By Lemma 2.1, the Dirichlet problem

is coercive and has a smooth solution. Testing by its negative part gives L\ge0. Hence -\Delta L=1+\mu L>0, and the boundary point lemma gives -\partial_\nu L>0. Green’s identity and the Neumann equation yield

If h_v=0, then v\in H^1_0(\Omega), and the Dirichlet inequality and \mu<\lambda_D force v=0. Positivity of \beta gives the last assertion.□

Proof. For each component U of a sign set, the function v_U=v\one_U belongs to H^1(\Omega) and has weak gradient \one_U\grad v. Indeed it is locally Lipschitz in interior balls: any transition between U and its complement crosses a zero of v, so the local Lipschitz bound for v also bounds the zeroed function. This gives the local weak-gradient identity, and the global L^2 bounds give global H^1 membership. Testing the eigenfunction equation by v_U shows

If at least three sign components existed, a nonzero linear combination of two restrictions could be chosen mean zero. Disjoint supports show that it attains the quotient \mu. It is therefore an eigenfunction, but vanishes on the omitted open component, contradicting interior analyticity. Both sign sets are nonempty, so each is connected.

Four alternating strict signs on the boundary would give a simple arc joining the two positive points, otherwise in the positive interior set, and another joining the negative points in the negative interior set. Such arcs exist by connectedness of open planar sign sets, using short interior end segments and then polygonal paths with loops removed. After applying \Phi^{-1}, their endpoints alternate on a circle. Planar separation forces the two arcs to intersect, contrary to their signs; see Figure 1.□

Proof. Suppose its dimension is at least three. Since the trace map is injective, choose 0\ne v\in\V whose trace h=h_v satisfies

It is already orthogonal to 1 by Lemma 2.3. Cut the circle at a strictly negative point and write its angles as (\theta_0,\theta_0+2\pi). If a and b are the infimum and supremum of the positive angles, then

There are no positive values outside [a,b]. Nor is there a negative value between a and b: positive values on either side of it, together with the cut point, would violate Lemma 2.4. The function

is strictly positive on (a,b) and negative outside [a,b] in the cut interval. Consequently Fh\ge0 everywhere and Fh>0 on a nonempty open set. Thus \int\beta Fh>0, contradicting the three orthogonalities.□

Proof. Solve

The Neumann compatibility follows from the tangent trace. The solutions are in H^2(\Omega) by smooth-boundary regularity [Sho94, Chapter III, Theorems 6.1 and 6.4]. The rotated gradient \grad^\perp j=(-\partial_2j,\partial_1j) is tangent. Thus

is tangent, curl-free, and divergence-free. Its components are harmonic in the interior, so Z is smooth there. Simple connectivity gives an interior potential r with Z=\grad r.

Here r\in H^1(\Omega), as can be seen without assuming its global integrability in advance. For M>0, its bounded truncation r_M=\max(-M,\min(r,M)) has local weak gradient of norm at most |Z|, hence belongs globally to H^1(\Omega). Fix an interior ball B. Its averages (r_M)_B are bounded by \sup_B|r|. The Poincaré inequality, with this fixed-ball average as anchor, gives

Fatou gives r\in L^2, and its weak gradient is Z. Testing \diver Z=0 against r, with its zero normal trace, yields \int|\grad r|^2=0. Thus

The summands are orthogonal in L^2, since the second is divergence-free and tangent. Moreover,

Integration by parts and the two Poincaré inequalities show

and likewise \norm{\Delta j}_2^2\ge\lambda_D\norm{\grad j}_2^2. Hence

By Lemma 2.1, this proves (3.1). Equality forces j=0 and equality in the Rayleigh inequality for v, so v\in\V. Conversely, gradients of functions in \V give equality directly.□

Proof. For smooth fields, the difference of the bulk integrands is 2\det\grad X. Its integral is the boundary pairing

This identity extends to H^1 fields by smooth approximation, since traces are in H^{1/2} and their tangential derivatives are in H^{-1/2}. If the trace is a\tau, the pairing is \int\kappa a^2. This follows first for smooth a and then by approximation in H^{1/2}. It gives (3.2).□

Proof. Write \dd q=a\dd x, with a bounded. For f\in L^2(q), the source af lies in L^2(\dd x), since \int|af|^2\le\norm a_\infty\int|f|^2\dd q. Its Dirichlet solution \psi=Tf satisfies

Combining this with (4.2) gives

hence the norm bound. Symmetry follows from the Green kernel, or by testing the two Poisson equations against each other. The Green formula can first be proved for smooth compactly supported sources and then passed to L^2(\dd x) by approximation; G(x,\cdot)\in L^2(\dd x) gives pointwise evaluation. These sections vary continuously in L^2 on compact interior sets, so this is the continuous interior representative. For \rho\le q, its density remains bounded and its weighted Dirichlet inequality follows from (4.2), proving the same result.□

Proof. The elementary estimates

give uniform bounds for P_s in L^{3/2}(\dd x) and for G(x,\cdot) in L^3(\dd x). Hölder and the bounded density imply

The section P_s need not belong to L^2(q), but one Green integration gives F_s=TP_s\in L^\infty(\D)\subset L^2(q). We can now apply the L^2(q) contraction estimate to the remaining iterates. For n\ge2,

The constants can be chosen uniformly in p and s. This proves the assertion about K.

For distinct s,t, the product P_sP_t is integrable: near either boundary pole, the other factor is bounded, and |y-s|^{-1} is locally integrable in two dimensions. The terms with n\ge1 in (4.7) are summable by the first smoothing estimate and (4.12), since \int P_s\dd q is uniformly bounded. Reversing the nonnegative chains proves symmetry. Tonelli’s theorem and \int P_s\dd s=1 give

which is (4.9). Diagonal values of R play no role in these boundary integrals. Expanding the Neumann series in (4.5) and using the positive kernels gives (4.10) and (4.11); the same bounds justify the exchanges for signed data by absolute domination.□

Proof. By Lemma 3.2 and conformal invariance of the scalar Dirichlet integral applied to each Cartesian component, the bulk terms of E are

The equation for W_k permits replacing W_h here by Ah, since their difference has zero trace. Harmonicity of Ah then replaces W_k by Ak in the gradient term. Since W_k=(I-T)^{-1}Ak in L^2(q), this gives

The boundary curvature term changes by the factor |\Phi'|, and (4.11) identifies the second term.

For the harmonic energy,

Indeed both sides diagonalize in Fourier modes. The Hermitian energy of s^m is 2\pi|m|, while

by a geometric sum, verifying the normalization. Smooth convolution approximation, with uniformly bounded Lipschitz constants and convergence in H^{1/2}, extends the identity to the stated data. This proves (4.14).□

Proof. Since E is positive semidefinite and E(g)=0, its bilinear form annihilates g: this follows by considering E(g+th) for real t. In particular, E(g,b^2g)=0. Subtract (4.14) for this pair from the formula for E(bg,bg). The curvature terms cancel. The harmonic term gives

in the double integral. Symmetry of R gives the same expression with R in place of N_0. Their sum is (4.18). All terms are integrable: the squared Lipschitz difference cancels the singularity of N_0, and R is integrable.□

Proof. Put

and f(h)=h+\tfrac12-L(h). After multiplying the x- and y-entries by 1-|x|^2 and 1-|y|^2, respectively, the kernel h+\tfrac12 becomes

This is a positive rank-one kernel minus a positive semidefinite kernel, so h+\tfrac12 has (P) on finite sets.

Since 1/G=4\pi L(h), the identity L=(h+\tfrac12)-f will be useful once we prove that f(h(x,y)) is positive semidefinite. For h>0,

For 0<\alpha<1, let c_\alpha normalize v^{\alpha-1}(1+v)^{-1}\dd v to a probability measure on (0,\infty), and set r=(1+v)^{-1}. Integration by parts gives

For h>0, scaling v in the normalized integral yields

The last equality remains true at h=0. Integrating in \alpha therefore gives

Now

is positive semidefinite. Indeed (1-x\overline y)^{-1} is a Gram kernel by its power series; multiplying it by its complex conjugate and the positive diagonal factors preserves positive semidefiniteness. For fixed 0<r<1,

where powers are entrywise. This series converges even when h=0. Each power is positive semidefinite, as follows by taking tensor products of Gram vectors. Thus 1/(h+r) is positive semidefinite, and the positive mixture (5.5) proves the claim. The mixture has finite entries by the displayed bound, including on the diagonal.

Define

Its reciprocal matrix has (P) by (5.2) and the positive semidefiniteness just proved. Its denominator is at least 2\pi\epsilon, because

It is positive and continuous, so C_\epsilon is bounded, continuous, and strictly positive. Since f\ge0, these kernels increase to G by (4.3), and 0<C_\epsilon\le G. On the diagonal C_\epsilon(x,x)=1/(2\pi\epsilon), as required.□

Proof. Positive diagonal congruence transforms the reciprocal of M into

Writing a=M_{zz}>0, its z-row and column are identically a. Put z last, and let H_0 denote the remaining principal block. Subtracting the last row from every other row, and doing the same to the columns, gives the congruence

Because a>0 and H has (P), the first block is nonpositive. Hence

Since H_{ii}>0, one has 0\le U_{ii}<a, and positive-semidefinite Cauchy–Schwarz gives |U_{ij}|<a.

Using the same diagonal normalization, the new reciprocal has entries H_{ij}/(1+cH_{ij}). The difference from the constant rank-one kernel is

The series converges absolutely because |cU_{ij}|<1+ca. Its coefficients are nonnegative and every entrywise power of U is positive semidefinite, even if U has negative entries. Thus the normalized reciprocal of \widetilde M is a positive rank-one kernel minus a positive semidefinite matrix, proving (P). For c=0 the identity reduces directly to U.□

Proof. First fix \epsilon>0 and replace G throughout (5.9) by C=C_\epsilon. Its integral operator T_C on L^2(\rho) satisfies

The absolute pointwise domination proves the norm bound without requiring a separate positivity assertion about its quadratic form.

Partition a compact integration set into finitely many measurable cells with uniformly vanishing diameters. For each cell of positive \rho-mass, choose a representative, and let \pi send that cell to the representative. Use its mass as the quadrature weight. Uniform continuity gives uniform convergence of

on the integration set up to null sets. The associated operators T_\pi on L^2(\rho) converge to T_C in operator norm; for example their norm difference is at most \rho(\D) times the uniform kernel error. Likewise C(x,\pi(\cdot))\to C(x,\cdot) in L^2(\rho) for each fixed evaluation point x. In particular, sufficiently fine quadratures have \norm{T_\pi}<1, and

converges to the regularized continuum resolvent evaluation. This follows from convergence of the source vectors and of the inverses in operator norm.

For clarity, if the nodes and weights are z_\alpha,w_\alpha, set

The normalized cell indicators form an orthonormal basis for the step functions. On this subspace T_\pi is represented by \mathsf T, and it vanishes on the orthogonal complement. Thus \norm{\mathsf T}<1, and the correction in (5.10) is \xi_x^T(I-\mathsf T)^{-1}\xi_y. Evaluation points enter only these finite vectors, including when they coincide with nodes.

We next prove (P) for each sufficiently fine quadrature. On the union of the evaluation points and quadrature nodes, start with M_{ij}=C(x_i,x_j). By Lemma 5.3, its reciprocal has (P). Expanding the finite resolvent sums all chains, multiplying kernel entries along each chain and the quadrature weight at every intermediate occurrence. These nonnegative sums are finite by the preceding matrix formula. Restricting the allowed intermediate nodes gives a subseries and hence remains finite.

Suppose some nodes have been allowed and their chain sum is M. Admit one more node z of weight w>0. Chains with exactly k\ge1 internal occurrences of z have total weight

All factors are positive. Finiteness of the full series forces wM_{zz}<1, and summing over k updates the matrix to

Lemma 5.4 therefore propagates (P) one node at a time. This counting remains valid when an endpoint is a quadrature node, since only internal occurrences receive weights. Coincident node locations may be merged by adding their weights. Thus the discrete kernel (5.10) has the required reciprocal property. The quadrature limit gives it for the regularized continuum kernel. If \rho=0, this conclusion follows directly from Lemma 5.3.

Finally let \epsilon\downarrow0. Every chain integral increases to its counterpart with G by monotone convergence, as does the sum of these nonnegative integrals. Off the diagonal, its limit is (5.9), and the correction is finite because

The direct diagonal term tends to infinity, so its reciprocal tends to zero. Finite entrywise limits preserve (P), completing the proof.□

Proof. Fix compactly supported 0\le\rho\le q. Apply Proposition 5.5 to the points r s_1,\ldots,r s_m,p, with r\uparrow1 and r>|p|. The explicit Green formula gives

The first convergence is uniform on compact interior sets. For the second, one may use

The bounded resolvent in (5.9) and the uniform convergence on \supp\rho therefore imply

Here the first resolvent correction tends to \ip{P_{s_i}}{(I-T_\rho)^{-1}P_{s_j}}_\rho=R^\rho(s_i,s_j). The second tends to P_{s_i}(p)+\ip{G(p,\cdot)}{(I-T_\rho)^{-1}P_{s_i}}_\rho=K^\rho(p,s_i). These pairings are legitimate on the compact support, where the Poisson kernels are bounded.

Scale the reciprocal matrix by positive diagonal congruence with factors 1-r at the moving points and factor 1 at p. Its limit is (5.12) with K^\rho,N^\rho, so it has (P). The zero diagonal stays zero throughout.

Now choose increasing disk cutoffs \dd\rho_n=\one_{\{|x|\le r_n\}}\dd q, with r_n\uparrow1. Every term of K^{\rho_n} and R^{\rho_n} is a nonnegative chain integral. Monotone convergence in each finite product integral and then in the series gives

These limits are finite by Lemma 4.2. Taking the reciprocals and the finite matrix limit proves (5.12). No L^2(q) boundary limit of an individual Poisson kernel has been used.□

Proof of Theorem 5.1. Use the matrix of Lemma 5.6. We first show that x^TBx\le0 whenever \ell^Tx=0. If instead x^TBx>0, choose y with \ell^Ty\ne0 and replace it by

Then x^TBy=0 and still \ell^Ty\ne0. In the bordered form, (x,0) and (y,c) are orthogonal, and c can be chosen so that

They would span a two-dimensional positive subspace, contradicting (P).

Finally, for coefficients with \sum_i a_i=0, put x_i=K(p,s_i)a_i. Then \ell^Tx=0 and

This proves the assertion on distinct points; repeated points are handled by combining their coefficients.□

Proof. Fix s_0\in\Sone. Conditional negativity makes the anchored kernel

positive semidefinite: append to any finite family of coefficients the coefficient at s_0 that makes their sum zero. Give formal finite linear combinations of symbols b(s) the bilinear form with Gram kernel \Gamma, quotient by its nullspace, and complete. Since D_p(s,s)=0, this realizes (6.1) and b(s_0)=0.

Positivity of K and finiteness and positivity of N off the diagonal give D_p(s,t)>0 for s\ne t, so b is injective. Also N\ge N_0 gives

Thus b is Lipschitz. Replace \HH by the closed span of b(\Sone); it is separable by continuity and separability of the circle.□

Proof of Theorem 1.1. Take an arbitrary 0\ne u\in\V, with g defined by (4.16). Suppose \Phi(p) is a critical point. Choose b from Lemma 6.1, and let b_j be its coordinates in a finite or countable orthonormal basis. Each b_j is real Lipschitz, so Proposition 4.4 applies. For partial sums, the absolute integrand is bounded by

The right side is bounded and integrable. Dominated convergence therefore gives

Every summand is nonnegative. By (4.13), for every j there is v_j\in\V whose boundary gradient, pulled back by \Phi, is b_jg. The trace equality holds pointwise because both sides are continuous. The boundary gradient of u is g.

Set S_g=\{s\in\Sone:g(s)\ne0\}. This is a nonempty open set. If it were empty, each Cartesian derivative of u would solve the Helmholtz equation with zero Dirichlet trace, contradicting \mu<\lambda_D unless both vanished. Then u would be a constant eigenfunction for a positive eigenvalue, hence zero.

The two possibilities for S_g lead respectively to multiplicity and boundary-uniqueness contradictions.

Case 1: S_g is dense. The set b(S_g) cannot lie in an affine line. Otherwise continuity would place b(\Sone) in that closed line, but a continuous injective circle cannot map into a line. Indeed its compact connected image would be an interval; deleting an interior point disconnects an interval but not a circle with one point removed.

Choose three points of S_g whose images are affinely independent. Their two difference vectors are linearly independent in \HH, so some two coordinates j,k have a nonzero 2\times2 minor. To see this, if all such minors vanished, a nonzero coordinate of the first difference vector would force every coordinate of the second to be proportional to it. Thus the functions 1,b_j,b_k are linearly independent on those three points. Since g\ne0 there, the vector traces g,b_jg,b_kg are linearly independent. They are gradient traces of three functions in \V, contradicting Proposition 2.5.

Case 2: S_g is not dense. There is a nonempty open arc I on which g=0. Since S_g is nonempty and open and b is injective, some coordinate b_j is nonconstant on S_g. Consequently g and b_jg are linearly independent, and so are their eigenfunctions u and v_j. Their gradients vanish on the boundary arc \Phi(I), so each has a constant trace there. Choose a nonzero linear combination w of them whose constant trace on that arc is zero. It also has zero normal derivative there.

Take a small neighborhood of an interior point of the arc that meets no other boundary portion, and extend w by zero to the exterior of \Omega in that neighborhood. This is a local version of the zero-extension argument in [Fil05, Lemma, p. 414]. Integration by parts against compactly supported tests shows that this extension solves (\Delta+\mu)w=0 distributionally: both its Dirichlet and normal traces on the intervening boundary vanish. Interior elliptic regularity and analyticity make the extension analytic. Since it is zero on an exterior open set, it is zero throughout a smaller neighborhood. Analytic continuation inside the connected domain gives w=0 on \Omega, contradicting independence of u and v_j. No analyticity of the boundary is used.

Both cases contradict the assumed critical point. Thus \grad u is nowhere zero in \Omega. The continuous function u attains both extrema on the compact closure. Any interior extremum would be critical, so both are attained on the boundary. An interior value equal to either boundary extremum would itself be a global extremum. This proves both strict inequalities in (1.2).□