==== TABLE OF CONTENTS ==== Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//figures/boundary-signs.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//main.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/01-introduction.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/02-preliminaries.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/03-vector-form.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/04-boundary-kernels.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/05-reciprocal-kernels.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/06-conclusion.tex Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//references/references.bib =========================== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//figures/boundary-signs.tex ==== \begin{figure}[htbp] \centering \begin{minipage}{.84\textwidth} \centering\small \begin{tikzpicture}[x=1cm,y=1cm,font=\small, line cap=round,line join=round] \draw[black!65,line width=.65pt] (0,0) circle[radius=1.25]; \node at (-.67,-.69) {$\D$}; \draw[line width=1.1pt] (-1.25,0) .. controls (-.8,.38) and (-.4,.22) .. (0,.22) .. controls (.48,.22) and (.78,.18) .. (1.25,0); \draw[line width=.9pt,dash pattern=on 3pt off 2.2pt] (0,1.25) .. controls (-.26,.91) and (-.20,.55) .. (0,.22) .. controls (.35,-.20) and (.30,-.90) .. (0,-1.25); \fill (0,.22) circle[radius=1.7pt]; \foreach \x/\y in {-1.25/0,1.25/0,0/1.25,0/-1.25} \fill (\x,\y) circle[radius=1.6pt]; \node[font=\large] at (-1.57,0) {$+$}; \node[font=\large] at (1.57,0) {$+$}; \node[font=\large] at (0,1.55) {$-$}; \node[font=\large] at (0,-1.55) {$-$}; \draw[line width=1.1pt] (2.10,.74) -- (2.72,.74); \node[anchor=west] at (2.89,.74) {Positive crosscut}; \draw[line width=.9pt,dash pattern=on 3pt off 2.2pt] (2.10,.10) -- (2.72,.10); \node[anchor=west] at (2.89,.10) {Putative negative path}; \fill (2.41,-.54) circle[radius=1.7pt]; \node[anchor=west] at (2.89,-.54) {Intersection forced}; \end{tikzpicture} \caption{Connectedness of the positive set gives a crosscut joining the $+$ points. It separates the two $-$ points, forcing any negative connecting path to meet it. The dashed path illustrates this contradiction; all four boundary signs are strict.} \label{fig:boundary-signs} \end{minipage} \end{figure} ==== END Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//figures/boundary-signs.tex ==== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//main.tex ==== \documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage{lmodern} \input{glyphtounicode} \input{glyphtounicode-cmex} \pdfgentounicode=1 \pdfglyphtounicode{negationslash}{0338} \usepackage[margin=1.08in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{microtype} \usepackage{enumitem} \usepackage{needspace} \usepackage{xcolor} \usepackage{tikz} \usetikzlibrary{arrows.meta} \usepackage[colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black]{hyperref} \usepackage[nameinlink,capitalise,noabbrev]{cleveref} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \AddToHook{env/theorem/before}{\Needspace{5\baselineskip}} \AddToHook{env/lemma/before}{\Needspace{5\baselineskip}} \AddToHook{env/proposition/before}{\Needspace{5\baselineskip}} \AddToHook{env/corollary/before}{\Needspace{5\baselineskip}} \AddToHook{env/remark/before}{\Needspace{5\baselineskip}} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\D}{\mathbb D} \newcommand{\Sone}{\mathbb S^1} \newcommand{\V}{\mathcal V} \newcommand{\HH}{\mathcal H} \newcommand{\Q}{\mathcal Q} \newcommand{\dd}{\,\mathrm d} \newcommand{\grad}{\nabla} \newcommand{\curl}{\operatorname{curl}} \newcommand{\diver}{\operatorname{div}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\ip}[2]{\langle #1,#2\rangle} \newcommand{\norm}[1]{\lVert #1\rVert} \newcommand{\one}{\mathbf 1} \newcommand{\supp}{\operatorname{supp}} \newcommand{\PSD}{\succeq0} \DeclareMathOperator{\spanop}{span} \setlist{topsep=5pt,itemsep=3pt} \setlength{\emergencystretch}{2em} \title{Strict hot spots and absence of interior critical points on smooth simply connected planar domains} \author{OpenAI} \date{September 24, 2026} \hypersetup{ pdftitle={Strict hot spots and absence of interior critical points on smooth simply connected planar domains}, pdfauthor={OpenAI} } \begin{document} \maketitle \begin{abstract} We prove the strict hot spots conjecture for smooth bounded simply connected planar domains. More precisely, every nonzero eigenfunction for the first positive Neumann eigenvalue has nonvanishing gradient in the interior, so all its global maxima and minima lie on the boundary. This holds even when the eigenvalue is multiple. \end{abstract} \input{sections/01-introduction} \input{sections/02-preliminaries} \input{sections/03-vector-form} \input{sections/04-boundary-kernels} \input{sections/05-reciprocal-kernels} \input{sections/06-conclusion} \bibliographystyle{alphaurl} \bibliography{references/references} \end{document} ==== END Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//main.tex ==== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/01-introduction.tex ==== \section{Introduction} The hot-spots problem arose from Rauch's discussion of the long-time behavior of heat flow with insulating boundary conditions \cite{Rauch1975}. Its spectral formulation asks whether the extrema of a first-positive Neumann eigenfunction occur on the boundary. BaƱuelos and Burdzy distinguished strict and nonstrict formulations, and assertions about every eigenfunction from the existence of one favorable eigenfunction \cite{BanuelosBurdzy1999}. These distinctions matter when the eigenvalue is multiple. Let $\Omega\subset\R^2$ be a nonempty bounded simply connected open set with $C^\infty$ boundary. In particular, $\Omega$ is connected. Set \begin{equation}\label{eq:mu} \mu=\mu_1(\Omega) =\inf_{\substack{0\ne v\in H^1(\Omega)\\ \int_\Omega v=0}} \frac{\int_\Omega|\grad v|^2}{\int_\Omega|v|^2}, \end{equation} and let $\V$ be the real Neumann eigenspace for $\mu$. We count only positive eigenvalues in this notation; sources that count the constant eigenfunction first write $\mu_2$ for this same eigenvalue. \begin{theorem}\label{thm:main} For every $0\ne u\in\V$, \[ \grad u(x)\ne0\qquad(x\in\Omega). \] In particular, \begin{equation}\label{eq:hotspots} \min_{y\in\partial\Omega}u(y)0$. Work temporarily over $\C$. For $k\in\R^2$ with $|k|^2=\lambda_D$, put \[ e_k(x)=e^{ik\cdot x},\qquad w_k=e_k-\frac{\int_\Omega e_k}{\int_\Omega\phi}\phi. \] 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. \end{proof} We also include the smooth conformal parametrization used throughout the proof; see \cite{Pommerenke1992} for the classical boundary-regularity theory of conformal maps. \begin{lemma}\label{lem:conformal} There is a conformal bijection $\Phi:\D\to\Omega$ extending to a smooth diffeomorphism of closures, with $\Phi'\ne0$ on $\overline\D$. \end{lemma} \begin{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 \[ m(z)=(z-a)\exp(-l(z)-i\widetilde l(z)) \] 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. Rouch\'e'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 $\Phi$. \end{proof} Throughout, $\Sone=\partial\D$ carries arclength $\dd s$, of total mass $2\pi$. To bound the dimension of $\V$, we transfer its mean-zero condition to the boundary and use connectedness of the positive and negative sets to control boundary signs. \begin{lemma}\label{lem:boundary-weight} There is a fixed smooth positive function $\beta$ on $\Sone$ such that every $v\in\V$ has boundary trace $h_v=v\circ\Phi$ satisfying \[ \int_{\Sone}\beta h_v\dd s=0. \] If $v\ne0$, its boundary trace is nonzero and takes both signs. \end{lemma} \begin{proof} By \Cref{lem:gap}, the Dirichlet problem \[ (-\Delta-\mu)L=1,\qquad L|_{\partial\Omega}=0 \] 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 \[ 0=\int_\Omega v=-\int_{\partial\Omega}v\,\partial_\nu L =\int_{\Sone}\beta(s)h_v(s)\dd s, \quad \beta=|\Phi'|\,(-\partial_\nu L)\circ\Phi. \] 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. \end{proof} \begin{lemma}\label{lem:nodal} For $0\ne v\in\V$, the sets $\{v>0\}$ and $\{v<0\}$ are connected. Its boundary trace cannot have four cyclically ordered points with strictly alternating signs. \end{lemma} \begin{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 \[ \int_\Omega|\grad v_U|^2=\mu\int_\Omega|v_U|^2. \] 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 \Cref{fig:boundary-signs}. \end{proof} \input{figures/boundary-signs} \Needspace{7\baselineskip} The following multiplicity bound is due to Nadirashvili \cite[Theorem~3, p.~227]{Nadirashvili1988}; see also \cite[Proposition~2.5]{BanuelosBurdzy1999}. The boundary-sign proof is included for completeness. \begin{proposition}\label{prop:multiplicity} The real eigenspace $\V$ has dimension at most two. \end{proposition} \begin{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 \[ \int_{\Sone}\beta h\cos\theta\dd s =\int_{\Sone}\beta h\sin\theta\dd s=0, \qquad s=e^{i\theta}. \] It is already orthogonal to $1$ by \Cref{lem:boundary-weight}. 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 \[ \theta_00$ on a nonempty open set. Thus $\int\beta Fh>0$, contradicting the three orthogonalities. \end{proof} ==== END Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/02-preliminaries.tex ==== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/03-vector-form.tex ==== \section{A variational principle for tangent vector fields}\label{sec:vector} For a real vector field $X\in H^1(\Omega;\R^2)$ with tangent trace, meaning $\tr X\cdot\nu=0$, set \[ \Q(X)=\int_\Omega\bigl(|\diver X|^2+|\curl X|^2\bigr), \qquad \curl X=\partial_1X_2-\partial_2X_1. \] The following is Rohleder's vector variational principle \cite[Theorem~1.1]{Rohleder2021}; see also \cite{Rohleder2024}. We include its derivation, in particular its equality case. Our curvature convention below is the opposite of his, so the corresponding boundary term has the opposite sign. \begin{proposition}\label{prop:vector} For every such $X$, \begin{equation}\label{eq:vector-inequality} \Q(X)\ge\mu\int_\Omega|X|^2, \end{equation} with equality exactly when $X=\grad v$ for some $v\in\V$. \end{proposition} \begin{proof} Solve \[ \Delta v=\diver X,\quad \partial_\nu v=0,\quad\int_\Omega v=0, \qquad \Delta j=\curl X,\quad j|_{\partial\Omega}=0. \] The Neumann compatibility follows from the tangent trace. The solutions are in $H^2(\Omega)$ by smooth-boundary regularity \cite[Chapter~III, Theorems~6.1 and~6.4]{Showalter1994}. The rotated gradient $\grad^\perp j=(-\partial_2j,\partial_1j)$ is tangent. Thus \[ Z=X-\grad v-\grad^\perp j \] 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 Poincar\'e inequality, with this fixed-ball average as anchor, gives \[ \norm{r_M}_{L^2(\Omega)} \le C\norm{\grad r_M}_{L^2(\Omega)}+C|(r_M)_B| \le C\norm Z_2+C\sup_B|r|. \] 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 \[ X=\grad v+\grad^\perp j. \] The summands are orthogonal in $L^2$, since the second is divergence-free and tangent. Moreover, \[ \Q(X)=\norm{\Delta v}_2^2+\norm{\Delta j}_2^2. \] Integration by parts and the two Poincar\'e inequalities show \[ \norm{\grad v}_2^2 \le\norm v_2\norm{\Delta v}_2 \le\mu^{-1/2}\norm{\grad v}_2\norm{\Delta v}_2, \] and likewise $\norm{\Delta j}_2^2\ge\lambda_D\norm{\grad j}_2^2$. Hence \[ \Q(X)\ge\mu\norm{\grad v}_2^2+\lambda_D\norm{\grad j}_2^2 =\mu\norm X_2^2+(\lambda_D-\mu)\norm{\grad j}_2^2. \] By \Cref{lem:gap}, this proves \eqref{eq:vector-inequality}. 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. \end{proof} For a differential-form interpretation of this divergence--curl energy, see \cite[Section~5.1]{FriesGoffengMiranda2026}. We will use the following boundary identity in the fixed Cartesian coordinates of the plane. Let $\tau$ be the counterclockwise unit tangent to $\partial\Omega$, and put \[ \kappa=\det(\tau,\partial_\tau\tau). \] Thus $\kappa$ is positive on a convex circle. \begin{lemma}\label{lem:curvature} For tangent $X\in H^1(\Omega;\R^2)$, \begin{equation}\label{eq:curvature} \Q(X)=\int_\Omega|\grad X|^2 +\int_{\partial\Omega}\kappa|X|^2. \end{equation} \end{lemma} \begin{proof} For smooth fields, the difference of the bulk integrands is $2\det\grad X$. Its integral is the boundary pairing \[ \int_{\partial\Omega} (X_1\partial_\tau X_2-X_2\partial_\tau X_1). \] 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 \eqref{eq:curvature}. \end{proof} ==== END Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/03-vector-form.tex ==== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/04-boundary-kernels.tex ==== \section{Boundary kernels and the multiplier identity}\label{sec:boundary} Our objective is a boundary formula for the nonnegative vector energy from \Cref{prop:vector}. We first construct scalar solution and interaction kernels for a general nonnegative subcritical potential on the disk. We then return to the conformal potential and derive the multiplier identity that will connect these kernels to an eigenfunction gradient. \subsection{A subcritical disk potential} The conformal map of \Cref{lem:conformal} defines a measure \begin{equation}\label{eq:q} \dd q(x)=\mu|\Phi'(x)|^2\dd x. \end{equation} It has a bounded positive density. A subscript $q$ on an inner product or norm denotes $L^2(q)$. Conformal invariance of the Dirichlet integral and the Dirichlet inequality on $\Omega$ give \begin{equation}\label{eq:subcritical} \norm\psi_{L^2(q)}^2\le d\int_\D|\grad\psi|^2, \qquad \psi\in H^1_0(\D),\qquad d=\frac\mu{\lambda_D}<1. \end{equation} Until the vector-form application in \Cref{subsec:energy}, only boundedness and nonnegativity of the density and \eqref{eq:subcritical} are needed. In particular, the kernel constructions apply to any measure $q$ with those properties. For such a general measure we fix a constant $00$ and compactly supported $0\le\rho\le q$; let $\epsilon\downarrow0$; send the moving evaluation points radially to the circle while fixing $p$; and finally increase $\rho$ to $q$. Compact support keeps the Poisson sections bounded during the boundary step. \subsection{A bounded regularization of the Green kernel} Direct quadrature of $G$ would encounter the infinite value $G(z,z)$ at every integration node. We therefore need a bounded replacement with finite positive diagonal that retains the reciprocal condition \textup{(P)}. The scalar integral used below is an equivalent form of the logarithmic-mean representations in \cite{QiZhangLi2014}. We derive it and then establish the positive-semidefinite disk-kernel property that we need. \begin{lemma}\label{lem:regularization} There are bounded continuous strictly positive kernels $C_\epsilon$ on $\D\times\D$, $0<\epsilon\le1$, such that $C_\epsilon\uparrow G$ as $\epsilon\downarrow0$, including on the diagonal, and every finite matrix $(1/C_\epsilon(x_i,x_j))_{i,j}$ has \textup{(P)}. \end{lemma} \begin{proof} Put \[ h(x,y)=\frac{|x-y|^2}{(1-|x|^2)(1-|y|^2)},\qquad L(h)=\frac1{\log(1+1/h)}\ (h>0),\quad L(0)=0, \] 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 \begin{equation}\label{eq:lorentz} \tfrac12(1+|x|^2)(1+|y|^2)-2x\cdot y. \end{equation} This is a positive rank-one kernel minus a positive semidefinite kernel, so $h+\tfrac12$ has \textup{(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 $h>0$, \begin{equation}\label{eq:log-mean} L(h)=\int_0^1h^{1-\alpha}(h+1)^\alpha\dd\alpha. \end{equation} 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 \[ \mathbb E_\alpha(1-r) =c_\alpha\int_0^\infty\frac{v^\alpha}{(1+v)^2}\dd v =\alpha. \] For $h>0$, scaling $v$ in the normalized integral yields \begin{align} h^{1-\alpha}(h+1)^\alpha &=c_\alpha\int_0^\infty \frac{h(h+1)}{(h+1)+vh}v^{\alpha-1}\dd v\notag\\ &=\mathbb E_\alpha\frac{h(h+1)}{h+r} =\mathbb E_\alpha\left[h+(1-r)-\frac{r(1-r)}{h+r}\right].\label{eq:mixture-algebra} \end{align} The last equality remains true at $h=0$. Integrating in $\alpha$ therefore gives \begin{equation}\label{eq:f-mixture} f(h)=\int_0^1\mathbb E_\alpha\frac{r(1-r)}{h+r}\dd\alpha, \qquad 0\le f(h)\le f(0)=\tfrac12. \end{equation} Now \[ k(x,y)=\frac1{1+h(x,y)} =\frac{(1-|x|^2)(1-|y|^2)}{|1-x\overline y|^2} \] 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 $00$, 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 \[ H\ \sim\ \begin{pmatrix}H_0-a\one\one^T&0\\0&a\end{pmatrix}. \] Because $a>0$ and $H$ has \textup{(P)}, the first block is nonpositive. Hence \[ U=a\one\one^T-H\PSD. \] Since $H_{ii}>0$, one has $0\le U_{ii}0$ and replace $G$ throughout \eqref{eq:M-rho} by $C=C_\epsilon$. Its integral operator $T_C$ on $L^2(\rho)$ satisfies \[ |T_Cf|\le T_\rho|f|,\qquad \norm{T_C}\le d<1. \] 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 \[ C(\pi(\cdot),\pi(\cdot))\longrightarrow C \] 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 \begin{equation}\label{eq:quadrature-resolvent} C(x,y)+\ip{C(x,\pi(\cdot))}{(I-T_\pi)^{-1}C(y,\pi(\cdot))}_\rho \end{equation} 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 \[ \mathsf T_{\alpha\beta} =\sqrt{w_\alpha}\,C(z_\alpha,z_\beta)\sqrt{w_\beta}, \qquad (\xi_x)_\alpha=\sqrt{w_\alpha}\,C(x,z_\alpha). \] 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 \eqref{eq:quadrature-resolvent} 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 \textup{(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 \Cref{lem:regularization}, its reciprocal has \textup{(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 \[ w^kM_{iz}M_{zz}^{k-1}M_{zj}. \] All factors are positive. Finiteness of the full series forces $wM_{zz}<1$, and summing over $k$ updates the matrix to \begin{equation}\label{eq:elimination} M_{ij}+\frac{w}{1-wM_{zz}}M_{iz}M_{zj}. \end{equation} \Cref{lem:update} therefore propagates \textup{(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 \eqref{eq:quadrature-resolvent} has the required reciprocal property. The quadrature limit gives it for the regularized continuum kernel. If $\rho=0$, this conclusion follows directly from \Cref{lem:regularization}. 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 \eqref{eq:M-rho}, and the correction is finite because \[ \big|\ip{G(x,\cdot)}{(I-T_\rho)^{-1}G(y,\cdot)}_\rho\big| \le\frac{\norm{G(x,\cdot)}_\rho\norm{G(y,\cdot)}_\rho}{1-d}. \] The direct diagonal term tends to infinity, so its reciprocal tends to zero. Finite entrywise limits preserve \textup{(P)}, completing the proof. \end{proof} \subsection{The boundary limit} For compactly supported $\rho$, write $K^\rho,R^\rho,N^\rho$ for the kernels in \eqref{eq:K}--\eqref{eq:N} with $q$ replaced by $\rho$. The boundary limit will be taken with this compact support fixed, before increasing $\rho$ to $q$. \begin{lemma}\label{lem:bordered} For distinct $s_1,\ldots,s_m\in\Sone$ and $p\in\D$, the matrix \begin{equation}\label{eq:bordered} \begin{pmatrix}B&\ell\\\ell^T&0\end{pmatrix}, \qquad B_{ij}=\begin{cases}1/N(s_i,s_j),&i\ne j,\\0,&i=j,\end{cases} \qquad \ell_i=1/K(p,s_i), \end{equation} has \textup{(P)}. \end{lemma} \begin{proof} Fix compactly supported $0\le\rho\le q$. Apply \Cref{prop:interior-reciprocal} to the points $r s_1,\ldots,r s_m,p$, with $r\uparrow1$ and $r>|p|$. The explicit Green formula gives \begin{align} \frac{G(rs,x)}{1-r}&\longrightarrow P_s(x),\label{eq:radial-one}\\ \frac{G(rs,rt)}{(1-r)^2}&\longrightarrow N_0(s,t)\qquad(s\ne t).\label{eq:radial-two} \end{align} The first convergence is uniform on compact interior sets. For the second, one may use \[ G(rs,rt)=\frac1{4\pi}\log\left(1+\frac{(1-r^2)^2}{r^2|s-t|^2}\right). \] The bounded resolvent in \eqref{eq:M-rho} and the uniform convergence on $\supp\rho$ therefore imply \[ \frac{M^\rho(rs_i,rs_j)}{(1-r)^2}\longrightarrow N^\rho(s_i,s_j) \quad(i\ne j),\qquad \frac{M^\rho(p,rs_i)}{1-r}\longrightarrow K^\rho(p,s_i). \] 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 \eqref{eq:bordered} with $K^\rho,N^\rho$, so it has \textup{(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 \[ K^{\rho_n}(p,s)\uparrow K(p,s),\qquad N^{\rho_n}(s,t)\uparrow N(s,t)\quad(s\ne t). \] These limits are finite by \Cref{lem:kernel-estimates}. Taking the reciprocals and the finite matrix limit proves \eqref{eq:bordered}. No $L^2(q)$ boundary limit of an individual Poisson kernel has been used. \end{proof} \begin{proof}[Proof of \Cref{thm:negative-kernel}] Use the matrix of \Cref{lem:bordered}. 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 \[ y-\frac{x^TBy}{x^TBx}x. \] 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 \[ (y,c)^T\begin{pmatrix}B&\ell\\\ell^T&0\end{pmatrix}(y,c) =y^TBy+2c\ell^Ty>0. \] They would span a two-dimensional positive subspace, contradicting \textup{(P)}. Finally, for coefficients with $\sum_i a_i=0$, put $x_i=K(p,s_i)a_i$. Then $\ell^Tx=0$ and \[ \sum_{i,j}a_i a_jD_p(s_i,s_j)=x^TBx\le0. \] This proves the assertion on distinct points; repeated points are handled by combining their coefficients. \end{proof} ==== END Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/05-reciprocal-kernels.tex ==== ==== BEGIN Strict-hot-spots-and-absence-of-interior-critical-points-on-smooth-simply-connected-planar-domains//sections/06-conclusion.tex ==== \section{Hilbert multipliers and the absence of critical points}\label{sec:conclusion} We now use the conditional negativity proved in \Cref{thm:negative-kernel} to construct all the scalar multipliers needed in \Cref{prop:multiplier} at once. The Hilbert-space construction is the classical squared-distance correspondence of Schoenberg \cite{Schoenberg1938}; we give the short Gram-kernel proof. \begin{lemma}\label{lem:hilbert} For every $p\in\D$ there is an injective Lipschitz map $b:\Sone\to\HH$, where $\HH$ is a separable real Hilbert space, such that \begin{equation}\label{eq:hilbert-distance} \norm{b(s)-b(t)}_\HH^2=D_p(s,t). \end{equation} \end{lemma} \begin{proof} Fix $s_0\in\Sone$. Conditional negativity makes the anchored kernel \[ \Gamma(s,t)=\tfrac12\bigl(D_p(s,s_0)+D_p(t,s_0)-D_p(s,t)\bigr) \] 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 \eqref{eq:hilbert-distance} 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 \[ \norm{b(s)-b(t)}_\HH^2 \le\pi\norm{K(p,\cdot)}_\infty^2|s-t|^2. \] Thus $b$ is Lipschitz. Replace $\HH$ by the closed span of $b(\Sone)$; it is separable by continuity and separability of the circle. \end{proof} \begin{proof}[Proof of \Cref{thm:main}] Take an arbitrary $0\ne u\in\V$, with $g$ defined by \eqref{eq:g}. Suppose $\Phi(p)$ is a critical point. Choose $b$ from \Cref{lem:hilbert}, and let $b_j$ be its coordinates in a finite or countable orthonormal basis. Each $b_j$ is real Lipschitz, so \Cref{prop:multiplier} applies. For partial sums, the absolute integrand is bounded by \begin{align*} N(s,t)\sum_{j\le m}|b_j(s)-b_j(t)|^2|g(s)||g(t)| &\le N(s,t)D_p(s,t)|g(s)||g(t)|\\ &=K(p,s)K(p,t)|g(s)||g(t)|. \end{align*} The right side is bounded and integrable. Dominated convergence therefore gives \begin{align} \sum_j E(b_jg) &=\frac12\iint K(p,s)K(p,t)g(s)\cdot g(t)\dd s\dd t\notag\\ &=\frac12\left|\int_{\Sone}K(p,s)g(s)\dd s\right|^2 =\frac12|W_g(p)|^2=0.\label{eq:sum-energy} \end{align} Every summand is nonnegative. By \eqref{eq:E-nullspace}, 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. \smallskip\noindent \emph{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 \Cref{prop:multiplicity}. \smallskip\noindent \emph{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 \cite[Lemma, p.~414]{Filonov2005}. 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. 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