# The absence of interior critical points *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This lesson proves the hot-spots theorem stated in [Neumann eigenfunctions and the hot-spots problem](neumann-eigenfunctions-and-the-hot-spots-problem.md) (Theorem 5.1 there): no eigenfunction of the first positive Neumann eigenvalue of a bounded simply connected planar domain with smooth boundary has an interior critical point. The conditionally negative kernel of [Reciprocal kernels and conditional negativity](reciprocal-kernels-and-conditional-negativity.md) is a squared distance in a Hilbert space; its coordinates are scalar multipliers to which the multiplier identity of [Boundary kernels and the multiplier identity](boundary-kernels-and-the-multiplier-identity.md) applies. At a critical point the multiplied boundary gradients are all gradients of eigenfunctions, and this contradicts either the multiplicity bound or unique continuation. - Squared distances in a Hilbert space (Section 1). - The proof of the theorem (Section 2). - Later developments (Section 3). Notation is that of the earlier lessons: \(\Phi\colon\mathbb D\to\Omega\) is the conformal map of [Conformal coordinates, Hopf's lemma and boundary signs](conformal-coordinates-hopfs-lemma-and-boundary-signs.md), \(q=\mu|\Phi'|^2dx\) is the conformal potential, \(K\), \(R\), \(N\) are its kernels, \(E\) is the boundary energy, and \(D_p(s,t)=K(p,s)K(p,t)/N(s,t)\). ## 1. Squared distances in a Hilbert space **Lemma 1.1** ([OpenAI-HS, Lemma 6.1]). For every \(p\in\mathbb D\) there are a separable real Hilbert space \(\mathcal H\) and an injective Lipschitz map \(b\colon\mathbb S^1\to\mathcal H\) with \[ \begin{gathered} \|b(s)-b(t)\|^2=D_p(s,t)\\ (s,t\in\mathbb S^1). \end{gathered} \tag{1.1} \] *Proof.* Fix \(s_0\in\mathbb S^1\) and put \[ \begin{aligned} \Gamma(s,t)&=\tfrac12\bigl(D_p(s,s_0)+D_p(t,s_0)\\ &\qquad-D_p(s,t)\bigr). \end{aligned} \] This kernel is positive semidefinite: given points \(s_1,\dots,s_n\) and real \(c_1,\dots,c_n\), adjoin \(s_0\) with \(c_0=-\sum_ic_i\); Theorem 2.1 of the previous lesson gives \[ \begin{aligned} 0&\ge\sum_{i,j\ge0}c_ic_jD_p(s_i,s_j)\\ &=\sum_{i,j\ge1}c_ic_jD_p(s_i,s_j)\\ &\quad-2\Bigl(\sum_jc_j\Bigr)\sum_ic_iD_p(s_i,s_0), \end{aligned} \] and the right side is \(-2\sum_{i,j\ge1}c_ic_j\Gamma(s_i,s_j)\). On the space of finitely supported real functions on \(\mathbb S^1\), \(\Gamma\) defines a positive semidefinite bilinear form; dividing by its null space and completing gives a real Hilbert space in which the class \(b(s)\) of the indicator of \(\{s\}\) satisfies \(\langle b(s),b(t)\rangle=\Gamma(s,t)\). Since \(D_p(s,s)=0\), \(\Gamma(s,s)=D_p(s,s_0)\), and \[ \begin{aligned} &\|b(s)-b(t)\|^2\\ &\quad=\Gamma(s,s)+\Gamma(t,t)-2\Gamma(s,t)\\ &\quad=D_p(s,t). \end{aligned} \] For \(s\ne t\), \(K(p,s)K(p,t)>0\) and \(N(s,t)<\infty\) (Lemma 4.1 of [Boundary kernels and the multiplier identity](boundary-kernels-and-the-multiplier-identity.md)), so \(D_p(s,t)>0\) and \(b\) is injective. Since \(N\ge N_0\), \[ \|b(s)-b(t)\|^2\le\pi\,\|K(p,\cdot)\|_\infty^2\,|s-t|^2, \] so \(b\) is Lipschitz. Replace \(\mathcal H\) by the closed span of \(b(\mathbb S^1)\); it is separable, being spanned by the images of a countable dense subset of the circle. \(\square\) Kernels of the form \(\|b(s)-b(t)\|^2\) are always conditionally negative semidefinite, as in Remark 2.2 of the previous lesson; the construction above shows the converse. ## 2. The proof of the theorem *Proof of Theorem 5.1 of the first lesson.* Let \(0\ne u\in V\), and let \(g(s)=(\nabla u)(\Phi(s))\) be its boundary gradient on the circle, as in (6.1) of [Boundary kernels and the multiplier identity](boundary-kernels-and-the-multiplier-identity.md). Suppose that \(\nabla u\) vanishes at the point \(\Phi(p)\), \(p\in\mathbb D\). *Step 1: the multipliers produce eigenfunctions.* Let \(b\) be as in Lemma 1.1 for this \(p\), let \((e_j)\) be a finite or countable orthonormal basis of \(\mathcal H\), and let \(b_j(s)=\langle b(s),e_j\rangle\). Each \(b_j\) is a real Lipschitz function, so Proposition 6.2 of [Boundary kernels and the multiplier identity](boundary-kernels-and-the-multiplier-identity.md) applies: \[ \begin{aligned} E(b_jg)&=\frac12\iint N(s,t)(b_j(s)-b_j(t))^2\\ &\qquad\cdot g(s)\cdot g(t)\,ds\,dt . \end{aligned} \] By Parseval's identity, \(\sum_j(b_j(s)-b_j(t))^2=D_p(s,t)\), so for every partial sum the absolute value of the integrand is at most \[ \begin{aligned} &N(s,t)D_p(s,t)|g(s)||g(t)|\\ &\quad=K(p,s)K(p,t)|g(s)||g(t)|, \end{aligned} \] which is bounded. By dominated convergence and (4.2) of that lesson, applied componentwise, \[ \begin{aligned} &\sum_jE(b_jg)\\ &\quad=\frac12\iint K(p,s)K(p,t)\\ &\qquad\cdot g(s)\cdot g(t)\,ds\,dt\\ &\quad=\frac12\Bigl|\int_{\mathbb S^1}K(p,s)g(s)\,ds\Bigr|^2\\ &\quad=\frac12|W_g(p)|^2 . \end{aligned} \] By Lemma 6.1 there, \(W_g(p)=\nabla u(\Phi(p))=0\). Every summand is nonnegative by (5.1) there, so \(E(b_jg)=0\) for every \(j\), and by (5.1) again there is \(v_j\in V\) with \(X_{b_jg}=\nabla v_j\). The function \((\nabla v_j)\circ\Phi\) lies in \(C^\infty(\overline{\mathbb D})\), and its trace equals the trace \(b_jg\) of \(W_{b_jg}\); both are continuous, so \[ \begin{gathered} (\nabla v_j)(\Phi(s))=b_j(s)g(s)\\ (s\in\mathbb S^1). \end{gathered} \tag{2.1} \] *Step 2: \(g\) does not vanish identically.* Let \(S_g=\{s\in\mathbb S^1:g(s)\ne0\}\), an open set. If it were empty, each \(w=\partial_iu\in C^\infty(\overline\Omega)\) would vanish on \(\partial\Omega\) and satisfy \(\Delta w+\mu w=0\); then \(w\in H^1_0(\Omega)\) and \(\int|\nabla w|^2=\mu\int w^2\), which forces \(w=0\) because \(\mu<\lambda_D\). So \(u\) would be constant, and \(\mu u=-\Delta u=0\) would give \(u=0\). *Step 3: the case where \(S_g\) is dense.* The set \(b(S_g)\) is not contained in an affine line of \(\mathcal H\). Otherwise, by continuity and density, \(b(\mathbb S^1)\) would lie in that line, and \(b\) would be a continuous injection of the circle into a copy of \(\mathbb R\). Its image would be a compact interval \([\alpha,\beta]\) with \(\alpha<\beta\), and \(b\) would be a homeomorphism onto it; but removing an interior point disconnects the interval, while the circle with one point removed is connected. Choose distinct \(s_1,s_2\in S_g\), and then \(s_3\in S_g\) with \(b(s_3)\) off the line through \(b(s_1)\) and \(b(s_2)\). The vectors \(y_1=b(s_2)-b(s_1)\) and \(y_2=b(s_3)-b(s_1)\) are linearly independent. Some coordinates \(j,k\) have \(\langle y_1,e_j\rangle\langle y_2,e_k\rangle-\langle y_1,e_k\rangle\langle y_2,e_j\rangle\ne0\): if all these \(2\times2\) minors vanished, a nonzero coordinate \(\langle y_1,e_j\rangle\) would make every coordinate of \(y_2\) proportional to that of \(y_1\). Subtracting the first row from the others shows \[ \det\begin{pmatrix}1&b_j(s_1)&b_k(s_1)\\1&b_j(s_2)&b_k(s_2)\\1&b_j(s_3)&b_k(s_3)\end{pmatrix}\ne0 . \] So \(1,b_j,b_k\) are linearly independent on \(\{s_1,s_2,s_3\}\). Since \(g(s_i)\ne0\), the functions \(g,b_jg,b_kg\) are linearly independent. By (2.1) they are the boundary gradients of \(u,v_j,v_k\in V\), which are therefore linearly independent. This contradicts \(\dim V\le2\) (Proposition 5.2 of [Conformal coordinates, Hopf's lemma and boundary signs](conformal-coordinates-hopfs-lemma-and-boundary-signs.md)). *Step 4: the case where \(S_g\) is not dense.* Then \(g=0\) on a nonempty open arc \(I\). Since \(S_g\) is nonempty and open, it has at least two points; \(b\) is injective, so by Parseval's identity some \(b_j\) is not constant on \(S_g\). Then \(g\) and \(b_jg\) are linearly independent: if \(\alpha g+\beta b_jg=0\), then \(\alpha+\beta b_j=0\) on \(S_g\), so \(\beta=0\) and \(\alpha=0\). Hence \(u\) and \(v_j\) are linearly independent. Their gradients vanish on the boundary arc \(\Phi(I)\), so each is constant on it. Choose \((\alpha,\beta)\ne(0,0)\) such that \(w=\alpha u+\beta v_j\) vanishes on \(\Phi(I)\). Then \(w\in V\), \(w\ne0\), and \(w=0\), \(\partial_\nu w=0\) on \(\Phi(I)\). Let \(x_0\in\Phi(I)\), and take a boundary chart at \(x_0\) whose rectangle \(U\) satisfies \(\partial\Omega\cap U\subseteq\Phi(I)\). Extend \(w\) by zero: \(\tilde w=w\) on \(\Omega\cap U\) and \(\tilde w=0\) on \(U\setminus\Omega\). For \(\phi\in C^\infty_c(U)\), Green's formula (2.2) of the first lesson, applied twice, gives \[ \begin{aligned} &\int_U\tilde w(\Delta\phi+\mu\phi)=\int_\Omega(\Delta w+\mu w)\phi\\ &\quad+\int_{\partial\Omega}(w\,\partial_\nu\phi-\phi\,\partial_\nu w)\,ds=0, \end{aligned} \] because \(\Delta w+\mu w=0\) and both \(w\) and \(\partial_\nu w\) vanish on \(\partial\Omega\cap U\). So \((\Delta+\mu)\tilde w=0\) in the sense of distributions on \(U\), and \(\tilde w\) is real analytic in \(U\) ([Anisotropic derivative classes and analyticity](course:constant-coefficient-equations-and-solvability/AN02-L031), Theorem 3.1). It vanishes on the nonempty open set \(U\setminus\overline\Omega\), and \(U\) is connected, so \(\tilde w=0\) on \(U\). Thus \(w\) vanishes on the nonempty open set \(\Omega\cap U\), and \(w=0\) by Proposition 3.1(4) of the first lesson, a contradiction. Both cases are impossible, so \(\nabla u\) does not vanish in \(\Omega\). \(\square\) The zero extension in Step 4 is a local form of a lemma of Filonov [Fil04, Lemma, p. 173]. No analyticity of the boundary is needed: only the vanishing of both Cauchy data on the arc. ## 3. Later developments Burdzy, and independently Bañuelos, reduced a non-strict hot-spots statement on general simply connected planar domains to the smooth theorem proved in this course [Bur26]. Let \(\Omega\) be a bounded simply connected planar domain, with no boundary regularity, such that the inclusion \(H^1(\Omega)\to L^2(\Omega)\) is compact and the first positive Neumann eigenvalue is simple. Assuming the smooth theorem, every nonzero eigenfunction \(u\) satisfies \[ \sup_\Omega u=\lim_{\varepsilon\to0}\ \sup_{\operatorname{dist}(x,\partial\Omega)<\varepsilon}u(x), \] and likewise for the infimum [Bur26, Theorem 2.1]; if \(u\) extends continuously to \(\overline\Omega\), its maximum and minimum over \(\overline\Omega\) are attained on \(\partial\Omega\) [Bur26, Corollary 2.2]. The argument approximates \(\Omega\) by the images of concentric discs under a conformal map and proves convergence of the eigenvalues and eigenfunctions. That preprint reports that it was written with ChatGPT and that its author has not checked every detail; its argument is not reproduced or checked here. Neither the strict conclusion nor the multiple-eigenvalue case is covered by that reduction. In higher dimensions the convex form of the conjecture fails in all sufficiently large dimensions [dDP24], and in the plane the simple connectivity assumption cannot be dropped [Bur05, Theorem 1.1]. ## 4. Exercises **4.1.** For \(q=0\) and \(p=0\), find an explicit map \(b\) in Lemma 1.1 with values in \(\mathbb R^2\). **4.2.** Show that the kernel \(\Gamma\) in the proof of Lemma 1.1 satisfies \(\Gamma(s,s_0)=0\), so \(b(s_0)=0\). **4.3.** In Step 4, suppose only \(w=0\) on the arc, not \(\partial_\nu w=0\). Compute the distribution \((\Delta+\mu)\tilde w\) on \(U\). **4.4.** Show that the theorem gives, for \(0\ne u\in V\), that every level set \(\{u=c\}\) with \(\min_{\partial\Omega}u