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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 (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 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 applies. At a critical point the multiplied boundary gradients are all gradients of eigenfunctions, and this contradicts either the multiplicity bound or unique continuation.

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, \(q=\mu|\Phi'|^2dx\) 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), 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. 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 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).

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, 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 \(\min_{\partial\Omega}u<c<\max_{\partial\Omega}u\) is, inside \(\Omega\), a one-dimensional smooth submanifold.

4.5. Show that a nonnegative Dirichlet eigenfunction for \(\lambda_D\) has an interior critical point. Which steps of the proof of the theorem use the Neumann condition?

5. Solutions

4.1. Here \(D_0(s,t)=\frac1{4\pi}|s-t|^2\) by Remark 2.2 of the previous lesson, so \(b(s)=s/(2\sqrt\pi)\in\mathbb R^2\) works (the construction of the proof gives \(b(s)-b(s_0)\) up to an isometry).

4.2. Since \(D_p(s_0,s_0)=0\), the definition gives \(\Gamma(s,s_0)=\frac12(D_p(s,s_0)-D_p(s,s_0))=0\), so \(\|b(s_0)\|^2=\Gamma(s_0,s_0)=0\).

4.3. The computation in Step 4 gives \(\int_U\tilde w(\Delta\phi+\mu\phi)=-\int_{\partial\Omega\cap U}\phi\,\partial_\nu w\,ds\), so \((\Delta+\mu)\tilde w\) is the measure \(-\partial_\nu w\,ds\) on the arc, which is not zero when \(\partial_\nu w\ne0\) there.

4.4. By the theorem \(\nabla u\ne0\) in \(\Omega\), so the implicit function theorem shows that \(\{u=c\}\cap\Omega\) is a smooth curve near each of its points.

4.5. Such a function \(\phi\) is continuous on \(\overline\Omega\), vanishes on \(\partial\Omega\) and is positive somewhere, so it attains its maximum in \(\Omega\), where \(\nabla\phi=0\). The Neumann condition makes the boundary gradient \(g\) tangent, which is needed to define \(E(bg)\) through tangent fields; and it gives \(E(g)=0\) through the equality case of Rohleder's principle, which is the starting point of the multiplier identity.

References