Conformal coordinates, Hopf's lemma and boundary signs

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The proof of the hot-spots theorem works on the unit disc \(\mathbb D=\{z\in\mathbb C:|z|<1\}\) with boundary circle \(\mathbb S^1\). This lesson constructs a conformal map from \(\mathbb D\) onto \(\Omega\) that is smooth up to the boundary, using Hopf's boundary point lemma and the argument principle. It then transfers the mean-zero condition of Neumann eigenfunctions to a weighted condition on the circle, shows that their sign sets are connected, and deduces that the first positive Neumann eigenvalue has multiplicity at most two.

Notation is that of Neumann eigenfunctions and the hot-spots problem: \(\Omega\) is a nonempty bounded simply connected open set with \(C^\infty\) boundary, \(\mu\) and \(V\) are the first positive Neumann eigenvalue and its eigenspace, \(\lambda_D\) is the first Dirichlet eigenvalue, and (SOB), (ER) are the named analytic facts. Proposition 3.1 there gives \(V\subseteq C^\infty(\overline\Omega)\), and Proposition 4.1 there gives \(\mu<\lambda_D\). The circle \(\mathbb S^1\) carries arclength \(ds\), of total mass \(2\pi\), and points \(s=e^{i\theta}\).

1. Interior discs and Hopf's lemma

Lemma 1.1 (interior discs). For every \(x_0\in\partial\Omega\) and every sufficiently small \(\rho>0\), the disc \(B\) of radius \(\rho\) with centre \(x_0-\rho\nu(x_0)\) satisfies \(B\subseteq\Omega\); its outward normal at \(x_0\in\partial B\) is \(\nu(x_0)\).

Proof. In a boundary chart centred at \(x_0\) (Section 1 of the first lesson), \(\Omega\) is the region \(x_2>\gamma(x_1)\) in a rectangle \((-a,a)\times(-b,b)\), with \(\gamma(0)=\gamma'(0)=0\), and \(\nu(x_0)=(0,-1)\). Let \(c\ge0\) bound \(|\gamma''|/2\), so \(\gamma(x_1)\le cx_1^2\). Take \(\rho<\min(a,b/2)\) with \(2c\rho<1\). A point of the disc of radius \(\rho\) centred at \((0,\rho)\) satisfies \(|x_1|<\rho\), \(0<x_2<2\rho\) and \[ \begin{aligned} x_2&>\rho-\sqrt{\rho^2-x_1^2}\\ &=\frac{x_1^2}{\rho+\sqrt{\rho^2-x_1^2}}\\ &\ge\frac{x_1^2}{2\rho}\ge cx_1^2\ge\gamma(x_1) \end{aligned} \] when \(x_1\ne0\), and \(x_2>0=\gamma(0)\) when \(x_1=0\). \(\square\)

Lemma 1.2 (Hopf's boundary point lemma). Let \(B\) be an open disc with centre \(y\) and radius \(\rho\), let \(x_0\in\partial B\) and \(n=(x_0-y)/\rho\). Let \(w\in C^2(B)\cap C^1(\overline B)\) satisfy \(\Delta w\ge0\) in \(B\) and \(w(x)<w(x_0)\) for all \(x\in B\). Then \(\partial_nw(x_0)>0\).

Proof. Let \(A\) be the annulus \(\rho/2<|x-y|<\rho\), fix \(\alpha>4/\rho^2\), and put \(v(x)=e^{-\alpha|x-y|^2}-e^{-\alpha\rho^2}\). With \(r=|x-y|\), in two dimensions \[ \begin{aligned} \Delta v&=e^{-\alpha r^2}(4\alpha^2r^2-4\alpha)\\ &=4\alpha e^{-\alpha r^2}(\alpha r^2-1)>0\quad\text{on }A, \end{aligned} \] since \(\alpha r^2>\alpha\rho^2/4>1\). On the inner circle, \(w-w(x_0)\le-\delta\) for some \(\delta>0\), by compactness. Choose \(\varepsilon>0\) with \(\varepsilon\le\delta\), and put \(z=w-w(x_0)+\varepsilon v\). Then \(\Delta z>0\) in \(A\), so \(z\) has no interior maximum in \(A\) (at an interior maximum the Hessian is negative semidefinite). On the inner circle \(z\le-\delta+\varepsilon\le0\); on the outer circle \(v=0\) and \(w\le w(x_0)\), so \(z\le0\). Hence \(z\le0\) on \(A\), while \(z(x_0)=0\). Approaching \(x_0\) along \(-n\) gives \(\partial_nz(x_0)\ge0\), that is, \[ \begin{aligned} \partial_nw(x_0)&\ge-\varepsilon\,\partial_nv(x_0)\\ &=2\varepsilon\alpha\rho e^{-\alpha\rho^2}>0.\qquad\square \end{aligned} \]

2. A conformal map that is smooth up to the boundary

We write \(z\) for the complex coordinate. For a bounded open set \(U\) with \(C^1\) boundary and a complex function \(F\in C^1(\overline U)\), (SOB 1) applied to the real and imaginary parts of \((F,0)\) and \((0,F)\) gives \(\int_{\partial U}F\nu\,ds=\int_U(\partial_xF+i\partial_yF)\), with \(\nu=\nu_1+i\nu_2\). Since \(\tau=i\nu\), \[ \begin{gathered} \int_{\partial U}F\,\tau\,ds=2i\int_U\partial_{\bar z}F,\\ \partial_{\bar z}=\tfrac12(\partial_x+i\partial_y). \end{gathered} \tag{2.1} \] In particular \(\int_{\partial U}F\tau\,ds=0\) if \(F\) is holomorphic in \(U\).

Proposition 2.1 ([OpenAI-HS, Lemma 2.2]). There is a holomorphic bijection \(\Phi\colon\mathbb D\to\Omega\) that extends to a homeomorphism \(\overline{\mathbb D}\to\overline\Omega\), with \(\Phi\in C^\infty(\overline{\mathbb D})\), \(\Phi^{-1}\in C^\infty(\overline\Omega)\), and \(\Phi'\ne0\) on \(\overline{\mathbb D}\).

Proof. Step 1: a harmonic function with logarithmic boundary values. Fix \(a\in\Omega\) and a smooth \(\chi\) on \(\mathbb R^2\) that vanishes near \(a\) and equals \(1\) near \(\partial\Omega\). Then \(E=\chi\log|z-a|\) is smooth on \(\mathbb R^2\). By (B10) and (B29) of Boundary energy, local inverses, and harmonic data, there is \(w\in H^1_0(\Omega)\cap C^\infty(\overline\Omega)\) with \(-\Delta w=\Delta E\). The function \(l=E+w\in C^\infty(\overline\Omega)\) is harmonic in \(\Omega\), and \(l=\log|z-a|\) on \(\partial\Omega\), because \(w\) is continuous with zero trace.

Step 2: a harmonic conjugate. The function \(F=l_x-il_y\) is holomorphic in \(\Omega\) by the Cauchy–Riemann equations, which hold because \(l\) is harmonic. By Lemma A.2 of Dirichlet series and Euler products, \(F\) has a primitive \(G\) in the simply connected domain \(\Omega\). Since \(\partial_x\operatorname{Re}G=\operatorname{Re}G'=l_x\) and \(\partial_y\operatorname{Re}G=-\operatorname{Im}G'=l_y\), adding a real constant to \(G\) gives \(G=l+i\tilde l\). All derivatives of \(\tilde l\) of positive order extend continuously to \(\overline\Omega\), because \(\nabla\tilde l=(-l_y,l_x)\). The function \(\tilde l\) itself extends continuously: in a boundary chart, two points \(p,q\) of \(\Omega\) near the boundary point can be joined inside \(\Omega\) by moving up a distance \(L|p-q|\), across, and down, where \(L=1+\sup|\gamma'|\); this path has length at most \((2L+1)|p-q|\), so \(\tilde l\) is Lipschitz there. Hence \(\tilde l\in C^\infty(\overline\Omega)\).

Step 3: the candidate inverse. Put \(m(z)=(z-a)e^{-G(z)}\). It is holomorphic in \(\Omega\), lies in \(C^\infty(\overline\Omega)\), and has exactly one zero, at \(a\), which is simple. The function \(\log|m|=\log|z-a|-l\) is harmonic in \(\Omega\setminus\{a\}\), continuous on \(\overline\Omega\setminus\{a\}\), zero on \(\partial\Omega\), and tends to \(-\infty\) at \(a\). We show \(|m|<1\) in \(\Omega\). Choose \(r>0\) with \(\overline{B(a,r)}\subseteq\Omega\) and \(\log|m|<-1\) on \(\overline{B(a,r)}\setminus\{a\}\), and let \(\Omega_r=\Omega\setminus\overline{B(a,r)}\). The maximum \(M\) of \(\log|m|\) over the compact set \(\overline{\Omega_r}\) is at least \(0\). If it is attained at a point of \(\Omega_r\), the harmonic maximum principle (Lemma A.3 of Dirichlet series and Euler products) makes \(\log|m|\) equal to \(M\) on the component \(C\) of \(\Omega_r\) containing that point. By continuity \(\log|m|=M\ge0\) on \(\partial C\), which is nonempty and contained in \(\partial\Omega\cup\partial B(a,r)\); since \(\log|m|<-1\) on \(\partial B(a,r)\), \(\partial C\subseteq\partial\Omega\). Then \(C\) is open and closed in \(\Omega\), so \(C=\Omega\), which is impossible because \(a\notin C\). Hence \(M\) is attained only on \(\partial\Omega_r\), and therefore only on \(\partial\Omega\), where \(\log|m|=0\). So \(M=0\), \(\log|m|<0\) on \(\Omega_r\), and \(|m|<1\) in \(\Omega\).

Step 4: \(m'\ne0\) on \(\partial\Omega\). Let \(x_0\in\partial\Omega\), and let \(B\) be an interior disc at \(x_0\) from Lemma 1.1, so small that \(a\notin\overline B\). The function \(\log|m|\) is \(C^1\) on \(\overline B\), harmonic in \(B\), negative in \(B\) and zero at \(x_0\). By Lemma 1.2, its normal derivative at \(x_0\) is positive. Near any point where \(m\ne0\), \(\log|m|\) is the real part of a local holomorphic logarithm of \(m\), so \(|\nabla\log|m||=|m'/m|\); by continuity this holds at \(x_0\), and \(m'(x_0)\ne0\).

Step 5: \(m\) maps \(\Omega\) bijectively onto \(\mathbb D\). Let \(w_0\in\mathbb D\). The zeros of \(m-w_0\) lie in the compact set \(\{|m|\le|w_0|\}\subseteq\Omega\); they are isolated (Lemma A.3 there), hence finitely many, say \(z_j\) with orders \(k_j\). Remove small closed discs \(\overline{B(z_j,\varepsilon)}\) from \(\Omega\) and apply (2.1) to \(F=m'/(m-w_0)\), which is \(C^1\) up to the boundary of the remaining set and holomorphic inside. On the circle around \(z_j\), oriented clockwise as part of that boundary, write \(m-w_0=(z-z_j)^{k_j}h\) with \(h\) zero-free near \(z_j\); then \(F=k_j/(z-z_j)+h'/h\), and the circle integral is \(-2\pi ik_j\), because \(h'/h\) has zero integral around the circle (Lemma A.2 there). Hence \[ \begin{aligned} N(w_0)&:=\sum_jk_j\\ &=\frac1{2\pi i}\int_{\partial\Omega}\frac{m'}{m-w_0}\,\tau\,ds . \end{aligned} \] On \(\partial\Omega\), \(|m-w_0|\ge1-|w_0|>0\), so the right side is continuous in \(w_0\in\mathbb D\). Being integer-valued on the connected set \(\mathbb D\), it is constant, equal to \(N(0)=1\). So every \(w_0\in\mathbb D\) has exactly one preimage in \(\Omega\), of multiplicity one; in particular \(m'\ne0\) in \(\Omega\), and \(m\colon\Omega\to\mathbb D\) is a bijection.

Step 6: the boundary. Every \(\zeta\in\mathbb S^1\) is a limit of \(m(z_n)\) with \(z_n\in\Omega\); a limit point \(z\) of \((z_n)\) satisfies \(m(z)=\zeta\), so \(z\in\partial\Omega\). Hence \(m(\partial\Omega)=\mathbb S^1\). Near \(x_0\in\partial\Omega\), flatten the boundary in a boundary chart by \((x_1,x_2)\mapsto(y_1,y_2)=(x_1,x_2-\gamma(x_1))\), write \(f\) for \(m\) in the coordinates \(y\), and put \(\hat f(y_1,y_2)=3f(y_1,-y_2)-2f(y_1,-2y_2)\) for \(y_2<0\). Values and first derivatives match at \(y_2=0\), so this gives a \(C^1\) map \(\hat m\) on a neighbourhood \(N\) of \(x_0\) that equals \(m\) on \(N\cap\overline\Omega\). Its Jacobian at \(x_0\) is \(|m'(x_0)|^2\ne0\). Suppose \(m(x_1)=m(x_2)=\zeta\) with \(x_1\ne x_2\) on \(\partial\Omega\). By the inverse function theorem there are disjoint neighbourhoods \(N_1,N_2\) of \(x_1,x_2\) and a disc \(U\) around \(\zeta\) such that \(\hat m\) maps each \(N_i\) homeomorphically onto \(U\). The preimage in \(N_i\) of the connected set \(U\cap\mathbb D\) is connected and contained in \(N_i\setminus\partial\Omega=(N_i\cap\Omega)\cup(N_i\setminus\overline\Omega)\), since \(\hat m\) maps \(\partial\Omega\) into \(\mathbb S^1\); it meets \(N_i\cap\Omega\), whose image contains points of \(\mathbb D\) near \(\zeta\). So it lies in \(\Omega\), and points of \(U\cap\mathbb D\) have two preimages in \(\Omega\), contradicting Step 5. Hence \(m\colon\overline\Omega\to\overline{\mathbb D}\) is a continuous bijection of compact spaces, and its inverse \(\Phi\) is continuous.

Step 7: smoothness of the inverse. At \(\zeta=m(x_0)\in\mathbb S^1\), the local \(C^1\) inverse of \(\hat m\) agrees with \(\Phi\) on \(\overline{\mathbb D}\) near \(\zeta\), by continuity of \(\Phi\) and injectivity of \(m\). So \(\Phi\) is \(C^1\) up to the boundary, holomorphic in \(\mathbb D\), and \(\Phi'=1/(m'\circ\Phi)\) is continuous and nonzero on \(\overline{\mathbb D}\). If all derivatives of \(\Phi\) of order at most \(k\) extend continuously to \(\overline{\mathbb D}\), the chain rule shows the same for \(m'\circ\Phi\), hence for \(\Phi'=1/(m'\circ\Phi)\), and so for the derivatives of \(\Phi\) of order \(k+1\). Thus \(\Phi\in C^\infty(\overline{\mathbb D})\), and \(\Phi^{-1}=m\in C^\infty(\overline\Omega)\). \(\square\)

From now on \(\Phi\) is fixed. It maps \(\mathbb S^1\) onto \(\partial\Omega\), and the arclength of \(\partial\Omega\) at \(\Phi(s)\) is \(|\Phi'(s)|\,ds\). For \(v\in V\) put \(h_v=v\circ\Phi\) on \(\mathbb S^1\).

3. A positive boundary weight

Lemma 3.1 ([OpenAI-HS, Lemma 2.3]). There is \(\beta\in C^\infty(\mathbb S^1)\) with \(\beta>0\) such that \[ \int_{\mathbb S^1}\beta h_v\,ds=0\quad(v\in V). \tag{3.1} \] If \(v\ne0\), then \(h_v\ne0\), and \(h_v\) takes both signs.

Proof. By Proposition 4.1 of the first lesson and Lemma 2.3 there, \(\langle\phi,\psi\rangle_\mu=\int\nabla\phi\cdot\nabla\psi-\mu\int\phi\psi\) is an inner product on \(H^1_0(\Omega)\) equivalent to the \(H^1\) inner product: \(\langle\phi,\phi\rangle_\mu\ge(1-\mu/\lambda_D)\|\nabla\phi\|^2\). The Riesz–Fréchet theorem gives \(L\in H^1_0(\Omega)\) with \(\langle L,\phi\rangle_\mu=\int\phi\) for all \(\phi\in H^1_0(\Omega)\). Thus \(\int\nabla L\cdot\nabla\phi=\int(1+\mu L)\phi\), and (ER D) applied repeatedly, with (SOB 4), gives \(L\in C^\infty(\overline\Omega)\), \(-\Delta L-\mu L=1\) in \(\Omega\) and \(L=0\) on \(\partial\Omega\).

\(L\ge0\). Let \(L^-=\max(-L,0)\). Approximating \(t\mapsto\max(-t,0)\) by \(\frac12(\sqrt{t^2+\delta^2}-\delta-t)\) as in Lemma 3.2 of the first lesson shows \(L^-\in H^1_0(\Omega)\) with \(\nabla L^-=-\mathbf 1_{\{L<0\}}\nabla L\) almost everywhere (the set where \(L=0\) and \(\nabla L\ne0\) is locally a smooth curve, of measure zero). Testing with \(\phi=L^-\) gives \[ -\int|\nabla L^-|^2+\mu\int(L^-)^2=\int L^-\ge0 . \] If \(L^-\ne0\), this gives \(\int|\nabla L^-|^2\le\mu\int(L^-)^2<\lambda_D\int(L^-)^2\), contradicting the definition of \(\lambda_D\). So \(L\ge0\).

\(L>0\) in \(\Omega\) and \(-\partial_\nu L>0\) on \(\partial\Omega\). At an interior zero of \(L\ge0\), \(L\) has a minimum and \(\Delta L\ge0\), whereas \(\Delta L=-1-\mu L=-1\) there. So \(L>0\) in \(\Omega\). At \(x_0\in\partial\Omega\) apply Lemma 1.2 to \(w=-L\) on an interior disc from Lemma 1.1: \(\Delta w=1+\mu L>0\) and \(w<0=w(x_0)\) in the disc. So \(-\partial_\nu L(x_0)>0\).

The identity. For \(v\in V\), Green's formula (2.2) of the first lesson applied twice gives \[ \begin{aligned} \int_\Omega v&=\int_\Omega v(-\Delta L-\mu L)\\ &=\int_\Omega(-\Delta v-\mu v)L\\ &\quad+\int_{\partial\Omega}(L\,\partial_\nu v-v\,\partial_\nu L)\,ds\\ &=-\int_{\partial\Omega}v\,\partial_\nu L\,ds, \end{aligned} \] since \(\Delta v+\mu v=0\), \(\partial_\nu v=0\) and \(L=0\) on \(\partial\Omega\). As \(\int_\Omega v=0\), the change of variables \(x=\Phi(s)\) gives (3.1) with \(\beta=|\Phi'|\,(-\partial_\nu L)\circ\Phi\), which is smooth and positive.

If \(h_v=0\), then \(v\in C^\infty(\overline\Omega)\) vanishes on \(\partial\Omega\), so \(v\in H^1_0(\Omega)\) and \(\mu\|v\|^2=\|\nabla v\|^2\ge\lambda_D\|v\|^2\), which forces \(v=0\). If \(h_v\ne0\), (3.1) and \(\beta>0\) force both signs. \(\square\)

4. Two paths with alternating endpoints meet

Lemma 4.1 (continuous logarithms). Let \(Q=[0,1]^2\) and let \(F\colon Q\to\mathbb C\setminus\{0\}\) be continuous. There is a continuous \(G\colon Q\to\mathbb C\) with \(e^G=F\).

Proof. Let \(m=\min_Q|F|>0\), and choose \(\delta>0\) with \(|F(x)-F(y)|<m\) whenever \(|x-y|\le\delta\). Fix an integer \(N>\sqrt2/\delta\) and a complex number \(g_0\) with \(e^{g_0}=F(0)\). For \(x\in Q\) put \(x_j=(j/N)x\), \(0\le j\le N\). Then \(|F(x_{j+1})/F(x_j)-1|<1\), so the principal logarithm \(\operatorname{Log}\) is defined and continuous at these quotients, and \[ G(x)=g_0+\sum_{j=0}^{N-1}\operatorname{Log}\frac{F(x_{j+1})}{F(x_j)} \] is continuous with \(e^{G(x)}=F(0)\prod_jF(x_{j+1})/F(x_j)=F(x)\). \(\square\)

For \(p\in\mathbb S^1\) let \(H_p=\{w\in\mathbb C:\operatorname{Re}(w\bar p)>0\}\), fix a real \(\varphi_p\) with \(p=e^{i\varphi_p}\), and define the continuous argument \(\theta_p(w)=\varphi_p+\operatorname{Arg}(w/p)\) on \(H_p\), where \(\operatorname{Arg}\) takes values in \((-\pi/2,\pi/2)\) on the right half-plane. For \(z\in\overline{\mathbb D}\setminus\{p\}\), \(\operatorname{Re}((p-z)\bar p)=1-\operatorname{Re}(z\bar p)>0\), so \(p-z\in H_p\). For \(q=e^{i\kappa}\in\mathbb S^1\setminus\{p\}\), with \(\kappa\) chosen in \((\varphi_p-2\pi,\varphi_p)\), \[ \begin{aligned} &p-q=2\sin\bigl(\tfrac{\varphi_p-\kappa}2\bigr)\\ &\qquad\cdot e^{i((\varphi_p+\kappa)/2+\pi/2)},\\ &\theta_p(p-q)=\tfrac{\varphi_p+\kappa}2+\tfrac\pi2, \end{aligned} \tag{4.1} \] because the sine is positive and \((\kappa-\varphi_p)/2+\pi/2\in(-\pi/2,\pi/2)\).

Lemma 4.2 (planar separation). Let \(a,b,c,d\in\mathbb S^1\) be distinct and in this counterclockwise order. Let \(\gamma_1,\gamma_2\colon[0,1]\to\overline{\mathbb D}\) be continuous paths from \(a\) to \(c\) and from \(b\) to \(d\), with \(\gamma_1(s),\gamma_2(s)\in\mathbb D\) for \(0<s<1\). Then \(\gamma_1(s)=\gamma_2(t)\) for some \(s,t\in(0,1)\).

Proof. Suppose not. Then \(F(s,t)=\gamma_1(s)-\gamma_2(t)\) never vanishes on \(Q\): interior points of the paths lie in \(\mathbb D\), their endpoints lie on \(\mathbb S^1\), and the four endpoints are distinct. Let \(G\) be as in Lemma 4.1; \(\operatorname{Im}G\) is a continuous argument of \(F\). Along the side \(t=0\), \(F(s,0)=-(b-\gamma_1(s))\) with \(b-\gamma_1(s)\in H_b\), so \(\operatorname{Im}G(s,0)-\pi-\theta_b(b-\gamma_1(s))\) is a continuous function with values in \(2\pi\mathbb Z\), hence constant. The same reasoning on the four sides, traversed counterclockwise around \(Q\), gives \[ \begin{aligned} 0&=[\theta_b(b-c)-\theta_b(b-a)]\\ &\quad+[\theta_c(c-d)-\theta_c(c-b)]\\ &\quad+[\theta_d(d-a)-\theta_d(d-c)]\\ &\quad+[\theta_a(a-b)-\theta_a(a-d)], \end{aligned} \] using \(F(1,t)=c-\gamma_2(t)\), \(F(s,1)=-(d-\gamma_1(s))\) and \(F(0,t)=a-\gamma_2(t)\). Write \(a,b,c,d=e^{i\alpha},e^{i\beta},e^{i\gamma},e^{i\delta}\) with \(\alpha<\beta<\gamma<\delta<\alpha+2\pi\), and take \(\varphi_a,\varphi_b,\varphi_c,\varphi_d=\alpha,\beta,\gamma,\delta\). By (4.1) the four brackets are \[ \begin{gathered} \frac{(\gamma-2\pi)-\alpha}2,\qquad\frac{(\delta-2\pi)-\beta}2,\\ \frac{\alpha-\gamma}2,\qquad\frac{(\beta-2\pi)-(\delta-2\pi)}2, \end{gathered} \] whose sum is \(-2\pi\ne0\). \(\square\)

5. Sign sets and the multiplicity bound

Lemma 5.1 ([OpenAI-HS, Lemma 2.4]). Let \(0\ne v\in V\).

  1. The sets \(\{v>0\}\) and \(\{v<0\}\) are nonempty and connected.
  2. There are no points \(s_1,s_2,s_3,s_4\in\mathbb S^1\), in this counterclockwise order, with \(h_v(s_1)>0\), \(h_v(s_2)<0\), \(h_v(s_3)>0\), \(h_v(s_4)<0\).

Proof. (1) Both sets are nonempty because \(v\) has mean zero. Let \(U\) be a connected component of \(\{v>0\}\). At a point \(z\in\partial U\cap\Omega\), \(v(z)=0\): if \(v(z)>0\), a small disc around \(z\) lies in \(\{v>0\}\) and meets \(U\), so it lies in \(U\); if \(v(z)<0\), a small disc around \(z\) misses \(U\). Choose smooth nondecreasing \(g_\varepsilon\) with \(g_\varepsilon=0\) on \((-\infty,\varepsilon]\) and \(g_\varepsilon=1\) on \([2\varepsilon,\infty)\), let \(G_\varepsilon(t)=\int_0^tg_\varepsilon\), and put \(w_\varepsilon=G_\varepsilon(v)\mathbf 1_U\). Near each point of \(\overline\Omega\), \(w_\varepsilon\) equals \(0\) or \(G_\varepsilon(v)\). This is clear near points of \(U\) and of \(\Omega\setminus\overline U\); near points of \(\partial U\cap\Omega\), and near boundary points where \(v\le0\), it holds because \(v<\varepsilon\) there; near a point of \(\partial\Omega\setminus\overline U\), because \(U\) stays away from it. Finally, if \(x_0\in\partial\Omega\cap\overline U\) and \(v(x_0)>0\), then in a boundary chart at \(x_0\) the set \(\{|x_1|<a',\ \gamma(x_1)<x_2<b'\}\) is connected, and for small \(a',b'\) it lies in \(\{v>0\}\) and meets \(U\), so it lies in \(U\). Hence \(w_\varepsilon\in C^\infty(\overline\Omega)\) with \(\nabla w_\varepsilon=g_\varepsilon(v)\mathbf 1_U\nabla v\). As \(\varepsilon\to0\), \(w_\varepsilon\to v\mathbf 1_U\) uniformly and \(\nabla w_\varepsilon\to\mathbf 1_U\nabla v\) in \(L^2\). So \(v_U=v\mathbf 1_U\in H^1(\Omega)\) with \(\nabla v_U=\mathbf 1_U\nabla v\). The same holds for components of \(\{v<0\}\), applied to \(-v\). Testing (3.1) of the first lesson with \(\phi=v_U\) gives \[ \begin{aligned} \int|\nabla v_U|^2&=\int_U|\nabla v|^2\\ &=\mu\int_Uv^2=\mu\int v_U^2 . \end{aligned} \] Suppose the two sign sets together have at least three components, among them \(U_1,U_2,U_3\). Choose \((c_1,c_2)\ne(0,0)\) with \(c_1\int v_{U_1}+c_2\int v_{U_2}=0\), and put \(w=c_1v_{U_1}+c_2v_{U_2}\). Since the supports are disjoint, \(\int|\nabla w|^2=\mu\int w^2\), and \(w\ne0\). By Proposition 3.1(2) of the first lesson \(w\in V\); it vanishes on the open set \(U_3\), so \(w=0\) by Proposition 3.1(4) there, a contradiction. Hence each sign set has exactly one component.

(2) Suppose such points exist, and let \(x_i=\Phi(s_i)\). By Lemma 1.1 and continuity of \(v\) on \(\overline\Omega\), there is a segment from \(x_1\) into \(\Omega\), along \(-\nu(x_1)\), on which \(v>0\); likewise at \(x_3\). Their inner endpoints lie in the open connected set \(\{v>0\}\), which is path-connected. This gives a path \(\sigma_1\) from \(x_1\) to \(x_3\), lying in \(\Omega\) except at its endpoints, with \(v>0\) on all of it. In the same way there is a path \(\sigma_2\) from \(x_2\) to \(x_4\) with \(v<0\) on all of it. The paths \(\Phi^{-1}\circ\sigma_1\) and \(\Phi^{-1}\circ\sigma_2\) satisfy the hypotheses of Lemma 4.2, so they meet, and so do \(\sigma_1\) and \(\sigma_2\), which is impossible. \(\square\)

Proposition 5.2 ([OpenAI-HS, Proposition 2.5]; [Nad88, Theorem 3]). \(\dim V\le2\).

Proof. Suppose \(\dim V\ge3\). The linear map \(v\mapsto\bigl(\int\beta h_v\cos\theta\,ds,\int\beta h_v\sin\theta\,ds\bigr)\) from \(V\) to \(\mathbb R^2\) has a nonzero kernel element \(v\). By Lemma 3.1, \(h=h_v\) is nonzero, takes both signs, and is orthogonal to \(\beta\), \(\beta\cos\theta\) and \(\beta\sin\theta\).

Choose \(\theta_0\) with \(h(e^{i\theta_0})<0\) and regard \(h\) as a function of \(\theta\in(\theta_0,\theta_0+2\pi)\). The set \(P\) of angles there where \(h>0\) is open and nonempty, and \(h<0\) near both ends of the interval; let \(a=\inf P\) and \(b=\sup P\). Then \(\theta_0<a<b<\theta_0+2\pi\), and \(h\le0\) outside \([a,b]\). There is no \(\theta'\in(a,b)\) with \(h(\theta')<0\): otherwise \(P\) has points \(p_1\in(a,\theta')\) and \(p_2\in(\theta',b)\), and the points \(e^{ip_1},e^{i\theta'},e^{ip_2},e^{i\theta_0}\) lie in counterclockwise order with signs \(+,-,+,-\), contrary to Lemma 5.1. So \(h\ge0\) on \([a,b]\). Put \(w=(b-a)/2\in(0,\pi)\), \(m=(a+b)/2\), and \[ F(\theta)=\cos(\theta-m)-\cos w . \] On \((a,b)\), \(|\theta-m|<w\) and \(F>0\). For \(\theta\) in \((\theta_0,a)\cup(b,\theta_0+2\pi)\), \(\theta-m\) lies in \((w-2\pi,-w)\cup(w,2\pi-w)\), where the cosine is less than \(\cos w\), so \(F<0\). Hence \(Fh\ge0\) everywhere and \(Fh>0\) on the nonempty open set \(P\). So \(\int\beta Fh\,ds>0\). But \(F=\cos m\cos\theta+\sin m\sin\theta-\cos w\) is a combination of \(1,\cos\theta,\sin\theta\), so this integral vanishes. \(\square\)

The bound is due to Nadirashvili [Nad88, Theorem 3], who proved it for the first positive Neumann eigenvalue of every two-dimensional Riemannian manifold diffeomorphic to a disc with nonpositive Gaussian curvature.

6. Exercises

6.1. Let \(\Omega=\mathbb D\) and \(\Phi(z)=z\). Show that the function \(L\) of the proof of Lemma 3.1 is radial, and deduce that \(\beta\) is constant.

6.2. Check (4.1) for \(p=1\) and \(q=i\).

6.3. Show that Lemma 4.2 fails when the endpoints do not alternate: give two disjoint chords of \(\overline{\mathbb D}\) joining \(1\) to \(i\) and \(-1\) to \(-i\).

6.4. In the proof of Proposition 5.2 take \(a=0\) and \(b=\pi\). Compute \(F\).

6.5. Let \(\Omega=\mathbb D\). Show that \(\dim V=2\).

7. Solutions

6.1. For a rotation \(R\), \(L\circ R\) solves the same problem, and the solution is unique (it is the Riesz–Fréchet representative). So \(L\) is radial, \(\partial_\nu L\) is constant on the circle, and \(\beta=|\Phi'|(-\partial_\nu L)\) is constant.

6.2. Here \(\varphi_p=0\) and \(q=e^{i\kappa}\) with \(\kappa=-3\pi/2\in(-2\pi,0)\). Formula (4.1) gives \(\theta_1(1-i)=-3\pi/4+\pi/2=-\pi/4\), the argument of \(1-i\).

6.3. The segments \([1,i]\) and \([-1,-i]\) lie in the half-planes \(x+y\ge1\) and \(x+y\le-1\), so they are disjoint. The points \(1,i,-1,-i\) are in counterclockwise order, and the paths join the first two and the last two, so the endpoints do not alternate.

6.4. \(m=\pi/2\), \(w=\pi/2\), and \(F(\theta)=\cos(\theta-\pi/2)=\sin\theta\).

6.5. Rotations \(R\) of the disc map \(V\) into itself, since (3.1) of the first lesson is rotation invariant. If \(V\) were spanned by one function \(v\), then \(v\circ R_\theta=c(\theta)v\) with \(c(\theta)=\pm1\) (the \(L^2\) norm is preserved) and \(c\) continuous with \(c(0)=1\), so \(c=1\) and \(v\) would be radial. Then \(h_v\) is constant, and (3.1) with constant \(\beta\) (Exercise 6.1) gives \(h_v=0\), so \(v=0\) by Lemma 3.1. Hence \(\dim V\ge2\), and Proposition 5.2 gives equality.

References