4 Deriving the geometric estimates

Proof. By scaling we may assume \(L(\sigma)=2\pi\). The symmetry implies that \(\kappa_\sigma\circ I=\kappa_\sigma\) and \(\mathbf{T}_\sigma\circ I=-\mathbf{T}_\sigma\). Hence, \(\mathcal{E}_S[\sigma, f]=\mathcal{E}_S[\sigma, f\circ I]\) and so, the variational characterization of the lowest eigenvalue implies that the lowest eigenfunction \(f\) must satisfy \(f\circ I=f\). As observed in [1], \[\mathcal{E}_S[\sigma,f]= \int_{\mathbb{S}^1} |\mathbf{y}'|^2-|\mathbf{y}|^2 \; \mathrm{d}\theta\] where \(\mathbf{y}=f \mathbf{T}_\sigma\). Moreover, \(\mathbf{y}(p)=(a \cos \theta(p)+b \sin \theta(p),c \cos \theta(p)+d \sin \theta(p) )\) if and only if \(\sigma\in \mathcal{O}\). As \(\mathbf{y}\circ I=-\mathbf{y}\), \[\int_{\mathbb{S}^1} \mathbf{y}\; \mathrm{d}\theta=0\] and the proposition follows from the one-dimensional Poincaré inequality.

Proof. Take \(f=\kappa_\sigma^{-1/2}\) in (1.2) and use Proposition 4.1.

Proof. As \(\int_{\mathbb{S}^1} \psi' \; \mathrm{d}\theta=2\pi\) and \(\psi'\) is continuous, there is a point \(p_0\), so that if \(\gamma_\pm\) are the components of \(\mathbb{S}^1\backslash \left\{p_0, I(p_0)\right\}\), then \(\int_{\gamma_\pm} \psi' \;\mathrm{d}\theta=\pi\). That is, \(p_0\) and \(I(p_0)\) are balance points. Expanding \(\psi'\) as a Fourier series, rotating so \(\theta(p_0)=0\) and abusing notation slightly, gives that \[\psi=C+\theta+\sum_{n=2}^\infty \left( a_n \sin n\theta + b_n \cos n\theta \right)=C+\theta+f(\theta)+g(\theta)\] where \(C\) is a constant, \(f\) are the remaining odd terms in the expansion and \(g\) are the remaining even terms. By construction, \(\psi(0)+\pi=\psi(\pi)\), \(f(\theta+\pi)=-f(\theta)\) and \(g(\theta+\pi)=g(\theta)\) and so \(f(0)=0=f(\pi)\) and all balance points of \(\psi\) in \(\gamma_+\) correspond to zeros of \(f\) in \((0,\pi)\). If \(f\) does not change sign on \((0,\pi)\), then either \(f\equiv 0\) and \(\psi\) has an infinite number of balance points, or \(\int_{0}^\pi f(\theta) \sin \theta \; \mathrm{d}\theta \neq 0\). However, as \(f(\theta+\pi)\sin (\theta+\pi)=f(\theta) \sin \theta\), this would imply \(\int_{0}^{2\pi} f(\theta) \sin \theta \; \mathrm{d}\theta \neq 0\) which is impossible. Hence, \(f\) must change sign and so \(f\) has at least one zero in \((0,\pi)\) which verifies the first claim.

To verify the second claim. We first scale so \(L(\sigma)=2\pi\). If \(\sigma\) is closed, then \(\int_{\mathbb{S}^1} \mathbf{T}_\sigma \; \mathrm{d}\theta=0\). That is, \(\int_{\mathbb{S}^1} \mathbf{T}_0 \circ \phi_\sigma \; \mathrm{d}\theta=0\). Changing variables, gives \[\int_{\mathbb{S}^1} (\phi_\sigma^{-1})' \mathbf{T}_0 \; \mathrm{d}\theta.\] Hence, \(n_B(\phi_\sigma^{-1})\geq 4\) and it is clear that \(n_B(\phi_\sigma)\geq 4\) as well.

Proof. Let \(B\) be the set of balance points of \(\phi\) and \(S\subset B\) be the set of stable balance points. It follows from the inverse function theorem that for each \(p\in S\), there is an open interval, \(I_p\), in \(\mathbb{S}^1\) so that \(B\cap I_p=\left\{p\right\}\). It is straightforward to show, after fixing smaller open intervals, \(I_p'\), satisfying \(p\subset I_p'\) and \(\bar{I}_p'\subset I_p\), that there are \(C^1\) neighborhoods, \(V_p\), of \(\phi\) in \(\mathrm{ Diff}_+^1(\mathbb{S}^1)\) so that all \(\psi \in V_p\) have only one stable balance point in \(I_p'\) and no unstable balance points.

If \(n_{SB}(\phi)> N\), then let \(S_N\subset S\) be some choice of \(N\) distinct points of \(S\), otherwise, let \(S_N=S\). As \(S_N\) is finite, \(V_N=\cap_{p\in S_N} V_p\) is a an open \(C^1\) neighborhood of \(\phi\) in \(\mathrm{ Diff}_+^1(\mathbb{S}^1)\) so that for any \(\psi\in V_N\), there are \(\min\left\{n_{SB}(\phi),N\right\}\) stable balance points in \(U'_N=\cup_{p\in S_N} I'_p\) and no unstable balance points. Hence, \(n_{SB}(\psi)\geq \min\left\{n_{SB}(\phi),N\right\}\) which completes the proof of the first claim. The second claim follows by taking \(N= n_{SB}(\phi)<\infty\). As \(n_{B}(\phi)=n_{SB}(\phi)\), there are no balance points in \(\mathbb{S}^1\backslash U'_N\) and so small \(C^0\) perturbations of \(\phi\) also have no balance points in \(\mathbb{S}^1\backslash U'_N\). In other words, by shrinking \(V_N\) one can ensure that \(n_{B}(\psi)=n_{SB}(\psi)=n_{SB}(\phi)=n_B(\phi)\) for all \(\psi\in V_N\).

Proof. If \(\phi \in \partial{\mathrm{ BDiff}}_+^{k, \alpha}(\mathbb{S}^1, 4)\) for \(k\geq 1\), then \(\phi \in \partial{\mathrm{ BDiff}}_+^{1}(\mathbb{S}^1, 4)\). Hence, we can restrict attention to the \(C^1\) setting. If \(n_B(\phi)=2\), then the two balance points must be unstable as otherwise Lemma 4.5 would imply that any \(C^1\) perturbation of \(\phi\) also has only two balance points – that is \(\phi \not\in \overline{\mathrm{ BDiff}}_+^{1}(\mathbb{S}^1, 4)\). If \(n_B(\phi)=\infty\), then it must have some unstable balance points, as the set balance points is closed while the set of stable balance points is discrete and so is a proper subset. Finally, if \(4\leq n_B(\phi)<\infty\), then Lemma 4.5 implies at least two of them are unstable. Otherwise, any \(C^1\) perturbation of \(\phi\) would continue to have at least four balance points, i.e., \(\phi\) is in the interior of \({\mathrm{ BDiff}}_+^{1}(\mathbb{S}^1, 4)\).

Proof. Fix \(p_0\in \mathbb{S}^1\), let \(\nu_{\epsilon}(p, p_0)\) be a family of \(C^\infty\) mollifiers with \(\nu_{\epsilon}(p,p_0)\geq 0\), \(\mathrm{supp}(\nu_{\epsilon}(\cdot, p_0))\subset B_{\epsilon}(p_0)\), \(\nu_{\epsilon}(p,p_0)=\nu_{\epsilon}(p_0,p)\), \(\nu_\epsilon(I(p_0),I(p))=\nu_\epsilon(p_0,p)\) and \(\int_{\mathbb{S}^1} \nu_{\epsilon}(p , p_0) \; \mathrm{d}\theta(p) =1\). That is, \(\lim_{\epsilon\to 0} \nu_\epsilon (p, p_0)= \delta_{p_0}(p)\) the Dirac delta with mass at \(p_0\). Set \[P_\epsilon=\int_{\mathbb{S}^1} \nu_{\epsilon} (\cdot , p ) \phi'(p)\; \mathrm{d}\theta (p)\in C^\infty(\mathbb{S}^1).\] Hence, \(\int_{\mathbb{S}^1} P_\epsilon \; \mathrm{d}\theta=2\pi\) and \(P_{\epsilon}\geq \min_{\mathbb{S}^1} \phi'>0\). It follows, that there are \(\phi_\epsilon \in \mathrm{ Diff}_+^\infty(\mathbb{S}^1)\) so that \(\phi_\epsilon(p_0)=\phi(p_0)\) and \(\phi_\epsilon'=P_\epsilon\). As \(\log \phi'\in H^1(\mathbb{S}^1)\), \(\phi'\in H^1(\mathbb{S}^1)\) and so \(P_\epsilon \to\phi'\) strongly in \(H^1(\mathbb{S}^1)\). This convergence together with the uniform lower bound on \(P_\epsilon\) and the Sobolev embedding theorem implies that \(\log P_\epsilon\) converge strongly in \(H^1(\mathbb{S}^1)\) to \(\log \phi'\) – that is, \(\phi_\epsilon\to\phi\) strongly in \(\mathrm{ HDiff}_+(\mathbb{S}^1)\).

Finally, we observe that if \(\phi\circ I=I\circ \phi\), then \(\phi'\circ I=\phi'\) and so \(P_\epsilon \circ I=P_\epsilon\). In particular, if \(\phi\circ I=I\circ \phi\), then \(\phi_\epsilon\circ I=I\circ \phi_\epsilon\).

Proof. By Lemma 4.7, there are a sequence of \(\phi_i\in \mathrm{ Diff}^\infty_+(\mathbb{S}^1)\), with \(\phi_i\circ I=I\circ \phi_i\) and \(\phi_i\to \phi\) strongly in \(\mathrm{ HDiff}_+(\mathbb{S}^1)\). In particular, \(\mathcal{E}_G^*[\phi_i]\to \mathcal{E}_G^*[\phi]\). Set \(\psi_i=\phi_i^{-1}\) and note that \(\psi_i\circ I=I\circ \psi_i\). Further, let \(\sigma_i\in \mathcal{D}^+_i\) have induced diffeomorphism \(\psi_i\). By (3.4), Proposition 3.1 and Corollary 4.2, \[\mathcal{E}_G^*[\phi_i]=\mathcal{E}_G^*[\sigma^*_i]=\mathcal{E}_G[\sigma_i]\geq 0,\] proving the desired inequality. If one has equality, then the inequality implies that \(\phi\) is critical with respect to variations preserving the symmetry. It follows that \(\phi\) is smooth and so \(\phi\in \mathrm{ M\ddot{o}b}(\mathbb{S}^1)\) by Corollary 4.2.

Proof. Let \(\phi \in \mathrm{ HDiff}^+(\mathbb{S}^1)\cap \partial{\mathrm{ BDiff}}_+^1(\mathbb{S}^1, 4)\). As \(\phi \in \mathrm{ Diff}_+^{1,\frac{1}{2}}(\mathbb{S}^1)\), Lemma 4.6 implies that \(\phi\) has at least one unstable balance point \(p_0\). Let \(\gamma_\pm\) be the two components of \(\mathbb{S}^1\backslash \left\{p_0, I(p_0)\right\}\). Up to relabeling, we may assume that \[\mathcal{E}_G^*[\phi]\geq 2 \left( \int_{\gamma_-} \frac{1}{4}(u')^2 -\mathrm{e}^{2u} \; \mathrm{d}\theta+2\pi\right)\] where \(u=\log \phi'\). Now define \[\tilde{u}(p)=\left \{ \begin{array}{cc} u(p) & p\in \bar{\gamma}_- \\ u(I(p)) & p \in \gamma_+ \end{array} \right.\] Here, \(\bar{\gamma}_-\) is the closure of \(\gamma_-\) in \(\mathbb{S}^1\). Clearly, \(\tilde{u}\) is continuous, \(\int_{\mathbb{S}^1} \mathrm{e}^{\tilde{u}} \; \mathrm{d}\theta=2\pi\) and \[\mathcal{E}_G^*[\phi]\geq E[\tilde{u}] +2\pi.\] Hence, there is a \(\tilde{\phi}\in \mathrm{ Diff}_+^{1,1}(\mathbb{S}^1)\subset \mathrm{ HDiff}_+(\mathbb{S}^1)\) so that \(\tilde{u}=\log\tilde{\phi}'\). By construction, \(\tilde{\phi}\circ I =I\circ \tilde{\phi}\) and so by Lemma 4.8 \[\mathcal{E}_G^*[\phi]\geq \mathcal{E}_G^*[\tilde{\phi}]\geq 0,\] with equality if and only if \(\tilde{\phi}\in \mathrm{ M\ddot{o}b}(\mathbb{S}^1)\).

In the case of equality for \(\phi\) we could reflect either \(\gamma_+\) or \(\gamma_-\), hence the preceding argument implies, \(\phi|_{\gamma_\pm}=\phi_\pm\) for \(\phi_\pm \in \mathrm{ M\ddot{o}b}(\mathbb{S}^1)\) which satisfy \(\phi_\pm(\gamma_+)=\gamma_+\). By precomposing with a rotation, we may assume that \(\left\{p_0,I(p_0)\right\}=\left\{\mathbf{e}_2,-\mathbf{e}_2\right\}\) and \(\theta(\gamma_+)=\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\). Taking \(\hat{\phi}=\phi_-\in \mathrm{ M\ddot{o}b}(\mathbb{S}^1)\), one has \(\phi\circ \hat{\phi}^{-1}\in \mathrm{ Diff}^1_+(\mathbb{S}^1)\) and is the identity map on \(\gamma_-\) and some element of \(\mathrm{ M\ddot{o}b}(\mathbb{S}^1)\) on \(\gamma_+\). This implies that \(\phi\circ\hat{\phi}^{-1}=\psi_\tau\) where \(\psi_\tau\) is of the form \(\eqref{eqn:BadFamily2}\) for some \(\tau\in \mathbb R\). That is, \(\phi=\psi_\tau \circ \hat{\phi}\).

Proof. It is straightforward to see that (4.5) has the conservation law \[\frac{1}{4}(u')^2-\alpha \mathrm{e}^{2u}+ 2 \beta \mathrm{e}^u = \eta.\] Integrating this we see that \[E[u]+(1-\alpha) \gamma + 4\pi \beta= 2\pi \eta.\] However, integrating (4.5) gives that \[-\alpha \gamma + 2\pi \beta=0\] and hence \[E[u]=2\pi \eta-\gamma- 2\pi \beta.\]

Now set \(U=\mathrm{e}^{-u}\) one has that \[\frac{1}{4}U'' =-\frac{1}{4}\mathrm{e}^{-u} u'' +\frac{1}{4}\mathrm{e}^{-u} (u')^2=-\alpha \mathrm{e}^{u}+\beta +\alpha \mathrm{e}^u -2\beta +\eta \mathrm{e}^{-u}=\eta U -\beta\] That is, \(U\) satisfies \[U''-4\eta U= -4\beta.\] As \(U\in C^\infty(\mathbb{S}^1)\), either \(U= \frac{\beta}{\eta}\), or \(4\eta=-n^2\) for some \(n\in \mathbb{Z}^+\) and \[U= \frac{\beta}{\eta}+C_1 \cos \sqrt{-4\eta} \theta + C_2 \sin \sqrt{-4\eta} \theta\] for some constants \(C_1, C_2\). In the first case, the constraints force \(\eta= \beta\) and so \(u=0\), \(\alpha=\beta=\eta\), \(\gamma=2\pi\) and \(E=-2\pi\).

In the second case, we first note that \(U>0\) and so \[\frac{\beta}{\eta}> \sqrt{C_1^2+C_2^2}.\] Using the calculus of residues, we compute that \[\begin{aligned} \int_{\mathbb{S}^1} \mathrm{e}^u \; \mathrm{d}\theta & = \int_{\mathbb{S}^1} \frac{1}{U} \; \mathrm{d}\theta \\ &= \int_{\mathbb{S}^1} \frac{1}{\frac{\beta}{\eta} +\frac{C_1}{2}(z^n+z^{-n})+\frac{C_2}{2i}(z^n-z^{-n})} \; \frac{\mathrm{d}z}{i z} \\&= \frac{2\pi}{\sqrt{\left(\frac{\beta}{\eta}\right)^2-C_1^2-C_2^2}}.\end{aligned}\] Keeping in mind that \(U>0\), the first constraint is satisfied if and only if \[\beta= \eta\sqrt{1+C_1^2+C_2^2}.\] Hence, \[u= -\log \left(\sqrt{1+C_1^2+C_2^2}+C_1 \cos \sqrt{-4\eta}\theta + C_2 \sin \sqrt{-4\eta} \theta\right).\] Plugging this into (4.5), shows that \(\alpha=\eta\). Hence, \[\gamma=2\pi \beta/ \alpha=2\pi \sqrt{1+C_2^2+C_2^2}.\] We conclude that, \[E[u]=2\pi \eta- \gamma -\eta \gamma=-2\pi \frac{n^2}{4} +(\frac{1}{4}n^2-1) \gamma.\] Hence, if \(n\geq 2\), then as \(\gamma\geq 2\pi\) \[E[u]\geq -2\pi \frac{n^2}{4} +2\pi (\frac{n^2}{4}-1)\geq -2\pi.\] with equality if and only if \(n=2\).

Proof. If inequality does not hold, then there is a \(\phi_0\in \mathrm{ HDiff}_+(\mathbb{S}^1)\cap \overline{\mathrm{ BDiff}}_+^1(\mathbb{S}^1, 4)\) so that \(\mathcal{E}_G^*[\phi_0]<0\). Let \(u_0=\log \phi_0'\) and set \(\gamma_0=\int_{\mathbb{S}^1} (\phi')^2 \; \mathrm{d}\theta=\int_{\mathbb{S}^1} \mathrm{e}^{2u_0} \; \mathrm{d}\theta\). The Cauchy-Schwarz inequality implies that \(\gamma_0\geq 2\pi\) with equality if and only if \(u_0\equiv 0\). Now consider the minimization problem \[E(\gamma)= \inf\left\{\mathcal{E}_G^*[\phi] \; \phi \in \mathrm{ HDiff}_+(\mathbb{S}^1)\cap \overline{\mathrm{ BDiff}}_+^1(\mathbb{S}^1, 4), \int_{\mathbb{S}^1} (\phi')^2\; \mathrm{d}\theta=\gamma\right\}.\] Clearly, our assumption ensures that \(E(\gamma_0)\leq \mathcal{E}_{G}^*[\phi_0]<0\). Notice without the constraint \(\int_{\mathbb{S}^1} (\phi')^2\; \mathrm{d}\theta\) the symmetry of Theorem 1.2 would imply that \(E\) is not coercive for the \(H^1\)-norm of \(u=\log \phi'\). However, with the constraint we are minimizing the Dirichlet energy of \(u\) and so the Rellich compactness theorem gives a \(u_{\min}\in H^1_{2\pi}(\mathbb{S}^1)\) satisfying \[E(\gamma_0)=\int_{\mathbb{S}^1}\frac{1}{4} (u'_{\min})^2 -\mathrm{e}^{2u_{\min}} \; \mathrm{d}\theta+2\pi=\int_{\mathbb{S}^1}\frac{1}{4} (u'_{\min})^2 \; \mathrm{d}\theta-\gamma_0+2\pi<0.\] and, hence, a \(\phi_{\min}\in \mathrm{ HDiff}_+(\mathbb{S}^1)\cap\overline{\mathrm{ BDiff}}_+^1(\mathbb{S}^1, 4)\) so that \(\log \phi'_{\min} =u_{\min}\). However, Proposition 4.9 implies that \(\phi_{\min} \in \mathring{\mathrm{ BDiff}}_+^1(\mathbb{S}^1, 4)\). This implies that \(u_{\min}\) is critical with respect to arbitrary variations in \(H^1(\mathbb{S}^1)\) which preserve the constraints \[\begin{array}{ccc} \int_{\mathbb{S}^1} \mathrm{e}^u \; \mathrm{d}\theta=2\pi& \mbox{and} & \int_{\mathbb{S}^1} \mathrm{e}^{2u} \; \mathrm{d}\theta=\gamma_0 . \end{array}\] Hence, \(u_{\min}\) weakly satisfies the Euler-Lagrange equation \[\frac{1}{4}u''_{\min} -\alpha \mathrm{e}^{2 u_{\min}}+\beta \mathrm{e}^{u_{\min}}=0.\] As this is a semi-linear ODE and \(u_{\min}\in C^{1/2}(\mathbb{S}^1\)) by Sobolev embedding, \(u_{\min}\in C^{2+\alpha}(\mathbb{S}^1)\) and satisfies this equation classically. Hence, \(u_{\min}\) is smooth by standard ODE theory. Notice, that if \(\phi_{\lambda}\) is one of the elements of (4.1), then \[u_{\lambda}=\log \phi_{\lambda}'=-\log\left( \frac{1}{2}(\lambda+\lambda^{-1})+\frac{1}{2}(\lambda-\lambda^{-1} )\cos \theta(p)\right).\] Applying, Proposition 4.10 to \(u_{\min}\) we see that, up to a rotation, if \(n=1\), then \(\phi_{\min}=\phi_{\lambda}\) for some \(\lambda\). As \(n_{B}(\phi_{\min})\geq 4\), this is impossible. Hence, \(n\geq 2\) and so \(E[u_{\min}]\geq 0\) which contradicts \(E(\gamma_0)< 0\) and proves the inequality.

Equality cannot hold for \(\phi\in \mathring{\mathrm{ BDiff}}_+^1(\mathbb{S}^1,4)\). If it did, \(\phi\) would be a critical point for \(\mathcal{E}_G^*\) with respect to arbitrary variations in \(\mathrm{ HDiff}_+(\mathbb{S}^1)\). Applying Proposition 4.10 to \(u=\log \phi'\) shows this is impossible. Hence, equality is only achieved on \(\partial \mathrm{ BDiff}_+^1(\mathbb{S}^1,4)\) and so the claim follows from Proposition 4.9 and the observation that, for \(\psi_\tau\) as in (4.2), \(\psi_\tau \in \mathrm{ Diff}_+^2(\mathbb{S}^1)\) or \(\mathrm{ BDiff}_+^1(\mathbb{S}^1, 4)\) if and only if \(\tau=0\).

Proof of Theorem 1.1. The natural scaling of the problem means that we may apply a homothety to take \(L(\sigma)=2\pi\). As \(\sigma\) is a smooth closed strictly convex curve, it is a smooth degree-one strictly convex curve. Let \(\phi_\sigma\in \mathrm{ Diff}^+(\mathbb{S}^1)\), be the induced diffeomorphism and let \(\psi_\sigma=\phi^{-1}_\sigma\). By Lemma 4.4, \(\phi_\sigma, \psi_\sigma\in \mathrm{ BDiff}_+^1(\mathbb{S}^1,4)\). The claim now follows from Propositions Proposition 4.12 and Proposition 3.1.