Hall’s Conjecture on Extremal Sets for Random Triangles

Abstract

In this paper, we partially resolve Hall’s conjecture (Hall in J Appl Probab 19(3):712–715, 1982) on random triangles. We consider the probability that three points chosen uniformly at random from a bounded convex region of the plane form an acute triangle. Hall’s conjecture states that this probability is maximized by the disk. This can be interpreted as a probabilistic version of the isoperimetric inequality. We first prove that the disk is a weak local maximum among bounded domains in ${\mathbb {R}}^2$ and that the ball is a weak local maximum in ${\mathbb {R}}^3$. In ${\mathbb {R}}^2$, we then prove a local $C^{2,\frac{1}{2}}$-type estimate on the probability in the Hausdorff topology. This enables us to prove that the disk is a strong local maximum in the Gromov–Hausdorff topology (modulo congruences). Finally, we give an explicit upper bound on the isoperimetric ratio for a region which maximizes the probability and show how this reduces verifying the full conjecture to a finite, though currently intractable, calculation.

Appendices

Appendix A

Until now, we have tried to avoid too many technical details so as to focus on the overall structure of the proof. However, for completeness we must prove the lemmas used in our theorems. For convenience, we will restate each of the lemmas at the beginning of each section. For the lemmas in Theorem 3, we found GeoGebra to be extremely useful for visualizing the geometry. We strongly recommend using this or some similar software to help gain familiarity with the approach, especially for the final parts. With a good picture, it is not too difficult to see what terms should be controlled and what general strategies should be used. The GeoGebra files that we used are available online [17].

Appendix B: The Proof of the Lemmas Used in Theorem 2

Lemma

(Properties of the Fourier series of an autocorrelation) Suppose that $\mu : {\mathbb {S}}^2 \rightarrow {\mathbb {R}}$ is a smooth function which decomposes as $\mu = \sum _{i,m} c_i^m Y_m^i$ where $Y_m^i$ are the spherical harmonics.

Consider the autocorrelation $R_\mu$ and write it in terms of spherical harmonics. That is to say, $R_\mu = \sum _{i,m} b_i^m Y_m^i$. Then the coefficients $b_i^m$ have the following properties.

• (1) For all i, m, $b_i^m \ge 0$. Furthermore, $b_i^m=0$ for all $i \ne 0.$

• (2) For any i and m, of $c_i^m \ne 0$ then $b_0^m > 0$.

• (3) The $c_i^m$ terms do not contribute to $b_j^n$ for $n \ne m$. In particular, none of the higher $c_i^m$ terms contribute to $b_0^0$ or $b_0^1$, which are the only positive terms in the Fourier series of $A_3(\theta )$.

Proof

We will give some indication of how to prove this lemma. The first two claims follow from Fourier analysis on SO(3) (see [25]). On SO(3), there is a correlation theorem, which follows from the convolution theorem. It states the following, where the overline denotes conjugation;

$$\begin{aligned} {\mathcal {F}}[R_\mu ]= \overline{ {\mathcal {F}}[\mu ]} \times {\mathcal {F}}[\mu ]. \end{aligned}$$

For our purposes, this immediately shows that $b_0^m \ge 0$ and that if $c_i^m \ne 0$ for any i, $b_0^m >0$, as desired. As in the two-dimensional case, it is worth noting that when we write $R_\mu = \sum _{i,m} b_i^m Y_m^i$ from $\mu = \sum _{i,m} c_i^m Y_m^i$, the terms $c_i^m$ with $i \ne 0$ contribute to $b_0^m$.

For the third claim, we must show that the $c_i^n$ contribute to $b_0^n$ but not to $b_0^m$ for $m \ne n$. There is a simple way to show this without calculating tedious integrals or diving into the correlation theorem. Instead, we can apply the Laplacian to the following integral expression:

$$\begin{aligned} \Delta _g \int _{x \in {\mathbb {S}}^2} Y_m^i(x) Y_n^j(xg) dS= & {} \int _{x \in {\mathbb {S}}^2} Y_m^i(x) \Delta _g Y_n^j(xg) dS \\= & {} \int _{x \in {\mathbb {S}}^2} Y_m^i(x) n(n+1) Y_n^j(xg) dS \\= & {} n(n+1) \int _{x \in {\mathbb {S}}^2} Y_m^i(x) Y_n^j(xg) dS. \end{aligned}$$

Using symmetry and a change of variables, this must be zero if $m \ne n$. From this, it follows that the $c_i^m$ terms do not contribute to $b_j^n$ for $n \ne m$ (as when we decompose $\mu$ in spherical harmonics, only the matching terms contribute to the Fourier series of the autocorrelation). $\square$

1.1 B.1: The Proof of the Lemma on Legendre Polynomials

Here we prove the following lemma which we used in Theorem 2. A Mathematica notebook with the calculations can be found online [17].

Lemma

Let $P_n(\cos (\theta ))$ be the n-th Legendre polynomial evaluated at $\cos (\theta )$. Then the following inequalities hold:

$$\begin{aligned} \int _0^\pi A_3(\theta ) P_n(\cos (\theta )) \sin {\theta } \,\mathrm{d} \theta \begin{array}{ll} >0 \text { for } n = 0,1 \\ \le 0 \text { otherwise}. \\ \end{array} \end{aligned}$$

Proof

Using an explicit formula for $P_n(\cos (\theta ))$, this reduces to the following:

$$\begin{aligned} \int _0^\pi \sum _{k=0}^n {n\atopwithdelims ()k} {-n-1\atopwithdelims ()k} \left( \sin (\theta /2) \right) ^k A_3(\theta ) \sin {\theta }\, \mathrm{d} \theta >0 \text { when n = 0,1} \\ \int _0^\pi \sum _{k=0}^n {n\atopwithdelims ()k} {-n-1\atopwithdelims ()k} \left( \sin (\theta /2) \right) ^k A_3(\theta ) \sin {\theta }\, \mathrm{d} \theta <0 \text { otherwise}. \\ \end{aligned}$$

For the three-dimensional ball, we can simplify 5 to obtain the following expression for $A_3(\theta )$:

$$\begin{aligned} A_3(\theta )= & {} - \frac{ 2 }{ 3 } \pi \left( 1 + \sin \left[ \frac{ \theta }{ 2 } \right] ^ { 3 } \right) \\&+ \pi \left( \frac{ 1 }{ 12 } \left( 9 \cos \left[ \frac{ \theta }{ 2 } \right] - \cos \left[ \frac{ 3 \theta }{ 2 } \right] \right) + \frac{ 1 }{ 6 } \left( 9 \sin \left[ \frac{ \theta }{ 2 } \right] + \sin \left[ \frac{ 3 }{ 2 } \right] \right) \right) . \end{aligned}$$

Thus, we have the following expression for $A_3(\theta ) \sin (\theta )$:

$$\begin{aligned}&A_3(\theta ) \sin (\theta ) \\&= \frac{ 1 }{ 24 } \left( 32 \pi \cos \left[ \frac{ \theta }{ 2 } \right] - 24 \pi \cos \left[ \frac{ 3 \theta }{ 2 } \right] - 8 \pi \cos \left[ \frac{ 5 \theta }{ 2 } \right] \right. \\&\quad \left. + 10 \sin \left[ \frac{ \theta }{ 2 } \right] + 64 \pi \sin [ \theta ] + 9 \sin \left[ \frac{ 3 \theta }{ 2 } \right] - \sin \left[ \frac{ 5 \theta }{ 2 } \right] \right) . \end{aligned}$$

For $n \ge 1$, we now want to compute the following:

$$\begin{aligned} \int _0^\pi \sum _{k=0}^n {n\atopwithdelims ()k} {-n-1\atopwithdelims ()k} \left( \sin (\theta /2) \right) ^k A_3(\theta ) \sin {\theta }\, \mathrm{d} \theta . \end{aligned}$$

(8)

In practice, we found trying to do this integral directly with Mathematica caused the computation to hang up. To avoid this problem, we derived the following intermediate identity so that the computation was tractable:

$$\begin{aligned}&\int _0^{\pi } \frac{1}{24} \left( \begin{array}{ll} 10 \pi \sin \left( \frac{\theta }{2}\right) +9 \pi \sin \left( \frac{3 \theta }{2}\right) -\pi \sin \left( \frac{5 \theta }{2}\right) +64 \pi \sin (\theta ) \\ + 32 \pi \cos \left( \frac{\theta }{2}\right) -24 \pi \cos \left( \frac{3 \theta }{2}\right) -8 \pi \cos \left( \frac{5 \theta }{2}\right) \\ \end{array} \right) \sin ^{2 k}\left( \frac{\theta }{2}\right) \, \mathrm{d}\theta \\&\quad =\frac{ ( 4 + 8 \mathrm { k } ) \pi }{ 15 + 31 \mathrm { k } + 20 \mathrm { k } ^ { 2 } + 4 \mathrm { k } ^ { 3 } } + \frac{ ( 2 + \mathrm { k } ) \pi ^ { 3 / 2 } \Gamma [ 1 + \mathrm { k } ] }{ \Gamma \left[ \frac{ 7 }{ 2 } + \mathrm { k } \right] }. \end{aligned}$$

Substituting this identity into the previous one (8) and switching the order of the summation and integration using Fubini’s Theorem, we find the following:

$$\begin{aligned}&\int _0^\pi \sum _{k=0}^n {n\atopwithdelims ()k} {-n-1\atopwithdelims ()k} \left( \sin (\theta /2) \right) ^k A_3(\theta ) \sin {\theta }\, \mathrm{d} \theta \nonumber \\&\quad = - \frac{ 8 \left( - 9 - 6 ( - 1 ) ^ { m } + 4 m + 2 ( - 1 ) ^ { m } m + 4 m ^ { 2 } + 2 ( - 1 ) ^ { m } m ^ { 2 } \right) \pi }{ ( - 3 + 2 m ) ( - 1 + 2 m ) ( 1 + 2 m ) ( 3 + 2 m ) ( 5 + 2 m ) }. \end{aligned}$$

(9)

We note that (9) is positive for $n=1$ and negative otherwise. For $n=0$, we find the following:

$$\begin{aligned} \int _0^\pi A_3(\theta ) \sin {\theta } \,\mathrm{d} \theta = \frac{4 \pi }{3} >0. \end{aligned}$$

Note that (9) is not equal to $\int _0^\pi A_3(\theta ) P_0(\cos (\theta )) \sin {\theta } \,\mathrm{d} \theta$ when $n=0$ due to a division by zero issue. $\square$

Appendix C: Existence of the Canonical Homotopy (Lemma 10)

Lemma

Let S be convex region whose Hausdorff distance from the disk is no more than $\epsilon$, with $\epsilon < \frac{1}{8\sqrt{2 \pi }}$. There is an embedding ${\bar{S}} = \{ re^{i \theta }~ |~ r \le 1 + {\bar{g}}(\theta ) \}$ with the following properties.

• (1) ${\bar{S}}$ is similar to S.

• (2) ${\bar{S}}$ can be obtained by translating S by no more than $3\epsilon$ and dilating S by a factor between $1-3\epsilon$ and $1+3\epsilon$. In other words, $d_{H.}(S, {\bar{S}}) < 7 \epsilon$.

• (3) The function ${\bar{g}}(\theta )$ satisfies the following:

  $$\begin{aligned} \int _0^{ 2 \pi } {\bar{g}}(\theta ) \mathrm{d} \theta = 0 \int _0^{ 2 \pi } {\bar{g}}(\theta ) e^{\ i \theta } \mathrm{d} \theta = 0. \end{aligned}$$

Proof

We want to find a similarity of S that satisfies the third condition, and show that it satisfies the second condition. Unfortunately, it is impossible to directly compute the integrals in the third condition for arbitrary S. Therefore, we need to find a work-around.

Heuristically, we expect that dilations should change the total integral of ${\bar{g}}(\theta )$ while leaving its first Fourier coefficients roughly constant. Similarly, we expect that translations do not change the total integral of ${\bar{g}}(\theta )$ much while changing the first Fourier coefficients. Unfortunately, this ansatz is not exactly true and so to salvage this intuition, we need to work topologically.

Before proving the estimates, we will explain the general strategy in more detail. Given a vector $w \in {\mathbb {R}}^2$ and a dilation factor $s \in R$, we can consider the map T:

$$\begin{aligned}&T: {\mathbb {R}}^2 \times {\mathbb {R}}\rightarrow {\mathbb {C}} \times {\mathbb {R}}, \\&T(w,s) = \left( \int _0^{ 2 \pi } g_{sS+w}(\theta ) e^{ i \theta } \, \mathrm{d} \theta , \int _0^{ 2 \pi } g_{sS+w}(\theta ) \, \mathrm{d} \theta \right) . \end{aligned}$$

We want to find a vector $w_0$ and a dilation $s_0$ with $|w_0|, |s_0-1|< 3 \epsilon$ for which T vanishes. To do so, assume that this is not the case. Then since T does not vanish on $B_{3\epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ]$, the following map is well defined and continuous:

$$\begin{aligned} {\bar{T}}: B_{3\epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ] \rightarrow {\mathbb {S}}^2 \\ {\bar{T}}(w,s) = \frac{T(w,s)}{|T(w,s)|.} \end{aligned}$$

We can restrict this map to the boundary of $B_{3\epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ]$. Topologically, this restriction is a continuous self-map of the sphere. The estimates here establish that the degree of this map is 1, after which we can use the degree theorem to obtain a contradiction. $\square$

1.1 C.1: The Estimates in Lemma 10

Since S is Hausdorff distance at most $\epsilon$ from D, S is completely contained within the disk of radius $1+\epsilon$ and completely contains the disk of radius $1-\epsilon$ . Therefore, the dilation of S by a factor of $(1+3 \epsilon )$ is completely contained in the disk of radius $1+ 4 \epsilon +3\epsilon ^2$ and completely contains the disk of radius $(1+2\epsilon -3\epsilon ^2)$.

Suppose we dilate S by a factor of $(1+3 \epsilon )$ and translate the S by some amount $3 \epsilon _T$ in a unit direction V, where $\epsilon _T$ is no greater than $\epsilon$. Then we can use the inner disk to obtain a lower estimate on $\int _0^{ 2 \pi } {\bar{g}}(\theta ) \mathrm{d} \theta$.

$$\begin{aligned} \int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon )S+3\epsilon _T V}(\theta ) \mathrm{d} \theta\ge & {} \int _0^{ 2 \pi } 2 \epsilon _T \cos (\theta ) + \sqrt{(1+2\epsilon -3\epsilon ^2)^2-9 \epsilon ^2_T \sin ^2(\theta ) }-1 \,\mathrm{d} \theta \\= & {} \int _0^{ 2 \pi } \sqrt{ 1+4\epsilon -4\epsilon ^2 - 12 \epsilon ^3+9 \epsilon ^4 - 9 \epsilon ^2_T \sin ^2(\theta )} -1 \,\mathrm{d} \theta \\\ge & {} \int _0^{ 2 \pi } 2\epsilon -O(\epsilon ^2) - O(\epsilon ^2_T) \, \mathrm{d} \theta \\= & {} 4 \pi \epsilon - O(\epsilon ^2). \end{aligned}$$

We can use the exact same argument to show that if we contract the disk by $1-3\epsilon$, and translate it by no more than $3 \epsilon$ in any direction, then $\int _0^{ 2 \pi } {\bar{g}}_{(1-3\epsilon )S+3\epsilon _T V}(\theta ) \mathrm{d} \theta < - 4 \pi \epsilon + O(\epsilon ^2)$.

Suppose now that we translate S by $3 \epsilon$ in some direction V, after dilating it by $1+3\epsilon _D$, with $\epsilon _D<\epsilon$. Using the inner disk, we obtain the following estimate:

$$\begin{aligned} \int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon _D)S+2\epsilon V}(\theta ) \mathrm{d} \theta\ge & {} \int _0^{ 2 \pi } 2 \epsilon \cos (\theta ) \\&+ \sqrt{((1-\epsilon )^2(1+3 \epsilon _D)^2-4 \epsilon ^2 \sin ^2(\theta ) }-1 \,\mathrm{d} \theta \\\ge & {} \int _0^{ 2 \pi } \sqrt{ 1-2\epsilon + 6 \epsilon _D - O(\epsilon ^2)} -1 \,\mathrm{d} \theta \\\ge & {} \int _0^{ 2 \pi } -\epsilon + 3 \epsilon _D -O(\epsilon ^2) \, \mathrm{d} \theta \\= & {} -2 \pi \epsilon + 6 \pi \epsilon _D + O(\epsilon ^2). \end{aligned}$$

Similarly, we want to estimate $\int _0^{ 2 \pi } {\bar{g}}(\theta ) e^{i \theta } \mathrm{d} \theta$ using the inner and outer disks. After performing a rotation, we can assume that V is the unit vector in the x-direction. For conciseness, we only consider the cosine term here. Suppose we translate S by $3 \epsilon$ in some direction, after dilating S by $1+3\epsilon _D$, with $|\epsilon _D| < \epsilon$. We can use the inner and outer disks to make the following estimates:

$$\begin{aligned}&\int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon _D)S+3\epsilon V}(\theta ) \cos (\theta ) \mathrm{d} \theta \\&\quad \ge \int _{-\pi /2}^{ \pi /2} \left( 3 \epsilon \cos (\theta ) + \sqrt{(1-\epsilon )^2(1+3 \epsilon _D)^2-9 \epsilon ^2 \sin ^2(\theta ) }-1 \right) \cos (\theta ) \,\mathrm{d} \theta \\&\qquad + \int _{\pi /2}^{ 3\pi /2} \left( 3 \epsilon \cos (\theta ) + \sqrt{(1+\epsilon )^2(1+3 \epsilon _D)^2-9 \epsilon ^2 \sin ^2(\theta ) }-1 \right) \cos (\theta ) \,\mathrm{d} \theta \\&\quad \ge 3 \pi \epsilon + \int _{-\pi /2}^{ \pi /2} \left( - \epsilon + 3 \epsilon _D - O(\epsilon ^2) \right) \cos (\theta ) \,\mathrm{d} \theta \\&\qquad + \int _{\pi /2}^{3 \pi /2} \left( \epsilon + 3 \epsilon _D + O(\epsilon ^2) \right) \cos (\theta ) \,\mathrm{d} \theta \\&\quad = 3 \pi \epsilon + 2 \left( - \epsilon + 3 \epsilon _D \right) - 2 \left( \epsilon + 3 \epsilon _D \right) - O(\epsilon ^2) \\&\quad = 3 \pi \epsilon - 4 \epsilon - O(\epsilon ^2). \end{aligned}$$

Note that we did not assume a sign for $\epsilon _D$, so this covers both the cases when the dilation expands or contracts S. Finally, we suppose now that we dilate S by $1+3\epsilon$ and translate it by $3 \epsilon _T$ in some direction V with $\epsilon _T<\epsilon$. Using the inner and outer disks, we obtain the following estimate. Note that we would obtain the same estimate if we dilated S by $1-3\epsilon$ instead.

$$\begin{aligned}&\int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon )S+3\epsilon _T V}(\theta ) \cos (\theta ) \mathrm{d} \theta \\&\quad \ge \int _{-\pi /2}^{ \pi /2} \left( 3 \epsilon _T \cos (\theta ) + \sqrt{(1-\epsilon )^2(1+3 \epsilon )^2-9 \epsilon _T^2 \sin ^2(\theta ) }-1 \right) \cos (\theta ) \,\mathrm{d} \theta \\&\qquad + \int _{\pi /2}^{ 3\pi /2} \left( 3 \epsilon _T \cos (\theta ) + \sqrt{(1+\epsilon )^2(1+3 \epsilon )^2-9 \epsilon _T^2 \sin ^2(\theta ) }-1 \right) \cos (\theta ) \,\mathrm{d} \theta \\&\quad \ge 3 \pi \epsilon _T + \int _{-\pi /2}^{ \pi /2} \left( 2 \epsilon - O(\epsilon ^2) \right) \cos (\theta ) \,\mathrm{d} \theta + \int _{\pi /2}^{3 \pi /2} \left( 4 \epsilon - O(\epsilon ^2) \right) \cos (\theta ) \,\mathrm{d} \theta \\&\quad = 3 \pi \epsilon _T + 2 \left( 2 \epsilon \right) - 2 \left( 4 \epsilon \right) - O(\epsilon ^2). \\&\quad = 3 \pi \epsilon _T - 4 \epsilon - O(\epsilon ^2) \end{aligned}$$

Now that we have proven the four estimates needed, we can finish the proof of Lemma 10. To prove that an embedding of S satisfying the previous conditions exists, we consider the continuous map T.

$$\begin{aligned}&T:B_{3\epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ] \rightarrow {\mathbb {R}}^3 \\&T(w,s) = \left( \int _0^{ 2 \pi } g_{sS+w}(\theta ) e^{ i \theta } \, \mathrm{d} \theta , \int _0^{ 2 \pi } g_{sS+w}(\theta ) \, \mathrm{d} \theta \right) . \end{aligned}$$

We want to show that there is a pair (w, s) so that T vanishes. Suppose, for the sake of contradiction, that T never vanishes. If so, it induces a continuous map:

$$\begin{aligned}&{\bar{T}}: B_{3 \epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ] \rightarrow {\mathbb {S}}^2 \\&{\bar{T}}(w,s) = \frac{T(w,s)}{|T(w,s)|}. \end{aligned}$$

The restriction of this map to the boundary (which we denote ${\bar{T}}|_{{\mathbb {S}}^2}$) is topologically a self-map of the sphere. We want to compute the degree of this map. To do so, let $x= (3 \epsilon _T V,3 \epsilon ) \in B_{3 \epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ]$, and observe we have the following estimate:

$$\begin{aligned} \langle {\bar{T}}|_{{\mathbb {S}}^2} (x), x\rangle\ge & {} 3 \epsilon _T \cdot \int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon _D)S+3\epsilon V}(\theta ) \cos (\theta ) \mathrm{d} \theta \\&+\, 3 \epsilon \int _0^{ 2 \pi } {\bar{g}}_{(1+3\epsilon )S+3\epsilon _T V}(\theta ) \mathrm{d} \theta \\\ge & {} 3 \epsilon _T( 3 \pi \epsilon _T - 4 \epsilon - O(\epsilon ^2)) + 3 \epsilon (4 \pi \epsilon - O(\epsilon ^2)) \\\ge & {} (12 \pi - 12) \epsilon ^2 - O(\epsilon ^3). \end{aligned}$$

If $\epsilon$ sufficiently small, this is positive. Repeating this argument for the other parts of the boundary of $B_{3\epsilon }(0) \times [1-3\epsilon , 1 + 3\epsilon ]$, we can show that ${\bar{T}}|_{{\mathbb {S}}^2}$ satisfies $\langle x, {\bar{T}}|_{{\mathbb {S}}^2}( x) \rangle > 0$ for all x. As a result, ${\bar{T}}|_{{\mathbb {S}}^2}(x)$ is contained in the same hemisphere as x. This implies the degree of the T is one, as we can construct a homotopy from ${\bar{T}}|_{{\mathbb {S}}^2}$ to the identity map. Therefore, the supposed map ${\bar{T}}$ retracts ${\bar{T}}|_{{\mathbb {S}}^2}$, which is impossible. As a result, T must vanish for some translation $w_0$ and some dilation $s_0$.

To finish the proof, we observe that since $|w_0| < 3 \epsilon$ and $|s_0|<3\epsilon$, we have that $d_{H.}(S, {\bar{S}})< 3 \epsilon (1+3 \epsilon )+3 \epsilon$. Using the assumption that $\epsilon < \frac{1}{8 \sqrt{2 \pi }}$, this implies that $d_{H.}(S, {\bar{S}}) < 7 \epsilon$. $\square$

In the previous calculations, it should be noted that the terms that were $O(\epsilon _D)$ and $O(\epsilon _T)$ canceled out, which shows that the effect of translations on the total integral of $g(\theta )$ is 0 to first order and that the effect of dilations on the first Fourier coefficients of $g(\theta )$ is also 0 to first order. In other words, our initial ansatz was correct to first order. We will also note that these estimates are rather finicky, and the factor of $3 \epsilon$ for the translations and dilations cannot be reduced much.

Appendix D: The Lipschitz Estimate of p(S(t)) Along the Canonical Homotopy

Notation: Before starting the proof, we introduce the following notation, which will help to write certain sets concisely. We will use $S \Delta D$ to denote the symmetric difference between the sets S and D:

$$\begin{aligned} S \Delta D= (S \backslash D) \cup (D \backslash S). \end{aligned}$$

In this section, we prove the following lemma.

Lemma

(The Lipschitz estimate on p(S)) There exist uniform constants $e, C > 0$ so that whenever $\epsilon < e$ and $d_{H.}(S,D) \le \epsilon$, then $\bigl |\bigl |p(S) - p(D) \bigl |\bigl |< C \epsilon$.

Proof

To prove this lemma, we suppose that $d_{H.}(S, D) < 8 \epsilon$ and assume that $\epsilon < \frac{1}{8 \sqrt{2 \pi }}$. This assumption implies that $\Vert g \Vert _{L^2} < 1$, which we use in the first inequality. The third inequality estimates the size of the set $(S \times S \times S) \Delta (D \times D \times D)$. The fourth inequality uses Taylor’s Theorem to control the first term.

$$\begin{aligned} | p(S) - p(D) | =&\Bigg | \frac{\int _S \int _S \int _S f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z}{V(S)^3} - \frac{\int _D \int _D \int _D f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z}{V(D)^3} \Bigg | \\ =&\Bigg | \frac{ \pi ^3 \int _S \int _S \int _S f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z - (\pi + \frac{1}{2} \Vert g(\theta ) \Vert _2^2)^3 \int _D \int _D \int _D f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z}{ \pi ^3(\pi + \frac{1}{2} \Vert g(\theta ) \Vert _2^2)^3} \Bigg | \\ \le&\pi ^3 \Bigg | \frac{ \int _S \int _S \int _S f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z - \int _D \int _D \int _D f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z}{ \pi ^6} \Bigg | \\&+ \frac{7}{2} \Vert g(\theta ) \Vert _2^2 \Bigg | \frac{ \int _D \int _D \int _D f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z}{\pi ^6} \Bigg | \\ \le&\frac{ \int _{(S \times S \times S) \Delta (D \times D \times D)} 1 \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z }{ \pi ^3} + \frac{7}{2} \Vert g(\theta ) \Vert _2^2 \frac{ \frac{4}{\pi ^2} - \frac{1}{8}}{\pi ^3} \\ \le&~ \frac{3}{\pi ^3} \left( \pi (1 + 8 \epsilon )^2 \right) ^2 \pi \left( (1 + 8 \epsilon )^2-(1 - 8 \epsilon )^2\right) + \frac{7}{2} \Vert g(\theta ) \Vert _2^2 \frac{ \frac{4}{\pi ^2} - \frac{1}{8}}{\pi ^3} \\ \le&~ 6(1 + 8 \epsilon )^5 8 \epsilon + 7 \pi (8 \epsilon )^2 \frac{ \frac{4}{\pi ^2} - \frac{1}{8}}{\pi ^3} \le 48 \epsilon +O(\epsilon ^2). \end{aligned}$$

As this upper bound is $O(\epsilon )$, the desired estimate holds.

$\square$

Appendix E: The Hölder Estimate on $\frac{\mathrm{d}^2}{\mathrm{dt}^2} M$

Here, we start the proof of the Hölder estimate by reducing it to three more manageable lemmas which we prove in the following subsections. Although this part of the proof does not require any abstract machinery, we have to carefully estimate each term using geometry.

Recall that we are proving the following lemma.

Lemma

(The Hölder-1 / 2 estimate on $\frac{\mathrm{d}^2M}{\mathrm{dt}^2}$) There exist uniform constants $e, C >0$ so that whenever $\epsilon < e$ and $d_{H.}(S,D) \le \epsilon$, then $\bigl |\frac{\mathrm{d}^2}{\mathrm{dt}^2} M (S) - \frac{\mathrm{d}^2}{\mathrm{dt}^2} M (D) \bigl |< C \epsilon ^{1/2}$.

The Hölder 1 / 2 estimate is natural, but it is likely possible to strengthen it with more effort. It seems that a full $C^3$ estimate does not hold for general convex regions near the disk, so the $C^{2,\alpha }$ estimate is likely optimal. The general strategy to prove this lemma is to expand $\frac{\mathrm{d}^2}{\mathrm{dt}^2} M (S)$ explicitly and to carefully control each term.

Proof

Recall that the second variation of M is the following:

$$\begin{aligned} \frac{\mathrm{d}^2}{\mathrm{dt}^2}M =&3 \int _0^{2 \pi } \int _S \int _S f(r( \theta _1, t) e^{i \theta _1},y,z) \frac{\mathrm{d}}{\mathrm{dt}}\mu (\theta _1, t) r(\theta _1, t) \,\mathrm{d}y \, \mathrm{d}z \, \mathrm{d} \theta _1 \\&+ 6 \int _0^{2 \pi } \int _0^{2 \pi } \int _S f(r(\theta _1, t) e^{i \theta _1},r(\theta _2, t) e^{i \theta _2},z) \mu (\theta _1, t)\\&\quad \mu (\theta _2, t) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \\&+ 3 \int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 r(\theta _1, t) \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \\&+ 3 \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1. \end{aligned}$$

$\square$

Since $\mu (\theta ,t)$ is time independent along the canonical homotopy, we write it as $\mu (\theta )$ and observe that the first term vanishes.

$$\begin{aligned} \frac{\mathrm{d}^2}{\mathrm{dt}^2}M =&6 \int _0^{2 \pi } \int _0^{2 \pi } \int _S f(r(\theta _1, t) e^{i \theta _1},r(\theta _2, t) e^{i \theta _2},z) \\&\quad \mu (\theta _1) \mu (\theta _2) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \\&+ 3 \int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 r(\theta _1, t) \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \\&+ 3 \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1. \end{aligned}$$

The rest of the proof is showing that each of these terms are uniformly Hölder continuous in $d_{H.}(D,S)$. So as to be more manageable, we split this into three lemmas for each of these terms.

Lemma 14

(Hölder continuity of the autocorrelation terms) There exist uniform constants $e ,C >0$ so that whenever $d_{H.}(S,D) \le \epsilon < e$, then

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _0^{2 \pi } \int _S f(r(\theta _1,t) e^{i \theta _1},r(\theta _2,t) e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \\&\quad - \int _0^{2 \pi } \int _0^{2 \pi } \int _D f(e^{i \theta _1}, e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$

Lemma 15

(Holder continuity of the terms involving $\frac{\partial f}{\partial r}$) There exist uniform constants $e, C >0$ so that whenever $d_{H.}(S,D) \le \epsilon < e$, then

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f(r(\theta _1,t)e^{i \theta _1},y,z) r(\theta _1,t) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta \\&\quad - \int _0^{2 \pi } \int _D \int _D \frac{\partial }{\partial r} f(e^{i \theta _1},y,z) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$

Lemma 16

(Lipschitz continuity of the arc-length term) There exist uniform constants $e, C >0$ so that whenever $\epsilon < e$ and $d_{H.}(S,D) \le \epsilon$,

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d}\theta _1\\&\quad -\int _0^{2 \pi } \int _D \int _D f(e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg | < C \epsilon . \end{aligned}$$

1.1 E.1: Hölder Continuity of the Autocorrelation Term

In this section, we prove Lemma 14, which we restate here.

Lemma

There exist uniform constants $e ,C >0$ so that whenever $d_{H.}(S,D) \le \epsilon < e$, then

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _0^{2 \pi } \int _S f(r(\theta _1) e^{i \theta _1},r(\theta _2) e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d}\theta _2 \\&\quad - \int _0^{2 \pi } \int _0^{2 \pi } \int _D f(e^{i \theta _1}, e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) \,\mathrm{d} z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \Bigg | < C \epsilon ^{1/2} \end{aligned}$$

Proof

To prove this, we note that it is sufficient to show that there is a C, uniform in $\theta _1$ and $\theta _2$, so that the following estimate holds:

$$\begin{aligned} \Bigg | \int _S f(r(\theta _1,t ) e^{i \theta _1},r(\theta _2,t) e^{i \theta _2},z) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z - \int _D f(e^{i \theta _1}, e^{i \theta _2},z) \, \mathrm{d}z \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$

If we can show this estimate, it immediately implies the following

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _0^{2 \pi } \int _S f(r(\theta _1,t) e^{i \theta _1},r(\theta _2,t) e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) r(\theta _1, t) r(\theta _2, t) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \\&\quad - \int _0^{2 \pi } \int _0^{2 \pi } \int _D f(e^{i \theta _1}, e^{i \theta _2},z) \mu (\theta _1) \mu (\theta _2) \, \mathrm{d}z \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 \Bigg | \\&\qquad < \int _0^{2 \pi } \int _0^{2 \pi } C \epsilon ^{1/2} \, \mathrm{d} \theta _1 \, \mathrm{d} \theta _2 = 4 \pi ^2 C \epsilon ^{1/2}. \end{aligned}$$

To do this, it is convenient to let $X=r(\theta _1) e^{i \theta _1}$, $Y=r(\theta _2) e^{i \theta _2}$, and use the triangle inequality to estimate the previous term.

$$\begin{aligned}&\Bigg | \int _S f(X,Y,z) |X| |Y| \, \mathrm{d}z - \int _D f\left( \frac{X}{|X|}, \frac{Y}{|Y|},z \right) \, \mathrm{d}z \Bigg | \\&\quad \le \Bigg | \int _S f(X,Y,z) \, \mathrm{d}z - \int _D f(X,Y,z) |X| |Y| \, \mathrm{d}z \Bigg | \\&\qquad +\Bigg | \int _D f(X,Y,z) |X| |Y| \, \mathrm{d}z - \int _D f\left( X,Y,z \right) \, \mathrm{d}z \Bigg | \\&\qquad +\Bigg | \int _D f(X,Y,z) \, \mathrm{d}z - \int _D f\left( \frac{X}{|X|}, \frac{Y}{|Y|},z \right) \, \mathrm{d}z \Bigg |. \end{aligned}$$

To prove this lemma, we show that all of these terms are $O(\epsilon ^{1/2})$.

For the first term, note that $f(X,Y,z) \le 1$ and that $d_{H.}(S,D) < 8 \epsilon$. Therefore, we have the following estimate:

$$\begin{aligned} \Bigg | \int _S f(X,Y,z) \, \mathrm{d}z - \int _D f(X,Y,z) \, \mathrm{d}z \Bigg | \le Vol(S \Delta D) < 16 \pi \epsilon + 64 \epsilon ^2. \end{aligned}$$

To control the second term, note $|1-|X||, |1-|Y|| < 8 \epsilon$, and that $\int _D f(X,Y,z) |X| |Y| \, \mathrm{d}z \le \pi ^2$. This yields the following estimate:

$$\begin{aligned} \Bigg | \int _D f(X,Y,z) |X| |Y| \, \mathrm{d}z \!- \!\int _D f\left( X,Y,z \right) \, \mathrm{d}z \Bigg | \!<\! ((1\!+\!8 \epsilon )^2\!-\!1) \pi ^2 \!=\! \pi ^2 \epsilon \!+\!64 \pi ^2 \epsilon ^2. \end{aligned}$$

Therefore, the first two terms are $O(\epsilon )$, and so $O(\epsilon ^{1/2})$ as well. To show that the final term is $O(\epsilon ^{1/2})$, we show that that $\int _D f(X, Y,z) \, \mathrm{d}z$ is Hölder 1 / 2 continuous in X and Y, when X and Y are near the boundary of the disk. Referring back to Fig. 1, $\int _D f(X, Y,z) \, \mathrm{d}z$ is the area between the two parallel lines outside of the smaller disk. Using Euclidean geometry, we can write $\int _D f(X, Y,z) \, \mathrm{d}z$ explicitly. In most cases, this is given by the following formula.

$$\begin{aligned} \int _D f(X, Y,z) \, \mathrm{d}z= & {} \frac{1}{2}\left( \arccos (\frac{X \cdot (X-Y)}{|X-Y|}\right) - \arccos \left( \frac{Y \cdot (X-Y)}{|X-Y|}\right) \\&- \sqrt{1 - (\frac{X \cdot (X-Y)}{|X-Y|})^2 } + \sqrt{1-(\frac{X \cdot (X-Y)}{|X-Y|})^2 } )\\&- r^2 \arccos (\frac{\mathrm{d}^2+r^2 -1}{2dr}) + \arccos (\frac{\mathrm{d}^2+1 -r^2}{2d})\\&- \frac{1}{2} \sqrt{(-d+r+1)(-d+r-1)(d-r+1)(d+r+1).} \end{aligned}$$

Here, $d = |X+Y|/2$ and $r = |X-Y|/2$. For a given pair X, Y, it is possible that there are fewer terms. For instance, if the small disk is contained entirely in the larger one or one of the lines does not intersect the unit disk, there will be fewer terms. By inspection, this is Hölder-1/2 continuous whenever $|X-Y|$ and $|X+Y|$ are not too small (say both greater than $48 \epsilon$).

We can use geometry to show that this expression is small whenever $|X-Y| < 64 \epsilon$ or $|X+Y|< 64 \epsilon$. To do so, note that that $\int _D f(X, Y,z) \, \mathrm{d}z < 2 |X-Y|$, as the disk has diameter 2 and the width of the parallel lines in Fig. 1 is $|X-Y|$. Therefore, when $|X-Y|$ is small, $\int _D f(X, Y,z) \, \mathrm{d}z$ is Lipschitz in $|X-Y|$. Furthermore, if $|X+Y|$ is small and X and Y lie on the boundary of S (which is at most $8 \epsilon$ from the boundary of D), then X and Y are nearly antipodes of the circle. In this case, very little of the unit disk will lie outside the disk defined by X and Y, as shown in Fig. 3.

Fig. 3

Bounding the error when $|X+Y|$ is small

To state this precisely, whenever $|X+Y|< 64 \epsilon$:

$$\begin{aligned} \int _D f(X, Y,z) \, \mathrm{d}z< \pi |1- |X|^2| + |1- |Y|^2| < 32\pi \epsilon + 128 \pi \epsilon ^2. \end{aligned}$$

$\square$

1.2 E.2: Holder Continuity of the Terms Involving $\frac{\partial f}{\partial r}$

In this section, we prove Lemma 15, which we restate here.

Lemma

There exist uniform constants $e, C >0$ so that whenever $d_{H.}(S,D) \le \epsilon < e$, then

$$\begin{aligned} \Bigg | \int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f((r(\theta _1) e^{i \theta _1},y,z) r(\theta _1) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta&\\ - \int _0^{2 \pi } \int _D \int _D \frac{\partial }{\partial r} f(e^{i \theta _1},y,z) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg |&< C \epsilon ^{1/2}. \end{aligned}$$

Proof

Earlier, we were content to interpret $\frac{\partial f}{\partial r}$ distributionally, without worrying about the formal definition. However, for this step, we write this out explicitly using classical integrals. To make the calculation more intuitive, we write $X=r(\theta _1) e^{i \theta _1}.$

$$\begin{aligned}&\int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f(X,y,z) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \\&\quad = \int _0^{2 \pi } \int _S \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot \frac{X}{|X|} |X| \mathrm{d}Z \,\mathrm{d} Y \, \mu (X)^2 \mathrm{d} \theta _1 \\&\qquad - \int _0^{2 \pi } \int _S \int _{ \{ Z \in S | \measuredangle (XYZ)= \pi /2 \} } \frac{(X-Y)}{|X-Y|} \cdot \frac{X}{|X|} |X| \mathrm{d}Z \,\mathrm{d} Y \, \mu (X)^2 \mathrm{d} \theta _1 \\&\qquad + \int _0^{2 \pi } \int _S \int _{ \{Z \in S | \measuredangle (XZY)= \pi /2 \}} \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot \frac{X}{|X|} |X| \mathrm{d}Z \,\mathrm{d} Y \, \mu (X)^2 \mathrm{d} \theta _1. \end{aligned}$$

We want to show that each of these three terms are Hölder continuous in Hausdorff distance. To do this, we use we use the triangle inequality on the inner two integrals as in the previous step. For conciseness, we will only write this out fully for the first term and we denote ${\bar{X}} = \frac{X}{|X|}$.

$$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \nonumber \\&\quad - \int _D \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot \frac{X}{|X|} \mathrm{d}Z \,\mathrm{d} Y \Bigg | \nonumber \\&\qquad \le \Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \nonumber \\&\qquad \quad -\int _S \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | \end{aligned}$$

(10)

$$\begin{aligned}&\qquad \quad + \Bigg | \int _S \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \nonumber \\&\qquad \quad -\int _S \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}}Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \Bigg | \end{aligned}$$

(11)

$$\begin{aligned}&\qquad \quad + \Bigg | \int _S \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}} Z)= \pi /2\} } \frac{( {\bar{X}}-Y)}{| {\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \nonumber \\&\qquad \quad -\int _D \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}} Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{| {\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \Bigg |. \end{aligned}$$

(12)

$\square$

1.3 E.3: Proving that (11) is $O(\epsilon ^{1/2})$

The title of this subsection is a bit of a misnomer. Not only must we bound (13), we need the same estimate on

$$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in D | \measuredangle (XYZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \\&\qquad -\int _S \int _{ \{ Z \in D | \measuredangle ( {\bar{X}} YZ) = \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \Bigg | \end{aligned}$$

and

$$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in D | \measuredangle (XZY)= \pi /2\} } \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}Z \,\mathrm{d} Y -\int _S \int _{ \{ Z \in D | \measuredangle ( {\bar{X}}ZY)= \pi /2\} } \\&\quad \frac{ Z -({\bar{X}}+Y)/2}{|Z -({\bar{X}}+Y)/2|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \Bigg | . \end{aligned}$$

Since X lies on the boundary of S, $|X-{\bar{X}}| < 8 \epsilon$, it is sufficient to show that (13) is uniformly Holder continuous in terms of X. To do this, we consider the innermost integral (e.g., $\int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \,\mathrm{d}Z$), and show the following three estimates:

• (1) Away from a small set of Y, this term is uniformly Hölder-1 / 2 continuous as a function of X.

• (2) The small set of Y’s where the uniform Hölder estimate fails (in terms of X) has size $O(\epsilon ^{1/2})$.

• (3) The innermost integrals are uniformly bounded.

With this overview out of the way, we now do this precisely.

• (1) We first show that the following terms are Hölder continuous in X, so long as Y avoids a small set:

  $$\begin{aligned}&\int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}z \\&\int _{ \{ Z \in D | \measuredangle (XYZ)= \pi /2 \} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}z \\&\int _{ \{Z \in D | \measuredangle (XZY)= \pi /2 \}} \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}z. \end{aligned}$$

  To do this, we show that both the integrand and bounds of the integral are uniformly Hölder continuous in X.

For the first two terms, the integrand is Hölder continuous in X, so long as $|X-Y|$ is not too small. Furthermore, if $|(X-Y)\cdot X| > \epsilon ^{1/2}$ and $|X-Y| > \epsilon ^{1/2}$, then the bounds of the first two integrals are Hölder continuous in X. To see this, note that the length of a chord through a point is uniformly Lipschitz in the angle that defines the chord, unless the point lies near the boundary on the disk and chord is nearly tangent to the boundary. Furthermore, the angle defining the chord is Hölder 1 / 2 continuous in X whenever $|X-Y| > \epsilon ^{1/2}$.

The integrand in the third term is uniformly Hölder continuous, except on the set where $|Z -(X+Y)/2|$ is small. Since Z lies on a semicircle with diameter $\overline{XY}$, this can only happen when $|X-Y|$ is small. It is worth noting that when Y is close to the line $y=0$, as the length of $\{Z \in D | \measuredangle (XZY)= \pi /2 \}$ is only Hölder continuous in X, not Lipschitz.

Combining all of this, we have uniform Hölder continuity of $\int _D \frac{\partial }{\partial r} f(X,Y,z) \mathrm{d}z$ for X near the boundary of D, so long as Y is away from where $|(X-Y)\cdot X| \le \epsilon ^{1/2}$ and $|X-Y| \le \epsilon ^{1/2}$.

• (2) We now show that the measure of the set where the Hölder estimate fails is $O(\epsilon ^{1/2})$. Note that the set $\{ Y | |X-Y| < \epsilon ^{1/2} \}$ has measure at most $\pi \epsilon$. The set $\{ Y \in D | |(X-Y)\cdot X| < \epsilon ^{1/2} \}$ is contained in a circular segment near X. We can find the area of this segment using Euclidean geometry, which shows that it is $O(\epsilon ^{3/4})$.

• (3) Given $X, Y \in {\mathbb {R}}^2$ with $|X| < 1+8 \epsilon$, we can make the following three uniform estimates:

  $$\begin{aligned}&\Bigg | \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}z \Bigg | \le 2 |X|\le 2(1+8 \epsilon ) \\&\Bigg | \int _{ \{ Z \in D | \measuredangle (XYZ)= \pi /2 \} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}z \Bigg | \le 2 |X|\le 2(1+8 \epsilon ) \\&\Bigg | \int _{ \{Z \in D | \measuredangle (XZY)= \pi /2 \}} \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}z \Bigg | \le 2 \pi |X| \le 2 \pi (1+8 \epsilon ). \end{aligned}$$

The preceding estimates imply that $\int _D \int _D \frac{\partial }{\partial r} f(X,y,z) d y \,d z$ is uniformly Hölder continuous in X when X is near the boundary of the disk. To see this, observe that if we move X by no more than $8 \epsilon$, the inner integral is Hölder continuous, except for a set of size $O(\epsilon ^{1/2})$. Since the inner integral is uniformly bounded, the integral on the bad set is also $O(\epsilon ^{1/2})$. As such, the change in the total integral is $O(\epsilon ^{1/2})$.

1.4 E.4: Proving that (10) is $O(\epsilon ^{1/2})$

We now consider (10) and its corresponding terms. We show the following three estimates:

• (1)
  $$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \\&\quad -\int _S \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$
• (2)
  $$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (XYZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y\\&\quad -\int _S \int _{ \{ Z \in D | \measuredangle (XYZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$
• (3)
  $$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (XZY)= \pi /2\} } \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \\&\quad -\int _S \int _{ \{ Z \in D | \measuredangle (XZY)= \pi /2\} } \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | < C \epsilon ^{1/2}. \end{aligned}$$

To do this, we use the triangle inequality again.

$$\begin{aligned}&\Bigg | \int _S \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \\&\qquad -\int _S \int _{ \{ Z \in D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | \\&\quad \le \Bigg | \int _S \int _{ \{ Z \in S \backslash D | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \,\mathrm{d} Y \Bigg | \\&\qquad +\Bigg | \int _S \int _{ \{ Z \in D \backslash S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}Z \mathrm{d} Y \Bigg |. \end{aligned}$$

We now want to show that both of these terms are $O(\epsilon ^{1/2})$. To do so, we use the same procedure as before. We first establish a Hölder estimate in terms of X away from a small set of Y values and then bound the size of the small set. Finally, we obtain a uniform estimate on the innermost integral. For the Hölder continuity away from a small set, we use the following transversality estimate. For conciseness, we omit the proof, which is a straightforward estimate using Euclidean geometry.

Lemma 17

(Transversality estimate) Suppose $\gamma _1$ and $\gamma _2$ are circles (or lines) which intersect transversally at an angle $\phi$. Suppose further that the curvatures of both $\gamma _1$ and $\gamma _2$ are less than some K. Then, given $\epsilon$ satisfying $\epsilon < \frac{1}{4K}$ and $\epsilon < \frac{\sin (\phi )}{4}$, then the length of the set $\{ x \in \gamma _1 | d(x, \gamma _2)< \epsilon \} < 8 \frac{\epsilon }{\sin (\phi )}$.

This lemma is useful because it allows us to bound the lengths of the loci of right triangles in $S \Delta D$. For instance, we can bound the length of the set $\{Z \in S \Delta D | \measuredangle (XZY)= \pi /2 \}$ by the length of the set $\{Z | \measuredangle (XZY)= \pi /2 \text { and } d(Z, D) < 8 \epsilon \}$.

Explicitly, this lemma shows that if $\{Z | \measuredangle (XZY)= \pi /2 \}$ intersects $\partial D$ at an angle $\phi$, then so long as the inverse curvatures of D and $\{Z \in D | \measuredangle (XZY)= \pi /2 \}$ and $\sin (\phi )$ are much larger than $\epsilon$, then we have the following estimate:

$$\begin{aligned} l(\{Z | \measuredangle (XZY)= \pi /2 \text { and } d(Z, D)< 8 \epsilon \} ) < 64 \frac{\epsilon }{\sin (\phi )}. \end{aligned}$$

This implies that $l(\{Z \in S \Delta D | \measuredangle (XZY)= \pi /2 \}) < 64 \frac{\epsilon }{\sin (\phi )}$.

Similarly, if $\{Z | \measuredangle (XYZ)= \pi /2 \}$ intersects $\partial D$ at an angle $\phi$, then so long as $\sin (\phi )$ is much larger than $\epsilon$, then we have the same estimate, and so $l(\{Z \in S \Delta D | \measuredangle (XYZ)= \pi /2 \}) < 64 \frac{\epsilon }{\sin (\phi )}$. The argument for the $\{Z \in S \Delta D | \measuredangle (ZXY)= \pi /2 \}$ is then exactly the same.

So long as the intersection angles are not too small and the curvature of $\{Z \in D | \measuredangle (XZY)= \pi /2 \}$ is not too large, this estimate shows that deforming from D to S does not change the integral $\int _{S} \frac{\partial }{\partial r} f(X,Y,z) \mathrm{d}z$ much. A good strategy for this step is to consider curves which intersect at an angle $\phi > \epsilon ^{1/2}$ as sufficiently transverse. This will give strong enough estimates to obtain uniform Hölder continuity. Meanwhile, the compliment of this set has size $O(\epsilon ^{1/2})$.

This transversality estimate will encounter problems in three cases.

• (1) If we apply a rotation so that X is close to the point (1, 0), an issue occurs when Y is close to the line $y=0$. When this occurs, the set $\{Z \in S | \measuredangle (XZY)= \pi /2 \}$ is nearly tangent to $\partial D$ at their intersection, so the length of $\{Z \in S | \measuredangle (XZY)= \pi /2 \}$ might differ greatly from $\{Z \in D | \measuredangle (XZY)= \pi /2 \}$. We discard the set of Y’s with y coordinate smaller that $\epsilon ^{1/2}$, which has measure at most 2$\epsilon ^{1/2}.$ Outside of this set, the length of the circle in $S \Delta D$ is $O(\epsilon ^{1/2}).$

• (2) Another problem occurs in a small segment containing the point $-X$, in which case the line $\{ Z \in D | \measuredangle (XYZ)= \pi /2 \}$ is nearly tangent to the disk at $-X$. The strategy to circumvent this is exactly the same as in the previous step, where we discarded the set of Y’s so that $|(X-Y)\cdot X| < \epsilon ^{1/2}$.

• (3) The final problem occurs when $|X-Y|$ is small, as the curvature of $\{Z \in D | \measuredangle (XZY)= \pi /2 \}$ blows up. However, the set where $|X-Y|<\epsilon ^{1/2}$ has size $O(\epsilon )$, and the set $\{Z \in S | \measuredangle (XZY)= \pi /2 \}$ has length less than $2 \pi \epsilon ^{1/2}$. Outside of this set, we can use the transversality estimate to obtain Hölder continuity.

To finish the proof, we establish the following uniform estimate.

$$\begin{aligned}&\Bigg | \int _{ \{ Z \in S | \measuredangle (YXZ)= \pi /2\} } \frac{(X-Y)}{|X-Y|} \cdot X \mathrm{d}z \Bigg | + \Bigg | \int _{ \{ Z \in S | \measuredangle (XYZ)= \pi /2 \} } \frac{(X-Y)}{|X-Y|} \cdot X\mathrm{d}z \Bigg | \\&\quad + \Bigg | \int _{ \{Z \in S | \measuredangle (XZY)= \pi /2 \}} \frac{ Z -(X+Y)/2}{|Z -(X+Y)/2|} \cdot X \mathrm{d}z \Bigg | \le 2 \pi (1+ 8 \epsilon ) + 4 (1+ 8 \epsilon ). \end{aligned}$$

1.5 E.5: Proving that (12) is $O(\epsilon ^{1/2})$

In fact, we can prove a stronger estimate and show (12) is $O(\epsilon )$. To do so, we must bound the following terms:

$$\begin{aligned}&\int _{S \Delta D} \int _{ \{Z \in D | \measuredangle ({\bar{X}}ZY)= \pi /2 \}} \frac{ Z -({\bar{X}}+Y)/2}{|Z -({\bar{X}}+Y)/2|} \cdot {\bar{X}} \mathrm{d}Z \, \mathrm{d}Y \\&\int _{S \Delta D} \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}} Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y \\&\int _{S \Delta D} \int _{ \{ Z \in D | \measuredangle ({\bar{X}} Y Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot {\bar{X}} \mathrm{d}Z \,\mathrm{d} Y. \end{aligned}$$

This term is quite a bit easier to estimate than the previous ones. For a fixed ${\bar{X}} \in \partial D$, for any $Y \in S \Delta D$, the length of the set of right triangles in D with vertices X and Y is at most $4+2 \pi$, (twice the diameter plus the circumference). Therefore, the inner integrals have a uniform estimate. Since $S \Delta D$ is contained entirely in the annulus of outer radius $1+8 \epsilon$ and inner radius $1-8 \epsilon$, $Vol(S \Delta D) \le 32 \pi \epsilon$, and hence (12) is bounded by $32 \pi (4+2 \pi ) \epsilon$.

1.6 E.6: Finishing the Proof of Lemma 15

From the previous three estimates in this lemma, we have shown that $\frac{\partial }{\partial r} \int _S \int _S f(X,Y,Z) |X| d Y \,d Z$ is Hölder continuous in X and S.

Integrating with respect to $\theta$ from 0 to $2 \pi$, this implies that

$$\begin{aligned} \int _0^{2 \pi } \int _S \int _S \frac{\partial }{\partial r} f(r(\theta _1) e^{i \theta _1},y,z) r(\theta _1) (\mu (\theta _1))^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \end{aligned}$$

is Hölder continuous in S, and so the proof of the lemma is complete. $\square$

1.7 E.7: Lipschitz Continuity of the Arc-Length Term

In this section, we prove Lemma 16, which we restate here. We refer to this as the arc-length term because it corresponds to the change in arc-length as the region evolves along the canonical homotopy.

Lemma

There exist uniform constants $e, C >0$ so that whenever $\epsilon < e$ and $d_{H.}(S,D) \le \epsilon$,

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1\\&\quad -\int _0^{2 \pi } \int _D \int _D f(e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg | < C \epsilon . \end{aligned}$$

Proof

To do this, we use the triangle inequality:

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1\\&\quad - \int _0^{2 \pi } \int _D \int _D f(e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg | \\&\quad \le \Bigg | \int _0^{2 \pi } \int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \\&\qquad - \int _0^{2 \pi } \int _D \int _D f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg | \\&\qquad + \, \Bigg | \int _0^{2 \pi } \int _D \int _D f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \\&\qquad - \int _0^{2 \pi } \int _D \int _D f(e^{i \theta _1},y,z) \mu (\theta _1)^2 \, \mathrm{d}y \,\mathrm{d} z \, \mathrm{d} \theta _1 \Bigg |. \end{aligned}$$

We can estimate the first term by estimating it as a single integral over $(S \times S) \Delta (D \times D)$. Using the same estimate as in the Lipschitz estimate of p(S), the volume of $(S \times S) \Delta (D \times D)$ is less than $16 \pi (1+8\epsilon )^3 \epsilon$. Furthermore, $|f(X,Y,Z)| \le 1$, which shows the following estimate:

$$\begin{aligned} \Big |\int _S \int _S f(r(\theta _1, t) e^{i \theta _1},y,z) \,\mathrm{d}y \,\mathrm{d} z -\int _D \int _D f(r(\theta _1, t) e^{i \theta _1},y,z) \, \mathrm{d}y \,\mathrm{d} z\Big | < C \epsilon . \end{aligned}$$

To estimate the second term, we consider the following:

$$\begin{aligned} \int _D \int _D f(r(\theta _1, t) e^{i \theta _1},y,z) \, \mathrm{d}y \,\mathrm{d} z. \end{aligned}$$

Note that this is the total mass of acute triangles with two vertices in a disk given a fixed third vertex (which may or may not be in the disk). We want to show that this mass is uniformly Lipschitz in the choice of fixed vertex. To see this, we can estimate $|\frac{\mathrm{d}}{\mathrm{dt}}\int _D \int _D f({\bar{X}}+tV,y,z) \, \mathrm{d}y \,\mathrm{d}z|$ for any point ${\bar{X}}$ and unit vector V.

We can use the triangle inequality and some estimates to bound this quantity.

$$\begin{aligned}&\Bigg | \frac{\mathrm{d}}{\mathrm{dt}}\int _D \int _D f({\bar{X}}+tV,y,z) \, \mathrm{d}y \,\mathrm{d}z \Bigg | \nonumber \\&\quad \le \Bigg | \int _D \int _{ \{Z \in D | \measuredangle ({\bar{X}}ZY)= \pi /2 \}} \frac{ Z -({\bar{X}}+Y)/2}{|Z -({\bar{X}}+Y)/2|} \cdot V \mathrm{d}Z \, \mathrm{d}Y \Bigg | \\&\qquad + \Bigg | \int _D \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}} Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot V \mathrm{d}Z \,\mathrm{d} Y \Bigg | \\&\qquad + \Bigg | \int _D \int _{ \{ Z \in D | \measuredangle ({\bar{X}} Y Z)= \pi /2\} } \frac{({\bar{X}}-Y)}{|{\bar{X}}-Y|} \cdot V\mathrm{d}Z \,\mathrm{d} Y \Bigg | \\&\quad \le \Bigg | \int _D \int _{ \{Z \in D | \measuredangle ({\bar{X}}ZY)= \pi /2 \}} 1~ \mathrm{d}Z \, \mathrm{d}Y \Bigg | \\&\qquad + \Bigg | \int _D \int _{ \{ Z \in D | \measuredangle (Y {\bar{X}} Z)= \pi /2\} } 1~ \mathrm{d}Z \,\mathrm{d} Y \Bigg | \\&\qquad + \Bigg | \int _D \int _{ \{ Z \in D | \measuredangle ({\bar{X}} Y Z)= \pi /2\} } 1~ \mathrm{d}Z \,\mathrm{d}Y \Bigg |. \end{aligned}$$

We can geometrically bound each of these.

$$\begin{aligned} \Bigg | \frac{\mathrm{d}}{\mathrm{dt}}\int _D \int _D f(X+tV,y,z) \, \mathrm{d}y \,\mathrm{d} z|\le & {} | \int _D 2 \pi \, \mathrm{d}Y |+ | \int _D 2 \,\mathrm{d} Y| + |\int _D 2 \,\mathrm{d} Y \Bigg | \\= & {} \pi (2\pi +4). \end{aligned}$$

This shows that the inner two integrals are uniformly Lipschitz. Integrating with respect to $\theta _1$, this implies that for $\epsilon$ small, we can find a uniform C so that

$$\begin{aligned}&\Bigg | \int _0^{2 \pi } \int _D \int _D f(r(\theta _1, t) e^{i \theta _1},y,z) \mu (\theta _1)^2 \mathrm{d}y\mathrm{d} z \mathrm{d} \theta _1\\&\quad -\int _0^{2 \pi } \int _D \int _D f(e^{i \theta _1},y,z) \mu (\theta _1)^2 \mathrm{d}y \mathrm{d} z \mathrm{d} \theta _1 \Bigg | < C \epsilon . \end{aligned}$$

This completes the proof of Lemma 16. $\square$

Since we have proven Lemmas 14–16, we have established the Hölder-1 / 2 continuity for each of the terms in $\frac{\mathrm{d}^2}{\mathrm{dt}^2} M$. This completes the proof of Theorem 3. $\square$