Appendix A. Some Geometric Results

Here, we assemble a few geometric statements that we used.

Lemma A.1

Let \(\Lambda \subset }}^\) be a piecewise \(\textsf^\) region for some \(0< \alpha <1\). Let \((\Psi _,i})_\) be a piecewise \(\textsf^\) atlas of \(\partial \Lambda \) and \(\Gamma \) as defined in Definition 2.1. Then, there is a constant C depending only on \(\Lambda \) such that for all unequal \(v_1\) and \(v_2\) in \(\partial \Lambda \), we have

$$\begin \Vert v_1 -v_2 \Vert \ge C \min \left\ \right|^} , }(v_1, \Gamma )\right\} . \end$$

(A.1)

Remark A.2

The normal vector \(n(v_1)\) is well defined if \(v_1 \not \in \Gamma \). In the case \(v_1 \in \Gamma \), the minimum on the right-hand side is meant to be \(0=}(v_1, \Gamma )\), which turns it into a trivial statement.

Proof

We begin with the case \(v_1 \in \Gamma \) or \(v_2 \in \Gamma \); \(v_1\in \Gamma \) is explained in the above remark. If \(v_2\in \Gamma \), then we trivially have \(\Vert v_1 - v_2 \Vert \ge }(v_1, \Gamma )\) and thus the claim holds for any \(C \le 1\).

Let us now consider the case that there is an \(i\in I\) such that both \(v_1\) and \(v_2\) are in \(\Psi _,i}((0,1)^d)\). As \(\Psi :=\Psi _,i}\) is injective, there are unique \(x_k\in (0,1)^d\) such that \(\Psi (x_k)=v_k\in }}^\) for \(k=1,2\). We observe that \(n(v_1) \cdot D\Psi (x_1)=0 \in }}^d\), as the image of the matrix \(D\Psi (x_1)\) is the tangent space to \(\partial \Lambda \) at \(v_1\) and hence is orthogonal to the outward normal vector \(n(v_1)\). Thus, using (2.13), we see

$$\begin |n(v_1) \cdot (v_1-v_2) |&= \big |n(v_1) \cdot \big (\Psi (x_1)-\Psi (x_2)\big ) \big |\end$$

(A.2)

$$\begin&\le \Vert x_1 - x_2 \Vert \sup _ \big |n(v_1) \cdot D\Psi (tx_1+ (1-t)x_2) \big |\end$$

(A.3)

$$\begin&= \Vert x_1 - x_2 \Vert \sup _ \big |n(v_1) \cdot (D\Psi (tx_1+ (1-t)x_2) - D\Psi (x_1) \big |\end$$

(A.4)

$$\begin&\le \Vert x_1 -x_2 \Vert \, C \Vert x_1-x_2 \Vert ^\alpha = C \Vert x_1-x_2 \Vert ^ . \end$$

(A.5)

As \(\Psi \) is bi-Lipschitz, we know that \(\Vert v_1- v_2 \Vert = \Vert \Psi (x_1)-\Psi (x_2) \Vert \ge C \Vert x_1 -x_2 \Vert \). Using this and dividing both sides by \(\Vert v_1-v_2\Vert \), we arrive at

$$\begin \left|n(v_1) \cdot \frac \right|\le C\Vert v_1-v_2 \Vert ^\alpha . \end$$

(A.6)

We are now in the remaining case that \(v_1\) and \(v_2\) lie in different sets \(\Psi _,i}((0,1)^d)\) since \(\partial \Lambda = \Gamma \cup \bigcup _ \Psi _,i}((0,1)^d)\).

Let \((\Psi _,j})_\) be a global Lipschitz atlas of \(\partial \Lambda \). As the boundary \(\partial \Lambda = \bigcup _ \Psi _,j}((0,1)^d)\) is a cover by (relatively) open sets and \(\partial \Lambda \) is a compact metric space, by Lebesgue’s number lemma, there is an \(\varepsilon >0\) such that for all \(v \in \partial \Lambda \) there is an \(j\in J\) with \(B_(v) \cap \partial \Lambda \subset \Psi _,j}((0,1)^d)\), where \(B_(v)\subset }}^\) is the open ball of radius \(\varepsilon \) at v.

If \(\Vert v_1- v_2 \Vert \ge \varepsilon \), we can choose \(C= \varepsilon \) to get the statement, as the first expression inside the minimum is at most 1.

Hence, we are left with the case \(\Vert v_1- v_2 \Vert < \varepsilon \). Now, we get an \(j \in J\) such that \(v_1,v_2 \in \Psi _,j}((0,1)^d)\). Again, we define \(y_k\) by \(\Psi _,j}(y_k)=v_k\) for \(k=1,2\). The image \(\gamma \) of the linear path from \(y_1\) to \(y_2\) is at most \(C \Vert y_1-y_2\Vert \le C \Vert v_1 -v_2 \Vert \) long. As \(v_1\) and \(v_2\) are in the images of two different \(\Psi _,i}\)’s, the path \(\gamma \) has to intersect some edge \(\Psi _,i}(\partial (0,1)^d)\) which implies \(\gamma \cap \Gamma \ne \emptyset \). Hence, we have

$$\begin }(v_1, \Gamma ) \le }^1}( \gamma ) \le C \Vert y_1- y_2 \Vert \le C \Vert v_1 -v_2 \Vert . \end$$

(A.7)

The last inequality follows since \(\Psi _,j}\) is bi-Lipschitz. This finishes the proof. \(\square \)

Lemma A.3

For \(d\ge 1\), let \(f :[0,1]^d \rightarrow }}^\) be a Lipschitz continuous function with Lipschitz constant \(C_}(f)\) and let \(\Lambda \subset }}^\) be a piecewise Lipschitz region. Then, for any \(r>0\), the \((d+1)\)-dimensional Lebesgue volume of the r-neighborhood (see (2.1)) of the set \(f([0,1]^d)\) in \(}}^\) satisfies

$$\begin |B_r (f([0,1]^d)) |\le )^d (C_}(f)^d+1) }(r+r^ ) , \end$$

(A.8)

and the set \(\partial \Lambda \) satisfies the bounds

$$\begin |B_r (\partial \Lambda ) |&\le }(\Lambda )}(r+r^ ) , \end$$

(A.9)

$$\begin }^d(\partial \Lambda )&\le }(\Lambda ) , \end$$

(A.10)

where \(}(\Lambda )\) is described as follows: Let \(}\) be the set of all piecewise Lipschitz atlases of \(\partial \Lambda \), as defined in Definition 2.1. Then, we define

$$\begin } (\Lambda ) :=\inf _,i})_ \in }} \sum _ (16 \sqrt)^d (C_}(\Psi _,i})^d+1) . \end$$

(A.11)

Proof

We consider the set

$$\begin A_r :=\left( \frac}(f)\sqrt} }\right) ^d \cap [0,1]^d . \end$$

(A.12)

The maximum distance a point in \([0,1]^d\) can have from \(A_r\) is less than \(\frac}(f)}\). For the cardinality \(\#A_r\) of \(A_r\), we observe

$$\begin & \#A_r \le \left( 1 +\frac}(f)\sqrt} \right) ^d\le 2^ \left( 1+ \frac}(f)^d \sqrt^d} \right) \nonumber \\ & \quad \le 2^ \sqrt ^d (1+C_}(f)^d )(1+r^) . \end$$

(A.13)

For any \(x \in [0,1]^d\), there is a \(z \in A_r\) such that \(\Vert x-z\Vert \le r /C_}(f)\) and thus \(\Vert f(x)- f(z) \Vert \le r\). This implies \(B_r(f(A_r)) \supset f([0,1]^d)\), which leads to \(B_(f(A_r)) \supset B_r(f([0,1]^d))\). Hence, we get

$$\begin |B_r (f([0,1]^d)) |&\le |B_(f(A_r)) |\le \, \# A_r (2r)^ \le 4^ \,\#A_r r^\\&\le (16 \sqrt)^d(1+C_}(f)^d ) (r+r^) . \end$$

This finishes the proof of the first statement. The second statement is trivially implied by the first one. Furthermore, as \(B_r(f(A_r)) \supset f([0,1]^d)\) due to the definition of the Hausdorff measure (see, e.g., [9, Definition 2.1]) we observe

$$\begin }^d(f([0,1]^d)\le & \lim _ |B_1^(0) |\#A_r r^d \le \lim _ (4 \sqrt)^d (1+C_}(f)^d )(1+r^) r^d \\= & (4 \sqrt)^d (1+C_}(f)^d ) \, . \end$$

The final statement is a corollary of this inequality. We want to note that \(}(\Lambda )<\infty \) for any piecewise Lipschitz region \(\Lambda \), as we require in this paper our atlases to be a finite collection of charts. \(\square \)

Lemma A.4

Let \(\Lambda \subset }}^3\) be a piecewise Lipschitz region with piecewise Lipschitz atlas \((\Psi _,i})_\). Let \(v_0 \in \partial \Lambda \) satisfy that there are \(i_0 \in I\) and \(x_0 \in (0,1)^2\) such that \(\Psi _,i_0}(x_0)=v_0\) and the Jacobi matrix \(D \Psi _,i_0}(x_0)\) exists. Then, the signed distance function \(}_\Lambda \) is differentiable at \(v_0\), the outward unit normal vector \(n(v_0)\) is well-defined, orthogonal to the image of \(D \Psi _,i_0}(x_0)\), and \(D}_\Lambda (v_0)=n(v_0)\).

To prove this statement, we need the following result from intersection theory.

Lemma A.5

Let \(R>0\), \(f_1 :[-1,1] \rightarrow \overline_R^ (0), f_2 :\overline_R^ (0) \rightarrow \overline_R^ (0)\) be continuous functions such that \(f_1( \pm 1 ) = \pm R e_3\) and \(f_2\) restricted to the boundary is the equatorial embedding, that is, \(f_2(x)=(x,0)\) for \(\Vert x \Vert =R\). Then, the images of \(f_1\) and \(f_2\) intersect.

Proof

Without loss of generality, we assume \(R=1\). Assume \(f_1\) and \(f_2\) were two such functions such that their images do not intersect. Let \(\eta _1 :}}^1 \rightarrow }}^3, t \mapsto (0,0,t)\) and \(\eta _2 :}}^2 \rightarrow }}^3, x \mapsto (x,0)\) be the natural orthogonal inclusions. The assumptions on \(f_j\) can now be stated as \(f_j(x_j)= \eta _j(x_j)\) for \(x_j \in }}^j\) with \(\Vert x_j \Vert =1\) for \(j\in \\). We extend the maps \(f_j\) to the \(}}^j\) by setting

$$\begin f_j(x_j) := f_j(x_j) &\quad \text \Vert x_j \Vert \le 1 \\ \eta _j(x_j) &\quad \text \Vert x_j \Vert >1 \end\right. } ,\quad x_j \in }}^j ,\,j\in \ . \end$$

(A.14)

Trivially, these extensions are still continuous and their images still do not intersect. As the images do not intersect and only get close to each other in the compact set \(\overline_1^(0)\), they have a positive distance. We can now mollify \(f_j\) by convolution with an appropriately chosen, compactly supported smooth function to get \(}_j\) such that the images of \(}_1\) and \(}_2\) still have positive distance and \(}_j (x_j)=\eta _j(x_j)\) for any \(x_j \in }}^j\) with \(\Vert x_j \Vert \ge 2\).

For \(d=1,2,3\), consider the sphere \(}^d= }}^d \cup \\). With the charts \(\text :}}^d \rightarrow }^d, x \mapsto x\) and \(\iota _d :}}^d \rightarrow }^d, x \mapsto x / \Vert x \Vert ^2, 0 \mapsto \infty \), it becomes a differentiable manifold. We now extend \(}_j\) to a function from \(}^j\) to \(}^3\) by setting \(}_j(\infty ):=\infty \) for \(j\in \\). The point \(\infty \) is now an intersection point of \(}_1\) and \(}_2\). We want to show that the extended functions are still smooth and that they intersect transversely at \(\infty \), see for instance [14, Page 113]. For \(j\in \\) and \(x_j \in }}^j\) with \(\Vert x_j \Vert < \frac\), we observe

$$\begin (\iota _3^ \circ }_j \circ \iota _j)(x_j)= (\iota _3^ \circ \eta _j )(x_j \Vert x_j \Vert ^)=\eta _j(x_j) . \end$$

(A.15)

Thus, in the charts \(\iota _j ,\iota _3\) the maps \(}_j\) are linear and orthogonal at 0 (which corresponds to \(\infty \in }^j\)). The maps \(}_1,}_2\) are therefore smooth and intersect transversely at \(\infty \). In conclusion, we have just constructed two smooth maps \(}_j :}^j \rightarrow \mathbb S^3\), which intersect transversely and have a unique intersection point. Thus, their oriented intersection number is equal to the local intersection number at this intersection point, which is \(+1\) or \(-1\) (in fact, it is \(+1\)). However, both maps are contractible (homotopic to a constant map) and thus, as oriented intersection numbers are homotopy invariant (see [14, Page 115]), they should have intersection number 0. This is a contradiction. Hence, the assumption that \(f_1\) and \(f_2\) do not intersect was wrong. \(\square \)

Proof of Lemma A.4

Let \(C_}\) be a bi-Lipschitz constant of \(\Psi _,i_0}\), as in footnote 1. Then, for any \(x \in }}^2\) with \(\Vert x \Vert =1\), we observe \(C_}^ \le \Vert D\Psi _,i_0}(x_0) x \Vert \le C_}\). This means that the Jacobi matrix is invertible. Thus, using affine linear transformations on \(}}^2\) and on \(}}^3\), we can transform the function \(\Psi _,i_0}\) into a function \(\Psi \) such that \(x_0,v_0\) are mapped to 0Footnote 4 and the Jacobi matrix turns into the standard inclusion \(J :}}^2 \rightarrow }}^3, x \mapsto (x,0)\). The function \(\Psi \) is now defined on some closed parallelogram P containing 0 in its interior \(P^}\). Let \(C_}\ge 2\) be a bi-Lipschitz constant for \(\Psi \). Let \(0<\varepsilon <1/2\). Then, there is an \(r>0\) such that

For any \(x \in B_}r}^(0)\), we have \(\Vert \Psi (x)-(x,0) \Vert \le \varepsilon \Vert x \Vert \), as \(D \Psi (0)=J\);

\(B_^(0)\cap \partial \Lambda \subset \Psi (P^})\), as \(\Psi ^}\) is relatively open in \(\partial \Lambda \);

The set has exactly two connected components, as \(\overline\) and are topological manifolds with common boundary \(\partial \Lambda \).

As \(\Psi \) is bi-Lipschitz, we observe

$$\begin B_^(0) \cap \partial \Lambda&\subset \Psi \left( B_}r}^(0) \right) \subset \bigcup _}r}^(0)} \overline_^ ((x,0)) \end$$

(A.16)

$$\begin&\subset \bigcup _}}^2 } \overline_^ ((x,0)) =\big \}}^3 :|v \cdot e_3 |\le \varepsilon \Vert v \Vert \big \} \, . \end$$

(A.17)

We define

$$\begin U_0&:=\big \}}^3 :|v \cdot e_3 |\le \varepsilon \Vert v \Vert \big \} , \end$$

(A.18)

$$\begin U_\pm&:=\big \}}^3 :\pm v \cdot e_3 > \varepsilon \Vert v \Vert \big \} . \end$$

(A.19)

The sets \(U_\pm \cap B_r^(0)\) are open, convex and do not intersect \(\partial \Lambda \) due to (A.17). We will now use Lemma A.5 to show that, up to a binary choice, we may assume \(U_- \cap B_r^(0) \subset \Lambda \) and . As has exactly two connected components, it is sufficient to prove that any (continuous) path \(p :[-1/2 , 1/2 ] \rightarrow B_r^(0)\) with \(p(\pm 1/2) \in U_\) intersects \(\partial \Lambda \).

We first use the convexity to extend p by an (affine) linear path at both ends to get a path \(f_1 :[-1,1] \rightarrow \overline_^(0)\) with \(f_1(\pm 1)= \pm 5 r e_3\). Then we define \(f_2 :\overline_^(0) \rightarrow }}^3\) by

$$\begin f_2(x) := \Psi (x) \quad & \text \Vert x \Vert< 2 r \\ \frac (x,0) + \frac \Psi (x) \quad & \text 2 r\le \Vert x \Vert < 3 r \\ (x,0) & \text 3r \le \Vert x \Vert \le 5 r \end\right. } . \end$$

(A.20)

We see that \(f_2 \) is Lipschitz continuous. The middle case is just a convex combination between the two other cases. Let \(x \in }}^2\) with \(\Vert x \Vert \le 3 r\). We observe

$$\begin \Vert f_2 (x) - (x,0) \Vert \le \sup _ \Vert t \Psi (x) + (1-t) (x,0) -(x,0) \Vert \le \sup _ t \Vert \Psi (x)- (x,0) \Vert < \varepsilon \Vert x \Vert . \end$$

(A.21)

This implies

$$\begin |f_2(x)\cdot e_3| = |f_2(x)\cdot e_3 -(x,0)\cdot e_3| \le \Vert f_2(x)-(x,0)\Vert <\varepsilon \Vert x \Vert , \end$$

that is, \(f_2\left( \overline_^(0)\right) \subset U_0\) and thus, by the definition of \(f_2(x)\) for \(\Vert x\Vert \ge 3r\), \(f_2 \left( \overline_^(0) \right) \subset U_0\). By the triangle inequality and with \(\varepsilon < \frac \) we obtain

$$\begin \frac \Vert x \Vert< \Vert f_2(x) \Vert <\frac \Vert x \Vert \, . \end$$

(A.22)

These inequalities yield \( f_2^ \left( B_^ (0) \right) \subset B_^(0)\) and \(f_2 \left( \overline_^ (0)\right) \subset B_^(0)\). The latter inclusion together with the definition of \(f_2\) outside \(B_^(0)\) implies \(f_2\left( B_^(0) \right) \subset B_^(0)\). Thus, \(f_1\) and \(f_2\) satisfy the assumptions of Lemma A.5 (with \(R :=5 r\)) and consequently, they have an intersection point \(s \in }}^3\). We have \(s \in f_2\left( \overline_^(0)\right) \subset U_0\) and \(f_1^(U_0) \subset (-\frac, \frac)\), which means s is in the image of the original path p. Thus, \(s \in B_r^(0)\), which implies that \(s \in f_2\left( B_^(0)\right) = \Psi \left( B_^(0)\right) \subset \partial \Lambda \). Therefore, the path p intersects \(\partial \Lambda \), which was our claim.

As a result, we know that the sets \(U_\pm \cap B_r^(0)\) lie on opposite sides of \(\partial \Lambda \). Without loss of generality, we assume \(U_- \cap B_r^(0) \subset \Lambda \) and . In terms of the signed distance function \(}_\Lambda \), this means that \(\pm }_\Lambda (v) >0\) for \(v \in U_\pm \cap B_r^(0)\).

We are left to estimate \(}(v, \partial \Lambda )\) for \(v \in B_r^(0)\). We start with the case \(v \in U_0 \cap B_r^(0)\). For that, we consider the map

$$\begin \Phi :B_r^(0) \times [-2 \varepsilon , 2 \varepsilon ] \rightarrow }}^3 , \quad (y,t) \mapsto (\sqrt y, t \Vert y\Vert ) . \end$$

(A.23)

We see that \(\Phi \left( B_r^(0) \times [- \varepsilon , \varepsilon ]\right) = U_0 \cap B_r^(0)\) and \(\Vert \Phi (y,t) \Vert = \Vert y \Vert \). Furthermore, for a fixed y, the map \(t \mapsto \Phi (y,t)\) defined on \([-2\varepsilon ,2\varepsilon ]\) is a path between \(U_+\) and \(U_-\) inside \(B_r^(0)\) and must thus intersect \(\partial \Lambda \). This path has a length of \(2 \Vert y \Vert \sin ^(2 \varepsilon ) \le 2 \pi \varepsilon \Vert y \Vert \). Hence, as each point \(v \in U_0 \cap B_r^(0)\) is on such a path for a y with \(\Vert y \Vert = \Vert v \Vert \), we get \(|}_\Lambda (v) |\le 2 \pi \varepsilon \Vert v \Vert \). Therefore, for \(v \in U_0 \cap B_r^(0)\), we get

$$\begin |}_\Lambda (v) - v \cdot e_3 |\le |}_\Lambda (v) |+ |v \cdot e_3 |\le (2\pi + 1 ) \varepsilon \Vert v \Vert . \end$$

(A.24)

For \(v \in U_ \cap B_r^(0)\), we know that \(\pm }_\Lambda (v) >0\) and only need upper and lower bounds for the distance to \(\partial \Lambda \). For the lower bound, as \(\partial \Lambda \cap B_^(0) \subset U_0\), we have

$$\begin |}_\Lambda (v) |\ge }(v,U_0) = \sqrt |v \cdot e_3 |- \varepsilon \Vert v^\perp \Vert \ge |v \cdot e_3 |- \varepsilon ^2 |v \cdot e_3 |- \varepsilon \Vert v^\perp \Vert \ge |v \cdot e_3 |- 2\varepsilon \Vert v \Vert . \end$$

(A.25)

For the upper bound, we just use

$$\begin |}_\Lambda (v) |\le |v \cdot e_3 |+ |}_\Lambda ((v^\perp ,0))| \le |v \cdot e_3 |+2 \pi \Vert v^\perp \Vert \le |v \cdot e_3 |+ 2 \pi \Vert v \Vert .\nonumber \\ \end$$

(A.26)

As the signs align, we finally get

$$\begin |}_\Lambda (v) - v \cdot e_3 |\le 2 \pi \Vert v \Vert . \end$$

(A.27)

Thus, (A.24) holds for all \(v \in B_^(0)\), which, by definition, says that \(}_\Lambda \) is differentiable at 0 and its differential is \(e_3\). We also see that \(e_3\) is orthogonal to the image of J and points toward , which means that it is the outward normal vector to \(\partial \Lambda \) at 0. This finishes the proof. \(\square \)

Lemma A.6

Let \(\Lambda \subset }}^3\) be a piecewise Lipschitz region. Then, the outward normal vector n(v) exists for \(}^2\) almost every \(v \in \partial \Lambda \) and the set

$$\begin } :=\left\}}^2 :\partial \overline\right) } \ne \left( \partial \Lambda \right) _ \right\} \end$$

(A.28)

is a (two-dimensional) Lebesgue null set, where \(\Lambda _\) and \(\left( \partial \Lambda \right) _\) are defined in Definition 3.1.

Proof

We observe

$$\begin \partial \overline\right) } \subset \partial \left( \Lambda _\right) \subset \left( \partial \Lambda \right) _ . \end$$

(A.29)

The first inclusion is trivial. The second inclusion can be seen as follows. Let \(t\in \partial \left( \Lambda _\right) \). Then for all \(r>0\), \(B_r^(t)\cap \Lambda _\not =\emptyset \) and . Therefore, \(B_r^(x^\perp ,t)\cap \Lambda \not =\emptyset \) and . Therefore, \((x^\perp ,t)\in \partial \Lambda \) and \(t\in \left( \partial \Lambda \right) _\).

Let \(\pi :}}^3 \rightarrow }}^2\) be the projection with \(\pi (e_3)=0\) and let \(\left( \Psi _,i}\right) _\) be a piecewise Lipschitz atlas of \(\partial \Lambda \). For \(i \in I\), we define the sets

$$\begin }_i :=\partial [0,1]^2 \cup \big \,i}(x) \text \big \} , \end$$

(A.30)

which are Lebesgue null sets due to Rademacher’s theorem.Footnote 5 Thus, the set \(\bigcup _\Psi _,i}(}_i)\) is an \(\mathcal H^2\) null set, see [9, Theorem 2.8(i)]. Combining this with Lemma A.4, we now know that the outward normal vector n(v) is well-defined for \(}^2\) every \(v \in \partial \Lambda \). As \(\pi \circ \Psi _,i} :[0,1]^2 \rightarrow }}^2\) is Lipschitz, this implies that \((\pi \circ \Psi _,i}) (}_i)\) is a Lebesgue null set [9, Lemma 3.2(iii)]. Furthermore, we define the sets

$$\begin }_i :=\big \,i})(x) \text \big \} . \end$$

(A.31)

By [9, Theorem 3.8], we know that \((\pi \circ \Psi _,i}) (}_i)\) is a Lebesgue null set. Let

$$\begin \Omega _i :=[0,1]^2 (}_i \cup \mathcal M_i) , \end$$

(A.32)

and

$$\begin } :=\partial \Lambda \bigcup _ \Psi _,i}(\Omega _i) . \end$$

(A.33)

Let \(v \in \partial \Lambda }\). Hence, there is an \(i \in I\) and a \(y \in \Omega _i\), such that \(v = \Psi _,i}(y)\). Thus, \(D\Psi _,i}(y)\) exists, has full rank and does not have \(e_3\) in its image. By Lemma A.4, we know that n(v) exists and that \(n(v) \cdot e_3 \ne 0\). Thus, the function \(p :}}\rightarrow }}\) given by \(p(t)=}_\Lambda (v+te_3)\) with \(}_\Lambda \) being the signed distance function to the boundary \(\partial \Lambda \) has non-vanishing differential at 0 and satisfies \(p(0)=0\). Hence, p changes sign at 0, which means that \(v^\parallel \in \partial \overline\right) }\). Conversely, this means that for \(v \in \partial \Lambda \), the property \(v^\parallel \not \in \partial \overline\right) }\) implies that \(v \in \mathcal M\). Thus, we have for the set \(}\) defined in (A.28),

$$\begin } \subset \pi \left( }\right) . \end$$

(A.34)

Finally, we observe

$$\begin } \subset \pi \left( \bigcup _ \Psi _,i}(}_i \cup }_i)\right) = \bigcup _ \left( (\pi \circ \Psi _,i}) (\mathcal N_i) \cup (\pi \circ \Psi _,i}) (}_i)\right) , \end$$

(A.35)

which shows that \(}\) is a Lebesgue null set.\(\square \)

Lemma A.7

Let \(\Lambda \subset }}^3\) be a piecewise Lipschitz region, \(\pi :}}^3 \rightarrow }}^2\) be the canonical projection and \(f :\partial \Lambda \rightarrow }}^+\) be measurable. Let \(n :\partial \Lambda \rightarrow }}^3\) be the outward normal vector field, which is defined almost everywhere (see Lemma A.6). Then, we have

$$\begin \int _ \textrm}^2}( v)\, f(v) |n(v) \cdot e_3 |= \int _}}^2}\textrmx \sum _(x) \cap \partial \Lambda } f(v) . \end$$

(A.36)

Proof

The proof is based on a quite general form of changing variables. We use the following area-formula (see [9, Theorem 3.9]) with one slight modification. To this end, let \(n,m \in }\) with \(n \le m\), \(U \subset }}^n\) be open, \(\Phi :U \rightarrow }}^m\) be Lipschitz continuous and let \(g :U \rightarrow }}^+\) be measurable. Then, we have the identity

$$\begin \int _U \textrmy\, g(y) |D\Phi (y) |= \int _}}^m} \textrm}^n}( x)\sum _(x)} g(y) , \end$$

(A.37)

where \(|D\Phi (y) |^2= \det (D\Phi (y)^* D\Phi (y))\). In [9], this is stated for \(g \in }^1(U)\). However, their proof also applies to positive, measurable functions g as an identity in \([0,\infty ]\).

We cannot apply this directly to \(\pi \), as it decreases the dimension and \(\pi \) does not have an inverse. Thus, we have to introduce a new map.

Let \((\Psi _,i})_\) be a piecewise Lipschitz atlas of \(\partial \Lambda \) so that \(\partial \Lambda = \bigcup _\Psi _,i}([0,1]^2)\). We may assume that \(}(f) \subset \Psi _,i} ((0,1)^2) \) for some \(i \in I\). For the remainder of this proof, we write \(\Psi \) for \(\Psi _,i}\).

Now, we can apply (A.37) with \(\Phi :=\pi \circ \Psi \) and \(g :=f\circ \Psi \). Thus, we see

$$\begin \int _ \textrmy\, f(\Psi (y)) |D(\pi \circ \Psi )(y) |&= \int _}}^2}\textrmx \sum _(x)} f(\Psi (y)) \end$$

(A.38)

$$\begin&= \int _}}^2} \textrmx \sum _(x)\cap \partial \Lambda } f(v) . \end$$

(A.39)

We used that \(\Psi \) is bijective. So, we already have the right-hand side of the claim. We will apply again (A.37) with the functions \(\Phi :=\Psi \) and g given by

$$\begin g(y):=f(\Psi (y))\frac ,\quad y\in (0,1)^2 . \end$$

(A.40)

Thus, using that \(\Psi \) is bijective and that the measure \(\mathcal H^2\) on \(\partial \Lambda \) is the 2-dimensional Hausdorff measure, we have

$$\begin \int _ \textrmy\,f(\Psi (y)) |D(\pi \circ \Psi )(y) |= \int _ \textrm}^2}(v)\, f(v) \frac(v)) |}(v))|} . \end$$

(A.41)

To conclude the proof, we only need to show that the quotient of the functional determinants is given by \(|n(v) \cdot e_3 |\) for almost every \(v \in \partial \Lambda \). Let \(v \in \partial \Lambda \) be such that \(B :=D\Psi (\Psi ^(v)) \in }}^\) is well defined. We identify the two column vectors of the \(3 \times 2\) matrix B as \(w_1\) and \(w_2\). The image of B is the tangent space to \(\partial \Lambda \) at v. As \(\Psi \) is bi-Lipschitz continuous, the matrix B has full rank. The normal vector n(v) is now orthogonal to the linear independent vectors \(w_1,w_2\). Thus, \(n(v) |w_1 \times w_2 |= \pm w_1 \times w_2\).

As \(\pi \) is linear, we have \(D(\pi \circ \Psi )(\Psi ^(v))= \pi B\). For the determinant of this \(2 \times 2\) matrix, we get \(\det (\pi B)=( w_ 1\times w_2) \cdot e_3\). For the denominator, we observe

$$\begin \det ( B^* B)= |w_1 |^2 |w_2 |^2 - (w_1 \cdot w_2)^2= |w_1 \times w_2|^2 . \end$$

(A.42)

In conclusion, we have

$$\begin \frac(v)) |}(v))|} = \frac =|n(v) \cdot e_3 |. \end$$

(A.43)

In combination with (A.39) and (A.41), we have proved the statement. \(\square \)

Corollary A.8

Let \(\Lambda \subset }}^3\) be a piecewise Lipschitz region. Then, for Lebesgue almost every \(x^\perp \in }}^2\), the set \(\Lambda _\) is a finite (possibly empty) union of intervals with disjoint closures.

Proof

As \(}^2( \partial \Lambda )\) is finite, see Lemma A.3, we have by Lemma A.7 (with \(f=1\))

$$\begin \int _}}^2} \textrm x^\perp \, \# (\partial (\Lambda _)) = \int _ \textrm }^2(v)\, |n(v) \cdot e_3 |\le }^2(\partial \Lambda ) < \infty . \end$$

(A.44)

This implies that the set \(\partial (\Lambda _)\subset }}\) is finite for almost every \(x^\perp \). Hence, \(\Lambda _\) is almost everywhere a finite union of intervals. If, for some \(x_0^\perp \in }}^2\), two different connected components of \(\Lambda _\) share a boundary point, t, then \(t \in \partial ( \Lambda _) \partial \overline\right) }\). Looking at Lemma A.6, we realize that this means \(x_0^\perp \in }\). Thus, we have proved the claim. \(\square \)

Lemma A.9

Let \(\Lambda \subset }}^3\) be a piecewise \(\textsf^\) region with \(\Gamma \) as in Definition 2.1. Then, there is a constant \(C< \infty \) such that

$$\begin \int _ \mathrm d }^2(w) |\ln (} (w, \Gamma ) ) |\le C . \end$$

(A.45)

Proof

We start with

$$\begin \int _ \textrm}^2(w)\,|\ln (}(w,\Gamma ))|&\le C \sum _}} (|k |+1) \cdot }^2\big ( \big \ \le } (w,\Gamma ) \le 2^k \big \} \big ) \end$$

(A.46)

$$\begin&\le C \sum _^}} (|k |+1)\cdot }^2\big ( \big \} (w,\Gamma ) \le 2^k \big \} \big ) . \end$$

(A.47)

For the first step, we just bound the integrand by a step function from above. As \(\Lambda \) is bounded, the associated set is empty for \(k>k_}\) with some finite \(k_}\). We are left to estimate the volume of these sets. Specifically, we will show that there is an \(r_0>0\) and a \(C < \infty \) such that for any \(r< r_0\), we have

$$\begin }^2( B_r(\Gamma ) \cap \partial \Lambda ) \le Cr . \end$$

(A.48)

We recall that \(B_r(\Gamma )\subset }}^3\) is the r-neighborhood of \(\Gamma \). By assumption, there is a global Lipschitz atlas \((\Psi _,j})_\) of \(\partial \Lambda \), as in Definition 2.1. For each \(i \in I\), the set \(U_j :=\Psi _,j} ((0,1)^d)\subset \partial \Lambda \) is a (relatively) open subset of the compact metric space \(\partial \Lambda \) and we have \(\partial \Lambda \subset \bigcup _ U_j \). Thus, by Lebesgue’s number lemma, there is a constant \(r_0>0\) such that any \(v\in \partial \Lambda \) there is a \(j \in J\) such that \(B_(v) \cap \partial \Lambda \subset U_j\).

Now, we need to understand the set \(\Gamma \). We recall its definition

$$\begin \Gamma :=\bigcup _ \Psi _,i}(\partial [0,1]^2) . \end$$

(A.49)

Let \(C_0\) be a Lipschitz constant for all \(\Psi _,i}\)’s which exists, as I is finite. As \(\partial [0,1]^2\) is just the boundary of the unit square, there is a surjective (piecewise linear) function \(\vartheta :[0,1] \rightarrow \partial [0,1]^2\) with Lipschitz constant 4. Let \(N\in }\) with \(N>4C_0/r_0\) and \(f_k :[0,1] \rightarrow [0,1]\) be the functions satisfying \(f_k(t)=\frac\). Now, for any \(1 \le k \le N\) and \(i \in I\), we define \(g_ :[0,1] \rightarrow \Gamma \) by \(g_ :=\Psi _,i} \circ \vartheta \circ f_k\) and observe

$$\begin C_}\left( g_ \right) \le 4 C_0 /N < r_0 . \end$$

(A.50)

Furthermore, \(\Gamma = \bigcup _ \bigcup _^N g_([0,1])\). By (A.50), we know \(g_([0,1]) \subset B_ (g_(0))\) and thus

$$\begin B_r(g_((0,1)) \subset B_(g_(0)) \end$$

(A.51)

for \(r \le r_0\). Hence, there is an \(j=j(i,k) \in J\), such that \( B_r(g_([0,1]) \cap \partial \Lambda \subset U_ \). For any \(r \le r_0\), we can estimate

$$\begin }^2( B_r(\Gamma ) \cap \partial \Lambda ) \le \sum _ \sum _^N }^2\big ( B_r(g_([0,1]) ) \cap \partial \Lambda \big ) . \end$$

(A.52)

As \(\Psi _,j(i,k)}\) is bi-Lipschitz, there is a constant C such that

$$\begin }^2\big ( B_r(g_([0,1]) ) \cap \partial \Lambda \big ) \le C \left|\Psi _,j(i,k)}^\big ( B_r(g_([0,1]) ) \cap \partial \Lambda \big ) \right|, \end$$

(A.53)

and

$$\begin \Psi _,j(i,k)}^ \left( B_r(g_([0,1]) ) \cap \partial \Lambda \right) \subset B_( \Psi _,j(i,k)}^(g_([0,1]))) . \end$$

(A.54)

We now apply Lemma A.3 with \(f=\Psi _,j(i,k)}^ \circ g_\) and \(d=1\) to obtain

$$\begin |B_( \Psi _,j(i,k)}^(g_([0,1]))) |\le C(r+r^2) \le Cr , \end$$

(A.55)

as \(r < r_0\).

In conclusion, as I is finite, we have

$$\begin }^2(B_r(\Gamma ) \cap \partial \Lambda )&\le \sum _ \sum _^N }^2\big ( B_r(g_([0,1]) ) \cap \partial \Lambda \big ) \end$$

(A.56)

$$\begin&\le C\sum _ \sum _^N \left|\Psi _,j(i,k)}^\big ( B_r(g_([0,1]) ) \cap \partial \Lambda \big ) \right|\end$$

(A.57)

$$\begin&\le C\sum _ \sum _^N |B_( \Psi _,j(i,k)}^(g_([0,1]))) |\end$$

(A.58)

$$\begin&\le C\sum _ \sum _^N Cr \le Cr . \end$$

(A.59)

For \(r_0<r<2^}}\), we trivially arrive at the same estimate as long as \(C \ge }^2( \partial \Lambda ) r_0^\), that is, \( }^2(B_r(\Gamma )\cap \partial \Lambda )\le C r\) also for “large” r.

Now, we are able to finish (A.47) and obtain for some (finite) constant C

$$\begin \int _ \textrm}^2(w)\,|\ln (}(w,\Gamma ))|\le C \sum _^}} (|k |+1) 2^k \le C , \end$$

(A.60)

which was the claim. \(\square \)

Appendix B. Proof of (3.15)

We observe

$$\begin \int _}}^}\prod _^\frac \mathrm d}^\parallel= & \int _}}^}\prod _^\frac \prod _^ \textrmy_j^\parallel \nonumber \\ = & \int _}}^ } \mathrm dx_1^\parallel \cdots \mathrm dx_^\parallel \prod _^ \frac^\parallel \rangle } , \end$$

(B.1)

where we switched back to the integration variables \(x_1, \dots , x_m\) and setFootnote 6\(x_0 :=x_m :=0\). As we can see, the last expression is the \((m-1)\)-fold convolution of \(\langle \, \cdot \, \rangle ^\) with itself evaluated at 0. This is a job for the Fourier transform. We use the convention

$$\begin } (f) (\xi ) :=\lim _ \int _^R \mathrm d t f(t) \,\textrm^ \xi t } , \quad \xi \in }}. \end$$

(B.2)

Thus, we have

$$\begin \int _}}^ } \mathrm dx_1^\parallel \cdots \mathrm dx_^\parallel \prod _^ \frac^\parallel \rangle } = \int _}}\mathrm d\xi \,\mathcal F(\langle \, \cdot \,\rangle ^ )(\xi )^m \, . \end$$

(B.3)

The Fourier transform of \(\langle \, \cdot \, \rangle ^\) can be expressed in terms of the modified Bessel function of the second kind \(K_0\), see [