Uniform maximal Fourier restriction for convex curves

Abstract

We extend the estimates for maximal Fourier restriction operators proved by Müller et al. (Rev Mat Iberoam 35:693–702, 2019) and Ramos (Proc Am Math Soc 148:1131–1138, 2020) to the case of arbitrary convex curves in the plane, with constants uniform in the curve. The improvement over Müller, Ricci, and Wright and Ramos is given by the removal of the ${\mathcal {C}}^2$ regularity condition on the curve. This requires the choice of an appropriate measure for each curve, that is suggested by an affine invariant construction of Oberlin (Michigan Math J 51:13–26, 2003). As corollaries, we obtain a uniform Fourier restriction theorem for arbitrary convex curves and a result on the Lebesgue points of the Fourier transform on the curve.

Appendix A. Compact convex curves

1.1 A.1. Proof of Theorem 2.4

First, for every compact convex curve $\Gamma$ we define a continuous parametrization $\gamma$ in Lemma A.5. We achieve this formalizing the following intuition, see Fig. 2. Let $x_0$ be a point in the bounded open convex set $K \subseteq {\mathbb {R}}^2$, whose boundary $\partial K$ is $\Gamma$. We parametrize $\Gamma$ by ${\mathbb {S}}^1$ via the unique intersection between $\Gamma$ and each positive half-line emanating from $x_0$. Moreover, we choose to parametrize ${\mathbb {S}}^1$ by $[0,2\pi )$ counterclockwise, hence $\Gamma$ too.

Fig. 2

The intuitive parametrization of $\Gamma =\partial K$

After that, we prove the rectifiability of every compact convex curve $\Gamma = \partial K$ claimed in Theorem 2.4. The main ingredient in the proof is the inequality between the perimeters of convex polygons A and B such that $A \subseteq B$ stated in Lemma A.11.

We begin with the definition of the continuous parametrization $\gamma$ for every compact convex curve $\Gamma = \partial K$ outlined above. We first state and prove three auxiliary lemmata.

Lemma A.1

Let $x \in K$, $y \in \partial K$. For every $0 < \lambda \le 1$ we have $\lambda x + (1-\lambda )y \in K$.

Proof

Fix $0< \lambda \le 1$. Since $y \in \partial K$, there exists a sequence $\{ y_n :n \in {\mathbb {N}}\} \subseteq K$ converging to y. Moreover, the sequence $\{ x_n :n \in {\mathbb {N}}\}$ defined by

$$\begin{aligned} x_n := x - \frac{1-\lambda }{\lambda } (y - y_n), \end{aligned}$$

converges to x. Therefore, there exists N such that $x_N \in K$, yielding

$$\begin{aligned} \lambda x +(1-\lambda ) y = \lambda x_N + (1-\lambda ) y_N \in K. \end{aligned}$$

$\square$

Lemma A.2

Let $x_0 \in K$. The function $T=T(x_0)$ defined by

$$\begin{aligned} T :{\mathbb {S}}^1 \rightarrow (0, \infty ), \qquad T(e) := \sup \Big \{ t \ge 0 :x_0+te \in K \Big \}, \end{aligned}$$

is well-defined. Moreover, for every $e \in {\mathbb {S}}^1$ we have

$$\begin{aligned} \partial K \cap \{ x_0+te :t \ge 0 \} = \{ x_0+T(e)e \}. \end{aligned}$$

Proof

Since K is open and bounded, for every $e \in {\mathbb {S}}^1$ we have $T(e) \in (0,\infty )$.

Next, by the definition of T(e), there exists an increasing sequence $\{t_n :n \in {\mathbb {N}}\} \subseteq (0,\infty )$ converging to T(e). Therefore, the sequence $\{ x_0+t_n e :n \in {\mathbb {N}}\} \subseteq K$ converges to $x_0+T(e)e$. Since for $T(e)<t< \infty$ the point $x_0+te \in {\mathbb {R}}^2 \setminus K$, then $x_0+T(e)e \in \partial K$.

To conclude, suppose there exists $t >0$, $t \ne T(e)$ such that $x_0+te \in \partial K$.

If $t< T(e)$, by Lemma A.1 we have $x_0+te \in K$, yielding a contradiction with $x_0+te \in \partial K$.

If $t> T(e)$, the same argument yields a contradiction with $x_0 +T(e)e \in \partial K$. $\square$

Lemma A.3

Let $x_0 \in K$. For $T=T(x_0)$ the function $\tau = \tau (x_0)$ defined by

$$\begin{aligned} \tau :{\mathbb {S}}^1 \rightarrow \partial K \subseteq {\mathbb {R}}^2, \qquad \tau (e) := x_0+T(e)e, \end{aligned}$$

is well-defined and bijective

Proof

The function is well-defined by Lemma A.2.

Injective. Suppose there exist $e_1,e_2 \in {\mathbb {S}}^1$, $e_1 \ne e_2$ such that

$$\begin{aligned} x_0+T(e_1)e_1=x_0+T(e_2)e_2. \end{aligned}$$

If $e_1 \ne -e_2$, they are two linearly independent vectors, hence $T(e_1)=T(e_2)=0$, yielding a contradiction with $T(e_1),T(e_2)>0$.

If $e_1=-e_2$, then $T(e_1)=-T(e_2)$. Since $T(e_1)>0$, then $T(e_2)<0$, yielding a contradiction with $T(e_2)>0$.

Surjective. Let $x \in \partial K$ and consider

$$\begin{aligned} e = \frac{x-x_0}{\left|{x-x_0}\right|} \in {\mathbb {S}}^1. \end{aligned}$$

Then $x \in \partial K \cap \{ x_0+te :t \ge 0 \}$. By Lemma A.2, we have $x = x_0+T(e)e$. $\square$

The remaining ingredient to define $\gamma$ is the following collection of parametrizations of ${\mathbb {S}}^1$.

Definition A.4

Let $e \in {\mathbb {S}}^1 \subseteq {\mathbb {R}}^2$. We define the the counterclockwise continuous parametrization $\Theta = \Theta (e)$ of the circle ${\mathbb {S}}^1$ with starting point e by

$$\begin{aligned} \Theta :[0,2\pi ) \rightarrow {\mathbb {S}}^1 \subseteq {\mathbb {R}}^2, \qquad \Theta (\theta ) := \begin{pmatrix} \cos \theta &{} -\sin \theta \\ \sin \theta &{} \cos \theta \end{pmatrix} e. \end{aligned}$$

In particular, for every $x_1 \in \partial K$ let $\Theta = \Theta (x_1)$ be the counterclockwise continuous parametrization of the circle ${\mathbb {S}}^1$ with starting point $\tau ^{-1}(x_1) \in {\mathbb {S}}^1$.

Lemma A.5

Let $x_0 \in K$, $x_1 \in \Gamma = \partial K$. For $\tau = \tau (x_0)$, $\Theta = \Theta (x_1)$ the function $\gamma = \gamma (x_0,x_1)$ defined by

$$\begin{aligned} \gamma :[0,2 \pi ) \rightarrow \Gamma = \partial K \subseteq {\mathbb {R}}^2, \qquad \gamma := \tau \circ \Theta , \end{aligned}$$

is well-defined, bijective and continuous.

Proof

The function is well-defined and bijective by Lemma A.3 and the definition of $\Theta$. The continuity of $\gamma$ follows from that of $\Theta$ and $T \circ \Theta$.

It is enough to prove that the function $T \circ \Theta$ is continuous. We argue by contradiction and we suppose that it has a discontinuity in $\theta$. Let $\{\theta _n :n \in {\mathbb {N}}\}$ be a sequence converging to $\theta$ such that $\{ T(\Theta (\theta _n)) :n \in {\mathbb {N}}\}$ does not converge to $T(\Theta (\theta ))$. In particular, there exists $\varepsilon > 0$ and a subsequence $\{ \theta _n :n \in M \subseteq {\mathbb {N}}\} \subseteq \{\theta _n :n \in {\mathbb {N}}\}$ such that

$$\begin{aligned} \inf \Big \{ \left|{ T(\Theta (\theta )) - T(\Theta (\theta _n))}\right| :n \in M \Big \} \ge \varepsilon . \end{aligned}$$

Since K is compact, there exists a subsequence $\{ \theta _n :n \in \widetilde{M} \subseteq M \} \subseteq \{\theta _n :n \in M \}$ such that the limit of $\{ T(\Theta (\theta _n)) :n \in \widetilde{M} \}$ exists and is $\widetilde{T} \ne T(\Theta (\theta ))$. We distinguish two cases.

Case I: $\widetilde{T} > T(\Theta (\theta ))$. Fix t such that $\widetilde{T}> t > T(\Theta (\theta ))$. The sequence

$$\begin{aligned} \Big \{ x_0+\frac{t}{\widetilde{T}} T(\Theta (\theta _n)) \Theta (\theta _n) :n \in \widetilde{M} \Big \} \subseteq K, \end{aligned}$$

converges to $x_0+t \Theta (\theta )$. Therefore, we have $x_0+t \Theta (\theta ) \in K \cup \partial K$. Then, by the convexity of K and Lemma A.1, we have $x_0+T(\Theta (\theta )) \Theta (\theta ) \in K$, yielding a contradiction with $x_0+T(\Theta (\theta )) \Theta (\theta ) \in \partial K$.

Case II: $T(\Theta (\theta )) > \widetilde{T}$. Fix t such that $T(\Theta (\theta ))> t > \widetilde{T}$. The sequence

$$\begin{aligned} \Big \{ x_0+\frac{t}{\widetilde{T}} T(\Theta (\theta _n)) \Theta (\theta _n) :n \in \widetilde{M} \Big \} \subseteq {\mathbb {R}}^2\setminus ( K \cup \partial K), \end{aligned}$$

converges to $x_0+t \Theta (\theta )$. Therefore, we have $x_0+t \Theta (\theta ) \in {\mathbb {R}}^2 {\setminus } K$. However, by Lemma A.1, we have $x_0+t \Theta (\theta ) \in K$, yielding a contradiction. $\square$

We continue with the proof that every compact convex curve $\Gamma = \partial K$ is rectifiable. We first recall the definition of rectifiability.

Definition A.6

Let $\gamma :I \rightarrow \Gamma \subseteq {\mathbb {R}}^2$ be a continuous parametrization of a curve, where $I \subseteq {\mathbb {R}}$ is a bounded interval of either of the following forms

$$\begin{aligned} I = [a,b], \qquad I = [a,b), \qquad I = (a,b], \qquad I = (a,b). \end{aligned}$$

Let $P=\{P_0,\dots ,P_k\}$ be a finite and strictly increasing collection of points in I, namely $P_0< P_1< \dots < P_k$. Let $\sigma _{\gamma (P)}$ be the polygonal curve given by the segments between $\gamma (P_{i})$ and $\gamma (P_{i+1})$. Let $\ell (\sigma _{\gamma (P)})$ be the length of $\sigma _{\gamma (P)}$ defined by

$$\begin{aligned} \ell (\sigma _{\gamma (P)}) := \sum _{i=0}^{k-1} \left|{\gamma (P_{i+1})- \gamma (P_i)}\right|. \end{aligned}$$

Let ${\mathcal {P}}$ be the set of all possible finite and strictly increasing collections of points in I. The curve $\gamma (I)$ is rectifiable if

$$\begin{aligned} \ell (\gamma (I)) := \sup \Big \{ \ell (\sigma _{\gamma (P)}) :P \in {\mathcal {P}} \Big \} < \infty , \end{aligned}$$

and we call $\ell (\gamma (I))$ the length of $\gamma (I)$.

Remark A.7

If $I = [a,b]$, without loss of generality we consider only finite and strictly increasing collections $\{P_0, \dots , P_k\}$ of points in I such that $P_0 = a$, $P_k = b$.

Now, for every parametrization $\gamma :[0, 2 \pi ) \rightarrow \Gamma = \partial K$ we define the parametrization $\widetilde{\gamma } :[0, 2 \pi ] \rightarrow \Gamma = \partial K$ by

$$\begin{aligned} \forall t \in [0, 2 \pi ), \widetilde{\gamma }(t) := \gamma (t), \qquad \widetilde{\gamma }(2 \pi ) := \gamma (0). \end{aligned}$$

In particular, for $\widetilde{\gamma }$ we can apply the observation made in Remark A.7. Moreover, it is straight-forward to observe that $\ell (\gamma ([0,2\pi ))) = \ell (\widetilde{\gamma }([0,2\pi ]))$. Therefore, with a slight abuse of notation, we denote by $\gamma$ also $\widetilde{\gamma }$.

Moreover, we introduce the auxiliary definition of convex hull we use in the remaining part of the Appendix.

Definition A.8

Let $Q = \{ Q_1, \dots , Q_k \}$ be a finite collection of points in ${\mathbb {R}}^2$. The open convex hull ${{\,\textrm{ch}\,}}(Q)$ is defined by

$$\begin{aligned} {{\,\textrm{ch}\,}}(Q) := \Big \{ \sum _{i=1}^k \alpha _i Q_i :( \alpha _1, \dots , \alpha _k ) \in (0,1)^k, \sum _{i=1}^k \alpha _i = 1 \Big \}. \end{aligned}$$

Next, we state and prove three auxiliary lemmata.

Lemma A.9

Let $x,y \in \Gamma = \partial K$, $x \ne y$. Let $\gamma =\gamma (x) :[0,2 \pi ] \rightarrow \Gamma = \partial K$ be the counterclockwise parametrization such that $\gamma (0)=x$. Let $s \in (0,2 \pi )$ be such that $\gamma (s)=y$. Then the two pieces $\gamma ((0,s))$ and $\gamma ((s,2\pi ))$ of the curve $\Gamma$ are in the closure of the distinct half-planes defined by the line l passing through x and y.

Fig. 3

The open subsets A, B, C in Case II

Proof

Let $x_0 \in K$. Let $l_x$ be the half-line emanating from $x_0$ and passing through x, and $l_y$ the half-line emanating from $x_0$ and passing through y. We distinguish three cases.

Case I: $s=\pi$. Then $l_x,l_y \subseteq l$, and the statement is satisfied.

Case II: $s<\pi$. In particular, $x_0 \notin l$. Let $H_0$ be the open half-plane such that $\partial H_0 = l$ and $x_0 \in H_0$. The piece $\gamma ((0,s))$ of the curve $\Gamma$ is in the section of the plane defined by the counterclockwise angle from $l_x$ to $l_y$. We claim that $\gamma ((0,s)) \subseteq H_0^c$. We argue by contradiction and we suppose that there exists $0<u<s$ such that $\gamma (u)$ belongs to the open subset $C = {{\,\textrm{ch}\,}}( x,y,x_0 ) \subseteq K$, see Fig. 3. Then $\gamma (u) \in K$, yielding a contradiction with $\gamma (u) \in \partial K$.

Let $\Pi$ be the open section of the plane defined by the counterclockwise angle from $l_y$ to $l_x$. Let A and B be the connected open subsets of the plane such that $A \cap B = \varnothing$, $A \cup B = \Pi \cap ( H_0 \cup \partial H_0)^c$, $x \in \partial A$ and $y \in \partial B$, see Fig. 3. The piece $\gamma ((s,2\pi ))$ of the curve $\Gamma$ is in the set $\Pi$. We claim that $\gamma ((s,2\pi )) \subseteq H_0 \cup \partial H_0$. We argue by contradiction and we suppose that there exists $s<u<2 \pi$ such that $\gamma (u)$ belongs to either of the subsets A and B. Without loss of generality, we assume $\gamma (u) \in A$. Then $x \in {{\,\textrm{ch}\,}}( \gamma (u),y,x_0 ) \subseteq K$, yielding a contradiction with $x \in \partial K$.

Case III: $s >\pi$. We proceed as in Case II, switching the arguments for the two subcases. $\square$

Lemma A.10

Let $\gamma :[0,2\pi ] \rightarrow \Gamma \subseteq {\mathbb {R}}^2$ be a parametrization of a compact convex curve $\Gamma$. Let $P = \{ P_0, \dots , P_k \}$ be a finite and strictly increasing collection of points in $[0,2\pi ]$ such that $P_0 = 0$, $P_k = 2 \pi$. Then the open convex hull ${{\,\textrm{ch}\,}}(\gamma (P))$ is an open convex polygon, and $\partial {{\,\textrm{ch}\,}}(\gamma (P)) = \sigma _{\gamma (P)}$.

Proof

Consider the segment between $\gamma (P_j)$ and $\gamma (P_{j+1})$. By Lemma A.9, all the points in $\gamma (P)$ are in the same closed half-plane defined by the line passing through $\gamma (P_j)$ and $\gamma (P_{j+1})$. Therefore, the open convex hull ${{\,\textrm{ch}\,}}(\gamma (P) )$ is in the same closed half-plane, and the segment between $\gamma (P_j)$ and $\gamma (P_{j+1})$ belongs to the boundary $\partial {{\,\textrm{ch}\,}}( \gamma (P) )$.

Lemma A.11

Let A, B be two convex polygons such that $A \subseteq B$. Then

$$\begin{aligned} \ell (\partial A) \le \ell (\partial B). \end{aligned}$$

Proof

We prove the claim by induction on the number n of sides of $\partial A$ that are not contained in $\partial B$. If $n=0$, then $A=B$ and the desired inequality is satisfied.

Next, suppose that there are $n \ge 1$ sides of $\partial A$ that are not contained in $\partial B$. We choose one, we draw the line l defined by it, and we let H be the closed half-plane defined by l containing A, see Fig. 4. Then $C= B \cap H$ is a convex polygon and, by triangle inequality, we have

$$\begin{aligned} \ell (\partial C) \le \ell (\partial B). \end{aligned}$$

We observe that there are $n-1$ sides of $\partial A$ that are not contained in $\partial C$. Therefore, by induction hypothesis, we obtain the desired inequality. $\square$

Fig. 4

The inductive step

Proof of Theorem 2.4

Let B(0, R) be a ball centred at the origin with radius R containing K. Let $\Delta$ be an equilateral triangle containing B(0, R).

By Lemma A.10, for every finite and strictly increasing collection $P = \{P_0, \dots , P_k\}$ of points in $[0,2\pi ]$ such that $P_0 = 0$, $P_k = 2 \pi$ the open convex hull ${{\,\textrm{ch}\,}}( \gamma (P) )$ is an open convex polygon contained in $\Delta$. Moreover, we have $\sigma _{\gamma (P)}= \partial {{\,\textrm{ch}\,}}( \gamma (P) )$.

By Lemma A.11, we have

$$\begin{aligned} \ell (\gamma (I)) := \sup \Big \{ \ell (\sigma _{\gamma (P)}) \le \ell (\partial \Delta ) :P \in {\mathcal {P}} \Big \} < \infty . \end{aligned}$$

$\square$

Remark A.12

Let $x_0 \in K$, $x_1 \in \Gamma = \partial K$. Let $\gamma = \gamma (x_0, x_1) :[0,2\pi ) \rightarrow \Gamma$ be the counterclockwise parametrization defined in Lemma A.5. Let $z = z(x_1) :[0,\ell (\Gamma )) \rightarrow \Gamma$ be the counterclockwise affine arclength parametrization defined by

$$\begin{aligned} z(0) = x_1. \end{aligned}$$

The function $\gamma ^{-1}\circ z$ is strictly increasing, because both $\gamma$ and z are counterclockwise parametrizations.

1.2 A.2. Proofs of Theorem 2.5, Theorem 2.6, and Theorem 2.7

We introduce two auxiliary functions $\theta _l$ and $\theta _r$ defined geometrically in every point of the convex curve $\Gamma =\partial K$ by the minimal cone centred at the point and containing the convex set K. These functions are strictly related to the left and right derivatives of the arclength parametrization z of $\Gamma$, and are helpful in proving the desired theorems.

Definition A.13

Let x be a point in $\Gamma = \partial K$. The cone $E_x$ is defined by

$$\begin{aligned} E_x := \Big \{ e \in {\mathbb {S}}^1 :\{ x+te :t > 0 \} \cap \partial K \ne \varnothing \Big \}. \end{aligned}$$

Fig. 5

Two instances of $E_x$

See Fig. 5.

Lemma A.14

For every $x \in \Gamma = \partial K$ we have ${\mathbb {S}}^1 {\setminus } E_x \ne \varnothing$.

Proof

We argue by contradiction and we suppose that $E_x= {\mathbb {S}}^1$. We fix any arbitrary counterclockwise parametrization $\Psi :[0, 2 \pi ) \rightarrow {\mathbb {S}}^1$ as in Definition A.4. Let $y_1,y_2,y_3 \in \partial K$ be the points corresponding to the directions $e_1 = \Psi (\pi /3)$, $e_2 = \Psi (\pi )$, and $e_3 = \Psi (5 \pi /3)$. Therefore, we have $x \in {{\,\textrm{ch}\,}}(y_1,y_2,y_3) \subseteq K$, yielding a contradiction with $x \in \partial K$. $\square$

The previous result guarantees that the following definition is meaningful. For every $x \in \Gamma = \partial K$ let $e_0 = e_0(x) \in {\mathbb {S}}^1 {\setminus } E_x$. Moreover, let $\Phi = \Phi (e_0) :[0, 2 \pi ) \rightarrow {\mathbb {S}}^1$ be the counterclockwise parametrization of the circle with starting point $e_0$ as in Definition A.4.

Lemma A.15

For every $x \in \Gamma = \partial K$ we have that $\Phi ^{-1}(E_x)$ is an interval with extremal points $a,b \in [0,2\pi )$ satisfying

$$\begin{aligned} a < b \le a +\pi . \end{aligned}$$

(A.1)

Proof

Let $\theta _1, \theta _2 \in \Phi ^{-1} (E_x)$ such that $\theta _1 < \theta _2$. We claim that for every $\theta \in [0, 2 \pi )$, $\theta _1< \theta < \theta _2$ we have $e = \Phi (\theta ) \in E_x$.

By the definition of $\Phi$, we have $\theta _1 \ne 0$ and $\theta _2 \ne 2 \pi$. Now, let $e_1, e_2 \in {\mathbb {S}}^1$ be defined by

$$\begin{aligned} e_1= \Phi (\theta _1), \qquad e_2= \Phi (\theta _2), \end{aligned}$$

and let $y_1,y_2 \in \partial K$ be defined by

$$\begin{aligned} y_1= \{ x+te_1 :t> 0 \} \cap \partial K, \qquad y_2= \{ x+te_2 :t > 0 \} \cap \partial K. \end{aligned}$$

We distinguish three cases.

Case I: $\theta _2>\theta _1+\pi$. We have

$$\begin{aligned} {{\,\textrm{ch}\,}}(y_1,y_2) \cap \{ x+te_0 :t > 0 \} \ne \varnothing , \end{aligned}$$

yielding a contradiction with $e_0 \notin E_x$.

Case II: $\theta _2<\theta _1+\pi$. We have

$$\begin{aligned} {{\,\textrm{ch}\,}}(y_1,y_2) \cap \{ x+te :t > 0 \} \ne \varnothing , \end{aligned}$$

hence $\theta \in \Phi ^{-1}(E)$.

Case III: $\theta _2=\theta _1+\pi$. By Case I, we have

$$\begin{aligned} \theta _1 = \inf \Big \{ \theta \in \Phi ^{-1}(E_x) \Big \}, \qquad \theta _2 = \sup \Big \{ \theta \in \Phi ^{-1}(E_x) \Big \}. \end{aligned}$$

Let $y \in K$. It belongs to one of the two half-planes defined by the line through $y_1,x,y_2$. Therefore, we have

$$\begin{aligned} \widetilde{\theta } := \Phi ^{-1} \Big ( \frac{y-x}{\left|{y-x}\right|} \Big ) \in E_x, \qquad \theta _1< \widetilde{\theta } < \theta _2 = \theta _1 + \pi , \end{aligned}$$

and we reduce to Case II for the couples $(\theta _1,\widetilde{\theta })$ and $(\widetilde{\theta },\theta _2)$.

Therefore, $\Phi ^{-1}(E_x)$ is an interval with extremal points $a,b \in [0, 2 \pi )$. By Case I, we obtain the desired relation between a, b described in (A.1). $\square$

In particular, $\Gamma = \partial K$ is contained in the closed section of the plane defined by the half-lines $\{ x+t \Phi (a) :t \ge 0 \}$ and $\{ x+t \Phi (b) :t \ge 0 \}$. Now, for every $x \in \Gamma = \partial K$ let $E_x$ be the cone as in Definition A.13 and let $e_0(x) \in {\mathbb {S}}^1 {\setminus } E_x$. Next, let $\Phi _{x} :[0, 2\pi ) \rightarrow {\mathbb {S}}^1$ be the counterclockwise parametrization of the circle with starting point in $e_0(x)$ as in Definition A.4. After that, let $x_1 \in \Gamma = \partial K$ and let the arclength parametrization $z = z(x_1) :J \rightarrow \Gamma$ be defined as in Remark A.12. Then, we choose the counterclockwise parametrization of the circle $\Upsilon = \Upsilon (x_1) :[0,2\pi ) \rightarrow {\mathbb {S}}^1$ with starting point

$$\begin{aligned} \Phi _{x_1} \Big ( \inf \Big \{ \theta \in [0,2\pi ) :\Phi _{x_1}(\theta ) \in E_{x_1} \Big \} \Big ). \end{aligned}$$

as in Definition A.4. Finally, we define the functions $\theta _l :(0,\ell (\Gamma )] \rightarrow [0, 2 \pi )$ and $\theta _r :[0,\ell (\Gamma )) \rightarrow [0,2 \pi )$ by

$$\begin{aligned} \theta _l (t)&:= \Upsilon ^{-1} \Big ( - \Phi _{z(t)} \Big ( \sup \Big \{ \theta :\theta \in \Phi _{z(t)}^{-1}(E_{z(t)}) \Big \} \Big ) \Big ), \\ \theta _r (t)&:= \Upsilon ^{-1} \Big ( \Phi _{z(t)} \Big ( \inf \Big \{ \theta :\theta \in \Phi _{z(t)}^{-1} (E_{z(t)}) \Big \} \Big ) \Big ). \end{aligned}$$

Lemma A.16

For all $s,t \in (0,\ell (\Gamma ))$, $s < t$ we have

$$\begin{aligned} \theta _r(s) \le \theta _l(t) \le \theta _r(t). \end{aligned}$$

(A.2)

Moreover, for every $s \in (0,\ell (\Gamma ))$ we have

$$\begin{aligned} \theta _r(0) \le \theta _l(s) \le \theta _r(s) \le \theta _l(\ell (\Gamma )). \end{aligned}$$

(A.3)

Proof

The first inequality in (A.2) follows from

$$\begin{aligned} \theta _r(s) \le \Upsilon ^{-1} \Big ( \frac{z(t) -z(s)}{\left|{z(t)-z(s)}\right|} \Big ) = \Upsilon ^{-1} \Big ( - \frac{z(s)-z(t)}{\left|{z(s)-z(t)}\right|} \Big ) \le \theta _l(t). \end{aligned}$$

(A.4)

The second inequality in (A.2) follows from Lemma A.15 and the definition of a counterclockwise parametrization of ${\mathbb {S}}^1$ in Definition A.4. The first and the third inequalities in (A.3) follow from the chain of inequalities in (A.4). $\square$

Lemma A.17

The functions $\theta _l$ and $\theta _r$ are increasing and have bounded variation. Moreover, they coincide m-almost everywhere.

Proof

By Lemma A.16, the functions $\theta _l$ and $\theta _r$ are increasing. Moreover, they take values in a bounded set, hence they have bounded variation.

Now, suppose that the functions $\theta _l$ and $\theta _r$ do not coincide m-almost everywhere. Therefore, there exists an uncountable collection $X \subseteq (0,\ell (\Gamma ))$ of points such that for every $x \in X$ we have

$$\begin{aligned} \lim _{t \rightarrow x^{-}} \theta _r(t) \le \theta _l(x) < \theta _r(x) \le \lim _{t \rightarrow x^{+}} \theta _r(t). \end{aligned}$$

Hence, we have

$$\begin{aligned} \lim _{t \rightarrow \ell (\Gamma )^{-}} \theta _r(t) \ge \sum _{x \in X} \Big ( \lim _{t \rightarrow x^{+}} \theta _r(t) - \lim _{t \rightarrow x^{-}} \theta _r(t) \Big ) = \infty , \end{aligned}$$

yielding a contradiction with $\theta _r([0,\ell (\Gamma )) \subseteq [0,2\pi )$. $\square$

Lemma A.18

Fix $s \in J$ and consider the function $\phi = \phi _s$ defined by

$$\begin{aligned} \phi :J \setminus \{s\} \rightarrow [0,2 \pi ), \qquad \phi (t) := {\left\{ \begin{array}{ll} \displaystyle \Upsilon ^{-1} \Big ( \frac{z(s)-z(t)}{\left|{z(s)-z(t)}\right|}\Big ), \qquad &{} \text {if}\ t < s, \\ \displaystyle \Upsilon ^{-1} \Big ( \frac{z(t)-z(s)}{\left|{z(t)-z(s)}\right|}\Big ), \qquad &{} \text {if}\ t > s. \end{array}\right. } \end{aligned}$$

Then, the function $\phi$ is increasing.

Proof

For all $t, u \in J {\setminus } \{ s\}$, $t < u$ we claim that

$$\begin{aligned} \phi (t) \le \phi (u). \end{aligned}$$

(A.5)

Let $x_0 \in K$, $x_1 \in \Gamma = \partial K$, and let $\gamma = \gamma (x_0,x_1)$ and $z = z(x_1)$ be the associated parametrizations as in Remark A.12. Moreover, we consider the points z(s), z(t), and z(u). By Remark A.12, we have

$$\begin{aligned} \gamma ^{-1} (z(t)) < \gamma ^{-1}(z(u)). \end{aligned}$$

We distinguish three cases according to the relation between s, t, and u.

Case I: $s< t < u$. We distinguish five additional subcases.

Case I.i. We assume

$$\begin{aligned} \gamma ^{-1}(z(t))< \gamma ^{-1}(z(s))+\pi < \gamma ^{-1}(z(u)). \end{aligned}$$

By Lemma A.15, the points z(t) and z(u) belong to distinct open half-planes defined by the line passing through z(s) and $\gamma (\gamma ^{-1}(z(s))+\pi )$. Moreover, let $e_0 \in {\mathbb {S}}^1$ be defined by

$$\begin{aligned} e_0 = \frac{z(s) - x_0}{\left|{z(s) - x_0}\right|}. \end{aligned}$$

In particular, we have

$$\begin{aligned} - e_0 = \frac{\gamma (\gamma ^{-1}(z(s))+\pi ) - x_0}{\left|{\gamma (\gamma ^{-1}(z(s))+\pi ) - x_0}\right|} \in E_{z(s)}. \end{aligned}$$

By Lemma A.15, we have that $-e_0$ belongs to the interior of $E_{z(s)}$, hence we have $e_0 \in {\mathbb {S}}^1 {\setminus } E_{z(s)}$. Let $\Phi _{z(s)} :[0,2\pi ) \rightarrow {\mathbb {S}}^1$ be the counterclockwise parametrization of the circle with starting point in $e_0$ as in Definition A.4. To prove the desired inequality in (A.5), it is enough to prove the inequality

$$\begin{aligned} \Phi _{z(s)}^{-1} \Big ( \frac{z(t)-z(s)}{\left|{z(t) - z(s)}\right|} \Big ) \le \Phi _{z(s)}^{-1} \Big ( \frac{z(u)-z(s)}{\left|{z(u) - z(s)}\right|}\Big ). \end{aligned}$$

(A.6)

To prove the desired inequality in (A.6), we argue by contradiction and we suppose that

$$\begin{aligned} \Phi _{z(s)}^{-1} \Big ( \frac{z(u)-z(s)}{\left|{z(u) - z(s)}\right|} \Big ) < \Phi _{z(s)}^{-1} \Big ( \frac{z(t)-z(s)}{\left|{z(t) - z(s)}\right|} \Big ) . \end{aligned}$$

Therefore, the points z(t) and z(u) belong to the same open half-plane defined by the line passing through z(s) and $\gamma (\gamma ^{-1}(z(s))+\pi )$, yielding a contradiction.

Case I.ii. We assume

$$\begin{aligned} \gamma ^{-1}(z(t))< \gamma ^{-1}(z(u)) < \gamma ^{-1}(z(s))+\pi . \end{aligned}$$

Let $l_s$ and $l_u$ be the half-lines emanating from $x_0$ and passing through z(s) and z(u) respectively. Since the parametrization z is counterclockwise, the point z(t) belongs to the open section of the plane defined by the angle strictly smaller than $\pi$ between $l_s$ and $l_u$. To prove the desired inequality in (A.5), we argue by contradiction and we suppose that $\phi (u) < \phi (t)$. Let $l'_u$ and $l'_s$ be the half-lines emanating from z(s) and passing through z(u) and $x_0$ respectively. Since $\phi (u) < \phi (t)$, the point z(t) belongs to the open section of the plane defined by the angle strictly smaller than $\pi$ between $l'_u$ and $l'_s$. Therefore, we obtain

$$\begin{aligned} z(t) \in {{\,\textrm{ch}\,}}( z(s),z(u),x_0 ) \subseteq K, \end{aligned}$$

yielding a contradiction with $z(t) \in \Gamma = \partial K$.

Case I.iii. We assume

$$\begin{aligned} \gamma ^{-1}(z(s))+\pi< \gamma ^{-1}(z(t)) < \gamma ^{-1}(z(u)) . \end{aligned}$$

We argue by contradiction and we suppose that $\phi (u) < \phi (t)$. Analogously to Case I.ii, we obtain

$$\begin{aligned} z(u) \in {{\,\textrm{ch}\,}}( z(s),z(t),x_0 ) \subseteq K, \end{aligned}$$

yielding a contradiction with $z(u) \in \Gamma = \partial K$.

Case I.iv. We assume

$$\begin{aligned} \gamma ^{-1}(z(t)) < \gamma ^{-1}(z(s))+\pi = \gamma ^{-1}(z(u)). \end{aligned}$$

The desired inequality in (A.5) follows from the fact that the parametrizations z, $\gamma$, and $\Upsilon$ are counterclockwise.

Case I.v. We assume

$$\begin{aligned} \gamma ^{-1}(z(t)) = \gamma ^{-1}(z(s))+\pi < \gamma ^{-1}(z(u)). \end{aligned}$$

We prove the desired inequality in (A.5) analogously to Case I.iv.

Case II: $t< u < s$. We distinguish five additional subcases and we prove the desired inequality in (A.5) analogously to Case I.

Case III: $t< s < u$. We prove the desired inequality in (A.5) by Case I applied to $\phi _t$ and Case II applied to $\phi _u$. $\square$

Lemma A.19

The function $\theta _r$ is right-continuous, and the function $\theta _l$ is left-continuous.

Proof

We focus on the case of the function $\theta _r$. The case of the function $\theta _l$ is analogous.

We want to prove that for every $s \in J$ we have

$$\begin{aligned} \theta _r(s) = \lim _{t \rightarrow s^{+}} \theta _r (t). \end{aligned}$$

We fix $s \in J$. By Lemma A.17, the limit is an infimum and it is enough to prove that for every $\varepsilon > 0$ there exists $t > s$ such that

$$\begin{aligned} \theta _r(t) \le \theta _r (s) + 2 \varepsilon . \end{aligned}$$

By the definition of $\theta _r$, there exists $u \in J$, $u > s$ such that

$$\begin{aligned} \theta _r(s) \le \Upsilon ^{-1}\Big (\frac{z(u)-z(s)}{\left|{z(u)-z(s)}\right|}\Big ) \le \theta _r(s)+ \varepsilon . \end{aligned}$$

(A.7)

By Lemma A.9, the piece z((s, u)) of the curve $\Gamma$ is in the closure of the half-plane defined by the line passing through z(s) and z(u). In particular, by the definition of $\theta _r$ and $\theta _l$, and Lemma A.18, for every $t \in J$, $s< t < u$ we have

$$\begin{aligned} \theta _r(s) \le \Upsilon ^{-1} \Big ( \frac{z(t) - z(s)}{\left|{z(t) - z(s)}\right|} \Big )\le \Upsilon ^{-1} \Big ( \frac{z(u) - z(t)}{\left|{z(u) - z(t)}\right|} \Big ) \le \theta _l (u). \end{aligned}$$

We distinguish two cases.

Case I. We assume

$$\begin{aligned} \Upsilon ^{-1}\Big (\frac{z(u)-z(s)}{\left|{z(u)-z(s)}\right|}\Big ) = \theta _r(s). \end{aligned}$$

Then, we have $\theta _l(u) = \theta _r(s)$. By Lemma A.16, for every $t \in J$, $s< t < u$ we have

$$\begin{aligned} \theta _r(t) = \theta _r(s). \end{aligned}$$

Case II. We assume

$$\begin{aligned} \Upsilon ^{-1}\Big (\frac{z(u)-z(s)}{\left|{z(u)-z(s)}\right|}\Big ) >\theta _r(s). \end{aligned}$$

Then, we have $\theta _l(u) > \theta _r(s)$. By Lemma A.15, there exists $t \in J$, $s< t < u$ such that

$$\begin{aligned} 0&\le \Upsilon ^{-1}\Big (\frac{z(u)-z(t)}{\left|{z(u)-z(t)}\right|}\Big ) -\Upsilon ^{-1}\Big (\frac{z(u)-z(s)}{\left|{z(u)-z(s)}\right|}\Big ) \nonumber \\&= \Phi _{z(u)}^{-1}\Big (\frac{z(t)-z(u)}{\left|{z(t)-z(u)}\right|}\Big ) - \Phi _{z(u)}^{-1}\Big (\frac{z(s)-z(u)}{\left|{z(s)-z(u)}\right|}\Big ) \le \varepsilon . \end{aligned}$$

(A.8)

Together with the definition of $\theta _r$, the inequalities in (A.7) and (A.8) yield

$$\begin{aligned} \theta _r (t) \le \Upsilon ^{-1}\Big (\frac{z(u)-z(t)}{\left|{z(u)-z(t)}\right|}\Big ) \le \theta _r(s) + 2\varepsilon . \end{aligned}$$

$\square$

Fig. 6

The obtuse triangle ${{\,\textrm{ch}\,}}(z(s),z(t),y(s,t))$ shaded in blue and the associated right-triangle in black

We turn now to the derivatives $z_l'$ and $z_r'$ and their relation with the functions $\theta _l$ and $\theta _r$.

Proof of Theorem 2.5

In the proof that the derivatives are well-defined, we focus on the case of the right derivative $z'_r$. The case of the left derivative $z'_l$ is analogous.

We want to prove that for every $s \in J$ the limit

$$\begin{aligned} z'_r(s) := \lim _{t \rightarrow s^{+}} \frac{z(t)-z(s)}{t-s} , \end{aligned}$$

is well-defined in ${\mathbb {S}}^1$.

We fix $s \in J$, we choose $\varepsilon >0$ such that $s + \varepsilon \in J$. First, we consider the function $\psi = \psi (s)$ defined by

$$\begin{aligned} \psi :[s, s+ \varepsilon ) \rightarrow [0,2 \pi ), \qquad \psi (t) := \Upsilon ^{-1} \Big ( \frac{z(t)-z(s)}{\left|{z(t)-z(s)}\right|} \Big ) -\theta _r(s). \end{aligned}$$

By the definition of $\theta _r$ and Lemma A.18, the following limit exists and we have

$$\begin{aligned} \lim _{t \rightarrow s^{+}} \Upsilon ^{-1} \Big ( \frac{z(t)-z(s)}{\left|{z(t)-z(s)}\right|} \Big ) \ge \theta _r(s). \end{aligned}$$

Moreover, by the definition of $\theta _r$, for every $\delta >0$ there exists $t \in J$, $t > s$ such that

$$\begin{aligned} \Upsilon ^{-1}\Big (\frac{z(t)-z(s)}{\left|{z(t)-z(s)}\right|}\Big ) \le \theta _r(s)+\delta . \end{aligned}$$

Therefore, by Lemma A.18 we have

$$\begin{aligned} \lim _{t \rightarrow s^{+}} \Upsilon ^{-1}\Big (\frac{z(t)-z(s)}{\left|{z(t)-z(s)}\right|}\Big ) = \theta _r(s). \end{aligned}$$

(A.9)

To conclude that $z_r'$ is well-defined in ${\mathbb {S}}^1$, it is enough to prove that

$$\begin{aligned} \lim _{t \rightarrow s^{+}} \frac{\left|{z(t)-z(s)}\right|}{t-s} = 1. \end{aligned}$$

(A.10)

By Lemma A.19, for $\rho > 0$ small enough we have

$$\begin{aligned} \theta _l(s+\rho ) \le \theta _r(s+\rho ) < \theta _r(s)+\frac{\pi }{2}. \end{aligned}$$

(A.11)

For every $t \in (s,s+\rho )$ let y(s, t) be the intersection between the half-line emanating from z(s) in the direction $\Upsilon ^{-1}(\theta _r(s))$ and the half-line emanating from z(t) in the direction $- \Upsilon ^{-1}(\theta _l(t))$. By the inequalities in (A.11), the arc z([s, t]) of the curve $\Gamma$ is contained in the closure of the open convex hull ${{\,\textrm{ch}\,}}(z(s),z(t),y(s,t))$, which is an obtuse triangle. This obtuse triangle is contained in a right-triangle with the segment between z(s) and z(t) as hypotenuse and a cathetus on the half-line emanating from z(s) in the direction $\Upsilon ^{-1}(\theta _r(s))$, see Fig. 6. By an argument analogous to that used to prove Theorem 2.4, we have

$$\begin{aligned} t-s \le \left|{z(t) - y(s,t)}\right| + \left|{y(s,t) - z(s)}\right| \le (\sin \psi (t)+ \cos \psi (t) ) \left|{z(t)-z(s)}\right|, \end{aligned}$$

where $t-s$ is the length of the arc z([s, t]) of the curve $\Gamma$. Therefore, we have

$$\begin{aligned} \lim _{t \rightarrow s^{+}} \frac{\left|{z(t)-z(s)}\right|}{t-s} \ge \lim _{t \rightarrow s^{+}} \frac{1}{\sin \psi (t)+ \cos \psi (t)} = 1. \end{aligned}$$

Together with $\left|{z(t)-z(s)}\right| \le t-s$, the inequality in the previous display yields the desired equality in (A.10).

In particular, by the equality in (A.9), we proved

$$\begin{aligned} z'_l = \Upsilon \circ \theta _l, \qquad z'_r = \Upsilon \circ \theta _r, \end{aligned}$$

(A.12)

Therefore, by Lemma A.17, the functions $z'_l$ and $z'_r$ coincide m-almost everywhere. $\square$

Finally, we recall a result about the differentiability of a function of bounded variation.

Theorem A.20

(Stein and Shakarchi [27], Theorem 3.4) Let $a,b \in {\mathbb {R}}$. If F is of bounded variation on [a, b], then F is differentiable almost everywhere.

Proof of Theorem 2.6

By Lemma A.17, the functions $\theta _l$ and $\theta _r$ have bounded variation. By Theorem A.20, they admit derivatives $\theta _l'$ and $\theta _r'$ well-defined m-almost everywhere.

Moreover, by Lemma A.16, the function $\theta _r - \theta _l$ is positive everywhere. By Lemma A.17, it has bounded variation and it is zero m-almost everywhere. By Theorem A.20, it admits a derivative m-almost everywhere, hence the derivative is zero m-almost everywhere. Therefore, the functions $\theta '_l$ and $\theta '_r$ coincide m-almost everywhere.

As we concluded in (A.12), we have

$$\begin{aligned} z_l'(t) = \begin{pmatrix} \cos \theta _l(t) \\ \sin \theta _l(t) \end{pmatrix} , \qquad z_r'(t) = \begin{pmatrix} \cos \theta _r(t) \\ \sin \theta _r(t) \end{pmatrix} , \end{aligned}$$

(A.13)

hence the functions $z_l''$ and $z_r''$ are well-defined m-almost everywhere by

$$\begin{aligned} z_l''(t) = \begin{pmatrix} - \sin \theta _l(t) \\ \cos \theta _l(t) \end{pmatrix} \theta _l'(t), \qquad z_r''(t) = \begin{pmatrix} - \sin \theta _r(t) \\ \cos \theta _r(t) \end{pmatrix} \theta _r'(t). \end{aligned}$$

(A.14)

In particular, they coincide m-almost everywhere. $\square$

Proof of Theorem 2.7

By Lemma A.16, the Borel measure $\sigma$ on J defined in (2.1) is positive. Now, by the equalities in (A.12), for all $a,b \in J$, $a \le b$ we have

$$\begin{aligned}&\sigma ((a,b)) = \max \{ 0, \theta _l(b) - \theta _r(a) \}, \qquad \sigma ((a,b]) = \theta _r(b) - \theta _r(a), \\&\sigma ([a,b)) = \theta _l(b) - \theta _l(a), \qquad \sigma ([a,b]) = \theta _r(b) - \theta _l(a). \end{aligned}$$

The metric density associated with the absolutely continuous part of $\sigma$ with respect to the Lebesgue measure m on J is $\kappa$.

Next, we define the Borel measure $\sigma _r$ on J as follows. For all $a,b \in J$, $a \le b$ we define

$$\begin{aligned}&\sigma _r((a,b)) = \max \{ 0, \lim _{t \rightarrow b^{-}}\theta _r(t) - \theta _r(a) \}, \qquad \sigma _r((a,b]) = \theta _r(b) - \theta _r(a), \\&\sigma _r([a,b)) = \lim _{t \rightarrow b^{-}}\theta _r(t) - \theta _r(a), \qquad \sigma _r([a,b]) = \theta _r(b) - \theta _r(a). \end{aligned}$$

The metric density associated with the absolutely continuous part of $\sigma _r$ with respect to the Lebesgue measure m on J coincides m-almost everywhere with $\theta _r'$.

For every $b \in J$ we consider the sequence of sets $\{ (b-\varepsilon , b] :\varepsilon > 0 \}$ that shrinks to b nicely as in Definition 2.1. On each of these sets, the Borel measure $\sigma - \sigma _r$ is zero. By Theorem 2.2, the metric density associated with the absolutely continuous part of $\sigma - \sigma _r$ with respect to the Lebesgue measure m on J is zero m-almost everywhere. Therefore, the functions $\kappa$ and $\theta '_r$ coincide m-almost everywhere. Analogously we prove that the functions $\kappa$ and $\theta '_l$ coincide m-almost everywhere. By Theorem 2.6 and the equalities in (A.13) and (A.14), for m-almost every $t \in J$ we have

$$\begin{aligned} \theta _l'(t) = \det \begin{pmatrix} z_l'(t)&z_l''(t) \end{pmatrix}, \qquad \theta _r'(t) = \det \begin{pmatrix} z_r'(t)&z_r''(t) \end{pmatrix}, \end{aligned}$$

yielding the desired result. $\square$