Proof. If the inferior limit of the curvature exponents is infinite, there is nothing to prove. Hence, assume that \(\liminf_{\varepsilon\to 0}N_\mathrm{curv}^\varepsilon<\infty\). This means that, for \(\varepsilon>0\) sufficiently small, \((\mathsf{X},\mathsf d_\varepsilon,\mathfrak m_\varepsilon)\) satisfies the \(\mathsf{MCP}(0,N_\mathrm{curv}^\varepsilon)\), cf.Remark 4. Let now \(N_0:=\liminf_{\varepsilon\to 0}N_\mathrm{curv}^\varepsilon\) and consider a sequence \(\{\varepsilon_n\}_{n\in\mathbb{N}}\) such that \(N^{\varepsilon_n}_{\mathrm{curv}}\to N_0\). Then, by stability of the measure contraction property under the pointed measured Gromov–Hausdorff convergence, \((\mathsf{X},\mathsf d_0,\mathfrak m_0)\) satisfies \(\mathsf{MCP}(0,N_0)\). This concludes the proof. ◻

Proof. Firstly, suppose that a norm \({\|\!\cdot\!\|}\) is induced by a scalar product. Let \(\{(a,0),(b_1,b_2)\}\) be an orthonormal basis of the dual norm \({\|\!\cdot\!\|}_\ast\), which is also induced by a scalar product. Without loss of generality, we can assume that \(ab_2>0\). Define a curve \(t \mapsto(x(t),y(t))\) by \[x(t)=a\cos(ct)+b_1\sin(ct),~~~~y(t)=b_2\sin(ct),\] where \(c:=1/(a b_2)\). Then the generalized trigonometric function \((\cos_{\Omega^\circ}(t),\sin_{\Omega^\circ}(t))\) associated to \({\|\!\cdot\!\|}_\ast\) coincides with \((x(t),y(t))\). Indeed, clearly the image of \((x(t),y(t))\) is the unit sphere \(\{\left\lVert v\right\rVert_*=1\}\) and the identity \(x\dot{y}-y\dot{x}\equiv 1\) holds by the choice of \(c\). By the formula \(\cos_{\Omega^\circ}^{\prime\prime}(t)=-C_\circ^\prime(t) \cos_{\Omega^\circ}(t)\), we have that \(C_\circ^\prime\equiv c^2\).

Conversely, suppose that \(C_\circ^\prime\equiv C^2>0\). Recall that the trigonometric functions \((x(t),y(t))=(\cos_{\Omega^\circ}(t),\sin_{\Omega^\circ}(t))\) are recovered by the differential equation \[\label{eq:trig95ode} \ddot{x}(t)=-C^2\cdot x(t),\qquad\ddot{y}(t) = -C^2\cdot y(t),\tag{16}\] with initial conditions given by \((x(0),y(0))=(x_0,0)\) and \((\dot{x}(0),\dot{y}(0))=(\dot{x}_0,\dot{y}_0)\). Then, the solution to 16 is given by \[\label{eq:normaltrigonometry} x(t)=A\cos(Ct)+B_1\sin(Ct),\qquad y(t)=B_2\sin(Ct),\tag{17}\] where the constants \(A,B_1,B_2\in\mathbb{R}\) are determined by the initial values \(x_0\), \(\dot{x}_0\) and \(\dot{y}_0\), and we can assume \(C>0\) by the symmetries of the (classical) trigonometric functions. Furthermore, the Pythagorean identity 9 , Proposition 5 and 17 imply the following identity \[\begin{align} 1&\equiv \cos_\Omega(t_\circ) \cos_{\Omega^\circ}(t) + \sin_\Omega(t_\circ)\sin_{\Omega^\circ}(t)=\dot{y}(t)x(t)-\dot{x}(t)y(t)\\ &=B_2C\cos(Ct)\left[A\cos(Ct)+B_1\sin(Ct)\right]-\left[-AC\sin(Ct)+B_1C\cos(Ct)\right]B_2\sin(Ct)\\ &=AB_2C, \end{align}\] showing that \(C = 1/(A B_2)\) and that the curve 17 is an ellipse. This implies that the norm is induced by a scalar product whose orthonormal basis is \(\{(A,0),(B_1,B_2)\}\). ◻

Proof. Observe that 10 implies that the function \[[0,2\pi_{\Omega^\circ})\ni \varphi\mapsto C_\circ(\varphi) - \frac{\pi_\Omega}{\pi_{\Omega^\circ}} \varphi\] is \(\pi_{\Omega^\circ}\)-periodic and non-constant. Thus it is not concave on \([0,2\pi_{\Omega^\circ})\). Consequently, \(C_\circ\) is not concave and we can find \(\varphi_0\) and \(\omega_0\) such that \(D_{\omega_0}^2 C_\circ(\varphi_0)=:H>0\). To prove this, we argue by contradiction, assuming that \(D_{\omega_0}^2 C_\circ(\varphi_0)\leq 0\), for every choice of angles \(\varphi_0\), \(\omega_0\). But this implies that \(C_\circ\) is midpoint concave, and thus concave.

Now, assume by contradiction that there exists \(h>0\) such that, for every \(\delta\in (0,h]\) and every \(\varphi\in \rinterval{0}{2 \pi_{\Omega^\circ}}\), we have \(D^2_\delta C_\circ (\varphi) < H\). From the very definition of \(D_\omega^2C_\circ(\varphi)\), we have that \[D_{\omega_0}^2C_\circ(\varphi_0)=\frac{1}{4}D^2_{\omega_0/2}C_\circ(\varphi_0)+\frac{1}{2}D_{\omega_0/2}^{2}C_\circ(\varphi_0+\omega_0/2)+\frac{1}{4}D_{\omega_0/2}^2C_\circ(\varphi_0+\omega_0).\] Iterating this identity, for every \(n\) we obtain \[\begin{align} D_{\omega_0}^2C_\circ(\varphi_0) = \frac{1}{2^{2n}}{}& \sum_{k=0}^{2^n-1} (k+1) D^2_{\omega_0/2^n} C_\circ(\varphi_0 + k\omega_0/2^n) \\ &+ \frac{1}{2^{2n}}\sum_{k=2^n}^{2^{n+1}-2} (2^{n+1}-1-k) D^2_{\omega_0/2^n}C_\circ (\varphi_0 + k\omega_0/2^n). \end{align}\] Then, taking \(n\) sufficiently large such that \(\omega_0/2^n<h\), we deduce that \[D_{\omega_0}^2C_\circ(\varphi_0) < \frac{H}{2^{2n}} \Bigg[\sum_{k=0}^{2^n-1} (k+1) + \sum_{k=2^n}^{2^{n+1}-2} (2^{n+1}-1-k) \Bigg]= \frac{H}{2^{2n}} \Bigg[ 2^n + 2 \sum_{k=1}^{2^{n}-1} k \Bigg] = H,\] finding a contradiction and concluding the proof. ◻

Proof. Given any \(h>0\), we apply Lemma 2 finding \(\delta\in (0,h]\) and \(\overline{\varphi} \in \rinterval{0}{2 \pi_{\Omega^\circ}}\) for which \[D^2_\delta C_\circ (\overline{\varphi}) \geq H.\] We set \(K := D_\delta C_\circ(\overline{\varphi})\) and define \[\begin{align} \varphi:={}& \sup \left\{ \theta \in \interval{\overline{\varphi}}{\overline{\varphi} + \delta} : C_\circ(\theta) - C_\circ(\overline{\varphi}) \geq \left(K + \frac{H \delta}{2}\right) (\theta - \overline{\varphi}) \right\}, \\ \omega :={}& \inf \left\{ t \in \linterval{0}{\overline{\varphi} + 2 \delta - \varphi} : \Delta_tC_\circ(\varphi) \geq \left(K + \frac{H \delta}{2}\right) t \right\}. \end{align}\] Observe that, by definition of \(K\), \[C_\circ(\overline{\varphi} + \delta) - C_\circ(\overline{\varphi}) = K \delta < \left(K + \frac{H \delta}{2}\right)\delta,\] which ensures that \(\varphi\) is well defined and that \(\varphi< \overline{\varphi} + \delta\). By continuity of \(C_\circ\), it actually holds that \[C_\circ(\varphi) - C_\circ(\overline{\varphi}) = \left(K + \frac{H \delta}{2}\right) (\varphi- \overline{\varphi}).\] Moreover, taking into account 20 , we note that \[\begin{align} C_\circ(\varphi+ (\overline{\varphi} + 2 \delta - \varphi)) - C_\circ(\varphi) ={}& C_\circ(\overline{\varphi} + 2 \delta) - C_\circ(\overline{\varphi}) + C_\circ(\overline{\varphi}) - C_\circ(\varphi) \\ \geq{}& 2 \delta K + H \delta^2 + \left(K + \frac{H \delta}{2}\right) (\overline{\varphi} - \varphi) \\ ={}& \left(K + \frac{H \delta}{2}\right) (\overline{\varphi} + 2 \delta - \varphi). \end{align}\] Thus, the set that determines \(\omega\) is not empty, which implies that \(\omega\) is well-defined. Furthermore, the very definition of \(\varphi\) guarantees that \(\omega > \overline{\varphi}+ \delta - \varphi> 0\). Then, since \(C_\circ\) is continuous, 22 follows from the definitions of \(\varphi\) and \(\omega\). Moreover, we observe that \[\begin{align} C_\circ(\overline{\varphi}+ \delta) - C_\circ(\varphi)= C_\circ(\overline{\varphi}+ \delta)- &C_\circ(\overline{\varphi}) - \big[C_\circ(\varphi)- C_\circ( \overline{\varphi})\big] \\ &= K \delta - \left(K + \frac{H \delta}{2}\right) (\varphi- \overline{\varphi}) \leq K (\overline{\varphi}+ \delta - \varphi) \end{align}\] and therefore we deduce that \[K \geq \mathop{\mathrm{ess\,inf}}_{t \in \rinterval{\varphi}{\overline{\varphi}+ \delta}} C_\circ'(t) \geq A \geq a,\] proving 21 . ◻

Proof. Given any \(h>0\), we apply Lemma 3 finding \(\delta \in \linterval{0}{h}\), \(\overline{\varphi}\in \rinterval{0}{2 \pi_{\Omega^\circ}}\), \(\overline{\omega} \in \linterval{0}{2 \delta}\), and \(K \geq A := \mathop{\mathrm{ess\,inf}}_{t \in \rinterval{\overline{\varphi}}{\overline{\varphi}+ \overline{\omega}}} C_\circ'(t) \geq a\) for which \[\label{eq:ddddd} \Delta_t C_\circ(\overline{\varphi})\leq \left(K + \frac{H \delta}{2}\right) t,\quad\forall t \in \interval{0}{\overline{\omega}} \qquad \text{and}\qquad \Delta_{\overline{\omega}} C_\circ(\overline{\varphi})=\left(K + \frac{H \delta}{2}\right) \overline{\omega}.\tag{24}\] Let \(\varepsilon> 0\) sufficiently small and pick any \(\varphi\in D(S_{A + \varepsilon}(\overline{\varphi}, \overline{\omega}))\), where we denote \[S_{A + \varepsilon}(\overline{\varphi}, \overline{\omega}) := \left\{ \theta \in \rinterval{\overline{\varphi}}{\overline{\varphi} + \overline{\omega}} : C_\circ'(\theta) \leq A + \varepsilon\right\} \subseteq S_{A + \varepsilon}.\] Note that, by definition of \(A\), this set is not empty and in addition \(\varphi=\overline{\varphi}+\overline{t}\), for some \(\overline{t}\in [0,\bar\omega]\). Now, we introduce \[\begin{align} B :={}& \frac{C_\circ(\overline{\varphi} + \overline{\omega}) - C_\circ(\varphi)}{\overline{\varphi} + \overline{\omega} - \varphi} - A = \frac{C_\circ(\overline{\varphi} + \overline{\omega}) - C_\circ(\overline{\varphi}) + C_\circ(\overline{\varphi}) - C_\circ(\varphi)}{\overline{\varphi} + \overline{\omega} - \varphi} - A \\ ={}&\frac{\Delta_{\overline{\omega}}C_\circ(\overline{\varphi}) -\Delta_{\overline{t}} C_\circ(\overline{\varphi})}{\overline{\varphi} + \overline{\omega} - \varphi} -A \geq \frac{(K + H \delta/2) (\overline{\varphi} + \overline{\omega} - \varphi)}{\overline{\varphi} + \overline{\omega} - \varphi} - A \geq \frac{H \delta}{2} > 0, \end{align}\] where the first inequality follows from 24 . Now, we take \[\omega := \inf \{t \in \linterval{0}{\overline{\varphi} + \overline{\omega} - \varphi} : \Delta_tC_\circ(\varphi) = C_\circ(\varphi+ t) - C_\circ(\varphi) \geq (A + B) t \}.\] The definition of \(B\) ensures that \[C_\circ(\varphi+ (\overline{\varphi} + \overline{\omega} - \varphi)) - C_\circ(\varphi) = C_\circ(\overline{\varphi} + \overline{\omega}) - C_\circ(\varphi) = (A + B) (\overline{\varphi} + \overline{\omega} - \varphi)\] and thus that \(\omega\) is well defined. Moreover, since \(\varphi\in D(S_{A + \varepsilon}(\overline{\varphi}, \overline{\omega}))\subseteq S_{A+\varepsilon}\), then \(C_\circ'(\varphi)\leq A+\varepsilon\) and we can deduce that \(\omega > 0\). Now, the lower bound of 23 is a consequence of the definition of \(A\), while the upper bound follows from the definitions of \(\varphi\) and \(\omega\). The equality \(\Delta_\omega C_\circ(\varphi) = (A + B) \omega\) is justified by the continuity of \(C_\circ\). ◻

Proof. Integrating by parts a first time we obtain \[\begin{align} P(\varphi, \omega)&= \frac{1}{2}\int_{0}^{ \omega} (t - \omega)^2 C_\circ'(\varphi+ t) C_\circ(\varphi+ \omega) \mathop{}\!\mathrm{d}t - \frac{1}{2}\int_{0}^{ \omega} t^2 C_\circ'(\varphi+ t) C_\circ(\varphi) \mathop{}\!\mathrm{d}t \\ &\quad +\int_{0}^{\omega} \int_{0}^{ \omega} (t - s) C_\circ'(\varphi+t) C_\circ(\varphi+s) \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t. \end{align}\] With another integration by parts and after standard computations, we have that: \[\begin{align} P(\varphi, \omega)&= - \omega^2 C_\circ(\varphi+\omega) C_\circ(\varphi) + 2 \int_{0}^{ \omega} (\omega - t) C_\circ(\varphi+\omega) C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t \\ &\quad + 2 \int_{0}^{ \omega} t \, C_\circ(\varphi) C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t -\int_{0}^{\omega} \int_{0}^{ \omega} C_\circ(\varphi+t) C_\circ(\varphi+s) \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t. \end{align}\] We add and remove the quantity \(\omega^2C_\circ(\varphi)^2+ 2 \omega C_\circ(\varphi) \int_{0}^{ \omega} C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t\), obtaining: \[\begin{align} P(\varphi, \omega) ={}& - \omega^2 C_\circ(\varphi)^2 - \omega^2 C_\circ(\varphi) [C_\circ(\varphi+\omega)- C_\circ(\varphi)] + 2 \omega C_\circ(\varphi) \int_{0}^{ \omega} C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t \\ &\quad + 2 [C_\circ(\varphi+\omega)- C_\circ(\varphi)] \int_{0}^{ \omega} (\omega - t) C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t - \bigg(\int_{0}^{ \omega} C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t \bigg)^2\\ ={}&[C_\circ(\varphi+\omega)-C_\circ(\varphi)] \bigg( \int_{0}^{ \omega} (\omega - t) C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t - \omega^2 C_\circ(\varphi) \bigg) \\ &\quad -\bigg(\int_{0}^{ \omega} C_\circ(\varphi+t) \mathop{}\!\mathrm{d}t- \omega C_\circ(\varphi) \bigg)^2 \end{align}\] Once we have this expression, the thesis simply follows from the definition of \(\Delta_\omega C_\circ(\varphi)\). ◻

Proof. If \({\|\!\cdot\!\|}\) is induced by a scalar product, then there is an isometry between \((\mathbb{H}, \mathsf d)\) and the standard sub-Riemannian Heisenberg group, and thus \((\mathbb{H}, \mathsf d,\mathscr{L}^3)\) has \(N_{\text{curv}}= 5\) (see [9]).

If \({\|\!\cdot\!\|}\) is not induced by a scalar product and it is neither strongly convex nor \(C^1\), then, according to [15], it does not satisfy \(\mathsf{MCP}(K, N)\) for any \(K \in \mathbb{R}\) and any \(N \in \ointerval{1}{+\infty}\) and thus \(N_{\mathrm{curv}} = +\infty\). For the rest of the proof, we assume that \({\|\!\cdot\!\|}\) is \(C^1\) and strongly convex and it is not induced by a scalar product, i.e. \(C_\circ\) is not an affine map, by Proposition 12. We are going to prove that \(N_{\mathrm{curv}} > 5\).

We place ourselves within the conclusions of Lemma 4, using the same notations except for \(S_{A + \varepsilon}\) that we denote by \(S\) for simplicity. Recall the definition of \(a\) in 18 and consider \(r \in \interval{0}{1}\).

3.0.0.1 Step 1: Finding an upper bound on \(P(\varphi, r \omega)\).

Let \(\varphi\in D(S_{A+\varepsilon})\) the angle identified by Lemma 4. By writing \(\interval{\varphi}{\varphi+r \omega}\) as the disjoint union of \(\interval{\varphi}{\varphi+r \omega} \cap S\) (on which we have \(C_\circ'(\theta) \leq A + \varepsilon\)) and \(\interval{\varphi}{\varphi+r \omega} \setminus S\) (on which we use the trivial estimate \(C_\circ'(\theta) \leq \mathsf {Lip}(C_\circ)\)), using the definition of \(P\) in ?? , we obtain

\[\begin{align} P(\varphi, r\omega) \leq \frac{1}{2} \Big({}& (A + \varepsilon)^2 \int_{\interval{\varphi}{\varphi+r \omega}\cap S} \int_{\interval{\varphi}{\varphi+r \omega}\cap S} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \\ & + 2 (A + \varepsilon) \mathsf {Lip}(C_\circ) \int_{\interval{\varphi}{\varphi+r \omega}\setminus S} \int_{\interval{\varphi}{\varphi+r \omega}\cap S} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \\ & + \mathsf {Lip}(C_\circ)^2\int_{\interval{\varphi}{\varphi+r \omega}\setminus S} \int_{\interval{\varphi}{\varphi+r \omega}\setminus S} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \Big). \end{align}\] We then use the trivial inclusion \(\interval{\varphi}{\varphi+r \omega} \cap S \subseteq \interval{\varphi}{\varphi+r \omega}\) to get \[\begin{align} P(\varphi, r\omega) \leq \frac{1}{2} \Big({} & (A + \varepsilon)^2 \int_{\interval{\varphi}{\varphi+r \omega}} \int_{\interval{\varphi}{\varphi+r \omega}} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \\ & + 2 (A + \varepsilon) \mathsf {Lip}(C_\circ) \int_{\interval{\varphi}{\varphi+r \omega}\setminus S} \int_{\interval{\varphi}{\varphi+r \omega}} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \\ & + \mathsf {Lip}(C_\circ)^2 \int_{\interval{\varphi}{\varphi+r \omega}\setminus S} \int_{\interval{\varphi}{\varphi+r \omega}\setminus S} (t - s)^2 \mathop{}\!\mathrm{d}s \mathop{}\!\mathrm{d}t \Big) \\ \leq \frac{1}{2} \Big({}& \frac{1}{6} (A + \varepsilon)^2 r^4 \omega^4 + \frac{4}{3} (A + \varepsilon) \mathsf {Lip}(C_\circ) r^3 \omega^3 \mathscr{L}^1(\interval{\varphi}{\varphi+ r \omega}\setminus S) \\ &+ \mathsf {Lip}(C_\circ)^2 r^2 \omega^2 \mathscr{L}^1(\interval{\varphi}{\varphi+ r \omega}\setminus S)^2 \Big). \end{align}\] Since \(\varphi\) is a density point of \(S\), we know that \[\frac{\mathscr{L}^1(\interval{\varphi}{\varphi+ r \omega}\setminus S)}{r \omega} \to 0, \qquad\text{as } r \to 0^+.\] In particular, we deduce that for every \(r > 0\) small enough, it holds \[\label{eq:Pupperbound} P(\varphi, r \omega) \leq \frac{1}{12} (A + 2 \varepsilon)^2 r^4 \omega^4.\tag{26}\]

3.0.0.2 Step 2: Finding a lower bound on \(P(\varphi, \omega)\).

Considering the function \(f_\varphi(t) := \Delta_t C_\circ(\varphi) - A t\), from 23 , we deduce that \[\label{eq:6061Gt6061} 0 \leq f_\varphi(t) \leq B t \;\;\text{ for all } t \in \interval{0}{\omega}, \text{ and } \;\;f_\varphi(\omega) = B \omega.\tag{27}\] To deduce the lower bound on \(P\), we use the expression 25 obtained in Lemma 5. Then, thanks to our choice of angles, we have \[\begin{align} \label{eq:PwithG} \begin{aligned} P(\varphi, \omega) ={}& 2 \omega (A + B) \left(\int_0^\omega (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t + \frac{1}{6} A \omega^3 \right) - \left( \frac{1}{2} A \omega^2 + \int_0^\omega f_\varphi(t) \right)^2 \\ ={}& \frac{1}{12} A^2 \omega^4 + \frac{1}{3} A B \omega^4 + A \omega \left( 2 \int_0^{\omega} (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t - \omega \int_0^{\omega} f_\varphi(t) \mathop{}\!\mathrm{d}t \right) \\ & + \left( 2 B \omega \int_0^{\omega} (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t - \left(\int_0^{\omega} f_\varphi(t) \mathop{}\!\mathrm{d}t\right)^2\right). \end{aligned} \end{align}\tag{28}\] We now estimate the two quantities within the parentheses. Firstly, we observe that \[2 \int_0^{\omega} (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t - \omega \int_0^{\omega} f_\varphi(t) \mathop{}\!\mathrm{d}t = \int_0^{\omega} (\omega -2 t) f_\varphi(t) \mathop{}\!\mathrm{d}t \geq \int_{\omega/2}^{\omega} (\omega - 2 t) Bt \mathop{}\!\mathrm{d}t = -\frac{5}{24} B \omega^3.\] Secondly, we claim that the third line of 28 is non-negative. To prove this, we note that 27 implies the existence of a unique \(s \in \interval{0}{\omega}\) satisfying \(\int_{s}^{\omega} B t \mathop{}\!\mathrm{d}t = \int_{0}^\omega f_\varphi(t) \mathop{}\!\mathrm{d}t\). Then, we have that \[\int_0^\omega (f_{\varphi}(t) - \chi_{\interval{s}{\omega}} Bt) \mathop{}\!\mathrm{d}t = 0,\quad \text{and thus}\quad \int_0^s f_\varphi(t) \mathop{}\!\mathrm{d}t = \int_s^\omega (Bt - f_{\varphi(t)}) \mathop{}\!\mathrm{d}t.\] This yields that \[\int_0^s (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t \geq \int_0^s (\omega - s) f_\varphi(t) \mathop{}\!\mathrm{d}t = \int_s^\omega (\omega - s) (B t - f_\varphi(t)) \mathop{}\!\mathrm{d}t \geq \int_s^\omega (\omega - t) (B t - f_\varphi(t)) \mathop{}\!\mathrm{d}t,\] and therefore \(\int_0^{\omega} (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t \geq \int_s^{\omega} (\omega - t) B t \mathop{}\!\mathrm{d}t\). Hence, we can estimate the third line of 28 as follows: \[\begin{gather} 2 B \omega \int_0^{\omega} (\omega - t) f_\varphi(t) \mathop{}\!\mathrm{d}t - \left(\int_0^{\omega} f_\varphi(t) \mathop{}\!\mathrm{d}t\right)^2\\ \geq 2 B \omega \int_s^{\omega} (\omega - t) B t \mathop{}\!\mathrm{d}t - \left(\int_s^{\omega} B t \mathop{}\!\mathrm{d}t\right)^2 = \frac{1}{12} B^2 (\omega - s)^3 (3 s + w) \geq 0. \end{gather}\] Finally, we conclude that \[\label{eq:Plowerbound} P(\varphi, \omega) \geq \frac{1}{12} A^2 \omega^4 + \frac{1}{8} A B \omega^4.\tag{29}\]

3.0.0.3 Step 3: Proving \(N_{\mathrm{curv}} > 5\) when \(a > 0\).

Using the estimates 26 and 29 , keeping in mind ?? , we deduce that for \(r\) sufficiently small, we have \[\begin{align} \frac{\mathcal{J}_R(\varphi, r \omega)}{\mathcal{J}_R(\varphi, \omega)} \leq{}& \frac{\frac{1}{12} (A + 2 \varepsilon)^2 r^4 \omega^4 + M r^6 \omega^6}{\frac{1}{12} A^2 \omega^4 + \frac{1}{8} A B \omega^4 - M \omega^6} = r^4 \frac{\frac{1}{12} (A + 2 \varepsilon)^2 + M r^2 \omega^2}{\frac{1}{12} A^2 + \frac{1}{8} A B - M \omega^2}. \end{align}\] Moreover, we observe that \[\label{eq:explimit} \lim_{r\to 0} \frac{\frac{1}{12} (A + 2 \varepsilon)^2 + M r^2 \omega^2}{\frac{1}{12} A^2 + \frac{1}{8} A B - M \omega^2} = \frac{(A+2 \varepsilon)^2}{A^2 + \frac{3}{2} AB - 12 M \omega^2} \leq \frac{(A+2 \varepsilon)^2}{A^2 + \frac{3}{4} H A \delta - 48 M \delta^2},\tag{30}\] where the inequality follows from the facts that \(B \geq H \delta/2\) and \(\omega\leq 2 \delta\). We finally choose \(h > 0\) and \(\varepsilon> 0\) in Lemma 4 so that the expression in 30 is smaller than 1. For example, we can take \[h \leq \frac{a H}{192 M}\qquad \text{and}\qquad \varepsilon\leq \frac{1}{2} \left( \sqrt{A^2 + \frac{1}{4} H a \delta} - A \right).\] Note that, since \(a > 0\), we may pick \(h, \varepsilon> 0\). With these choices, recalling that \(A\geq a\) and \(\delta\leq h\), we get \[\label{eq:finalkick} A^2 + \frac{3}{4} H A \delta - 48 M \delta^2 \geq A^2 + \frac{3}{4} H a \delta - 48 M \delta^2 \geq A^2 + \frac{1}{2} H a \delta > (A+2 \varepsilon)^2.\tag{31}\]

Putting everything together, we deduce that there exists \(r\in \interval{0}{1}\) (close to \(0\)) such that \[\frac{\mathcal{J}_R (\varphi, r \omega)}{\mathcal{J}_R (\varphi, \omega)} < r^4.\] According to Proposition 10, this is sufficient to prove that the metric measure space \((\mathbb{H}, \mathsf d, \mathscr{L}^3)\) does not satisfy \(\mathsf{MCP}(0,5)\), and then \(N_{\text{curv}}> 5\), thanks to Remark 1.

3.0.0.4 Step 4: Proving \(N_{\mathrm{curv}} > 5\) when \(a = 0\).

In this case, we can provide a similar and simpler argument. Since \(a = 0\), the set \(S_\varepsilon\) is non-empty for all \(\varepsilon> 0\). Therefore, repeating the argument from Step 1 with \(A = 0\), we obtain that if \(\varphi\in D(S_\varepsilon)\) and \(\omega > 0\), then for all \(r \in \interval{0}{1}\) sufficiently small \[P(\varphi, r \omega) \leq 3 \varepsilon^2 r^4 \omega^4.\] Recalling Proposition 8, this yields that \[\label{eq:failMCPa610} \lim_{r \to 0} \frac{\mathcal{J}_R(\varphi, r \omega)}{r^4} \leq \lim_{r \to 0} \frac{1}{r^4} \left( 3 \varepsilon^2 r^4 \omega^4 + M r^2 \omega^5 \right) = 3 \varepsilon^2 \omega^4.\tag{32}\] As we are focusing on \(C^1\) and strongly convex reference norms, the reduced Jacobian \((\varphi,\omega) \mapsto \mathcal{J}_R (\varphi,\omega)\) is continuous and nonnegative when \(\omega\) is not a multiple of \(2\pi_{\Omega^\circ}\), see Remark 3.12 and Proposition 3.16 in [15]. In particular, by periodicity, we have that \[u:=\min_{\theta \in [0,2\pi_{\Omega^\circ})} \mathcal{J}_R (\theta,\pi_{\Omega^\circ}) >0.\] Finally, we choose \(\varepsilon< \sqrt{u/(3 \pi_{\Omega^\circ}^4)}\) and \(\omega = \pi_{\Omega^\circ}\) and we get that for \(r\in[0,1]\) sufficiently small it holds that \[\mathcal{J}_R (\varphi, r \pi_{\Omega^\circ}) < r^4 u\leq r^4 \mathcal{J}_R (\varphi, \pi_{\Omega^\circ}).\] As in the fist case, this is sufficient to conclude. ◻

Proof. Recall that the map \(P\) is bi-Lipschitz, cf.[15]. Moreover, if \({\|\!\cdot\!\|}\) is of class \(C^k\), then \(\partial\Omega=\{\left\lVert v\right\rVert=1\}\) is a \(C^k\) curve, as \(1\) is a regular value of \(f(v):=\left\lVert v\right\rVert^2\). Let \(\gamma:[0,1]\to \partial\Omega\) be a \(C^k\) parametrization of \(\partial\Omega\), with \(\gamma(0)=P(0)\) and constant (Euclidean) speed. Then, there exists a \(C^k\) function \(\theta:[0,1]\to [0,2\pi_\Omega]\) such that \[\gamma(t)=P(\theta(t)),\qquad\forall\, t\in [0,1],\] where, by definition, \(\theta(0)=0\) and \(\theta(1)=2\pi_\Omega\). Indeed, using Stokes’ theorem, we see that the function \(\theta(\cdot)\) satisfies \[\label{eq:def95theta} \theta(t)= \int_0^t (\gamma_1(s)\dot{\gamma}_2(s)-\gamma_2(s)\dot{\gamma}_1(s))\, \mathrm ds,\tag{34}\] and thus is \(C^k\). Now, we compute \[\dot{\theta}(t)=\gamma_1(t)\dot{\gamma}_2(t)-\gamma_2(t)\dot{\gamma}_1(t)= \gamma(t)\cdot\dot{\gamma}(t)^\perp,\] where \(\cdot\) denotes here the Euclidean scalar product and \(v^\perp\) is the oriented Euclidean orthogonal complement of \(v\). Since \(\dot{\gamma}(t)\neq 0\) for every \(t\in [0,1]\), we deduce that also \(\dot{\theta}(t)\neq 0\) for every \(t\in [0,1]\). Indeed, we have \[0=\frac{1}{2}\frac{\, \mathrm d}{\, \mathrm dt}|\dot{\gamma}(t)|^2= \gamma(t)\cdot\dot{\gamma}(t),\qquad\forall\,t\in [0,1],\] therefore \(\dot{\theta}(t)= 0\) would imply that \(\dot{\gamma}(t)\) is parallel and orthogonal to \(\gamma(t)\), meaning that \(\dot{\gamma}(t)=0\) and thus giving a contradiction. Therefore, \(\theta\) is a \(C^k\)-diffeomorphism and thus \(P=\gamma\circ\theta^{-1}\) is \(C^k\)-diffeomorphism as well. ◻

Proof. By assumptions, we have that \(\gamma^\varepsilon(s)\to \gamma(s)\) for every \(s\in[0,1]\) and \(\{f=1\}, \{f_\varepsilon=1\}\) are compact. Thus, it suffices to show the local \(C^k\)-convergence. Let \(v\in\{f=1\}\). Since \(1\) is a regular value for \(f\), \(\nabla f(v)\neq 0\), and w.l.o.g we may assume that \(v\) is such that \(\partial_yf(v)\neq 0\). Let \(O\subseteq\mathbb{R}^2\) be a neighborhood of \(v\) such that \(\partial_yf\neq 0\). Since \(\nabla f_\varepsilon\to\nabla f\) uniformly, for \(\varepsilon\) sufficiently small \(\partial_yf_\varepsilon\neq 0\) on \(O\) as well. For every \(\varepsilon\), let \(v_\varepsilon\in O\cap\{f_\varepsilon=1\}\), such that \(v_\varepsilon\to v\). With the natural identification \(v=v_0\), for every \(\varepsilon\geq 0\), there exist open sets \(U_{v_\varepsilon},V_{v_\varepsilon}\subseteq\mathbb{R}\) and a function \(g_\varepsilon:U_{v_\varepsilon}\subseteq\mathbb{R}\to \mathbb{R}\) such that \(v_\varepsilon\in U_{v_\varepsilon}\times V_{v_\varepsilon}\subseteq O\), \[\{f_\varepsilon=1\}\cap U_{v_\varepsilon}\times V_{v_\varepsilon}=\{(t,g_\varepsilon(t)):t\in U_{v_\varepsilon}\},\] and \[\label{eq:der95g} g_\varepsilon'(t)=-\frac{\partial_xf_\varepsilon}{\partial_yf_\varepsilon}(t,g_\varepsilon(t)).\tag{35}\] Moreover, we may choose \(U_{v_\varepsilon}=U_{v}\) and \(V_{v_\varepsilon}=V_v\) and, up to an affine reparametrization, we may assume that \(v_\varepsilon=(0,g_\varepsilon(0))\). Clearly \(g_\varepsilon\to g\) uniformly on \(U_v\). Thanks to 35 , \(g_\varepsilon\to g\) in \(C^1(U_v)\) and a simple iterative argument, using 35 , shows that \(g_\varepsilon\to g\) in \(C^k(U_v)\). At this point, consider \(\eta_\varepsilon:U_v\to\mathbb{R}^2\); \(\eta_\varepsilon(t):=(t,g_\varepsilon(t))\). Then, there exists \(I_v\subseteq[0,1]\) such that \(\gamma^\varepsilon(I_v)=\eta_\varepsilon(U_v)\), up to possibly restricting \(U_v\). Then, we can find \(\tilde{I}_v=[0,\alpha]\) and \(\tilde{\gamma}^\varepsilon:\tilde{I}_v\to \mathbb{R}^2\) for every \(\varepsilon\geq 0\), such that \(\tilde{\gamma}^\varepsilon:\tilde{I}_v\to \mathbb{R}^2\) is a \(C^k\) arc-length (affine) reparametrization of a sub-arc of \({\gamma^\varepsilon}_{|I_v}\). By further restricting \(U_v\) to \(\tilde{U}_v\), and defining the \(C^k\)-diffeomorphic embedding \(\varphi_\varepsilon: \tilde{U}_v\to \mathbb{R}\) by \[\varphi_\varepsilon(t):=\int_0^t|\dot{\eta}_\varepsilon(\tau)|\, \mathrm d\tau\] we have \(\varphi_\varepsilon\to \varphi_0\) in \(C^k(\tilde{U}_v)\). Furthermore, note that \(\tilde{I}_v\subseteq\bigcap_{\varepsilon>0} \varphi_\varepsilon(\tilde{U}_v)\) and \(\dot{\varphi}_\varepsilon\) is uniformly bounded below for small \(\varepsilon>0\) on \(\tilde{U}_v\). Thus, we have \(\varphi_\varepsilon^{-1}\to \varphi^{-1}\) in \(C^k(\tilde{I}_v)\). Thus, since \(\eta_\varepsilon=\tilde{\gamma}^\varepsilon\circ \varphi_\varepsilon\), we conclude that \(\tilde{\gamma}^\varepsilon\to\tilde{\gamma}\) in \(C^k(\mathcal{I}_v)\), and also \(\gamma_\varepsilon\to\gamma\) in \(C^k_{loc}\) since their reparametrizations are affine. ◻

Proof. For every \(\varepsilon\geq 0\), let \(\gamma^\varepsilon:[0,1]\to \mathbb{R}^2\) be a constant-speed \(C^k\) parametrization of \(\partial\Omega_\varepsilon\). Assume that \(\gamma^\varepsilon(0)=\gamma^\varepsilon(1)=P^\varepsilon(0)\) for every \(\varepsilon\geq 0\). Then, as a consequence of Lemma 7, \(\gamma^\varepsilon\to\gamma^0\) in the \(C^k\)-topology. Reasoning as in Lemma 6, for every \(\varepsilon\geq 0\), we find a \(C^k\)-diffeomorphism \(\theta_\varepsilon:[0,1]\to [0,2\pi_{\Omega_\varepsilon}]\) such that \[\label{eq:parametrization} \gamma^\varepsilon(t)=P^\varepsilon(\theta_\varepsilon(t)),\qquad\forall\,t\in[0,1].\tag{37}\] Since \(\theta_\varepsilon\) is defined by 34 , with \(\gamma^\varepsilon\) in place of \(\gamma\) in the left-hand side, we deduce that \(\theta_\varepsilon\to\theta_0\) in \(C^k([0,1])\). Now, for every \(\varepsilon\geq 0\), consider the function \(\varphi_\varepsilon:[0,1]\to [0,1]\) defined by \(\varphi_\varepsilon(t):=\theta_\varepsilon^{-1}(2\pi_{\Omega_\varepsilon}t)\). Since \(2\pi_{\Omega_\varepsilon}\to 2\pi_{\Omega_0}\) and \(\theta_\varepsilon\to\theta_0\) in \(C^k([0,1])\), we have that \(\varphi_\varepsilon\to \varphi_0\) in \(C^k([0,1])\), implying 36 as \(P^\varepsilon(2\pi_{\Omega_\varepsilon}t)=\gamma^\varepsilon(\varphi_\varepsilon(t))\), for every \(t\in [0,1]\). ◻

Proof. Recall that, for every \(\varepsilon\geq 0\), \(C_\circ^\varepsilon=(P^\varepsilon)^{-1}\circ\mathrm{d}({\|\!\cdot\!\|}_\varepsilon)_*\circ Q^\varepsilon\) by 11 , and observe that, by Lemma 6, \(C_\circ^\varepsilon\) is a \(C^{k-1}\) function. Now, thanks to our assumption and by Lemma 8, \(\mathrm{d}({\|\!\cdot\!\|}_\varepsilon)_*\circ Q^\varepsilon\) converges to \(\mathrm{d}({\|\!\cdot\!\|}_0)_*\circ Q^0\) as \(\varepsilon\to 0\) in the \(C^{k-1}\)-topology. Thus, we are left to check that \((P^\varepsilon)^{-1}\to (P^0)^{-1}\) as \(\varepsilon \to 0\) in the \(C^k\)-topology. Define \(f_\varepsilon:\mathbb{R}^2\setminus\{0\}\to \mathbb{R}\) as \[f_\varepsilon(x):=(P^\varepsilon)^{-1}\left(\frac{x}{\left\lVert x\right\rVert_\varepsilon}\right),\qquad\forall\,x\in\mathbb{R}^2\setminus\{0\},\;\varepsilon\geq 0.\] We claim that \(f_\varepsilon\to f\) in \(C^k_{loc}(\mathbb{R}^2\setminus\{0\})\) as \(\varepsilon\to 0\). Indeed, let \(K\subseteq\mathbb{R}^2\setminus\{0\}\) be compact, then for \(\varepsilon_0\) sufficiently small the family \(\{f_\varepsilon\}_{\varepsilon\in (0,\varepsilon_0)}\) is uniformly bounded in \(K\): \[|f_\varepsilon(x)|\leq 2\pi_{\Omega_\varepsilon}\leq 4\pi_{\Omega_0}, \qquad\forall\,x\in K.\] In addition, it is equi-Lipschitz on \(K\), as for \(x,y\in K\) and \(\varepsilon\in (0,\varepsilon_0)\) we have \[|f_\varepsilon(x)-f_\varepsilon(y)|\leq \mathsf {Lip}((P^\varepsilon)^{-1})\left|\frac{x}{\left\lVert x\right\rVert_\varepsilon}-\frac{y}{\left\lVert y\right\rVert_\varepsilon}\right|_{\mathbb{R}^2}\leq \mathsf {Lip}((P^\varepsilon)^{-1})\sup_{z\in K} \mathrm{d}_z\left(\frac{\cdot}{{\|\!\cdot\!\|}_\varepsilon}\right)|x-y|_{\mathbb{R}^2},\] and the constant in the right-hand side is uniformly bounded since \(\{{\|\!\cdot\!\|}_\varepsilon\}_{\varepsilon >0}\) \(C^k\)-strongly converges to \({\|\!\cdot\!\|}_0\) as \(\varepsilon\to 0\) and \(\mathsf {Lip}((P^\varepsilon)^{-1})\) is bounded by the Euclidean distance between \(\partial\Omega^\varepsilon\) and the origin, cf.[15]. Then, given any sequence \(\{\varepsilon_n\}_{n \in \mathbb{N}}\) such that \(\varepsilon_n \to 0\) as \(n\to \infty\), using Arzelà-Ascoli theorem we can extract a subsequence \(\{\varepsilon_{n(m)}\}_{m \in \mathbb{N}}\) such that \(\{f_{\varepsilon_{n(m)}}\}_{m \in \mathbb{N}}\) converges uniformly as \(m \to \infty\). Thanks to Lemma 8, we must have that \(\{f_{\varepsilon_{n(m)}}\}_{m \in \mathbb{N}} \to f_0\) uniformly as \(m \to \infty\). Computing the differential of \(f_{\varepsilon_{n(m)}}\), we see that \[\mathrm d_xf_{\varepsilon_{n(m)}}=\frac{1}{(P^{\varepsilon_{n(m)}})'\circ f_{\varepsilon_{n(m)}}}\,\mathrm{d}_x\left(\frac{\cdot}{{\|\!\cdot\!\|}_{\varepsilon_{n(m)}}}\right).\] Since \((P^{\varepsilon_{n(m)}})'\) is uniformly bounded away from \(0\) and converges uniformly to \((P^0)'\), we deduce that \(\mathrm d_xf_{\varepsilon_{n(m)}}\to \mathrm d_xf_0\) on \(K\) as \(m\to \infty\), implying that \(f_{\varepsilon_{n(m)}}\to f_0\) in \(C^1_{loc}(\mathbb{R}^2\setminus\{0\})\) as \(m\to \infty\). With an inductive argument, we conclude that \((P^{\varepsilon_{n(m)}})^{-1}\to (P^0)^{-1}\) as \(m \to \infty\) in \(C^k_{loc}(\mathbb{R}^2\setminus\{0\})\). As this holds for every sequence \(\{\varepsilon_n\}_{n \in \mathbb{N}}\), we deduce that \((P^\varepsilon)^{-1}\to (P^0)^{-1}\) in \(C^k_{loc}(\mathbb{R}^2\setminus\{0\})\) as \(\varepsilon \to 0\), concluding the proof. ◻

Proof. In order to prove the continuity of the curvature exponent, we need to address the convergence of the reduced Jacobians associated with the norms. To this aim, it is convenient to introduce the following parametrization: for every \(\varepsilon\geq 0\), define the map \[\widetilde{\mathcal{J}}_R^\varepsilon:[0,1)\times (0,1)\to [0,\infty); \qquad \widetilde{\mathcal{J}}_R^\varepsilon(s,r):=\mathcal{J}_R^\varepsilon(2\pi_{\Omega^\circ_\varepsilon}s,-2\pi_{\Omega^\circ_\varepsilon}(1-r)+2\pi_{\Omega^\circ_\varepsilon}r).\] Then, recalling the explicit expression of the reduced Jacobian 15 , as a consequence of Lemma 8 and Corollary 1, we easily obtain that \[\label{eq:uniform95convergence95red95jactilde} \widetilde{\mathcal{J}}_R^\varepsilon\xrightarrow{\varepsilon\to 0}\widetilde{\mathcal{J}}_R^0,\qquad\text{in }C^{k-1}([0,1)\times (0,1)).\tag{38}\] In addition, since the norms of the sequence are \(C^1\) and strongly convex, [15] implies that the reduced Jacobians are always non-negative and \[\label{eq:zeros95redjac} \lim_{r\to r^\star}\widetilde{\mathcal{J}}_R^\varepsilon(s,r)=0\qquad\Longleftrightarrow\qquad r^\star=0,1,\frac{1}{2}.\tag{39}\] According to Proposition 11, after reparametrization, we have that \[\label{eq:tha95sup} N_{\mathrm{curv}}^\varepsilon = \sup_{(s,r)\in [0,1)\times (0,1)} \widetilde{N}^\varepsilon(s,r)\tag{40}\] where, for every \((s,r)\in [0,1)\times (0,1)\), \[\widetilde{N}^\varepsilon(s,r):=1+\frac{(r-\frac{1}{2})\partial_r \widetilde{\mathcal{J}}_R^\varepsilon(s,r)}{\widetilde{\mathcal{J}}_R^\varepsilon(s,r)}.\] Firstly, observe that for every \(s\in [0,1)\), \(\mathcal{J}_R^0(s,\cdot)\) is strictly positive on \((0,1)\setminus\{\frac{1}{2}\}\) and 39 holds. Therefore, there exists \(\delta>0\) sufficiently small such that \((r-\frac{1}{2})\partial_r\mathcal{J}_R^0(s,r)<0\) in \([0,1)\times ((0,\delta)\cup (1-\delta,1))\). Since \(\partial_r \widetilde{\mathcal{J}}_R^\varepsilon(s,r)\to\partial_r \widetilde{\mathcal{J}}_R^0(s,r)\) uniformly as \(\varepsilon\to 0\), then there exists \(\varepsilon_0>0\) such that \[\Big(r-\frac{1}{2}\Big)\partial_r \widetilde{\mathcal{J}}_R^\varepsilon(s,r)<0,\qquad\forall\, (s,r)\in [0,1)\times ((0,\delta)\cup (1-\delta,1)),\,\varepsilon< \varepsilon_0.\] This means that the supremum in 40 can be taken on the set \([0,1)\times [\delta,1-\delta]\). Secondly, we claim that \(\widetilde{N}^\varepsilon\to \widetilde{N}^0\) uniformly on \([0,1)\times [\delta,1-\delta]\), this would conclude the proof.

To prove the claim, first of all, by 38 and 39 , we immediately deduce that \(\widetilde{N}^\varepsilon\to \widetilde{N}^0\) locally uniformly on \([0,1)\times ((0,1)\setminus \{\frac{1}{2}\})\). Second of all, by periodicity and compactness, the convergence is actually uniform as \(s\in [0,1)\), therefore we are left to check that we have a uniform convergence around \(r=\frac{1}{2}\). To do so, we use ?? to obtain a simplified expression for \(N^\varepsilon\): after reparametrization, we get the following: \[\label{eq:Ntilde95epsilon} \widetilde{N}^\varepsilon(s,r)=1+ \frac{(r-\frac{1}{2})\partial_r(I^\varepsilon(s,r)+\widetilde{R}^\varepsilon(s,r))}{I^\varepsilon(s,r)+\widetilde{R}^\varepsilon(s,r)}=1+\frac{(r-\frac{1}{2})\partial_rI^\varepsilon(s,r)}{I^\varepsilon(s,r)}+\mathcal{R}^\varepsilon(s,r),\tag{41}\] where \(\widetilde{R}^\varepsilon\) denotes the reparametrization of \(R^\varepsilon\) defined in ?? (for the norm \({\|\!\cdot\!\|}_\varepsilon\)) and \[\label{eq:def95I} I^\varepsilon(s,r):=\frac{1}{2} \int_{1/2}^{r} \left( \int_{1/2}^{r} (t - \tau)^2 (C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}(s-1+2t)) (C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}(s-1+2\tau)) \mathop{}\!\mathrm{d}\tau \right) \mathop{}\!\mathrm{d}t.\tag{42}\] In 41 , the term \(\mathcal{R}^\varepsilon(s,r)\) is defined as the following difference \[\widetilde{N}^\varepsilon(s,r)-1-\frac{(r-\frac{1}{2})\partial_rI^\varepsilon(s,r)}{I^\varepsilon(s,r)}.\] The proof now follows from Lemma 9 and Lemma 10. ◻

Proof. Before proving the claim, let us observe that \[\label{eq:expression95I95eps} \partial_rI^\varepsilon(s,r)=(C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}(s-1+2r))\underbrace{\int_{1/2}^{r} (t - r)^2 (C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}(s-1+2t)) \, \mathrm dt}_{=:J^\varepsilon(s,r)}.\tag{43}\] The function \(J^\varepsilon(s,r)\) has a Taylor expansion with uniform (in \(\varepsilon\)) remainder at \(r=\frac{1}{2}\), indeed: \[\label{eq:expansion95J95eps} \begin{align} J^\varepsilon(s,r) &=\frac{(C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}s)}{3}\Big(r-\frac{1}{2}\Big)^3+\int_{1/2}^r \frac{(t-\frac{1}{2})^3}{3}(C_\circ^\varepsilon)''(2\pi_{\Omega^\circ_\varepsilon}(s-1+2t))\, \mathrm dt\\ &=\frac{(C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}s)}{3}\Big(r-\frac{1}{2}\Big)^3+O\left(\Big(r-\frac{1}{2}\Big)^4\right),\qquad\text{as }r\to\frac{1}{2}, \end{align}\tag{44}\] where the big-\(O\) term is uniform in \(\varepsilon\) and depends on a uniform bound for \(\mathsf {Lip}((C_\circ^\varepsilon)'')\). From this expansion, we deduce an expansion for the function \(I^\varepsilon(s,r)\) at \(r=\frac{1}{2}\), using 43 and a uniform bound on \(\mathsf {Lip}((C_\circ^\varepsilon)')\). More precisely, we have: \[\label{eq:expansion95I95eps} I^\varepsilon(s,r)=\int_{1/2}^r \partial_rI^\varepsilon(s,\tau)\, \mathrm d\tau=\frac{(C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}s)^2}{12}\Big(r-\frac{1}{2}\Big)^4+O\left(\Big(r-\frac{1}{2}\Big)^5\right),\qquad\text{as }r\to\frac{1}{2},\tag{45}\] where the big-\(O\) term is once again uniform in \(\varepsilon\).

We intend to apply Arzelà-Ascoli theorem to the functions \(\mathcal{Q}^\varepsilon\). Firstly, by 45 and 43 , there exists constants \(C>c>0\), not depending on \(\varepsilon\), such that \[\label{eq:uniform95bds95I} c \Big|r-\frac{1}{2}\Big|^4\leq |I^\varepsilon|\leq C \Big|r-\frac{1}{2}\Big|^4,\qquad |\partial_rI^\varepsilon|\leq C\Big|r-\frac{1}{2}\Big|^3.\tag{46}\] This means that the functions \(\mathcal{Q}^\varepsilon\) are uniformly bounded by a constant not depending on \(\varepsilon\). Secondly, we find a uniform bound for the gradient of \(\mathcal{Q}^\varepsilon\). For the derivative in the \(s\)-direction, the estimates from the first part of the proof, together with a uniform bound on \(\mathsf {Lip}((C_\circ^\varepsilon)')\), allows to conclude. For the derivative in the \(r\)-direction, it is convenient to write: \[\begin{gather} \label{eq:expression95Q95eps} \partial_r \mathcal{Q}^\varepsilon(s,r)=4\pi_{\Omega^\circ_\varepsilon} (C_\circ^\varepsilon)''(2\pi_{\Omega^\circ_\varepsilon}(s-1+2r)) \frac{(r-\frac{1}{2})J^\varepsilon(s,r)}{I^\varepsilon(s,r)}\\+ (C_\circ^\varepsilon)'(2\pi_{\Omega^\circ_\varepsilon}(s-1+2r))\partial_r\left( \frac{(r-\frac{1}{2})J^\varepsilon(s,r)}{I^\varepsilon(s,r)}\right) \end{gather}\tag{47}\] The first term can be uniformly estimated reasoning as in the first part of the proof, using the expansion 44 and also a uniform bound on \(\mathsf {Lip}((C_\circ^\varepsilon)')\). For controlling the second term, we need a uniform bound on \[\partial_r\left( \frac{(r-\frac{1}{2})J^\varepsilon(s,r)}{I^\varepsilon(s,r)}\right)=\frac{(J^\varepsilon(s,r)+(r-\frac{1}{2})\partial_rJ^\varepsilon(s,r))I^\varepsilon(s,r)-(r-\frac{1}{2})J^\varepsilon(s,r)\partial_r I^\varepsilon(s,r)}{I^\varepsilon(s,r)^2}.\] Note that, a priori, the numerator has order \(7\) at \(r=\frac{1}{2}\), while the denominator has order \(8\). However, replacing the Taylor expansions 45 and 44 , after a routine computation, one obtains that: \[(J^\varepsilon(s,r)+(r-\frac{1}{2})\partial_rJ^\varepsilon(s,r))I^\varepsilon(s,r)-(r-\frac{1}{2})J^\varepsilon(s,r)\partial_r I^\varepsilon(s,r)=O\left(\Big(r-\frac{1}{2}\Big)^8\right),\qquad\text{as }r\to\frac{1}{2}\] where the big-\(O\) is uniform in \(\varepsilon\). This shows the claimed uniform bound for the second term of 47 . Finally, we have proven that \(\{\mathcal{Q}^\varepsilon\}_{\varepsilon\geq 0}\) is uniformly bounded and equi-Lipschitz as \(\varepsilon\to 0\). Therefore, since we also know that \(\mathcal{Q}^\varepsilon(s,r)\to\mathcal{Q}^0(s,r)\) pointwise as \(\varepsilon\to 0\), we conclude the proof of the lemma. ◻

Proof. Firstly observe that, for the norm \({\|\!\cdot\!\|}_\varepsilon\), the remainder term ?? , denoted by \(R^\varepsilon\), can be bounded by \(M_\varepsilon \omega^6\) around \(\omega=0\), where \(M_\varepsilon\) depends on the \(L^\infty\)-norms of \(P^\varepsilon\), \(Q^\varepsilon\) and \((C_\circ^\varepsilon)'\) which are all uniformly bounded in \(\varepsilon\). Indeed: \[\begin{align} |R^\varepsilon(\varphi,\omega)| &\leq \mathsf {Lip}(C_\circ^\varepsilon)^3\int_\varphi^{\varphi+\omega}\int_\varphi^t\int_\varphi^s (t-s)(s-u)[2\|Q^\varepsilon\|_{L^\infty}\|P^\varepsilon\|_{L^\infty}(u-\varphi)+(t-\varphi)]\, \mathrm du \, \mathrm ds \, \mathrm dt\\ &\leq 2\mathsf {Lip}(C_\circ^\varepsilon)^3(2\|Q^\varepsilon\|_{L^\infty}\|P^\varepsilon\|_{L^\infty}+1)\int_\varphi^{\varphi+\omega}\int_\varphi^t\int_\varphi^s (t-s)(s-u)(t-\varphi)\, \mathrm du \, \mathrm ds \, \mathrm dt\\ &\leq 2\mathsf {Lip}(C_\circ^\varepsilon)^3(2\|Q^\varepsilon\|_{L^\infty}\|P^\varepsilon\|_{L^\infty}+1)\int_\varphi^{\varphi+\omega}\int_\varphi^t\int_\varphi^s (t-s)(s-u)(t-\varphi)\, \mathrm du \, \mathrm ds \, \mathrm dt\\ &\leq \frac{1}{72}\mathsf {Lip}(C_\circ^\varepsilon)^3(2\|Q^\varepsilon\|_{L^\infty}\|P^\varepsilon\|_{L^\infty}+1)\omega^6, \qquad\forall\, \varphi,\omega. \end{align}\] Hence, after reparametrization, there exists \(M_1>0\) such that \[\label{eq:contazzi1} |\widetilde{R}^\varepsilon(s,r)|\leq M_1\left|r-\frac{1}{2}\right|^6,\qquad\text{on } [0,1)\times(0,1).\tag{48}\] Reasoning as above, one can obtain a similar estimate for the \(r\)-derivative, namely there exists \(M_2>0\) such that \[\label{eq:contazzi2} |\partial_r\widetilde{R}^\varepsilon(s,r)|\leq M_2\left|r-\frac{1}{2}\right|^5,\qquad\text{on } [0,1)\times(0,1).\tag{49}\] Now, by definition of \(\mathcal{R}^\varepsilon(s,r)\), we see that \[\mathcal{R}^\varepsilon(s,r)=\frac{(r-\frac{1}{2})\left(I^\varepsilon(s,r)\partial_r\widetilde{R}^\varepsilon(s,r)-\widetilde{R}^\varepsilon(s,r)\partial_rI^\varepsilon(s,r))\right)}{I^\varepsilon(s,r)(I^\varepsilon(s,r)+\widetilde{R}^\varepsilon(s,r))}.\] The right-hand side of the above can be uniformly bounded combining 48 , 49 and 46 . Thus, we conclude that there exists a uniform constant \(M_3>0\) such that \[|\mathcal{R}^\varepsilon(s,r)|\leq M_3\left|r-\frac{1}{2}\right|^2,\qquad\text{on } [0,1)\times(0,1).\] With completely similar arguments, one can show that there exits a constant \(M_4>0\), depending on the uniform bounds of the \(L^\infty\)-norms of \(P^\varepsilon\), \(Q^\varepsilon\), \((C_\circ^\varepsilon)'\) and \((C_\circ^\varepsilon)''\), such that \[|\partial_r\mathcal{R}^\varepsilon(s,r)|\leq M_4\left|r-\frac{1}{2}\right|,\qquad\text{on } [0,1)\times(0,1).\] Thus, applying Arzelà-Ascoli, we conclude the proof. ◻

Proof. Let \(p\in (1,2)\) and let \(q\in (2,\infty)\) be its conjugate exponent. Choose \(p\) sufficiently close to \(1\) so that \(q\geq \min\{k,3\}\) and \(2q+1\geq N^\star\). Consider the curve of norms \[\label{eq:construction} [0,1]\ni t\mapsto {\|\!\cdot\!\|}^t:=f^t_*,\qquad\text{with }f^t:=t\, \ell^q+(1-t)\ell^2,\tag{50}\] where \(\ell^q\) denotes the standard \(\ell^q\)-norm and \(\ell^2\) is the classical Euclidean norm. Then, for every \(t\in [0,1)\), \(f^t\) is strongly convex and of class \(C^{\lfloor q \rfloor}\) and \(t\mapsto f^t\) is a continuous curve of norms with respect to the topology of \(C^{\lfloor q \rfloor}\)-strongly convergence (with \(\lfloor q \rfloor\geq \min\{k,3\}\)). Therefore, according to Remark 15, \(t\mapsto {\|\!\cdot\!\|}^t\) is a continuous curve of strongly convex and \(C^{\lfloor q \rfloor}\) norms on \([0,1)\). Moreover, the endpoints of \(t\mapsto{\|\!\cdot\!\|}^t\) are respectively the \(\ell^2\)-norm at \(t=0\) and the \(\ell^p\)-norm at \(t=1\). By Theorem 16 (which applies since \(\lfloor q \rfloor\geq 3\)) and by Lemma 1, we have \[\label{eq:crazy95continuities} \lim_{t\to t_0}N_{\mathrm{curv}}^t=N_{\mathrm{curv}}^{t_0},\quad\forall\, t_0\in (0,1)\qquad\text{and}\qquad N_ {\mathrm{curv}}^1\leq\liminf_{t\to 1}N_{\mathrm{curv}}^t.\tag{51}\] But now, since at time \(t=1\), \({\|\!\cdot\!\|}^1=\ell^p\), by [13], \(N_{\mathrm{curv}}^1>2q+1\geq N^\star\). Hence, from 51 , we have that \(t\mapsto{\|\!\cdot\!\|}^t\) is a curve of norms whose curvature exponents connect continuously \(N_{\mathrm{curv}}^0=5\) and \(2q+1\). Thus, there exists \(t_0\) such that \(N_{\mathrm{curv}}^{t_0}=N^\star\), and setting \({\|\!\cdot\!\|}:={\|\!\cdot\!\|}^{t_0}\), we find a \(C^k\) and strongly convex norm with the desired curvature exponent. ◻

Proof. We repeat the same argument of Theorem 17, replacing the \(\ell^q\)-norm in 50 with the smooth and strongly convex norm \({\|\!\cdot\!\|}^h_{*}\) of Remark 18, with \(h\) sufficiently big. Note that the norm \({\|\!\cdot\!\|}\), found with the construction, is \(C^\infty\) and strongly convex, as it is the dual of a \(C^{\infty}\) and strongly convex norm, cf.Remark 15. ◻