Gehring-Hayman Inequality for Meromorphic Univalent Mappings

Theorem 1. The following inequalities hold for each $p \in (0,1)$ : (1.3) $$\frac{(1+p)^2\pi}{4p} \le A_p \le \tilde{A}_p < \frac{1+p^2}{p} \left(1+\sqrt{2}+\frac{20}{3p}\right)^2 \log 2.$$ Actually, we will give a better but much more complicated bound in Theorem 2 below. We mention here that the proof of Theorem 2 will be accomplished by employing a similar technique adopted by Gehring and Hayman in [4, Theorem 1]. The structure of the paper is as follows. Section 2 is devoted to several lemmas, which will be used to prove our main results. In Section 3, Theorem 2 will be presented together with its proof. Then Theorem 1 will be derived from it. Finally, as an application of the proof of Theorem 2, we will extend Theorem B to the meromorphic case (see Theorem 3).

Lemma 1. Let $\beta = [a, b]$ and $\beta' = [ia, ib]$ for positive numbers a, b with a < b. Then (2.5) $$\frac{\operatorname{arccot} M_p(b/a)}{\pi} \le \omega(z, \beta, \Omega_1) \le \frac{1}{2}, \quad z \in \beta',$$ where $\Omega_1$ is the hyperbolic half-plane defined above and <span id="page-3-3"></span>(2.6) $$M_p(q) = \frac{q+1}{q-1} + \frac{(1-p^2)^2(1+q^2)}{2p(q-1)(4p\sqrt{q}+(1+q)(1+p^2))}.$$

Lemma 2. Let k be a constant with 0 < k < 1. Then the function $G(x) = \cot(k \operatorname{arccot} x)$ is strictly convex and increasing on $0 < x < +\infty$ . Moreover, $G(0^+) = \cot(k\pi/2)$ , $G'(0^+) = k/\sin^2(k\pi/2)$ and the following inequalities hold: $$\cot \frac{k\pi}{2} + \frac{kx}{\sin^2(k\pi/2)} < G(x) < \cot \frac{k\pi}{2} + \frac{x}{k}, \quad 0 < x < +\infty.$$ Proof. First we compute $G(0^+) = \lim_{x\to 0^+} G(x) = \cot(k\pi/2)$ because $\operatorname{arccot}(0^+) = \pi/2$ . Next we have $$G'(x) = \frac{k}{(1+x^2)\sin^2(k \operatorname{arccot} x)} > 0,$$ which implies that G(x) is (strictly) increasing and gives the value of $G'(0^+)$ in the assertion. Since $\tan \theta$ is strictly convex on $0 < \theta < \pi/2$ and satisfies $\tan 0 = 0$ , the inequality $\tan(k\theta) < k \tan \theta$ ; equivalently, $\cot \theta < k \cot(k\theta)$ , holds for $0 < \theta < \pi/2$ . Letting $x = \cot \theta$ , we obtain $x < k \cot(k \operatorname{arccot} x)$ for $0 < x < +\infty$ . Hence, we observe that $$G''(x) = \frac{2k(k\cot(k\operatorname{arccot} x) - x)}{(1+x^2)^2\sin^2(k\operatorname{arccot} x)} > 0$$ for $0 < x < +\infty$ . Thus we have checked strict convexity of G. We next find the asymptotic value by using the L'Hôpital's rule as follows: $$\lim_{x \to +\infty} \frac{G(x)}{x} = \lim_{x \to +\infty} G'(x) = \lim_{x \to +\infty} \frac{k}{(1+x^2)\sin^2(k \operatorname{arccot} x)}$$ $$= \lim_{\theta \to 0^+} \frac{k}{(1+\cot^2\theta)\sin^2(k\theta)} = \lim_{\theta \to 0^+} \frac{k\sin^2\theta}{\sin^2(k\theta)} = \frac{1}{k}.$$ The strict convexity of G yields the inequalities $$G(0^+) + G'(0^+)(x - 0) < G(x) < G(0^+) + \frac{x(G(x_1) - G(0^+))}{x_1 - 0}$$ for $0 < x < x_1 < +\infty$ . In particular, the left-hand inequality follows. We now let $x_1 \to +\infty$ to obtain the right-hand inequality.

Theorem 2. Let $p \in (0,1)$ . Then (3.1) $$\frac{(1+p)^2\pi}{4p} \le A_p \le \tilde{A}_p \le \inf_{q \in (1,\infty)} N_p(q),$$ where $$N_p(q) = \frac{(1+p^2)\log q}{2p}\cot^2\left(\frac{1}{4}\cot^{-1}\left(\frac{q+1}{q-1} + \frac{(1-p^2)^2(1+q^2)}{2p(q-1)(4p\sqrt{q}+(1+q)(1+p^2))}\right)\right).$$ Proof. The proof of this theorem is similar to that of [3, Theorem 1]. But, for the sake of completeness, we will repeat all the necessary details. Let $f \in \tilde{\mathcal{M}}_p$ for a given number 0 . In other words, <math>f is a univalent analytic function on $\mathbb{D} \setminus \overline{\Delta}_p$ which is continuous up to the (open) left half $\mathbb{T}^-$ of the unit circle. Let $g: \mathbb{D} \to \mathbb{H}$ be the map defined in (2.3) and let $\alpha$ be the number defined in (2.1). Then $h = f \circ g^{-1}$ is analytic and univalent on $\Omega_1 = g(\Omega)$ and remains continuous on $\mathbb{R}^+$ . In particular, the image $D = h(\Omega_1) = f(\Omega)$ is a simply connected domain in $\mathbb{C}$ . Choose $q \in (1, \infty)$ arbitrarily and fix it. Let $\beta_n = [q^n, q^{n+1}]$ and $\beta'_n = [iq^n, iq^{n+1}]$ for each integer n. Let $\ell_n$ and $\ell'_n$ denote the lengths of the images of these segments under the mapping h. Then Lemma 1 yields the inequalities <span id="page-7-1"></span>(3.2) $$\frac{\xi}{\pi} \le \omega(h^{-1}(w), \beta_n, \Omega_1) = \omega(w, h(\beta_n), D) \le 1/2$$ at each point $w \in h(\beta'_n)$ , where $\xi = \cot^{-1}(M_p(q))$ and $M_p(q)$ is given in (2.6). Let $w_0$ be the point on $h(\beta_n)$ that divides its arclength in half. Then $h(\beta_n)$ lies in the disk $|w - w_0| \le \ell_n/2$ , and thus Lemma A together with (3.2) implies (3.3) $$\delta_D(w) \le \frac{\ell_n}{2} \cot^2(\pi \omega(w, h(\beta_n), D)/4) \le \frac{\ell_n}{2} \cot^2(\xi/4)$$ for $w \in h(\beta'_n)$ . We now recall that $\phi_\alpha$ maps $\Omega$ conformally onto the unit disk $\mathbb{D}$ and that $g = g^{-1}$ maps $\Omega$ onto $\Omega_1$ . Thus $\phi_\alpha \circ g : \Omega_1 \to \mathbb{D}$ is conformal. Since $h : \Omega_1 \to D$ is conformal, too, we have the relations <span id="page-7-2"></span> $$\lambda_{D}(h(z))|h'(z)| = \lambda_{\Omega_{1}}(z) = \lambda_{\mathbb{D}}(\phi_{\alpha}(g(z)))|(\phi_{\alpha} \circ g)'(z)|$$ $$= \frac{2|\phi'_{\alpha}(w)|}{|z+i|^{2}(1-|\phi_{\alpha}(w)|^{2})}, \quad z \in \Omega_{1},$$ where we put w = g(z). By the well-known estimate $1/\lambda_D \leq 4\delta_D$ on D, we obtain $$\delta_D(h(z)) \ge \frac{1}{4\lambda_D(h(z))} = \frac{|h'(z)|(1-|\phi_\alpha(w)|^2)|z+i|^2}{8|\phi'_\alpha(w)|}, \quad w = g(z) \in \Omega.$$ We now take $z = iy \in \mathbb{I}^+$ and compute w = g(iy) = i(1-y)/(1+y) = iv, where $v \in (-1,1)$ . Hence, in conjunction with (3.3), we get $$|h'(iy)| \le \frac{\ell_n}{2} \cot^2(\xi/4) \frac{8|\phi'_{\alpha}(iv)|}{(1-|\phi_{\alpha}(iv)|^2)(y+1)^2}.$$ Noting $dv/dy = -2/(y+1)^2$ , we estimate $$\ell'_{n} = \int_{\beta'_{n}} |h'(z)| |dz| = \int_{q^{n}}^{q^{n+1}} |h'(iy)| dy$$ $$\leq \frac{\ell_{n}}{2} \cot^{2}(\xi/4) \int_{q^{n}}^{q^{n+1}} \frac{8|\phi'_{\alpha}(iv)|}{(1 - |\phi_{\alpha}(iv)|^{2})(y+1)^{2}} dy$$ $$= 2\ell_{n} \cot^{2}(\xi/4) \int_{v_{n+1}}^{v_{n}} \frac{|\phi'_{\alpha}(iv)|}{1 - |\phi_{\alpha}(iv)|^{2}} dv,$$ where we put $v_n = (1 - q^n)/(1 + q^n)$ for $n \in \mathbb{Z}$ . Using the expression of $\phi_{\alpha}$ , we have the inequality $$\frac{|\phi_{\alpha}'(iv)|}{1 - |\phi_{\alpha}(iv)|^2} = \frac{\sqrt{(1 - v^2)^2 + 4\alpha^2 v^2}}{\alpha(1 - v^4)} \le \frac{\sqrt{(1 - v^2)^2 + 4v^2}}{\alpha(1 - v^4)} = \frac{1}{\alpha(1 - v^2)}$$ for $v \in (-1,1)$ and $0 < \alpha < 1$ . Hence, using (2.1), we obtain $$\ell'_n \le 2\ell_n \cot^2(\xi/4) \int_{v_{n+1}}^{v_n} \frac{dv}{\alpha(1-v^2)} = \ell_n \cot^2(\xi/4) \frac{\log q}{\alpha} = N_p(q)\ell_n$$ for each $n \in \mathbb{Z}$ . Summing up for $n \in \mathbb{Z}$ , we finally get $$\ell(f(\mathbb{I}_1)) = \ell(h(\mathbb{I}^+)) = \sum_{n=-\infty}^{\infty} \ell'_n \le N_p(q) \sum_{n=-\infty}^{\infty} \ell_n = N_p(q)\ell(h(\mathbb{R}^+)) = N_p(q)\ell(f(\mathbb{T}^-)).$$ The assertion of the theorem is now clear. We now deduce Theorem 1 from Theorem 2. Proof of Theorem 1. By Theorem 2 with the choice q = 4, we obtain $A_p \leq N_p(4)$ for 0 , where $$N_p(4) = \frac{(1+p^2)\log 2}{p} \cot^2 \left(\frac{1}{4} \operatorname{arccot} \left(\frac{5}{3} + \frac{17(1-p^2)^2}{6p(8p+5(1+p^2))}\right)\right).$$ We now apply Lemma 2 with k = 1/4 to get $$N_p(4) < \frac{(1+p^2)\log 2}{p} \left(\cot \frac{\pi}{8} + 4\left(\frac{5}{3} + \frac{17(1-p^2)^2}{6p(8p+5(1+p^2))}\right)\right)^2$$ Consider the function $$H(p) = 6p \left( \frac{5}{3} + \frac{17(1-p^2)^2}{6p(8p+5(1+p^2))} \right) = \frac{17p^4 + 50p^3 + 46p^2 + 50p + 17}{5p^2 + 8p + 5}.$$ Since $$H(1) - H(p) = \frac{(1-p)(17p^3 + 67p^2 + 63p + 33)}{5p^2 + 8p + 5} > 0$$ for 0 , we obtain <math>H(p) < H(1) = 10. Noting the formula $\cot(\pi/8) = 1 + \sqrt{2}$ , we now get $$N_p(4) < \frac{(1+p^2)\log 2}{p} \left(\cot \frac{\pi}{8} + \frac{2H(p)}{3p}\right)^2 < \frac{(1+p^2)\log 2}{p} \left(1 + \sqrt{2} + \frac{20}{3p}\right)^2.$$ Table 1 presents a comparison between ranges of $A_p$ by listing the upper bounds (UB) and lower bounds (LB) for selected values of $p \in (0,1)$ with the previously established results from [3, Table 1]. As evident from the following table, the newly obtained upper bounds for $A_p$ show an improvement. | | | UB of $A_p$ | UB of $A_p$ | UB of $A_p$ | |-------|-------------|----------------|----------------|----------------| | p | LB of $A_p$ | from Theorem C | from Theorem 1 | from Theorem 2 | | 0.999 | 3.141 | 73.421 | 114.486 | 73.251 | | 0.99 | 3.141 | 74.995 | 116.025 | 73.259 | | 0.9 | 3.150 | 95.491 | 134.471 | 74.212 | | 0.8 | 3.180 | 135.733 | 164.134 | 77.634 | | 0.7 | 3.242 | 221.807 | 210.271 | 84.837 | | 0.6 | 3.351 | 471.016 | 287.415 | 98.455 | | 0.5 | 3.534 | 1984.431 | 429.726 | 124.383 | | 0.4 | 3.848 | | 731.847 | 178.045 | | 0.3 | 4.424 | | 1528.574 | 310.577 | | 0.2 | 5.654 | | 4605.973 | 775.275 | | 0.1 | 9.503 | | 33408.930 | 4608.760 | <span id="page-9-0"></span>Table 1. Comparison between ranges of $A_p$ for various values of $p \in (0,1)$ It is plausible that the quantity $A_p$ is decreasing in $0 but it seems difficult to show it rigorously. Instead, we will show that <math>\tilde{A}_p$ is monotone.

Lemma 3. The quantity $\tilde{A}_p$ is non-increasing in 0 . In particular, the quantity $$A_p^+ = \sup_{p \le p' < 1} A_{p'}$$ satisfies the inequalities $$A_p \le A_p^+ \le \tilde{A}_p$$ for 0 . Proof. It is enough to show the inclusion relation $\Delta_{p_1} \supset \Delta_{p_2}$ for $0 < p_1 < p_2 < 1$ . Here, we recall that $\overline{\Delta}_p = \{\zeta : |\zeta - \alpha^{-1}| \le \sqrt{\alpha^{-2} - 1}\}$ and $\alpha = 2p/(1 + p^2)$ for $0 . To this end, we need to check the inequality <math>\alpha_1^{-1} - \alpha_2^{-1} + \sqrt{\alpha_2^{-2} - 1} \le \sqrt{\alpha_1^{-2} - 1}$ for $\alpha_k = 2p_k/(1 + p_k^2)$ , k = 1, 2. Indeed, it follows from the fact that the function $x - \sqrt{x^2 - 1}$ is decreasing in $1 < x < +\infty$ . Remarks. (1) It is clear that $A_1 \leq \tilde{A}_p$ for each $0 , where <math>A_1$ is the best possible constant appearing in Theorem A. In particular, we have from Theorem 2 $$A_1 \le \tilde{A}_{1^-} \le \lim_{p \to 1^-} N_p(q) = \log q \cot^2 \left(\frac{1}{4} \operatorname{arccot} \frac{q+1}{q-1}\right)$$ for each $1 < q < +\infty$ . With the help of Mathematica, we choose q = 5.55465 to obtain $\tilde{A}_{1^-} \leq 73.2502105$ . This agrees with the upper bound in Theorem A. (2) In Theorems 1 and 2, we consider the configuration $\overline{\Delta}_p$ and $\mathbb{I}_1$ only. However, we can translate $\overline{\Delta}_p$ by a disk automorphism of the form $T_a(z) = (z+ia)/(1-iaz)$ for -1 < a < 1 while leaving $\mathbb{I}_1$ invariant. Hence, we will have the same result by replacing $\overline{\Delta}_p$ by $T_a(\overline{\Delta}_p)$ , where $$T_a(\overline{\Delta}_p) = \left\{ \zeta : \left| \zeta - \frac{1 - a^2}{\alpha(1 + a^2)} - \frac{2ai}{1 + a^2} \right| \le \frac{\sqrt{\alpha^{-2} - 1}(1 - a^2)}{1 + a^2} \right\}.$$ Finally, we will generalize Theorem B. In order to enhance the difference, we rephrase Theorem B in terms of the hyperbolic geometry of a simply connected domain D as follows. Let $\Gamma$ be the hyperbolic geodesic that joins two points $w_1$ and $w_2$ in a simply connected domain $D \subset \mathbb{C}$ . Then, for any Jordan arc J joining $w_1$ and $w_2$ in D, the inequality $$(3.4) \ell(\Gamma) \le A_1 \ell(J)$$ holds, where $A_1$ is the constant in Theorem A. We will generalize this to the case when D is a simply connected hyperbolic subdomain of $\widehat{\mathbb{C}}$ containing $\infty$ . In this case, the bound should depend on a distance from $\infty$ to the geodesic $\Gamma$ . Our result can be stated in the form similar to that of Theorem B. We denote by hull(C) for a closed curve C in $\mathbb{C}$ the compact set $K = \mathbb{C} \setminus U_{\infty}$ , where $U_{\infty}$ is the (unique) unbounded connected component of $\mathbb{C} \setminus C$ , and denote by hull(C) the interior of the set hull(C). Here, we also note that the Euclidean length $\ell(J)$ of a Jordan arc J is the same as the (suitably normalized) 1-dimensional Hausdorff measure $\mathcal{H}^1(J)$ (see [8, p. 56]).

Theorem 3. Let J be a (closed) Jordan arc in $\mathbb{D}$ with endpoints $\zeta_1$ and $\zeta_2$ and let $\gamma$ be the (closed) hyperbolic line segment joining $\zeta_1$ and $\zeta_2$ in $\mathbb{D}$ . Let $\tilde{\gamma}$ be the set $C \setminus C_0$ , where C is the Euclidean circle (or Euclidean line attached with $\infty$ ) containing $\gamma$ and $C_0$ is the connected component of $C \setminus J$ containing a point outside $\mathbb{D}$ . Let f be a meromorphic univalent function on $\mathbb{D}$ with a pole at $s \in \mathbb{D}$ . Suppose that $s \notin \text{hull}(\tilde{\gamma} \cup J)$ , then $$\ell(f(\gamma)) \le \begin{cases} A_1 \ell(f(J)) & \text{if } s \notin \text{hull}^{\circ}(J \cup \hat{J}) \\ A_{\tau}^{+} \ell(f(J)) & \text{if } s \in \text{hull}^{\circ}(J \cup \hat{J}), \end{cases}$$ where $\hat{J}$ denotes the reflection of J with respect to the circle C and $$\tau = \tanh d_{\mathbb{D}}(s, \tilde{\gamma}).$$ Here, $A_1$ is the same as defined in Theorem A and $A_p^+$ is defined in Lemma 3 for $0 . If <math>A_p$ is non-increasing, we would have $A_p^+ = A_p$ . Proof. By our assumption, $\tau > 0$ . Translating by a disk automorphism if necessary, we can assume that $\zeta_1$ and $\zeta_2$ lie on $\mathbb{I}_1 = (-i, i)$ . Then $\hat{J}$ is the reflection of J in the imaginary axis. Let $\mathcal{V}$ be the set of all connected components V of $J \setminus \tilde{\gamma}$ . Then each $V \in \mathcal{V}$ is an open subarc of J with endpoints in $\tilde{\gamma}$ . We denote by $I_V$ the open line segment sharing two endpoints with V. Let $I = \bigcup_{V \in \mathcal{V}} I_V$ . Then I is an open subset of $\tilde{\gamma}$ . We set $\gamma_0 = \tilde{\gamma} \cap J$ and show that $$\gamma \subset \gamma_0 \cup I$$ and $J = \gamma_0 \cup \left(\bigcup_{V \in \mathcal{V}} V\right)$ . Since $\bigcup_{V\in\mathcal{V}}V=J\setminus\tilde{\gamma}$ , the latter equality is clear. We will verify the inclusion relation $\gamma\subset\gamma_0\cup I$ . Choose an arbitrary point $\zeta_0\in\gamma\setminus I$ and we will show the claim $\zeta_0\in J$ . We denote by $\tilde{\zeta}_m$ (m=1,2) the endpoints of $\tilde{\gamma}$ so that $\tilde{\zeta}_1,\zeta_1,\zeta_2,\tilde{\zeta}_2$ lie in $\mathbb{I}_1$ in this order. Note that $\tilde{\zeta}_m$ may be equal to $\zeta_m$ . Let $\tilde{\gamma}_m=[\tilde{\zeta}_m,\zeta_0]$ for m=1,2. For convenience, we parametrize J by a homeomorphism $j:[0,1]\to J$ such that $j(0)=\zeta_1$ and $j(1)=\zeta_2$ . Let $t_1=\sup\{t\in[0,1]:j(t)\in\tilde{\gamma}_1\}$ , that is, $t_1$ is the "final exit time" of j from $\tilde{\gamma}_1$ . In particular, $j(t)\neq\zeta_0$ for $t>t_1$ . It suffices to show that $\zeta_0=j(t_1)$ $(\in J)$ . To the contrary, we suppose that $j(t_1)\neq\zeta_0$ . We next let $t_2=\inf\{t\in(t_1,1]:j(t)\in\tilde{\gamma}_2\}$ . By continuity of j, we see that $\zeta_m':=j(t_m)\in\tilde{\gamma}_m$ and $j(t_m)\neq\zeta_0$ for m=1,2. In particular, $t_1< t_2$ . Since $j((t_1,t_2))$ does not meet $\tilde{\gamma}$ by construction, $V=j((t_1,t_2))\in\mathcal{V}$ and $I_V=(\zeta_1',\zeta_2')$ contains the point $\zeta_0$ , which contradicts the assumption $\zeta_0\in\gamma\setminus I$ . Hence, we conclude that $\zeta_0=j(t_1)\in J$ . Thus, the claim has been verified and $\zeta_0\in\gamma\cap J\subset\gamma_0$ follows. We now fix a Jordan arc $V \in \mathcal{V}$ . We denote by $D_V$ the Jordan domain bounded by $J_V = \overline{V} \cup I_V$ and denote by $\widetilde{D}_V$ the union of $D_V$ , $I_V$ and $\widehat{D}_V$ which stands for the reflection of $D_V$ in $\mathbb{I}_1$ . Let $g_V : \mathbb{D}_- \to D_V$ be a conformal homeomorphism whose continuous extension to $\partial \mathbb{D}_-$ maps $\mathbb{I}_1$ onto $I_V$ . Here, we recall that $\mathbb{D}_-$ is the left-half of the unit disk $\mathbb{D}$ . Schwarz's reflection principle now allows us to extend $g_V$ to a conformal homeomorphism of $\mathbb{D}$ onto $\widetilde{D}_V$ , which will be denoted by the same symbol. We next consider the following two cases. Case I: $s \notin D_V$ . In this case, $f \circ g_V : \mathbb{D} \to \mathbb{C}$ is analytic and univalent. Thus, by Theorem A, we obtain $$\ell(f(I_V)) = \ell((f \circ g_V)(\mathbb{I}_1)) \le A_1 \ell((f \circ g_V)(\mathbb{T}^-)) = A_1 \ell(f(V)).$$ Case II: $s \in \widetilde{D}_V$ . In this case, since $s \notin D_V \cup I_V$ by assumption, we may further normalize the conformal map $g_V : \mathbb{D} \to \widetilde{D}_V$ so that $p = g_V^{-1}(s) \in (0,1)$ . Since $\widetilde{D}_V \subset \mathbb{D}$ , the domain monotonicity of hyperbolic metric yields arth $$p = d_{\mathbb{D}}(p, 0) = d_{\widetilde{D}_V}(s, g_V(0)) \ge d_{\widetilde{D}_V}(s, I_V) \ge d_{\mathbb{D}}(s, I_V) \ge d_{\mathbb{D}}(s, \tilde{\gamma}) = \operatorname{arth} \tau$$ , and therefore, $p \ge \tau$ . By the definition of $A_p$ , we have $$\ell(f(I_V)) = \ell((f \circ q_V)(\mathbb{I}_1)) < A_n \ell((f \circ q_V)(\mathbb{T}^-)) = A_n \ell(f(V)) < A_{\tau}^+ \ell(f(V)).$$ Finally, we show the inequality in the assertion. When $s \in \text{hull}^{\circ}(J \cup \hat{J})$ , we have $\ell(f(I_V)) \leq A_{\tau}^{+}\ell(f(V))$ for every $V \in \mathcal{V}$ because $A_1 \leq A_{\tau}^{+}$ . Since $\gamma \subset \gamma_0 \cup I$ and J is the disjoint union of $\gamma_0$ and V over $\mathcal{V}$ , we obtain $$\ell(f(\gamma)) = \mathcal{H}^{1}(f(\gamma)) \leq \mathcal{H}^{1}(f(\gamma_{0})) + \mathcal{H}^{1}(f(I))$$ $$\leq \mathcal{H}^{1}(f(\gamma_{0})) + \sum_{V \in \mathcal{V}} \mathcal{H}^{1}(f(I_{V}))$$ $$\leq A_{\tau}^{+} \mathcal{H}^{1}(f(\gamma_{0})) + A_{\tau}^{+} \sum_{V \in \mathcal{V}} \mathcal{H}^{1}(f(V))$$ $$= A_{\tau}^{+} \mathcal{H}^{1}(f(J)) = A_{\tau}^{+} \ell(f(J)).$$ When $s \notin \text{hull}^{\circ}(J \cup \hat{J})$ , only Case I occurs. Therefore, we can show the corresponding inequality in a similar way. Statements and Declarations: Nil. Competing interests: We have nothing to declare. Conflict of interest: Nil.