On the generalized Zalcman functional $λa_n^2-a_{2n-1}$ in the close-to-convex family

Theorem 1. Suppose that $f \in \mathcal{C}$ as in the form (1) and $n \geq 3$ . (1) If $$\lambda \ge \frac{3n+\sqrt{5n^2-4n}}{n^2+n}$$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \lambda n^2 - (2n-1),$$ where the equality holds if f(z) is the Koebe function $z/(1-z)^2$ . (2) If $\frac{2n}{n^2-n+1} < \lambda < \frac{3n+\sqrt{5n^2-4n}}{n^2+n}$ , then (2) If $$\frac{2n}{n^2-n+1} < \lambda < \frac{3n+\sqrt{5n^2-4n}}{n^2+n}$$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \frac{\lambda \left[ 4n(n+1) - (3n^2 + 1)\lambda \right] + 4n(2-\lambda)\sqrt{n\lambda(2-\lambda)}}{\lambda \left[ 8n - \lambda(n+1)^2 \right]}.$$ (3) If $0 < \lambda \leq \frac{2n}{n^2-n+1}$ , then we have $$|\lambda a_n^2 - a_{2n-1}| \le 2n - 1.$$ Substituting $\lambda = 1$ in Theorem 1(1), it follows easily that $|a_n^2 - a_{2n-1}| \le (n-1)^2$ for all $n \ge 4$ and $f \in \mathcal{C}$ . It is worth to state the cases n = 3, 4 explicitly.

Corollary 1. Suppose that $f \in \mathcal{C}$ as in the form (1). (1) If $\lambda \geq 1$ , then $$|\lambda a_4^2 - a_7| \le 16\lambda - 7,$$ where the equality holds if f(z) is the Koebe function $z/(1-z)^2$ . (2) If $\frac{13}{8} < \lambda < 1$ , then $$|\lambda a_4^2 - a_7| \le \frac{\lambda(80 - 49\lambda) + 32(2 - \lambda)\sqrt{\lambda(2 - \lambda)}}{\lambda(32 - 25\lambda)}.$$ (3) If $0 < \lambda \le \frac{8}{13}$ , then we have $|\lambda a_4^2 - a_7| \le 7$ .

Corollary 2. Suppose that $f \in C$ as in the form (1). (1) If $\lambda \geq \frac{9+\sqrt{33}}{12}$ , then $$|\lambda a_3^2 - a_5| \le 9\lambda - 5,$$ where the equality holds if f(z) is the Koebe function $z/(1-z)^2$ . (2) If $\frac{6}{7} < \lambda < \frac{9+\sqrt{33}}{12}$ , then $$|\lambda a_3^2 - a_5| \le \frac{\lambda(12 - 7\lambda) + 3(2 - \lambda)\sqrt{3\lambda(2 - \lambda)}}{\lambda(6 - 4\lambda)}.$$ (3) If $0 < \lambda \le \frac{6}{7}$ , then we have $|\lambda a_3^2 - a_5| \le 5$ . Clearly, Corollary 2(2) gives that if $f \in \mathcal{C}$ is given by (1), then $$|a_3^2 - a_5| \le \frac{5 + 3\sqrt{3}}{2} \approx 5.098.$$ Proof of Theorem 1 rely on a number of lemmas. In Section 2, we present three important lemmas which play vital role in the formulation of several lemmas in Section 3. In Section 3, we state and prove several lemmas based on different interval range values of $\lambda$ . The proof of Theorem 1 will be given in Section 4.

Lemma 1. For $g(z) = z + \sum_{n=2}^{\infty} b_n z^n$ is analytic in $\mathbb{D}$ , consider $$J(g) = \lambda \left( \operatorname{Re} b_n \right)^2 - \operatorname{Re} b_{2n-1},$$ where $\lambda > 0$ . Then J is a real-valued, continuous and convex functional on $\mathcal{HC}$ . Proof. Let $h(z) = z + \sum_{n=2}^{\infty} c_n z^n$ be analytic in $\mathbb{D}$ , $0 \le t \le 1$ and G = tg + (1-t)h. Then, by the definition of J, we have $$J(G) = \lambda \left[ \text{Re} \left( tb_n + (1 - t)c_n \right) \right]^2 - \text{Re} \left[ tb_{2n-1} + (1 - t)c_{2n-1} \right]$$ which may be easily rearranged as $$J(G) = tJ(g) + (1-t)J(h) - \lambda t(1-t)[\operatorname{Re} b_n - \operatorname{Re} c_n]^2.$$ This gives $J(tg + (1-t)h) \le tJ(g) + (1-t)J(h)$ and the desired conclusions follow. Since C is compact, for the functional J defined as in Lemma 1, Theorem 4.6 in [7] yields the following. Lemma B. $\max\{J(f): f \in \mathcal{HC}\} = \max\{J(f): f \in \mathcal{C}\} = \max\{J(f): f \in \mathcal{EHC}\}.$ By using Lemmas A, 1 and B, we derive the following lemma.

Lemma 2. We have $4\max\{J(f): f \in \mathcal{C}\} - 4n \leq F_{n,\lambda}(u,v)$ , where $F_{n,\lambda}(u,v) =: F(u,v)$ is given by $$F(u,v) = \left[ (n+1)^2 \lambda - 8n \right] u^2 - 2(n-1) \left[ (n+1)\lambda - 2 \right] uv + 4(n-1)\sqrt{1-u^2} \sqrt{1-v^2} + (n-1)^2 \lambda v^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right] u^2 + 2 \left[ (n+1)^2 \lambda - 8n \right]$$ Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{EHC}$ . Then Lemma A and (3) (with $x = e^{is}$ and $y = e^{it}$ ) give that $$a_n = \frac{n+1}{2}e^{i(n-1)t} - \frac{n-1}{2}e^{i[s+(n-2)t]}$$ where $t, s \in [0, 2\pi)$ and $t \neq s$ . This representation quickly yields $$J(f) = \lambda \left\{ \frac{n+1}{2} \cos(n-1)t - \frac{n-1}{2} \cos(s+(n-2)t) \right\}^{2}$$ $$-n \cos 2(n-1)t + (n-1) \cos(s+(2n-3)t).$$ By the identity $\cos 2\theta = 2\cos^2 \theta - 1$ and the addition formula for $\cos(s + (n-2)t + (n-1)t)$ , we may rewrite the last expression as $$J(f) = \left(\frac{(n+1)^2}{4}\lambda - 2n\right)\cos^2(n-1)t$$ $$-(n-1)\left(\frac{n+1}{2}\lambda - 1\right)\cos(n-1)t\cos(s+(n-2)t)$$ $$-(n-1)\sin(n-1)t\sin(s+(n-2)t) + \frac{(n-1)^2}{4}\lambda\cos^2(s+(n-2)t) + n.$$ If we set $\cos(n-1)t = u$ and $\cos(s+(n-2)t) = v$ and use the identity $\sin \theta = \pm \sqrt{1-\cos^2 \theta}$ , then the above equation reduces to $$4J(f) - 4n < F(u, v),$$ where $u, v \in [-1, 1]$ . Lemmas A, 1 and B show that the proof is completed.

Lemma 3. Suppose that $n \geq 3$ and $\lambda \geq \frac{10n-2}{(n+1)^2}$ . Then $$F(u, v) \le 4\lambda n^2 - 12n + 4$$ for $(u, v) \in R$ , where the equality holds if and only if (u, v) = (1, -1) or (u, v) = (-1, 1).

Lemma 4. Suppose that $n \geq 3$ and $\frac{6n-2}{n^2+n} \leq \lambda < \frac{10n-2}{(n+1)^2}$ . Then $$F(u, v) \le 4\lambda n^2 - 12n + 4$$ for $(u, v) \in R$ , where the equality holds if and only if (u, v) = (1, -1) or (u, v) = (-1, 1).

Lemma 5. Suppose that $n \geq 3$ and $0 < \lambda \leq \frac{2n}{n^2 - n + 1}$ . Then $G(u, v) \leq 0$ for $(u, v) \in R$ , where G(u, v) is defined by (5).

Lemma 6. Suppose that $n \geq 3$ , $\frac{2n}{n^2-n+1} < \lambda < \frac{6n-2}{n^2+n}$ and $F(u,v) := F_{n,\lambda}(u,v)$ is given by Lemma 2. Then $$\max \{F(u,v): (u,v) \in \partial R\} \le A_{n,\lambda},$$ where (14) $$A_{n,\lambda} = \begin{cases} \frac{4(n-1)^2 \left[\lambda(n-1)+1\right]}{8n-(n+1)^2 \lambda} & \text{for } \frac{2n}{n^2-n+1} < \lambda \le \frac{5n-1}{n^2+n} \\ 4\lambda n^2 - 12n + 4 & \text{for } \frac{5n-1}{n^2+n} < \lambda < \frac{6n-2}{n^2+n}. \end{cases}$$ Proof. We continue to use the chain of inequalities given by (12) and as before, the proof is divided into four cases. Case 1: u = 1. For this case, $\lambda > \frac{2n}{n^2 - n + 1} > \frac{2}{n + 1}$ , the function $\Psi(v) := F(1, v)$ defined by $$\Psi(v) = [(n+1)^2\lambda - 8n] - 2(n-1)[(n+1)\lambda - 2]v + (n-1)^2\lambda v^2$$ attains its maximum at v = -1. This observation yields that $$\Psi(v) \le \Psi(-1) = 4\lambda n^2 - 12n + 4.$$ Case 2: u = -1. By similar reasoning as in Case 1, we conclude that $F(-1, v) \leq 4\lambda n^2 - 12n + 4$ . Case 3: v = 1. In this case, we need to consider the function $\Phi(u) := F(u,1)$ defined by $$\Phi(u) = [(n+1)^2\lambda - 8n]u^2 - 2(n-1)[(n+1)\lambda - 2]u + (n-1)^2\lambda.$$ Clearly, the only solution $u_0$ to the equation $\Phi'(u) = 0$ is given by $$u_0 = \frac{(n-1)[\lambda(n+1) - 2]}{\lambda(n+1)^2 - 8n}.$$ Moreover, $|u_0| < 1$ if and only if $$[(n^2 + n)\lambda - (5n - 1)][(n + 1)\lambda - (3n + 1)] \ge 0.$$ Also, we have $$\Phi(u_0) = F(u_0, 1) = \frac{4(n-1)^2 \left[\lambda(n-1) + 1\right]}{8n - (n+1)^2 \lambda}$$ and $$\Phi(-1) = F(-1,1) = F(1,-1) = 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2 - 12n + 4\lambda n^2$$ $\Phi(-1) = F(-1,1) = F(1,-1) = 4\lambda n^2 - 12n + 4.$ Since $\lambda < \frac{6n-2}{n^2+n} < \frac{3n+1}{n+1}$ and $\frac{5n-1}{n^2+n} < \frac{6n-2}{n^2+n}$ , the above facts show that $$|u_0| \left\{ \begin{array}{l} \leq 1 & \text{ for } \frac{2n}{n^2 - n + 1} < \lambda \leq \frac{5n - 1}{n^2 + n} \\ > 1 & \text{ for } \frac{5n - 1}{n^2 + n} < \lambda < \frac{6n - 2}{n^2 + n}. \end{array} \right.$$ It follows that $F(u,1) \leq A_{n,\lambda}$ , where $A_{n,\lambda}$ is given by (14). Case 4. v = -1. Again, by similar reasoning as in Case 3, we can prove that $F(u,-1) \leq A_{n,\lambda}$ . Moreover, by a computation, for $\lambda < \frac{6n-2}{n^2+n}$ , we have $$\Phi(-1) - \Phi(u_0) = 4\lambda n^2 - 12n + 4 - \frac{4(n-1)^2 \left[\lambda(n-1) + 1\right]}{8n - (n+1)^2 \lambda} = -\frac{4 \left[n(n+1)\lambda - (5n-1)\right]^2}{8n - (n+1)^2 \lambda} \le 0.$$ The desired conclusion follows if we use the above facts and combine the four cases.

Lemma 7. Suppose that $n \geq 3$ , $\frac{2n}{n^2-n+1} < \lambda < \frac{6n-2}{n^2+n}$ , and $F(u,v) := F_{n,\lambda}(u,v)$ is given by Lemma 2. Then the critical points of F(u,v) are (0,0) and (u,v), where (u,v) satisfies (15) $$v^2 = \frac{\left[\sqrt{n\lambda} + \sqrt{2-\lambda}\right]^2 \left[(n-1)\lambda - \sqrt{\lambda n(2-\lambda)}\right]}{(n-1)^2 \lambda^2}$$ (16) $$uv = \frac{[(n+1)\lambda - 2][\lambda(n-1) - \sqrt{\lambda n(2-\lambda)}]}{(n-1)\lambda[\lambda(n-1) - 2\sqrt{\lambda n(2-\lambda)}]}$$ and (17) $$u^{2} = \frac{\left[\lambda(n-1) - 2\sqrt{\lambda n(2-\lambda) + 2\right]\left[(n-1)\lambda - \sqrt{\lambda n(2-\lambda)}\right]}{\left[\lambda(n-1) - 2\sqrt{\lambda n(2-\lambda)}\right]^{2}}.$$

Lemma 8. Suppose that $F(u,v) := F_{n,\lambda}(u,v)$ is given by Lemma 2. Then we have the following: - If n ≥ 3 and 3n+√5n²-4n / n²+n ≤ λ < 6n-2 / n²+n, then the only critical point of F(u, v) for (u, v) ∈ (-1,1) × (-1,1) is (0,0), and that F(0,0) = 4(n-1).</li> If n ≥ 3 and 2n / n²-n+1 < λ < 3n+√5n²-4n / n²+n, then there are three critical points for F(u, v) on (-1,1) × (-1,1), namely, (0,0) and (u_i, v_i) (i = 1,2) which satisfy (15), (16) and (23) $$F(u_1, v_1) = F(u_2, v_2) = \frac{4\lambda(n-1)\left[\lambda(n^2+1) - 4n\right] + 16n(2-\lambda)\sqrt{n\lambda(2-\lambda)}}{\lambda\left[8n - \lambda(n+1)^2\right]},$$ and $F(u_1, v_1) > F(0, 0).$

Lemma 9. Let $n \geq 3$ and $R = [-1, 1] \times [-1, 1]$ . Then we have the following: (1) If $$\frac{3n+\sqrt{5n^2-4n}}{n^2+n} \le \lambda < \frac{6n-2}{n^2+n}$$ , then $F(u,v) \le 4\lambda n^2 - 12n + 4$ for $(u,v) \in R$ . (2) If $\frac{2n}{n^2-n+1} < \lambda < \frac{3n+\sqrt{5n^2-4n}}{n^2+n}$ , then $F(u,v) \le F(u_1,v_1)$ . (2) If $$\frac{2n}{n^2-n+1} < \lambda < \frac{3n+\sqrt{5n^2-4n}}{n^2+n}$$ , then $F(u,v) \le F(u_1,v_1)$ . Here $F(u,v) := F_{n,\lambda}(u,v)$ is given by Lemma 2 and $F(u_1,v_1)$ is given by (23).