The second and third Hankel determinants for starlike MA--Minda subclass associated to quadratic polynomials

Lemma 3.2. [3] Let $A, B, C \in \mathbb{R}$ and define $$Y(A, B, C) := \max_{z \in \overline{\mathbb{D}}} (|A + Bz + Cz^2| + 1 - |z|^2).$$ (i) If AC > 0, then $$Y(A, B, C) = \begin{cases} |A| + |B| + |C|, & \text{if } |B| \ge 2(1 - |C|), \\ 1 + |A| + \frac{B^2}{4(1 - |C|)}, & \text{if } |B| < 2(1 - |C|). \end{cases}$$ (ii) If AC < 0, then $$Y(A,B,C) = \begin{cases} 1 - |A| + \frac{B^2}{4(1-|C|)}, & \text{if } -4AC\left(C^2 - 1\right) \leq B^2 \text{ and } |B| < 2\left(1 - |C|\right), \\ 1 + |A| + \frac{B^2}{4\left(1 + |C|\right)}, & \text{if } B^2 < \min\left\{4(1 + |C|)^2, -4AC\left(C^2 - 1\right)\right\}, \\ R(A,B,C), & \text{otherwise,} \end{cases}$$ where. $$R(A, B, C) = \begin{cases} |A| + |B| + |C|, & \text{if } |C| (|B| + 4|A|) \le |AB|, \\ -|A| + |B| + |C|, & \text{if } |AB| \le |C| (|B| - 4|A|), \\ (|A| + |C|) \sqrt{1 - \frac{B^2}{4AC}}, & \text{otherwise.} \end{cases}$$ The next well-known lemma by Prokhrov and Szynal [14] will be crucial for our result.

Lemma 3.3. Let $w(z) = \sum_{n=1}^{\infty} c_n z^n$ be a Schwarz function, that is, w is analytic in $\mathbb{D}$ , w(0) = 0, and |w(z)| < 1 for all $z \in \mathbb{D}$ . If $c_1 \geq 0$ , then there exist complex numbers $\gamma, \eta, \rho$ with $|\gamma| < 1$ , $|\eta| < 1$ , $|\rho| < 1$ , such that $$c_{2} = (1 - c_{1}^{2})\gamma,$$ $$c_{3} = (1 - c_{1}^{2})(\eta(1 - |\gamma|^{2}) - c_{1}\gamma^{2}),$$ $$c_{4} = (1 - c_{1}^{2})(c_{1}^{2}\gamma^{3} - (1 - |\gamma|^{2})(2c_{1}\gamma\eta + \overline{\gamma}\eta^{2}) + (1 - |\gamma|^{2})(1 - |\eta|^{2})\rho).$$ First, we establish a bound for the second Hankel determinant and show that the bound is sharp.

Theorem 3.1. Let $f \in \mathcal{S}^*(\varphi)$ , then $|H_2(2)| \leq 1/4$ . The result is sharp.

Theorem 3.2. Let $f \in \mathcal{S}^*(\varphi)$ , then $|H_3(1)| \leq 1/9$ , the result is sharp. Proof. Let f ∈ S^∗ (φ) be given by $$z \frac{f'(z)}{f(z)} \prec \varphi(z)$$ , for $z \in \mathbb{D}$ . Thus, there exists a Schwarz function w(z) in the form of a power series w(z) = P^∞ ⁿ=1 cnz n such that $$z\frac{f'(z)}{f(z)} = \varphi(w(z)), \quad z \in \mathbb{D}.$$ (3.7) As in the proof of Theorem 3.1, by replacing f with it's rotations $$f_{\theta}(z) := e^{-i\theta} f(e^{i\theta}z),$$ we do not change the value of |H3(1)|, and the corresponding Schwarz function becomes $$w_{\theta}(z) = w(e^{i\theta}z) = \sum_{n=1}^{\infty} c_n e^{in\theta} z^n.$$ Choosing θ = − arg c1, we may assume without loss of generality that c¹ ≥ 0. The third Hankel determinant is defined by $$H_3(1) = a_3(a_2a_4 - a_3^2) - a_4(a_4 - a_2a_3) + a_5(a_3 - a_2^2).$$ (3.8) Putting the values of (3.2) in (3.8), we obtain $$144 H_3(1) = -\left(1 - 3t + 9t^2 + 9t^3\right) c_1^6 + \left(3 - 2t + 21t^2\right) c_1^4 c_2 + 9\left(-1 + 2t\right) c_2^3 + 4\left(2 - 3t + 9t^2\right) c_1^3 c_3 - 4\left(-6 + 7t\right) c_1 c_2 c_3 - 16c_3^2 + 18c_2 c_4 + c_1^2 \left(\left(-9 + 3t - 46t^2\right) c_2^2 + 18\left(-1 + t\right) c_4\right).$$ $$(3.9)$$ Now using the parametric relation in Lemma 3.3 in (3.9), we obtain $$144 H_{3}(1) = -\left(1 - 3t + 9t^{2} + 9t^{3}\right) c_{1}^{6} + \left(3 - 2t + 21t^{2}\right) \gamma c_{1}^{4} (1 - c_{1}^{2}) + 4\left(2 - 3t + 9t^{2}\right) c_{1}^{3} \left(\eta(1 - |\gamma|^{2}) - \gamma^{2}c_{1}\right) (1 - c_{1}^{2}) - 4\left(-6 + 7t\right) \gamma c_{1} \left(\eta(1 - |\gamma|^{2}) - \gamma^{2}c_{1}\right) (1 - c_{1}^{2})^{2} - 16\left(\eta(1 - |\gamma|^{2}) - \gamma^{2}c_{1}\right)^{2} (1 - c_{1}^{2})^{2} + 9\left(-1 + 2t\right) \gamma^{3} (1 - c_{1}^{2})^{3} + 18\gamma(1 - c_{1}^{2})^{2} \left(\gamma^{3}c_{1}^{2} - (1 - |\gamma|^{2})\left(-\rho(1 - |\eta|^{2}) + 2\gamma\eta c_{1} + \eta^{2}\overline{\gamma}\right)\right) + c_{1}^{2} \left(\left(-9 + 3t - 46t^{2}\right) \gamma^{2} (1 - c_{1}^{2})^{2} + 18\left(-1 + t\right) (1 - c_{1}^{2})\left(\gamma^{3}c_{1}^{2} - (1 - |\gamma|^{2})\left(-\rho(1 - |\eta|^{2}) + 2\gamma\eta c_{1} + \eta^{2}\overline{\gamma}\right)\right)\right),$$ $$(3.10)$$ where |γ| ≤ 1, |η| ≤ 1, and |ρ| ≤ 1. Next, we convert the above expression in the following form $$144 H_3(1) = A_1 + B_1 \eta + C_1 \eta^2 + D_1 \rho, \tag{3.11}$$ where, $$A_{1} = -\left(1 - 3t + 9t^{2} + 9t^{3}\right)c_{1}^{6} - \left(3 - 2t + 21t^{2}\right)\gamma c_{1}^{4}\left(-1 + c_{1}^{2}\right)$$ $$+ 2\gamma^{4}c_{1}^{2}\left(-1 + c_{1}^{2}\right)^{2} - \gamma^{2}c_{1}^{2}\left(-1 + c_{1}^{2}\right)\left(-9 + c_{1}^{2} + 2t^{2}\left(-23 + 5c_{1}^{2}\right) + t\left(3 + 9c_{1}^{2}\right)\right)$$ $$- \gamma^{3}\left(-1 + c_{1}^{2}\right)\left(-3\left(3 + 2c_{1}^{2} + c_{1}^{4}\right) + 2t\left(9 - 4c_{1}^{2} + 4c_{1}^{4}\right)\right),$$ $$B_{1} = 4\left(-1 + |\gamma|^{2}\right)c_{1}\left(-1 + c_{1}^{2}\right)\left(\left(2 - 3t + 9t^{2}\right)c_{1}^{2} + \gamma^{2}\left(-1 + c_{1}^{2}\right)\right)$$ $$+ \gamma\left(6 - 7t + 3c_{1}^{2} - 2tc_{1}^{2}\right),$$ $$C_{1} = -2\left(-1 + |\gamma|^{2}\right)\left(-1 + c_{1}^{2}\right)\left(8 - 8c_{1}^{2} + 8|\gamma|^{2}\left(-1 + c_{1}^{2}\right) + 9\left(-1 + t\right)c_{1}^{2}\overline{\gamma}\right)$$ $$- 9\gamma\left(-1 + c_{1}^{2}\right)\overline{\gamma},$$ $$D_{1} = 12\left(-1 + c_{1}^{2}\right)\left(-1 + c_{1}^{2}\right)\left(-1 + c_{1}^{2}\right)\left(-1 + c_{1}^{2}\right)\left(-1 + c_{1}^{2}\right)^{2}$$ $$D_1 = -18 \left( -1 + |\gamma|^2 \right) \left( -1 + |\eta|^2 \right) \left( -1 + c_1^2 \right) \left( \gamma + (-1 + t) c_1^2 - \gamma c_1^2 \right).$$ Now taking modulus on both sides of (3.11), writing $p_1 := c_1$ , using $$|\gamma| = x, \qquad |\eta| = y, \qquad |\rho| \le 1,$$ and applying the triangle inequality, we obtain $$H(p_{1}, x, y, t) := (1 - 3t + 9t^{2} + 9t^{3}) p_{1}^{6} + x ((3 - 2t + 21t^{2}) p_{1}^{4} (1 - p_{1}^{2})) + x^{2} (p_{1}^{2} (1 - p_{1}^{2}) (9 - p_{1}^{2} + 2t^{2} (23 - 5p_{1}^{2}) - t(3 + 9p_{1}^{2}))) + x^{3} (3(3 + 2p_{1}^{2} + p_{1}^{4}) + 2t(-9 + 4p_{1}^{2} - 4p_{1}^{4})) + x^{4} (2p_{1}^{2} (1 - p_{1}^{2})^{2}) + y (4(1 - x^{2})p_{1}(1 - p_{1}^{2})((2 - 3t + 9t^{2})p_{1}^{2} + x^{2}(1 - p_{1}^{2})) + x(6 - 7t + 3p_{1}^{2} - 2tp_{1}^{2}))) + y^{2} (2(1 - x^{2})(1 - p_{1}^{2})(8(1 - x^{2})(1 - p_{1}^{2}) + 9(x(1 - p_{1}^{2}) + (1 - t)p_{1}^{2})x)) + 18(1 - x^{2})(1 - y^{2})(1 - p_{1}^{2})(x(1 - p_{1}^{2}) + (1 - t)p_{1}^{2}).$$ (3.12) It is easy to see that the coefficient of the coefficient of the term linear in y, precisely $$4(1-x^2)p_1(1-p_1^2)\left((2-3t+9t^2)p_1^2+x^2(1-p_1^2)+x(6-7t+3p_1^2-2tp_1^2)\right) \ge 0$$ is non-negative which implies we can replace $y^1 = 1$ , and maxima will increase. We define the new function where $y^1 = 1$ as $H_1(p_1, x, y, t)$ , which can be written as $$H_{1}(p_{1}, x, y, t) := (1 - 3t + 9t^{2} + 9t^{3}) p_{1}^{6} + x ((3 - 2t + 21t^{2}) p_{1}^{4} (1 - p_{1}^{2})) + x^{2} (p_{1}^{2} (1 - p_{1}^{2}) (9 - p_{1}^{2} + 2t^{2} (23 - 5p_{1}^{2}) - t(3 + 9p_{1}^{2}))) + x^{3} (3(3 + 2p_{1}^{2} + p_{1}^{4}) + 2t(-9 + 4p_{1}^{2} - 4p_{1}^{4})) + x^{4} (2p_{1}^{2} (1 - p_{1}^{2})^{2}) + 4(1 - x^{2})p_{1}(1 - p_{1}^{2})((2 - 3t + 9t^{2})p_{1}^{2} + x^{2}(1 - p_{1}^{2}) + x(6 - 7t + 3p_{1}^{2} - 2tp_{1}^{2})) + y^{2} (2(1 - x^{2})(1 - p_{1}^{2})(8(1 - x^{2})(1 - p_{1}^{2}) + 9(x(1 - p_{1}^{2}) + (1 - t)p_{1}^{2})x)) + 18(1 - x^{2})(1 - y^{2})(1 - p_{1}^{2})(x(1 - p_{1}^{2}) + (1 - t)p_{1}^{2}).$$ (3.13) Now our aim is to maximize the function $H_1(p_1, x, y, t)$ over the region $\mathcal{D} := \{(p_1, x, y, t) \in [0, 1] \times [0, 1] \times [0, 1] \times [0, 1/2]\}.$ It is easy to see that the function $H_1$ is a quadratic polynomial in the variable y, since $H_1$ does not have the term with y, only constant term and term with $y^2$ , the maximum occurs at one of the end points either y = 0 or y = 1. We denote the function $R_1(p_1, x, t) :=$ H1(p1, x, 1, t) and R2(p1, x, t) := H1(p1, x, 0, t). $$\begin{split} R_1(p_1,x,t) &:= p_1^6 (1-3t+9t^2+9t^3) + p_1^4 (1-p_1^2)(3-2t+21t^2)x \\ &+ p_1^2 (1-p_1^2) \left(9-p_1^2-(3+9p_1^2)t+2(23-5p_1^2)t^2\right)x^2 \\ &+ (1-p_1^2) \left(3(3+2p_1^2+p_1^4)-2(9-4p_1^2+4p_1^4)t\right)x^3 \\ &+ 2p_1^2 (-1+p_1^2)^2 x^4 + 4p_1 (1-p_1^2)(1-x^2)(p_1^2(2-3t+9t^2)+(6(1-t)+2p_1^2(1-t)+t)x + (1-p_1^2)x^2) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) \left(9x(p_1^2(1-t)+t)x+(1-p_1^2)x^2\right) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x^2) + 2(1-p_1^2)(1-x$$ We claim that $$\max \left\{ R_1(p_1, x, t) : 0 \le p_1 \le 1, \ 0 \le x \le 1, \ 0 \le t \le \frac{1}{2} \right\} = 16.$$ Taking, ^p := ^p1, u := 2t, ^Re(p, x, u) := ^R1(p, x, u/2). We have ⁰ ^≤ ^p ^≤ ¹, ⁰ ^≤ ^x ^≤ ¹, ⁰ ^≤ ^u ^≤ ¹, and ^R^e is a polynomial of tri-degree (6, ⁴, 3) on the unit cube. We first apply the Bernstein method on the whole cube. Write $$\widetilde{R}(p, x, u) = \sum_{i=0}^{6} \sum_{j=0}^{4} \sum_{k=0}^{3} \beta_{i,j,k} B_i^6(p) B_j^4(x) B_k^3(u),$$ where $$B_r^N(s) = \binom{N}{r} s^r (1-s)^{N-r}.$$ By the basic Bernstein enclosure property, $$\widetilde{R}(p, x, u) \le \max_{0 \le i \le 6, \ 0 \le j \le 4, \ 0 \le k \le 3} \beta_{i,j,k}.$$ We now calculate these coefficients. First we write $$\widetilde{R}(p, x, u) = \sum_{r=0}^{6} \sum_{s=0}^{4} \sum_{\ell=0}^{3} \alpha_{r,s,\ell} p^{r} x^{s} u^{\ell}.$$ Next, for each monomial, use the standard conversion formula from the monomial basis to the Bernstein basis $$p^{r} = \sum_{i=r}^{6} \frac{\binom{i}{r}}{\binom{6}{r}} B_{i}^{6}(p), \qquad x^{s} = \sum_{j=s}^{4} \frac{\binom{j}{s}}{\binom{4}{s}} B_{j}^{4}(x), \qquad u^{\ell} = \sum_{k=\ell}^{3} \frac{\binom{k}{\ell}}{\binom{3}{\ell}} B_{k}^{3}(u).$$ Substituting these identities into the monomial expansion of ^R^e and computing like Bernstein terms, we obtain $$\widetilde{R}(p, x, u) = \sum_{i=0}^{6} \sum_{j=0}^{4} \sum_{k=0}^{3} \beta_{i,j,k} B_i^6(p) B_j^4(x) B_k^3(u),$$ where $$\beta_{i,j,k} = \sum_{r=0}^{i} \sum_{s=0}^{j} \sum_{\ell=0}^{k} \alpha_{r,s,\ell} \frac{\binom{i}{r}}{\binom{6}{r}} \frac{\binom{j}{s}}{\binom{4}{s}} \frac{\binom{k}{\ell}}{\binom{3}{\ell}}.$$ For example, the coefficient β1,1,⁰ receives contributions only from the constant term 16 and the term 24px. Hence, $$\beta_{1,1,0} = 16 + 24 \frac{\binom{1}{1}}{\binom{6}{1}} \frac{\binom{1}{1}}{\binom{4}{1}} = 16 + 24 \cdot \frac{1}{6} \cdot \frac{1}{4} = 17.$$ In the same way we compute all the Bernstein coefficients βi,j,k. For convenience, write $$M_k = (\beta_{i,j,k})_{0 \le i \le 6, \ 0 \le j \le 4}, \qquad k = 0, 1, 2, 3,$$ where the rows correspond to i = 0, 1, . . . , 6 and the columns correspond to j = 0, 1, . . . , 4. The exact Bernstein coefficients are as follows. For k = 0, $$M_0 = \begin{pmatrix} 16 & 16 & \frac{41}{3} & \frac{45}{4} & 9 \\ 16 & 17 & \frac{142}{9} & \frac{163}{12} & 9 \\ \frac{208}{15} & \frac{97}{6} & \frac{503}{30} & \frac{467}{30} & \frac{143}{15} \\ 10 & \frac{137}{10} & \frac{33}{2} & \frac{84}{5} & \frac{53}{5} \\ \frac{88}{15} & \frac{637}{60} & \frac{1363}{90} & \frac{333}{20} & \frac{169}{15} \\ \frac{8}{3} & \frac{85}{12} & \frac{23}{2} & \frac{53}{4} & \frac{29}{3} \\ 1 & 1 & 1 & 1 & 1 \end{pmatrix}$$ For k = 1, we have $$M_{1} = \begin{pmatrix} 16 & 16 & \frac{41}{3} & \frac{21}{2} & 6 \\ 16 & \frac{607}{36} & \frac{31}{2} & \frac{113}{9} & 6 \\ \\ \frac{208}{15} & \frac{2851}{180} & \frac{2899}{180} & \frac{853}{60} & \frac{611}{90} \\ \\ \frac{99}{10} & \frac{1567}{120} & \frac{919}{60} & \frac{181}{12} & \frac{251}{30} \\ \\ \frac{82}{15} & \frac{1711}{180} & \frac{1201}{90} & \frac{523}{36} & \frac{143}{15} \\ \\ 2 & \frac{23}{4} & \frac{19}{2} & \frac{395}{36} & \frac{71}{9} \\ \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ For k = 2, we have $$M_2 = \begin{pmatrix} 16 & 16 & \frac{41}{3} & \frac{39}{4} & 3 \\ 16 & \frac{301}{18} & \frac{137}{9} & \frac{415}{36} & 3 \\ \\ \frac{208}{15} & \frac{698}{45} & \frac{8363}{540} & \frac{2339}{180} & \frac{43}{10} \\ \\ \frac{199}{20} & \frac{377}{30} & \frac{5179}{360} & \frac{553}{40} & \frac{69}{10} \\ \\ \frac{17}{3} & \frac{6497}{720} & \frac{13291}{1080} & \frac{3217}{240} & \frac{329}{36} \\ \\ \frac{7}{3} & \frac{89}{16} & \frac{211}{24} & \frac{1459}{144} & \frac{277}{36} \\ \\ \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} \end{pmatrix}$$ For k = 3, we have $$M_3 = \begin{pmatrix} 16 & 16 & \frac{41}{3} & 9 & 0 \\ 16 & \frac{199}{12} & \frac{269}{18} & \frac{21}{2} & 0 \\ \\ \frac{208}{15} & \frac{911}{60} & \frac{671}{45} & \frac{119}{10} & \frac{31}{15} \\ \\ \frac{203}{20} & \frac{489}{40} & \frac{329}{24} & \frac{521}{40} & \frac{31}{5} \\ \\ \frac{97}{15} & \frac{2201}{240} & \frac{4331}{360} & \frac{3187}{240} & \frac{121}{12} \\ \\ \frac{11}{3} & \frac{313}{48} & \frac{75}{8} & \frac{515}{48} & \frac{109}{12} \\ \\ \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} \end{pmatrix}$$ Hence, by direct inspection of the above coefficients, we obtain $$\max_{0 \le i \le 6, \ 0 \le j \le 4, \ 0 \le k \le 3} \beta_{i,j,k} = 17, \quad \text{attained at } \beta_{1,1,0} = 17.$$ Thus the whole-cube Bernstein step gives only $$\widetilde{R}(p, x, u) \le 17,$$ which is not yet sufficient as per our claim. We therefore subdivide only in the p- and x-directions. For 0 ≤ i, j ≤ 7, let $$Q_{ij} := \left[\frac{i}{8}, \frac{i+1}{8}\right] \times \left[\frac{j}{8}, \frac{j+1}{8}\right] \times [0, 1].$$ On each Qij , perform the affine change of variables $$p = \frac{i+\xi}{8}, \qquad x = \frac{j+\eta}{8}, \qquad u = \zeta, \qquad (\xi, \eta, \zeta) \in [0, 1] \times [0, 1],$$ and expand the transformed polynomial in the Bernstein basis of degrees (6, 4, 3), $$\widetilde{R}\left(\frac{i+\xi}{8}, \frac{j+\eta}{8}, \zeta\right) = \sum_{a=0}^{6} \sum_{b=0}^{4} \sum_{c=0}^{3} \beta_{a,b,c}^{(ij)} B_a^6(\xi) B_b^4(\eta) B_c^3(\zeta).$$ Take, $$M_{ij} := \max_{a,b,c} \beta_{a,b,c}^{(ij)}.$$ For convenience, we list all the values Mij explicitly, $$\begin{array}{llllllllllllllllllllllllllllllllllll$$ $$M_{70} = \frac{2498276837}{536870912}, \quad M_{71} = \frac{180167627}{33554432}, \quad M_{72} = \frac{3222265757}{536870912}, \quad M_{73} = \frac{3494483}{524288},$$ $M_{74} = \frac{3828389717}{536870912}, \quad M_{75} = \frac{973980383}{134217728}, \quad M_{76} = \frac{242567447}{33554432}, \quad M_{77} = \frac{3696095117}{536870912}.$ Equivalently, each entry in the (i, j)-position of the above list is precisely the number Mij . Again by the consequence of Bernstein method, we have $$\widetilde{R}(p, x, u) \le M_{ij}$$ on $Q_{ij}$ . Now the exact rational computation gives $$M_{00} = \frac{1025}{64},$$ and, for all (i, j) ̸= (0, 0), $$M_{ij} < 16.$$ In fact, $$\max_{(i,j)\neq(0,0)} M_{ij} = \frac{83341001}{5242880} < 16.$$ Therefore the Bernstein method already proves $$\widetilde{R}(p, x, u) \le 16$$ on the union of all Qij except $$Q_{00} = \left[0, \frac{1}{8}\right] \times \left[0, \frac{1}{8}\right] \times [0, 1].$$ So it remains only to show that $$\widetilde{R}(p, x, u) \le 16$$ for $(p, x, u) \in Q_{00}$ . Set, $$F(p, x, u) := 16 - \widetilde{R}(p, x, u).$$ We write F as a polynomial in x: $$F(p, x, u) = a_0(p, u) + a_1(p, u)x + a_2(p, u)x^2 + a_3(p, u)x^3 + a_4(p)x^4,$$ where $$a_0(p, u) = \frac{p^2}{8} \left( 256 - 64p - 128p^2 + 64p^3 - 8p^4 + 48pu - 48p^3u + 12p^4u - 72pu^2 + 72p^3u^2 - 18p^4u^2 - 9p^4u^3 \right),$$ $$a_1(p,u) = -\frac{p(1-p^2)}{4} \Big( 96 - 40u + 72p - 36pu + 32p^2 - 16p^2u + 12p^3 - 4p^3u + 21p^3u^2 \Big),$$ $$a_2(p, u) = 14 - 4p - 37p^2 + \left(\frac{3}{2}u - \frac{23}{2}u^2\right)p^2 + (16 - 6u + 9u^2)p^3 + (24 + 3u + 14u^2)p^4 + (-12 + 6u - 9u^2)p^5 + \left(-1 - \frac{9}{2}u - \frac{5}{2}u^2\right)p^6,$$ $$a_3(p, u) = -9 + 9u + (24 - 10u)p + (21 - 22u)p^2 + (-16 + 6u)p^3$$ $$+(-15+17u)p^4+(-8+4u)p^5+(3-4u)p^6,$$ $$a_4(p) = 2 + 4p - 6p^2 - 8p^3 + 6p^4 + 4p^5 - 2p^6.$$ We now estimate these coefficients on 0 ≤ p ≤ 1/8, 0 ≤ x ≤ 1/8, 0 ≤ u ≤ 1. For a0, write $$a_0(p, u) = \frac{p^2}{8}A(p, u),$$ where A(p, u) = 256+p(−64+48u−72u 2 )−128p ²+p 3 (64−48u+72u 2 )+p 4 (−8+12u−18u ²−9u 3 ). For 0 ≤ u ≤ 1, $$-64 + 48u - 72u^{2} \ge -88,$$ $$64 - 48u + 72u^{2} \ge 56,$$ $$-8 + 12u - 18u^{2} - 9u^{3} \ge -23.$$ Hence $$A(p, u) \ge 256 - 88p - 128p^2 + 56p^3 - 23p^4 := O(p).$$ The right-hand side is decreasing on -0, 1/8 , because its derivative is $$O'(p) = -88 - 256p + 168p^2 - 92p^3 < 0.$$ Therefore $$A(p,u) \ge 256 - \frac{88}{8} - \frac{128}{64} + \frac{56}{512} - \frac{23}{4096} = \frac{995753}{4096} > 240.$$ So, we have a0(p, u) ≥ 30p 2 . For a1, let $$B(p,u) := 96 - 40u + 72p - 36pu + 32p^2 - 16p^2u + 12p^3 - 4p^3u + 21p^3u^2.$$ Since 0 ≤ u ≤ 1 and 0 ≤ p ≤ 1/8, $$B(p,u) \le 96 + 72p + 32p^2 + 33p^3 \le 96 + 9 + \frac{1}{2} + \frac{33}{512} < 108.$$ Hence $$a_1(p,u) \ge -\frac{108}{4}p = -27p.$$ For a2, we use the bounds $$\frac{3}{2}u - \frac{23}{2}u^2 \ge -\frac{23}{2}, \qquad 16 - 6u + 9u^2 \ge 15,$$ also $$24 + 3u + 14u^2 \ge 24$$ , $-12 + 6u - 9u^2 \ge -15$ , $-1 - \frac{9}{2}u - \frac{5}{2}u^2 \ge -8$ . Therefore $$a_2(p,u) \ge 14 - 4p - \frac{97}{2}p^2 + 15p^3 + 24p^4 - 15p^5 - 8p^6 := L(p).$$ The right-hand side is decreasing on -0, 1/8 , since its derivative is $$L'(p) = -4 - 97p + 45p^2 + 96p^3 - 75p^4 - 48p^5 < 0.$$ Hence $$a_2(p,u) \ge 14 - \frac{4}{8} - \frac{97}{2 \cdot 64} + \frac{15}{512} + \frac{24}{4096} - \frac{15}{32768} - \frac{8}{262144} = \frac{26167}{2048} > \frac{51}{4}.$$ For a3, observe that $$a_3(p, u) + 9 = u(9 - 10p - 22p^2 + 6p^3 + 17p^4 + 4p^5 - 4p^6) +$$ $$p(24 + 21p - 16p^2 - 15p^3 - 8p^4 + 3p^5).$$ Now $$9 - 10p - 22p^{2} + 6p^{3} + 17p^{4} + 4p^{5} - 4p^{6} \ge 9 - 10p - 22p^{2} > 9 - \frac{10}{8} - \frac{22}{64} = \frac{237}{32} > 0,$$ and $$24 + 21p - 16p^2 - 15p^3 - 8p^4 + 3p^5 \ge 24 - 16p^2 - 15p^3 - 8p^4 > 24 - \frac{1}{4} - \frac{15}{512} - \frac{1}{64} > 0.$$ Thus a3(p, u) ≥ −9. Finally, $$a_4(p) = 2 + 4p - 6p^2 - 8p^3 + 6p^4 + 4p^5 - 2p^6 \ge 2 + 4p - 6p^2 - 8p^3.$$ Since $$4p - 6p^2 - 8p^3 = 2p(2 - 3p - 4p^2) \ge 0$$ for $0 \le p \le 1/8$ , we get a4(p) ≥ 2. Combining the coefficient bounds, we obtain on Q00, $$F(p, x, u) \ge 30p^2 - 27px + \frac{51}{4}x^2 - 9x^3 + 2x^4.$$ As 0 ≤ x ≤ 1/8, we have −9x ³ ≥ −9/8x 2 , and 2x ⁴ ≥ 0, and hence $$F(p, x, u) \ge 30p^2 - 27px + \left(\frac{51}{4} - \frac{9}{8}\right)x^2 = 30p^2 - 27px + \frac{93}{8}x^2 := V(p, x).$$ For each fixed x, this is a quadratic polynomial in p with leading coefficient 30, which is a positive quantity and the discriminant is (−27x) 2 (−15) · 93x ² = −666x ² ≤ 0. Hence V (p, x) ≥ 0, thus, $$F(p, x, u) \ge 0 \quad \text{ on } Q_{00}.$$ So $$\widetilde{R}(p, x, u) \le 16$$ on $Q_{00}$ . We have thus proved $$\widetilde{R}(p, x, u) \le 16$$ for all $(p, x, u) \in [0, 1] \times [0, 1] \times [0, 1]$ . Returning to t = u/2, we conclude that $$R_1(p_1, x, t) \le 16$$ for $0 \le p_1 \le 1, \ 0 \le x \le 1, \ 0 \le t \le 1/2$ . To see that this bound is indeed sharp, evaluate at the point p¹ = 0, x = 0. Then $$R_1(0,0,t) = 16$$ for every $0 \le t \le 1/2$ . Hence the maximum is exactly 16. Now we consider the part where y = 0, and the function is defined as R2(p1, x, t) := H1(p1, x, 0, t). $$R_{2}(p_{1}, x, t) := p_{1}^{6}(1 - 3t + 9t^{2} + 9t^{3}) + p_{1}^{4}(1 - p_{1}^{2})(3 - 2t + 21t^{2})x$$ $$+ p_{1}^{2}(1 - p_{1}^{2})\left(9 - p_{1}^{2} - (3 + 9p_{1}^{2})t + 2(23 - 5p_{1}^{2})t^{2}\right)x^{2}$$ $$+ (1 - p_{1}^{2})\left(3(3 + 2p_{1}^{2} + p_{1}^{4}) - 2(9 - 4p_{1}^{2} + 4p_{1}^{4})t\right)x^{3}$$ $$+ 2p_{1}^{2}(-1 + p_{1}^{2})^{2}x^{4}$$ $$+ 18(1 - p_{1}^{2})\left(p_{1}^{2}(1 - t) + (1 - p_{1}^{2})x\right)(1 - x^{2})$$ $$+ 4p_{1}(1 - p_{1}^{2})(1 - x^{2})\left(p_{1}^{2}(2 - 3t + 9t^{2})\right)$$ $$+ (6(1 - t) + 2p_{1}^{2}(1 - t) + t)x + (1 - p_{1}^{2})x^{2}\right).$$ For this case we seek an upper bound in the region 0 ≤ p¹ ≤ 1, 0 ≤ x ≤ 1, 0 ≤ t ≤ 1/2. Let us take, $$p := p_1, \quad u := 2t, \quad \widetilde{R}_2(p, x, u) := R_2(p, x, \frac{u}{2}).$$ Then R := {(p, x, u) ∈ R : (p, x, u) ∈ [0, 1] × [0, 1] × [0, 1]}, and $$\begin{split} \widetilde{R}_2(p,x,u) &= \frac{9}{8} p^6 u^3 + \frac{5}{2} p^6 u^2 x^2 - \frac{21}{4} p^6 u^2 x + \frac{9}{4} p^6 u^2 + 4 p^6 u x^3 + \frac{9}{2} p^6 u x^2 + p^6 u x \\ &- \frac{3}{2} p^6 u + 2 p^6 x^4 - 3 p^6 x^3 + p^6 x^2 - 3 p^6 x + p^6 + 9 p^5 u^2 x^2 - 9 p^5 u^2 \\ &- 4 p^5 u x^3 - 6 p^5 u x^2 + 4 p^5 u x + 6 p^5 u - 4 p^5 x^4 + 8 p^5 x^3 + 12 p^5 x^2 \\ &- 8 p^5 x - 8 p^5 - 14 p^4 u^2 x^2 + \frac{21}{4} p^4 u^2 x - 8 p^4 u x^3 - 12 p^4 u x^2 \\ &- p^4 u x + 9 p^4 u - 4 p^4 x^4 - 21 p^4 x^3 + 8 p^4 x^2 + 21 p^4 x - 18 p^4 - 9 p^3 u^2 x^2 \\ &+ 9 p^3 u^2 - 6 p^3 u x^3 + 6 p^3 u x^2 + 6 p^3 u x - 6 p^3 u + 8 p^3 x^4 + 16 p^3 x^3 \\ &- 16 p^3 x^2 - 16 p^3 x + 8 p^3 + \frac{23}{2} p^2 u^2 x^2 + 13 p^2 u x^3 + \frac{15}{2} p^2 u x^2 - 9 p^2 u \\ &+ 2 p^2 x^4 + 33 p^2 x^3 - 9 p^2 x^2 - 36 p^2 x + 18 p^2 + 10 p u x^3 - 10 p u x - 4 p x^4 \\ &- 24 p x^3 + 4 p x^2 + 24 p x - 9 u x^3 - 9 x^3 + 18 x. \end{split}$$ We first expand ^Re² in the tensor-product Bernstein basis of degree (6, ⁴, 3): $$\widetilde{R}_2(p, x, u) = \sum_{i=0}^6 \sum_{j=0}^4 \sum_{k=0}^3 \beta_{i,j,k} B_i^6(p) B_j^4(x) B_k^3(u),$$ where $$B_r^N(s) = \binom{N}{r} s^r (1-s)^{N-r}.$$ Write $$M_k = (\beta_{i,j,k})_{0 \le i \le 6, \ 0 \le j \le 4}, \qquad k = 0, 1, 2, 3.$$ The rows correspond to i = 0, 1, . . . , 6, and the columns correspond to j = 0, 1, . . . , 4. For k = 0, $$M_0 = \begin{pmatrix} 0 & \frac{9}{2} & 9 & \frac{45}{4} & 9 \\ 0 & \frac{11}{2} & \frac{100}{9} & \frac{163}{12} & 9 \\ \frac{6}{5} & \frac{71}{10} & \frac{1181}{90} & \frac{467}{30} & \frac{143}{15} \\ 4 & \frac{19}{2} & \frac{149}{10} & \frac{84}{5} & \frac{53}{5} \\ \frac{38}{5} & \frac{241}{20} & \frac{159}{10} & \frac{333}{20} & \frac{169}{15} \\ \frac{26}{3} & \frac{139}{12} & \frac{27}{2} & \frac{53}{4} & \frac{29}{3} \\ 1 & 1 & 1 & 1 & 1 \end{pmatrix}.$$ For k = 1, $$M_1 = \begin{pmatrix} 0 & \frac{9}{2} & 9 & \frac{21}{2} & 6 \\ 0 & \frac{193}{36} & \frac{65}{6} & \frac{113}{9} & 6 \\ 1 & \frac{298}{45} & \frac{2231}{180} & \frac{853}{60} & \frac{611}{90} \\ \frac{33}{10} & \frac{1009}{120} & \frac{811}{60} & \frac{181}{12} & \frac{251}{30} \\ \frac{31}{5} & \frac{917}{90} & \frac{413}{30} & \frac{523}{36} & \frac{143}{15} \\ 7 & \frac{19}{2} & \frac{67}{6} & \frac{395}{36} & \frac{71}{9} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$ For k = 2, $$M_2 = \begin{pmatrix} 0 & \frac{9}{2} & 9 & \frac{39}{4} & 3 \\ 0 & \frac{47}{9} & \frac{95}{9} & \frac{415}{36} & 3 \\ \frac{4}{5} & \frac{553}{90} & \frac{6323}{540} & \frac{2339}{180} & \frac{43}{10} \\ \frac{11}{4} & \frac{112}{15} & \frac{4459}{360} & \frac{553}{40} & \frac{69}{10} \\ \frac{27}{5} & \frac{6449}{720} & \frac{13387}{1080} & \frac{3217}{240} & \frac{329}{36} \\ \frac{19}{3} & \frac{137}{16} & \frac{81}{8} & \frac{1459}{144} & \frac{277}{36} \\ \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} \end{pmatrix}$$ . For k = 3, $$M_3 = \begin{pmatrix} 0 & \frac{9}{2} & 9 & 9 & 0 \\ 0 & \frac{61}{12} & \frac{185}{18} & \frac{21}{2} & 0 \\ & \frac{3}{5} & \frac{17}{3} & \frac{166}{15} & \frac{119}{10} & \frac{31}{15} \\ & \frac{47}{20} & \frac{267}{40} & \frac{1381}{120} & \frac{521}{40} & \frac{31}{5} \\ & \frac{26}{5} & \frac{401}{48} & \frac{4243}{360} & \frac{3187}{240} & \frac{121}{12} \\ & \frac{20}{3} & \frac{421}{48} & \frac{83}{8} & \frac{515}{48} & \frac{109}{12} \\ & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} \end{pmatrix}$$ By the Bernstein enclosure property, $$\widetilde{R}_2(p, x, u) \le \max_{0 \le i \le 6, \ 0 \le j \le 4, \ 0 \le k \le 3} \beta_{i,j,k}.$$ From the matrices above, the largest coefficient is $$\beta_{3,3,0} = \frac{84}{5}.$$ Hence $$\widetilde{R}_2(p, x, u) \le \frac{84}{5}$$ for all $(p, x, u) \in [0, 1] \times [0, 1] \times [0, 1]$ . Returning to u = 2t, we obtain $$R_2(p_1, x, t) \le \frac{84}{5}$$ for all $0 \le p_1 \le 1, \ 0 \le x \le 1, \ 0 \le t \le 1/2$ . We now subdivide the (p1, x)-rectangle into four parts and apply the Bernstein method again on each part, keeping 0 ≤ t ≤ 1/2 unchanged. We have the following four subregions of the region R. $$Q_{11} = \left[0, \frac{1}{2}\right] \times \left[0, \frac{1}{2}\right] \times [0, 1], \qquad Q_{12} = \left[0, \frac{1}{2}\right] \times \left[\frac{1}{2}, 1\right] \times [0, 1],$$ $$Q_{21} = \left[\frac{1}{2}, 1\right] \times \left[0, \frac{1}{2}\right] \times [0, 1], \qquad Q_{22} = \left[\frac{1}{2}, 1\right] \times \left[\frac{1}{2}, 1\right] \times [0, 1].$$ The corresponding affine changes are $$Q_{11}: \quad p = \frac{P}{2}, \quad x = \frac{X}{2},$$ $$Q_{12}: \quad p = \frac{P}{2}, \quad x = \frac{1+X}{2},$$ $$Q_{21}: \quad p = \frac{1+P}{2}, \quad x = \frac{X}{2},$$ $$Q_{22}: \quad p = \frac{1+P}{2}, \quad x = \frac{1+X}{2}.$$ For each subbox we write the transformed polynomial in the Bernstein basis of degree (6, 4, 3), $$\widetilde{R}_{2}^{(\nu)}(P, X, U) = \sum_{i=0}^{6} \sum_{j=0}^{4} \sum_{k=0}^{3} \beta_{i,j,k}^{(\nu)} B_{i}^{6}(P) B_{j}^{4}(X) B_{k}^{3}(U),$$ where ν ∈ {11, 12, 21, 22}. For Q11, the coefficient matrices M (11) ^k = (β (11) i,j,k) are $$M_0^{(11)} = \begin{pmatrix} 0 & \frac{9}{4} & \frac{9}{2} & \frac{207}{32} & \frac{63}{8} \\ 0 & \frac{5}{2} & \frac{361}{72} & \frac{691}{96} & \frac{139}{16} \\ \\ \frac{3}{10} & \frac{119}{40} & \frac{8167}{1440} & \frac{15361}{1920} & \frac{143}{15} \\ \\ \frac{19}{20} & \frac{297}{80} & \frac{1039}{160} & \frac{5683}{640} & \frac{3329}{320} \\ \\ \frac{77}{40} & \frac{2999}{640} & \frac{4291}{576} & \frac{75247}{7680} & \frac{4321}{384} \\ \\ \frac{37}{12} & \frac{737}{128} & \frac{9685}{1152} & \frac{16325}{1536} & \frac{9185}{768} \\ \\ \frac{265}{64} & \frac{3401}{512} & \frac{4653}{512} & \frac{22745}{2048} & \frac{1571}{128} \end{pmatrix}$$ $$M_1^{(11)} = \begin{pmatrix} 0 & \frac{9}{4} & \frac{9}{2} & \frac{51}{8} & \frac{15}{2} \\ 0 & \frac{355}{144} & \frac{89}{18} & \frac{4037}{576} & \frac{197}{24} \\ \frac{1}{4} & \frac{257}{90} & \frac{351}{64} & \frac{22099}{2880} & \frac{3211}{360} \\ \frac{63}{80} & \frac{6619}{1920} & \frac{11773}{1920} & \frac{4283}{512} & \frac{18469}{1920} \\ \frac{127}{80} & \frac{1349}{320} & \frac{823}{120} & \frac{69503}{7680} & \frac{19709}{1920} \\ \frac{81}{32} & \frac{5815}{1152} & \frac{17413}{2304} & \frac{22175}{2304} & \frac{24775}{2304} \\ \frac{433}{128} & \frac{731}{128} & \frac{8199}{1024} & \frac{20231}{2048} & \frac{5571}{512} \end{pmatrix}$$ $$M_2^{(11)} = \begin{pmatrix} 0 & \frac{9}{4} & \frac{9}{2} & \frac{201}{32} & \frac{57}{8} \\ 0 & \frac{175}{72} & \frac{39}{8} & \frac{491}{72} & \frac{371}{48} \\ \frac{1}{5} & \frac{197}{72} & \frac{45791}{8640} & \frac{42359}{5760} & \frac{11983}{1440} \\ \frac{1}{60} & \frac{3073}{960} & \frac{66767}{11520} & \frac{15151}{1920} & \frac{17083}{1920} \\ \frac{53}{40} & \frac{29347}{7680} & \frac{219659}{34560} & \frac{9695}{1152} & \frac{13577}{1440} \\ \frac{413}{192} & \frac{20801}{4608} & \frac{31817}{4608} & \frac{40793}{4608} & \frac{11323}{1152} \\ \frac{747}{256} & \frac{10409}{2048} & \frac{3715}{512} & \frac{9241}{1024} & \frac{635}{64} \end{pmatrix}$$ , $$M_3^{(11)} = \begin{pmatrix} 0 & \frac{9}{4} & \frac{9}{2} & \frac{99}{16} & \frac{27}{4} \\ 0 & \frac{115}{48} & \frac{173}{36} & \frac{1273}{192} & \frac{29}{4} \\ \\ \frac{3}{20} & \frac{157}{60} & \frac{737}{144} & \frac{6761}{960} & \frac{743}{96} \\ \\ \frac{83}{160} & \frac{1903}{640} & \frac{4213}{768} & \frac{19091}{2560} & \frac{659}{80} \\ \\ \frac{91}{80} & \frac{8967}{2560} & \frac{13691}{2304} & \frac{30319}{3840} & \frac{8393}{960} \\ \\ \frac{373}{192} & \frac{6385}{1536} & \frac{29713}{4608} & \frac{133}{16} & \frac{147}{16} \\ \\ \frac{1415}{512} & \frac{9779}{2048} & \frac{7017}{1024} & \frac{8767}{1024} & \frac{1205}{128} \end{pmatrix}$$ It is easy to see that $$\max_{i,j,k} \beta_{i,j,k}^{(11)} = \frac{1571}{128},$$ and this occurs in M (11) , last row and last column, that is, at $$\beta_{6,4,0}^{(11)} = \frac{1571}{128}.$$ Hence $$\widetilde{R}_2^{(11)}(P, X, U) \le \frac{1571}{128},$$ so equivalently $$R_2 \le \frac{1571}{128} \approx 12.2734375$$ on $Q_{11}$ . Therefore $$R_2(p, x, t) \le \max_{i,j,k} \beta_{i,j,k}^{(11)}$$ for all $(p, x, t) \in Q_{11}$ . For Q12, the coefficient matrices M (12) ^k = (β (12) i,j,k) are $$M_0^{(12)} = \begin{pmatrix} \frac{63}{8} & \frac{297}{32} & \frac{81}{8} & \frac{81}{8} & 9 \\ \frac{139}{16} & \frac{977}{96} & \frac{395}{36} & \frac{257}{24} & 9 \\ \frac{143}{15} & \frac{21247}{1920} & \frac{4249}{360} & \frac{5431}{480} & \frac{137}{15} \\ \frac{3329}{320} & \frac{7633}{640} & \frac{1007}{80} & \frac{1907}{160} & \frac{47}{5} \\ \frac{4321}{384} & \frac{32531}{2560} & \frac{76429}{5760} & \frac{3987}{320} & \frac{1169}{120} \\ \frac{9185}{768} & \frac{6805}{512} & \frac{3955}{288} & \frac{1231}{96} & \frac{241}{24} \\ \frac{1571}{128} & \frac{27527}{2048} & \frac{1761}{128} & \frac{3277}{256} & \frac{323}{32} \end{pmatrix}$$ , $$M_3^{(12)} = \begin{pmatrix} \frac{27}{4} & \frac{117}{16} & \frac{27}{4} & \frac{9}{2} & 0 \\ \frac{29}{4} & \frac{1511}{192} & \frac{1049}{144} & \frac{39}{8} & 0 \\ \frac{743}{96} & \frac{8099}{960} & \frac{1423}{180} & \frac{1319}{240} & \frac{31}{60} \\ \frac{659}{80} & \frac{4617}{512} & \frac{33047}{3840} & \frac{813}{128} & \frac{31}{20} \\ \frac{8393}{960} & \frac{2455}{256} & \frac{107491}{11520} & \frac{3761}{512} & \frac{2837}{960} \\ \frac{147}{16} & \frac{161}{16} & \frac{45841}{4608} & \frac{12721}{1536} & \frac{853}{192} \\ \frac{1205}{128} & \frac{10513}{1024} & \frac{10509}{1024} & \frac{18257}{2048} & \frac{2909}{512} \end{pmatrix}$$ Therefore, $$R_2(p, x, t) \le \max_{i,j,k} \beta_{i,j,k}^{(12)}$$ for all $(p, x, t) \in Q_{12}$ , thus, we have $$\max_{Q_{12}} R_2 \le \max_{i,j,k} \beta_{i,j,k}^{(12)} = \frac{1761}{128} \approx 13.7578125.$$ For Q21, the coefficient matrices M (21) ^k = (β (21) i,j,k) are $$M_0^{(21)} = \begin{pmatrix} \frac{265}{64} & \frac{3401}{512} & \frac{4653}{512} & \frac{22745}{2048} & \frac{1571}{128} \\ \frac{499}{96} & \frac{1927}{256} & \frac{22507}{2304} & \frac{35585}{3072} & \frac{9667}{768} \\ \\ \frac{1477}{240} & \frac{329}{40} & \frac{11719}{1152} & \frac{14987}{1280} & \frac{1601}{128} \\ \\ \frac{269}{40} & \frac{169}{20} & \frac{3207}{320} & \frac{3589}{320} & \frac{3771}{320} \\ \\ \frac{389}{60} & \frac{1243}{160} & \frac{1067}{120} & \frac{9313}{960} & \frac{803}{80} \\ \\ \frac{29}{6} & \frac{89}{16} & \frac{37}{6} & \frac{631}{96} & \frac{161}{24} \\ \\ 1 & 1 & 1 & 1 & 1 \end{pmatrix}$$ $$M_1^{(21)} = \begin{pmatrix} \frac{433}{128} & \frac{731}{128} & \frac{8199}{1024} & \frac{20231}{2048} & \frac{5571}{512} \\ \frac{271}{64} & \frac{7343}{1152} & \frac{38965}{4608} & \frac{93379}{9216} & \frac{6341}{576} \\ \frac{799}{160} & \frac{19781}{2880} & \frac{99703}{11520} & \frac{7247}{720} & \frac{7759}{720} \\ \frac{217}{40} & \frac{891}{128} & \frac{1609}{192} & \frac{72703}{7680} & \frac{3833}{384} \\ \frac{207}{40} & \frac{907}{144} & \frac{10501}{1440} & \frac{46177}{5760} & \frac{11987}{1440} \\ \frac{15}{2} & \frac{35}{4} & \frac{235}{8} & \frac{235}{48} & \frac{3023}{576} & \frac{773}{1444} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ $$M_2^{(21)} = \begin{pmatrix} \frac{747}{256} & \frac{10409}{2048} & \frac{3715}{512} & \frac{9241}{1024} & \frac{635}{64} \\ \frac{1415}{384} & \frac{52079}{9216} & \frac{35053}{4608} & \frac{5297}{576} & \frac{11537}{1152} \\ \frac{4217}{960} & \frac{70213}{11520} & \frac{268199}{4608} & \frac{6991}{68} & \frac{49}{5} \\ \frac{4217}{960} & \frac{70213}{11520} & \frac{268199}{34560} & \frac{6991}{68} & \frac{49}{5} \\ \frac{4217}{240} & \frac{70213}{5760} & \frac{268199}{8640} & \frac{6991}{13520} & \frac{49}{640} \\ \frac{1129}{240} & \frac{32867}{5760} & \frac{57173}{8640} & \frac{2807}{384} & \frac{3671}{480} \\ \frac{85}{24} & \frac{787}{192} & \frac{439}{96} & \frac{2827}{576} & \frac{727}{144} \\ \frac{3}{3} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} \end{pmatrix}$$ $$\begin{pmatrix} \frac{1415}{512} & \frac{9779}{2048} & \frac{7017}{1024} & \frac{8767}{1024} & \frac{1205}{128} \\ \frac{2753}{768} & \frac{16567}{3072} & \frac{1045}{144} & \frac{4511}{512} & \frac{617}{64} \\ \frac{8489}{1920} & \frac{22943}{3840} & \frac{2903}{3840} & \frac{1067}{120} & \frac{9263}{960} \\ \frac{1920}{480} & \frac{380}{1920} & \frac{4117}{576} & \frac{7547}{960} & \frac{3979}{480} \\ \frac{480}{1920} & \frac{12043}{576} & \frac{4117}{768} & \frac{7547}{596} & \frac{3979}{480} \\ \frac{480}{1920} & \frac{229}{3804} & \frac{364}{96} & \frac{96}{96} & \frac{101}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{587}{96} & \frac{101}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{587}{96} & \frac{101}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{587}{96} & \frac{101}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{587}{96} & \frac{101}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{53}{96} & \frac{106}{16} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96} & \frac{339}{96} & \frac{339}{96} & \frac{339}{96} \\ \frac{229}{48} & \frac{339}{64} & \frac{553}{96}$$ Therefore $$R_2(p, x, t) \le \max_{i,j,k} \beta_{i,j,k}^{(21)}$$ for all $(p, x, t) \in Q_{21}$ . Thus, we have $$\max_{Q_{21}} R_2 \le \frac{9667}{768} \approx 12.5872395833.$$ For Q22, the coefficient matrices M (22) ^k = (β (22) i,j,k) are $$M_0^{(22)} = \begin{pmatrix} \frac{1571}{128} & \frac{27527}{2048} & \frac{1761}{128} & \frac{3277}{256} & \frac{323}{32} \\ \frac{9667}{768} & \frac{13917}{1024} & \frac{7939}{576} & \frac{4907}{384} & \frac{487}{48} \\ \\ \frac{1601}{128} & \frac{17033}{1280} & \frac{77009}{5760} & \frac{2969}{240} & \frac{199}{20} \\ \\ \frac{3771}{320} & \frac{3953}{320} & \frac{787}{64} & \frac{1819}{160} & \frac{93}{10} \\ \\ \frac{803}{80} & \frac{9959}{960} & \frac{819}{80} & \frac{303}{32} & \frac{79}{10} \\ \\ \frac{161}{24} & \frac{219}{32} & \frac{161}{24} & \frac{299}{48} & \frac{16}{3} \\ \\ 1 & 1 & 1 & 1 & 1 \end{pmatrix}$$ , , $$M_{1}^{(22)} = \begin{pmatrix} \frac{5571}{512} & \frac{24337}{2048} & \frac{12305}{1024} & \frac{1387}{128} & \frac{251}{32} \\ \frac{6341}{576} & \frac{36511}{3072} & \frac{18373}{1536} & \frac{347}{32} & \frac{1165}{144} \\ \\ \frac{7759}{720} & \frac{919}{80} & \frac{14719}{1280} & \frac{20137}{1920} & \frac{1457}{180} \\ \\ \frac{3833}{384} & \frac{80617}{7680} & \frac{20047}{1920} & \frac{9193}{960} & \frac{917}{120} \\ \\ \frac{11987}{1440} & \frac{16573}{1920} & \frac{767}{90} & \frac{3767}{480} & \frac{2323}{360} \\ \\ \frac{773}{144} & \frac{3161}{576} & \frac{43}{8} & \frac{715}{144} & \frac{151}{36} \\ \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ $$M_2^{(22)} = \begin{pmatrix} \frac{635}{64} & \frac{11079}{1024} & \frac{5553}{512} & \frac{19527}{2048} & \frac{815}{128} \\ \frac{11537}{1152} & \frac{65}{6} & \frac{50141}{4608} & \frac{9989}{1024} & \frac{2017}{288} \\ \frac{49}{5} & \frac{40309}{3840} & \frac{364571}{34560} & \frac{36991}{3840} & \frac{21287}{2880} \\ \frac{17473}{1920} & \frac{18497}{1920} & \frac{111317}{11520} & \frac{215}{24} & \frac{1163}{160} \\ \frac{3671}{480} & \frac{5111}{640} & \frac{13771}{1728} & \frac{14327}{1920} & \frac{455}{72} \\ \frac{727}{144} & \frac{2989}{576} & \frac{493}{96} & \frac{2783}{576} & \frac{38}{9} \\ \frac{3}{3} & \frac{3}{3} & \frac{3}{3} & \frac{3}{3} & \frac{3}{3} & \frac{3}{3} \end{pmatrix}$$ $$M_3^{(22)} = \begin{pmatrix} \frac{1205}{128} & \frac{10513}{1024} & \frac{10509}{1024} & \frac{18257}{2048} & \frac{2909}{512} \\ \frac{617}{64} & \frac{5361}{512} & \frac{12185}{1152} & \frac{29329}{3072} & \frac{5315}{768} \\ \frac{9263}{960} & \frac{333}{32} & \frac{6777}{640} & \frac{7585}{768} & \frac{5063}{640} \\ \\ \frac{297}{32} & \frac{5071}{512} & \frac{7759}{768} & \frac{309}{32} & \frac{2663}{320} \\ \\ \frac{3979}{480} & \frac{8369}{960} & \frac{25517}{2880} & \frac{5499}{640} & \frac{249}{32} \\ \\ \frac{101}{16} & \frac{625}{96} & \frac{629}{96} & \frac{409}{64} & \frac{287}{48} \\ \\ \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} & \frac{23}{8} \end{pmatrix}$$ Therefore $$\max_{Q_{22}} R_2(p, x, t) \le \max_{i, j, k} \beta_{i, j, k}^{(22)} = \frac{7939}{576} \approx 13.7829861111.$$ Thus the Bernstein upper bounds on the four subregions are $$Q_{11}: \frac{1571}{128}, \qquad Q_{12}: \frac{1761}{128}, \qquad Q_{21}: \frac{9667}{768}, \qquad Q_{22}: \frac{7939}{576}.$$ Taking the largest of these four numbers, we conclude that $$R_2(p_1, x, t) \le \frac{7939}{576}$$ for all $0 \le p_1 \le 1, \ 0 \le x \le 1, \ 0 \le t \le 1/2,$ precisely, 7939/576 ≈ 13.7829861111. Exhausting both the cases y = 0 and y = 1 for H1(p1, x, y, t), we obtain $$144 |H_3(1)| \le 16$$ Hence, $$|H_3(1)| \le \frac{1}{9}.$$ Now we prove that we obtain the extremal function for H3(1) by taking w(z) = z 3 , thus, we have $$\varphi(z^3) = 1 + z^3 + \frac{m}{n}z^6.$$ Hence $$\frac{zf'(z)}{f(z)} = \varphi(z^3) = 1 + z^3 + \frac{m}{n}z^6,$$ so $$\frac{f'(z)}{f(z)} = \frac{1}{z} + z^2 + \frac{m}{n}z^5.$$ Integrating, we obtain $$\log f(z) = \log z + \frac{z^3}{3} + \frac{m}{6n}z^6 + C.$$ Therefore $$f(z) = C z \exp\left(\frac{z^3}{3} + \frac{m}{6n}z^6\right).$$ Since f(0) = 0 and f'(0) = 1, we must have C = 1. Thus $$f(z) = z \exp\left(\frac{z^3}{3} + \frac{m}{6n}z^6\right).$$ Set, $\alpha = m/6n$ . Using the exponential series, $$e^{\frac{z^3}{3} + \alpha z^6} = 1 + \left(\frac{z^3}{3} + \alpha z^6\right) + \frac{1}{2}\left(\frac{z^3}{3} + \alpha z^6\right)^2 + \frac{1}{6}\left(\frac{z^3}{3}\right)^3 + \cdots$$ Now $$\frac{1}{2} \left( \frac{z^3}{3} + \alpha z^6 \right)^2 = \frac{z^6}{18} + \frac{\alpha}{3} z^9 + \cdots,$$ and $$\frac{1}{6} \left( \frac{z^3}{3} \right)^3 = \frac{z^9}{162}.$$ Therefore $$f(z) = z + \frac{1}{3}z^4 + \left(\alpha + \frac{1}{18}\right)z^7 + \left(\frac{\alpha}{3} + \frac{1}{162}\right)z^{10} + \cdots$$ Substituting $\alpha = m/6n$ , we get $$f(z) = z + \frac{1}{3}z^4 + \frac{n+3m}{18n}z^7 + \frac{n+9m}{162n}z^{10} + \cdots$$ Thus, we have $$a_2 = 0,$$ $a_3 = 0,$ $a_4 = \frac{1}{3},$ $a_5 = 0.$ Hence, $$|H_3(1)| = \frac{1}{9},$$ which implies that $w(z) = z^3$ provides the extremal function, which proves the theorem.