In our first theorem we determine a sharp upper bound for the coefficient body *T*_{2}(2).

#### Theorem 2.1

*Let f given by (1) be in the class* *B*(*β*); 0 ≤ *β* ≤ 1. *Then we have the sharp bound*

$$|{T}_{2}(2)|=\left|{a}_{3}^{2}-{a}_{2}^{2}\right|\le \frac{4}{(\beta +2{)}^{2}}max\left\{1,\left|\frac{(-{\beta}^{4}-6{\beta}^{3}-12{\beta}^{2}-6\beta +5)}{(\beta +1{)}^{4}}\right|\right\}.$$

#### Proof

First note that by equating the corresponding coefficients in the equation

$$\frac{{z}^{1-\beta}{f}^{\mathrm{\prime}}(z)}{{\left[f(z)\right]}^{1-\beta}}=h(z)$$

we obtain

$${a}_{2}=\frac{{p}_{1}}{(\beta +1)},$$(2)

$${a}_{3}=\frac{{p}_{2}}{\beta +2}-\frac{(\beta -1){p}_{1}^{2}}{2(\beta +1{)}^{2}}.$$(3)

In view of (2) and (3), a simple computation leads to

$${a}_{3}^{2}-{a}_{2}^{2}=\frac{{p}_{2}^{2}}{(\beta +2{)}^{2}}+\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{4(\beta +1{)}^{4}}-\frac{{p}_{1}^{2}{p}_{2}(\beta -1)}{(\beta +2)(\beta +1{)}^{2}}-\frac{{p}_{1}^{2}}{(\beta +1{)}^{2}}.$$(4)

Note that, by Lemma 1.1, we may write 2*p*_{2} = *p*^{2} + *x*(4 − *p*^{2}) where without loss of generality we let 0 ≤ *p*_{1} = *p* ≤ 2. Substituting this into the above equation we obtain the following quadratic equation in terms of *x*.

$${a}_{3}^{2}-{a}_{2}^{2}=\frac{(4-{p}^{2}{)}^{2}}{4(\beta +2{)}^{2}}{x}^{2}+\frac{(\beta +3)(4-{p}^{2}){p}^{2}}{2(\beta +1{)}^{2}(\beta +2{)}^{2}}x+\frac{\left[({\beta}^{2}+6\beta +9){p}^{2}-4(\beta +1{)}^{2}(2+\beta {)}^{2}\right]{p}^{2}}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}.$$

Using the triangle inequality we obtain

$$\left|{a}_{3}^{2}-{a}_{2}^{2}\right|\le \frac{(4-{p}^{2}{)}^{2}}{4(\beta +2{)}^{2}}+\frac{(\beta +3)(4-{p}^{2}){p}^{2}}{2(\beta +1{)}^{2}(\beta +2{)}^{2}}+\frac{\left[({\beta}^{2}+6\beta +9){p}^{2}+4(\beta +1{)}^{2}(2+\beta {)}^{2}\right]{p}^{2}}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}=\mathrm{\Phi}(p,\beta ).$$

Differentiating *Φ*(*p*,*β*) with respect to *p* we obtain

$$\frac{\mathrm{\partial}\left(\mathrm{\Phi}(p,\beta )\right)}{\mathrm{\partial}p}=\frac{p\left[{p}^{2}\left({\beta}^{3}-3\beta +2\right)-2{\beta}^{3}+4{\beta}^{2}+14\beta +8\right]}{(\beta +1{)}^{4}(\beta +2)}.$$

Setting $\frac{\mathrm{\partial}\left(\mathrm{\Phi}(p,\beta )\right)}{\mathrm{\partial}p}=0$ yields either *p* = 0 or

$${p}^{2}=\frac{2{\beta}^{3}-4{\beta}^{2}-14\beta -8}{{\beta}^{3}-3\beta +2}.$$

But 2*β*^{3} − 4*β*^{2} − 14*β* − 8 < 0 for 0 ≤ *β* ≤ 1. Therefore, the maximum of $|{a}_{3}^{2}-{a}_{2}^{2}|$ is attained at the end points *p*_{1} = *p* ∈ [0,2].

For *p*_{1}=0, we have *p*_{2} = 2*x*. Therefore, from (4),

$$\left|{a}_{3}^{2}-{a}_{2}^{2}\right|=\frac{4{\left|x\right|}^{2}}{(\beta +2{)}^{2}}\le \frac{4}{(\beta +2{)}^{2}}.$$

For *p*_{1}=2 we have ${a}_{2}=\frac{2}{\beta +1}$ and ${a}_{3}=\frac{2}{\beta +2}-\frac{2(\beta -1)}{(\beta +1{)}^{2}}$ which yields

$$\left|{a}_{3}^{2}-{a}_{2}^{2}\right|\le \left|\frac{4(-{\beta}^{4}-6{\beta}^{3}-12{\beta}^{2}-6\beta +5)}{(\beta +1{)}^{4}(\beta +2{)}^{2}}\right|.$$

The result is sharp for the functions given by

$$\frac{{z}^{1-\beta}{f}^{\mathrm{\prime}}(z)}{{\left[f(z)\right]}^{1-\beta}}=\frac{1+z}{1-z}.$$ □

In our next theorem, we determine an upper bound for the coefficient body *T*_{2}(3).

#### Theorem 2.3

Let *f* given by (1), be in the class *B*(*β*), 0 ≤ *β* ≤ 1. Then

$$\left|{T}_{2}(3)\right|=\left|{a}_{4}^{2}-{a}_{3}^{2}\right|\le max\left\{\frac{|N(\beta )|}{9(\beta +3{)}^{2}(\beta +1{)}^{6}(\beta +2{)}^{2}},\frac{4}{(2+\beta {)}^{2}}\right\}.$$

where

$$N(\beta )=4{\beta}^{8}+40{\beta}^{7}+152{\beta}^{6}+88{\beta}^{5}-1312{\beta}^{4}-5096{\beta}^{3}-8024{\beta}^{2}-4248\beta +2268.$$

#### Proof

By equating the corresponding coefficients in the equation,

$$\frac{{z}^{1-\beta}{f}^{\mathrm{\prime}}(z)}{{\left[f(z)\right]}^{1-\beta}}=h(z)$$

we obtain

$${a}_{2}=\frac{{p}_{1}}{(\beta +1)},$$(5)

$${a}_{3}=\frac{{p}_{2}}{\beta +2}-\frac{(\beta -1){p}_{1}^{2}}{2(\beta +1{)}^{2}}\text{, and}$$(6)

$${a}_{4}=\frac{{p}_{3}}{\beta +3}-\frac{(\beta -1){p}_{1}{p}_{2}}{(\beta +1)(\beta +2)}+\frac{(\beta -1)(2\beta -1){p}_{1}^{3}}{6(\beta +1{)}^{3}}.$$(7)

In view of (6) and (7) and applying Lemma 1, denoting *X* = 4 − *p*^{2} and *Y* = (1 − |*x*|^{2})*ζ*, where 0 ≤ *p* ≤ 2 and |*ζ*| < 1 we get,

$$\begin{array}{rl}{a}_{4}^{2}-{a}_{3}^{2}& =\frac{{\beta}^{8}+10{\beta}^{7}+47{\beta}^{6}+148{\beta}^{5}+383{\beta}^{4}+778{\beta}^{3}+1153{\beta}^{2}+1368\beta +1296}{144(\beta +3{)}^{2}(\beta +1{)}^{6}(\beta +2{)}^{2}}{p}_{1}^{6}\\ & \phantom{\rule{2em}{0ex}}-\frac{{\beta}^{2}+6\beta +9}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}{p}_{1}^{4}+\frac{{X}^{2}{Y}^{2}}{4(\beta +3{)}^{2}}-\frac{{p}_{1}{x}^{2}{X}^{2}Y}{4(\beta +3{)}^{2}}\\ & \phantom{\rule{2em}{0ex}}+\frac{(\beta +5)x{X}^{2}{p}_{1}Y}{2(\beta +3{)}^{2}(\beta +2)(\beta +1)}+\frac{{\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36}{12(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}{p}_{1}^{3}XY\\ & \phantom{\rule{2em}{0ex}}+\frac{{p}_{1}^{2}{X}^{2}{x}^{4}}{16(\beta +3{)}^{2}}-\frac{\beta +5}{4(\beta +3{)}^{2}(\beta +2)(\beta +1)}{p}_{1}^{2}{x}^{3}{X}^{2}\\ & \phantom{\rule{2em}{0ex}}+\frac{{\beta}^{2}+10\beta +25}{4(\beta +3{)}^{2}(\beta +2{)}^{2}(\beta +1{)}^{2}}{p}_{1}^{2}{X}^{2}{x}^{2}-\frac{{x}^{2}{X}^{2}}{4(\beta +2{)}^{2}}\\ & \phantom{\rule{2em}{0ex}}-\frac{{\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36}{24(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}X{x}^{2}{p}_{1}^{4}\\ & \phantom{\rule{2em}{0ex}}+\frac{{\beta}^{5}+10{\beta}^{4}+36{\beta}^{3}+74{\beta}^{2}+131\beta +180}{12(\beta +3{)}^{2}(\beta +2{)}^{2}(\beta +1{)}^{4}}Xx{p}_{1}^{4}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{(\beta +3)}{2(\beta +2{)}^{2}(\beta +1{)}^{2}}Xx{p}_{1}^{2}.\end{array}$$

As in the proof of Theorem 1, without loss of generality, we can write *p*_{1} = *p*, where 0 ≤ *p* ≤ 2. Then an application of triangle inequality gives,

$$\begin{array}{rl}\left|{a}_{4}^{2}-{a}_{3}^{2}\right|& \le \frac{(2-p{)}^{2}}{16(\beta +3{)}^{2}}|x{|}^{4}+\frac{(\beta +5)({p}^{2}-2p)(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}(\beta +2)(\beta +1)}|x{|}^{3}\\ & \phantom{\rule{2em}{0ex}}+\left[\frac{{\beta}^{2}+10\beta +25}{4(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{2}}{p}^{2}(4-{p}^{2})\right.\\ & \phantom{\rule{2em}{0ex}}+\frac{{p}^{3}(p-2)(4-{p}^{2})({\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36)}{24(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}\\ & \left.\phantom{\rule{2em}{0ex}}-\frac{({\beta}^{2}+2\beta -1)(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}(\beta +2{)}^{2}}+\frac{p(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}}\right]|x{|}^{2}\\ & \phantom{\rule{2em}{0ex}}+\left[\frac{(\beta +3)}{2(\beta +2{)}^{2}(\beta +1{)}^{2}}(4-{p}^{2}){p}^{2}+\frac{(\beta +5)(4-{p}^{2}{)}^{2}p}{2(\beta +3{)}^{2}(\beta +2)(\beta +1)}\right.\\ & \left.\phantom{\rule{2em}{0ex}}+\frac{({\beta}^{5}+10{\beta}^{4}+36{\beta}^{3}+74{\beta}^{2}+131\beta +180){p}^{4}(4-{p}^{2})}{12(\beta +3{)}^{2}(\beta +1{)}^{4}(\beta +2{)}^{2}}\right]|x|\\ & \phantom{\rule{2em}{0ex}}+\left|{N}_{1}(\beta ){p}^{6}-{N}_{2}(\beta ){p}^{4}\right|+\frac{(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}}\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36){p}^{3}(4-{p}^{2})}{12(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}\\ & =\mathrm{\Psi}(p,|x|)\end{array}$$

where

$$\begin{array}{rcl}{N}_{1}(\beta )& =& \frac{{\beta}^{8}+10{\beta}^{7}+47{\beta}^{6}+148{\beta}^{5}+383{\beta}^{4}+778{\beta}^{3}+1153{\beta}^{2}+1368\beta +1296}{144(\beta +3{)}^{2}(\beta +2{)}^{2}(\beta +1{)}^{6}},\\ {N}_{2}(\beta )& =& \frac{{\beta}^{2}+6\beta +9}{4(\beta +2{)}^{2}(\beta +1{)}^{4}}.\end{array}$$

We need to find the maximum value of *Ψ*(*p*, |*x*|) on [0,2] × [0,1]. First, assume that there is a maximum at an interior point *Ψ*(*p*_{0}, |*x*_{0}|) of [0,2] × [0,1]. Differentiating *Ψ*(*p*, |*x*|) with respect to |*x*| and equating it to 0 implies that *p* = *p*_{0} = 2, which is a contradiction. Thus for the maximum of *Ψ*(*p*, |*x*|), we need only to consider the end points of [0,2] × [0,1].

For *p* = 0 we obtain

$$\mathrm{\Psi}(0,|x|)=\frac{4}{(\beta +3{)}^{2}}|x{|}^{4}-\frac{4({\beta}^{2}+2\beta -1)}{(\beta +3{)}^{2}(\beta +2{)}^{2}}|x{|}^{2}+\frac{4}{(\beta +3{)}^{2}}\le \frac{4}{(\beta +2{)}^{2}}.$$

For *p*=2 we obtain

$$\mathrm{\Phi}(2,|x|)=\left|64{N}_{1}(\beta )-16{N}_{2}(\beta )\right|.$$

For |*x*|=0 we obtain

$$\begin{array}{rl}\mathrm{\Psi}(p,0)& =\left|{N}_{1}(\beta ){p}^{6}-{N}_{2}(\beta ){p}^{4}\right|\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36){p}^{3}(4-{p}^{2})}{12(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}+\frac{p(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}},\end{array}$$

which has the maximum value |*N*_{1}(*β*)*p*^{6} − *N*_{2}(*β*)*p*^{4}| on [0,2].

For |*x*| = 1 we obtain

$$\begin{array}{rl}\mathrm{\Psi}(p,1)& =\frac{(2-p{)}^{2}(4-{p}^{2}{)}^{2}}{16(\beta +3{)}^{2}}+\frac{(\beta +5)({p}^{2}-2p)(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}(\beta +2)(\beta +1)}\\ & \phantom{\rule{2em}{0ex}}+\left[\frac{({\beta}^{2}+10\beta +25){p}^{2}(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{2}}+\frac{p(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}}\right.\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36){p}^{3}(p-2)(4-{p}^{2})}{24(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}\\ & \left.\phantom{\rule{2em}{0ex}}-\frac{({\beta}^{2}+2\beta -1)(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}(\beta +2{)}^{2}}\right]\\ & \phantom{\rule{2em}{0ex}}+\left[\frac{(\beta +3)}{2(\beta +2{)}^{2}(\beta +1{)}^{2}}(4-{p}^{2}){p}^{2}+\frac{(\beta +5)(4-{p}^{2}{)}^{2}p}{2(\beta +3{)}^{2}(\beta +2)(\beta +1)}\right.\\ & \left.\phantom{\rule{2em}{0ex}}+\frac{({\beta}^{5}+10{\beta}^{4}+36{\beta}^{3}+74{\beta}^{2}+131\beta +180){p}^{4}(4-{p}^{2})}{12(\beta +3{)}^{2}(\beta +2{)}^{2}(\beta +1{)}^{4}}\right]\\ & \phantom{\rule{2em}{0ex}}+\left|{N}_{1}(\beta ){p}^{6}-{N}_{2}(\beta ){p}^{4}\right|+\frac{(4-{p}^{2}{)}^{2}}{4(\beta +3{)}^{2}}\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{4}+5{\beta}^{3}+11{\beta}^{2}+19\beta +36){p}^{3}(4-{p}^{2})}{12(\beta +3{)}^{2}(\beta +2)(\beta +1{)}^{3}}\end{array}$$

which has the maximum values |64*N*_{1}(*β*)−16*N*_{2}(*β*)| for *p* = 2 and $\frac{4}{(2+\beta {)}^{2}}$ for *p* = 0. □

#### Theorem 2.5

*Let f given by (1) be in the class B(β), (0 ≤ β ≤ 1; β ≠ β*_{0}), then

$$\left|{T}_{3}(2)\right|=\left|\left(\left|\begin{array}{ccc}{a}_{2}& {a}_{3}& {a}_{4}\\ {a}_{3}& {a}_{2}& {a}_{3}\\ {a}_{4}& {a}_{3}& {a}_{2}\end{array}\right|\right)\right|\le \left\{\begin{array}{ll}max\left\{|{M}_{1}(\beta ){M}_{2}(\beta )|,\text{\hspace{0.17em}}\frac{8\text{\hspace{0.17em}}|{M}_{1}(\beta )|}{(\beta +2{)}^{2}}\right\};& \text{if\hspace{0.17em}}\beta \ne {\beta}_{0},\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\\ max\left\{|{M}_{2}(\beta ){M}_{3}(\beta )|,\text{\hspace{0.17em}}\frac{8\text{\hspace{0.17em}}|{M}_{3}(\beta )|}{(\beta +2{)}^{2}}\right\};& \text{if\hspace{0.17em}}\beta ={\beta}_{0},\end{array}\right.$$

where *β*_{0} ≈ 0.3676 is the positive root of the polynomial

$$4{\beta}^{4}+32{\beta}^{3}+80{\beta}^{2}+64\beta -36=0,$$

$${M}_{1}(\beta )=\frac{4{\beta}^{4}+32{\beta}^{3}+80{\beta}^{2}+64\beta -36}{3(\beta +1{)}^{3}(\beta +2)(\beta +3)},{M}_{2}(\beta )=\frac{4(4{\beta}^{5}+34{\beta}^{4}+108{\beta}^{3}+140{\beta}^{2}+32\beta -54)}{3(\beta +1{)}^{4}(\beta +2{)}^{2}(\beta +3)}$$

and

$${M}_{3}(\beta )=\frac{8{\beta}^{4}+52{\beta}^{3}+124{\beta}^{2}+140\beta +108}{3(\beta +1{)}^{3}(\beta +2)(\beta +3)}.$$

#### Proof

Write

$$\begin{array}{rcl}\left|{T}_{3}(2)\right|& =& \left|{a}_{2}^{3}-2{a}_{2}{a}_{3}^{2}+2{a}_{3}^{2}{a}_{4}-{a}_{2}{a}_{4}\right|\\ & =& \left|({a}_{2}-{a}_{4})({a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4})\right|.\end{array}$$

Using the same techniques as in Theorem 2, one can obtain with simple computations that

$$|{a}_{2}-{a}_{4}|\le \left|{M}_{1}(\beta )\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{f}\mathrm{o}\mathrm{r}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta \ne {\beta}_{0}.$$

We need to show that

$$\left|{a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4}\right|\le |{M}_{2}(\beta )|.$$

In view of (2), (3) and (7) and Lemma 1, where we denote *X* = 4 − *p*^{2} and *Y* = (1 − |*x*|^{2})*ζ*, where 0 ≤ *p* ≤ 2 and |*ζ*| < 1, one may easily get,

$$\begin{array}{rl}\left|{a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4}\right|& =\left|\frac{{p}_{1}^{2}}{(\beta +1{)}^{2}}-2\left(\frac{{p}_{2}^{2}}{(\beta +2{)}^{2}}+\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{4(\beta +1{)}^{4}}-\frac{(\beta -1){p}_{1}^{2}{p}_{2}}{(\beta +1{)}^{2}(\beta +2)}\right)\right.\\ & \phantom{\rule{2em}{0ex}}\left.+\frac{{p}_{1}}{(\beta +1)}\left(\frac{{p}_{3}}{\beta +3}-\frac{(\beta -1){p}_{1}{p}_{2}}{(\beta +1)(\beta +2)}+\frac{(\beta -1)(2\beta -1){p}_{1}^{3}}{6(\beta +1{)}^{3}}\right)\right|\\ & =\left|\frac{{p}_{1}^{2}}{(\beta +1{)}^{2}}-2\frac{{p}_{2}^{2}}{(\beta +2{)}^{2}}-\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{2(\beta +1{)}^{4}}+2\frac{(\beta -1){p}_{1}^{2}{p}_{2}}{(\beta +1{)}^{2}(\beta +2)}\right.\\ & \phantom{\rule{2em}{0ex}}\left.+\frac{{p}_{1}{p}_{3}}{(\beta +1)(\beta +3)}-\frac{(\beta -1){p}_{1}^{2}{p}_{2}}{(\beta +1{)}^{2}(\beta +2)}+\frac{(\beta -1)(2\beta -1){p}_{1}^{4}}{6(\beta +1{)}^{4}}\right|\\ & =\left|\frac{{p}_{1}^{2}}{(\beta +1{)}^{2}}-\frac{1}{2(\beta +2{)}^{2}}\left[{p}_{1}^{4}+{X}^{2}{x}^{2}+2{p}_{1}^{2}Xx\right]\right.\\ & \phantom{\rule{2em}{0ex}}-\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{2(\beta +1{)}^{4}}+\frac{(\beta -1){p}_{1}^{2}}{2(\beta +1{)}^{2}(\beta +2)}\left[{p}_{1}^{2}+Xx\right]\\ & \phantom{\rule{2em}{0ex}}+\frac{{p}_{1}}{4(\beta +1)(\beta +3)}\left[{p}_{1}^{3}+2{p}_{1}Xx-{p}_{1}X{x}^{2}+2XY\right]\\ & \phantom{\rule{2em}{0ex}}\left.+\frac{(\beta -1)(2\beta -1){p}_{1}^{4}}{6(\beta +1{)}^{4}}\right|.\end{array}$$

Applying the triangle inequality and assuming that *p*_{1} = *p*, where 0 ≤ *p* ≤ 2 we obtain

$$\begin{array}{rl}\left|{a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4}\right|& \le \left[\frac{1}{4(\beta +1)(\beta +3)}\left({p}^{2}-2p\right)+\frac{1}{2(2+\beta {)}^{2}}(4-{p}^{2})\right](4-{p}^{2})|x{|}^{2}\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{2}+5\beta +8)}{2(\beta +1{)}^{2}(\beta +2{)}^{2}(\beta +3)}{p}^{2}(4-{p}^{2})|x|\\ & \phantom{\rule{2em}{0ex}}+\left|\frac{{p}^{2}}{(\beta +1{)}^{2}}-\frac{(90-{\beta}^{5}-7{\beta}^{4}-15{\beta}^{3}+13{\beta}^{2}+88\beta ){p}^{4}}{12(\beta +1{)}^{4}(\beta +2{)}^{2}(\beta +3)}\right|\\ & \phantom{\rule{2em}{0ex}}+\frac{p}{2(\beta +1)(\beta +3)}(4-{p}^{2})\\ & =\mathrm{\Omega}(p,|x|)\end{array}$$

We need to find the maximum value of *Ω*(*p*,|*x*|) on [0,2] × [0,1]. First, assume that there is a maximum at an interior point *Ω*(*p*_{0},|*x*_{0}|) of [0,2] × [0,1]. Differentiating *Ω*(*[*,|*x*|) with respect to |*x*| and equating it to zero implies that *p* = *p*_{0} = 2, which is a contradiction. Thus for the maximum of Ω(*p*,|*x*|), we need only to consider the end points of [0,2] × [0,1].

For *p* = 0 we obtain

$$\mathrm{\Omega}(0,|x|)=\frac{8}{(\beta +2{)}^{2}}|x{|}^{2}\le \frac{8}{(\beta +2{)}^{2}}.$$

For *p* = 2 we obtain

$$\mathrm{\Omega}(2,|x|)=\frac{4(4{\beta}^{5}+34{\beta}^{4}+108{\beta}^{3}+140{\beta}^{2}+32\beta -54)}{3(\beta +1{)}^{4}(\beta +2{)}^{2}(\beta +3)}={M}_{2}(\beta ).$$

For |*x*| = 0 we obtain

$$\begin{array}{rl}\mathrm{\Omega}(p,0)& =\left|\frac{{p}^{2}}{(\beta +1{)}^{2}}-\frac{(90-{\beta}^{5}-7{\beta}^{4}-15{\beta}^{3}+13{\beta}^{2}+88\beta )}{12(\beta +1{)}^{4}(\beta +2{)}^{2}(\beta +3)}{p}^{4}\right|\\ & \phantom{\rule{2em}{0ex}}+\frac{p}{2(\beta +1)(\beta +3)}(4-{p}^{2})\end{array}$$

which has maximum value *Ω*(*p*,0) = *M*_{2}(*β*) attained at the end point *p* = 2.

For |*x*| = 1 we obtain

$$\begin{array}{rl}\mathrm{\Omega}(p,1)& =\left|\frac{{p}^{2}}{(\beta +1{)}^{2}}-\frac{(90-{\beta}^{5}-7{\beta}^{4}-15{\beta}^{3}+13{\beta}^{2}+88\beta )}{12(\beta +1{)}^{4}(\beta +2{)}^{2}(\beta +3)}{p}^{4}\right|\\ & \phantom{\rule{2em}{0ex}}+\frac{1}{2(\beta +2{)}^{2}}(4-{p}^{2}{)}^{2}+\frac{1}{4(\beta +1)(\beta +3)}{p}^{2}(4-{p}^{2})\\ & \phantom{\rule{2em}{0ex}}+\frac{({\beta}^{2}+5\beta +8)}{2(\beta +1{)}^{2}(\beta +2{)}^{2}(\beta +3)}{p}^{2}(4-{p}^{2}),\end{array}$$

which has maximum value $\mathrm{\Omega}(p,1)=\frac{8}{(\beta +2{)}^{2}}$ at *p* = 0 and *Ω*(*p*,1) = *M*_{2}(*β*) at *p* = 2. Hence

$$\left|{a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4}\right|\le max\left\{|{M}_{2}(\beta )|,\frac{8}{(\beta +2{)}^{2}}\right\}.$$

Thus

$$\left|{T}_{3}(2)\right|=\left|({a}_{2}-{a}_{4})({a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4})\right|\le max\left\{|{M}_{1}(\beta ){M}_{2}(\beta )|,\frac{8\text{\hspace{0.17em}}|{M}_{1}(\beta )|}{(\beta +2{)}^{2}}\right\}.$$

For the case *β* = *β*_{0}, we compute |*a*_{2}−*a*_{4}| as follows

$$|{a}_{2}-{a}_{4}|=\left|\frac{{p}_{1}}{\beta +1}-\frac{{p}_{3}}{\beta +3}+\frac{(\beta -1){p}_{1}{p}_{2}}{(\beta +1)(\beta +2)}-\frac{(\beta -1)(2\beta -1){p}_{1}^{3}}{6(\beta +1{)}^{3}}\right|.$$

Since, each |*p*_{i}| ≤ 2, an application of triangle inequality shows that

$$|{a}_{2}-{a}_{4}|\le |{M}_{3}(\beta )|=\frac{8{\beta}^{4}+52{\beta}^{3}+124{\beta}^{2}+140\beta +108}{3(\beta +1{)}^{3}(\beta +2)(\beta +3)}.$$

Therefore

$$\left|{T}_{3}(2)\right|=\left|({a}_{2}-{a}_{4})({a}_{2}^{2}-2{a}_{3}^{2}+{a}_{2}{a}_{4})\right|\le max\left\{|{M}_{2}(\beta ){M}_{3}(\beta )|,\frac{8\text{\hspace{0.17em}}|{M}_{3}(\beta )|}{(\beta +2{)}^{2}}\right\}.$$

This completes the proof of Theorem 2.5. □

#### Theorem 2.7

*Let f given by (1), be in the class B(β), 0 ≤ β ≤ 1. Then*

$$\left|{T}_{3}(1)\right|=\left|\left(\left|\begin{array}{ccc}1& {a}_{2}& {a}_{3}\\ {a}_{2}& 1& {a}_{2}\\ {a}_{3}& {a}_{2}& 1\end{array}\right|\right)\right|\le max\left\{1+\frac{1}{4(\beta +2{)}^{2}},|{M}_{4}(\beta )|\right\}$$

*where*

$${M}_{4}(\beta )=\frac{{\beta}^{6}+8{\beta}^{5}+18{\beta}^{4}-4{\beta}^{3}-51{\beta}^{2}-20\beta +32}{(\beta +1{)}^{4}(\beta +2{)}^{2}}.$$

#### Proof

Expanding the determinant by using equations (2) and (3) and applying Lemma 1.1, we have

$$\begin{array}{rl}{T}_{3}(1)& =1+2{a}_{2}^{2}({a}_{3}-1)-{a}_{3}^{2}\\ & =1+\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}\left(\frac{{p}_{2}}{(\beta +2)}-\frac{(\beta -1){p}_{1}^{2}}{2(\beta +1{)}^{2}}-1\right)\\ & \phantom{\rule{2em}{0ex}}-\frac{{p}_{2}^{2}}{(\beta +2{)}^{2}}-\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{4(\beta +1{)}^{4}}+\frac{{p}_{2}{p}_{1}^{2}(\beta -1)}{(\beta +1{)}^{2}(\beta +2)}\\ & =1+\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}(2+\beta )}\left(\frac{{p}_{1}^{2}+Xx}{2}\right)-\frac{(\beta -1){p}_{1}^{4}}{(\beta +1{)}^{4}}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}-\frac{(\beta -1{)}^{2}{p}_{1}^{4}}{4(\beta +1{)}^{4}}\\ & \phantom{\rule{2em}{0ex}}-\frac{1}{4(\beta +2{)}^{2}}\left[{p}_{1}^{4}+{X}^{2}{x}^{2}+2{p}_{1}^{2}Xx\right]+\frac{{p}_{1}^{2}(\beta -1)}{(\beta +1{)}^{2}(\beta +2)}\left(\frac{{p}_{1}^{2}+Xx}{2}\right)\\ & =1+\frac{{p}_{1}^{4}}{(\beta +1{)}^{2}(\beta +2)}+\frac{{p}_{1}^{2}Xx}{(\beta +1{)}^{2}(\beta +2)}-\frac{(\beta -1){p}_{1}^{4}}{(\beta +1{)}^{4}}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}\\ & \phantom{\rule{2em}{0ex}}-\frac{{p}_{1}^{4}}{4(\beta +2{)}^{2}}-\frac{{X}^{2}{x}^{2}}{4(\beta +2{)}^{2}}-\frac{{p}_{1}^{2}Xx}{2(\beta +2{)}^{2}}-\frac{(1-\beta {)}^{2}{p}_{1}^{4}}{4(\beta +1{)}^{4}}\\ & \phantom{\rule{2em}{0ex}}-\frac{{p}_{1}^{4}(1-\beta )}{2(\beta +1{)}^{2}(\beta +2)}-\frac{{p}_{1}^{2}Xx(1-\beta )}{2(\beta +1{)}^{2}(\beta +2)}\\ & =1+{p}_{1}^{4}\left[\frac{1}{(\beta +1{)}^{2}(2+\beta )}+\frac{(1-\beta )}{(\beta +1{)}^{4}}-\frac{1}{4(\beta +2{)}^{2}}-\frac{(1-\beta {)}^{2}}{4(\beta +1{)}^{4}}\right.\\ & \left.\phantom{\rule{2em}{0ex}}-\frac{(1-\beta )}{2(\beta +1{)}^{4}(\beta +2)}\right]-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}+{p}_{1}^{2}Xx\left[\frac{(1+\beta )(\beta +2)-(\beta +1{)}^{2}}{2(\beta +1{)}^{2}(\beta +2{)}^{2}}\right]\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{{X}^{2}{x}^{2}}{4(\beta +2{)}^{2}}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}\\ & =1+\frac{{p}_{1}^{4}}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}\left[4(\beta +1{)}^{2}(\beta +2)+4(1-\beta )(\beta +2{)}^{2}-(\beta +1{)}^{4}\right.\\ & \left.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-(1-\beta {)}^{2}(\beta +2{)}^{2}-2(1-\beta )(\beta +1{)}^{2}(\beta +2)\right]\\ & \phantom{\rule{2em}{0ex}}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}+{p}_{1}^{2}Xx\left[\frac{(1+\beta )(\beta +2)-(\beta +1{)}^{2}}{2(\beta +1{)}^{2}(\beta +2{)}^{2}}\right]-\frac{{X}^{2}{x}^{2}}{4(\beta +2{)}^{2}}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}\\ & =1+\frac{3{\beta}^{2}+14\beta +15}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}{p}_{1}^{4}-\frac{2{p}_{1}^{2}}{(\beta +1{)}^{2}}+\frac{1}{2(\beta +1)(\beta +2{)}^{2}}{p}_{1}^{2}xX-\frac{1}{4(\beta +2{)}^{2}}{x}^{2}{X}^{2}.\end{array}$$

Without loss of generality, we let 0 ≤ *p*_{1} = *p* ≤ 2. Now substituting this into the above equation and applying the triangle inequality we obtain the following quadratic equation in terms of *x*.

$$\begin{array}{rcl}\left|{T}_{3}(1)\right|& \le & \frac{(4-{p}^{2}{)}^{2}}{4(\beta +2{)}^{2}}{\left|x\right|}^{2}+\frac{{p}^{2}(4-{p}^{2})}{2(\beta +1)(\beta +2{)}^{2}}\left|x\right|+\left[1+\frac{\left|(3{\beta}^{2}+14\beta +15){p}^{2}-8(\beta +1{)}^{2}(\beta +2{)}^{2}\right|}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}{p}^{2}\right]\\ & \le & \frac{(4-{p}^{2}{)}^{2}}{4(\beta +2{)}^{2}}+\frac{{p}^{2}(4-{p}^{2})}{2(\beta +1)(\beta +2{)}^{2}}+\left[1+\frac{\left|(3{\beta}^{2}+14\beta +15){p}^{2}-8(\beta +1{)}^{2}(\beta +2{)}^{2}\right|}{4(\beta +1{)}^{4}(\beta +2{)}^{2}}{p}^{2}\right]\\ & =& \mathrm{\Gamma}(p,\beta ).\end{array}$$

Differentiating Γ(*p*, *β*) with respect to *p* we obtain

$$\frac{\mathrm{\partial}\left(\mathrm{\Gamma}(p,\beta )\right)}{\mathrm{\partial}p}=\frac{p\left[{p}^{2}\left({\beta}^{4}+2{\beta}^{3}+3{\beta}^{2}+12\beta +14\right)+4{\beta}^{3}+13{\beta}^{2}+14\beta +15\right]}{(\beta +1{)}^{4}(\beta +2{)}^{2}}.$$

Setting ∂(Γ(*p*,*β*))/∂*p* = 0 yields either *p* = 0 or

$${p}^{2}=\frac{-4{\beta}^{3}-13{\beta}^{2}-14\beta -15}{{\beta}^{4}+2{\beta}^{3}+3{\beta}^{2}+12\beta +14}.$$

But −4*β*_{3} − 13*β*_{2} − 14*β* − 15 < 0 for 0 ≤ *β* ≤ 1. Therefore, the maximum of |*T*_{3}(1)| is attained at the end points *p*_{1} = *p* ∈ [0,2].

For *p*_{1} = 0 we have *a*_{2} = 0 and ${a}_{3}={\displaystyle 1-\frac{{x}^{2}}{4(\beta +2{)}^{2}}}$ which yields\newline

$$\left|{T}_{3}(1)\right|=1+\frac{{\left|x\right|}^{2}}{4(\beta +2{)}^{2}}\le 1+\frac{1}{4(\beta +2{)}^{2}}.$$

For *p*_{1} = 2 we obtain

$$\left|{T}_{3}(1)\right|\le \left|1+\frac{4(3{\beta}^{2}+14\beta +15)}{(\beta +1{)}^{4}(\beta +2{)}^{2}}-\frac{8}{(\beta +1{)}^{2}}\right|\le \left|{M}_{4}(\beta )\right|.$$

where

$${M}_{4}(\beta )=\frac{{\beta}^{6}+8{\beta}^{5}+18{\beta}^{4}-4{\beta}^{3}-51{\beta}^{2}-20\beta +32}{(\beta +1{)}^{4}(\beta +2{)}^{2}}.$$

This completes the proof of Theorem 2.7. □

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