Although we have discussed the Hankel determinant problem in the paper, the first two problems are specifically related with the Fekete-Szegö functional, which is a special case of the Hankel determinant.

**Theorem 2.1** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then* $|{a}_{3}-\mu {a}_{2}^{2}|\le \{\begin{array}{ll}\frac{1}{16}(1-4\mu ),& \mu <-\frac{3}{4},\\ \frac{1}{4},& -\frac{3}{4}\le \mu \le \frac{5}{4},\\ \frac{1}{16}(4\mu -1),& \mu >\frac{5}{4}.\end{array}$

*Furthermore*,

*for* $-\frac{3}{4}<\mu \le \frac{1}{4}$,

$|{a}_{3}-\mu {a}_{2}^{2}|+\frac{1}{4}(4\mu +3)|{a}_{2}{|}^{2}\le \frac{1}{4},$

*and for* $\frac{1}{4}<\mu \le \frac{5}{4}$,

$|{a}_{3}-\mu {a}_{2}^{2}|+\frac{1}{4}(5-4\mu )|{a}_{2}{|}^{2}\le \frac{1}{4}.$

*These results are sharp*.

*Proof* If

$f\in {\mathit{SL}}^{\ast}$, then it follows from (1.3) that

$\frac{z{f}^{\prime}(z)}{f(z)}\prec \varphi (z),$

(2.1)

where

$\varphi (z)=\sqrt{1+z}$. Define a function

$p(z)=\frac{1+w(z)}{1-w(z)}=1+{p}_{1}z+{p}_{2}{z}^{2}+\cdots .$

It is clear that

$p\in P$. This implies that

$w(z)=\frac{p(z)-1}{p(z)+1}.$

From (2.1), we have

$\frac{z{f}^{\prime}(z)}{f(z)}=\varphi (w(z)),$

with

$\varphi (w(z))={\left(\frac{2p(z)}{p(z)+1}\right)}^{\frac{1}{2}}.$

Now

${\left(\frac{2p(z)}{p(z)+1}\right)}^{\frac{1}{2}}=1+\frac{1}{4}{p}_{1}z+[\frac{1}{4}{p}_{2}-\frac{5}{32}{p}_{1}^{2}]{z}^{2}+[\frac{1}{4}{p}_{3}-\frac{5}{16}{p}_{1}{p}_{2}+\frac{13}{128}{p}_{1}^{3}]{z}^{3}+\cdots .$

Similarly,

$\frac{z{f}^{\prime}(z)}{f(z)}=1+{a}_{2}z+[2{a}_{3}-{a}_{2}^{2}]{z}^{2}+[3{a}_{4}-3{a}_{2}{a}_{3}+{a}_{2}^{3}]{z}^{3}+\cdots .$

Therefore

${a}_{2}=\frac{1}{4}{p}_{1},$

(2.2)

${a}_{3}=\frac{1}{8}{p}_{2}-\frac{3}{64}{p}_{1}^{2},$

(2.3)

${a}_{4}=\frac{1}{12}{p}_{3}-\frac{7}{96}{p}_{1}{p}_{2}+\frac{13}{768}{p}_{1}^{2}.$

(2.4)

This implies that

$|{a}_{3}-\mu {a}_{2}^{2}|=\frac{1}{8}|{p}_{2}-\frac{1}{8}(4\mu +3){p}_{1}^{2}|.$

Now, using Lemma 1.1, we have the required result. □

The results are sharp for the functions

${K}_{i}(z)$,

$i=1,2,3,4$, such that

$\begin{array}{c}\frac{z{K}_{1}^{\prime}(z)}{{K}_{1}(z)}=\sqrt{1+z}\phantom{\rule{1em}{0ex}}\text{if}\mu -\frac{3}{4}\text{or}\mu \frac{5}{4},\hfill \\ \frac{z{K}_{2}^{\prime}(z)}{{K}_{2}(z)}=\sqrt{1+{z}^{2}}\phantom{\rule{1em}{0ex}}\text{if}-\frac{3}{4}\mu \frac{5}{4},\hfill \\ \frac{z{K}_{3}^{\prime}(z)}{{K}_{3}(z)}=\sqrt{1+\mathrm{\Phi}(z)}\phantom{\rule{1em}{0ex}}\text{if}\mu =-\frac{3}{4},\hfill \\ \frac{z{K}_{4}^{\prime}(z)}{{K}_{4}(z)}=\sqrt{1-\mathrm{\Phi}(z)}\phantom{\rule{1em}{0ex}}\text{if}\mu =\frac{5}{4},\hfill \end{array}$

where $\mathrm{\Phi}(z)=\frac{z(z+\eta )}{1+\eta z}$ with $0\le \eta \le 1$.

**Theorem 2.2** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then for a complex number* *μ*,

$|{a}_{3}-\mu {a}_{2}^{2}|\le \frac{1}{4}max\{1;|\mu -\frac{1}{4}\left|\right\}.$

*Proof* Since

$|{a}_{3}-\mu {a}_{2}^{2}|=\frac{1}{8}|{p}_{2}-\frac{1}{8}(4\mu +3){p}_{1}^{2}|,$

therefore, using Lemma 1.2, we get the result. This result is sharp for the functions

$\frac{z{f}^{\prime}(z)}{f(z)}=\sqrt{1+z}$

or

$\frac{z{f}^{\prime}(z)}{f(z)}=\sqrt{1+{z}^{2}}.$

□

For $\mu =1$, we have ${H}_{2}(1)$.

**Corollary 2.3** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then* $|{a}_{3}-{a}_{2}^{2}|\le \frac{1}{4}.$

**Theorem 2.4** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then* $|{a}_{2}{a}_{4}-{a}_{3}^{2}|\le \frac{1}{16}.$

*Proof* From (2.2), (2.3) and (2.4), we obtain

$\begin{array}{rcl}{a}_{2}{a}_{4}-{a}_{3}^{2}& =& \frac{1}{48}({p}_{1}{p}_{3}-\frac{7}{8}{p}_{1}^{2}{p}_{2}+\frac{13}{64}{p}_{1}^{4})-{(\frac{1}{8}{p}_{2}-\frac{3}{64}{p}_{1}^{2})}^{2}\\ =& \frac{1}{48}{p}_{1}{p}_{3}-\frac{1}{64}{p}_{2}^{2}-\frac{5}{768}{p}_{1}^{2}{p}_{2}+\frac{25}{12\text{,}288}{p}_{1}^{4}\\ =& \frac{1}{12\text{,}288}(256{p}_{1}{p}_{3}-192{p}_{2}^{2}-80{p}_{1}^{2}{p}_{2}+25{p}_{1}^{4}).\end{array}$

Putting the values of

${p}_{2}$ and

${p}_{3}$ from Lemma 1.3, we assume that

$p>0$, and taking

${p}_{1}=p\in [0,2]$, we get

$\begin{array}{rl}|{a}_{2}{a}_{4}-{a}_{3}^{2}|=& \frac{1}{12\text{,}288}|64{p}_{1}\{{p}_{1}^{3}+2(4-{p}_{1}^{2}){p}_{1}x-(4-{p}_{1}^{2}){p}_{1}{x}^{2}+2(4-{p}_{1}^{2})(1-|x{|}^{2})z\}\\ -48{\{{p}_{1}^{2}+x(4-{p}_{1}^{2})\}}^{2}-40{p}_{1}^{2}\{{p}_{1}^{2}+x(4-{p}_{1}^{2})\}+25{p}_{1}^{4}|.\end{array}$

After simple calculations, we get

$\begin{array}{rl}|{a}_{2}{a}_{4}-{a}_{3}^{2}|=& \frac{1}{12\text{,}288}|41{p}^{4}-8(4-{p}^{2}){p}^{2}x-128(4-{p}^{2})(1-|x{|}^{2})z\\ +{x}^{2}(4-{p}^{2})(64{p}^{2}+48)(4-{p}^{2})|.\end{array}$

Now, applying the triangle inequality and replacing

$|x|$ by

*ρ*, we obtain

$\begin{array}{rcl}|{a}_{2}{a}_{4}-{a}_{3}^{2}|& \le & \frac{1}{12\text{,}288}[41{p}^{4}+128(4-{p}^{2})+8(4-{p}^{2}){p}^{2}\rho +{\rho}^{2}(4-{p}^{2})(16{p}^{2}+64)]\\ =& F(p,\rho )\phantom{\rule{1em}{0ex}}\text{(say).}\end{array}$

Differentiating with respect to

*ρ*, we have

$\frac{\partial F(p,\rho )}{\partial \rho}=\frac{1}{12\text{,}288}[8(4-{p}^{2}){p}^{2}+2\rho (4-{p}^{2})(16{p}^{2}+64)].$

It is clear that

$\frac{\partial F(p,\rho )}{\partial \rho}>0$, which shows that

$F(p,\rho )$ is an increasing function on the closed interval

$[0,1]$. This implies that maximum occurs at

$\rho =1$. Therefore

$maxF(p,\rho )=F(p,1)=G(p)$ (say). Now

$G(p)=\frac{1}{12\text{,}288}[17{p}^{4}-96{p}^{2}+768].$

Therefore

${G}^{\prime}(p)=\frac{1}{12\text{,}288}[68{p}^{3}-192p]$

and

${G}^{\u2033}(p)=\frac{1}{12\text{,}288}[204{p}^{2}-192]<0$

for

$p=0$. This shows that maximum of

$G(p)$ occurs at

$p=0$. Hence, we obtain

$\begin{array}{rcl}|{a}_{2}{a}_{4}-{a}_{3}^{2}|& \le & \frac{768}{12\text{,}288}\\ =& \frac{1}{16}.\end{array}$

This result is sharp for the functions

$\frac{z{f}^{\prime}(z)}{f(z)}=\sqrt{1+z}$

or

$\frac{z{f}^{\prime}(z)}{f(z)}=\sqrt{1+{z}^{2}}.$

□

**Theorem 2.5** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then* $|{a}_{2}{a}_{3}-{a}_{4}|\le \frac{1}{6}.$

*Proof* Since

$\begin{array}{c}{a}_{2}=\frac{1}{4}{p}_{1},\hfill \\ {a}_{3}=\frac{1}{8}{p}_{2}-\frac{3}{64}{p}_{1}^{2},\hfill \\ {a}_{4}=\frac{1}{12}{p}_{3}-\frac{7}{96}{p}_{1}{p}_{2}+\frac{13}{768}{p}_{1}^{2}.\hfill \end{array}$

Therefore, by using Lemma 1.3, we can obtain

$|{a}_{2}{a}_{3}-{a}_{4}|\le \frac{1}{768}\{7{p}^{3}+8p\rho (4-{p}^{2})+32(4-{p}^{2})+16{\rho}^{2}(p-2)(4-{p}^{2})\}.$

Let

${F}_{1}(p,\rho )=\frac{1}{768}\{7{p}^{3}+8p\rho (4-{p}^{2})+32(4-{p}^{2})+16{\rho}^{2}(p-2)(4-{p}^{2})\}.$

(2.5)

We assume that the upper bound occurs at the interior point of the rectangle

$[0,2]\times [0,1]$. Differentiating (2.5) with respect to

*ρ*, we get

$\frac{\partial {F}_{1}}{\partial \rho}=\frac{1}{768}\{8p(4-{p}^{2})+32\rho (p-2)(4-{p}^{2})\}.$

For

$0<\rho <1$ and fixed

$p\in (0,2)$, it can easily be seen that

$\frac{\partial {F}_{1}}{\partial \rho}<0$. This shows that

${F}_{1}(p,\rho )$ is a decreasing function of

*ρ*, which contradicts our assumption; therefore,

$max{F}_{1}(p,\rho )={F}_{1}(p,0)={G}_{1}(p)$. This implies that

${G}_{1}^{\prime}(p)=\frac{1}{768}\{21{p}^{2}-64p\}$

and

${G}_{1}^{\prime \prime}(p)=\frac{1}{768}\{42p-64\}<0$

for $p=0$. Therefore $p=0$ is a point of maximum. Hence, we get the required result. □

**Lemma 2.6** *If the function* $f(z)=\stackrel{\mathrm{\infty}}{\sum _{n=1}}{a}_{n}{z}^{n}$ *belongs to the class* ${\mathit{SL}}^{\ast}$,

*then* $|{a}_{2}|\le 1/2,\phantom{\rule{2em}{0ex}}|{a}_{3}|\le 1/4,\phantom{\rule{2em}{0ex}}|{a}_{4}|\le 1/6,\phantom{\rule{2em}{0ex}}|{a}_{5}|\le 1/8.$

*These estimations are sharp*. *The first three bounds were obtained by Sokół* [3]*and the bound for* $|{a}_{5}|$ *can be obtained in a similar way*.

**Theorem 2.7** *Let* $f\in {\mathit{SL}}^{\ast}$ *and of the form* (1.1).

*Then* $|{H}_{3}(1)|\le \frac{43}{576}.$

*Proof* Since

${H}_{3}(1)={a}_{3}({a}_{2}{a}_{4}-{a}_{3}^{2})-{a}_{4}({a}_{4}-{a}_{2}{a}_{3})+{a}_{5}({a}_{1}{a}_{3}-{a}_{2}^{2}).$

Now, using the triangle inequality, we obtain

$|{H}_{3}(1)|\le |{a}_{3}||{a}_{2}{a}_{4}-{a}_{3}^{2}|+|{a}_{4}||{a}_{2}{a}_{3}-{a}_{4}|+|{a}_{5}||{a}_{1}{a}_{3}-{a}_{2}^{2}|.$

Using the fact that

${a}_{1}=1$ with the results of Corollary 2.3, Theorem 2.4, Theorem 2.5 and Lemma 2.6, we obtain

$|{H}_{3}(1)|\le \frac{43}{576}.$

□