数理解析研究所講究録
第 1878 巻 2014 年 100-111

100

THE SHARP GROWTH ESTIMATE FOR$\mathcal{U}(\lambda)$

Hiroshi Yanagihara
Department of Applied Science
Faculty of Engineering, Yamaguchi University
Tokiwadai, Ube 755-8611, Japanbe the class of analytic
ABSTRACT. For $0<\lambda\leq 1$ let
functions in the unit disk with $f(O)=f’(O)-1=0$ satisfying $|f’(z)(z/f(z))^{2}-1|<\lambda$ in D. Then it is known that every
is univalent in . In the present article we shall prove
estimates $|f"(0)|\leq 2(1+\lambda)$ and $|z|/\{(1+|z|)(1+\lambda|z|)\}\leq$the$|f(z)|\leq|z|/\{(1-|z|)(1-\lambda|z|)\}$ . As an application we shall also
give the sharp covering theorems.
$\mathcal{U}(\lambda)$$\mathbb{D}$$\mathbb{D}$$f\in \mathcal{U}(\lambda)$$sha\iota p$

1. INTRODUCTION
We denote the complex plane by and the extended complex plane. For
and $r>0$ let $\mathbb{D}(c, r)=\{z\in \mathbb{C}:|z-c|<r\}$by: $r<|z|\leq$
let
.
and
and
denote the space of analytic functions in andLet$\{f\in \mathcal{A}(\{\mathbb{D}\}) : f(O)=f’(O)-i=0\}$ . Here we regard
as atopological vector space endowed with the topology of uniform converfunction is said to be univalent ingence over compact subsets ofa domain if it is one-to-one in . Let denote the class of univalent.
functions insatisfying
be the class of functions
For $0<\lambda\leq 1$ let$\mathbb{C}$$\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$$c\in \mathbb{C}$$\mathbb{D}=\mathbb{D}(0,1)$$Simi1_{\dot{c}}u\cdot 1y$$\Delta_{t}=\hat{\mathbb{C}}\backslash \overline{\mathbb{D}}(0,r)=\{z\in\hat{\mathbb{C}}$$\Delta=\Delta_{1}.$$\infty\}$$A_{0}(\{\mathbb{D}\})=$$\mathbb{D}$$\mathcal{A}(\{\mathbb{D}\})$$A(\{\mathbb{D}\})$$\mathbb{D}.$

$A$

$f$

$D$

$D$

$S$

$\mathcal{A}_{0}(\{\mathbb{D}\})$$f\in \mathcal{A}_{0}$$\mathcal{U}(\lambda)$(1.1)$|f’(z)( \frac{z}{f(z)})^{2}-1|<\lambda$. The boundedness of $f’(z)(z/f(z))^{2}$ forcesthat $f(z)\neq 0$. Hence $f’(z)\neq 0$ holds in and is locally univalent in
is univalent in , i.e.,
Moreover it is known thatinin$\mathbb{D}$$\mathbb{D}$$\mathbb{D}\backslash \{0\}$$f\in \mathcal{U}(\lambda)$$f\in \mathcal{U}(\lambda)$$\mathbb{D}.$$f$$\mathbb{D}$$\mathcal{U}(\lambda)\subset S.$

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In the present article we shall prove the sharp inequalities$|a_{2}(f)|=2^{-1}|f^{l/}(0)|\leq 1+\lambda,$$\frac{|z|}{(1-\lambda|z|)(1-|z|)}\leq|f(z)|\leq\frac{|z|}{(1-\lambda|z|)(1-|z|)}$for. To this end we introduce three classes of meromorphic
functions in closely related to. For $0<\lambda\leq i$ let
be theclass of meromorphic functions in which ha a Laurent expansionof a form$f\in \mathcal{U}(\lambda)$$\Delta$$\mathcal{U}(\lambda)$

$g$

(1.2)

$\mathcal{M}(\lambda)$$\Delta$$g(w)=w+c_{0}+ \frac{c_{1}}{w}+\frac{c_{2}}{w^{2}}+\cdots, 1<|w|<\infty$

and satisfying$|g’(w)-1|<\lambda.$

Now for$f\in \mathcal{U}(\lambda)$put$T(f)(w)= \frac{1}{f(\frac{1}{w})}, w\in\Delta.$

Then we have$\frac{d}{dw}T(f)(w)=\frac{f’(1/w)}{w^{2}f^{2}(\frac{1}{w})}=f’(z)(\frac{z}{f(z)})^{2}, z=\frac{1}{w}$and hence. Thus is a transformation which maps
injectively into
. The image
is a proper subset of
and it is easy to see that$Tf\in \mathcal{M}(\lambda)$$\mathcal{M}(\lambda)$$T$$T(\mathcal{U}(\lambda))$$\mathcal{M}(\lambda)$$\mathcal{M}_{0}(\lambda)=\{g\in \mathcal{M}(\lambda);g(w)\neq 0 in \Delta\}=T(\mathcal{U}(\lambda))$Notice that $a_{2}(f)=-c_{0}(T(f))$ hold for$\mathcal{U}(\lambda)$$f\in \mathcal{U}(\lambda)$.

. Moreover let$\tilde{\mathcal{M}}(\lambda)=\{g\in \mathcal{M}(\lambda):c_{0}(g)=0\}.$In the next section we shall show that every
in . Thus the relation$0$$g\in\tilde{\mathcal{M}}(\lambda)$

satisfies $g(w)\neq$$\Delta$$T(\tilde{\mathcal{U}}(\lambda))=\tilde{\mathcal{M}}(\lambda)\subset T(\mathcal{U}(\lambda))=\mathcal{M}_{0}(\lambda)\subset \mathcal{M}(\lambda)$

holds.
In Section 2 we shall derive an integral representation ofand prove existence of the boundary limitforeach. Further we shall precisely study the boundary values ofand obtain the sharp estimate $|g(\eta)|\leq 1+\lambda$ forand, which is equivalent to the sharp upper bound $|a_{2}(f)|\leq$for.
$g\in \mathcal{M}(\lambda)$$g( \eta)=\lim_{\Delta\ni warrow\eta}g(w)$$\sim\eta\in\partial\Delta$$g\in \mathcal{M}(\lambda)$$\eta\in\partial\Delta$$1+\lambda$$f\in \mathcal{U}(\lambda)$$g\in\tilde{\mathcal{M}}(\lambda)$

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we shall treat the sharp
In Section 3 for each fixed. The result has an immediateforestimate on
and we shall derive the sharp growth estimate andcounterpart in.
the sharp covering theorem for$w_{0}\in\Delta\backslash \{\infty\}$$g\in \mathcal{M}_{0}(\lambda)$$|g(w_{0})|$$\mathcal{U}(\lambda)$$\mathcal{U}(\lambda)$

2. INTEGRAL REPRESENTATIONFor$g\in \mathcal{M}(\lambda)$

let

(2.1)

$b_{g}(w)= \frac{w^{2}}{\lambda}(1-g’(w)) , w\in\Delta,$(2.2)$\beta_{9}(z)=b_{9}(1/z) , z\in \mathbb{D}.$and that $g’(1/z)$ is analytic
Note that $g’(w)-1=O(w^{-2})$ as
in . Applying the maximum modulus principle to $g’(1/z)-1$ we have$warrow\infty$$\mathbb{D}$$|g’(w)-1| \leq\frac{\lambda}{|w|^{2}}, 1<|w|<\infty.$

and for any $w,$
$1=-\lambda b_{g}(w)w^{-2}$ we obtainHence$\beta_{g}\in H_{1}^{\infty}(\mathbb{D})$$w_{0}\in\Delta\backslash \{\infty\}$

by integrating $g’(w)-$$g(w)-w=g(w_{0})-w_{0}- \lambda\int_{w0}^{w}\frac{b_{g}(\zeta)}{\zeta^{2}}d\zeta.$

Since$\lim_{w_{0^{arrow\infty}}}(g(w_{0})-w_{0})=c_{0}$, we have$g(w)=w+c_{0}- \lambda\int_{\infty}^{w}\frac{b_{g}(\zeta)}{\zeta^{2}}d\zeta=w+c_{0}+\lambda\int_{0}^{1/w}\beta_{g}(\zeta)d\zeta.$

Converse is also true and we have the following.
Theorem 2.1. For a meromorphic function insuch thatandonly if there esist$g$$\beta\in H_{1}^{\infty}(\mathbb{D})$$\Delta,$$g\in \mathcal{M}(\lambda)$

if and$c\in \mathbb{C}$$g(w)=w+c+ \lambda\int_{0}^{1/w}\beta(\zeta)d\zeta.$

Corollary 2.2. Each(2.3)

$g\in \mathcal{M}(\lambda)$

is Lipschitz continuous and satisfies$(1- \frac{\lambda}{|w_{0}w_{1}|})|w_{1}-w_{0}|\leq|g(w_{1})-g(w_{0})|\leq(1+\frac{\lambda}{|w_{0}w_{1}|})|w_{1}-w_{0}|$for$w_{0},w_{1}\in\Delta$. Particularly(i) The limit $g( \eta)=\lim_{\Delta\ni warrow\eta}g(w)$ の ists for every(ii)
is univalent in . Furthermore if $0<\lambda<1$ , then$\eta\in\partial\Delta.$$\Delta$$g$

univalent on$\overline{\Delta}.$

$g$

is

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Proof. Inequality (2.3) easily follows from Theorem 2.1 and$| \int_{1/w_{0}}^{1/w1}\beta(\zeta)d\zeta|\leq|\frac{1}{w_{1}}-\frac{1}{w_{0}}|=\frac{|w_{i}-w_{0}|}{|w_{0}w_{1}|}.$By (2.3) the function is Lipschitz continuous inaild from thisthe boundary limit $g( \eta)=\lim_{\Delta\ni warrow\eta}g(w)$ exists for each
Inequality (2.3) also shows that is univalent in . Since (2.3) stillholds on by continuity, is univalent on , when $0<\lambda<1.$$\Delta$$g$$\eta\in\partial\Delta.$$\Delta$$g$$\overline{\Delta}$$\overline{\Delta}$$\square$

$g$

Each
Corollary 2.2.

$f\in \mathcal{U}(\lambda)$

is univalent in$\mathbb{D}$, since $Tf$ is univalent inCorollary 2.3. For each $0<\lambda\leq 1$ the inclusion relation
holds.

$\Delta$by$\tilde{\mathcal{M}}(\lambda)\subset$$\mathcal{M}_{0}(\lambda)$

Proof. For$g\in\tilde{\mathcal{M}}(\lambda)(\subset \mathcal{M}(\lambda))$we have by Theorem 2.1$|g(w)|=|w+ \lambda\int_{0}^{1/w}\beta(\zeta)d\zeta|\geq|w|-\frac{\lambda}{|w|}>0, |w|>1.$

Thus 9 has no zeros in

For$g\in \mathcal{M}(\lambda)$$\Delta$and hence$g\in \mathcal{M}_{0}(\lambda)$.

$\square$let $E(g)$ be the omitted set of , i.e.,$g$$E(g)=\hat{\mathbb{C}}\backslash g(\Delta)$.

For $R>1$ the image
is an analytic Jordan curve andis the domain outside. Let
be the domain bounded by. Then $\{D_{R} : 1<R\}$ is a 1-parameter family of increasing
domains in and$g(\partial\Delta_{R})$$g(\partial\Delta_{R})$$g(\Delta_{R})$$D_{R}$$g(\partial\Delta_{R})$$\mathbb{C}$$E(g)= \bigcap_{R>1}D_{R}.$In particular when $0<\lambda<1$ , by Corollary 2.2 $E(g)$ is a closed Jordan
domain bounded by. Notice that for any
with $0<$
and
there exists
such that $a=g(\eta)$ .
$g(\partial\Delta)$$\lambda\leq 1$$a\in\partial E(g)$Theorem 2.4. Let
and$\beta\in H_{i}^{\infty}(\mathbb{D})$(2.4)$g\in \mathcal{M}(\lambda)$$\eta\in\partial\Delta$

and $g(w)=w+c_{0}+ \int_{0}^{1/w}\beta(\zeta)d\zeta$ with. Thenif and only if$g\in \mathcal{M}(\lambda)$$c_{0}\in \mathbb{C}$$g\in \mathcal{M}_{0}(\lambda)$$-c_{0}\in E(\tilde{g})$

,

where
$\tilde{g}(w)=w+\int_{0}^{1/w}\beta(\zeta)d\zeta\in\tilde{\mathcal{M}}(\lambda)$

.

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Proof. For

if and only if $g(w)=$, by definition,. This is equivalent to (2.4).
for all$g\in \mathcal{M}_{0}(\lambda)$$g\in \mathcal{M}(\lambda)$$\square$$w\in\Delta$$\tilde{g}(w)+c_{0}\neq 0$

the coefficient $c_{0}=q(g)$ in the expansion (1.2)
For
is called the conformal center of the set $E(g)$ . For more details on
conformal center we refer to [8].
andholds for any
Notice that.
This implies that there are no upper bound on $|q(g)|$ for, it is not difficult to get the
However concerning with the classsharp estimate.
$g\in \mathcal{M}_{0}(\lambda)$$c\in \mathbb{C}.$$g\in \mathcal{M}(\lambda)$$9+c\in \mathcal{M}(\lambda)$$g\in \mathcal{M}(\lambda)$$\mathcal{M}_{0}(\lambda)$

Theorem 2.5. Let$\lambda\in(0,1]$. Then,the sharp estimate $1-\lambda\leq|g(\eta)|\leq 1+\lambda$ holds
on . Furthermore equahty $|g(\eta)|=1-\lambda$ at someif$|g(\eta)|=1+\lambda$
only
,
and
and
and only ifif(a) For$g\in\tilde{\mathcal{M}}(\lambda)$$\eta\in\partial\Delta$$\partial\Delta$$g(w) \equiv w-\lambda\prime\int^{2}/w$

if$g(w)\equiv w+\lambda\eta^{2}/w.$(b) Inequality $|q(g)|\leq 1+\lambda$ holds for$g\in \mathcal{M}_{0}(\lambda)$

and only ifwith equality if$g(w)=w(1+ \frac{\lambda e^{i\theta}}{w})(1+\frac{e^{i\theta}}{w}) , w\in\Delta.$

for some real$\theta.$(c) Inequahty $|a_{2}(f)|\leq 1+\lambda$ holds for $f\in u(\lambda)$ with equality if
and only if$f(z)= \frac{z}{(1+\lambda e^{i\theta}z)(1+e^{i\theta}z)}, z\in \mathbb{D}$for some real
Proof. Let$g\in\tilde{\mathcal{M}}(\lambda)$

(2.5)

For

$\theta.$and put$\beta\in H_{1}^{\infty}(\mathbb{D})$as in (2.2). Then$g(w)=w+ \lambda\int_{0}^{1/w}\beta(\zeta)d\zeta.$$\eta\in\partial\Delta$

we consider$\int_{0}^{1/\eta}\beta(\zeta)d\zeta$

as the Lebesgue integral along

and contained in except for the end-path connecting to. Then the integral does not depend on choice of path. Thuspoint
and we have(2.5) still holds fora$C^{1}$$0$$1/\eta$$\mathbb{D}$$1/\eta$$\eta\in\partial\Delta$$|g( \eta)|\leq|\eta|+\lambda|\int_{0}^{1/\eta}\beta(\zeta)d\zeta|\leq 1+\lambda\int_{[0,1/\eta]}|\beta(\zeta)||d\zeta|\leq 1+\lambda,$

105
where $[0,1/\eta]$ is the radial line segment connecting and. If$|g(\eta_{0})|=1+\lambda$ at some, then by the maximum modulus
theoremfor someand$0$$1/\eta$$\eta_{0}\in\partial\Delta$$\epsilon\in\partial \mathbb{D}$$\beta=\epsilon$$\frac{\lambda\int_{0}^{1/\eta}\beta(\zeta)d\zeta}{\eta}=\frac{\lambda\epsilon}{\eta^{2}}>0.$

Therefore$\epsilon=\eta^{2}$

and $g(w)\equiv w+\lambda\eta^{2}/w$ . Similarly$|g( \eta)|\geq|\eta|-\lambda|\int_{0}^{1/\eta}\beta(\zeta)d\zeta|\geq 1-\lambda\int_{[0,1/\eta]}|\beta(\zeta)|\downarrow d\zeta|\geq 1-\lambda$, i.e.,
with equality if and only if
and$g(w)\equiv
w-\lambda\eta^{2}/w$
and therefore. This completes the proof of (a).
$\beta=\epsilon\in\partial \mathbb{D}$$\lambda\epsilon/\eta^{2}<0$$\epsilon=-\eta^{2}$To show (b) let. Then can be expressed aswith. By (a) and $E( \tilde{g})=\bigcap_{R>1}D(R)$ , where$D(R)$ is a domain bounded by the Jordan curve. Hence wehave
and $|c_{0}(g)|\leq 1+\lambda.$
Suppose now that $|c_{0}(g)|=1+\lambda$ . Combining $-c_{0}(g)\in E(\tilde{g})\cap$, we haveand. By
Lipschitz continuity of there exists $\eta\in\partial\Delta with-c_{0}(g)=g(\eta)$ . Since$|\tilde{g}(\eta)|=|-c_{0}(g)|=1+\lambda$ , it follows from (a) that $\tilde{g}(w)=w+\lambda\eta^{2}/w$
and hence$g\in \mathcal{M}_{0}(\lambda)$$g=\tilde{g}+c_{0}$$g$$\tilde{g}\in\tilde{\mathcal{M}}(\lambda)and-c_{0}\in E(\tilde{g})$$\tilde{g}(\partial\Delta_{R})$$E(\tilde{g})\subset\overline{\mathbb{D}}(0,1+\lambda)$$\partial \mathbb{D}(0,1+\lambda)$$E(\tilde{g})\subset\overline{\mathbb{D}}(0,1+\lambda)$$-c_{Q}(g)\in\partial E(\tilde{g})$$\tilde{g}$$g(w)= \tilde{g}(w)+c_{0}=\tilde{g}(w)-\tilde{g}(\eta)=w(1+\frac{\lambda\eta}{w})(1+\frac{\eta}{w})$By letting$e^{i\theta}=-\eta$.

we obtain (b).

Since $a_{2}(f)=-c_{0}(T(f))$ holds for $f\in u(\lambda),$ $(c)$ follows directlyfrom (b).
$\square$

3. GROWTH ESTIMATES
Let $0<\lambda\leq 1$ and $1<|w_{0}|<\infty$ . Combimng Theorem 2.1 andTheorem 2.5 (b) it is easily seen that ifthen$g\in \mathcal{M}_{0}(\lambda)$(3.1)$|g(w_{0})| \leq|w_{0}|+\lambda\int_{[0,i/w_{0}]}|\beta(\zeta)||d\zeta|+|c_{0}|$$\leq|w_{0}|+\frac{\lambda}{|w_{0}|}+1+\lambda=|w_{0}|(1+\frac{\lambda}{|w_{0}|})(1+\frac{1}{|w_{0}|})$

106
if and only if $g(w)\equiv w(1+\lambda e^{i\theta}w^{-i})(1+$. Similarly the lower estimate

with equality at$e^{i\theta}w^{-1})$$w_{0}=Re^{i\theta}$$|g(w_{0})| \geq|w_{0}|-\frac{\lambda}{|w_{0}|}-(1+\lambda)$

holds foris negative for ffi. Clearly it is not sharp, since the right hand side$g\in \mathcal{M}(\lambda)$

sufficiently close to 1.

$r$

In this section we deal with the region of variability, i.e.,
when varies on$V_{\lambda}(w_{0})$of$g(w_{0})$,$\mathcal{M}_{0}(\lambda)$$g$$V_{\lambda}(w_{0})=\{g(w_{0}):g\in \mathcal{M}_{0}(\lambda)\}.$

is a closed Jordan domain bounded by a
We shall show that
simple closed curve and give a parameterization of the boundary curve.
Using these ideas we will obtain the sharp lower estimate on.
when
is a compact subset of . Indeed by (3.1)
First we notice thatwhich
is a family of analytic functions in
it is clear thatis locally uniformly bounded and hence normal. Moreover if a sequence, then it isconverge to locally uniformly ininis a compact family. Thusnot difficult to see thatwith respect to the topology of locally uniform convergence. Sincewith respect to the continuous mappingis the image ofis also an compact subset offor. Thenlet
Next for. From this it follows thatany
$V_{\lambda}(w_{0})$$|g(w_{0})|$$g\in \mathcal{M}_{0}(\lambda)$$\mathbb{C}$$V_{\lambda}(w_{0})$$\Delta\backslash \{\infty\}$$\mathcal{M}_{0}(\lambda)$$\mathcal{M}_{0}(\lambda)$$\{g_{n}\}_{n=1}^{\infty}$$\Delta\backslash \{\infty\}$$g$$g\in \mathcal{M}_{0}(\lambda)$$\mathcal{M}_{0}(\lambda)$$\mathcal{M}_{0}(\lambda)$$V_{\lambda}(w_{0})$$\mathcal{M}_{0}(\lambda)\ni g\mapsto g(w_{0})\in \mathbb{C},$$g\in \mathcal{M}_{0}(\lambda)$$\mathbb{C}.$$V_{\lambda}(w_{0})$$g_{\theta}(w)=e^{-i\theta}g(e^{i\theta}w)$$g_{\theta}\in \mathcal{M}_{0}(\lambda)$$\theta\in \mathbb{R}$$V_{\lambda}(Re^{i\theta})=e^{i\theta}V_{\lambda}(R)$for $1<R<\infty$ . Similarly for
and it suffices to determineisand hence. Thenlet
symmetric with respect to$V_{\lambda}(R)$$\overline{g}(w)=\overline{g(\overline{w})}$$g\in \mathcal{M}_{0}(\lambda)$$\overline{g}\in \mathcal{M}_{0}(\lambda)$$V_{\lambda}(R)$$\mathbb{R}.$$Th\infty rem3.1$(i) For. Let$0<\lambda\leq 1$. Then$g\in \mathcal{M}_{0}(\lambda)$$|w|(1- \frac{\lambda}{|w|})(1-\frac{1}{|w|})\leq|g(w)|\leq|w|(1+\frac{\lambda}{|w|})(1+\frac{1}{|w|})$

for $1<|w|<\infty$ with equality$g(w)=w(1- \frac{\lambda c^{i\theta_{0}}}{w})(1-\frac{e^{i\theta_{0}}}{w})$respectively.

or$w_{0}=R_{0}e^{i\theta\eta}$,

if and only if$g(w)=w(1+ \frac{\lambda e^{i\theta_{0}}}{w})(1+\frac{e^{i\theta_{0}}}{w})$,107(ii) For$f\in \mathcal{U}(\lambda)$$\frac{|z|}{(1+|z|)(1+\lambda|z|)}\leq|f(z)|\leq\frac{|z|}{(1-|z|)(1-\lambda|z|)},$

with equality at$z=r_{0}e^{i\theta_{0}}$if and only ifor$f(z)= \frac{z}{(1+\lambda e^{i\theta_{0}}z)(1+e^{i\theta_{0}}z)}$$0<|z|<1$$f(z)= \frac{z}{(1-\lambda e^{i\theta_{0}}z)(1-e^{i\theta_{0}}z)}$respectively.

Theorem 3.2. Let$f\in \mathcal{U}(\lambda)$with $0<\lambda\leq 1$ . Then$\mathbb{D}(0, \frac{1}{2(1+\lambda)})\subset f(\mathbb{D})$Funhermore$\frac{e^{t\theta_{0}}}{2(1+\lambda)}\not\in f(\mathbb{D})$.

holds if and only if$f(z)= \frac{z}{(1+\lambda e^{-i\theta 0}z)(1+e^{-1\theta_{0}}z)}.$Now we define auxiliary functions. For$\epsilon\in\overline{\mathbb{D}}$let$\tilde{G}_{\lambda,\vee=}(w)=w+\frac{\lambda\epsilon}{w}$and$E_{\lambda}=\{\begin{array}{ll}\{u+iv:(u/(1+\lambda))^{2}+(v/(1-\lambda))^{2}\leq 1\}, 0<\lambda<1{[}-2, 2], \lambda=1.\end{array}$

Notice that

$E(\tilde{G}_{\lambda,c^{i\theta}})=e^{i\theta/2}E_{\lambda}$

for$\theta\in \mathbb{R}.$Proposition 3.3. Let. Ifwith $|\epsilon|=|\eta|=1$ , such that$g\in \mathcal{M}_{0}(\lambda)$$\epsilon,$$g(R)\in\partial V_{\lambda}(R)$$g=\tilde{G}_{\lambda,\epsilon}-\tilde{G}_{\lambda,c}(\eta)$$\eta$

Proof. By Theorem 2.4$g$.

, then there enists

can be decomposed as $g=\tilde{g}+q$ , where$\tilde{g}(w)=w+\lambda\int_{0}^{1/w}\beta_{g}(\zeta)d\zeta\in\tilde{M}(\lambda)and-c_{0}\in E(\tilde{g})$rem 2.4. Again by Theo-$g(R)=\tilde{g}(R)+c_{0}\in\tilde{g}(R)-E(\tilde{g})\subset V_{\lambda}(R)$.

Thus, otherwise $g(R)$ is an
cannot be an interior point ofinterior point of. Hence. By Lipschitz continuity$that-c_{0}=\tilde{g}(\eta)$of there exists
such. Therefore$E(\tilde{g})$$-c_{\theta}$$V_{\lambda}(R)$$\tilde{g}$- $c_{0}\in\partial E(\tilde{g})$$\eta\in\partial\Delta$$g(R)= \tilde{g}(R)-\tilde{g}(\eta)=R-\eta+\lambda\int_{1/\eta}^{1/R}\beta_{g}(\zeta)d\zeta.$

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Notice that

$R\neq\eta$, since $g(R)\neq 0$ . Then we have$|l_{/\eta}^{1/R} \beta_{g}(\zeta)d\zeta|\leq|\frac{1}{R}-\frac{1}{\eta}|$with equality if and only if
Suppose that$\beta_{9}=\epsilon$

for some$\epsilon$

with$|\epsilon|=1.$$|l_{/\eta}^{1/R} \beta_{g}(\zeta)d\zeta|<|\frac{1}{R}-\frac{1}{\eta}|=|\frac{\eta-R}{R\eta}|.$Put$\epsilon_{0}=\frac{R\eta}{\eta-R}\int_{1/\eta}^{1/R}\beta_{g}(\zeta)d\zeta(\in\mathbb{D})$.

Then$\tilde{G}_{\lambda.e0}(R)-\tilde{G}_{\lambda,\epsilon 0}(\eta)=R-\eta+\lambda\epsilon_{0}(\frac{1}{R}-\frac{1}{\eta})$$=R-\eta+\lambda l_{/\eta}^{1/R}\beta_{g}(\zeta)d\zeta=\tilde{g}(R)-\tilde{g}(\eta)=g(R)$

.

, we
for
. The mappingfor. Sinceis an analytic function of, the mapping is not constant and hence it is an open mapping.
, whichis an interior point of
Thusfor. Therefore
contradict the assumption that
.
and
some

On the other hand since
have$\tilde{G}_{\lambda,c}-\tilde{G}_{\lambda,c}.(\eta)\in \mathcal{M}_{0}(\lambda)$$c\in\overline{\mathbb{D}}$$\mathbb{D}\ni c\mapsto$$c\in \mathbb{D}$$\tilde{G}_{\lambda,c}(R)-\tilde{G}_{\lambda,c}(\eta)\in V_{\lambda}(R)$$c\in \mathbb{D}$$\tilde{G}_{\lambda},,(R)-\tilde{G}_{\lambda,r}.(\eta)\in V_{\lambda}(R)$$R\neq\eta$$g(R)=\tilde{G}_{\lambda,\epsilon 0}(R)-\tilde{G}_{\lambda,\epsilon 0}(\eta)$$V_{\lambda}(R)$$g(R)\in\partial V_{\lambda}(R)$$\beta_{g}=\epsilon$$\square$$g=\tilde{G}_{\lambda,\epsilon}-\tilde{G}_{\lambda,\epsilon}(\eta)$$\epsilon\in\partial \mathbb{D}$Proof of Theorem 3.1. Since (ii) follows directly from (i), it suffices toshow (i). From compactness ofsuch that$\mathcal{M}_{0}(\lambda)$

it follows that there exist 91,$g_{2}\in$$\mathcal{M}_{0}(\lambda)$$|g_{1}(R)|= \min_{g\in \mathcal{M}_{0}(\lambda)}|g(R)|$

Then clearlyfact thatand. Also$|g_{2}(R)|= \max_{g\in \mathcal{M}_{0}(\lambda)}|g(R)|.$

follows from thewith. Thus by Proposition 3.3 there exist
for $j=1,2$ . Sincesuch that$g_{2}(R)\in\partial V_{\lambda}(R)$$g_{1}(R)\in\partial V_{\lambda}(R)$$0\not\in V_{\lambda}(R)$$|\epsilon_{j}|=|\eta_{j}|=1$$\epsilon_{j},$$g_{j}\cdot=\tilde{G}_{\lambda,e_{j}}-\tilde{G}_{\lambda,e_{j}}(\eta_{j})$$g_{j}(R)= \tilde{G}_{\lambda,\epsilon_{j}}(R)-\tilde{G}_{\lambda,\epsilon_{j}}(\eta_{j})=(R-\eta_{j})(1-\frac{\lambda\epsilon_{j}}{R\eta_{j}})$,$\eta_{j}$109
we have$|g_{1}(R)|= \min_{g\in \mathcal{M}o(\lambda)}|g(R)|\leq\tilde{G}_{\lambda,1}(R)-\tilde{G}_{\lambda,1}(1)$$=(R-1)(1- \frac{\lambda}{R})$$\leq|(R-\eta_{i})(1-\frac{\lambda\epsilon_{1}}{R\eta_{1}})|=|g_{1}(R)|.$

Thus$\eta_{1}=\epsilon_{1}=1$

and hence$91(w) \equiv\tilde{G}_{\lambda.1}(w)-\tilde{G}_{\lambda,1}(1)=w+\frac{\lambda}{w}-(1+\lambda)$

We have shown that for.

$g\in \mathcal{M}(\lambda)$$(R-1)(1- \frac{\lambda}{R})\leq|g(R_{1})|$

with equality if and only if $g(w)=w+\lambda w^{-1}-(1+\lambda)$ . Applying thisto
we have for
and$w=Re^{i\theta}$$g_{\theta}(w)=e^{-i\theta}g(e^{i\theta}w)$$g\in \mathcal{M}_{0}(\lambda)$$(|w|-1)(1- \frac{\lambda}{|w|})=(R-1)(1-\frac{\lambda}{R})\leq|g_{\theta}(R)|=|g(w)|$with equality $g_{\theta}(w)=w+\lambda w^{-1}-(1+\lambda)$ , i.e.,$g(w)=w+ \lambda e^{2i)}w^{-1}-(1+\lambda)e^{i\theta}=w(1-\frac{\lambda e^{i\theta}}{w})(1-\frac{e^{i\theta}}{w})$In the same manner we can treat the rest of the proof of (i).

Proof of Theorem 3.2. For$f(\mathbb{D})$$f\in\cdot \mathcal{U}(\lambda)$

directly follows from 3.1 (ii).

the relation.
$\square$$\mathbb{D}(0, (2(1+\lambda))^{-1})\subset$

Suppose that
. Then $2(1+\lambda)e^{-i\theta_{0}}\in E(g)=$$E(\tilde{g})+c_{0}(g)$ , where $g=Tf=\tilde{g}+c_{0}(g)$. Sincewithand $|c_{0}(g)|\leq 1+\lambda$ by
Theorem 2.5, we have$e^{i\theta_{0}}\{2(1+\lambda)\}^{-1}\not\in f(\mathbb{D})$$\tilde{g}\in\tilde{\mathcal{M}}(\lambda)$$2(1+\lambda)e^{-i\theta_{O}}-c_{0}(g)\in E(\tilde{g})\subset\overline{\mathbb{D}}(0,1+\lambda)$$1+\lambda\leq 2(1+\lambda)-|c_{0}(g)|\leq|2(1+\lambda)e^{-t}\prime 0_{0}-c_{0}(g)|\leq 1+\lambda.$Thus$c_{0}(g)=(1+\lambda)e^{-i\theta_{0}}$. By Theorem 2.5 (b)$e^{i\theta}w^{-1})=w+(1+\lambda)e^{i\theta}+\lambda e^{2i\theta}w^{-1}$for some$g(w)=w(1+\lambda e^{i\theta}w^{-1})(1+$$\theta\in \mathbb{R}$. Therefore$e^{i\theta}=e^{-i\theta_{0}}$and $g(w)=w(1+\lambda e^{-i\theta_{0}}w^{-1})(1+e^{-i\theta_{0}}w^{-1})$ . This implies$f(z)= \frac{z}{(1+\lambda e^{-i\theta_{0}}z)(1+e^{-i\theta_{0}}z)}.$$\square$110
Proposition 3.3 implies that$\partial\partial V_{\lambda}(R)$is contained in$V_{\lambda}^{*}(R)= \{(R-\eta)(1-\frac{\lambda\epsilon}{R\eta}):|\epsilon|=|\eta|=1\}.$FIGURE 1.

$V_{0.5}^{*}(2)$and$V_{0.9}^{*}(1.1)$

andconsists of two Jordan curves
One can proveis contained inside ofwhich are starlike with respect to and.
, and thatis a closed Jordan domain surrounded by
For details see forthcoming paper [11].
$J_{e}(R)$$\partial V_{\lambda}^{*}(R)$$R$$J_{e}(R)$$J_{i}(R)$$J_{i}(R)$$J_{e}(R)$$V_{\lambda}(R)$

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class$\mathcal{U}(\lambda)$