数理解析研究所講究録 1341 巻 2003 年 31-44

31
SOME FAMLIES OF MEROMORPHIC MULTIVALENT FUNCTIONSINVOLVING CERTAIN LINEAR OPERATOR

Jin-Lin Liu
Deparment Mathematics
Yangzhou University
Yangzhou 225002, Jiangsu
People Republic of China
E-mail jlliucn@yahoo.com.cn$:$

Shigeyoshi Owa
Deparment Mathematics
Kinki University

Higashi-Osaka Osaka 577-8502
Japan
E-mail : owa@math.kIndai.ac.jP
AbstractLet$\mathrm{f}(\mathrm{z})=z^{-p}+\sum_{k-0}^{\infty}a_{k}z^{k}$punctured disc$\sum_{p}$denotethe$(p\in N=\{1,2,3,\cdots\})$$D=\{z:0<|z |<1\}$ .

class

of functions

which are analytic andof the$p$formvalent in the

We introduce and study some new families ofmeromorphic multivalent functions defined by certain linear operator. Anumber of
useful characteristics of fimctions in these families are obtained. In particular, someproperties of neighborhoods of functions in these families are given.
Key words and phrases. Meromorphic, neighborhood, operator, partial sum.
;Secondary
2000 Mathematics Subject Classification. Primary$33\mathrm{C}20$$30\mathrm{C}45$.

,$30\mathrm{D}30$

1. Introduction
Let$\sum_{p}$denote the class of$\mathrm{f}\mathrm{i}\mathrm{m}\mathrm{c}\dot{\mathrm{b}}\mathrm{o}\mathrm{n}\mathrm{s}$$f(z)=z^{-p}+ \sum_{k4}^{\infty}a_{k}z^{k}$which are analytic and$p$ofthe form$(p\in N=\{1,2,3,\cdots\})$(1.1)-valent in the punctured disc $D=\{z:0<|z|<1\}$ . Define a

linear operator as followin$D^{0}f(z)=f(z)$32$D^{1}f(z)=z^{-p}+(p+1)a_{0}+(p+2)a_{1}z+(p+3)a_{2}z^{2}+\cdots$$= \frac{(z^{p+1}f(z))’}{z^{p}}$$D^{2}f(z)=D(D^{1}f(z))$

and for $n=1,2,\cdots$$D^{n}f(z)=D(D^{n-1}f(z))=z^{-p}+ \sum_{k-0}^{\infty}(p+k+1)^{n}a_{k}z^{k}$$= \frac{(z^{p+1}D^{n-1}f(z))’}{z^{p}}$

.

(1.2)

It is easy to see that$z(D^{n}f(z))’=D^{n+1}f(z)-(p+1)D^{n}f(z)$Whenoperator$D^{n}$$p=,\mathrm{U}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{d}\mathrm{i}$$D^{n}$.

(1.3)

and Somanatha[20] investigated certain properties of the. Recendy, Aouf and Hossen [2] showed some results of the operator

for $p\in N=\{1,2,3,\cdots\}$ .

Let $-1\leq B<A\leq 1$ . Afunction$f(z)=z^{-p}+ \sum_{k\cdot 0}^{\infty}a_{k}z^{k}\in\sum_{p}$is said to be in theclass $T_{n}(A,B)$ if it satisfies the condition(1.4)

for all $z\in E=\{z :|z|<1\}$ .
Furthermore, afunction$T_{n}^{\cdot}(A,B)$$f(z)=z^{-p}+ \sum_{k-p}^{\infty}|a_{k}|z^{k}\in\sum_{p}$is said to be in the class

if it satisfies the condition (1.4).

It should be remarked in passing that the definition (1.4) is motivated essentially by
the recent work of Mogra [13] and Liu and Srivastava [12]. The special class
TO(AyB) was studied by Mogra [13]. Another subclass associated with the linear

operator
was considered recently by Liu and Srivastava[12].
In recent years, many importantand characteristics ofvarious interestin$D^{n}$$\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{n}\cdot \mathrm{e}\mathrm{s}$33subclasses of the class$\sum_{p}$of meromorphically$p$-valent fimctions were

investigated extensively by (among others) Aouf et al. ([2] and [3]), Joshi and, Liu and Srivastava ([10],[11] and [12]), Mogra ([13]Srivastava [8], Kulkarni etand [14] , Owa et a1.[15], Srivastava et a1.[18], Uralegaddi and Somanatha ([20] and[21] , and Yang ([22]). The main object of this paper is to present several inclusion$\mathrm{a}\mathrm{i}.[9]$

$)$

$)$

and other properties of functions in the classes $T_{n}(A,B)$ andwhich we$T_{n}^{\cdot}(A,B)$

have introduced here. We also apply the familiar concept of neighborhoods of analytic

ffinctions (see [6] and [16]) to meromorphically$p$-valent ffinctions in the class$\Sigma_{p}$2. Properties of the class $T_{n}(A,B)$
In proving our results, we shall need the following lemma which is due to Jack [7].

Lemma.
$|w(z)|$

Let

be non-constant analytic in $E=\{z:|z|<1\}$ , $w(0)=0$ . If

$w(z)$

attains its maximum value on the circle $|z|=r<1$ at$z_{0}w’(z_{0})=Mz_{0})$ ,

where

$k$

is real number and$k\geq 1$$z_{0}$

we have,.

Theorem 2.1. Let $1+B\geq p(A-B)$ , then $T_{n+1}(A,B)\subset T_{n}(A,B)$ .

Proof. Let $f(z)\in T_{n+1}(A,B)$ . Suppose that$\frac{z(D^{n}f(z))’}{D^{n}f(z)}=-p\frac{1+Aw(z)}{1+Bw(z)}$

(2.1)

,

where $w(0)=0$ .
By using (2.1) and (1.3), we have.

$\frac{z(D^{n+1}f(z))’}{D^{n+1}f(z)}=-p\frac{1+Aw(z)}{1+Bw(z)}-\frac{p(A-B)zw’(z)}{(1+Mz))\{1+[B-p(A-B)]w(z)\}}$

Suppose now that, for$z_{0}w’(z_{0})=\mathrm{w}(\mathrm{z})$$z_{0}\in E$$k\geq 1$,$\max|w(z)|\triangleleft \mathrm{d}z_{\Phi}||-\dashv w(z_{0})|=1$. Writing$\mathrm{w}(\mathrm{z}0)=e^{l\theta}$$z(D^{n+1}f(z))’+pD^{n+1}f(z)2$$z(D^{n+1}f(z))’$ $+pD^{n+1}f(z)$$Bz(D^{n+1}f(z))’$$Bz(D^{n+1}f(z))’+ApD^{n*1}f(z)$$+ApD^{n*1}f(z)$$z\cdot z_{\mathrm{Q}}$

-1

(2.2)

. Applying Lemma, we haveand putting $z=z_{0}$ in (2.2), we obtain34$=| \frac{(k+1)+[B-p(A-B)]e^{j\theta}}{1-[B(k-1)+p(A-B)]e^{i\theta}}|^{2}-1$$= \frac{H_{1}(\cos\theta)}{|1-[B(k-1)+p(A-B)]e^{l\theta}|^{2}}$

(2.3)

,

where
$H_{1}(\cos\theta)=k^{2}(1-B^{2})+2k[1+B^{2}-pB(A-,B)]+2k[2B-p(A-B)]\cos\theta$.

Since $1+B\geq p(A-B)$ , then$H_{1}(1)=k^{2}(1-B^{2})+2k(1+B)[(1+B)-p(A-B)]\geq 0$

and$H_{1}(-1)=k^{2}(1-B^{2})+2k(1-B)[(1-B)+p(A-B)]\geq 0$ .

This shows that $H(\cos\theta)\geq 0$ for all$\theta(0\leq\theta<2\pi)$

and it follows that (2.3)

contradicts the hypothesis $f(z)\in T_{n+1}(A,B)$ . Hence $|w(z)|<1$ for all $z\in E$ and(2.1) shows that $f(z)\in H\mathrm{n}(A,B)$ .

Theorem 2.2

Let. If $f(z)$${\rm Re} \lambda\geq pa\frac{1+A}{1+B}$$a>0$$\square$and let !be acomplex number such that$\in T_{n}(\mathrm{i}4,5)$, then the fimction$\mathrm{g}\{\mathrm{z}$) defined by$D^{n}g(z)= \{\frac{\lambda-pa}{z^{\lambda}}\int_{0}^{z}t^{\lambda-1}(D^{n}f(t))^{a}dt\}^{\frac{1}{a}}$(2.4)

also belongs to $T_{n}(A,B)$ .
Proof. Put$\frac{z(D^{n}g(z))}{D^{n}g(z)},$$=-p \frac{1+Aw(z)}{1+Bw(z\rangle}$

,

(2.5)

where $w(0)=0$ .

Using (2.4) and (2.5) together with some computations, it follows that$\frac{z(D^{n}f(z))’}{D^{n}f(z)}=-p\frac{1+Aw(z)}{1+Bw(z)}-\frac{p(A-B)zw’(z)}{(1+Bw(z))[(\lambda-pa)+(B\lambda-Apa)w(z)]}$(2.6)

The remaining part of the proof of Theorem 2.2 is similar to that of Theorem 2.1
and so is omitted.
$\square$

3. Properties of the class$T_{n}^{\cdot}(A,B)$35
In this section, we assume that $A+B\leq 0$ .

Theorem 3.1

Let$/( \mathrm{z})=z^{-p}+\sum_{k=p}^{\infty}|a_{k}|z^{k}$$D=\{z:0<|z|<1\}$ . Then$f(z)\in T_{n}^{*}(A,B)$

be analytic and$p$-valent in

if and only if$\sum_{k\cdot p}^{\infty}[p(1-A)+k(1-B)](p+k+1)^{n}|a_{k}|\leq p(A-B)$

.

(3.1)

The result is sharp for the fimction $f(z)$ given by$f(z)=z^{-p}+ \frac{p(A-B)}{(p+k+1)^{n}[p(1-A)+k(1-B)]}z^{k}$

Proof. Let$f(z)=z^{-p}+ \sum_{k\cdot p}^{\infty}|a_{k}|z^{k}\in T_{n}^{\cdot}(A,B)$

$<1$

Since

$|{\rm Re} z|\leq|z|$

for any$(k\geq p)$

.

(3.2)

. Then.

(3.3)$z,\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}z$

to be real and letting$zarrow 1^{-}$

ffiroug realvalues, (3.3) yields$\sum_{k\cdot p}^{\infty}(p+k\mathrm{X}p+k+1)^{n}|a_{k}|\leq p(A-B)+\sum_{k-p}^{\infty}(Ap+Bk)(p+k+1)^{n}|a_{k}|$,which gives (3.1).
On the other hand, we have that$<1$This shows that$/(\mathrm{z})$.

$T_{n}^{\cdot}(A,B).\square$

Next, we prove the following growth and distortion property for the class$T_{n}^{\cdot}(A,B)$.

Theorem 3.2 Let 05

$m$

$<p$

. If

$/(\mathrm{z})\in T_{n}^{l}(A,B)$, then for $0<|z|=r<1$ ,36$\{\frac{(p+m-1)!}{(p-1)!}-\frac{(A-B)p!}{(2-(A+B))(2p+1)^{n}(p-m)!}r^{2p}\}r^{-(p+m)}$1$f^{(m)}(z)|$$\leq\{\frac{(p+m-1)!}{(p-1)!}+\frac{(A-B)p!}{(2-(A+B)\mathrm{X}2p+1)^{n}(p-m)!}r^{2p}\}r^{\prec p+m)}$

The results ore sharp for the function $f(z)$ given by$f(z)=z^{-p}+ \frac{A-B}{(2-(A+B)\mathrm{X}2p+1)^{n}}z^{p}$

Proof. Let $f(z)\in T_{n}^{\cdot}(A,B)$ . Then we find fiom Theorem 3.1 that$\frac{p(2-(A+B)\mathrm{X}2p+1)^{n}}{p!}\sum_{k-p}^{\infty}k!|a_{k}|\leq\sum_{k\cdot p}^{\infty}[p(1-A)+k(1-B)](p+k+1)^{\hslash}|a_{k}$$\leq p(A-B)$ ,which yields(3.4)$\sum_{k\cdot p}^{\infty}k!|a_{k}|\leq^{j}(2-(A+B))(2p+1)^{n}(A-B)p^{1}$

Now by differentiating $f(z)m$ times, we have
(3.5)$f^{(m)}(z)=(-1)^{n} \frac{(p+m-1)!}{(p-1)!}z^{-p-m}+\sum_{k-p}^{\infty}\frac{k!}{(k-m)!}|a_{k}|z^{k- m}$Hence Theorem 3.2 would follow ffom (3.4) and (3.5).

$\square$Finally, we detennine the radius of meromorphicallyconvexity for functions in the class
Theorem 33 Let(i)$T_{n}^{\cdot}(A,B)$$\mathrm{f}(\mathrm{z})\mathrm{e}T_{n}^{*}(A,B)$$f(z)$ is meromorphically$p$${\rm Re} \{\frac{zf’(z)}{f(z)}\}<-p\delta$

where $0\leq\delta<1$ and$p$-valent starlikeness and.

. Thenvalent starlike oforder$(|z|<r_{1})$$\delta$in$|z|<r_{1}$, that is

(3.6)

37
$r_{1}= \inf_{k\geq p}\{\frac{(1-\delta)[p(1-A)+k(1-B)](p+k+1)^{n}}{(A-B)(k+p\delta)}\}^{\overline{k+p}}-$(ii)$f(z)$ is meromorphically$p$-valent convex oforder${\rm Re} \{1+\frac{zf^{l}(z)}{f’(z)}\}<-p\delta$

where$0\leq\delta<1$$(|z|<r_{2})$$\delta$in $|z|<r_{2}$ , that is
(3.7)

,

and

$r_{2}= \inf_{k\geq p}\{\frac{p(1-\delta)[p(1-A)+k(1-B)](p+k+1)^{n}}{k(k+p\delta \mathrm{X}A-B)}\}^{\frac{1}{k+p}}$

Eachthese results is sharp for the fimction $f(z)$ given by (3.2).

Proof, (i) From Theorem 3.1, we have$\sum_{k-p}^{\infty}\frac{k+p\delta}{p(1-\delta)}|a_{k}||z|^{k+p}<\sum_{k-p}^{\infty}\frac{[p(1-A)+k(1-B)](p+k+1)^{n}}{p(A-B)}|a_{k}|$$(|z|<r_{1})$$\leq 1$

Therefore for.

$|z|<r_{1}$$zf’(z)/f(z)+p$$\leq\frac{\sum_{k\cdot p}^{\infty}(k+p)|a_{k}||z|^{k+p}}{\infty}$$ff’(z)/f(z)-p(1-2\delta)$$2p(1- \delta)-\sum_{k\cdot p}[k-p(1-2\delta)]|a_{k}||z|^{k+p}$

$<1$

,

which shows that (3.6) is true,(ii) It follows from Theorem 3.1 that$\sum_{k\cdot p}^{\infty}\frac{k(k+p\delta)}{p^{2}(1-\delta)}|a_{k}||z|^{k+p}<\sum_{k\cdot p}^{\infty}\frac{[p(1-A)+k(1-B)](p+k+1)^{n}}{p(A-B)}|a_{k}$51Thus for$|z|<r_{2}$.we obtain$<1$

which shows that (3.7) is true,$(|z|<r.1$$(|z|<r_{2})$.

38
Sharpness can be verified easily.

$\square$

4. Neighborhoods and partial sums
Following the earlier works (based upon the familiar concept of neighborhoods ofanalytic functions) by Goodman[6] and Ruscheweyh[16], and(more recently) by
Altintas and Owa [1] and Liu and Srivastava ([10] and [12]), we begin by introducing
here the -neighborhood of afunctionof the form (1.1) by means of the$f \in\sum_{p}$$\delta$definition:$N_{\delta}(f)= \{g(z)=z^{-p}+\sum_{k-0}^{\infty}b_{k}z^{k}\in\sum_{p}$:$\sum_{k\cdot 0}^{\infty}\frac{[p(1-A)+k(1-B)]}{p(A-B)}(p+k+1)^{n}|b_{k}-a_{k}|4,-1\leq B<A\leq 1;\delta\geq 0\}$.

Making use ofthe definition’ we now prove
Theorem 4.1

Let$\delta>0\mathrm{a}\mathrm{n}\mathrm{d}-1<A\leq 0$.If$f(z)=z^{-p}+ \sum_{k4}^{\infty}a_{k}z^{k}\in\sum_{p}$satisfies

the condition(4.1)$\frac{f(z)+\epsilon z^{-p}}{1+\epsilon}\in T_{n}(A,B)$

for any complex number$e$such that$|\epsilon|<\delta$, then $N_{\delta}(f)\subset Tn(A,B)$ .

Proof. It is obvious ffom (1.4) that $g(z)\in T_{n}(A,B)$ if and only if for any complex

number$\sigma$with$|\sigma|=1$$\frac{z(D^{n}g(z))’+pD^{n}g(z)}{Bz(D^{n}g(z))+ApD^{n}g(z)},\neq\sigma$$(z\in E)$,

which is equivalent to$\frac{g(z)*h(z)}{z^{-p}}\neq 0$$(z\in E)$

,

(4.2)

where
$h(z)=z^{-p}+ \sum_{k=0}^{\infty}c_{k}z^{k}$$=z^{-p}+ \sum_{k-0}^{\infty}\frac{[(p+k)-\sigma(pA+kB)]}{p\sigma(B-A)}(p+k+1)^{n}z^{k}$

(4.3)

39
From (4.3), we have$|c_{k}|=| \frac{[(p+k)-\sigma(pA+kB)]}{p\sigma(B-A)}(p+k+1)^{n}|$$\leq\frac{p(1-A)+k(1-B)}{p(A-B)}(p+k+1)^{n}$Ifsatisfies the condition (4.1), then (4.2) yields$f(z)=z^{-p}+ \sum_{k4}^{\infty}a_{k}z^{k}\in\sum_{p}$$f(z)$$\mathrm{s}$$h(z)$$(z\in E)$$f(z)*h(z)z^{-p}\geq\delta$$z^{-p}$Now let$\mathrm{p}\{\mathrm{z}$)$=z^{-p}+ \sum_{k\triangleleft}^{\infty}b_{k}z^{k}\in N_{\delta}(f)$$\exists$

.

(4.4)

, then$z| \sum_{k4}^{\infty}\frac{[p(1-A)+k(1-B)]}{p(A-B)}(p+k+1)|b_{k}-a_{k}|$$<\delta$

Thus for any complex number$\sigma$.
such that$|\sigma|=1$$(z \in E)$$\frac{p(z)*h(z)}{z^{-p}}\neq 0$, we have,

which implies that $p(z)\in T_{n}(A,B).\square$
Theorem 42Let

and$-1<A\leq 0$$sm(z)=z^{-p}+ \sum_{k-0}^{m-2}a_{k}z^{k}$. Let$(m\geq 2)$$f(z)=z^{-p}+ \sum_{k\cdot 0}^{\infty}a_{k}z^{k}\in\sum_{p}$where$c_{k}= \frac{p(1-A)+k(1-B)}{p(A-B)}(p+k+1)^{n}$(i)$f(z)\in T_{n}(A,B)$;$s_{1}(z)=z^{-p}$. Suppose that$\sum_{k\underline{-}0}^{\infty}c_{k}|a_{k}|\leq 1$

Then we have

,

(4.5)

40
(ii)

(4.6)

${\rm Re} \{\frac{f(z)}{s_{m}(z)}\}>1-\frac{1}{c_{m- 1}}$and${\rm Re} \{\frac{s_{m}(z)}{f(z)}\}>\frac{c_{l\hslash- 1}}{1+c_{m- 1}}$

.

(4.7)

The estimates are sharp.
Proof,(i) It is obvious that $z^{-p}\in T_{n}(A,B)$ . Thus from Theorem 4.1 and thecondition (4.5), we have$(\mathrm{i}\mathrm{i})\mathrm{I}\mathrm{t}$$N_{1}(z^{-p})\subset T_{n}(A,B)$. This gives$\mathrm{f}(\mathrm{z})$$T_{n}(A,B)$.

is easy to see that $c_{k+1}>c_{k}>1$ . Thus$. \sum_{k\approx 0}^{-2}|a_{k}|+c_{m-1}\sum_{k=n-1}^{\infty}|a_{k}|\leq\sum_{k=0}^{\infty}c_{k}|a_{k}|\leq 1$

.

(4.8)

Let
$h_{1}(z)=c_{m-1} \{\frac{f(z)}{s_{n}(z)}-(1-\frac{1}{c_{n\vdash 1}})\}$$=1+ \frac{c_{n1-1}\sum_{k\cdot m-1}^{\infty}a_{k}z^{k+p}}{n-2}$$1+ \sum_{k\cdot 0}a_{k}z^{k+p}$

It follows from (4.8)$(z\in E)$.

From this we obtain the inequality (4.6).

If we take$f(z)=z^{-p}- \frac{z^{m-1}}{c_{m-1}}$

,

(4.9)

then
$\frac{f(z)}{\mathrm{s}_{n}(z)}=1-\frac{z^{p+m- 1}}{c_{m- 1}}arrow 1-\frac{1}{c_{n- 1}}$as$zarrow 1^{-}$

This shows that the bound in (4.6) is best possible for each
Similarly, if we take

$m$

.

41
$h_{2}(z)=(1+c_{m-1}) \{\frac{s_{m}(z)}{f(z)}-\frac{c_{m- 1}}{1+c_{m- 1}}\}$$=1- \frac{(1+c_{m-1})\sum_{k=m-1}^{\infty}a_{k}z^{k+p}}{1+\sum_{k*0}^{\infty}a_{k}z^{l+p}}$,

then we deduce that$h_{2}(z)-1\leq\underline{(1+c_{m-1})\sum^{\infty}|a_{k}|k-m-1}$$h_{2}(z)-1$$h_{2}(z)+1$$2-2 \sum_{k4}^{n-2}|a_{k}|+(1-c_{m-1})\sum^{\infty}|a_{k}|k\cdot m-1$

51$(z\in E)$ ,which yields (4.7). The estimate (4.7) is sharp with the extremal function $f(z)$ givenby (4.9).
For$\square$$\delta$$\geq 0,$$-1\leq B<A\leq 1$

neighborhood of$/(\mathrm{z})$and$\mathrm{f}(\mathrm{z})=z^{-p}+\sum_{k\cdot 0}^{\infty}|a_{k}|z^{k}\in\sum_{p}$by$N_{\delta}^{l}(f)= \{g(z)=z^{-p}+\sum_{k\cdot p}^{\infty}|b_{k}|z^{k}\in\sum_{p}$:$\sum_{k\cdot p}^{\infty}\frac{[p(1-A)+k(1-B)]}{p(A-B)}(p+k+1)^{n}|b_{k}|-|a_{k}||\ovalbox{\tt\small REJECT}$

Theorem 43$N_{\delta}^{\cdot}(f)\subset T_{n}^{\cdot}(A,B)$

we define,Let$A+B\leq 0$, where. If $f(z)$$\delta=\frac{2p}{1+2p}$

$\}$

.

$=z^{-p}+ \sum_{k-p}^{\infty}|a_{k}|z^{k}\in T_{n+1}^{*}(4B)$. The result is sharp.

Proof. Using the same method as in Theorem 4.1 we would have$h(z)=z^{-p}+ \sum_{k\approx p}^{\infty}c_{k}z^{k}$

Under the hypothesis $A+B\leq 0$ , we obtain that

,

then

42
$| \frac{f(z)*h(z)}{z^{-p}}|=|1+\sum_{k\overline{-}p}^{\infty}c_{k}|a_{k}|z^{k+p}|$$\geq 1-\frac{1}{1+2p}\sum_{k\cdot p}^{\infty}\frac{[p(1-A)+k(1-B)]}{p(A-B)}(p+k+1)^{n+1}|a_{k}|$.

From Theorem 3.1, we get

The remaining part ofthe proof is similar to that of Theorem 4.1.
To show the sharpness, we consider the$\mathrm{f}\mathrm{i}\mathrm{m}\mathrm{c}\dot{\mathrm{b}}\mathrm{o}\mathrm{n}$$f(z)=z^{-p}+ \frac{A-B}{(2-(A+B))(1+2p)^{n*1}}z^{p}\in T_{n+1}^{*}(A,B)$

and$g(z)=z^{-p}+[ \frac{A-B}{(2-(A+B))(1+2p)^{n+1}}+\frac{(A-B)\delta’}{(2-(A+B)\mathrm{X}1+2p)^{n}}]z^{p}$

where$\delta’>\frac{2p}{1+2p}$. Then the fimction $g(z)$ belongs to$N_{\delta}^{\cdot},(f)$.

On the other hand, we find ffom Theorem 3.1 that $g(z)$ is not in
the proof is complete.

,$T_{n}^{\cdot}(A,B)$. Now$\square$Acknowledgements
The authors warmly thank the referee for his suggestions and criticisms which haveessentiaUy improved our original paper

43
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