# A Fundamental Inequality of Algebroidal Function

Authors: Yingying Huo; Daochun Sun
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By using a new mapping of Ahlfors covering surfaces, a fundamental inequality in the angular domain for the algebroidal function is obtained.


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## Body

## 1. Introduction and Main Results

In the field of valued distribution, the fundamental inequality is an important tool. For example, it can be used to investigate the singular direction [1]. Using geometric theory, Tsuji firstly obtained the second fundamental theory in an angular domain and proved the existence of Borel direction [2]. The value distribution theory of meromorphic functions was extended to algebroidal functions last century [3]. In 1983, Lv and Gu proved an inequality of algebroidal function for an angular domain [4]. By the inequality, some results of singular direction are obtained; see [5, 6]. In [7], the authors obtained a more accurate inequality for angular domain. In this paper, we will use a new method to simplify and extent an inequality of Tsuji to algebroidal functions.First, we recall some definitions from [3].Suppose thatAv(z),…,A0(z) are analytic functions with no common zeros in the complex plane. Ψ(z,W) is a bivariate complex function and satisfies
(1.1)Ψ(z,W)=Av(z)Wv+Av-1(z)Wv-1+⋯+A0(z)=0.
For all z in the complex plane, the equation Ψ(z,W)=0 has v complex roots w1(z),w2(z),…,wv(z). Then, (1.1) defines a v-valued algebroidal function W(z); see [3, 8]. If Av(z)=1, then W(z) is called v-valued integral algebroidal function. If Ψ(z,W) is irreducible, correspondingly W(z) is called v-valued irreducible algebroidal function (note that W(z) is a meromorphic function, if v=1). Now we suppose that W(z) is an irreducible algebriodal function defined by (1.1).IfAv(z0)≠0, and the k-degree equation Ψ(z0,W)=0 and its partial derivative (∂Ψ/∂W)(z0,W)=0 have no common roots (i.e., z0 is not a multiple root of Ψ(z0,W)=0), then z0 is said to be a regular point. The set of all regular points is called the regular set, denoted by TW. Its complementary set SW : ={z∣|z|<∞}-TW is called the critical set. Obviously, Sw includes all branch points of W(see [3]).The domain of av-valued irreducible algebriodal function W is a connected Riemann surface [8], and its single-valued domain is denoted by R~z. A point in R~z is z~ and sets lying over |z|<r and {ϕ1<argz<ϕ2}(ϕ1<ϕ2) are |z~|<r and Ω~(ϕ1,ϕ2). Let n(r,W=a) and n(Ω(ϕ1,ϕ2),r,W=a) be the number of zeros, counted according to their multiplicities, of W=a in {|z~|<r} and {|z~|<r}⋂Ω~(ϕ1,ϕ2), respectively. Let n¯(r,W=a) be the number of distinct zeros in {|z~|<r}, and let n(r,R~z) be the number of branch points in {|z~|<r}. Similarly, we can define n¯(Ω(ϕ1,ϕ2),r,W=a) and n(Ω(ϕ1,ϕ2),r,R~z). Let
(1.2)S(r,W)=1π∬|z~|≤r(|W′(z)|1+|W(z)|2)2dWz=reiθ,S(Ω(ϕ1,ϕ2),r,W)=1π∬{|z~|≤r}⋂Ω~(ϕ1,ϕ2)(|W'(z)|1+|W(z)|2)2dW,T(r,W)=1v∫0rS(t,W)tdt,T(Ω(ϕ1,ϕ2),r,W)=1v∫0rS(Ω(ϕ1,ϕ2),t,W)tdt,N(r,W=a)=1v∫0rn(t,W=a)-n(0,W=a)tdt+n(0,W=a)vlnr,N(r,R~z)=1v∫0rn(t,R~z)-n(0,R~z)tdt+n(0,R~z)vlnr,m(r,W)=12πv∑k=1v∫02πln+|wk(reiθ)|dθ.Similarly, we can defineN(Ω(ϕ1,ϕ2),r,W=a), N¯(Ω(ϕ1,ϕ2),r,W=a) and N(Ω(ϕ1,ϕ2),r,R~z). From [3], we know that T(r,w)=N(r,W=∞)+m(r,W)+O(1) and N(r,R~z)≤2(v-1)T(r,W)+O(1).In this paper, we will prove the main theorem.Theorem 1.1.
LetW(z) be a v-valued algebroidal function in region Ω(ϕ1,ϕ2)≜{|z|∣ϕ1<argz<ϕ2}(ϕ1<ϕ2). a1,a2,…,aq(q⩾3) are q different complex numbers on the sphere with radius δ∈(0,1/2). For ϕ,ε*,ε(0<ε*<ε,ϕ1<ϕ-ε<ϕ-ε*<ϕ+ε*<ϕ+ε<ϕ2), and R>R*>2, we have
(1.3)(q-2)S(Ω(ϕ-ε*,ϕ+ε*),R*,W)⩽n(Ω(ϕ-ε,ϕ+ε),R,R~z)+∑j=1qn¯(Ω(ϕ-ε,ϕ+ε),R,W=aj)+256vπ24lnRδ38(ε-ε*)(lnR-lnR*)+(q-2)S(Ω(ϕ-ε*,ϕ+ε*,1R*,W)).By the inequality in Theorem1.2, we will immediately have the following.Theorem 1.2.
For a meromorphic functionW (a 1-valued algebroidal function with no branch points) defined by (1.1) satisfying
(1.4)limR→∞¯T(r,W)ln2R=∞,
it has at least one Nevanlinna direction, that is, there exists argz=ϕ0, such that ∑a∈C⋃{∞}δ(a,ϕ0)≤2 holds for any finitely many deficient value a, where
(1.5)δ(a,ϕ0)=1-limε→0+¯limR→∞¯N¯(Ω(ϕ0-ε,ϕ0+ε),R,W=a)T(Ω(ϕ0-ε,ϕ0+ε),R,W)>0.


## 2. Some Lemmas

First, it is easy to prove the following.Lemma 2.1.
Suppose thata,b>0, then there exists p,q>0, such that
(2.1)p+iq=1a-bi,
where
(2.2)p=a2+b2+a2(a2+b2),q=a2+b2-a2(a2+b2).Lemma 2.2.
Suppose thatA(z)=∫0z(dt/t(1-t2)) and B(z)=((1-z)/(1+z))i, then (1)
the mappingG1=A∘B(z) maps the unit disc U to the square Q≜{z=x+iy∣0<x<2h,0<y<2h}, where h is constant;(2)
G1 maps {|z|<r}, where 0<r<1, into a symmetrical convex region in Q;(3)
h in Lemma 2.2 satisfies
(2.3)h=∫2-11dtt(1-t2)=∫02-1dtt(1-t2)>1.Proof.
Obviously, (1) holds. By the definition ofB(z), we have B(1)=0,B(-1)=∞,B(i)=1,B(-i)=-1.
In order to computeh, first we prove that G1 maps d(0), d(π/2), d(π/4), d(-π/4) to four symmetry axes of Q, where d(θ)={reiθ;-1<r<1}. Let z=reiθ (θ is fixed), r∈(-1,1),
(2.4)G1(z)=A∘B(z)=∫0Bdtt(1-t2)=-(1+i)∫1reiθdr1-r4ei4θ.
Hence, whenθ=0, π/4, π/2, -π/4, argG1(reiθ)=π/4,π/2,-π/4,0, respectively, see Figure 1.
Then, we computeh. Since z=0 is the only intersection point of the lines d(0) and d(π/4). The center of the square Q, h+hi, is that of the curves G1(d(0)) and G1(d(π/4)). Then, G1 conforms 0 onto h+hi.
Hence,(2.5)h=G1(eiπ/4)=A(2-1)=∫02-1dtt(1-t2)=∫2-11dtt(1-t2)>∫2-11dtt>1.
(2) At last we prove thatG1({|z|<r}) is a convex region, see Figure 1.
For a fixedr∈(0,1), by (2.4),
(2.6)G1(reiθ)=∫1z-(1+i)dz1-z4=(1+i)∫r1dr1-r4+∫0θ(1-i)dθ(r-2-r2)cos2θ-i(r-2+r2)sin2θ.
SetG1(reiθ)=x(θ)+iy(θ). When θ∈(0,π/4), cos2θ, sin2θ>0, by Lemma 2.1,
(2.7)∂G1∂θ=1-i(r-2-r2)cos2θ-i(r-2-r2)sin2θ=(1-i)(p+iq)=(p+q)-(p-q)i≜x'+y'i,
where p>q>0. 							(2.8)∂2G1∂θ2=(1-i)[(r-2-r2)sin2θ+i(r-2+r2)cos2θ][(r-2-r2)cos2θ-i(r-2+r2)sin2θ]3=(1-i)(p+qi)[(r-2-r2)sin2θ+i(r-2+r2)cos2θ](r-2-r2)cos2θ-i(r-2+r2)sin2θ=(p-q)(r-4-r4)-2(p+q)sin4θ(r-2-r2)2cos22θ+(r-2+r2)2sin22θ+i[2(p-q)sin4θ+(p+q)(r-4-r4))(r-2-r2)2cos22θ+(r-2+r2)2sin22θ=x′′+y′′i.
Hence,(2.9)dydx=y'x'=-p-qp+q,d2ydx2=y′′x′-y′x′′x'2=2(r-4-r4)(p2+q2)x'2[(r-2-r2)cos22θ+(r-2+r2)2sin2θ]>0.
Therefore, the image ofL(r)={reiθ;0<θ<π/4} is a descending convex curve. By the symmetry of square, G1({|z|=r}) is a smooth curve, and G1({|z|<r<1}) is a convex symmetric figure in the square Q.
Then, we obtain Lemma2.2.Figure 1Lemma 2.3.
Suppose that mappings(2.10)C(z)≜h+i(z-h)(wherehisdefinedinLemma  2.2),D(z,z¯)≜xlnRh+iyεh=logR+ε2hz+logR-ε2hz¯,H(z)≜ez,G2(z)≜C-1∘D-1∘H-1(z).
Then, the mapping G2 maps the region E≜{1/R<|z|<R}⋂{|argz|⩽ε} into the square Q, and G1-1∘G2(E*)⊂{|z|<r}, where E*≜{1/R*<|z|<R*}⋂{|argz|⩽ε*}, R*∈(2,R), ε*∈(0,ε), and
(2.11)0<1-r<min{(ε-ε*)22πε2,(ε-ε*)(lnR-lnR*)4πεlnR}.Proof.
The conclusion is equivalent toG2(E*)⊂G1({|z|<r}). We prove the lemma by two cases.
Case I. When
(2.12)ε*ε⩾lnR*lnR,
then
(2.13)0⩽argG2(h-hε*ε+i(h-hlnR*/lnR))=arg[h(ε-ε*ε+ilnR-lnR*lnR)]⩽π4.G1({|z|<r<1}) is a convex symmetric figure if there exists a θ0∈(0,π/4], such that
(2.14)ReG1(reiθ0)<hε-ε*ε,ImG1(reiθ0)<hlnR-lnR*lnR,
then Lemma 2.3 holds, see Figure 2.
In fact, for anyr∈(0,1), θ∈(0,π/4],
(2.15)G1(reiθ)=-∫1z(1+i)dz1-z4=∫1eiθ(1+i)dz1-z4+∫eiθreiθ(1+i)dz1-z4≜α+β,
where
(2.16)α=∫1eiθ(1+i)dz1-z4=∫0θ-(1+i)ieiθ1-ei4θdθ=∫0θdθsin2θ<π2∫0θdθθ=πθ,β=∫eiθreiθ(1+i)dz1-z4=(1+i)∫r1dr(1-r4)cos2θ-i(1+r4)sin2θ≜(1+i)∫r1dra-bi,
where
(2.17)a=(1-r4)cos2θ<1,b=(1+r4)sin2θ,a2+b2=1+r8+2r4(sin22θ-cos22θ)=(1-r4)+4r4sin22θ>4r4sin2θ.
By Lemma2.1, we have
(2.18)β=∫r1(1+i)(p+iq)dr=∫r1[(p-q)+(p+q)i]dr≜ξ+iη,
where
(2.19)ξ=∫r1(p-q)dr=∫r1p2-q2p+qdr<∫r1p2+q2pdr<∫r12dra2+b2<∫r1drr22sin2θ<1-r2rsin2θ<1-r2rπθ,η=∫r1(p+q)dr<2∫r1pdr<2∫r1dra2+b24<1-rrπ2θ.
Let(2.20)θ0=(ε-ε*)24πε2<π4.
Combing (3.1) (notethat by (3.1), we have r>2/2),
(2.21)ReG1(reiθ0)=α+ξ<πθ0+1-r2rπθ0<h(ε-ε*)ε,ImG1(reiθ0)=η<1-rrπ2θ<12lnR-lnR*lnR<hlnR-lnR*lnR.
Therefore, a vertex ofG2(E*)(h-ε*h/ε,h-hlnR*/lnR)∈G1({|z|<r}). By Lemma 2.2, G2(E*)⊂G1({|z|<r}).
Case II. When
(2.22)ε*ε≤lnR*lnR,
since G1({|z|<r<1}) is a convex symmetric figure, we also have Lemma 2.3, see Figure 3.Figure 2Figure 3For the convenience of readers, we prove the following lemma again, it can be found in [9].Lemma 2.4.
(1) LetG(z)=G2-1∘G1(z)eiϕ, then
(2.23)|Gz|+|Gz¯|⩽lnRε(|Gz|-|Gz¯|).
(2) Puts(x,y)=ReG, t(x,y)=ImG, then
(2.24)sx2+sy2+tx2+ty2⩽lnRε(sxty-sytx).Proof.
(1) Sincef≜H-1∘G1(z)eiϕ and C-1 are holomorphic functions, then
(2.25)|f¯z|=|fz¯|=|(C-1)D-1¯|=0,|fz¯¯|=|fz|.
For(2.26)D(z)=xlnRh+iyεh
then
(2.27)D-1(f,f¯)=xhlnR+iyhε=h(f+f¯)2lnR+h(f-f¯)ε.
Hence,
(2.28)max{|Df-1|+|Df¯-1||Df-1|-|Df¯-1|}=|lnR+ε|+|lnR-ε||lnR+ε|-|lnR-ε|=lnRε.
Thus,(2.29)|Gz|+|Gz¯|=|(C-1)D-1Df-1fz|+|(C-1)D-1Df¯-1f¯z¯|⩽|(C-1)D-1fz|lnRε(|Df-1|-|Df¯-1|)=lnRε(|(C-1)D-1Df-1fz|-|(C-1)D-1Df¯-1f¯z¯|).
(2) By(2.30)sx=sz+sz¯sy=sz-isz¯,
we have
(2.31)sz=sx-isy1-i,sz¯=sy-isx1-i.
Similarly,
(2.32)tz=tx-ity1-i,tz¯=ty-itx1-i.
Then,(2.33)|Gz|2=12[(ty+sx)2+(tx-sy)2],|Gz¯|2=12[(tx+sy)2+(ty-sx)2].
Therefore,(2.34)sx2+sy2+tx2+ty2=|Gz|2+|Gz¯|2⩽(|Gz|+|Gz¯|)2(by(2.23))⩽lnRε(|Gz|2-|Gz¯|2)⩽lnRε(sxty-sytx).Lemma 2.5 (see [10]).
LetF be a connected covering surface on F0, F0 is bounded by q different points with radius δ∈(0,1/2), then
(2.35)max{0,ρ(F)}⩾(q-2)S-225π11δ-19L,
where L is the length of F and ρ(F) is Euler characteristic of F, |F| is the area of F and
(2.36)S=|F||F0|.Lemma 2.6 (see [10]).
LetV be a sphere with radius 1/2, F0 be bounded by q different points with radius δ∈(0,1/2) and Fr=W∘G(F0) then
(2.37)S=|Fr||F0|=1π∑k=1v∫0r∫02π|wk′|2(sxty-sytx)r(1+|wk∘G|2)2drdθ,
where s(x,y)=ReG, t(x,y)=ImG.Proof.
Suppose thatwk=u+iv. Then
(2.38)|Fr|=∑k=1v∬G(|z~|<r)1(1+|wk|2)2dudv,
where
(2.39)du=(ussx+uttx)dx+(ussy+utty)dy,dv=(vssx+vttx)dx+(vssy+vtty)dy.
Hence, by the Jacobian determinant, we have(2.40)dudv=[(us2+vs2)(sxty-sytx)]dxdy,|Fr|=∑k=1v∬|z~|<r(us2+vs2)(sxty-sytx)(1+|wk∘G|2)2dxdy=∑k=1v∫0r∫02π|wk′|2(sxty-sytx)r(1+|wk∘G|2)2drdθ.
By|F0|=π, we have Lemma 2.6.


## 3. Proof of Theorem1.1

Proof.
SetG(z)=G2-1∘G1(z)eiϕ. It conforms the unit disc U={|z|<1} to the sector E={1/R<|W|<R}⋂{|argz-ϕ|<ε} and the interior of U*={|z|<r} to E*={1/R*<|W|<R*}⋂{|argz-ϕ|<ε*}, where 2<R*<R, 0<ε*<ε, and
(3.1)0<1-r<min{(ε-ε*)22πε2,(ε-ε*)(lnR-lnR*)4πεlnR}.
Hence W∘G conforms {|z|<r} to the sphere V.
PutDr~={|z~|<r}. Then by M. Hurwite Formula, we have
(3.2)ρ(Dr~)=n(r,R~z)-v.
Put Dr=Dr~-{z∣∏j=1q(w∘G(z)-aj)=0} and Fr=W∘G(F0). Then
(3.3)ρ(Fr)=ρ(Dr)=n(r,R~z)-v+∑j=1qn¯(r,W∘G=aj)⩽n(1,R~z)-v+∑j=1qn¯(1,W∘G=aj)≜N.
By Lemma 2.5, it follows that
(3.4)N⩾ρ(Fr)⩾(q-2)S(r,W∘G)-227π11δ-19L(r).
Now we will prove
(3.5)L2(r)⩽24vπrlnRεdS(r,W∘G)dr.
For any r∈(0,1), k=1,2,…,v and ε>0, we have
(3.6)||wk∘G(reiθ)|-|wk∘Gj||<ε,
where θj=jπ/n(j=1,2,…,2n), θ∈[θj-1,θj], |wk∘Gj|=min{|wk∘G(reiθ)|,θj-1⩽θ⩽θj}.
By (3.6), for any θ∈[θj-1,θj], we have
(3.7)|wk∘G(reiθ)|2⩽|wk∘Gj|2+2ε(ε+|wk∘Gj|).
Therefore
(3.8)1+|wk∘G(reiθ)|21+|wk∘Gj|2⩽1+2ε21+|wk∘Gj|2+2ε|wk∘Gj|1+|wk∘Gj|2⩽2.
Put wk=uk(s,t)+ivk(s,t), G=s(r,θ)+it(r,θ). Hence
(3.9)Lk(r)≜limn→∞∑j=12n|wk∘G(reiθj)-wk∘G(reiθj-1)|1+|wk∘G(reiθj)|21+|wk∘G(reiθj-1)|2⩽limn→∞∑j=12n{[∫θj-1θj((uk)ssθ+(ut)stθ)dθ]2+[∫θj-1θj((vk)ssθ+(vk)ttθ)dθ]2}1/21+|wk∘Gj|2⩽2limn→∞∑j=12n∫θj-1θj[((uk)ssθ+(uk)ttθ)2+((vk)ssθ+(vk)ttθ)2]1/2dθ1+|wk∘Gkj|2=2∫02π|wk′|(sθ2+tθ2)1/2dθ1+|wk∘Gk|2.
By(3.10)sθ=-sxrsinθ+syrcosθ,tθ=-txrsinθ+t-tyrsinθ,
where
(3.11)x=-rcosθ,y=rsinθ,
we obtain
(3.12)sθ2+tθ2⩽2r2(sx2+sy2+tx2+ty2),L2(r)=(∑k=1vLk(r))2⩽8[∑k=1v∫02π|wk′|(sx2+sy2+tx2+ty2)1/2r1+|wk∘G|2dθ]2(by(2.24))⩽8lnRε[∑k=1v∫02π|wk′|(sxty-sytx)1/2r1+|wk∘G|2dθ]2(Cauchyinequality)⩽8vlnRε∑k=1v[∫02π|wk′|(sxty-sytx)1/2r1+|wk∘G|2dθ]2(Schwarzinequality)⩽8vlnRε∑k=1v[∫02π|wk′|2(sxty-sytx)r(1+|wk∘G|2)2dθ][∫02πrdθ]⩽16vπrlnRε∑k=1v∫02π‍|wk′|2(sxty-sytx)r(1+|wk∘G|2)2dθ(byLemma2.6)⩽16vπrlnRεdS(r,W∘G)dr.
(1) Ifforallr'∈(r,1)(3.13)S(r,W∘G)⩾Nq-2,
then by (3.4)
(3.14)(S(r′,W∘G)-Nq-2)2⩽250π22(q-2)2δ38L(r′)(by(3.5))⩽254vπ23r'lnR(q-2)2δ38εdS(r′,W∘G)dr',
that is,
(3.15)dr'⩽254vπ23lnR(q-2)2δ38εdS(r′,W∘G)(S(r′,W∘G)-N/(q-2))2.
Hence,
(3.16)1-r=∫r1dr′⩽254vπ23lnR(q-2)2δ38ε∫r1dS(r′,W∘G)(S(r′,W∘G)-N/(q-2))2⩽254vπ23lnR(q-2)δ38ε(1S(r,W∘G)-N/(q-2)-1S(R,W∘G)-N/(q-2))<254vπ23lnRδ38ε1(q-2)S(r,W∘G)-N.
Therefore
(3.17)(q-2)S(r,W∘G)⩽254vπ23lnRδ38ε(1-r)+N⩽n(1,R~z)+∑j=1qn¯(1,W∘G=aj)+254vπ23lnRδ38ε(1-r).
(2) If there is ar'∈(r,1), such that
(3.18)(q-2)S(r′,W∘G)-N<0,
then
(3.19)(q-2)S(r,W∘G)<(q-2)S(r′,W∘G)<N,
Equation (3.17) holds.
By (3.17) and Lemma 2.3, we have
(3.20)(q-2)S(Ω(ϕ-ε*,ϕ+ε*,R*,W))-(q-2)S(Ω(ϕ-ε*,ϕ+ε*,1R*,W))⩽(q-2)S(r,W∘G)(by(3.1))⩽n(Ω(ϕ-ε,ϕ+ε),R,R~z)+∑j=1q‍n¯(Ω(ϕ-ε,ϕ+ε),R,W=aj)+256vπ24lnRδ38(ε-ε*)(lnR-lnR*).


## 4. Proof of Theorem1.2

Proof.
By the hypothesis of Theorem1.2, there exists an increasing sequence Rn (Rn→∞, when n→∞), such that
(4.1)limn→∞T(Rn,W)ln2Rn=+∞.
Then, there exist someϕ0∈[0,2π], such that for arbitrary ε∈(0,ϕ0),
(4.2)limn→∞¯T(Rn,ϕ0-ε,ϕ0+ε,W)ln2Rn=+∞
holds. We claim that argz=ϕ0 is the Nevanlinna direction.
Otherwise, for a positive numberε0, there exist some a1,a2,…,aq(q≥3), such that
(4.3)∑j=1qδ(aj,ϕ0)>2+3ε0.
By the definition of δ(aj,ϕ0), we have
(4.4)limε→0+¯limR→∞¯∑j=1qN¯(Ω(ϕ0-ε,ϕ0+ε),R,W=aj)T(Ω(ϕ0-ε,ϕ0+ε),R,W)<q-2-3ε0.
There exists ε1>0, such that for any ε∈(0,ε1),
(4.5)limR→∞¯∑j=1qN¯(Ω(ϕ0-ε,ϕ0+ε),R,W=aj)T(Ω(ϕ0-ε,ϕ0+ε),R,W)<q-2-2ε0.
Hence, for{Rn} defined earlier, when n is sufficiently large,
(4.6)∑j=1qN¯(Ω(ϕ0-ε,ϕ0+ε),Rn,W=aj)<(q-2-ε0)T(Ω(ϕ0-ε,ϕ0+ε),Rn,W).
By Theorem 1.1, we have
(4.7)(q-2)S(Ω(ϕ-ε2,ϕ+ε2),R,W)-(q-2)S(Ω(ϕ-ε2,ϕ+ε2,1,W))⩽∑j=1qn¯(Ω(ϕ-ε,ϕ+ε),2R,W=aj)+252π24ln2Rδ38εln2.
Hence,
(4.8)(q-2)T(Ω(ϕ-ε2,ϕ+ε2),Rn,W)⩽∑j=1qN¯(Ω(ϕ-ε,ϕ+ε),Rn,W=aj)+254π24ln22Rnδ38(ε-ε*)ln2+O(1)<(q-2-ε0+O(1))ln2Rn.
Hence,
(4.9)limn→∞¯T(Ω(ϕ-ε/2,ϕ+ε/2),Rn,W)ln2Rn<O(1),
which contradicts (4.2). Therefore, Theorem 1.2 holds.



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*Source: 570324-2012-12-03.xml*