Radial length, radial John disks and $K$-quasiconformal harmonic mappings

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by  Shaolin Chen

RADIAL LENGTH, RADIAL JOHN DISKS AND
K-QUASICONFORMAL HARMONIC MAPPINGS
7
1
SHAOLIN CHEN AND SAMINATHAN PONNUSAMY
0
2

n Abstract. Inthisarticle,wecontinueourinvestigationsoftheboundarybehav-
a ior of harmonic mappings. We first discuss the classical problem on the growth
J
of radial length and obtain a sharp growth theorem of the radial length of K-

6
quasiconformal harmonic mappings. Then we present an alternate characteriza-
2
tionofradialJohndisks. Inaddition,weinvestigatethelinearmeasuredistortion

] andthe Lipschitz continuity onK-quasiconformalharmonic mappings ofthe unit
V diskontoaradialJohndisk. Finally,usingPommerenkeinteriordomains,wechar-
C acterize certain differential properties of K-quasiconformalharmonic mappings
.
h
t
a
m
1. Introduction and statement of main results
[

1 This paper continues the study of previous work of the authors [6] and is mainly
v
motivatedbythearticlesofBeardonandCarne[3],CarrollandTwomey[4], Chuaqui
7
et al. [10], Pommerenke [29], and the monograph of Pommerenke [30]. In order to
3
5 state our first result concerning the growth of the radial length of K-quasiconformal
7
harmonic mappings (see Theorem 1), we need to recall some basic definitions and
0
. some results which motivate the present work.
1
0 Let f be a complex-valued and continuously differentiable function defined in the
7 unit disk D = z : z < 1 and let ℓ (θ,r) be the length of the f-image (with
f
1 { | | }
counting multiplicity) of the radial line segment [0,z] from 0 to z = reiθ D, where
:
v ∈
θ [0,2π] is fixed and r [0,1). Then (cf. [5])
Xi ∈ ∈
r r
r ℓ (θ,r) := ℓ f([0,z]) = df(ρeiθ) = f (ρeiθ)+e 2iθf (ρeiθ) dρ.
a f z − z
Z0 Z0
In [21], Keogh sh(cid:0)owed tha(cid:1)t if f is(cid:12)(cid:12)a bound(cid:12)(cid:12)ed, anal(cid:12)(cid:12)ytic and univalent func(cid:12)(cid:12)tion in D,
then, for each θ [0,2π],
∈
1/2
(1.1) ℓ (θ,r) = O log(1/(1 r)) as r 1 .
f −
− →
(cid:16) (cid:17)
Throughout the discussion, we l(cid:0)et (cid:1)
1/2
(1.2) ψ(r) = log(1/(1 r)) for 0 < r < 1.
−
(cid:0) (cid:1)
File: Ch-P-Kqc-2016_final.tex, printed: 27-1-2017, 1.42
2010 Mathematics Subject Classification. Primary: 30C62, 30C75; Secondary: 30C20, 30C25,
30C45, 30F45, 30H10.
Key words and phrases. K-quasiconformal harmonic mapping, radial John disk, radial length,
Pommerenke interior domain.
This second author is on leave from IIT Madras.
1
2 S.L. Chen and S. Ponnusamy
Keogh also gave some examples to show that the exponent 1/2 in (1.1) can not be
decreased. Jenkins improved on these examples in [16], and Kennedy [20] presented
further examples by showing that
ℓ (θ,r) = O(µ(r)ψ(r)) as r 1
f −
→
isfalseingeneralforeverypositivefunctionµin[0,1)satisfying µ(r) 0asr 1 .
−
→ →
In [4], Carroll and Twomey established certain refinements and extension of these
results without the boundedness condition in the following form.
Theorem A. Suppose that f(z) = a z + a z2 + is univalent in D. Then, for
1 2
···
any fixed θ [0,2π], there is a constant C > 0 such that
1
∈
(1.3) ℓ (θ,r) C max f(ρeiθ) ψ(r) for r (0.5,1).
f 1
≤ ρ [0,r]| | ∈
∈
If, further, f(reiθ) = O(1) as r 1 , then (1.1) holds.
−
→
Later, Beardon and Carne [3] gave a relatively simple argument to Theorem A in
hyperbolic geometry and provided with further examples. It is worth pointing out
here two results which strengthened (1.3) and was inspired by the work of Sheil-
Small [33] and Hall [14]. If f is starlike, i.e. f(D) contains the line segment
∈ S
[0,w] whenever it contains w, then (see [19])
ℓ (θ,r) f(reiθ) (1+r) < 2 f(reiθ) for r (0,1)
f
≤ | | | | ∈
and the inequality of course is not sharp for all r, but the bound 2 sharp as the
Koebe function k(z) = z/(1 z)2 shows and is attained when r approaches 1 (see
−
[14, 33]). Later in 1993, Balasubramanian et al. [1] showed that if f is convex,
i.e. f(D) is a convex domain, then ∈ S
ℓ (θ,r) f(reiθ) r 1arcsinr for r (0,1)
f −
≤ | | ∈
and the inequality is sharp as the convex function f(z) = z/(1 z) shows. Note
−
that ϕ(r) = r−1arcsinr is increasing on (0,1) and ϕ(r) limr 1−ϕ(r) = π/2 and
≤ →
thus, the conjecture of Hall [15] was settled (see also [2]).
The first aims of this paper is to extend Theorem A for the case of harmonic
quasiconformal mappings (see Theorem 1 below). We need some preparation to
state this result.
For a real 2 2 matrix A, we use the matrix norm A = sup Az : z = 1
and the matrix×function l(A) = inf Az : z = 1 . Forkz =k x+iy{| C,| th|e |forma}l
{| | | | } ∈
derivative of the complex-valued function f = u+iv is given by the Jacobian matrix
u u
x y
D = ,
f
vx vy !
so that
D = f + f and l(D ) = f f ,
f z z f z z
k k | | | | | |−| |
where fz = (1/2) fx ify and fz = (1/2) fx + if(cid:12)y . Let Ω(cid:12) be a domain in C,
− (cid:12) (cid:12)
with non-empty boundary. A sense-preserving homeomorphism f from a domain Ω
(cid:0) (cid:1) (cid:0) (cid:1)
Radial length, radial John disks and K-quasiconformal harmonic mappings 3
onto Ω, contained in the Sobolev class W1,2(Ω), is said to be a K-quasiconformal
′ loc
mapping if, for z Ω,
∈
D (z) 2 K detD (z) , i.e., D (z) Kl D (z) ,
f f f f
k k ≤ k k ≤
where K 1 and detD is t(cid:12)he determi(cid:12)nant of D (cf. [18, 2(cid:0)2, 35, 3(cid:1)6]).
≥ f (cid:12) (cid:12) f
Let denote the family of sense-preserving planar harmonic univalent mappings
H
f = hS+g in D, with the normalization h(0) = g(0) = 0 and h(0) = 1. Recall that
′
f is sense-preserving if the Jacobian J of f given by
f
J := detD = f 2 f 2 = h 2 g 2
f f z z ′ ′
| | −| | | | −| |
is positive. Thus, f is locally univalent and sense-preserving in D if and only if
J (z) > 0 in D; or equivalently if h = 0 in D and the dilatation ω = g /h has the
f ′ ′ ′
property that ω(z) < 1 in D (see [611, 12, 23]). The family together with a
H
| | S
few other geometric subclasses, originally investigated in detail by [11, 34], became
instrumental in the study of univalent harmonic mappings (see [12, 31]) and has
attracted the attention of many function theorists. If the co-analytic part g is
identically zero in the decomposition of f = h+g, then the class reduces to the
H
S
icnlaDss.icIaflfa0m=ilySf =ofhal+lngormalize:dga(n0a)l=yti0cu,ntihveanletnhtefufanmctiiloyns0h(izs)b=otzh+nor∞nm=a2laannzdn
SH { ∈ SH ′ } SH P
compact. See [11] and also [8, 6, 12, 31].
Theorem 1. For K 1, let f be a K-quasiconformal harmonic mapping.
H
≥ ∈ S
Then, for any fixed θ [0,2π], there is a constant C > 0 such that
2
∈
ℓ (θ,r) C max f(ρeiθ) ψ(r) for r (0.5,1).
f 2
≤ ρ [0,r]| | ∈
∈
If, further, f(reiθ) = O(1) as r 1 , then
−
→
ℓ (θ,r) = O(ψ(r)) as r 1 ,
f −
→
and the exponent1/2 in ψ(r) definedby (1.2) cannotbe replacedbya smallernumber.
First we remark that if K = 1, then Theorem 1 coincides with Theorem A. Sec-
ondly, the proof of Theorem 1 is substantially harder than the proof of Theorem A.
This is because Beardon and Carne’s argument of Theorem A in [3] is not applicable
in the proof of Theorem 1.
We need further notation and terminology before stating our second result. Let
d (z) be the Euclidean distance from z to the boundary ∂Ω of Ω. If Ω = D, then
Ω
we set d(z) := dD(z).
Definition 1. A bounded simply connected plane domain G is called a c-John disk
for c 1 with John center w G if for each w G there is a rectifiable arc γ,
0 1
≥ ∈ ∈
called a John curve, in G with end points w and w such that
1 0
σ (w) cd (w)
ℓ G
≤
for all w on γ, where γ[w ,w] is the subarc of γ between w and w, and σ (w) is the
1 1 ℓ
Euclidean length of γ[w ,w] (see [6, 13, 17, 28, 30]).
1
4 S.L. Chen and S. Ponnusamy
Remark 1. If f is a complex-valued and univalent mapping in D, G = f(D) and,
for z D, γ = f([0,z]) in Definition 1, then we call c-John disk a radial c-John disk,
∈
where w = f(0) and w = f(z). In particular, if f is a conformal mapping, then we
0
call c-John disk a hyperbolic c-John disk. It is well known that any point w G
0
∈
can be chosen as a John center by modifying the constant c if necessary. When we
do not wish to emphasize the role of c, then we regard the c-John disk simply as a
John disk in the natural way (cf. [6, 13, 17, 28]).
Unless otherwise stated, throughout the discussion we consider the following ter-
minology. Denote by (K) if f and is a K-quasiconformal harmonic mapping
in D, where K 1. AFlso, we de∈noFte by (K,Ω) if f (K) and f maps D onto
≥ F ∈ F
Ω. We prove several results mainly when equals one of , 0 , and , and Ω
F SH SH SH2
equals either radial John disk or Pommerenke interior domain.
Further, for z D, we define
∈
(1.4) B(z) := ζ : z ζ < 1, argz argζ π(1 z ) .
{ | | ≤ | | | − | ≤ −| | }
In the following, we continue our previous work of [6], and give an another charac-
terization of the radial John disk.
Theorem 2. Let f 0 (K). Then the following are equivalent:
∈ SH
(i) Ω = f(D) is a radial John disk.
(ii) There is an x (0,1) such that
∈
(1 ρ2) D (ρζ) x+r
sup sup − k f k < 1 for ρ = .
(1 r2) D (rζ) 1+xr
ζ=1r (0,1) f
| | ∈ − k k
f(z) f(w)
(iii) sup | − | < .
z D, w B(z) (1 z 2) Df(z) ∞
∈ ∈ −| | k k
Next, we establish the linear measure distortion on K-quasiconformal harmonic
mappings of D onto a radial John disk.
Theorem 3. Let f = h+g 0(K,Ω), where Ω is a radial John disk. Then, for
a ,a D with B(a ) B(a∈),SthHere is a positive constant C such that
1 2 1 2 3
∈ ⊂
diamf(B(a )) ℓ(B(a ) ∂D) α
1 1
C ∩ .
diamf(B(a )) ≤ 3 ℓ(B(a ) ∂D)
2 (cid:18) 2 ∩ (cid:19)
where α = supf H |h′′2(0)| and B(z) is defined by (1.4).
∈S
knWowenre(smeear[k6,t9h,a1t22,≤34]α).=Wseupdfis∈cSuHss|h′t′2(h0e)|L<ip∞sch,ibtuztcothnetinshuaitrypovnalKue-qoufaαsiicsosntfiollrmuna-l
harmonic mappings of D onto a radial John disk, which is as follows.
Theorem 4. Let f = h+g 0(K,Ω), where Ω is a radial John disk. Then, for
z D with z 1 and ζ ,ζ∈ SHB(z), there are constants δ (0,1) and C > 0
∈ | | ≥ 2 1 2 ∈ 1 ∈ 4
such that
ζ ζ δ1
1 2
f(ζ ) f(ζ ) C d (f(z)) | − | .
1 2 4 Ω
| − | ≤ 1 z
(cid:18) −| | (cid:19)
Radial length, radial John disks and K-quasiconformal harmonic mappings 5
Let f (K,G), where G is domain. For 0 < r < 1, let D = z : z < r and
H r
∂D deno∈teSthe boundary of D . Now, for w ,w f (∂D ), let γ{ be|t|he sm}aller
r r 1 2 r r
subarc of f (∂D ) between w and w , and let ∈
r 1 2
d (w ,w ) = infdiamΓ,
Gr 1 2
Γ
where Γ runs through all arcs from w to w that lie in G = f(D ) except for their
1 2 r r
endpoints. If
ℓ γ [w ,w ]
r 1 2
(1.5) sup sup < ,
d (w ,w ) ∞
0<r<1(w1,w2∈γr (cid:0)Gr 1 2(cid:1))
then we call G a Pommerenke interior domain (cf. [6, 29]). In particular, if G is
bounded, then we call G as a bounded Pommerenke interior domain.
Given a sense-preserving harmonic mapping f = h+g in D, fix ζ D andperform
∈
a disk automorphism (also called Koebe transform F of f) to obtain
f z+ζ f(ζ)
1+ζz −
(1.6) F(z) = =: H(z)+G(z).
h(cid:16)(ζ)(1(cid:17) ζ 2)
′
−| |
A calculation gives,
H (0) 1 h (ζ)
′′ = (1 ζ 2) ′′ 2ζ .
2 2 −| | h(ζ) −
(cid:26) ′ (cid:27)
Now, we consider the class of all harmonic mappings f = h+g satisfying
SH2 ∈ SH
h (z)
(1.7) sup (1 z 2) ′′ 2z < 4.
z D(cid:12) −| | h′(z) − (cid:12)
∈
(cid:12) (cid:12)
Thisinequality obviouslyholds(cid:12)ifh andhisnot(cid:12)theKoebefunctionz/(1 eiθz)2,
θ R. Note that for the Koeb(cid:12)e fun∈ctiSon the suprem(cid:12)um turns out to be 4. O−ur next
∈
two results are extension of [29, Theorem 3].
Theorem 5. Let f (K,G), where G is a bounded Pommerenke interior do-
main. If there are po∈sitSivHe2 constants δ (0,1) and C such that, for each ζ ∂D
2 5
∈ ∈
and for 0 ρ ρ < 1,
1 2
≤ ≤
1 ρ2 δ2−1
(1.8) D (ρ ζ) C − D (ρ ζ) ,
f 2 5 f 1
k k ≤ 1 ρ k k
(cid:18) − 1(cid:19)
then
1 D (ξ) 1 ζ 2
f
sup k k −| | dξ < .
ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞
We remark that if K = 1, then Theorem 5 coincides with [29, Theorem 3].
By using similar reasoning as in the proof of Theorem 5, one can easily get the
following result which replaces the assumption f by a more general condition
∈ SH2
f and thus, we omit its proof.
H
∈ S
6 S.L. Chen and S. Ponnusamy
Theorem 6. Let f (K,G), where G is a bounded Pommerenke interior do-
H
∈ S
main. If there are constants C > 0, C > 0, δ > 0 and δ (0,1) such that, for
6 7 3 4
each ζ ∂D and for 0 ρ ρ < 1, ∈
1 2
∈ ≤ ≤
1 ρ1 δ3−1 1 ρ2 δ4−1
C − D (ρ ζ) D (ρ ζ) C − D (ρ ζ) ,
6 f 1 f 2 7 f 1
1 ρ k k ≤ k k ≤ 1 ρ k k
(cid:18) − 2(cid:19) (cid:18) − 1(cid:19)
then
1 D (ξ) 1 ζ 2
f
sup k k −| | dξ < .
ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞
Also, the following result easily follows from Theorem 5 and [6, Theorem 1].
Corollary 1. For K 1, let f 0 be a K-quasiconformal harmonic
mapping from D onto a≥bounded Po∈mmSHer2e∩nkSeHinterior domain G. If G is a radial
John disk, then
1 D (ξ) 1 ζ 2
f
sup k k −| | dξ < .
ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞
The proofs of Theorems 1-5 will be presented in Section 2.
2. The proofs of the main results
Let λD stand for the hyperbolic distance (or Poincar´e distance) on the unit disk
D. We have
dz z z
λD(z1,z2) = inγfZγ 1−| |z||2 = tanh−1(cid:12)11−−z1z22(cid:12),
where the infimum is taken over all smooth curves γ (cid:12)(cid:12)in D conn(cid:12)(cid:12)ecting z1 D and
z D (cf. [30]). In [34], Sheil-Small proved that if f =(cid:12) h+g (cid:12) , then ∈
2 H
∈ ∈ S
(1 z )α 1 (1+ z )α 1
− −
(2.1) −| | h(z) | |
′
(1+ z )α+1 ≤ | | ≤ (1 z )α+1
| | −| |
and
h (0)
′′
α := sup | | < .
2 ∞
f H
∈S
Unless otherwise stated, the number α will be used throughout the discussion and
is indeed called the order of the linear invariant family (see [34]).
H
S
Lemma 1. Suppose that f (K). Then, for z ,z D,
H 0 1
∈ S ∈
1 f(z ) f(z ) K
1 e 2αλD(z1,z0) | 1 − 0 | e2αλD(z1,z0) 1 .
−
α(1+K) − ≤ (1 z 2) f (z ) ≤ α(1+K) −
0 z 0
−| | | |
(cid:2) (cid:3) (cid:2) (cid:3)
In particular,
1 K
(2.2) 1 e 2αλD(z,0) f(z) e2αλD(z,0) 1 , z D.
−
α(1+K) − ≤ | | ≤ α(1+K) − ∈
(cid:2) (cid:3) (cid:2) (cid:3)
Radial length, radial John disks and K-quasiconformal harmonic mappings 7
Proof. By assumption f = h+ g is a K-quasiconformal harmonic mapping,
H
where h and g are analytic in D. T∈huSs, by (2.1), we have
2K 2K (1+ z )α 1
−
(2.3) D (z) h(z) | |
f ′
k k ≤ K +1| | ≤ K +1(1 z )α+1
−| |
and thus, for z D, we obtain
∈
f(z) D (ζ) dζ
f
| | ≤ k k| |
Z[0,z]
α
2K z (1+ρ)α 1 K 1+ z
| | −
(2.4) dρ = | | 1 .
≤ K +1 (1 ρ)α+1 α(K +1) 1 z −
Z0 − " −| |! #
On the other hand, let Γ be the preimage under f of the radial segment from 0 to
f(z). Again, because
2 2 (1 z )α 1
−
l(D (z)) h(z) −| | ,
f ≥ K +1| ′ | ≥ K +1(1+ z )α+1
| |
it follows that
α
1 1 z
(2.5) f(z) l(D (ζ)) dζ 1 −| | .
f
| | ≥ | | ≥ α(K +1) − 1+ z
ZΓ " | |! #
Let z = z1 z0 so that z = z+z0 , where z ,z D. Then, by assumption,
1 −z0z1 1 1+z0z 0 1 ∈
−
f(z) f(z )
0
F(z) = −
(1 z 2)h(z ) ∈ SH
0 ′ 0
−| |
and is a K-quasiconformal harmonic mapping, i.e. F (K). Applying (2.4) and
H
∈ S
(2.5) to F gives us the desired result if we take into account of the fact that
z z e2λD(z1,z2) 1
1 2
1 −z z = e2λD(z1,z2) −+1 = tanhλD(z1,z2).
(cid:12) − 1 2(cid:12)
(cid:12) (cid:12) (cid:3)
The proof of the lemma is complete.
(cid:12) (cid:12)
(cid:12) (cid:12)
Lemma 2. Assume that f (K). Then
H
∈ S
C f(z)
D (z) z 8| | for z D,
f
k k| | ≤ 1 z ∈
−| |
where
z (1+ z )α 1
−
(2.6) C = 2αKsup | | | | K.
8 z∈D (cid:26)[(1+|z|)α −(1−|z|)α](cid:27) ≥
Proof. Suppose that f = h+g (K), where h and g are analytic in D. Next, for
H
fixed ζ D, consider the Koeb∈eStransform F of f given by (1.6). By assumption,
∈
F and is also a K-quasiconformal harmonic mapping. By letting z = ζ in
H
∈ S −
(1.6) and applying (2.2) to F, we obtain (since f(0) = 0)
f(ζ) 1 [(1+ ζ )α (1 ζ )α]
F( ζ) = | | | | − −| |
| − | (1 ζ 2) h(ζ) ≥ α(1+K) (1+ ζ )α
′
−| | | | | |
8 S.L. Chen and S. Ponnusamy
which gives
h(ζ) (1+ ζ )α 1 α
′ −
| | | | .
(1+K) f(ζ) ≤ [(1+ ζ )α (1 ζ )α] · 1 ζ
| | | | − −| | −| |
Since this follows for each ζ D, by the first inequality in (2.3), we easily have
∈
D (z) z 2K h(z) z C
f ′ 8
k k| | | || | ,
f(z) ≤ K +1 f(z) ≤ 1 z
| | | | −| |
where C is given by (2.6). (cid:3)
8
Lemma 3. Let f (K) and, for any fixed θ [0,2π], set
H
∈ S ∈
m (r,θ) = max f(ρeiθ) ,
f
ρ [0,r]| |
∈
where r [0,1). Then, for 0 < ρ r < 1 and 0 ρ r, there is a constant
0
∈ ≤ ≤ ≤
C > 0 which depends only on ρ such that
9 0
f(ρeiθ)
(2.7) | | C m (r,θ),
9 f
ρ ≤
where ρ is a constant.
0
Proof. Without loss of generality, we assume that θ = 0. Clearly, (2.2) yields that
f(ρ) K e2αλD(ρ,0) 1 2K
lim | | lim − = ,
ρ 0+ ρ ≤ α(1+K) ρ 0+ ρ 1+K
→ →
which implies that f(ρ)/ρ is bounded in [0,ρ ], where ρ is a constant such that
0 0
0 < ρ r < 1. Hence there is a constant C > 0 such that
0 10
≤
f(ρ)
(2.8) C m (r,0) for ρ [0,ρ ],
10 f 0
ρ ≤ ∈
where r [ρ ,1). For r [0,1), let
0
∈ ∈
K
T(r) = 1 e 2αλD(r,0) .
−
α(1+K) −
(cid:2) (cid:3)
Then T is increasing in [0,1), which, together with (2.2), yields that
f(ρ) f(ρ) 1
(2.9) 0 < T(ρ ) T(ρ) f(ρ) m (r,0) for ρ [ρ ,r],
0 f 0
≤ ≤ ≤ ρ ≤ ρ ≤ ρ ∈
0 0
where r [ρ ,1). Therefore, (2.7) follows from (2.8) and (2.9). (cid:3)
0
∈
Lemma 4. For r (0,1), let Ω be the Stolz-type domain consisting of the interior
r
of the convex hull ∈of the point r and the disk D . Then, for z = ρeiη Ω D ,
r/4 r r/4
∈ \
4π 4π
η (r ρ) < (1 ρ).
| | ≤ r√15 − r√15 −
Radial length, radial John disks and K-quasiconformal harmonic mappings 9
C
E = ρ eiη
1
z = ρeiη
η
O A
D
B
Figure 1. Stolz-type domain
Proof. Assume without loss of generality that η 0. Let A, D, and E represent the
points r, r/4, and ρ eiη (see Figure 1), respecti≥vely. As ∠OCA = π/2, it is clear
1
that
sin∠COA = √15, cos∠COA = 1, sin∠COE = ρ21 − 1r62 and cos∠COE = r .
4 4 q ρ 4ρ
1 1
Then, because (sinη)/η 2/π for η < π/2, it follows that for η 0
| | ≥ | | ≥
2η
sinη = sin(∠COA ∠COE)
π ≤ −
= sin∠COAcos∠COE cos∠COAsin∠COE
−
√15r2 16ρ2 r2
= − 1 −
16ρ
p1
r2 ρ2
= − 1
ρ √15r + 16ρ2 r2
1 1 −
Note that r < ρ < ρ < r and(cid:16), becausep (cid:17)
4 1
r2 ρ2 4
− 1 < r2 ρ2 < 8(r ρ ) < 8(r ρ),
ρ r − 1 − 1 −
1
(cid:0) (cid:1)
the last above relation clearly implies that
2η 8(r ρ)
< −
π r√15
which gives the desired conclusion. Observe that 8 (r ρ) is less than 6/√15 from
r√15 −
which we also deduce that η < 3π/√15. (cid:3)
| |
10 S.L. Chen and S. Ponnusamy
Proof of Theorem 1. Assume without loss of generality that θ = 0. For r (0,1),
∈
we use Ω to denote the Stolz-type domain, where Ω is same as in Lemma 4. Let
r r
z = ρeiη Ω D . Then, by Lemma 4, there is a constant C > 0 which depends
r r/4 11
∈ \
only on r such that
(2.10) η < C (1 ρ).
11
| | −
Suppose that f = h+g (K). By calculations, we get
H
∈ S
f(ρeiη) f(ρ) η ρeith(ρeit) ρe itg (ρeit)
(2.11) log log = i ′ − − ′ 1 dt.
ρeiη − ρ f(ρeit) −
Z0 !
Taking real part of (2.11) on both sides, and then using (2.10), (2.11) and Lemma
2, we see that there is a constant C such that
12
f(ρeiη) f(ρ) η ρ D (ρeit)
f
log | | log | | k k dt
ρ − ρ ≤ f(ρeit)
Z0 | |
η dt
C C C ,
12 11 12
≤ 1 ρ ≤
Z0 −
which gives that
(2.12) f(z) = f(ρeiη) eC11C12 f(ρ) .
| | | | ≤ | |
By (2.2), weseethat(2.12)alsoholdsforz D . Then, by(2.12), thereisconstant
r/4
C such that f(Ω ) is contained in D ∈ , which yields that
13 r C13mf(r,0)
(2.13) J (ζ)dA(ζ) C2 m2(r,0),
f ≤ 13 f
ZΩr
where ζ = x+iy, dA = dxdy/π and m (r,θ) is defined as in Lemma 3.
f
By [24, Theorem 2], there is a constant C such that
14
1
(2.14) (1 ρ) H (ρ) 2dρ C H (z) 2dA(z),
′ 14 ′
− | | ≤ | |
Z0 ZΩ1
where H(z) is analytic in D. For r (0,1), let H(z) = h(rz) for z D. Then, by
∈ ∈
(2.14), we obtain
r
(r ρ) h(ρ) 2dρ C h(z) 2dA(z),
′ 15 ′
− | | ≤ | |
Z0 ZΩr
which implies that
r
(r ρ) D (ρ) 2dρ C K J (z)dA(z),
f 15 f
− k k ≤
Z0 ZΩr
(2.15) C2 C Km2(r,0) (by (2.13)),
≤ 13 15 f
where C > 0 is a constant.
15
By Lemmas 2 and 3, for r (1/2,1) and ρ [0,r], there is a constant C such
16
∈ ∈
that
r r m2(r,0) r
(2.16) D (ρ) 2dρ C f dρ = C m2(r,0) .
k f k ≤ 16 (1 ρ)2 16 f 1 r
Z0 Z0 − −