balasubramanian. 77KB Jun 04 2011 12:08:23 AM

An alternate proof of Hall’s theorem on a
conformal mapping inequality
R. Balasubramanian

S. Ponnusamy∗

In this note we give a different and direct proof of the following result of Hall
[2], which actually implies the conjecture of Sheil-Small [3]. For details about the
related problems we refer to [1, 3].
THEOREM. Let f be regular for |z| < 1 and f(0) = 0. Further, let f be starlike
of order 1/2. Then
Z r
π
|f ′ (ρeiθ )|dρ < |f(reiθ )|
2
0
for every r < 1 and real θ.
Proof. As in [2, p.125] (see also [1]), to prove our result it suffices to show that
J = I(t, τ ) + I(τ, t) < π − 2 for 0 < t < τ < π

(1)

where
I(t, τ ) =

Z

0

1

2| sin(t/2)|
√
1 − 2ρ cos t + ρ2

(

)

1
1 − ρ cos τ

√
−
dρ.
1 − 2ρ cos τ + ρ2 1 − 2ρ cos τ + ρ2

To evaluate these integrals we define k by
sin2 (τ /2) − sin2 (t/2)
k =
cos2 (t/2) sin2 (τ /2)
2

∗

This work has been done with the support of National Board for Higher Mathematics.
Received by the editors May 1995.
Communicated by R. Delanghe.
1991 Mathematics Subject Classification : Primary:30C45; Secondary:30C35.
Key words and phrases : Starlike and conformal mappings.

Bull. Belg. Math. Soc. 3 (1996), 209–213

210

R. Balasubramanian – S. Ponnusamy

so that
cos t =
and

cos τ + k 2 sin2 (τ /2)
1 − k 2 sin2 (τ /2)

(2)

(1 − k 2 ) sin2 (τ /2)
sin (t/2) =
.
1 − k 2 sin2 (τ /2)
2

Further we let

ρ=

π−τ
sin θ
, 0≤θ≤
.
sin(θ + τ )
2

(3)

(The idea of change of variables already occurs in [2, 3]). Then, from (2) and (3),
we easily have
sin τ
dθ
dρ =
2
sin (θ + τ )

|1 − ρeiτ |2 =

sin2 τ
= 1 − 2ρ cos τ + ρ2
sin2 (θ + τ )

|1 − ρeit|2 =
and
1 − ρ cos t =
After some work we find that

sin2 τ [1 − k 2 sin2(θ + τ /2)]
sin2 (θ + τ )[1 − k 2 sin2(τ /2)]

sin τ [cos θ − k 2 sin(τ /2) sin(θ + τ /2)]
.
sin(θ + τ )[1 − k 2 sin2 (τ /2)]

J = I(t, τ ) + I(τ, t) =

Z

(π−τ )/2

0

H(k, τ, θ)dθ

(4)

where
v
u

u 1 − k 2 sin2 (τ /2)
1
cos θ − k 2 sin(τ /2) sin(θ + τ /2)
t
H(k, τ, θ) =
−

+
cos(τ /2)
1 − k 2 sin2 (θ + τ /2)
1 − k 2 sin2 (θ + τ /2)
q

We put

(1 − k 2 )(1 − cos θ)



.
+q
2
2
1 − k sin (θ + τ /2)

π−τ
=λ

2

to obtain
J =

Z

=

Z

(π−τ )/2

0

where F and G are defined by
F (k, λ, θ)

0

λ

H(k, τ,

π−τ
− θ)dθ
2

[F (k, λ, θ) + G(k, λ, θ)]dθ

(5)

211

An alternate proof of Hall’s theorem on a conformal mapping inequality

s



1  1 − k 2 cos2 λ cos(λ − θ) − k 2 cos λ cos θ 
=
−
sin λ
1 − k 2 cos2 θ
1 − k 2 cos2 θ

(1 − k 2) sin2 (λ − θ)
1
√
√
=
sin λ (1 − k 2 cos2 θ)[ 1 − k 2 cos2 λ 1 − k 2 cos2 θ + cos(λ − θ) − k 2 cos λ cos θ]
"

and

√
1 − k 2(1 − cos(λ − θ))
√

G(k, λ, θ) =
.
sin λ 1 − k 2 cos2 θ

Therefore
∂J
=
∂λ

#

Z

0

λ

∂F (k, λ, θ)
dθ +
∂λ

λ

Z

0

∂G(k, λ, θ)
dθ + F (k, λ, λ) + G(k, λ, λ).
∂λ

Since F (k, λ, λ) = 0 = G(k, λ, λ) the above becomes
∂J
=
∂λ

Z

0

λ

∂F
dθ +
∂λ

Z

0

λ

∂G
dθ.
∂λ

A simple calculation shows that

[

s

(1 − k 2 ) sin(λ − θ)
1 − k 2 cos2 θ
∂F
=
{cos(λ − θ) sin λ + sin θ
∂λ
(1 − k 2 cos2 θ)X 2
1 − k 2 cos2 λ
o
−k 2 cos λ sin(λ + θ) − k 2 cos θ sin(λ + θ) + cos(λ − θ) sin θ + sin λ

]

where X is the denominator of the second expression in F (k, λ, θ).
Since the right hand side of the above expression is a decreasing function of k
for each k in (0, 1) and since the square bracketed term in this expression for k = 1
is positive, we have

∂F
∂F 
= 0.
(6)

≥
∂λ
∂λ k=1
Similarly we have



∂G
∂G 

≥
= 0.
∂λ
∂λ k=1

Thus to prove (1), by (5), (6) and the above, it is sufficient to prove that
Z

0

π/2

[F (k, π/2, θ) + G(k, π/2, θ)]dθ ≤ π − 2,

or equivalently
Z π/2
Z π/2 √
√
1 − sin θ
1 − k 2 cos2 θ − sin θ
2
√
dθ ≤ π − 2.
dθ
+
1
−
k
1 − k 2 cos2 θ
0
0
1 − k 2 cos2 θ

(7)

(Note that this corresponds
to τ = 0 in (4) or (5) ). In the first of the above integrals
√
2
we put tan θ = y 1 − k so that it becomes
√
Z ∞
1 + y2 − y
q
dy
0
(1 + y 2) 1 + (1 − k 2 )y 2

212

R. Balasubramanian – S. Ponnusamy

which, by substituting y = tan θ, yields
Z

π/2

√

0

Put

1 − sin θ

1 − k 2 sin2 θ

dθ.

√
#
1 − k2
1
L(k, θ) = (1 − sin θ) √
.
+√
1 − k 2 cos2 θ
1 − k 2 sin2 θ
"

Therefore (7) is now equivalent to
Z

π/4

L(k, θ)dθ =

Z

=

Z

π/4

π/2

0

L(k, θ)dθ +

0

0

Z

π/2

π/4

L(k, θ)dθ

[L(k, θ) + L(k, π/2 − θ)]dθ

≤ π − 2.
Note that
Z

0

π/2

L(0, θ)dθ = π − 2.

It is therefore suffices to prove that
∂
∂
L(k, θ) +
L(k, π/2 − θ) ≤ 0,
∂k
∂k
for all 0 ≤ θ ≤ π/4. For this we easily find that
∂L(k, θ) ∂L(k, π/2 − θ)
+
∂k
∂k
√
= k(1 − sin θ)(1 − cos θ)[(1 − k 2 sin2 θ)−3/2 { 1 − k 2 (1 + cos θ) − (1 + sin θ)}
√
+ (1 − k 2 cos2 θ)−3/2 { 1 − k 2 (1 + sin θ) − (1 + cos θ)}](1 − k 2 )−1/2 .
Since the inequalities
(1 − k 2 cos2 θ)−3/2 ≥ (1 − k 2 sin2 θ)−3/2
and
1 + cos θ −

√

1 − k 2 (1 + sin θ) ≥ −(1 + sin θ) +

√

1 − k 2 (1 + cos θ)

hold for 0 ≤ θ ≤ π/4, (8) follows easily. This finishes the proof of the theorem.

(8)

An alternate proof of Hall’s theorem on a conformal mapping inequality

213

References
[1] R. BALASUBRAMANIAN, V. KARUNAKARAN and S. PONNUSAMY, A
proof of Hall’s conjecture on starlike mappings, J. London Math. Soc. (2)
48(1993), 278-288.
[2] R.R. HALL, A conformal mapping inequality for starlike functions of order 1/2,
Bull. London Math. Soc. 12 (1980) 119-126.
[3] T. SHEIL-SMALL, Some conformal mapping inequalities for starlike and convex
functions, J. London Math. Soc. (2) 1 (1969) 577-587.

R. Balasubramanian
Institute of Mathematical Sciences
C.I.T. Campus, Madras -600 113, INDIA.
S. Ponnusamy
Department of Mathematics
PO. BOX 4 (Yliopistonkatu 5)
00014 University of Helsinki, Finland.
e-mail:samy@geom.helsinki.fi