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 order1/2. Then
Z r
0 |f
′
(ρeiθ)|dρ < π 2|f(re
iθ) |
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 1
0
2|sin(t/2)|
√
1−2ρcost+ρ2
(
1
√
1−2ρcosτ +ρ2 −
1−ρcosτ 1−2ρcosτ +ρ2
)
dρ.
To evaluate these integrals we define k by
k2= sin
2(τ /2)
−sin2(t/2) cos2(t/2) sin2(τ /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.
1991Mathematics Subject Classification : Primary:30C45; Secondary:30C35.
Key words and phrases : Starlike and conformal mappings.
so that
Further we let
ρ = sinθ
sin(θ+τ), 0≤θ ≤ π−τ
2 . (3)
(The idea of change of variables already occurs in [2, 3]). Then, from (2) and (3), we easily have
dρ= sinτ
After some work we find that
J =I(t, τ) +I(τ, t) =
= 1
A simple calculation shows that
∂F whereX 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 eachk in (0,1) and since the square bracketed term in this expression for k = 1 is positive, we have
∂F
Similarly we have
∂G
Thus to prove (1), by (5), (6) and the above, it is sufficient to prove that
Z π/2
which, by substitutingy= tanθ, yields
Z π/2
0
1−sinθ
√
1−k2sin2θdθ.
Put
L(k, θ) = (1−sinθ)
"
1
√
1−k2sin2θ +
√
1−k2
√
1−k2cos2θ
#
.
Therefore (7) is now equivalent to
Z π/2
0 L(k, θ)dθ =
Z π/4
0 L(k, θ)dθ+
Z π/2
π/4 L(k, θ)dθ
=
Z π/4
0 [L(k, θ) +L(k, π/2−θ)]dθ
≤π−2.
Note that
Z π/2
0 L(0, θ)dθ =π−2.
It is therefore suffices to prove that
∂
∂kL(k, θ) + ∂
∂kL(k, π/2−θ)≤0, (8)
for all 0≤θ ≤π/4. For this we easily find that ∂L(k, θ)
∂k +
∂L(k, π/2−θ) ∂k
=k(1−sinθ)(1−cosθ)[(1−k2sin2θ)−3/2
{√1−k2(1 + cosθ)−(1 + sinθ)}
+ (1−k2cos2θ)−3/2
{√1−k2(1 + sinθ)−(1 + cosθ)}](1−k2)−1/2
.
Since the inequalities
(1−k2cos2θ)−3/2
≥(1−k2sin2θ)−3/2
and
1 + cosθ−√1−k2(1 + sinθ)≥ −(1 + sinθ) +√1−k2(1 + cosθ)
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)
