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2.3. REVERSED TRIANGLE INEQUALITY 25

With (2.2.9) established, it is quite easy to finish the proof in the special case.

In fact diam(t2) :S: alz1 - z2I in this case. To see this, let z , w E -y 2. If both wand z

belong to [z1, z 2] or to some -y 1 (u 1 , v 1 ) , there is nothing to prove. If w E [z 1 , z 2 ] and

z '1-[z1, z2J, then z E 'Y 1 ( Uj, Vj) for some j. Now either z 1 ::=; w ::=; Uj or v 1 ::=; w ::=; z 2.

We will assume the former, but a similar argument works for the second situation.

Then

lw - z l :S: lw - u1I + lui - zl

:S: lw - u1I + alu1 - v1I

:S: a(lw-u1I + lu1 - vjl)

:S: al z1 - z2I-

Finally, suppose that w E -y 1 ( Uj, Vj) and z E -y 1 (Uk, vk), assuming Vj ::=; Uk. Then

lw - z l :S: lw - Vj I + lv1 - uk I + luk - z l

:S: alu1 - v1I + lv1 - ukl + al ·uk - vkl

:S: alu1 - Vk I

:S: al z1 - z2I·

In the general case let [zlk , z 2 k] denote t he closures of the components of [z 1 , z 2 ]\
'¥ 1 , and let 'Ylk = -y 1 (zlk,z2k) C -y 1. For each 'Ylk we are in the situation we have
treated already, and
lz1k - z l :S: a lz1k - z2k I
for each z E 'Ylk· Finally z E -y 1 must be in [z 1 , z 2 ] or in 'Ylk for some k , in which
case we have

lz1 - z l :S: lz1 - zlkl + lz1k - z l

:S: lz1 - zlk l +a lzlk - Z2kl

:S: a lz1 - z2I·

By applying the triangle inequality we obtain (2.2.3) with the constant a replaced

by 2a. D

2.3. Reversed triangle inequality

Inequality (2.2.3) is not quantitatively sharp since for a = 1 it does not imply
that D is a disk or half-plane. An alternative formulation of this condition as a
reversed triangle inequality, or sum of two cross ratios, is better in this respect.
The cross ratio of four points z 1 , z 2 , z 3 , z 4 E R^2 is defined by

(z1 - z3) (z2 - Z4)

(z1 - z4) (z2 - z3) ·

When z 4 = oo, the cross ratio reduces to

and similarly when z 1 , z 2 , or z 3 lies at oo. See Ahlfors [3]. We will sometimes write
-2
[z1,z2,z3,z4] for the cross ratio of z1,z 2 ,z3,Z 4 ER.