# Mobius-induced inequality on the unit disk.

• Sep 27th 2009, 06:37 PM
subfallen
Mobius-induced inequality on the unit disk.
I need to prove that, for r ∈ (0, 1) and z in the unit disk:

|(z - r) / (1 - rz)| ≤ |z| + r

Even the smallest hint will be greatly appreciated! Thanks guys.
• Sep 27th 2009, 06:57 PM
Quote:

Originally Posted by subfallen
I need to prove that, for r ∈ (0, 1) and z in the unit disk:

|(z - r) / (1 - rz)| ≤ |z| + r

Even the smallest hint will be greatly appreciated! Thanks guys.

Use the Triangle Inequality.

| (z - r) / (1 - rz) | = | z / (1 - rz) - r / (1 - rz) | ≤ | z / (1 - rz)| + |(-r)/(1-rz)| ...
• Sep 27th 2009, 07:31 PM
subfallen

I've tried that for hours on end, but can never make it work. For if Re(z) > 0 then |1 - rz| < 1, and |z / (1 - rz)| > |z|.

(|z| + r) seems to be an extremely tight bound on the LHS; every time I try to use a basic inequality, it blows up in my face. :/
• Sep 27th 2009, 07:37 PM
subfallen
I can't even prove the inequality when it's restricted to a, b ∈ (0, 1):
|(a - b)/(1 - ab)| ≤ a + b
It seems there must be a way to turn the LHS into the form:
|X(a + b)|
where |X| ≤ 1. Otherwise I just can't imagine how it can be done. :/
• Sep 27th 2009, 07:41 PM
hmm... so either the Triangle Inequality is not the right approximation for this problem or it is the case that the LHS is greater than or equal to RHS.
• Sep 27th 2009, 08:01 PM
subfallen
Suppose we use the notation ϕr(z) = (z-r) / (1-rz). Then it is a fact from complex analysis that:
arctanh|ϕr(z)| ≤ arctanh|z| + arctanh|r|
(This holds because Mobius transformations are isometries of the disk under the Poincaré metric.) Do you have any idea how to go from that inequality to the one desired in my original post?
r(z)| ≤ |z| + r
• Sep 27th 2009, 08:08 PM
I'm sorry, in that aspect I cannot help you. I'm not at a high level of mathematics yet; I'm in Analysis I.

However, this looks a lot like...
what do we know about inverse hyperbolic tan?
is it a function such that f(a+b) = f(a) + f(b)?

In that case tanh-1|z| + tanh-1r = tanh-1(|z| + r)

so then
tanh-1|ϕr(z)| ≤ tanh-1(|z| + r)

so it follows that
|ϕr(z)| ≤ |z| + r
• Sep 27th 2009, 08:22 PM
subfallen
Well, if we can show that:
[1] a ≤ b implies tanh(a) ≤ tanh(b)
[2] tanh(a + b) ≤ tanh(a) + tanh(b)
for a, b ∈ (0, 1) then we can just take tanh of both sides of the inequality
arctanh|ϕr(z)| ≤ arctanh|z| + arctanh|r|
and we'll be done. Clearly #1 holds (a graph here: Hyperbolic Tangent -- from Wolfram MathWorld). But I'm not sure about #2. Hmm...maybe I'll try to talk to a teacher. Meh. This problem is breaking my will to live.