I've seen this a couple of times used in proofs, and it's apparently so obvious it doesn't need to be proved itself. It's not obvious to be though...
write dD for the boundary of the unit circle in C. take f to be any polynomial. then:
sup_dD |z* - f(z)| >= 1
where z* is the complex conjugate, and >= means 'greater than or equal to'.
complex numbers, polynomials Watch
- Thread Starter
- 17-01-2010 23:12
- 18-01-2010 00:52
I genuinely don't know if this would work, but how about the reverse triangle ineqality?
sup_dD |z* - f(z)| >= |z* - f(z)| >= | ||z*|| - ||f(z)|| | = | ||z|| - ||f(z)|| | = | 1 - ||f(z)|| |>= ... ?Last edited by JohnnySPal; 18-01-2010 at 00:58.
- 18-01-2010 01:30
I'm not seeing anything terribly obvious, but I think this works:
On dD, z* = 1/z. So |z*-f(z)| = |1/z - f(z)| = |z||1/z - f(z)| = |1 - z f(z)|.
Let g = 1-zf. g is a poly, so analytic. And |g(0)| = 1. Now apply the maximum modulus principle to g.
(There's probably a slightly more direct approach using Schwarz' Lemma).
- 18-01-2010 01:33
(Original post by SimonM)
- 18-01-2010 19:38
I don't know how JohnnySPal's proof will work ...