proof of Hadamard three-circle theorem

Let f be holomorphic on a closed annulus 0<r1|z|r2. Let


Let M(r)=Mf(r)=||f||r=max|z|=r|f(z)|. Then we have to prove that


For this, let α be a real number; the function αlog|z|+log|f(z)| is harmonic outside the zeros of f. Near the zeros of f the above function has values which are large negative. Hence by the maximum modulus principle this function has its maximum on the boundary of the annulus, specifically on the two circles |z|=r1 and |z|=r2. Therefore


for all z in the annulus. In particular, we get the inequalityMathworldPlanetmath


Now let α be such that the two values inside the parentheses on the right are equal, that is


Then from the previous inequality, we get


which upon substituting the value for α gives the result stated in the theorem.


Lang, S. Complex analysis, Fourth edition. Graduate Texts in Mathematics, 103. Springer-Verlag, New York, 1999. xiv+485 pp. ISBN 0-387-98592-1

Title proof of Hadamard three-circle theorem
Canonical name ProofOfHadamardThreecircleTheorem
Date of creation 2013-03-22 15:56:02
Last modified on 2013-03-22 15:56:02
Owner Simone (5904)
Last modified by Simone (5904)
Numerical id 5
Author Simone (5904)
Entry type Proof
Classification msc 30C80
Classification msc 30A10