# proof of Hadamard three-circle theorem

Let $f$ be holomorphic on a closed annulus $0. Let

 $s=\frac{\log r_{1}-\log r}{\log r_{2}-\log r_{1}}.$

Let $M(r)=M_{f}(r)=||f||_{r}=\max_{|z|=r}|f(z)|$. Then we have to prove that

 $\log M(r)\leq(1-s)\log M(r_{1})+s\log M(r_{2}).$

For this, let $\alpha$ be a real number; the function $\alpha\log|z|+\log|f(z)|$ 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 $|z|=r_{1}$ and $|z|=r_{2}$. Therefore

 $\alpha\log|z|+\log|f(z)|\leq\max(\alpha\log r_{1}+\log M(r_{1}),\alpha\log r_{% 2}+\log M(r_{2}))$

for all $z$ in the annulus. In particular, we get the inequality  $\alpha\log r+\log M(r)\leq\max(\alpha\log r_{1}+\log M(r_{1}),\alpha\log r_{2}% +\log M(r_{2})).$

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

 $\alpha=\frac{\log M(r_{2})-\log M(r_{1})}{\log r_{1}-\log r_{2}}.$

Then from the previous inequality, we get

 $\log M(r)\leq\alpha\log r_{1}+\log M(r_{1})-\alpha\log r,$

which upon substituting the value for $\alpha$ gives the result stated in the theorem.

## References

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 ProofOfHadamardThreecircleTheorem 2013-03-22 15:56:02 2013-03-22 15:56:02 Simone (5904) Simone (5904) 5 Simone (5904) Proof msc 30C80 msc 30A10