proof of Hadamard three-circle theorem

Let $f$ be holomorphic on a closed annulus . 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

$$\log M(r)\le (1-s)\log M(r_1)+s\log M(r_2).$$

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

$$\alpha \log |z|+\log |f(z)|\le \max (\alpha \log r_1+\log M(r_1),\alpha \log r_2+\log M(r_2))$$

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

$$\alpha \log r+\log M(r)\le \max (\alpha \log r_1+\log M(r_1),\alpha \log r_2+\log M(r_2)).$$

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

$$\log M(r)\le \alpha \log r_1+\log M(r_1)-\alpha \log r,$$

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