TITLE: Hadamard's three circle theorem
QUESTION [11 upvotes]: Hadamard's three circle theorem is given as follows:
$$A(a,b)=\{z:a<|z|< b\},$$ and $f$ is a holomorphic function in this annulus $A$.
Let $$M(r) = \max_{|z|=r}|f(z)| $$for $a < r < b$. Then 
$$\log\left(\frac ba\right)\log(M(r)) \le\ \log\left(\frac br\right)\log(M(a)) + \log\left(\frac ra\right)\log(M(b))$$
For which non-constant $f$ does equality hold in this inequality?

REPLY [5 votes]: Let $\lambda=\frac{\log(b/r)}{\log(b/a)}$. Then $1-\lambda=\frac{\log(r/a)}{\log(b/a)}$. Dividing both sides of your equation by $\log(b/a)$ gives:
$$\log(M(r))\leq \lambda \log(M(a))+(1-\lambda)\log(M(b)).$$
Notice that $a^{\lambda}b^{1-\lambda}=\exp(\lambda\log(a)+(1-\lambda)\log(b))=r$  (verify this!). Thus,
$$\log(M(a^\lambda b^{1-\lambda})\leq \lambda\log(M(a))+(1-\lambda)\log(M(b))$$
which is saying that $M$ is log-convex, in that $\log(M(\exp(z))$ is convex in $z$. It is not hard to show that the only times you get strict equality for convex functions is iff the function is affine (linear): $\log(M(\exp(z))=Az+B$. Thus, $M(\exp(z))=Ce^{Az}$, or $M(z)=Cz^A$ 
Addendum: to show that all strictly equal convex functions are affine, write:
$f''(z)=\lim_{h\rightarrow 0}\frac{f(z+h)+f(z-h)-2f(z)}{h^2}=\lim_{h\rightarrow 0}\frac{2f(z)-2f(z)}{h^2}=0$
where we used $z=\frac{1}{2}(z+h)+\frac{1}{2}(z-h)$ and equal convexity.