Possible generalizations of Hadamard's three line lemma Let $f$ be an analytic function on a sector 
$$
S=\left\{re^{i\theta}:0<r<\infty,\; 0<\theta<\gamma<\frac{\pi}{2}\right\}
$$
with opening angle $\gamma$ at the origin. Suppose $f$ is bounded on the boundary; that is,
$$
|f(r)|\leq M_1
$$
and 
$$
|f(re^{i\gamma})|\leq M_2
$$
for all $r>0$, and $f$ satisfies some mild growth condition in the interior (such as polynomial growth); then Hadamard's three line lemma says that $\log\sup_{r>0}|f(re^{i\theta})|$ is a convex  function of $\theta$. In another words, we have 
$$
\sup_{r>0}|f(re^{i\theta})|\leq M_1^{1-\frac{\theta}{\gamma}}\times M_2^{\frac{\theta}{\gamma}}.
$$
I want to know if the assumptions of the lemma are sharp. For example, if we change the second condition to the following
$$
|f(re^{i\gamma})|\leq M_2(1+r)^{a},
$$
for some $a>0$ (now $f$ has polynomial growth ), and $f$ is still bounded on the positive real line, and satisfy the same mild growth condition on the interior, then can we also obtain the same conclusion? If not, is there any counterexample to this? 
Thanks very much for any comment.
 A: The theorem is sharp.
We will only consider the case $M_1=M_2=1$. In this case Hadamard's theorem is very similar to the maximum principle. In general, $f(z)$ can be replaced by $\frac{f(z)}{M_1}\cdot c_1 z^{ic_2}$ with some suitable  constants $c_1$ and $c_2$.
Instead of the condition $|f(re^{iγ})|\le (1+r)^a$, we will use a stronger one: $|f|\le1$ along the entire boundary of $S$, except for a small segment $AB$; along $AB$ we prescribe $|f|<1+\varepsilon$ with a small constant $\varepsilon$.
So, let $AB$ be an arbitrary segment $AB$ on the boundary of $S$ and let $\varepsilon>0$ be also arbitrary.
Take a moon-shaped set $M$ like in the rightmost part of the picture, between the circles with radii $1$ and $1+\varepsilon$ such that $M$ has an arc $B'A'$ on the unit circle.

By Riemann's mapping theorem, there is a conformal map from $S$ to $M$ that maps $A$ and $B$ to $A'$ and $B'$, respectively. This map and its inverse are continuous along the boundary ($S$ must be compactified by adding the infinite point). The green and the red parts of the boundary are mapped to the green and red parts, respectively.
As can be seen,
(1) All values of $f$ are in $|z|<1+\varepsilon$. In particular, $|f|<1+\varepsilon$ along the segment $AB$.
(2) We have $|f|=1$ along the boundary of $S$, except for the segment $AB$.
(3) Inside $S$ we have $|f|>1$.
This function shows that Hadamard's theorem becomes false if we have a weaker constraint along any segment.
Remark: The function $f(z)$ can easily be constructed directly, without using the mapping theorem.
