Suppose that $f(z)$ is analytic for $\Re(z)>0$ and it is positive on the real axis. Suppose that for some $y\_0$ we have
$$
\left| {f\left( {x + iy} \right)} \right| \le f\left( x \right)
$$
for any $x>0$ and $y>y\_0$. Does it follow that this inequality holds for any $y\geq 0$? This question is related to my earlier question <https://mathoverflow.net/questions/138596/inequality-for-complex-hankel-function>.



Inequality for certain analytic functions
The answer is no. For example, consider $f(z)=1/(z^2+1)$. Then for every $y \geq \sqrt{2}$, we have
$$
|f(x+iy)|\leq f(x)
$$
for all $x>0$. Indeed, if you write everything out, you get that this inequality is equivalent to
$$
(x^2+1)^2 \leq (x^2-y^2+1)^2 + 4x^2y^2,
$$
which after simplification is seen to be equivalent to
$$
0 \leq y^2(y^2 -2 + 2x^2),
$$
which is true for all $x$ provided that $y \geq \sqrt 2$.


However, the inequality fails for $y=1$. Indeed, $|f(x+i)|\to \infty$ as $x \to 0$, whereas $f(x)\to1$.


