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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 Inequality for complex Hankel function.

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up vote 8 down vote accepted

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 $f(x)\to1$.

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+1, this is very nice! – Kalim Aug 8 '13 at 17:55

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