The $M-L$ estimation lemma inequality states:
$$\left |\int_\Gamma f(z) dz\right| < ML(\Gamma)$$
Where $M = \max|f(z)|$ and $L(\Gamma)$ is the arc length of $\Gamma$.
Here: Wikipedia: Estimation Lemma
But how do we find the maximum of the modulus of $f(z)$?
Take the simpler example:
$$f(z) = \frac{\log(z)}{z^2 + 1}$$
How do we find the maximum using a semi-circle contour, with large radius $R$? Such that the parameter is:
$$z = Re^{i\theta}$$
I am confused about this "maximum" system, I can find a bound, but the lemma requires the maximum, which is what is the confusion here.
Thanks. Any help is welcome.
You could use:
$$\left| \int_\Gamma f(z) \text{d} z\right| \leqslant \int_\Gamma |f(z)| |\text{d} z| \leqslant \operatorname*{max}_{z\in \Gamma}|f(z)| \int_\Gamma |\text{d} z| = \operatorname*{max}_{z\in \Gamma}|f(z)| \ell(\Gamma)$$
Where $\ell(\Gamma)$ is the length of the contour.
Example
In the case of $f(z) = \frac{\log z}{z^2+1}$.
On the semi-circle $z(\theta) = Re^{i\theta}$ with $\theta: 0\to \pi$
$$\begin{align} \left|\frac{\log z}{z^2+1}\right| &= \left|\frac{\log Re^{i\theta}}{R^2e^{2i\theta}+1}\right| \\ &= \left|\frac{\log{R} +i\theta}{R^2e^{2i\theta}+1}\right| \\ &\leqslant\frac{\log{R} +\theta}{R^2-1}\\ &\leqslant\frac{\log{R} +\pi}{R^2-1} \end{align}$$
Such that if $R$ big enough (and $\ell(\Gamma) = R\pi$)
$$\left| \int_\Gamma f(z) \text{d} z\right| \leqslant \frac{\log{R} +\pi}{R^2-1}\cdot R\pi$$