Arc length in ML estimation

Question

There are a lot of proofs in Complex Analysis where we use the following „Estimation lemma” (ML inequality):

$$\left|\int_{\Gamma}f\left(z\right)dz\right|\leq M\cdot l\left(\Gamma\right),$$

where $M=\sup_{z\in\Gamma}\left|f\left(z\right)\right|$, $l\left(\Gamma\right)=\int_{a}^{b}\left|\gamma'\left(t\right)\right|dt$ and $\gamma:\left[a,b\right]\rightarrow\Gamma$ is a parametrization of the $\Gamma$ curve (Check for example: https://en.wikipedia.org/wiki/Estimation_lemma).

In this inequality $l\left(\Gamma\right)$ looks like a kind of „total variation” on interval $\left[a,b\right]$ of $\gamma$, however in most of the proofs we calculate it as a kind of „Euclidean length” of the curve.

E.g. we give $2r\pi$ circumference of a circle, instead of $4r$. This Euclidean length shouldn't be calculated as

$$\int_{a}^{b}\sqrt{1+\left(\gamma'\left(t\right)\right)^{2}}dt?$$

This doesn't really bother the correctness of the proofs since in most cases these lengths are only constants. However I still don't understand why we „mix them”.

Or is it because in Complex Analysis $\left|.\right|$ stands for the „modulus” of the $z\in\mathbb{C}$ complex number and not just a simple "absolute value"?

Although, I see in my example that

$$\int_{0}^{2\pi}\left|\left(re^{it}\right)'\right|dt=\int_{0}^{2\pi}\left|\left(rie^{it}\right)\right|dt=\int_{0}^{2\pi}\left|r\right|\left|i\right|\left|e^{it}\right|dt=\int_{0}^{2\pi}rdt=2r\pi,$$

so what we give is actually correct since it is the circumference of the circle. It is maybe trivial, but I don't really see the intuition behind it how they can coincide. Is it because in Complex Analysis the $\left|.\right|$ modulus is basically the Euclidean norm if we imagine it in $\mathbb{R}^{2}$?

Answer 1

You seem to be confusing the formulas for arclengths of

1. a parametrized curve $\gamma:[a,b]\to\Bbb{R}^n$, in which case the arclength is $\int_a^b\|\gamma’(t)\|_{\Bbb{R}^n}\,dt$.
2. the graph of a function $f:[a,b]\to\Bbb{R}$. The formula here can be deduced from the previous case by considering $\gamma_f:[a,b]\to\Bbb{R}^2$, $\gamma_f(t)=(t,f(t))$, in which case the arclength is \begin{align} \int_a^b\|\gamma_f’(t)\|_{\Bbb{R}^2}\,dt=\int_a^b\|(1,f’(t))\|_{\Bbb{R}^2}\,dt=\int_a^b\sqrt{1+|f’(t)|^2}\,dt. \end{align}

Here, $\|x\|_{\Bbb{R}^n}:=\sqrt{\sum_{j=1}^n|x_j|^2}$ is the usual $\ell^2$-norm . In complex analysis, you parametrize $\Gamma$ by $\gamma:[a,b]\to\Bbb{C}$, so you use the formula (1) with $n=2$ (keeping in mind of course that the $\ell^2$-norm of a vector $(a,b)\in\Bbb{R}^2$ is the same as the modulus of the corresponding complex number $a+ib$, since both are equal by definition to $\sqrt{a^2+b^2}$).