Estimation Lemma

Lemma

Let $D \subseteq \C$ be open.

Let $f : D \to \C$ be continuous.

Let $\gamma : [a,b] \to D$ be a path in $D$.

Then:

$\displaystyle \left| \int_\gamma f(z)\ dz \right| \leq \int_\gamma |f(z)|\cdot |dz| \leq M L(\gamma) $

where $L(\gamma)$ is the length of $\gamma$:

$\displaystyle M = \sup_{z \in \gamma} |f(z)| $

and

$\displaystyle \int_\gamma |f(z)|\cdot |dz| = \int_a^b |f(\gamma(t))|\cdot |\gamma'(t)|\ dt$

Proof of Estimation Lemma

We have

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert \int_\gamma f(z)\ dz \right\vert\)	\(\)	\(\displaystyle \left\vert \int_a^b f(\gamma(t))\gamma'(t)\ dt \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \int_a^b \vert f(\gamma(t))\vert \cdot \vert \gamma'(t) \vert \ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the Absolute Value of Complex Integral (this proves the first inequality)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle M \int_a^b \vert\gamma'(t)\vert\ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle ML(\gamma)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the definition of Definition:Length (Complex Plane)length

$\blacksquare$