proof of estimating theorem of contour integral
WLOG consider $g(t):\mathbb{R}\to \u2102$ a parameterization of the $\gamma $ curve along which the integral^{} is evaluated with ${g}^{\prime}(t)=1$. This amounts to a canonical parameterization and is always possible. Since the integral is independent of reparameterization^{1}^{1}apart from a possible sign change due to exchange of orientation of the path the result will be completely general.
With this in mind, the contour integral can be explicitly written as
$${\int}_{\gamma}f(z)\mathit{d}z={\int}_{0}^{L}f(g(t)){g}^{\prime}(t)\mathit{d}t$$  (1) 
where $L$ is the arc length^{} of the curve $\gamma $.
Consider the set of all continuous functions^{} $[0,L]\to \u2102$ as a vector space^{}^{2}^{2}axioms are trivial to verify, we can define an inner product^{} in it via
$$\u27e8f,g\u27e9={\int}_{0}^{L}f(t)\overline{g}(t)\mathit{d}t$$  (2) 
The axioms are easy to verify:

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$\u27e8{k}_{1}{a}_{1}+{k}_{2}{a}_{2},{a}_{3}\u27e9={\int}_{0}^{L}({k}_{1}{a}_{1}(t)+{k}_{2}{a}_{2}(t))\overline{{a}_{3}}(t)\mathit{d}t={k}_{1}\u27e8{a}_{1},{a}_{3}\u27e9+{k}_{2}\u27e8{a}_{2},{a}_{3}\u27e9$

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$\u27e8a,b\u27e9={\int}_{0}^{L}a(t)\overline{b}(t)\mathit{d}t={\int}_{0}^{L}\overline{b(t)\overline{a}(t)}\mathit{d}t=\overline{{\int}_{0}^{L}b(t)\overline{a}(t)\mathit{d}t}=\overline{\u27e8b,a\u27e9}$
 •
With all this in mind, equation 1 can be written as
$${\int}_{\gamma}f(z)\mathit{d}z=\u27e8f\circ g,{\overline{g}}^{\prime}\u27e9$$  (3) 
Where by definition $$ is the norm associated with the inner product defined previously.
Using CauchySchwarz inequality we can write that
$$\u27e8f\circ g,{\overline{g}}^{\prime}\u27e9\le \parallel f\circ g\parallel \parallel {\overline{g}}^{\prime}\parallel $$  (4) 
But since by assumption the parameterization $g$ is canonic, $\parallel {\overline{g}}^{\prime}\parallel =\parallel {g}^{\prime}\parallel =\sqrt{{\int}_{0}^{L}1\mathit{d}t}=\sqrt{L}$.
On the other hand $\parallel f\circ g\parallel =\sqrt{{\int}_{0}^{L}f(g(t))\overline{f}(g(t))\mathit{d}t}\le \sqrt{{\int}_{0}^{L}{M}^{2}\mathit{d}t}=M\sqrt{L}$, where $f(g(t))\le M$ for every point on $\gamma $.
The previous paragraphs imply that
$$\left{\int}_{\gamma}f(z)\mathit{d}z\right\le ML$$  (5) 
which is the result we aimed to prove.
CauchySchwarz inequality says more, it also says that $\u27e8a,b\u27e9=\parallel a\parallel \parallel b\parallel \iff a=\lambda b$ where $\lambda $ is a constant.
So if $\u27e8f\circ g,{\overline{g}}^{\prime}\u27e9=\parallel f\circ g\parallel \parallel {\overline{g}}^{\prime}\parallel $ then $f\circ g=\lambda {\overline{g}}^{\prime}$, where $\lambda \in \u2102$ is a constant. If $g$ is a canonical parameterization ${g}^{\prime}=1$ and we get the absolute modulus^{} $\lambda =f\circ g$ (which must be constant) and all that remains is to find the phase of $\lambda $ which must also be constant.
Title  proof of estimating theorem of contour integral 

Canonical name  ProofOfEstimatingTheoremOfContourIntegral 
Date of creation  20130322 15:46:02 
Last modified on  20130322 15:46:02 
Owner  cvalente (11260) 
Last modified by  cvalente (11260) 
Numerical id  22 
Author  cvalente (11260) 
Entry type  Proof 
Classification  msc 30E20 
Classification  msc 30A99 