proof of estimating theorem of contour integral

WLOG consider $g(t):ℝ\to ℂ$ 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 re-parameterization the result will be completely general.

With this in mind, the contour integral can be explicitly written as

$${\int }_{\gamma }f(z)𝑑z={\int }_0^{L}f(g(t))g^{\prime }(t)𝑑t$$(1)

where $L$ is the arc length of the curve $\gamma$.

Consider the set of all continuous functions $[0,L]\to ℂ$ as a vector space, we can define an inner product in it via

$$⟨f,g⟩={\int }_0^{L}f(t)\overline{g}(t)𝑑t$$(2)

The axioms are easy to verify:

• $⟨k_1{a}_1+k_2{a}_2,a_3⟩={\int }_0^L(k_1{a}_1(t)+k_2{a}_2(t))\overline{a_3}(t)𝑑t=k_1⟨a_1,a_3⟩+k_2⟨a_2,a_3⟩$

• $⟨a,b⟩={\int }_0^{L}a(t)\overline{b}(t)𝑑t={\int }_0^L\overline{b(t)\overline{a}(t)}𝑑t=\overline{{\int }_0^{L}b(t)\overline{a}(t)𝑑t}=\overline{⟨b,a⟩}$

• $⟨a,a⟩={\int }_0^{L}a(t)\overline{a}(t)𝑑t={\int }_0^L{|a(t)|}^2𝑑t\ge 0$ since the integrand is a non-negative (real) function, and $0$ iff ${|a|}^2=0$ everywhere in the interval, that is: $⟨a,a⟩=0\iff a=0$

With all this in mind, equation 1 can be written as

$${\int }_{\gamma }f(z)𝑑z=⟨f\circ g,{\overline{g}}^{\prime }⟩$$(3)

Where by definition is the norm associated with the inner product defined previously.

Using Cauchy-Schwarz inequality we can write that

$$|⟨f\circ g,{\overline{g}}^{\prime }⟩|\le \parallel f\circ g\parallel \parallel {\overline{g}}^{\prime }\parallel$$(4)

But since by assumption the parameterization $g$ is canonic, $\parallel {\overline{g}}^{\prime }\parallel =\parallel g^{\prime }\parallel =\sqrt{{\int }_0^{L}1𝑑t}=\sqrt{L}$.

On the other hand $\parallel f\circ g\parallel =\sqrt{{\int }_0^{L}f(g(t))\overline{f}(g(t))𝑑t}\le \sqrt{{\int }_0^L{M}^2𝑑t}=M\sqrt{L}$, where $|f(g(t))|\le M$ for every point on $\gamma$.

The previous paragraphs imply that

$$\left|{\int }_{\gamma }f(z)𝑑z\right|\le ML$$(5)

which is the result we aimed to prove.

Cauchy-Schwarz inequality says more, it also says that $|⟨a,b⟩|=\parallel a\parallel \parallel b\parallel \iff a=\lambda b$ where $\lambda$ is a constant.

So if $|⟨f\circ g,{\overline{g}}^{\prime }⟩=\parallel f\circ g\parallel \parallel {\overline{g}}^{\prime }\parallel$ then $f\circ g=\lambda {\overline{g}}^{\prime }$, where $\lambda \in ℂ$ 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.