proof of estimating theorem of contour integral

WLOG consider $g(t):\mathbb{R}\to\mathbb{C}$ 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-parameterization11apart 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

 $\int_{\gamma}f(z)dz=\int_{0}^{L}f(g(t))g^{\prime}(t)dt$ (1)

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

Consider the set of all continuous functions $[0,L]\to\mathbb{C}$ as a vector space22axioms are trivial to verify, we can define an inner product in it via

 $\langle f,g\rangle=\int_{0}^{L}f(t)\bar{g}(t)dt$ (2)

The axioms are easy to verify:

• $\langle k_{1}a_{1}+k_{2}a_{2},a_{3}\rangle=\int_{0}^{L}(k_{1}a_{1}(t)+k_{2}a_{% 2}(t))\bar{a_{3}}(t)dt=k_{1}\langle a_{1},a_{3}\rangle+k_{2}\langle a_{2},a_{3}\rangle$

• $\langle a,b\rangle=\int_{0}^{L}a(t)\bar{b}(t)dt=\int_{0}^{L}\overline{b(t)\bar% {a}(t)}dt=\overline{\int_{0}^{L}b(t)\bar{a}(t)dt}=\overline{\langle b,a\rangle}$

• $\langle a,a\rangle=\int_{0}^{L}a(t)\bar{a}(t)dt=\int_{0}^{L}|a(t)|^{2}dt\geq 0$ since the integrand is a non-negative (real) function, and $0$ iff $|a|^{2}=0$ everywhere in the interval, that is: $\langle a,a\rangle=0\iff a=0$

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

 $\int_{\gamma}f(z)dz=\langle f\circ g,\bar{g}^{\prime}\rangle$ (3)

Where by definition $\|f\|=\sqrt{}$ is the norm associated with the inner product defined previously.

Using Cauchy-Schwarz inequality we can write that

 $|\langle f\circ g,\bar{g}^{\prime}\rangle|\leq\|f\circ g\|\|\bar{g}^{\prime}\|$ (4)

But since by assumption the parameterization $g$ is canonic, $\|\bar{g}^{\prime}\|=\|g^{\prime}\|=\sqrt{\int_{0}^{L}1dt}=\sqrt{L}$.

On the other hand $\|f\circ g\|=\sqrt{\int_{0}^{L}f(g(t))\bar{f}(g(t))dt}\leq\sqrt{\int_{0}^{L}M^% {2}dt}=M\sqrt{L}$, where $|f(g(t))|\leq M$ for every point on $\gamma$.

The previous paragraphs imply that

 $\left|\int_{\gamma}f(z)dz\right|\leq ML$ (5)

which is the result we aimed to prove.

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

So if $|\langle f\circ g,\bar{g}^{\prime}\rangle=\|f\circ g\|\|\bar{g}^{\prime}\|$ then $f\circ g=\lambda\bar{g}^{\prime}$, where $\lambda\in\mathbb{C}$ 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.

Title proof of estimating theorem of contour integral ProofOfEstimatingTheoremOfContourIntegral 2013-03-22 15:46:02 2013-03-22 15:46:02 cvalente (11260) cvalente (11260) 22 cvalente (11260) Proof msc 30E20 msc 30A99